IIT JEE Advanced Comprehensive Mathematics

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IIT JEE eBooks: www.crackjee.xyz

IIT JEE eBooks: www.crackjee.xyz

JEE ADVANCED Comprehensive

2019

Mathematics

IIT JEE eBooks: www.crackjee.xyz

IIT JEE eBooks: www.crackjee.xyz

JEE ADVANCED

2019

Comprehensive

Mathematics Ravi Prakash

Retired Associate Professor, Rajdhani College University of Delhi, Delhi

Ajay Kumar

Professor of Mathematics University of Delhi, Delhi

Usha Gupta

Retired Associate Professor, Rajdhani College University of Delhi, Delhi

McGraw Hill Education (India) Private Limited CHENNAI

McGraw Hill Education Offices Chennai New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San  Juan Santiago  Singapore Sydney Tokyo Toronto

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McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited, 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur,Chennai - 600 116, Tamil Nadu, India Comprehensive Mathematics—JEE Advanced Copyright © 2018 by McGraw Hill Education (India) Private Limited No Part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers. McGraw Hill Education (India) Private Limited 1 2 3 4 5 6 7 8 9 D102542 22 21 20 19 18

ISBN (13) : 978-93-87572-59-1 ISBN (10) : 93-87572-59-5 Information contained in this work has been obtained McGraw Hill Education (India), from sources believed to be reliable. However, neither, McGraw Hill nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought.

at Printers, 312 EPIP, HSIDC, Kundli, Sonepat, Haryana Cover Design: Neeraj Dayal Visit us at: www.mheducation.co.in

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Preface Although the first edition of our book was published twenty five years ago in 1992, the purpose and goal of the book remains the same, i.e., to provide the students with material that enables them to crack IIT Advanced Question Papers of Mathematics. With this purpose and goal in mind the book has been revised several times in the last twenty five years to meet the exacting standards of IIT Advanced Exam. This edition has been streamlined with the present day trends of the question paper. The book has 28 chapters. Each chapter begins with important definitions, formulae, tips, tricks and techniques required to deal with the problems in the chapter.

Questions in each chapter are split into four groups: ∑ Solved Examples ∑ Exercise (Level 1) Questions ∑ Exercise (Level 2) Questions ∑ Past Years IIT Advanced Questions (onwards 1978)

Solved Examples, Level 1, Level 2 and Past Years Questions are further split in to following five categories:  Single Correct Answer Type Questions  Multiple Correct Answers Type Questions  Assertion–Reason Type Questions  Comprehension Type Questions  Integer Answer Type Questions However, for the sake of completeness, Fill in the Blanks Type Questions and Subjective Type Questions from Past Years are also included in the Past Years Questions. We hope the book meets the needs of the students for which it is written. Over the last 25 years we have received valuable comments and suggestions from many teachers and students towards the improvement of the book. We are thankful to all of them. We hope that readers will continue to be kind enough to give the feedback for further improvement of the book. Although we hope the book is free from errors, but experience suggests otherwise. If you happen to find an error (typo or factual) please do point it to us to make the book error free for next generation readers. Dr. Ravi Prakash Prof. Ajay Kumar

Mrs. Usha Gupta

IIT JEE eBooks: www.crackjee.xyz

IIT JEE eBooks: www.crackjee.xyz

Syllabus Algebra: Algebra of complex numbers, addition, multiplication, conjugation, polar representation, properties of modulus and principal argument, triangle inequality, cube roots of unity, geometric interpretations. Quadratic equations with real coefficients, relations between roots and coefficients, formation of quadratic equations with given roots, symmetric functions of roots. Arithmetic, geometric and harmonic progressions, arithmetic, geometric and harmonic means, sums of finite arithmetic and geometric progressions, infinite geometric series, sums of squares and cubes of the first n natural numbers. Logarithms and their properties. Permutations and combinations, Binomial theorem for a positive integral index, properties of binomial coefficients. Matrices as a rectangular array of real numbers, equality of matrices, addition, multiplication by a scalar and product of matrices, transpose of a matrix, determinant of a square matrix of order up to three, inverse of a square matrix of order up to three, properties of these matrix operations, diagonal, symmetric and skew-symmetric matrices and their properties, solutions of simultaneous linear equations in two or three variables. Addition and multiplication rules of probability, conditional probability, Bayes Theorem, independence of events, computation of probability of events using permutations and combinations. Trigonometry: Trigonometric functions, their periodicity and graphs, addition and subtraction formulae, formulae involving multiple and sub-multiple angles, general solution of trigonometric equations. Relations between sides and angles of a triangle, sine rule, cosine rule, half-angle formula and the area of a triangle, inverse trigonometric functions (principal value only). Analytical geometry (2 dimensions): Cartesian coordinates, distance between two points, section formulae, shift of origin. Equation of a straight line in various forms, angle between two lines, distance of a point from a line; Lines through the point of intersection of two given lines, equation of the bisector of the angle between two lines, concurrency of lines; Centroid, orthocentre, incentre and circumcentre of a triangle. Equation of a circle in various forms, equations of tangent, normal and chord. Parametric equations of a circle, intersection of a circle with a straight line or a circle, equation of a circle through the points of intersection of two circles and those of a circle and a straight line. Equations of a parabola, ellipse and hyperbola in standard form, their foci, directrices and eccentricity, parametric equations, equations of tangent and normal. Locus Problems. Analytical geometry (3 dimensions): Direction cosines and direction ratios, equation of a straight line in space, equation of a plane, distance of a point from a plane. Differential calculus: Real valued functions of a real variable, into, onto and one-to-one functions, sum, difference, product and quotient of two functions, composite functions, absolute value, polynomial, rational, trigonometric, exponential and logarithmic functions. Limit and continuity of a function, limit and continuity of the sum, difference, product and quotient of two functions, L’Hospital rule of evaluation of limits of functions. Even and odd functions, inverse of a function, continuity of composite functions, intermediate value property of continuous functions. Derivative of a function, derivative of the sum, difference, product and quotient of two functions, chain rule, derivatives of polynomial, rational, trigonometric, inverse trigonometric, exponential and logarithmic functions.

IIT JEE eBooks: www.crackjee.xyz viii  Comprehensive Mathematics—JEE Advanced

Derivatives of implicit functions, derivatives up to order two, geometrical interpretation of the derivative, tangents and normals, increasing and decreasing functions, maximum and minimum values of a function, Rolle’s Theorem and Lagrange’s Mean Value Theorem. Integral calculus: Integration as the inverse process of differentiation, indefinite integrals of standard functions, definite integrals and their properties, Fundamental Theorem of Integral Calculus. Integration by parts, integration by the methods of substitution and partial fractions, application of definite integrals to the determination of areas involving simple curves. Formation of ordinary differential equations, solution of homogeneous differential equations, separation of variables method, linear first order differential equations. Vectors: Addition of vectors, scalar multiplication, dot and cross products, scalar triple products and their geometrical interpretations.

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Contents Preface

v

Syllabus vii Format of Questions in this Book

xi

1. Complex Numbers

1.1–1.81

2. Quadratic Equations

2.1–2.62

3. Progressions

3.1–3.57

4. Logarithm

4.1–4.27

5. Permutations and Combinations

5.1–5.46



6.1–6.49

6. Binomial Theorem

7. Matrices

7.1–7.41

8. Determinants

8.1–8.59

9. Probability

9.1–9.65

10. Trigonometry

10.1–10.57

11. Trigonometric Equations

11.1–11.22

12. Solution of Triangles and Applications of Trigonometry

12.1–12.51

13. Inverse Trigonometric Functions

13.1–13.30

14. Cartesian System of Coordinates and Locus

14.1–14.24

15. Straight Lines

15.1–15.52

16. Circles and Systems of Circles

16.1–16.60

17. Parabola

17.1–17.43

18. Ellipse

18.1–18.42

19. Hyperbola

19.1–19.33

20. Three Dimensional Geometry

20.1–20.45

21. Functions

21.1–21.38

22. Limits and Continuity

22.1–22.49

23. Differentiability and Differentiation

23.1–23.52

24. Applications of Derivatives

24.1–24.52

25. Indefinite Integration

25.1–25.66

26. Definite Integrals

26.1–26.70

IIT JEE eBooks: www.crackjee.xyz x  Comprehensive Mathematics—JEE Advanced

27. Differential Equations

27.1–27.44

JEE (Advanced)—2017

PYQ.1–PYQ.7

JEE (Advanced)—2017

PYQ.9–PYQ.14

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Format of Questions in this Book Single Correct Answer Type Questions This section contains multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which only one answer is correct.

Multiple Correct Answers Type Questions This section contains multiple correct answer type questions. Each question has four choices (a), (b), (c) and (d), out of which one or more answers are correct.

Matrix-Match Type Questions Each question in this section contains statements given in two columns, which have to be matched. Statements in Column I are labelled as a, b, c and d whereas statements in Column II are labelled as p, q, r and s. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are a – q, a – r, b – p, b – s, c – s and d – q, then the correctly bubbled matrix will look like the following. p q r s a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Assertion-Reason Type Questions Each question in this section has four choices (a), (b), (c) and (d) out of which only one is correct. Mark your choices as follows: (a) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1 (b) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1 (c) STATEMENT-1 is True, STATEMENT-2 is False (d) STATEMENT-1 is False, STATEMENT-2 is True

Comprehension-Type Questions Each set in this section contains a paragraph followed by questions. Each question has four choices (a), (b), (c) and (d), out of which one or more answers are correct.

Integer-Answer Type Questions Each question in this section has an answer which is a single digit (0 to 9)

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1 Complex Numbers 1.1

DEFINITIONS

A number of the form a + ib where a, b ΠR, the set of real numbers, and i = -1 , is called a complex number. ordered pair of real numbers a and b, and may be written as (a, b), real part and the second number denotes the imaginary part. If z = a + ib, then the real part of z is denoted by Re(z) and the imaginary part of z is denoted by Im(z). A complex number z is said to be purely real if Im(z) = 0 and is said to be purely imaginary if Re(z) = 0. Note that the complex number 0 = 0 + i0 is both purely real and purely imaginary. It is the only complex number with this property. We denote the set of all complex numbers by C. That is, C = {a + ib| a, b ΠR}. Two complex numbers z1 = a1 + ib1 and z2 = a2 + ib2 are said to be equal if a1 = a2 and b 1 = b 2. Very Important The main advantage of complex number is to use one symbol z for an ordered pair (x, y) of real numbers. We lose this advantage whenever we replace z by x + iy. As far as possible do not write z = x + iy while solving the problems. Putting z = x + iy should be the last resort

5. Quotient: If at least one of c, d is non-zero, then quotient of a + ib and c + id is given by a + ib (a + ib) (c - id ) (ac + bd ) + i(bc - ad ) = = c + id (c + id ) (c - id ) c2 + d 2 = 1.3

ac + bd c +d 2

2

+i

bc - ad c2 + d 2

CONJUGATE OF COMPLEX NUMBER

Let z = a + ib of z, denoted by z to be the complex number a – ib. That is, if z = a + ib, then z = a – ib. Properties of Conjugate of a Complex Number

(i) z1 = z2 ¤ z1 = z2 (ii) ( z ) = z (iii) z + z = 2 Re (z) (iv) z – z = 2i Im(z) (v) z = z ¤ z is purely real (vi) z + z = 0 ¤ z is purely imaginary (vii) z z = [Re (z)]2 + [Im(z)]2 (viii) z1 + z2 = z1 + z2 (ix) z1 - z2 = z1 - z2

1.2

ALGEBRAIC OPERATIONS WITH COMPLEX NUMBERS

1. Addition: (a + ib) + (c + id ) = (a + c) + i(b + d ) 2. Subtraction: (a + ib) – (c + id ) = (a – c) + i(b – d ) 3. Multiplication: (a + ib) (c + id ) = (ac – bd ) + i(ad + bc) 4. Reciprocal: If at least one of a, b is non-zero then the reciprocal of a + ib is given by 1 a - ib a b = = 2 -i 2 2 a + ib (a + ib) (a - ib) a +b a + b2

(x) z1z2 = z1 z2 z Êz ˆ (xi) Á 1 ˜ = 1 if z2 π 0 Ë z2 ¯ z2 (xii) If P(z) = a0 + a1 z + a2 z 2 + … + an zn where a0, a1, … an and z are complex number, then P ( z ) = a0 + a1 ( z ) + a2 ( z )2 +  + an ( z )n = P( z ) where P ( z ) = ao + a1z + a2 z 2 +  + an z n

IIT JEE eBooks: www.crackjee.xyz 1.2 Comprehensive Mathematics—JEE Advanced

P (z) (xiii) If R(z) = where P (z) and Q (z) are Q( z ) polynomials in z, and Q(z) π 0, then R (z) =

(x) |z1 + z2|2 + |z1 – z2|2 = 2(|z1|2 + |z2|2) (xi) If a and b are real numbers and z1, z2 are complex numbers, then |az1 + bz2|2 + |bz1 – az2|2 = (a2 + b2) (|z1|2 + |z2|2)

P( z ) Q( z )

(xii) If z1, z2 π 0, then |z1 + z2|2 = |z1|2 + |z2|2 ¤ z1/z2 is purely imaginary.

Illustration 1 (xiii) Triangle Inequality. If z1 and z2 are two complex numbers, then |z1 + z2| £ | z1| + |z2|.

Ê z + 3z 2 ˆ z + 3z 2 = ÁË z - 1 ˜¯ z -1 a1 (xiv) If z = b1 c1

a2 b2 c2

a3 a1 b3 , then z = b1 c3 c1

The equality holds if and only if z1 z2 ≥ 0.

a2 b2 c2

In general, |z1 + z2 + ... + zn | £ | z1| + |z2| + ... + |zn| and the equality sign holds if and only if the ratio of any two non-zero terms is positive.

a3 b3 c3

(xiv) |z1 – z2| £ |z1| + |z2| (xv )

where ai, bi, ci (i = 1, 2, 3) are complex numbers. 1.4

z1 - z2 £ |z1| + |z2|

(xvi) |z1 – z2| ≥ ||z1| – |z2||

MODULUS OF A COMPLEX NUMBER

Let z = a + ib

modulus

or the absolute value of z to be the real number and denote it by | z|. Note that | z "zŒC

a 2 + b2

(xvii) If a1, a2, a3 and a4 are four complex numbers, then | z – a1 | + | z – a2 | + | z – a3 | + | z – a4 | ≥ max {| a1 – al | + | am – an|: l, m, n are distinct integers lying in {2, 3, 4} and m < n}. 1.5

Properties of Modulus

GEOMETRICAL REPRESENTATION OF COMPLEX NUMBERS

If z is a complex number, then A complex number z = x + iy can be represented by a point P whose Cartesian co-ordinates are (x, y) referred to rectangular axes Ox and Oy, usually called the real and imaginary axes respectively. The plane is called the Argand plane, complex plane or Gaussian plane. The point P (x, y) is called the image of the complex number z and z is said to be the or complex co-ordinate of point P.

(i) |z| = 0 ¤ z = 0 (ii) |z| = | z | = |– z| = |– z | (iii) – | z| £ Re(z) £ |z| (iv) – |z| £ Im(z) £ |z| (v) z z = |z|2 In particular, note that if z π 0, then

()

Im ( z ) 1 Ê 1 ˆ Re ( z ) == and Im . 2 Ë z¯ z | z |2 |z| If z1, z2 are two complex numbers, then

Re

z 1 = 2 so that z |z|

Note All purely real numbers lie on the real axis and all purely imaginary numbers lie on the imaginary axis. The complex number 0 = 0 + i0 lies at the origin O.

(vi) |z1 z2| = |z1| |z2| (vii)

z1 z = 1 , if z2 π 0 z2 z2

(viii) |z1 + z2|2 = |z1|2 + |z2|2 + z1 z2 + z1 z2 = |z1|2 + |z2|2 + 2Re( z1 z2 ) (ix) |z1 – z2|2 = |z1|2 + |z2|2 – z1z2 - z1 z2 = |z1|2 + |z2|2 – 2Re( z1 z2 )

We have OP = 1.6

x 2 + y 2 = |z| . Thus, |z| is the length of OP.

ARGUMENT OF A COMPLEX NUMBER

If z is a non-zero complex numbers represented by point P in the complex plane, then argument of z is the angle which OP makes with the positive direction of the real axis. See Fig. 1.1.

IIT JEE eBooks: www.crackjee.xyz Complex Numbers 1.3 Y

Note P (x + iy)

y q

x

O

OP = | z | arg (z) = q

Argument of a non-zero complex number is not unique, since, if q is a value of the argument, then 2np + q where n Œ I, the set of integers, are also values of the argument of z. The value q inequality – p < q £ p is called the principal value of the argument or principal argument.

X

M

Fig. 1.1

Principal Value of the Argument for Different Positions of z in the Complex Plane 1. When z

2. When z lies in the second quadrant Y z = x + iy

y q

X

x

O

Ê yˆ arg (z) = tan–1 Á ˜ Ë x¯

Ê yˆ arg (z) = p - tan -1 Á ˜ Ë x¯

Fig. 1.2 (i)

Fig. 1.2 (ii)

3. When z lies in the third quadrant

Ê yˆ arg (z) = -p + tan -1 Á ˜ Ë x¯ Fig. 1.2 (iii)

4. When z lies in the fourth quadrant

Ê yˆ arg (z) = - tan -1 Á ˜ Ë x¯ Fig. 1.2 (iv)

Thus, if z = x + iy, then Ïtan ( y / x ) Ô -1 Ôtan ( y / x ) + p Ô arg ( z ) = Ìtan -1 ( y / x ) - p Ô Ôp / 2 ÔÓ- p /2 -1

Common Mistake if x > 0 if x < 0, y ≥ 0 if x < 0, y < 0 if x = 0, y > 0 if x = 0, y < 0

A usual mistake committed by the students is to take the argument of z = x + iy as tan–1 (y/x) irrespective of the values of x and y. Kindly remember that tan–1 (y/x) lies in the interval (– p/2, p/2) whereas the principal value of argument of z lies in the interval (– p, p ].

IIT JEE eBooks: www.crackjee.xyz 1.4 Comprehensive Mathematics—JEE Advanced 1.7

POLAR FORM OF A COMPLEX NUMBER

Let z be a non-zero complex number, then we can write z = r (cos q + i sin q ) where r = |z| and q = arg (z).

and – 1 – i =

= r12 + r22 + 2r1r2 cos (q1 - q 2 ) |z1 + z2|2 = |z1|2 + |z2|2 ¤ cos (q1 – q2) = 0

Illustration 2 –1 + i =

where r1 = |z1|, r2 = |z2| and q1 = arg (z1), q2 = arg (z2), then (vii) |z1 + z2|2 = |z1|2 + |z2|2 + 2|z1| |z2| cos (q1 – q2)

3p 3p ˆ Ê 2 Á cos + i sin ˜ Ë 4 4¯

{

Ê 3p ˆ Ê 3p ˆ 2 cos Á - ˜ + i sin Á - ˜ Ë 4¯ Ë 4¯

¤ Oz1 is at right angles to Oz2. (viii) |z1 – z2|2 = |z1|2 + |z2|2 – 2|z1| |z2| cos (q1– q2) = r12 + r22 - 2r1r2 cos (q1 - q 2 )

}

In fact, if z = r (cos q + i sin q ), then z is also given by z = r [cos (2kp +q ) + i sin (2kp +q)] where k is any integer. Euler’s Formula

The complex number cos q + i sin q is denoted by e or cis q. That is e iq = cis q = cos q + i sin q iq

Some Important Results Involving Argument

If z, z1 and z2 are non-zero complex numbers, then (i) arg ( z ) = – arg (z) (ii) *arg (z1 z2) = arg (z1) + arg (z2) In fact, arg (z1 z2) = arg (z1) + arg (z2) + 2kp where Ï0 if - p < arg ( z1) + arg ( z2 ) £ p Ô k = Ì1 if - 2p < arg ( z1) + arg ( z2 ) £ - p Ô-1 if p < arg ( z ) + arg ( z ) £ 2p Ó 1 2 (iii) *arg (z1 z2 ) = arg (z1) – arg (z2) Êz ˆ (iv) *arg Á 1 ˜ = arg (z1) – arg (z2) Ë z2 ¯ In fact,

|z1 – z2|2 = |z1|2 + |z2|2 ¤ Oz1 is perpendicular to Oz2. 1.8

VECTOR REPRESENTATION OF COMPLEX NUMBERS

We can also represent  the complex number z = x + iy by O of the complex using the vector OP plane to the point P(x, y),instead of using   point P itself. The length of the vector OP , that is, OP is the modulus of z. The angle between the positive real axis and the vector OP , more exactly, the angle through which the positive real axis must  be rotated to cause it to have the same direction as OP (considered positive if the rotation is counter-clockwise and negative otherwise) is the argument of the complex number z. 1.9

GEOMETRICAL REPRESENTATION OF ALGEBRAIC OPERATIONS ON COMPLEX NUMBERS

Let z1 and z2 be two complex numbers represented by the points P1(x1, y1) and P2 (x2, y2) respectively. Sum

z1 + z2 should be represented by the point (x1 + x2, y1 + y2). This point is nothing but the vertex P which the origin with z1 and z2 Y

Êz ˆ arg Á 1 ˜ = arg (z1) – arg (z2) + 2kp Ë z2 ¯ where k arg (z1) and arg (z2 ) replaced by – (v) |z1 + z2| = |z1 – z2| ¤ arg (z1) – arg (z2) = p/2 (vi) |z1 + z2| = |z1| + |z2| ¤ arg (z1) = arg (z2) If z1 = r1 (cos q1 + i sin q1) and z2 = r2 (cos q2 + i sin q2) * L.H.S. and R.H.S. may differ by a multiple of 2p.

P (x1 + x2, y1 + y 2)

P2 (x1, y 2)

y2 P1 (x1, y1) M

y1 O

x2

N

L Fig. 1.3

K

X

IIT JEE eBooks: www.crackjee.xyz Complex Numbers 1.5

Note that the addition of two complex numbers z1 and z2 follows the same law of addition as that of vectors, represented both in magnitude and direction by the line segments z1 and z2, for OP1 + OP2 = OP1 + P1P = OP Difference

z2 by P ¢2 so that P2P ¢2 is bisected at O. Complete the parallelogram OP1 PP ¢2. Then, it can be easily seen that P represents the difference z1 – z2. See Fig. 1.4. As OP1 PP¢2 is a parallelogram so P1P = OP¢2. Using vector notation, we have z1 – z2 = OP1 – OP2 = OP1 + P2O = OP1 + OP¢2 = OP1 + P1P = OP = P2P1

Let E be a point on the x-axis such that OE = 1 unit. (See Fig 1.5.) Complete the triangle OP1E. Now, taking OP2 as the base, construct a triangle OPP2 similar to OP1E so that OP : OP1 = OP2 : OE [∵ OE = 1] i.e., OP = OP1 ◊ OP2 Also – P2OP = –EOP1 = –XOP1 = q1 Thus – XOP = q1 + q2 Hence P represents the complex number for which the modulus is r1 r2 and the argument is q1 + q2. That is, it represents the complex number z1 z2.

Y P2 (x2, y2)

P1 (x1, y1) X

O

Fig. 1.5

Quotient P (x1 – x2, y1 – y 2)

P¢2 (– x2, – y2) Fig. 1.4

Note the complex number z1 – z2 is represented by the vector P2P1, where the points P2 and P1 represent the complex number z2 and z1 respectively. Note that arg(z1 – z2) is the angle through which OX must be rotated in the anticlockwise direction so that it becomes parallel to P2P1.

Let z1 = r1 (cos q1 + i sin q1) and z2 = r2 (cos q2 + i sin q2). We take z2 π 0, so that r2 π 0. Now z1 r (cos q1 + i sin q1 ) = 1 z2 r2 (cos q 2 + i sin q 2 ) =

r1 {cos (q1 – q2) + i sin (q1 –q2)} r2

We shall use this to get a geometrical interpretation of the quotient of a complex number by a non-zero complex number. Y P2

Product (Multiplication)

= r1 r2 {cos(q1 + q2) + i sin (q1 + q2)} Thus |z1 z2 | = r1 r2 and arg(z1 z2) = q1 + q2 This shows that modulus of the product of two complex numbers is the product of their moduli, and the argument of the product of any two complex numbers is the sum of their arguments. Using this, we shall derive a geometrical interpretation of the product of two complex numbers.

r2

Let z1 = r1(cos q1 + i sin q1) and z2 = r2 (cos q2 + i sin q2). Then z1 z2 = r1 r2 (cos q1 + i sin q1) (cos q2 + i sin q2)

q2 O

P1

r1 q1 E

q2

X

P Fig. 1.6

IIT JEE eBooks: www.crackjee.xyz 1.6 Comprehensive Mathematics—JEE Advanced Y

Let P1 and P2 represent z1 and z2 respectively. On OP1 construct the triangle OPP1 similar to OEP2, where E lies on the x-axis and OE = 1 unit. (See Fig. 1.6.) fi

OP : OE = r1 : r2

Now,

– XOP = q1 – q2

Also

z4

r OP = 1 r2

z2 q = arg

z1

The point P thus represents the quotient z1/z2, since its modulus is r1/r2 and its arguments is q1 – q2.

Êz -z ˆ

1 Interpretation of arg Á 3 Ë z2 - z1 ˜¯

If z1, z2, z3 are the vertices of a triangle ABC described in the counter-clockwise sense, then

– z2 3 – z4

1

I JK

z3

Remark Note that if q1 and q2 are the principal values of arg z1 and arg z2 then q1 + q2 is not necessarily the principal value of arg(z1 z2), nor is q1 – q2 necessarily the principal value of arg(z1/z2).

Fz GH z

O

X Fig. 1.8

Corollary

z4 and z3 z2 and z1 if Ê z - z2 ˆ arg Á 1 =± Ë z3 - z4 ˜¯ i.e., if z1 – z2 = ± ik(z3 – z4), where number. (Fig. 1.9).

is inclined at 90° to p 2 k is a non-zero real

Êz -z ˆ (i) arg Á 3 1 ˜ = –BAC = a (say ), and Ë z2 - z1 ¯ (ii)

z3 - z1 CA = (cos a + i sin a) z2 - z1 BA

See Fig. 1.7. C (z3)

Y

Q (z 3 – z1)

a

B (z2)

Fig. 1.9

A (z1)

1.10 a

P (z 2 – z1) X

O Fig. 1.7

Corollary The points z1, z2, z3 will be collinear if and only z -z if angle a = 0 or p, i.e., if and only if 3 1 is purely real. z2 - z1 Êz -z ˆ

2 Interpretation of arg Á 1 Ë z3 - z4 ˜¯

Let z1, z2, z3 and z4 be four complex numbers. Then the line z2 and z1 at z4 and z3 the following angle: Ê z - z2 ˆ arg Á 1 Ë z3 - z4 ˜¯

SOME IMPORTANT GEOMETRICAL RESULTS AND EQUATIONS

1. Distance Formula Distance between A(z1) and B(z2) is given by AB = |z2 – z1| B (z2)

A (z1) Fig. 1.10

2. Section Formula The point P(z AB in the ratio m : n is given by

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z=

mz2 + nz1 m+n

Fig. 1.11

3. Mid-point Formula Mid-point M(z) of the segment AB is given by 1 z = (z1 + z2) 2

(iii) Square ¤

(a) the diagonals AC and BD bisect each other z1 + z3 = z2 + z4

instance, AD = AB ¤ |z4 – z1| = |z2 – z1| (c) the two diagonals are equal, that is, AC = BD ¤ |z3 – z1| = |z4 – z2|

Fig. 1.15 Fig. 1.12

4. Condition(s) for four non-collinear A(z1), B(z2), C(z3) and D(z4) to represent vertices of a (i) Parallelogram The diagonals AC and BD must bisect each other ¤ ¤

(iv) Rectangle ¤

(a) the diagonals AC and BD bisect each other z1 + z3 = z2 + z4

(b) the diagonals AC and BD are equal ¤ |z3 – z1| = |z4 – z2|.

1 1 (z1 + z3) = (z2 + z4) 2 2 z1 + z3 = z2 + z4

Fig. 1.16 Showing that four points, no three of which are collinear, form a Parallelogram/Rhombus/Square/Rectangle Fig. 1.13

(ii) Rhombus (a) the diagonals AC and BD bisect each other ¤ z1 + z3 = z2 + z4, and

¤

instance, AD = AB |z4 – z1| = |z2 – z1|

Fig. 1.14

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5. Centroid, Incentre, Orthocentre and Circumcentre of a Triangle Let ABC be a triangle with vertices A(z1), B(z2) and C(z3), (i) Centroid G (z) of the DABC is the point of concurrence of medians of DABC and is given by 1 (z1 + z2 + z3) z= 3

Also

z=

z1 (sin 2 A) + z2 (sin 2 B) + z3 (sin 2C ) sin 2 A + sin 2 B + sin 2C

Fig. 1.19

(iv) Orthocentre H(z) of the DABC is the point of concurrence of altitudes of DABC and is given by

z12

Fig. 1.17

z22

(ii) Incentre I(z) of the DABC is the point of concurrence of internal bisectors of angles of DABC and is given by z=

z=

z32

az1 + bz2 + cz3 a+b+c

z1 2

z1 1

z2 1 + z2

2

z2 1

z3 1 z1 z2 z3

2

z3 1

z1 1

z3

z1 1 z2 1 z3 1

or z =

(tan A) z1 + (tan B) z2 + (tan C )z3 tan A + tan B + tan C

or z =

(a sec A) z1 + (b sec B)z2 + (c sec C )z3 a sec A + b sec B + c sec C z

z

Fig. 1.18

(iii) Circumcentre S(z) of the DABC is the point of concurrence of perpendicular bisectors of sides of DABC and is given by z=

( z2 - z3 ) z1 2 + ( z3 - z1 ) z2 2 + ( z1 - z2 ) z3 2 z1 ( z2 - z3 ) + z2 ( z3 - z1 ) + z3 ( z1 - z2 )

z=

z1

2

z1 1

z2

2

z2 1

z3 z1 z2 z3

2

z3 z1 z2 z3

1 1 1 1

z

z

Fig. 1.20

Remark In case circumcentre of DABC is at the origin, then orthocentre of triangle is given by z1 + z2 + z3.

Important Note It is not necessary to remember formulae for circumcentre and orthocentre of a triangle.

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Euler’s Line

The centroid G the orthocentre H and the circumcentre S of the triangle. G H and S in the ratio 2 : 1.

p/3

p/3

B (z2)

C (z3) Fig. 1.23

Fig. 1.21

8. Equation of a Straight Line (i) Non-parametric form

Thus, zG =

1 ( z + 2 zS ) 3 H

points A(z1) and B(z2) is z z1 z2

6. Area of a Triangle Area of DABC with vertices A(z1), B(z2) and C(z3) is given by z1 1 D = | z2 4i z3 =

z1 1 z2 1 | z3 1

or or

z 1 z1 1 = 0 z2 1 z - z1 z - z1 = z2 - z1 z2 - z1

z( z1 - z2 ) - z ( z1 - z2 ) + z1 z2 - z2 z1 = 0

1 Im ( z1z2 + z2 z3 + z3 z1 ) 2 Fig. 1.24

(ii) Parametric form

Fig. 1.22

7. Condition for Triangle to be Equilateral Triangle ABC with vertices A(z1), B(z2) and C(z3) is equilateral if and only if 1 1 1 + + =0 z2 - z3 z3 - z1 z1 - z2 ¤ ¤

z 21 + z22 + z 23 = z2 z3 + z3 z1 + z1 z2 z1 z2 = z2 z3 = z3 z1

¤

z12 = z2 z3 and z22 = z1z3 1 z2 1 z3 1 z1

¤

z - z2 z2 - z1 = 3 z3 - z2 z1 - z3

¤ ¤

z3 z1 = 0 z2

1 1 1 + + =0 z - z1 z - z2 z - z3

where

1 z = (z1 + z2 + z3) 3

points A(z1) and B(z2) is z = tz1 + (1 – t)z2 where t is a real parameter. (iii) General Equation of a Straight Line The general equation of a straight line is az + az + b = 0 where a is a non-zero complex number and b is a real number. 9. Complex Slope of a Line If A(z1) and B(z2) are two distinct points in the complex plane, then complex slope of AB to be z1 - z2 z1 - z2 Two lines with complex slopes m1 and m2 are (i) parallel, if m1 = m2 (ii) perpendicular, if m1 + m2 = 0 m=

The complex slope of the line az + az + b = 0 is given by -(a / a )

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10. Length of Perpendicular from a Point to a Line Length of perpendicular of point A(w) from the line az + az + b = 0 (a ΠC Р{0}, b ΠR) is given by

(z – z1) ( z – z2 ) + ( z – z1 ) (z – z2) z + i k z1 z2

p

az

b z+

=0

+a

where k is real a parameter.

Fig. 1.25

(v) Equation of a circle passing through three non-collinear points. Let three non-collinear points be A(z1), B(z2) and C(z3). Let P(z) be any point on the circle. Then either –ACB = –APB [when angles are in the same segment] –ACB + –APB = p [when angles are in the opposite segment]

aw + aw + b p= 2a 11. Some Results on Circle (i) Equation of a circle An equation of a circle with centre at z0 and radius r is |z – z0| = r iq or z = z0 + re , 0 £ q < 2p (parametric form) or

z 1 z1 1 = 0 z2 1

zz - z0 z - z0 z + z0 z0 - r 2 = 0

Fig. 1.26

Fig. 1.28

(ii) General equation of a circle General equation of a circle is (1)

Êz -z ˆ Ê z - z2 ˆ =0 arg Á 3 2 ˜ - arg Á Ë z - z1 ˜¯ Ë z3 - z1 ¯ or

where a is a complex number and b is a real number such that aa - b ≥ 0

Êz -z ˆ Ê z - z1 ˆ =p arg Á 3 2 ˜ + arg Á Ë z - z2 ˜¯ Ë z3 - z1 ¯

zz + az + az + b = 0



aa - b .



(iii) Diameter form of a circle An equation of the circle one of whose diameter A(z1) and B(z2) is (z – z1) ( z – z2 ) + ( z – z1 ) (z – z2) = 0

ÈÊ z - z ˆ Ê z - z1 ˆ ˘ arg ÍÁ 3 2 ˜ Á ˜˙ = 0 ÎË z3 - z1 ¯ Ë z - z2 ¯ ˚

or

ÈÊ z - z1 ˆ Ê z3 - z2 ˆ ˘ arg ÍÁ ˜Á ˜˙ = p ÎË z - z2 ¯ Ë z3 - z1 ¯ ˚

Centre of (1) is – a and its radius is

Êz ˆ [using arg Á 1 ˜ = arg (z1) – arg (z2) and Ë z2 ¯ arg (z1 z2) = arg (z1) + arg (z2)]

P (z) A (z 1)

B (z 2)

Fig. 1.27

(iv) An equation of circle through two points An equation of the circle passing through two points A (z1) and B (z2) is

( z - z1 ) ( z3 - z2 ) In any case, we get is purely ( z - z2 ) ( z3 - z1 ) real. ¤

( z - z1 ) ( z3 - z2 ) ( z - z1 ) ( z3 - z2 ) = ( z - z2 ) ( z3 - z1 ) ( z - z2 ) ( z3 - z1 ) (vi) Condition for four points to be concyclic. Four points z1, z2, z3 and z4 will lie on the (z - z ) (z - z ) same circle if and only if 4 1 3 2 ( z4 - z2 ) ( z3 - z1 ) is purely real.

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( z1 - z3 ) ( z2 - z4 ) is purely real. ( z1 - z4 ) ( z2 - z3 )

¤

1.11

RECOGNIZING SOME LOCI BY INSPECTION

(i) arg (z) = a (– p < a < p) arg (z) = a represents a ray starting at the origin (excluding the origin) and making an angle a with the positive direction of the real axis. See Fig. 1.29.

k > 0, k π 1 is a

(iv) If z1 and z2 real number, then z - z1 =k z - z2

represents a circle. For k = 1, it represents perpendicular bisector of the A(z1) and B(z2). (v) |z – z1| + |z – z2| = k k be a Let z1 and z2 positive real number. (a) If k > |z1 – z2|, then |z –z1| + |z – z2| = k represents an ellipse with foci k. at A(z1) and B(z2 See Fig. 1.32

C

A(z1)

B(z2)

Fig. 1.29

(ii) arg (z – z0) = a (– p < a < p) arg (z – z0) = a point z0 (excluding the point z0) and making an angle a with the positive direction of the real axis.

D CD = k

Fig. 1.32

(b) If k = |z1 – z2|, then |z – z1| + |z – z2| = k z1 and z2. (c) If k < |z1 – z2|, then |z – z1| + |z – z2| = k does not represent any curve in the Argand plane. (vi) |z – z1| – |z – z2 | = k k be a positive real Let z1 and z2 number. (a) If k < |z1 – z2|, then ||z – z1| – |z – z2|| = k represents a hyperbola with foci at A(z1) and B(z2). See Fig 1.33

Fig. 1.30

(iii) If z1 and z2 |z – z1| = |z – z2| represents the perpendicular bisector of the A(z1) and B(z2).

A(z1)

z

|

z–

|z

| z1

–z

2|

A (z1)

Fig. 1.33

B (z2) Fig. 1.31

B(z2)

(b) If k = |z1 – z2|, then ||z – z1| – |z – z2|| = k A(z1) and B(z2) but excluding the segment AB. See Fig. 1.34.

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A(z1)

P(z)

B(z2)

p/2

Fig. 1.34

A(z1)

Remark

B(z2)

If k > |z1 – z2|, then ||z – z1| – |z – z2|| = k does not represent any curve in the Argand plane.

p/2 P(z)

(vii) If z1 and z2

Fig. 1.37

|z – z1|2 + |z – z2|2 = |z1– z2|2

(c) If a = p, then

represent a circle with z1 and z2 as the extremities of a diameter. See Fig. 1.35. z

Ê z - z1 ˆ arg Á = a (= p ) Ë z - z2 ˜¯

B(z2)

A(z1) and B(z2) but excluding the segment AB. See Fig. 1.38

A(z1)

A(z1)

B(z2)

Fig. 1.38

Fig. 1.35

(d) If a = 0, then

Ê z - z1 ˆ =a (viii) arg Á Ë z - z2 ˜¯ Let z1 and z2 number such 0 £ a £ p. (a) If 0 < a < p and a π p /2, then

Ê z - z1 ˆ arg Á = a ( = 0) Ë z - z2 ˜¯

a be a real

A(z1) and B(z2) see Fig. 1.39.

Ê z - z1 ˆ =a arg Á Ë z - z2 ˜¯

A(z1)

represents a segment of the circle passing through A(z1) and B(z2). See Fig. 1.36.

a

B(z2)

A(z1) Fig. 1.36

(b) If a = p/2, then Ê z - z1 ˆ Ê pˆ = a Á= ˜ arg Á ˜ Ë 2¯ Ë z - z2 ¯ represents a circle with diameter as the segment A(z1) and B(z2). See Fig. 1.37.

B(z2) Fig. 1.39

1.12

GEOMETRIC INTERPRETATION OF MULTIPLYING A COMPLEX NUMBER BY eia.

Let z be a non-zero complex number. We can write z in the polar form as follows: z = r(cos q + i sin q) = reiq where r = |z| and arg (z) = q We have zeia = reiq eia = rei(q + a ) Thus, ze ia represents the complex number whose modulus is r and argument is q + a. Geometrically, ze ia can be obtained O and P(z) through an angle a in the anticlockwise direction. See Fig. 1.40

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Q(zeia )

Illustration 3 P(z)

r a

r

q Fig. 1.40

In particular iz is obtained by rotating OP through an angle of p/2 in the anticlockwise direction. Corollary If A(z1) and B(z2) are two complex number such that angle –AOB = q, then we can write z2 z eiq z1 1 z1 = r1 eia and z2 = r2 eib, z2 =

|z1| = r1, |z2| = r2.

where

z2 r2 eib r2 i ( b -a ) = = e z1 r1eia r1

then Note that

b–a=q

(

)

˘ 32 + 42 - 3 i ˚

1

1.13

DE MOIVRE’S THEOREM AND ITS APPLICATIONS

(a) De Moivre’s Theorem for integral index. If n is an integer, then (cos q + i sin q)n = cos (nq) + i sin (nq) (b) De Moivre’s Theorem for rational index. If n is a rational number, then value of or one of the values of (cosq + isinq)n is cos (nq) + isin (nq). In fact, if n = p/q where p, q Œ I, q > 0 and p,q have no factors in common, then (cos q + i sin q)n has q distinct values, one of which is cos (nq) + i sin (nq).

The values of (cos q + i sin q) p/q where p, q ΠI, q > 0, hcf (p,q) =1 are given by

z2 r2 iq = e z1 r1

Èp Èp ˘ ˘ cos Í (2kp + q ) ˙ + i sin Í (2kp + q )˙ q q ˚ ˚ Î Î where k = 0, 1, 2,…, q–1.

z2 z1eiq z1

z2 =

32 + 42 + 3 +

Note

Thus,



1 È Î 2

[2 2 + 2i ] = ± (2 + i) 2 Alternatively we can use De Moivre’s Theorem. = ±

O

Suppose

3 + 4i = ±

nth Roots of Unity

By an nth root of unity we mean any complex number z

Fig. 1.41

Square Roots of a Complex Number

Let z = x + iy. If a + ib is a square root of z, then (a + ib)2 = x + iy fi



a2 – b2 = x, 2ab = y

Also, a2 + b2 = Thus,

zn = 1 (1) Since, an equation of degree n has n roots, there are n values of z which satisfy the equation (1). To obtain these n values of z, we write 1 = cos (2kp) + i sin (2kp) where k ΠI and

x 2 + y2

a= ±

1

b= ±

1

2

where x 2 + y2 + x

x +y -x 2 If y > 0, then a and b both have the same signs. If y < 0, then a and b have opposite signs. 2

2

Ê 2 kp ˆ Ê 2 kp ˆ + i sin Á z = cos Á Ë n ˜¯ Ë n ˜¯ [using the De Moivre’s Theorem] k = 0, 1, 2, …, n –1.

Note We may use any n consecutive integral values to k. For instance, in case of 3, we may take – 1, 0 and 1 and in case of 4, we may take – 1, 0, 1 and 2 or – 2, – 1, 0 and 1.

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Notation

Ê 2p ˆ Ê 2p ˆ Let w = cos Á ˜ + i sin Á ˜ Ë n¯ Ë n¯

By using the De Moivre’s theorem, we can write the nth roots of unity as 1, w, w 2, …, w n –1.

Some Results Involving Complex Cube Root of Unity (w)

(i) w3 = 1, w = w 2 and

1 = w2 w

Sum of the Roots of Unity is Zero

(ii) 1 + w + w2 = 0 (iii) x3 – 1 = (x – 1) (x – w) (x – w2)

We have

(iv) w and w2 are roots of x2 + x + 1 = 0

1- wn 1-w n But w = 1 as w is a nth root of unity. \ 1 + w + … + w n –1 = 0 Also, note that 1 + w + … + wn – 1 =

n -1

nx 1 1 1 + …+ = n n -1 x -1 x -w x -w x -1

(v) a3 – b3 = (a – b) (a – bw) (a – bw2) = (a – b) (aw – bw2) (aw2 – bw) (vi) a2 + b2 + c2 – bc – ca – ab = (a + bw + cw2) (a + bw2 + cw) (vii) a3 + b3 + c3 – 3abc = (a + b + c) (a + bw + cw2) (a + bw2 + cw) (viii) x3+1 = (x + 1) (x + w) (x + w2)

Writing nth Roots of Unity When n is Odd

(ix) a3 + b3 = (a + b) (a + bw) (a + bw2)

If n = 2m + 1, then nth roots of unity are also given by

(x) Cube roots of real number a are a1/3, a1/3 w, a1/3 w2.

Ê 2 kp ˆ Ê 2 kp ˆ z = cos Á + i sin Á Ë n ˜¯ Ë n ˜¯ where k = – m, – (m – 1), …, – 1, 0, 1, 2, …m. Since

Ê 2 kp ˆ Ê 2 kp ˆ cos Á = cos Á Ë n ˜¯ Ë n ˜¯

and

Ê 2 kp ˆ Ê 2 kp ˆ sin Á = - sin Á Ë n ˜¯ Ë n ˜¯

we may take the roots as Ê 2 kp ˆ Ê 2 kp ˆ ± i sin Á 1, cos Á ˜ Ë n ¯ Ë n ˜¯ where k = 1, 2, …, m. In terms, w we may take nth roots of unity to be 1, w+1, w+2, … w ± m. Writing nth Roots of Unity When n is Even

If n = 2m, then nth roots of unity are given z = + 1, + w, + w2,…+ w m –1 Êpˆ Êpˆ Ê 2p ˆ Ê 2p ˆ = cos Á ˜ + i sin Á ˜ + i sin Á where w = cos Á Ë m¯ Ë m¯ Ë 2m ˜¯ Ë 2m ˜¯ Cube Roots of Unity

Cube roots of unity are given by 1, w, w2, where -1 - 3i Ê 2p ˆ Ê 2p ˆ -1 + 3i w = cos Á ˜ + i sin Á ˜ = and w 2 = Ë 3¯ Ë 3¯ 2 2

To obtain cube roots of a real number a, we write x3 = a as y3 = 1 where y = x/a1/3. y3 = 1 are 1, w, w2.

Solution of

x = a1/3, a1/3 w, a1/3 w2.

\ Illustration 4

3 Ê xˆ To obtain cube roots of –27, we write Ë ¯ = 1 -3



x = – 3, – 3w, – 3w2

nth Roots of a Complex Number

Let z π 0 be a complex number. We can write z in the polar form as follows: z = r (cos q + i sin q) where

r = |z| and q = arg (z). Recall – p < q £ p.

The nth root of z has n values one of which is equal to Ê arg z ˆ È Ê arg z ˆ ˘ z0 = n | z | Ícos Á ˜¯ + i sin ÁË ˜ and is called as the Ë n n ¯ ˙˚ Î principal value of we write z as

n

| z | . To obtain other values of

z = r [cos (q + 2kp) + i sin (q + 2kp)] Ê q + 2 kp ˆ Ê q + 2 kp ˆ ˘ È fi z1 / n = r1 / n Ícos Á ˜ + i sin ÁË ˜ Ë n ¯ n ¯ ˙˚ Î = z0 w k where k = 0, 1, 2, …, n –1.

n

|z| ,

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2p 2p + i sin is a complex nth root of unity. n n Thus, all the nth roots of z can be obtained by multiplying the principal value of n | z | by different roots of unity.

Ê 2018 ˆ Ê 2018 ˆ ( ) a b  k k ÁË ˜¯ ÁË Â (ak + bk )˜¯ k =1 k =1

and w = cos

Rational Power of a Complex Number

If z is a complex number and m/n is a rational number such that m and n are relatively prime integers and n z m/n =

(n z )

m

E=

(a) (b) (c) (d)

is is is is

Ê 2018 ÁË Â | zk k =1

ˆ |˜ ¯

2

at least 2018 at least 1009 always less than 2 always greater than 2

Ans. (c) For a, b Œ R, a + ib π 0

Solution:

Thus, z m/n has n distinct values which are given by

( )

(a + b)2 + (a – b)2 = 2(a2 + b2)

mÈ Êm ˆ ˆ˘ Êm z m/n = n z Ícos ÁË n (q + 2kp )˜¯ + i sin ÁË n (q + 2k p )˜¯ ˙ ˚ Î



where k = 0, 1, 2, …, n – 1.

Note that both |a + b| and |a – b| cannot be equal to

2 a 2 + b 2 = 2 | a + ib |

|a + b|, |a – b| £

2 a2 + b2 Now,

SOLVED EXAMPLES

2018

2018

k =1

k =1

k =1

 (ak ± bk ) £  | ak ± bk | £ 2  | zk |

SINGLE CORRECT ANSWER TYPE QUESTIONS

Therefore,

All the solutions of

Example 1 z 2018 +

2018

1 2018

z lie on the curve

+ z 2016 +

1 z

2016

=0

(a) |z| = 1

(b) |z| = 1/2018

(c) |z| = 1/2016

(d) |z| = 1/4034

E
0, and the equation |z – a2| +

Example 15

|z – 2a| = 3 represents an ellipse, then a lies in (a) (1, 3)

(b) ( 2 ,

(c) (0, 3)

(d) (1,

3)

3)

Ans. (c) Solution: The equation |z – a2| + |z – 2a| = 3 will represent an ellipse if [See Theory] |a2 – 2a| < 3 ¤ –3 < a2 – 2a < 3 ¤ –2 < (a – 1)2 < 4 ¤ (a – 1)2 < 4 ¤ –2 < a – 1 < 2 ¤ –1 < a < 3 fi a Œ (0, 3) [∵ a > 0]

fi fi

z 21 + z 22 + 2z1 z2 cos q = 0, then triangle with vertices 0, z1 and z2 is (a) equilateral (b) right angled (c) isosceles (d) none of these. Ans. (c) Solution: z 21 + z 22 + 2z1 z2 cos q = 0

| z|2 £ 2| z | + 4



|z | – 1 £

z-



z1 = – cos q ± i sin q z2



z1 z2



| z1 | = | z 2 |

=

( - cos q )2 + sin 2 q = 1 fi | z1 – 0 | = | z2 – 0|



(| z | – 1)2 £ 5



|z| £

5 +1

5 + 1,

=2 5 + 1 which is

a z + a z = 0 in the

real axis is (a) a z + az = 0

(b)

a z = a z

(c) (a + a ) (z + z ) = 0 (d) none of these Ans. (a) Solution:

Let a = a + ib and z = x + iy, then a z + a z = 0

Ê-aˆ x. becomes a x + by = 0 or y = Á Ë b ˜¯ x-axis is y=

2

2

4 |z|

Example 18

Ê z1 ˆ Ê z1 ˆ ÁË z ˜¯ + 2 ÁË z ˜¯ cos q + 1 = 0 2 2

Êz ˆ fi Á 1 + cos q ˜ = - (1 - cos2 q) = – sin2 q Ëz ¯

4 z

5

5 +1

4 4 4 4 + £ z+ z z z |z|

Therefore, the greatest value of | z| is attained when z = 5 + 1

2



z=

Also, for

If q is real and z1, z2 are connected by

Example 16

|z | £ 2 +

2

or or

a x b

or

a x – by = 0

Ê a + aˆ Ê z + zˆ Ê a - aˆ Ê z - zˆ =0 ˜ÁË ˜Á 2 ¯ Ë 2 ¯ ÁË 2 i ˜¯ ÁË 2 i ˜¯ az + a z = 0

Example 19 If z1 and z2 are two non-zero complex numbers such that |z1 + z2| = |z1| + |z2|, then arg (z1) – arg (z2) is equal to (a) – p (b) – p/2 (c) p/2 (d) 0 Ans. (d)

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Solution:

Let z1 = r1(cos q1 + i sin q1) and

z2 = r2(cos q2 + i sin q2) where r1 = |z1|, r2 = |z2|, q1 = arg (z1) and q2 = arg (z2). We have |z1 + z2|2 = r 12 + r22 + 2r1r2 cos (q1 – q2)

From right DOAC, cos ( fi

p OA 20 –q)= = 2 OC 25

p/2 – q = cos–1 (4/5)

Now, for |z – 25i| £ 15 Y

= (r1 + r2)2 + 2r1r2 {cos (q1 – q2) – 1} Now, |z1 + z2| = |z1| + |z2| ¤

cos (q1 – q2) = 1 ¤ q1 – q2 = 0 ¤ q1 = q2.

Alternatively, ¤ ¤

| z1 + z2 | = |z1 | + |z2 |

C 25i

z1/z2 is a non-negative real number arg(z1) = arg(z2)

25

B

Example 20 If A(z1), B(z2) and C(z3) are the vertices of a triangle such that |z1| = |z2| = |z3| > 0, then orthocentre of DABC is (b) z1 z2 z3 (a) z1 + z2 + z3 (c) 0 (d) 1 Ans. (a) Solution: As |z1| = |z2| = |z3|, circumcentre of DABC is 1 zS = 0. Also, centroid of DABC is zG = (z1 + z2 + z3) 3



1 (zH + 2zS) 3

1 1 (z1 + z2 + z3) = (zH + 2(0)) 3 3



zH = z1 + z2 + z3

q

If |z – 25i| £ 15, then

|maximum arg (z) – minimum arg (z)| equals –1

(a) 2 cos (3/5) (b) 2 cos–1 (4/5) (c) p/2 + cos–1 (3/5) (d) sin–1 (3/5) – cos–1 (3/5) Ans. (b) Solution: If |z – 25i| £ 15, then z lies either in the interior and or on the boundary of the circle with centre at C (25i) and radius equal to 15. From the Fig. 1.43, it is clear that argument is least for point A and argument is greatest for point B.

X

Fig. 1.45

|maximum (arg z) – minimum (arg z)| Êp ˆ = |arg (B) – arg (A)| = –BOA = 2 Á - q ˜ Ë2 ¯ = 2 cos–1 (4/5) Example 22 If |z1| = |z2| = |z3| = 1 and z1 + z2 + z3 = 0, then area of the triangle whose vertices are z1, z2, z3 is (a) 3 3 /4 (b)

3 /4

(c) 1

(d) 2

Ans. (a) Solution:

Example 21

A

O

If zH denotes the orthocentre of DABC, then zG =

15

15



|z1 + z2|2 + |z1 – z2|2 = 2(|z1|2 + |z2|2)

|– z3|2 + |z1 – z2|2 = 2(1 + 1)



|z1 – z2|2 = 3



|z1 – z2| =

|z2 – z3| = |z3 – z1| =

Similarly,

3

3

Thus, D whose vertices are z1, z2 and z3 is equilateral and its side is of length 3 . Therefore, area of triangle = Example 23

3 4

( 3)

2

=

3 3 4

Shaded region is given by

(a) |z + 2| ≥ 6, 0 £ arg (z) £

p 6

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(b) |z + 2| ≥ 6, 0 £ arg (z) £

p 3

(c) |z + 2| £ 6, 0 £ arg (z) £

p 2

Example 25

If k >1, |z1| < k and

(a) |z2| < k (c) z2 = 0 Ans. (d)

(d) none of these.

k - z1 z2 = 1, then z1 - k z2

(b) |z2| = k (d) |z2| = 1

k - z1 z2 =1 z1 - k z2

Solution: ¤

|k – z1 z 2|2 = |z1 – k z2|2

¤

k2 + |z1 z 2|2 – kz1 z 2 – k z 1z2 = |z1 |2 + k2|z2|2 – kz1 z 2 – k z 1z2

Fig. 1.44

Ans. (b) Solution: Note that AB = 6 and 1 + 3 3i = – 2 + 3 + 3 3i

(c)

(d) –

p 2

Solution: Let arg (z) = q, then arg z = – q and arg (– z ) = – p – q. See Fig. 1.45 arg ( z ) – arg (– z ) = – q – (– p – q) = p

|z2 | = 1 z-w k + zw 2

(b) k/2 (d) 2k/3 z -w k2 + z w

k2 k2 z w = w-z = – a a = 2 2 k k zw + k 2 k2 + z w fi

a+ a =0

Example 27

fi Re(a) = 0

Roots of the equation are

(z + 1)5 = (z – 1)5 are Ê 2p ˆ Êpˆ (a) ± i tan Á ˜ , ± i tan Á ˜ Ë 5¯ Ë 5¯ Ê 2p ˆ Êpˆ (b) ± i cot Á ˜ , ± i cot Á ˜ Ë 5¯ Ë 5¯ Ê 2p ˆ Êpˆ (c) ± i cot Á ˜ , ± i tan Á ˜ Ë 5¯ Ë 5¯

Fig. 1.45

or

z z = w w = k2, thus,

But

Ans. (a)

Thus,

| z2|2 – 1 = 0

If k > 0, |z| = |w | = k, and a =

a =

p , then arg z – 2

p 2

we get

Ans. (a)

0 £ arg z £ p /3

If – p < arg (z) < –

(b) –p

| z1 | < k,

Solution:

arg (– z ) is (a) p

(|z1 |2 – k2) (|z2 |2 – 1) = 0

then Re (a) equals (a) 0 (c) k

\ –BAC = p /3 Thus, shaded region is given by

Example 24



Example 26

p pˆ Ê = – 2 + 6 Á cos + i sin ˜ Ë 3 3¯

and

|z1 |2 |z2|2 – |z1 |2 – k2|z2|2 + k2 = 0

As

Ê1 3 ˆ =–2+6 Á + i 2 ˜¯ Ë2

|z + 2| ≥ 6



(d) none of these Ans. (b)

(1)

,

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For z π 1, (1) can be written as

Solution:

=

5

Ê z + 1ˆ 2kp i ÁË z - 1˜¯ = 1 = e , k Œ I

=

z +1 = e2kp i/5, z -1



=

If we denote this value of z by zk, then zk =

e e

2 kp i /5 2 kp i /5

+1 -1

=

e e

kp i/5

+e

kp i/5

- e- kp i/5

- kp i/5

=

Ê kp ˆ = – i cot Á ˜ , k = –2, –1, 1, 2 Ë 5¯ Ê 2p ˆ Êpˆ ± i cot Á ˜ , ± i cot Á ˜ Ë 5¯ Ë 5¯

Example 29

Example 28 Let z1, z2, z3 be three distinct complex numbers lying on a circle with centre at the origin such that z1 + z2z3, z2 + z3z1 and z3 + z1z2 are real numbers, then z1z2z3 equals (a) Р1 (b) 0 (c) 1 (d) i Ans. (c) Solution: As z1, z2, z3 lie on a circle with centre at the origin, |z1| = |z2| = |z3| = r(say). z1 + z2z3 ΠR, z1 + z2z3 = z1 + z2 z3 =

r2 r4 + z1 z2 z3

=

z3 - 1 z3 - r 2

( z2 - 1) - ( z1 - 1) ( z2 - r 2 ) - ( z1 - r 2 ) z2 - z1 =1 z2 - z1 r2 = 1.

If the points z, – iz and 1 are collinear

then z lies on (a) a straight line (c) an ellipse Ans. (b)

(b) a circle (d) a pair of straight lines.

Solution: As z, – iz and 1 are collinear z z 1 z z 1 0 = – iz iz 1 = (– i) z - z i 1 1 1 1 1 1 [taking – i common from R2] z -1 z -1 1 z - i –z - i i = 0 0 0 1



fi (z – 1) ( z + i) + (z – i ) ( z – 1) = 0 fi z z – z + iz – i + z z – i z – z + i = 0 fi 2z z – (1 – i)z – (1 + i) z = 0



z + z2 z3 r2 = 1 z1 z2 z3 z2 z3 + r 2 z1

(1)



zz -

Similarly,

z + z3 z1 r2 = 2 z1 z2 z3 z3 z1 + r 2 z2

(2)



z-

and

z +z z r2 = 3 1 22 z1 z2 z3 z1 z2 + r z3

z z2 z3 z + z3 z1 z +z z r2 = 1 = 2 = 3 1 22 2 z1 z2 z3 z2 z3 r z1 z3 z1 + r z2 z1 z2 + r z3

( z1 - z2 )( z3 - r ) 2

[using C1 Æ C1 – C3 and C2 Æ C2 – C3]

r 2 ( z2 z3 + r 2 z1 ) = z1 z2 z3

From (1), (2) and (3), we get

( z1 - z2 )( z3 - 1)

fi z3 – 1 = z3 – r2 fi Hence, z1 z2 z3 = 1

\ Roots of (1) are

As

z2 z3 + r 2 z1 - ( z3 z1 + r 2 z2 )

z -1 z -1 z -1 r2 = 3 2 = 2 2 = 1 2 z1 z2 z3 z3 - r z2 - r z1 - r

\

where k = –2, –1, 1, 2 [For k = 0, we get 1 = – 1.]

z1 + z2 z3 - ( z2 + z3 z1 )

1- i 1+ i zz =0 2 2

1+ i 2

=

1 2

Thus, z lies on a circle. (3)

Example 30

If | z – i| = 1 and arg (z) = q where

q Π(0, p /2), then cot q Рequals

2 z

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(b) – i

(a) 2i

(c) i

(d) 1 + i

Ans. (c) Solution: |z – i| = 1 represents a circle with centre at i and radius 1. As 0 < q < p /2, q quadrant. See Fig. 1.46. As –OPA = p/2

and

–POA = p /2 – q, we get

(c) 4 2 Ans. (a)

Solution: Let z = a + bw + cw2 Then |z|2 = z z = (a + bw + cw2) (a + bw2 + cw) = a2 + b2 + c2 – bc – ca – ab

OP Êp ˆ cos Á - q ˜ = Ë2 ¯ OA fi

sin q =

=

|z| 2

x = |z| + | z | = 2|z|

2i OP È Êp ˆ Êp ˆ˘ = cos Á - q ˜ + i sin Á - q ˜ ˙ Ë2 ¯ Ë2 ¯˚ z OA ÍÎ



-

1 [sin q + i cos q ] = 1 + i cot q sinq

2 2 = i – cot q fi cot q – =i z z A 2i P (z)

O

q

=

Therefore minimum value of x is Example 33

x3 – a3) |(f (x) – f (a)) for each positive real number a, then f (x) is of the form (b) xp(x3) (a) p(x3) 2 3 (d) a constant (c) x p(x ) where p(x Ans. (a) Solution

Let f (x) = a0 + a1x +a2x2 + a 3x 3 + a 4x 4 + a 5x 5 + a 6x 6 + a 7x 7 + a 8x 8

If |z1| = |z2 | = |z3 | = 1 and z1 + z2 + z3 =

purely imaginary purely real positive real number none of these

+… = p1(x3) + xp2 (x3) + x2p3(x3)

(1)

where p1(x), p2(x) and p3(x) are polynomials with real As

(x3 – a3)| (f (x) – f (a)) f (x) – f (a) = (x3 – a3) p(x)

2

= |z1 + z2 + z3|2

Solution:

2 +i



3 = |z1 |2 + |z 2|2 + |z3|2 + 2 Re(z2 z 3 + z3 z 1 + z1 z 2)



12 = 2 3

Let f (x) be a polynomial with real coef-

2 + i, then the complex number z2 z 3 + z3 z 1 + z1 z 2 is (a) (b) (c) (d) Ans. (a)

2 [(b – c)2 + (c – a)2 + (a – b)2]1/2

As a, b, c are distinct integers, minimum value of (b – c)2 + (c – a)2 + (a – b)2 is 12 + 12 + 22 = 6.

Fig. 1.46

Example 31

1 [(b – c)2 + (c – a)2 + (a – b)2] 2

Thus,

Also, since –POA = p /2 – q

=

(d) 2

Re(z2 z 3 + z3 z 1 + z1 z 2) = 0

Example 32 If a, b, c are distinct integers and w π 1 is a cube root of unity then minimum value of x = |a + bw + cw2 | + |a + bw2 + cw| (a) 2 3

(b) 3

where p(x Now, f (x) – f (a) = (x – a) (x – aw) (x – aw2) p(x) (2) where w π 1 is cube root of unity. From (1) and (2) p1(a3) + ap2(a3) + a2p3(a3) = f (a)

(3)

p1(a ) + awp2(a ) + a w p3(a ) = f (a)

(4)

p1(a3) + aw2p2(a3) + a2wp3(a3) = f (a)

(5)

3

3

2 2

3

Adding above equations, we get 3p1(a3) = 3f (a) fi p1(a3) = f (a)

(6)

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From (3), (4), (6) we get p2(a3) + ap3(a3) = 0

(7)

p2(a3) + awp3(a3) = 0

(8)

Example 35 Let rk > 0 and zk = rk (cos ak + i sin ak ) for k = 1, 2, 3 be such that 1 1 1 =0 + + z1 z2 z3

Solving (7), (8) we get

Let Ak be the point in the complex plane given by wk =

p3(a3) = 0, p2(a3) = 0 " a > 0 Thus, p2(x3) = p3(x3) = 0 " x ΠR Hence, f (x) = p(x3) for some polynomial p(x) with real

cos 2a k + i sin 2a k for k = 1, 2, 3. The origin, O is the zk (a) (b) (c) (d)

Example 34 Suppose m, n, Œ R – {0} and m π n. If incentre of triangle with vertices A(m, 0), B(0, n) and C(a, – a) is the origin, then a is equal to (a)

mn n-m

(b)

incentre of D A1 A2A3 orthocentre of DA1 A2 A3 circumcentre of D A1A2 A3 centroid of D A1 A2 A3

Ans. (d)

mn m-n

Solution

wk =

mn (d) |m-n|

(c) 0

zk2 1 = zk zk zk zk

w1 + w2 + w3 =

1 1 1 + + z1 z2 z3

A(m), B(ni), C(a – ai) a = BC = |a – ai – ni| =

a 2 + (a + n) 2

b = CA = |a – ai – m| =

(a - m) + a 2

\ Thus, origin O is the centroid of D A1A2A3.

m2 + n2

Example 36 If a and b are two real numbers and z1 and z2 are two non zero complex numbers such that a|z1 |=

az1 + bz2 + cz3 = 0 fi

b| z2 |, then z =

a + (a + n) (m) + (a - m) + a (ni ) 2

2

2



m a 2 + (a + n) 2 + a m 2 + n 2 = 0

(1)

Ans. (a)

and

n (a - m) 2 + a 2 - a m 2 + n 2 = 0

(2)

Solution:

m a 2 + (a + n) 2 + n (a - m) 2 + a 2 = 0 mn < 0 and m2[a2 + (a + n)2] = n2[(a – n)2 + a2] = 0 2(m2 – n2) a2 + 2mna (m + n) = 0



2a (m + n) [a(m – n) + mn] = 0



a=

mn n-m

Let w =

(b) z is purely imaginary (d) none of these

az1 a | z1 | , then |w| = =1 bz2 b | z2 |



From (1), (2)



az1 bz2 + bz2 az1

(a) z is purely real (c) |z| = a/b

+ m 2 + n 2 (a – ai) = 0



rk2 zk

Ê1 1 1ˆ = Á + + ˜ =0 Ë z1 z2 z3 ¯

2

As incetre of DABC is at the origin

2

zk2

We have

Use complex coordinates:

c = AB = |ni – m| =

=

zk

=

Ans. (a) Solution:

(cos a k + i sin a k )2

(∵ a π 0)

w = eiq for some q, 0 £ q < 2p 1 Now, z = w + = e iq + e–iq = 2 cos q w fi z is purely real.

Example 37 z1 and z2 lie on a circle with centre at the origin. The point of intersection z3 of the tangents at z1 and z2 is given by (a)

1 ( z1 + z2 ) 2

(b)

2z1 z2 z1 + z2

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(c)

1Ê1 1ˆ + ˜ Á 2 Ë z1 z 2 ¯

(d)

È1 - z 30 ˘ = Im Í ˙ Î z-z ˚

z1 + z2 z1 z2

Ans. (b)

È -1 ˘ = Im Í {1 - cos 30 q - i sin 30 q }˙ Î 2i sin q ˚

Solution: Refer Fig. 1.47. As DOAC is a right triangle with right angle at A, | z1 | 2 + |z3 – z1| 2 = |z3 |2 fi 2|z1 |2 – z 3 z1 – z 1z3 = 0 fi

2 z1 - z3 -

z1 z3 = 0 z1

(1)

Similarly,

2 z2 - z3 -

z2 z3 = 0 z2

(2)

Example 39 Let z = x + iy be a complex numbers where x and y are integers. Then area of the rectangle whose vertices are roots of z z 3 + z z 3 = 738

Subtracting (2) from (1) we get 2 ( z2 - z1 ) fi

2r 2 ( z1 - z2 ) z1 z2

is

Êz z ˆ = - z3 Á 1 - 2 ˜ Ë z1 z2 ¯

(a) 80 Ans. (a)

Ê z2 - z2 ˆ = - z3 r 2 Á 2 2 21 ˜ Ë z1 z2 ¯



z3 =

2z1 z2 z2 + z1

O

C (z3) B (z 2 )

(d) 96



(x2 + y2) [(x – iy)2 + (x + iy)2] = 738



(x2 + y2) (2x2 – 2y2) = 738



x4 – y4 = 369 fi x4 ≥ 369 fi |x| > 4



r

(c) 56

z z ( z 2 + z 2 ) = 738

x =± 5

Let us try

A (z1)

(b) 48

z z 3 + z z 3 = 738

Solution: [\ |z1 |2 = |z2|2 = r 2]



1 1 = 2sin q 2sin 3∞

=

625 – 369 = y4 fi y = ± 4

Thus, roots of z z 3 + z z 3 = 738 are ± 5 ± 4i. Area of the rectangle ABCD whose vertices are ± 5 ± 4i is (10) (8) = 80 y – 5 + 4i B

Fig. 1.47

Example 38

Let z = cos q + i sin q. Then value of 15

 Im ( z

2 m -1

)

C – 5 – 4i

at q = 3° is

D 5 – 4i y¢

(a) – 1

(b) 0

1 2sin 3∞

(d)

-1 2sin 3∞

Ans. (c) 15

Solution

x



m =1

(c)

5 + 4i A

Â

m =1

Ê 15 ˆ Im (z2m – 1) = Im Á Â z 2 m - 1 ˜ Ëm =1 ¯ È z (1 - ( z 2 )15 ) ˘ = Im Í ˙ 2 Î 1- z ˚

Fig. 1.48

Example 40 A man walks a distance of 3 units from the origin towards the north-east (N 45° E) direction. From there he walks a distance of 4 units towards the north-west (N 45° W) direction to reach a point B. Then position of B in the argand plane is (b) (3 – 4i)eip/4 (a) 3eip/4 + 4i (c) (4 + 3i)eip/4 Ans. (d)

(d) (3 + 4i)eip/4

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Solution: Suppose the man reaches A after walking 3 units in N 45° E and B after walking a distance of 4 units in N 45°W. Position of A in the Argand diagram is 3eip/4. Let position of B be z. Since –OAB = p/2, we have

2 D C z1 3 5 A z (1 + 2i) B 0

E z2

Ê 0 - 3eip /4 ˆ p arg Á = ip /4 ˜ 2 Ë z - 3e ¯ fi fi fi fi

- 3eip /4 z - 3eip /4 - 3eip /4 z - 3eip /4

0

p p OA È = cos + i sin ˙˘ Í OB Î 2 2˚

Fig. 1.50

A, B Example 42 If P(w where A(a), B(b) lie on the unit circle |z| = 1, then w is equal to

3i = 4

z – 3eip/4 = 4ieip/4 z = (3 + 4i)eip/4 N

B

x

(a)

a + b -w a+b

(b)

a + b -w ab

(c)

w 1 ( a + b) 2 ab

(d)

1 (a + b + w) ab

z

Ans. (b) Solution:

p/2

4) ip/

A

W

Suppose P(w) A, B in the ratio k : 1, then

(3e

3 p/4

O

w= E

S Fig. 1.49

Example 41

k=

w -a b -w

Now,

w=

a + kb k +1

1 w - a Ê 1ˆ + Á ˜ a b - w Ë b¯ = w -a +1 b -w

where i = - 1 origin by 5 units and then vertically away from the origin by 3 units to reach a point z1. From z1 the particle moves 2 units in the direction of ˆi + ˆj and then it moves through an angle p/2 in the anticlockwise direction on a circle with centre at origin, to reach point z2. The point z2 is given by (a) 6 + 7i (b) – 7 + 6i (c) 7 + 6i (d) – 6 + 7i

=

b 2 - bw + aw - a 2 ab(b - a)

=

(b - a)(b + a - w ) ab(b - a )

=

a +b -w ab

Ans. (d) Solution B is (1 + 2i) + 5 = 6 + 2i C is (6 + 2i) + 3i = 6 + 5i D is (6 + 5i) +

1+ iˆ 2 Ê = 7 + 6i Ë 2¯

E is (7 + 6i)epi/2 = (7 + 6i)i = – 6 + 7i

a + kb k +1



A particle P starts from the point z0 = 1 + 2i,

Example 43

and is

k

A(a)

1 P(w)

B(b)

O

Fig. 1.51

[∵ aa = 1, bb = 1]

Number of solution of

|z – 7 – i| = 3 2

(1)

|z – 1 – 7i| £ 3 2

(2)

Ans. (b)

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Solution: |z – 7 – i| = 3 2 represents a circle C1 with centre at A(7 + i) and radius r1 = 3 2 Also, |7 – 1 – 7i| = 3 2 represents a circle C2 with centre at B(1 + 7i) and radius r2 = 3 2 . AB = |(7 + i) – (1 + 7i)|

As

1 =1 +



b2 + c2 (sin 2a) = 0 fi b2 + c2 = 0 2bc



= |6 – 6i| = 6 2 = r1 + r2

b = ± ic

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

C1 and C2 touch each other externally at exactly one point viz. mid-point (4 + 4i) of AB. Thus, (1) and (2) have exactly one solution.

b2 + c2 (sin 2a) 2bc

Example 46

Let x1, x2, x3, x4 ΠC be such that

x1(x1 – x2) (x1 – x3) (x1 – x4) = 4 A

x2(x2 – x1) (x2 – x3) (x2 – x4) = 4

B

x3(x3 – x1) (x3 – x2) (x3 – x4) = 4 x4(x4 – x1) (x4 – x2) (x4 – x3) = 4

Fig. 1.52

Example 44

then

È 7 + 5i 3 + 3i ˘ Let Z = Í ˙ Î 8 + 5i 5 + 2i ˚

then det ( ZZ ) , (where Z is the matrix obtained by taking Z) is equal to (a) 256 Ans. (a) Solution: and

(b) 121

(c) 36

(d) 25

(a) (b) (c) (d)

Ans. (a) (b) (c) Solution: Note that none of xi is zero and no two of xi’s are equal.

Let z1 = 7 + 5i, z2 = 3 + 3i, z3 = 3 + 8i

Let

z4 = 8 + 5i, then È z1 ZZ = Í Î z3

z2 ˘ È z1 z4 ˙˚ ÍÎ z3

= |z1z4 – z2z3|

f (t) = (t – x1) (t – x2) (t – x3) (t – x4) f ¢(t) = (t – x2) (t – x3) (t – x4)

z2 ˘ z4 ˙˚

+ (t – x1) (t – x3) (t – x4) + (t – x1) (t – x2) (t – x4)

fi det( ZZ ) = det ( Z ) det ( Z )

+ (t – x1) (t – x2) (t – x3)

2

2

= 16 = 256 Example 45 Let a, b, c be three points lying on the circle |z| = 1 and suppose a Œ (0, p /2) be such that a + b cos a + c sin a = 0, then (a) b = ± ic (c) a2 + b2 + c2 = 0

(b) 2a2 + b2 + c2 = 0 (d) b = c

Ans. (a) Solution:

x1 + x2 + x3 + x4 = 0 x 1x 2x 3x 4 = – 1 x21 + x22 + x23 + x24 = 0 x 1x 2x 3x 4 = 0

1 = |– a|2 = |b cos a + c sin a|2

= |b|2 cos2 a + |c|2 sin2 a + (bc + bc ) cos a sin a b c = 1 + ÊÁ + ˆ˜ cos a sin a Ë c b¯

4 for k = 1, 2, 3, 4 xk

\

f ¢(xk) =



x1, x2, x3, x4 are zeros of tf ¢(t) – 4

Also, tf ¢(t) – 4 is a polynomial of degree 4 with leading \ fi

4f (t) = tf ¢(t) – 4 4 f ¢(t ) = f (t ) + 1 t



f (t) + 1 = at4 for some constant a



f (t) = at4 – 1 t4, we get a = 1

\

f (t) = t4 – 1

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fi x1, x2, x3, x4 are 4th roots of unity viz 1, – 1, i, – i Thus, x1 + x2 + x3 + x4 = 0 fi

x21 + x22 + x23 + x24 = 0

and

x 1x 2x 3x 4 = – 1

Example 48 Suppose r, s Œ R, r π s and rs π 0. If incentre of triangle with vertices A(r, 0), B(0, s) and C(a, – a) is (0, 0), then (a) a =

Example 47 Suppose a ΠR, b < 0, and four solutions of the equation z4 + 2az2 + b = 0 form four vertices of a quadrilateral Q in the complex plane. If area (Q) < 1, then

(c) ar < 0

We have a = BC = |a – ai – si| = b = CA = |a – ai – r| = c = BC = |r – si| =

Ans. (b) Solution:

As b < 0, D = 4a – 4b = 4(a – b) > 0.

\

-2a ± 2 a - b =–a± 2

2

2

2

0= a2 - b fi

(a - r )2 + a 2

r 2 + s2

az1 + bz2 + cz3 a+b+c

(r ) a 2 + (a + s ) 2 + ( si ) (a - r )2 + a 2 + (a - ai ) r 2 + s 2 = 0

2

we get z + 2az + b = 0 has two real and two purely imaginary solutions. z1 = A, z2 = – A, z3 = Bi, z4 = – Bi where

a 2 + (a + s )2

As incentre of DABC is 0,

a 2 - b > |a| > a, 4

Let

(d) as > 0

Ans. (a), (b), (c), (d)

(c) Р1 < b < 0, a ΠR 1 (d) - < b < 0, a > 0 2

As

(b) rs < 0

Solution: Vertices of triangle ABC are A(r), B(si) and C(a – ai). Note that a π 0.

(a) Р1 < b < 0, a > 0 1 (b) - < b < 0 , a ΠR 2

z2 =

rs s-r

a 2 - b – a, B =

A=

a2 - b + a

y Bi



r a 2 + (a + s )2 + a r 2 + s 2 = 0

and

s (a - r )2 + a 2 - a r 2 + s 2 = 0



r a 2 + (a + s ) 2 + s (a - r )2 + a 2 = 0



ar < 0, sa > 0, rs < 0

and

r2[a2 + (a + s)2] = s2 [(a – r)2 + a2]

We have Q -A

A -Bi

Area (Q) =

r2{2a2 + 2as + s2} = s2{2a2 – 2ar + r2}

x

1 (2A) (2B) 2



(r2 – s2) (2a2) = – 2a(r + s)rs



a=

Example 49

2

= 2(a – b – a ) = – 2b 1 As area (Q) < 1, we get – 2b < 1 or b > - . 2 1 Thus, - < b < 0 and a is any real number. 2

Suppose x, y ΠR and

Ê 1 ˆ 5 x Á1 + 2 ˜ = 12 Ë x + y2 ¯

= 2AB 2

rs - rs = r-s s-r

Ê 1 ˆ 5 y Á1 - 2 ˜ = 4, Ë x + y2 ¯ then

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(a) x = 2, y = 1 (c) x =

1 2 ,y= 5 5

(b) x = – 2, y = – 1

That is, z1/z2 lies on Re(z) = -

2 1 (d) x = - , y = 5 5

Let

Ans. (a), (c)

Solution:

and

Let z = x + iy, then

Ê 1 ˆ 12 Re(z) Á1 + 2 ˜ = Ë |z| ¯ 5

(1)

Ê 1 ˆ 4 Im(z) Á1 - 2 ˜ = Ë |z| ¯ 5

(2)

z1 1 = - + ai , then z2 2 -

TIP As x > 0, answers (b) and (d) can be ruled out.

a2 = 3/4 fi a = ± 3 / 2

\

z1 1 = ( -1 ± 3i ) = w, w2 z2 2

(c)

fi fi

5z2 – (12 + 4i)z + 5 = 0



z=

\

1

(1 + i )

2

Solution Put w = cos

(d) S – S = 0

2p 2p + i sin , 11 11

so that for 1 £ k £ 10 2p k 2p k - i cos 11 11

=

1 [6 + 2i ± 7 + 24i ] 5

=

1 [6 + 2i ± (4 + 3i)] 5

2p k 2p k ˆ Ê + i sin = - i Á cos ˜ Ë 11 11 ¯

=

1 1 (10 + 5i), (2 – i) 5 5

= – iw k

sin

10

S = - i Âw k = k =1

If z, z2 are non-zero complex numbers

Solution:

(b) w

But \ fi

then z1/z2 can be (a) 1 Ans. (b), (c)

(c) w2

(d) – 1

z z We have 1 = 1 = 1 + 1 z2 z2

z1 lie on the perpendicular bisector of the segment z2 A(– 1 + 0i) and 0(0 + 0i)

[De Moivre’s theorem]

thus,

1 2 x = 2, y = 1 or x = , y = - . 5 5

|z1| = |z2| = |z1 + z2|



S =±

(b) S S = 1

Ans. (a), (b), (c)

12 + 4i ± (12 + 4i ) 2 - 100 10

Example 50 such that

-1/2

10 2p k ˆ Ê 2p k - i cos If S = Â Á sin ˜ then 11 11 ¯ k =1 Ë

(a) S + S = 0

z 1 = (12 + 4i ) zz 5

O

A



Example 51

1 1 z + = (12 + 4i ) z 5

-1

1 1 + ai = 1 fi + a 2 = 1 2 4

Multiply (2) by i and add to (1), then z+

1 2

i (w - w 11 ) i w (1 - w 10 ) = 1-w 1- w

w11 = cos 2p + i sin 2p = 1 + i0 = 1 S= i S + S = 0, S S = 1 S =±

and

1 2

(1 + i)

If |z1| = |z2| = |z3| = 1

Example 52

2

and

S = |z2 – z3| + |z3 – z1| + |z1 – z2|2 Let and then

2

a = minimum value of S b = maximum value of S (a) a = 0

(b) b = 9

(c) b = 3

(d) a = 2

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Ans. (a), (b) Solution: Thus, Since

|z1 – z3 |2 = |z1 |2 + |z3|2 – z1 z 3 – z 1z3

Also,

Note that S ≥ 0 and S = 0 for z1 = z2 = z3 a= 0

= 2 – z1 (– z 1) – z 2 (–z2) = 4 fi

|z1 – z3| = 2

|a – b|2 = |a|2 + |b|2 – ab - ab , we get

Similarly

|z2 – z4 | = 2

|z2 – z3|2 + | z3 – z1|2 + |z1 – z2|2

Thus, z1, z2, z3 z4 are vertices of a rectangle.

= 6 – (z2 z 3 + z 2z3 + z3 z 1 + z 3z1 + z1 z 2 + z 1z2)

Example 54 If z1 and z2 are two non-zero complex numbers such that |z1 + z2 |2 = |z1|2 + |z2 |2, then (a) z1 z 2 is purely imaginary (b) z1/z2 is purely imaginary (c) z1 z 2 + z 1z2 = 0 (d) O, z1, z2 are the vertices of a right triangle.

= 6 – z 1(z2 + z3) – z 2 (z3 + z1) – z 3 (z1 + z2)

(1)

0 £ |z1 + z2 + z3|2

Also

= |z1|2 + |z2|2 + |z3|2 + z 1 (z2 + z3) + z 2 (z3 + z1) + z 3 (z1 + z2) fi

Ans. (a), (b), (c), (d)

z 1 (z2 + z3) + z 2 (z3 + z1) + z3 (z1 + z2) ≥ –3

(2)

Thus, from (1) and (2), we get |z2 – z3|2 + |z3 – z1|2 + |z1 – z2|2 £ 9 The value 9 is attained when z2 = w z1 = 1

z3 = w

2

where w π 1 is a cube root of unity. Example 53 Suppose z1 + z2 + z3 + z4 = 0 and |z1| = |z2| = |z3| = |z4| = 1. If z1, z2, z3 z4 are the vertices of a quadrilateral, then the quadrilateral can be a (a) parallelogram (b) rhombus (c) rectangle (d) square Ans. (a), (c) Solution:

z1 + z2 + z3 + z4 = 0



z1 + z2 + z3 + z4 = 0



1 1 1 1 =0 + + + z1 z2 z3 z4



(1)



|z1|2 + z2 z 1 + z1 z 2 + |z2 |2 = |z1|2 + |z2 |2



z2 z 1 + z1 z 2 = 0 fi z1 z2 = z 1z2 = – z1 z 2



z1 z 2 is purely imaginary.



z1 |z2 |2 is purely imaginary. z2

Since |z2|2 is real, we get z1/z2 is purely imaginary As z1/z2 is purely imaginary, arg(z1/z2) = p /2 Thus, O, z1, z2 forms a right triangle.

(2 – w) (2 – w 2) ... (2 – w n–1) (a) 2n – 1 (b) nC1 + nC2 + ... + nCn–1 + nCn (c) [2n+1C0 + 2n+1C1 + ... + 2n+1Cn]1/2 – 1

(2)

|z1 z2 z3z4| = 1

(d)

The equation whose roots are z1, z2, z3 and z4 is (z – z1) (z – z2) (z – z3) (z – z4) = 0 4

2

z + bz + d = 0

If 1, w, w2, ..., w n–1 are the nth roots of

Example 55 unity, then equals

z 2 z 3z 4 + z 1z 3z 4 + z 1z 2z 4 + z 1 z 2 z 3 = 0

Also,

or

Solution: |z1 + z2 |2 = |z1|2 + |z2 |2 fi (z1 + z2) ( z 1 + z 2) = |z1 |2 + |z2 |2

(3)

where

b = (z1 + z2) (z3 + z4) + z1z2 + z3z4

and

d = z 1 z 2z 3z 4

fi z1, z2, z3, z4 are vertices of a parallelogram. (Similarly for other pairs)

Cn

Ans. (a), (b), (c) Solution: We have zn–1 = (z – 1) (z – w) (z – w2) ... (z – w n–1)

[use (1) and (2)]

Clearly, if a is a root of (4), then – a is a root of (4). Thus, root of (3) are of the form ± a, ± b. If z1 + z3 = 0 then z2 + z4 = 0

2n

(z – w) (z – w2) ... (z – w n–1) =

fi Putting fi

zn - 1 z -1

z = 2, we get

(2 – w) (2 – w2) ... (2 – w n–1) =

zn - 1 = 2n – 1 z -1

IIT JEE eBooks: www.crackjee.xyz Complex Numbers 1.31

( 1 1 (d) mix = ( 2 |z| 1 1 = z ŒSa | z | 2

Example 56 If a, b, c are real numbers and z is a complex number such that a2 + b2 + c2 = 1 and b + ic = 1+ i z (1 + a)z, then equals 1- i z (a)

b

ic ia

(b)

a + ib 1+ c

a2 = z +

Solution:

Ans. (b), (c) Solution: We have z =

b + ic 1+ a



z =

b +c

and

1- a

b - ic 1+ a



2

2bi 1 - a 1+ a 1+ a 1 + a + 2bi - 1 + a = = 2c 1 - a 1 + a + 2c + 1 - a 1+ + 1+ a 1+ a 1+

a + ib 2 ( a + i b) = = 2 (1 + c) 1+ c

(1)

(2)

From (1) and (2) we get (b) and (c) are true. Let a be a positive real number and let

(a 1 (b) mix |z| = ( a 2 z ŒSa

1 2

2

2

z ŒSa

) + 4 + a) +4-a

z2 + z 2 | z |2

(1)

1 z

= a},

=

1È (2 + a 2 ) ± a 4 + 4a 2 ˘ ˚ 2Î

=

1 [a2 + (a2 + 4) ± 2a 4

=

1 4

|z| Œ

or

1- c 1- c = a - ib (1 + c) ( a - i b)

(a) min |z| =

| z|

+

(2 + a 2 ) ± (2 + a 2 ) 2 - 4 2

(

a2 + 4 ± a

( È1 ÍÎ 2 (

È1 | z|2 Œ Í Î4

2

then

2

| z|

z z + z z

x=

From (1) we get

a2 + b2 a + ib a - ib ¥ = = 1+ c a - ib (1 + c) ( a - i b)

Sa = {z Œ C : z π 0 and z +

1

+

2

Roots of x2 – (2 + a2)x + 1 = 0 are

1+ i z 1+ i z 1+ i z 1 + i (z + z ) - z z ¥ = = 1- i z 1 - i (z - z ) + z z 1- i z 1+ i z

Example 57

1

= |z|2 +

|z|4 – (2 + a2) |z|2 + 1 £ 0

We have

=

2

Ê 1 ˆ fi |z|2 Á a 2 - | z |2 - 2 ˜ = (z + z )2 – 2|z|2 ≥ – 2 |z|2 Ë |z| ¯

1- a zz = = = 2 2 1+ a (1 + a) (1 + a) 2

1 z

= |z|2 +

2b 2i c fi z+ z = ,z– z = , 1+ a 1+ a 2

a2

z ŒSa

Ans. (a), (b), (c), (d)

1+ a (d) b + ic

1- c (c) a - ib

) + 4 + a)

a2 + 4 - a

(c) min

)

a2 + 4 ]

2

) 14 ( a + 4 + a) ˙˚˘ 1 + 4 - a ) , ( a + 4 + a )˙˘ 2 ˚ 2

a2 + 4 - a ,

2

a2

2

2

Note that the minimum value of |z| is obtained when 1 z= a 2 + 4 - a = z1 (say) and the maximum value of 2 1 |z| is obtained when z = a 2 + 4 + a = z2 (say). 2

(

)

(

Next, note that z Œ Sa ¤ Since, we get

1 2

(

)

)

1 ΠS a. z

a 2 + 4 - a £ |z| £

1 , 2

(

)

a 2 + 4 + a " z ΠSa

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2

£

a2 + 4 + a or

1 2

(

)

a2 + 4 - a £

1 £ |z|

2 a2 + 4 - a

1 1 £ |z| 2

(

(c) a lies on or outside the unit circle |z| = 1 (d) a lies inside the unit circle | z| = 1. Ans. (a), (c).

" z ΠSa

)

a 2 + 4 + a " z ΠSa

These extreme values are obtained when z = z2 and z = z1. Example 58 If the roots of (z – 1)n = 2w (z + 1)n (where n ≥ 3 and w is a complex cube root of unity) are plotted in the argand plane, then these roots lie on (a) straight line (b) circle (c) ellipse (d) rectangular hyperbola. Ans. (b) Solution:

Solution: Suppose | a| < 1. We have | – 3| = |a5 + a3 + a | £ |a|5+ |a|3 + |a| < 1 + 1 + 1 fi 3 < 3. A contradiction fi

Example 60 The equation |z – z1 |2 + |z – z2 |2 = k (where k is a real number) will represent a circle if. (a) k = |z1 – z2 | (b) k £ |z1 – z2|

n

z -1 z +1



= 2w

(c) k ≥



= |2w| = 2

Solution:

sin q1 z3 + sin q 2 z2 + sin q3 z + sin q 4 = 3

(1)

(b) |z | > 2/3 (d) |z| > 1/2

Solution: If z z π 0 for otherwise, sin q4 = 3. Also 3 = |3| = |sin q1 z3 + sin q2 z2 + sin q3 z + sin q4 | £ |sin q1 | | z|3 + |sin q 2 | | z|2 + |sin q3 | | z| + |sin q4| [∵ |sin q | £ 1]

£ |z|3 + | z|2 + | z| + 1

1 £ 1 + | z| + | z | +| z| + | z| + ... £ 1 - | z| 3

| z| > 2/3 and hence | z| > 1/2 If a is a root of z5 + z3 + z + 3 = 0, then

(a) | a| ≥ 1 (b) | a| < 1

1

+ z 2)] = k – (|z1 |2 + |z2 |2)

È Ê z + z2 ˆ ˘ 1 2 |z|2 – Re Í z Á 1 ˙ + |z1 + z2| ˜ 4 ÍÎ Ë 2 ¯ ˙˚



=

k 1 + [|z1 + z2 |2 – 2 |z1 |2 – 2|z2 |2] 2 4

{

2

}



z +z 1È 1 2 2 ˘ z - 1 2 = Ík z1 + z2 - 2 Re ( z1 z2 ) ˙ 2 2Î 2 ˚



z + z2 z- 1 2

2

=

1 2

1 2˘ È ÍÎk - 2 z1 - z2 ˙˚

This will represent a circle with centre at 12 (z1 + z2) and 1 2 radius 2k - z1 - z2 provided k ≥ 12 |z1 – z2 |2. 2 Example 62 a, b Œ R,

4

fi 1 – | z| < 1/3 fi | z| > 2/3

Example 60

|z – z1 |2 + |z – z2 |2 = k implies

fi 2|z|2 – 2Re[z( z

Example 59 If q1, q2, q3 and q4 are four real numbers, then any root of the equation

2

(d) k = |z1 + z2|

|z|2 + |z1|2 – 2Re(z z 1) + |z|2 + |z2|2 – 2Re(z z 2) = k

As 21/n > 1, we get z lies on a circle. [See section 1.11 in Theory]

lying inside the unit circle | z (a) |z| < 2/3 (c) |z| < 1/2 Ans. (b), (d)

1 |z1 – z2|2 2

Ans. (c)

n

z -1 = 21/n z +1

Thus,

a lies either on the or outside the unit circle | z| = 1.

If z is a root of (z – 1)n = 2w (z + 1)n, then

Ê z - 1ˆ ÁË z + 1˜¯

| a| ≥ 1.

Thus,

(a) (b) (c) (d) Ans. (a)

For any two complex numbers z1, z2 and |az1 + bz2|2 + |bz1 – az2|2 =

(a2 + b2) (| z1|2 + | z2 |2) (a2 – b2) (| z1|2 – | z2 |2) a2 + b2 2(a2 + b2)

Solution: We have | az1 + bz2 |2 + | bz1 – az2|2

IIT JEE eBooks: www.crackjee.xyz Complex Numbers 1.33

(

= |az1|2 + |bz2 |2 + 2R az1 bz2 + bz1 2

2

2

+ az2

2

2

(

)

- 2 R bz1 az2

Ans. (b), (d). Solution: We have

)

A + B cos 4x =

2

= a |z1 | + b |z2 | + 2abR(z1 z 2) + b2|z1 |2 + a2|z2|2 – 2abR(z1 z 2)

=

[∵ a and b are real numbers] = (a2 + b2) (|z1|2 + |z2 |2) Example 63 If w π 1 is a cube root of unity and w and w 2 satisfy the equation 1 1 1 1 2 = + + + a+x b+x c+x d+x x then value of

(1)

1 1 1 1 is + + + a +1 b +1 c +1 d +1

(a) 1 (c) 3 Solution: We can write (1) as

S x(b + x) (c + x) (d + x) = 2(a + x) (b + x) (c + x) (d + x) fi S x{x3 + (b + c + d)x2 + (bc + cd + bd)x + bcd} = 2{x4 + (a + b + c + d)x3 + (ab + ac + ad + bc + bd + cd)x2 + (abc + abd + acd + bcd)x + abcd} 4 fi 2x + (a + b + c + d)x3 + 0x2 – (abc + abd + acd + bcd)x – 2abcd = 0 This is a fourth degree equation whose two roots are w and w2. Let a, b be the two other roots, then (a + b) (w + w ) + ab + w ◊ w = 0 fi fi fi

=

1 [6(e4ix + e–4ix) + 20] 32

=

1 [(6) (2 cos 4x) + 20] 32

=

3 5 cos 4x + 8 8

2

[sum of the products taken two at a time] (a + b ) (–1) + ab + 1 = 0 (1 – a ) (1 – b) = 0 a = 1 or b = 1

B = 3/8

Let z1, z2, z3 be complex numbers such z2 z12 z2 + 2 + 3 = –1, then z2 z3 z3 z1 z1 z2

value of |z1 + z2 + z3| can be (a) 1 (c) 3

(b) 2 (d) 4

Ans. (a), (b) Solution: We have z13 + z23 + z33 = –z1z2z3. fi

– 4z1z2z3 = z13 + z23 + z33 – 3z1z2z3 = (z1 + z2 + z3) (z12 + z22 + z32 – z2z3 – z3z1 – z1z2) = (z1 + z2 + z3) [(z1 + z2 + z3)2 – 3(z2z3 + z3z1 + z1z2)]



z3 – 3z(z2z3 + z3z1 + z1z2) + 4z1z2z3 = 0

where fi

z = z1 + z2 + z3 È Ê1 1 1ˆ ˘ z3 = z1z2z3 Í3z Á + + ˜ - 4˙ Î Ë z1 z2 z3 ¯ ˚ = z1z2z3 [3z ( z1 + z2 + z3 ) – 4 ]

\ Thus,

and

that |z1| = |z2| = |z3| = 1 and

Ans. (b)

2

1 6 [ C1 (eix)5 (e–ix) + 6C3(eix)3 (e–ix)3 32 + 6C5 (eix) (e– ix)5]

A = 5/8

Thus

Example 65

(b) 2 (d) 0

1 [(eix + e– ix)6 – (eix – e – ix)6] 64

[∵ 1 1 1 1 = 2. + + + a +1 b +1 c +1 d +1

Example 64

|z1| = |z2| = |z3| = 1]

= z1z2z3 [3zz – 4] fi

If sin6 x + cos6 x = A + B cos 4x, then

(a) A =

5 3

(b) B =

1 8

(c) A =

5 8

(d) B =

3 8



3

|z| = |z1| |z2| |z3| |3|z|2 – 4| |z|3 – |3|z|2 – 4| = 0

If |z| ≥ 2/ 3 , we get |z|3 – 3|z|2 + 4 = 0 fi

(|z| – 2) (|z|2 – |z| – 2) = 0



(|z| – 2)2 (|z| + 1) = 0

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|z| – 2 = 0 or |z| = 2

Example 68 If w π 1 is a cube root of unity, then sum of the series S = 1 + 2w + 3w2 + ... + 3n w3n–1 (n Œ N) is

If |z| < 2/ 3 , we get |z|3 + 3|z|2 – 4 = 0 fi

(|z| – 1) (|z|2 + 4|z| + 4) = 0



|z| – 1 = 0 fi |z| = 1.

(a)

cos g = 0, then (a) cos 2a + cos 2b + cos 2g = 0 (b) sin 2a + sin 2b + sin 2g = 0 (c) cos (b +g ) + cos (g + a) + cos (a + b ) = 0 (d) sin (b + g ) + sin (g + a) + sin (a + b ) = 0 Let a = cos a + i sin a, b = cos b + i sin b and c = cos g + i sin g,

then

a+b+c =0

and

1 1 1 + + =0 a b c



bc + ca + ab = 0

Solution:

Put

=

d d È x - x 3n + 1 ˘ ÈÎ x + x 2 + º + x3n ˘˚ = Í ˙ dx dx Î 1 - x ˚

=

(1 - x) (1 - (3n + 1) x3n ) + ( x - x3n + 1 ) (1 - x)2

x = w, we get

(1)

(1 - w ) (1 - 3n - 1) + (w - w ) - 3n = 2 (1 - w ) (1 - w )

=

(

=

(w – 1) (w

i [sin (b + g) + sin (g + a) + sin (a + b)] = 0 From this (c) and (d) follow. Similarly, (a), (b) follow from (3). If sin a + sin b + sin g = cos a + cos b +

cos g = 0 then (a) cos 4a + cos 4b + cos 4g = 0 (b) sin 4a + sin 4b + sin 4g = 0 (c) cos 3a + cos 3b + cos 3g = 3 cos (a + b + g)

(a), (b) follows from (a) and (b) of Example 66.

= (a + b + c) (a2 + b2 + c2 – bc – ca – ab) =0 From this identity both (c) and (d) follow.

(

)

Ans. (a), (b), (c), (d) Solution:

Centroid of DABC is zG =

1 (z1 + z2 + z3), 3

circumcentre of DABC is zS = 0. As zG zS and zH (orthocentre) in the ratio 1 : 2, we get zG =

Ans. (a), (b), (c), (d)

a3 + b3 + c3 – 3abc

)

–1

= n w2 – 1

Example 69 Suppose A(z1), B(z2) and C(z3) are vertices of a triangle lying on the unit circle |z| = 1. AD is altitude of the DABC meeting the unit circle in E. (a) orthocentre of DABC is z1 + z2 + z3 E is –z2z3/z1 2 (c) if z1 = z2z3 and z22 = z3z1, then DABC is equilateral. (d) if z2 + z3 = 0, then DABC is a right angled.

(d) sin 3a + sin 3b + sin 3g = 3 sin (a + b + g)

Using notation of Example 66,

2

(3)

cos (b + g) + cos (g + a) + cos (a + b) +

Solution:

)

3n w 2 – 1

(2)

From (2)

Example 67

1 + 2x + 3x2 + . . . + 3nx3n–1

1 + 2w + 3w2 + . . . + 3n w3n–1

From (1) and (2) a2 + b2 + c2 = 0

(d) 1

Ans. (a), (b)

Ans. (a), (b), (c), (d) Solution

(b) n(w 2 – 1)

(c) 0

If sin a + sin b + sin g = cos a + cos b +

Example 66

3n w –1



2 zs + 1z H 1 = zH 3 3

1 1 (z1 + z2 + z3) = zH fi zH = z1 + z2 + z3 3 3

As AD passes through H, its equation is z z1 z1 + z2 + z3

1 z 1 =0 z1 z1 + z2 + z3 1

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z - z1 z1 z2 + z3



z - z1 z1 z2 + z3

0 1 =0 0

Lastly, if z2 + z3 = 0, zH = z1, that is, orthocentre of DABC is at A fi DABC is right angled at A. A(z1)

[use R1 Æ R1 – R2 and R3 Æ R3 – R2] fi

(z – z1) ( z2 + z3 ) - ( z - z1 ) (z2 + z3) = 0



Ê 1 1ˆ 1ˆ Ê (z – z1) Á + ˜ – Á z - ˜ (z2 + z3) = 0 Ë Ë z2 z3 ¯ z1 ¯



( z + z3 ) 1ˆ Ê (z – z1) 2 – Á z - ˜ (z2 + z3) = 0 Ë z2 z3 z1 ¯ z - z1 1 = zz2 z3 z1



As E lies on the circle |z| = 1, we have z = 1/z of E. Thus, z - z1 z - z1 1 1 = – =– z2 z3 z z1 z1 z fi

D ABC is equilateral.

B(z2)

H D

C(z3) E

Fig. 1.53

Example 70 Let z and w be two complex numbers such that |z| £ 1, |w| £ 1, |z + iw| = |z – i w | = 2, then z can be (a) 1 (b) – 1 (c) i (d) – i Ans. (a), (b) Solution: Note that z, w, – iw and i w lie inside or on the boundary of the disc | z| £ 1. Since |z + i w| = |z – i w | = 2, we get – iw, i w are the other end points of the diameter having one end point as z. Thus, – iw = i w = – z. Therefore, |z + z| = 2 fi |z| = 1.

z z z =– 2 3 z1

Next, z 12 = z 2z 3, z 22 = z 3z 1 2 2 fi z1 z2 = z1z2z32 fi z32 = z1z2 Thus, z12 + z22 + z32 = z2z3 + z3z1 + z1z2 fi

O

Now, – z = – iw = i w fi w = – iz and w = iz w = i z and w = iz \ \ i z = iz fi z = z fi

z is real.

As |z| = 1, we get z = 1, – 1.

MATRIX-MATCH TYPE QUESTIONS Example 71

Match the statements in Column-I with those in Column-II. Column I

(a) The set of points z satisfying | z – i | z | | = | z + i | z | | is contained in or equal to (b) The set of points z satisfying | z + 4 | + | z – 4 | = 10 is contained in or equal to 1 (c) If | w | = 2, then the set of points z = w – w is contained in or equal to 1 (d) If | w | = 1, then the set of points z = w + w is contained in or equal to

Column II

(p) an ellipse with eccentricity

4 5

(q) the set of points z satisfying Im z = 0 (r) the set of points z satisfying | Im z |

1

(s) the set of points z satisfying | Re z |

2

(t) the set of points z satisfying | z |

3

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We have 2ae = 8 and 2a = 10 e = 4/5 (c) | ω | = 2 ω = 2eiθ = 2(cosθ + i sinθ) 1 1 = 2eiθ – e–iθ z=ω– w 2

Ans.

x= Solution:

(a) z

x2 y2 + =1 9/4 25/4

| z – i | z| | = | z + i | z| | For z 0, we can write z z -i = +i |z| |z| z is equidistant from –i and i. |z| z lies on the real axis. |z| z lies on the real axis. Im(z) = 0 and | Im(z) | 1 (b) | z + 4 | + | z – 4 | = 10 represents an ellipse whose foci are – 4 and 4

3 5 cos θ, y = sin θ 2 2

This represents an ellipse and its eccentricity is 1-

b2 a

2

= 1-

(d) | ω | = 1

9 4 = . 25 5

ω = eiθ

Now, z = ω +

1 = eiθ + e–iθ = 2cos θ w

As z is real, Im(z) = 0, | Im(z) | Also, Re(z) = 2cos θ | Re(z) | = | 2cos θ | Finally, z = 2cos θ

Example 72

R.

|1 - z1| |1 - z2 | 10

|1 - z9 |

equals

2k p ˆ 9 S. 1 -  k = 1 cos ÊÁ equals Ë 10 ˜¯ P Q R S (a) 1 2 4 3 (c) 1 2 3 4 Ans. (c) 2p 2p k + i sin ˆ˜ Solution: P. For 1 £ k £ 9, zk = ÊÁ cos Ë 10 10 ¯ Now

2 |z|

⎛ 2k π ⎞ ⎛ 2k π ⎞ + i sin ⎜ k = 1, 2, ..., 9 Let zk = cos ⎜ ⎝ 10 ⎟⎠ ⎝ 10 ⎟⎠

List I P. For each zk there exists a zj such that zk . zj = 1 Q. There exists a k {1,2, , 9} such that z1 z = zk has no solution for z in the set of complex numbers.

zk . z10–k 2p 2p k 2p 2p 10- k + i sin ˆ˜ ÊÁ cos + i sin ˆ˜ = ÊÁ cos Ë 10 10 ¯ Ë 10 10 ¯

1.

List II 1. True 2. False

3. 1 4. 2 P (b) 2 (d) 2

Q 1 1

R 3 4

S 4 3

2p 2p 10 = ÁÊ cos + i sin ˜ˆ Ë 10 10 ¯ 2p 2p = cos ÁÊ10 ˜ˆ + i sin ÁÊ10 ˜ˆ Ë 10 ¯ Ë 10 ¯ = cos (2π) + i sin (2π) = 1

2

3.

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this statement is true Q. For 1

k

9 Ê 2kp ˆ fi 1 -  cos Á = 1 - cos p = 2 Ë 10 ˜¯ k =1

9, z1 z = zk

z = zk/z1 = zk–1

Number solutions of

Example 73

2p 2p k -1 = ÊÁ cos + i sin ˆ˜ Ë 10 10 ¯

Column 1 2

this statement is false R. We have 10

z –1 = (z – 1) (z – z1)(z – z2)...(z – z9)

(a) z + |z|

=0

(p) 1

(b) z2 + z 2

=0

(q) 3

(c) z2 + 8z

=0

(r) 4

(d) |z and |z – 1| = 2

(z – z1) (z – z2) ... (z – z9) = z9 + z8 +  + z + 1

Column 2

Ans.

p

q

r

s

Putting z = 1, we get

a

p

q

r

s

(1 – z1)(1 – z2)...(1 – z9) = 10

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

1 1 - z1 1 - z2 ... 1 - z9 = 1 10 S. Now, 2k p ˆ 1 - Â cos ÊÁ ˜ Ë 10 ¯ k =1 9

Solution:

2p 18p ˆ 4p 1 - ÈÍcos ÁÊ ˜ˆ + cos ÁÊ + cos ÁÊ ˜ˆ Ë 10 ˜¯ Ë 10 ¯ Î Ë 10 ¯ 16p ˆ 6p 14p ˆ + cos ÊÁ + cos ÊÁ ˆ˜ + cos ÊÁ Ë 10 ˜¯ Ë 10 ¯ Ë 10 ˜¯ = 8p 12p ˆ + cos ÊÁ ˆ˜ + cos ÊÁ Ë 10 ¯ Ë 10 ˜¯ 10p ˆ ˘ + cos ÁÊ Ë 10 ˜¯ ˚˙ We have cos

2p 18p ˆ + cos ÊÁ Ë 10 ˜¯ 10

2p 2p p = cos ÊÁ ˆ˜ + cos ÊÁ 2p - ˆ˜ = 2 cos ÊÁ ˆ˜ ; Ë 10 ¯ Ë Ë 5¯ 10 ¯ 4p 16p ˆ 2p Similarly cos ÊÁ ˆ˜ + cos ÊÁ = 2 cos ÊÁ ˆ˜ ; ˜ Ë 10 ¯ Ë 10 ¯ Ë 5¯ 6p 14p ˆ 3p and cos ÊÁ ˆ˜ + cos ÊÁ = 2 cos ÊÁ ˆ˜ Ë 10 ¯ Ë 10 ˜¯ Ë 5¯ 2p = -2 cos 5 8p 12p ˆ 8p and cos ÊÁ ˆ˜ + cos ÊÁ = 2 cos ÊÁ ˆ˜ Ë 10 ¯ Ë 10 ˜¯ Ë 10 ¯ 4p Êpˆ = 2 cos ÊÁ ˆ˜ = –2cos Á ˜ Ë 5¯ Ë 5¯

(a) z2 = – |z| fi

|z|2 = |z| fi |z| = 0

or

|z| = 1

If

|z| = 0,

then

z=0

If

|z| = 1,

then

z2 = – 1 fi z = ± i

\ z2 + |z| = 0 has 3 solutions. (b) For z π 0, z2 = – z

2

2



Ê zˆ ÁË ˜¯ = – 1 z



z = ± iz



z = ±i z

Each complex number of the form a(1 ± i) where aŒR (c) |z2| = |– 8 z | fi

|z|2 = 8|z| fi |z| = 0

If

|z| = 8 fi z z = 64

\

z2 = – 8 z = – 83/z



z3 = – 83



or

|z| = 8

z = – 8, – 8w, – 8w2

Thus, z2 + 8 z = 0 has 4 solutions. (d) Circles |z – 2| = 1 and |z – 1| = 2 touch each other at z = 3. Thus, |z – 2| = 1 solution.

and

|z

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Example 74 If a, b, c are distinct integers, and w π 1 is a cube root of unity, then value of Column 1 (a) |a + bw + cw 2|2

Column 2 (p) 1 (q) ≥ 1

22

(b) |a + bw + cw |

+|a + bw2 + cw|2 (c)

a + bw + cw 2 a + bw 2 + cw p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

(s) 2

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

fi fi

= (a + bw + cw2) (a + bw2 + cw) = a2 + b2 + c2 – bc – ca – ab

– 3 – 4i and – 5 + 2i. Thus, its centre to – 4 – i.

1 [(b – c)2 + (c – a)2 + (a – b)2] 2 As a, b, c are distinct integers the least value of

Modulus of complex number whose Example 76 reciprocal is Column 1 Column 2

=

(b – c)2 + (c – a)2 + (a – b)2 ≥ 2

(a)

1 1 + a b + ic

(p)

(b)

1 1 a - ib a - ic

(q)

(c)

b c + a + ib a - ic

(r)

(d)

1 a + ib + ic

(s)

|a + bw 2 + cw |2 = |a + bw 2 + cw |2 = |a + bw + cw 2|2

TIP a + bw + cw2| = |a + bw2 + cw| we can guess all the answers. Example 75

Centre of circle

Column 1 2

3z z – 5z – 5z + 3 = 0

Its circle with centre at 5/3 + 0i (c) It represents a circle with centre at 1 – i

= (a + bw + cw2) (a + bw + cw 2 )

Also,

4 (z z – z – z + 1) = zz + z + z + 1

|a + bw + cw2|2

Solution:

q

Solution (a) The given equation represents a circle with the i as diameter. Therefore, its centre is 1 + 2i (b) 4 |z – 1|2 = |z + 1|2

(r) ≥ 2

(d) |a + bw2 + cw |

p

Ans.

Column 2 2

(a) |z – 2| + |z – 4i| = 20

(p) 1 – i

z -1 1 = (b) 2 z +1

(q) 5/3 + 0i

(c) z z – (1 + i)z

(r) – 4 – i

– (1 – i) z + 7 = 0 Ê z + 3 + 4i ˆ p (d) arg Á = Ë z + 5 - 2i ˜¯ 2

(s) 1 + 2i

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

Solution: (a)

1 a + b + ic = ab + iac z

a 2 + b2

a2 + c2

|b - c| a 2 + (b + c) 2 | a | b2 + c2 ( a + b) 2 + c 2 a 2 + b2

a2 + c2

| a | |b + c|

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( a + b) 2 + c 2 1 = | z| | a | b2 + c2





(b)

(c)

|z| =

1 - cos q Êqˆ = tan Á ˜ Ë 2¯ 1 + cos q Êqˆ | z| = – tan Á ˜ Ë 2¯

\

( a + b) 2 + c 2

1 i (b - c) = (a - ib) (a - ic) z fi

1 = | z|



|z| =

|b - c | a + b2

a2 + c2

2

a 2 + b2 a 2 + c2 |b - c |

|z| =

1 = z

a + (b + c)



a 2 + b2 a 2 + c2 | a | |b + c|

(d) a 2 + (b + c) 2

when

Column 1

Column 2

(a) – p < q < 0

(p) 0

(b) q = 0

Ê1 ˆ (q) tan Á q ˜ Ë2 ¯

(c) 0 < q < p

Ê1 ˆ (r) – tan Á q ˜ Ë2 ¯

(d) q = 2p /3

(s)

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution

|z| =

if q = 0

Êqˆ = tan Á ˜ Ë 2¯

if 0 < q < p

(1 + cos q ) 2 + sin 2 q

1- z

4

z4 - 1 z3 + 1 q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(q)

p a + 2 2

(r)

3a 2

(s)

p a 2 2

Solution (a) 1 + z3 = 1 + cos 3a + i sin 3a 3a 3a 3a + 2 i sin cos = 2 cos2 2 2 2 Ê 3a ˆ È Ê 3a ˆ Ê 3a ˆ ˘ = 2 cos Á ˜ Ícos Á ˜ + i sin Á ˜ ˙ Ë 2¯Î Ë 2¯ Ë 2 ¯˚

3

(1 - cos q ) 2 + sin 2 q

1 + z3

p

Ans.

1 - cos q - i sin q , Modulus of z = 1 + cos q - i sin q

Example 77

Ans.

(c)

2

|z| =

=0

(b) 1 – z4

1 2

if – p < q < 0

Example 78 If z = cos a + i sin a, 0 < a < p/6, then the argument of Column 1 Column 2 p (p) 2a – (a) 1 + z3 2

1 a(b + c) = (a + ib) (a - ic) z fi

(d)

| a | b2 + c2

=



arg (1 + z3) = 3a/2

(b) 1 – z4 = 1 – cos 4a – i sin 4a = 2 sin2 2a – 2i cos 2a sin 2a = 2 sin 2a [cos (2a – p /2) + i sin (2a – p/2)] fi (c)

arg (1 – z4) = 2a – p/2

1 + z 3 cos (3a /2 ) È Êp aˆ Ê p aˆ˘ = Ícos ÁË - ˜¯ + i sin ÁË - ˜¯ ˙ 4 sin (2a ) Î 2 2 2 2 ˚ 1- z

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fi (d)

Ê 1 + z3 ˆ p a arg Á ˜= Ë 1 - z4 ¯ 2 2

Column 1

È Êa pˆ Êa p ˆ˘ Ícos ÁË - ˜¯ + i sin ÁË - ˜¯ ˙ 2 2 2 2 ˚ Î

=

sin (2a ) cos (3a /2)

=

sin (2a ) È Êp aˆ Ê p a ˆ˘ Ícos ÁË + ˜¯ + i sin ÁË + ˜¯ ˙ cos (3a /2) Î 2 2 2 2 ˚

+

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

p

q

r

s

Ans.

(s) ellipse

q

r

s

a

p

q

r

s

d

b

p

q

r

s

Solution

c

p

q

r

s

d

p

q

r

s

Solution:

For (a) and (d) see theory. (b) can be written as

z - 3/2 =1 z-i

z1 z2 , |z1 | = 3,

|z2| = 1 (s) 10 (d) z1 – z2 + z3 – z4 if z1, z2, z3, z4 represent vertices of a parallelogram.

p

Ans.

(r) 4

1 ( z1 + z2 ) - w 2

if w =

(q) hyperbola

z-6 =3 z - 2i

1 ( z1 + z2 ) + w 2

(c)

(r) straight line

(q) 1

if z z = 1

Example 79 Match the equation on the left with the curve they represent on the right Column 1 Column 2 (a) |z – 3| + |z – i| = 10 (p) circle

(c) z2 + z 2 = 5

(p) 0

2

Ê z - 1ˆ p a Thus, arg Á 3 ˜= + Ë z + 1¯ 2 2

2z - 3 =2 z-i

4

1 + |2 – z|2 (b) 2 + z

4

(d)

Column 2

(a) 4 a (b – a ) if a + ib, b π 0 is a root of z5 = 1 4

z4 - 1 z3 + 1

(b)

Value of

Example 80

(a) Let z = a + ib, then z5 = 1. Also,

|z|5 = 1 fi |z| = 1 fi z = 1/z

Now, 4a (b 4 – a 4) = 4a(b2 – a2) |z|2 1 (z + z ) [(z – z )2 + (z + z )2] 2 1 1ˆ Ê = – (z + z ) (z2 + z 2) = - Á z 3 + z + + 3 ˜ Ë z z ¯

= -

fi z lies on the perpendicular bisector of the segment i. (c) Let z = x + iy fi z2 = x2 – y2 + 2ixy Thus, z2 + z 2 = 5 fi 2(x2 – y2) = 5 fi x2 – y2 = 5/2 Thus, z lies on a hyperbola.

= -

Ê z8 - 1 ˆ Ê 1 - z3 ˆ 1 Ê z 8 - 1ˆ = = Á ˜ Á ˜ Á ˜ z 3 Ë z 2 - 1¯ Ë z5 - z3 ¯ Ë 1 - z3 ¯ [∵ z5 = 1]

=1 (b) |2 + z |2 + |2 – z|2 = 2{22 + |z|2} = 10 (c)

1 1 z1 + z2 ) + w + ( z1 + z2 ) - w ( 2 2

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=

1 2

(

z1 + z2

)

2

+

1 2

(

z1 - z2

)

Ê z3 - z1 ˆ (b) Re Á =0 Ë z3 - z2 ˜¯

2

= |z1| + |z2| = 4

Ê z3 - z1 ˆ (c) Re Á 0, then |z – z1| = |z – z2| represents the z1 and z2. Ans. (b) Solution: We have

numbers and w π 1 is a cube root of unity, then a + bw + cw 2 aw 2 + b + cw Statement-2: If z π 0, then

\

Statement-1: Two nonzero complex Example 88 numbers z1 and z2 lie on a straight line through origin if and only if z1 z2 is real. [∵ w 3 = 1]

Statement-2: Two non-zero complex numbers z1 and z2 always lie on a straight line passing through origin if and only if z1 z2 is real.

[∵ |w | = 1]

Ans. (a)

z = 1 " z π 0. z

Statement-1: a, b, c are three non-zero

real numbers such that a + b + c = 0 and z1, z2, z3 are three complex numbers such that az1 + bz2 + cz3 = 0, then z1, z2 and z3 lie on a circle. Statement-2: If z1, z2 and z3 are collinear then z1 z2 z3

z1 1 z2 1 = 0 z3 1

Ans. (d) Solution a + b + c = 0 fi c = – (a + b) \ az1 + bz2 + cz3 = 0 fi az1 + bz2 – (a + b)z3 = 0

least value of y is 2 which is attained at z = 1

For statement-2, see Theory.

z = 1. z

Solution: We have |a + bw + cw 2 | = |aw3 + bw + cw 2| = |w (aw2 + b + cw)| = |w| |aw2 + b + cw| = |aw2 + b + cw| 2 a + bw + cw =1 fi aw 2 + b + cw

Example 86

y = |z | + |2 – z| + |z – 1| ≥ |z + (2 – z)| + |z – 1| = 2 + |z – 1|

=1

Ans. (b)

Also, since | z | = |z|, we get

Statement-1: If z ΠC, least value of

Solution: z1, z2 will lie on a straight line through the origin if the origin O z1, z2 is some ratio. fi 0=

z1 + kz2 for some k ΠR. 1+ k

z1 = – k Œ R fi z1 z2 = – k |z2|2 Œ R. z2 Next z1 z2 Œ R ¤ z1 z2 Œ R. fi

Example 89 Statement-1: If A, B, C, D are four points in the complex plane, then AC 2 + BD2 + AD 2 + BC2 ≥ AB2 + CD2 Statement-2: If z1, z2, z3, z4 are four complex numbers then |z1 + z2 – z3 – z4|2 ≥ 0 Ans. (a) Solution: Let z1, z2, z3 and z4 represent the points A, B, C and D in the complex plane, then AC2 + BD2 + AD2 + BC2 – AB2 – CD2

IIT JEE eBooks: www.crackjee.xyz Complex Numbers 1.43

= |z1 – z3|2 + |z2 – z4|2 + |z1 – z4|2 + |z2 – z3|2– |z1 – z2|2 – |z3 – z4|2

(a) bz1 + bz2 is a real number (b) bz1 + bz2 is a non-real complex number

= 2(|z1|2 + |z2|2 + |z3|2 + |z4|2 – Re(z1 z3 ) – Re(z2 z4 )

(c) bz1 + bz2 = 0

– Re(z1 z4 ) – Re(z2 z3 ) – (|z1|2 + |z2|2 + |z3|2 +

(d) z1 z2 + |b|2 = c

|z4|2 – 2Re(z1 z2 ) – 2Re(z3 z4 )

Ans. 91 (a) 92 (a)

= |z1|2 + |z2|2 + |z3|2 + |z4|2 + 2Re[z1(z3 z4 )] – 2Re[z2( z3 + z4 ) + 2Re(z3 z4 ) = |z1 + z2 – z3 – z4|2 ≥ 0. Example 90

bz + bz = c

Statement-1. Let z1, z2 ΠC РR.

(1)

L P(z)

then PA = PB

Êz ˆ Êz ˆ then arg Á 1 ˜ + arg Á 1 ˜ = 0 Ëz ¯ Ëz ¯ 2

A(z1)

Solution: 91. Let P(z) be any point on the line

2

Statement-2. arg(z) + arg ( z ) = 0 for each non-zero z ΠC



PA2 = PB2



|z – z1|2 = |z – z2|2



|z|2 – zz1 - zz1 + |z1|2

B(z2)

= | z |2 - zz2 - zz2 + | z2 |2

Ans. (c) Solution: Statement-2 is false. For instance, if z = – 2, then arg(z) + arg ( z ) . = p + p = 2p



We have bz1 + bz2 =

Ê | z |2 ˆ Êzz ˆ = arg Á 1 1 ˜ = arg Á 1 2 ˜ = 0 Ëz z ¯ Ë| z | ¯

COMPREHENSION-TYPE QUESTIONS

c | z2 | - | z1 |2 2

c

2

=

| z2 | - | z1 | 2

2

{( z2 - z1 ) z1 + ( z2 - z1 ) z2 }

{|z2|2 – |z1|2} = c

We have c = bz + bz

Paragraph for Questions Nos. 91 and 92



Let bz + bz = c , b π 0 be the line L in the complex plane.

fi c and hence bz1 + bz2 is a real number.

Example 91 If z1 through the line L, then

z 2,

(a) bz1 + bz2 is a real number (b) bz1 + bz2 is a non-real complex number (c) bz1 + bz2 = 0 (d) z1 z2 + | b | = c

z1, then

(2)

b b c = = 2 z2 - z1 z2 - z1 | z2 | - | z1 |2

Êz ˆ Êz ˆ arg Á 1 ˜ + arg Á 1 ˜ z Ë 2¯ Ë z2 ¯

2 2

z ( z2 - z1 ) + z ( z2 - z1 ) = |z2|2 – |z1|2

As (1) and (2) are the same line

Statement-1 is true as

Example 92

bz + bz = c

If foot of perpendicular from z2 to L is

c = bz + bz = c

92. Let b1 = b1 + ib2, z1 = x1 + iy1 and z2 = x2 + iy2 As BA ^ L, Ê b ˆ Ê y - y1 ˆ =–1 then Á - 1 ˜ Á 2 Ë b2 ¯ Ë x2 - x1 ˜¯

z2

B

L bz + bz = c



b1 x2 - x1 = b2 y2 - y1



b1 + ib2 ( x2 - x1 ) + i ( y2 - y1 ) = b1 - ib2 ( x2 - x1 ) - i ( y2 - y1 )

z1

A

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b z2 - z1 = b z2 - z1

So there cannot exist distinct x, z such that c = 0 As

a, b, c ≥ 0, a + b + c = 0



bz2 - bz1 = bz2 - bz1



a = 0, b = 0, c = 0



bz2 + bz1 = bz1 + bz2

But

b = 0 fi a, c > 0. A contradiction.

94.

a = |xu – y|,

fi bz1 + bz2 is a real numbers.

b = |y – vz|

c = |x – wz|

a + b = |xu – y| + |y – vz|

Now,

Paragraph for Questions Nos. 93 to 94

≥ |(xu – y) + (y – vz)|

Let x, y, z be three real numbers, and let

= |xu – uz|

and

a = x2 –

3xy + y2

b = y2 –

2 yz + z2

|xu – vz|2

But

= |xu|2 + |vz|2 – xz (uv + u v ) Ê 5p ˆ = x2 + z2 – 2cos Á ˜ xz Ë 12 ¯

c = x2 – xz + z2

Example 93 (a) (b) (c) (d)

If x, y, z are distinct, then

b = 0 fi a, c > 0 there exist distinct number x, z such that c = 0 there exist x, y, z such that a + b + c = 0 none of these

Example 94

If x, y, z > 0, then

> x2 + z2 – 2cos(p/3)xz = |x – wz| \

a+ b> c

Also,

a + c = |x – uy| + |wz – x| ≥ |wz – uy| |wz – uy|2 = y2 + z2 – 2yz cos(7p/12)

But

(a)

a + b = c for some x, y, z

= y2 + z2 + 2cos(p/12)yz

(b)

a + b > c for each x, y, z

> y2 + z2 – 2cos(p/4)yz = b2

(c)

a + c = b for some x, y, z

(d)

a + c < b for each x, y, z

Ans. 93. (d) 94. (b) Solution:

93. Let u = epi/6, v = epi/4 and w = epi/3

\

a+ c> b

Paragraph for Questions Nos. 95 to 99. If a0, a1, a2, ..., an–1 are root nth roots of unity, then ak = cos

We have a = x2 – 2 cos(p/6)xy + y2 2

= |xu – y| = |x – uy| b = |y – vz|

2

x – 1 = (x – a0) (x – a1) ... (x – a n – 1) n

2

Now, b = 0 fi y = vz If z π 0, then v = y/z, that is, y/z is a non-zero complex number. A contradiction Thus, b = 0 fi y = z = 0 As x, y, z are distinct, x π 0 a = |xu|2 = |x|2 = c > 0

Note that c = 0 fi x = z = 0

(b) (–1)n (d) 1 + (–1)n–1

(a) 3 (c) 0

Value of (1 – a1) ... (1 – a n–1) is

Example 96 (a) n (c) (–1) Example 97

(1)

Value of (1 + a 0) (1 + a1) . . . (1 + a n–1) is

Example 95

c = |x – wz| = |wz – x|

\

0 £ k £ n – 1. Also

where

2 2

2k p 2k p + i sin n n

(b) n – 1 n

(d) 0 Value of

Êpˆ Ê 2p ˆ Ê n -1 ˆ 2n – 1 sin Á ˜ sin Á ˜ º sin Á p ˜ is Ë n¯ Ë n ¯ Ë n ¯

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(a) n (c) n – 1

(b) –1 (d) n/3

2p Ê ˆ z5 + 1 = (z + 1) Á z 2 + 2 cos z + 1˜ Ë ¯ 5 4p Ê 2 ˆ z + 1˜ (3) ÁË z + 2 cos ¯ 5 Now use cos (4p/5) = – cos (p/5)

If n = 5 in (1), then value of

Example 98

p 2p Ê ˆÊ ˆ + 1˜ is (z + 1) Á z 2 - 2 z cos + 1˜ Á z 2 + 2 z cos Ë ¯ Ë ¯ 5 5 (a) z5 – 1 (c) z5 + 1 Value of sin

Example 99

99.

(b) z5 (d) 0

(a) 1/4

2p ˆ Ê 2 = 23 Á1 + cos ˜ Ë 5 ¯

p p cos is 10 5

(b) 1/2

(c) 1/3

(d)

Put x = –1 in (1) so that

p 2p cos2 5 5

2 = 25 cos2



p 1 Êp p ˆ = cos cos Á - ˜ Ë 2 10 ¯ 5 4



(–1) – 1 = (–1 – a0) (–1 – a1) ... (–1 – an–1) n

4p ˆ Ê ÁË1 + cos ˜¯ 5

fi 1/ 3

Ans. 95. (d), 96. (a), 97. (a), 98. (c), 99. (a) Solution 95.

In (3) put z = 1, so that

p p 1 = cos sin 5 10 4

fi (–1)n [1 + (–1)n – 1] = (–1)n (1 + a0) ... (1 + an – 1)

Paragraph for Questions Nos. 100 to 102

fi (1 + a0) (1 + a1) ... (1 + an–1) = 1 + (–1)n–1 96.

If n n as follows:

Note that a0 = 1 and (1) can be written as

(x – a1) (x – a2) .. (x – a n–1) =

xn - 1 x -1

cos nq = Pn (cos q ) For example, P2(x) = 2x2 – 1 and P3(x) = 4x3 – 3x. 1 [Pn+1 (x) + Pn–1 (x)] equals Example 100 2x

= 1 + x + ... + xn–1 Put

x = 1, so that (1 – a1) (1 – a2) ... (1 – a n–1) = n

(2)

2k p ˆ Ê |1 – ak | = Á1 - cos ˜ Ë n ¯

2

+ sin2

2k p n

2k p ˆ kp Ê = 2 Á1 - cos = 4 sin2 ˜ Ë n ¯ n

(x + Example 101 (a) Pn (x) (c) 2Pn (x) Example 102

From (2)

(a) (b) (c) (d)

|(1 – a1) (1 – a2) ... (1 – a n – 1)| = n fi

Êpˆ Ê 2p ˆ Ê n -1 ˆ 2n–1 sin Á ˜ sin Á ˜ ... sin Á p˜ = n Ë n¯ Ë n ¯ Ë n ¯

98.

For n = 5, note that a3 = e6pi/5 = a 2 and a4 = a1

Thus, (1) can be written as x5 – 1 = (x – 1) (x – a1)(x – a1 ) (x – a2) (x – a 2 ) = (x – 1) [x2 – (a1 + a1 ) x + 1] [x – (a2 + a 2 )x + 1] 2

Putting

x = – z, we get

(b) Pn–1 (x) + Pn (x) (d) Pn+1 (x) – Pn (x)

(a) Pn+2 (x) (c) Pn (x)

97. We have for 1 £ k £ n – 1 2

Pn(x) of degree

Ans.

x 2 - 1 )n + (x – x 2 - 1 )n equals (b) Pn+1 + Pn–1 (x) (d) –3Pn ( x )

P6(x) equals

36x6 – 45x4 + 18x2 – 8 32x6 – 48x4 + 18x2 – 1 36x6 – 48x4 + 18x2 – 5 none of these

100. (c), 101. (c), 102. (b)

Solution 100. We have P1(x) = x and cos (n + 1) q + cos (n – 1)q = 2 cos (nq) cos q fi Pn+1 (x) + Pn–1 (x) = 2x Pn(x) fi 101.

Pn (x) =

1 [Pn+1 (x) + Pn–1 (x)] 2x

Put x = cos q so that

x 2 - 1 = i sin q

IIT JEE eBooks: www.crackjee.xyz 1.46 Comprehensive Mathematics—JEE Advanced

Now,

x 2 - 1 )n + (x –

(x +

(c) z1 z2 + z1 ¥ z2 = 0

x 2 - 1 )n

= (cos q + i sin q)n + (cos q – i sin q)n = 2 cos (nq) [using DeMoivre’s theorem] = 2Pn (cos q) = 2Pn(x) Thus, (x +

x 2 - 1 )n + (x –

x - 1 ) + (x –

2P6(x) = (x +

102.

6

6

4

Example 106 and OQ is

x 2 - 1 )n = 2Pn (x). 6

2

(d) none of these

2

6

x -1) 2

2

6

2

The area of parallelogram with sides OP

(a) |z1 z2|

(b) |z1 ¥ z2|

(c) |z1 z2| + |z1 ¥ z2|

(c) |z1 z2| |z1 ¥ z2 | OP on OQ is

Example 107 2

= 2[x + C2 x (x – 1) + C4 x (x – 1) + 6C6 (x2 – 1)3] 6 4 2 P6(x) = 32x – 48x + 18x – 1

Paragraph for Questions Nos. 103 to 107. For non-zero complex numbers z1 and z2 product and the cross product of z1 and z2 as follows: Dot product: z1 z2 = |z1| |z2| cos q ∑

= x1 x2 + y1 y2 = Re( z1 z2) Cross product: z1 ¥ z2 = |z1| |z2| sin q = x1 y2 – x2 y1 = Im( z1 z2)

(a) |z1 z2|/|z1| (c) |z1 z2|

(b) |z1 z2|/|z2| (d) |z1 ¥ z2|

Ans. 103. (c), 104. (a), 105. (b), 106. (b), 107. (b) Solution 103. z1 z2 + i(z1 ¥ z2) = (x1 x2 + y1 y2) + i(x1 y2 – x2 y1) = (x1 – iy1) (x2 + iy2) = z1 z2 104.

z1 z2 = 0

¤

Ê 2 z ˆ ¤ Re Á z1 2 ˜ = 0 z1 ¯ Ë

Re( z1 z2) = 0 z2 is purely imaginary z1

¤

105.

OP coincides with OQ

¤

q = 0 ¤ z1 ¥ z2 = 0

106.

Area of parallelogram with sides OP and OQ = 2 [area (DOPQ)] = ||z1| |z2| sin q| = |z1 ¥ z2| OP and OQ

107. = ||z1| cos q | =

| z1 | | z2| cos q | z2 |

=

| z1 ◊ z2 | | z2 |

Paragraph for Questions Nos. 108 to 110 Fig. 1.54

Example 103

z1 z2 + i(z1 ¥ z2) equals

(a) z1 z2

(b) z1 z2

(c) z1 z2

(d) z1 z2

Example 104

Let a, b, g be three real numbers such that a 2 + b 2 + g 2 – g a + ib = 0 and z = . 1- g



z1 z2 = 0 is equivalent to say

Example 108 (a) g (c)

(a) z2/z1 is purely imaginary (b) z1/z2 is purely real (c) z1 = 0 or z2 = 0 (d) none of these Example 105

OP coincides with OQ if and only if

(a) z1 z2 = 0 (b) z1 ¥ z2 = 0

|z|2 equals (b) 1 – g

g 1- g

Example 109

(d)

1- g g

( z - z )i 2(1 + | z |2 )

a equals

(a)

z+z 2(1 + | z |2 )

(b)

(c)

( z - z )i 2(1 + | z |2 )

(d) none of these

IIT JEE eBooks: www.crackjee.xyz Complex Numbers 1.47

b equals

Example 110 (a)

z-z 2(1 + | z |2 )

(c) (b)

( z - z )i 2(1 + | z |2 )

Example 113 (a) (b) (c) (d)

2z (d) 1 + | z |2

z (c) 2(1 + | z |2 ) Ans. 108. (c), 109. (a), 110. (b) Solution 108.

109. From Example 108,



1 + |z|2 =

g 1 = –1 + 1- g 1- g 1 | z |2 fig= 1- g 1 + | z |2



2a = (z + z ) (1 – g) =

z+z 1 + | z |2

Solution 111. Let l = ia where a ΠR. Let z1, z2, z3 and w1, w2, w3 be vertices of two triangles ABC and PQR respectively, such that

2ib =z– z 1- g

110.

z3 - z1 w - w1 = 3 = ia z2 - z1 w2 - w1

z-z 2ib = (z – z ) (1 – g) = 1 + | z |2

fi fi

b=

( z - z )i 2(1 + | z |2 )



| z3 - z1 | | w3 - w1 | = = |a| | z2 - z1 | | w2 - w1 |



AC PR = = |a| AB PQ

Paragraph for Questions Nos. 111 to 115 Let l ΠC and z1, z2, z3 ΠC are such that

Also,

z3 - z1 =l z2 - z1



If l is a given purely imaginary number Example 111 then all the triangles with vertices z1, z2, z3 are (a) similar (b) acute angled (c) equilateral (d) obtuse angled Value of l for which triangle with Example 112 vertices z1, z2, z3 is equilateral, is (a)

(

1 1 - 3i 2

)

lie on a circle are vertices of a right triangle lie on a straight line lie on a parabola

Ans. 111. (a), 112. (b), 113. (c), 114. (c), 115. (a)

z+z a= 2(1 + | z |2 )



If l ΠR, then z1, z2, z3

If l = eit (t ΠR) and z2, z3 Example 115 locus of z1 is (a) a straight line (b) an ellipse (c) a circle (d) a pair of straight lines

2a =z+ z 1- g

Now,

(d) – 3i

If l = it (t ΠR) and z2, z3 Example 114 locus of z1 is (a) a straight line (b) an ellipse (c) a circle (d) a pair of straight lines

a2 + b2 g - g 2 g |z| = = = 2 2 1- g (1 - g ) (1 - g ) 2

|z|2 =

3i

(b)

(

1 1 + 3i 2

)

–BAC = –QPR = p/2

l=

z3 - z1 | z3 - z1 | Ê p pˆ = ÁË cos + i sin ˜¯ z2 - z1 | z2 - z1 | 3 3 =

113. fi fi

(2)

p Ê z - z1 ˆ Ê w - w1 ˆ arg Á 3 = arg Á 3 = ˜ ˜ 2 Ë z2 - z1 ¯ Ë w2 - w1 ¯

From (2) and (3) we get DABC and DPQR are similar. 112.

(1)

(

1 1 + 3i 2

)

z3 – z1= l(z2 – z1) z3 = (1 – l) z1 + lz2 z3 lies on the line joining z1 and z2.

(3)

IIT JEE eBooks: www.crackjee.xyz 1.48 Comprehensive Mathematics—JEE Advanced

114. As

y

z3 - z1 is purely imaginary, angle between the z2 - z1 z3z1 and z2z1 is a right angle.

fi z1 diameter. 115. fi fi

P

z2z3 as

z

÷2

A

z3 - z1 = |eit| = 1 z2 - z1 z1 is equidistant from z2 and z3 z1 lies the perpendicular bisector of the segment z3 and z3.

Paragraph for Questions Nos. 116 to 119 Let A, B, C follows: A = {z : Im(z) ≥ 1} B = {z : |z – 2 – i | = 3} C = {z : Re((1 – i)z) = 2 }.

O

117. Points A and B are end points of a diameter of |z – (2 + i A is – 1 + i and that of B is 5 + i. |z + 1 – i|2 + |z – 5 – i|2 = PA2 + PB2 = AB2 = 36 where lies between 35 and 39. 118. ||z| – |w|| £ |z – w| Here z z and w would be maximum for diametrically opposite points.

A « B « C is (a) 0 (c) 2 Example 117

(b) 1 (d) • Let z be any point in A « B « C. Then,

x

÷2 Fig. 1.55

The number of elements in the set

Example 116

B

1

\

||z| – |w || £ |z – w| < 6



– 6 < |z| – |w | < 6 fi – 3 < |z| – |w| + 3 < 9

Ê 5 + i – zˆ p = . 119. As –APB = p /2, arg Á Ë i – 1 – z ˜¯ 2

2

|z + 1 – i| + |z – 5 – i| 2 lies between (a) 25 and 29 (b) 30 and 34 (c) 35 and 39 (d) 40 and 44 Let z be any point in A « B « C and let Example 118 w be any point satisfying |w – 2 – i| < 3. Then, |z| – |w| + 3 lies between (a) – 6 and 3 (b) – 3 and 9 (c) – 6 and 6 (d) – 3 and 6 Example 119

If z is any point A « B « C, then

INTEGER-ANSWER TYPE QUESTIONS If a = w + w2 + w4 and b = w2 + w5 + w6 then (a + 1) (b + 1) is equal to Ans. 1 6

Ans.

(b) p/3 (d) p/4

116. (b), 117. (c), 118. (b), 119. (a)

Solution 116. A consists all point in the half plane Imz ≥ B all points on the circle |z – (2 + i)| = 3 and C all the points on the line x + y =

2 . These three curve intersect

a + b = Â wk =

Solution:

k =1

Ê 5 + i – zˆ equals arg Á Ë i – 1 – z ˜¯ (a) p /2 (c) 2p/3

Let w = cos(2p/7) + i sin (2p/7)

Example 120

=

w -w 1- w

w (1 - w 6 ) 1- w

7

But

w7 = cos(2p) + i sin (2p) = 1



a+b=

w -1 =–1 1- w

Also, ab = (w + w2 + w4) (w3 + w5 + w6) 7

= Â wk + 2 k =1

=0+2=2

[using w7 = 1]

IIT JEE eBooks: www.crackjee.xyz Complex Numbers 1.49

Now, (1 + a) (1 + b) = 1 + (a + b) + ab = 1 + (– 1) + 2 = 2 Example 121

Now, |z1| = |z2| fi z1 z1 = z2 z2 2

Also, |z1| = |z1 – z2|

If x ΠC, then

= |z1|2 + |z2|2 – z1 z2 - z1 z2

min (|z –1| + |z + 1| + |z – 3|) is



|z2|2 – z1 z2 - z1 z2 = 0



z2 z1 z2 = ◊ +1 z1 z2 z1



z1 z1 z1 = ◊ + 1 [use (1)] z2 z2 z2



z21 + z22 – z1z2 = 0

Ans. 4 Solution:

min (|z –1| + |z + 1| + |z – 3|)

≥ min {|1 + 1| + |1 – 3|, |– 1 – 1| + |– 1 – 3|, |3 – 1| + |3 + 1|} = 4 Example 122

Let a, b, c ΠC such that a + b + c =

0. If |a| = |b| = |c| = 1, then 3

3

Ê kp ˆ Ê kp ˆ ak = cos Á ˜ + i sin Á ˜ , Ë 7 ¯ Ë 7 ¯

Ans. 0

then the value of the expression

We have

12

|b – c|2 + |b + c|2 = 2(|b|2 + |c|2) fi

|b – c|2 + |– a|2 = 2(1 + 1)



|b – c| =

 | ak +1 - ak |

S=

k =1

is

3

 | a4k -1 - a4k - 2 |

3

k =1

Similarly, |c – a| = |a – b| = \

For any integer k, let

Example 125

3

|a – b| + |b – c| + |c – a| – 3 |a – b| |b – c| |c – a| is equal to

Solution:

(1)

2

Ans. 4

3

|a – b|3 + |b – c|3 + |c – a|3 – 3|a – b| |b – c| |c – a| = 3 3 + 3 3 + 3 3 - 3(3 3 ) = 0

Note that ak = eikp/7 " k ΠN

Solution: fi

ak + 1 – ak = eikp/7 (eip/7 – 1)



|ak + 1 – ak| = |eikp/7| |eip/7 – 1|

If (a, b) lies on the unit circle |z| = 1 Example 123 2 2 and |z1| + |z2| = 3, then |az1 – bz2|2 + |bz1 + az2|2 is equal to

= |eip/7 – 1| Thus, S =

12 | eip / 7 - 1 | 3 | eip / 7 - 1 |

=4

Ans. 3 Solution:

If a1, a2, º, an are real numbers and Example 126 cos a + i sin a is a root of

|az1 – bz2|2 + |bz1 + az2|2

= |az1|2 + |bz2|2 – 2Re (ab z1 z2 ) 2

2

+ |bz1| + |az2| – 2Re (ba z1 z2 ) = (a2 + b2) (|z1|2 + |z2|2) = (1) (3) = 3

[∵ (a, b) lies on |z| = 1]

Let z1, z2 be two distinct non-zero Example 124 complex numbers such that the origin, z1 and z2 form an equilateral triangle, then z21 + z22 – z1z2 is equal to Ans. 0 Solution:

|z1| = |z2| = |z1 – z2|

(1)

then value of |a1 cos a + a2 cos 2a + º + an cos (na)| is Ans. 1. Solution: As a1, a2, º an are real numbers and z = cos a + i sin a

is a root of (1),

z = cos a – i sin a

is also a root of (1).

1 1 = is a root of Thus, cos a – i sin a = cos a + i sin a z (1). Therefore,

We have

|0 – z1| = |z2 – 0| = |z1 – z2| fi

x n + a1 xn – 1 + a2 xn – 2 + º + an – 1 x + an = 0

a a 1 + n 1- 1 + n -2 2 + º + an = 0 n z z z

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fi fi

1 + a1 z + a2 z2 + º + an zn = 0 1 + a1 (cos a + i sin a) + a 2 (cos 2a + i sin 2a) + º + an (cos na + i sin na) = 0 [using DeMoivre’s Theorem] Equating real parts, we get 1 + a1 cos a + a2 cos 2a + º + an cos na = 0 fi | a1 cos a + a2 cos 2a + º + an cos na | = 1.

If the complex number z is such that Example 129 |z – 1| £ 1 and |z – 2| = 1 then maximum possible value |z|2 is Ans. 3

value of 2(z + z ) – |z|2 is

Solution: |z – 1| £ 1 represent the interior and boundary of the circle with centre at 1 + 0i and radius 1 and |z – 2| = 1 represent circle with centre at 2 + 0i and radius 1. Clearly the points z satisfying |z – 1| £ 1 and |z – 2| = 1 lie on the arc DAC. \ OA £ |z| £ OC (= OD)

Ans. 3

As

Solution: We have



OC =

Thus,

|z| £ 3

If z π 0 and 2 + cos q + i sin q = 3/z, then

Example 127

9 = (2 + cosq )2 + sin2 q = 5 + 4 cos q 2 | z|

(1)

–OCB = p/2, OC2 = OB2 – BC2 = 4 – 1 = 3 3

2

and 3 3 = 4 + 2 cos q + z z

(2)

Eliminating q from (1) and (2), we get 9 Ê 1 1ˆ - 6Á + ˜ = – 3 2 Ëz z¯ | z| fi

3 = 2( z + z) – |z|2 Fig. 1.56

If a, b ΠR Example 128 2 3 plex numbers a + ib for which (a + ib) = (a Рib) . Ans. 6

Example 130

If

z - 25 = 5, then |z| is equal to z -1

Ans. 5

Solution: as

Let z = a + ib, the given equation can be written Solution: z2 = z 3

z - 25 =5 z -1



|z|2 = | z |3



|z – 25|2 = 25 |z – 1|2



|z|2 = |z|3



|z|2 – 25z – 25 z + 625 = 25 {|z|2 – z – z + 1}



|z| = 0 or |z|4 = 1



|z| = 0 or |z| = 1



24|z|2 = 600

If |z| = 0, then z = 0



|z| = 5

If |z| = 1, we get z = 1/z and the equation becomes 2

z = fi

1 z3

5

z =1

Example 131

fi |z|2 = 25

If 0 £ arg z £ p/4, then least value of

2 2 |z – i| is Ans. 2 0 £ arg (z) £ p/4 represent the region of the

This equation has 5 roots.

Solution:

Thus, there are 6 complex numbers satisfying the given equation.

the x-axis and the line y = x. The least value of |z – i| is the length of perpendicular from (0, 1) to y = x, which is

1 2

.

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Thus, least value of

2 |z – i| = 1 fi 2 2 | z - i | = 2 fi

w w 1 a 1/a 1 = 0 b 1/b 1



Ê a bˆ Ê 1 1ˆ w Á - ˜ + w(b - a) + Á - ˜ = 0 Ë b a¯ Ë a b¯



w a+b =0 + wab ab



w=

Fig. 1.57

Suppose A1(a1), A2(a2), A3(a3) and Example 132 A4(a4) are four non-collinear points. A1A2 is perpendicular to A3A4 if and only if

a2 - a1 a4 - a3 + =0 a2 - a1 a4 - a3

Ans. 0 Solution:

Let aj = xj + iyj (j = 1, 2, 3, 4)

y - y3 y - y1 and slope of A3A4 is 4 Slope of A2A1 is 2 x2 - x1 x4 - x3 A2A 1 ^ A 3A 4 ¤

Ê y2 - y1 ˆ Ê y4 - y3 ˆ ÁË x - x ˜¯ ÁË x - x ˜¯ = – 1 2 1 4 3

¤

x - x3 i ( y2 - y1 ) = 4 x2 - x1 i ( y4 - y3 )

¤

( x2 - x1 ) + i ( y2 - y1 ) i ( y4 - y3 ) + ( x4 - x3 ) = ( x2 - x1 ) - i ( y2 - y1 ) i ( y4 - y3 ) - ( x4 - x3 )

¤

a - a3 a2 - a1 =- 4 a4 - a3 a2 - a1

¤

a+b-w ab

Also, as PQ ^ AB a-b w- z = 0 [see Example 132] + a -b w- z w- z =0 w- z



– ab +



ab( w - z ) = w – z



Ê a + b - wˆ - abz = w – z ab Á Ë ab ˜¯



a + b – w – abz = w – z



w=



4w =2 a + b + z - abz

1 (a + b + z - abz ) 2

Let a, b, c be three distinct complex Example 134 numbers such that |a| = |b| = |c| = |b + c – a| = 1, then 3 – |b + c| is equal to Ans. 3 Solution:

a2 - a1 a4 - a3 + =0 a2 - a1 a4 - a3

1 = |b + c – a|2

= |b|2 + |c|2 + |a|2 + bc + bc - ab - ab - ac - ac

Let A(a) and B(b) be two distinct Example 133 points on the unit circle. If Q(w) is the foot of perpendicular from point P(z A and B, then 4w is a + b + z - abz P

Solution: As A, Q, B are collinear

c b a b a c + - - - - =0 b c b a c a



2+



2abc + ac2 + ab2 – a2c – b2c – a2b – bc2 = 0



Ans. 2

w w 1 a a 1 =0 b b 1

[∵ aa = bb = 1]

(abc + ac2 – a2b – a2c) + (abc + ab2 – b2c – bc2) = 0



a(c – a) (b + c) + (a – c) b(b + c) = 0



(a – b) (c – a) (b + c) = 0

As a, b, c are distinct, we get b+c=0 A

B Fig. 1.57

Q

\

|b + c| = 0



3 – |b + c| = 3

IIT JEE eBooks: www.crackjee.xyz 1.52 Comprehensive Mathematics—JEE Advanced

Example 135

If iz3 + 3z2 – 9z + 27i = 0, then 3|z|

is equal to Ans. 9 Solution: Divide by i to obtain z3 – 3iz2 + 9iz + 27 = 0 fi z2(z – 3i) + 9i(z – 3i) = 0 fi (z2 + 9i) (z – 3i) = 0 fi z2 = – 9i or z = 3i fi |z| = 3 fi 3|z| = 9

EXERCISE LEVEL 1

SINGLE CORRECT CHOICE TYPE QUESTIONS 1. Let z1, z2 be two non-zero complex numbers such that (|z1| – |z2|)2 + |z1z2| = z1 z2 + z1 z2 then

z2 lies in the interval z1

1 È1 ˘ (a) Í (3 - 5 ), (3 + 5 ) ˙ 2 2 Î ˚ 1 È1 ˘ (b) Í ( 5 - 1), ( 5 + 1) ˙ 2 2 Î ˚

1 f (z1, z2, z3) = sgn z1 z1

|z2 + z3|2 + |z3 + z1|2 + |z1 + z2|2 equal (a) 2|z1 + z2 + z3|2 (b) |z1 + z2 + z3|2 (c) 2|z1 + z2 + z3|2 – |z1|2 – |z2|2 – |z3|2 (d) none of these 7. If |z| = 1, |a| π 1, and x = z/(z – a) (1 – z a ), then (a) x < 0 (b) x > 0 (c) x ≥ 1 (d) 0 < x < 1 8. Let z1, z2 Œ C and x = |z1z2| – Re(z1z2) – +

È1 ˘ (d) Í (3 - 5 ), 1˙ Î2 ˚

(a) x < 0 (c) x ≥ 1

2. Suppose A(a), B(b) lie on the unit circle |z| = 1. If w b + a -w is a complex number such that w = , then ab P(w) lies on (a) chord AB (b) perpendicular bisector of chord AB (c) unit circle |z| = 1 (d) the circle |z| = 1/2 z1 - z2 1 - z1 z2

(a) |z2| = 1 (c) |z2| > 1 4. Let z1, z2, z3 ΠC and

1 z3 z3

x where sgn x = if x π 0 and sgn (0) = 0, then for | x| a π 0, (a) f (z1, z2, z3) = f (az1, az2, az3) (b) f(z1, z2, z3) = f (az2, az1, az3) (c) f(z1, z2, z3) = f(az3, az2, az1) (d) none of these 5. If f is as in Question 4, then for a Œ C, (a) f (z1, z2, z3) = f (z1 + a, z2 + a, z3 + a) (b) f (z1, z2, z3) = |a| f (z1, z2, z3) (c) f(z1, z2, z3) = |a| f (a1, a2, a3) where a1, a2, a3 is a permutation of z1, z2, z3. (d) none of these 6. If z1, z2, z3 are three complex numbers, then

È 1 ˘ (c) Í1, (3 + 5 ) ˙ Î 2 ˚

3. If |z1| π 1,

1 z2 z2

= 1, then (b) |z2| = 0 (d) 0 < |z2| < 1

1 | z1 – z2|2 2

1 (|z2| – |z1|)2 then 2

9. If the ratio

(b) x = 0 (d) 0 < x < 1 1- z is purely imaginary, then 1+ z

(a) |z| £ 1/2 (b) |z| = 1 (c) |z| > 1 (d) 0 < |z| < 1 10. If |a| π |b|, then the equation az + b z + c = 0 represents (a) a circle (b) an ellipse (c) a straight line (d) a point 11. If |a| = |b| and ac π bc , then the equation az + bz + c = 0 has (a) no solution (b) exactly one solution

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12. If |a| = |b| π 0 and ac = bc , then az + bz + c = 0 represents (a) a circle (b) an ellipse (c) a straight line (d) a point 13. The equation z = z 0 + A(z – z0), where A is a constant, represents (a) straight line (b) a circle (c) a point (d) none of these 14. If m is slope of the straight line in Question 13, then (a) A =

1 + im 1 - im

(b) A =

1 - im 1 + im

(c) A =

i+m 1 - mi

(d) A =

i-m 1 + mi

15. If q is the angle which the line of Question 13 makes with the positive direction of real axis, then (b) A = e–2iq (a) A = e2iq (c) A = eiq (d) A = e–iq 16. The equation z = a+

r2 , z-a

r>0

represents (a) an ellipse (b) a parabola (c) a circle (d) a straight line through point a 17. If a π b, then the equation z z + az + bz + c = 0 represents (a) a circle (b) an ellipse (c) a straight line C 2 18. If a = b, c Œ R and |b| > c, then zz + az + b z + c = 0 represents (a) a circle (b) a parabola (c) a straight line 19. If a = b, c Œ R, and |b|2 < c, then the equation z z + az + b z + c = 0 (a) has no solution (b) exactly two solutions (d) none of these 20. If z1 + z2 + z3 = 0, |z1 – z2|2 equals

then

|z2 – z3|2 + |z3 – z1|2 +

(a)

1 |z1|2 + 2|z2|2 + 2|z3|2 3

(b)

2 (|z1|2 + |z2|2 + |z3|2) 3

(c) 2 (|z1|2 + |z2|2 + |z3|2) (d) 3 (|z1|2 + |z2|2 + |z3|2) 21. If |z1| = |z2| = |z3| = |z4| = 1, and z1 + z2 + z3 + z4 = 0, then least value of the expression E = |z1 – z2|2 + |z2 – z3|2 + |z3 – z4|2 + |z4 – z1|2 is (a) 6 (b) 8 (c) 10 (d) 12 22. If z1 + z2 + z3 + z4 = 0, then the expression |z1 – z2|2 + |z2 – z3|2 + |z3 – z4|2 + |z4 – z1|2 – 2(|z1|2 + |z2|2 + |z3|2 + |z4|2) is equal to 0, if and only if, (a) z1 = – z3 and z4 = – z2 (b) z1 = – z4 and z2 = – z3 (c) z1 = z2 and z3 = z4 (d) z1 = z3 and z2 = z4 u-v 23. If |u| = |v| = 1, uv π – 1, and z = , then 1 + uv (a) |z| = 1 (c) Im(z) = 0

(b) Re(z) = 0 (d) Re(z) = Im(z)

u-v , then least value of 24. If |u| < 1, |v| < 1, and z = - uv 1 |z| is (a)

|u| - | v| 1 + |u| |v|

(b)

|u| + | v| 1 - |u| |v|

(c)

| |u| - | v| | 1 - |u| | v|

(d) none of these

25. Let L be the straight line a z + a z + c = 0, a π 0, c Œ R, and OA O and A(a), then (a) L || OA (b) L ^ OA (c) L makes an angle of p /4 with OA (d) none of these 26. Locus of point z so that z, i, and iz are collinear, is (a) a straight line (b) a circle (c) an ellipse (d) a rectangular hyperbola 27. The number of roots of the equation |z2| – 5|z| + 1 = 0 is (a) 0 (b) 2

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28. The set of points in C satisfying the inequality |arg (z) – p /2| < p /2 is given by (a) {z : Re(z) > 0} (b) {z : In(z) < 0} (c) {z : Im(z) > 0} (d) {z : Re(z) = Im(z)} 29. If n ≥ 3 and 1, a1, a2, ... an–1 are nth roots of unity, then value of (a) 0 (c) –1

Â

ai a j 1£ i < j £ n -1

equation |z| = 2 and |z| = |z – 1| is (a) 0 (b) 1

is

(b) 1 (d) (–1)n

0, z, z + iz and iz is (a) parallelogram (b) rhombus (c) rectangle (d) square 31. If |z – 2| = 2|z – 1|, then 3|z|2 – 4Re(z) equals (a) 0 (b) –1 (c) 1 (d) none of these 32. Sum of the common roots of z2006 + z100 + 1 = 0 and z3 + 2z2 + 2z + 1 = 0 is (a) 0 (b) – 1 (c) 1 (d) 2 33. If z1 and z2 lie on the same side the line a z + a z +b = 0, where a Œ C, a π 0, b Œ R, then the ratio a z1 + a z1 + b is a z2 + a z2 + b (a) purely imaginary (b) a positive real number (c) a negative real number (d) none of these 34. If z1 and z2 are the roots of the equation z2 + az + b = 0, then prove that the origin, z1 and z2 form an equilateral triangle if and only if (b) b2 = 3a (a) a2 = 3b (c) a2 + 3b = 0 (d) b2 + 3a = 0 35. If a > 0 and z|z| + az + 2a = 0 then z must be (a) purely imaginary (b) a positive real number (c) a negative real number (d) 0 3 = a + ib, then (a – 2)2 + b2 36. If 2 + cos q + i sin q equals (a) 0 (c) – 1

37. If z = 4 + i 7 , then value of z3 – 4z2 – 9z + 91 equals (a) 0 (b) 1 (c) – 1 (d) 2

(b) 1 (d) 2

39. If z =

mz1 + z2 , then distance of point z from the m +1

line joining z1 + 1 and z2 + 1 is (a) 0 (b) 1 (c)

2m m +1

(d)

m m +1

40. If z1, z2, z3 – 7z2 + 3z3 = 0, then z1, z2, z3 are (a) vertices of a scalane triangle (b) vertices of a right triangle (c) points on a circle (d) collinear points

z1

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS z has modulus 1 and argument p /3, then z2 + z (a) is purely imaginary (b) has modulus 3 (d) none of these 42. If z1 = a + ib and z2 = c + id numbers such that | z1| = | z2 | = 1 and Re (z1 z2) = 0, w1 = a + ic and w2 = b + id satisfy (a) |w1| = 1 (b) |w2| = 1 (d) Re( w1 w2) = 0 (c) |w1 w2 | = 1 43. If 2 cos q = x + (a) xn + (b)

1 1 and 2 cos j = y + , then y x

1 = 2 cos (nq) xn

x y + = 2 cos (q – j) y x

(c) x y +

1 = 2 cos (q + j) xy

(d) none of these

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44. If z1, z2 relation

50. Which of the following value(s) of k, the equation z1 + z2 z1 - z2

then

=1

z (a) 1 is purely imaginary z2 z1 >0 z2 (c) z1 = ikz2

(b)

(d) –z1 O z2 = p/2 45. Points A, B z1, z2 and (1 – i) z1 + iz2 are the vertices of (a) an isosceles triangle (b) an equilateral triangle (c) a right triangle (d) an obtuse angled triangle 46. If | z1 – z2 | = | z1| + | z2|, then Êz ˆ p (a) arg Á 1 ˜ = 2 Ë z2 ¯ Êz ˆ (b) arg Á 1 ˜ = p Ë z2 ¯ (c) z1 z2 + z1 z2 £ 0 (d) z1, O, z2 lie on a straight line. 47. If z1 lies on | z| = 1 and z2 lies on |z| = 2, then (a) 3 £ | z1 – 2z2| £ 5 (b) 1 £ | z1 + z2 | £ 3 (c) | z1 – 3z2 | ≥ 5 (d) |z1 – z2| ≥ 1 48. If z

z – 1 – 2i | £ 1, then

(a) min (arg (z)) = tan

–1

Ê 3ˆ ÁË ˜¯ 4

p z)) = 2 (c) min (|z|) = z|) =

3z - 1 - i =k z+2+i

5 –1 5 +1

49. If O, z1 and z2 are vertices of an equilateral triangle, then (a) |z1| = |z2| = |z1 – z2| (b) (z1 + z2)2 = 3z1z2 (c) |arg (z1) – arg (z2)| = p /3 (d) |z1 + z2| = 2|z1| + |z2|

represents a circle? (a) 3 (b) 2 (c) 5 (d) p 51. Let z Œ C and x = |z – 1| + |z| + |z + 1|, then (a) minimum value of x is 2 (b) minimum value is attained on at least three points (c) minimum value of x number of points (d) x is attained on the circle | z| = 2 52. If z1, z2 and z3 are vertices of an equilateral triangle whose orthocentre is, the origin, then (a) z1 + z2 + z3 = 0 (b) |z1| = |z2| = |z3| (c) z2z3 + z3z1 + z1z2 = 0 (d) O z1 is perpendicular to z2z3. z2 through 53. If the point z1 the line b z + b z = c, where b π 0, then bz1 + bz2 is (a) proportional to |z1|2 – |z2|2 (b) equal to c (c) inversely proportional to c (d) none of these and touching 2 |z| = 1 is (a) x + y = 1 (b) x – y = – 1 (c) x – y = 1 (d) x + y = – 1 55. If z1, z2 Œ C are such that z1 - z2 = 1, 1 - z1 z2 then (a) |z1| = 1 (b) |z2| = 1 (d) z2 = e if, f Œ R (c) z1 = eiq, q Œ R 56. If z1 and z2 |z1 + z2|2 = |z1|2 + |z2|2, then (a) z1 z2 is purely imaginary (b) z1/z2 is purely imaginary (c) z1 z2 + z1 z2 = 0 (d) O, z1, z2 are vertices of a right triangle 57. If z2/(z – 1) is always real, then z can lie on (c) a circle

(d) a parabola

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58. If z1 and z2 c > 0, then |z1 – z2|2 ≥ (1 – c) |z1|2 + (1 – 1/c) |z2|2 is (a) true for all z1, z2 Œ C (b) true if z1 = z2 (c) false for all z1, z2 Œ C (d) true if z1/z2 is real 59. If z Œ C, the minimum value of |z| + |z – 5| is attained at (a) z = 0 (b) z = 5 (c) z = 5/2 (d) for all z Œ [0, 5] 60. The curve represented by 3 ,qŒR z= 2 + cos q + i sin q

z| as 3 (d) has minimum value of |z| as 1 z + 1| = z + 2 + 2i (a) 1 + i (c)

1 – 2i 2

1 (b) - – i 2 (d) 0

z1, z2, z3 are the vertices of a z which makes the triangle a parallelogram is given by (b) z1 + z3 – z2 (a) z1 + z2 – z3 (c) z2 + z3 – z1 (d) z1 + z2 + z3 Ê n n zj ˆ 63. If z1, z2, …, zn lie on |z| = r and Re Á Â Â ˜ = 0, Ë j = 1 k = 1 zk ¯ then n

(a)

Â

n

zj = 0

j =1 n

(c)

Â

j =1

(b)

 zj

=0

j =1

1 =0 zj

(d) none of these

64. Let S z such that log1/3 (log 1/2 (| z|2 + 4|z| + 3)) < 0, then S is contained in (a) (0, 1) (b) {z | Re(z) > 0} (c) {z | Re(z) > 3} (d) f 65. If z ΠC and log|z + i| |3 + 4i| < log|z + i| |5 + 12i|, then z lies. (a) inside a circle passing through the origin (b) outside a circle passing through the origin (c) inside the disc |z + i| < 5 (d) outside the curve z = Рi + eiq, q ΠR

MATRIX-MATCH TYPE QUESTIONS 66. Match the curve on the right on which the roots of the equation in on the left lie. Column 1 Column 2 n n (p) x = 0 (a) z = (1 – z) n n (b) (1 – z) = (1 + z) (q) x = 1/2 (c) zn = 2 + 21 i

(r) circle with radius 5

(d) (z – i)n = 21 + 2i

(s) circle with centre at z = i 67. a, b are cube roots of i which are not purely imaginary. Column 1 Column 2 2 2 (p) – 1 (a) a + b (b) a4 + b 4 (q) 2 (c) a6 + b6 (r) 1 12 12 (d) a + b (s) – 2 68. If a, b Œ R and a2 – 4b ≥ 0, the equation z4 + az2 + b = 0 has Column 1 Column 2 (a) four real roots if (p) a = 0, b = 0 (b) two real and two (q) a > 0, b > 0 imaginary roots if (c) four imaginary roots if (r) a £ 0, b ≥ 0 (d) four equal roots

(a) (b) (c) (d)

Column 1 Column 2 –1 (p) z π 1 sin |z| (q) 0 < |z| £ 1 tan–1 |z| (r) |z| £ 1 log |z| + cos–1 |z| log |z – 1| + sec–1 |z| (s) |z| ≥ 1

70. Centre of circle Column 1 (a)

(s) a £ 0

z−9 =3 z −1

⎛ 1 + it ⎞ (b) z = 5 + 3i + 2 ⎜ , ⎝ 1 − it ⎟⎠ tŒR (c) passing through

Column 2 (p) 5 + 3i (q) 1 + 4i

(r) 0

3, – 3i , 3i (d) having join of 3 + 5i and – 1 + 3i as diameter

(s) 1

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71. If a, b are roots of z2 sin2 q – z sin 2q + 1 = 0 where 0 < q < p/2 and n is an integer, then value of Column 1 Column 2 (p) 2n cotn q (a) (ab)n (q) cosec2n q (b) (a + b )n n n (r) 2 cosecn q cos (nq) (c) a + b (d) |a|n + |b|n (s) 2 cosecn q z x 72. For z = x + iy e = e (cos y + i sin y). The set z such that Column 1 (a) ez = 2 (b) ez = 1 – 3i

Column 2 (p) {2 + 2npi | n ΠI} (q) {log 2 + i(2n Р1/3)p | n ΠI} (r) {(2n + 1)pi | n ΠI} (s) {log 2 + 2np i | n ΠI}

(c) ez = e2 (d) ez = – 1

ASSERTION-REASON TYPE QUESTIONS 73. Circum-centre of D ABC with vertices A(z1), B(z2) and C(z3) is S(z0). A to side BC meets the circumcircle at E(z4). A(z1)

B(z2)

S(z0) D

E(z4) Fig. 1.58

z0 –

76. Statement-1: If z 2z + 3 = 3z + 1

that

5 , then z lies on a circle. 3

Statement-2: If a, b k > 0, k π 1, then z+a =k z+b represents a circle. 77. Statement-1: If z 2 – z + 1 = 0 and n is a natural number, then n

Â(z

k

k =1

+ z–k

)

2

Èn˘ = n + 3Í ˙ Î3˚

where [x] denotes the greatest integer £ x. Statement-2: If w π 1 is a cube root of unity, then Ï0 if k is not a multiple of 3 (a) w k + w k = Ì Ó2 if k is a multiple of 3 78. Statement-1: Two lines az + az + b = 0 = 0, a1 z + a1 z + b1 = 0 (where a, a1 Œ C, a, a1 π 0 and a b, b1 Œ R) are parallel if and only if is real. a1

C(z3)

Statement-1: z1 + z2 + z3 – 2z0. Statement-2: z4 is given by

Statement-2: If z is any point on the circle with A1A2 as diameter, then ( z – z1 )( z – z2 ) + ( z – z1 )( z – z2 ) = 0

DABC is

( z0 – z2 )( z0 – z3 ) z0 – z1

74. Statement-1: Let a, b, c Œ R and z1, z2 Œ are comaz1 z1 + b(z1 z2 + z1 z2) + cz2 z2 is purely real. Statement-2: If a, b, c Œ R and a œ R is a root of ax2 + bx + c = 0, then a(a + a ) + b = 0 and aa a = c. 75. Let A1 and A2 represent z1 and z2. Statement-1: Equation of circle with A1A2 as diameter is |2z – z1 – z2| = |z1 – z2|

Statement-2: Two lines az + az + b = 0 , a1 z + a1 z + b1 = 0 (where a, a1 Œ C, a, a1 π 0 and b, a b1 Œ R) are perpendicular if and only if is purely a1 imaginary. 79. Statement-1: The minimum value of f (q) =

2i 3 – ieiq

is

1 2

Statement-2: f (q) =

2i is 1. 3 – ieiq

80. Statement-1: Let m, n be two natural numbers such that m = a2 + b2 and n = c2 + d2 for some a, b, c, d ΠI e, f ΠI such that mn = e2 + f 2 Statement-2: If z1, z2 ΠC, then |z1 + z2| = |z1| + |z2| if and only if z1z2 > 0.

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COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 81 to 85 For z = x + iy, x, y Œ R ez = ex (cos y + i sin y), 1 (eiz – e–iz) sin z = 2i cos z =

and

(b) |z| (d) none of these

90. log (–i) equals (a) pi/2 (c) – p i/2

(b) pi (d) 0

Paragraph for Question Nos. 91 to 95

1 iz (e + e–iz) 2

(a) 0

z for which ez = 0 is (b) 1

(a) 0

z for which sin z = 0 is (b) 1 z for which cos z = 0 is (b) 1

(a) 0

89. elog z equals (a) z (c) z + 2kp i, k ΠI

If a1, a2, ..., a m are the roots of polynomial equation f (x) = a0 xm + a1 xm–1 + ... + am–1 x + am = 0, then f (x) = a0(x – a1) ... (x – a m). f ¢( x) 1 = + f ( x) x - a1

and

1+

1 1 + + 2 - w2 2-w

+

1 2 - w n -1

equals (d) none of these (a)

z

85. e z1 = e 2 if and only if z1 and z2 differ by (a) np (b) 2np (c) (2n + 1)p /2 (d) none of these where n ΠI.

Paragraph for Question Nos. 86 to 90 zπ log z = log |z| + i (arg z) where – p < arg (z) £ p i.e. arg (z) stands for the principal argument of z. For

87. For z, z¢ π 0, log (zz¢) and log z + log z¢ differ by a multiple of (a) np (b) 2np (c) 0 (d) (2n + 1)p /2, where n Œ I. (b) x + iy (d) none of these

n ( 2n - 1 ) 2n + 1

(c) 0

(b)

n ( 2n - 1 ) 2n - 1

(d) 1

Êpˆ Êpˆ 92. If a = cos Á ˜ + i sin Á ˜ , then Ë n¯ Ë n¯ (x – 1) (x + 1) (x – a) (x – a ) ... (x – a n–1 ) (x – a equals (b) xn + 1 (a) xn – 1 (d) x2n – 1 (c) x2n + 1 93. ( x 2 - 2 x cos

86. log z = 1 if and only if z equals (a) e (b) e, – e (c) e + 2kpi, k Œ I (d) ± e + 2kpi, k Œ I

88. zlog (ex + iy) equals (a) ex + arg (x + iy) (c) x + iy + 2kp, k ΠI

1 x - am

Ê 2p ˆ Ê 2p ˆ 91. If w = cos Á ˜ + i sin Á ˜ , then numerical value Ë n ¯ Ë n ¯ of

84. Solution set of sin (iy) = 0 is (a) {np | n Œ I} (b) {0} p Ï ¸ (c) Ì(2n - 1) , n Œ I ˝ 2 Ó ˛

+

n–1

)

p 2p + 1) ( x 2 - 2 x cos + 1) n n n -1 º ( x 2 - 2 x cos p + 1) equals n

1 + x + . . . + xn–1 1 – x + x2 – . . . + (–1)n–1 xn–1 1 + x + ... + x2n–2 0 p 2p 94. 2n–1 (cos q - cos ) (cos q - cos ) n n (a) (b) (c) (d)

. . . (cos q - cos

n -1 p ) equals n

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sin q sin(nq )

(a)

(b)

sin n q sin q

(d) – n cos q

(c) 1 95. 2n–1 sin

p 2p sin n n

sin

n -1 p equals n

(a) n (c) 2n

(b) 0 (d) –n

and

and

a = w + w 2 + w4

b = w 3 + w 5 + w 6.

Ê 99 ˆ ˆ is sin Á p Ë 100 ˜¯ ˜¯

107. If 8iz3 + 12z2 – 18z + 27i = 0, then 4|z|2 is 108. If |z + 4| = 3, then greatest possible value of |z – 7|/7 is

1

and

a +w

2

+

1 b+w

2

+

1 c+w

2

=

2 w2

then the value of

96. a + b equals (a) 0 (b) –1 (c) –2 (d) 1 97. ab equals (a) –1 (b) 0 (c) 1 (d) 2 98. a and b are roots of the equations (b) x2 + x + 2 = 0 (a) x2 + x + 1 = 0 2 (c) x + 3x + 5 = 0 (d) none of these 99. 2a equals

1 1 1 is + + a +1 b +1 c +1 110. If a, b are the roots of x2 – 2x + 4 = 0, then the value of (a 6 + b 6)/64 is 111. If the points a(cos a + i sin g), b(cos b + i sin b) and c(cos g + isin g ) are collinear, then value of |z| where z = bc sin (b – g ) + ca sin (g – a) + ab sin (a – b) + 3i is

(a) –1 + 7i

(b) –1 – 7i

112. If z1, z2, z3, z4 lie on the circle |z| = 3, then

(c) 1 + 7i

(d) 1 – 7i

2 + |z1 + z2 + z3 + z4| – 9

6

100.

1 Ê 100 Ê p ˆ Ê 2p ˆ 2 sin Á sin Á Ë 100 ˜¯ Ë 100 ˜¯ 50 ÁË

109. If w π 1 is a cube root of unity, and 1 1 1 2 + + = a +w b+w c+w w

Paragraph for Question Nos. 96 to 100 ⎛ 2π ⎞ ⎛ 2π ⎞ Let w = cos ⎜ ⎟ + i sin ⎜ ⎟ ⎝ 7 ⎠ ⎝ 7 ⎠

106. The value of

∑ωk

2

equals

k =0

LEVEL 2

(a) i

(b)

7i

(c) – i

(d) – 7i

INTEGER-ANSWER TYPE QUESTIONS 101. If

3iz2 3 z + 7 z2 is purely real, then value of 5 1 is 5 z1 3z1 - 7 z2

102. If –3 + x2yi and x2 + y + 4i are conjugate of each other, then |x| + |y| is 103. If z + 2 |z + 1| + i = 0, then |z|2 is 104. If |z| < 1/2 and 4|(1 + i)z3 + iz| < k, then the least integral value of k is 105. If z = cos q + i sin q is a root of n

x + p1 x where

1 1 1 1 + + + is equal to z1 z2 z3 z4

n–1

+ p2 x

n–2

+ ... + pn = 0,

p1, p2, ..., pn ΠR,

then the value of 7 (p1 sin q + p2 sin 2q + ... + pn sin (nq)) is

SINGLE CORRECT ANSWERS TYPE QUESTIONS 1. If |z1| = |z2| = |z3| = 1 and z1 + z2 = 0, then z1, z2, z3 are the vertices of a triangle which is (a) right angled (b) isosceles (c) obtuse angled (d) equilateral 2. If z1, z2, z3 are the vertices of an isosceles triangle with right angle at z2, then value of (z1 – z2)2 + (z2 – z3)2 is (a) 0 (c) –1

(b) 1 (d) 3i

3. If |z z + 2|, |z – 4|}, then z lies on (a) circle (b) a straight line (c) a pair of circles (d) a pair of straight lines

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4. If roots of the equation z2 + a z + b = 0 (a, b Œ C) are real, then value of (Im b)2 + (Im a) (Im (a β )) is (a) 0 (b) 1 (c) –1 (d) i 5. If roots of the equation z2 + az + b = 0 (a, b Œ C) are purely imaginary, then value of (Im b)2 – (Re a) (Re (a β )) is (a) 0 (b) 1 (c) –1 (d) i 2 6. If roots of the equation z + az + b = 0 lie on |z| = 1, then (a) 2 |Im a| = 1 – |b |2 (b) 2 |Im a| = |b|2 – 1 (c) Im a = 0 (d) none of these 7. If z1 = 1 + i and z2 = 1 – i are the two vertices of an (a) ± 3

(b) 1 ± 3

3 ±1

(c)

(d) none of these

8. If a π 0, |a| π 1, then all the circles passing through a and

1 , α

(a) are concentric with the circle |z| = 1 (c) are orthogonal to the circle |z| = 1 (d) none of these 9. If Re(z) is a positive integer, then value of the |1 + z + ... + z n| cannot be less than 1 1 (a) |z|n – (b) |z|n + | z| | z| (c) n|z|n

(d) n|z|n + 1

(a)

1+ x (x Œ R) 1− x

(b)

(c)

i+x (x Œ R) 1− x

(d) none of these

1 + ix (x Œ R) 1 − ix

11. If x2 + x + 1 = 0, then value of 3

3

1⎞ ⎛ 2 1 ⎞ ⎟ + ⎜ x + 2 ⎟⎠ + x⎠ ⎝ x

is (a) 100 (c) 50

1 ⎞ ⎛ + ⎜ x100 + 100 ⎟ ⎝ x ⎠ (b) 197 (d) 97

(a) 7 (b) 5 (c) 3 (d) 1 13. If w π 1 is a cube root of unity, then value of (x + y)2 + (xw + yw2)2 + (xw2 + yw)2 is (a) 3xy (c) 4xy

3

(b) 6xy (d) xy

14. If z ΠC, then

2z − i = m, m Œ R represents a 5z + 1

straight line if (a) m = 2 (b) m = 5 (c) m = 2.5 (d) m = 0.4 15. If w π 1 is a cube root of unity, then roots of (x – 2i)3 + i = 0 are (a) i, i, i (b) 3i, 2i + iw, 2i + iw2 (c) 2i, 2i, 2i (d) i, i + w, i + w2 16. The minimum value of |z – 3| + |z + 5| is (a) 1 (b) 4 (c) 6 (d) 8 Ê 3 iˆ 17. If z = Á + ˜ Ë 2 2¯

2009

Ê 3 iˆ +Á - ˜ Ë 2 2¯

2009

, then

(a) Im(z) = 0 (b) Re(z) > 0 (c) Im(z) > 0 (d) Re(z) < 0, Im(z) > 0 18. If |z| = 3, then

10. If |z| = 1, z π i then z can be written in the form

⎛ ⎜⎝ x +

12. If |z1| = |z2| = |z3| = 1 and |z2 – z3|2 + |z3 – z1|2 + |z1 – z2|2 = 9, then value of 12z1 + 3z2 + 2z3| is

(a) z

9+ z equals 1+ z (b) z

(c) 3z (d) z + z 19. If p, q, r are positive integers, then remainder when x3p + x3q +1 + x3r +2 is divided by x2 + x + 1 is (a) 0 (b) 1 (c) x + 1 (d) x – 1 20. If z lies on the curve |z – 3 – 4i| = 3, then least value of |z| is (a) 2 (b) 3 (c) 8 (d) none of these

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MULTIPLE CORRECT ANSWERS TYPE QUESTIONS (z – 1) (cos a – i sin a) + (z – 1)–1 (cos a + i sin a) is zero, then (a) |z| = 1 (c) arg (z) = a

(b) |z – 1| = 1 (d) arg (z – 1) = a

22. Two lines az + az + c = 0, bz + bz + d = 0 where a, b ΠC Р{0} and c, d ΠR, are (a) parallel iff ab = ab (b) parallel iff

a is real b

25. Let C be the circle |z – z1| = k|z – z2| where z1, z2 Œ C, z1 π z2,

k > 0, k π 1. Ê z - z1 ˆ =0 (a) centre of C lies on the line Re Á Ë z2 - z1 ˜¯

(b) radius of C is

and

k | z2 - z1 | |1 - k 2 |

(c) mid point of the segment joining z1 z2 lies on the circle (d) none of these 26. Let c > 0. For the ellipse |z – ci| + |z + ci| = 4c (a) eccentricity is 1/2 z) = 4ci

(c) perpendicular iff ab + ba = 0 (d) perpendicular iff

a is imaginary b

23. Let z1, z2 Œ C, z1 π z2. (a) Equation of straight line through z1 and z2 is Ê z - z1 ˆ Im Á =0 Ë z2 - z1 ˜¯ (b) Equation of circle with z1 z2 as diameter is |2z – z1 – z2| = |z1 – z2| (c) Equation of any circle passing through z1 and z2 Ê z - z1 ˆ |2z – z1 – z2| – |z1 – z2| + l Im Á =0 Ë z2 - z1 ˜¯ where l Œ R. (d) none of these 24. Let two given circles be C1 : zz + az + az + c = 0 and

C2 : zz + bz + bz + d = 0

where a, b ΠC

and

c, d ΠR.

(a) Radius of circle C1 is | a |2 - c (b) C1 and C2 |a – b| = | a |2 - c + | b |2 - d (c) C1 and C2 will intersect orthogonally iff ab + ab = c + d C1 and C2 is

27. ABC is an equilateral triangle whose circumcentre is at the origin. Let the point A be represented by –1 +

3i .

(a) length of side of DABC is 2 3 (b) centroid of DABC is –1 + i (c) orthocentre of DABC is O (d) in-radius of DABC is 1. 28. If z1 and z2 z so that z - z1 = e iq, z - z2 is (a) (b) (c) (d)

q ΠR,

a circle a semicircle a straight line perpendicular bisector of segment joining z1 and z2.

29. Which of the following represent a straight line through z1 and z2? (a) z = z1 + t (z2 – z1), t Œ R Ê z - z1 ˆ (b) Im Á =0 Ë z2 - z1 ˜¯ (c)

z z1 z2

(d)

z - z1 = e iq, q ΠR z - z2

(a − b ) z + (a – b) z + c – d = 0

z 1 z1 1 = 0 z2 1

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(c) G and L are the roots of t 2 – 10t + 36 = 0.

⎛ z⎞ 30. If a π 0, then Re ⎜ ⎟ = 1 represents ⎝ a⎠

(r) G – L = 2

(d) G geometric series

(a) a straight line (b) rectangular hyperbola

5+

(c) a hyperbola with eccentricity 2 . (d) part of circle with centre at a.

31. If w π 1 is a cube root of unity, a, b Œ R then value of Column 2

(a) (a + b) (aw + bw2)

(p) 6ab

(aw2 + bw) (b) (a + b)2 + (aw + bw2)2 + (q) a3 + b3 + (aw2 + bw)2

1 – 3ab

(c) (a + b + 1) (a + bw + w ) 2

(a + bw2 + w)

(s)

ASSERTIONS-REASON TYPE QUESTIONS 34. Statement-1: Let ABC be an equilateral triangle and P be a point on the circumcircle of DABC, then PA2 + PB2 + PC2 is independent of the position of point P. Statement-2: Let A, B, C be three points in a plane such that CA2 + CB2 = AB2/2, then DCAB is an isosceles triangle. 35. Statement-1: Centre of circle

z +1 5 = 2 is z –1 3

Statement-2: Radius of circle

z –1 4 = 2 is z +1 3

(r) a3 + b3

(d) (a + bw + cw 2)3 +

(s) (2a – b – c)

(a + bw2 + cw)3

G 3 = L 2

and L = G – 2.

MATRIX-MATCH TYPE QUESTIONS

Column 1

5 5 5 + 2 + 3 + 6 6 6

(2b – c – a)

36. Let ABC be a triangle and P be any point in the

(2c – a – b) 32. If w = −

Statement-1: BC2 + CA2 + AB2

1 3i , then value of + 4 4

Column 1

£ 3(PA2 + PB2 + PC2) Column 2

(

)

(a) 1 – w + w2 – w3 + ...

(p)

1 5 + 3i 7

(b) 1 – w2 + w4 – w6 + ...

(q)

2 9 − 3i 21

(c) 1 + w + w2 + w3 + ...

(r) 1 −

(d) 1 + w2 + w4 + ...

(s)

Column 1 33. (a) G is the greatest and L be the least value of |z – 2| if z |z + 2 + 3i| £ 1 (b) G is the greatest and L is least value of |z | when z |z – 5i| £ 1

(

1 3

= 3(PA2 + PB2 + PC2) ¤ P is centroid of D ABC.

)

i

(

2 7 + 3i 13

)

L + G = 10

LG = 24

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 37 to 40

Column 2

(p)

Statement-2: BC2 + CA2 + AB2

For

z = x + iy, x, y ΠR, ez = ex (cos y + i sin y),

1 (e z – e–z) 2 1 (e z + e–z ) cosh z = 2 37. For z Œ C, conjugate of e z equals sinh z =

(a) e–z

(b) e z

(c) e− z (d) ez 38. (cosh z)2 + (sinh z)2 equals (a) 1 (b) –1 (c) 0 (d) cosh (2z)

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39. cosh z = sinh z holds if z equals (a) (2n Р1)p /2 (b) 2np (c) (2n + 1)pi (d) none of these 40. sinh z = 0 if z equals (a) npi (b) (2n Р1)pi (c) (2n Р1)pi (d) (2n Р1)pi/2, n ΠI

Paragraph for Question Nos. 41 to 44 If z1, z2

A AB as m =

and B

and b ΠR is (a)

a a

a (c) a 42. If m and m¢ lines, then (a) mm¢ = –1 (c) m + m¢ = 0

z1 − z2 . z1 − z2

az + az + b = 0 where a ΠC

(b)

a a

(b) mm¢ = –1

(c) mm¢ = 1

(d) m = m¢ eip /2

44. If a line makes an angle q (π p/2) with the positive direction of the x (a) e2iq (c) ie2iq

(b) e–2iq (d) none of these

INTEGER-ANSWER TYPE QUESTIONS gate of their own cube is 46. If w π 1 is a cube root of unity, and a + b = 11, a3 + b3 = 1001, then value of (aw2 + bw) (aw + bw2)/13 is 47. If |z1| = |z2| = |z3| and arg (z2/z1) = 60∞, then the de-

a (d) – a

(b) m = m¢ e (d) mm¢ = 1

43. If m and m¢ then (a) m = m¢

Êz -z ˆ gree measure of arg Á 3 1 ˜ 10 is Ë z2 - z1 ¯ 48. Let eccentricity of ellipse with focii z1 and z2 be 1/3. If |z1 – z2| = 5 and z lies on the ellipse, then z – z1|/2 is

p i/2

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS

(a) The x (b) The straight line y = 5 (c) The circle passing through the origin

1. If the cube roots of unity are 1, ω, ω then the roots of the equation (x – 1)3 + 8 = 0 are (a) – 1, 1 + 2ω, 1 + 2ω2 (b) – 1, 1 – 2ω, 1 – 2ω2 (c) – 1, – 1, – 1 Ê1 + iˆ 2. The smallest positive integer n for which Á Ë 1 - i ˜¯ = 1 is (a) 8

n

(b) 16 [1980] z = x + iy, which satisfy the

z – 5i = 1, lie on equation z + 5i

5

5

Ê 3 iˆ Ê 3 iˆ 4. If z = Á + ˜ +Á – ˜ then 2¯ 2¯ Ë 2 Ë 2 (a) (b) (c) (d)

Re(z) = 0 Im(z) = 0 Re(z) = 0, Im(z) > 0 Re(z) > 0, Im(z) < 0.

[1982]

5. The inequality |z – 4| < |z – 2| represents the region given (a) Re(z) 0 (b) Re(z) < 0 (c) Re(z) > 0 (d) none of these. [1982] 6. If z = x + iy and w =

1 – iz then |w| = 1 implies that, z–i

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(a) z (b) z (c) z lies on the unit circle (d) none of these. [1983] 7. The points z1, z2, z3, z4 vertices of a parallelogram if and only if (a) z1 + z4 = z2 + z3 (b) z1 + z3 = z2 + z4 (c) z1 + z2 = z3 + z4 (d) none of these [1983] 8. If a, b, c; u, v, w senting the vertices of two triangles such that c = (1 – r) a + rb, w = (1 – r)u + rv where r is a (a) have the same area (c) are congruent

(b) are similar (d) none of these. [1985]

9. If z1 and z2 that |z1 + z2| = |z1| + |z2|, then Arg z1 – Arg z2 is equal to (a) –π (b) –π/2 (c) 0 (d) π/2 (e) π [1987] 2p k 2p k ˆ Ê ÁË sin 7 – i cos 7 ˜¯ k =1

Â

(b) 0 (c) – i (e) none of these [1987] x + i cos 2x and cos x – i sin 2x are conjugate to each other, for (a) x = nπ (b) x = 0 (c) x = (n + 1/2)π (d) no value of x [1988] 12. Let α and β be the roots of the equation x2 + x + 1 = 0. The equation whose roots are α19, β 7 is (a) – 1 (d) i

(a) x2 – x – 1 = 0 (c) x2 + x – 1 = 0

(b) x2 – x + 1 = 0 (d) x2 + x + 1 = 0 [1994] 13. If ω is an imaginary cube root of unity, then the value of

}

w2

–1 w2 – 1 – i +w –1 –1

= 1– i –i equals (a) 0 (c) i

(b) 1 (d) ω

[1995]

18. For positive integers n1 and n2 the values of the z = (1 + i ) n1 + (1 + i 3 )n1 + (1 + i 5 )n2 + –1 is real if and only if

(b) n1 = n2 – 1 (d) n1 > 0, n2 > 0 [1996] 19. If ω is an imaginary cube roots of unity, then (1 + ω – ω2)7 equals (a) 128 ω (c) 128 ω 2

(a) - 3 2

(b) -1

(c) 1

(d)

(b) –128 ω (d) –128 ω2 13

20. The value of the sum

 (i

n

+ i n + 1 ), where

–1 equals

i=

(a) i (c) – i 21. If

6i – 3i = 4 3i 20 3

(b) i – 1 (d) 0

[1998]

1 –1 = x + iy, then i (b) x = 1, y = 1 (d) x = 0, y = 0 [1998]

2

3 2

[1998]

n =1

(a) x = 3, y = 1 (c) x = 0, y = 3

sin (w 10 + w 23 )p – p / 4 is

2

1+ i + w2

1

[1995]

(a) n1 = n2 + 1 (c) n1 = n2

6

{

(a) 1 or i (b) i or – i (c) 1 or – 1 (d) i or – 1 17. If ω ( 1) is a cube root of unity, then

(1 + i 7 ) n2 when i =

10. The value of S=

15. Let z and w that |z| = |w| and Arg z + Arg w = π, then z equals (a) w (b) –w (d) – w [1995] (c) w 16. Let z and w z| 1, |w| 1 and |z + iw| = |z – i w | = 2 then z equals

[1994]

14. If ω ( 1) is a cube root of unity and (1 + ω)7 = A + Bω then A and B are respectively (a) 0, 1 (b) 1, 1 (c) 1, 0 (d) – 1, 1 [1995]

22. Let A0 A1A2A3A4A5 in a circle of unit radius. Then the product of the lengths of the line segments A0A1, A0A2 and A0A4 is (a) 3/4

(b) 3 3

(c) 3

(d) 3 3 4

[1998]

IIT JEE eBooks: www.crackjee.xyz Complex Numbers 1.65

– 1 , then

23. If i =

(b) –

(a) 0

Ê –1 i 3ˆ z = 4 + 5Á + 2 ˜¯ Ë 2

334

Ê –1 i 3ˆ + 3Á + 2 ˜¯ Ë 2

365

is equal to

(a) 1 – i 3

(b) – 1 + i 3

(c) i 3

(d) – i 3

[1999]

(b) – π (d) π/2.

[2000]

25. If z1, z2, z3 are complex numbers such that |z1| = |z2| = |z3| = 1,

1 1 1 + + = 1, then z1 z2 z3

[2003]

(b) 3 (d) 6

[2004]

QAR is an 32. PQ and PR arc. The set of points lying in the shaded region excluding the boundary is

(b)

(b) less than 1 (d) equal to 3 [2000]

2 | z + 1|2

(d)

31. If ω ( 1) be a cube root of unity and (1 + ω2)n = (1 + ω4)n, then the least positive value of n is

(a)

|z1 + z2 + z3| is (a) equal to 1 (c) greater than 3

z 1 ◊ z + 1 | z + 1|2

(a) 2` (c) 5

24. If arg (z) < 0, then arg (– z) – arg (z) equals (a) π (c) – π/2

(c)

1 | z + 1|2

(c) (d)

26. Let z1 and z2 be nth roots of unity which subtend a right angle at the origin, then ‘n’ must be of the form

{ z : z + 1 > 2; arg( z + 1) < p /2} { z : z + 1 < 2; arg( z + 1) < p /4} { z : z – 1 > 2; arg( z – 1) < p /4} { z : z – 1 < 2; arg( z – 1) < p /2}

[2005]

,

(a) 4k + 1 (b) 4k + 2 (c) 4k + 3 (d) 4k [2001] 27. The complex number z1, z2 and z3 satisfying z1 – z3 1 – i 3 = are the vertices of the triangle z2 – z3 2 which is (a) of area zero (b) right angled isosceles (c) equilateral (d) obtuse angled isosceles 28. Let ω = – 1

= 1 –1–w 1 (a) 3ω (c) 3ω2

,

[2001]

1 3 +i . Then the value of 2 2 1 w

2

1 2

3

(d) 1/2

34. If the complex number u =

4

(b) 3ω (ω – 1) (d) 3ω (1 – ω) [2002]

29. For all complex numbers z1, z2 satisfying |z1| = 12 and |z2 – 3 – 4i| = 5, the minimum value of |z1 – z2| is (a) 0 (c) 7

33. The minimum value of |a + bω + cω2|, where a, b and c are integers, not all equal and (ω 1) is a cube of root of unity, is (a) 1 (b) 0 (c)

w2 w

A

(b) 2 (d) 17

z –1 (where z 30. If |z| = 1 and w = z +1 Re (w) is

[2002] –1), then

where w = α + iβ, β values of z is (a) {z : |z| = 1} (c) {z : z 1}

0 and z

[2005]

w – wz is purely real 1– z 1, then the set of the

(b) {z : z = z } (d) {z : |z| = 1, z 1}. [2006] 35. A man walks a distance of 3 units from the origin towards the north-east (N 45ºE) direction. From there, he walks a distance of 4 units towards the north-west (N 45ºW) direction to reach a point P. Then the position of P in the Argand plane is

IIT JEE eBooks: www.crackjee.xyz 1.66 Comprehensive Mathematics—JEE Advanced

(a) 3eiπ/4 + 4i (c) (4 + 3i)eiπ/4

(b) (3 – 4i)eiπ/4 (d) (3 + 4i)eiπ/4

respectively. If z0 = x0 + iy0 2|z0|2 = r2 + 2, then | | =

[2007] z 36. If |z| = 1 and z 1, then all the values of ω = 1 – z2 lie on (a) a line not passing through the origin (b) |z| = 2 (c) the x y 37. A particle P starts from the point z0 = 1 + 2i, where i = -1 by 5 units and then vertically away from origin by 3 units to reach a point z1. From z1 the particle moves 2 units in the direction of the vector i + j and p then it moves through an angle in anticlockwise 2 direction on a circle with centre at origin, to reach a point z2. The point z2 is given by (a) 6 + 7i (b) –7 + 6i (c) 7 + 6i (d) –6 + 7i [2008] 38. Let z = cos θ + i sin θ. Then the value of 15

 Im ( z 2m – 1 )

m =1

1 sin 2∞

(b)

1 (c) 2 sin 2∞

1 3 sin 2∞

1 (d) 4 sin 2∞

[2009]

39. Let z = x + iy x and y are integers. Then the area of the rectangle whose vertices are the roots of the equation zz + zz = 350 3

(d)

3 4

[2012]

1 lie on circles a (x – x0)2 + (y – y0)2 = r2 and (x – x0)2 + (y – y0)2 = 4r2, α and

2 1 7

(b)

1 2

(d)

1 3

[2013]

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 1. If z1 = a + ib and z2 = c + id such that |z1| = |z2| = 1 and Re( z1 z2 ) = 0, then the w1 = a + ic and w2 = b + id (a) |w1| = 1

(b) |w2| = 1

(c) Re( w1w2 ) = 0

(d) none of these [1985]

2. Let z1 and z2 z1 z2 and |z1| = |z2|. If z1 has positive real part and z2 has z + z2 may be negative imaginary part, then 1 z1 – z2 (b) real and positive (d) purely imaginary [1986]

3. Let z1 and z2 let z = (1 – t) z1 + tz2 for some real number t with 0 < t < 1. If Arg (w) denotes the principal argument w, then (a) | z – z1 | + | z – z2 | = | z1 – z2 | (b) Arg (z – z1) = Arg (z – z2)

3

is (a) 48 (b) 32 (c) 40 (d) 80 [2009] 40. Let z part of z is nonzero and a = z2 + z + 1 is real. Then a cannot take the value. 1 (a) – 1 (b) 3 1 (c) 2

(c)

1

(a) zero (c) real and negative (e) none of these

at θ = 2° is (a)

(a)

(c)

z – z1 z2 – z1

z – z1 z2 – z1

=0

(d) Arg (z – z1) = Arg (z2 – z1) 5. Let w = Further

[2010]

3+i and P = {wn : n = 1, 2, 3, …}. 2 H1

=

1¸ Ï Ì z Œ : Re z > ˝ and 2˛ Ó

H2

=

-1 ¸ Ï Ì z Œ : Re z < ˝ , where  is the set of all 2˛ Ó z1 P H1, z2 P H2 and O represents the origin, then z1Oz2 =

IIT JEE eBooks: www.crackjee.xyz 1.67

p p (b) 2 6 2p 5p (c) (d) [2013] 3 6 6. Let a, b Œ R and a2 + b2 π 0. Suppose S = 1 ¸ Ï , t Œ R, t π 0˝ , where i = -1 . Ìz ŒC : z = a + ibt ˛ Ó If z = x + iy and z ŒS, then (x, y) lies on (a)

(a) the circle with radius

1 Ê 1 ˆ and centre Á , 0˜ Ë 2a ¯ 2a

for a > 0, b π 0 (b) the circle with radius -

1 Ê 1 ˆ and centre Á - , 0˜ Ë 2a ¯ 2a

for a < 0, b π 0 a π 0, b = 0 a = 0, b π 0

(c) the x (d) the y

[2016]

MATRIX-MATCH TYPE QUESTIONS 1. z Column 1 (a) Re z = 0 π (b) Arg z = 4

Column 2 (p) Re z2 = 0 (q) Im z2 = 0 (r) Re z2 = Im z2

[1992]

2. Match the statements in Column-I with those in Column-II. [Note: Here z the real part of z.]

z and Re z denote, respectively, the imaginary part and Column I

(a) The set of points z satisfying | z – i | z | | = | z + i | z | | is contained in or equal to (b) The set of points z satisfying | z + 4 | + | z – 4 | = 10 is contained in or equal to 1 (c) If | w | = 2, then the set of points z = w – w is contained in or equal to 1 (d) If | w | = 1, then the set of points z = w + w is contained in or equal to

Column II (p) an ellipse with eccentricity

4 5

(q) the set of points z satisfying Im z = 0 (r) the set of points z satisfying | Im z |

1

(s) the set of points z satisfying | Re z |

2

(t) the set of points z satisfying | z | ⎛ 2k π ⎞ ⎛ 2k π ⎞ ; k = 1, 2, ..., 9 + i sin ⎜ 3. Let zk = cos ⎜ ⎝ 10 ⎟⎠ ⎝ 10 ⎟⎠ List I P. For each zk zj such that zk . zj = 1 Q. k {1,2, , 9} such that z1 z = zk has no solution z R.

|1 - z1| |1 - z2 | 10

|1 - z9 |

equals

2k p ˆ 9 S. 1 -  k = 1 cos ÊÁ equals Ë 10 ˜¯ P Q R S (a) 1 2 4 3 (c) 1 2 3 4

3 [2010]

List II 1. True 2. False

3. 1 4. 2 P (b) 2 (d) 2

Q 1 1

R 3 4

S 4 3

[2014]

IIT JEE eBooks: www.crackjee.xyz 1.68 Comprehensive Mathematics—JEE Advanced

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 1 to 3. Let A, B and C below. A = {z : Imz 1] B = {z : |z – 2 – i| = 3

Then the value of

C = {z : Re ((1 – i)z) = 2 }. 1. The number of elements in the set A B C is (a) 0 (b) 1 (c) 2 (d) 2. Let z be any point in A B C. Then |z + 1 – i|2 + |z – 5 – i|2 lies between (a) 25 and 29 (b) 30 and 34 (c) 35 and 39 (d) 40 and 44 3. Let z be any point in A B C and let w be any point satisfying |w – 2 – i| < 3. Then, |z| – |w| + 3 lies between (a) –6 and 3 (b) –3 and 6 (c) –6 and 6 (d) –3 and 9 [2008]

S2

S3, where

È z - 1 + 3i ˘ ÔÏ Ô¸ S1 = {z ∈  : | z | < 4}, S2 = Ì z Œ : Im Í ˙ > 0˝ ÔÓ Ô˛ Î 1 - 3i ˚ and S3 = {z

4. Area of S = 10p 3 16p (c) 3 5. min zŒS |1 - 3i - z | =

(b) (d)

20p 3 32p 3

(a)

2- 3 2

(b)

2+ 3 2

(c)

3- 3 2

(d)

3+ 3 2

| a |2 + | b |2 | + | c |2

is

[2011]

⎛ kπ ⎞ ⎛ kπ ⎞ 3. For any integer k, let α k = cos ⎜ ⎟ + i sin ⎜ ⎟ , ⎝ 7⎠ ⎝ 7⎠ where i =

–1.

12

∑ α k +1 – α k

k =1 3

is

[2015]

∑ α 4k –1 – α 4k –2

k =1

FILL

IN THE

z=

BLANKS TYPE QUESTIONS

sin( x /2) + cos( x /2) – i tan( x) 1 + 2i sin( x /2)

z1, z2 and any real numbers a and b, |az1 – bz2|2 + |bz1 + az2|2 = _______ [1988] 3. If a and b are real numbers between 0 and 1 such that the points z1 = a + i, z2 = 1 + bi, z3 = 0 form an equilateral triangle, then a = _______ and b = _______ [1989]

 : Re z > 0}.

(a)

| x |2 + | y |2 | + | z |2

is real, then the set of all possible values of x is _______ [1987]

Paragraph for Questions 4 and 5 Let S = S1

a+b+c =x a+b +c 2 =y a + b 2 + c = z.

4. If α, β, γ are the cube roots of p, p < 0, then for any x, y and z, xa + y b + zg = _______ xb + yg + za

[2013]

INTEGER-ANSWER TYPE QUESTIONS 1. If z z – 3 – 2i | 2, then the minimum value of | 2z – 6 + 5i | is [2011] * i /3 2. Let w = e , and a, b, c, x, y, z numbers such that * There is a misake in the question. However it has been solved by taking = e2 i/3

[1989]

5. ABCD is a rhombus. Its diagonals AC and BD intersect at the point M and satisfy BD = 2AC. If the point D and M i and 2 – i respectively, then A number _______ or _______ [1993] 6. Suppose z1, z2, z3 are the vertices of an equilateral triangle inscribed in the circle |z | = 2. If z1 = 1 + i 3 then z2 = _______ and z3 = _______ [1994] 1(2 – ω) (2 – ω2) + 2 (3 – ω) (3 – ω2) + ... + (n – 1) (n – ω) (n – ω2), where ω is an imaginary cube root of unity, is _______ [1996]

IIT JEE eBooks: www.crackjee.xyz 1.69

TRUE/FALSE TYPE QUESTIONS we write z1

z12 + z22 – z1z2 = 0.

z1 = x1 + iy1 and z2 = x2 + iy2, z2, if x1 x2 and y1 y2. Then for all z with 1

z, we have

1– z « 0. 1+ z [1981]

z1, z2 and z3 represent the vertices of an equilateral triange such that |z1| = |z2| = |z3| then z1 + z2 + z3 = 0. [1984]

4. The cube roots of unity when represented on Argand diagram form the vertices of an equilateral triangle. [1988]

SUBJECTIVE-TYPE QUESTIONS 1. If x = a + b, y = aα + bβ, z = aβ + bα where α, β are xyz = a3 + b3. [1978] 1 in the form A + iB. (1 – cos q ) + 2i sin q [1978] 3. If x + iy =

a + ib a 2 + b2 prove that (x2 + y2)2 = 2 . c + id c + d2 [1978]

4. Find the real values of x and y for which the following (1 + i ) x – 2i (2 – 3i ) y + i =i [1980] + 3+i 3–i 5. It is given that n is an odd integer greater than 3, but n is not a multiple of 3. Prove that x3 + x2 + x is a factor of f(x) = (x + 1)n – xn – 1. [1980] z1, z2 and z3 be the vertices of an equilateral triangle. If z0 be the circumcentre 2 2 2 2 of the triangle, then prove that z1 + z2 + z3 = 3 z0 . [1981] 7†. A relation R z1Rz2 if and only if

z1 – z2 is real z1 + z2

show that R is an equivalence relation.

[1982]

z1, z2 and the origin form an equilateral triangle only if † The word non-zero was missing from the Original Question.

[1983]

9. If 1, a1, a2, .... an – 1 are the n roots of unity then show that (1 – a1) (1 – a2) .... (1 – an – 1) = n. [1984] 10. Show that the area of the triangle on the Argand z, iz and 1 2 z + iz is |z| . [1986] 2 z1, z2, z3 are the vertices A, B, C respectively of an isosceles right angled triangle with right angle at C show that [1986] (z1 – z2)2 = 2(z1 – z3) (z3 – z2) 12. Let z1 = 10 + 6i and z2 = 4 + 6i. If z number such that the argument of (z – z1)/(z – z2) is π/4, then prove that |z – 7 – 9i| = 3 2

[1990]

13. Prove that k

S=

 (– 3)

r – 1 3n

( C2 r – 1 ) = 0

r =1

where k = (3n/2) and n is an even positive integer. [1993] 3

2

14. If iz + z – z + i = 0 then show that |z| = 1 [1995] 15. Let z and w |z| 1, |w| 1 then show that |z – w|2

(|z| – |w|)2 + |Arg z – Arg w|2

[1995] z satisfying [1996]

z = iz2.

17. Let z1 and z2 be roots of the equation z2 + pz + q = 0, p and q bers. Let A and B represent z1 and z2 plane. If AOB = α 0 and OA = OB, where O is the origin, prove that p2 = 4q cos2 (α/2). [1997] 18. Let bz + bz = c, b plane, where b b, if a point z1 z2 through the line, then show that c = z1b + z2b . [1999] n –1

19. Prove that

 (n – k ) cos

k =1

an integer.

2k p n = , where n n 2

3 is

[1997]

IIT JEE eBooks: www.crackjee.xyz 1.70 Comprehensive Mathematics—JEE Advanced

z and w, prove that |z|2w – |w|2z = z – w if and only if z = w or zw = 1. [1999] α, α 1 be a root of the equation z – z – z + 1 = 0, when p, q are distinct primes. Show that either 1 + α + α + ... α p–1 = 0 or 1 + α + α2 + ... α q–1 but not both together. p+q

p

q

[2002] 22. If z1 and z2 |z1| < 1 < |z2| then prove that

49. 51. 53. 55. 57. 59. 61. 63. 65.

1 – z1 z2 < 1. [2003] z1 – z2 z such

24. If k > 0, k

1, show that

25. |z – 1| =

[2003]

z –a = k represents a z–b

circle. Find its centre and radius.

[2004]

Answers

(a) (d) (a) (c) (b) (b) (a) (b)

2. 7. 12. 17. 22. 27. 32. 37.

(a) (b) (c) (d) (a) (d) (b) (c)

3. 8. 13. 18. 23. 28. 33. 38.

(a) (b) (a) (a) (b) (c) (b) (c)

4. 9. 14. 19. 24. 29. 34. 39.

(a) (b) (b) (a) (c) (b) (a) (b)

5. 10. 15. 20. 25. 30. 35. 40.

(a) (d) (b) (d) (b) (d) (c) (d)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 41. 43. 45. 47.

(a), (a), (a), (a),

(b), (c) (b), (c) (c) (b), (c), (d)

42. 44. 46. 48.

(a), (a), (b), (a),

(b), (c), (c), (b),

(c), (d) (d) (d) (c), (d)

(c), (d) (c), (d)

(b), (c) (d)

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

68.

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 6. 11. 16. 21. 26. 31. 36.

50. 52. 54. 56. 58. 60. 62. 64.

q

67.

2 is a circle inscribed in a square whose i 3 . Find the remaining vertices. [2005]

LEVEL 1

(c) (c)

p

66.

Â

1 ar z r = 1 where |ar| < 2. and 3 r =1

(b), (b), (b) (b), (c) (b),

(b), (a), (a), (a), (a) (a), (a), (d)

(c), (b), (b), (b),

(d) (c), (d) (c), (d) (c), (d)

(b), (c), (d) (b), (c)

MATRIX-MATCH TYPE QUESTIONS

n

that |z|
0 we get

8. x = |z1 z2| −

z=

L.H.S. = 0 and 12. az + bz + c = 0

2

1 z2 + a z2 + a

az +bz+c =0

(|a|2 – |b|2)z = bc − ac

= 1 + | z1 |2 |z2|2 – z1 z2 – z1z2

1 5. D = z1 + a z1 + a

(1)

11. From problem 10, we get

|z1|2 + |z2|2 – z1 z2 – z1 z2

1 az2 a z2

10. az + bz + c = 0

we get

3. |z1 – z2|2 = |1 – z1 z2|2

1 4. az1 a z1

2(1 - | z |2 ) (1 − z ) (1 + z ) + (1 − z ) (1 + z ) = |1 + z |2 (1 + z ) (1 + z )

Multiplying (1) by a and (2) by b and subtracting,

fi w lies on chord AB. fi

1− z 1− z + 1+ z 1+ z

1 (z1z2 + z1 z2 ) 2

1 ⎡| z1 |2 + | z2 |2 − z1 z2 − z1 z2 2⎣

Az + Az + α = 0 where A = a + b and a = c + c Œ R and thus represents a straight line. 13. A =

z - z0 z - z0

fi | A| = 1

Also, Az – z + z0 – Az0 = 0 fi A( z0 - A z0 ) z - ( z0 - Az0 ) z + z0 - A z0 2 = 0 fi ( Az0 - z0 ) z + ( Az0 - z0 ) z + z0 - A z0

2

= 0 (1)

As it is of the form αz + αz + c = 0 where a = Az0 - z0 and c Œ R, the given equation represents a straight line. 14. If m is slope of the line (1) in problem 13, then – m=

Re(a ) Im(a)



im = a + a a -a

IIT JEE eBooks: www.crackjee.xyz 1.74 Comprehensive Mathematics—JEE Advanced

24. Let u = r1 e if, v = r2 eiq

A z0 - z0 1 + im α 1 =− = = z0 - A z 0 1 − im α A



1 – u v = 1 – r1 r2 e

i(q – f)

|1 – u v|2 = (1 – r1 r2 cos (q – f))2

[∵ | A| = 1] 15. A =

r1 < 1, r2 < 1.

where

1 - i tan q cos q - i sin q = 1 + i tan q cos q + i sin q

+ r12 r22 sin 2 (q - f ) = 1 – 2r1r2 cos (q – f) + r12 r22

= (cos q – i sin q)2 = e–2iq

£ (1 – r1r2)2 = (1 – |u| |v|)2

16. As |z – a|2 = r2 or | z – a| = r fi z lies on a circle with centre at a and radius r.



|1 – u v| £ 1 – |u| |v|

17. zz + az + bz + c = 0



1 1 £ |1 - u v | 1 - |u| |v|

(1)

zz + a z + bz + c = 0



(2)

Also,

||u| – |v|| < |u – v|

Thus,

u-v || u | - | v || £ 1 – uv 1 - | u || v |

Subtracting (2) from (1) we get (a − b ) z + (b − a ) z + c − c = 0 fi

z =

25. Let a = a + ib, z = x + iy, so that

(c − c ) − ( a − b ) z b−a

az + az + c = 0

Substituting this in (1) we get a quadratic equation in z. 18. Putting a = b and c = c in (2) above we get zz + bz + bz + c = 0 fi

This is a circle with centre at – b and radius 19. If |b| < c,

(3) has no solution.

20. |z2 – z3|2 + |z3 – z1|2 + |z1 – z2|2 = 2[|z1|2 + |z2|2 + |z3|2] – z1( z2 + z3 ) – z2 ( z3 + z1 ) – z3 ( z1 + z2 ) = 2[|z1|2 + |z2|2 + |z3|2] + z1 z1 + z2 z2 + z3 z3 = 3(|z1|2 + |z2|2 + |z3|2) 21. E = 2(|z1|2 + |z2|2 + |z3|2 + |z4|2) – z1 (z2 + z4) – z2 (z1 + z3) – z3 (z2 + z4) – z4 (z1 + z3) = 8 + |z1 + z3|2 + |z2 + z4|2 ≥ 8 E – 2(|z1|2 + |z2|2 + |z3|2 + |z4|2) = |z1 + z3|2 + |z2 + z4|2 = 0 z1 + z3 = 0

β α

(3) | b |2 − c .

As m1 m2 = – 1, OA ^ L z z 1 z −i z +i 1 0 1 26. 0 = i −i 1 = 0 iz −i z 1 iz − i −i z + i 1 fi (z – i) ( z – 1) + (z – 1) (z + i) = 0 fi 2z z – (1 – i)z – (1 + i) z = 0 fi z lies on a circle 27. |z|2 – 5|z| + 1 = 0 fi

|z| =



z=

5 ± 21 2 1 (5 ± 21)eiθ 2

where q ΠR. 28. 0 < arg (z) < p

22. From problem 21,

¤

Let

(1)

α m 1 = slope of (1) = − β

Also, m2 = slope OA =

|z + b|2 = |b|2 – c

2

becomes 2(a x + by) + c = 0

and

z2 + z4 = 0

u-v u -v 23. 2Re(z) = + 1 + uv 1 + u v v-u u-v = + =0 1 + uv 1 + u v

fi Im(z) > 0 29. As 1, a1, a2, . . . a n–1 are roots of xn – 1 = 0, sum of the products taken two at a time is 0. fi

n -1

 1(a k ) +  a ia j = 0

k =1



i< j

Ê n -1

ˆ

 a ia j = - Á  a k ˜ = 1 Ë k =1 ¯ i< j

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30. The point O(0), A(z), B(z + iz), C(iz) are vertices of a square as (i) the mid point of OC = the mid point of AB (ii) OB = AC = 2 |z| (iii) OA = OC = |z| 31. |z – 2|2 = 4|z – 1|2 fi

|z|2 – 2z – 2 z + 4 = 4(|z|2 – z – z + 1)



3|z|2 – 2(z + z ) = 0



3|z|2 – 4Re(z) = 0



38. |z| = |z – 1| represent the straight z = 1/2. This line meets the circle |z 1 15 ± i. 2 2

39. The line joining z1 + 1 and z2 + 1 is parallel to the line joining z1 and z2 and is at distance 1 from it.

2

(z + 1) (z + z + 1) = 0 z = –1, w, w2

\ z is at a distance one from the line joining z1 + 1 and z2 + 1

out of these w and w satisfy 2

z2006 + z100 + 1 = 0

40. As z2 =

33. Produce the join of z1 and z2 to meet az + az + b = 0 in z. Suppose z divides the join of z1 and z2 nally in the ratio k : 1where k > 0, so that z - kz2 z= 1 1- k As z lies on az + az + b = 0,

4 z1 + 3z3 7

z2 divides the join of z1 and z3 in the ratio 3 : 4. 41. z = cos (p /3) + i sin (p/3) z2 = cos (2p /3) + i sin (2p /3) = −

az1 + az1 + b = k >0 az2 + az2 + b

z2 = cosf + i sinf

and

a = cos q, b = sin q c = cos f, d = sin f z1 z2 = cos (q – f) + i sin (q – f)

z1, z2 form an equilateral triangle ¤

02 + z12 + z22 = 0(z1 + z2) + z1z2

Re( z1 z2 ) = 0 fi

2

(z1 + z2) = 3z1z2

¤ a = 3b 35. z(|z| + a|) = – 2a − 2a fi z= | z| + a 2

=

cos (q – f) = 0

q – f = p /2

\

z1 = – sin f + i cos f



a = – sin f = – d, w1 = a + ic = – sin f + i cos f w2 = b + id = cos f + i sin f |w1| = |w2| = 1

Also,

(1 + 2 cos q ) + 4 sin q 2

( a - 2)2 + b2 =



b = cos f = c

As a > 0, |z| ≥ 0, z is real and negative. (1 + 2 cos θ ) + 2i sin θ 36. (a – 2) + ib = − (2 + cos θ ) + i sin θ fi

)

42. Let z1 = cos q + i sin q

34. z1 + z2 = – a, z1 z2 = b ¤

(

1 1 − 3i 2

\ z2 + z = 3i

a ( z1 - kz2 ) + a( z1 - kz2 ) + b(1 - k ) = 0 fi

Dividing z3 – 4z2 – 9z + 91 by z2 – 8z + 23 we get z3 – 4z2 – 9z + 91 = (z + 4) (z2 – 8z + 23) –1 = –1

viz.

32. (z + 1) (z2 – z + 1) + 2z(z + 1) = 0 fi

37. (z – 4)2 = – 7 fi z2 – 8z + 23 = 0

2

(2 + cos q )2 + sin 2 q 5 + 4 cos q =1 5 + 4 cos q



|w1w2| = 1

Re( w1 w2) = 0

43. 2 cos q = x + 1/x fi

x = cos q ± i sin q

Similarly

y = cos f ± i sin f

xn +

1 xn

= cos (nq) + i sin (nq) + cos (nq) – i sin (nq)

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= 2 cos (nq) x = cos (q – f) + i sin (q – f) y \ Lastly, fi

48. Refer Fig. 1.59

x y + = 2 cos (q – f) y x xy = cos (q + f) + i sin (q + f) xy +

1 = 2 cos (q + f) xy Fig. 1.59

44. |z1/z2 + 1| = |z1/z2 – 1| fi z1/z2 lies on the perpendicular bisector of –1 and 1 fi z1/z2

Min (arg z) = a where y = (tan a) x is tangent to the circle (x – 1)2 + (y – 2)2 = 1

fi z1 = ikz2 for some k ΠR

\

Also, the join of O and z1 makes an angle of p /2 with the join of O and z2.



45. AC = |– iz1 + iz2| = |z1 – z2| = AB Also,

BC = |(1 – i) (z1 – z2)| =

2 | AB.

|z1 – z2|2 = |z1|2 + |z2|2 + 2|z1| |z2| fi z1z2 + z1 z2 £ 0 Also, since (1)

we get z1, z2, O are collinear. In view of (1), z1 and z2 lie on a straight line on a straight line opposite side of O and Êz ˆ arg Á 1 ˜ = p Ë z2 ¯

Also, Min (|z|) = OD z|) = OB where ODB pass through the centre of (1). 5-1

and OB = OP + 1 = 5 + 1 49. (a) is true as sides an equilateral of triangle are equal. For (b) note that 02 + z12 + z22 = 0(z1 + z2) + z1z2 fi (z1 + z2)2 = 3z1z2 As –z1 Oz2 = p /3, (c) is true. 50. For all values of k k = 3. In case k = 3, the equation represents a straight line. 51. x = |1 – z | + | z + 1| + |z| ≥ |(1 – z) + (z + 1)| + |z | ≥ 2 + |z| ≥ 2

z2 = 2(cos f + i sin f)

This value is attained at z = – 1, 0, 1.

fi |z1 – 2z2| = (cos q – 4 cos f) + (sin q – 4 sin f) 2

2

= 1 + 16 – 8 cos (q – f) fi

(tan a – 2)2 = sec2a

z) = p /2

47. z1 = cos q + i sin q 2

=1

OD = OP – 1 =

fi |z1|2 + |z2|2 – z1 z2 – z1 z2 = |z1|2 + |z2|2 + 2|z1z2|

|z1 – z2| = |z1 – 0| + |z2 – 0|

1 + tan 2 a

fi tan2a – 4 tana + 4 = sec2a fi tana = 3/4 \ Min (arg z) = tan–1 (3/4)

Thus, A, B, C are vertices of an isosceles right triangle. 46.

tan a - 2

(1)

9 £ |z1 – 2z2|2 £ 25

Similarly, 1 £ |z1 + z2| £ 3, |z1 – z2| ≥ 1 and |z1 – 3z2| ≥ 5.

52. In an equilateral triangle orthocentre, circumcentre and centroid coincide. For (c), recall for an equilateral triangle (z1 + z2 + z3)2 = 3(z2z3 + z3z1 + z1z2) 53. If z is any point bz + bz = c, then |z – z1| = |z – z2| fi

( z2 - z1 ) z + ( z2 - z1 ) z + | z1 |2 - | z2 |2 = 0

This line is identical to the line

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63. Let

bz + bz - c = 0

n

z2 - z1 z2 - z1 | z2 |2 - | z1 |2 = = b b c

\

Ê n ˆ Ê n zj ˆ = Á Â zk ˜ Á Â 2 ˜ Ë k =1 ¯ Ë j =1 r ¯

55. |z1 – z2|2 = |1 – z1 z2|2 ¤ ¤

|z1| + | z2| – z1 z2 - z1 z2 = 1 + | z1 z2 | - z1 z2 - z1 z2 2

2

2

(|z1| – 1) (|z2| – 1) = 0 2

56. |z1 + z2| = |z1| + |z2|

2

¤

z1 z2 + z1 z2 = 0

¤

z1 |z2|2 is purely imaginary z2

¤ z1 z2 is purely imaginary

(1)

Ê 1 ˆ c | z1 | Á |z | Ë c 2 ˜¯

)

|z1 – z2| ≥ (1 – c) |z1| + (1 – 1/c) |z2| 2

= 0fi

n

r2

= 0fi

n

1

Âz

=0

k =1 k

|z|2 + 4|z

65. log |z + i| (5) < log |z + i| (13) fi 66. (a)

|z + i| > 1 |z|n = |1 – z|n



|z| = |1 – z|

fi z lies on x = 1/2 (c) |z|n = 5



|z| = 51/n

(d) |z – i| = 51/n 67.

z3 = i = (–i)3 fi z = –i, – iw, –iw2 Let a = –iw, b = – iw2

68. (a) Both the roots of x2 + ax + b = 0 are non-negative if a £ 0, b ≥ 0

=1

fi 4|z| – 6z – 6z + 9 = |z|

=0

k =1

Âz



2

2

2

n

 zk

Similarly for (b)

59. Since |z| + |5 – z| ≥ |z + (5 – z)| = 5, minimum value of |z| + |z – 5| is 5. 3 60. cos q + i sin q = –2 z | z |2

=0

64. log1/2 (|z|2 + 4|z| + 3) > 1

Thus, from (1)



n

 zk

k =1 k

£ c|z1|2 + (1/c) |z2|2

3 - 2z

n

 zk

2

58. |z1 – z2|2 = |z1|2 + |z2|2 – 2Re(z1 z2 )

2

¯

k =1



zz (z – z ) – (z – z ) = 0 (zz – z – z ) (z – z ) = 0 z lies on circle z z – z – z = 0 or z lies on

(

j =1

k =1

k =1

But Re(z1 z2 ) £ |z1 z2 | =

ˆ

Âz j ˜

 zk

r2

Re S = S = 0 ¤



z2 ( z – 1) = z 2 (z – 1)

¤ ¤ ¤

n

Thus, S is a real number, so

z2 z2 57. = z -1 z -1

2

[∵ z z j = r]

2

n

1

=

z As 1 is purely imaginary, Oz1z2 is a right triangle. z2

¤

1 Ê n ˆÊ Á  zk ˜ Á r 2 Ë k =1 ¯ Ë

=

2

2

Ê n ˆÊ n 1ˆ = Á  zk ˜ Á  ˜ Ë k =1 ¯ Ë j =1 z j ¯

zj

j =1 k =1 k

54. Each of the given line is tangent to x2 + y2 = 1/2

2

n

ÂÂz

S=

(b) x2 + ax + b = 0 has one non-negative and one negative root if b £ 0

2

fi |z|2 – 2z – 2z + 3 = 0 fi |z – 2|2 = 1 or |z – 2| = 1 61. As L.H.S. is real, Im(z) = – 2 62. Complete parallelogram taking BC, CA and AB as diagonals.

(c) x2 + ax + b = 0 has both negative roots if a > 0, b > 0 69.

(a) sin–1 |z (b) tan

–1

|z

z with |z| £ 1. z Œ C.

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z| > 0 and cos–1 |z| for |z|

(c) log |z £1

z – 1| > 0 and

(d) log |z sec–1 |z| 70.

|z| ≥ 1.

for

z-9 =3 z -1

(a)

As AD ^ BC,

¤

(b) |z – (5 + 3i)| =



z4 – z1 z – z1 + =0 z2 – z3 z2 – z3



w4 – w1 w – w1 + =0 w2 – w3 w2 – w3

|z| = 3 2 1 + t2

=2

z4 – z1 is purely imaginary. z2 – z3

(c) 3, – 3i and 3i are equidistant from 1.

where wi = zi – z0 for i = 1, 2, 3, 4 As z1, z2, z3, z4 lie on a circle with centre at z0 |zi – z0| = r

(d) Mid point of the diameter is the centre



1 + t2

a, b = cot q ± i

71.

for i = 1, 2, 3, 4.

ab = cosec2q,

Also,

wi wi = r 2

a + b = 2 cot q w – w1 r 2 w – r 2 w1 + 2 =0 w2 – w3 r w2 – r 2 w3

a + b = 2 cosec q cos (nq) n

n

n

|a| n + |b|n = 2 cosecnq

and

72. (a) e x cos y = 2, e x sin y = 0 fi sin y = 0 and cos y > 0 fi Also, ex = 2 fi x = log2 Thus, z = log2 + 2npi, n ΠI

y = 2np, n ΠI



w – w1 w2 – w3



w= –

(b) ex cos y = 1, exsin y = - 3 Square and add to obtain e2x = 4 fi ex = 2





y = 2np – p /3

\

z = log2 ± i (2np – p/3), n Œ I.

For statement-2 see Theory in chapter 2. 75. Use Pythagoras theorem and statement-2. 5 2z + 3 2 z + 3 2 = = 3 3z + 1 3 z + 1 3

76.

(c) ex cos y = e2, ex sin y = 0 Squaring and adding e

2x

=e fix=2

\ cos y = 1, sin y = 0 Thus,



y = 2np

z = 2 + 2np i, n ΠI

x

(d) e cos y = –1, ex sin y = 0 Square and add to obtain x = 0 cos y = –1, sin y = 0



y = (2n + 1)p, n ΠI

1 73. Centroid G of D ABC is (z1 + z2 + z3). Let Ortho3 centre H of DABC be z. As G divides the join of H and S in the ratio 2 : 1, we get 2z + z 1 (z1 + z2 + z3) = 0 2 +1 3 fi

z = z1 + z2 + z3 – 2z0.

z+3 2 3 5 15 = = π1 z +1 3 2 3 2



4

w2 w3 w1

az1 z1 + b(z1 z2 + z1 z2) + cz2 z2 is real.

x = log2

\ cos y = 1/2, sin y = - 3/2

È w2 w3 ˘ Í1 + ˙ =0 ww1 ˚ Î

represents a circle. For statement-2, see Theory. 77. For statement-2, note that w k = w k = 1 if k is a multiple of 3, and if k is not a multiple of 3, then one of w k, w k equals w and other equals w2. z2 – z + 1 = 0 are – w and – w = – w – 1/w 2

(z k + z – k)2 = [(– w)k + (– w )k ]2 Ï1 if k is not a multiple of 3 = Ì Ó4 if k is a multiple of 3 78. Let a = a + ib and a1 = a1 + ib1

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2(ax + by) + b = 0 2(a 1x + b1 y) + b1 = 0

and

(1) (2)

The lines (1) and (2) are parallel if and only if –

a a a a1 =– 1 ¤ = ib ib b b1

\

e





x = np / 2, n ΠI.



n = 2m

cos (ny) = 1

(–1)n e–2y = 1 e

–2y

=1 fi

84. sin (iy) = 0 fi e–2y = 1 85. ex1 cos y1 = ex2 cos y2 fi

tan y1 = tan y2



y2 = np + y1

As in problem 82, y2 = 2np + y1.

¤

a – b1 = ib ia1

¤

i (a1 + ib1 ) a + ib – b1 + ia1 = = a – ib – b1 – ia1 ( – i ) (a1 – ib1 )

¤

a a a =– 1 ¤ is purely imaginary a a1 a1 4

(3 + sin q )

2

4 = 2 + cos q 10 + 6sin q

2 5 + 3sinq

\ f (q) q when cosq = 1. Thus, 1/2 £ f(q) £ 1. 80. m = |a + ib|2 and n = |c + id|2, mn = (|a + ib| |c + id |)2 = |(ac – bd) + i(ad + bc)]

2

= e2 + f 2 where e = ac Рbd, f = ad + bc ΠI. Statement-2 is false, since 2

|z1 + z2| = (|z1| + |z2|)

Re(z1 z2 ) = |z1 z2 |

¤

z1 z2 ≥ 0.

\

z1 = z2 + 2np

86. log z = 1

¤

x1 = x2

log |z| = 1, arg z = 0

87. Can be checked easily by taking z = – 1, z¢ = – 2 88. |e x + iy| = ex 89. e logz = e log|z| ei(arg z) = |z|eiarg(z) wn = 1 and 1, w, w2, ..., w n–1 are roots of x – 1 = 0. Thus, n

1 1 + + x -1 x - w

+

1 nx n -1 = x - w n -1 x n - 1

Put x = 2 a, a , . . . an–1, a of x – 1 = 0. Thus,

n -1

are the 2n roots

2n

x2n – 1 = (x – 1) (x + 1) (x – a) (x – a ) ... (x – an–1) (x – a n–1) 93. Use (x – a) (x – a ) = x2 – 2x cos

= |z1| + |z2| + 2|z1| |z2| ¤

e x1 = e x2

2

|z1|2 + |z2|2 + 2Re(z1 z2 ) 2



Thus,

90. log (–i) = log |–i| + i arg (–i) = – p i/2

2

2

y=0

e x1 sin y1 = ex2 sin y2

1

¤

sin 2x = 0 –2y

83. Similar to 82.

Ê a ˆ Ê a1 ˆ ÁË – b ˜¯ ÁË – b ˜¯ = –1

=

e2iz = 1

e–2y cos 2x = 1, e –2y sin 2x = 0

and

and only if

79. f (q)2 =



82. sin z = 0 fi

a Ê aˆ a = ¤ is real a1 ÁË a1 ˜¯ a1

¤

As ex π 0, cos y = sin y fi

a + ib a1 + ib1 a a = ¤ = 1 a – ib a1 – ib1 a a1

¤

81. ez = 0 fi ex cos y = 0, ex sin y = 0

and

x 2n -1 x2 - 1

p + 1 etc. n

= x2n–2 + x2n–3 + ... + x + 1

94. In 93, divide both the sides by x n–1 to obtain 1 pˆ Ê ÁË x + - 2 cos ˜¯ x n

1 2p ˆ Ê ÁË x + - 2 cos ˜¯ x n

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n - 1 ˆ xn - x-n 1 Ê p˜ = ÁË x + - 2 cos x n ¯ x - x -1 95. From 94, p sin(nq ) n -1 p) = 2n -1 (cos q - cos ) (cos q - cos sin q n n Take limit as q Æ 0 to obtain 2p ˆ Ê p ˆÊ Ê 2 n -1 ˆ n = 2n –1 Á 2sin 2 ˜ Á 2sin 2 p˜ ˜ Á 2sin Ë 2n ¯ Ë 2n ¯ Ë 2n ¯ p 2p n -1 ˆ Ê n-2 ˆ Ê p ˜ Á 2sin p˜ = 2n–1 Á 2sin sin sin Ë ¯ Ë 2n 2n 2n 2n ¯ pˆ n -1 Ê p sin ˜ ÁË 2sin 2n n¯ kp Ên-k ˆ Ê kp ˆ 2sin sin Á p ˜ = sin Á ˜ Ë 2n ¯ Ë n¯ 2n n

w (1 - w 6 ) 96. a + b = wk = = -1 1-w k =1

Â

as w7 = 1 97. ab = (w + w2 + w4) (w3 + w5 + w6 ) = 3 + (a + b ) = 2 98. Desired equation is x2 – (a + b )x + ab = 0 1 99. a = ( - 1 ± 7i ) 2 p 2p 4p + sin - sin > 0 7 7 7

1 a = ( -1 + 7i ) 2

\ 6

100. (b) Â wk = 1 + 2a = 7i k=0

where

Thus,

2

3z1 + 7 z2 3 + 7ai = 5 =5 3z1 - 7 z2 3 - 7 ai

– 4 + y2 = –3y

fi fi

y = 1, –4 x2 = –4,



x=±1

|x| + |y| = 5 x + iy + 2( x + 1)2 + 2 y 2 + i = 0



x + 2( x + 1)2 + 2 y 2 = 0, y = –1

\

x + 2( x + 1)2 + 2 = 0



2(x + 1)2 + 2 = x2



x2 + 4x + 4 = 0



x = –2

2

|z| = 4 + 1 = 5

Thus,

104. 4|(1 + i)z + iz| £ 4|1 + i| |z|3 + 4|i| |z|3 3

Ê 1ˆ < 4 2Á ˜ + 2< 3 Ë 8¯ 105. As p1, p2, ... pn Œ R, cos q – i sin q = 1/z a root of the given equation. fi 1 + p1 z + p2 z2 + ... + pn zn = 0 fi p1 sin q + p2 sin2 q + ... + pn sin (nq) = 0 106. 100th roots of unity are a0, a1, ..., a99 Ê 2p j ˆ Ê 2p j ˆ where a j = cos Á + i sin Á Ë 100 ˜¯ Ë 100 ˜¯ Thus, (x – a1) ... (x – a99) = x99 + ... + x + 1 Putting x = 1, (1 – a1) (1 – a2) ... (1 – a99) = 100 fi

(1 – a1 ) (1 – a 2 ) ... (1 – a 99 ) = 100

\

(1 – a1) (1 – a1 ) (1 – a2) (1 – a 2 ) ... (1 – a99 ) (1 – a 99 99 ) = 104

But

(1 – a k) (1 – a k ) 2k p ˆ Ê 2 2 Ê kp ˆ = 2 Á1 - cos ˜ ˜¯ = 2 sin ÁË Ë 100 ¯ 100

Thus, Ê 99p ˆ Ê p ˆ ... sin2 Á = 104 2198 sin2 Á Ë 100 ˜¯ Ë 100 ˜¯

2

102. – 3 + x yi = x + y – 4i fi x2 + y = –3, x2y = –4



103. Let z = x + iy

kŒR

z 5i fi 2 = - k = ai ( say ) 3 z1 5

4 + y = –3 y

y2 + 3y – 4 = 0 y= 1 not possible. y = – 4 fi x2 = 1

2

3iz2 101. Let =k 5 z1

-



x = cosq + i sin q

But Im a = sin





Ê 99p ˆ Ê p ˆ 299 sin Á ... sin Á = 100 Ë 100 ˜¯ Ë 100 ˜¯

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From (1), we get

107. The given equation can be written as 3

2

8z – 12iz + 18iz + 27 = 0 fi

4z (2z – 3i) + 9i(2z – 3i) = 0



(4z2 + 9i) (2z – 3i) = 0



z2 = – 9i/4



|z|2 = 9/4

1 1 1 + + =2 a +1 b +1 c +1

2

or

p pˆ Ê 110. x = 1 ± 3i = 2 Á cos ± i sin ˜ Ë 3 3¯

z = 3i/2

a6 + b 6 = 27 cos 2p = 128

108. If |z + 4| = 3, z lies on a circle with centre at – 4 and radius 3. Furtherst point on this circle from 7 is z – 7| is 14 2 w, w are roots of 1 1 1 2 + + = a+x b+x c+x x fi

a cos a 111. b cos b c cos g

a sin a 1 b sin b 1 = 0 c sin g 1

fi bc sin (g – b) + ca sin (a – g) + ab sin (b – a) = 0 (1)

x3 + (bc + ca + ab)x + 2abc = 0

If a is the third root of this equation, then a + w + w 2 = 0 or a = 1.

\

|z| = 3

112. zz = 9 fi

9 =z. z

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2 Quadratic Equations 2.1 QUADRATIC EQUATIONS

An equation of the form ax 2 + bx + c = 0

complex numbers, then equation (1) becomes an identity, that is a = b = c = 0. (1)

where a π 0, a, b, c Œ C, the set of complex numbers, is called a quadratic equation. A root of the quadratic equation (1) is a complex number a such that

2.3 RELATION BETWEEN ROOTS AND COEFFICIENTS

If a and b are the roots of the quadratic equation (1), then a+b=–

aa2 + ba + c = 0 The quantity D = b2 – 4ac is known as the discriminant of the equation (1). The roots of equation (1) are given by the quadratic formula x=

ab =

c a

Note that a quadratic equation whose roots are a and b is given by (x – a) (x – b) = 0

-b ± D 2a

2.2 NATURE OF ROOTS

1. If a, b, c Œ R and a π 0. Then the followings hold good: (a) The equation (1) has real and distinct roots if and only if D > 0. (b) The equation (1) has real and equal roots if and only if D = 0. (c) The equation (1) has complex roots with non-zero imaginary parts if and only if D < 0. (d) p + iq (p, q Œ R, q π 0) is a root of equation (1) if and only if p – iq is a root of equation (1). 2. If a, b, c Œ Q and D is a perfect square of a rational number, then equation (1) has rational roots. 3. If a, b, c Œ Q and p +

and

b a

q (p, q ΠQ) is an irrational

root of equation (1) then p Рq is also a root of equation (1). 4. If a = 1, b, c ΠI, the set of integers, and the roots of equation (1) are rational numbers, then these roots must be integers.

2.4 QUADRATIC EXPRESSION AND ITS GRAPH

Let f (x) = ax2 + bx + c, where a, b, c Œ R and a π 0. We have

b c˘ È f (x) = a Í x 2 + x + ˙ a a˚ Î È b b2 c b2 ˘ = a Í x2 + x + 2 + - 2 ˙ a a 4a ˚ 4a Î ÈÊ b ˆ 2 4ac - b2 ˘ = a ÍÁ x + ˜ + ˙ 2a ¯ 4a2 ˚ ÎË

(2)

When is a Quadratic Expression always positive Or always negative? It follows from (2) that f (x) > 0 (< 0) "x Œ R if and only if a > 0 (< 0) and D = b2 – 4ac < 0. See Fig. 2.1 and (Fig. 2.2). Also, it follows from (2) that f (x) ≥ 0 (£ 0) "x Œ R if and only if a > 0 (< 0) and D = b2 – 4ac = 0. In this case f (x) > 0 (< 0) for each x Œ R, x π – b/2a. Also, in this case the graph of y = f (x) will touch the x-axis at x = – b/2a.

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If D = b2 – 4ac > 0 and a < 0 then Ï< 0 for x < b or x > a Ô f (x) Ì> 0 for b < x < a Ô= 0 for x = a , b Ó See Fig. 2.4

Fig. 2.1

Fig. 2.4

Note that if a > 0, then f (x) attains the least value at x = – b/2a. This least value is given by 2 Ê b ˆ 4 ac - b f Á- ˜ = Ë 2a ¯ 4a

Fig. 2.2

If a < 0, then f (x) attains the a greatest value at x = – b/2a. This greatest value is given by

Sign of a Quadratic Expression If D = b2 – 4ac > 0, then (2) can be written as ˘ È b ˆ 2 Ê b2 - 4ac ˆ ˙ Ê f (x) = a ÍÁ x + ˜ - Á ˜ ÍË 2a ¯ 2a Ë ¯ ˙˚ Î ÈÊ b - b2 - 4ac ˆ ˘ b + b2 - 4ac ˆ Ê = a ÍÁ x + ˜˙ ˜ Áx + 2a 2a ÍÎË ¯ ˙˚ ¯Ë = a (x – a) (x – b) 2

where a = If

- b - b2 - 4ac 2a

and b =

D = b2 – 4ac > 0 and a > 0 then Ï> 0 for x < a or x > b Ô f (x) Ì< 0 for a < x < b Ô= 0 for x = a , b Ó See Fig. 2.3

- b + b2 - 4ac 2a

2 Ê b ˆ 4 ac - b f Á- ˜ = Ë 2a ¯ 4a

2.5 POSITION OF ROOTS OF A QUADRATIC EQUATION

Let f (x) = ax2 + bx + c, where a, b, c Œ R be a quadratic expression and let k be a real number. Conditions for Both the Roots to be more than a Real Number k. If a > 0 and b2 – 4ac > 0 then the parabola y = ax2 + bx + c opens upwards and intersect the x-axis in a and b where - b ± b2 - 4ac 2a In this case both the roots a and b will be more than k if k lies to left of both a and b. See Fig. 2.5 a, b =

Y

Fig. 2.3

Fig. 2.5

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From the Fig. 2.5, we note that both the roots are more than k if and only if -b (iii) f (k) > 0 (i) D > 0 (ii) k < 2a In case a < 0, both the roots will be more than k (see Fig. 2.6) if and only if (i) D > 0

(ii) k
0 (ii) af (k) < 0

(iii) f (k) < 0 Fig. 2.7

Fig. 2.6

Fig. 2.8

Combining the above two sets, we get both the roots of ax2 + bx + c = 0 are more than a real number k if only if (ii) k


If a > 0, then exactly one root of f (x) = ax2 + bx + c = 0 lies in the interval (k1, k2) if and only if f (k1) > 0 and f (k2) < 0. Also, exactly one root lies in the interval (k1, k2) if and only if f (k1) < 0 and f (k2) > 0. See Fig. 2.9. Thus, if a > 0, exactly one root of f (x) = ax2 + bx + c = 0 lies in the interval (k1, k2) if and only if f (k1) f (k2) < 0.

-b (iii) af (k) > 0 2a Conditions for Both the Roots to be less than a Real Number k Both the roots of ax2 + bx + c = 0 are less than a real number k if and only if (i) D > 0

(i) D > 0

-b 2a

(iii) af (k) > 0

Note 1. Both the roots of ax2 + bx + c = 0 are positive if and only if c b D ≥ 0, - > 0 and > 0 a a 2. Both the roots of ax2 + bx + c = 0 are negative if and only if c b D ≥ 0, - < 0 and > 0 a a

Fig. 2.9

Similarly, if a < 0, exactly one of the roots of f (x) = ax2 + bx + c = 0 lies in the interval (k1, k2) if f (k1) f (k2) < 0.

2.6 CONDITIONS FOR A NUMBER k TO LIE BETWEEN THE ROOTS OF A QUADRATIC EQUATION

A real number k lies between the roots of the quadratic equation f (x) = ax2 + bx + c = 0 if and only if a and f (k) are of opposite signs, that is, if and only if (i) a > 0

(ii) D > 0

(i) a < 0

(ii) D > 0

(iii) f (k) < 0 [See Fig. 2.7]

Fig. 2.10

(iii) f (k) > 0 [See Fig. 2.8]

Hence, if f (k1) f (k2) < 0, then exactly one root of f (x) = 0 lies in the interval (k1, k2).

or

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0, then 1 is a root of the quadratic equation.

If a > 0, both the roots of f (x) = ax2 + bx + c = 0 lie in the interval (k1, k2) if and only if b (i) D > 0 (ii) k1 < < k2 2a (iii) f (k1) > 0 and f (k2) > 0

Illustration 1 1 is a root of (b – c) x2 + (c – a) x + a – b = 0 as (b – c) + (c – a) + (a – b) = 0.

Finding Range of a Rational Function

See Fig. 2.11

Let f (x) = ax2 + bx + c, g(x) = px2 + qx + r where a, b, c, p, q, r Œ R and one of a, p π 0.

R(x) =

Fig. 2.11

y=

In case a < 0, the conditions read as (ii) k1 < -

(i) D > 0

f ( x) , x ΠR, put g( x )

ax 2 + bx + c px 2 + qx + r

,

fi (a – py)x2 + (b – qy)x + (c – r y) = 0. Since x is real, (b – qy)2 – 4(a – py) (c – r y) ≥ 0 We use this inequality to obtain range of R(x).

b < k2 2a

(iii) f (k1) < 0 and f (k2) < 0 See Fig. 2.12

Illustration 2 Find range of

y= Fig. 2.12

Some Useful Tips Let quadratic equation be f (x) = ax2 + bx + c = 0 where a, b, c Œ R, a π 0. Let roots of (1) be a, b, and D = b2 – 4ac. 1. Both the roots are positive ¤ D ≥ 0, -

c b = ab > 0 = a + b > 0, a 2a

(1)

x2 - 2 x + 5 x2 + 2 x + 7

fi (x2 + 2x + 7)y = x2 – 2x + 5 fi ( y – 1)x2 + 2x(y + 1) + 7y – 5 = 0 As x is real, 4(y + 1)2 – 4(y –1) (7y – 5) ≥ 0 fi (y2 + 2y + 1) – (7y2 – 12y + 5) ≥ 0 fi 3y2 – 7y + 2 £ 0 fi (3y – 1) (y – 2) £ 0 fi 1/3 £ y £ 2

Some Useful Tips

2. Both the roots are negative ¤ D ≥ 0, -

c b = a + b < 0, = ab > 0 a 2a

3. Both the roots lie in the interval (p, q). b 0, af (q) > 0 ¤ D ≥ 0, p < 2a

4. If one of a, b is real, then other must be real.

Suppose range of

ax 2 + bx + c px 2 + qx + r

all real values, then range of also S.

is the set S and h(x) takes a(h ( x ))2 + b (h( x )) + c p(h ( x ))2 + q(h ( x )) + r

is

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Eliminating a we get

Illustration 3

(a ¢ c - ac ¢ )2 tan x - 2 tan x + 5 2

Range of

tan x + 2 tan x + 7 2

However, range of

is

1 £ y £ 2. 3

sin 2 x - 2 sin x + 5 sin 2 x + 2 sin x + 7

is not

(ab ¢ - a ¢b) fi

2

=

bc ¢ - b ¢c a b ¢ - a ¢b

(a ¢c - ac ¢ )2 = (bc ¢ - b ¢c) (ab ¢ - a ¢b)

This is the required condition for two quadratic equation to have a common root. See Fig. 2.15

[1/3, 2] as – 1 £ sin x £ 1 will curtail its range. For instance try y = 2. 2.9 CONDITIONS FOR A QUADRATIC EQUATION TO HAVE A REPEATED ROOT

The quadratic equation f (x) = ax2 + bx + c = 0, a π 0 has a as a repeated if and only if f (a) = 0 and f ¢(a) = 0. In this case f (x) = a(x – a)2. In fact a = – b/2a. See Fig. 2.13 and Fig. 2.14.

Fig. 2.15

Tip

Fig. 2.13

How to Obtain the Common Root? Make coefficients of x2 in both the equations same and subtract one equation from the other to obtain a linear equation in x. Solve it for x to obtain the common root.

2.11 CONDITION FOR TWO QUADRATIC EQUATIONS TO HAVE THE SAME ROOTS

Fig. 2.14

2.10 CONDITION FOR TWO QUADRATIC EQUATIONS TO HAVE A COMMON ROOT

Suppose that the quadratic equations ax2 + bx + c = 0 and a¢ x2 + b¢ x + c¢ = 0 (where a, a ¢ π 0 and ab¢ – a¢b π 0) have a common root. Let this common root be a. Then aa2 + ba + c = 0 and a¢a2 + b¢a + c¢ = 0 Solving the above equations, we get a2 a 1 = = bc ¢ - b ¢c a ¢ c - ac ¢ ab ¢ - a ¢b fi

a2 =

bc ¢ - b ¢c a ¢ c - ac ¢ and a = ab ¢ - a ¢b a b ¢ - a ¢b

Two quadratic equations ax2 + bx + c = 0. and a¢x2 + b¢x + c¢ = 0 have the same roots if and only if a b c = = a¢ b¢ c¢ 2.12 EQUATIONS OF HIGHER DEGREE

The equation f (x) = a0 xn + a1xn – 1 + a2xn– 2 + … + an–1 x + an = 0 where a0, a1, …, an – 1, an Œ C, the set of complex numbers, and a0 π 0, is said to be an equation of degree n. An equation of degree n has exactly n roots. Let a1, a2, …, an Œ C be the n roots of (1). Then f (x) = a0(x – a1) (x – a2) … (x – an) Also

a1 + a2 + … + an = -

and

a1a2 … an = (–1)n

an . a0

a1 a0

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If a, b, g are the roots of ax3 + bx2 + cx + d = 0, then b c d a + b + g = – , bg + g a + ab = and a b g = – . a a a 4 3 Also, if a, b, g, d are the roots of the equation ax + bx + cx2 + dx + e = 0, then –b c , (a + b ) (g + d ) + ab + g d = a a –d e ab (g + d) + gd ( a + b) = , abgd = . a a

a+b+g+d=

2.14 TRANSFORMATION OF EQUATIONS

We now list some of the rules to form an equation whose roots are given in terms of the roots of another equation. Let given equation be (1) a0 xn + a1 xn–1 + … + an–1 x + an = 0 Rule 1: To form an equation whose roots are k(π 0) times x roots of the equations in (1), replace x by k in (1). Rule 2: To form an equation whose roots are the negatives of the roots in equation (1), replace x by – x in (1). n–1

n–3

n–5

of x , x , x , … etc. in (1). Rule 3: To form an equation whose roots are k more than the roots of equation in (1), replace x by x – k in (1). Rule 4: To form an equation whose roots are reciprocals 1 in (1) of the roots in equation (1), replace x by x and then multiply both the sides by xn. Rule 5: To form an equation whose roots are square of the roots of the equation in (1) proceed as follows: Step 1

Replace x by

x in (1)

Step 2 Collect all the terms involving x on one side. Step 3 Square both the sides and simplify. For instance, to form an equation whose roots are squares of the roots of x3 + 2x2– x + 2 = 0, replace x by x to obtain x x + 2x – x + 2 = 0 fi

x (x – 1) = –2(x + 1)

Squaring we get or

x(x – 1)2 = 4(x + 1)2 x3 – 6x2 – 7x – 4 = 0

Rule 6: To form an equation whose roots are cubes of the roots of the equation in (1) proceed as follows: Step 1 Replace x by x1/3. Step 2 Collect all the terms involving x1/3 and x2/3 on one side. Step 3 Cube both the sides and simplify. 2.15 DESCARTES RULE OF SIGNS FOR THE ROOTS OF A POLYNOMIAL

Rule 1: The maximum number of positive real roots of a polynomial equation f(x) = a0 xn + a1 xn–1 + a2 xn–2 + … + an–1 x + an = 0 from positive to negative and negative to positive. For instance, in the equation x3 + 3x2 + 7x – 11 = 0 +++– As there is just one change of sign, the number of positive roots of x3 + 3x2 + 7x – 11 = 0 is at most 1. Rule 2: The maximum number of negative roots of the polynomial equation f (x) = 0 is the number of changes from positive to negative and negative to f (– x) = 0. 2.16 SOME HINTS FOR SOLVING POLYNOMIAL EQUATIONS

1. To solve an equation of the form (x – a)4 + (x – b)4 = A a+b 2 In general to solve an equation of the form (x – a)2n + (x – b)2n = A put

y=x–

where n is a positive integer, we put y = x – 2. To solve an equation of the form a0 f (x)2n + a1(f (x))n + a2 = 0 n

a+b . 2 (1)

2

we put (f (x)) = y and solve a0 y + a1y + a2 = 0 to obtain its roots y1 and y2. Finally, to obtain solutions of (1) we solve, (f (x))n = y1 and

(f (x))n = y2

3. An equation of the form (ax2 + bx + c1) (ax2 + bx + c2) … (ax2 + bx + cn) = A where c1, c2, …, cn , A Œ R, can be solved by putting ax2 + bx = y.

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4. An equation of the form (x – a) (x – b) (x – c) (x – d) = Ax2 where ab = cd, can be reduced to a product of two ab quadratic polynomials by putting y = x + . 2 5. An equation of the form (x – a) (x – b) (x – c) (x – d) = A where a < b < c < d, b – a = d – c can be solved by change of variable y=

( x - a ) + ( x - b) + ( x - c) + ( x - d )

we divide the equation by x2, to obtain 1ˆ 1ˆ Ê Ê a Á x2 + 2 ˜ + b Á x + ˜ + c = 0 Ë Ë ¯ x¯ x 1 = y. x 6. To solve equation of the type (x + a) (x + b) (x + c) (x + d) + k = 0 where a + b = c + d, put x2 + (a + b)x = y 7. To solve equation of the type and then put

4

ax + b = cx + d

1 = x - (a + b + c + d) 4 6. A polynomial f (x, y) is said to be symmetric if f (x, y) = f (y, x) ⁄ x, y. A symmetric polynomials can be represented as a function of x + y and xy.

Equations Reducible to Quadratic 1. To solve an equation of the type ax4 + bx2 + c = 0, put x2 = y. 2. To solve an equation of the type a (p(x))2 + bp(x) + c = 0 (where p(x) is an expression in x) put p(x) = y. 3. To solve an equation of the type ap(x) +

b +c=0 p ( x)

where p(x) is an expression in x, put p(x) = y. This reduces the equation to ay2 + cy + b = 0. 4. To solve an equation of the form 1ˆ Ê Ê a Á x2 + 2 ˜ + b Á x + Ë Ë ¯ x

x+

1ˆ ˜ +c=0 x¯

or ax 2 + bx + c = dx + e square both the sides and solve for x. 8. To solve equation of the type ax + b ± cx + d = e proceed as follows. Step 1 Transfer one of the radical to the other side and square both the sides. Step 2 Keep the expression with radical sign on one side and transfer the remaining expression on the other side. Step 3 Now solve as in 7 above. 2.17 USE OF CONTINUITY AND DIFFERENTIABILITY TO FIND ROOTS OF A POLYNOMIAL EQUATION

Let f (x) = a0 xn + a1 xn – 1 + a2xn – 2 + … + an – 1 x + an, then f is continuous on R. Since f is continuous on R, we may use the intermediate f has a real root in an interval (a, b). If there exist a and b such that a < b and f (a) f (b) < 0, then there exists at least one c Œ (a, b) such that f (c) = 0. Also, if lim f ( x ) and f (a) are of opposite signs, then at xÆ - •

1 =y x+ x

put and to solve

1ˆ Ê Ê a Á x2 + 2 ˜ + b Á x Ë Ë x ¯ put

1ˆ ˜ +c=0 x¯

x–

1 =y x

5. To solve reciprocal equation of the type ax4 + bx3 + cx2 + bx + a = 0, a π 0

least one root of f (x) = 0 lies in the interval (– •, a). Also, if f (a) and lim f ( x ) are of opposite signs, then at least one xÆ •

root of f (x) = 0 lies in the interval (a, •). Result 1 If f (x) = 0 has a root a of multiplicity r (where r > 1), then we can write f (x) = (x – a)r g(x) where g(a) π 0. Also, f ¢(x) = 0 has a as a root with multiplicity r – 1. Result 2 If f (x) = 0 has n real roots, then f ¢(x) = 0 has (n – 1) real roots.

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Dividing (2) by (1), we get

It follows immediately using Result 1 and Rolle’s Theorem. Result 3

n

Â

k =1

(3)

2 x 2 + 2 px - p 2 = 1 + 4px

where a1, a2, … an are n distinct roots of f (x) = 0. We also have f ¢ (x) = f (x)

x 2 + 2 px - p 2 = 4px

Adding (1) and (3) we get

f (x) = a0 (x – a1) (x – a2) … (x – an)

1 x - ak



4[x2 + 2px – p2] = 1 + 8px + 16p2x2



4x2 (1 – 4p2) = 1 + 4p2

(4)

As p > 1/2, (4) has no real solution, therefore, (1) has no solution. If a, b, c are distinct real numbers, then Example 3 number of solutions of

SOLVED EXAMPLES

SINGLE CORRECT ANSWER TYPE QUESTIONS

is

Example 1

Number of real solutions of the equation

(

)

5

x 2 + 2 px - p 2 –

If f (x) = 0 has n distinct real roots, we can write

1 + x + 1 - x = 6 x + 8 1 - x2

(1)

x+a x+b x+c + + =3 x+b x+c x+a

(1)

Ans. (b) Solution: Write (1) as Ê x+a ˆ Ê x+b ˆ Ê x+c ˆ - 1˜ + Á - 1˜ + Á - 1˜ = 0 ÁË x+b ¯ Ë x+c ¯ Ë x+a ¯

is Ans. (a) Solution: Note that 1 + x ≥ 0, 1 – x ≥ 0 ¤ –1 £ x £ 1.



x = cos 2q, 0 £ q £ p/2.



(a – b) (x + c) (x + a) + (b – c) (x + a) (x + b) + (c – a) (x + b) (x + c) = 0



[(a – b) + (b – c) + (c – a)]x2 + [(a – b) (a + c) + (a + b) (b – c) + (c – a) (b + c)]x + ac (a – b) + ba (b – c) + bc (c – a) = 0



x=

Put

Now (1) becomes 5 2 (cosq + sinq) = 6 cos2q + 8 sin2q fi 5 2 2 cos (q – p/4) =

62 + 82 cos (2q – a)

where cosa = 3/5, sina = 4/5 fi cos(q – p/4) = cos(2q – a)

If p > 1/2, the number of real solutions

of the equation x 2 + 2 px - p 2 + x 2 - 2 px - p 2 = 1

Number real solutions of the equation

2019

 k 2 | x 2 + (k + 3) x - k - 4| = 0

is (1)

k =1

Ans. (b) Solution: x2 + (k + 3)x – k – 4 = (x – 1) (x + k + 4)

is Ans. (a) Solution: For each x ΠR, we have (x2 + 2px Рp2) Р(x2 Р2px Рp2) = 4px

a 2 + b 2 + c 2 - ab - bc - ca

Example 4

x = cos (2q) = sin (2a) = 24/25.

Example 2

3abc - a 2 c - ab 2 - bc 2

[As a, b, c are distinct, a2 + b2 + c2 – bc – ca – ab π 0]

fi q – p/4 = 2q – a fi 2q = 2a – p/2 \

a-b b-c c-a =0 + + x+b x+c x+a

\ (1) can be written as (2)

2019

| x - 1| Â k 2 | x + k + 4| = 0 k =1

(1)

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.9



x=1 È 2019 2 ˘ Í∵  k | x + k + 4| > 0 " x Œ R ˙ Î k =1 ˚

Example 5 1 |f -1

If Ú

Suppose a, b ΠR. Let f (x) = 3x2 + 2ax + b

Ans. (a) Solution: As 1 – 2x ≥ 0, 1 + 2x ≥ 0, Put

1 - 2 x = a,

¤

a b + b a ab (a + b) = a2 + b2 = 2

( x)| dx > 2 , then f (x) = 0 has

1 1 £x£ . 2 2

1 + 2 x = b, then (1) can be written as

a+b=

(a) distinct real roots (b) equal roots (c) purely imaginary roots (d) nature of roots depend on values of a, b Ans. (a)

[∵ a2 + b2 = (1 – 2x) + (1 + 2x) = 2]

Solution: Let D = 4a2 – 4(3)b = 4(a2 – 3b) If a2 – 3b £ 0, then b ≥ 0 and f (x) = 3x2 + 2ax + b ≥ 0 xŒR

Now (2) becomes



Suppose a + b = m, then a2 + b2 + 2ab = m2 fi 2ab = m2 – 2 1 (m2 – 2)m = 2 2

1

1

2 Ú-1| f ( x)| dx = Ú-1 (3x + 2ax + b)dx



m3 – 2m – 4 = 0

A contradiction.



(m – 2) (m2 + 2m + 2) = 0

Thus, a2 – 3b > 0, that is D > 0



(m – 2) [(m + 1)2 + 1] = 0

Hence, f (x) = 0 has two distinct real roots.



m = 2.

\

2ab = 22 – 2 = 2 fi ab = 1.

= 2(1 + b) < 2 fi b < 0.

If [x] = greatest integer £ x, then number Example 6 of solutions of the equation Ê1 1 ˆ (x – [x]) Á + ˜ = 2 Ë x [ x] ¯

is

(1)

Ans. (d) Solution: We can write (1) as [ x] x 1+ -1 = 2 x [ x] fi

1 x t - - 2 = 0 where t = [ x] t



t2 – 2t – 1 = 0



t = (1 ± 2 )



x = (1 ± 2 )[ x] , [x] π 0.

It is easy to verify that (1 ± 2 )m , where m is a non-zero integer, is a solution of (1) Example 7

Number of real solutions of the equations

1 - 2x + 1 + 2x =

1 - 2x 1 + 2x (1) is + 1 + 2x 1 - 2x

(2)

As a + b = 2, ab = 1, we get a = b = 1. fi

1 - 2x =

1 + 2 x = 1 fi x = 0.

Let p and q be real numbers such that

Example 8

p π 0, p3 π q and p3 π – q. If a and b are nonzero complex numbers satisfying a + b = –p and a3 + b3 = q, then a a b quadratic equation having and as its roots is b a (a) (p3 (b) (p3 (c) (p3 (d) (p3 Ans. (b)

+ + – –

q)x2 q)x2 q)x2 q)x2

– – – –

(p3 + 2q)x + (p3 + q) = 0 (p3 – 2q)x + (p3 + q) = 0 (5p3 – 2q)x + (p3 – q) = 0 (5p3 + 2q)x + (p3 – q) = 0

Solution: q = a3 + b3 = (a + b)3 – 3ab (a + b) = –p3 + 3abp fi

ab =

p3 + q 3p

We have a b a2 + b2 (a + b ) 2 - 2ab + = = b a ab ab

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(a + b ) 2 p2 = -2 -2 = ab ( p3 + q) /3 p = and

3 p 3 - 2 p 3 - 2q p3 + q

=

p3 + q

or

p +q 3

|x|3/5 = 33

x +1 = 0

The quadratic equation p(x) = 0 with real

(b) – 1



(y – 6) (y + 1) = 0



(x + 3) (x – 2) = 0



Example 12

fi p(p(x)) = 0 cannot have real or purely imaginary roots.

(1)

is

(a)

x = – 3, 2

If tan A and tan B are the roots of

p2 p2 + q2 p (1 - q)2

Solution: Put |x|3/5 = y [See Theory], so that (1) becomes y2 – 26y – 27 = 0

(b)

p2 ( q + p)2

(d)

p2 (1 - q) 2 + p 2

Ans. (d) Solution: We have tan A + tan B = p, and tan A tan B = q. \

tan (A + B) =

tan A + tan B p = 1- q 1 - tan A tan B

Now, sin2 (A + B) =

=

tan 2 ( A + B) sec2 ( A + B)

p2 (1 - q)2 1 + p2 (1 - q)2

Example 13 (c) – 312/5 (d) – 321/5

y = – 1 or 27

or

the quadratic equation x2 – px + q = 0, then value of sin2 (A + B) is

The product of real roots of the equation



y = 6, – 1.

x = w, w2 and their sum is –1.

q(x) = p(p(x)) > 0

(y + 1) (y – 27) = 0



When y = – 1, we get x + x + 1 = 0

If x is real or purely imaginary then p(x) is real and hence

(b) – 312

(d) 6

2

(c) 1 –

|x|6/5 – 26 |x|3/5 – 27 = 0

(c) – 6

y2 – 5y – 6 = 0

q(x) = p(p(x)).

(a) – 310 Ans. (a)

(1)

When y = 6, we get x2 + x – 6 = 0

This shows that p(x) is real for all real x and purely imaginary x.



Sum of the non-real roots of



p(x) π 0 if x π ± bi

Example 10

x = ± 35

Solution: Put x2 + x = y [See Theory], so that the equation (1) becomes (y – 2) (y – 3) = 12

Also, p(x) = 0 if x = ± bi





product of roots of (1) is (35) (– 35) = – 310.

(a) 1 Ans. (b)

Solution: As p(x) is quadratic and p(x) = 0 has purely imaginary roots, p(x) must be of the form p(x) = a(x2 + b) where a π 0, b > 0. Assume that a > 0, then p(x) > 0 x Œ R.

Let

|x| = 35

is

p(p(x)) = 0 has (a) only purely imaginary roots (b) all real roots (c) two real and two purely imaginary roots (d) neither real nor purely imaginary roots Ans. (d)

and



(x2 + x – 2) (x2 + x – 3) = 12

(p3 + q)x2 – (p3 – 2q)x + (p3 + q) = 0

Example 9



Example 11

Thus, required quadratic equation is ( p 3 - 2q )

y = |x|3/5 ≥ 0 " x, we get y = 27

\

p 3 - 2q

a b ◊ = 1. b a

x2 -

As

=

p2 (1 - q)2 + p2

The real value of a for which the sum of

the squares of the roots of the equation x2 – (a – 2)x – a – 1 = 0 assume the least value is (a) 0 Ans. (b)

(b) 1

(c) 2

Solution: Discriminant of the equation is (a – 2)2 + 4(a + 1)

(d) 3

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.11

= a2 Р4a + 4 + 4a + 4 = a2 + 8 > 0 as a ΠR. \ Roots of the given equation are real. Let these roots be a and b.

Solution: Let f (x) = ax3 + bx2 + cx. Note that f is continuous and derivable on R. Also f (0) = 0 and f (1) = a + b + c = 0. By the Rolle’s theorem, there exists at least one a Œ (0, 1) such that f ¢(a) = 0 fi 3aa2 + 2ba + c = 0

Then a + b = a – 2, ab = – (a + 1).

Thus, 3ax2 + 2bx + c = 0 has at least one root in [0, 1].

We have

Example 17 Let f (x) = ax2 + bx + c, a, b, c Œ R and a π 0. Suppose f (x) > 0 for all x Œ R. Let g(x) = f (x) + f ¢(x) + f ¢ ¢(x). Then (a) g (x) > 0 " x Œ R (b) g (x) < 0 " x Œ R (c) g (x) = 0 " x Œ R (d) g(x) = 0 has real roots. Ans. (a)

a2 + b2 = (a + b)2 – 2ab = (a – 2)2 + 2 (a + 1) = a2 – 4a + 4 + 2a + 2 = a2 – 2a + 6 = (a – 1)2 + 5 a2 + b2 is least when a = 1.

Thus,

Example 14 If p and q are distinct primes and x2 – px + q = 0 has distinct positive integral roots, then p + q equals (a) 5 (b) 7 (c) 19 (d) 40 Ans. (a) Solution. Let a and b be two roots of x2 – px + q = 0 and assume a < b. We have a + b = p, ab = q As q is prime, we must have a = 1, b = q. Thus, p = 1 + q fi p – q = 1. As p and q are prime, this is possible if p = 3, q = 2. Example 15 The real values of a for which the quadratic equation 3x2 + 2(a2 + 1)x + (a2 – 3a + 2) = 0 possesses roots of opposite signs lie in (a) (– •, 1) (b) (– •, 0) (c) (1, 2)

Ê3 ˆ (d) Á , 2˜ Ë2 ¯

Ans. (c) Solution: The quadratic equation 3x2 + 2(a2 + 1) x + a2 – 3a + 2 = 0 will have two roots of opposite signs if it has real roots and the product of the roots is negative, that is, if 4(a2 + 1)2 – 12 (a2 – 3a + 2) ≥ 0 and

a2 - 3a + 2 < 0. 3

Both of these conditions are met if a2 Р3a + 2 < 0 i.e. if (a Р1) (a Р2) < 0 or 1 < a < 2. Example 16 If a, b, c ΠR and a + b + c = 0, then the quadratic equation 3ax2 + 2bx + c = 0 has (a) at least one root in [0, 1] (b) at least one root in [1, 2] (c) at least one root in [3/2, 2] (d) none of these. Ans. (a)

Solution Since f (x) > 0 " x Œ R, a > 0 and b2 – 4ac < 0. We have f ¢(x) = 2ax + b and f ¢¢ (x) = a. Thus g(x) = ax2 + bx + c + 2ax + b + 2a = ax2 + (2a + b)x + (2a + b + c) We have a > 0 and D = (2a + b)2 – 4a(2a + b + c) = 4a2 + 4ab + b2 – (8a2 + 4ab + 4ac) = b2 – 4ac – 4a2 < 0, since b2 – 4ac < 0. Thus g(x) > 0 " x Œ R. Example 18 If b < 0, then the roots x1 and x2 of the equation 2x2 + 6x + b = 0, satisfy the condition (x1/x2) + (x2/x1) < k where k is equal to. (a) – 3 (b) – 5 (c) – 6 (d) – 2 Ans. (d) Solution: The discriminant of the quadratic equation 2x2 + 6x + b = 0 is given by D = 36 – 8b > 0. Therefore, the given equation has real roots. We have

=

x1 x2 x 2 + x22 ( x1 + x2 )2 - 2 x1 x2 + = 1 = x2 x1 x1 x2 x1 x2 ( -3)2 - 2(b / 2) 18 –2 0 (c) a > 0 (d) none of these Ans. (a) 2

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Solution: Let f (x) = ax2 + bx + c. Since f (x) has no real zeros, either f (x) > 0 or f (x) < 0 for all x ΠR. Since f (0) = c < 0, we get f (x) < 0 " x ΠR. Therefore, a < 0 as the parabola y = f (x) must open downwards.



3 ( a – 1) > 2 a – 1

\

a > 1 and

Alternative solution ax2 + bx + c = 0 has no real roots fi b2 – 4ab < 0



a>1+



2 11 = 9 9

4ac > b2 > 0 fi a < 0 as c < 0. Example 22 If x is real, then the maximum value of

Example 20

(

y = 2(a – x) x + x + b 2

(a) a2 + b2

) is

(b) a2

a2 + b2

(c)

2

(d) a a2 + b2

Solution:

\

Let t = x +

1 1 = = t x + x 2 + b2

t-

b2 = 2x t

and

(

t+

Thus 2(a – x) x + x 2 + b2

)

x 2 + b2 - x

b2 = 2 x 2 + b2 t Ê b2 ˆ = Á 2a - t + ˜ ◊(t ) t ¯ Ë

= 2at – t2 + b2 = a2 + b2 – (a2 – 2at + t2)

which is a real number This is against the hypothesis. Therefore, a = 0.

) £ a + b . Hence, 2

If both the roots of the equation

x2 – 6ax + 2 – 2a + 9a2 = 0 exceed 3, then (a) a < 1 (b) a > 11/9 (c) a > 3/2 (d) a < 5/2 Ans. (b)

has real roots that are equal in magnitude but opposite in sign is (b) b2 = 2m2 (a) b2 = m2 (c) 2b2 = m2 (d) none of these Ans. (b) Solution: Clearly x = m is a root of the equation. Therefore, the other root must be – m. That is, 1 1 1 1 + = + -m -m+b m m+b

Solution: We can write the given equation (x – 3a)2 = 2a – 2 Note that a ≥ 1 and



x = 3a ± 2a – 2 Both the roots will exceed 3 if smaller of the two roots exceed 3, that is, if 3a – 2a – 2 > 3

The condition that the equation

1 1 1 1 + = + x x+b m m+b

2

the maximum value of y is a2 + b2. This value is attained at t = a or x = (a2 – b2)/2a. Example 21

4 ac - b2 Ô¸ ÔÏ = a Ì- 1 + ˝ 4 a2 Ô˛ ÔÓ

Example 23

= a2 + b2 – (a – t)2

(

Suppose a π 0. We rewrite f (x) as follows:

2 ÏÔÊ b 4 ac - b2 ¸Ô bˆ ˆ Ê b +i+ f Á+ i ˜ = a ÌÁ + ˝ Ë 2a ¯ 2 a ˜¯ Ë 2a 4 a2 Ô ÓÔ ˛

b2

Therefore, y = 2(a – x) x + x 2 + b2

Let f (x) = ax2 + bx + c, a, b, c ΠR. If f (x)

2 ÏÔÊ 4 ac - b2 ¸Ô bˆ f (x) = a ÌÁ x + ˜ + ˝ 2a ¯ 4 a2 Ô˛ ÔÓË

x 2 + b2



(4)

takes real values for real values of x and non-real values for non-real values of x, then. (a) a = 0 (b) b = 0 (c) c = 0 (d) nothing can be said about a, b, c. Ans. (a) Solution:

Ans. (a)

a –1> 2 3

fi fi

1 1 2 = b-m b+m m b+m-b+m b -m 2

2

=

2 m

2 m2 = 2b2 – 2m2

or

2m2 = b2.

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If x is real, and

Example 24 k=

x2 - x + 1 x2 + x + 1

then

(a) 1/3 £ k £ 3 (c) k £ 0 Ans. (a) Solution We have k = fi

x2 - x + 1 x2 + x + 1

(b) k ≥ 5 (d) 2/3 £ k £ 1

(a) no root in (0, 1) (b) at least one root in (1, 2) (c) a double root in (0, 1) (d) none of these. Ans. (b)

fi kx + kx + k = x – x + 1 2

2

As x is real, the discriminant D = (k + 1)2 – 4(k – 1)2 ≥ 0 fi

(3k – 1) (– k + 3) ≥ 0



(k - 1 3) ( k - 3) £ 0

fi 1 3£ k £3

These extreme values are attained at x = 1 and x = –1. Example 25 Let a > 0, b > 0 and c > 0. Then both the roots of the equation ax2 + bx + c = 0 (a) are real and negative (b) have negative real parts (c) are rational numbers (d) none of these. Ans. (b) Solution: We have D = b2 – 4ac. If D ≥ 0, then the roots of the equation are given by -b ± D 2a

As D = b2 – 4ac < b2 (∵ a > 0, c > 0), it follows that the roots of the quadratic equation are negative. In case D < 0, then the roots of the equation are given by



1

2

0

0

1

which have negative real parts.

2

Ú f ( x ) dx = 0

1

2

If f (x) > 0 (< 0) " x Œ [1, 2], then Ú f ( x ) dx > 0 (< 0). 1

Thus, f (x) = (e–x + ex) (ax2 + bx + c) must be positive for some value of x in [1, 2] and must be negative for some value of x in [1, 2]. As e–x + ex ≥ 2, it follows that if g(x) = ax2 + bx + c, then there exists some a, b Œ [1, 2] such that g(a) > 0 and g(b) < 0. Since g is continuous on R, there exists some g between a and b such that g(g) = 0. Thus, ax2 + bx + c = 0 has at least one root in (1, 2). Example 27

The equation

x + 3 - 4 x -1 + x + 8 - 6 x -1 = 1 has (a) no solution (b) only one solution (c) only two solution (d) more than two solutions. Ans. (d) Solution: Put x - 1 = t equation becomes

or x = t2 + 1 so that the given

t2 + 4 - 4 t + t2 + 9 - 6 t = 1

-b ± i - D x= 2a Example 26 that

2

We have Ú f ( x ) dx = Ú f ( x ) dx + Ú f ( x ) dx

(k – 1)x2 + (k + 1)x + k – 1 = 0

x=

Let f (x) = (e–x + ex) (ax2 + bx + c)

Solution:

or

(t - 2)2 + (t - 3)2 = 1 or

t -2 + t -3 = 1

Let a, b, c be non-zero real numbers such 2 and 3, i.e., 2 £ t £ 3.

t lying between

1

-x x 2 Ú (e + e ) (ax + bx + c)dx

0

2

= Ú (e - x + e x ) (ax2 + bx + c) dx 0

Then the quadratic equation ax2 + bx + c = 0 has

x lying between 5 and 10. Example 28 The equation 3x – 1 + 5x – 1 = 34 (a) no solution (b) one solution (c) two solutions

has

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(d) more than two solutions Ans. (b) Solution: It is quite clear that x equation we now show that there is no other solution. Note that y = 3x – 1 and y = 5x – 1 are both increasing functions of x (exponential functions with base greater than 1). Therefore their sum y = 3x – 1 + 5x – 1 is also an increasing

Example 31 Let (a1, a2, a3, a4, a5) denote a rearrangement of (3, – 5, 7, 4, – 9), then the equation a1 x4 + a2 x3 + a3 x2 + a4 x + a5 = 0 has (a) at least two real roots (b) all four real roots (c) only imaginary roots (d) none of these Ans. (a) Solution:

x = 1 is always a root of the equation.

Example 32

If three distinct real number a, b and c

satisfy a2(a + p) = b2(b + p) = c2(c + p) Fig. 2.16

function of x. This means for x < 3, 3x – 1 + 5x – 1 < 34 and for x > 3, 3x – 1 + 5x – 1 > 34. Thus, the equation has no other solution. (Fig. 2.16) If the harmonic mean between roots of

Example 29

2 )x2 – bx + 8 + 2 5 = 0

(5 +

(1)

where p ΠR, then value of bc + ca + ab is (a) Рp (b) p (c) 0 (d) p2/2 Ans. (c) Solution: If value of each relation is k, then a, b, c are roots of x3 + px2 Рk = 0. Example 33 equation

is 4 then b equals (b) 4 –

5

is

(c) 3 Ans. (d)

(d) 4 +

5

(a) 0 Ans. (b)

4= fi

2ab a +b

b 8+2 5 = 5+ 2 2 5+ 2



(

)

fi fi

If a £ 0, then number of real roots of

x = ± 2 as x is an integer. Let S denote the set of all values of the

parameter a for which

is

x+ Ans. (b) Solution: Put x – a = t and rewrite (1) as (a + t)2 + 2a|t| + 3a2 = 0 t2 + 2a (t + |t|) + 4a2 = 0

(d) 8

y = 16, 29.

Example 34 x2 + 2a |x – a| + 3a2 = 0



(c) 4

But y = 29 is not possible and x4 = y = 16

5.

Example 30

(b) 2

Solution: Put x4 = y and write y + 20 = (22 – y)2

Let a, b be the roots of (1), then

b=4+

x 4 + 20 = 22

x4 +

(a) 2

Solution:

The number of integral roots of the

(2)

x2 + a = a

(1)

has no solution, then S equals (a) (– 1, 1) (b) (– •, – 1) (c) [– 1, •) (d) (0, •) Ans. (b)

Not possible. Therefore t ≥ 0.

Solution: If a = 0, (1) holds for all x £ 0. Write (1) as x2 + a = a2 – 2ax + x2

Thus, (2) gives t2 + 4at + 4a2 = 0



fi fi

Substituting this (1) we get a – 1 + |a + 1| = 2a

2

2

If t < 0, we get t + 4a = 0.

t = – 2a. x = – a.

2ax = a(a – 1) fi x = (a – 1)/2

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.15

This holds if a ≥ – 1.

¤

\ The equation has no solution if a < – 1.

(a – 2) (1 + a) > 0

The number of roots of the equation

Example 35

1 1 35 + = x 1 - x 2 12

a (7a + 16) £ 0,

(1)

and

(2a + 1) (1 + a) > 0

¤ and

16 £ a £ 0; (a < –1 or a > 1/2 7 a < – 1 or a > – 1/2)

¤

-

-

is (a) 0 Ans. (d) Solution:

(b) 1 Put x = 1/u,

(c) 2

(d) 3

1 - x 2 = 1/v

(2)

u + v = 35/12, u2 + v2 = u2v2

When uv = 25/12, we get x = 3/5, 4/5

Example 36

73 )/14.

Let S denote the set of all real values of a

for which the roots of the equation x2 – 2ax + a2 – 1 = 0

a (a + 1) > 0 ¤ a Œ (–•, –1) » (0, •)

Example 39

Now, 5 < a ± 1 < 10 4 < a < 9, 6 < a < 11

\

a Π(6, 9). Let S denote the set of all values of a for

which the roots of the equation (1 + a)x2 – 3ax + 4a = 0 exceed 1, then S equals (a) [– 16/7, 0) (c) (– 1, 2) Ans. (b)

(1)

(b) (– 16/7, – 1) (d) (– 1, 3)

Solution: Clearly a π 0, – 1. Both the roots of (1) will exceed 1 if and only if 9a2 – 4(4a) (1 + a) ≥ 0, 3a >2 1+ a (1 + a) [(1 + a) – 3a + 4a] > 0

Let a, b, p, q ΠQ and suppose that

f (x) = x2 + ax + b = 0 and g(x) = x3 + px + q = 0

Solution: We can write (1) as (x – a)2 = 1 fi x = a ± 1

and

2a2 – 2(2a + 1)a + a(a + 1) < 0 ¤

Ans. (d)

Example 37

Solution: As (2a + 1)2 Р2a(a + 1) > 0 " a ΠR the required a must satisfy the inequality

(1)

lie between 5 and 10, then S equals (a) (– 1, 2) (b) (2, 9) (c) (4, 9) (d) (6, 9)



Let S denote the set of all values of S for

has one root less than a and other root greater than a, then S equals (a) (0, 1) (b) (– 1, 0) (c) (0, 1/2) (d) none of these Ans. (d)

uv = 25/12, – 49/12

When uv = – 49/12, we get x = – (5 +

Example 38

which the equations 2x2 – 2(2a + 1)x + a(a + 1) = 0

(1) and (2) give fi

16 £a 0 1 1 1 fi a Œ ÊÁ - , ˆ˜ Ë 2 2¯ 4



a2
1



1 ˆ Ê 1 Ê ˆ a Œ Á -•, - ˜ » Á , •˜ Ë ¯ Ë 5 5 ¯

t3 – (x + y + z + p + q)t2 + [(p + q)x + qy + pz]t – pqx = 0

From here the answers can be given easily. Let a be a repeated root of

Example 43 (1)

Next, as |x1 – x2| < 1 fi

¤

3

2

p(x) = x + 3ax + 3bx + c = 0, then (a) a is a root of x2 + 2ax + b = 0 c - ab (b) a = 2 ( a 2 - b) (c) a =

2

ab - c a2 - b

(d) a is a root of ax2 + 2bx + c = 0 Ans. (a), (b), (d)

(2)

Solution: Now

a is also a root of p¢(x) = 0 or x2 + 2ax + b = 0

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.17

a3 + 3aa2 + 3ba + c = 0 fi

a2 + 2aa + b = 0

(1)

aa2 + 2ba + c = 0

(2)

Eliminating a2 from (1) and (2), we get 2(a2 – b)a + ab – c = 0 fi

a=

c - ab 2 ( a 2 - b)

Example 44

pa - qa = pa – q a 2



a=

(y + q)3 = – p3y



y3 + 3qy2 + (3q2 + p3)y + q3 = 0

1 (y + q) p

As a is a root of the given equation 1 p q (y + q)3 – 2 (y + q)2 + (y + q) – r = 0 3 p p p fi

(y + q)3 – p2(y + q)2 + p2q (y + q) – p3r = 0



y3 + (3q – p2)y2 + q(3q – p2)y + q3 – p3r = 0

Next y = b3 + g3 = – 3q – a3 fi

a3 = – 3q – y

As a3 is a root of (1) (– 3q – y)3 + 3q (– 3q – y)2 + (3q2 + p3) (– 3q – y) + q3 =0 fi

y3 + 6qy2 + (p3 + 12q2)y + 3p3q + 8q3 = 0

Lastly, a2 + b2 + g2 = (a + b + g)2 – 2(bg + ga + ab) = – 2p a4 + b4 + g4 = (a2 + b2 + g2)2 – 2(b2g2 + g2a2 + a2b2) b 2g 2 + g 2a 2 + a 2b 2

But

= (bg + ga + ab)2 – 2abg(a + b + g) = p2 Example 46

If the equations x2 + bx + c = 0 and

bx2 + cx + 1 = 0 have a common root then (a) b + c + 1 = 0 (b) b2 + c2 + 1 = bc (c) (b – c)2 + (b – 1)2 + (c – 1)2 = 0 (d) b + c + 1 = bc Ans. (a), (c) Solution:

If a is a common root of the two equations,

a + ba + c = 0, ba2 + ca + 1 = 0 2

b and subtracting we get

If q3 – p3 r = 0, then y = 0 is a root of the above equation. Example 45 Let a, b, g be roots of x3 + px + q = 0, then (a) an equation whose roots are a3, b3, g3 is x3 + 3qx2 + (3q2 + p3)y + q3 = 0 (b) a3 + b3 + g3 = 3abg (c) an equation whose roots are b3 + g3, g3 + a3, a3 + b3 is x3 + 6qx2 + (p3 + 12q2)x + 3p3q + 8q3 = 0 (d) a4 + b4 + g4 = 2p2 Ans. (a), (b), (c), (d) Solution: y = a3 fi a = y1/3 \ y + py1/3 + q = 0

(1)

a3 + b3 + g3 = – 3q = 3abg

Let a, b, g be roots of

x3 – px2 + qx – r = 0, then (a) equation whose a2 – bg, b2 – ga, g2 – ab is x3 + (3q – p2)x2 + q(3q – p2)x + q3 – p3r = 0 (b) a permutation of a, b, g is in G.P. if q3 = p3 r (c) Square of one of the roots will be additive inverse of the product of the other two if q3 = p3 r (d) none of these Ans. (a), (b) a3 - r r = Solution: y = a2 – bg = a 2 a a =



(b2 – c)a + bc – 1 = 0 fi a = Thus,

1 - bc b2 - c

(1 – bc)2 + b(1 – bc) (b2 – c) + c(b2 – c2) = 0 fi

b3 + c3 + 1 – 3bc = 0



(b + c + 1) (b2 + c2 + 1 – bc – b – c) = 0



1 (b + c + 1) [(b – c)2 + (b – 1)2 + (c – 1)2] = 0. 2

Example 47

Let a and b be two distinct real numbers

and p(x) be a quadratic polynomial such that p(a) = a and p(b) = b,

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then (a) p(p(x)) Рx = 0 has at least two real roots. (b) a and b are roots of p(p(x)) Рx = 0 (c) p(p(x)) = x for each x ΠR (d) none of these. Ans. (a), (b) Solution: and

q2 - pr r2

Solution: a, b, g, d are in A.P. fi b–a=d–g fi

p(p(b)) = p(b) = b Suppose m ΠR. The quadratic equation

x2 – (m – 3)x + m = 0

c2

=

(d) none of these Ans. (a), (b), (c)

Use p(p(a)) = p(a) = a

Example 48

b2 - ac

(c) H.P. then

(1)

has (a) real distinct roots if and only if m Œ (– •, 1) » (9, •) (b) both positive roots if and only if m Œ [9, •) (c) both negative roots if and only if m Œ (0, 1] (d) at least one positive root if and only if m Œ (9, •) Ans. (a), (b), (c)



(b + a)2 – 4ba = (d + g)2 – 4dg b2 - ac a2

=

q2 - pr p2

Next, a, b, g, d are in G.P. fi

a g = b d



a -b g -d = a +b g +d



1-

Solution: (1) will have real and distinct roots ¤ D = (m – 3)2 – 4m > 0



b2 q2 = ac pr

¤

m2 – 10m + 9 > 0

¤

m Œ (– •, 1) » (9, •)

Finally, 1/a, 1/b are roots of cx2 + 2bx + a = 0 and 1/g, 1/d are roots of rx2 + 2qx + p = 0.

Roots of (1) will be both positive if D ≥ 0, m – 3 > 0, m > 0 fi

m ≥ 9, m > 3.

\

m Œ [9, •)

Roots of (1) will be negative if

Next, 1/a, 1/b, 1/g, 1/d are in A.P. \

b2 - ac q2 - pr = . c2 r2

a(p + r)2 + 2bpr + c = 0, then (a) q + r = -

2( a + b) p a

(b) qr = p2 +

c a

m £ 1, 0 < m < 3

Thus, m Π(0, 1] For m < 0, equation has one positive and one negative root. Example 49

(c) |q – r| =

2 (2a + b) bp2 - ac |a|

(d) |q – r| =

2 p2 - 4ac |a|

Let a, b be the roots of

ax2 + 2bx + c = 0 and g, d be the roots of px2 + 2qx + r = 0. If a, b, g, d are in (a) A.P. then

b2 - ac q2 - pr = a2 p2

(b) G.P. then

b2 q2 = ac pr

If a(p + q)2 + 2bpq + c = 0 and

Example 50

D ≥ 0, m – 3 < 0, m > 0 fi

4g d 4ab = 12 (a + b ) (g + d )2

Ans. (a), (b), (c) Solution: q and r are roots of a(p + x)2 + 2bpx + c = 0 or

ax2 + 2(a + b)px + ap2 + c = 0

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.19

\

q+r= -

2( a + b) p c , qr = p2 + a a

Next (q – r)2 = (q + r)2 – 4qr 4 = 2 [(a + b)2p2 – a2p2 – ac] a 4

=

a

2

[(2ab + b2) p2 – ac] Let a, b, c be three distinct real numbers

such that each of the expression ax2 + bx + c, bx2 + cx + a and cx2 + ax + b is positive for each x ΠR and bc + ca + ab let a = 2 , then a + b2 + c 2 (a) a < 4 (c) a > 1/4 Ans. (b), (c)

(b) a < 1 (d) a > 1

a2 + b2 + c2 < 4(bc + ca + ab)



1 < a. 4

Suppose |f (x)| £ 1 " x Œ [0, 1], then (a) |a| £ 8 (b) |b| £ 8 (c) |c| £ 1 (d) |a| + |b| + |c| £ 17 Ans. (a), (b), (c), (d) Putting x = 0, 1, 1/2, we obtain

|c| £ 1, |a + b + c| £ 1, fi

1 1 a + b + c £1 4 2

– 1 £ c £ 1, – 1 £ a + b + c £ 1,



– 4 £ 4a + 4b + 4c £ 4

and

– 4 £ – a – 2b – 4c £ 4 – 8 £ 3a + 2b £ 8

Also – 8 £ a + 2b £ 8 \

– 16 £ 2a £ 16 fi |a| £ 8

Since – 1 £ – c £ 1, – 8 £ – a £ 8,

Also a2 + b2 + c2 – (bc + ca + ab) 1 [(b – c)2 + (c – a)2 + (a – b)2] > 0 2

a < 1.

Example 52

Let f (x) = ax2 + bx + c, where a, b, c ΠR.

Adding we get

\

Suppose a, b, c are positive integers and

2

f (x) = ax – bx + c = 0 has two distinct roots in (0, 1), then (a) a ≥ 5 (b) b ≥ 5 (c) abc ≥ 25 (d) abc ≥ 250 Ans. (a), (b), (c) Solution:

Example 53

– 4 £ a + 2b + 4c £ 4

Solution: According to the given conditions a > 0, b2 < 4ac; b > 0, c2 < 4ab; c > 0, a2 < 4bc



Next as a ≥ 5, b ≥ 5, c ≥ 1 we get abc ≥ 25.

Solution:

Example 51

=

Also, b2 ≥ 4ac ≥ 20 fi b ≥ 5.

f (0) and f (1) are of the same sign, thus,

c (a – b + c) > 0 fi c (a – b + c) ≥ 1. Suppose f (x) = a(x – a) (x – b) \

a2 ab(1 – a) (1 – b) ≥ 1

But

a(1 – a) = 1/4 – (1/2 – a)2 £ 1/4

Thus, a2a (1 – a) b(1 – b) < a2/16 as a, b are distinct As a2/16 > 1, we get a > 4 As a Œ I, a ≥ 5.

we get – 16 £ 2b £ 16 fi |b| £ 8 Thus, |a| + |b| + |c| £ 17. Example 54

Let a, b, c > 0 and a, b, c be in G.P. and

w π 1 be a cube root unity. The equation ax2 + bx + c = 0 has (a) real roots (b) imaginary roots (c) roots with negative real parts (d) roots are aw2, aw2 where a = b/a Ans. (b), (c), (d) Solution: x=

b2 = ac and -b ± b2 - 4ac - ac ± i 3ac = 2a 2a

Ê -1 ± 3i ˆ c b b = Á = w, w 2 ˜ 2 ¯ a a a Ë Example 55 If the equation ax2 + bx + c = 0 (a > 0) has two roots a and b such that a < – 2 and b > 2, then (a) b2 – 4ac > 0 (b) c < 0 (c) a + | b | + c < 0 (d) 4a + 2| b | + c < 0.

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Thus d –2 –1

1

2

a

Ans. (a), (b), (c), (d) Solution: Since the equation has two distinct roots a and b, the discriminant b2 – 4ac > 0, we must have f (x) = ax2 + bx + c < 0 for a < x < b. Since



2e 2e d f 2e c + = = = a c c a ac b



d e f , , a b c

b

Fig. 2.17

a < 0 < b we must have, f (0) = c < 0.

c c - 2e + f =0 a a

are in A.P. If ax + by = 1, cx2 + dy2 = 1 have only one

Example 57 solution, then (a)

a2 b2 + =1 c d

(c) y =

b d

(d)

Also, as a < –1, 1 < b we get, f (–1) = a – b + c < 0 and f (1) = a + b + c < 0.

Ans. (a), (b), (c)



Solution:

a + | b | + c < 0.

We have

f (2) = 4a + 2b + c < 0, i.e., 4a + 2| b | + c < 0. cx2 +

If a, b, c are in G.P., then the equations

ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 have a common root then a b c , , are in H.P. (a) d e f a b c , , , are in G.P. (b) d e f



d (1 – ax)2 = 1 b2

(b2c + a2d)x2 – 2adx + d – b2 = 0 D = 4a2d2 – 4(b2c + a2d) (d – b2) = 0



a2d2 + (b2c + a2d)b2 – b2cd – a2d2 = 0



b2[b2c + a2d – cd] = 0 b2c + a2d = cd fi

d e f , , , are in A.P. a b c



(d)

a b c , , , are in A.P. d e f

Also, in this case



x=–

c ,a

(

a x+ c

)

2

=0

c a

If the two given equations have a common root, then this root must be –

(c / a ) .

b2 a2 + =1 d c

x=

2ad ad a = = 2 2 (b c + a d ) cd c

y=

1 - ax 1 Ê a2 ˆ 1 b2 b = Á1 - ˜ = ◊ = b bË c¯ b d d

Ans. (a), (c) Solution: As a, b, c are in G.P., b2 = ac. Now, the equation ax2 + 2bx + c = 0 can be written as

(1)

This quadratic equation will have equal roots if

(c)

ax2 + 2 ac x + c = 0 fi

a b + =1 c d

ax + by = 1 fi y =

f (–2) = 4a – 2b + c < 0

Example 56

a c

1 - ax b Putting this value in the second equation, we get

Next, since a < –2, 2 < b,

and

(b) x =

2

Example 58

For real x, the expression

( x - a) ( x - b) x-c

will assume all real values provided (a) a > b > c (b) a < b < c (c) a > c > b (d) a < c < b Ans. (c), (d)

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Solution: fi

Let y =

Thus, two roots of (1) are 1, 1.

( x - a) ( x - b) x-c

\ product of roots of (2) = 1

x2 – (a + b + y)x + ab + cy = 0



a-b =1 b-c



a, b, c are in A.P.

Since x is real, the discriminant (a + b + y)2 – 4(ab + cy) ≥ 0 fi

y2 + 2(a + b – 2c)y + (a + b)2 – 4ab ≥ 0



y2 + 2(a + b – 2c) y + (a – b)2 ≥ 0

Example 61



a–b=b–c

If b2 ≥ 4ac for the equations ax4 + bx2 + c

4(a + b – 2c)2 – 4(a – b)2 < 0

= 0 then all the roots of the equation will be real if (a) b > 0, a < 0, c > 0 (b) b < 0, a > 0, c > 0 (c) b > 0, a > 0, c > 0 (d) b > 0, a < 0, c < 0 Ans. (a), (b)



(a + b – 2c + a – b) (a + b – 2c – a + b) < 0

Solution:



4(a – c) (b – c) < 0 fi (a – c) (b – c) < 0

(1)

Since y takes all real values, (1) is possible if and only if

f (y) = ay2 + by + c = 0

This is possible if c lies between a and b, that is, if a < c < b or a > c > b.

(1)

The given equation will have four real roots if (1) has two non-negative roots. This can happen if -b ≥ 0, af (0) ≥ 0, b2 – 4ac ≥ 0 a

If sin a, cos a are the roots of the equation

Example 59

Put x2 = y. The given equation becomes

ax2 + bx + c = 0, (c π 0) then (b) (a + c)2 = b2 + c2 (a) a2 – b2 + 2ac = 0 (c) a2 + b2 – 2ac = 0 (d) (a – c)2 = b2 + c2. Ans. (a), (b)

Thus a and b must have opposite sign whereas a, – b and c must have the same sign.

Solution:





We have

sin a + cos a = –

b c , sin a cos a = a a

a > 0, b < 0, c > 0

Example 62 4

Now, 1 = sin a + cos a 2

2

= (sin a + cos a) – 2sin a cos a 1=

b 2 2c a a2



a2 = b2 – 2ac fi a2 – b2 + 2ac = 0



a2 + c2 + 2ac

Example 60

= b2 + c2 fi (a + c)2 = b2 + c2

If x2 + mx + 1 = 0

(b – c)x2 + (c – a)x + (a – b) = 0

have both the roots common, then (a) m = – 2 (b) m = – 1 (c) a, b, c are in A.P. (d) a, b, c are in H.P. Ans. (a), (c) Solution: \ x

Note x

Thus 1 + m + 1 = 0



m=–2

Therefore, (1) becomes x2 – 2x + 1 = 0



(x – 1)2 = 0.

[∵ b2 – 4ac ≥ 0 is given]

or

a < 0, b > 0, c < 0.

For x ΠR, consider the equation

x + 5 x + 243 = 3

(1)

(a) (1) has at least one solution. (b) x = 0 is the only solution of (1) (c) (1) has exactly 4 solutions (d) (1) has exactly 5 solutions. Ans. (a), (b)

2

and

– ab ≥ 0, ac ≥ 0

(1) (2)

Solution x≥0 1/5 5 1/5 fi (x + 243) = (x + 3 ) ≥ 3 \x For x > 0, x1/4 + (x + 35)1/5 > 3. Thus x = 0 is the only solution of (1) Example 63

The polynomial P(x) has remainder x + 8

when P(x) is divided by Q(x) = x3 – 7x2 + 14x – 8 and P(x) P(16) = 3p(2). Let R(x) be the remainder when P(x2) is divided by Q(x), then (a) R(x) is a linear polynomial (b) R(x) is a quadratic polynomial (c) R(x) = 0 has real roots

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(d) R(x) = 0 does not have real roots. Ans. (b), (d) Solution: Q(x) = (x – 1) (x – 2) (x – 4) Suppose P(x) = A(x) Q(x) + (x + 8) As

P(2) = 10, we get P(16) = 30.

Next assume P(x2) = B(x) Q(x) + (ax2 + bx + c) a = P(1) = a + b + c

(1)

2

and

12 = P(4) = P(2 ) = 4a + 2b + c

(2)

30 = P(16) = P(42) = 16a + 4b + c

(3)

g, d be the roots of x2 – 36x + B = 0. If a, b, g, d forms an increasing G.P., then (a) B = 81 A (b) A = 3 (c) B = 243 (d) A + B = 251 Ans. (a), (b), (c) Solution: Let common ratio be r, then a(1 + r) = a + b = 4, a r2 (1 + r) = 36 \

r2 = 9 fi r = 3. Thus, a = 1

Therefore A = ab = a2r = 3 and

From (1), (2), (3) we get

B = g d = a2r5 = 243

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

a = 2, b = –3, c = 10. 2

Thus, R(x) = 2x – 3x + 10 and

Let a, b be the roots of x2 – 4x + A = 0 and

Example 65

R(x) = 0

Let -

Example 66

does not have real roots. If a, b are the roots of the equation

Example 64

ax2 + 2bx + c = 0 and a + h, b + h are the roots of the equations Ax2 + 2Bx + C = 0, then (a) h =

b B a A

(c) h =

Ac + aC A+ a

(b)

b2 - a c B 2 - AC b - ac

=

2

(d)

B - AC 2

=

a2 A2 a A

Ans. (a), (b)

Suppose a1 and b1 be the roots of the equation x2 – 2xsecq + 1 = 0 and a2 and b2 be the roots of equation x2 + 2xtanq – 1 = 0. Suppose a1 > b1 and a2 > b2. Column 1 Column 2 (a) a1 + b2

(p) 2(sec q – tan q)

(b) a2 + b1

(q) 2sec q

(c) a1 + a2

(r) –2tan q (s) 0

(d) b1 + b2

Solution: We have 2b a+b=– a and fi

2 B 2b 2h = + A a



b B h= a A

(a – b)2 = (a + b)2 – 4ab =



4 b2 a

2

-

4c 4 (b2 - ac) = a a2

(a – b) = [(a + h) – (b + h)] = 2

4 (b - ac)

2

2

Thus,

a2

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution:

Also

and

p

Ans.

2B (a + h) + (b + h) = – A

4 ( B - AC ) 2

=

b2 - a c a2 = A2 B 2 - AC

A2

4 ( B 2 - AC ) A2

p p 0, tanq < 0. therefore, a1 = sec q – tan q b1 = sec q + tan q Also, x2 + 2xtanq – 1 = 0 fi

(x + tanq)2 = 1 + tan2q = sec2q



x = –tanq ± secq.

As –p/6 < q < –p/12, tanq < 0, secq > 0 therefore a2 = – tanq – secq

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.23

b2 = – tanq + secq \

a1 + b2 = 2(secq – tanq) a2 + b1 = 0 a1 + a2 = – 2tanq b1 + b2 = 2secq

Example 67 Match the equations in Column 1 with the properties in Column 2. Column 1 Column 2 (a) a < b < c < d and (p) real roots equation is (x – a) (x – c) + p(x – b) (x – d) = 0 (b) a > 0, a + b + c < 0 (q) distinct real roots and equation is ax2 + bx + c = 0 (c) b, c, Œ I and the (r) integral roots equation x2 + bx + c = 0 has rational roots (d) a, b, c, d Œ R are (s) discriminant ≥ 0 in G.P. and equation is (a2 + b2 + c2)x2 + 2(ab + bc + ca)x + b2 + c2 + d2 = 0 p q r s Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: (a) Let P(x) = (x – a) (x – c) + p(x – b) (x – d) Note that P is continuous on R and P(a) > 0, P(b) < 0, P(c) < 0, P(d) > 0 \ P(x) = 0 has a root in (a, b) and a root in (c, d). (b) As a > 0, the parabola y = ax2 + bx + c open upwards. Also, as y(1) < 0, y = ax2 + bx + c meets the x–axis twice. \ ax2 + bx + c = 0 has distinct real roots. (c) x2 + bx + c = 0 has integral roots. See Theory page 2.1. (d) If r is the common ratio of the G.P., we can write the given equation as (x2 + 2rx + r2) (1 + r2 + r4) = 0 fi (x + r)2 = 0 fi x = – r, – r.

Example 68 Match the equations in Column 1 with the properties in Column 2. Column 1 Column 2 (a) a, b, c Œ R (p) real roots are such that a + b + c = 0. Equation is 3ax2 + 2bx + c = 0 (b) a, b, c Œ R are (q) discriminant > 0 such that 2a + 3b + 6c = 0. Equation is ax2 + bx + c = 0 (c) a, b, c Œ R are (r) at least one root in such that (0, 1) (a + c)2 < b2. Equation is ax2 + bx + c = 0 (d) a, b, c Œ R (s) a root in (– •, – 1) are such that or a root in (1, •) |b| c + < 0. a a Equation ax2 + bx + c = 0 p q r Ans. a p q r 1+

s s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: (a) Consider the polynomial P(x) = ax3 + bx2 + cx and use Rolle’s Theorem on [0, 1]. (b) Consider the polynomial 1 1 1 P(x) = ax3 + bx2 + cx 3 2 6 and use Rolle’s theorem on [0, 1]. (c) Let P(x) = ax2 + bx + c Note that P(–1) P(1) < 0 fi P(x) vanishes at least once in (–1, 1) fi P(x) = 0 has real roots. (d) Let P(x) = ax2 + bx + c b c and Q(x) = x2 + x + a a

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As x Œ R, (y + 3)2 + 4(y – 1) (6y + 2) ≥ 0

b c + < 0, we get As 1 + a a Q(–1) or Q(1) < 0 Since Q(x) Æ• as x Ʊ•, we get Q(x) = 0 or P(x) = 0 has a root in (–1, •) or (1, •).

fi 25y2 – 10y + 1 = (5y – 1)2 ≥ 0 However, y = 1 fi x = 2 for which expression 0 . Therefore, y π 1. Thus, range of the 0 given expression R – {1}. becomes

If x ΠR, match each expression with its

Example 69 range.

Column 1 (a)

Column 2

x2 – 2x + 3

(d) y =

(p) (– •, 3] » [4, •)

x2 – 2x – 8

dy =1+ dx

2

(b)

x – 12 2x – 7

(q) R – {1}

x – 3x + 2

3

( x – 1)2

>0"xπ1

\ y increases as x increases. Thus, range is R.

2

(c)

x 2 – 3x – 1 3 =x–2– x –1 x –1

(r) R

x2 + x – 6

Let a, b, g be three numbers such that

Example 70

x – 3x + 1 x –1 2

(d)

(s) (– •, – 2/9] » (1, •)

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

1 1 1 1 1 1 1 9 + + = , 2+ 2+ 2 = , a b g 2 a 4 b g and a + b + g = 2. Column 1 abg bg + ga + ab a2 + b2 + g2 a3 + b3 + g3

(a) (b) (c) (d)

Solution: (a) y =

x – 2x + 3 x – 2x – 8

fi (x – 2x – 8) y = x – 2x + 3 2

fi (y – 1) x2 – 2(y – 1) x – (8y + 3) = 0 As x Œ R, 4(y – 1)2 + 4 (y – 1) (8y + 3) ≥ 0 fi (y – 1) (9y + 2) ≥ 0 fi y £ – 2/9 or y ≥ 1. But y = 1 fi 3 = – 8. Not possible. Thus, y Œ (– •, – 2/9] » (1, •) x 2 – 12 2x – 7

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

fi y2 – 7y + 12 ≥ 0 fi (y – 3) (y – 4) ≥ 0 fi y Œ (– •, 3] » [4, •) x2 + x – 6

Ê 1 1 1ˆ Ê 1 1 1 ˆ 2Á = Á + + ˜ + + ˜ Ëa b g ¯ Ë bg ga ab ¯

\

2

Ê 1 1 1ˆ – Á 2 + 2 + 2˜ b g ¯ Ëa

2(a + b + g ) = – 2 fi abg = – 2 abg

Also,

As x Œ R, 4y2 – 4 (7y – 12) ≥ 0

x 2 – 3x + 2

Solution:



fi x2 – 2yx + 7y – 12 = 0

(c) y =

q

2

2

(b) y =

p

Ans. 2

(p) (q) (r) (s)

Column 2 6 8 –2 –1

bg + ga + ab 1 = abg 2



bg + ga + ab = – 1.

a2 + b2 + g2 = (a + b + g)2 – 2(bg + ga + ab) =6

fi (y – 1) x2 + (y + 3) x – (6y + 2) = 0

Lastly a + b + g – 3abg 3

3

3

= (a + b + g) (a2 + b2 + g2 – bg – ga – ab)

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.25

The set of value(s) of k ΠR for which

Example 71

Column 1 (a) kx2 – (k + 1)x + 2k – 1 = 0 has no real roots (b) x2 – 2(4k – 1)x + 15k2 – 2k – 7 > 0 for each x (c) Sum of the roots of x2 + (2 – k – k2)x – k2 = 0 is zero (d) The roots of x2 + (2k – 1)x + k2 + 2 = 0 are in the ratio 1 : 2 p q r s Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Column 2 (p) {1, – 2}

(q) (– •, – 1/7) » (1, •) (r) {– 4} (s) (2, 4)

Solution: (a) Any rational root of x3 – px2 + 1 = 0 must be an integer. But for a Œ I, a2(p – a) = 1 is not possible if p > 2. (b) As in (i) any root of x10 – x9 – 2 = 0 must be an integer. Clearly x For, x π – 1, x Œ I, x9(x – 1) = 2 is not possible. (c) The given equation can be written as (x2 + 5x + 4) (x2 + 5x + 6) = 24 Put x2 + 5x = t to obtain t2 + 10t = 0 fi t = 0, t = – 10. For t = 0, x2 + 5x = 0 fi x = 0, x = – 5 For t = – 10, x2 + 5x = – 10 does not have rational roots. (d) Put log3 x = t to obtain 1- t t2 + = 1 fi t3 + t2 – 2t = 0 1+ t fi t(t + 2) (t – 1) = 0 fi t = 0, 1, – 2 This gives x = 1, 3, 1/9.

Solution: (a) (k + 1)2 – 4k (2k – 1) < 0, k π 0 fi 7k2 – 6k – 1 > 0 fi (7k + 1) (k – 1) > 0 fi k < – 1/7 or k > 1 (b) (4k – 1)2 – (15k2 – 2k – 7) < 0 fi k2 – 6k + 8 < 0 fi (k – 2) (k – 4) < 0 fi 2 0 and g, d (g < d) are the roots of ax2 + bx + c = 0, then graph of the parabola y = ax2 + bx + c, for g < x < d lies below the x-axis. Ans. (a) a, b are roots of ax2 + bx + c = 0 ¤ a, b are b c x+ = 0. roots of x2 + a a x2 > 0, graph of the parabola y = x2 + b c x+ for a < x < b lies below the x-axis. As –1, 1 Œ a a b c b c (–1) + < 0 and 1 + + 0, bc + ca + ab ¤

a > 0, b > 0, c > 0.

a2 + b2 .

Ans. (c)

As q p and hcf (p, q) = 1 we get q = 1 Thus,

Equation whose roots are a, b, c is (x – a) (x – b) (x – c) = 0

Statement-2: If a, b ΠR, then |ab|
0.

Example 84

roots of x2 + ax + b = 0 are rational then these roots must be integers.

Solution:

Solution: Statement-2 is true as x < 0 implies x3 – px2 + qx – r < 0 [∵ p, q, r > 0]

As this equation cannot have negative roots a, b, c > 0.

which is real. This contradicts our assumption that f (x) is non-real for non-real x. Therefore, a = 0. Statement-2 is false since – i is a root of x2 + x + 1 + i = 0 but i is not a root of x2 + x + 1 + i = 0. Example 82

Statement-2: If p, q, r > 0, then the equation x3 – px2 + qx – r = 0 cannot have negative roots.

fia+b+c=0

Also, x1 = b and x3 = a. Equation whose roots are x1 and x3 is x2 – (a + b)x + ab = 0 or x2 + cx + ab = 0. Statement-2 is false can be checked by taking a = b = 0.

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.29

Example 86

Statement-1: The equation x x2 x3 x 4 + + + =0 1 2! 3! 4 !

f (x) = 1 +

has two pairs of repeated roots. Statement-2: Polynomial equation P(x) = 0 has a repeated root a if P(a) = 0 and P¢(a) = 0. Ans. (d) Solution: If a is a repeated root of f (x) = 0, then f (a) = 0 and f ¢(a) = 0. a a a a + + + =0 1 2! 3! 4 ! 2

3

4

\

1+

and

a a2 a3 1+ + + =0 1 2! 3!



a4 =0 4!



a = 0.

Solution: By multiplying a, b, c by an appropriate natural number, we may assume that a, b, c are integers. We may further assume that a, b, c have no factor in common. Now, a + b21/3 + c22/3 = 0 fi

a3 = – 2 (b + 21/3 c)3 = – 2(b3 + 2c3 + 3abc)



2|a3

fi 2|a

fi a = 2a1 for some a1 ΠI.

Thus, 4a13 = – (b3 + 2c3 – 6a1bc) fi

b3 = – 2(2a13 + c3 – 3a1bc)



2|b3

fi 2|b

fi b = 2b1 for some b1 ΠI.

Therefore 4b13 = – (2a13 + c3 – 6a1b1c) fi

c3 = – 2(a13 + 2b13 – 3a1b1c)



2|c3



2|c

But a = 0 does not satisfy f (x) = 0.

This is a contradiction.

For truth of statement-2, see Theory.

Statement-2 is false as 21/3 is irrational but is a root of x3 – 2 = 0.

Example 87

Statement-1: If all the four roots of x4 – 4x3 + ax2 – bx + 1 = 0

are positive, then a = 6 and b = 4. Statement-2: If polynomial equation P(x) = 0 has four positive roots, then the polynomial equation P¢(x) = 0 has 3 positive roots. Ans. (b) Solution: Let x1, x2, x3, x4 be four positive roots of x4 – 4x3 + ax2 – bx + 1 = 0, then x1 + x2 + x3 + x4 = 4 and x1 x2 x3 x4 = 1. fi

1 (x1 + x2 + x3 + x4) = (x1 x2 x3 x4)1/4 4

That is, A.M. = G.M., but this is possible if and only if x1 = x2 = x3 = x4 = 1. Thus, the given equation becomes (x – 1)4 = 0



a = 6, b = 4

Example 89

a, b are the roots of equation x2 + 2px + q = 0 and a, 1/b are roots of the equation ax2 + 2bx + c = 0, where b2 œ {–1, 0, 1}. Statement-1: (p2 – q) (b2 – ac) ≥ 0 and Statement-2: b π pa or c π qa Ans. (a) Solution:

Statement-1: If a, b, c ΠQ and 21/3

a + bx + cx2 = 0, then a = 0, b = 0, c = 0. Statement-2: cients cannot have irrational roots. Ans. (c)

Suppose a ΠR.

As a, b, c, p, q ΠR, the other root of x2 + 2px + q = 0 and ax2 + 2bx + c = 0 must be a . Thus a = b = 1/b fi b2 = 1. But

b2 œ {– 1, 0, 1}.

\

aŒR



D1 = 4(p2 – q) ≥ 0 and D2 = 4 (b2 – ac) ≥ 0



b, 1/b ΠR

Next, let b = pa and c = qa, then ax2 + 2bx + c = a (x2 + 2px + q)

Statement-2 follows immediately from the Rolle’s theorem. Example 88

Let a, b, c, p, q be real numbers. Suppose



both roots of the two equations are common.



b = 1/b



b2 = 1. A contradiction.

Statement-2 is true but is not the correct explanation for the statement-1.

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Example 90

Statement-1: Let a, b, c be real numbers,

a π 0. If a is a root of a2x2 + bx + c = 0, b is a root of a2x2 – bx – c = 0 and 0 < a < b, then the equation a2x2 + 2bx + 2c = 0 has a root g a < g < b.

(a)

( x - x1 )(2 x0 - x1 - x ) ( x - x0 )( x - x0 - t ) y0 + y1 ( x1 - x0 )( x1 - x0 - t ) ( x0 - x1 )( x0 - x1 + t )

(b)

( x - x1 )(2 x0 - x1 - x ) Ï ( x1 - x0 ) y1 ¸ Ì y0 + ˝ x1 - x0 - t ˛ ( x0 - x1 )( x0 - x1 - t ) Ó

(c)

( x - x0 )(2 x0 - x1 - x ) Ï ( x0 - x1 ) y1 ¸ ˝ Ì y0 + ( x0 - x1 )( x0 - x1 + t ) Ó x1 + t - x0 ˛

Statement-2: If P(x) is a polynomial such for some a < b, P(a) P(b) < 0, then there exists a g such that P(g) = 0. Ans. (a) Solution: Statement-2 is true [See Theory] Next, note that

and

(d) none of these

a2 a2 + ba + c = 0

(1)

a2 b2 – bb – c = 0

(2)

where 0 < a < b, and a π 0. Let

f (x) = a2x2 + 2bx + 2c

Since f is a polynomial function, f is continuous on R and hence on [a, b]. Also, f (a) = a2a 2 + 2ba + 2c = 2(0) – a2a 2 = – a2a 2 < 0

[using (1)]

f (b) = a b + 2bb + 2c 2 2

= 3a2 b2 – 2(a2b 2 – bb – c) = 3a2 b2 – 2(0) = 3a2 b2 > 0

[using (2)]

Since f is continuous on [a, b] and f (a) < 0, f (b) > 0, f must vanish at least once in (a, b). Thus, statement-1 is true and statement-1 is correct explanation for it.

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 91 to 95 A polynomial p(x) of degree two or less which takes values y0, y1, y2 at three distinct values x0, x1, x2 respectively, is given by p(x)* =

( x - x0 )( x - x2 ) ( x - x1 )( x - x2 ) y0 + y1 ( x1 - x0 )( x1 - x2 ) ( x0 - x1 )( x0 - x2 ) +

Example 91

Let p be a polynomial of degree 2 and

a π 0, 1, then (a) p(x) = x[(x – 1) p(0) + xp(a)] + x(x – 1) p(a) (b) p(x) = x(x – a) p(0) + ap(a) {x(x – a) + x2} È x ( x - a) ˘ (c) p(x) = Í1 p(0) 1 - a ˙˚ Î x (1 - x ) p( a) - p(0) x ( x - a) + p(1) 1- a a 1- a (d) none of these. +

= 2(a2a2 + ba + c) – a2a2

and

Example 92

( x - x0 )( x - x1 ) y2 ( x2 - x0 )( x2 - x1 )

A polynomial of degree 2 which takes

Example 93

In Example 67 if a Æ 0, then p(x) is

given by (a) (1 – x2) p(0) + x(1 – x) p¢(0) + x2 p(1) (b) x2 p(0) + xp¢(0) + p(1) (c) x2 p(0) + xp¢(0) + xp(1) + p¢(1) (d) none of these Example 94

A polynomial of degree 2 which takes

values – 1, 0 and 1 at – 1, 0 and 1 respectively is (a) x2 – x + 1 (b) x2 – x + 3 2 (d) none of these. (c) x + x – 1 Example 95

Condition under which p(x) will be a

polynomial of degree 2 is (a)

(c)

x0 x1 x2

y0 1 y1 1 π 0 y2 1

x02

y1 1

x12

y1 1 = 0

x22

y2 1

(b)

(d)

x0

y02 1

x1

y12 1 π 0

x2

y22 1

x02

y02 1

x12

y12 1 = 0

x22

y22 1

values y0, y0, y1 at points x0, x0 + t, x1 (t π 0) is given by Ans. 91. (d), 92. (c), 93. (a), 94. (d), 95. (a) Solution: 91. * It is called as Lagrange’s interpolation polynomial.

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.31

p(x) =

( x - x0 - t )( x - x1 ) ( x - x0 )( x - x1 ) y0 + y0 ( -t )( x0 - x1 ) (t )( x0 + t - x1 ) ( x - x0 )( x - x0 - t ) + y1 ( x1 - x0 )( x1 - x0 - t )

( x - x1 )(2 x0 - x1 - x + t ) ( x - x0 )( x - x0 - t ) = y0 + y1 ( x1 - x0 )( x1 - x0 - t ) ( x0 - x1 )( x0 - x1 + t ) 92.

x ( x - 1) ( x - a)( x - 1) p(a) + p(0) a ( a - 1) a x ( x - a) + p(1) 1- a

p(x) =

=

x(1 - x ) È p( a) - p(0) ˘ x( x - a) ˙˚ + 1 - a p(1) a 1 - a ÍÎ + a (x, a) p(0)

( x - 1) x ( x - 1)( x - a) + where a (x, a) = a ( a - 1) a =

x ( x - a) x -1È x ˘ + x - a˙ = 1 – Í 1- a a Îa -1 ˚

93. Take limit as a Æ 0 in (c) of Example 92. 94. p(x) is a polynomial of degree 1. In fact p(x) = x. 95. If point (x0, y0), (x1, y1), (x2, y2) are not collinear, then a parabola can pass through these points.

Paragraph for Question Nos. 96 to 100 Let x0, x1 and x2 polynomials l0(x), l1(x), l2(x) and l(x) as follows:

and

l0(x) =

( x - x1 )( x - x2 ) ( x0 - x1 )( x0 - x2 )

l1(x) =

( x - x0 )( x - x2 ) ( x1 - x0 )( x1 - x2 )

l2(x) =

( x - x0 )( x - x1 ) ( x2 - x0 )( x2 - x1 )

l(x) = (x – x0) (x – x1) (x – x2) Example 96

p(x) = l0(x) + l1(x) + l2(x) equals (b) x (d) 1 + x + x2

(a) 1 (c) x2 Example 97

p(x) = (x1 + x2) l0(x) + (x2 + x0) l1(x) + (x0

+ x1) l2(x) then p(x) equals (a) x

1 (b) x – (x0 + x1 + x2) 2

(c) x0 + x1 + x2 – x

(d) none of these

Example 98

The Polynomial

l ( x) l ( x) l ( x) + + ( x - x1 ) l ¢( x1 ) ( x - x0 ) l ¢( x0 ) ( x - x2 ) l ¢( x2 ) equals (b) x (d) none of these

(a) 1 (c) x2 + x 2

Example 99

 lk(x) x2 equals

k =0

(b) x (d) none of these

(a) 0 (c) x2 Example 100

If

A0 A A2 1 + 1 + = then x - x0 x - x1 x - x2 l( x)

Ai(for i = 0, 1, 2) equals l ¢( xi ) xi

(a)

1 l ¢( xi )

(b)

(c)

xi l ¢( xi )

(d) l¢(xi)

Ans. 96. (a), 97. (c), 98. (a), 99. (c), 100. (a) Solution: 96.

p(x) –1 = 0 for x = x0, x1, x2.

Thus, p(x) = 1 " x. 97. p(x) – (x0 + x1 + x2 – x) = 0 for x = x 0, x 1, x 2. \ p(x) = x0 + x1 + x2 – x " x 98. l¢(x) = (x – x1) (x – x2) + (x – x0) (x – x2) + (x – x0) (x – x1) This gives l ( x) = l0(x) etc. ( x - x0 ) l ¢( x0 ) Now, use Example 96. 99. Similar to Example 96. 100. Ai can be obtained by putting x = xi in

1 after deletl( x)

ing the factor (x – xi) of l(x).

Paragraph for Question Nos. 101 to 105 A polynomial p(x) = a0 xn + a1 xn–1 + ... + an–1 x + an, a0 π 0 is said to be a reciprocal equation if ai = an–i for 0 £ i £ [n/2] where [x] denote the greatest integer £ x. Example 101

If p(x) is a reciprocal polynomial of odd

degree, then one of the roots of p(x) = 0 is (a) – 1 (b) 1 (c) 0 (d) (n + 1)/2

IIT JEE eBooks: www.crackjee.xyz 2.32 Comprehensive Mathematics—JEE Advanced

If p(x) is a reciprocal equation and a π

Example 102

0 is a root of p(x) = 0, then which one of the following is also a root of p(x) = 0? (a) – a (b) 1/a (c) a – 1 (d) none of these Sum of the rational roots of

Example 103

133x - 78 x5 = 133 - 78 x

is

(a) 2/9 (c) 13/6

Now

(x10 – 1) + (x5 – 1) + 5 = (x5 – 1) q(x) + 5 where q(x) = (x15 + x10 + x5 + 1)

(1)

\ the required remainder is 5. 105. p(x) =

(b) x – 1 (d) 5

(

Let m, n ΠN and p(x) = 1 + x + ... + xm. p(x) will divide p(xn) if (a) hcf (m, n) = 1 (b) hcf (m + 1, n) = 1 (c) hcf (m, n + 1) = 1 (d) hcf (m + 1, n + 1) = 1

Paragraph for Question Nos. 106 to 110

Ê 1ˆ p(x) = xn p Á ˜ Ë x¯

A polynomial p(x) = a0 xn + a1 xn–1 + ... + an–1 x + an where a0, a1, ... an Œ I, a0 π 0 is said to be irreducible over I if p(x) cannot be written as product of two non-constant polynomials of degrees < n. For example, p(x) = x3 + x2 + x + 1 is not irreducible over I as p(x) = (x2 + 1) (x + 1) where as q(x) = x2 + x + 1 is irreducible over I.

p(a) = 0 ¤ p(1/a) = 0

Eisenstein’s Irreducibility Criterion

Ans. 101. (a), 102. (b), 103. (c), 104. (d), 105. (b) Solution: 101 As n is odd (– 1)i and (– 1)n–i are of opposite sign. Now use the fact that ai = an – i. 102. As p(x) is a reciprocal polynomial

103. We can write (1) as 78x6 – 133x5 + 133x – 78 = 0

(2)

Let p be a prime and a0, a1, ... an Œ I be such that p | ai for 1 £ i £ n, p a0 and p2 an, then the polynomial a0 xn + a1 xn–1 + ... + an–1 x + an

Clearly 1 and – 1 are root of (2)

is irreducible over I.

Dividing (2) by x2 – 1, we get quotient as 78x4 – 133x3 + 78x2 – 133x + 78 = 0

(3)

Example 106

2

Dividing (3) by x and putting x + 1/x = y we get 78y2 – 133y – 78 = 0 y = 13/6, – 6/13

Now, x + 1/x = 13/6 fi x = 2/3, 3/2 and

)

1 - x ( m+1) n (1 - x ) p( x n ) = p( x ) (1 - x n )(1 - x m+1 ) is a polynomial. Since both 1 – xn and 1 – xm + 1 divide 1 – x(m + 1)n and since factors of 1 – x(m + 1)n are distinct, it is necessary xm + 1 – 1 and xn – 1 have no factor in common except x – 1, that is, hcf (m + 1, n) = 1.

Example 105



1 - x m+1 . 1- x m, n so that

Remainder when p(x5) is divided by

p(x) = x4 + x3 + x2 + x + 1 is (a) x + 1 (c) x3 + 1

p(x5) = (x20 – 1) + (x15 – 1) +

+ (x10 + x5 + 1) + (x5 + 1) + 1

(b) 9/2 (d) 6/13

Example 104

\

104. (x – 1) p(x) = x5 – 1

x + 1/x = – 6/13 gives imaginary roots.

\ sum of rational roots is 1 + (– 1) + 2/3 + 3/2 = 13/6.

(a) (b) (c) (d)

Let a, b ΠI. The polynomial

p(x) = (x – a)2 (x – b)2 + 1. p(x) is irreducible over I p(x) has a rational zero p(x) has two integral none of these

Example 107

Let p(x) = x5 – 5x4 + 25x – 5, then

(a) p(x) is irreducible over I (b) p(x) has at least one rational zeros

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.33

(c) p(x) has three integral zeros (d) none of these. Let p be a prime and

Example 108

p(x) = x then (a) (b) (c) (d)

p–1

+x

p–2

+ ... + x + 1

Least degree of the polynomial with a = 31/2007 is a zero is (b) 669 (d) none of these

(a) 223 (c) 2007

If a, b, c and d are different integers then

Example 110

Ci =

As p is prime, p

p(x) = 0 has no real roots p(x) has at least one rational root p(x) is irreducible over I none of these

Example 109

(a) (b) (c) (d)

p

p(x) = (x – a) (x – b) (x – c) (x – d) – 1 is irreducible over I has at least two rational roots has four real roots none of these.

Ans. 106. (a), 107. (a), 108. (c), 109. (c), 110. (a)

p! ( p - 1)!i ! (p – i)! and p

i!

1 £ i £ p – 1. Also, p | p!

for

Thus, p | pCi for 1 £ i £ p – 1. 1 and p2

As p

p

C p -1 , p(x + 1) and hence p(x) is

irreducible over I by the Eisenstein’s irreducible over I. 109. Clearly a is zero of f (x) = x2007 – 3. Also by the Eisenstein’s criterion f (x) is irreducible over I. If g(x) is the desired polynomial of the least degree then by the division algorithm there exist q(x) and r(x) such that f (x) = g(x) q(x) + r(x) where

r(x) = 0 or deg r(x) < deg g(x)

Let

r(x) π 0, then since r(a) = f (a) – g(a) q(a) = 0 deg r(x) < deg g(x), we reach a contradiction.

and

r(x) = 0 and f (x) = g(x) q(x)

Thus,

and f (x) is irreducible over I and deg g(x) ≥ 1, q(x) must be a constant. Therefore, f (x) is the desired polynomial. Let p(x) = q(x) r(x) where q(x), r(x) are non-constant

110.

Solution: 106. Suppose p(x) = q(x) r(x) where q(x) and r(x) are non-constant polynomials of degrees < 4. We have q(a) = r(a) = q(b) = r(b) = ± 1

Also degree of q(x), r(x) and hence of q(x) + r(x) < 4.

fi both p(x) ± 1 and q(x) ± 1 are divisible by

Since

(x – a) (x – b). As deg q(x) + deg r(x) = 4, we have q(x) ± 1 = r(x) ± 1 = (x – a) (x – b) \ fi fi

q(x) r(x) = [(x – a) (x – b) ± 1]

2

p(x) = p(x) ± 2(x – a) (x – b)

q(a) = – r(a) = ± 1 for a = a, b, c, d

we get

p(x) = – (q(x))2 x4 on the L.H.S. is positive

and on the R.H.S. it is negative.

INTEGER-ANSWER TYPE QUESTIONS

107. Use Eisenstein’s Irreducibility criterion with p = 5.

Example 111

x p -1 108. p(x) = x -1

If x, y, z ΠR, x + y + z = 4 and x2 + y2 +

z2 = 6, then maximum possible value of z is Ans. 2 Solution: x + y = 4 – z, x2 + y2 = 6 – z2,

( x + 1) p - 1 x

\

= xp–1 + pC1 x p – 2 + ...

2xy = (x + y)2 – (x2 + y2) = (4 – z)2 – (6 – z2) = 2z2 – 8z + 10

+ pC p - 2 x + pC p -1 For1 £ i £ p – 1,

q(x) + r(x) ∫ 0.

\

(x – a) (x – b) = 0. A contradiction.

fi p(x + 1) =

q(a) + r(a) = 0 for a = a, b, c, d,

The quadratic equation whose roots are x and y is t2 – (x + y)t + xy = 0

IIT JEE eBooks: www.crackjee.xyz 2.34 Comprehensive Mathematics—JEE Advanced

t2 – (4 – z)t + z2 – 4z + 5 = 0



m2 – 4m + 4 < m2 – 4

and

16 – 8m + m2 < m2 – 4

(4 – z)2 – 4(z2 – 4z + 5) ≥ 0



2 < m and m > 5/2

¤

3z – 8z + 4 £ 0 ¤ (3z – 2) (z – 2) £ 0

\ the least integral value of m is 3.

¤

2/3 £ z £ 2.

or

Since x and y are real 2

Example 114

That z can take value 2 can be checked by taking x = y = 1. The number of rational roots of

Example 112

satisfying the equation x2 + |x – 1| = 1 is Solution:

Ans. 2

For x ≥ 1, (1) becomes

x2 + x – 1 = 1

p(x) = (2x + 3) (x97 + x96 + ... + 1). = (2x + 3) (x + 1) (x96 + x94 + ... + x2 + 1)

The sum of all the integral roots of

2

(log5 x) + log5x (5/x) = 1 is

Example 115 (1)

Ans. 6 Solution:

Clearly x > 0 and x π 1/5 log5x (5/x) =

log5 5 - log5 x log5 5 + log5 x

Putting log5 x = t, then equation (1) becomes t2 + ¤

1- t =1 1+ t

¤

t(t – 1) (t + 2) = 0

t3 + t2 – 2t = 0 ¤

The least integral value of m for which

mˆ m2 Ê x < –1 ˜ ÁË 2¯ 4

m – 2



m2 m m2 -1 < x < + -1 2 4 4

m2 - 4 and 4 – m
0 fi a < 9/4 Also, a – 2 product of roots of (1) ≥ 0. Thus, 2 £ a < 9/4. Example 116

The sum of all the real roots of the (1)

is Ans. 8 Put 4 – x = t, so that (1) becomes

(t – 1)5 – (t + 1)5 = – 32 fi

10t4 + 20t2 – 30 = 0



(t2 + 3) (t2 – 1) = 0

\

m2 - 4

t = ± 1 fi x = 3, 5.

Example 117 x

m2 – 4 m m2 < 1 and 2 < + -1 2 4 4

m–2
0

We must have m 2

Solution:

Clearly m > 0. Now x2 – mx + 1 < 0

Thus, m2 – 4 > 0 or m > 2 and

x=1

has non-negative roots is Ans. 2

Solution: 2

¤



equation (3 – x)5 + (x – 5)5 = – 32

every solution of the inequality 1 £ x £ 2 is a solution of the inequality x2 – mx + 1 < 0 is Ans. 3 Solution:

(x + 2) (x – 1) = 0

which the equation (2 – x) (x + 1) = a

t = 0, 1, – 2

So integral roots of (1) are 1 and 5. Example 113

¤

For x < 1, (1) becomes x2 + 1 – x = 1 fi x = 0.

Also, x96 + x94 + ... + x2 + 1 > 1 " x ΠR. Example 112

(1)

Ans. 1

p(x) = 2x98 + 3x97 + 2x96 + ... + 2x + 3 = 0 is Solution:

The sum of all the real numbers

2007

The degree of the remainder when

– 1 is divided by (x2 + 1) (x2 + x + 1) is

Ans. 3. Solution: x

2007

Write as

– 1 = q(x) (x2 + 1) (x2 + x + 1) + ax3 + bx2 + cx + d

and substitute x = w, w2, i, – i to show a = 1.

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.35

Let p(x) be a polynomial with integral

Example 118

and

f≤ (x) = 12x2 – 24x + 24 = 12[(x – 1)2 + 1] > 0

a, b, c be three distinct integers such that

x ΠR.

\ f ¢ is increases on R.

p(a) = p(b) = p(c) = – 1. The number of integral roots of p(x) is

Thus, y = f ¢ (x) intersects the x-axis exactly once.

Ans. 0

\ f (x) = 0 cannot have four distinct real roots for otherwise

Solution:

y = f ¢ (x) will meet the x-axis three times.

a, b, c are roots of p(x) + 1 = 0.

Let p(x) + 1 = (x – a) (x – b) (x – c) q(x)

[By the Rolle’s Theorem]

(1)

where q(x

EXERCISE

Let m be an integral root of p(x). Then (1) gives LEVEL 1

1 = (m – a) (m – b) (m – c) q(m) As a, b, c are distinct at least one of m – a, m – b, m – c is different from 1 and – 1. This is a contradiction. The smallest value of k for which both

Example 119

the roots of the equation x2 – 8kx + 16(k2 – k + 1) = 0 are real and distinct and have least value 4, is Ans. 2 Solution:

As x2 – 8kx + 16(k2 – k + 1) = 0

(1)

has real and distinct roots, (–8k)2 – (4) (16) (k2 – k + 1) > 0



k–1>0

As k is an integer, we get k ≥ 2. If a, b are roots of (1), then a, b ≥ 4. fi

a + b ≥ 8 and ab ≥ 16



8k ≥ 8 and 16(k2 – k + 1) ≥ 16



k ≥ 1 and k(k – 1) ≥ 0



k≥1

Thus, smallest value of k is 2 Example 120

The number of distinct real roots of x4

– 4x3 + 12x2 + x – 1 = 0 is Ans. 2 Solution: We have

Let f (x) = x4 – 4x3 + 12x2 + x – 1

f (–1) = 15 > 0, f (0) = –1 < 0, f (1) = 9 > 0 As f (x) is a polynomial and hence continuous, f (x) = 0 has a root in (–1, 0) and a root in (0, 1). Thus, f (x) = 0 has a least two distinct real roots. We have f ¢ (x) = 4x3 – 12x2 + 24x + 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. The number or real roots of (x + 3)4 + (x + 5)4 = 16 is (a) 0 (b) 2 (c) 3 (d) 4 2 2. Let a and b be the roots of x – 6x – 2 = 0, with a > b. If an = an – bn for n ≥ 1, then the value of a10 - 2a8 is 2a9 (a) 3 (b) 2 (c) 1 (d) 4 3. A value of b for which the equations x2 + bx – 1 = 0 x2 + x + b = 0 have one root in common is (a) - 2

(b) i 5

(c) i 3

(d)

4. If x ΠR, the number of solutions of 2 x - 1 = 1 is

2 2x + 1 –

5. If l, m, n are real, l + m π 0, then the roots of the equation (l + m)x2 – 3(l – m)x – 2(l + m) = 0 are (a) real and unequal (b) complex (c) real and equal (d) purely imaginary 6. If the equation x + 1 - x = a has a solution, then (a) 0 < a < 1 (b) a > 1 (c) 0 < a £ 1 (d) a £ 1 7. Let a, b be the roots of the equation x2 – ax + b = 0 and An = an + bn. Then

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An+1 – aAn + bAn–1 is equal to. (a) – a (b) b (c) a – b (d) 0 8. If a, b, g are such that a + b + g = 2, a2 + b2 + g2 = 6, a3 + b3 + g 3 = 8, then a 4 + b 4 + g 4 is (a) 5 (b) 18 (c) 12 (d) 36 9. In a triangle PQR, –R = p/4. If tan (P/3) and tan (Q/3) are the roots of the equation ax2 + bx + c = 0, then (a) a + b = c (b) b + c = 0 (c) a + c = b (d) b = c 10. The product of the roots of 3 8 + x + 3 8 - x = 1 is (a) – 21 (b) – 189 (c) – 9 (d) –5. 3 11. If all the roots of x + px + q = 0 p, q Œ R, q π 0 are real, then (a) p < 0 (b) p = 0 (c) p > 0 (d) p > q 12. The number of irrational roots of the equation 3 4x 5x = + 2 2 2 x + x + 3 x - 5x + 3 is (a) 0 (b) 1 (c) 2 (d) 3 13. If a, b, g, d are the roots of (x2 + x + 4)2 + 3x(x2 + x + 4) + 2x2 = 0 then |a| + |b| + |g| + |d| is equal (a) 6 (b) 8 (c) 12 (d) 25 14. x = 1 is a root of (x2 – x + 1)4 – 6x2 (x2 – x + 1)2 + 5x4 = 0 of multiplicity (a) 2 (b) 3 (c) 4 (d) 6 15. The number of irrational roots of the equation (x2 – 3x + 1) (x2 + 3x + 2) (x2 – 9x + 20) = – 30 is (a) 0 (b) 2 (c) 4 (d) 6 16. The number of roots of the equation 3x 2 + 6 x + 7 + 5 x 2 + 10 x + 14 = 4 – 2x – x2 is (a) 4 (b) 3 (c) 2 (d) 1 17. The number of real values of x which satisfy the equation

x2 - 6 x + 7 = 1. x2 + 6 x + 7 (a) 0

(b) 1

18. The number of real values of x which satisfy the equation x x2 + |x| = x -1 | x -1| (a) 1

(b) 2

19. Sum of all the real values of x which satisfy the equation 2- x 2- x = 2+ x 2+ x (a) 0 (b) 2 (c) 7.5 (d) 11.5 20. The set of real values of a for which the equation 2a 2 + x 2 2x 1 =0 + 3 3 2 2 x-a a -x ax + a + x has a unique solution is (a) (– •, 1) (b) (– 1, •) (c) (– 1, 1) (d) R – {0} 21. The set of real values of a for which sum of the roots of the equation 1 1 1 1 + - 2 = x a a x + a - a2 is less than a3/4 is (a) (0, 2) » (2, •) (b) (3, •) (c) (– 1, 0) » (3, •) (d) (2, •) x 2 - ax - 2 >–1 22. The real values of a for which 2 x - 3x + 4 for each x Œ R, is (a) (– 1, 2) (b) (0, 7) (c) (– 7, 1) (d) (2, 7) 23. If a, b, c Œ R are distinct, then the condition(s) on a, b, c for which the equation 1 1 1 1 =0 + + + x-a x-b x-c x-b-c+a has real roots is (a) a + b + c π 0, a π 0 (b) a – b – c π 0, a π 0 (c) 2a = b + c (d) none of these

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24. If f (x) = ax2 + bx + c, f (– 1) < 1, f (1) > – 1 and f (– 3) < – 4 then (a) a = 0 (b) a < 0 (c) a > 0 (d) sign of a cannot be determined 25. The set of all values of k for which the equation x2 + 2(k – 1)x + (k – 5) = 0 has at least one non-negative root is (a) [1, •) (b) [– 1, 1] (c) (– •, – 5] (d) [5, •) 26. If 3p/4 < a < p, then the set of values of a for which 1 (sin a + cos a) (sin a)x2 + (2 cos a)x + 2 is square of a linear polynomial is Ï 5p 11p ¸ (a) Ì , ˝ Ó 6 12 ˛

Ï11p ¸ (b) Ì ˝ Ó 12 ˛

Ï 5p ¸ (c) Ì ˝ Ó 6 ˛

(d) f

27. If 0 £ f £ 3p, then the set of all values of f for which sum of the squares of the roots of the equation 1 cos2 f = 0 x2 + (sin f – 1)x – 2 is greatest is ¸ Ï 3p , 2p , 3p ˝ (a) Ìp , 2 Ó ˛

(b) {p, 3p}

5p ¸ Ï (c) Ì2p , ˝ 4 ˛ Ó

Ï 3p ¸ (d) Ì ˝ Ó2 ˛

28. The set of values of b for which 2log1/36 (bx + 28) = – log6 (12 – 4x – x2) has exactly one solution is (a) (– •, – 14) » [14/3, •) (b) [14/9, •) (c) (– •, – 14] » {4} » [14/3, •] (d) {4} 29. If 2x2 + 3x + 4 = 0 and ax2 + bx + c = 0, where a, b, c Œ N have a common root, then the least value of a + b + c is (a) 10 (b) 9 (c) 8 (d) 6

30. Let p, q, r, s be real numbers such that pr = 2(q + s). Consider quadratic equations x2 + px + q = 0 and x2 + rx + s = 0. Then (a) none of these has real roots (b) both have real roots (c) at least one has real roots (d) at most one has real roots 31. Let a < 0, a π – 2. The equation x2 + a|x| + 1 = 0 has (a) no real roots (b) at least two real roots (c) exactly four real roots or no real roots (d) none of these 32. Let f (x) = x3 + 3x2 + 6x + 2009 and 1 2 3 . g(x) = + + x - f (1) x - f (2) x - f (3) The number of real solutions of g(x) = 0 is (a) 0 (b) 1 33. Let a, b Œ R, b π 0 and a + ib be a root of x3 + qx + r = 0 where q, r Œ R. A cubic equation with a, is (b) x3 + 2qx + r = 0 (a) x3 – qx + r = 0 (d) none of these (c) x3 + 4qx – 8r = 0 34. Suppose a, b are complex numbers and x3 + ax + b = 0 has a pair of complex conjugate roots, then (a) a is real and b is imaginary (b) a is imaginary and b is real (c) both a and b are real (d) none of these 35. The number of solution of Ê 65 ˆ 102/x + 251/x = Á ˜ (501/x) is Ë 8¯ (a) 0

(b) 2 x

9 Ê 10 ˆ has 36. The equation Á ˜ = – 3x2 + 2x – Ë 9¯ 11 (a) no solution (b) exactly one solution (c) exactly two solutions (d) none of these 37. Let a, b, c Œ R and a π 0 be such that (a + c)2 < b2, then the quadratic equation ax2 + bx + c = 0 has

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(a) imaginary roots (b) real roots (c) two real roots lying between (– 1, 1) (d) none of these 38. The number of real solution of 4x + 1.5 + 9x + 0.5 = (10) (6x) is (a) zero (b) one 39. If roots of the equation x2 – 2mx + m2 – 1 = 0 lie in the interval (– 2, 4), then (a) – 1 < m < 3 (b) 1 < m < 5 (c) 1 < m < 3 (d) – 1 < m < 5 40. The number of solutions of the equation Ê px ˆ = x2 – 2 3 x + 4 is sin Á Ë 2 3 ˜¯ (a) 1

(b) 2

41. If x is real, then the least value of the expression x2 - 6x + 5 is x2 + 2x + 2 (a) –1 (b) –1/3 (c) –1/2 (d) none of these. 42. If a and b (a < b) are the roots of the equation x2 + bx + c = 0, where c < 0 < b, then (a) a < 0 < b < |a| (b) b > |a| (c) |a| + |b| < 2 (d) none of these x2 + 2 x + c takes all real val43. If x is real, and y = 2 x + 4x +3 ues then (a) 0 < c < 2 (b) 0 £ c £ 1 (c) – 1 < c < 1 (d) – 3 £ c £ 1 44. The smallest integer x for which the inequality x-5 x 2 + 5 x - 14 (a) –6 (b) –5 (c) –4 (d) –3 45. If |a ± b| > c and a π 0, then the quadratic equation a2 x2 + (b2 + a2 – c2) x + b2 = 0 (a) has two real roots (b) both positive roots (c) cannot have real roots (d) none of these

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 46. If abc < 0, then the equation ax2 + 2(b + c – a)x + bc = 0, has (a) real roots (b) one positive and one negative root (c) both positive roots (d) both negative roots 47. The equation 1 + 1 - x 4 - x 2 = x, x Œ R, has (a) only positive solutions (b) exactly one solution (c) at least two solutions 48. The equation x + x + x + 2 + x 2 + 2 x = 3 has (a) no solution (b) at least one solution (c) only positive solutions 49. Let a, b, c be the sides of an obtuse angled triangle with –C > p/2. The equation a2x2 + (b2 + a2 – c2)x + b2 = 0 has (a) two positive roots (b) one positive and one negative root (c) two real roots (d) two imaginary roots. 50. The equation x - 1 + x + 3 + 2 ( x - 1)( x + 3) = 4 – 2x has (a) exactly one integral solution (b) all its solutions in [1, 2] (c) sum of all the solutions is 1 (d) no solution 51. The equation x3 + 1 = 2(2x – 1)1/3 has (a) one rational solution (b) two irrational solutions (c) sum of roots equal to zero (d) product of the roots as – 1. 52. Let P(x) be a polynomial such that P(x + 2) = x2 + 3x – 2. Which one of the followings are true statement?

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(a) There are two polynomials satisfying the above relation. (b) P(x) must be of degree 2 (c) P(x) = 0 has two irrational roots (d) Product of the roots of P(x) = 0 is – 4. 53. Let P(x) be a polynomial such that P(x2 + 2) = x17 – 3x5 + x3 – 3, then (a) P(x) = 0 has at least 34 roots. (b) P(x) = 0 has exactly 17 roots (c) Sum of the roots of P(x) = 0 is – 193. (d) none of these. 54. Let P(x) = a0 + a1x + ... + anxn, an π 0 be such that P(x2) = P(x)2, then (a) a0 = 1 (b) a0 = a1 = ... = an–1 = 0 (c) an = 1 (d) an = – 1 55. Let p, q be non-zero, real numbers and x1, x2, x3 be the roots of x3 + px + q = 0 such that 1 1 + , then x3 = x1 x2 2

(a) q is a root of x + px + q = 0 (b) q < 0 (c) p < 0 (d) x1 + x2 < 0 56. Let n Œ N, n > 1 and a Œ R be such that sin a π 0, and let p(x) = xn sin a – x sin na + sin (n – 1)a then (a) cos a ± i sin a are zeros of p(x) (b) p(x) is divisible by x2 – 2x cos a + 1 (c) p(x) is divisible by x2 + 2x cos a + 1 (d) sum of all the roots of p(x) = 0 is 0. 57. Let n Œ N, n > 1 and a Œ R be such that sin a π 0 and let p(x) = (x sin a + cos a)n – x sin (na) – cos (na), then (a) ± i are zeros of p(x) (b) p(x) is divisible by x2 + 1 (c) ± 1 are zeros of p(x) (d) p(x) is divisible by x2 – 1 58. Let a, b, c > 0 and for all p, q Œ R with p + q = 1 pa2 + qb2 > pqc2, then (a) a + b > c (b) a + c > b (c) b + c > a (d) a + b = c

59. Let a, b be two distinct roots of x4 + x3 – 1 = 0, and p(x) = x6 + x4 + x3 – x2 – 1 (a) ab is a root of p(x) = 0 (b) a + b is a root of p(x) = 0 (c) both a + b and ab are roots of p(x) = 0 (d) none of ab, a + b is a root of p(x) = 0 60. If x2 + px + 1 is a factor of ax3 + bx + c, then (a) ax + c is also a factor of ax3 + bx + c (b) ap + c = 0 (c) a2 – c2 = ab (d) (2a – b)2 = b2 + 4c2 61. Let a, b, c, d and p be integers such that p π 0. If x + p is a factor of ax3 + bx2 + cx + d, then (a) p|d (b) p2|(pc – d) (c) d = 0 fi p is a root of ax2 – bx + c = 0 (d) none of these. 62. The equation log x 2 +6 x +8 (log2 x 2 + 2 x +3 ( x 2 - 2 x )) = 0 has (a) no solution (b) exactly one solution (c) at least one negative solution (d) at least two solutions. 63. Let – 1 £ p £ 1 and p(x) = 4x3 – 3x – p (a) p(x) = 0 has at least one root in [1/2, 1] (b) p(x) is strictly increasing on [1/2, 1] Ï1 ¸ (c) a = cos Ì cos -1 ( p) ˝ is a root of p(x) = 0 Ó3 ˛ (d) p(x) = 0 has real roots. 64. Let f (x) = ax2 + bx + c where a, b, c Œ R. Suppose f (x) is an integer whenever x is an integer, then which of the following are integers? (a) 2a

(b) a + b

(c) a – b

(d) c

65. Let ab < 0, a + b π 0 and p(x) = (a + b)x2 – 4abx + ab(a + b). (a) p(x) = 0 has real roots (b) p(x) = 0 has roots of opposite signs (c) p(x) > 0 " x if a + b > 0 (d) p(x) < 0 " x if a + b < 0

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MATRIX-MATCH TYPE QUESTIONS 66. Let a < b < c < d. Match with the which root of equation lies Column 1 (a) (x – a) (x – c) (p) + 2009(x – b) (x – d) = 0 (b) 2009 (x – a) (x – c) (q) – 2008 (x – b) (x – d) = 0 (c) (x – a) (x – b) (x – c) + (x – d) = 0 (r) (d) (x – a) (x – b) + (x – c) = 0 (s)

interval(s) in Column 2 (c, d) (a, b)

(– •, a) (b, c)

67. Let a, b, g be the roots of x + px + q = 0, then value of Column 1 Column 2 1 1 1 + + b +g g +a a + b

(b) a4 + b4 + g4 (c) a–3 + b–3 + g –3 (d) a–2 + b–2 + g –2 68. Let a, b, c Œ R be such b2 – ax2 + 2bx + c = 0 has Column 1 (a) two positive roots (p)

p

(p) (q) (r) (s) ac ≥

2

q2 – (p3 + 3q2)/q3 2p2 p/q 0. The equation

x2 + 2(a + 1)x + 9a – 5 = 0

(b) (q)

x – 2 x –1 =1 (d)

1 + 5 – 4 log9 x 4 =3 1 + log9 x

(s) 1

Column 1 (a) ax2 + (2a + b)x +a+b+c=0 (b) (a – b + c)x2 +

Column 2

(p)

a b , a +1 b +1

a -1 b -1 , (b – 2c)x + c = 0 (q) a +1 b +1 2 (c) (a – b + c)x + 2(a – c)x +a+b+c=0 (r) a – 1, b – 1 2 (d) ax + (b – 2a)x + a–b+c=0 (r) a + 1, b + 1

ASSERTION-REASON TYPE QUESTIONS 72. If P(x), Q(x) and R(x) are polynomials in x such that

Column 2 a, – c are of the same sign (b) two negative roots (q) a, – b, c are of the same sign (c) one positive and (r) a, b, c are of the one negative root same sign (d) equal roots (r) a, b, c are in G.P. 69. Consider the quadratic equations

Column 1 (a) a < 1 (b) a > 6

x + 3 – 4 x –1 +

71. a, b are roots of ax2 + bx + c = 0. Match equation with its roots.

3

(a)

(c)

Column 2 real and distinct roots one positive and one negative root imaginary roots both negative roots.

(c) 1 < a < 6 (r) (d) 2 < a < 5 (s) 70. The number of real roots of Column 1 (a) x4 + x2 + 2 = log0.75 (2009) (b) (14)2/x + 491/x = (4.25) (98)1/x

Column 2 (p) 0 (q) 2

P(x3) + xQ(x3) = (x2 + x + 1) R(x) Statement-1: 1 is a root of P(x) = 0 Statement-2: 1 is a root of Q(x) = 0 73. Let p, q, r, s be four real numbers such that pr = 2(q + s). Statement-1: At least one of x2 + px + q = 0 and x2 + rx + s = 0 has real roots. Statement-2: x2 + px + q = 0 has real roots if and only if x2 + rx + s = 0 has imaginary roots. 74. Statement-1: If a, b, c Œ R and 3c < 4a + 4b and the equation ax2 + 2bx – 3c = 0 has no real roots, then c < 0. Statement-2: If a, b, c Œ R, the equation ax2 + bx + c = 0, has no real roots if b2 – 4ac < 0. 75. Statement-1: Let f: R Æ R f (x) = 2 a x + 6x – 8 . If f is onto, then a Œ [2, 14]. a + 6 x – 8x2 Statement-2: If the expression y =

a x 2 + 3x – 4 a + 3x – 4 x 2

x ΠR assumes all real values, then a Π[1, 7].

,

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76. Let a, b, p, q Œ R, p π q, b π 0 and the ratio of the roots of equation ax2 + bx + b = 0 is p : q. Statement-1:

p + q

q b + =0 p a

Statement-2: a(p + q)2 + bpq = 0

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 77 to 81 Let a, b, g be roots of the equation ax3 + bx2 + cx + d = 0. To obtain the equation whose roots f (a), f (b), f (g) where f is a function, we put y = f (a) and obtain a = f–1 (y). Now, we use the fact that a is root of the equation ax3 + bx2 + cx + d = 0 to obtain the desired equation. Note Desired equation may require some manipulation after substitution. For example if a, b, g are roots of ax3 + bx2 + cx + d a, 1/b, 1/g, we put 1 1 = y fi a = . As a is a root of ax3 + bx2 + cx + d = y a 0, we get a y

3

+

b y

2

+

c + d = 0 or a + by + cy2 + dy3 = 0 y

This is the desired equation. The same result holds for all polynomial equations. 77. If a, b are roots of ax2 + bx + c = 0, the roots of a(x – 1)2 + b(x – 1) (x – 2) + c(x – 2)2 = 0 are (a)

2a - 1 2b - 1 , a -1 b -1

(b)

a -1 b -1 , a -2 b -2

(c)

a +1 b +1 , a -1 b -1

(d)

3a - 1 3b - 1 , a -1 b -1

78. If a, b are roots of px2 + qx + r = 0, then the equation whose roots of a2 + r/p and b2 + r/p is (a) p3x2 + pq2x + r = 0 (b) px2 – qx + r2 – p = 0 (c) p3x2 – pq2x + q2r = 0 (d) none of these 79. If a, b, g are the roots of x3 – 13x + 11 = 0 then equation whose roots are b + g, g + a, a + b is (a) x3 – 13x – 11 + 0 (b) x3 + 13x – 11 = 0 (c) x3 + 13x + 11 = 0 (d) x3 – 13x + 11 = 0 80. If a, b, g are the roots of x3 + 3px + q = 0, then the equation whose roots are (b – g)2, (g – a)2 and (a – b)2 is

(a) (b) (c) (d)

y3 – 3(p – q)y + pq = 0 y3 + 3(p – q)y – pq = 0 y3 + 3(p + q)y – pq = 0 none of these

81. If a, b, g are roots of x3 + 3x + 2 = 0, then an equation whose roots are (a – b) (a – g), (b – g) (b – a), (g – a) (g – b) is (a) y3 – 6y2 + 216 = 0 (b) y3 + 9y2 – 216 = 0 (c) y3 + 3y2 – 128 = 0 (d) y3 + 6y2 + 184 = 0

Paragraph for Question Nos. 82 to 86 Let f (x) = ax2 + bx + c. If x1, x2, x3 are three distinct real numbers such that f (x1) = f (x2) = f (x3) = 0, then f (x) ∫ 0, i.e. f (x) = 0 " x Œ R. 82. If f and x1, x2, x3 are as in the above paragraph, then (a) a π 0, b π 0, c = 0 (b) a = b π 0, c = 0 (c) a = b = c = 0 (d) a π 0, b = c = 0. 83. If x1, x2, x3 are three distinct real numbers and g(x) =

( x - x2 )( x - x3 ) ( x - x3 )( x - x1 ) + ( x1 - x2 )( x1 - x3 ) ( x2 - x3 )( x2 - x3 ) +

( x - x1 )( x - x2 ) ( x3 - x1 )( x3 - x2 )

then g(x) is identically equal to (a) 0 (b) 1 (c) (x – x1) (x – x2) (x – x3) (d) none of these. 84. If g(x) = x1 + x2

( x - x2 )( x - x3 ) ( x1 - x2 )( x1 - x3 )

( x - x3 )( x - x1 ) ( x - x1 )( x - x2 ) + x3 ( x2 - x1 )( x2 - x3 ) ( x3 - x1 )( x3 - x2 )

then g(x) is identically equal to (a) 0 (b) 1 (c) x (d) (x – x1 – x2 – x3)2 + (x12 + x22 + x32)x – x1x2x3

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85. If g(x) = x1(x1 – 1) + x2(x2 – 1) + x3 (x3 – 1)

( x - x2 )( x - x3 ) ( x1 - x2 )( x1 - x3 )

( x - x3 )( x - x1 ) ( x2 - x1 )( x2 - x3 ) ( x - x1 )( x - x2 ) ( x3 - x1 )( x3 - x2 )

then g(x) is identically equal to (a) x (b) x2 – x 3 2 (d) x3 – x2 + x + 1 (c) x – x x2 x3 ( x - x1 )2 x3 x1 ( x - x2 )2 86. If g(x) = + ( x1 - x2 )( x1 - x3 ) ( x2 - x3 )( x2 - x1 ) +

x1 x2 ( x - x3 )2 ( x3 - x1 )( x3 - x2 )

then g(x) is identically equal to (a) 1 (b) x 2 (d) x2 – x + 1 (c) x

Paragraph for Question Nos. 87 to 91 By the Rolle’s theorem between any two real zeros of a polynomial p(x) lies a zero of p¢(x). 87. If p, q, r Œ R and x3 + px2 + qx + r = 0 has three real roots, then (b) p2 £ 3q (a) p2 < 3q (d) none of these. (c) p2 ≥ 3q 88. The number of real roots of x3 – 3x + 1 = 0 lying in the interval (0, 1) is (a) 0 (b) 1 (c) 2 (d) 3 89. Let f (x) be a polynomial such that a, b with a < b are two consecutive zeros of f ¢(x). The number of zeros of f (x) lying in the interval [a, b] is (a) at most 1 (b) at most 2 (c) at least 1 (d) at least 2 90. If a, b, c Œ R and 3b2 – 8ac < 0, then the equation f (x) = ax4 + bx3 + cx2 + 2009x + 2010 = 0 has (a) all real roots (b) all imaginary roots (c) cannot have all real roots (d) two real and two imaginary roots 91. If a + b + c = 0, then equation 3ax2 + 2bx + c = 0 has (a) at least one root in [0, 1] (b) no real roots (c) one positive and one negative root (d) none of these.

Paragraph for Question Nos. 92 to 96 Descrate’s Rule of Signs

Let p(x) be a polynomial. (a) Positive roots The number of positive roots of the equation p(x) = 0 cannot exceed the number of changes of sign from positive to negative and negap(x). (b) Negative roots The number of negative roots p(x) = 0 cannot exceed the number of changes in the signs of p(– x). 92. The least number of imaginary roots of the equation P(x) = x9 – x5 + x4 + x2 + 1 = 0 is (a) 2 (b) 8 (c) 4 (d) 6 5 3 93. The equation p(x) = x + x – 8x – 5 = 0 has (a) exactly 3 real roots (b) exactly one real root (c) at least one root in (0, 1) (d) none of these. 94. The equation p(x) = x5 – x + 16 = 0 has (a) no imaginary roots (b) exactly two imaginary roots (c) exactly four imaginary roots (d) at least one root in (– •, – 2) 95. The number of imaginary roots of p(x) = x9 – x5 + x4 + x2 + 1 = 0 is (a) 2 (b) 4 (c) 6 96. If a, b, c > 0, then the equation

(d) 8

p(x) = x4 – ax3 – bx – c = 0 has (a) two negative roots (b) all the four imaginary roots (c) one positive, one negative and two imaginary roots (d) two positive and two negative roots.

INTEGER-ANSWER TYPE QUESTIONS 97. If a, b, c > 0, a2 = bc and a + b + c = abc, then the least possible value of a2 is 98. If x1 + x2 + x3 = 1, x2x3 + x3x1 + x1x2 = 1 and x1x2x3 = 1, then value of |x1| + |x2| + |x3| is 99. The number of real roots of P(x) = x100 – 2x99 + 3x98 – ... – 100x + 101 = 0 is

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100. The number of positive integral solutions x4 – y4 = 3879108 is 101. The number of solutions of Ê xˆ 2 cos2 Á ˜ = (0.2)x + (0.2)–x is Ë 2¯ 102. Let f (x) = ax2 + 2bx + c, a, b, c Œ R. If f (x) takes real values for real values of x and imaginary value for imaginary values of x, then the value of a is 103. The real value of – a for which the roots x1, x2, x3 of x3 – 6x2 + ax – a = 0 satisfy (x1 – 3)3 + (x2 – 3)3 + (x3 – 3)3 = 0 is 104. Sum of all the real values of x for which 2- x +

4 = 2 is 2- x +3

105. Let a and b be two integers such that 10a + b = 5 and p(x) = x2 + ax + b. Let integer n be such that n p(10) p(11) = p(n) then value of is. 23 106. If x1 and x2 are roots of the equation x2 + x + c = 0, c π 0, and x13 x3 1 + 2 = - , 2 + x2 2 + x1 2 the value of – 4c is 107. If x1, x2, x3 are such that x1 + x2 + x3 = 2 x12 + x22 + x32 = 6 x13 + x23 + x33 = 8 then value of – (x2 – x3) (x3 – x1) (x1 – x2) is 108. Let f (x) =

x2 - 2 x + 4 x + 2x + 4 2

,xŒR

The maximum integral value of x tying in the range of f (x) is LEVEL 2

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. The number of cubic polynomials p(x) with integral a, b and c such that p(a) = b, p(b) = c and p(c) = a is

(a) 0 (b) 1 (c) dependent on a, b, c 2. Let p(x such that p(0) and p(1) are both odd integers, then the number of integral roots of p(x) = 0, is (a) 0 (b) 1 (c) 2 (d) none of these 3. If n ΠN, then number of real roots of 1 2 1 x + ... + x2n = 0 1+x+ (2n)! 2! is (a) n (b) 2 (c) 0 (d) none of these 4. The number of real values of x satisfying 641/x Р2(3x + 3)/x + 12 = 0 is (a) 0

(b) 1

5. Let S denote the sum of all the roots of the equation 27(16x) – 30(36x) + 8(81x) = 0 then S equals (a) 2/3 (b) 1 (c) 3/2 (d) 5/2 6. The number of real values of a for which 4x – a(2x) – a + 3 = 0 has at least one solution is (a) 0 (b) 1

7. The solution set of

1 4 + =3 5 - 4 log 4 x 1 + log 4 x

is (a) {4, 8} (b) {2, 8} (c) {2, 4} (d) none of these log34 + 4log3x = 64 is 8. The solution set of 3x (a) {9} (b) {27} (c) {81} (d) {16} 9. Let S denote the set of all values of a for which 2

2

4 x + 2(2a + 1)2 x + 4a2 - 3 > 0 x, then S equals (a) (– •, – 1) (c)

( – 1,

3 2

)

(b)

(

3 2, •

)

(d) none of these

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10. The number of integral solutions of 5 - xˆ Ê 5 - xˆ Ê x+ xÁ ˜ ˜ =6 Á Ë x + 1¯ Ë x + 1¯ is (a) 0 (b) 1 (c) 2 (d) 3 11. The number of subsets of the set of real solutions of 11x - 6 x4 = 6 x - 11 is (a) 4 (b) 8 (c) 16 (d) 32 12. Let S denote the set of integral values of x for which x2 + 1 < 3x – x2 – 3 then the number of elements in S contains is (a) 0 (b) 1 13. The equation (2.4)x = (2.6)x – 1 has (a) no solution (b) exactly one solution (c) at least two solutions

then the equation 2x3 – Ax2 + Bx – C = 0 has (a) exactly one real root (b) three positive roots (c) three negative roots (d) none of these 17. Let A, B, C be the angles of triangle and tan A, tan B, tan C be the three roots of x4 – ax3 + bx2 – cx + d = 0, then the fourth root of this equation is also a root of (b) x2 – ax – c = 0 (a) x2 + bx + d = 0 (d) none of these (c) x2 – ax + d = 0 2 18. The equation x – ax + b = 0 and x2 – cx + d = 0 have a common root and x2 – cx + d = 0 has equal roots, then b + d equals (a) ac (b) ac/2 (c) 2ac

(d)

ac

1ˆ Ê 19. If 4 cosec2a x2 + x + Á b 2 - b + ˜ = 0 has real Ë 2¯ roots, then sin2 a + (cos (sin–1 b))2 equals (a) 3/4 (b) 7/4 (c) 11/4 (d) 15/4 20. Suppose a1, a2, ..., an–1 ≥ 0 and the equation

(3x) (8x/(x+1)) = 36

P(x) = xn + a1xn–1 + ... + an–1 x + 1 has n real roots, then least value of P(2) is

then the number of elements in S is (a) 0 (b) 1

(a) 3n (c) 1

14. Let S denote the set of all irrational solutions of

15. If a, b, c Œ R and (b – 1)2 < ac, then the system of equations ax12 + 2bx1 + c = 2x2 ax22 + 2bx2 + c = 2x3 .................................. .................................. .................................. ax2n–1 + 2bxn–1 + c = 2xn ax2n + 2bxn + c = 2x1 has (a) no solution (b) exactly one solution (d) exactly n solutions. 16. Let x1 > x2 > x3 > x4 > x5 > x6 > 0, and A = x1 + x2 + ... + x6 B = x 1 x 3 + x 1x 5 + x 3 x 5 + x 2 x 4 + x 2 x 6 + x 4 x 6 C = x 1 x 3 x 5 + x 2 x 4 x 6,

(b) 2n (d) none of these.

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 21. The equation x (3 log3 x + 7 / 2) (log3 x ) = 3 has (a) a rational root (b) two irrational roots (c) sum of all the roots of the equation 7/2 (d) product of the roots of the equation is less than 1. 22. If a, b, c ΠR and (a + c)2 < b2 then the equation ax2 + bx + c = 0 has (a) real roots (b) at least one root in (Р1, 1) (c) one of the two roots lie in the region given by |x| > 1 (d) both the roots lie in (Р1, 1) 23. If S denotes the solution set of the equation 2

1ˆ Ê 2 2 Ê p xˆ ÁË x + ˜¯ = 4 cos (px) sin ÁË ˜ , then 2 ¯ x

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.45

(a) S (b) P(S), the power set of S has exactly four elements (c) S (d) S Õ I, the set of integers. 24. If a, b, g Œ C are such that a + b + g = 2, a2 + b2 + g2 = 6, a3 + b3 + g3 = 8, then (a) abg = – 2 (b) a4 + b 4 + g 4 = 18 (c) a5 + b 5 + g 5 = 32 (d) a5 + b 5 + g 5 = 34 25. Suppose p/2 < q < p and b > 0. If a and b (a < b) are the roots of (sin q)x2 + bx + cos q = 0, then (a) 0 < a < b (b) a < 0 < b < |a| (c) a < b < 0 (d) a < 0 < a < |b| 26. Let a1, a2, a3, ... be a sequence of non-zero real numbers which are in A.P. For k Œ N, let fk(x) = ak x2 + 2ak + 1 x + ak + 2, (a) fk(x) = 0 has real roots for each k Œ N. (b) Each of fk(x) = 0 has one root in common. (c) non-common roots of f1(x) = 0, f2(x) = 0, f3(x) = 0, ... form an A.P. (d) none of these. 27. Let p(x a, b be two distinct real numbers. The remainder when p(x) is divided by (x – a) (x – b) is (a) 0 if a and b are roots of p(x) = 0 p( a) (x – b) if b is a root of p(x) = 0 a-b 1 [x(p(a) – p(b)) + ap(b) – bp(a)] (c) a-b 1 [x(p(b) – p(a)) + bp(a) – ap(b)] (d) a-b

(b)

28. If p, q, r, s are positive real numbers and the equation x4 – px3 + qx2 – rx + s = 0 has four positive real roots, then (b) r4 ≥ 256s3 (a) p4 ≥ 256s (c) pr ≥ 16s (d) q2 ≥ 36s 29. If a, b are the roots of l (x2 + x) + x + 5 = 0 and l1, l2 are the two values of l for which a, b are a b connected by the relation + = 4, then the value b a l l of 1 + 2 is l2 l1

(a) 254 (b) 482 (c) 780 (d) 782 30. Suppose a, b, c Œ R, a > 0 be such that a + b + c ≥ 0, a – b + c ≥ 0, a > c, and let x1, x2 be the roots of ax2 + bx + c = 0, then (b) |x1|, |x2| £ 1 (a) |x1|, |x2| > 0 (c) |b| £ 2a (d) |x1 x2| £ 1

MATRIX-MATCH TYPE QUESTIONS 31. The value of k for which the equation x3 – 3x + k = 0 has Column 1 Column 2 (a) all three real roots (p) |k| > 2 (b) two equal roots (q) k = – 2, 2 (c) exactly one real root (r) |k| £ 2 (d) three equal roots (s) no value of k 32. Range of the function Column 1 Column 2 (a)

(b)

(c)

x2 – x + 1

(p) (– •, 0]

x2 + x + 1 x 2 – 3x + 6 x –1

(q)

x+ x

È1 ˘ (r) Í , 3˙ Î3 ˚

1 + x2

(d) (3 – x) [x +

[0, 1]

x2 - 9 ]

(s) (– • – 5] » [3, •)

33. Suppose a, b, c Œ R, abc π 0 and b2 – 4ac ≥ 0. Let a, b be the roots of ax2 + bx + c = 0, then roots of Column 1 Column 2 Ê b2 ˆ (a) x2 + Á 2 – ˜ x + 1 = 0 ac ¯ Ë

(p)

È (a + c )2 ˘˙ x + 1 = 0 (b) x2 + Í2 – ac ˙ ÍÎ ˚

(q) ab ,

È 4b 2 ˘ b4 (c) x2 + Í – 2 2 – 2˙ x + 1 = 0 ˚ Î ac a c

2 (a + c ) ac

x=0

1 ab

(r)

a b , b a

(s)

1 ab , 2ab 2

2

(d) (2x + 1)2 –

a2 b2 , b2 a2

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34. The value of k for which the equation kx3 – 3x2 + 1 = 0 has Column 1 Column 2 (a) all three real roots (p) |k| > 2 (b) two equal roots (q) k = – 2, 2 (c) exactly one real roots (r) |k| < 2 (d) all three equal roots (s) no value of k 35. Polynomial p(x) which satisfy Column 1 Column 2 (a) p(x + 3) = (p) no such polynomial x2 + 7x + 12 (b) p(x – 1) + 2p (x + 1) (q) x2 + x = 3x2 – 7x (r) x2 – 3x (c) p(x2 + 1) 5 3 =x +x –1 (s) x3 – 7x2 (d) p(x2 + 1) = x7 – x5 + 3

ASSERTION-REASON TYPE QUESTIONS 36. Let p, q Œ R be such that p2 – 4q ≥ 0. Statement-1: If the quadratic equation x2 – px + q = 0 has distinct positive roots, then the equation can be written as (ax + b)2 = x for some real numbers a and b. Statement-2: If the quadratic equation x2 – px + q = 0 has distinct positive roots, then q > 0 and p > 2 q . 37. Statement-1: If m, n are positive integers b and c are distinct, xm(bn – cn) + bm (cn – xn) + cm (xn – bn) is divisible by x2 – (b + c)x + bc. Statement-2: If a, b are different complex numbers and r(x) is the remainder when polynomial p(x) is divided by (x – a) (x – b), then r(x) =

(x – a)n divdies p(x) but (x – a)n+1 does not divide p(x). For instance 2 is a root of multiplicity 3 of (x – 2)3 (x – 5) (x2 + 1) = 0 If a is a root of multiplicity n of p(x) = 0, then a is a root of multiplicity n – r of pr(x) = 0. 39. Values of the parameter a for which the equation p(x) = 3x4 + 4x3 – 6x2 – 12x + a = 0 has a multiple root. (a) 11, – 5 (b) 11, – 7 (c) – 11, 5 (d) – 11, 7 40. If q π 0, the relation between p and q for which the equation x5 + px4 + q = 0 has a root of multiplicity two is (b) 3125q + 256p5 = 0 (a) 3125q = 256p5 5 (d) none of these (c) 256p = 1025q 3 2 41. If the equation ax + 3bx + 3cx + d = 0 has a root of multiplicity 2, then this root is given by 2 bc - ad b (b) (a) 2 ac + b 2 ( ac - b2 ) (c)

then B = A . Statement-2: If a, b, c ΠR, and ax2 + bx + c = 0 has no real roots, then b2 Р4ac < 0.

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 39 to 43 A root a of the polynomial equation p(x) = 0 is said to be of multiplicity n ΠN if

(d)

d ab - c 2

42. If q π 0 and the equation xn – npx + (n – 1)q = 0 has a root of multiplicity 2, then (b) pn + qn–1 = 0 (a) pn = qn–1 (d) pn + nqn–1 = 0 (c) np + qn = 0 43. If a is a root of xn + p1 xn–1 + p2 xn–2 + ... + pn = 0 of multiplicity 2, then a must be a root of (a) (b) (c) (d)

p (a ) – p ( b ) a p ( b ) – b a (a ) x+ a –b a –b

38. Statement-1: Let a, b, c p, q Œ R and suppose ax2 + bx + c = 0 has no real roots. Let a, a be the roots of ax2 + bx + c = 0 and px + q A B + = 2 x –a x –a ax + bx + c

bc ac - b2

p1 xn–1 + p2 xn–2 + ... + pn = 0 p1 xn–1 + 2p2 xn–2 + ... + npn = 0 np1 xn–1 + (n – 1) p2 xn–2 + ... + pn = 0 none of these.

Paragraph for Question Nos. 44 to 48 Let f (x) = ax2 + 2bx + c, a π 0, b2 – ac ≥ 0. Both the roots of f (x) = 0 will lie in the interval (x1, x2) if b < x2 a a f (x1) > 0, a f (x2) > 0

x1 < – and

If x1 = –•, we drop the condition a f (x1) > 0 and in case x2 = •, we drop the condition a f (x2) > 0. 44. If the equation (1 – a2)x2 + 2ax – 1 = 0 has both the roots lying in (0, 1), then (a) a > 2 (b) – 1 < a < 1 (c) a < – 1 (d) 1 < a < 2

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.47

45. If |p| < 1, p2 > 4q and both the roots of x2 + px + q = 0 satisfying the inequality |x| < 1, then (a) 1 + p + |q| > 0 (b) 1 + |p| + q > 0 (c) p + q > 1 (d) |p + q| < 1 46. If a > 0, a + |b| + c > 0 and |c| < a, then both the roots of the equation ax2 + bx + c = 0 lie in the interval (a) (0, 1) (b) (– 1, 0) (c) (– 1, 1) (d) none of these. 47. If both the roots of the equation x2 – 2(a – 1)x + 2a + 1 = 0 are positive, then (a) a > 2 (b) 1 < a < 2 (c) a ≥ 4 (d) 1 £ a £ 4

51. Let f (x) = ax2 + bx + c. Given f (x) = x has no real roots, then number of number of real roots of f ( f (x)) = x is 52. Let f (x) be a monic polynomial with integral coefa, b, c, d be four distinct integers such that f (a) = f (b) = f (c) = f (d) = 5. The number of integral values of k for which f (k) = 8 is 53. If three distinct real numbers a, b, c ΠR satisfy a(a2 + p) = b(b2 + p) = c(c2 + p) for some p ΠR, then the value of a + b + c is 54. The numbers of values of x ΠR for which 641/x + 481/x = 801/x is

2

48. If b – ac > 0, ab < 0, ac > 0, then the equation ax4 + 2bx2 + c = 0 has (a) four real roots (b) two negative and two imaginary roots (c) two positive and two imaginary roots (d) four imaginary roots

1 1 1 1 55. If a, b, c Œ R satisfy + + = a b c a + b+c value of Ê 1 1 1ˆ ÁË + + ˜¯ a b c

2019

-

2019 a

2019

+ b2019 + c 2019 +

INTEGER-ANSWER TYPE QUESTIONS 49. Let a, b Œ R. If for |x| £ 1, |ax2 + bx + c| £ 1, then the maximum possible value of |2ax + b| for x Œ [– 1, 1] is 8 50. Let a, b, c Œ R. If |ax2 + bx + c| £ 1 and a2 + 2b2 3 is maximum, then |a + c| is equal to

2018 is ( a + b + c)2019

56. The value of the parameter a ΠR such that the roots a, b, g of x3 + 6x2 + ax Рa = 0 satisfy the relation (a + 3)3 + (b + 3)3 + (g + 3)3 = 0 is

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. If l, m are real, l

m, then the roots of the equation

2

(l – m)x – 5(l + m)x – 2(l – m) = 0 are (a) real and equal (c) real and unequal

(b) complex (d) none of these [1979] 2. The entire graph of the equation y = x2 + kx – x + 9 is strictly above the x-axis if and only if (a) k < 7 (c) k > – 5

(b) – 5 < k < 7 (d) none of these [1979]

3. Let a > 0, b > 0 , and c > 0. Then both the roots of the equation ax2 + bx + c = 0 (a) are real and negative (b) have negative real parts (c) have positive real parts (d) none of these. [1980] 4*. Both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are always (a) positive (b) negative (c) real (d) none of these * It is to be assumed that a, b and c are real.

[1980]

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5. If x2 + px + 1 is a factor of ax3 + bx + c, then (b) a2 – c2 = – ab (a) a2 + c2 = – ab (d) none of these (c) a2 – c2 = ab [1980] 6. The equation Ê xˆ 2 cos2 Á ˜ sin2x = x2 + x–2, 0 < x π/2 Ë 2¯ has (a) no real solution (b) one real solution (c) more than one real solution. [1980] 7. Let a, b, c be non-zero real numbers such that 1 0

(d) α < γ < β.

[1989] 13. Let f (x) be a quadratic expression which is positive for all real x. If g(x) = f (x) + f (x) + f (x), then for all real x, (a) g(x) < 0 (b) g(x) > 0 (c) g(x) = 0 (d) g(x) 0 [1990]

Then the roots of the equation (x – α) (x – β) + c = 0 are

2

= Ú (1 + cos8 x ) ( ax 2 + bx + c) dx 0

Then the quadratic equation ax2 + bx + c = 0 has (a) no root in (0, 2) (b) at least one root in (0, 2) (c) a double root in (0, 2) (d) two imaginary roots. [1980] 8. The number of real solutions of the equation |x|2 – 3|x| + 2 = 0 is (b) 1 (d) 2

(c) γ = α

14. Let α, β be the roots of the equation (x – a) (x – b) = c, c 0.

8 2 Ú (1 + cos x ) (ax + bx + c)dx

(a) 4 (c) 3

12. Let a, b, c be real numbers, a 0. If α is a root of a2x2 + bx + c = 0, β is the root of a2x2 – bx – c = 0 and 0 < α < β, then the equation a2x2 + 2bx + 2c = 0 has a root γ 1 1 (a) γ = (α + β) (b) γ = α + β 2 2

[1982]

(a) a, c (c) a, b

(b) b, c (d) a + c, b + c. [1992] 2 15. If p and q are roots of x + px + q = 0, then (a) p = 1 (b) p = 1 or 0 (c) p = – 2 (d) p = – 2 or 0 [1995] 16. If p, q, r are positive and are in A.P., the roots of quadratic equation px2 + qx + r = 0 are all real for (a)

9. If a + b + c = 0, then the quadratic equation 3ax2 + 2bx + c = 0 has (a) at least one root in (0, 1) (b) one root in (2, 3) and the other in (– 2, – 1) (c) imaginary roots (d) none of these. [1983] 2 2 =1– has: 10. The equation x – x –1 x –1 (a) no root (b) one root (c) two equal roots

r –4 ≥4 3 p

(c) all p and r

2

2

ax + 2bx + c = 0 and dx + 2ex + f = 0 have a common root if (a) A.P. (c) H.P.

d e f are in , , a b c (b) G.P. (d) none of these [1985]

p –7 ≥4 3 r

(d) no p and r [1995]

17. The equation x +1 – (a) (b) (c) (d)

x –1 =

4 x – 1 has

no solution one solution two solutions more than two solutions.

18. In a triangle PQR, 11. If a, b, c are in G.P., then the equations

(b)

R=

[1997]

p Ê pˆ Ê Qˆ , if tan Á ˜ , tan Á ˜ Ë 2¯ Ë 2¯ 2

are the roots of the equation ax2 + bx + c = 0, (a 0), then (a) a + b = c (b) b + c = a (c) a + c = b (d) b = c 2 19. The roots of the equation x – 2ax + a2 + a – 3 = 0 are real less then 3, then

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.49

(a) a < 2 (b) 2 < a 3 (c) 3 a 4 (d) a > 4. [1999] 20. If α and β (α < β), are the roots of the equation x2 + bx + c = 0, where c < 0 < b, then (a) 0 < α < β (b) α < 0 < β < |α| (c) α < β < 0 (d) α < 0 < |α| < β. [2000] 21. If b > a, then the equation (x – a) (x – b) – 1 = 0 has (a) both roots in (a, b) (b) both roots in (– , a) (c) both root in (b, ) (d) one root in (– , a) and other in (b, ) [2000] 22. For the equation 3x2 + px + 3 = 0, p > 0, if one of the roots is square of the other, then p is equal to (a) 1/3 (b) 1 (c) 3 (d) 2/3 [2000] 23. Let α, β be the roots of x2 – x + p = 0 and γ, δ be the roots of x2 – 4x + q = 0. If α, β, γ, δ are in G.P, then the integral value of p and q respectively, are (a) – 2, – 32 (b) – 2, 3 (c) – 6, 3 (d) – 6, – 32. [2001] 24. Let f (x) = (1 + b2)x2 + 2bx + 1 and let m(b) be the minimum value of f (x). As b varies, the range of m(b) in (a) (0, 1] (b) (0, 1/2] (c) (1/2, 1] (d) [0, 1] [2001] 25. The set of all real numbers x for which x2 – |x + 2| + x > 0, is (a) (– ,– 2)

(2,

)

(d) ( 2 , •) . Ê pˆ ÁË 0, ˜¯ then 2

[2002] x2 + x +

greater than or equal to (a) 2tanα (b) 1

tan 2 a x2 + x

is always

(d) sec2 α. [2003] 2 2 2 27. If f (x) = x + 2bx + 2c and g(x) = – x – 2cx + b2 are such that min f (x) > max g(x), then relation between b and c, is (a) no relation (b) 0 < c < b/2 (c) |c|


2 |b| [2003]

[2006]

32. Let α, β be the roots of the equation x2 – px + r = 0 and α/2, 2β be roots of the equation x2 – qx + r = 0. Then the value of r is 2 2 (a) (p – q) (2q – p) (b) (q – p) (2p – q) 9 9 (c)

(b) ( -•, - 2 ) » ( 2 , •) (c) (– ,– 1) (1, )

26. If α

28. For all x, x2 + 2ax + 10 – 3a > 0, then the interval in which a lies is (a) a < – 5 (b) |a| < 5 (c) – 5 < a < 2 (d) 2 < a < 5. [2004] 29. If one root is square of the other root of the equation x2 + px + q = 0, then the relation between p and q is (a) p3 – q(3p – 1) + q2 = 0 (b) p3 – q(3p + 1) + q2 = 0 (c) p3 + q(3p – 1) + q2 = 0 [2004] (d) p3 + q(3p + 1) + q2 = 0 30. Let α, β be the roots of the quadratic equation ax2 + bx + c = 0 and = b2 – 4ac. If α + β, α2 + β2, α3 + β3 are in G.P., then (a) =0 (b) 0 (c) b = 0 (d) c = 0 [2005] 31. Let a, b, c be the sides of a triangle where a c and λ R. If the roots of the equation x2 + 2(a + b + c)x + 3λ(ab + bc + ca) = 0 are real, then

2 (q – 2p)(2q – p) 9

(d)

2 (2p – q)(2q – p) 9 [2007]

33. Let p and q be real numbers such that p 0, p3 q and p3 – q. If α and β are nonzero complex numbers satisfying α + β = – p and α3 + β3 = q, then a a b quadratic equation having and as its roots is b a (a) (p3 + q)x2 – (p3 + 2q)x + (p3 + q) = 0 (b) (p3 + q)x2 – (p3 – 2q)x + (p3 + q) = 0 (c) (p3 – q)x2 – (5p3 – 2q)x + (p3 – q) = 0 (d) (p3 – q)x2 – (5p3 + 2q)x + (p3 – q) = 0 [2010] 34. Let α and β be the roots of x2 – 6x – 2 = 0, with α > β a - 2a8 If an = an – bn for n 1, then the value of 10 2a9 is (a) 1 (c) 3

(b) 2 (d) 4

[2011]

IIT JEE eBooks: www.crackjee.xyz 2.50 Comprehensive Mathematics—JEE Advanced

35. A value of b for which the equations x2 + bx – 1 = 0 x2 + x + b = 0 have one root in common is (a)

- 2

(b) -i 3

(c)

i 5

(d)

2

1 ˆ Ê (c) Á 0, ˜ Ë 5¯

[2015]

MATRIX-MATCH TYPE QUESTIONS [2011]

36. The quadratic equation p(x has purely imaginary roots. Then the equation p( p (x)) = 0 has (a) only purely imaginary roots (b) all real roots (c) two real and two purely imaginary roots (d) neither real nor purely imaginary roots [2014] p p 37. Let - < q < - . Suppose a1 and b1 are the roots 6 12 of the equation x2 – 2x secq + 1 = 0 and a2 and b2 are the roots of the equation x2 + 2x tanq – 1 = 0. If a1 > b1 and a2 > b2, then a1 + b2 equals (a) 2(secq – tanq) (b) 2 secq (c) – 2 tanq (d) 0 [2016]

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 1. For real x, the function all real values provided (a) a > b > c (c) a > c > b

( x - a) ( x - b) will assume x-c (b) a < b < c (d) a < c < b.

[1984] 2. If α and β are the roots of x2 + px + q = 0 and α4, β4 are roots of x2 – rx + s = 0 then the equation x2 – 4qx + 2q2 – r = 0 has always (a) two real roots (b) two positive roots (c) two negative roots (d) one positive and one negative root. [1989] 3. Let S be the set of all non-zero real numbers α such that the quadratic equation αx2 – x + α = 0 has two distinct real roots x1 and x2 satisfying the inequality |x1 – x2| < 1. Which of the following intervals is(are) a subset(s) of S ? 1 ˆ Ê 1 (a) Á – , – ˜ Ë 2 5¯

Ê 1 ˆ (b) Á – , 0˜ Ë 5 ¯

Ê 1 1 ˆ (d) Á , Ë 5 2 ˜¯

1. Let f (x) =

(a) (b) (c) (d)

x2 – 6x + 5 x2 – 5x + 6

.

Column 1 Column 2 If – 1 < x < 1, then (p) 0 < f (x) < 1 f (x If 1 < x < 2, then (q) f (x) < 0 f (x If 3 < x < 5, then (r) f (x) > 0 f (x If x > 5, then (s) f (x) < 1 f (x

ASSERTION-REASON TYPE QUESTIONS 1. Let a, b, c, p, q be real numbers. Suppose α, β are 1 the roots of the equation x2 + 2px + q = 0 and α, b are the roots of the equation ax2 + 2bx + c = 0, where β2 {– 1, 0, 1}. Statement–1: (p2 – q) (b2 – ac) 0 Statement–2: b pa or c qa [2008]

INTEGER-ANSWER TYPE QUESTIONS 1. The smallest value of k, for which both the roots of the equation x2 – 8kx + 16(k2 – k + 1) = 0 are real, distinct and have value at least 4, is [2009] 4

2. The number of distinct real roots of x – 4x + 12x2 + x – 1 = 0 is [2011]

FILL

IN THE

3

BLANK TYPE QUESTIONS

1. If 2 + 3 i is a root of the equation x2 + px + q = 0, where p, q R, then (p, q) = _______ [1982] 2. If the product of the roots of the equation x2 – 3kx + 2e2lnk – 1 = 0 is 7, then roots are real for k = _______ [1984] 3. If the quadratic equations x2 + ax + b = 0 and x2 + bx + a = 0, (a b) have a common root, then the numerical value of a + b is _______ [1986]

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.51

1 x 1 x = and (x + y) = – , y 2 y 2 then x = _______ and y = _______ [1990]

4. If x < 0, y < 0, x + y +

5. The sum of all the real roots of the equation |x – 2|2 + |x – 2| – 2 = 0 is _______ [1997]

TRUE/FALSE TYPE QUESTIONS 1. The equation 2x2 + 3x + 1 = 0 has an irrational root. [1983] 2. If x – r is a factor of the polynomial f (x) = a0xn + a1xn – 1 + .... + an repeated m times, (1 – c is ( a - c + b - c )2

[1979]

5. Show that the equation esin x – e– sin x – 4 = 0 has no real solution.

[1982]

2

6. If one root of the equation ax + bx + c = 0 is equal to the nth power of the other, then show that 1 n n +1

( ac )

+ ( a n c)

1 n +1

+b =0

[1983]

(5 + 2 6 ) x

2

-3

+ (5 - 2 6 ) x

0. [1983]

2

-3

= 10.

[1985]

9. For a 0, determine all real roots of the equation [1986] x2 – 2a |x – a| – 3a2 = 0 10. Let α1, α2, β1, β2 be the roots of ax2 + bx + c = 0 and px2 + qx + r = 0 respectively. If the system of equations α1y + α2z = 0 and β1y + β2z = 0 and has non-trivial solution, then prove that

b2 ac = q2 pr [1987]

11. Solve

|x2 + 4x + 3| + 2x + 5 = 0

[1988]

12. Let a, b, c R and α, β be the roots of ax2 + bx + c = 0 such that α < – 1 and β > 1 then show that b c 1 + + < 0. [1995] a a 13. A function f : R

R, where R is the set of real a x2 + 6x – 8 . Find f (x) = a + 6 x – 8x2 the interval of values of α for which f is onto. Is the function one-to-one for α = 3? Justify your answer. [1996]

14. Find the set of all solutions of the equation [1997] 2|y| – |2y – 1 – 1| = 2y – 1 + 1 15. Let f (x) = Ax2 + Bx + C where A, B, C are real numbers. Prove that if f (x) is an integer whenever x is an integer, then the numbers 2A, A + B and C are all integers. Conversely, prove that if the numbers 2A, A + B and C are all integers then f (x) is an integer whenever x is an integer. [1998] 16. Let a, b, c, d be real numbers in G.P. If u, v, w satisfy the system of equations u + 2v + 3w = 16 4u + 5v + 6w = 12 6u + 9v = 4 then show that the roots of the equation Ê1 1 1ˆ 2 2 2 2 ÁË + + ˜¯ x + [(b – c) + (c – a) + (d – b) ]x u v w + u + v + w = 0 and 20x2 + 10(a – d)2 x – 9 = 0 are reciprocal of each other. [1999] 17. If α, β are the roots of ax2 + bx + c = 0, (a 0) and α + δ, β + δ are the roots of Ax2 + Bx + C = 0, (A 0) for some constant δ, then prove that b2 – 4 ac

7. Find all real values of x which satisfy x2 – 3x + 2 > 0 and x2 – 3x – 4

8. Solve for x.

a2

=

B 2 – 4 AC A2

[2000]

IIT JEE eBooks: www.crackjee.xyz 2.52 Comprehensive Mathematics—JEE Advanced

18. Let – 1 p 1. Show that the equation 4x3 – 3x – p = 0 has a unique root in the interval [1/2, 1] and identify it. [2001]

MATRIX-MATCH TYPE QUESTIONS 66.

p

q

r

s

19. Let a, b, c be real numbers with a 0 and let α, β be the roots of the equation ax2 + bx + c = 0. Express the roots of a3x2 + abcx + c3 = 0 in terms of α, β. [2001]

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

20. If x2 + (a – b)x + (1 – a – b) = 0 where a, b R then a for which equation has unequal real roots for all values of b. [2003]

d

p

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p

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a

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b

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c

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b

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67.

2

21. Let a and b be the roots of the equation x – 10cx – 11d = 0 and those of x2 – 10ax – 11b = 0 are c, d a + b + c + d, assuming that a, b, c, d are distinct. [2006] 68.

Answers LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 6. 11. 16. 21. 26. 31. 36. 41.

(b) (c) (a) (d) (a) (d) (c) (a) (d)

2. 7. 12. 17. 22. 27. 32. 37. 42.

(a) (d) (c) (b) (c) (d) (c) (b) (a)

3. 8. 13. 18. 23. 28. 33. 38. 43.

(c) (b) (b) (d) (d) (d) (d) (c) (d)

4. 9. 14. 19. 24. 29. 34. 39. 44.

(b) (a) (b) (b) (d) (b) (c) (a) (a)

5. 10. 15. 20. 25. 30. 35. 40. 45.

69. (a) (b) (d) (d) (d) (c) (b) (a) (a)

70.

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 46. 49. 51. 53. 55. 57. 59. 61. 63. 65.

(a), (d) (a), (d) (a), (a), (a) (a), (a), (a),

(b) (b), (c) (b) (b), (b), (b)

47. (a), (b) 48. 50. (c), (d) 52. 54. 56. 58. 60. (c) 62. (c), (d) 64.

(b), (a), (b), (b), (a), (a), (a), (b), (a),

(c) (b), (c), (c) (b) (b), (b), (c) (b),

71. (c) (d)

(c) (c), (d) (c), (d)

ASSERTION-REASON TYPE QUESTIONS 72. (b) 76. (d)

73. (c)

74. (a)

75. (a)

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.53

COMPREHENSION-TYPE QUESTIONS 77. 81. 85. 89. 93.

(a) (b) (b) (a) (a)

78. 82. 86. 90. 94.

(c) (c) (c) (c) (c)

79. 83. 87. 91. 95.

(a) (b) (c) (a) (d)

80. 84. 88. 92. 96.

(d) (c) (b) (d) (c)

INTEGER-ANSWER TYPE QUESTIONS 97. 3 101. 1 105. 5

98. 3 102. 0 106. 5

99. 0 103. 9 107. 6

p

q

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33.

34.

100. 0 104. 1 108. 3

LEVEL 2

35.

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17.

(a) (c) (d) (b) (c)

2. 6. 10. 14. 18.

(a) (d) (c) (b) (b)

3. 7. 11. 15. 19.

(c) (c) (b) (a) (b)

4. 8. 12. 16. 20.

(c) (a) (a) (b) (a)

MULTIPLE ANSWERS TYPE QUESTIONS 21. 23. 25. 27. 29.

(a), (b), (d) (a), (b), (d) (b) (a), (b), (c) (d)

22. 24. 26. 28. 30.

(a), (a), (a), (a), (b),

(b), (b), (b) (b), (c),

(c) (c) (c), (d) (d)

MATRIX-MATCH TYPE QUESTIONS p

q

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31.

32.

ASSERTION-REASON TYPE QUESTIONS 36. (a)

37. (b)

38. (b)

COMPREHENSION-TYPE QUESTIONS 39. (a) 43. (b) 47. (c)

40. (a) 44. (a) 48. (a)

41. (b) 45. (b)

42. (a) 46. (c)

INTEGER-ANSWER TYPE QUESTIONS 49. 4 53. 0

50. 1 54. 1

51. 0 55. 0

52. 0 56. 9

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17. 21. 25.

(c) (c) (a) (b) (a) (d) (b)

2. 6. 10. 14. 18. 22. 26.

(b) (a) (a) (c) (a) (c) (a)

3. 7. 11. 15. 19. 23. 27.

(b) (b) (a) (b) (a) (a) (d)

4. 8. 12. 16. 20. 24. 28.

(c) (a) (d) (b) (b) (a) (c)

IIT JEE eBooks: www.crackjee.xyz 2.54 Comprehensive Mathematics—JEE Advanced

29. (a) 33. (b) 37. (c)

30. (d) 34. (c)

31. (a) 35. (b)

32. (d) 36. (d)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 1. (a), (b)

2. (a), (d)

3. (a), (d)

MATRIX-MATCH TYPE QUESTIONS p

q

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b

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1.

1. (b)

INTEGER-ANSWER TYPE QUESTIONS 2. 2

FILL

IN THE

BLANKS TYPE QUESTIONS

1. (– 4, 7) 3. – 1 5. 1, 3

2. 6 4. – 1/4, – 1/4

TRUE/FALSE TYPE QUESTIONS 1. False

2. False

LEVEL 1

3+5 =x+4 2 The equation becomes (y – 1)4 + (y + 1)4 = 16 fi 2{y4 + 6y2 + 1} = 16 fi y4 + 6y2 – 7 = 0 fi (y2 + 7) (y2 – 1) = 0 fi y2 = –7 or y2 = 1

1. Put y = x +



ASSERTION-REASON TYPE QUESTIONS

1. 2

Hints and Solutions

3. True

4. True

y = ± 7i or y = ±1

fi x = -4 ± 7i or x = – 4 ± 1 = – 5 or 3. Thus, the given equation has two real roots. 2. a2 – 6a – 2 = 0, b 2 – 6b – 2 = 0 Now, a10 – 2a8 = a10 + a10 – 2(a8 + b8) = a8 (a2 – 2) + b8(b 2 – 2) = a8 (6a) + b8(6b) = 6a9 fi

a10 - 2a8

=3 2a 9 3. If a is a common root of x2 + bx – 1 = 0 and x2 + x + b = 0, we get a2 + ba – 1 = 0, a 2 + a + b = 0. fi ba – a – 1 – b = 0, fi a(b – 1) = b + 1 b +1 fi a= b -1 2 As a + ba – 1 = 0, we get

SUBJECTIVE-TYPE QUESIONS

2

Ê b + 1ˆ Ê b + 1ˆ ÁË ˜¯ + b ÁË ˜ -1 = 0 b -1 b - 1¯

1. 5/4 3. (q – s)2 + q(q – s) (r – p) + s(r – p)2 = 0 7. [–1, 1) (2, 4]

fi fi

(b + 1)2 + b(b + 1) (b – 1) – (b – 1)2 = 0 4b + b(b2 – 1) = 0 fi b2 + 3 = 0

8. ± 2, ± 2



b = ± 3i

9. (1 - 2 )a, ( 6 - 1)a

x+1≥0 and 2x + 1 ≥ 0 fi 2x ≥ – 1 and 2x ≥ 1 fi x ≥ 1/2. We rewrite the equation as

11. {– 4, – (1 + 3 )} 13. 2 α 14 14. {–1} [1, ) Ï1 ¸ 18. cos Ì cos -1 p ˝ Ó3 ˛ 20. a > 1

2x + 1 = 1 + 2x - 1 Squaring both the sides, we get 19. α2β, αβ2 21. 1210

2x + 1 = 1 + 2x – 1 + 2 2 x - 1 fi

1 = 2 2 x - 1 fi 2x – 1 = 1/4 fi x = 5/8

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.55

5.

6.

7.

8.

Since x equation has just one solution. Discriminant of the given equation is D = 9(l – m)2 + 8 (l + m)2 As l + m π 0, (l + m)2 > 0. Also, (l – m)2 ≥ 0. Thus, D > 0. Hence, roots of the given equation are real and unequal. 1 a= x +1 + x Thus, 0 < a £ 1. We have a + b = a and ab = b. Now, aAn – bAn–1 = (a + b) (a n + b n) – ab (an–1 + bn–1) = an+1 + anb + abn + bn+1 – anb – abn = an+1 + bn+1 = An+1 We have (a + b + g)2 = a2 + b2 + g 2 + 2(bg + ga + ab) fi 4 = 6 + 2(bg + ga + ab) fi bg + ga + ab = –1. Also, a3 + b3 + g 3 – 3abg = (a + b + g) (a2 + b2 + g2 – bg – ga – ab) fi 8 – 3abg = 2(6 + 1) fi 3abg = 8 – 14 = – 6 or abg = – 2. Now, (a2 + b2 + g 2)2 = Â a4 + 2 Â b2 g 2 2 = Â a 4 + 2 È(Â bg ) - 2abg (Â a )˘ Î ˚



c Ê Pˆ Ê Qˆ tan Á ˜ tan Á ˜ = Ë 3¯ Ë 3¯ a Now, –R = p/4 fi P + Q = 3p/4 tan( P / 3) + tan(Q / 3) Êpˆ 1 = tan Á ˜ = Ë 4¯ 1 - tan( P / 3)(Q / 3) - b /a -b = = 1 - c /a a-c fi a + b = c. 10. Cubing we get 8 + x + 8 – x + 3(64 – x2)1/3 = 1 fi (64 – x2)1/3 = –5 fi 64 – x2 = – 125 fi

2p = (a + b + g)2 – (a2 + b2 + g 2) < 0 12. x = 0 is not a root. Divide both the numerators and denominators by x and put x + 3/x = y to obtain 3 4 5 + = - fi y = – 5, 3 y +1 y - 5 2 x + 3/x = – 5 has two irrational roots and x + 3/x = 3 has imaginary roots. 13. (x2 + x + 4 + x) (x2 + x + 4 + 2x) = 0

(

fi x = -1 ± 3 i , -3 ± 7 i

x2 = 189 fi x = ±3 21

11. If a, b, g are the roots of

)

2

14. The given equation gives (x2 – x + 1)2 = x2, 5x2 15. Use x2 + 3x + 2 = (x + 1) (x + 2), and x2 – 9x + 20 = (x – 4) (x – 5) to obtain (x2 + 3x + 2) (x2 – 9x + 20) = (x2 – 3x – 4) (x2 – 3x – 10) Now, put x2 – 3x = y and rewrite the equation as (y + 1) (y – 4) (y – 10) = – 30 Roots of this equation are 5, 4 ± 30 . 16. Write the equation as 3( x + 1)2 + 4 +

 a4 = 36 – 2[(–1)2 – 2(–2) (2)] = 18

b Ê Pˆ Ê Qˆ 9. tan Á ˜ + tan Á ˜ = Ë 3¯ Ë 3¯ a

\

x3 + px + q = 0, then a + b + g = 0, bg + ga + ab = p, abg = – q As q π 0, none of a, b, g is zero. Now, use

5( x + 1)2 + 9 = 5 – (x + 1)2

and observe that LHS > 5 and RHS < 5 for x π –1. 17.

x2 - 6 x + 7 = ±1 fi x = 0 or x2 + 7 = 0. 2 x +6 x +7

18. Since

x x2 +x = , we must have x and x -1 | x - 1|

x x2 ≥0 are of the same sign i.e. x -1 x -1 fi x = 0 or x –1 > 0 19. Clearly – 2 < x £ 2. Also, x For – 2 < x < 2, we get 2- x = 1 or x = 0. 2+ x 20. For a π 0, x = a/2. For a = 0, no solution. 21. Roots are – a and a2. Now, a2 – a < a3/4.

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Clearly a cannot be negative or 0. Thus,

= p2 + r2 – 2pr = (p – r)2 ≥ 0

a2 – 4a + 4 > 0 fi (a – 2)2 > 0

fi At least one of D1, D2 ≥ 0.

22. As x – 3x + 4 > 0 " x, 2x2 – (a + 3)x + 2 > 0 " x 2

¤ (a + 3)2 < 16 ¤ – 4 < a + 3 < 4 ¤ – 7 < a < 1. 1 1 ˆ ˆ Ê 1 Ê 1 + 23. Á =0 + + Ë x - a x - b - c + a ˜¯ ÁË x - b x - c ˜¯ 2x - b - c 2x - b - c + =0 fi ( x - a)( x - b - c + a) ( x - b)( x - c) 1 (b + c) and 2x2 – 2(b + c)x + ab + ac 2 + bc – a2 = 0 The last equation has real roots. 24. That sign of a cannot be determined may be checked by taking examples. For instance, f (x) = x2 – 4x – 5 and f (x) = 1 – x2. fi x=

25. Find values of k for which both the roots are negative. (k – 1)2 – (k – 5) ≥ 0, k – 1 > 0, k – 5 > 0



k > 5.

26. 4 cos2 a – 2 sin a (sin a + cos a) = 0 fi

1 + 3 cos 2a = sin 2a.

For 3p/4 < a < p, left side is positive and right side is negative. 27. a2 + b2 = (sin f – 1)2 + 2 cos2 f = 3 – 2 sin f – sin2 f = 4 – (sin f + 1)2 28. bx + 28 = 12 – 4x – x2 fi x2 + (4 + b) x + 16 = 0 For unique solution (4 + b)2 – 64 = 0 fi b = 4, – 12, b = – 12 is not possible. 29. 2x2 + 3x + 4 = 0 has imaginary roots. Thus, the given equations have identical roots. \

a b c = = = k (say) 2 3 4

fi a + b + c = 9k. For least value take k = 1. 30. Let D1 = p2 – 4q, D2 = r2 – 4s Now, D1 + D2 = p2 + r2 – 4(q + s)

31. D = a2 – 4 > 0 if a < – 2 and D < 0 if – 2 < a < 0. If – 2 < a < 0, the equation has no real roots. If a < – 2, the equation has exactly four real roots. 32. f ¢(x) = 3(x + 1)2 + 3 > 0 " x. Thus, f (x) increases on R. Let a = f (1), b = f (2) and c = f (3), then a < b < c and g(x) = (x – b) (x – c) + 2(x – a) (x – c) + 3(x – a) (x – b) As g(a) > 0, g(b) < 0 and g(c) > 0, g(x) = 0 has exactly two real solutions. 33. As q, r Œ R, a – ib must be a root of x3 + qx + r = 0. Let the third root be g, then a + ib + a – ib + g = 0 fi g = – 2a As g is a root of x3 + qx + r = 0, – 8a3 – 2qa + r = 0 \ required equation is 8x3 + 2qx – r = 0. 34. Let a + ib, a – ib, g where b π 0 be roots of x3 + ax + b = 0. As a + ib + a – ib + g = 0, we get g is real. \ a = (a + ib + a – ib) (– 2a) + (a + ib) (a – ib) = – 3a2 + b2 is real. Also, b = 2a (a + ib) (a – ib) = 2a (a2 + b2) is real. 35. Write the given equation as Ê 100 ˆ ˜ ÁË 50 ¯

1x

Ê 25 ˆ +Á ˜ Ë 50 ¯

1x

=

65 8

fi 21/x = 8, 1/8 fi x = –1/3, 1/3 36. – 3x2 + 2x –

9 =–3 11

2 ÈÊ 1ˆ 16 ˘ ÍÁ x - ˜ + ˙ < 0 " x 3¯ 99 ˚˙ ÍÎË

and LHS > 0 " x. 37. (a + c)2 < b2 fi (a – b + c) (a + b + c) < 0 Thus, ax2 + bx + c = 0 has at least one root in (– 1, 1). 38. Write the given equation as 3t2 – 10t + 8 = 0 where t = (3/2)x fi t = 2, 4/3. 39. (x – m)2 = 1 fi x = m ± 1 \ – 2 < m – 1, m + 1 < 4 fi – 1 < m < 3.

(

)

+1

Ê px ˆ fi sin Á = 1 and x = Ë 2 3 ˜¯

3.

40. x2 – 2 3 x + 4 = x - 3

2

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.57

41. Let y =

Put

x2 + 2x + 2

fi y = 2 as y ≥ 0. This gives x = 1. 51. Put (2x – 1)1/3 = y or y3 = (2x – 1).

fi (y – 1)x + 2(y + 3)x + (2y – 5) = 0 2

As x is real, (y + 3)2 – (y – 1) (2y – 5) ≥ 0

The original equation becomes

fi y2 – 13y – 4 £ 0

\ x3 – y3 = – 2(x – y)

x3 = 2y – 1 fi x – y = 0 or x2 + y2 + xy = – 2

42. a + b = – b < 0, ab = c < 0

fi x = y as other equation is not possible

As a < b, we get a < 0, b > 0 and b < |a|

\ 2x – 1 = x3 or x3 – 2x + 1 = 0

2

43. x (y – 1) + 2x (2y – 1) + 3y – c = 0

fi x = 1, – (1 + 5 ) 2 , – (1 - 5 ) 2 52. Replace x by x – 2 to obtain

As x is real, (2y – 1)2 – (y – 1) (3y – c) ≥ 0 fi y2 + (c – 1)y – (c – 1) ≥ 0 " y Œ R fi (c – 1)2 + 4(c – 1) £ 0 fi –3£c£1 44.

x - 1 + x + 3 = y to obtain y2 + y – 6 = 0

x2 - 6x + 5

P(x) = x2 – x – 4. 53. As P(x2 + 2) is of even degree, no such polynomial exists.

x-5 >0 ( x + 7)( x - 2)

54. Clearly, an2 = an fi an = 1

fi – 7 < x < 2 or x > 5

Let k be the largest non-negative integer < n such that ak π 0, then

Thus, the smallest integer for which the inequality

P(x) = a0 + a1x + ... + ak xk + an xn.

2

2

2 2

2 2

45. D = (b + a – c ) – 4a b

= (a2 + b2 – 2ab – c2) (a2 + b2 – c2 + 2ab)

55. None of x1, x2, x3, is 0. Also,

= [(a – b)2 – c2] [(a + b)2 – c2] > 0

x1 + x2 + x3 = 0, x1 x2 x3 = – q.

46. D = 4 (b + c – a)2 – 4abc > 0 – ( b + c - a) ± D , one is positive and other a, b = 2a is negative. 47. As LHS ≥ 1, x ≥ 1. Write the given equation as 1–

x 4 - x 2 = (x – 1)2

As x3 =

1 1 + , we get x1 x2 = – 1, x3 = q. x1 x2

Thus q3 + pq + q = 0 fi q2 + p + 1 = 0 fi p < 0. 56. sin a (cos a ± i sin a)n – (cos a ± i sin a) sin (na) + sin (n – 1)a = 0 ¤ sin a cos (na) – cos a sin (na) + sin (n – 1)a ± i [sin a sin (na) – sin a sin (na)] = 0

fi x2 (2 – x)2 = x2(x2 – 1) fi x = 5/4. 48. Put y =

Comparing coefficient of xn+k on both sides of P(x2) = P(x)2, we obtain 0 = 2anak fi ak = 0. A contradiction.

which is true.

x + x + 2 fi y = 2x + 2 + 2 x + 2 x . 2

2

The given equation becomes y2/2 – 1 + y = 3. As y > 0, we get y = 2. \ x + 2 = 4 – 4 x + x fi x = 1/4

From here it also follows that p(x) is divisible by x2 Р2x cos a + 1. 57. Similarly to Question 56. 58. Putting q = 1 Рp in the given inequality we obtain p2c2 + (a2 Рc2 Рb2)p + b2 > 0. It holds for each p ΠR if and only if (a2 Рc2 Рb2)2 Р4b2c2 < 0

49. See solution of Question number 45.

¤ (a2 – c2 – b2 – 2bc) (a2 – c2 – b2 + 2bc) < 0

50. For equation to be meaningful, x ≥ 1, x ≥ – 3, x £ 2. Thus, x Œ [1, 2]

¤ (a + b + c) (a – b – c) (a + b – c) (a – c + b) < 0

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59. Let a = a + b, b = ab. Now, a3 (a + 1) = 1, and b3 (b + 1) = 1 gives 1 = (ab)3 (a + b + ab + 1) fi 1 = b3 (a + b + 1)

exactly one root in [1/2, 1]. Let this root be a. Put a = cos q. As 1/2 £ a £ 1, 0 £ q £ p/3 Now, 0 = p(a) = cos (3q) – p

Also, a4 + a3 = b4 + b3, a π b (a3 + a2) + (b3 + b2) + ab (a + b) + ab = 0 fi

1 1 + + ab + b = 0 a b



a (1 + b2) + b2 = 0



Ê 1 ˆ 2 2 Á 3 – b – 1˜ (1 + b ) + b = 0 Ëb ¯



b6 + b4 + b3 – b2 – 1 = 0.

1 cos–1 (p). 3

64. f (0), f (1), f (–1) are all integers fi c, a + b + c, a – b + c are integers. fi c, a + b, a – b are integers fi c, a + b, a – b, 2a are integers. 65. D = 4a2b2 – ab(a + b)2 > 0 as ab < 0 and product of roots < 0.

60. Write

66. (a) f (a) > 0, f (b) < 0

ax3 + bx + c = a(x2 + px + 1) (x + l)

(b) f (x) Æ • as x Æ – • and f (a) < 0

where l is some constant.

(c) f (c) < 0, f (d) > 0



0 = a (l + p), b = a (pl + 1), c = al

Thus,

0 = c + ap

and

b = a(1 – p2) fi ab = a2 – c2

(d) f (b) < 0, f (c) > 0 (∵ c = al)

Also, (2a – b)2 = b2 + 4c2 61. – p is a root of ax3 + bx2 + cx + d fi

fiq=

– ap3 + bp2 – cp + d = 0 fi p|d

Also, cp – d = p2 (b – ap) fi p2|(cp – d)

67. (a) Use a + b + g = 0 (b) For large values of x. 1 1 a -1 = ÊÁ1 - ˆ˜ xË x¯ x -a =

If d = 0, p(ap2 – bp + c) = 0. 62. x2 + 6x + 8 > 0, x2 + 6x + 8 π 1

ˆ 1 Ê a a2 a3 a4 1 + + 2 + 3 + 4 + ...˜ Á xË x x x x ¯

Thus, a4 + b4 + g4

2x + 2x + 3 > 0, 2x + 2x + 3 π 1 2

2

f ¢( x ) 1 in the expansion of . 5 f ( x) x

log2 x 2 + 2 x +3 ( x - 2 x ) = 1 2

fi x2 + 6x + 8 > 0, x2 + 6x + 7 π 0 2x2 + 2x + 3 > 0, x2 + x + 1 π 0 and x2 – 2x = 2x2 + 2x + 3 fi x = – 1, – 3. But x = – 3 does not satisfy x2 + 6x + 8 > 0 63. p¢(x) = 12x2 – 3 = 12(x2 – 1/4) > 0 if 1/2 < x < 1. 1 3 Ê 1ˆ Also p Á ˜ = - - p £ 0 Ë 2¯ 2 2 and p(1) = 4 – 3 – p ≥ 0 Thus p(x) has at least one root in [1/2, 1] As p(x) is strictly increasing on [1/2, 1], p(x) has

In the present case f ¢(x) = 3x2 + p Therefore f ¢( x ) 3x 2 + p 3 2 p 3q 2 p2 = 3 = - 3 - 4 + 5 + ... f ( x) x x x + px + q x x \ a4 + b4 + g4 = 2p2 1 1Ê xˆ (c) and (d) = - Á1 - ˜ Ë a a¯ x -a = -

ˆ x x2 1Ê + + 2 + ...˜ 1 Á aË a a ¯

Thus, a–3 + b–3 + g–3

-1

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.59

f ¢( x ) x in the expansion of f ( x) 2

But Ê p3 3 ˆ - f ¢( x ) - p p2 = + 2 x - Á 3 + ˜ x 2 + ... f ( x) q q¯ q Ëq

71. (a) a(x2 + 2x + 1) + b(x + 1) + c = 0 fi x + 1 = a, b fi x = a – 1 b – 1 (b) ax2 – bx (x – 1) + c(x – 1)2 = 0 2

x ˆ Ê +b fi a ÁË x - 1˜¯

Ê p3 3 ˆ Thus, a–3 + b–3 + g –3 = - Á 3 + ˜ q¯ Ëq and a–2 + b–2 + g –2 = p2/q2 2b c 68. (a) > 0, > 0 fi a, – b, c are of the same sign a a (b) -



69. Discriminant of the equation is D = 4(a + 1)2 – 4(9a – 5) = 4 (a – 1) (a – 6) Equation will have real roots if a £ 1 or a ≥ 6 and imaginary roots if 1 < a < 6.

2

Ê x + 1ˆ Ê x + 1ˆ –b Á +c=0 fia Á Ë x - 1˜¯ Ë x - 1˜¯ fi–

fi x – 1 = a, b

Roots will positive if D ≥ 0; – (a + 1) > 0 and 9a – 5 > 0.



P(1) = 0 and Q(1) = 0

73. Let D1 = p2 – 4q and D2 = r2 – 4s Now, D1 + D2 = p2 + r2 – 4(q + s) = p2 + r2 – 2pr [\ 2(q + s) = pr] = (p – r)2 ≥ 0

70. (a) Use log0.75 x < 0 if x > 1. (b) (196)1/x + (49)1/x = (4.25) (98)1/x Ê 1ˆ + Á ˜ Ë 2¯

1x

9 fi 2 = ¤ 21/x = 3/2, 2/3. 4 Thus, there are two values of x. (c) Put

x – 1 = t or x = 1 + t2 to obtain t 2 – 4t + 4 + t 2 – 2t + 1 = 1



|t–2|+|t–1|=1fi1£t£2

\



x = 3, 9



at least one of D1, D2, is non-negative.

2

74. ax + 2bx – 3c = 0 has no real roots fi

4b2 + 12ac < 0



ac < 0

If c > 0, then a < 0 In this case ax2 + 2bx Р3c < 0 " x ΠR fi

a(22) + 2b(2) – 3c < 0



a+b
6.

1/x



72. Let w π 1 be a cube root of unity. Putting x = w, w2 we get P(1) + w Q(1) = 0 and

¤

x +1 a -1 b -1 , = a, b fi x = . a +1 b +1 x -1

(d) a(x2 – 2x + 1) + b(x – 1) + c = 0

No such value of a

Roots will be negative if D ≥ 0; – (a + 1) < 0 and 9a – 5 > 0

-x a b , = a, b fi x = a +1 b +1 x -1

(c) a(x + 1)2 – b(x2 – 1) + c(x – 1)2 = 0

2b c < 0, > 0 fi a, b, c are of the same sign a a

c (c) < 0 fi a, – c are of the same sign. a

x ˆ Ê ˜ +c=0 ÁË x - 1¯



t = 1/2, 1

75. y =

a x 2 + 3x – 4 a + 3x – 4 x 2

fi (a + 3x – 4x2)y = ax2 + 3x – 4 fi (a + 4y) x2 + 3(1 – y)x – (4 + ay) = 0

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As x Œ R, 9(1 – y)2 + 4(a + 4y) (4 + ay) ≥ 0 fi (9 + 16a) y2 + 2 (2a2 + 23) y + 16a + 9 ≥ 0 This is true for each y Œ R. fi 9 + 16a > 0 and (2a2 + 23)2 – (16a + 9)2 £ 0 fi (2a2 + 23 + 16a + 9) (2a2 + 23 – 16a – 9) £ 0 fi 9 + 16a > 0 and 4(a + 4)2 (a – 1) (a – 7) £ 0 fi a Œ [1, 7] a x + 6x – 8 2

The function f (x) =

a + 6 x – 8x

(a 2) x + 3x – 4 (a 2) + 3x – 4 x 2

2

will be onto if the

fi a=

Thus

3q 3p + y 27q3

(3 p + y )

3

+

9 pq +q=0 3p + y

fi 27q2 + 9p(3p + y)2 + q(3p + y)3 = 0. 81. y = (a – b) (a – g) = a2 – (ab + ag) + bg = a2 – (ab + ag + bg) + 2bg

2

expression

takes all real values.

y = a2 – 3 +

fi a/2 Π[1, 7] or a Π[2, 14]. = p , q

76. Expression

q b have meanings if p/q ≥ 0 , p a

and b/a ≥ 0.

2abg a

-6 (a + 1) a 3 - 3a - 4 = a a

fia=

-6 . 6+ y 3

But

p , q

q b =0 , p a

p q b = 0, = 0, = 0 q p a



or y3 + 9y2 – 216 = 0

This is not possible. pa + qa =

Ê -6 ˆ Ê -6 ˆ Thus, Á +3 Á +2=0 ˜ Ë 6 + y¯ Ë 6 + y ˜¯

–b a



a= –

b a ( p + q)

2

Now, use pa is a root of ax + bx + c = 0. 2

Ê x - 1ˆ Ê x - 1ˆ + bÁ 77. a Á +c=0 Ë x - 2 ˜¯ Ë x - 2 ˜¯ 2a - 1 2b - 1 x -1 , fi = a, b fi x = . a -1 b -1 x-2

82. f (x2) – f (x1) = 0 fi a(x1 + x2) + b = 0. Similarly, a(x2 + x3) + b = 0 Subtracting, we get a(x3 – x1) = 0 fi a = 0. \ b = 0, c = 0. 83. g(x) = 1 for x = x1, x2, x3. 84. g(x) = x for x = x1, x2, x3. 85. g(x) = x(x – 1) for x = x1, x2, x3.

r y- . p

r 78. Put y = a + or a = p 2

Ê rˆ \p Áy- ˜ +q Ë p¯

y-

r +r=0 p

fi p y – pq y + q r = 0. 3 2

2

2

79. As a + b + g = 0, b + g = – a. The desired equation can be obtained by changing x to – x. 80. y = (b – g)2 = (b + g)2 – 4bg = a2 + 4q/a ˆ Êq = 3 Á - p˜ ¯ Ëa

86. g(x) = x2 for x = x1, x2, x3. 87. 3x2 + 2px + q = 0 has two real roots. 88. f (0) > 0, f (1) < 0 and f ¢(x) < 0 for 0 < x < 1. 89. If a £ a £ b £ b and a, b are zeros of f (x), then $ c Œ (a, b) such that f ¢(c) = 0. Contradicts a, b are consecutive zero of f ¢(x). 90. If f (x) = 0 has four real roots then f ¢¢(x) = 12ax2 + 6bx + 2c = 0 has two real roots. This implies 9b2 – 24ac ≥ 0. Contradiction. 91. Let f (x) = ax3 + bx2 + cx. Apply Rolle’s theorem to f (x) on [0, 1]. 92. p(x) = 0 can have at most two positive and one negative root.

IIT JEE eBooks: www.crackjee.xyz Quadratic Equations 2.61

93. p(x) = 0 can have at most one positive and two negative roots. Also, p(– 2) < 0, p(– 1) > 0 and p(0) < 0. 94. The equation p(x) = 0 has two variation of signs, thus it can have at most two positive roots. Also, p(–x) = 0 has just one variation of sign, so p(x) = 0 can have at most one negative root. As p¢(x) = 0 fi x = ± (1/5)1/4 and p¢¢((1/5)1/4) > 0, p(x) has a local minimum at x = (1/5)1/4. As p(x) > 0 " p(0) = 16 > 0, p((1/5)1/4 x ≥ 0. Thus p(x) = 0 has no positive root. Also, as p(– 2) < 0 and p(– 1) > 0, p(x) = 0 has one negative root lying in (– 2, – 1). \ p(x) = 0 has four imaginary roots.

=

x (1 + x101 ) + 101 > 0 1+ x

fi P(x) > 0 " x > 0. 100. Both x and y must be odd or both must be even. In both the cases LHS is divisible by 8 but the RHS is not divisible by 8. 101. For x π 0, (0.2)x + (0.2)–x > 2. For x = 0, LHS = 2 = RHS. 2 4ac - b2 Ô¸ bˆ ÔÏÊ 102. If a π 0, f (x) = a ÌÁ x + ˜ + ˝. Ë 2a ¯ 4a2 ˛Ô ÓÔ

For x = -

b + i , f (x) is real. 2a

A contradiction. Thus, a = 0.

95. For x > 1, p(x) > 0. For 0 £ x < 1,

103. Let Sk = x1k + x2k + x3k.

p(x) = x9 + x4(1 – x) + x2 + 1 > 0. Thus, p(x) = 0 cannot have a positive root. As p(– x) = 0 has just one variation of sign, p(x) = 0 can have at most one negative root. Therefore, p(x) = 0 must have eight imaginary root. [It cannot have nine imaginary roots.] 96. As there is just one variation in signs of both p(x) and p(– x), the equation can have at most one positive and at most one negative root. Also, as p(– •) = •, p(0) = – c < 0 and p(•) = •, the equation has exactly one positive and one negative root. The remaining two roots are imaginary. 97. b and c are roots of

S1 = 6, S2 = S12 – 2(x2x3 + x3x1 + x1x2) = 36 – 2a S3 = 6S2 – aS1 + 3a = 216 – 21a. Now, 0 = (x1 – 3)3 + (x2 – 3)2 + (x3 – 3)3 = S3 – 9S2 + 27S1 – 243 = – 27 – 3a fi – a = 9. 104. Put

2 - x + 3 = t to obtain

t–3+

4 = 2 fi t2 – 5t + 4 = 0 t

fi t = 4, 1 As t ≥ 3,

x2 – (a3 – a)x + a2 = 0

2- x = 1

fi x = 1.

As b, c are real

105. Let p(x) = (x – a) (x – b)

(a3 – a)2 – 4a2 ≥ 0

p(10) p(11) = (10 – a) (11 – b) (11 – a) (10 – b)

fi a2 (a2 + 1) (a2 – 3) ≥ 0 fi a2 ≥ 3.

= [110 – 10 (a + b) – a + ab]

98. x1, x2, x3 are roots of

[110 – 10 (a + b) – b + ab]

x3 – x2 + x – 1 = 0 fi (x – 1) (x2 + 1) = 0 fi x = 1, i, – i

= (110 + 10a + b – a) (110 + 10a + b – b)

\ |x1| + |x2| + |x3| = 3.

= (115 – a) (115 – b) = p(115)

99. Clearly x £ 0 cannot be a solution. For x > 0, xP(x) = x

101

– 2x

100

99

2

+ 3x + ... – 100x + 101x

Adding xP(x) and P(x), we obtain (x + 1) P(x) = x101 – x100 + x99 ... + x + 101

106. Given relation implies 4(x13 + x23) + 2(x14 + x24) + 4 + 2(x1 + x2) + x1x2 = 0 But x1 + x2 = – 1, x1x2 = c gives 4c2 + 5c = 0. As c π 0, – 4c = 5.

IIT JEE eBooks: www.crackjee.xyz 2.62 Comprehensive Mathematics—JEE Advanced

107. 2(x2x3 + x3x1 + x1x2) = (x1 + x2 + x3)2 – (x12 + x22 + x32) = – 2

108. Let y =

2 x 2 - 2 x + 4 ( x - 1) + 3 = > 0 for each x ΠR 2 x 2 + 2 x + 4 ( x + 1) + 3

x13 + x23 + x33 – 3x1 x2 x3

fi (y – 1) x2 + 2(y + 1) x + 4(y – 1) = 0

= (x1 + x2 + x3) (x12 + x22 + x32 – x2x3 – x3x1 – x1x2)

As x is real

= (2) (6 + 1) = 14 fi x1 x2 x3 = – 2.

4(y + 1)2 – 16(y – 1)2 ≥ 0

fi (y + 1 + 2y – 2) (y + 1 – 2y + 2) ≥ 0

Equation whose roots are x1, x2, x3 is

fi 3(y – 1/3) (y – 3) £ 0

x – 2x – x + 2 = 0 fi x = 2, 1, – 1.

fi 1/3 £ y £ 3

Thus, – (x2 – x3) (x3 – x1) (x1 – x2) = 6.

fi Required integer is 3

3

2

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3 Progressions 3.1 SEQUENCES

Illustration 1

A sequence is a function whose domain is the set of natural numbers. If the range of a sequence is a subset of real numbers (complex numbers), then it is called a real sequence (complex sequence). If f : N Æ C, is a sequence, we usually denote it by · f (1), f (2), f (3), Ò = · f (n) Ò If tn = f (n), then the sequence is written as · t1, t2, t3,

Ò

3.2 ARITHMETIC PROGRESSION

A sequence of numbers ·tnÒ is said to be in arithmetic progression (A.P.) when the difference tn – tn–1 is a constant for all n Œ N, n > 1. This constant is called the common difference of the A.P., and it is usually denoted by the letter d. If a d the common difference of the A.P., then its nth term tn is given by tn = a + (n – 1)d tn = tm + (n – m)d n terms of such an A.P. is given by The sum Sn n n [2a + (n – 1)d] = (a + l) 2 2 where l is the last term, that is, the nth term of the A.P. Sn =

Some Facts About A.P.

term of a given A.P., then the resulting sequence is also an A.P., and it has the same common difference as that of the given A.P.

term of the A.P. If d is the common difference of the A.P., then (25 – 20)d = a25 – a20 = 15 – 25 fi d = – 2. d=6 Also a – a20 fi a = 25 + 6 = 31 5. If we have to take three terms in A.P. whose sum is known, it is convenient to take them as a – d, a, a + d. In general, if we have to take (2r + 1) terms in A.P., we take them as a – rd, a – (r – 1)d, , a – d, a, a + d, , a + rd. 6. If we have to take four terms in A.P., whose sum is known, we take them as a – 3d, a – d, a + d, a + 3d. arithmetic mean A of any two numbers a and b is given by (a + b)/2. It is to be noted that the sequence a, A, b is in A.P. If a1, a2, , an are n numbers, then the arithmetic mean A, of these numbers is given by 1 (a1 + a2 + + a n) n 8. n numbers A1, A2, , An are said to be n arithmetic means between a and b if a, A1, A2, , An, b is an A.P. Here a b is the (n + 2)th term of the A.P. If d is the common difference of this A.P., then A=

b = a + (n + 2 – 1)d = a + (n + 1)d fid=

then the resulting sequence is also an A.P. and b1, b2, b3, are two arithmetic 3. If a1, a2, a3, progressions, then a1 + b1, a2 + b2, a3 + b3, is also an A.P. 4. If a1, a2, a3, º are in A.P., then am – an = (m – n)d where d is the common difference of the A.P.

b-a n +1

Thus, A1 = a +

b-a 2(b - a) , A2 = a + , n +1 n +1 An = a +

n(b - a ) n +1

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Tip Sometimes, it is useful to use An = b – d etc. n Sum of n Arithmetic means A1 + A2 + . . . + An = (a + b) 2

3.3 GEOMETRIC PROGRESSION

A sequence is said to be in geometric progression (G.P.) terms is r times the preceding term, where r r is known as the common ratio no term of a G.P. is zero. If a r its common ratio, then its nth term, tn, is given by tn = ar n – 1 The sum Sn n terms of such a G.P. is given by Ï a(r n - 1) Ô Sn = Ì r - 1 Ô na Ó

if r π 1

.

if r = 1

If –1 < x < 1, then lim x n = 0.

Thus (1 – x)–1

=

1 . 1- x

= 1 + x + x2 + x3 +

(–1 < x < 1) (1) Differentiating both the sides w.r.t. x, we get (–1 < x < 1) (1 – x)–2 = 1 + 2x + 3x2 + 4x3 + Differentiating (1) (m – 1) times w.r.t. x, we get for – 1< x < 1 (m + 1)! 2 (m – 1)! (1 – x)–m = (m – 1)! + m!x + x 2! (m + 2)! 3 + x + 3! fi (1 – x)–m = 1 + mC1 x + m +1C2 x2 + m +2C3 x3 + (2) for – 1 < x < 1 Also, if n is any real number, then (1 – x)–n = 1 + nx + +

is also a G.P. and b1, b2, b3, are two geometric 2. If a1, a2, a3, progressions, then the sequence a1b1, a2b2, a3b3, is also a G.P. 3. The geometric mean G of two positive numbers a and b is given by a b . It is to be noted that a, G and b are in G.P. If a1, a2, , an are n positive numbers, then their geometric mean is given by G = (a1 a2 an)1/n 4. n numbers G1, G2, . . . Gn are said to be n geometric means between two positive real numbers a and b if a, G1, G2, . . ., Gn, b forms a G.P. Let r be the common ratio of this G.P., then b = (n + 2)th term of the G.P. = ar n + 2 – 1 = ar n + 1 1 ( n + 1) Ê bˆ fir= Ë ¯ a 1 ( n + 1) 2 ( n + 1) Ê bˆ Ê bˆ Thus, G1 = a Ë ¯ , G2 = a Ë ¯ a a n ( n + 1)

nƕ

1 + x + x2 +

1. If each term of a G.P. is multiplied (divided) by a

n (n + 1) 2 n (n + 1)(n + 2) 3 x + x 2! 3!

Ê bˆ . . ., Gn = a Ë ¯ a Also, note that G1 G2 . . . Gn = (ab)n / 2 5. If we have to take three terms in G.P., we take them as a/r, a, ar and four terms as a/r3, a/r, ar, ar3. is a G.P. (ai > 0 " i), then log a1, 6. If a1, a2, a3, log a2, log a3, is an A.P. In this case the converse also holds. Suppose a1, a2, a3, is a G.P. Let ai = AR i – 1, where A R is the common ratio of the G.P. Then log ai = log A + (i – 1) log R is an This shows that log a1, log a2, log a3, A and the common difference is log R. Conversely, assume that is an A.P. Suppose that log a1, log a2, log a3, d is log ai = a + (i – 1)d, where a the common difference of the A.P. Then ai = a a + (i – 1)d = a a (a d)i – 1 where a is the base of the logarithm. This shows a a and that a1, a2, a3, is a d common ratio a .

(–1 < x < 1)

If – 1 < r

3.4 ARITHMETICO-GEOMETRIC SEQUENCE

a + ar + ar2 +

=

a 1- r

Suppose a1, a2, , an, is an A.P., and b1, b2, is a G.P. Then the sequence a1 b1, a2 b2, , an bn,

, b n, is said

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Progressions

to be an arithmetico-geometric sequence. An arithmeticogeometric sequence is of the form ab, (a + d)br, (a + 2d)br 2, (a + 3d)br 3, Sum to n term of an Arithemtico-Geometric series. Let Sn = ab + (a + d)br + (a + 2d)br2 + + [a + (n – 2)d]br n–2 + [a + (n – 1)d]br n –1 (1) We multiply each term by r below the second, the second below the third and so on. rSn = abr + (a + d)br2 + + [a + (n – 3)d]br n–2 + [a + (n – 2)d]br n – 1 + [a + (n – 1)d]br n Subtracting, we get + dbr n – 2 + dbrn – 1 (1 – r)Sn = ab + dbr + dbr2 + – [a + (n – 1)d]br n = ab + fi Sn =

in H.P., that is, if 1/a, 1/H1, 1/H2, , 1/Hn, 1/b are in A.P. Let d be the common difference of this A.P. Then 1 1 + (n + 2 – 1)d = b a fid=

Thus 1 1 a-b = + , H1 a (n + 1)ab 1 H2

dbr (1 - r n - 1 ) – [a + (n – 1)d]br n 1- r

a b dbr (1 - r n - 1 ) [ a + (n - 1)d ]br n + ,rπ1 1- r 1- r (1 - r )2

lim r n = 0

1 2(a - b) , + a (n + 1) ab

=

1 1 n( a - b ) = + H n a (n + 1) ab

,

From here, we can get the values of H1, H2,

n

and

1.

lim nr n = 0

Âk n

series in (1) is

Â

k2 = 12 + 22 +

1 n(n + 1) (2n + 1) 6

+ n2 =

k =1

ab dbr + 1 - r (1 - r )2

1 n (n + 1) 2

+n=

=1+2+3+

k =1

nƕ

2. S=

, Hn.

3.6 SUMMATION OF SOME SERIES OF NATURAL NUMBERS

If –1 < r < 1, then nÆ•

1 Ê 1 1ˆ a-b ÁË - ˜¯ = (n + 1) ab n +1 b a

Ê n ˆ + n3 = Á Â k ˜ Ë k =1 ¯

n

(–1 < r < 1)

3.

Â

k3 = 13 + 23 +

k =1

+ n)2

= (1 + 2 + 3 +

3.5 HARMONIC PROGRESSION

2

2 1 2 È1 ˘ = Í n ( n + 1)˙ = n (n + 1)2 4 Î2 ˚

The sequence a1, a2, , an, where ai π 0 for each i is said to be in harmonic progression (H.P.) if the sequence is in A.P. Note that an, the nth term 1/a1, 1/a2, , 1/an, of the H.P., is given by

n

n

Â

k4 and

k =1

1 1 1 1 an = where a = and d = a + (n - 1) d a1 a2 a1

5

Â

k5 we can use the identities:

k =1

5

4

k – (k – 1) = 5k –10k3 + 10k2 – 5k + 1 and k6 – (k – 1)6 = 6k5 – 15k4 + 20k3 – 15k2 + 6k – 1

Some Facts about H.P. n

Â

1. If a and b are two non-zero numbers, then the harmonic mean of a and b is a number H such that the sequence a, H, b is a H.P. We have 2 ab 1 1 Ê 1 1ˆ or H= = Á + ˜ a+b H 2 Ë a b¯ If a1, a2, , an are n non-zero numbers, then the harmonic mean H of these numbers is given by 1 1Ê 1 1 = Á + + H n Ë a1 a2

+

1ˆ an ˜¯

2. n numbers H1, H2, , Hn are said to be harmonic means between a and b if a, H1, H2, , Hn, b are

n

{k5 – (k – 1)5} = 5

k =1

n

Â

k4 – 10

k =1



n5 + 10

{

1 2 2 n ( n + 1) 4

n

= 5Â k 4 k =1

}

Â

n

k2 – 5

k =1

– 10

{

k3 +

k =1

n

10

Â

Â

k+n

k =1

}

1 n ( n + 1) (2n + 1) 6 1 +5 n (n + 1) - n 2

{

}

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Similarly we can use 2nd identity to obtain sum of

 k5 .

(a)

k =1

256 41

128 41

(b)

64 41

(c)

(d)

32 41

Ans. (a) 3.7 SUM OF THE PRODUCTS OF TWO TERMS OF A SEQUENCE

To obtain the sum

Â

Solution:

Â

ak = =

ai aj = (a1 + a2 + º + an) – 2

(a 21

+

a 22

+º+

a 2n )

i< j

=

3.8 METHOD OF DIFFERENCE FOR SUMMATION OF SERIES

=

If possible express rth term as difference of two terms as follows: tr = f(r) – f(r ± 1).

(k 2 + 1)2 (k 2 + 2) 2 - (2k )2 (k 2 + 1)2 (k 2 - 2k + 2)(k 2 + 2k + 2) (k 2 + 1)2 [(k - 1) 2 + 1][(k + 1) 2 + 1]

For n ŒN a 1a 2 an =

Illustration 2

22

then tr = (2r – 1) (2r + 1) (2r + 3) and tr +1 = (2r + 1) (2r + 3) (2r + 5) fi (2r + 5)tr = (2r – 1)tr +1 fi (2r – 3)tr + 8tr = (2r – 1)tr +1 fi 8tr = f (r + 1) – f (r) where f (r) = (2r – 3)tr = (2r – 3) (2r – 1) (2r + 1) (2r + 3) r2 – 1) = (4r2 This gives 8 Â tr = r =1

n

Â

(22 + 1) 2

Â

8

=

2(n 2 + 1) (n + 1) 2 + 1

Next a 1a 2

an–1 =

a 1a 2

an–2 =

{f(r + 1) – f (r)} = f(n + 1) – f (1)

tr = [4(n + 1)2

n + 1)2 – 1]

n

tr = 2(n + 1)4 – 5(n + 1)2 + 3

r =1

SOLVED EXAMPLES

SINGLE CORRECT ANSWER TYPE QUESTIONS Example 1 2

2[(n - 1) 2 + 1] n2 + 1 2[(n - 2) 2 + 1] (n - 1)2 + 1

r =1

= 16(n + 1)4 – 40(n + 1)2

Â

(n 2 + 1)2

[(n - 2)2 + 1](n 2 + 1) [(n - 1) 2 + 1][(n + 1) 2 + 1]

a 1a 2 =

r =1



(32 + 1) 2



[(n - 1) 2 + 1]2

n



a18a2



(1)(22 + 1) (12 + 1)(32 + 1) (22 + 1)(42 + 1)

º

n

k 4 + 4k 2 + 4 - 4k 2

ai aj, we use the identity

i< j

2

(k 2 + 1)2

Let ak =

a a8 is equal to

(k 2 + 1)2 k4 + 4

, k ΠN, then value of

2(22 + 1) 32 + 1

22 5 Multiplying the above equations, we get a1 =

a1n a2n -1

an2-1an =

2n +1 (n + 1)2 + 1

For n = 8, we get a18a2

a 2a 8 =

29 92 + 1

=

28 256 = 41 41

Suppose a, b, c are positive real numbers Example 2 and that a, b, c are pth, qth, rth terms of a G.P., then least value of E = aq–r + br–p + cp–q is

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Ans. (a) Solution: As A.M. ≥ G.M., E = aq–r + br–p + cp–q≥ 3[ aq–rbr–pcp–q]1/3 (1) Let A R be the common ratio of the G.P., then b AR q -1 = = Rq–p a AR p -1 Ê bˆ ÁË ˜¯ a



q-r

Ê bˆ q- p q-r = (R ) = Á ˜ Ë c¯

Let a2, a2,

(k 2 - 1)ak + k 2 + 2k

k =1

ak2 + ak + 1

Â

k =1

=

(b)

1 n  ak n + 1 k =1

an > 0 be such that  ak =

(c)

n n  ak n + 1 k =1

(d)

n n +1 Â ak n + 1 k =2

n

k =1

Ans. (a) Solution:

For a, b > 0,

is

a 2 - ab + b 2 ≥

k =1

k =1

k =1

¤ ¤ n

2 2 Thus, Â ak - ak ak +1 + ak +1 ≥ k =1

k =1

n



 bk = 0

n

=

n n (2a + (n - 1)d ) + d 2 2

=

n (2a + nd ) 2

=

n n +1 ◊ (2a + nd ) n +1 2

=

n n +1 Â ak n + 1 k =1

k =1

Consider (k 2 - 1)ak + k 2 + 2k ak2 + ak + 1 (ak + bk ) 2 ak - ak + (ak + bk ) 2 + 2(ak + bk ) ak2 + ak + 1 (ak3 + ak2 + ak ) + 2bk (ak2 + ak + 1) + bk2 (ak + 1) ak2 + ak + 1

n

Â

k =1

bk2 (ak + 1) ak2 + ak + 1

bk2 (ak + 1) ak2 + ak + 1

(k 2 - 1)ak + k 2 + 2k ak2

+ ak + 1

1 n  (ak + ak +1 ) 2 k =1

n 1 = Â ak + (an +1 - a1 ) 2 k =1

 k =  (ak + bk ) =  ak +  bk

1 1 n(n + 1) = n(n + 1) + Â bk 2 2 k =1

1 ( a + b) 2

4a2 – 4ab + 4b2 ≥ a2 + b2 + 2ab 3(a2 – 2ab + b2) ≥ 0 ¤ (a – b)2 ≥ 0

n





1 n(n + 1) 2

n n +1 Â ak n + 1 k =1

n

= k + bk +

1 n(n + 1) + 0 2

(a)

n

= ak + 2bk +



(2)

n

=

+ ak + 1

2 k =1 ak

k =1

1 2 1 n(n + 1) n (b) 2 2 (c) 2n (d) n2 – 1 Ans. (b) Solution: Let k = ak + bk for k = 1, 2, n.

=

k =1

bk2 (ak + 1)

n

(a)



n

 ak2 - ak ak +1 + ak2+1 is

1 n(n + 1) , then least value of 2 n

n

Suppose a, d Œ (0, •) and an = a + Example 4 (n – 1)d ∀ n Œ N. Least value of

q- p

fi 1 = aq–rbq–p–q+rcp–q fi aq–rbr–pcp–q = 1 From (1) and (2), E ≥ 3. Example 3

n

= Â k + Â bk + Â

3.5

Example 5

Let S denote the set of all real values of

x for which (x2020 + 1) (1 + x2 + x4 + + x2018) = 2020x then the number of elements in S is (a) 0 (b) 1 Ans. (b) Solution: We can write 1 ˆ Ê 2 4 ÁË x + 2019 ˜¯ (1 + x + x + x

+ x2018) = 2020

,

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¤ 2020 = x + x3 + x5 +

+x

+

1 x

2019

+

1 x

2017

+ ... +

1 x

1ˆ Ê 3 1 ˆ 1 ˆ Ê Ê 2019 + 2019 ˜ = Á x + ˜ + Á x + 3 ˜ + ... + Á x Ë Ë ¯ x¯ Ë x ¯ x ≥ 2 + 2 + ... + 2 = 2 × 1010 1010 times

1 1 = 2, x3 + 3 = 2, x x x ¤x=1 \ S consists of just one element.

¤x+

+

1 x

2019

=2

n

Example 6

1 , then for n ≥ 2 k =1 k ( n + 1 - k )

Let an = Â

(a) an + 1 > an

(b) an + 1 < an

(c) an + 1 = an Ans. (b) Solution:

(d) an + 1 – an = 1/n

We have

2 n 1 Â n + 1 k =1 k

For n ≥ 2, 1 1 n 1 1 n +1 1 (an – an + 1) =   n + 1 k =1 k n + 2 k =1 k 2 1 ˆ n 1 1 Ê 1 = Á Ë n + 1 n + 2 ˜¯ k (n + 1)(n + 2) =1 k n

=

1 1 Â >0 (n + 1)(n + 2) k = 2 k

fi an > an + 1 If a, b, c are three unequal numbers such Example 7 that a, b, c are in A.P. and b – a, c – b, a are in G.P., then ratio a : b : c is equal (a) 1 : 2 : 3 (c) 2 : 3 : 4 Ans. (a)

Let d be the common difference. Then logy x = 1 + d fi x = y1+d logz y = 1 + 2d fi y = z1 + 2d and – 15 logx z = 1 + 3d fi z = x–(1 + 3d)/15 \ x = y1 + d = z(1 + 2d) (1 + d) = x – (1 + d) (1 + 2d) (1 + 3d)/15 fi (1 + d) (1 + 2d) (1 + 3d)= – 15 fi 6d3 + 11d2 + 6d + 16 = 0 fi (d + 2) (6d2 – d + 8) = 0 fi d = – 2 [Note that 6d2 – d + 8 = 0 has complex roots] \ x = y1 + d = y– 1, y = z1 – 4 = z–3 and x = (z–3)–1 = z3. Example 9

If log 2, log (2x – 1) and log (2x + 3) are

in A.P., then x is equal to 5 (a) (b) log25 2

(c) log32 (d)

3 . 2

Ans. (b)

1 1 ˆ Ê1 an = Â ÁË + ˜ n + 1 k =1 k n + 1 - k ¯ n

=

Ans. (a) Solution:

(b) 1 : 3 : 4 (d) 1 : 2 : 4

Solution: By the hypothesis, b – a = c – b and (c – b)2 = a(b – a) ( b π a) fi (b – a)2 = a(b – a) fi b – a = a fi b = 2a and c = 3a \ a : b : c = 1 : 2 : 3. If 1, logy x, logz y, – 15 logx z are in A.P., Example 8 then (b) x = y–2 (a) z3 = x (c) z–2 = y (d) x = y

Solution: 2, 2x – 1, 2x + 3 are in G.P. fi (2x – 1)2 = 2(2x + 3) 2x fi 2 – 4 ¥ 2x – 5 = 0 fi (2x – 5) (2x + 1) = 0 As 2x cannot be negative, we get 2x – 5 = 0 fi 2x = 5 or x = log2 5. Example 10

The sum Sn to n terms of the series 1 3 7 15 + + + + 2 4 8 16

is equal to (a) 2n – n – 1 (c) 2– n + n – 1 Ans. (c)

(b) 1– 2– n (d) 2n – 1.

Solution: Let tr denote the rth term of the series. Then tr = 1 – (1/2r). We have Ê Sn = Á 1 Ë

1ˆ Ê ˜ + Á1 2¯ Ë

Ê1 1 =n – Á + 2 + Ë2 2 = nExample 11

1ˆ Ê 1ˆ ˜¯ + ÁË1 - 3 ˜¯ + 4 2 +

1ˆ Ê + Á1 - n ˜ Ë 2 ¯

1ˆ ˜ 2n ¯

(1/2)[1 - (1/2n )] = n - 1 + 2- n . 1 - (1/2) If 12 + 22 + 32 +

+ 20032

and (1) (2003) + (2) (2002) + (3) (2001) + = (2003) (334) (x), then x equals (a) 2005 (b) 2004 (c) 2003

+ (2003) (1)

(d) 2001

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Ans. (a)

Example 14

Solution: Let S = (1) (2003) + (2) (2002) + (3) (2001) + + (2003)(1) 2 2 2 2 + 2003 and K =1 + 2 + 3 + Adding, we get S + K = (2004) [1 + 2 + 3 + + 2003] fi

= (2004) (1002) (2003)

Solution: a + c = 2b fi 3b = a + b + c = 3/2 fi b = 1/2 Thus, a + c = 1 and a2 c2 = b4 = 1/16 fi a2(1 – a)2 = 1/16 fi a(1 – a) = ± 1/4 fi a2 – a ± 1/4 = 0

x = 2005. Sum of the series

Example 12 1

1

+

1

+

fi 4a2 – 4a ± 1 = 0 fi (2a – 1)2 = 0 or (2a – 1)2 = 2

+ upto 2n terms is 4 -1 6 -1 n 1 1 4 (a) (b) (c) (d) 2n + 1 2n + 1 n +1 2n - 1 Ans. (a) 2 -1 2

2

Solution:

tk =

= fi

2

1 (2k ) - 1 2

=

fi a=

1 (2k - 1)(2k + 1)

Example 15

=

n

1

1

Let a1, a2,

, a10 be in A.P. and h1, h2,

,

h10 be in H.P. If a1 = h1 = 2 and a10 = h10 = 3, then a5 h6 is (a) 2 (b) 3 (c) 5 (d) 6 Ans. (d)

a5 = 2 + 4d =

22 9

Ê 22 ˆ Ê 27 ˆ Hence, a5 h6 = Á ˜ Á ˜ = 6. Ë 9 ¯ Ë 11 ¯



h6 =

,

2 an + 1 fi

27 11

an +1 1 = . an 2

Thus, a1, a2, a2, ratio 1/ 2 . Therefore,

(

an = 10 1/ 2

)

n-1

Area (Sn) = a2n < 1

fi [10 (1/ 2 )n – 1]2 < 1 fi 100 < 2n – 1 fi smallest possible value of n is 8. Example 16

1 1 1 1 = = +9D fi D = 3 h10 2 54 1 1 11 = + 5D = h6 h1 27

a n=

Also,

Next, let D be the common difference of 1/h1,1/h2, 1/h10. Then

\

be squares such that for

Solution: Let an denote the length of a side of Sn. We are given: Length of a side of Sn = Length of a diagonal of Sn+1 fi

Solution: Let d be the common difference of the A.P., then 3 = a10 d fi d \

Let S1, S2,

Ans. (b)

n 1Ê 1 ˆ ÁË1 ˜¯ = 2 2n + 1 2n + 1

Example 13

1 1 1 , we get a = . 2 2 2

each n ≥ 1, the length of a side of Sn equals the length of a diagonal of Sn +1. If the length of a side of S1 is 10 cm, then the smallest value of n for which Area (Sn) < 1 is

Ê ˆ Â tk = Â ÁË ˜ 2 k =1 2k - 1 2k + 1¯ k =1 1

1 1 1 or a = ± 2 2 2

As a < b < c and b =

1È 1 1 ˘ 2 ÍÎ 2k - 1 2k + 1 ˙˚

n

Suppose a, b, c are in A.P. and a2, b2, c2

are in G.P. If a < b < c and a + b + c = 3/2, then the value of a is 1 1 (b) (a) 2 2 2 3 1 1 1 1 (c) (d) 2 2 2 3 Ans. (d)

(2003) (334) (x



3.7

If a, b, c are in G.P., and log a – log 2b,

log 2b – log 3c and log 3c – log a are in A.P., then a, b, c are the lengths of the sides of a triangle which is (a) acute-angled (b) obtuse-angled (c) right-angled Ans. (b) Solution: We have b2 = ac and

(d) equilateral.

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2(log 2b – log 3c) = log a – log 2b + log 3c – log a 2a 4a and c = fi b2 = ac and 2b = 3c fi b = 3 9 Since a + b = 5a/3 > c, b + c = 10a a and c + a = 13a b, therefore a, b, c are the sides of a triangle. Also, as a A of D ABC. cos A =

b2 + c2 - a 2 29 =< 0. 2 bc 48

Hence, D ABC is an obtuse-angled triangle. If x1, x2,

Example 17

xn are n non-zero real numbers

such that + + + x2n – 1) (x22 + x23 + + x2n) £ 2 (x1 x2 + x2 x3 + + xn –1 xn) then x1, x2, , xn are in (x21

x22

(a) A.P. (c) H.P. Ans. (b)

=

= (a1b2 – a2b1)2 + (a1b3 – a3b1)2 + + (am – 1 bm – am bm – 1)2 Thus,

(x21 + x22 + + x2n – 1) (x22 + x32 + + x 2n) – (x1 x2 + x2 x3 + + xn – 1 xn)2 £ 0



(x1x3 – x2x2)2 + (x1x4 – x3x3)2 + + (xn – 2 xn – xn – 1 xn – 1)2 £ 0

As x1, x2, x1 x3 – fi

sin (a2 - a1 ) sin (a3 - a2 ) + + sin a1 sin a2 sin a2 sin a3

= x2 x4 –

x1 x2 x3 = = = x2 x3 x4

fi x 1, x 2,

x23 =

=

= xn – 2 xn –

x2n – 1

S = 3 + 75 + 243 + 507 + equal to (a) 14

sin an - 1 sin an

is 435 3 , then n is (b) 15

Ans. (b) Solution: 3(1) + 3 52 + 3 92 + 3 132 + upto n terms

S=

=0

xn xn - 1 \

(c > a) are in H.P., then log (a + c) + log (a – 2b + c) is equal to (a) 2 log (c – b) (b) 2 log (a + c) (c) 2 log (c – a) (d) log a + log b + log c. Ans. (c)

fi fi fi

upto n terms]

=

3

=

Ê nˆ 3 Á ˜ (2(1) + ( n - 1)( 4) ) Ë 2¯

, xn are in G.P.

3n (2n - 1)

435 3 = 2

2n – n – 435 = 0 n – 15) = 0

(2n

n = 15 as n ΠN.

Example 21

Sum to n terms of the series

1 1 1 + + + 1◊ 2 ◊ 3◊ 4 2 ◊ 3◊ 4 ◊ 5 3◊ 4 ◊ 5 ◊ 6

Solution: a, b, c are in H.P., b = (2ac)/(a + c). We have log (a + c) + log (a – 2b + c) È 4ac ˘ = log (a + c) + log Í (a + c) ˙ a + c˚ Î

sin (an - an - 1 )

n terms of the series

Example 20

If three positive real numbers a, b, c

Example 18

+

= (cot a1 – cot a2) + (cot a2 – cot a3) + + (cot an – 1 – cot an) = cot a1 – cot an.

, xn are real, this is possible if and only if x22

, an are in A.P. with common

difference d π 0, then the sum of the series sin d [cosec a1 cosec a2 + cosec a2 cosec a3 + + cosec an – 1 cosec an] is (b) cosec a1 – cosec an (a) sec a1 – sec an (c) cot a1 – cot an (d) tan a1 – tan an. Ans. (c) Solution: We have = a n – a n – 1. d = a2 – a1 = a3 – a2 = Thus, + sin d [cosec a1 cosec a2 + cosec a2 cosec a3 + cosec an – 1 cosec an]

(b) G.P. (d) A.G.P.

Solution: We shall make use of the identity + a2m) (b21 + b22 + + b2m) – (a21 + a22 + (a1b1 + a2b2 + + am bm)2

If a1, a2,

Example 19

is (a)

n3 3(n + 1) (n + 2) (n + 3)

(b)

n 2 + 6n 2 - 3 n 6(n + 2) (n + 3) (n + 4)

= log [(a + c)2 – 4ac] = log (c – a)2 = 2 log (c – a)

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Progressions

(c)

15 n 2 + 7 n 4n(n + 1) (n + 5)

a(n) = 1 +

n + 6 n + 11 n (d) . 18(n + 1) (n + 2) (n + 3) 3

2

1Ê 1 ˆ 1Ê1 1 ˆ 1Ê1 1 ˆ fi  t r = Á1 – ˜¯ – 3 ÁË 2 – n + 2 ˜¯ + 6 ÁË 3 – n + 3 ˜¯ n + 6 1 Ë r =1 n3 + 6 n 2 + 11 n = . 18 (n + 1) ( n + 2) ( n + 3)

. Then

Example 22 positive integers taken two at a time is 1 (n – 1)n (n + 1) (3n + 2) (a) 24 1 (n – 2) (n – 1)n2 48 1 (c) n (n + 1) (n + 2) (n + 5) 6

(b)

2

r b101 (b) s > t and a101 < b101 (c) s < t and a101 > b101 (d) s < t and a101 < b101 Ans. (d) Solution: logeb1, logeb2 …, logeb101 are in A.P. with common difference loge 2, b1, b2 …, b101 forms a G.P. with common ratio 2. We have t = b1 + b2 + … + b51 b1 (251 - 1) = b1 (251 – 1) 2 -1 s = a1 + a2 … + a51 51 51 (a1 + a51 ) = (b1 + b1 (250 )) = 2 2

Ans. (b) Solution: We have S= \ fi

=

1/3 1 = 1 - 1/3 2

(1/ 2)

log (1/2) log y = log (0.64) log ( 0.25) =



=

1 1 1 + 2 + 3+ 3 3 3

log y = (0.64) 0.25

=

=

y = 0.8. Let ax2 +

a > 0 and b be less than (a) 4c3 (c) 8c3 Ans. (a)

b ≥ c for all positive x, where x ab2 cannot (b) 4c2 (d) c3

b bˆ 1Ê 2 b bˆ Ê + ˜ ≥ Á ax 2 ◊ ◊ ˜ ÁË ax + Ë ¯ 3 2x 2x 2x 2x ¯ Ê ab 2 ˆ 1 Ê 2 bˆ + ax ≥ ÁË ˜ Á ˜ 3Ë 4 ¯ x¯

250 + 1

Ê ab 2 ˆ 3 ÁË ˜ 4 ¯

1/3

≥c

ab2 ≥ 4c3

>1

t>s = 2a51 – a1 2b51 – b1 2(b1250) – b1 b1 (251 – 1) < b1 (2100) = b101 n

If Sn = Â tr = r =1

1 n(2n2 6

n + 13), then

n

1/3

 tr equals

r =1

1 n(n + 1) 2 1 (c) n(n + 3) 2 (a)

1/3

But the least value of ax2 +



fi Also, a101 = = =

2(250 + 250 - 1)

Example 38

Solution: As A.M. ≥ G.M., we get



b1 50 (2 + 1) 2

t 2(251 - 1) = s 250 + 1

- log 2 1 log (0.64) = log (0.64) - log 4 2

Example 36

Let bi > 1 for i = 1, 2, …, 101. Suppose

b is c, therefore x

(b) (d)

1 n(n + 2) 2 1 n(n + 5) 2

Ans. (c) Solution: We have tn = Sn – Sn–1 " n ≥ 2 \

tn =

1 [2(n3 – (n – 1)3 6

n2 – (n – 1)2) + 13(n – n + 1)]

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1 [6n2 – 6n 6

=

1 (6n2 + 12n + 6) = (n + 1)2 6

n – 1) + 13]

n

n

r =1

r =1

1 n(n + 3). 2

=

If 0 < ,

Example 39 •



n =0

n =0

< /2 and



1

=

1 - sin q 2

1 cos q 2

(b) xyz – 1 = yz + zx (d) xyz + 1 = yz + zx

,y=

1 1 - cos f 2

=

1 sin 2 f

Also, cos (q + f) cos (q – f) = cos As, 0 < cos2 –1 < cos

2

2

– sin

– sin



–1
1,

b+c c+a a+b - 1, - 1, - 1 are in A.P. a b c [subtracting 1 from each term]



Now, ( x )n - 1

3.15

n



1 Ê 1ˆ (2n + 5) Á ˜ cn = (n + 1) (n + 2) (n + 3) Ë 3¯ 3 –

1 n(n + 1) (n + 2) 3

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cn =

(n + 1)(n + 2)3n "n ≥ 2 2n + 5

For n≥ 2 1 2n + 5 1 fi = cn (n + 1)(n + 2) 3n

\

=

(n + 1)3n -1

-

• 1 1 1 +Â = c1 n = 2 cn n=1 cn •

• È ˘ 7 1 1 +ÂÍ = n -1 n˙ 24 n = 2 Î (n + 1)3 (n + 2)3 ˚

7 1 21 + 8 29 + = = 24 9 72 72 Let a1 = 1,

+ a n.

Ê 1ˆ (a) an = 2 - Á ˜ Ë 2¯ n

"n ≥ 1

Ans. (a), (b), (c), (d) Solution: For n ≥ 1, 1 1 2n + 1 - 2 an+1 = an + 2 2 (n + 1) n (n + 1)2 =

Ê 1 1 1 1 ˆ an + -Á 2 ˜ 2 2 (n + 1) Ën (n + 1) 2 ¯

=

1 2 1 an + - 2 2 2 (n + 1) n 2 (n + 1)2

Let bn = an -

2 n

2

=

∀n≥1

1 , a = 0, and an > 0 ∀ n ≥ 5 36 4 [ 2n > n2 ∀ n ≥ 5]

35 35 = S4 and Sn > ∀n≥5 36 36 35 Sn is least when n = 3, 4 and min (Sn) = 36 n+ 1)th terms of an

Example 51

A.P.; G.P. and H.P. of positive terms are equal and their (n + 1)th terms are a, b and c respectively, then (a) a ≥ b ≥ c

(b) ac = b2

(c) a, b, c are in G.P.

(d) a + c = b

D be the common Solution: Let A difference and B be the (2n + 1)th term of the A.P., then B-A . B = A + 2nD fi D = 2n

n -1

(b) Sn is minimum for n = 3 (c) Sn is minimum for n = 4 35 (d) Min (Sn) = 36

fi an+1 –

n -1

Ê 1ˆ -Á ˜ 2 Ë 2¯ n 2

Ans. (a), (b), (c)

n 2 - 2n - 1 1 "n ≥ 1 an+1 = an + 2 2 n (n + 1) 2

2

n -1

S1 = 1, S2 = 1, S3 = fi

Â

and Sn = a1 + a2 + Then

Ê 1ˆ = -Á ˜ Ë 2¯

Now,

Next,

Example 50

n -1

(n + 2)3n

lim

=

Ê 1ˆ bn = -1Á ˜ Ë 2¯

fi a1 = 1, a2 = 0, a3 = -

1

1 =0–0=0 nÆ• cn



1 bn ∀ n ≥ and b1 = –1 2

Thus, an =

3 1 1 1 = n n +1 3 n + 2 3n 1

bn+1 =

1Ê 2ˆ ÁË an - 2 ˜¯ ∀ n ≥ 1 2 n

∀ n ≥ 1, then

A+ B . 2 2AB AB and c = . A+ B

Also,

a = A + (n + 1 – 1)D =

Similarly,

b=

Clearly, we have b2 = ac fi a, b, c are in G.P. Also, as

A.M. ≥ G.M. ≥ H.M, we get a ≥ b ≥ c.

Example 52

For a positive integer n, let

a(n) = 1 +

1 1 1 1 + + +º+ n . Then 2 3 4 (2 ) - 1

(a) a(n) < n (c) a(2n) > n Ans. (a), (b), (c), (d) Solution: We have

n 2 (d) a(2n) < 2n (b) a(n) >

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1ˆ Ê 1 1ˆ Ê 1 1ˆ Ê 1 a(n) = 1 + Á + ˜ + Á + º + ˜ + Á + º + ˜ + Ë 2 3¯ Ë 4 7¯ Ë 8 15 ¯

When n = 2m – 1, then S = m(2m + 1)2 – 2(2m)2

Ê 1 1 1 ˆ º + Á n -1 + n -1 +º+ n ˜ 2 (2 ) - 1¯ +1 Ë2 2 4 8 2n - 1 < 1 + + + + º + n -1 = n 2 4 8 2 Thus, a (n) < n fi Next, 1 Ê1 a(n) = 1 + + Á + 2 Ë3

n 1 1 + = = 2 2 2

(a) a = b = c 1 c are in G.P. 2 (c) a, b, c are in H.P.

(b) a, b, –

1 a, b, c are in G.P. 2

(d) –

Ans. (a), (b), (d) Solution:

From the given condition, 2b = a + c and b2 =

a(n) >

Example 53

n fi a(2n) > n. 2 Sum to n terms of the series

S = 12 + 2(2)2 + 32 + 2(42) + 52 + 2(62) + is 1 n (n + 1)2 when n is even (a) 2 1 2 n (n + 1) when n is odd (b) 2 1 2 n (n + 2) when n is odd (c) 4 1 n (n + 2)2 when n is even. (d) 4 Ans. (a), (b) Solution:

Ê a + cˆ ÁË ˜ 2 ¯

2

=

2 a2c2 a2 + c2

(a2 + c2)2 + 2ac (a2 + c2) – 8a2c2 = 0



[(a + c)2 + 2ac] (a – c)2 = 0



Ê 2 1 ˆ 4 Á b + ac˜ (a – c)2 = 0 Ë 2 ¯

1 ac 2 Note that a = c, ¤ a = b = c. 1 1 and b2 = – ac, if a, b, – c are in G.P. 2 2 fi

a – c = 0 or b2 = –

or –

1 a, b, c are in G.P. 2

Example 55

For the nth row of the triangle 1

+ (2m)2] + [22 + 42 + + (2m)2]

1 = (2m) (2m + 1) (4m + 1) + 6 4 m (m + 1) (2m + 1) 6 1 m(2m + 1) [4m + 1 + 2m + 2] = 3 1 = n(n + 1)2 2

a2 + c2



Let n = 2m, then

S = [12 + 22 + 32 +

2 a2c2

Eliminating b, we get

n times

Thus,

1 2 n (n + 1). 2

a, b, c are in A.P., and a2, b2, c2 are in H.P., if

Example 54

1ˆ Ê 1 1ˆ ˜¯ + ÁË + º + ˜¯ 4 5 8

Ê 1 2 4 2n - 1 1 ˆ + + + + º + - n˜ 1 > Á 2 4 8 2n 2 ¯ Ë 1 1 ≥ + + 2 2

= m(2m – 1)2 =

a(2n) < 2n

Ê 1 1ˆ 1 +º + n ˜ - n + º + Á n -1 +1 2 ¯ 2 Ë2

3.17

2 4

3 5

6

..................................................... ..................................................... (a) Last term =

1 n (n + 1) 2

(b) First term =

1 (n2 – n + 2) 2

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= [(10 )12 + (10 )11 + + 10 + 1] 6 ¥ [10 + 105 + + 10 + 1] fi a is not prime.

1 n (n2 + 1) 2 1 2 n (n + 1) (d) Sum = 2 (c) Sum =

Example 58

Ans. (a), (b), (c) Solution:

Last term of nth row

1 n(n + 1) 2 As terms in the nth row forms an A.P. with common difference 1, n – 1) (1) 1 = n (n + 1) – n + 1 2 1 (n2 – n + 2) = 2 +n=

=1+2+3+

1 1 2 È1 2 ˘ n Í (n - n + 2) + (n + n) ˙ 2 2 2 Î ˚ 1 = n (n2 + 1) 2

Sum

=

Example 56

1 1 1 If S = + 2 + 3 + ... upto • , then 3 3 3

(a) (0.25)log2 (S) = 4 (b) (0.008)log5 (S) = 8 (d) (0.25)log2 (S) = 8 (c) (0.008)log5 (S) = 4 Ans. (a), (b) 13 1 = Solution: We have S = 1-1 3 2

A.P., then (a) common difference of the A.P. is a rational number (b) all the terms of the A.P. must be rational. (c) all the terms of the A.P. must be integers. must be rational. Ans. (a), (b), (d) Solution:

Let d be the common difference and am = 2, 29 an = 31, then (n – m)d = 2a fi d = which is rational. n-m Also fi

Example 57

Let an = (111

Example 59

is not prime.

Suppose A1, A2 are two arithmetic means; G1, G2 are two geometric means and H1, H2 are two harmonic means between a and b; then G1 G2 A + A2 (a) = 1 H1 + H 2 H1 H 2

(c) a480 is not prime

(d) a

is not prime. is not prime.

Solution: As a , a and a480 are divisible by 3, none of them is prime. For a , we have 1 1 (10 – 1) a = ( 99 9) = 9 9



(c)

H1 + H 2 9ab = A1 + A2 (2a + b) (a + 2b)

È (107 )13 - 1 ˘ È107 - 1 ˘ – 1] = Í ˙Í ˙ 7 Î 10 - 1 ˚ Î 10 - 1 ˚

1 2 (b – a), A2 = b + (b – a) 3 3

A1 + A2 = a + b

Ê bˆ Similarly, G1 = a Á ˜ Ë a¯ fi and

91 times

1 [(10 )13 9

G1 G2 5 2 Ê a bˆ - = Á + ˜ H1 H 2 9 9 Ë b a¯

Solution: A1 = a +

Ans. (a), (b), (c), (d)

=

(b)

G1 G2 H + H2 = 1 H1 H 2 A1 + A2 Ans. (a), (b), (c)

1) , then (b) a

Let a and b be two positive real numbers.

(d)

n times

(a) a

ar – am= (r – m)d ar is rational.

must be rational.

Thus, (0.25)log2 (S) = (0.5)2 log2 (1/2) = (0.5)–2 = 4, (0.008)log5 (S) = (0.2)3log5 (1/2) 1 = - 3 log 2 = 8. 5 5

If 2 and 31 appear as two terms in an

Now,

1

3

Ê bˆ , G2 = a Á ˜ Ë a¯

2

3

G1G2 = ab 1 1 1 Ê 1 1ˆ 1 2 Ê 1 1ˆ = + Á - ˜ , H2 = + Á - ˜ Ë ¯ a 3 Ë b a¯ H1 a 3 b a 1 1 1 1 + = + H1 H 2 a b

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H1 + H 2 A + A2 a+b = = 1 H1 H 2 G1 G2 ab



G1 G2 A + A2 = 1 H1 + H 2 H1 H 2

a(2n)
0 1 1 fi r < - ( 5 + 1) or r > ( 5 - 1) 2 2 Thus, 2 sin 18° < r < 1 1 1 n -1 1 Example 61 Let a(n) = 1 - + - + (-1) n 2 3 1 1 1 + + + (a) = a(2n) n +1 n + 2 2n (b) a(2n) < 1 " n (c) a(2n) ≥ 0.5 " n (d) 0.5 < a(n) < 1 " n. As r > 1, 1 < r
1, pn is odd and qn is even. Example 64

Solution: Let 1, 5, 25 be the pth, qth, rth terms of an A.P. with common difference d, then

p p p – c1 , – c2 , – c3 ... are in H.P. 2 2 2

(b) a2, b2, c2 are in G.P.

Ans. (a), (b), (d) 1 n +3 1 1 x n +3 ˘ x tan -1 x ˙ – dx In+2 = Ú n+3 n + 3 0 1 + x2 ˚0 1

Solution:

fi(n + 3)In+2 =

n +3 p 1 x – Ú0 dx 4 1 + x2

Similarly, n +1 p 1 x - Ú0 dx 4 1 + x2 (n + 3) In+2 + (n + 1)In

(n + 1)In = \

p p 1 1 - Ú0 x n +1 dx = = 2 2 n+2 p 1 \ an = n + 3, bn = n + 1, cn = . 2 n +1 Example 65 The numbers 1, 5, 25 can be three terms (not necessarily consecutive) of (a) at least one A.P. (b) at least one G.P.

(c) a2, b2, c2 are in H.P.

(d) a + b + c =

3 2

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

Solution:

(p) a = b = c 1 (q) - a , b, c are in 2 G.P. 1 (r) a, b, - c are in 2 G.P. 1 (s) b = 2

(a) We have 2b = a + c and 2b2 = a2 + c2

fi (a + c)2 = 2a2 + 2c2 fi a2 + c2 + 2ac = 2a2 + 2c2 fi a2 + c2 – 2ac = 0 fi (a – c)2 = 0 fi a = c \ a=b=c

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3.21

(b) a2, b2, c2 are in G.P. fi (b2)2 = a2 c2 fi b2 = ac or b2 = – ac If b2 = ac, we get a, b, c are in G.P. Also, a, b, c are in A.P. \ a=b=c 2

(c) b =

2a 2 c 2 a2 + c2

fi (2b)2 =

8a 2 c 2

a2 + c2

fi a2 + c2 – 2ac = 0 or (a + c)2 + 2ac = 0 fi (a – c) = 0 or 4b = – 2ac 2

2

Example 67 Match the sequence a1, a2, a3, ... whose nth term is given in Column 1 with properties of the sequence in Column 2. Column 1 Column 2 p / 2 1 - cos 2nx 0 1 - cos 2 x

(a) an = Ú

(p) a1, a2, a3, ... are A.P. (q) a1, a2, a3, ... are G.P.

(r) a1, a2, a3, ... are in H.P.

–1 2

"n≥1

nx

p / 2 sin

dx sin x and an = bn – bn – 1 " n ≥ 1 2

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s p /2

(a) a1 = Ú0

dx =

p ,a =p 2 2

p / 2 2 cos

(2n + 2) x - cos (2n + 4) x - cos (2nx) dx 1 - cos 2 x

p / 2 2 cos

(2n + 2) x - 2 cos (2n + 2) x cos 2 x dx 1 - cos 2 x

= Ú0

p /2

= Ú0

2 cos (2n + 2)x dx

˘ 2 sin (2n + 2) x ˙ = 2n + 2 ˚0

=0

\ a1, a2, a3, ... are in A.P. (b) In + 1 + In + 3 p /4

= Ú0

tan n + 1 x sec2 x dx p /4

˘ 1 tan n + 2 x ˙ = n+2 ˚0

=

1 n+2

\ an = n + 2 fi a1, a2, ... are in A.P. (c) In – In – 1 = Ú0 p / 2 sin

= Ú0

2

(nx) - sin 2 (n - 1) x dx sin x

(2n - 1) x sin x dx sin x p /2

˘ 1 cos (2n - 1) x ˙ = 2n - 1 ˚0

=

1 = an 2n - 1

fi a1, a2, a3, ... are in H.P.

2

nx dx sin x

"n≥0 and an = In – In (d) bn = Ú0

dx

1 + In + 3

p / 4 sin

(c) In = Ú0

p

p / 2 sin

tan n x dx

In + 1

a

p /2

a fi a = c or - , b, c are in G.P. 2 1 or a, b, - c are in G.P. 2 If a = c then b = c. 3 1 fib= (d) a + c = 2b, a + b + c = 2 2

an =

s

= Ú0

fi (a2 + c2 – 2ac) (a2 + c2 + 4ac) = 0

p /4

r

Also, an + 2 + an – 2an + 1

8a 2 c 2

fi (a2 + c2)2 + 2ac (a2 + c2) – 8a2c2 = 0

(b) In = Ú0 and

q

Solution:

a2 + c2

fi (a + c)2 =

p

Ans.

p /21 -

(d) bn = Ú0

p " n. 2 \ a1, a2, a3, ... are in A.P., G.P. and H.P. Also a1, a2, a3 ... is a constant sequence. an =

(s) a1, a2, a3, ... is a constant sequence.

np cos 2nx [see (a)] dx = 2 1 - cos 2 x

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Sum of the series upto n terms

Example 68

Column 1 (a) 13 + (1.5)3 + 23 + (2.5)3 + ... 2

Column 2 (p) n (3n2 + 6n + 1)

fi  tr =

1 (q) n(n + 1) 12 ¥ (n + 2) (3n + 5)

(b) 1(2 ) + 2(3 ) + 3(42) + ...

1 (n + 1)2 32 1 ¥ (n + 2)2 – 8 1 2 (s) n (n – 1)2 4

(c) (n2 – 1)2 + 2(n2 – 22) + 3(n – 3 ) + ... (d) (2) (5) + (5) (8) + (8) (11) + ...

r2 + 3r – 2

r =1

9 3 n(n + 1) (2n + 1) + n(n + 1) – 2n 6 2

3 = n(n + 1) (2n + 1 + 1) – 2n 2 = 3n (n2 + 2n + 1) – 2n

(r)

2

1 2 2 n (n – 1). 4

(d) tr =(3r – 1) (3r n

2

2

=

= n (3n2 + 6n + 1). Example 69 Match the value of x in Column 1 with the value in Column 2. Column 1 Column 2 2 4 6 2x –28 (p) 3 log3 5 (a) 5 5 5 ... 5 = (0.04)

p

q

r

s

a

p

q

r

s

(b) x2 = ( 0.2)

b

p

q

r

s

c

p

q

r

s

(c) x = ( 0.16)

d

p

q

r

s

Ans.

1 (r + 1)3 8 n

fi  tr = r =1

1 n 1 n +1 1 Â (r + 1)3 = Â r3 – 8 r =1 8 r =1 8

(b) tr = r (r + 1)2 = r3 + 2r2 + r n 1 1 fi  tr = n2(n + 1)2 + n(n + 1) (2n + 1) 4 3 r =1 +

=

1 n(n + 1) 2

1 n(n + 1) (3n2 + 11n + 10) 12

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: (a) 52 + 4 + 6 + ... + 2x = (25)28 fi 5x (x + 1) = 556 fi x2 + x – 56 = 0 fi x (b)

2 log5 x = log = log

1 n(n + 1) (n + 2) (3n + 5). 12 2

= -

2

(c) tr = r(n – r ) n Ê n -1 ˆ n -1 3 fi  t r = n2 Á  r ˜ –  r Ë r =1 ¯ r =1 r =1

1 n2 (n – 1) n – n2 (n – 1)2 2 4 1 = n2 (n – 1) (2n – n + 1) 4 =

Ê1 1 1 ˆ ÁË + + +....˜¯ 4 8 16

Ê1 1 1 ˆ log 2.5 Á + 2 + 3 +....˜ Ë3 3 3 ¯

p

Ans.

1 1 = (n + 1)2 (n + 2)2 – 32 8

=

5

(d) 3x–1 + 3x–2 + 3x–3 1 1 Ê 2 ˆ = 2 Á 5 + 5 + 1 + + 2 + ......˜ Ë ¯ 5 5

Solution: (a) tr =

log



5

5

x>0

Ê 14 ˆ ÁË 1 - 1 2 ˜¯ log5 (0.2)

(1 2) log5 ÊÁË

log5 (1 2) log5 5

1ˆ ˜ 5¯

= log5 4

x= 2

Ê 13 ˆ log (0.16) (c) log x = log2.5 Á Ë 1 - 1 3 ˜¯ = log5/2 (1/2) log (2/5)2 = log 4

(q) 4 (r) 2

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fi x = 4. x (d) 3



(1 3) 1-1 3

=

2 (5 ) 1-1 5

1-

1 x 1 (3 ) = (53) 2 2



x = Â (sin q )

2n

n =0

xy 2 - 1



(b) Â tan2nq

(q)

y x- y

(r)

xy xy - 1

n =0 •

(c) Â sin2nq cos4nq n =0

(s)

q

r

s

a

p

q

r

s

x2 y

b

p

q

r

s

x y -1

c

p

q

r

s

a

p

q

r

s

d

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution:

Ans.

2

n =0

s

Solution: Let tr = log10 ar, then t1, t2, t3, ... are in A.P. Let its common difference be d. Now, 1 1 tm – tn = n m

1

1 x= fi cos q = , 2 x 1 - sin q y= •

1 - cos 2 q

(a) Â sin n =0

2n

fi (m – n)d =

2

1

fi sin2 q =

q cos

2n

q= =

1 . y

1 1xy

1

n =0

1 - tan q 2

=

=

1 x 1y

=

1

n =0

1 - sin q cos 4 q 2



d=

1 . mn

r 1 mn n

tr = r/mn ar = 10r/mn.

ASSERTION-REASON TYPE QUESTIONS

xy . xy - 1



(c) Â sin2n q cos4n q =

fi fi

1 - sin 2 q cos 2 q 1

( m - n) mn

Now, tr – tm = (r – m)d = 1



(b) Â tan2n q =

.

Let a1, a2, a3, ... be a geometric progression

p



(d) Â cos2nq sin4nq r

.

1 1 and log10(an) = n m m and n, with m < n, then Column 1 Column 2 (p) 101/m–1/n (a) a2m+n (b) amn (q) 10 (r) 102/n+1/m (c) am+n (d) an–m (s) 101/n+1/m

xy 2

(p)

n =0

q

xy 2 - 1

x2 y - 1

log10 (am) =

Column 2



Ans.

2n

n =0

Sum of the series Column 1

p

n =0

x2 y

such that



, y = Â (cosq )

(a) Â sin2nq cos2nq

=

x2 y xy 2

Example 71

For 0 < q < p/4, let

1



(d) Â cos2n q sin4n q =

fi x = 3 log3 5. Example 70

1

=

2

3.23

y . y-x

Example 72 Suppose four distinct positive numbers a1, a2, a3, a4 are in G.P. Let b1 = a1, b2 = b1 + a2, b3 = b2 + a3 and b4 = b3 + a4. Statement-1: The numbers b1, b2, b3, b4 are neither in A.P. nor in G.P. Statement-2: The numbers b1, b2, b3, b4 are in H.P. Ans. (c)

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Solution: Let a r be the common ratio of the G.P. As four numbers are distinct and positive a > 0, r > 0 and r π 1. b1 = a, b2 = a(1 + r), b3 = a(1 + r + r2) and b4 = a(1 + r + r2 + r3). As b2 – b1 = ar π ar2 = b3 – b2 fi b1, b2, b3, b4 are not in A.P. b2 1 + r + r 2 b3 = =1+rπ b1 1+ r b2 b1, b2, b3, b4 are not in G.P.



fi 35 is rational. This is a contradiction Hence, 3, 5 and 7 cannot be three terms of an A.P. Statement-1:

Example 74

+

(

)

b1, b2, b3, b4 cannot be in H.P. Example 73

Statement-1: There exists no A.P. whose

three terms are 3, 5 and 7 . Statement-2: If tp, tq and tr are three distinct terms of an tr - t p A.P., then is a rational number. tq - t p

n2 n(n + 1) = (2n - 1) (2n + 1) 2(2n + 1)

1 1 1 + +...+ (1) (3) (3) (5) (2n - 1) (2n + 1)

Statement-2:

1 1 –r – r2 1 1 – = – π = 2 b2 b1 a (1 + r ) a (1 + r ) 1 + r + r b3 b2

As

12 22 + + (1)(3) (3)(5)

= Ans. (c) Solution:



Let tr =

r2 (2r - 1) (2r + 1)

4tr =

4r 2 - 1 + 1 (2r - 1) (2r + 1)

Ans. (a)

=1+

Solution: Suppose 3 , 5 and 7 are the pth, qth and rth terms of an A.P. whose common difference is d, then tr – tp = (r – p) d

1 (2r - 1) (2r + 1)

=1+

1Ê 1 1 ˆ Á 2 Ë 2r - 1 2r + 1˜¯

tq – tp = (q – p) d tr - t p r-p fi = which is a rational numbers. tq - t p q- p

and

7- 3

fi ¤ ¤

5- 3

(

7 - 3) ( 5 + 3) is rational 5-3 35 – 15 + 21 – 3 is rational 2

15 – 21 =

35 – r

Squaring both sides, we get 15 + 21 – (2) (6) 35 = 35 + r2 – 2r 35 fi

r =1

1Ê 1 ˆ 1Á 2Ë 2n + 1˜¯

=n+

n 2n (n + 1) = 2n + 1 2n + 1

35 =

1- r 2 12 + 2r

n

 tr =



r =1

1 n Ê 1 1 ˆ  2 r = 1 ÁË 2r - 1 2r + 1˜¯

=n+

is a rational numbers.

¤ 35 - 15 + 21 is rational, say r. Now, 35 - 15 + 21 = r fi

n

4 Â tr = n +



1 2n + 1

n(n + 1) 2(2n + 1)

\ Statement-1 is true. But statement-2 is false since fi

1 1 n Ê 1 1 ˆ = ÂÁ 2 r = 1 Ë 2r - 1 2r + 1˜¯ r = 1 (2r - 1) (2r + 1) n

Â

=

n 1Ê 1 ˆ 1= Á ˜ 2Ë 2 n + 1¯ 2n + 1

Example 75 Statement-1: For k ≥ 0, let Sk(n) = 1k + 2k + ... + nk, 1 n(n + 1) (2n + 1) (3n2 + n + 1) then S4(n) = 30

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Ê k + 1ˆ Ê k + 1ˆ Statement-2: Á Sk(n) + Á S (n) + ... + ˜ 1 Ë ¯ Ë 2 ˜¯ k – 1

Ê 2m ˆ = Á [(2m + 1) (8m + 5) – 36 m] Ë 6 ˜¯

Ê k + 1ˆ Ê k + 1ˆ k+1 ÁË k ˜¯ S1(n) + ÁË k + 1˜¯ S0(n) = (n + 1) – 1 Ans. (d) Solution: We have (r + 1)

k+1

–r

k+1

Ê k + 1ˆ k Ê k + 1ˆ k–1 = Á r + Á r Ë 1 ˜¯ Ë 2 ˜¯ Ê k + 1ˆ Ê k + 1ˆ 0 +...+ Á r+ Á r ˜ Ë k + 1˜¯ Ë k ¯



=

1 m [16m2 – 18m + 5] 3

=

1 m (2m – 1) (8m – 5) 3

For n = 2m – 1, we get 12 + (3) (22) + 32 + (3) (42) + 52 + . . . upto n terms =

1 Ê n + 1ˆ 1 n(n + 1) (4n – 1) Á ˜ (n) (4n – 1) = 3Ë 2 ¯ 6 Statement-1: If x

Example 77

n

(n + 1)k + 1 – 1 = Â {(r + 1)k + 1 – rk + 1}

terms of

r =1

Ê k + 1ˆ S (n) + = Á Ë 1 ˜¯ k

2

Ê k + 1ˆ ÁË 2 ˜¯ Sk – 1(n) + . . .

Ê k + 1ˆ Ê k + 1ˆ + Á S1(n) + Á S (n) ˜ Ë k + 1˜¯ 0 Ë k ¯ \ Statement-2 is true. That statement-1 is false can be checked by putting n = 4 in the statement -2. In fact 1 S4(n) = n(n + 1) (2n + 1) (3n2 + 3n – 1) 30 Example 76

3.25

Statement-1: If n is odd,

12 + (3) (22) + 32 + (3) (42) + 52 + ... upto n terms =

1 n(n + 1) (4n – 1) 6

Statement-2: If n is even, 12 + (3) (22) + 32 + (3) (42) + 52 + ... upto n terms 1 = n(n + 1) (4n + 5) 6 Ans. (a) Solution: If n = 2m, then statement-2 becomes 12 + (3) (22) + 32 + (3) (42) + 52 + . . . + 3(2m)2 1 = (2m) (2m + 1) (8m + 5) 6 fi 12 + (3) (22) + 32 + (3) (42) + 52 + . . . + 3(2m – 1)2 1 = (2m) (2m + 1) (8m + 5) – 3(2m)2 6

3

1ˆ 1ˆ 1ˆ Ê Ê Ê 1 + 3 Á1 – ˜ + 5 Á1 – ˜ + 7 Á1 – ˜ + Ë Ë Ë x¯ x¯ x¯ is 2x2 – x Statement-2: If 0 < y < 1, the sum of the series 1+ y 1 + 3y + 5y2 y3 + is (1 – y ) 2 Ans. (a) Solution:

Let y3 + S = 1 + 3y + 5y2 2 yS = y + 3y + 5y3 + Subtracting (2) from (1) we get (1 – y)S = 1 + 2y + 2y2 + 2y3 + 2y = 1+ 1– y 1 2y 1+ y + = 2 1 – y (1 – y ) (1 – y ) 2



S=

Putting

y = 1– S=

(1) (2)

1 , we get x

1+1–1 x

(1 x )

2

= 2x2 – x

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 78 to 82 Sum of the following three series is given 1 1 1 1 1 - + - + - ........ = log 2 2 3 4 5 1+

p 1 1 1 1 1 - - + + - ........ = 3 5 7 9 11 2 2

(1) (2)

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p 1 1 1 1 1 - + - + ........ = 4 3 5 7 9

(3) tr =

80.

Ê 1 ˆ 1 1 + + + ......˜ 1–2 Á Ë (3) (5) (7) (9) (11) (13) ¯ is (a) p/2

(b) p/2 – 1 (c) p/4

1–

1Ê 1 1 1 1ˆ + - ˜ ÁË 2 4r - 3 4r - 2 4r - 1 4r ¯



4 2

-1

(d)

r =1





4 2

 tr =

r =1

1 1 1 + + + ..... upto • 1.2.3 5.6.7 9.10.11 is (b)

tr =

log 2



 tr =

\ (d) log 4

Sum of the series 1 1 1 1 1 1 - + + - - + .... upto • 2 3 4 5 6 7

1 1 1 log 2 – log 2 = log 2. 2 4 4

81. Multiply (1) by 82.

1 log 2 4 1 (c) log 2 3

1 n Ê 1 1ˆ - ˜ Â ÁË 4 r =1 2r - 1 2r ¯

Taking limit as n Æ •, we get

p

Sum of the series

(a)

1Ê 1 1ˆ - ˜ ÁË 4 2r - 1 2r ¯

1 n Ê 1 1 1 1ˆ + - ˜ ÂÁ 2 r =1 Ë 4r - 3 4r - 2 4r - 1 4r ¯

 tr =

\

p (1 + 2) (b) 8

p

1-

=



1 1 1 1 + - + ...... upto • 7 9 15 17

p (a) 2

Example 81

1Ê 1 2 1 ˆ + ÁË ˜ 2 4 r - 3 4 r - 2 4 r - 1¯

(d) p/4 + 1

is

Example 80

=

Sum of the series

Example 79

(c)

(4r - 3) (4r - 2) (4r - 1)

Sum of the series

Example 78

r =1

1 and subtract from (3). 2

1 1 1 1Ê 1 1ˆ - ˜ = Á Ë 2 2r - 1 2r ¯ 2r - 1 4r - 2 4r 1 log 2 2

Paragraph for Question Nos. 83 to 87 Given a sequence t1, t2 f(r) such that tr = f(r + 1) – f(r) then

is

n

(a)

p – log 2 4

(c) p – log 2 Example 82

(b)

p + log 2 4

(d) p + log 2

Sum of the series

1ˆ Ê 1 1 ˆ Ê 1 1 1ˆ Ê 1 1 ÁË1 - - ˜¯ + ÁË - - ˜¯ + ÁË - - ˜¯ +... upto • 2 4 3 6 8 5 10 12 is 1 1 log 2 (b) log 2 (a) 2 3 (c) log 4 (d) p – log 2 Ans. Solution: p Ê1 1 1 1 1 1 ˆ 1 – Á - + - + - + .....˜ = Ë 3 5 7 9 11 13 ¯ 4

 tr = f(n + 1) – f(1)

r =1

Example 83 •

Â

r =1 4r

1 2

-1

Sum of the series

is

(a) 2 Example 84

(b) 1

(c) 1/2

(d) 1/4

If u1 = 1, un+1 = 2un + 1, then un+1 equals

(a) 2n + 1

(b) 2n+1 – 1

(c) 2n – 2

(d) 2n+1 – 2 •

Example 85 (a) 1

1 is r =1 r ( r + 1) ( r + 2)

Sum of the  (b) 1/2

(c) 1/4

(d) 1/8

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n

Sum of the series  r (r + 1) (r + 2) is

Example 86

r =1

(a) 4 (

n+3

(b) 6 (n+3C4)

C 2)

(c) n2 – n

(a)

Sum of the series  r(r + 3) (r + 6) is

n

4 Â tr = ntn+1 = 4! (n+3C4).



r =1

n

Tip: In general  r(r + 1) ..... (r + k – 1) r =1

r =1

=

1 n(n + 3) (n 3

(b) n4 n2 + 20n 1 n(n + 3) (n + 5) (n (c) 4 (d) none of these Ans.

1 n(n + 1) ..... (n + k) k

tr = r (r + 3) (r + 6) = r [r2

r + 18]

= r [(r + 1) (r + 2) + 6 (r + 1) + 10] = r (r + 1) (r + 2) + 6r (r + 1) + 10r Now, use the above tip.

Solution:

83.

1 4r - 1 2

=

1Ê 1 1 ˆ Á ˜ 2 Ë 2 r - 1 2 r + 1¯

1Ê 1 ˆ = Á1 Â 2 ˜ 2 Ë 2 n + 1¯ r =1 4r - 1 n



1

Now take limit as n Æ • un+1 – un = 2(un – un–1)

84.



(r + 3)tr – r tr+1 = 0 4tr = rtr+1 – (r – 1)tr

(d) n2 + n n

Example 87

fi fi

3.27

un+1

= = = = = = = = =

85. Let fi fi fi fi fi 86. Let

rtr =

22 (un–1 – un–2) 23 (un–2 – un–3) ...................... 2n–1 (u2 – u1) = 2n un + 2n = un–1 + 2n–1 + 2n un–2 + 2n–2 + 2n–1 + 2n ...................... u1 + 2 + ........ + 2n–1 + 2n 2n+1 – 1. 1 tr = r (r + 1) (r + 2)

Paragraph for Question Nos. 88 to 92 For k, n ∈ N B(k, n) = 1.2.3 ... k + 2.3.4 ... (k + 1) + ... + n(n + 1) ... (n + k – 1), S0 (n) = n and Sk(n) = 1k + 2k + ... + nk To obtain value of B(k, n), we rewrite B(k, n) as follows:

( ) + ( ) + ( ) + ..... + ( ) ˘˚ n (n + 1).......(n + k ) = k! ( ) = . k +1

B(k, n) = k! ÈÎ

n -1

2 Â tr+1 = t1 – ntn r =1

n

2 Â tr = r =1 •

2 Â tr = r =1

1 1 2 (n + 1) (n + 2) 1 1 -0 = . 2 2

tr = r(r + 1) (r + 2)

k +2 k

n + k -1 k

n+k k +1

Example 88

S2(n) + S1(n) equals

(a) B(2, n) (c)

(d) none of these

S3(n) + 3S2(n) equals

(a) B(3, n) (c) B(3, n) – 2B(1, n)

(b) B(3, n) – 2B(2, n) (d) B(3, n) + 2B(1, n)

( ) S (n) + ( ) S ( ) S (n) + ( ) S (n) k +1 1

Example 90 k +1 k

1

1 B(2, n) 2

(b)

1 B(2, n) 6

Example 89

1 = (r + 3)tr+1 (r + 1) (r + 2)

rtr – (r + 1)tr+1 = 2tr+1

k +1 k

k k

k +1 k +1

k

k +1 2

k–1(n)

+ ... +

0

equals (a) (n + 1)k

(b) (n + 1)k – 1

(c) nk – (n – 1)k

(d) (n + 1)k – (n – 1)k

n

Example 91

 k 4 equals

k =1

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(a)

1 n(n + 1) (2n + 1) (3n2 + 2n – 1) 6

= (n + 1)5 – 1 –

(b)

1 n(n + 1) (2n + 1) (3n2 + 3n – 1) 6



(c)

1 n(n + 1) (2n + 1) (3n2 + 1) 6

(d)

n3 ( n + 1)( n + 2) 6 n

=

k =1

Alternative Solution Write k4 = k(k + 1) (k + 2) (k + 3) + ak (k + 1) (k + 2) + bk(k + 1) + ck Put k = – 1, – 2, – 3 to obtain c = – 1, b a=–6 Thus, S4(n) = B(4, n) – 6B(3, n B(2, n) – B(1, n) 1 = n(n + 1) (n + 2) (n + 3) (n + 4) 5 6 – n(n + 1) (n + 2) (n + 3) 4 7 1 + n(n + 1) (n + 2) – n(n + 1) 3 2

1 (a) n(n + 1) (n + 2) (n + 3) (4n + 15) 20 (b)

1 n(n + 1) (n + 2) (n + 3) (3n + 12) 20

(c)

1 n(n + 1) (n + 2) (n + 3) (2n + 13) 20

(d)

1 n(n + 1) (n + 2) (n + 3) (n + 14) 20

Ans. Solution: n

n

k =1

k =1

k =1

88. S2(n) + S1(n) = Â k2 + Â k = Â k(k + 1)

1 n(n + 1) [6(n + 2) (n + 3) (n + 4) 30 – 45 (n + 2) (n n + 2) – 15] 1 n(n + 1) (6n3 n2 + n – 1) = 30 =

= B(2, n). S3(n) + 3S2(n) n

n

n

k =1

k =1

k =1

= Â k2(k + 3) = Â k(k + 1) (k + 2) – 2 Â k = B(3, n) – 2B(1, n)

( ) S (n) + ( ) S k +1 1

k +1 2

k

k–1(n)

+ ... +

= Â ÈÎ r =1

( )r + ( )r k +1 1

n

k

= Â [(r + 1)

k +1 2

k -1

=

( ) S (n) k +1 k

1

( ) S (n) + ..... + ( ) r + ( ) ˘ ˚ +

n

n (6n4 + 15n3 + 10n2 – 1) 6 n (n + 1) (2n + 1) (3n2 + 3n – 1). 6

=

n

k +1 k +1

k +1 k

0

k +1 k +1

1 n(n + 1) (2n + 1) (3n2 + 3n – 1). 30

k(k + 1) (k + 2)2 = k(k + 1) (k + 2) (k + 3) – k(k + 1) (k + 2) n

fi  k(k + 1) (k + 2)2 = B(4, n) – B(3, n) k =1

= k+1

–r

10 5 n(n + 1) (2n + 1) – n(n + 1) – n 6 2

È 4 5 3 5 2 1˘ = n Ín + n + n - ˙ 2 3 6˚ Î

 k(k + 1) (k + 2)2 equals

Example 92

k+1

] = (n + 1)

k+1

–1

1 n(n + 1) (n + 2) (n + 3) (n + 4) 5

r =1

– n

 k4 = S4(n)

k =1

5S4(n) = (n + 1)5 – 1 –

10 2 n (n + 1)2 4

=

( ) S (n) – ( ) S (n) – ( ) S (n) – n 5 2

3

5 3

2

5 4

1

1 n(n + 1) (n + 2) (n + 3) 4

1 n(n + 1) (n + 2) (n + 3) (4n + 11) 20

Paragraph for Question Nos. 93 to 97 Let Vr

r terms of an arithmetic r and the common

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difference is (2r – 1). Let Tr = Vr + 1 – Vr – 2; Qr = Tr + 1 – Tr, Wr =

+

1 1 + and Tr 4 ( r + 1)

1 n(n + 1)[3n(n + 1) – (2n + 1) + 3] 12 1 n(n + 1) (3n2 + n + 2) = 12

(a) (b) (c) (d)

1 1 (2r + 1) + –2 2 2 = 3r2 + 2r – 1 = (3r – 1) (r + 1) \ Tr is always a composite number. Qr = Tr + 1 – Tr = 3{(r + 1)2 – r2} + 2{(r + 1) – r}

an odd number an even number a prime number a composite number

= 3(2r + 1) + 2 = 5 + 6r \ Q 1, Q 2, 6.

Example 95 Which one of the following is a correct statement? (a) Q1, Q2, Q3, are in A.P. with common difference 5. are in A.P. with common dif(b) Q1, Q2, Q3, ference 6. are in A.P. with common dif(c) Q1, Q2, Q3, ference 11. (d) Q1 = Q2 = Q3 = are in Example 96 W1, W2, W3, (a) A.P. (c) H.P. Example 97 X1, X2, X3, (a) A.P. (c) H.P. Ans Solution

Vr = =

\

(b) (d) are in (b) (d)

G.P. A.G.P. G.P. none of these.

r [2(r) + (r – 1) (2r – 1)] 2 1 (2r3 – r2 + r) 2 n

1 n 1 n  Vk =  k –  k 2 +  k 2 k =1 2 k =1 k =1 k =1 n

=

1 [(r + 1)2 – r2] 2 1 + [(r + 1) – r] – 2 2

= (3r2 + 3r + 1) –

Tr is always:

Example 94

Tr = Vr + 1 – Vr – 2 = [(r + 1)3 – r3] –

1 (b) (n + 1)(3n2 + n + 2) 12 1 n(2n2 – n + 1) (c) 2 1 (d) (2n3 – 2n + 3) 3

1 n(n + 1) 4

=

Xr = 3Qr for r = 1, 2, + Vn is: Example 93 The sum V1 + V2 + 1 n(n + 1) (3n2 – n + 1) (a) 12

3.29

3

1 2 1 n (n + 1)2 – n(n + 1) (2n + 1) 4 12

Wr = =

forms an A.P. with common difference

1 1 1 1 + = + Tr 4 ( r + 1) (3r – 1)( r + 1) 4 ( r + 1) 3 4 (3r – 1)

\ W 1, W 2, W 3

are in H.P.

Q1, Q2, Q3, are in A.P. X 1, X 2, X 3

are in G.P.

Paragraph for Question Nos. 98 to 100 Let A1, G1, H1 denote the arithmetic, geometric and harmonic means, respectively of two distinct positive numbers. For n ≥ 2, let An – 1 and Hn – 1 has arithmetic, geometric and harmonic means as An, Gn, Hn respectively. Example 98 Which of the following statements is a correct statement? (a) G1 > G2 > G3 > (b) G1 < G2 < G3 < (c) G1 = G2 = G3 = and G2 > G4 > G6 > (d) G1 < G3 < G5 < Example 99 correct? (a) A1 (b) A1 (c) A1 (d) A1

Which of the following statements is > < >
< A3 < > A5 > < A5
A4 > A6

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Example 100 correct? (a) H1 > (b) H1 < (c) H1 > (d) H1 < Ans

Which of the following statements is H2 H2 H3 H3

> < >
< >
A2 > H2 > H1



A1 > A2 > A3 > H3 > H2 > H1

Ê x3 - 1ˆ Ê x11 - 1ˆ Ê x 7 - 1ˆ Á x -1 ˜ Á x -1 ˜ = Á x -1 ˜ Ë ¯Ë ¯ Ë ¯

And so on. Thus,

A1 > A2 > A3 >

and

H1 < H2 < H3
H4 > H6 >

Let two distinct positive numbers be a and b. 1 2ab A1 = (a + b), G1 = ab , H1 = 2 a+b

Hn =

4 -4 = 475 25 ¥ 19

We have

Solution:

An =

d=

(1)

2

fi x14 – x3 – x11 + 1 = x14 – 2x + 1 fi x11 – 2x + x3 = 0 fi x8 – 2x4 + 1 = 0 [ fi (x4 – 1)2 = 0 fi x4 = 1 fi x = –1 Thus, x + 5 = 4 Example 104 x101 Ans. 3 Solution: x2n = – =– \ x101 = – Example 105

x π 0]

If xn = 12 – 22 + 32 – ... + (–1)n–1 n2

(1 + 2 + ...... + 2n) n(2n + 1) 50(101) + 1012 = 5151. If xn = 12 + (2)(22) + 32 + (2)(42) + ... = n(n + 1)2/2 if n is even, then x51 is (13)(512 )

Ans. 2 Solution x51 = x50 + 512 1 = (50) (51)2 + 512 2 = (26)(512).

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The sum of

Example 106

Example 109 Let be an arithmetic progression with common difference d (d π 0) and be a geometric progression with common ratio q, where q is a rational a 2 + a22 + a32 number and 0 < q < 1. If a1 = d, b1 = d 2, and 1 b1 + b2 + b3 is a positive integer, then 6q is

1 1 Ê 1 ˆ + + + ..... upto •˜ 12 Á Ë 1.3.5 3.5.7 5.7.9 ¯ Ans. 1. Solution:

1 (2r - 1) (2r + 1) (2r + 3)

Let tr =

1 1 1 1 + = 8 2r - 1 4 (2r + 1) 8(2r + 3) 1Ê 1 1 ˆ 1Ê 1 1 ˆ Á ˜- Á ˜ 8 Ë 2 r - 1 2 r + 1¯ 8 Ë 2 r + 1 2 r + 3 ¯

= n

 tr

r =1

1Ê 1 ˆ 1Ê1 1 ˆ = Á1 ˜¯ - ÁË ˜ Ë 8 2n + 1 8 3 2n + 3 ¯

n

nƕ r =1

Example 107

Ans. 3 Solution: Note that ak = kd and bk = d2 qk – 1 " k ∈ N. a 2 + a22 + a32 d 2 (12 + 22 + 32 ) 14 Now, 1 = = . b1 + b2 + b3 d 2 (1 + q + q 2 ) 1 + q + q2 As

14

is a positive integer, say, m we get

1 + q + q2

1 1 1 = . 8 24 12

lim  tr =

1 + q + q2 = q= -

If S sum to 50 terms of

Ans. 5

r=1 25

= Â

r2 – 2)

S Let

an =

2 -1 3 -1 4 -1 n -1 . 3 . 3 ....... 3 , 3 2 +1 3 +1 4 +1 n +1 3

3

3

then lim 3an is nƕ

Ans. 2 Solution:

an =

(2 - 1) (3 - 1)........(n - 1) × (2 + 1) (3 + 1)........(n + 1) (n 2 + n + 1)

(22 - 2 + 1) (32 - 3 + 1)

(n 2 - n + 1)

Since (k + 1)2 – (k + 1) + 1 = k2 + k + 1, we get



lim an =

nƕ

2 . 3

14 3 3 - < m 4 2

0 < q < 1,



1 14 3 9 14 - < fi1< < 0 and a, b, c be in A.P. If a2, b2, c2 are in G.P., then (a) a = b = c (b) a2 + b2 = c2 2 2 2 (c) a + c = 3b (d) none of these

100 digits

1313.....13 100 digits

(c)

n

Sn = Â t r =

dx , then a1, a2, a3, ... are in

2323.....23

(b)

33.....3

1 3 5 7 + + + + ... is 3 (3)(7) (3)(7)(11) (3)(7)(11)(15)

100 digits

(a) 1/2

(d) none of these

100 digits

Sn (a) = 1 + 2(1 – a) + 3(1 – a) (1 – 2a) + 4(1 –a) (1 – 2a) (1 – 3a) + upto n terms, then Ê 1ˆ S100 Á ˜ equals Ë 99 ¯

(b) 1/3 (c) 1/6 (d) 1/4 4 44 444 + + + upto •. Then S is equal to 36. Let S = 19 192 193

a, b, c, ... are in G.P., and a1/x = b1/y = c1/z = ..., then x, y, z, ... are in (a) H.P. (b) G.P. (c) A.P. (d) none of these

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38. If ax = by = cz = du and a, b, c, d are in G.P., then x, y, z, u are in (a) A.P. (b) G.P. (c) H.P. (d) none of these F0 = 1, F1 = 1 and Fn + 1 = Fn + Fn – 1 " n ≥ 1. • Fn is Sum of the series  n =1 Fn -1 Fn +1 (a) 1 (c) 1/2

(b) 2 (d) 2/3

40. If the sum to n terms of an A.P. is 3n2 + 5n, while Tm = 164, then value of m is 41. If G1 and G2 are two geometric means and A is the arithmetic mean inserted between two positive numG2 G 2 bers a and b then the value of 1 + 2 is G2 G1 (a) A

(b) 2A

(c) A/2

(d) 3A/2

42. If A1, A2 be two arithmetic means and G1, G2 be two geometric means between two positive numbers A + A2 is equal to a and b, then 1 G1 G2 (a)

a b + b a

a b + b a 43. Suppose for each n ΠN. (c)

(b)

1 1 + a b

(d)

ab a+b

(12 – a1) + (22 – a2) + ... + (n2 – an) = then an equals (a) n (b) n – 1

1 n (n2 – 1) 3

(c) n + 1 (d) 2n

44. Let A1, A2 be two arithmetic means, G1, G2 be two geometric means, and H1, H2 be two harmonic means between two positive numbers a and b. The G G H + H2 is value of 1 2 . 1 H1 H 2 A1 + A2 (a) 1/2

(b) 1

(c) 3/2

(d) 2

45. Sum of the series S = (n) (n) + (n – 1) (n + 1) + (n – 2) (n + 2) + ... + 1(2n + 1) is 1 (b) n(n + 1) (n + 2) (a) n3 6 1 3 n – n2 (d) none of these (c) 3

3.35

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 46. Let a, d Œ (0, •) and ar = a + (r – 1) d ∀ r Œ N. If Sk k 1 n k =  , then  cannot exceed i =1 ai k =1 S k (a)

n (3a1 + an) 4

(b) n(3a1 + an)

(c)

1 n  ak 2 k =1

(d) Â ak

n

k =1

3 1 5 1+ 2 7 1+ 2 + 3 + . + ..... . + . 2 12 2 12 + 22 2 12 + 22 + 32 upto n terms, then Sn cannot exceed (a) n (b) 2n (c) 3n (d) 4n 48. Let x and y be two positive real numbers. Let P be the rth mean when n arithmetic means are inserted between x and y and Q be the rth harmonic mean between x and y when n harmonic means are inserted P y is independent of between x and y, then + x Q Sn =

(a) n (b) r (c) both n, r (d) none of these N = 2 + 22 + 222 + ... upto 30000 terms. (a) last four digits of N are 8580 (b) last three digits of N are 580 (c) last four digits of N (d) last digit of N is 0. 2

2

2

xˆ Ê xˆ Ê xˆ Ê 50. Let (1 + x)2 ÁË1 + ˜¯ Á1 + 2 ˜ Á1 + 3 ˜ 2 Ë 2 ¯ Ë 2 ¯ = a0 + a1 x + a2 x2 + upto •, then (b) a2 = 20/3 (a) a1 = 4 (d) a3 (c) a2 = 16/3 51. If a, b, c, d are in G.P., then value of

upto •

(a – c)2 + (c – b)2 + (b – d)2 – (d – a)2 is independent of (a) a (b) b

(c) c

(d) d

p, 1, 1/p, 1/p2 then value of p is 53. If a, b, c are in A.P., and a2, b2, c2 are in H.P., then (a) a = b = c (b) 2a2 = b2 = c2 (c) a, b, c are in G.P. (d) – a, 2b, 2c are in G.P.

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54. Let Sn denote the sum to n terms of the series 2

2

2

2

2

1 + 2(2 ) + 3 + 2(4 ) + 5 + ... then Sn equals (a)

1 n (n + 1)2 if n is even 2

3 (d) 2 : 3 62. If 1, log3 (3 – 2), 2 log (3x – 8/3) are in A.P., the value of x can be

1 2 (b) n (n + 1) if n is odd 2

x

1 n (n + 1)2 if n is even 2 1 2 n (n + 1) if n is odd (d) 6 are in A.P. with common difference 55. If a1, a2, a3, n Ê ˆ d –1 d, then  tan Á ˜ equals r =1 Ë 1 + ar ar + 1 ¯ (c)

Ê an + 1 – a1 ˆ –1 (a) tan Á ˜ Ë 1 + an + 1a1 ¯ Ê ˆ nd –1 (b) tan Á ˜ Ë 1 + an + 1a1 ¯ (c) tan – 1an + 1 – tan – 1(a1) (d) p/2 xr in (1 + x) (1 + 2x) 56. Let ar n–1 (1 + 2 x), then (a) a1 = 2n – 1 1 (b) a2 = (2n – 2) (2n – 1) 3 (d) an = 2n (c) an = 2n(n – 1)/2 n > 1, the values of the positive integer m for which nm + 1 divides a = 1 + n + n2 + + n63 are (a) 8 (b) 16 (c) 32 (d) 64 58. If p, q, r are positive and are in A.P., the roots of the quadratic equation px2 + qx + r = 0 are all real for (a)

r -7 ≥4 3 p

(c)

p -7 ≥4 3 r 3

(b)

q -4 ≥2 3 p

(d)

p 3 -1 ≥ q 2

2

x + bx + cx + d = 0 are bc d=0 (a) in A.P. then 2b3 3 3 (b) in G.P. then b d = c d3 bcd2 – 4c3d (d) equal then c3 = b3 + 3bc 1 1+ 2 1+ 2 +º+ n +º+ 3 60. If Sn = 3 + 3 then 3 1 1 +2 1 + 23 + º + n3 for, n > 1 Sn cannot exceed (a) 1 (b) 3/2 (c) 2

61. If the arithmetic mean of two positive numbers a and b (a > b) is twice their geometric mean, then a : b is 3 :1 (a) 2 + 3 : 2 – 3

(d) 5/2

Ê 4ˆ (a) log Á ˜ Ë 3¯

(b)

(c) 1

(d) log3 4

log 4 log 3

n

63. If  r(r + 1) (2r + 3) r =1

= an4 + bn3 + cn2 + dn + e, then (a) a + c = b + d (b) e = 0 (c) a, b – 2/3, c – 1 are in A.P. (d) c/a is an integers. 64. Let the H.M. and the G.M. between two positive numbers a and b be in the ratio 4 : 5, then (a) a : b = 4 : 1 (b) b : a = 4 : 1 (c) A.M. : G.M. = 5 : 4 (d) (G.M.)2 = (A.M.) (H.M.) 65. Let Sn = (1) (5) + (2) (52) + (3) (53) + ... + (n) (5n) 1 [(4n – 1)5a + b], 16 then (a) a = n + 1 (c) b = 5

=

(b) a = n (d) b = 25

MATRIX-MATCH TYPE QUESTIONS 1 1 1 + 2 + 3 + ...... , then 3 3 3 Column 1 Column 2

66. Let S =

log (S ) (a) (0.64) 0.25

(b) 5

log 2 ( S )

(c) [log8 (1/S)]

(q) 0.8 3

(d) [log0.125 (S)]

(r) 0.2

3

y = exp {(sin2 x + sin4 x + sin6 x + ...) loge 2} satisfy the equation x2 x + 16 = 0 and 0 < x < p/2 Column 1 Column 2 2 sin 2 x (p) 1 (a) 1 + cos 2 x (b)

2sin x sin x + cos x

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(c) a logp c, b logp b, c logp a are in (d) qloga pa, qlogb pb, qlogc pc are in



n (c) Â (cot x)

(r) 2/3

n =1 •

2n (d) Â n (cot x)

(s) 4/3

n =1

2n

2n

r=0

r=0

68. Suppose  ar (x – 2)r =  br (x – 3)r, then value of bn when for each r ≥ n Column 1 Column 2 (p) n + 1 (a) ar = 1 1

(b) ar =

r

(q)

Cn r

(c) ar =

r

r

(d) ar =

r

(r)

Cn

2n+1

2

Cn

Cn

1 1

1

1 1

2

(q)

1

3

3

1 n(n2 + 1) 2

1

3 7 13

5 9

(r) 2n–1

11

15

17

19

2

1 2

12

1

(d)

12 12

22 32

(s)

12 32

r ≥ 1, and x π 1. Let tr = 1 + 2x + 3x2 +

(

n 1 + xn + 1

2n–2

Cn–2

12

a, b, c, p > 1 and q > 0. Suppose G.P. Column 1 (a) logp a, logp b, logp c are in (b) loga p, logb p, logc p are in

+

(1 – x ) 2

a, b, c are in Column 2 (p) G.P. (q) A.G.P.

+ tn is

) – 2 x (1 – x ) n

(1 – x )3

Statement-2: For r ≥ 1, and x π 1 1 + x + x2 + + xr – 1 = and 1 + 2x + 3x2 +

+ rxr – 1 =

1 – xr

(1 – x ) 2



1 – xr , 1– x rx r 1– x

a, b, c be three positive real numbers which are in H.P. a+b c+b + ≥4. Statement-1: 2 a – b 2c – b Statement-2: If x > 0, then x +

1 (c)

ASSERTION-REASON TYPE QUESTIONS rx Statement-1: Sum of t1 + t2 +

n > 1. Sum of all the terms in the nth row of the triangle Column 1 Column 2 1 2 3 (a) (p) n3 4 5 6 7 8 9 10

(b)

(r) positive (s) non-negative

r–1

3 n(n + 1) 2

(s)

(r) H.P. (s) A.P.

a, b, g be distinct real numbers which are in A.P. If a, b, g are the roots of x3 + bx + c = 0, then Column 1 Column 2 (a) b (p) 0 (b) c (q) negative (c) b2 – - c (d) b2 + c2

1 n(n + 1) (14n + 1) 6

3.37

1 ≥4. x

Statement-1: If a, b, c are three positive real numbers such that a + c π b and 1 1 1 1 + + + =0 a a–b c c–b then a, b, c are in H.P. Statement-2: If a, b, c are distinct positive real numbers such that a(b – c)x2 + b(c – a)xy + c(a – b)y2 is a perfect square, then a, b, c are in H.P. a, b, c be three distinct non-zero real numbers. Statement-1: If a, b, c are in A.P. and b, c, a are in G.P., then c, a, b are in H.P. Statement-2: If a, b, c are in A.P. are b, c, a are in G.P., then a : b : c = 4 : 2 : – 1

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COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 76 to 80 A man invests Rs Pi in a bank at the end of ith year for n years. The rate of interest is Rs r per rupee per year and the interest is compounded annually. Pi = P " i, the amount the man will receive at the end of n years is P (a) [(1 + r)n – 1] r (b)

P [(1 + r)n + 1] r

(c)

P [(1 + r)n + 1 – 1 – r] r

r = 1/2, Pi = P "i, the smallest value of n in which the man will get at least 10 P, is (a) 5 (b) 6 (c) 18 (d) 20 P , then liability of the bank 80. If r = 1 and Pk = (n - k )! cannot exceed (a) Pe (b) Pe2 1 2 Pe 2

(d)

(c)

1 ( 5 + 3) 2

(d)

1 ( 3 + 5) 3

2 , then a2 equals

83. If r = (a)

1 (2 ± 3) 7

(b)

1 (11 ± 7) 4

(c)

1 (11 ± 6 2) 7

(d)

1 (11 ± 6 2) 5

84. If a = 1/2, S term is

(d) P(rn – 1) Pi = iP " i, the amount received by man at the end of n years is P (a) 2 [(1 + r)n+2 – (1 + r)2] r P n P(1 + r) (b) 2 [(1 + r)n+2 – (1 + r)2] – r r P n (1 + r)] (c) 2 [(1 + r)n+2 – r r (d) none of these Pi = P and r = 1, the smallest value of n in which the man receives at least 2000 P, is

(c)

82. If a2 = 2, then value of r equals 1 1 (5 - 3) (3 + 5) (b) (a) 2 2

Pe2 n!

Paragraph for Question Nos. 81 to 85

85. If we drop the condition that the G.P. is strictly increasing and take a2 = 3, then common ratio is given by (a) ± 2

(b) ±1

(c) 0

(d) ± 3

Paragraph for Question Nos. 86 to 90 Let ABCD is a unit square and 0 < a < 1. Each side of the square is divided in the ratio a : 1 – a as shown in Fig. 3.2. These points are connected to obtain another square. The sides of new square are divided in the ratio a : 1 – a and points are joined to obtain another square. The process is an denote the length of side and An the area of the nth square.

Fig. 3.2

86. If a = 1/3, the least value of n for which An < 1/10 is a= (a) 2

1 • , Â A n equals 2 n =1 (b) 4

(c) 8

(d) 8/3 8 88. The value of a for which  A n = is 3 n =1 •

The sum of three terms of a strictly increasing G.P. is aS and sum of the squares of these terms is S2. 81. a2 lies (a) (1/3, 2) (b) (1, 2) (c) (1/3, 3) (d) (1/3, 1) » (1, 3)

(a) 1/3, 2/3

(b) 1/4, 3/4

(c) 1/5, 4/5

(d) 1/2

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a for which side of nth square equals the diagonal of (n + 1)th square is (a) 1/3

(b) 1/4

(d) 1

(c) 1/2

LEVEL 2

2

SINGLE CORRECT ANSWER TYPE QUESTIONS

a = 1/4 and Pn denotes the perimeter of the nth •

square then  Pn equals n =1

(a) 8/3 (c) 16/3

(b) 32/3 (d) none of these

INTEGER-ANSWER TYPE QUESTIONS a, x, y, z and b are in A.P. when x + y + z = 15, and a, a, b, g, b are in H.P. when 1/a + 1/b + 1/g = 5/3. If a > b, then a is equal to a, b, c between 2 and 18 be such that (i) their sum is 25, (ii) the number 2, a, b are in A.P. and (iii) the numbers b, c, 18 are in G.P. then c n

an = Â 1 + k =1

1 k

2

+

1 (k + 1) 2

then 6a5 – 33 is

y, logz x, logy z are in G.P., xyz = 64 and x , y , z3 are in A.P., then x + y is 3

3

x

x1, x2, ... ∈ (0, p) denote the of values of x satisfying the equation 27 (1+ |cos x |+ cos

2

x + |cos x |3 +........upto • )

then the value of tk =

k

,

1 (x1 + x2 + ...) p n

and an = Â tk , then

k +4 a4 – 3/26)/325 is 4

3

k =1

a x25 in (1 + x + 2x2 + 25 2 + 25x ) then a/325 is equal to ABC right angled at C are in A.P., then 5 (sin A + sin B) is its second, tenth and thirty-fourth terms form a G.P., then the common difference is

1. Sum of the series 4 5 4 5 S= - 2 + 3 - 4+ 7 7 7 7

upto • is

(a) 23/24 (b) 23/48 (c) 4/43 (d) 3/11 2. Sum to n terms of the series 32 52 7 2 + + + is 2 22 23 1 (a) 34 - n - 1 (4n2 + 12n 2 1 (b) 34 - n - 1 (2n2 + 3n + 28) 2 1 (c) 34 - n (4n2 + 3n 2 (d) none of these 3. Sum to n terms of the series 13 – (1.5)3 + 23 – (2.5)3 +... is 1 1 (n + 1)2 (n + 2)2 – (a) 6 4 1 1 (b) (n + 1)2 (n + 2)2 – 32 8 2 2 (c) (n + 1) (n + 2) – 35 (d) none of these 4. Sum to n terms of the series 12 +

(16) (32) +... (a) (b) (c) (d)

2

n

(n – n + 4)2 – 4 (n2 + n + 2)2n – 4 (n2 – 2n + 5)2n – 4 none of these (k 2 ) (4k ) , then S equals k = 1 ( k + 1) ( k + 2) n

5. If S = Â time it rebounds, it rises to 2/3 of the height it has fallen through. If s is the total distance travelled by the ball before it comes to rest then s/500 is

3.39

(a)

˘ 1 È 2n + 1 Ê n - 1 ˆ + 2˙ Í2 Á ˜ 3Î Ë n + 2¯ ˚

(b)

2 È 2n + 1 Ê n - 1 ˆ ˘ + 1˙ Í2 3Î ËÁ n + 2 ¯˜ ˚

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(c)

13. If each ai > 0 and a1, a2, ... an are in harmonic progression, then expression

2 È 2n Ê n - 1 ˆ ˘ + 1˙ Í2 Á 3Î Ë n + 2 ˜¯ ˚

a1 a2

1 È 2 n Ê n - 1ˆ ˘ + 1˙ (d) Í 2 Á 3Î Ë n + 1˜¯ ˚

a1 + a2

6. Let a, b, c be three non-zero real numbers such that |a|, |b| ≥ 1 2 . If the circles

2

7.

8.

9.

10.

11.

2

2

touch each other, then a , b , c are in (a) A.P. (b) G.P. (c) H.P. (d) none of these If a1 = 50 and a1 + a2 + + an = n2 an " n ≥ 1, then a100 equals (a) 1/100 (b) 1/101 (c) 1/50 (d) 1/51 Let a, b, c, d and p be 5 real numbers such that their arithmetic mean is 2. If arithmetic mean of a2, b2, c2, d2 and p2 is 4, then p is equal to (a) 2 (b) 3 (c) 4 (d) 5 First three terms of an arithmetic-geometric progression are 1, 3, 8. If the common difference of the A.P. is positive, then sum of the next three terms of the series is (a) 120 (b) 140 (c) 160 (d) 180 Suppose 27, 8, 12 are pth, qth and rth term of a G.P. and let p(x) = px2 + 2qx – 2r, then p(x) = 0 has at least one root in (a) (– 1, 0) (b) (0, 1) (c) (1, •) (d) (– •, 0) The sum to n terms of the series 1 7 37 175 + + + + 4 16 64 256 is (a) n – 3–n + 1

(b) n – 4–n + 3–n

Ê Ê 3ˆ n ˆ (c) n – 3 Á1 - Á ˜ ˜ Ë Ë 4¯ ¯

ÊÊ 4ˆ n ˆ (d) n – Á Á ˜ - 1˜ ËË 3¯ ¯

12. If n 1ˆ 1ˆ Ê Ê 1 + 2 Á1 - 2 ˜ + 3 Á1 - 2 ˜ Ë ¯ Ë n n ¯

2

3

1ˆ Ê + 4 Á1 - 2 ˜ + ... Ë n ¯

is (a) n4

(b) n2

(c) n3

(d) n –

a2 a3 a2 + a3

+

+

(a)

(c)

(1 - n) a1 a2 a1 + an (1 - n) a1 an a1 + an

(b)

an - 1 + an

(d)

(n - 1) a1 an a1 + an n a1 + an

14. If a, b, c ∈ R and 225a2 + 25b2 + 9c2 – 75ab – 15bc – 45ca = 0 then a, b, c are in (a) A.P. (b) G.P. (c) H.P. (d) none of these 15. Sum to (n + 1) terms of the series x x ( x + a1 ) x ( x + a1 ) ( x + a2 ) + + +... is 1 + a1 a1 a2 a1 a2 a3 (a)

x ( x + a1 ) ( x + an ) a1 a2 an

(b)

( x + a1 ) ( x + an ) a1 a2 an

(c)

( x + a1 ) ( x + an ) -x a1 a2 an

(d) none of these 16. If a > b > c > 0 and 2 log (a – c), log (a2 – c2), log (a2 + 2b2 + c2) are in A.P., then (a) a, b, c are in A.P. (b) a2, b2, c2 are in A.P. (c) a, b, c are in G.P. (d) a2, b2, c2 are in H.P. 17. Sum of the series 2 3 4 2 3 4 + + + + + + 5 52 53 54 55 56 is 69 71 (a) (b) 124 124 35 75 (d) (c) 124 124 •

n

an - 1 an

equals

2|z|2 – 4a Re(z) + c2 = 0 2|z|2 – 4b Im(z) + c2 = 0

and

+

n3 - 1 equals 3 n=2 n + 1 (a) 1/3 (c) 1/10

18. ’

(b) 2/3 (d) 2/7

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d(n) denote the number of divisors of n, then d d(10) are in (a) A.P. (b) G.P. (c) H.P. (d) A.G.P. 1 1 1 20. If an = 1 + 2 + 2 + + 2 , and [x] denotes the 2 3 n greatest integer £ x, then [a100] equals d

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 21. If three positive real numbers a, b, c are the pth, qth and rth term of a G.P., then (q – r) log a + (r – p) log b + (p – q) log c (a) is independent of p, q, r (b) is independent of a, b, c (c) equals a constant (d) depends on the base of log 22. Let An stand for the expressive n

–2 (1 + 3–1) (1 + 3–2) (1 + 3–4) ... (1 + 3 ) nÆ•

b 2 - 4 ac q 2 - 4 pr

=

c2 r2

(b)

a -b 1 Ê b qˆ = Á - ˜ ab 4Ë c r¯

(c)

r -d 1 Ê b qˆ = Á - ˜ gd 4Ë c r¯

(d)

b 2 - 4 ac q 2 - 4 pr

=

a2 p2

28. Sixteen players P1, P2, ..., P16 participate in a tennis

16

16

i =1

i =1



(b) Â (wi

16

2

i =1

1 are removed from the series  . Sum n =1 n •

1 of the remaining terms of the series  cannot n =1 n exceed (c) 100 (d) none of these 24. Let s(n) denote the sum of all the divisors of n, then s(2 32) is divisible by (a) 3 (b) 11 (c) 13 (d) 31 25. Let f(n) = number of natural numbers £ n, which are relatively prime to n, then f(125) is divisible by (a) 2 (b) 5 (c) 50 (d) 100 Ê 2 x 4 x6 ˆ + + ˜ tan–1 Á x 2 4 Ë ¯

(a)

(a) Â wi = Â li = 120



2 and

then x equals (a) –1 (b) 1 (c) 1/2 (d) –1/2 2 a, b be the roots of ax + bx + c = 0 and g, d be the roots of px2 + qx + r = 0. If a, b, g, d are in H.P., then

(b) L < 2 (d) L > 1

1 term 1/n from the series  n =1 n

x

Ê 4 x8 x12 ˆ p + + ˜ = + cot–1 Á x , 2 4 2 Ë ¯

a game with each of the remaining players. Let (wi, li) denote the (number of wins, number of losses) of the ith player, then

If L = lim An, then (a) 2L is prime (c) L is prime 23. If the decimal form of n

3.41

16

= Â (li

2

i =1

16

16

i =1

i =1

(c) Â wi2 = Â li2 16

16

i =1

i =1

(d) Â wi2 = 240 + Â li2 ax3 + bx2 – x + 1 = 0 are real and are in H.P., then (a) b ∈ (– •,1/3] (b) a∈ •) a + 4b = 2 (d) a + b = 5 30. Let a, b, c > 0 be such that a, b, c are in G.P. and log a – log b – log 2, log b – log c – log (1.5), log 3 + log c – log a are in A.P., then (a) a, b, c are the sides of a triangle, say ABC. (b) DABC is obtuse angled triangle (c) cos B (d) cos C = 101/108

MATRIX-MATCH TYPE QUESTIONS 31. Let an = integer £ x.

n ≥ 1 and [x] denotes the greatest

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Column 1

Column 2

È1 ˘ È1 ˘ (a) Í + a1 ˙ + Í + a3 ˙ + 50 50 Î ˚ Î ˚

È1 ˘ Í 50 + a101 ˙ (p) 2 Î ˚

(b) [a1 + a100] + [a2 + a ] + ... + [a100 + a1]

(q) 1

forms an Statement-1: If tr = ar/2r, then t2, t3, t4 A.P. Statement-2: If Tr = r/ar, then T1, T2, T3, forms a G.P.

COMPREHESION-TYPE QUESTIONS

(c) [a1 + a2 + ... + a11]

(r) 3

Paragraph for Question Nos. 31 to 40

(d) [a1 + a2 + ... + a11 – 3]

(s) 0

Suppose x1, x2 be the roots of ax2 + bx + c = 0 and x3, x4 be the roots of px2 + qx + r = 0.

32. Let 0 < a < 1. Column 1

Column 2

89

(a) Â loga (sin k°)

(p) positive

89

(q) zero

k =1 89

(c) Â loga (sec k°)

(r) negative

k =1

(a)

1Êb qˆ 2 ÁË a p ˜¯

(b)

1Êb qˆ 3 ÁË a p ˜¯

(c)

1Ê c qˆ 3 ÁË a p ˜¯

(d)

1⎛b q⎞ − 4 ⎜⎝ a p ⎟⎠

89

(d) Â loga (cosec k°)

(s) non-negative

k =1

33. Let a, b, c be in H.P. Column 1

Column 2

a b c , , (a) (p) A.P. b + c - 2 a c + a - 2b a + b - 2 c b+c c+a a+b , , a b c b b b (c) a - , , c 2 2 2 b 1 b , ,1 (d) 1 2a 2 2c

(b)

(q) G.P. (r) H.P. (s) A.G.P.

ASSERTION-REASON TYPE QUESTIONS 34. Let a0, a1, a2, a3, be an arithmetic progression. Statement-1: sin a2 + sin a4 + + sin a2n =

=

38. If a, b/2, c are in G.P. as well as x1, x2, x3, x4 are in G.P., then p, q/2, r are in (a) A.P. (b) G.P. (c) H.P. (d) A.G.P. x1, x2, x3, x4 are in G.P., then its common ratio is Ê ar ˆ (a) Á ˜ Ë c p¯ (c)

2 sin ( a2 – a1 )

2 sin ( a2 – a1 )

35. Let a1, a2, a3, be a sequence of real numbers such that sum to n terms of the sequence is (n – 2)2n + 1 " n ∈ N

cr ap

Ê cr ˆ (b) Á ˜ Ë a p¯ (d)

1/3

ap bq

a b - 4 ad 4 a d 2 - 2bd + c (a) = = q r p (b)

a b - ad ad 2 - bd + c = = q r p

(c)

a b - 2 ad ad 2 - bd + c = = q r p

+ cos a2n

sin a2 n + 1 – sin a1

1/4

40. If x1, x2, x3, x4 are in A.P., with common difference d, then

cos a1 – cos a2 n + 1

Statement-2: cos a2 + cos a4 +

1 1 b 2 - 4 ac , are in A.P., then 2 equals x3 x4 q - 4 pr

(a) a2/r2 (b) b2/q2 (c) c2/p2 (d) a2/p2 x1, x2, x3, x4 are in A.P., then common difference of this A.P. is

k =1

(b) Â loga (cot k°)

36. If x1, x2,

(d) none of these Paragraph for Question Nos. 36 to 40

A man is to receive Rs Pi at the end of ith year for n years. Assume that the rate of interest is Rs r per

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rupee per year and the interest is compounded annually. 41. If Pi = P " i, then present value of income stream is P [1 – (1 + r)–n] (a) r P [1 + r – (1 + r)–n] (b) r (c) P[1 – (1 + r)–n] (d) p/r 42. If Pi = P " i and n Æ •, then present value of the income stream is (a) P (b) 2P (c) P/r (d) P + P/r 43. If Pi = iP " i, then present value of the income stream is P (1 + r ) (a) [1 – (1 + r)–n] r2 P (1 + r ) nP [1 – (1 + r)–n] – (b) 2 r r (1 + r ) n P (1 + r ) nP (c) [1 – (1 + r)–n] – 2 r (1 + r ) n (d) p/r 44. If Pi = iP " i and n Æ •, then present value of the income stream is (a)

P (1 + r )

(b)

r2

P

r2 (d) r(1 – r)P

(c) rP

3.43

45. If Pi = iP " i and n Æ •, then value of r for which present value of income stream is 2P is

INTEGER-ANSWER TYPE QUESTIONS 46. Three numbers a, b, c are chosen from the numbers 1, 2, 3, ..., 100. If N is the number of ways of choosing, a, b, c so that these are in A.P. then N/350 is terms are 2/5 and 12/23 is 48. The value of 2551 S 50 k is S=  4 + k k2 + 1 k =1 a = maximum value of the function 1 1 - cos x f(x) = (6x2 – x3 – 16) and r = lim . nÆ• 8 x2 If S = a + ar + ar2 + ..., then •

50. If a = Â

1

n =1 n

n

S is

, then [a], where [x] denoted the great-

est integer £ a is 51. Let an = 11 1 . The number of primes in the sen times

quence a2, a3, a4, a5, a6. 52. The sum of the series •

Â

6k

k k =1 (3

- 2k ) (3k +1 - 2k +1 )

is

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. If x, y and z are the pth, qth and rth terms respectively of an A.P. and also of a G.P., then xy – z yz – x zx – y is equal to (a) xyz

(b) 0

(c) 1

(d) none of these

2. The third term of geometric progression is 4, the (a) 43 (c) 44

(b) 45 (d) none of these

3. The rational number, which equals the number 2.357 with recurring decimal is (a)

2355 1001

(b)

2379 997

(c)

2355 999

(d) none of these

4. If a, b, c, d and p are distinct real numbers such that E = (a2 + b2 + c2) p2– 2 (ab + bc + cd) p + (b2 + c2 + d2) £ 0 then a, b, c, d (a) are in A.P. (c) are in H.P.

(b) are in G.P. (d) satisfy ab = cd

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n terms of the series S=

1 3 7 15 + + + + ......... is equal to 2 4 8 16

(a) 2n – n + 1 (c) n + 2–n – 1

(b) 1 – 2–n (d) 2n

6. Let n (> 1) be a positive integer. Then the largest integer m such that (nm + 1) divides S = 1 + n + n2 + ...... + n

is

n terms of the A.P. 2, 5, 8 ....... n n equals, (a) 100 (b) 12 (c) 11 (d) 13. [2001] 14. Suppose a, b, c are in A.P. and a2, b2, c2 are in G.P. If a < b < c and a + b + c = 3/2 then the value of a is (a)

1 2 2

1 1 + (c) 2 2 Tr the rth term of an A.P. for r = 1, 2, 3, ............... If for some positive integers m, n we have Tm = 1/n and Tn = 1/m, then Tmn equals, 1 mn

(a)

(b)

1 1 + m n

8. Let a1, a2, ..... a10 be in A.P. and h1, h2, .... h10 be in H.P. If a1 = h1 = 2 and a10 = h10 = 3, then a4 h is (a) 2

(b) 3

2 )x2 – (4 +

(5 + (a) 2

(b) 4

(c) 5

(d) 6.

5 )x + 8 + 2 5 = 0 is (c) 6

(d) 8

a and common ratio r. If its sum is 4 and the second term is 3/4, then (a) a r (c) a = 3/2, r = 1/2 11. If

a = 2, r = 3/8 (d) a = 3, r = 1/4. [2000]

Ê x 2 x3 ˆ + ... sin–1 Á x – 2 4 ˜¯ Ë Ê 2 x 4 x6 ˆ p + ...˜ = + cos–1 Á x – 2 2 4 ¯ Ë x

(a)

1 2

(c) –

1 2

1 2 3

1 1 – (d) 2 2

[2002]

x and sum 5, then (b) – 10 < x < 0 (d) x > 10 [2004] 2 n terms of an A.P. is cn , then the sum of squares of these n terms is (a) x < – 10 (c) 0 < x < 10

(a)

n ( 4n 2 – 1) c 2 6

(b)

n ( 4n 2 + 1) c 2 3

(c)

n ( 4n 2 – 1) c 2 3

(d)

n ( 4n 2 + 1) c 2 6

a1, a2, a3,... be in harmonic progression with a1 = 5 and a20 = 25. The least positive integer n for which an < 0 is (a) 22 (b) 23 (c) 24 (d) 25 [2012] 18. Let bi > 1 for i = 1, 2, …, 101. Suppose loge b1, loge b2 , …, loge b101 are in Arithmetic Progression (A.P) with the common difference loge 2. Suppose a1, a2, …, a101 are in A.P. such that a1 = b1 and a51 = b51. If t = b1 + b2 + … + b51 and s = a1 + a2 + … + a51, then (a) s > t and a101 > b101 (b) s > t and a101 < b101 (c) s < t and a101 > b101 (d) s < t and a101 < b101 [2016]

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

2 , then x equals (b) 1

(b)

(d) – 1.

[2001] 12. Let the positive numbers a, b, c, d be in A.P. then abc, abd, acd, bcd, are (a) Not in A.P./G.P./H.P. (b) in A.P. (c) in G.P. (d) in H.P. [2001]

n – 1)th term of an A.P., a G.P and an H.P are equal and their nth terms are a, b and c respectively then (a) a = b = c (b) a ≥ b ≥ c (c) a + c = b (d) ac – b2 • • p , if x = Â cos2n f, y = Â sin2n f 2. For 0 < f < 2 n =0 n =0 •

and z = Â cos2n f sin2n f, then n =0

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(a) xyz = xz + y (b) xyz = xy + z (c) xyz = x + y + z (d) xyz = yz + x 3. Let S1, S2, ..... be squares such that for each n ≥ 1, the length of a side of Sn equals the length of a diagonal of Sn + 1. If the length of a side of S1 is 10 cm, then the values of n for which Area (Sn) < 1 is

4. For a positive integer n, let 1 1 1 1 a(n) = 1 + + + + ...... + n . Then 2 3 4 (2 ) – 1 (a) a(100) £ 100 (c) a(200) £ 100 4n

5. Let Sn = Â ( -1) 1056 1120

(b) 1088 (d) 1332

[2013]

1. Suppose four distinct positive numbers a1, a2, a3, a4 are in G.P. Let b1 = a1, b2 = b1 + a2, b3 = b2 + a3 and b4 = b3 + a4. Statement-1: The numbers b1, b2, b3, b4 are neither in A.P. nor in G.P. Statement-2: The numbers b1, b2, b3, b4 are in H.P. [2008]

INTEGER-ANSWER TYPE QUESTIONS 1. Let a1, a2, a3, , a11 be real numbers satisfying a1 a2 > 0 and ak = 2ak–1 – ak–2 for k = 3, 4, , 11. 2 a12 + a22 + ... + a11 11

a1 + a2 + ... + a11 is equal to [2010] 11 2. Let a1, a2, a3 ...., a100 be an arithmetic progression with p

a1 = 3 and Sp = Â ai , 1 £ p £ 100. For any integer n i =1

with 1 £ n £ 20, let m = 5n. If on n, then a2 is

4. Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is. [2015]

FILL

IN THE

BLANKS TYPE QUESTIONS

k . Then Sn can take value(s)

ASSERTION-REASON TYPE QUESTIONS

If

the arithmetic mean of a, b, c is b + 2, then the value a 2 + a – 14 of is [2014] a +1

x in the polynomial P(x) = (x – 1) (x – 2) ..... (x – 100) is _____. 2. The sum of integers between 1 and 100 that are

(b) a(100) > 100 (d) a

k ( k +1) 2 2

k =1

(a) (c)

3.45

Sm does not depend Sn

[2011] b 3. Let a, b, c be positive integers such that is an a integer. If a, b, c are in geometric progression and

3. If a, b, c are in A.P., then the straight line ax + by + c ______, ______) coordinates are ( n terms of the series 12 + n (n + 1)2 , when 2(2 2)+ 32 + 2(42) + 52 + 2(62) + ......... is 2 n is even. When n 5. Let the harmonic mean and geometric mean of two positive numbers be in the ratio 4 : 5.Then the two 6. For any odd integer n ≥ 1, n3 – (n – 1)3 + ......... + (– 1)n – 1 13 p and q be roots of the equation x2 – 2x + A = 0 and let r and s be the roots of the equation x2 – 18x + B = 0. If p < q < r < s are in arithmetic progression, then A = _____ and B 8. Let x be the arithmetic mean and y, z be two geometric means between any two positive numbers. Then

y3 + z3 xyz

SUBJECTIVE-TYPE QUESTIONS 1. A sequence of numbers a0, a1, a2, a3 2 relation an – an – 1 an + 1 = (– 1)n " n ≥ 2. Find a3 if a0 = 1 and a1 2. The value of x + y + z is 15 if a, x, y, z, b are in A.P., 1 1 1 5 while the value of + + is if a, x, y, z, b y x z 3 are in H.P. Find a and b 3. The harmonic mean between two numbers is 4. The arithmetic mean A and the geometric mean G satisfy the relation 2A + G2

IIT JEE eBooks: www.crackjee.xyz 3.46 Comprehensive Mathematics—JEE Advanced

4. The interior angles of a polygon are in A.P. The smallest angle is 120° and the common difference is 5°. Find the number of sides of the polygon. 5. mn squares of equal size are arranged to form a rectangle of dimension m by n, where m and n are natural numbers. Two squares will be called ‘neighbours’ if they have exactly one common side. A natural number is written in each square such that the number written in any square is the arithmetic mean of the numbers written in its neighbouring squares. Show that this is possible only if all the 6. If a1, a2, .... an are in A.P where ai > 0 for all i, 1 1 + + ...... + show that a1 + a2 a2 + a3 1 an –1 + an

=

n –1

14. The real number x1, x2, x3 satisfying the equation x3 – x2 + bx + g = 0 are in A.P. Find the intervals in which b and g 15. The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive terms of it, prove that the resulting sum is square of an integer. [2000] 16. Let a1, a2, ... be positive real number in G.P. For each n, let An, Gn, Hn be respectively, the arithmetic mean, geometric mean, harmonic mean of a1, a2 ... an. Find an expression for the geometric mean of G1, G2, .... Gn in terms of A1, A2, ... An, H1, H2, ... Hn. [2001] a and b be positive real numbers. If a, A1, A2, b are in arithmetic progression a, G1, G2, b are in G.P. and a, H1, H2, b are in H.P. show that G1G2 A + A2 (2a + b) (a + 2b) = 1 = H1 + H 2 H1 H 2 9ab

a1 + an

as three of its terms? If it exists, how many such 8. Find three numbers a, b, c between 2 and 18 such that (i) their sum is 25 (ii) 2, a, b are consecutive terms of an A.P (iii) the numbers b, c, 18 are consecutive x ∈ (–p, p) which satisfy the 2x 3x x equation 8 = 43 10. The sum of the squares of three distinct real numbers, which are in G.P. is S2. If their sum is a S, show that a2 ∈ (1/3, 1) » 11. Let p n arithmetic means between two numbers and q n harmonic means between the same numbers. Show that q does not lie 2

Ê n + 1ˆ p between p and Á Ë n – 1˜¯

18. If a, b are c are in arithmetic progression and a2. b2 and c2 are in harmonic progression, then prove that either a = b = c or a, b and – c/2 are in geometric progression. [2003] n (> 1) matches and scores k (2n + 1 – k) runs in his kth match (1 £ k £ n). If the total runs scored by him in the n matches is 1 (n + 1) (2n + 1 – n n. 4 [2005] 2

20. If an =

bn = 1 – an that bn > an " n ≥ n0.

n and whose common ratios are

3

3 Ê 3ˆ Ê 3ˆ – Á ˜ + Á ˜ + ...... (– 1)n – 1 Ë 4¯ 4 Ë 4¯

n

Ê 3ˆ ÁË ˜¯ and 4 n0 such [2005]

• Ê 1 ˆ 21. If q = Â tan–1 Á 2 ˜ Ë2k ¯ k =1

12. If S1, S2, S3, ..... Sn

[2002]

q

[2006]

Answers

1 1 1 1 , , , 2 3 4 n +1

LEVEL 1

S12 + S22 + S32 + ...... + S22 n – 1 13. If y = exp {(sin2x + sin4x + sin6x + ......) ln 2} x2 x p cos x of ,0 0 and b2 + c2 = b2 > 0

tr = n

 tr =

r =1

=

1 – xr

(1 – x ) 2 (1 – x )

2

1

(1 – x ) 2

n

(1 – x )3

2ac b 2c b 2a fi = and = a+c a a+c c a+c

a + b 1 + b a a + 3c = = 2a – b 2 – b a 2a and

c + b 1 + b c 3a + c = = 2c – b 2 – b c 2c

Thus,

a+b c+b + 2 a – b 2c – b ≥1+

3 Ê c aˆ 3 ÁË + ˜¯ ≥ 1 + ( 2) = 4 2 a c 2

1 1 1 1 + + + =0 a c–b c a–b fi

a+c–b a+c–b + =0 a (c – b) c ( a – b)

fi a(c – b) = c(b – a) fi b =

2ac a+c

For statement-2, put t = x/y, and write the expression as y2[a(b – c)t2 + b(c – a)t + c(a – b)]

a + c = 2b, ab = c2 2bc c ( a + c ) ac + ab = = =a Now b+c b+c b+c \ c, a, b are in H.P.

(1)

Eliminating a from two expressions in (1) we get c2/b + c = 2b fi c2 + bc – 2b2 = 0

rx r – 1– x

 (1 – x r ) – n

1

(1 – x ) 2

) – 2 x (1 – x )

t perfect square other zero must be 1.

a + b + g = 0 fi 3b = 0 fi b = 0 \c=0

Thus, b2 –

b=

(

n 1 + xn + 1

r =1

(

È x 1 – xn Ín – Í 1– x Î

(

1 n r  rx 1 – x r =1

) ˘˙ ˙ ˚

)

n È ˘ 1 Íx 1– x nx n + 1 ˙ – – 1 – x Í (1 – x ) 2 1– x ˙ Î ˚

(c – b) (c + 2b) = 0 fi c = – 2b [ c π b] Thus, a = 4b. Now, a : b : c = 4 : 1 : – 2. Pi invested at the end of i years fetches interest for (n – i) years and becomes Ai = Pi(1 + r)n–i The amount A is given by + An A = A1 + A2 + P = ((1 + r)n – 1) if Pi = P " i. r

IIT JEE eBooks: www.crackjee.xyz Progressions

But a2 = 1 fi a = 0. Not possible. 82. Putting a2 = 2 in (4) (r – 1/r)2 = 5

77. A = P(1 + r)n–1 + 2P(1 + r)n–2 + + nP A nP + (n – 1)P + = P(1 + r)n–2 + 1+ r 1+ r Subtracting we obtain r A = P(1 + r)n–1 + P(1 + r)n–2 + 1+ r +P– = fi A=

fi r – 1/r = nP 1+ r

nP P [(1 + r)n – 1] – 1+ r r P (1 + r ) r2

5 +3 2 2 in (4) and solve.

10 1 ˆ Ê = Á r˜ +3≥3 Ë a r¯

Ê 1 2ˆ Also, a2 Á 2 + 1 + r ˜ = 202 Ër ¯ fi a £ 20 3 Thus, a £ 10/3. 85. From (4) in Solution of 81

A = A1 + A2 + + An 2 Ê 2 2 2n - 1 ˆ + + = P Á1 + + (n - 1)!˜¯ Ë 1! 2!

2

1ˆ Ê ÁË r - ˜¯ = 0 fi r = ± 1. r

< Pe2.

86. a1 = 1 and

Ê1 ˆ a Á + 1 + r ˜ = aS Ër ¯

(1)

Ê 1 2ˆ a2 Á 2 + 1 + r ˜ = S2 Ër ¯

(2)

Divide (2) by (1) to obtain S Ê1 ˆ a Á -1 + r ˜ = Ër ¯ a



fi a £ 10/3

80. If the investments are made for n years, then P 2n - k P Ak = (1 + 1)n–k = (n - k )! (n - k )!

a2n+1 = [(1 – a)2 + a2] an2 5 2 Ê 5ˆ a = Á ˜ Ë 9¯ 9 n

=

Now, An = an2 < (3)

From (2) and (3)

Ê 9ˆ fi Á ˜ Ë 5¯

n

1 10

n -1

> 10

fi 9n–1 > 2(5n) fi n ≥ 5.

Ê a 2 - 1ˆ 1ˆ Ê 2a = S Á a - ˜ = S Á ˜ Ë a¯ Ë a ¯

87. If a =

Putting this in (2) we get (a 2 - 1)2 Ê 1 2ˆ Á + 1 + r ˜¯ = 1 4a 2 Ë r 2

1 2 1 2 Ê 1ˆ , a n+1 = a = Á ˜ 2 2 n Ë 2¯

Ê 1ˆ \ A1 = 1 and An = Á ˜ Ë 2¯

2

¤ 3a4 – 10a2 + 3 < 0 ¤ (3a2 – 1) (a2 – 3) < 0 ¤ 1/3 < a2 < 3

fi r=

2

n so that

fi 3n –1 ≥ 2n+1 fi n ≥ 5.

1ˆ 4a 2 Ê fi Ár - ˜ + 3 = Ë r¯ (a 2 - 1)2

5r – 1 = 0

Ê1 ˆ Ê 1ˆ 84. a Á + 1 + r ˜ = Á ˜ (20) Ër ¯ Ë 2¯

nP [(1 + r) – 1] – . r

79. 2P [(3/2)n – 1] ≥ 10 P

81.

5.

fi r2 –

83. Put r =

n

78. In Solution of 76, put r A > 2000P.

3.55

(4)

n -1



fi  An = 2. n =1

88. a1 = 1, a2n+1 = (2a2 – 2a + 1)n •

\ Â An = n =1

1 2a - 2a

2

=

8 3

n

[

r > 1]

[

r > 1]

IIT JEE eBooks: www.crackjee.xyz 3.56 Comprehensive Mathematics—JEE Advanced

fi 2a – 2a2 = 3/8 fi 16a2 – 16a + 3 = 0 fi a = 1/4, 3/4.

1 k

=

2 an+1

an =



2

an = 2a

=

2 n+1 2

fi an = 2(2a – 2a + 1)an fi a = 1/2. P1 = 4, Pn = 4an = 4(1 – 2a + 2a2)(n–1)/2 2

Ê 5ˆ =4 Á ˜ Ë 8¯ •

8- 5

n =1

2

(

8 2 =

(

8+ 5

8-5

)

)

+

k 2 (k + 1)2 (k 2 + k + 1)2 k 2 (k + 1)2 1

k (k + 1) + 1 k (k + 1)

=

(k + 1)

2

=1 +

1 1 k k +1

n

an = Â 1 +

1

k =1

k

=n +

n n +1

2

+

1 (k + 1)

2

=n+1–

1 n +1

3

3

Ê xˆ Ê zˆ y =x +z fi Á ˜ +Á ˜ =2 Ë y¯ Ë y¯ 3

3y = 15 or y = 5

\ a + b = 10 Since a, a, b, g and b are in H.P., we have

(1)

3

3

(1)

Since logx y, logz x, logy z are in G.P., log y z log z x (log x)2 (log z ) 2 = fi = log z x log x y log y log z (log y ) (log x) fi (log x)3 = (log z)3 fi log x = log z fi x = z From (1) we get

1 1 1 1 2 + = + = b a b a g Also

2

k 2 (k + 1)2 + (k + 1)2 + k 2

5ˆ Ê \ 6a5 – 33 = 6 Á 5 + ˜ – 33 = 2 Ë 6¯

Also x + y + z = 15 fi 2y + y = 15 fi

k



8 4 + 10 . 3 a, x, y, z and b are in A.P., we have a + b = x + z = 2y =

1

fi 1+

( n - 1) 2

4 8

\ Â Pn =

(k + 1)2

2 an+1

n + 1)th square = Now

1

+

2

1 1 1 5 3 5 1 5 + + = fi = or = 3 b b a b g 3 9

Ê xˆ 2Á ˜ Ë y¯

1 1 10 b+a 10 + = fi = 9 ab a b 9 fi ab From (1) and (2) we get a b = 1. as a > b.

Thus

a + b + c = 25 2a = b + 2 and c2 = 18b

(1) (2) (3)

From (1) and (2) we have b + 2 + 2b + 2c = 50 fi 3b + 2c = 48 Using (3) we get

(4)

Ê 3c 2 ˆ 2 Á 18 ˜ + 2c = 48 fi c + 12c – 288 = 0 Ë ¯ fi (c + 24) (c – 12) = 0 fi c = – 24 or c = 12 As a, b, c lie between 2 and 18 we must have c = 12.

3

3

Ê xˆ =2 fi Á ˜ = 1 fi x = y Ë y¯

Therefore, x = y = z. Since xyz = 64, we get x = y = z = 4. Thus, x + y = 8. x ∈ (0, p x 2 3 S• x x x • x x2 x 3 + ... up to • 1 = 1 - cos x \ 33/(1 fi fi

x

x

= 36 fi

3 =6 1 - cos x

p 2p 1 1 fi cos x = ± fi x = , 3 3 2 2

1 (x1 + x2) = 1. p

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tk =

k k4 + 4 = =

a + d)2 = a2 + (a – d)2

k

=

(k 2 + 2)2 - 4k 2 k

(k + 2k + 2)(k 2 - 2k + 2) 2

1È 1 1 ˘ 4 ÍÎ (k - 1) 2 + 1 (k + 1)2 + 1 ˙˚

fi sin A =

a-d 3 = a+d 5

sin B = cos A =

n

an = Â t k



4 . 5

k =1

=

1È 1 1 1 ˘ 1+ - 2 Í 2 4 Î 2 n + 1 (n + 1) + 1 ˙˚

=

n(n + 2) ˘ 1 È n2 + Í 2 ˙ 4 ÎÍ n + 1 2 ÈÎ (n + 1)2 + 1˘˚ ˚˙

P(x) = (1 + x + 2x2 +

+ 25x25) (1 + x + 2x2 + 25x25)

\a x25 (1) (25) + (2) (24) + + (24) (2) + (25) (1) 25

= Â k(26 – k) = k =1



a+d a a-d = = sin A cos A 1

a 325

26 ¥ 25 ¥ 26 25 ¥ 26 ¥ 51 2 6

Fig. 3.3

d)2 = (3 + d) (3 + 33d) fi d = 1. 2 È Ê 2ˆ Ê 2ˆ Í900 Á ˜ + 900 Á ˜ + Ë 3¯ Ë 3¯ ÍÎ

˘ ˙ = 4500. ˙˚

3.57

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4 Logarithm 4.1

a xy aπ aŒR

EXPONENTIAL FUNCTION

For x ŒR 2

ex = 1 + x +

3

x x + + 2! 3!

+



n

ex Properties ex ex xŒR e0 a, b Œ R, a < b x e × e y = ex + y

xa = ea

ex) y = e x y

x, y ΠR

a

x =

a, c π a



= Â ( a log e x ) n

log e x log e a a

a, b, c b=

log c x log c a

xa

ey = x

x

a

x 0, a π a a = 1, a > 0, a π x e ex = x " x ee = x " x Œ R ex x, y a xy a x a y a > 0, a π x a a x " x > 0, a > 0, a π a

ex

Fig. 4.2

x

x

x

4.2

y x

x y

ex

x =y x " x > 0, y Œ R (Change of base formula) a > 0, a π

e 0, a > 0,

y

b

ex = ex – y ey

a

n=0

ex

xŒR a

x

x

n

x x += Â n! n= 0 n!

xŒR

a

(

(

x

ax
0 j x

f x) ≥

j x)

g x)

¤ 0 < f x) £ g x) j x

j x)

¤ f x) ≥ aj

Example 2 e

j x)

ex

is

=x

Ans.

f x) ≥ a

Solution:

x)

j x

j x)

¤ 0 < f x) £ aj

f x) ≥ a

x ex

f x) = e

x)

– x, x > 0

e e- x -1 = x x f¢ x xe

f ¢ x) =

Characteristics and Mantissa characteristic mantissa



f x e, •) fi

e

f x) < f e

x 1 is

x

x

x

y) =2

y

x0 is 1 6 1 2

x



1 3

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 27

a+c

is

a–c

x (3/4) (log 2 x )

2.

x > 1, y > 1, z 1 1 1 , , 1 + log x 1 + log y 1 + log z

x

= 4x – 1

=

2 has

2

x=

2 log3 2 2 log3 2 - 1

2 2 - log 2 3

1 1 - log 4 3

2 log 2 3 2 log 2 3 - 1

INTEGER-ANSWER TYPE QUESTIONS Ê Á 1 Á ÁË 3 2

x

+ log 2 x - 5/4

a – 2b + c

a, b, c a2, b2, c2 a, b, c a, b, c

4

2

x – 3) is

ˆ

4- 1 4- 1 4- 1 ... ˜˜ is 3 2 3 2 3 2 ˜¯

IIT JEE eBooks: www.crackjee.xyz 4.22

FILL

BLANKS TYPE QUESTIONS

IN THE

y = 10x

y

10

MATRIX-MATCH TYPE QUESTIONS

–1 Ê

x2 ˆ log Á 2 2˜ Ë ¯

f x

7

5

x+5 + x )=0

TRUE/FALSE TYPE QUESTIONS a 0. Let f (x) = elnx – x, x > 0

IIT JEE eBooks: www.crackjee.xyz 4.24 Comprehensive Mathematics—JEE Advanced

e e- x -1 = x x

f ¢(x) =

t 2 – 3t + 3 –1 0 if 0 < x < e Ô x=e fi f ¢(x) = Ì= 0 if Ô< 0 if x>e Ó fi f (x) increases on (0, e] and decreases on [e, •) \ f (x) < f (e) = 0 for 0 < x < e and f (x) < f (e) = 0 for x>e fi e lnx – x < 0 fi xe < ex

xπe

xπe

\ only solution of the equation is x = e. 4.

1 1 4 + = log a x log a y log a xy fi

5.

15.

16. 17.

(loga x – loga y)2 = 0

(5 x – 2) (4) =3fix=1 3x + 1

18. 19.

6. Use change of base. 7. (log (0.6) (0.6)3) log5(5 – 2x) £ 0 fi 5 – 2x £ 1 fi x ≥ 2 Also, 5 – 2x > 0 8. 2 + log 2

x + 1 > 1 – log2

4 – x2

20.

=

log5 34 1 – log5 33 + 3 2

=

1 4 – 3 log5 3 + log5 3 2 3

=

1 5 – log53 2 3

= log5 \

(

x=

5 / 35/3 5 10/3

)

is an irrational number

3 12. a log1/2 (x) + b log1/2 ( y) + c log 1/2 (z) = 0 13. xy = 100, x – y = 15 fi x = 20, y = 5 14. Put loge x = t and rewrite as

t 2 – 3t + 3 – t + 1 22 = 4 fi 4 – 5x < – 4 or 4 – 5x > 4 fi 5x > 8 or – 5x > 0 Use (a + 2b)2 = 16ab x2 – 16 £ 4x – 11 fi x2 – 4x – 5 £ 0 fi (x – 5) (x + 1) £ 0 fi –1£x£5 But x > 4. log 2 + 2log 3 3log 3 + log 2 ,b= a= 2 log 2 + log 3 3log 2 + log 3 Now ab + a – b = (a – 1) (b + 1) + 1

As x ΠN, 4 Рx 2 > 0 fi x = 1 9. log10(x + 1) log10(x + 1) = 2 + log10(x + 1) fi log10(x + 1) = 2, Р1 fi x + 1 = 100, 1/10 10. Use change of base. 1 + log1/5 27 + log125 81 11. 2



Ê log 3 – log 2 ˆ È 4(log 3 + log 2) ˘ = Á Í ˙ +1 Ë 2 log 2 + log 3 ˜¯ Î 3log 2 + log 3 ˚ π 1, 2, log12(24) x+5 +

21.

x =7

But (x + 5) – x = 5 \

x+5 –

x =

5 7

Thus, 2 x = 7 – 5/7 = 44/7 fi x = 484/49 22. If 0 < x < 1, then LHS < 0. For x > 1 1 1 3 + >1fi >1 log 2 x 2 log 2 x 2 log 2 x fi

0 < log 2 x < 3/2

Ê 9x – 1 + 7 ˆ 23. log 2 Á x – 1 ˜ =2 +1¯ Ë3 fi

32x – 2 + 7 = 4(3x–1 + 1)

IIT JEE eBooks: www.crackjee.xyz Logarithm 4.25

fi (3x–1)2 – 4(3x–1) + 3 = 0 fi (3x–1 – 1) (3x–1 – 3) = 0 fi x – 1 = 0 or x – 1 = 1 fi x = 1, 2 24. log x + log y + log z = 0, a log x + b log y + c log z = 0 Also (b + c) log x + (c + a) log y + (a + b) log z = 0 25. Taking logarithm of both the sides, [(log 2 x)2 – 6 log2 x + 11] log2 x = 6 fi log 2 x = 1, 2, 3 26. log 2(4x2 – x – 1) – log2 (x2 + 1) > 0 fi

27. 28. 29. 30. 31.

32.

2

4x – x – 1 x +1 2

> 1 fi 3x2 – x – 2 > 0

fi (3x + 2) (x – 1) > 0 fi x < – 2/3 or x > 1 Use a = x – 1, b = x, c = x + 1 x + 1 = log a bc + 1 = log(abc)/loga etc. 100.3 < 2 < 100.30103 fi 10 30 < 2100 < 1030.103 < 1031 Use change of base. [log10 x] = 1 fi 1 £ log10 x < 2 fi 10 £ x < 100. For x ≥ 1, let f (x) = x2lnx. For x ≥ y, x < y, by the Lagrange’s theorem there exist a Œ (x, y) such that f ( x) - f ( y ) = f ¢(a) = 2a ln + a x- y

34. (a) 2 + fi

1 log 2 x + log2 x = 7 4 5 log2 x = 5 or x = 16 4

(b) Put log2 x = t to obtain Ê 1 ˆ Ê1+ tˆ 1 t (1 + t) + Á =0fit=2fix=4 6 Ë 1 + t ˜¯ ÁË –1 ˜¯ (c) 3 log 4 (x – 150) = 3 fi x = 154 (d)

x2 ( x + 1/ 2) 2 = fi x = 1 as x > 0. x – 1/ 2 x + 1/ 8

35. (a) Taking log (x 2 – 6x + 8) log10(x – 2) > 0 fi

(x – 2) (x – 4) log10(x – 2) > 0

As x – 2 > 0, x – 4 > 0, log10(x – 2) > 0 or x – 4 < 0, log10 (x – 2) < 0 fi

x > 4 or 2 < x < 3

(b) Put 27x = t,

t 12 –5£0 + 2 t





t 2 – 10 t + 24 £ 0



4 £ t £ 6 fi log 2 4 £ 7x £ log2 6



2 1 £x£ log2 6 7 7

(c) 23x/2 + 23x/2 – 2(22x) ≥ 0

fi Let g(x) = 2x lnx + x, x ≥ 1, g¢(x) = 2lnx + 2 + 1 > 0

(d) Use change of base, | x| = 16.

x>1

Thus, g(x) is strictly increasing on [1, •). Now, there exist a1 Œ (a, b), a2 Œ (b, c), a3 Œ (c, d), a4 Œ (b, d), such that D(a, b) = g(a1), D(b, c) = g(a2) D(c, d) = g(a3), D(b, d) = g(a4)

fi fi

2(23x/2) (1 – 2x/2) ≥ 0 fi 2 x/2 £ 1 x£0 x2

(d) 2 < 22x+3 fi x2 < 2x + 3 fi

(x – 1)2 < 4 fi – 2 < x – 1 < 2



– 1 < x < 3.

36. Clearly x > 3. The given equation gives

As a < a1 < b < a2 < c < a3 < d, and g is increasing on [1, •) we get

Ê 2ˆ 2t – 3 Á ˜ = – 1 where t = log2 x Ët¯

D(a, b) < D(b, c) < D(c, d) < D(b, d)



2t2 + t – 6 = 0

Also, D(a, b) < D(c, d)



t = 3/2, – 2

33. (a) log 7 | x + 9| = 1 fi x = – 2, – 16 (b) log 8 (– 4x) = 2 fi x = – 16 (c) Use t + 1/t ≥ 2 or £ – 2

37. Using sin we get





(2t – 3) (t + 2) = 0

x = 2 2 , 1/4

p 3p 2p 4p = sin and sin = sin 5 5 5 5

IIT JEE eBooks: www.crackjee.xyz 4.26 Comprehensive Mathematics—JEE Advanced

p 2p x = 4 log5 2 + log5 sin sin2 5 5 2 = 4 log5 2 + log5 (1 – cos 36°) (1 – cos2 72°) = 4 log5 2 + log5 (1 – cos2 36°) (1 – sin2 18°) = 4 log5 2 + 2

1 È log5 Í1 – Î 16

(

)

2˘ È 1 5 + 1 ˙ Í1 – ˚ Î 16

(

)(

(

)

2˘ 5 –1 ˙ ˚

)

È 10 – 2 5 10 + 2 5 ˘ ˙ = 4 log5 2 + log5 Í Í ˙ 256 Î ˚ Ê 80 ˆ Ê 5ˆ = 4 log 5 2 + log 5 Á = 4 log 5 2 + log 5 Á 4 ˜ = 1 Ë 256 ˜¯ Ë2 ¯ Ê x + 1ˆ 38. log 0.2 Á ≥ log0.2 (0.2) Ë x ˜¯ x +1 1 x +1 x +1 ¤0< £ 0.2 ¤ >0 – £ 0 and x 5 x x ¤

x +1 4x + 5 > 0 and £0 x 5x

5ˆ Ê ¤ x π 0; x Á x + ˜ £ 0 and x (x + 1) > 0 Ë 4¯ ¤ – 5/4 £ x < – 1. 39. Rewrite the equation as log 5 (x4 + 5) (x2 + 25) = 3/2 and note that LHS ≥ 3. 40. Rewrite the equation as log3 (3 + x ) (1 + x2) = 0 and note that LHS ≥ 1. 41. x > 0, x π 1. Put log x 5 = t, to obtain t 3 – t2 – 6t = 0 fi t(t2 – t – 6) = 0 fi t (t – 3) (t + 2) = 0 fi t = 0, 3, – 2 fi 5 = x0, x 3, x–2 But 5 = x0 is not possible. 42. Rewrite the equation as 2t + 2t – t = 9 where t = log2 x fi t = 3 or x = 8. 43. Put log5 x = t and rewrite the equation as 1– t = 1 fi t2 + t3 + 1 – t = 1 + t t2 + 1+ t

fi t 3 + t2 – 2t = 0 fi t(t – 1) (t + 2) = 0 fi x = 1, 5, 1/25. 44. x > 0, x π 1 and (x + 3) 2 = 16 fi x = 1, – 7. 45. x > 0 and use xlog5 2 = 2log5 x to obtain 2log5 x = 16 fi log5 x = 4 fi x = 625 46. x > 0. Take log to obtain Ê 1ˆ log2 Á ˜ + (log2 Ë 4¯ fi

x ) (log2 x) =

(log 2 x)2 = 8 fi x = 2 – 2

2

1 (log2 x) 2 4

or 22

2

.

4

47. x < 0 and log6 (– 4x) = log2 (log7 7 ) = 2 fi – 4x = 62 = 36 fi x = – 9. 48. x > 0 and (log 4 x) (log 2 x) = (log4 x + 3) log 2 (8) Put log 2 x = t to obtain fi

1 2 Ê1 ˆ t = Á t + 3˜ (3) Ë2 ¯ 2

fi t2 = 3t + 18 fi t = 6 – 3 fi x = 26, 2–3 = 64, 1/8. 49. x > 0. Put log 2 x = t to obtain 2t2 + 2(1 – fi

2 )t – 2 2 = 0

(t + 1) (t –

2 ) = 0 fi x = 1/2, 2

2

.

50. x < 1/2. Rewrite equation as 1 log (1 – 2x) 2 – log (280 + x2) = log (1 – 2x) 2 fi fi

2 log (1 – 2x) = log (280 + x 2) 1 – 4x + 4x2 = 280 + x2 fi x = – 9. x –1 51. x 2 + 3x – 4 > 0, >0 x+4 fi

(x – 1) (x + 4) > 0,



x < – 4 or x > 1

Also, x 2 + 3x – 4 =

x –1 >0 x+4

x –1 x+4

fi (x + 4) 2 = 1 fi x = – 5, as x < – 4. 52. 9 – 2x > 0, 3 – x > 0 fi x < 3 Also, 9 – 2 x = 23 – x fi x = 0, 3. 53. ax = bc fi x log a = log bc fi

1 log a = etc. x log (bc )

IIT JEE eBooks: www.crackjee.xyz Logarithm 4.27

54. Note that x > 0. Put log1/7 x = t to obtain |4 + t| = 2 + |2 + t| fi t ≥ – 2 fi x £ 49. 55. Note that x > 0. Rewrite the equation as 1 log 2 log2 x + log2 log2 x – 1 = 2. 2 fi log 2 (log2 x) = 2 fi x = 16. 56. For x > 0, put log2 x = y, log3 x = z to obtain yz = 3z + 2y – 6 fi ( y – 3) (z – 2) = 0 fi x = 8, 9. 57. For x > 0, rewrite the equation as 1–t + t2 = 1 where t = log 3 x 1+ t fi

1–t = 1 – t2 1+ t



t = 0, 1, – 2

\ x = 1, 3, 1/9 fi P = 1/3. 58. 2x2 – 1 > 0, x2 – 2/3 > 0, 2x2 – 1 π 1 Rewrite the equation as log3 ( x 2 – 2 / 3) 2

log3 (2 x – 1) fi

=2–

1 log3 (2 x 2 – 1)

log3 ( x 2 – 2 / 3) + 1 log3 (2 x 2 – 1)

=2



x 2 = 3/4 as x2 π 1



1 3 2 P = – 3/4 fi 4|P| = 3.

\

x=±

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5 Permutations and Combinations 5.1 FUNDAMENTAL PRINCIPLES OF COUNTING The Sum Rule

Suppose that A and B are two disjoint events (mutually exclusive); that is, they never occur together. Further suppose that A occurs in m ways and B in n ways. Then A or B can occur in m + n ways. This rule can also be applied to more than two mutually exclusive events. The Product Rule

Suppose that an event X can be decomposed into two stages, A and B. Let stage A occur in m ways and suppose that these stages are unrelated, in the sense that stage B occurs in n ways regardless of the outcome of stage A. Then event X occurs in mn ways. This rule is applicable even if event X can be decomposed in more than two stages.

We now introduce some terminology. If the order is important, then we say that our sample consists of arrangements. If the order is not important then we have selections. If repetition is allowed, then we are using replacement. An arrangement of r objects with replacement is called an r-sequence (or simply sequence). Vehicle numbers are examples of sequences. An r-permutation (or simply permutation) is an arrangement without replacement. The different orderings of a deck of cards are examples of permutations. A selection of r objects with replacement is called an r-multiset. The letters used to form the word MATHEMATICS can be thought of as a multiset with two Ms, two As, two Ts, one H, one E, one I, one C and one S. An r-combination (or simply combination) is a selection of r objects without replacement. In Table 5.1 we illustrate the four different problems and their solutions. We take two objects from {x1, x2, x3}. Table 5.1

5.2 PERMUTATIONS AND COMBINATIONS

Suppose we have a set of n distinct objects {x1, x2, , xn} from which we have to take a sample of r (1 £ r £ n) objects. In how many ways can this be done? However, phrase ‘take a sample’ is somewhat ambiguous. 1. Is the order in which we take the sample important? For example, if r x1 and then x2 to be different from picking x2 and then x1? 2. Do we allow repetition of objects? For example, do we allow a sample to consists of two x1? Since each question has two answers, the product rule says that our original problem is really made up of four separate problems. 1. How many such samples are there if the order is important and we allow repetitions? 2. How many such samples are there if the order is important and we do not allow repetitions? 3. How many such samples are there if the order is not important and we allow repetitions? 4. How many such samples are there if the order is not important and we do not allow repetitions?

Repetition x1 x1 x1 x2 x1 x3 x2 x1 x2 x2 r-sequence x2 x3 x3 x1 x3 x2 x3 x3 {x1, x1} {x1, x2} {x1, x3} {x2, x2} r-multiset {x2, x3} {x3, x3}

No repetition x1 x2 x1 x3 x2 x1 x2 x3 r-permutation x3 x1 x3 x2

{x1, x2} {x1, x3} {x2, x3}

r-combination

What happens when r = 0? We may view the r-combination as a subset of r elements, and Thus, a 0-combination may be viewed as the empty set. Since there is exactly one empty set, we may say that there is exactly one 0-combination. Mathematicians also say that there is exactly one 0-sequence, one 0-permutation, and

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terms sequence and permutation interchangeably and call a multiset, a combination. 5.3 SOME IMPORTANT RESULTS

1. The number of sequences (that is, permutations) of n distinct objects taken r at a time, when repetition of objects is allowed, is nr (r > 0). 2. The number of permutations of n distinct objects taken r (0 £ r £ n) at a time, when repetition is not allowed, is given by n! n Pr = (n - r )! Conventionally, nPr = 0 for r > n. 3. The number of permutations of n objects taken all together, when p1 of the objects are alike and of one kind, p2 of them are alike and of the second kind, , pr of them are alike and of the rth kind, where p1 + p2 + + pr = n is given by n! p1 ! p2 !

Remark 1. Note that even if one object is lying on the circle, then the circle has to be treated as a row, so far as arranging the objects around the circle is concerned. 2. Note that if no distinction is to be made between clockwise and counter-clockwise arrangements, then the number of arrangements is equal to (n – 1)!/2. 3. The number of ways of arranging r identical objects (n - 1)! . and (n – r) distinct objects along a circle is r!

Illustration 1 In how many ways can 8 ladies and 5 men can be seated around a round table so that no two men are together. Solution: 8 ladies can be arranged arround a round table in (8 – 1)! ways. After the ladies have been arranged, there are 8 places for men. We can place 5 men at these places in 8P5 ways. Thus, required number of ways is (7!) (8P5).

pr !

L

4. The number of combinations of n distinct objects taken r (0 £ r £ n) at a time is given by n! Ê nˆ n ÁË r ˜¯ = Cr = (n - r )! r !

L

5.

6.

7.

8. 9.

n

C0 < nC1 < nC2 … < nCm = nCm+1

Cm+1 > nCm+2 > nCm+3 > … > nCn 10. If n = 2m, then n C 0 < nC 1 < nC 2 … < nC m n Cm > nCm+1 > nCm+2 … > nCn n

5.4 CIRCULAR PERMUTATION

The number of ways of arranging n distinct objects around a circle is (n – 1)!.

L

M

M

M

M

L

L

n

Cr = 0 if r > n. The number of ways of selecting zero or more objects p2 identical obout of p1 jects of second kind, …, pr identical objects of rth kind is (p1 +1) (p2 +1) … (pr +1). The number of combinations of n distinct objects taken r (£ n) at a time, when k (0 £ k £ r) particular objects always occur, is n – kCr – k. The number of combinations of n distinct objects taken r at a time, when k (1 £ k £ n) never occur, is n–k C r. If n > 1, nP0 < nP1 < nP2 < . . . < nPn – 1 = nPn If n = 2m +1, then

M

M

L

M

M

L

L

5.5 SOME IMPORTANT IDENTITIES

(i) (ii) (iii) (iv) (v) (vi)

(vii) (viii) (ix)

n

C0 = 1 = nCn. n Cr = nCn – r (0 £ r £ n). n n n+1 Cr – 1 + Cr = Cr. n Cr = nCs implies r = s or r + s = n. n - r +1 n n Cr = Cr – 1 (1 £ r £ n). r If n is even, then the greatest value of nCr is nCm, where m = n/2. If n is odd, then the greatest value of n Cr is nCm, where m = (n – 1)/2 or (n + 1)/2. n C0 + nC1 + nC2 + + nCn = 2n. n n n C0 + C2 + = C1 + nC3 + = 2n – 1. 2n + 1 C0 + 2n + 1C1 + + 2n + 1Cn = 22n. 2n + 1 2n + 1 Cn + 1 + Cn + 2 + + 2n+1C2n + 1 = 2 2n.

5.6 DIVISION OF IDENTICAL OBJECTS

The number of ways of distributing n identical objects among r persons giving zero or more to each is equal to the

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number of ways of arranging n identical object of one kind and (r – 1) identical objects of second kind (separators) in a row (n + r - 1)! n + r – 1 = = Cr – 1 n!(r - 1)! Illustration 2 To distribute 10 identical toys among 3 children, we may use two separators as follows: 1st child

1st child

2nd child

3rd child

2nd child

1st child

2nd child

3rd child

3rd child

etc. The required number of ways

=

(10 + 2)! = 10! 2!

12

C2 = 66.

Alternative method

1. The number of distributing n identical objects among r persions giving zero or more to each xn in (1 – x) – r xn in (1 + x + x2 + . . .)r 2. The number of ways of distributing n identical object among r persons giving at least one to each. = the number of ways of distributing remaining (n – r) identical objects (after giving one to each of r persons) among r persons giving zero or more to each = (n – r) + (r – 1)Cr – 1 = n – 1Cr – 1

(iii). If wish to distribute n identical object among r persons so that kth person (1 £ k £ r) does not get more than ak, then the desired number of ways = Coeffcient of xn in (1 + x + . . . + xa1) (1 + x + . . . + x a2) . . . (1 + x + . . . + x ar) Illustration 4 Find the number of ways of distributing 10 identical toys among 3 children so as give ai (1 £ i £ 3) to the ith children, where 2 £ a1 £ 4, 3 £ a2 £ 5, 1 £ a3 £ 7. The required number of ways = coefficient of x10 in (x2 + x3 + x4) (x3 + x4 + x5) (x + . . . + x7) = coefficient of x10 in x6 (1 + x + x2)2 (1 + + x6) 2

Ê 1 - x3 ˆ Ê 1 - x 7 ˆ = coefficient of x in Á Ë 1 - x ˜¯ ÁË 1 - x ˜¯ 4

= coefficient of x4 in (1 – 2x3) (1 – x)– 3 [ignore any power that is more than 4] = coeffcient of x4 in (1 – 2x3) (1 + 3C1 x + 4C2 x2 + 5 C3 x3 + 6C4 x4 + . . .) =

6

C4 – 2(3C1) = 15 – 6 = 9

Finding Number Integral Solutions of a Linear Equation

The number of non-negative integral solution of x1 + x2 + . + xr = n, where n Œ N » {0}, and r Œ N = the number of ways of distributing n identical objects among r persons giving zero or more to each = n + r – 1C r – 1. 5.7 DERANGEMENT

Illustration 3 (i). To distribute 10 toys among 3 children giving at of the child and distribute the remaining 7 toys among 3 children giving zero or more to each in 7+3–1 C3 – 1 = 9C2 = 36 ways. (ii). However if we wish to give at least two to

to the three children as desired and distribute the remaining 4 toys among three children giving zero or more to each. This can be done in 4 + 3 – 1C3 – 1 = 6C2 = 15 ways.

Number of ways arranging n objects numbered 1, 2, 3, ... , n at n places numbered 1, 2, 3, ... , n so that no object goes to the place corresponding to its number is 1 1 1 1 n ! ÊÁ - + - + ( -1)n -1 ˆ˜ Ë 2! 3! 4! n !¯ 5.8 DISTRIBUTION INTO GROUPS

1. The number of ways in which n distinct objects can be split into three groups containing respectively r, s and t objects (where r, s and t are distinct and r + s + t = n) is

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( nC r) ( n – rC s) ( n – r – sC t) n! (n - r )! (n - r - s)! = r !(n - r )! (n - r - s)! s ! (n - r - s - t )! t !

The last integer amongst 1, 2, by p is

Èn ˘ È [ n / p] ˘ Í p ˙ p=Í 2˙ p Î ˚ Îp ˚

n! = r ! s! t !

2. Suppose mn distinct objects are to be divided into m groups, each containing n objects and the order of the groups is not important. Then the number of ways of doing this is given by

Èn˘ È n ˘ Ep(n!) = Í ˙ + Í 2 ˙ + E p Î p˚ Î p ˚

Ê ÁË 1◊ 2

È n ˘ˆ Í 2 ˙˜ Î p ˚¯

m

If, however, the order of groups is important, then the number of ways is given by

Èn˘ È n ˘ Ep(n!) = Í ˙ + Í 2 ˙ + Î p˚ Î p ˚

Èn˘ +Í s˙ Îp ˚

where ps £ n < ps + 1.

(m n)! (n !)

Since the remaining integers are not divisible by p, we get

Continuing in this way, we get

(m n)! m !(n !)

, [n/p] which is divisible

m

Illustration 5 5.9 SELECTION

1. The number of ways of selecting one or more objects out of n distinct objects is 2n – 1. 2. Suppose we have n distinct objects, and p like objects of one kind, q like objects of a second kind, r like objects of a third kind, etc. Then the number of ways of selecting one or more objects from these objects is – 1. 2n (p + 1) (q + 1) (r + 1) 5.10 EXPONENT OF PRIME p IN n!

Let Ep(m) denote the exponent of the prime p in the prime factorization of positive integer m. We have (n – 1)◊n) Ep(n!) = Ep (1◊2◊3◊4 The last integer amongst 1, 2, 3, , (n – 1), n which is divisible by p is [n/p]p, where [x] denotes greatest integer £ x. Therefore, Ê Ep(n!) = Ep Á p ◊ 2 p ◊ 3 p Ë

Èn˘ ˆ Í p ˙ p˜¯ Î ˚

because the remaining integers from the set {1, 2, 3, (n – 1), n} are not divisible by p. Ê Ep(n!) = Ep Á p ◊ 2 p ◊ 3 p Ë Èn˘ Ê = Í ˙ + E p Á 1◊ 2 ◊ 3 Ë p Î ˚

Èn˘ ˆ Í p ˙ p˜¯ Î ˚ È n ˘ˆ Í p ˙˜¯ Î ˚

,

Find exponent of 3 in 32! As 3 divides 3, 6, 9 , 30 amongst 1, 2, . . . 31, 32, E3(32!) = E3(3 ¥ 6 ¥ 9 ¥ . . . ¥ 30) = E3(310 ¥ 1 ¥ 2 ¥ . . . ¥ 10) = 10 + E3(3 ¥ 6 ¥ 9) [Ignore 1, 2, 4, 5, 7, 8, 10 as 3 does not divide these.] = 10 + E3(33 ¥ 1 ¥ 2 ¥ 3) = 10 + 3 + E3(3) = 10 + 3 + 1 = 14. È 31 ˘ Note, 10 = Í ˙ , 3 = Î3˚

È 31 ˘ and 1 = ÍÎ 32 ˙˚

È 31 ˘ ÍÎ 33 ˙˚

Some Tips

1. Number of diagonal of a polygon of n sides is 1 n C2 – n = n(n – 3). 2 2. Number of circular permutations of n things taken r 1 at a time = (nPr). r 3. The number of permutations of n(>1) distinct objects taken r at a time when repetition of objects is n(n r - 1) allowed is n + n2 + n3 + nr = . n -1 4. Sum of the numbers formed by taking all the given n digits (excluding 0) when repetition of digits is not allowed.

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((n - 1)!) ¥ (11...1) ¥ (sum of all the digits)  n times

For, instance, let us sum of all the numbers formed by permuting 4 digits 1, 2, 3, 4. Let k Œ{1, 2, 3, 4} k occurs at units place 3! times k occurs at ten place 3! times k occurs at hundreds place at 3! times k occurs at thousands place at 3! times Thus, sum of the numbers due to k is 3!(1 + 10 + 100 + 1000)k Thus, sum of the 4! numbers is 3!(1 + 10 + 100 + 1000) (1 + 2 + 3 + 4) = 6(111) (24). 5. Sum of all the numbers formed by taking all the given n digits (including 0) is 11....1) [((n – 1)!)¥ (    n times

-((n - 2)!) ¥ (11....1) ] (sum of all the n digits)     ( n -1) times

6. Let there be n points no three of which are collinear except k points which lie on a straight line (i) Number of straight lines formed by these points is nC2 – kC2 + 1 (ii) Number of triangles formed by these points is nC3 – kC3. 7. n straight lines drawn in the plane such that no two of them are parallel and no three of them are concur1 rent divide the plane into n + 1C2 + 1 = (n2 + n + 2) 2 different regions. SOLVED EXAMPLES

For x = 2, LHS of (1) = 8 and RHS of (1) = 4 For x ≥ 3, LHS of (1) is even and RHS of (1) is odd. Thus, x = 1 is the only solution of the given equation. Let n = 2019. The least positive integer k Example 2 for which k(n2) (n2 – 12) (n2 – 22) (n2 – 32) … (n2 – (n – 1)2) = r! for some positive integer r is (a) 2018 (b) 2019 (c) 1 (d) 2 Ans. (d) Solution: We can rewrite the given expression as k(n2) (n – 1) (n + 1) (n – 2) (n + 2) (n – 3) (n + 3)… (n + n – 1) (n – n + 1) = r! fi kn (1) (2)… (n – 1) n(n + 1) (n + 2)… (2n – 1) = r! fi kn (2n – 1)! = r! \ To convert L.H.S. to a factorial, we require, k = 2 which will convert it into (2n)! If 0 < r < s £ n and nPr = nPs, then value of Example 3 r + s is (a) 2n – 2 (b) 2n – 1 (c) 2 (d) 1 Ans. (b) Solution:

n

P r = nP s fi

n! n! = (n - r )! (n - s )!

fi (n – r)! = (n – s)! As r < s, n – r > n – s. But the only two different factorials which are equal are 0! and 1!. Thus, n – r = 1 and n – s = 0 fi r = n – 1 and s = n. fi r + s = 2n – 1. Example 4 value of x is (a) – 7

If E =

1 2 3 4 . . . 4 6 8 10

(b) – 9

30 31 . = 8x, then 62 64

(c) – 10

(d) – 12

Ans. (d)

SINGLE CORRECT ANSWER TYPE QUESTIONS Example 1 2x

(x! + x)

We have E=

Number of solutions of

= (x! + x) , x ŒN is

Ans. (b) Solution: 2xxxx! = = x! fi 2x = For x

Solution:

x

2 (32!)

=

1 31

2 (32)

=

1 236

= 2–36 = (23)–12 = 8–12 Thus, x = –12. Example 5

For x ΠN, we can write the given equation as [x(1 + (x Р1)!)]x xx[1 + (x Р1)!]x [1 + (x Р1)]x (1)

31! 31

m Ê10ˆ Ê 20 ˆ Ê pˆ (where Á ˜ = 0 The sum  Á ˜ Á ˜ Ë q¯ i =0 Ë i ¯ Ë m - i¯

if (p < q) is maximum where m is (a) 5 (b) 10 (c) 15 Ans. (c)

(d) 20

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Solution:

Ans. (c)

m Ê10ˆ Ê 20 ˆ We have  Á ˜ Á ˜ i =0 Ë i ¯ Ë m - i¯

= the number of ways of choosing m persons out of 10 men and 20 women = the number of ways of choosing m out of 30 persons = 30Cm But 30Cm is maximum for m = 15.

Solution: A rational number of the desired category is of the form 2018. x1 x2 … xk where 1 £ k £ 9 and 9 ≥ x1 > x2 > … > xk ≥ 1. We can choose k digits out 9 in 9Ck ways and arrange them in decreasing order in just one way. Thus, the desired number of rational numbers is 9 C1 + 9C2 + …+ 9C9 = 29–1.

Let Tn denote the number of triangles Example 6 which can be formed by using the vertices of a regular polygon of n sides. If Tn +1 – Tn = 21, then n equals (a) 5 (b) 7 (c) 6 (d) 4

The number of functions f from the set A Example 9 = {0, 1, 2} in to the set B = {0, 1, 2, 3, 4, 5, 6, 7} such that f (i) £ f (j) for i < j and i, j ∈ A is (a) 8C3 (b) 8C3 + 2(8C2) (c) 10C3 (d) 9C3

Ans. (b)

Ans. (c)

Solution: The number of triangles that can be formed by using the vertices of a regular polygon is nC3. That is, Tn = nC 3 Now, Tn +1 – Tn = 21 fi n +1C3 – nC3 = 21

Solution: A function f : A Æ B such that f (0) £ f (1) £ f (2) falls in one of the following four categories. Case 1 f (0) < f (1) < f (2) There are 8C3 functions in this category. Case 2 f (0) = f (1) < f (2) There are 8C2 functions in this category. Case 3 f (0) < f (1) = f (2) There are 8C2 functions in this category. Case 4 f (0) = f (1) = f (2) There are 8C1 functions in this category. Thus, the number of desired functions is

fi nC2 + nC3 – nC3 = 21

[

n+1

Cr= nCr–1 + nCr]

1 n(n – 1) = 21 fi n = – 6 or 7. 2 As n is a positive integer, n = 7.



An eight digits number divisible by 9 is Example 7 to be formed by using 8 digits out of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 without replacement. The number of ways in which this can be done is (a) 9! (b) 2(7!) (c) 4(7!) (d) (36) (7!) Ans. (d) Solution: We have 0 + 1 + 2 + 3 … + 8 + 9 = 45 To obtain an eight digits number exactly divisible by 9, we must not use either (0, 9) or (1, 8) or (2, 7) or (3, 6) or (4, 5). [Sum of the remaining eight digits is 36 which is exactly divisible by 9]. When, we do not use (0, 9), then the number of required 8 digit number is 8!. When one of (1, 8) or (2, 7) or (3, 6) or (4, 5) is not used, the remaining digits can be arranged in 8! – 7! ways as 0 cannot be at extreme left. Hence, there are 8! + 4(8! – 7!) = (36) (7!) numbers in the desired category. The number of rational numbers lying in Example 8 the interval (2018, 2019) all whose digits after the decimal point are non-zero and are in decreasing order is 9

9 (a) Â Pi i =1

10

9 (b) Â Pi i =1

(c) 29–1

(d) 210–1

8

C 3 + 8C 2 + 8C 2 + 8C 1 = 9C 3 + 9C 2 =

10

C 3.

Example 10 The number of positive integral solutions of the equation x1 x2 x3 x4 x5 = 1050 is (a) 1800 (b) 1675 (c) 1400 (d) 1875 Ans. (d) Solution: Using prime-factorization of 1050, we can write the given equation as x1 x2 x3 x4 x5 = 2 × 3 × 52 × 7 We can assign 2, 3 or 7 to any of 5 variables. We can assign entire 52 to just one variable in 5 ways or can assign 52 = 5 × 5 to two variables in 5C2 ways. Thus, 52 can be assigned in 5 C1 + 5C2 = 5 + 10 = 15 ways Hence, required number of solutions = 5 × 5 × 5 × 15 = 1875. Example 11 (a) 0 (c) 3 Ans. (a)

The exponent of 7 in 100C50 is (b) 2 (d) 4

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Solution:

We have 100C50 =

100! . 50!50!

(a) 32400 Ans. (a)

(b) 16200 (c) 14000 (d) 18000

The exponent of 7 in 50! is È 50 ˘ È 50 ˘ Í 7 ˙ + Í 7 2 ˙ = 7 + 1 = 8, Î ˚ Î ˚ and the exponent of 7 in 100! is È100 ˘ È100 ˘ Í 7 ˙ + Í 7 2 ˙ = 14 + 2 = 16 Î ˚ Î ˚ Thus, exponent of 7 in 100C50 is 16 – 2(8) = 0. Example 12 In a certain test there are n questions. In this test 2k students gave wrong answers to at least (n – k) questions, where k = 0, 1, 2, …, n. If the total number of wrong answers is 4095, then value of n is (a) 11 (b) 12 (c) 13 (d) 15 Ans. (b) Solution: The number of students answering at least r questions incorrectly is 2n– r. \ the number of students answering exactly r (1 £ r £ n–1) questions incorrectly is 2n–r – 2n – (r +1). Also, the number of students answering all questions wrongly is 20 = 1. Thus, the total number of wrong answers is 1(2n–1 – 2n –2) + 2(2n – 2 – 2n –3) + 3(2n – 3 – 2n– 4) + … + (n – 1) (21 – 20) + n (20)

Solution: There are 12 horizontal and 12 vertical lines on a 11 ¥ 11 grid. We can choose the middle rectangle in (10C2) ¥ (10C2) ways, and the remaining four rectangles by choosing exactly one of the two dotted lines from the four corners. This can be done in 24 ways. Thus, required number of ways = (45)2 (16) = 32400. n

n

r =1

s =1

For n ≥ 2, let a =  r ! s ! s log10 ( s )

Example 15 n

n

r =1

s =1

and b =  r ! r  s !log10 ( s ) , then (a) a = b Ans. (c) Solution: a – b

(b) ab = 1 (c) a > b (d) a < b

n

n

n

n

r =1

s =1

r =1

s =1

=  r ! s ! s log10 ( s ) -  r ! r  s !log10 ( s ) n

= Â r !r ! r log10 (r ) + Â r ! s ! s log10 ( s ) r =1

= 2n–1 + 2n–2 + … + 20 = 2n – 1.

s >r

+ Â r ! s ! s log10 ( s ) s 1 and n divides (n – 1)! + 1, then (a) n must be prime (b) n must be divisible by exactly two primes (c) n must be a composite number (d) n must be divisible by p2 wher p is primes Ans. (a)

- Â r !r ! r log10 (r ) - Â r ! rs !log10 ( s ) r =1

- Â r ! rs !log10 ( s ) s r

Solution: If n is not prime, then there exists r ∈ N such that 2 £ r £ n – 1 and r|n. As r|n and n|[(n – 1)! + 1], we get r|[(n – 1)! + 1] As 2 £ r £ n – 1, r|(n – 1)!, therefore r|1. A contradiction. Example 14 A 11 ¥ 11 grid is divided into five rectangles along the grid. The number of ways to divide the grid into five rectangles so that one of the rectangles generated by the division doesn’t have any edge on the periphery is

s >r

+ Â r ! s !( s - r ) log10 ( s ) s r

+ Â r ! s !( r - s ) log10 (r ) s >r

= Â r ! s !( s - r ) (log10(s) – log10(r)) s >r

For s > r,

s > r and log10(s) > log10(r) \a–b>0fia>b

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Example 16

2 Sum of the series  (r + 1) (r !) is

2bn = Â

r =1

(a) (n + 1)! (c) n(n + 1)!

(b) (n + 2)! – 1 (d) n(n + 2)!

Ans. (c) Solution:

n



r =0



bn =

We can write r2 + 1= (r + 2) (r + 1) – 3(r + 1) + 2

n

(a)

r =1

n

= Â [(r + 2) (r + 1) - (r + 1) - 2 {(r + 1) - 1}] r ! r =1 n

n

r =1

r =1

= Â [(r + 2)! - (r + 1)!] - 2 Â {(r + 1)!- r !} = (n + 2)! – 2! – 2{(n + 1)! – 1} = n (n + 1)! Example 17 In a group of 8 girls, two girls are sisters. The number of ways in which the girls can sit so that two sisters are not sitting together is (a) 48200 (b) 14100 (c) 28300 (d) 30240 Ans. (d) Solution:

The required number of ways = the number of ways in which 8 girls can sit – the number of ways in which two sisters are together = 8! – (2) (7!) = 30240.

Example 18 The number of words that can be formed by using the letters of the word MATHEMATICS that start as well as end with T is (a) 80720 (b) 90720 (c) 20860 (d) 37528 Ans. (b) Solution: The word MATHEMATICS contains 11 letters viz. M, M, A, A, T, T, H, E, I, C, S. The number of words that begin with T and end with T is 9! = 90720. 2! 2! n

Example 19 (a)

n an 2

n

r equals If an =  n , then  n r = 0 Cr r = 0 Cr (b)

n an 4

1

(c) nan

(d) (n–1)an

Ans. (a) n

Solution:

Let bn = Â

r =0

r n

Cr

n



r =0

n-r n

Cn - r

n



r =0

n-r n

n

Cr

n

= nÂ

r =0

1 n

Cr

= nan

n an . 2

Example 20 m men and w women are to be seated in a row so that no two women sit together. If m > w, then the number of ways in which they can be seated is

 (r 2 + 1) (r !)

Thus,

r + (n - r )

Cr

(c)

m ! (m + 1)! (m - w + 1)! m+w

(b)

Cm (m – w)!

m

Cm–w (m – w)!

(d) none of these

Ans. (a) Solution: m men. This can be done in m! ways. After m men have taken their seats, the women must choose w seats out of (m + 1) seats marked with X below. X M X M X M X …X M X 1st 2nd 3rd mth They can choose w seats in m+1Cw ways and take their seats in w! ways. Thus, the required number of arrangements is m! (m +1Cw) (w!) =

m ! (m + 1)! w! m ! (m + 1)! = w! (m + 1 - w)! (m + 1 - w)!

Example 21 The number of subsets of the set A = {a1, a2, …, an} which contain even number of elements is (b) 2n –1 (c) 2n –2 (d) 2n (a) 2n–1 Ans. (a) Solution: n – 1) elements a1, a2, …, an–1 we have two choices: either ai (1 £ i £ n–1) lies in the subset or ai doesn’t lie in the subset. For the last element we have just one choice. If even number of elements have already been taken, we do not include an in the subset, otherwise (when odd number of elements have been added), we include it in the subset. Thus, the number of subsets of A = {a1, a2,…, an} which contain even number of elements is equal to 2n–1. Example 22 The number of ways in which we can post 5 letters in 10 letter boxes is (a) 50 (b) 510 (c) 105 (d) none of these Ans. (c) Solution: second letter in 10 ways and so on. Thus, the number of ways of posting 5 letters in 10 letter boxes is 10 ¥ 10 ¥ 10 ¥ 10 ¥ 10 = 105.

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Example 23 numbers having at least one of their digits repeated is (a) 90000 (b) 100000 (c) 30240 (d) 69760 Ans. (d) Solution: which can be formed using the digits 0, 1, 2, , 9 is 105 numbers which have none of their digits repeated is 10P5 = 30240. Thus, the required number of telephone numbers is 105 – 30240 = 69760. Example 24 (a) 1

The ten’s digit of 1! + 2! + 3! + + 49! is (b) 2 (c) 3 (d) 4

Ans. (a) Solution: We have 1! + 2! + 3! + 4! = 33. Also 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 326880. Thus, the ten digits of 1! + 2! + + 9! is 1. Also, note that n! is divisible by 100 for all n ≥ 10, so that ten’s digit of 10! + 11! + + 49! is zero. Therefore, ten’s digit of 1! + 2! + + 49! is 1. Example 25 Four dice are rolled. The number of possible outcomes in which at least one die shows 2 is (a) 1296 (b) 625 (c) 671 (d) 1023 Ans. (c) Solution: The total number of possible outcomes is 64. The number of possible outcomes in which 2 does not appear on any die is 54. Therefore, the number of possible outcomes in which at least one die shows a 2 is 64 – 54 = 1296 – 625 = 671. Example 26 A set contains (2n + 1) elements. The number of subsets of the set which contain at most n elements is (a) 2n (b) 2n + 1 (c) 2n – 1 (d) 22n Ans. (d) Solution: Let N = the number of subsets of the set which contain at most n elements + 2n + 1Cn = 2n + 1C0 + 2n + 1C1 + 2n + 1C2 + We have + 2n + 1Cn) 2N = 2(2n + 1C0 + 2n + 1C1 + 2n + 1C2 + 2n + 1 2n + 1 2n + 1 C0 + C2n + 1) + ( C1 + 2n + 1C2n) =( 2n + 1 Cn + 2n + 1Cn + 1) + +( ( nC r = nC n – r) 2n + 1 2n + 1 2n + 1 C0 + C1 + C2 + + 2n + 1C2n + 1 = 2n + 1 =2 fi N = 22n. Example 27 A is a set containing n elements. A subset P of A is chosen. The set A is reconstructed by replacing

the elements of P. A subset Q of A is again chosen. The number of ways of choosing P and Q so that P « Q = f is (c) 2n – 1 (d) 3n (a) 22n – 2nCn (b) 2n Ans (d) Solution: Let A = {a1, a2, , an}. For ai ∈ A, we have the following choices: (i) ai ∈ P and ai ∈ Q (ii) ai ∈ P and ai ∉Q (iii) ai ∉ P and ai ∈ Q (iv) ai ∉ P and ai ∉ Q Out of these only (ii), (iii) and (iv) imply ai ∉ P « Q. Therefore, the number of required subsets is 3n. Example 28

Let E = {1, 2, 3, 4} and F = {a, b}. Then

the number of onto functions from E to F is (a) 14 (b) 16 (c) 12 (d) 32 Ans. (a) Solution: E to F. We can assign 1 to one of the two values a or b. Similarly, we can assign 2, 3 and 4 to one of the two values a or b. Thus, there are 2 ¥ 2 ¥ 2 ¥ 2 = 16 functions from E to F. Out of these functions the two constant functions: f (x) = a " x ∈ E and g(x) = b " x ∈ E are not onto functions. \ there are 16 – 2 = 14 onto functions from E to F. Example 29

The number of ways of arranging letters of

the word HAVANA so that V and N do not appear together is (a) 40 (b) 60 (c) 80 (d) 100 Ans. (c) Solution: We can arrange the letters H, A, A, A in 4! = 4 ways. 3! If one possible arrangement is XXXX Then we can arrange V, N at any of the two places marked with O in the following arrangement. OXOXOXOXO This, we can arrange V and N in = 5P2 = 20 ways Thus, the number of ways in which letters can be arranged is 4 ¥ 20 = 80. Example 30 The number of integral points (integral point means both the coordinates should be integer) that lie exactly in the interior of the triangle with vertices. (0, 0), (0, 21), (21,0) is (a) 133 (b) 190 (c) 233 (d) 105 Ans. (b)

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Solution: The integral points lying on the line x = 1 which also lie in the interior of DOAB are (1, 1), (1, 2), , (1, 19) \ there are 19 points lying on the line x = 1. B (0, 21)

A (21, 0)

(0, 0)

Example 31 The number of ways of arranging letters of the word RACHIT so that the vowels are in alphabetical order is (a) 120 (b) 240 (c) 360 (d) 480 Ans. (c) Solution: We can arrange letters of the word RACHIT in 6! ways. Out of these exactly half, that is, there are 1 (6!) = 360 ways to arrange the letters of the word 2 RACHIT so that vowels are in alphabetical order.

(a) [ - 3,

C r = (K 2 – 3) ( nC r

3]

(c) (2, •)

3 or

1 > 0] n

3 n! + 1 as n≥ 2. =

Example 52 A class has 30 students. The following prizes are to be awarded to the students of this class. First and second in Mathematics; first and second in Physics first in Chemistry and first in Biology. If N denote the number of ways in which this can be done, then (a) 400 ÍN (b) 600 ÍN (c) 8100 ÍN (d) N is divisible by four distinct prime numbers. Ans. (a), (b), (c), (d) Solution: First and second prizes in Mathematics (Physics) can be awarded in 30P2 (30P2) ways. First prize

in Chemistry (Biology) can be awarded in 30 (30) ways. Therefore, N = (30P2)2 (302) = 304292 = 24 ◊ 34 ◊ 54 ◊ 292. Since 400= 24 ◊ 52, 600 = 23 ◊ 3 ◊ 52 and 8100 = 22 ◊ 34 ◊ 52 we get N is divisible by each of 400, 600 and 8100. Also, N is divisible by four distinct prizes, viz. 2, 3, 5 and 29. A letter lock consists of three rings Example 53 marked with 15 different letters. If N denotes the number of ways in which it is possible to make unsuccessful attempts to open the lock, then (a) 482 ÍN (b) N is product of 2 distinct prime numbers (c) N is product of 3 distinct prime numbers (d) N is product of 4 distinct primes Ans. (a),(c) Solution: Since each ring has 15 positions, the total number of attempts that can be made to open the lock is 153. Out of these, there is just one attempt in which the lock will open. Therefore, N = 153 – 1 = (15 – 1) (152 + 15 + 1) = 2 ¥ 7 ¥ 241 Clearly, 482 ÍN and N is product of three distinct prime numbers. Example 54 2nPn is equal to (a) (n + 1) (n + 2) (2n) (2n – 1)] (b) 2n [1 ◊ 3 ◊ 5 (c) (2) ◊ (6) ◊ (10) (4n – 2) 2n (d) n! ( Cn) Ans. (a), (b), (c), (d) Solution:

We have (2 n)! (2n)! 2n = Pn = (2n - n)! n! = =

(2n) (2n - 1) (2n - 2) (2n - 3) n! 2n [n(n - 1) (n - 2)

= 2n [1 ◊ 3 ◊ 5

2 ◊1] [1◊ 3◊ 5 n!

4 ◊ 3◊ 2 ◊1

(1)

(2n - 1)]

(2n – 1)]

= 2◊6◊10 (4n – 6) (4n – 2) Also, from (1) 2n Pn = (2n) (2n – 1) (n + 2) (n + 1) (2n)! (2n)! 2n Lastly, n! (2nCn) = (n!) = = P n. n! n! n! Example 55 If n objects are arranged in a row, then the number of ways of selecting three of these objects so that no two of them are next to each other is

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1 (n – 2) (n – 3) (n – 4) 6 (b) n – 2C3 (c) n – 3C3 + n – 3C2 (d) n – 3C2 (a)

Ans. (a), (b), (c)

ways of choosing the requisite triplets. Alternatively, triplets with x = y < z, x < y < z, y < x < z can be chosen in n + 1C2, n + 1C3, n + 1C3 ways. \ there are n + 1C2 + 2(n + 1C3) = n +2C3 + n + 1C3 = 2(n + 2C3) – Example 58

Solution:

Let x0 be the number of objects to the left of the x1 the number of objects between the x2 the number of objects between the second and the third and x3 the number of objects to the right of the third object. We have x0, x3 ≥ 0, x1, x2 ≥ 1 and x0 + x1 + x2 + x3 = n – 3 (1) Put x1 = y1 + 1 and x2 = y2 + 1. Then (1) reads as (2) x0 + y1 + y2 + x3 = n – 5 where x 0, y 1, y 2, x 3 ≥ 0 The number of non-negative integral solution of (2) is n–5+3

C3 = n – 2C3

We have 1 C3 = (n – 2) (n – 3) (n – 4) 6

n–2

Also,

(a)

n+1

C3 +

(c) 12 + 22 +

n+2

C3 + n2

1 n (n + 1) (2n + 1) 6 (d) 2(n + 2C3) – n + 1C2

(b)

Ans. (a), (b), (c), (d) Solution: When z = n + 1, we can choose x, y from {1, 2, , n}. Thus, when z = n + 1, x, y can be chosen in n2. When z = n, x, y can be chosen in (n – 1)2 ways and so on. Thus, there are 1 + 12 = n (n + 1) (2n + 1) n2 + (n – 1)2 + 6

C2 ways.

If 100! = 2 3 5 7d

g

, then

1 (a – 1) 2 1 (d) d = b 3 (b) b =

1 b 2 Ans. (a), (b), (c), (d)

Solution: We have È100 ˘ È100 ˘ È100 ˘ È100 ˘ È100 ˘ È100 ˘ a= Í ˙+Í ˙+Í ˙+Í ˙+Í ˙+Í ˙ Î 2 ˚ Î 4 ˚ Î 8 ˚ Î 16 ˚ Î 32 ˚ Î 64 ˚ = 97 È100 ˘ È100 ˘ È100 ˘ È100 ˘ b= Í ˙+Í ˙+Í ˙+Í ˙ Î 3 ˚ Î 9 ˚ Î 27 ˚ Î 81 ˚ = 33 + 11 + 3 + 1 = 48 =

C3 + n – 3C2 = n – 2C3

Solution: Let x5 be such that x1 + x2 + x3 + x4 + x5 = n. We now seek the non-negative integral solutions of x1 + x2 + x3 + x4 + x5 = n. The number of required solutions = n + 4C n = n + 4C 4 Example 57 The number of ways of choosing triplets (x, y, z) such that z ≥ max {x, y} and x, y, z ∈ {1, 2, , n, n + 1} is

b

(c) g =

n–3

Ans. (b), (d)

a

(a) a = 97

Similarly,

The number of non-negative integral Example 56 solutions of x1 + x2 + x3 + x4 £ n (where n is a positive integer) is (c) n + 5C5 (d) n + 4Cn (a) n + 3C3 (b) n+4C4

n+1

g = 24 =

Example 59

1 (a – 1) 2

1 1 b and d = 16 = b. 2 3

The number of triangles that can be

formed with the sides of lengths a, b and c where a, b, c are integers such that a £ b £ c is 1 (c + 1)2 when c is odd (a) 4 1 c (c + 1) when c is odd (b) 2 1 c (c + 2) when c is even (c) 4 1 2 c when c is even (d) 4 Ans. (a), (c) Solution:

Let c = 2m + 1 where m is a positive integer.

1 1 c=m+ . 2 2 Thus, b can take values from m + 1 to 2m + 1. If b = 2m + 1, a can be 1, 2, , (2m + 1). Thus, there are (2m + 1) values. If b = 2m, a can be 2, 3, 2m. In this case a can take (2m – 1) values. And so on. When b = m + 1, a can take just one value viz. (m + 1). Thus, there are Since c < a + b and a £ b, we get c < 2b fi b >

(2m + 1) + (2m – 1) +

+ 1 = (m + 1)2

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1 (c + 1)2 4 such triangles. Next, let c = 2m When b = 2m, a can be 1, 2, 2m When b = 2m – 1, a can be 2, 3, 2m – 1 When b = 2m – 2, a can be 3, 4, , 2m – 2 And so on. When b = m + 1, a can be m, m + 1. For b = m and any value £ m, for a will not work. Thus, required number of triangles is 2m + (2m – 2) + +2 c 1 Ê ˆ Êc ˆ È ˘ = 2 Í m (m + 1) ˙ = Á ˜ Á + 1˜ Ë 2¯ Ë 2 ¯ Î2 ˚ =

Example 60 equal to

k

If k is odd then Cr is maximum for r

1 (k – 1) 2 (c) k – 1

1 (k + 1) 2 (d) k

(b)

(a)

Ans. (a), (b) Solution: Let k = 2n + 1, then 2n +1Cr is maximum when r = n. Also 2n + 1Cn = 2n + 1Cn + 1. Thus, kCr is maximum 1 1 when r = (k – 1) or r = (k + 1). 2 2 Example 61

Let

È1 1 ˘ È1 2 ˘ E = Í + ˙ + Í + ˙ + … upto 50 terms, then Î 3 50 ˚ Î 3 50 ˚ (a) (b) (c) (d)

E E E E

is divisible by exactly 2 primes is prime ≥ 30 £ 35

Ans. (b), (d) Solution:

For 1 £ x £ 33,

1 1 x < + < 1, 3 3 50

È1 x ˘ ÍÎ 3 + 50 ˙˚ = 0 for 1 £ x £ 33 and for 34 £ x £ 50, 1 x 4 < 1< + 3 50 3



È1 x ˘ fi Í + ˙ = 1 for 34 £ x £ 50 Î 3 50 ˚ \

E = 17

Ans. (a), (d) Solution: n! N = (n + 1) (n + 2) (2n) Since n < p < 2n, p|(n + 1) (2n) fi p|n! N But as p > n, p n! fi p|N. If fi As

p2|N, then p2|(n + 1) (2n) p|(n + 1) (p - 1) (p + 1) (2n) (p - 1) (p + 1) n < p < 2n, p (n + 1)

we get p

1 c (c + 2) 4

=

Example 62 If n < p < 2n and p is prime and N = 2nCn, then N (a) p|N (b) p 2 2 (d) p N (c) p |N

2

(2n),

N

Example 63 A is a set containing n elements. A subset P1 of A is chosen. The set A is reconstructed by replacing the elements of P1. Next a subset P2 of A is chosen and again the set is reconstructed by replacing the elements Pm of A are of P2. In this way m(> 1) subsets P1, P2, chosen. The number of ways of choosing P1, P2, Pm so that Pi « Pj = f for i π j, is (a) (m + 1)n (b) 2m - mCn n

(c) Â

k =0

m

Ck mk

(d) (2m - 1)n

Ans. (a), (c) Solution: Let the set A be {a1, a2, , an each ak can belong to at most one of the subsets, for if it belongs to two of them, say Pr and Ps for r π s, then ak ∈ Pr « Ps, implying Pr « Ps π f. Thus, for each ak we have m + 1 choices: it does not belong to any subset; it belongs to P1; it belongs to P2; ; or it belongs to Pm. For the n elements, therefore, we have (m + 1)n choices in this case. Pm as in Example 63, Example 64 With P1, P2, the number of ways of choosing P1, Pm so that P1 « P2 « « Pm = f is (a) (2m - 1)n (b) (1 + 2 + + 2m-1)n (c) (m + 1)n

(d) 2m (mCn)

Ans. (a), (b) Solution: For each ai (1 £ i £ n) we have either ai ∈ Pj or ai ∉ Pj (1 £ j £ m). That is, there are 2m choices in which ai (1 £ i £ n) may belong to the Pj’s. Out of these, there is only one choice, viz., ai ∈ Pj for j = 1, 2, , m, « Pm to be f. which is not favourable for P1 « P2 « Thus, ai ∉ P1 « P2 « « Pm in 2m - 1 ways. Since there are n elements in the set A, the total number of choices is + 2m-1. (2m - 1)n. Also, 2m - 1 = 1 + 2 +

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Example 65 If N is the number of positive integral solutions of x1 x2 x3 x4 = 770, then (a) N is divisible by 4 distinct primes (b) N is a perfect square (c) N is a perfect fourth power (d) N is a perfect 8th power Ans. (b), (c), (d) Solution: 770 = (2) (5) (7) (11). We can assign each of 2, 5, 7, 1 in 4 ways. Thus, required number of solutions is 44 = 256 = 162 = 28. Example 66 The number of ways in which we can arrange the 2n letters x1, x2, , xn, y1, y2, , yn in a line x and those of the letters y are respectively in ascending order of magnitude is + ( nC n) 2 (a) (nC0)2 + (nC1)2 + (b) 2nCn (c) 2n [1.3.5 (2n - 1)]/n! (d) 2nCn - 1

(3n + 2) (3n + 1) (n + 1) n -3 2 2 1 = [9n2 + 9n + 2 - 3n2 - 3n] 2 1 (6n2 + 6n + 2) = 3n2 + 3n + 1 = 2 fi N - 1 = 3n2 + 3n = 3n (n + 1). Clearly 3|(N - 1), n|(N - 1), (n + 1)|(N - 1) and 3n(n + 1)|(N - 1) =

Example 68

2n

and

Cn =

( 2n ) ! n !n !

=

[2 ◊ 4 ◊ 6

( 2n ) ] ÈÎ1 ◊ 3 ◊ 5

=

2n ( n !) ÈÎ1 ◊ 3 ◊ 5

Then (a) 2n E is divisible by 4nC2n (b) 2n E is divisible by n!

n !n !

=

2 ÎÈ1 ◊ 3 ◊ 5 ( 2n - 1) ˚˘ . n!

Example 67 Let N denote the number of ways in which 3n letters can be selected from 2n A¢s, 2n B¢s and 2n C¢s. Then (a) 3|(N - 1) (b) n|(N - 1) (c) (n + 1)|(N - 1) (d) 3n(n + 1)|(N - 1) Ans. (a), (b), (c), (d) Solution: have \

N

Let x1 A’s, x2 B’s and x3 C’s be selected. We

2n E is a positive integer n!

(d)

2n E is not an integer (4 n)!

Solution: We have E = (2n + 1) (2n + 3) (2n + 5) (4n - 3) (4n - 1) ( 2n ) !( 2n + 1)( 2n + 2)( 2n + 3)( 2n + 4) º ( 4n - 1) ( 4n ) = ( 2n ) !( 2n + 2)( 2n + 4) º ( 4n ) =

n !n !

n

(c)

Ans. (a), (b), (c), (d)

( 2n - 1) ˘˚

( 2n - 1) ˘˚

( 4n ) !n ! ( 2n ) !2n n !( n + 1)( n + 2) º ( 2n )

=

1 2n



( 4n ) !n ! ( 2n ) ! ( 2n ) !

fi 2n E = (4nC2n) (n!) This shows that 2n E is divisible by 4nC2n and also by n!. 2n E = 4nC2n, a positive integer. n! Also, note that fi

2n E n! = is not an integer as n > 1. (2 n)! (2 n)! (4 n)! Example 69 Let N denote the number of ways in which n boys can be arranged in a line so that 3 particular boys are separated. Then (a) 3!|N (b) (n - 2)!|N (c)

x1 + x2 + x3 = 3n +t2n )3 t3n in (1 + t + t2 + 3n 2n + 1 3 t in (1 - t ) (1 - t)-3 t3n in (1 - 3t2n + 1) (1 +3C1 t 4 2 5 + C2 t + C3 t3 + ) = 3n + 2C3n - 3 ( n + 1Cn - 1)

(4n - 3) (4n - 1)

E = (2n + 1) (2n + 3) (2n + 5)

Ans. (a), (b), (c) Solution: Out of 2n places we just have to choose n places for x’s and at the remaining places we arrange y’s. This can be done in 2nCn ways. Also ( nC 0) 2 + ( nC 1) 2 + + (nCn)2 = 2nCn

For n > 1, let

n-2

C3|N

(d) (n - 3)2 (n - 4)2|N

Ans (a), (b), (c), (d) Solution n - 3) remaining boys. This can be done in (n - 3)! ways. Then three particular boys must now be arranged at (n - 2) places marked with box . ¥ ¥ ¥ ¥ (X denote one of the (n - 3) boys)

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N = (n - 3)! (n - 2C3) (3!) = (n - 3)!

\

(n - 2)! (n - 5)!

= (n - 3) (n - 4) (n - 2)! Thus, 3!|N, (n - 2)!|N, n - 2C3|N and (n - 3)2 (n - 4)2|N. The number of permutations of the set

Example 70

consecutive integers (increasing order) (a) is a prime (b) is a composite number divisible by 3 (c) does not exceed 17 (d) is at most 19 Ans. (a), (c), (d) Solution: It means the pattern 12, 23, 34 do not appear. Let Si denote the set that i(i number of elements in Si, we have to permute 2 numbers and the entity i(i + 1), this can be done in 3! ways. Also, (12), [or (23), (34)] can occur together in 3! ways, and (12), (34) can occur together in 2! ways. Next, 12, 23 and 34 can occur together (in that order) in just one way. Thus, the required number of ways = 4! - 3(3!) + 2(3!) + 2! – 1! = 19 ways.

MATRIX-MATCH TYPE QUESTIONS Let m and n be two positive integers such

Example 71

that m ≥ n. The number of ways of Column 1 (a) distributing m distinct books among n children (b) arranging n distinct books at m places (c) selecting m persons out of n persons so that two particular persons are not selected (d) number of functions from {1, 2, 3, n} to {1, 2, 3, m} p q r s Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Column 2 (p) 0

(c) When two particular persons are excluded the number of persons become n - 2. Since m ≥ n > n - 2, it is impossible to choose m persons out of n - 2. (d) For each k (1 £ k £ n) we have m choices. Example 72 The value of Column 1 Column 2 (a) (1)1! + (2)2! + (3)3! (p) (n + 2)2n-1 - (n + 1) + + (n).n! (q) 2nCn (b) n.nC1 + (n - 1) ◊ nC2 + (n- 2) ◊ nC3 + + 1 ◊ nC n 2 3 4 (c) C2 + C2 + C2 + + nC2(r) (n + 1)! - 1 n

(d) Â (nCr)2 r =0

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: (a) We have for k ≥ 1 (k) ◊ k! = [(k + 1) - 1} k! = (k + 1)! - k! Thus, 1 ◊ 1! + 2 ◊ 2! + 3 ◊ 3! + + n ◊ n! = (2! - 1) + (3! - 2!) + (4! – 3!) + + ((n + 1)! - n!) = (n + 1)! - 1. (b) Let S = (n) nC1 + (n - 1) nC2 + (n - 2) nC3

Let (r) n

p

Ans.

(q) mn m

+ + (1) nCn = (n - 1) ◊ nC1 + (n - 2) ◊ nC2 + + + nC n) 1 ◊ nC n - 1 + ( nC 1 + nC 2 + + C = (n - 1) ◊ nC1 + (n - 2) ◊ nC2 +

1 ◊ nCn-1 (1) We use Cr = Cn-r and re-write the terms in the reverse order, to obtain + (n - 1) ◊ nCn-1 (2) C = 1 ◊ nC 1 + 2 ◊ nC 2 + Add (1) and (2) and simplify. 2 C 2 + 3C 2 + 4C 2 + + nC 2 (c) = 3C 3 + 3C 2 + 4C 2 + + nC 2 4 4 n = C3 + C2 + + C2 = = n+1C3 (d) We have n

(s) (mCn) (n!)

Solution: (a) Each book can be given in n ways. Since, there are m books, the number of ways is nm. (b) We can choose n places out of m in mCn ways and then can arrange n books at these places in n! ways. Thus, the required number of ways in (mCn) (n!).

(s) n + 1C3

n

n

n

r =0

r =0

 ( nC r) 2 = Â

(nCr) (nCn-r)

= the number of ways of selecting n persons out of n men and n women = 2nCn Example 73 Let m = 100! Column 1 (a) The largest exponent k for which 3k divides m

Column 2 (p) 1

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(b) the number of zeros at the end of m (c) the largest exponent m k for which 11k divides 50!50! (d) The largest exponent k for which 61 for which 61k m divides 55! 45! p q r s Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(q) 2 (r) 24

(s) 48

= The total number of subsets - The number of subsets having only even numbers = 29 - 24 = 496 (b) The number of ways = 12C3 + 12C4 + 12C5 + + 12C12 12 12 12 12 = 2 - ( C0 + C1 + C2) = 4096 - (1 + 12 + 66) = 4017. (c) The prime factorization of 210 is 2 ◊ 3 ◊ 5 ◊ 7. We can assign 2 to any of x1, x2, x3 or x4. That is, 2 can be assigned in 4 ways. Similarly, 3(5) [7] can be assigned in 4 ways. A variable which is not assigned any prime will be assigned value 1. Thus, the number of solutions of the given equation is (4) (4) (4) (4) = 256. (d) Note that we cannot use 0 at any place. The required of ways of selecting 5 digits out of 9

Solution

9

Exponent of 3 in m is 48 Exponent of 2 in m is 97 and that of 5 is 24 Exponent of 11 in m is 9 and in 50! is 4 Proceed as in (c)

(a) (b) (c) (d)

Example 74 Column 1

Column 2

(a) The number of subsets of (p) 4017 the set {1, 2, 3, , 9} containing at least one odd number. (b) The number of ways of (q) 496 choosing at least 3 students from a class of 12. (c) The number of solution (r) 256 x1 x2 x3 x4 = 210 numbers in which every digit exceeds the immediately preceeding digit on the left p q r s Ans. a p q r s b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: (a) The required number of subsets.

C5 =

9¥8¥7¥6 = 126. 4¥3¥ 2

Example 75 Consider all possible permutations of the letters of the word R A C H I T I H C A R Column 1 Column 2 (a) The number of words (p) 56700 containing the word ACHIT is (b) The number of words (q) 630 beginning with RA and ending with AR is (c) The number of words (r) 45360 in which vowels occur at the odd places is (d) The number of words (s) 2520 in which the word IIT appears is p q r s Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: (a) We can permute ACHIT, R, R, I, H, C, A in 7! = 2520 ways. 2! (b) We can permute C, C, H, H, I, I, T at 7 places in 7! = 630 ways. 2!2!2!

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(c) We can choose 4 places out of 6 odd places in 4! 6 C4 ways and arrange A, A, I, I in . There2!2! fore, vowels can be arranged in

4! ˆ ( C ) ÊÁË 2!2! ˜¯ 6

4

ways. Remaining letters R, R, C, C, H, H, T 7! ways. Therefore, the can be arranged 2!2!2! desired number of ways 4! 7! = (15) = 56700 2!2! 2!2!2! (d) The number of words in which IIT appears = number of ways of arranging IIT, R, R, A, A, C, C, H, H 9! = 45360 = 2!2!2!2!

ASSERTION-REASON TYPE QUESTIONS Example 76

Statement-1: The expression Ê 40ˆ Ê 60ˆ Ê 40 ˆ Ê 60ˆ ÁË r ˜¯ ÁË 0 ˜¯ + ÁË r - 1˜¯ ÁË 1 ˜¯ + ◊◊◊◊

attains maximum value when r = 50. Ê 2nˆ Statement-2: Á ˜ is maximum when r = n. Ë r¯ Ans. (a) Solution: We have Ê 2n ˆ ÁË r ˜¯ (2n)! (r + 1)! (2n - r - 1)! = r ! (2n - r )! (2n)! Ê 2n ˆ ÁË r + 1˜¯ =

r +1 2n - r

Since for 0 £ r £ n – 1, Ê 2n ˆ Ê 2n ˆ ÁË 0 ˜¯ < ÁË 1 ˜¯ < Also, as

r +1 < 1, we get 2n - r

Ê 2nˆ < Á ˜ Ë n¯

Ê 2n ˆ Ê 2nˆ ÁË r ˜¯ = ÁË 2n - r ˜¯

Ê 2n ˆ Ê 2n ˆ ÁË 2n˜¯ < ÁË 2n - 1˜¯
1 2

(d) Sn < 1 " n

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24. Let p be an odd prime and N = 2pCp – 2, then (a) p|N (b) p2|N (c) 2p|N (d) 2p2|N 25. Let m, n, p, q be positive integers, and N = mn + pq, then N is divisible by (b) (n!)m (q!)p (a) (m!)n (p!)q (c) (m!)p (q!)n (d) (m!)q (n!)p 26. Let X and Y f π S Õ X × Y. For x ∈ X, let Ax = {y ∈ Y | (x, y) ∈ S} By = {x ∈ X | (x, y) ∈ S}, then (a) |Ax| = |By|

(b) Â |Ax| = Â

(c) Â |Ax| = |S|

(d) none of these

x Œx

y Œy

(c) 2C2 + 3C2 + 4C2 n

(d) Â (nCr)2 r =0

1 1 = d 2 d /n

(c) Â

(b) Â d = N d /n

1 =2 d d /n

(d) Â

29. Let N = 1! + 2! + + 2007!, then (a) unit’s digit of N is 3 (b) ten’s digit of N is 1 (c) N is a perfect square (d) N is a perfect cube 30. The number 1000C500 is divisible by (a) 7 (b) 13 (c) 191 (d) 201

MATRIX-MATCH TYPE QUESTIONS 31. The value of Column 1 (a) 1.1! + 2.2! + 3.3! + + n.n! n (b) n ◊ C1 + (n - 1) ◊ nC2 + (n - 2) ◊ nC3 + + 1 ◊ nC n

Column 2 (p) (n + 2)2n-1- (n + 1) (q) 2nCn

(s) n+1C3

32. Let m = 100! Column 1 (a) The largest exponent k for which 3k divides m (b) The number of zeros at the end of m

Column 2 (p) 24 (q) 48

|By|

27. For n ≥ 2, let h = hcf (n! + 1, (n + 1)! + 1), then (a) h = 1 " n ≥ 2 (b) h is a prime if n is prime (c) h = n " n ≥ 2 (d) none of these 28. Let N = 2n - 1 (2n - 1) where 2n - 1 is a prime, then d /N

+ C2

+

in which every digits exceeds the immediately preceding digit (d) The number of four digit numbers (s) 758 that can be formed by using the digits 1, 1, 1, 2, 2, 3, 3, 4, 5, 6.

x Œx

(a) Â d = 2N

(r) (n + 1)! - 1

n

33. Column 1 Column 2 (a) The last digit of (p) 30 500 (1! + 2! + … + 2009!) (b) Number of positive integral (q) 3 solutions of xyz = 24 (c) If nCr – 1 = (k2 – 8) (n + 1Cr) (r) – 3 then integral value of k (d) The number of elements (s) 1 7–x in the range of Px – 3 34. Suppose a set A consists of 10 distinct elements x1, x2, … x10 The number of subsets of A which contain Column 1 Column 2 (p) 512 (a) none of x1, x2, x3 (q) 896 (b) each of x1, x2, x3 (r) 128 (c) at least one of x1, x2, x3 (s) 256 (d) at most one of x1, x2, x3

ASSERTION-REASON TYPE QUESTIONS Ê nˆ 35. Let n ∈ N and Cr = Á ˜ Ë r¯ Statement-1: Sum of the series (C0 + C1) + (C1 + C2) + … + (Cn – 1 + Cn) is 2(2n – 1). Statement-2: C0 + C1 + … + Cn = 2n

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36. Statement-1: The number of onto function f from A = {1, 2, 3, 4, 5, 6} to B = {7, 8, 9} such that f(i) £ f(j) " i < j is 6. Statement-2: The number of permutations of 7, 8, 9 taken 3 at a time is 6.

COMPREHESION-TYPE QUESTIONS Paragraph for Question Nos 37 to 40 If #(A A and A1, A2, n Ê n ˆ # Á Ai ˜ = Â #(Ai) – Â #(Ai « Aj) + Ë i = 1 ¯ i =1 i< j

Â

i< j n then show that the number of ways in which they can be seated is m !(m + 1)! . [1983] (m – n + 1)! 5. Relatives of a man comprises 4 ladies and 3 gentlemen and his wife has also 7 relatives : 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of man’s relative and 3 of the wife’s relatives? [1985] 6. A box contains two white balls, three black balls, and four red balls. In how many ways can three balls be drawn from a box if at least one black ball is to be included in the draw? [1986] 7. A student is allowed to select at most n books from a collection of (2n + 1) books. If the total number of ways in which he can select at least one book is 63, n. [1987] 8. Eighteen guests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on the other side. Determine the number of ways in which the sitting arrangements can be made. [1991] 9. A committee of 12 is to be formed from 9 women and 8 men. In how many ways this can be done if at

IIT JEE eBooks: www.crackjee.xyz Permutations and Combinations 5.39

( n )! is Using permutation or otherwise, prove that 2

In how many ways in these committees (i) the women are in majority? (ii) the men are in majority?

14. [1994]

( n!) n

an integer, where n is a positive integer.

10. In how many ways three girls and nine boys can be seated in two vans each having numbered seats, 3 in front and 4 at the back? How many seating arrangements are possible if 3 girls sit together in a back row on adjacent seats? Now, if all the seating arrangements are equally likely, what is the probability of 3 girls sitting together in a back row on adjacent seats? [1996] 11. Let p be a prime and m be a positive integer. By mathematical induction on m, or otherwise, prove that whenever r is an integer such that p does not divide r, p divides mpCr. [1998] 12. Let n be any positive integer. Prove that Ê 2n – k ˆ m Á Ë k ˜¯ (2n – 4k + 1) n –2 k 2  k = 0 Ê 2n – k ˆ (2n – 2k + 1) ÁË n ˜¯ Ê nˆ ÁË m˜¯ 2n –2 m = n m 2 – 2 Ê ˆ ÁË n - m ˜¯

Answers LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17. 21. 25. 29. 33. 37. 41. 45.

(b) (b) (b) (d) (c) (d) (a) (a) (a) (b) (d) (c)

Ê pˆ Here Á ˜ = pCq. Ë q¯

46. 48. 50. 52. 54. 56. 58. 60. 62. 64.

[1999]

13. For any positive integer m, n (with n ≥ m), Ê nˆ let Á ˜ = nCm. Prove that Ë m¯ Ê mˆ + Á ˜ Ë m¯ Ê n + 1ˆ = Á Ë m + 1¯˜

(a) (a) (d) (d) (a) (d) (a) (d) (a) (a) (b)

3. 7. 11. 15. 19. 23. 27. 31. 35. 39. 43.

(d) (c) (b) (b) (b) (d) (a) (c) (d) (a) (c)

4. 8. 12. 16. 20. 24. 28. 32. 36. 40. 44.

(a) (b) (a) (c) (b) (a) (c) (d) (a) (c) (c)

Ê n - 1ˆ Ê n - 2ˆ Ê nˆ T = Á ˜ +2 Á + 3Á + ˜ Ë m ¯ Ë m ˜¯ Ë m¯

(b) (b), (c), (d) (d) (c) (b), (c), (d) (b), (c), (d) (c) (b), (c)

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

66.

[2000]

(a), (a), (a), (a), (a), (a), (a) (a), (a) (a),

47. 49. 51. 53. 55. 57. 59. 61. 63. 65.

(d) (a), (a), (a), (a), (a), (b), (b) (a), (a)

(b) (b), (c), (d) (d) (b) (b) (c), (d) (b), (c)

MATRIX-MATCH TYPE QUESTIONS

Hence or otherwise prove that

Ê n + 2ˆ Ê mˆ + (n – m + 1) Á ˜ = Á Ë m + 2¯˜ Ë m¯

2. 6. 10. 14. 18. 22. 26. 30. 34. 38. 42.

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

for each non-negative integer m £ n.

Ê n - 1ˆ Ê n - 2ˆ Ê nˆ S= Á ˜ + Á + Á + ˜ Ë m ¯ Ë m ˜¯ Ë m¯

[2004]

IIT JEE eBooks: www.crackjee.xyz 5.40 Comprehensive Mathematics—JEE Advanced

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

p

q

a

p

b

67.

68.

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 21. 23. 25. 27. 29.

(b), (d) (b), (c), (d) (b)

22.(a), (b), (c), (d) 24. (a), (b), (c), (d) 26. (b), (c) 28. (a), (d) 30. (b), (c)

(b)

MATRIX-MATCH TYPE QUESTIONS p

q

r

s

a

p

q

r

s

s

b

p

q

r

s

r

s

c

p

q

r

s

q

r

s

d

p

q

r

s

p

q

r

s

p

q

r

s

c

p

q

r

s

a

p

q

r

s

d

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

69.

31.

32.

ASSERTION-REASON TYPE QUESTIONS 70. (c)

71. (b)

72. (d)

73. (c)

COMPREHENSION-TYPE QUESTIONS 74. (d) 78. (d) 82. (d)

75. (b) 79. (a) 83. (b)

76. (d) 80. (c)

84. 2 88. 4 92. 2

85. 8 89. 9

86. 3 90. 5

87. 2 91. 3

LEVEL 2

SINGLE CORRECT ANSWER TYPE QUESTIONS (d) (c) (b) (c) (b)

2. 6. 10. 14. 18.

(d) (c) (a) (b) (a)

3. 7. 11. 15. 19.

(d) (b) (a) (b) (c)

33.

77. (c) 81. (b)

INTEGER-ANSWER TYPE QUESTIONS

1. 5. 9. 13. 17.

(a), (a), (a), (a) (a),

4. 8. 12. 16. 20.

34.

ASSERTION-REASON TYPE QUESTIONS (c) (a) (a) (c) (c)

35. (a)

36. (b)

COMPREHENSION-TYPE QUESTIONS 37. (c)

38. (b)

39. (a)

40. (a)

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INTEGER-ANSWER TYPE QUESTIONS 41. 7 45. 2

42. 8 46. 2

43. 4

44. 3

3. 150 5. 485 6. 64 8. 11C5(9!)2 9. 6062, (i) 2702 10. 12! ; 1/91

PAST YEARS IIT QUESTIONS

Hints and Solutions

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17. 21. 25.

(c) (d) (b) (c) (c) (c) (b)

2. 6. 10. 14. 18. 22. 26.

(a) (a) (c) (b) (b) (c) (c)

3. 7. 11. 15. 19. 23. 27.

(b) (c) (a) (a) (d) (c) (a)

LEVEL 1

4. 8. 12. 16. 20. 24.

(c) (c) (c) (a) (b) (d)

MATRIX-MATCH TYPE QUESTIONS

1. For each place we have two choices. 2. Use rPr = r(r!) = (r + 1)! - r! n - 2r

n

3. Â

n

r =0

n

= Â

Cr

r =0

n-r n

Cr

n

- Â

r =0

6. S = 2n + 1Cn + 1 + 2n + 1Cn + 2 +

r

s

a

p

q

r

s

S + 2n + 1C2n + 1 = 22n

b

p

q

r

s

7. (8C3) (7C4) - (7C2) (6C3)

c

p

q

r

s

8. 1.3.5.

d

p

q

r

s

2. (d)

INTEGER-ANSWER TYPE QUESTIONS 1. 7 5. 5

2. 4 6. 5

FILL

IN THE

3. 5

4. 7

BLANKS TYPE QUESTIONS

n

1. Â ak

2. 205

5. 9

6. pCk – 1 where p = n –

3. nn, n!

4. 35

k =1

1 2 (k – k + 2) 2

TRUE/FALSE-TYPE QUESTIONS

=

(2n)! n

2 (n !)

(2n - 1) = =

1 2n

+ 2n + 1C2n

S = 22n - 1 = 63 1.2.3.4.5......(2n - 1) (2n) (2) (4).....(2n)

(2nCn) (nPn)

9. Let x1 = x + 5, y1 = y + 5, z1 = z + 5, so that x1 + y1 + z1 = 15. The number of required solutions = 17C2 = 136 10. For each book we have (p + 1) choices. + x 6) 3 x11 in (x + x2 + 8 6 3 x in (1 - x ) (1 - x)-3 x8 in (1 - 3x6) (1 + 3C1 x + 4C2 x2 + ) 10 4 = C8 - 3( C2) = 27 12. Arrange m white and n red counters on one side of the central mark. (2n + 1) ºº (4n - 1)(4n) 1 (4n)! n ! = =y 13. x = (2n)! 2n (2n)! (2n + 2) ºº (4n) 14. (pC2) (qC2) + (qC2) (rC2) + (rC2) (pC2)

1. True

SUBJECTIVE-TYPE QUESTIONS 1. 26

Cr

n

q

COMPREHENSION-TYPE QUESTIONS

r n

Now, use Cr = Cn-r. for 0 £ r £ n. 4. First treat m men as just one object. We can permute (w + 1) objects in (w + 1)! and the m men is m! ways. 5. Use sum of digits must be divisible by 3 and the last digit should be an even number. n

p

1. (b)

7. 3 (ii) 1008

2. (i)

52!

(13!)

4

(ii)

1 52! 3! (17!)3

of the next four 10 choices. At this stage add the numbers already written and the choose digit for the unit’s place is 5 ways. 16. Ak is obtained by putting x = - k every where in the expression except in x + k. Therefore

IIT JEE eBooks: www.crackjee.xyz 5.42 Comprehensive Mathematics—JEE Advanced

m! Ak = (- k ) º (- k + k - 1) (- k + k + 1) º (- k + m) =

( - 1)k m ! = (- 1)k (mCk) (k !) (m - k )!

17. Consider the product (1 + 1) (1 + 2) (1 + n) 18. Consider the product Ê (1 + 1) ËÁ1 +

1ˆ Ê ˜ º Á1 + 2¯ Ë

1ˆ ˜ n¯

19. (M, M), (A, A), (T, T), E, H, I, C, S We can arrange 4 distinct letters in 8P4 ways, 2 iden4! 3 7 C2 tical and 2 distinct in C1 ways and 2 2!

( )(

)

4! pairs of identical letters in ( C2) ways 2! 2! 3

20. (mnCn) (mn-nCn) 21. nC2 - n = 90

( nC n)

30

C4 - 27

5

Cn Cn

+

4

Cn

6

Cn

fi n=2

36. 5C1 + 5C2 + 5C3 = 25

38. 39.

n

26. 2(nC5) = nC4 + nC6 fi 2 =

n

C4 C5

+

n

C6

n

C5

27. The letters of SURITI can be arranged in 6!/2! = 720/2 = 360 ways. The alphabetic order is I, R, S, T and U. The number of words beginning I is 5P5 = 5! = 120, and those beginning R number 5!/2! = 120/2 = 60. Then come words beginning SI, numbering 4! = 24, and SR, numbering 4!/2! = 12. We then get 3P3 = 3! = 6 words beginning SUI, after which come the words SURIIT and SURITI. Thus, the rank of SURITI is 120 + 60 + 24 + 12 + 6 + 2 = 224. 28. The number of ways the candidate can choose questions under the given conditions is enumerated below. Group 1 Group 2 No. of ways

4

35. 1 =

37. If n is odd, nCr is maximum when r =

22. (10 + 1) (8 + 1) (6 + 1) - 1 23. Choose two elements nC2 ways, and rest of the (n - 2) elements in 3n-2 ways. 24. 105 - 10P5 25.

29. Total number of functions = 3n Number of functions with exactly two elements in range is (3C2) (2n - 2) and with exactly one element in range is 3. 1 1 30. a = m(m - 1) and aC2 = a(a - 1) 2 2 31. The balls must be of alternate colours. 32. For a square of size k × k, we have to choose (k + 1) consecutive horizontal and (k + 1) consecutive vertical lines from the chess board, and this can be done in (9 - k) × (9 - k) ways. numbers is 94 and for any other four place it is 8 × 93. 34. Unit’s digit of 172009 - 72009 is 0.

4 2 ( 5C 4) ( 5C 2)

3 3 ( 5C 3) ( 5C 3)

2 4 ( 5C 2) ( 5C 4)

From this we see that the required number is ( 5C 4) ( 5C 2) + ( 5C 3) ( 5C 3) + ( 5C 2) ( 5C 4) = (5) (10) + (10) (10) + (10) (5) = 200.

40. 41. 42. 43. 44.

1 (n - 1) or 2

1 (n + 1) 2 BAAA can be arranged in 4 ways and two N’s can be adjusted in 5C2 ways. Use = 840 = 23 × 3 × 5 × 7 2’s can be assigned in 5C3 + 5C2 + 5C1 ways and rest of the three digits in 5 × 5 × 5 ways. Total number of ways - number of ways in which exactly four boxes remain empty. (x + 3)2 + y2 = 13 fi x + 3 = ± 2, y = ± 3 or x + 3 = ± 3, y = ± 2 20! = 218 × 38 × 54 × 72 × 11 × 13 × 17 × 19 All the words beginning with CC, CH, CI and CN occur before COCHIN. 5! ( 3C 1) = 60 3!

45. 100! = 297 ¥ 348 ¥ 524 ¥ 46.

an n +1 = < 1 for 1 £ n £ 8, an + 1 10 an an = 1 for n = 9 and > 1 for n ≥ 10. an + 1 an + 1

Thus, an is maximum for n = 9 and n = 10. 47. We have r

Cr2 Ê C ˆ = Á r r ˜ Cr Cr - 1 Ë Cr - 1 ¯

IIT JEE eBooks: www.crackjee.xyz Permutations and Combinations 5.43

But r

Cr n! (r - 1)! (n - r + 1)! ¥ =r Cr - 1 r ! (n - r )! n! =n-r+1 + 1Cn nC1 + (n - 1)C2 + Cn–r we can write + nCn-1 1C0 + 2C1 + 3C2 + 1 + (n + 2) (C1 + C2 + + Cn-1) + 1 2 + (n + 2) (2n - 2) 1 + (n + 2) (2n-1 - 1)

Thus, S = Using Cr = S= fi 2S = = fi S=

48. As r(r!) = (r + 1)! - r!, therefore S = (n + 1)! 49. Choose n place out of 2n for xi’s. 50. Use Cr = C2n + 1 - r " r 51. We can choose r objects out of n distinct objects in n Cr ways remaining (n - r) can be chosen from the n identical objects. 52. The number k (1 £ k £ n - 2) is minimum in n-kC2 sets. Thus, n-2

N= Â n-2

= Â

n-2

= (n + 1) Â

k =1

n-2

But Â

n-k

k =1

(n-kC2) - 3

Â

(n-k+1C3)

k =1

+ n-1C2

= 3C 3 + 3C 2 +

+ n-2C2

= 4C 3 + 4C 2 +

+ n-2C2

k =1

n-k+1

C 3 = 3C 3 +

+ nC3 = n+1C4

N = (n + 1) (nC3) - 3(n+1C4) = 4(n+1C4) - 3(n+1C4) = n+1C4 53. N = 2nCn =

(1) (2) (3) º (2n - 1) (2n)

= (1) (3)

2n (2n - 1) (1 × 2 ×

× n)

56. 0 cannot be used. Out of remaining 9 digits, choose 3. Therefore N = 9C3 = 84. 57. Total number of ways is 10! and out of these exactly half are such that A1 is ranked above A2. 58. Choose n place out of (p + 1) for negative signs. 59. nP2 = 1640 fi n = 41

62. Use nC0 = n +1C0, nCr + nCr – 1 = n+1Cr to obtain x=

= nC 3 n-3

=

n-2

C 2 = 2C 2 + 3C 2 +

Similarly, Â

vowels in 2! ways. Vowels do not occur together in (360 - 120) = 240 ways (2n)! 55. N = (2nC2) (2n-2C2) ( 4C 2) ( 2C 2) = 2n

61. (3C2) (9C1) + (4C2) (8C1) + (5C2) (7C1) + (3) (4) (5)

[n + 1 – (n – k + 1)] (n-kC2)

k =1

4! ways and two 2!

60. 9C3 - 7C3

k(n-kC2)

k =1

S, D, S, H can be permuted in

(2n)! (n + 1) (n + 2) º (2n) = n! n! (1) (2) º (n)

As n > p and p < 2n, p occurs exactly once in the numerator but does not occur in the denominator. 5! 54. Vowels are together in (2!) ways. Letters of 2! 6! = 360 ways out 2! of which in exactly half of them are in alphabetical order. SUDESH can be arranged in

n+r+1

Cr

63. For f(i) < f(j) whenever i < j, is equivalent to choose 3 numbers out of {1, 2, 3, 4, 5, 6, 7} in 7C3. For f(i) £ f(j) whenever i < j, the number of ways is 7 C3 + 2(7C2) + 7C1 = 84 For (c), see (a). 64. See solution to Question 15. 65. Sum of the seven digits use must be a multiple of 3. As 1 + 2 + + 9 = 72, we must not use one of the following pairs: (1, 2), (1, 5), (1, 8), (2, 4), (2, 7), (3, 6), (3, 9), (4, 5), (4, 8), (5, 7), (6, 9), (7, 8) Thus, required number of ways is (12) (7!) 66. (a) The given equation can be written as C3 - n+3C3 = 15(n + 2) fi n+3C2 + n+3C3 - n+3C3 = 15(n + 2) [n+1Cr = nCr-1 + nCr] (n + 3)! fi = 15(n + 2) (n + 1)! 2! n+4

fi (n + 3) (n + 2) = 30(n + 2) or n + 3 = 30 or n = 27. (b) 11 (nP4) = (20)(n-2P4)

IIT JEE eBooks: www.crackjee.xyz 5.44 Comprehensive Mathematics—JEE Advanced



11

fi 11

n! (n - 2)! = 20 (n - 6)! (n - 4)! n (n - 1) = 20 (n - 4) (n - 5)

fi 11(n2 - n) = 20(n2 - 9n + 20) fi n = 23/9 or 16. As n cannot be in fraction, n = 16. (c) 2nC3 = 11nC3 fi

(2n)! n! = 11 (2n - 3)!3! (n - 3)!3!

fi (2n) (2n - 1) (2n - 2) = 11n(n - 1) (n - 2) fi 4(2n - 1) = 11(n - 2) fi 3n = 18 or n = 6. (d) 2(14Cn–1) = 14Cn–2 + 14Cn fi n = 6, 10 67. Let A = {a1, a2, , an}. For each ai Œ A, we have following four choices (1) ai ∈ P, ai ∈ Q (2) ai ∈ P, ai ∉ Q (3) ai ∉ P, ai ∈ Q (4) ai ∉ P, ai ∉ Q For P » Q = A or P « Q = j the number of choice is 3n. The singleton can be chose in nC1 ways, rest of the elements in 3n-1 ways. P » Q = j in just one way. 1 1 1 68. (a) + = fi 7(x + y) = xy x y 7 fi (x – 7) (y – 7) = 49 = 72 x – 7 = 1, y – 7 = 72 fi x = 8, y = 56 x – 7 = 7, y – 7 = 7 fi x = 14, y = 14 x – 7 = 72, y – 7 = 1 fi x = 56, y = 8 (b) Similar to (a) (c) Note that x + y > x – y \ x + y = 7, x – y = ± 5 fi (x, y) = (6, 1), (1, 6) 1 1 1 1 1 (d) – = fi > or y > x x y 3 x y Also,

1 1 1 – = fi 3(y – x) = xy x y 3

fi (x – 3) (y + 3) = – 32 fi x – 3 = – 1, y + 3 = 9 fi x = 2, y = 6 x – 3 = – 3, y + 3 = 3. Not possible x – 3 = – 9, y + 3 = 1 Not possible 69. (a) The number of permutations of ENDEA N, O, E, L is 5!

to permute 2N’s 1E, 1A, 1D, 1O and 1L 7! 7 ¥ 6 = This can be done (5!) = (21) (5!) 2! 2 ways. 4! ways. The remaining letters can be per2! 5! ways. muted in 3! 5! (d) A, E, E, E, O can occur at odd places in ways 3! 4! and the remaining letters can be arranged in 2! ways. \ required number of ways Ê 5!ˆ Ê 4!ˆ = Á ˜ Á ˜ = 2(5!) Ë 3!¯ Ë 2!¯ in

70. The number of divisors 10m = 2m 5m is (m + 1)2 \ Number of divisors which divide 102009 but 102008 is 20102 – 20092 = 4019. Èx˘ Èx˘ 71. Suppose Í ˙ = Í ˙ = m (say). Î19 ˚ Î 21 ˚ For m = 0, x = 1, 2, 3, …, 18, m = 1, x = 21, 22, …, 37 m = 2, x = 42, 43, …, 56 m = 3, x = 63, 64, …, 75 m = 4, x = 84, 85, …, 94 m = 5, x = 105, 106, …, 113 m = 6, x = 126, 127, …, 132 m = 7, x = 147, 148, …, 151 m = 8, x = 168, 169, 170 m = 9, x = 189 Thus, the number of solutions is 18 + (17 + 15 + … + 3 + 1) = 99 72. Required number of ways = number of non-negative integral solutions of x1 + x2 + x3 + x4 + x5 = 6 10! = 6 + 5 – 1C5 – 1 = 10C4 = 4!6! = The number of different ways of arranging 6A’s and 4B’s. 73. The man will be just two steps forward from his initial position (in at most 10 steps) if his movement is as follows:

IIT JEE eBooks: www.crackjee.xyz Permutations and Combinations 5.45

2F or 3F; 1B or 4F, 2B or 5F, 3B or 6F, 4B Thus, required number of ways = 2C0 + 4C1 + 5C2 + 8C3 + 10C4 = 286 \ Statement-2 is false. Also, the number of ways in which the man is just two steps backwards = 286. \ required number of ways = 2(286) = 572. 74. Use < x + m> = for each integer m. 75. As RHS is an integer, and 0 £ < 1, we get =0 fi x is an integer. Therefore, 0 = x3 + x2 + x + 1 = (x2 + 1) (x + 1) fix =-1 76. Write x = [x] + to obtain 2[x] = [2[x] + 2] As 0 £ 2 < 1 and 2[x] ∈ I, k we get = where k ∈ {1, 2, , 2[x] - 1}. 2 [ x] For each of the values of [x] = 1, 2, , 9, must take exactly 2[x] different values and for [x] = 10, we have = 0. Thus, the number of solutions is 2+4+ + 2(9) + 1 = 91 77. As 3[x] - [x2] is an integer, = 0 or = 1/2. When = 0, we get 3x = x2 fi x = 0 or x = 3 When = 1/2, let x = m + 1/2 to obtain 3m = m2 + m + 1 fi m = 1 Thus, x = 0, 3/2, 3 78. = x for 0 £ x < 1 79.

1000

C500 =

1000! (500!) 2

È1000 ˘ È1000 ˘ E11(1000) = Í = 98 + Î 11 ˙˚ ÍÎ 121 ˙˚ È 500 ˘ È 500 ˘ E11(500) = Í = 49 + Î 11 ˙˚ ÍÎ 121 ˙˚

E2(n) = (a0 2 + (a0 2 +

k-2

k-1

+ a1 2

+ a1 2

k-3

+

+ ak-1 (2) + ak k-2

+

+ ak-1)

+ ak-2)

+ (2a0 + a1) + a0

= a0 (2k - 1) + a1(2k-1 - 1) + =n-s

+ ak-1

(200) (199) º (101) (1) (2) º (100) All the two digit numbers occur in the denominator. Thus, the exponent of the largest two digit prime 200

C100 =

the required largest two digit prime is 61. • n Èn˘ • Ê nˆ 83. E5(n) = Â Í k ˙ < Â Á k ˜ = 4 k =1 Î 5 ˚ k =1 Ë 5 ¯

fi n > 4E5(n) = 104 But for n £ 109, E5(n) < 26 \ n ≥ 110 The desired numbers are 110, 111, 112, 113, 114. 84. We must have x ≥ 1 and y ≥ 2, since x is less than y. That is, if y = j with 2 £ j £ 9, then x can take the j -1 values from 1 to j - 1, and z can take the j values from 0 to j - 1. For each value y = j, therefore, xyz can take j(j - 1) values. We can now get the answer by summing over all values of y: 9

9

9

9

j=2

j =1

j =1

j =1

 j(j - 1) =  j(j - 1) =  j2 -  j

fi m = 285 - 45 = 240 85. A number lying strictly between 100 and 1000 is of the form xyz where x π 0, and if x = 1, then either y or z is different from zero. Therefore x can be chosen from the given digits in 5 ways, and y and z in 6 ways each, giving 5 × 6 × 6 = 180 numbers. However, one of these numbers is 100 (corresponding to x = 1, y = 0 and z = 0), and must be excluded. Hence the answer is 179 fi m – 171 = 9 86. The number of solutions of x1 + x2 + x3 = 5 is (7C2) and that of x4 + x5 = 15 is (16C1). Thus, number of required solutions is (21) (16) = 336. 87. Let t ≥ 0 be such that x + y + z + t = 29. Put x = x1 + 1, y = y1 + 2, z = z1 + 3 where x1, y1, z1 ≥ 0, the equation becomes x1 + y1 + z1 + t = 23. Its number of solutions is 26C3 = 2600. 88.

80. Use E2(50) = 47 and E5(50) = 12 81. n = a0 2k + a1 2k-1 +

82.

15 + 6-1

C6-1 = 20C5 = 15504

89. The number of nine-digit tenary sequences beginning 210 is 36; the number ending 210 is also 36; and the number both beginning 210 and ending 210 is 33. Thus, the number of sequences either beginning 210 or ending 210 is 2(36) - 33 = 2(729) - 27 = 1431.

IIT JEE eBooks: www.crackjee.xyz 5.46 Comprehensive Mathematics—JEE Advanced

90. Number of non-negative integral solutions of x + y + z = 8, x, y, z ≥ 0 91. Number of positive integral solutions of x + y + z = 8, x, y, z ≥ 1. 92. Five consecutive cards can be chosen out of 13

(spades, hearts, diamonds and clubs). The second and subsequent cards, too, can be chosen in 4 ways. Thus, there are 45 that the total number of straights is 9(45) = 9(1024) are not from the same suit is 45 - 4 = 1020, so the number of straights in this case is 9(1020) = 9180. fi m/4590 = 2

IIT JEE eBooks: www.crackjee.xyz

6 Binomial Theorem 6.1 BINOMIAL THEOREM (FOR A POSITIVE INTEGRAL INDEX)

If n is a positive integer and x, y are two complex numbers, then (x + y)n = nC0 xn + nC1 xn–1 y + nC2 xn–2y2 + ... + nCr xn–r yr +…+ nCn–1 xyn–1 + nCnyn (1) n n n C0, C1, , Cn are called binomial .

Illustration 1 3 2 10 Find the middle term of Ê x - yˆ Ë2 5 ¯ As n = 10 is even, middle term of 2 ˆ 10 Ê10 ˆ Ê3 x y is Ë + 1¯ th or 6th term. Ë2 5 ¯ 2 It is given by 10 - 5

Ê3 ˆ T6 = T5 + 1 = 10C5 Ë x ¯ 2

6.2 PROPERTIES OF THE BINOMIAL EXPANSION

5 Ê 2 ˆ y Ë 5 ¯

5

1. There are (n + 1) terms in the expansion (1). 2. In any term of (1), the sum of the exponents of x and y is always n. Cr = nCn–r

Ê 3ˆ = –252 Ë ¯ x5 y5 5 Illustration 2

n

(0 £ r £ n). 4. The term nCr xn–r yr is the (r + 1)th term from the beginning of the expansion. It is usually denoted by of the expansion. Tr+1 and is called the 6.3 MIDDLE TERM(S)

1. n is even If n is even, then the expansion (1) has just one middle n term, viz. ÊÁ + 1ˆ˜ th term. It is given by Ë2 ¯

2 ˆ 11 Ê1 x Find the middle terms of Ë x + 2 3 ¯ As n = 11, is odd, the binomial expansion of 2 ˆ 11 Ê11 + 1 ˆ Ê1 x + x ¯ has two middle terms viz. Ë th Ë2 3 2 ¯ Ê11 + 1 ˆ and Ë + 1¯ th terms, that is, 6th and 7th terms. 2 Ê1 ˆ T6 = T5 + 1 = 11C5 Ë x ¯ 2

Cn/2 xn/2 yn/2.

11 - 5

Ê2 ˆ Ë 3 x¯

5

n

2. n is odd If n is odd, then the expansion (1) has two middle terms, Ê n + 1ˆ Ê n+1 ˆ th term and Á viz. Á + 1˜ th term. These are Ë 2 ˜¯ Ë 2 ¯ given by n

and

n

C( n -1) 2 x ( n -1) 2 y( n +1) 2 C( n +1) 2 x

( n +1) 2

y

( n -1) 2

=

11! 1 25 6+5/2 ◊ ◊ x 5! 6! 26 35

=

77 17 2 x 81

Ê1 ˆ and T7 = T6 + 1 = 11C6 Ë x ¯ 2 =

11- 6

Ê2 ˆ Ë 3 x¯

11! 1 26 5 + 3 308 8 = x ◊ ◊ x 243 5! 6! 25 36

6

IIT JEE eBooks: www.crackjee.xyz 6.2 Comprehensive Mathematics—JEE Advanced The Greatest Coefficient

Some Other Useful Expansions

If n (x + y)n is nCn/2. If n is odd expansion of (x + y)n. These are nC(n – 1)/2 and nC(n + 1)/2.

In the following expansions Cr stands for nCr 1. (x + y)n + (x – y) n = 2 [C0 x n + C2 x n – 2 y2 + C4 x n – 4 y4 + . . . .] 2. (x + y)n – (x – y)n

The Greatest Term

(x + y)n, x, y > 0, n ΠN put (n + 1) y a= x+y If a is not an integer, then the expansion of (x + y)n has just one greatest term and it is given by k = [a] where [a] denotes the greatest integer < a. If a is an integer, then expansion of (x + y)n has two greatest terms viz. ath and (a + 1)th terms.

= 2 [C1 xn – 1 y + C3 xn – 3 y3 + C5 xn – 5 y5 + . . . .] 3. n (n–1) (n –2)… (n– k + 1) (x + y) n–k n

= Â r (r - 1) (r - 2)º(r - k + 1)Cr x r - k y n -r r =k

4. (1 + x)n + (1 – x)n = 2 [C0 + C2 x2 + C4 x4 + . . . .] 5. (1 + x)n – (1 – x)n = 2 [C1 x + C3 x3 + . . . .] 6. C1 + 2C2x + 3C3x2 + . . . . + nCn xn = n(1 + x)n – 1 7. C0 x + C1

TIP If we are given that tr is numerically the greatest term, we use |tr–1| < | tr | and |tr+1| < | tr | to obtain two inequalities from which some other results can be obtained.

Illustration 3 Find numerically greatest term in the expansion of (5 – 3x)7 when x = 2/3. We have Tr + 1 = 7Cr 57 – r (–3x)r r T 7! (r - 1)! (8 - r )! 57 - r ( -3 x ) fi r +1 = Tr r ! (7 - r )! 7! 57 - r + 1 ( - 3 x )r -1

=

8-r 1 (–3x) r 5

For x = 2/3, Tr +1 (8 - r ) (2) = Tr 5r Similarly

Tr + 2 Tr + 1

=

(7 - r ) (2 ) 5 (r + 1)

For (r + 1)th term to be numerically greatest, |Tr| < |Tr + 1| and |Tr + 2| < |Tr + 1| fi 5r < 16 – 2r and 14 – 2r < 5r + 5 fi 9 < 7r < 16 fi r = 2. Thus, 3rd term is numerically greatest.

– 1

x2 x3 x n +1 + C2 + .... + Cn 2 3 n +1 x

= Ú (1 + t )n dt 0

6.4 SOME PROPERTIES OF THE BINOMIAL COEFFICIENTS

1. 2. 3. 4.

C0 + C1 + C2 + . . . . + Cn = 2n C0 + C2 + C4 + . . . . = C1 + C3 + C5 + . . . . = 2n – 1 C0 – C1 + C2 – . . . . + (–1)n Cn = 0 For n > 1, C1 – 2C2 + 3C3 – . . . . + (–1)n – 1 nCn = 0

5. If Pn = C0 C1 C2 . . . . Cn, then

Pn + 1 Pn

=

(n + 1)n n!

6.5 REVERSING TECHNIQUE

If a0, a1, a2, . . . ., an are in A.P., then sum of the series a0 C0 + a1 C1 + . . . . + an Cn below. Let S = a0 C0 + a1 C1 + a2 C2 + . . . . + an

+ an Cn (1) Using Cr = Cn – r and reversing the order in which terms are written above, we obtain S = anC0 + an – 1 C1 + an – 2 C2 + . . . . + a1Cn – 1 + a0 Cn (2) Adding (1) and (2) we get 2S = (a0 + an) C0 + (a1 + an – 1) C1 + (a2 + an – 2) C2 + . . . . + (an – 1 + a1) Cn – 1 + (an + a0) Cn As a0, a1, a2, …an are in A.P., we have a0 + an = a1 + an – 1 = a2 + an – 2 = . . . ., –1

Cn

–1

IIT JEE eBooks: www.crackjee.xyz Binomial Theorem 6.3

so that

fi e.g.

2S = (a0 + an) (C0 + C1 + C2 + . . . . + Cn) = (a0 + an)2n. 1 S = (a0 + an)2n = (a0 + an)2n – 1. 2 (i) C0 + 2C1 + 3C2 + . . . . + (n + 1)Cn = (1 + n + 1)2n – 1 = (n + 2)2n – 1 (ii) C1 + 2C2 + . . . . + nCn = 0C0 + 1C1 + 2C2 + . . . . + nCn = (0 + n)2n – 1 = n(2n – 1)

6.6 MULTINOMIAL THEOREM

If we wish to expand (x + y + z)n (where n is a positive integer and x, y, z are complex numbers), we may take y + z = a, expand the binomial (x + a)n, and then expand (y + z)r in each term of the expansion of (x + a)n. In general, we have the following result: n! n1 n2 nm (a1 + a2 + ...+ am)n = Â n ! n !  n ! a1 a2  am 1 2 m (1) where the summation is taken over all non-negative integers n1, n2, ..., nm such that n1 + n2 +  + nm = n. The number of distinct terms in the expansion (1) is n+m–1 C m – 1. 6.7 SOME USEFUL TIPS AND TRICKS

a + b r to a – b r . p,

q are irrational, then often it

is preferable to simplify ( p + q )2n ( p + q )2 = p + q+ 2 pq . n

C r – 1, 1 n Cr, nCr + 1 are in A.P., then r = (n + n + 2 ) 2 be in A.P. be in G.P. or H.P.

For | x | < 1 1. (1 + x)–1 = 1 – x + x2 – x3 + x4 – . . . . 2. (1 – x)–1 = 1 + x + x2 + x3 + x4 + . . . . 3. (1 + x)–2 = 1 – 2x + 3x2 – 4x3 + 5x4 – . . . . 4. (1 – x)–2 = 1 + 2x + 3x2 + 4x3 + 5x4 + . . . . 5. (1 + x)–3 = 1 – 3x + 6x2 – 10x3 + 15x4 – . . . . 6. (1 – x)–3 = 1 + 3x + 6x2 + 10x3 + 15x4 + . . . . 7. (1 – x)–n = 1 + nC1 x + n +1C2 x2 + n +2C3 x3 + . . . . where if n is a natural number. Expansion when x is very small

If x is very small so that x2 and higher powers of x can be neglected, then (1 + x)n = 1 + nx. If x is very small so that x3 and higher powers of x can be neglected, then 1 (1 + x)n = 1 + nx + n (n –1) x2 2 SOLVED EXAMPLES

SINGLE CORRECT ANSWER TYPE QUESTIONS Example 1

x n is divided by y, try to express x or some power of x as ky ± 1. For instance, 200 is divided by 50, we 2 write 7 = 50 – 1 and then use binomial theorem. 2. If a, b, r Œ Q and r is an irrational number, then sometimes it is useful to switch from 3. If p, q Œ Q and

6.8 SOME PARTICULAR EXPANSIONS

Let

2019 ÈÊ ˘ 2018 ˆ 2017 x + ˆ Ê ÍÁ Á ˙ 2018 + ˜ ˙ Í Á Ë 2018 ˜¯ ˜ + 2019 ˙ ÍË ¯ 2019 f ( x) = Í ˙ 2020 Î ˚

2020

and suppose a, b, c > 0 with abc = 1, then least value of f(a) + f(b) + f(c) is (a) 1 (b) 2017 (c) 3 (d) 20172 . (c) : By using binomial theorem we expand the above expression to obtain f(x) as a polynomial as follows: f(x) = a0 + a1x + a2x2 + … + amxm where ai > 0 for 0 £ i £ m. We have f(a) + f(b) + f(c) = 3a0 + a1(a + b + c) + a2(a2 + b2 + c2) +  + am (am + bm + cm) ≥ 3a0 + 3a1(abc)1/3 + 3a2(abc)2/3 +  + 3am(abc)m/3 [∵ AM ≥ G.M] = 3(a0 + a1 +  + am) [∵ abc = 1] = 3f(1) = 3(1) = 3

IIT JEE eBooks: www.crackjee.xyz 6.4 Comprehensive Mathematics—JEE Advanced

Suppose m and n are positive integers

Example 2 and let n

S = Â ( -1)

k

k =0 m

and

T = Â ( -1)

k

k =0

1 ( n Ck ) k + m +1 1 ( m Ck ) , k + n +1

then S – T

r Ê 5 - 1ˆ Ê1ˆ Â Cr Á ˜ = Â s Cr Á ˜ Ë bs ¯ s = 0 Ë 2 ¯ s =0 r

\

r

Ê 2 ˆ = Á Ë 5 - 1˜¯

Ê 1 k +m ˆ k n S = Â ( -1) ( Ck ) Á Ú0 x dx˜ ¯ Ë k =0 n

k n k +m ˆ = Ú Â ( -1) ( Ck )x ˜¯ dx k =0

Thus,

= Ú

m

m

k m n+k = Â ( -1) ( Ck ) Ú0 x dx 1

k =0 m

k m = Â ( -1) ( Ck ) k =0

1 =T n + k +1

Thus, S – T = 0 Let n

n

r =0

Cr p

t

t =0

pˆ Ê s : Let b = Â Ct Á 2 sin ˜ Ë 10 ¯ t =0 s

n

1 Êp ˆ ( 5 + 1) = 2 cos Á ˜ Ë 5¯ 2

Let P(x) be a polynomial with real

Example 4

1

m Ú0 x P(1 – x)dx = 0 " m

coef

t

N

{0}, then (a) (b) (c) (d) (d)

P(x) P(x) P(x) P(x)

= = = ∫

xn (1 – x)n for some n N (1 – x)2n for some n N 1 – xm (1 – x)n for some m, n 0

1 m

Ú0 x

1

P (1 - x ) dx = Ú (1 - x ) P ( x ) dx m

0

k Let P(x) = Â ak x where ak

R

k =0 n

= Â ak [1 – (1 – x)]k k =0

(b) 2cos(p/10) (d) 2sin(p/5)

r

n

n

Cr

ˆ Ê Â Ct ÁË 2 sin ˜¯ 10

s =0 s s

Then S1/n (a) 2cos(p/5) (c) 2sin(p/10) . (a)

S1/n =

: 0= s

r

Â



x) dx

m 1 nÊ k m kˆ = Ú0 x Á Â ( -1) ( Ck ) x ˜ dx Ë k =0 ¯

S= Â

= br

n

1

Example 3

r

Ê Ê 5 + 1ˆ 5 - 1ˆ =Á = Á1 + ˜ ˜ 2 ¯ Ë Ë 2 ¯

1

1 n x (1 0

r

n Ê 5 - 1ˆ Ê1ˆ n S = Â Cr Á ˜ = Â Cr Á ˜ Ë br ¯ r = 0 Ë 2 ¯ r =0 n

n m = Ú0 (1 - x) x dx n m = Ú0 (1 - (1 - x)) (1 - x) dx

s

s

Ê Ê 5 + 1ˆ 5 - 1ˆ =Á = Á1 + ˜ ˜ 2 ¯ Ë Ë 2 ¯

:

n

s

s

Ê 5 + 1ˆ Ê 2 ˆ =Á = Á ˜ Ë 5 - 1˜¯ Ë 2 ¯

(a) 0 (b) nm – mn m n (c) (n + 1) – (m + 1) (d) (1 – n)m – (1 – m)n (a)

1Ê 0Á Ë

t

Ê 5 - 1ˆ Ê 5 - 1ˆ = Á1 + =  Ct Á ˜ 2 ˜¯ Ë 2 ¯ Ë t =0 s

n Ê k k k jˆ = Â ak Á Â (-1) ( C j )(1 - x) ˜ Ë j =0 ¯ k =0 n

= Â bk (1 - x)

k

k =0 n

k j where bk = Â ( -1) ( Ck )a j j =k

N

IIT JEE eBooks: www.crackjee.xyz Binomial Theorem 6.5

Ê n 1 1 2 kˆ Now Ú0 ( P( x)) dx = Ú0 P( x) Á Â bk (1 - x) ˜ dx Ë k =0 ¯ n

a n - 4b4



1

k = Â bk Ú0 (1 - x) P( x)dx = 0

a n - 5 b5

2 25

[n C0 x 2 ] + [n C1 x 2 ] +  + [n Cn x 2 ]

(a)

1 2 x 2

(b) x2

(c) 2x2 (b)

(d) 4x2

: We know x – 1 < [x] £ x " x

R, therefore

Ck x – 1 < [ Ck x ] £ Ck x

n

n

2

fi fi

2

n

2

n

n

n

k =0

k =0

k =0

2n - 2

< £

1

n

2

x 2 (2n )

2n - 2 •, and using sandwich theorem, we

Taking limit as n n 1 È n Ck x 2 ˘ = x 2 get lim n  ˚ nÆ• 2 - 2 k = 0 Î

n-4 5

5 (c) n-4

2 (n - 4) 5

5 (d) 2 (n - 4)

. (b) T5 = nC4 an – 4 (– 2b)4 T6 = nC5 an – 5 (– 2b)5 T5 + T6 = 0, we get n C 4 2 4 a n – 4 b 4 = nC 5 2 5 a n – 5 b 5 :

and As

) (1 +

47

(b) 1 + (d) 1

we can ignore 25, 45 and 47 in (1) = 25C25 + 25C5 = 1 +

)

(1)

25

25

C5 +

C7

50

25

C5

The expression

(

x5 - 1 + x

) ( 7

-

x5 - 1 - x

(c) 20

)

7

(d) 27

. (a) : Using (a + b)7 – (a – b)7 = 2[7C1 a6 b + 7C3 a4 b3 + 7C5 a2 b5 + 7C7 b7], P(x) = 7C1 (x5 – 1)3 x + 7C2 (x5 – 1)2 x3 + 7C3 (x5 – 1)x5 + 7C7 x7

n

(b)

in

50

we get

If in the expansion of (a – 2b) , the sum of a 5th and 6th term is 0, then value of is b (a)

) (1 +

45

is a polynomial of degree (a) 16 (b) 18

2n - 2 k = 0

Example 6

40

. (a)

P(x) =

 [ Ck x ] n

(1 + ) (1 + ) (1 + is (a) 1 + 25C5 (c) 1 + 25C7

Example 8

 (n Ck x 2 - 1) <  [n Ck x 2 ] £  n Ck x 2

x 2 (2n ) - (n + 1)

25

, we shall ignore all the term with exponent more than 50. Thus, we can write (1) as (1 + 25C1 2 + ... + 25C25 50) ¥ (1 + 25 + 40 + 45 + 47)

2n - 2

nƕ

50

Example 7

Suppose [x] denote the greatest integer Example 5 £ x, and n Œ N, then lim

4!(n - 4)! n !25 ◊ 5!(n - 5)! n !24

a 2 (n - 4) = b 5



k =0

As P(x) is a polynomial, P(x) ∫ 0 " x Œ [0, 1] fi P(x) ∫ 0

=

which is a polynomial of degree 16. The expression

Example 9

(

2 x2 + 1 + 2 x2 - 1

)

6

Ê ˆ 2 +Á ˜ ÁË 2 x 2 + 1 + 2 x 2 - 1 ˜¯

is a polynomial of degree (a) 6 (b) 8

(c) 10

(d) 12

(a). :

We have 2

2 x2 + 1 + 2 x2 - 1

2 =

(

2 x2 + 1 - 2 x2 - 1

(2x

2

) (

)

+ 1 - 2 x2 - 1

)

6

IIT JEE eBooks: www.crackjee.xyz 6.6 Comprehensive Mathematics—JEE Advanced

(a) 22n – 1 (c) 2nCn

2x + 1 - 2x - 1 2

=

2

Thus, the given expression can be written as

(

2 x2 + 1 + 2 x2 - 1 6

) +( 6

6

6

2 x2 + 1 - 2 x2 - 1 6

4 2

6

)

(c) 6

: We have n

2 4

C02 + C12 + C22 +  + Cn2 = Â (nCr) (nCr)

6

But (a + b) + (a – b) = 2[a + C2a b + C4a b + b ]

r=0

Therefore,

(

(b) 2n (2nCn) (d) none of these

n

2 x2 + 1 + 2 x2 - 1 2

3

) +(

2

6

2 x2 + 1 - 2 x2 - 1 2

)

= Â ( nC r) ( nC n – r)

6

= number of ways of choosing n persons out of n men and n women = number of ways of choosing n persons out of 2n persons

2

= 2[(2x + 1) + 15(2x + 1) (2x – 1) + 15(2x2 + 1) (2x2 – 1)2 + (2x2 – 1)3] which is a polynomial of degree 6. Example 10

The expression C0 + 2C1 + 3C2 + … +

(n +1)Cn is equal to (a) 2n –1 (c) n(2n –1) + 2n

= 2nCn

n–1

(b) n(2 ) (d) (n + 1)2n

(c) :

Example 11

Value of the expression

Example 13

+ 2C12 +  + (n + 1) Cn2 is

C02

Let E = C0 + 2C1 + 3C2 + … + nCn – 1 + (n + 1)Cn (1) Using Cr = Cn – r, we can rewrite (1) as E = (n + 1) C0 + nC1 + (n – 1)C2 + … (2) + 2Cn –1 + Cn Adding (1) and (2), we get 2E = (n + 2)C0 + (n + 2)C1 + (n + 2)C2 + … + (n + 2)Cn = (n + 2) {C0 + C1 +… + Cn} = (n +2)2n fi E = (n + 2)2n –1 If n > 1 then value of the expression is

(a) (2n + 1) (2n Cn)

(b) (2n – 1) (2n Cn)

Ên ˆ (c) Á + 1˜ (2n Cn) Ë2 ¯

Ên ˆ (d) Á + 1˜ (2n – 1 Cn) Ë2 ¯

(c) :

Let

S = C02 + 2C12 + … + nCn2 – 1 + (n + 1)Cn2

S = (n + 1) C02 + nC12 + … + 2Cn2 – 1 + Cn2 (2) Adding (1) and (2), we get 2S = (n + 2) [C02 + C12 + … + Cn2] [Example 12] = (n + 2) (2nCn) Ên ˆ S = Á + 1˜ (2nCn). Ë2 ¯



If n ∈ N, then value of

Example 14 (a) – 1 (c) 1

(b) 0 (d) none of these

n

S = Â (– 1) r =0

. (b)

: We have C0x + C1x2 + C2x3 + … + Cnxn + 1 = x(1 + x)n Differentiating both the sides, we get C0 + 2C1 x + 3C2 x2 +  + (n + 1)Cn xn = (1 + x)n + nx(1 + x)n – 1 (1) Putting x = – 1, we get C0 – 2C1 + 3C2 +  + (n + 1) (– 1)n Cn = 0 Example 12

Value of the expression

C02 + C12 + C22 +  + Cn2 is

(1)

Using Cr = Cn – r, we write (1) in the reverse order to obtain

C0 – 2C1 + 3C2 – 4C3 + … + (– 1)n (n + 1)Cn is

[∵ nCr = nCn – r]

r=0

( C) n

r

(

r

r+2

Cr

1 n+2 (c) n + 2 . (b) (a)

)

is

2 n+2 (d) n + 1

(b)

: n

Cr

r+2

Cr

= =

n! r ! 2! ◊ (n - r )! r ! (r + 2) ! 2! (n + 2)! (n + 1) (n + 2) (r + 2)! (n - r )!

IIT JEE eBooks: www.crackjee.xyz Binomial Theorem 6.7

= \

2! (n (n + 1) (n + 2)

+ 2

Cr

+ 2)

S=

n 2 Â (– 1)r (n + 1) (n + 2) r = 0

=

Èn + 2 2 r Í Â (- 1) (n + 1) (n + 2) ÍÎ r = 0

(n

+ 2

(

n+2

+ 2

Cr

n+2

-

Cr

2k

ak =

and

k

" k ≥ n, then bn

Cn

(a) 2n(2n + 1 – 1) (c) 2n(2n – 1)

+ 2)

)

. (a) Put x – 101 = , so that

:

C0 +

n+2

C1 ˘˚

2n

 br

r

\

2n

= Â ar ( + 1)r n

bn n

= Cn an +

For n ∈ N, let n

S(n) = Â (– 1)r r =0 n

Value of S = Â (– 1)r

2n

= Â

Cr r+2

r=0

(a) S(n + 2) –

+ ... +

1 n

n

Cr

1 (n + 1)2 2

n

Cr

+1

2n

k=n

– n

= 2n (2n

+ 1

– 1)

C02 –

1 2 1 2 C1 + C – 2 3 2

+

(- 1)n C2 n +1 n

where Cr = nCr, is (a)

(c) 1

n+2

If n is even, then value of the expression

Example 17

From Example 14 (n + 1) (n + 2) = 2

Cn an

k=n

. (d) Cr

on the R.H.S. of (1)

Cn ak = Â 2k

= 2n  2k

is

(c) 0 (d) none of these

r+2

k

n+1

Cn a2n

2n

1 (b) S(n + 2) + (n + 1)2 2

:

2n

k=n

Cr

(1)

r=0

r=0

2 2 = [0 + (n + 1)] = (n + 1) (n + 2) n+2 Example 15

(b) 2n(2n + 1) (d) 2n + 1(2n – 1)

(- 1)n n ! (n + 1) (n / 2)!2 -1 (n + 1) (n /2)!2

(b)

(d)

(- 1)n - 1 (n !) (n + 1) (n /2)!2 (- 1)n /2 n ! (n + 1) (n /2)! 2

. (d)

Cr + 2

: We have

Thus, n

S=

1 (n + 1)(n + 2) Â (– 1)r 2 r=0

Èn + 2 1 r = (n + 1)(n + 2) Í Â (- 1) 2 ÍÎ r = 0 -

1 n+2

C0 –

Cr + 2

Cr

1 n+2

C0

+

1 n+ 2

˘ ˙ C1 ˙˚

È 1 1 ˘ (n + 1) (n + 2) Í S (n + 2) - 1 + ˙ 2 n + 2˚ Î 1 1 = (n + 1) (n + 2) S(n + 2) – (n + 1)2 2 2 =

Example 16

2n

2n

r=0

r=0

1 [1 – (1 – x)n + 1] (1) ( n + 1) x n 1 1 Ê 1ˆ + C2 2 + ... + Cn Á ˜ and C0 + C1 Ë x¯ x x n = (1 + 1/x) (2) Note that 1 1 1 C 12 + C22 – ... + (– 1)n C2 S = C 02 – n +1 n 2 3 =

1 n+2

1 1 1 C1 x + C2 x2 – ... + (– 1)n C xn n +1 n 2 3

If  ar(x – 100)r =  br(x – 101)r

È 1 1 2 ÍC0 - C1 x + C2 x 2 3 Î 1 1 È ¥ ÍC0 + C1 + C2 2 + x x Î

+ (- 1) n + Cn

˘ 1 Cn x n ˙ n +1 ˚

1˘ x n ˙˚

IIT JEE eBooks: www.crackjee.xyz 6.8 Comprehensive Mathematics—JEE Advanced

xr in the

:

expansion of (1 + x)m is mCr. Thus, A = 2nCn and B =

n 1 È1 - (1 - x)n + 1 ˘ Ê 1ˆ + 1 = Í ˙Á n +1 Î x x ˜¯ ˚Ë

x in 1 [(1 + x)n – (1 – x2)n (1 – x)] n +1 1 = n +1

x

n +1

= =

in (1 – x ) (1 – x)

x2m in (1 – x2)2m

xk in the expansion of

(c)

=

(- 1)n /2 n (- 1)n /2 n ! ( Cn/2) = n +1 (n + 1) (n /2)!2

(b) 3(nCm) (d) 3nCm

(d)

1 - (1 + x) n +1 1 = ÈÎ(1 + x) n +1 - 1˘˚ 1 - (1 + x) x xk in E xk+1 in [(1 + x)n + 1 – 1]

=

n +1

Ck

+1

Example 21 The number of irrational terms in the expansion of (51/6 + 21/8)100 is (a) 96 (b) 97 (c) 98 (d) 99 (b)

:

We have (1 + x)n (1 + y)n (1 + z)n

:

Ê n n ˆÊ n n ˆÊ n n ˆ =Á Cr x r ˜ Á Cs y s ˜ Á Ct z t ˜ Ë r =0 ¯ Ë s =0 ¯ Ë t =0 ¯

Â

Â

Â

=

0£ r , s , t £ n

Â

( n Cr )( n Cs )( n Ct ) x r y s z t m, we must have

r+ + =m where r, , are integers with r, ,

≥ 0.

 ( C )( C )( C ) n

n

r

n

s

t

r , s ,t ≥ 0 r + s +t = m

= the number of ways of choosing a total number of m balls out of n black, n white and n green balls = 3nCm xn in the Example 19 If A and B 2n 2n – 1 expansions of (1 + x) and (1 + x) respectively, then A/B (a) 1 (b) 2 (c) 1/2 (d) 1/n (b)

(2n)! (n !) (n - 1)! 2n =2 = n n ! n ! (2n - 1)!

E = 1 + (1 + x) + (1 + x)2 + … + (1 + x)n is (b) n +1Ck (a) nCk (c) n +1Ck +1 (d) nCk + 1

Example 18 Sum of the coefficients of the terms of degree m in the expansion of (1 + x)n (1 + y)n (1 + z)n is

=

Cn

=

C n.

: We have E = 1 + (1 + x) + (1 + x)2 + … + (1 + x)n

1 (– 1)m (2mCm) 2m + 1

(a) (nCm)3 (c) nC3m

Cn

2 n -1

Example 20 2 n

As n is even, let n = 2m, then 1 S= 2m +1

A = B

We have

n + 1

2n

2n – 1

(r +1)th in the expansion of (51/6 + 21/8)100 is

Cr (51/6)100 – r (21/8)r As 5 and 2 are relatively prime, r +1 will be rational r 100 - r if and are both integers. i.e. if 100 – r is a 8 6 multiple of 6 and r is a multiple of 8. As 0 £ r £ 100, multiples of 8 upto 100 and corresponding value of 100– r are r = 0, 8, 16, 24, …, 88, 96 100 – r = 100, 92, 84, 76, …, 12, 4 Out of values of 100 – r, multiples of 6 are 84, 60, 36, 12. \ there are just four rational terms. fi number of irrational terms is 101 – 4 = 97 r+1 =

100

Example 22 Sum of the last 30 coefficients in the expansion of (1 + x)59, when expanded in ascending powers of x, is (a) 229

(b) 228

(c)

60

C30219

(d) 258

(d) : There are 60 terms in the expansion of (1 + x)59 S = 59C30 + 59C31 + … + 59C58 + 59C59

IIT JEE eBooks: www.crackjee.xyz Binomial Theorem 6.9



C1 + 59C0 [using nCr = nCn – r] Adding the above two expansions, we get 2S = 59C0 + 59C1 + … + 59C59 = 259 fi S = 258 S=

59

C29 +

59

The sum of the rational terms in the

Example 23

(

1/5 expansion of 2 + 3

)

=

20

20

Cr 24–r/5 (3)r/2

-

Example 24 m

x is

Also,

Ê 2 1ˆ Now, (r + 1)th term of Á x + ˜ Ë x¯

S= 1¥2¥3¥4+2¥3¥4¥5 + ... + n (n + 1) (n + 2) (n + 3) is 1 n(n + 1) (n + 2) (n + 3) (n + 4) 5 1 n+ 3 ( C 5) (b) 5!

– r

Ê 1ˆ ÁË x ˜¯

m

is

r

= mCr x2m

1 n+ 4 ( C 4) 5 1 n ( C 4) (d) 5 . (a) (c)

– 3r

:

For this to be independent of x is 2m – 3r = 0 fi r = 6 \ term independent of x is 9C6 = 84. Example 25 (a)

2n

Cn

Thus,

(c)

2n

Cn

Value of  (Ck) (Ck – 1) is k =1

(b) +2

(d)

1 2n ( 2 2n

+ 2

Cn + 1

k(k + 1) (k + 2) (k + 3) = 4!

n

. (b)

Value of

(a)

: We are given C0 + mC1 + mC2 = 46 fi 2m + m(m – 1) = 90 m2 + m – 90 = 0 fi m = 9 as m > 0

Cr(x2)m

(1 + x)n x

1 2n + 2 ( Cn + 1) – 2nCn = 2nCn – 1 2

Example 26

m

m

n

= 2nCn–1

. (a)



1ˆ ˜ x¯

xn–1 in (1 + x)2n

is 46,

(d) 106

+ Cn x n + 1 ˘˚

Ê = constant term in Á1 + Ë

r/2

(c) 98

(2)

– 1)

¥ ÈÎC0 x + C1 x 2 +

As 2 and 3 are relatively prime 2 and 3 will be rational if both 4– r/5 and r/2 are integers, i.e. if r/5 and r/2 are integers. This is possible if r = 0, 10, 20. Thus, the sum of rational terms is 20 C0(24) (30) + 20C10 (22) (35) + 20C20 (20) (310) As this is more than 71, 85 or 97, the answer is (d).

(b) 92

n

= constant term in 1 1˘ È ÍC0 + C1 x + + Cn x n ˙ Î ˚

is

Ê 2 1ˆ ond and third terms of the expansion of Á x + ˜ Ë x¯

1ˆ x ˜¯

n

20

4 – r/5

(1)

Now,

(b) 85 (d) none of these

)

= x(1 + x)n

1 1 Ê + ... + Cn n = Á1 + Ë x x

k =1

Cr (21/5)20 – r (31/2 ) r =

(a) 84

C0 + C1

is

(

+ 1

and

 (Ck) (Ck

1/5 (r + 1)th term of 2 + 3

:

: We have C0 x + C1 x2 + ... + Cn xn

20

(a) 71 (c) 97 (d)

r+1

59

C28 + … +

Cn

+ 1)



2n

Cn

(k + 3)! = 4! (k + 3C4) (k - 1)!4!

S = 4! (4C4 + 5C4 + ... + n + 3C4) x4 in (1 + x)4 + (1 + x)5 + ... + (1 + x)n + 3 ] x4 in (1 + x )n - 1˘˙ (1 + x)4 1+ x -1 ˙ ˚

IIT JEE eBooks: www.crackjee.xyz 6.10 Comprehensive Mathematics—JEE Advanced

x5 in (1 + x)n + 4]

Also, we have

= 4! [n + 4C5] 1 = n(n + 1) (n + 2) (n + 3) (n + 4) 5

Ê 1 ˆ ÁË 5/3 ˜¯ 3

log3 8

=

n

Ê 3 1ˆ If in the expansion of Á x - 2 ˜ , Ë x ¯ x5 and x10 is 0, then value

Example 27 n∈N of n is (a) 5 (c) 15

(b) 10 (d) 20

(-1)n

Thus,

Now,

5

=

4+1

(r + 1)th term in the expansion of

= =

Ê 3 1ˆ ÁË x - 2 ˜¯ is x n

(3 )

(-1)n 2n /2

10

Cr(x3)n – r

= nCr(– 1)r x3n–5r 3n - 5 5

x10, we set

=p 3n – 5r = 10 fi r =

3n - 10 = q (say) 5

Note that p – q = 1. We are given

( )

1 ˆ ÁË ˜ 2¯

C4 2

( )

10! 1/3 2 4!6!

6

Cp(– 1)p + nCq (–1)q = 0



n

Cp(– 1)p + nCp –1 (–1)p –1 = 0



n

Cp = nCp –1 fi n – p = p – 1



n +1 n = 2p – 1 fi p = 2

Thus,

n +1 3n - 5 = fi 5n + 5 = 6n – 10 2 5



15 = n

(a) (c) . (b) : fi (1

( -1)4 ( 2-1/2 )

x20 in (1 – x + x2)20

a=b a b (d) a + b = 0

Ê1 ˆ ÁË + 1 + px˜¯ x But

Ê1 ˆ ÁË + 1 + px˜¯ x +

\

20

20

=1+

Ê1 ˆ C2 Á + px˜ Ëx ¯

a = b20 ] p = 1

Thus,

n

Ê 1 ˆ ÁË ˜ 2¯

2

+ .... +

20

C20

Ê1 ˆ ÁË + px˜¯ x

20

+ .... + (20C20) (20C10)p10 and b = b20 ] p =–1

Example 30 terms in the expansion of (1 + x)n are in A.P., then value of n is (a) 5 (b) 7 (c) 11 (d) 14 (b)

Ê 1/3 1 ˆ : Last term of Á 2 ˜ is Ë 2¯ n-n

Ê1 ˆ C1 Á + px˜ Ëx ¯

a > b.

Hence,

(b) 420 (d) 35

20

20

b20 = 1 + (20C2) (2C1) p + (20C4) (4C2)p2

, then the 5th term from the

(a)

( )

= 210

Dividing both the sides by x20, we get b20

log3 8

1/3 n n+1 = Cn 2

4

Let (1 – x + x2)20 = a0 +a1x + .... + a40x40 + x + x2)20 = a0 – a1x + a2x2 – .... + a40x40 x20 in (1 – x + x2)20 x20 in (1 + x + x2)20 Therefore, we consider the expansion

If the last term in the binomial expansion n

4

(1 + x + px2)20 = b0 + b1x +b2x2 + .... + b40x40

n

Ê 1/3 1 ˆ Ê 1 ˆ of Á 2 is Á 5/3 ˜ ˜ Ë3 ¯ Ë 2¯ beginning is (a) 210 (c) 105

25

and in (1 + x – x2)20 are respectively a and b, then

r

x5, we get 3n – 5r = 5 fi r =

Example 28

(-1)10

1/3 10 - 4 Ê

Example 29 Ê 1ˆ ÁË - 2 ˜¯ x

=

n/2 = 5 fi n = 10

(c) :

= 3- (5/3) log3 2 = 2–5 3

5/3 3 log3 2

-5 = 2 fi

2n /2



1

n

=

(-1)n 2n /2

:

Note that n ≥ 4. As C1, C2 and C3 are in A.P.,

IIT JEE eBooks: www.crackjee.xyz Binomial Theorem 6.11

2C2 = C1 + C3 C C 2= 1 + 3 C 2 C2

fi fi

Also I + F + f =

2 n–2 + fi n = 7 as n ≥ 4 n –1 3

2=

Example 31 of x in the expansion of

)

2 +1

6

+

(

)

2 -1

6

= 2[6C0 ◊ 23 + 6C2 ◊ 22 + 6C4 ◊ 2 + 6C6] = 2(8 + 60 + 30 + 1) = 198 Hence, F + f = 198 – I is an integer. But 0 < F + f < 2. Therefore, F + f = 1, and thus, I = 197. Example 33

Ê3 2 1 ˆ (1 + x + 2x3) Á x - ˜ Ë2 3x ¯

(

9

Sum to (n + 1) terms of the series C0 C1 C2 C3 + + 2 3 4 5

is

is (a) 1/3 (c) 17/54 (c)

(b) 19/54 (d) 1/4

: The (r + 1)th term in the expansion of [(3/2)x2 – (1/3x)]9 is given by Ê 3 2ˆ Tr + 1 = Cr Á x ˜ Ë2 ¯

9-r

9

r = Cr (- 1) 9

Ê 1ˆ ÁË - ˜¯ 3x

39 - 2 r 29 - r

(b)

(c)

1 n (n + 1)

(d)

1 n+2 1

( n + 1)( n + 2)

: We have (1 – x)n = C0 – C1x + C2x2 – C3x3 +  x (1 – x)n = C0 x – C1x2 + C2 x3 – C3x4 + 

fi (1)

1 n +1

(d).

r

x18 - 3r

(a)

1

1

0

0

n fi Ú x(1 - x) dx = Ú (C0x – C1x2 + C2x3 – ) dx

(1) independent of x in the expansion of Ê3 2 1 ˆ (1 + x + 2x3) Á x - ˜ Ë2 3x ¯

1 0

n LHS of (1) = Ú (1 - x) x d x

9

È Î

(2)

a

a

Ú0 f ( x) d x = Úo f (a - x) dx

x0, x–1 and x–3 in the expansion of [(3/2)x – (1/ 3x)] . For x0, r must be 6 in (1); for x–1, there is no value of r; and for x–3, r must be 7

Ê xn + 1 xn + 2 ˆ ˘ = Á ˙ n + 2 ˜¯ ˙˚ Ë n +1 0

of x in (2) is

=

2

1 ◊ 9C6 (–1)6 ◊ =

1

9

39 - 12 2 9-6

+ 2 ◊ 9C7 (- 1)7 ◊

39 - 14

Ê C x 2 C x3 C x 4 RHS of (1) = Á 0 - 1 + 2 3 4 Ë 2

2 9-7

9 ◊ 8 ◊ 7 3-3 9 ◊8 3-5 7 2 17 ◊ 3 + 2◊ (- 1) ◊ 2 = = 1◊ 2 ◊ 3 2 1◊ 2 18 27 54 2

C0 C1 C2 + 2 3 4 C0 C1 C2 1 + - = 2 3 4 (n + 1) (n + 2) =

Example 32

(

)

2 +1

Thus

6

is

(a) 196

(b) 197

. (b). : Let 0 £ F < 1. Let

f = 2 –1=

(

(

)

2 +1

)

(c) 198

(d) 199

= I + F, where I is an integer and

2 - 1 . We have 1 2 +1

Example 34

6

6

fi0
j (ix) lower triangular matrix if m = n and aij = 0 "i 0 (b) bc = 0 Example 4

Ê 1 ˆ (c) bc > min Ë0, ad ¯ 2

(d) a = 0

Ans. (c) As A – a I is invertible for all a Œ R. det (A – a I) π 0 " a Œ R. fi (a – a) (d – a) – bc π 0 " a Œ R. 2 " a Œ R. fi a – (a + d)a + ad – bc π 0 Therefore (a + d)2 – 4(ad – bc) < 0 fi (a – d)2 + 4 bc < 0 Therefore, bc < 0. Also, a2 + d2 – 2ad + 4bc < 0 fi 0 £ a2 + d2 < 2ad – 4bc 1 fi bc < ad. 2 Solution:

Ê 1 ˆ bc < min Ë0, ad ¯ 2

Thus, Example 5

If A + B is a non-singular matrix, then

A – B – A (A + B)– 1 A + B(A + B)– 1 B equals (a) O Ans. (a)

(b) I

(c) A

(d) B

Solution: A – B – A(A + B)– 1 A + B (A + B)– 1 B = A[I – (A + B)– 1 A] – [I – B(A + B)– 1] B = A(A + B)– 1 [A + B – A] – [A + B – B] (A + B)– 1 B –1 A(A + B) B – A (A + B)– 1 B =O Example 6

If P is a 3 ¥ 3 matrix such that

P ¢ = 2P + I, then there exists a column matrix Ê xˆ X = Á y˜ π O such that Á ˜ Ë z¯ (a) PX = O (c) PX = 2X

(b) PX = X (d) PX= –X

Solution: We have P = (P¢)¢ = (2P + I )¢ = 2P¢ + I P = 2(2P + I ) + I fi P = –I Thus, there exists X π 0 such that PX = –X Example 7 3

3

Let P and Q be 3 ¥ 3 matrices with P π Q 2

If P = Q and P Q = Q2P, then determinant of(P2 + Q2) is equal to (a) 1 (b) 0 (c) – 1 (d) –2 Ans. (d) Solution: P3 = Q3 and P2Q = Q2P gives fi fi

P 3 – P 2Q = Q 3– Q 2P P2(P – Q) = – Q2(P – Q) (P2 + Q2) (P – Q) = 0

If det (P2 + Q2) π 0, then P2 + Q2 is invertible and hence P = Q. Therefore, det (P2 + Q2) = 0.

Example 8

È 1 0 0˘ Let P = Í 9 1 0˙ and Q = ÈÎqij ˘˚3¥3 Í ˙ ÍÎ27 9 1˙˚

be such that P5 – Q = I, then

q21 + q31 is equal to q32

(a) 22 (b) 33 (c) 44 Ans. (a) Solution: Write P = I + R where È 0 0 0˘ R = Í 9 0 0˙ Í ˙ ÍÎ27 9 0˙˚

È 0 0 0˘ R = Í 0 0 0 ˙ , R3 = 0 Í ˙ ÎÍ81 0 0˙˚ 2

fi Rk = 0 " k ≥ 4. We have P5 = (I + R)5 = I + 5R + 10R2 fi Q = P5 – I = 5R + 10R2

0 0˘ È 0 Í = 45 0 0˙ Í ˙ ÍÎ945 45 0˙˚ q21 + q31 45 + 945 = = 22 q32 45

(d) 55

IIT JEE eBooks: www.crackjee.xyz Matrices 7.9

Let aij = ( 2 + 3 ) + ( 2 - 3 ) Example 9 j £ 3 and let A = (aij) 3 ¥ 3, then det (A) is equal to i+ j

(a) 1 (c) Ans. (d) Solution: =

(b)

(2 - 3 )

9

i+ j

, 1 £ i,

( 2 + 3 )9

(d) 0

For n ΠN, an + an + 2

(2 + 3 )

n

È1 + ( 2 + 3 ) Î ˚ 2˘

( 2 + 3 ) n (8 + 4 3 ) + ( 2 - 3 ) n (8 - 4 3 )

= 4 an + 1 Applying R1 Æ R1 + R3 – 4R2, we get det (A) = 0 Alternative Solution Let a = 2 + 3 then 2 - 3 = 1/a = a – 1 we are write A as Ê a2 Á A = Á a3 Á 4 Ëa



a -2 a -3 a -4

0ˆ Ê 1 a a2 ˆ ˜Á ˜ 0˜ Á 1 1/a 1/a 2 ˜ ˜Á 0 ˜¯ 0¯ Ë 0 0

We have

AA–1 = A–1 A = In



(AA–1)¢ = (A–1 A)¢ = (In)¢



(A–1)¢ A¢ = A¢(A–1)¢ = In



(A–1)¢ A = A(A–1)¢ = In (A–1)¢ = A–1

equal to (a) Aa + b (c) Aa – b Ans. (a) Solution. We have

Ans. (b) Solution: We have A¢ = – A Now, AA–1 = A–1 A = In fi

(AA–1)¢ = (A–1 A)¢ = (In)¢ (A–1)¢ A¢ = A¢(A–1)¢ = In

fi (A–1)¢ (–A) = (– A) (A–1)¢ = In

sin a ˘ , then Aa Ab is cos a ˙˚ (b) Aab (d) none of these

sin a ˘ cos a ˙˚

È cos a cos b - sin a sin b = Í Î- sin a cos b - cos a sin b

È cos b Í- sin b Î

[inverse of a matrix is unique]

Example 12 The inverse of a skew symmetric matrix (if it exists) is (a) a symmetric matrix (b) a skew symmetric matrix (c) a diagonal matrix (d) none of these

Thus,

È cos a If Aa = Í Î- sin a

È cos a Aa Ab = Í Î- sin a

Let A be an invertible symmetric matrix.



det (A) = 0

Example 10

Solution:



n 2 + ( 2 - 3 ) ÈÎ1 + ( 2 - 3 ) ˘˚

=

Ans. (a)

sin b ˘ cos b ˙˚

cos a sin b + sin a cos b ˘ - sin a sin b + cos a cos b ˙˚

È cos (a + b ) sin (a + b ) ˘ = Í ˙ = Aa + b Î- sin (a + b ) cos (a + b ) ˚ Example 11 The inverse of a symmetric matrix (if it exists) is (a) a symmetric matrix (b) a skew symmetric matrix (c) a diagonal matrix (d) none of these

(A–1)¢ = – (A–1) [inverse of a matrix is unique]

Example 13 The inverse of a skew symmetric matrix of odd order is (a) a symmetric matrix (b) a skew symmetric matrix (c) diagonal matrix (d) does not exist Ans. (d) Let A be a skew symmetric, matrix of order n. fi fi \

A–1

A¢ = – A fi |A¢| = |– A| |A| = (– 1)n |A| fi |A| = – |A| [ 2|A| = 0 fi |A| = 0 does not exist.

Example 14

n is odd]

If A is an orthogonal matrix, then |A| is

(a) 1 (b) – 1 (c) ± 1 (d) 0 Ans. (c) As A is an orthogonal matrix, A¢ A = AA¢ = In fi |A¢ A| = |AA¢| = |In| fi |A¢| |A| = 1 fi |A| |A| = 1 fi |A|2 = 1 fi |A| = + 1

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Example 15 If A is a 3 ¥ 3 non-singular matrix, then adj (adj A) is equal to (a) |A| A (b) |A|2 A (d) 0 (c) |A|–1 A Ans. (a) 1 As A is a non-singular matrix A–1 = (adj A) A

È - x 14 x 7 x ˘ Í 1 0 ˙˙ B= Í 0 ÍÎ x -4 x -2 x ˙˚ is (a)

-1 = A A (|A| A–1)–1

= |A|3 |A–1| |A|–1 (A–1)–1 [using scalar multiple property of determinants] 1 1 . A = |A|A = |A|3 A A If A and B are two square matrices such

–1

that B = – A BA, then (A + B)2 is equal to (a) O (b) A2 + B2 (c) A2 + 2AB + B2 (d) A + B Ans. (b) –1

Solution: As B = – A BA, we get AB = – BA or AB + BA = O. Now, (A + B)2 = (A + B) (A + B) = A2 + BA + AB + B2 = A 2 + O + B 2 = A 2 + B 2. Èa b ˘ 2 Example 17 If A = Í ˙ is such that A = I, then g a Î ˚

Thus,

(b) 1 – a2 – bg = 0 (d) 1 + a2 – bg = 0

ab - ab ˘ ˙ bg + a 2 ˙˚

A2 = I fi a2 + bg = 1, or 1 – a2 – bg = 0.

Example 18

(c)

1 4

(d)

1 5

0 0˘ È5 x È1 0 0 ˘ Í0 ˙ 1 0 ˙ = ÍÍ0 1 0˙˙ AB = Í ÍÎ 0 10 x - 2 5 x ˙˚ ÍÎ0 0 1˙˚

adj (adj A) = adj (B) = |B|B–1

È a 2 + bg A2 = Í ÍÎ ag - ag

1 3

We have

fi adj A = |A| A = B (say). Now,

(a) 1 + a2 + bg = 0 (c) 1 – a2 + bg = 0 Ans. (b) We have

(b)

Ans. (d)

–1

Example 16

1 2

The value of x for which the matrix

È2 0 7˘ Í ˙ A = Í 0 1 0 ˙ is inverse of ÎÍ 1 -2 1 ˙˚



x = 1/5

È cos a sin a 0 ˘ Í ˙ Example 19 If A(a, b) = Í- sin a cos a 0 ˙ , Í 0 0 eb ˙˚ Î –1 then A(a, b) is equal to (a) A (– a, b) (b) A (– a, – b) (c) A (a, – b) (d) A (a, b) Ans. (b) We have Èeb cos a 1 Í b A (a, b)–1 = b Í e sin a e Í 0 ÍÎ = A (– a, – b)

- eb sin a eb cos a 0

0˘ ˙ 0˙ ˙ 1˙ ˚

Example 20 If A is a 3 ¥ 3 skew-symmetric matrix, then trace of A is equal to (a) 1 (b) 3 (c) – 1 (d) |A| Ans. (d) As A is a skew symmetric matrix A¢ = – A fi aii = 0 " i fi trace (A) = 0 Also |A| = |A¢| = |– A| = (– 1)3 |A| fi 2|A| = 0 fi |A| = 0 \ trace (A) = |A| Example 21

Ê 1 If A(q) = Á Ë - tan q

then (sec2 q) B is equal to (a) A (q) (c) A (q/2) Ans. (b)

tan q ˆ and AB = I, 1 ˜¯

(b) A (– q) (d) A (– q/2)

IIT JEE eBooks: www.crackjee.xyz Matrices 7.11

As AB = I, we get B = A–1 \

Èa b ˘ 2 If A = Í ˙ and A = b a Î ˚

Example 22 (a) (b) (c) (d)

a a a a

= = = =

a 2 + b 2, b = a 2 + b 2, b = 2ab, b= a 2 + b 2, b =

Èa b ˘ Í b a ˙ , then Î ˚

2ab a2 – b2 a2 + b2 ab

We have Èa 2 + b 2 Èa b ˘ Èa b ˘ Í = A2 = Í ˙ Í ˙ Îb a ˚ Îb a ˚ ÍÎ 2ab 2 2 fi a = a + b , b = 2ab

2ab ˘ Èa b ˘ ˙=Í ˙ b 2 + a 2 ˙˚ Î b a ˚

We have |A| = (a + ib) (a – ib) – (– c + id) (c + id) = a2 + b2 + c2 + d2 =1 È a - ib - c - id ˘ A–1 = Í ˙ Îc - id a + ib ˚

Example 26 The number of values of k for which the system of equations (k + 1)x + 8y = 4k, kx + (k + 3)y = 3k – 1 has no solution is (a) 0 (b) 3 (c) 2 (d) 1 For the system of equations to have no solution, we must have k +1 8 4k π k k +3 3k - 1 k2 + 4k + 3 = 8k fi k2 – 4k + 3 = 0

Ans. (d) As |A| = 0, we get ad – bc = 0 Also, A – (a + d)A = A[A – (a + d)I] Ê a bˆ Ê - d b ˆ Ê - ad + bc ab - ba ˆ = Á = Á ˜ Á ˜ c d c a ¯ Ë - cd + cd bc - ad ˜¯ Ë ¯ Ë 2

Ê 0 0ˆ = Á Ë 0 0˜¯

Example 24 equal to (a) ± 3



k = 1, 3. For k = 1, k = 3,

Example 27

Èa 2 ˘ 3 If A = Í ˙ and |A | = 125, then a is 2 a Î ˚ (b) ± 2

(c) ± 5 3

È a + ib If A = Í Î- c + id

3

c + id ˘ and a - ib ˚˙

a2 + b2 + c2 + d2 = 1, then A–1 is equal to

8 4k π k + 3 k -1 If È1 1˘ Í0 1˙ Î ˚

and Q = PAP¢ then P¢ Q2019 P is equal to

(d) 0

We have 125 = |A | = |A| fi |A| = 5 fi a2 – 4 = 5 fi a2 = 9 fi a= ± 3

8 k +1 = and for k +3 k

È cos (p /6) sin (p /6) ˘ P= Í ˙, A= Î- sin (p /6) cos (p /6) ˚

– kI = 0 fi k = 0

Ans. (a)

Example 25

(d) none of these.

Ans. (d)

Ê a bˆ is such that |A| = 0 and Example 23 If A = Á Ë c d ˜¯ A2 – (a + d)A + kI = O, then k is equal to (a) b + c (b) a + d (c) ab + cd (d) 0

Thus,

c - id ˘ a + ib ˚˙

Ans. (c)

Thus,

Ans. (a)

- c + id ˘ È a - ib (b) Í a - ib ˚˙ Î- c - id

È a - ib - c - id ˘ (c) Í ˙ Îc - id a + ib ˚

- tan q ˆ = A(– q) 1 ˜¯

Ê 1 (sec2 q) B = Á Ë tan q



È a + ib (a) Í Î- c + id

- tan q ˆ 1 ˜¯

Ê 1 B= Á 2 1 + tan q Ë tan q 1

È1 3 2˘ (a) Í ˙ Î0 2019 ˚

È1 2019˘ (b) Í 1 ˙˚ Î0

È 3 2 2019˘ (c) Í ˙ 1 ˚ Î 0

È 3 2 - 1/2 ˘ (d) Í ˙ 2019˚ Î 1

Ans. (a) Solution Now, fi

Note that P¢ = P–1. Q = PAP¢ = PAP–1 Q2019 = PA2019 P–1

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\

P¢ Q2019P = P–1 (PA2019 P–1) P = A2019 = (I + B)2019 Ê 0 1ˆ where B = Á . As B2 = O, we get Br = O " r ≥ Ë 0 0˜¯ 2. Thus, by binomial theorem,

Example 28

\

Ê 1 kˆ , k Œ N. Then (S2)n (Sk)–1 Let Sk = Á Ë 0 1˜¯

(where n ΠN) is equal to: (a) S2n + k

c1 ˘ ˙ c2 ˙ and |A| π 0, then the c3 ˙˚

b1 b2 b3

system of equations a1 x + b1 y + c1 z = 0, a2 x + b2 y + c2 z = 0 and a3 x + b3 y + c3 z = 0 has

(c) S2n + k -1

Solution:

Ê 1 2ˆ =I+B S2 = Á Ë 0 1˜¯

Ans. (a) Note that A is invertible as |A| π 0. Now, AX = O fi A–1 (AX) = O fi (A–1 A)X = O or IX = O or X = O \ x = y = z = 0 is the only solution of the system of equations. If

- tan q ˆ Ê 1 1 ˜¯ ÁË - tan q

tan q ˆ 1 ˜¯

-1

Èa - b ˘ = Í ˙, Îb a ˚

then a a a a

= = = =

b=1 cos 2q, b = sin 2q sin 2q, b = cos 2q 1, b = sin 2q

Ans. (b) We have È 1 Í- tan q Î

tan q ˘ 1 ˙˚

where

Ê 1 0ˆ Ê 0 2ˆ and B = Á I= Á ˜ Ë 0 1¯ Ë 0 0˜¯ n

(a) (b) (c) (d)

(d) S2n - k

Note that B2 = O, therefore, Br = O

(c) no solution

Ê 1 ÁË tan q

(b) S2n – k

Ans. (b)

(a) only one solution

Example 29

- sin 2q ˘ cos 2q ˙˚

a = cos 2q, b = sin 2q

Example 30

Ê 1 2019ˆ . A2019 = I+2019 B = Á 1 ˜¯ Ë0 È a1 Í If A = Ía2 ÍÎ a3

Ècos 2q = Í Î sin 2q

-1

=

È 1 Í 1 + tan q Î tan q

Èa - b ˘ È 1 \ Í = cos2 q Í ˙ Îb a ˚ Î tan q

1

2

- tan q ˘ 1 ˙˚

- tan q ˘ È 1 1 ˙˚ ÍÎ tan q

È1 - tan 2 q - 2 tan q ˘ ˙ = cos2 q Í 2 ÎÍ 2 tan q 1 - tan q ˚˙

- tan q ˘ 1 ˙˚

" r ≥ 2.

n

Thus, (S2) = (I + B) = I + nB Ê 1 2 nˆ = Á Ë 0 1 ˜¯ Also,

Ê 1 -k ˆ Sk-1 = Á Ë 0 1 ˜¯

Ê 1 2 nˆ Ê 1 - k ˆ \ (S2)n(Sk)–1 = Á Ë 0 1 ˜¯ ÁË 0 1 ˜¯ Ê 1 2n - k ˆ = S2 n - k = Á Ë0 1 ˜¯

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS Example 31 Let A, B, C be three 3 ¥ 3 matrices with entries from R such that ABC + AB + BC + AC + A + B + C = O. (1) (a) If A commutes with B + C then A commutes with BC (b) If A commutes with BC then A commutes with B + C (c) A + I is invertible (d) B + I is invertible Ans. (a), (b), (c), (d) Solution: We can write (1) as A(BC + B + C + I) + BC + B + C + I = I fi A(B + I) (C + I) + (B + I) (C + I) = I fi (A + I) (B + I) (C + I) = I

IIT JEE eBooks: www.crackjee.xyz Matrices 7.13

Let aij = (i, j)the element of P2, then



A + I is invertible and (A + I) – 1 = (B + I) (C + I) Post multiplying by A + I, we get fi (B + I) (C + I) (A + I) = I fi B + I is invertible A commutes with B + C ¤ A(B + C) = (B + C)A ¤

¤ ¤

(3)

(From (3) and (2)) (BC)A + BA + CA + BC + B + C + A + I = A(BC) + AB + BC + AC + A + B + C + I (BC)A = A(BC) BC commutes with A.

Example 32 Suppose A, B are two distinct 2 ¥ 2 matrices such that A2 – 5A + 6I = 0 B2 – 5B + 6I = 0 Then (a) A3 – B3 = 19(A – B) (b) A4 – B4 = 65 (A – B) (c) A5 – B5 = 211(A – B) (d) A6 – B6 = 665(A – B) Ans. (a),(b), (c), (d) Solution: A3 – B3 = (5A2 – 6A) – (5B2 – 6B) = 5(A2 – B2) – 6(A – B) = 5[5A – 6I – 5B + 6I] – 6(A – B) = 19(A – B) Next A4 – B4 = 5(A3 – B3) – 6(A2 – B2) = 5(19)(A – B) – 6 (5)(A – B) = 65(A – B) A5 – B5 = 5(A4 – B4) – 6(A3 – B3) = 5(65)(A – B) – 6(19)(A – B) = 211(A – B) A6 – B6 = 5(A5 – B5) – 6(A4 – B4) = 5(211)(A – B) – 6(65)(A – B) = 665(A – B) Example 33 Let w π 1 be cube root of unity and let P = pij be n ¥ n matrix such that pij = w i + j. Then P2 π O when n is (a) 57 (b) 55 (c) 58 (d) 56 Ans. (b), (c), (d) Solution: È w2 Í 3 Íw P= Í Í ÍÎw 1+ n

w3 w4 w 2+ n

w 1+ n ˘ ˙ w 2+ n ˙ ˙ ˙ w n + n ˙˚

Ï0 if n = 3k Ô if n = 3k + 1 aij = Ìw Ô 2 Ó-w if n = 3k + 2 In fact aij = 0 for each i, j if n = 3k. Thus, P2 = O when n = 3k, and P2 π O when n = 3k + 1, 3k + 2 P2 π O for n = 55, 58, 56. Example 34 Let A = (aij)3 ¥ 3, B = (bij)3 ¥ 3 be two real symmetric matrices such that A2 + B2 = O, then (a) AB = O (b) A = O (c) B = O (d) det(A) + det(B) + tr(A) + tr(B) = 0 Ans. (a), (b), (c), (d) n

2

Solution: (i, j) th element of A =

 aik a jk k =1

n

= Â aik a jk

[

A is symmetric]

k =1

n



(i, i)th element of A2 =

 aik2 k =1

n

Thus (i, i)th element of A2 + B2 =

 (aik2 + bik2 ) k =1

As A2 + B2 = O, we get n

 (aik2 + bik2 ) = 0 for 1 £ i £ n k =1



aik = 0, bik = 0 for 1 £ I , k £ n Therefore, A = O, B = O

Example 35 Let A be 4 ¥ 4 matrix with real entries such that determinant of every 2 ¥ 2 submatrix is 0, then (a) adj(A) = O (b) det(A) = 0 (c) A = O (d) AX = O Ans. (a), (b), (d) Solution: As determinant of every 2 ¥ 2 submatrix is 0. fi determinant of every 3 ¥ 3 submatrix is 0. fi adj(A) = 0 Also (det(A))3 = det(adjA) = 0 fi det(A) = 0 Thus, AX = O

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È3 -1 -2˘ Í ˙ Example 36 Let P = Í2 0 a ˙ , where a ∈ R, ÍÎ3 -5 0 ˙˚ Suppose Q = [qij] is a matrix such that PQ = kI, where k ∈ k R, k π 0 and I is the identity matrix of order 3. If q23 = 8 2

k , then 2 (a) a = 0, k = 8 (b) 4a – k + 8 = 0 (c) det(P adj(Q)) = 29 (d) det(Q adj(P)) = 213 Ans. (b), (c) Solutions: PQ = kI fi Q = (kI) P–1 = kP–1 Since det P.det Q = k π 0, so P is invertible and cofactor of (i, j)th element of P are P12 = 3a, P13 = –10 P11 = 5a, P22 = 6 P23 = 12 P21 = 10, P32 = – 3a – 4 P33 = 2 P31 = – a, det(P) = 3P11 + (– 1) P12 + (– 2)P13 and det(Q) =

=

= k3 (12a + 20) = 43 (8) = 2623 = 29 Example 37

Let M be a 2 × 2 symmetric matrix with

integer entries. Then M is invertible if M is the transpose of the second row of M (b) the second row of M is the transpose of the M (c) M is a diagonal matrix with nonzero entries in the main diagonal (d) the product of entries in the main diagonal of M is not the square of an integer Ans. (c), (d) Solution: As M is a 2 × 2 symmetric matrix, Èa b ˘ M= Í ˙ Îb c ˚

= 15a – 3a + 20 = 12a + 20 Now,

k ( -3a - 4) 1 Ê ˆ P32 = q23 = k Á ˜ Ë 12a + 20 ¯ 12a + 20 –



k ( -3a - 4) k = 8 12a + 20



If

a=–1 det(Q) = k3 det(P–1) =

fi fi



k3 12a + 20

k2 k3 = fi 6a + 10 = k 2 12a + 20 k=4

Now, det(P(adj Q)) = det (P) det (adj Q) = det (P) (det (Q))2 6

=

6

k k = 12 20 12 + + 20 a a ( )

46 212 = 3 = 29 8 2

Next, det (Q (adj (P))) = det Q det (adj (P)) = (det Q) (det (P))2

Èa ˘ M is Í ˙ and second row of M is Îb ˚ Èa ˘ Èb ˘ Íb ˙ = [b c]¢ = Íc ˙ Î ˚ Î ˚

a=b=c and hence M is not invertible The second row of M is [b c

Èa ˘ Íb ˙ . Î ˚

Èa ˘¢ If [b c] = Í ˙ = [a b], we get a = b = c and hence M Îb ˚ is not invertible. If M is a diagonal matrix with non zero entries, then

\ 4a – k + 8 = – 4 – 4 + 8 = 0

= (12a + 20)

where a, b, c ∈ I.

[b c].

12a + 20 = 24a + 32

k3 (12a + 20)2 (12a + 20)

Èa 0˘ M= Í ˙ and det(M) = ac π 0 Î0 c ˚ ∴ M is invertible. If the product of main diagonal elements is not a perfect square, then det(M) = ac – b2 π 0 and hence M is invertible. Example 38 Let X and Y be two arbitrary, 3 × 3, nonzero, skew-symmetric matrices and Z be an arbitrary 3 × 3, non-zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric?

IIT JEE eBooks: www.crackjee.xyz Matrices 7.15

(a) Y3Z4 – Z4Y3 (c) X4Z3 – Z3Y4 Ans. (c), (d)

(b) X44 + Y44 (d) X23 + Y23

Solution: X' = –X, Y ¢ = –Y, Z' = Z. Let A = Y3Z4 – Z4Y3, then A ¢ = (Y3Z4) ¢ – (Z4Y3) ¢ = (Z4) ¢ (Y3) ¢ – (Y3) ¢ (Z4) ¢ = (Z ¢)4 (Y ¢)3 – (Y ¢)3 (Z ¢)4 = (Z)4 (–Y)3 – (–Y)3 (Z)4

Example 40 Let A and B be two matrices different from I such that AB = BA and An – Bn is invertible for some positive integer n. If An – Bn = An + 1 – Bn + 1 = An + 2 – Bn + 2, then (a) I – A is singular (b) I – B is singular (c) A + B = AB + I (d) (I – A) (I – B) is non-singular Ans. (a), (b), (c) n+2

= – Z4Y 3 + Y 3Z4 = A

– Bn + 2 = (A + B) (An + 1 – Bn + 1)

fi A is a symmetric matrix.

– AB(An – Bn)

Let B = X44 + Y44 fi B ¢ = (X ) ¢ + (Y )' = (X ¢) + (Y ¢) 44

44

44

44

44

= (–X) + (–Y)

44

44

44

=X +Y =B

fi B is a symmetric matrix.



An – Bn = (A + B) (An – Bn) – AB(An – Bn)



I = A + B – AB



(I – A) (I – B) = O.

I – A and I – B are signular matrices.

C ¢ = (X4Z3 – Z3X4)' = (Z3) ¢ (X4) ¢ – (X4) ¢ (Z3)' = (Z ¢)3 (X ¢)4 – (X ¢)4 (Z ¢)3 = Z3 (–X)4 – (–X)4 (Z)3 = Z3X4 – X4Z3 = – (X4Z3 – Z3X4) =–C

Example 41 If A is matrix of size n ¥ n such that A2 + A + 2I = O, then (a) A is non-singular (b) A π O 1 (c) |A| π 0 (d) A–1 = – (A + I). 2 Ans. (a), (b), (c), (d) We have A(A + I) = – 2I

fi C is a skew symmetric matrix. Finally, let D = X23 + Y23 fi

D ¢ = (X23 + Y23) ¢ = (X23) ¢ + (Y23) ¢ = (X ¢)23 + (Y ¢)23 = (–X)23 + (–Y)23 = –(X 23 + Y23) = – D

D is a skew symmetric matrix. Example 39 Let A, B be two 2 ¥ 2 real commutative matrices, such that det (A2 + AB + B2) = 0, then (a) det (A – wB) = 0 (b) det (A – w2B) = 0 (c) det (A + B) = 0 (b) det (A – B) = 0 Ans. (a), (b) Solution: As A, B commute. A2 + AB + B2 = (A – wB) (A – w2B) where w π 1 cube root of unity. fi A2 + AB + B2 = (A – wB) ( A - w B ) fi 0 = det(A2 + AB + B2) = det(A – wB) det ( A - w B ) = |det(A – wB)|2 fi fi

det(A – wB) = 0 fi det ( A - w B ) = 0. det(A – w2B) = 0

An – Bn is invertible]

As A, B π I, we get

Next, let C = X4Z3 – Z3X4 = (X4Z3) ¢ – (Z3X4)'

[



|A(A + I)| = |– 2I| fi |A| |A + I| = (–2)n π 0.

Ï 1 ¸ Thus, |A| π 0. Also, A Ì- ( A + I ) ˝ = I 2 ˛ Ó fi

A–1 = -

1 (A + I). 2

Clearly A π O for otherwise |A| = 0. Example 42

Let a = p/5 and

È cos a A= Í Î- sin a

sin a ˘ , then cos a ˙˚

B = A + A2 + A3 + A4 is (a) singular (c) skew-symmetric Ans. (b), (c) We have

(b) non-singular (d) |B| = 1

IIT JEE eBooks: www.crackjee.xyz 7.16 Comprehensive Mathematics—JEE Advanced

Ê cos 2 a A2 = Á Ë - sin 2 a

sin 2 a ˆ Ê cos 3a , A3 = Á cos 2 a ˜¯ Ë - sin 3a

Ê cos 4 a and A4 = Á Ë - sin 4 a

sin 3a ˆ cos 3a ˜¯

t 1 -1 + t 1 1 1 = D= t -1 3 1

sin 4 a ˆ cos 4 a ˜¯

[using R1 Æ R1 + R2] = (2t + 1) (– 4) = 0 fi t = – 1/2

We have cos a + cos 2a + cos 3a + cos 4a = cos a + cos 2a + cos (p – 2a) + cos (p – a)

Example 44

The system of equations

Ï a¸ p Ê 3a ˆ Ê 3p ˆ = 2 Ì2 sin Á ˜¯ cos ˝ = 4 sin ÁË ˜¯ cos Ë 2 2 10 10 Ó ˛

(aa + b)x + ay + bz = 0 (ba + c)x + by + cz = 0 (aa + b)y + (ba + c)z = 0 has a non-trivial solution, if (a) a, b, c are in A.P. (b) a, b, c are in G.P. (c) a, b, c are in H.P. (d) a is a root of ax2 + 2bx + c = 0 Ans. (b), (d)

p Êp pˆ = 4 sin Á - ˜ cos Ë 2 5¯ 10

The given system of equations will have a nontrivial solution if

[

5a = p]

= cos a + cos 2a – cos 2a – cos a = 0 sin a + sin 2a + sin 3a + sin 4a

and

= sin a + sin 2a + sin (p – 2a) + sin (p – a) = 2 [sin a + sin 2a]

= 4 cos

p p cos = a (say) 5 10

D=

Ê 0 aˆ Thus, B = Á Ë – a 0˜¯ \ B is skew-symmetric. |B| = a2 = 16 cos2

Also,

2t + 1 0 0 t +1 -1 1 t -1 3 1

p p cos2 >0 5 10

\ B is non-singular. Example 43

The system of homogeneous equations

tx + (t + 1)y + (t – 1)z = 0 (t + 1)x + ty + (t +2)z = 0 (t – 1)x + (t + 2)y + tz = 0 has non-trivial solutions for (a) exactly three real values of t

Applying R3 Æ R3 – aR1 – R2, we get aa + b a b ba + c b c =0 D= - (a a 2 + 2 b a + c) 0 0 ¤ ¤ ¤

– (aa2 + 2ba + c) (ac – b2) = 0 aa2 + 2ba + c = 0 or ac – b2 = 0 a is a root of ax2 + 2bx + c = 0 or a, b, c are in G.P.

Example 45

The system of equations

ax + by + cz = q - r bx + cy + az = r - p cx + ay + bz = p - q

(b) exactly two real values of t (c) exactly one real value of t

is t.

Ans. (c) For the system of equations to have non-trivial solution, t t +1 t -1 t t+2 D = t +1 t -1 t + 2 t

aa + b a b ba + c b c =0 0 aa + b ba + c

=0

Applying C2 Æ C2 – C1, C3 Æ C3 – C1, we get

(a) consistent if p = q = r (b) inconsistent if a = b = c, and p, q, r are distinct (c) consistent if a, b, c are distinct and a + b + c π 0. (d) none of these Ans. (a), (c) Solution: If p = q = r, then x = 0, y = 0, z = 0 is a solution of the system of equation. If a, b, c matrix is

IIT JEE eBooks: www.crackjee.xyz Matrices 7.17

1 (a + b + c) [(b - c)2 + (c - a)2 + (a - b)2] π 0 2

(a) (b) (c) (d)

and hence the system of equations has a unique solution.

MATRIX-MATCH TYPE QUESTIONS Let ak = nCk for 0 £ k £ n and

Example 46

Èak - 1 Ak = Í Î 0 n -1

Èa 0˘ Ak◊Ak+1 = Í ˙, Î0 b˚

B= Â

and

0˘ ˙ for 1 £ k £ n ak ˚

k =1

Column 1

Column 2 2n 2n ( C n) n +1 (q) 0 (r) 2nCn+1

(a) a

(p)

(b) a - b (c) a + b a (d) b

(s) 1

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

a =1 b 2 (2n)! (n + 1)! (n - 1)!

2n (2n)! 2n 2n = = ( C n) n + 1 n! n! n +1 Example 47

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

È 1 I-A= Í Î- tan a

Solution:

È 1 (I - A)-1 = cos2 a Í Î tan a Thus (I + A) (I - A)-1 È 1 = cos2a Í Î tan a

Ècos 2 a = Í Î sin 2 a

a = a0 a1 + a1a2 + ... + an-2 an-1 = ana1 + an-1a2 + ... + a2 an-1 = number of ways of selecting (n + 1) persons out of n men and n women = 2nCn+1 Similarly b = 2nCn+1

a + b = 2(2nCn+1) =

a

(p) (q) (r) (s)

Column 2 A – B(a) B(2a) B(–2a) AB(-a)

tan a ˘ 1 ˙˚ - tan a ˘ 1 ˙˚

- tan a ˘ È 1 1 ˙˚ ÍÎ tan a

- tan a ˘ 1 ˙˚

È1 - tan 2 a - 2 tan a ˘ ˙ = cos2 a Í 1 - tan 2 a ˚˙ ÎÍ 2 tan a

Solution:

\ a - b = 0 and

Ans.

Column 1 (I + A) (I - A)-1 (I - A) (I + A)-1 B(a)2 B(a)–2 p q r s

È 0 Let A = Í Î tan a Ècos a B(a) = Í Î sin a

- tan a ˘ , 0 ˙˚ - sin a ˘ cos a ˙˚

- sin 2 a ˘ = B(2a) = B(a)2 cos 2 a ˙˚

fi (I - A) (I + A)-1 = B(a)-2 = B(-2a) Let A and B be two non-singular matrices

Example 48 k

such that (AB) = Ak Bk for three consecutive positive integral values of k. Column 1 Column 2 (p) A2 (a) ABA-1 (b) BAB-1 (q) B 2 –1 (r) A (c) AB A (s) B2 (d) BA2B–1 p q r s Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: As A and B are invertible matrices A-1, B-1 both exist. Also, for every positive integer, An and Bn are invertible.

IIT JEE eBooks: www.crackjee.xyz 7.18 Comprehensive Mathematics—JEE Advanced

Suppose (AB)n = An Bn holds for three consecutive positive integers m, m + 1 and m + 2. We have (AB)m = Am Bm (AB)m+1 = Am+1 Bm+1 and (AB) From (2), we have

m+2

=A

m+2

B

(1) (2)

m+2

(3)

Am+1 Bm+1 = (AB)m+1 = (AB)m (AB) fi Am ABm B = Am Bm AB [using (1)] m Since A and B are invertible matrices, (4) ABm = Bm A Similarly, using (2) and (3), we can show that ABm+1 = Bm+1 A

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

Solution:

(a) Put x + 2 = t, then t > 0. Also x2 + 2 x + 4 (t – 2) 2 + 2(t – 2) + 4 = x+2 t

(5)

=

We have (AB)Bm = ABm+1 = Bm+1 A

2ˆ Ê = Á t– ˜ Ë t¯

[using (5)]

= B(Bm A) = B(ABm) = (BA)Bm [using (4)] Thus, (AB)Bm = (BA)Bm As B is an invertible matrix, we can cancel Bm from both the sides to obtain m

fi and

\ least value of

k satisfying 1 < 2 (– k + 3–a) < 2, must be less than

x2 + 2 x + 4 is 2 x+2

(b) (A + B) (A + B) = (A + B) (A – B) fi AB = BA. \ (– 1)k = – 1 fi k = 1, 3. (c) 1 < 2(– k + 3–a) < 2 fi 0 < – k + 3–a < 1 fi k < 3–a < k + 1. fi k < (log32)–1 < k + 1 fi k < log23 < k + 1 fi 2k < 3 < 2k + 1 fi k = 1.

(p) 0 (q) 1

(d) k2 – 3 =

= But 0
–2 x+2

t 2 – 2t + 4 4 =t+ –2 t t

(s) 3

Let t, q, f ∈ R be such that

- sin q ˆ Ê et cos q ˜¯ ÁË 0

0 ˆ Ê cos f ˜Á e-t ¯ Ë sin f

- sin f ˆ Ê 2 3ˆ = cos f ˜¯ ÁË 1 2˜¯

(1)

Match the entries in column 1 with that of column 2.

IIT JEE eBooks: www.crackjee.xyz Matrices 7.19

Ans.

Column 1 (a) t is equal to

Column 2 (p) R

(b) q lies in

(q) ln( 5 + 2)

(c) f lies in (d) et lies in p q r s

(r) ln( 5 – 2) (s) [0, 2p]

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution:

Now, A2 = I fi det(A2) = 1 fi





a = d, b = – b, c = – c fi a = d, b = c = 0. Ê a 0ˆ In this case A = Á . Ë 0 a˜¯

-et sin f ˆ Ê cos q ˜ =Á e-t cos f ¯ Ë - sin q

- sin q ˆ -1 Ê 2 3ˆ cos q ˜¯ ÁË 1 2˜¯

e =9±4

5 = ( 5 ± 2)

|A| = 1 fi a2 = 1 fi a = ± 1. \

A contradiction.

Now (2) (3) (4) (5)

\

Ê a bˆ Ê d – bˆ Ê – d ÁË c d ˜¯ = – ÁË – c a ˜¯ = ÁË c



a = – d fi tr(A) = a + d = 0.

\

Statement-1: is true and statement-2 is false.

Example 52 (6)

bˆ – a˜¯

Let

Ê cos q + sin q A(q) = Á Ë – 2 sin q

(7)

2 sin q ˆ ˜ cos q – sin q ¯

Statement-1: A(p/3)3 = – I Statement-2: A(q) A(f) = A(q + f) Ans. (a) 2



et =

5 ±2



t = ln( 5 ± 2) = ± ln( 5 + 2) = ln ( 5 + 2) , ln ( 5 - 2)

ASSERTION-REASON TYPE QUESTIONS Example 51

A = I or A = – I.

Thus, det(A) = – 1.

sin q ˆ Ê 2 3ˆ cos q ˜¯ ÁË 1 2˜¯

et cosf = 2cosq + sinq e–t sinf = –2sinq + cosq – et sinf = 3cosq + 2sinq e–tcosf = – 3sinq + 2cosq Square and add (2), (3) e2t cos2f + e–2t sin2f = 5 Square and add (4), (5) e2t sin2f + e–2t cos2f = 13 Adding (6), (7) we get e2t + e–2t = 18 e4t – 18e2t + 1 = 0 2t

Ê a bˆ 1 Ê d – bˆ ÁË c d ˜¯ = det A ÁË – c a ˜¯

If detA = 1, then

0 ˆ Ê cos f - sin f ˆ Ê cos q ˜ =Á ˜Á e-t ¯ Ë sin f cos f ¯ Ë sin q

Ê et cos f fiÁ Ë e-t sin f

(detA)2 = 1 fi detA = ± 1. Also, A2 = I fi A = A–1

Solution: Write (1) as Ê et Á Ë0

Ê a bˆ Let A = Á . Ë c d ˜¯

Let A be a 2 ¥ 2 matrix with real entries.

Let I be the 2 ¥ 2 identity matrix. Denote by tr(A), the sum of diagonal entries of A. Assume that A2 = I. Statement-1: If A π I and A π – I, then det(A) = – 1. Statement-2: If A π I and A π – I, then tr(A) π 0. Ans. (c)

Solution:

Ê a bˆ A(q) A(f) = Á Ë – b c˜¯

where a = (cos q + sin q) (cos f + sin f) – 2 sin q sin f = (cos q cos f – sin q sin f) + (sin q cos f + cos q sin f) = cos (q + f) + sin (q + f); b=

2 [sin f(cos q + sin q) + sin q(cos f – sin f)]

= 2 sin (q + f ) and c = – 2 sin q sin f + (cos q – sin q) (cos f – sin f) = cos q cos f – sin q sin f – (sin q cos f + cos f sin q) = cos (q + f) – sin (q + f) Thus, A(q) A(f) = A(q + f) fi A(q)2 = A(2q)

IIT JEE eBooks: www.crackjee.xyz 7.20 Comprehensive Mathematics—JEE Advanced

A(q)3 = A(2q) A(q) = A(3q) A(p/3)3 = A(p) = – I

\

Example 53

Ê a bˆ Suppose X = Á Ë c d ˜¯

Statement-1: If X¢AX = O for each X, then A must be a symmetric matrix. satisfies the

Statement-2: If A is symmetric and X¢AX = 0 for each X, then A = O Ans. (b)

equation X2 – 4X + 3I = O. Statement-1: If a + d π 4, then there are just two such matrices X. X,

Statement-2: satisfying X2 – 4X + 3I = O. Ans. (b) Solution:

X2 – 4X + 3I = O fi (X – I) (X – 3I) = O

b(a + d – 4) ˆ Ê (a – 1) (a – 3) + bc Ê 0 0ˆ = Á fi Á ˜ (d – 1) (d – 3) + bc¯ Ë c(a + d – 4) Ë 0 0˜¯ If a + d π 4, then b = 0, c = 0. (a – 1) (a – 3) = 0, (d – 1) (d – 3) = 0. fi a = 1, 3, d = 1, 3 As a + d π 4, a = 1, d = 1 or a = 3, d = 3.

X2 – 4X + 3I = O Ê2 0 ˆ Ê1 0 ˆ and B = Á Suppose A = Á ˜ Ë 0 -2˜¯ Ë 0 –1¯

Let X be a 2 ¥ 2 matrices such that X¢AX = B. Statement-1: X is non-singular and det(X) = ± 2 Statement-2: X is a singular matrix. Ans. (c) Solution: If X = O, then X¢AX = O fi B = O. A contradiction. det(X) = a, then det(X¢) = a

fi a(– 1)a = – 4 fi a = ± 2 As det(X) π 0, X cannot be a singular matrix.

È y1 ˘ Y= Í ˙ Î y2 ˚

Ê a11 Let A = Á Ë a21

Ê 0 Thus, A = Á Ë – a12

a12 ˆ Èx ˘ , X = Í 1 ˙, ˜ a22 ¯ Î x2 ˚

a12 ˆ is a skew symmetric matrix. 0 ˜¯

Ï ¸ Èa b ˘ Tp = Ì A = Í ˙ : a, b, c Œ{0, 1, 2, ..., p – 1}˝ c a Î ˚ ÓÔ ˛Ô Example 56 The number of A in Tp such that A is either symmetric or skew-symmetric or both, and det (A) divisible by p is (a) (p – 1)2 (b) 2 (p – 1) 2 (d) 2p – 1 (c) (p – 1) + 1 Example 57 The number of A in Tp such that the trace of A is not divisible by p but det (A) is divisible by p is (a) (p – 1) (p2 – p + 1) (b) p3 – (p – 1)2 (c) (p – 1)2 (d) (p – 1) (p2 – 2) Example 58 The number of A in Tp such that det (A) is not divisible by p is (a) 2p2

\ det(X¢AX) = det(B)

Example 55

Next, (E1 + E2)¢ A(E1 + E2) = O fi a12 + a21 = 0

Paragraph for Questions Nos 56 to 58 Let p be an odd prime number and Tp be the following set of 2 ¥ 2 matrices:

0 Ê (a – 1) (a – 3) + bc ˆ = 0. ÁË 0 (1 – a) (3 – a) + bc˜¯

Let

X¢AX = O " X, E1¢AE1 = O fi a11 = 0 E2¢AE2 = O fi a22 = 0.

COMPREHENSION-TYPE QUESTIONS

If a + d = 4, we get

Example 54

As

È1 ˘ È0 ˘ Let E1 = Í ˙ , E2 = Í ˙ . Î0 ˚ Î1 ˚

In case A is symmetric, a12 = a21 \ 2a12 = 0 fi a12 = 0. Thus, A = O, in this case.

Ê 1 0ˆ Ê 3 0ˆ or X = Á Thus, X = Á ˜ Ë 0 1¯ Ë 0 3˜¯

\

Solution:

(b) p3 – 5p

(c) p3 – 3p (d) p3 – p2 Solution: Ans. 56. (d), 57. (c), 58. (d) 56. A symmetric matrix is of the form Èa b ˘ A = Íb a ˙ Î ˚

IIT JEE eBooks: www.crackjee.xyz Matrices 7.21

and skew symmetric matrix is of the form È 0 b˘ A = Í -b 0 ˙ Î ˚ For symmetric matrix, det (A) = a2 – b2 = (a – b) (a + b) Note that p | det (A) fi p | (a – b) or p | (a + b) As 0 £ a, b £ (p – 1), we get – (p – 1) £ a – b £ p – 1 and 0 £ a + b £ 2p – 2 Thus, p | (a – b) fi a – b = 0 or a = b. and p | (a + b) fi a + b = p. There are p symmetric matrices for which a = b and (p – 1) symmetric matrices for which a + b = p. Thus, there are (2p – 1) symmetric matrices in Tp for which det (A) is divisible by p. When A is skew symmetric, det (A) = b2 Now, p | det (A) fi p | b2 fi p | b fi b = 0 fi A is symmetric. The only matrix in Tp which is both symmetric and skew È0 0 ˘ symmetric is Í0 0˙ , which has already been counted. Î ˚ 57.

Èa b ˘ Let A = Í c a ˙ Î ˚

tr (A) = 2a and det (A) = a2 – bc Note that p | 2a fi p | a fi a = 0. Now, p 2a means a π 0 and hence a can be chosen in (p – 1) ways. As p |det (A), we get p| (a2 – bc). Since p a, we get b π 0. Thus, b can be chosen in (p – 1) ways. After a and b have been chosen c can be chosen in just one way so that p | (a2 – bc). Thus, there are (p – 1)2 matrices such that p 2a but p | (a2 – bc). 58. Note that Tp contains p3 elements. We now p | det (A), that is, p | (a2 – bc) Following two cases arise. Case 1. a = 0 In this case p | (a2 – bc) fi p | bc fi p | b or p | c fi b = 0 or c = 0. There are 2p – 1 such matrices.

Case 2. a π 0. We have already seen in Example 56 that, there are (p – 1)2 matrices such that p a but p | det (A). Thus, the number of matrices A such that p | det (A), is (2p – 1) + (p – 1)2 = p2 Hence, the number of matrices A in Tp such that p det (A) is p3 – p2. Paragraph for Question Nos. 59 to 61 Let a, b and c be three real numbers satisfying È1 9 7 ˘ [a b c] ÍÍ8 2 7 ˙˙ = [0 0 0] ÍÎ7 3 7 ˙˚ Example 59

(E)

If the point P(a, b, c), with reference to

(E), lies on the plane 2x + y + z = 1, then the value of 7a + b + c is (a) 0 (b) 12 (c) 7 (d) 6 Example 60

Let w be a solution of x3 – 1 = 0 with

Im(w) > 0. If a = 2 with b and c satisfying (E), then the 3 1 3 value of u = a + b + c is equal to w w w (a) – 2 (c) 3 Example 61

(b) 2 (d) – 3 Let b = 6, with a and c satisfying (E). If α

and b are the roots of the quadratic equation ax2 + bx + c = Ê 1 1ˆ 0, then  Á + ˜ b¯ n=0 Ë a •

n

is

(a) 6 (b) 7 (c) 6/7 (d) • Ans. 59. (d), 60. (a), 61. (b) Solution: 59. We have a + 8b + 7c = 0 9a + 2b + 3c = 0 7a + 7b + 7c = 0 Eliminating c from (1) and (3), we get 6a – b = 0 fi b = 6a. Also, c = – a – b = – a – 6a = – 7a. Thus,

a b c = = = k (say) 1 6 -7

...(1) ...(2) ...(3)

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fi a = k, b = 6k, c = – 7k As (a, b, c) lies on 2x + y + z = 1, 2k + 6k – 7k = 1 fi k = 1 ∴ a = 1, b = 6, c = – 7 Thus, 7a + b + c = 7 + 6 – 7 = 6 60. As a = 2, we get b = 12, c = – 14 Thus,

3 w2

+

1 w 12

+

3 w -14

3

1 2 = 2 + + 3w = –2 1 w

61. As b = 6, we get a = 1, c = – 7 -b -6 -7 ∴a+b= = = -7 , ab = a 1 1 n

• Ê 1 • Êa + bˆ 1ˆ Now  Á + ˜ =  Á ˜ b¯ n=0 Ë a n = 0 Ë ab ¯ •

Example 66

If A and B are two non-singular matrices,

then (a) A is similar to B (b) AB is similar to BA (c) AB is similar to A-1B (d) none of these Ans. 62 (a), 63. (a), 64. (a), 65. (b), 66. (b) Solution: As A and B are similar matrices there exists a non-singular matrix P such that A = P-1 BP fi det (A) = det (P-1 BP) = det (P-1) det (B) det (P) =

n

n

1 Ê 6ˆ  ÁË ˜¯ = 6 = 7 7 = n=0 17

1 det (B) (det P) det ( P)

= det B Thus, det (A) = 0 ¤ det (B) = 0 and det (A) = 1 ¤ det (B) = 1 Next, A = P-1 BP, B = Q-1 C Q, fi A = P-1 (Q-1 C Q)P = (QP)-1 C QP

Paragraph for Question Nos. 62 to 66

Thus, A is similar to C

Two n ¥ n square matrices A and B are said to be similar if there exists a non-singular matrix P such that P-1 A P = B

Also, as AB = B-1(BA)B, we get AB is similar to BA.

Example 62 If A and B are two singular matrices, then (a) det (A) = det (B) (b) det (A) + det (B) = 0 (c) det (AB) π 0 (d) det (AB) = 0 Example 63

If A and B are similar matrices such that

det (AB) = 0, then (a) det (A) = 0 and det (B) = 0 (b) det (A) = 0 or det (B) = 0 (c) A = O and B = O (d) A = O or B = O Example 64

If A and B are similar matrices such that

det (A) = 1, then (a) det (B) = 1 (c) det (B) = - 1

(b) det (A) + det (B) = 0 (d) det (B) = 0

Example 65 If A and B are similar and B and C are similar, then (a) AB and BC are similar (b) A and C are similar (c) A + C and B are similar (d) AC is similar to B2

Paragraph for Question Nos. 67 to 71 Let A be a square matrix of order n ¥ n. A constant l is said to be characteristic root of A if there exists a n ¥ 1 matrix X such that AX = lX Example 67

If l is a characteristic root of A, then

(a) A - lI = O

(b) A - lI is singular

(c) A - lI is non-singular(d) det (A) = l Example 68 If a constant l is such that A - lI is nonsingular, then (a) l = 0 (b) l π 0 (c) l is not a characteristic root of A (d) none of these Example 69

If 0 is a characteristic root of A, then

(a) A is non-singular (c) A = O Example 70

(b) A is singular (d) A = In

If l is a characteristic root of A and n ∈

N, then l is a characteristic root of (a) An (b) An-1 (c) A-n (d) A - An + A-n n

IIT JEE eBooks: www.crackjee.xyz Matrices 7.23

Example 71 Let P be a non-singular matrix, then which of the following matrices have the same characteristic roots. (a) A and AP (b) A and PA (c) A and P-1 AP (d) none of these Ans. 67. (b), 68. (c), 69. (b), 70. (a), 71. (c) Solution: Since X π O is such that (A - lI)X = O, |A - lI| = 0 ¤ A - lI is singular. If A - lI is non-singular then the equation (A - lI)X = O fi X = O If l = 0, we get |A| = 0 fi A is singular. A2 X = A(AX) = A(lX) = l(AX)

We have

= l2X, A3X = A(A2X) = A(l2X) = l2(AX) = l2(lX) = l3X Continuing in this way, we obtain A X =l X n

Also,

|P

-1

" n ∈ N.

n

AP - lI| = |P

-1

(A - lI)P|

-1

= |P | |A - lI| |P| = |A - lI| Paragraph for Question Nos. 72 to 75 Let A be the set of all 3 × 3 symmetric matrices all whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0. Example 72 (a) 12 (c) 9

The number of matrices in A is (b) 6 (d) 3

Example 73 The number of elements in A for which the system of linear equations È x ˘ È1 ˘ A ÍÍ y ˙˙ = ÍÍ0˙˙ ÍÎ z ˙˚ ÍÎ0˙˚ has a unique solution, is (a) less than 4 (b) at least 4 but less than 7 (c) at least 7 but less than 10 (d) at least 10. Example 74 The number of matrices A in A for which the system of linear equations È x ˘ È1 ˘ A ÍÍ y ˙˙ = ÍÍ0˙˙ ÍÎ z ˙˚ ÍÎ0˙˚ is inconsistent is

(a) 0 (c) 2 Example 75 |A| = 0, is (a) 4 (c) 8

(b) more than 2 (d) 1 The number of matrices A in A such that (b) 6 (d) none of these

Ans. 72 (a), 73. (b), 71. (b), 75. (b) Solution: Let x be the number of 1's on the main diagonal and y be the number of 1¢s above the main diagonal, then x + 2y = 5 fi

x = 1, y = 2 or x = 3, y = 1.

When x = 1, the main diagonal can be chosen in 3 ways, and the elements above the main diagonal in 3 ways. Therefore, there are 9 such matrices. These are È1 1 1˘ A1 = ÍÍ1 0 0˙˙ , ÍÎ1 0 0˙˚

È0 1 1 ˘ A2 = ÍÍ1 1 0˙˙ ÍÎ1 0 0˙˚

È0 1 1 ˘ A3 = ÍÍ1 0 0˙˙ , ÍÎ1 0 1˙˚

È1 0 1 ˘ A4 = ÍÍ0 0 1˙˙ ÍÎ1 1 0˙˚

È0 0 1 ˘ A5 = ÍÍ0 1 1˙˙ , ÍÎ1 1 0˙˚

È0 0 1˘ A6 = ÍÍ0 0 1˙˙ ÍÎ1 1 1˙˚

È1 1 0 ˘ A7 = ÍÍ1 0 1˙˙ , ÎÍ0 1 0˙˚

È0 1 0 ˘ A8 = ÍÍ1 1 1 ˙˙ ÎÍ0 1 0˙˚

È1 0 0 ˘ A9 = ÍÍ0 1 1˙˙ ÍÎ0 1 1˙˚ When x = 3, the main diagonal can be chosen in 1 way; and the element above the main diagonal in 3 ways. Therefore, there are 3 such matrices. È1 1 0 ˘ A10 = ÍÍ1 1 0˙˙ , ÍÎ0 0 1˙˚

È1 0 1 ˘ A11 = ÍÍ0 1 0˙˙ ÍÎ1 0 1˙˚

È1 0 0 ˘ A12 = ÍÍ0 1 1˙˙ ÍÎ0 1 1˙˚ For example 73, note that |A2| π 0, |A3| π 0, |A4| π 0, |A5| π 0, |A7| π 0, |A9| π 0, there are six matrices A such that

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Ans. 2

È x ˘ È1 ˘ A ÍÍ y ˙˙ = ÍÍ0˙˙ ÍÎ z ˙˚ ÍÎ0˙˚ has a unique solution.

9

Solution: a + b = Â (ak + bk) k =1

9

= 10 Â (10Ck) = 10(210 - 2)

È x˘ For example 74, let X = ÍÍ y ˙˙ and B = ÍÎ z ˙˚

È1 ˘ Í0˙ , and note that Í ˙ ÍÎ0˙˚

k =1

= 10220 Example 79

If a, b are as in Example 78, then

ab

A1X = B and A12X = B AX = B has no solution when A = A6, A8, A10, A11 For example 75, note that |A| = 0 for six matrices.

INTEGER-ANSWER TYPE QUESTIONS

(

)

È10 29 - 1 ˘ Î ˚

2

is equal to

Ans. 1 Solution:

a

9

10

k =1

k =1

= Â k(10Ck) = Â k(10Ck) - 10(10C10)

= 10 (2 ) - 10 = 10(29 - 1) Similarly, b = 10(29 - 1) Thus, ab = 100(29 - 1)2 9

Example 76 Let S denote the sum of all the values of l for which the system of equations (1 + l)x1 + x2 + x3 = 1 x1 + (1 + l)x2 + x3 = l x1 + x2 + (1 + l)x3 = l2 is inconsistent. then number of elements in S is

Ans. 8

Ans. 2

Solution:

Solution: If A |A| = l2(l + 3) For l = 0 and l = - 3, the system is inconsistent. Example 77 Let S denote the set of all values of l for which the system of equation lx1 + x2 + x3 = 1 x1 + lx2 + x3 = 1 x1 + x2 + lx3 = 1 is inconsistent, then  | l | is l ŒS

Ans. 2 Solution: If A |A| = l3 + 2 - 3l = (l - 1)2 (l + 2) For l For l = - 2, the system has no solution. Example 78 and

If

Let ak = k(10Ck), bk = (10 - k) (10Ck) Èak Ak = Í Î0

0˘ bk ˙˚

9 Èa 0˘ A = Â Ak = Í ˙, k =1 Î0 b˚

then (a + b)/5110 is

Example 80 |A| = 4,

Suppose A is a 4 ¥ 4 matrix such that

adj A is equal to |adj A| = |A|n-1 = 43 = 64

Example 81

È0 a ˘ Let A = Í ˙ and Î0 0 ˚

Èa b ˘ (A + I)50 - 50A = Í ˙ , then a + b + c + d is Îc d ˚ Ans. 2 Solution: As A2 = O, Ak = O " k ≥ 2. Thus, (A + I)50 = I + 50A fi (A + I)50 - 50A = I \ a = 1, b = 0, c = 0, d = 1 Example 82 3

If A and B are two distinct matrices such

3

that A = B and A2B = B2A, then det(A3 + B3) is Ans. 0 Solution: We have (A2 + B2) (A – B) = A3 – A2B + AB2 – B3 = O. If det (A2 + B2) π 0, then A2 + B2 is invertible and we shall get A – B = 0 fi A = B. \ det (A2 + B2) = 0 Example 83

Suppose a matrix A satisfies A2 - 5A + 7I

= O. If A8 = aA + bI, then a/253 is Ans. 5

IIT JEE eBooks: www.crackjee.xyz Matrices 7.25

A4 = (5A - 7I)2 = 25A2 - 70A + 49I = 25(5A - 7I) - 70A + 49I = 55A - 126I 8 A = 3025A2 - 13860A + 15876I = 3025(5A - 7I) - 13860A + 15876I = 1265A - 5299I a/253 = 5

Solution:



Example 84 Suppose A and B are two non singular matrices such that B π I, A6 = I and AB2 = BA. If k is least value for which Bk = I, then k – 120 is Ans. 7 Use A-1 BA = B2 Ê a bˆ and b π 0. If A3 = I, then Example 85 Let A = Á Ë c d ˜¯

Solution:

a + d + 4 is equal to Ans. 3 Solution: A3 = I fi |A3| = 1 fi |A3| = 1 fi |A| = 1 π 0 \ A is invertible. Also A – 1 = A2 fi

Ê a 2 + bc b(a + d )ˆ Ê d - bˆ = Á ˜ ËÁ -c a ¯˜ Ë c(a + d ) d 2 + bc ¯



– b = b(a + d) fi a + d π – 1



a+d+4=3

EXERCISE LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. Let A, B be two 2 ¥ 2 matrices. Let a = det(A) + det(B) – det(A + B) and b = tr(AB) – (trA)(trB), then (a) a = b (b) a + b = 0 (d) a = b2 (c) a2 = b 2. Let A be a 3 ¥ 3 matrix with real entries such that sum of the entries in each column of A is 1, then sum of the entries of A2018 is (a) 1 (b) 3 (c) 2018 (d) 6054 0 1 2 È ˘ Í1 ˙ 0 3 3. If A = Í ˙ , then A + 2A¢ equals ÍÎ- 2 - 3 0˙˚

(a) A (c) - A¢

(b) A¢ (d) 2A

3 2 ˘ È1 Í2 5 t ˙˙ , then the value(s) of t for 4. Let At = Í ÍÎ4 7 - t - 6˙˚ which inverse of At does not exist. (a) - 2, 1 (b) 3, 2 (c) 2, - 3 (d) 3, - 1 5. Let a, b, c ∈ R be such that a + b + c > 0 and abc = 2. Let Èa b c ˘ Í ˙ A = Íb c a ˙ ÍÎ c a b ˙˚

If A2 = I, then value of a3 + b3 + c3 is (a) 7 (b) 2 (c) 0 (d) – 1. 6. If a is a 3 ¥ 3 skew summertic matrix with real entries and trace of A2 equals zero, then (a) A = O (b) 2A = I (c) A is orthogonal (d) none of these Èi 0 ˘ 7. If Í ˙ +X= Î3 -i ˚ È0 -1˘ (a) Í ˙ Î3 i ˚

2 ˘ Èi Í3 4 + i ˙ – X, then X is equal to Î ˚ 1 ˘ È0 (b) Í ˙ Î0 2 + i ˚

0 ˘ È1 (c) Í ˙ Î0 2 - i ˚ È0 -i ˘ 8. If A = Í ˙, B = Îi 0 ˚

(d) none of these

È1 0 ˘ Í0 -1˙ , then AB + BA is Î ˚ (a) null matrix (b) unit matrix (c) invertible matrix (d) none of these È1 2 3˘ Í ˙ 9. A = Í 1 2 3 ˙ , then A is a nilpotent matrix of ÎÍ-1 -2 -3˚˙ index (a) 2 (c) 4

(b) 3 (d) 5

( (

È 1 ix - ix Í2 e + e 10. If A = Í Í 1 eix - e-ix ÎÍ 2 (a) (b) (c) (d)

) 12 (e ) 12 (e

) )

˘ - e-ix ˙ ˙ then A–1 exists ix - ix ˙ +e ˚˙

ix

for all real x for positive real x only for negative real x only none of these

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È ab 11. If A = Í 2 ÎÍ- a

b2 ˘ ˙ , then A2 is equal - ab ˚˙

(a) O (c) – I

(b) I (d) none of these

12. If A is 2 × 2 matrix such that A2 = O, then tr (A) is (a) 1 (b) – 1 (c) 0 (d) none of these a b È ˘ 13. If A = Í ˙ such that A c d Î ˚ A2 – (a + d)A = O, then inverse of A is (a) I (b) A (c) (a + d)A (d) does not exist È3 2˘ –3 14. If A = Í ˙ , then A is 0 1 Î ˚ (a)

(c)

1 È1 -26 ˘ 27 ÍÎ0 -27 ˙˚

(b)

1 È1 -26˘ 27 ÍÎ0 27 ˙˚

(d)

1 È-1 -26˘ 27 ÍÎ 0 -27 ˙˚ 1 È-1 26 ˘ 27 ÍÎ 0 -27 ˙˚

15. If A is a skew Hermitian matrix, then the main diagonal elements of A are all (a) zero (b) positive (c) negative (d) none of these È1 2 1 ˘ Í ˙ 16. If A = Í0 1 -1˙ , then A3 – 3A2 – A – 9I is equal ÍÎ3 -1 1 ˙˚ to (a) O

(b) I (c) A (d) A2 Ê w -w ˆ Ê 1 -1ˆ 17. If A = Á and B = Á , then A9 ˜ Ë -w w ¯ Ë -1 1 ˜¯ equals (a) 64 B (b) 32 B (c) 16 B (d) 256 B 3 - i -i ˆ Ê 2 Á 18. If A = 3 + i - 5 7 + i˜ , then A is Á ˜ 7-i e ¯ Ë i (a) symmetric (b) Hermitian (c) skew Hermitian (d) none of these 19. If A and B are two square matrices of the same order and m is a positive integer, then (A + B)m = mC0 Am + mC1 Am - 1 B + m C2 Am - 2 B2 + + mCm–1 ABm - 1 + mCm Bm if (a) AB = BA (c) Am = O, Bm = O

(b) AB + BA = O (d) none of these.

Ê cos q - sin q 0ˆ cos q 0˜ , then A(q)3 will be a 20. If A(q) = Á sin q Á ˜ 0 0 0¯ Ë null matrix if and only if (a) q = (2k + 1)p/3, k ∈ I (b) q = (4k - 1)p/3 k ∈ I (c) q = (3k - 1)p/4, k ∈ I (d) none of these 21. If A and B are two non-singular matrices such that AB = C, then |B| is equal to (a)

|C | | A|

(b)

| A| |C |

(c) |C| (d) none of these 22. If the system of equations ax + y= 3, x + 2y = 3, 3x + 4y = 7 is consistent, then value of a is given by (a) 2 (b) 1 (c) – 1 (d) 0 23. If the system of equations x + 2y – 3z = 1, (p + 2)z = 3, (2p + 1) y + z = 2 is consistent, then the value of p is (a) – 2 (b) – 1/2 (c) 0 (d) 2 24. The system of linear equations x + y + z = 2, 2x + y – z = 3, 3x + 2y + kz = 4 has a unique solution if (a) k π 0 (b) – 1 < k < 1 (c) – 2 < k < 2 (d) k = 0 25. If A, B and A + B are non-singular matrices, then (A – 1 + B– 1) [A – A (A + B)– 1 A] (a) O (b) I (c) A (d) B

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 26. If w π 1 is cube root of unity, and È 1 w w2 ˘ Í ˙ 2 1 ˙ is A= Í w w Í 2 ˙ 1 w ˙˚ ÎÍ w (a) symmetric (b) skew symmetric (c) singular (d) orthogonal 27. Suppose A and B are two 3 ¥ 3 non-singular matrices such that (AB)k = AkBk for k = 2017, 2018, 2019, then (b) A–1B–2A = B2 (a) AB–1A–1 = B–1 (c) AB = BA (d) A–2BA2 = (A–1BA)2

IIT JEE eBooks: www.crackjee.xyz Matrices 7.27

28. If A and B are 3 ¥ 3 matrices and |A| π 0, then (a) |AB| = 0 fi |B| = 0 (b) |AB| π 0 fi |B| π 0 (c) |A–1| = |A|–1 (d) |2A| = 2|A| 29. If A and B are two symmetric matrices of the same order, then (a) A + B is symmetric (b) A – B is symmetric (c) AB is symmetric (d) BAB is symmetric 30. If A and B are two matrices of the same order, then (a) A2 – B2 = (A + B) (A – B) (b) A2 = I ¤ (A + I) (A – I) = O (c) (A¢)¢ = A (d) A + A¢ is symmetric 31. If A = (aij)3 ¥ 3 is a skew symmetric matrix, then (a) aii = 0 " i (b) A + A¢ is a null matrix (c) |A| = 0 (d) A is not invertible. 32. If A, B and C are three square matrices of the same order, then AB = AC fi B = C if (a) |A| π 0 (b) A is invertible (c) A is orthogonal (d) A is symmetric 33. If a, b, g are three real numbers and È 1 cos (a - b ) cos (a - g ) ˘ Í ˙ 1 cos (b - g ) ˙ A = Í cos (b - a ) ÍÎ cos (g - a ) cos (g - b ) ˙˚ 1 then (a) A is symmetric (b) A is orthogonal (c) A is singular (d) A is not invertible. n 34. Let Ck = Ck for 0 £ k £ n and Ê Ck2 - 1 0 ˆ Ak = Á ˜ for k ≥ 1, and A1 + A2 + ÁË 0 Ck2 ˜¯ Ê k1 = Á Ë0

+ An

0ˆ , then k2 ˜¯

(a) k1 = k2 (b) k1 + k2 = C2n +1 (d) k2 = 2nCn + 1 (c) k1 = 2nCn – 1 Ê sin q i cos q ˆ 35. If A(q) = Á , then Ë i cos q sin q ˜¯ A(q) is invertible for all q ∈ R A(q)–1 = A(– q) A(q)–1 = A(p – q) A(q) + A(p + q) is a null matrix

È1 - 1 1 ˘ Í ˙ 38. If A = Í2 - 1 0˙ , then ÍÎ1 0 0˙˚ (a) A3 = I (b) A-1 = A2 n (c) A = A " n π 4 (d) none of these 39. If D1 and D2 are two 3 ¥ 3 diagonal matrices, then (a) D1 D2 is diagonal matrix (b) D1 D2 = D2 D1 (c) D12 + D22 is a diagonal matrix (d) D12 + D22 is a diagonal matrix È1 0˘ 40. Let A = Í ˙ then Î1 1 ˚ È 1 0˘ (a) A-n = Í ˙ " n ∈ N. Î- n 1˚ È 1 0˘ (b) A- n = Í ˙ " n ŒN În 1˚ È 0 0˘ 1 -n A = Í ˙ n Î- 1 0˚ È0 0 ˘ 1 (d) lim 2 A-n = Í ˙ nÆ• n Î0 0 ˚ 41. Let A, B and C be 2 ¥ 2 matrices with entries from (c) lim

nƕ

A * B= 2n

(a) (b) (c) (d)

36. D is a 3 ¥ 3 diagonal matrix. Which of the following statements is not true? (a) D¢ = D (b) AD = DA for each matrix of size 3 ¥ 3 (c) D-1 if exists is a diagonal matrix (d) D2 is a diagonal matrix 37. If A is an invertible matrix, then which of the followings are true: (a) A π O (b) Adj. A π O (c) |A| π 0 (d) A-1 = |A| Adj.A.

1 (AB + BA), then 2

(a) A * B = B * A (b) A * A = A2 (c) A * (B + C) = A * B + A * C (d) A * I = A 42. With A, B, C o as follows: 1 AoB = (AB - BA) then 2 (a) AoA = O (b) AoI = O (c) Ao(B + C) = AoB + AoC (d) AoB = BoA

IIT JEE eBooks: www.crackjee.xyz 7.28 Comprehensive Mathematics—JEE Advanced

0 Ê Á a sin 43. Let A = Á Ë - sin a sin b then (a) (b) (c) (d)

sin a 0 - cos a cos b

sin a sin b ˆ cos a sin b ˜ , ˜ 0 ¯

|A| is independent of a and b A-1 depends only on a A-1 depends only on b A is skew symmetric matrix

44. Let A and B are two matrices such that AB = BA, then for every n ∈ N, (a) AnB = BAn (b) (AB)n = AnBn (c) (A + B)n = nC0An + nC1 An-1 B + n C2 An-2 B + ... + nCn Bn. 2n 2n n (d) A - B = (A - Bn) (An + Bn) 45. If A and B are square matrices of the same order such that A2 = A, B2 = B, AB = BA = O, then (a) AB2 = O (c) (A - B)2 = A - B

(b) (A + B)2 = A + B (d) (A – B)2 = A + B

MATRIX-MATCH TYPE QUESTIONS 46. Let p, q, r be the roots of x3 + x2 + k π 0, with k π 0. Denote the matrix on the left side by A. Column 1 Column 2 È 0 Í 3 (a) Í(q – p ) Í 3 ÎÍ (r – p) Èp Í (b) Í q ÎÍ r

q r p

( p – q )3 0 ( r – q )3 r˘ p ˙˙ q ˙˚

( p – r )3 ˘ ˙ (q – r )3 ˙ (p) A is ˙ singular 0 ˚˙

(q) A is nonsingular

È p 0 0˘ Í ˙ (c) Í 0 q 0˙ ÍÎ 0 0 r ˙˚

(r) |A| = 0

È0 0 Í (d) Í0 q ÍÎ r 0

(s) |A|2 = k2

p˘ 0 ˙˙ 0 ˙˚

47. The system of equations x +y+z=3 x + 2y + 3z = 6 x + 3y + lz = m

Column 1 Column 2 (a) has unique solution (p) l = 5, m = 9 (b) has no solution (q) l = 5, m π 9 l π 5, m π 9 of solutions (d) is consistent (s) l π 5. 48. With each matrix A on the left associate values(s) of l for which there exists l π 0 such that AX = lX. Column 1 Column 2 2 2˘ È 3 Í 2 4 1 ˙˙ (p) 0 (a) Í ÍÎ – 2 – 4 – 1˙˚ È – 3 – 7 – 5˘ Í 4 3 ˙˙ (b) Í 2 2 2 ˙˚ ÎÍ 1

(q) 1

6 6˘ È4 Í1 3 2 ˙˙ (c) Í ÍÎ –1 – 5 – 2˙˚

(r) 2

È11 – 4 – 7 ˘ Í ˙ (d) Í 7 – 2 – 5˙ ÍÎ10 – 4 – 6˙˚

(s) 3

49. Suppose a, b, c are three distinct real numbers and f (x) is a real quadratic polynomial such that È 4a 2 Í Í 4b 2 Í 2 ÍÎ 4c

4a 1˘ È f (–1) ˘ È3a 2 + 3a ˘ ˙ Í ˙ 4b 1˙ ÍÍ f (1) ˙˙ = Í 3b 2 + 3b ˙ ˙ Í ˙ 4c 1˙˚ ÎÍ f (1) ˚˙ ÍÎ 3c 2 + 3c ˙˚

Column 1 (a) x-coordinate(s) of the point of intersection of y = f(x) with the x-axis. (b) Area bounded by y =

Column 2

(p) 4

3 f(x) 2

and the x-axis. (c) Maximum value of f(x) (d) Length of the intercept made by y = f(x) on the x-axis.

(q) 2 (r) 1 (s) – 2

Ê -2 -1ˆ 50. Let A = Á Ë 3 1 ˜¯ Match the enteries in column 1 with in the enteries in column 2

IIT JEE eBooks: www.crackjee.xyz Matrices 7.29

Column 1 (a) A2020

Column 2 Ê 1 0ˆ (p) Á Ë 0 1˜¯

(b) A2019

Ê -2 -1ˆ (q) Á Ë 3 1 ˜¯

(c) A2018

Ê 1 1ˆ (r) Á Ë -3 -2˜¯

(d) A2017

Ê 2017 0ˆ (s) Á Ë 2019 1˜¯

ASSERTION-REASON TYPE QUESTIONS 51. Statement-1: If A and B are two 3 ¥ 3 matrices such that AB = O, then A = O or B = O. Statement-2: If A, B and X are three 3 ¥ 3 matrices such that AX = B, |A| π 0, then X = A–1 B. È1 p ˘ È1 100p ˘ 52. Statement-1: If A = Í , then A100 = Í . ˙ 1 ˙˚ Î0 1 ˚ Î0 Statement-2: If B is a 2 ¥ 2 matrix such that B2 = O, then (I + B)n = I + nB for each n ∈ N. È2 1 1 ˘ Í ˙ 53. Statement-1: As A = Í0 1 1 ˙ ÍÎ1 1 2˙˚ 3 2 tion x – 5x + 7x – 3 = 0, therefore A is invertible. Statement-2: If a square matrix A n n – 1 tion a0x + a1x + ... an – 1x + an = 0, and an π 0, then A is invertible. 54. Statement-1: If a, b, c are in A.P. the system of equations 3x + 4y + 5z = a (1) 4x + 5y + 6z = b (2) 5x + 6y + 7z = c (3) is consistent. Statement-2: If |A| π 0, the system of equations AX = B is consistent. È a11 55. Let A = Í Îa21

a12 ˘ Èx ˘ Èy ˘ , X = Í 1 ˙, y = Í 1 ˙ ˙ a22 ˚ Î x2 ˚ Î y2 ˚

Statement-1: If A is symmetric, then X¢AY = Y ¢AX for each pair of X and Y. Statement-2: If X¢AY = Y¢AX for each pair of X and Y, then A is symmetric.

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 56 to 59 A square matrix A of size n ¥ n is said to be unimodular if det (A) = 1. 56. If A and B are two unimodular matrices, then which one of the following is a unimodular matrix? (a) A + B (b) A - B (c) AB (d) A – B–1 57. If A is unimodular, then which of the following is not necessarily unimodular. (a) - A (b) A-1 (c) adj A (d) wA, where w is cube root of unity 58. If A and B are unimodular matrices, then adjoint of APB is (a) A (adj P) B (b) B (adj P) A (d) A-1 (adj P) B-1 (c) B-1 (adj P) A-1 59 If A is a matrix such that A¢A = In, then (a) A is unimodular (b) - A is unimodular (c) I + A is unimodular (d) none of these Paragraph for Question Nos. 60 to 63

60.

61.

62.

63.

A square matrix A is said to be a nilpotent matrix of degree r, if r is the least positive integer such that Ar = O. If A and B are two nilpotent matrices then AB will be a nilpotent matrix if (b) AB = BA (a) AB = In (c) AB = A (d) ABA = 0 If A and B are nilpotent matrices then A + B will be a nilpotent matrix if (a) A + B = AB (b) AB = BA (c) A - B = AB (d) none of these If A is a nilpotent matrix, then I + A is (a) singular (b) non-singular (c) symmetric (d) skew symmetric If A is nilpotent matrix, then A - I is (a) singular (b) non-singular (c) symmetric (d) skew-symmetric

Paragraph for Question Nos. 64 to 67 A square matrix A is said to be orthogonal if A¢A = AA¢ = In 64. If A and B are orthogonal matrices, of the same size, then which one of the following is an orthogonal matrix

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(a) AB (b) A + B (c) A + iB (d) i(A + B) 65. If k ∈ C and both A = (aij) n ¥ n and kA are orthogonal matrices, then (a) k is nth root of unity (b) k is imaginary (c) k must be a positive real number (d) k = ± 1 66. If A is real skew-symmetric matrix is such that A2 + I = 0, then (a) A is orthogonal matrix (b) A is orthogonal matrix of odd order (c) A is orthogonal matrix of even order (d) none of these 1 1 I and A + I are orthogonal matri67. If both A 2 2 ces, then (a) A is skew-symmetric matrix of even order (b) A is orthogonal 3 I (c) A2 = 4 (d) none of these

INTEGER-ANSWER TYPE QUESTIONS È1 2 x ˘ Í ˙ 68. If A = Í0 1 0 ˙ and B = ÎÍ0 0 1 ˙˚

È1 - 2 y ˘ Í0 1 0 ˙ Í ˙ and ÎÍ0 0 1 ˙˚

AB = I3, then x + y is equal to È 2a b c ˘ Í ˙ 69. Suppose a, b, c ∈ R and abc = 1. If A = Í b 2c a ˙ ÍÎ c a 2b ˙˚ is such that A¢A = 41/3I and |A| > 0, then the value of a3 + b3 + c3 is 70. Let A be 3 ¥ 3 matrix such that A¢A = I and |A| = 1, then the value of |A - I| is È2 q ˘ È x y q˘ 71. Let A = Í and A8 = Í ˙ ˙ , then x - y is Î0 1 ˚ Î0 1 ˚ 72. Let A and B be two non-singular matrices such that A π I, B3 = I and AB = BA2, then the least value of k such that Ak = I is 73. Suppose A is a matrix such that A2 = A and (I + A)10 = I + kA, then k/341 is È 2 3˘ -1 74. Let A = Í ˙ . If A = xA + yI, then x + 2y is Î - 1 5˚ equal to

Èa b ˘ 3 75. Let A = Í ˙ be such that A = O, then sum of Îc d ˚ all the elements of A2 is 76. Let S denote the set of all values of t such that the system of homogeneous equations tx + (t + 1)y + (t - 1)z = 0 (t + 1)x + ty + (t + 2)z = 0 (t - 1)x + (t + 2)y + tz = 0 has non-trivial then 4 Â (- t) t Œs

77. The number of values of l for which the homogeneous system of equations (a - l)x + by + cz = 0 bx + (c - l)y + az = 0 cx + ay + (b - l)z = 0 has a non-trivial solution is È0 1 1 ˘ Í ˙ 78. Let A = Í1 0 1˙ and X π O be a column matrix ÍÎ1 1 0˙˚ such that AX = lX, then sum of distinct values of l is 79. Let A, B be two 3 ¥ 3 matrices with real entries. Let C = AB – BA, and suppose C commutes with both A and B, then AB4 – B4A = kB3A, then k is equal to È2017 ˘ Í ˙ 80. If A = Í 2018˙ and B = [2017 ÍÎ 2019˙˚

2018

2019],

then det (AB)/(2017) (2018) (2019) is equal to LEVEL 2

SINGLE CORRECT ANSWER TYPE QUESTIONS È0 1 0 ˘ Í ˙ 1. Let E = Í0 0 1˙ , F = ÍÎ0 0 0˙˚ (E + xF2)3 equals (a) xI (c) xE

È0 0 0 ˘ Í1 0 0 ˙ Í ˙ and x ∈ R, then ÍÎ0 1 0˙˚ (b) O (d) xF

IIT JEE eBooks: www.crackjee.xyz Matrices 7.31

Èa b ˘ 2. Let A = Í ˙ , a, b, c, d π 0, then B = AA¢ - A¢A Îc d ˚ equals (a) (ad - bc)I (b) (ac - bd)I (c) O (d) none of these 2 2˘ È6 Í- 2 3 - 1˙ 3. If A = Í ˙ , then |xI - A| equals ÍÎ 2 - 1 3 ˙˚ (a) x3 + 12x2 - 44x + 12 (b) x3 - 12x2 + 36x - 12 (c) x3 - 12x2 - 36x + 12 (d) x3 – 12x2 + 44x – 8 4. Let A be as in question 3 and f(x) = |xI - A|, then f(A) equals (a) O (b) A (c) 2A (d) - A Èa b ˘ Èa c ˘ , a, b, c, d ∈ C, put Aq = Í 5. For A = Í ˙ ˙. Îc d ˚ Îb d ˚ We say matrix A is a Hermitian matrix, if A = Aq. Suppose A is Hermitian matrix such that A2 = O, then (a) A = A¢ (b) A = - A¢ (c) A = O (d) A = I 6. Let A be a square matrix such that A2 = I, then at least one of I - A and I + A is (a) singular (b) symmetric (c) skew symmetric (d) none-singular 7. Let A be a square matrix such that A2 = A and |A| π 0, then (a) A = A¢ (b) A = - A¢ (c) A¢ = –I (d) A = - I 8. Let A be as in problem 7 and B = 2A - I where I is the identity matrix, then (b) B2 = I (a) B2 = 2B 2 (d) B2 + B = O (c) B = O 9. Suppose A and B are two nonsingular matrices such that AB = BA2 and B5 = I, then (b) A31 = I (a) A32 = I (d) A50 = I (c) A30 = I 10. Let A = (aij) 3 ¥ 3 where aij = i2 + j2. Then A is (a) orthogonal (b) skew Hermitian (c) singular (d) none of these 11. Suppose A and B two matrices. Let C, D, E be the matrices obtained by interchanging ith and jth rows of A, B and AB respectively then

(a) E = CD (b) E = CB (c) E = AD (d) E = BC 12. Suppose A and B are two matrices. Let C, D and E be the matrices obtained by interchange ith and jth columns of A, B and AB respectively, then (a) E = CD (b) E = CB (c) E = AD (d) E = DC

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 13. Let A and B problem 5. Suppose A2 + B2 = O, then (a) A = O (b) B = O (c) A = ± iB (d) A = - B = iC for some matrix C 14. The number of matrices A = (aij)n¥n and B = (bij)n¥n be such that AB - BA = I is (a) 0 (b) nn (c) nnn (d) n2 Èa b ˘ 15. Let A = Í ˙ and f(l) = det (A - lI), then Îc d ˚ (a) f(l) = l2 - (a + d)l + ad - bc (b) f(A) = 0 (c) A2 = O implies Ar = 0 " r ≥ 2 (d) none of these 16. Let A be as in problem 15 and Ak = O for some k ≥ 3; then (a) A = O (b) A2 = O (c) A is singular (d) A is non-singular 17 If a2 + b2 + c2 = – 2, and È 1 + a 2 x (1 + b 2 ) x (1 + c 2 ) x ˘ Í ˙ 2 2 2 A = Í(1 + a ) x 1 + b x (1 + c ) x ˙ , then A is Í 2 2 2 ˙ ÎÍ(1 + a ) x (1 + b ) x 1 + c x ˚˙ non-singular if (a) x = 0 (b) – 1 (c) 1 (d) 2 18. Let A, B and C be 2 ¥ 2 matrices with entries from A * B= (a) (b) (c) (d)

1 (AB¢ + BA¢) 2

A*B=B*A A * A = A2 A * (B + C) = A * B + A * C 2A * I = A + A¢

IIT JEE eBooks: www.crackjee.xyz 7.32 Comprehensive Mathematics—JEE Advanced

19. With A, B, C follows:

o as

1 (AB¢ - BA¢), then 2 (a) AoA = O (b) AoI = O (c) Ao(B + C) = AoB + AoC (d) none of these 20. Let A be a symmetric matrix such that A5 = O and B = I + A + A2 + A4, then B is (a) symmetric (b) singular (c) non-singular (d) skew symmetric AoB =

MATRIX-MATCH TYPE QUESTIONS 21. Match the matrix on the left with the equations satColumn 1 Ê 0 c bˆ (a) Á c 0 a˜ Á ˜ Ë b a 0¯

Column 2 (p) x3 – (a2 + b2 + c2)x – 2abc = 0

Ê 0 a cˆ (b) Á a 0 b˜ Á ˜ Ë c b 0¯

(q) x3 + (a2 + b2 + c2)x = 0

Ê 0 b aˆ (c) Á b 0 c ˜ Á ˜ Ë a c 0¯

(r) x2 + (a2 + b2 + c) = 0

c – bˆ Ê 0 Á a ˜ (s) x4 – (a2 + b2 + c2)x2 – (d) – c 0 Á ˜ 2abcx = 0 Ë b –a 0 ¯ 22. Match each matrix on the left with the condition(s) on the right to make the matrix invertible. Column 1 Column 2 c c ˘ Èa + b Í a b+c a ˙˙ (a) Í (p) abc π 0 ÍÎ b b c + a ˙˚ È( a 2 + b 2 ) / c ˘ c c Í ˙ 2 2 a a (b + c ) / a ˙ (b) Í Í ˙ b b (c 2 + a 2 ) / b ˙˚ ÍÎ (q) a + b + c π 0 c c ˘ Èa + b – c Í a b+c –a a ˙˙ (c) Í ÍÎ b b c + a – b ˙˚

(r) bc + ca + ab π 0 a + b + c a b 2 È ˘ Í ˙ c b + c + a c 2 (d) Í ˙ ÍÎ c a c + a + 2b ˙˚ (s) a3 + b3 + c3 π 0

ASSERTION-REASON TYPE QUESTIONS Ê 1 0ˆ Ê 0 1ˆ 23. Let a, b ∈ R, and I = Á and J = Á . ˜ Ë 0 1¯ Ë –1 0˜¯ Statement-1: Inverse of aI + bJ is cI + dJ if and only if ac – bd π 0 and ad + bc = 0. Statement-2: (aI + bJ) (cI + dJ) = (ac – bd)I + (ad + bc)J c – bˆ Ê 0 Á a ˜ and t ∈ C. 24. Let A = – c 0 Á ˜ Ë b –a 0 ¯ Statement-1: adj(tI – A) = t2I + tA + A2 + (a2 + b2 + c2)I Statement-2: adjA = A2 + (a2 + b2 + c2)I 25. Let Ai, Bi, i = 0, 1, 2, . . ., k be 2k matrices each of the size n ¥ n. Statement-1: A0 + A1l + . . . + Aklk = B0 + B1l + . . . + Bklk for each l ∈ R if and only if Ai = Bi for i = 0, 1, 2 . . . k. Statement-2 l ∈ R, l π 0 such that A 0 + A 1l + . . . + A kl k = B 0 + B 1l + . . . + B kl k Ê a bˆ Ê p qˆ 26. Let A = Á and B = Á . ˜ b a Ë ¯ Ë r s ˜¯ Statement-1: If b = 0, then AB = BA Statement-2: If b π 0, then AB = BA ¤ p = s and q = r. 27. Suppose a, b, g are real numbers such that cos (b – g ) + cos(g – a) + cos (a – b ) = –3/2 (1) Statement-1: The matrix Ê cos a sin a -2ˆ A = Á cos b sin b 1 ˜ Á ˜ Ë cos g sin g 1 ¯ is a singular matrix. Statement-2: cos a + cosb + cos g = 0 and sin a + sin b + sin g = 0

IIT JEE eBooks: www.crackjee.xyz Matrices 7.33

28. Let A = (aij)3¥3 be a 3 × 3 matrix with aij Œ R for 1 £ i £ 3, 1 £ j £ 3. Let X be a 3 × 1 matrix. Statement-1: If A is skew-symmetric, then I + A is non-singular. Statement-2: If AX = O implies X = O, then A is non-singular.

COMPREHESION-TYPE QUESTIONS Paragraph for Question Nos. 29 to 33 A square matrix A is said to be an idempotent matrix if A2 = A. 29. If A is a non-singular idempotent matrix, then (b) A = O (a) A = In (c) A + A¢ = O

(d) A = – In

30. If A and B are idempotent matrices and AB = BA, then which of the following is an idempotent matrix (a) A + B

(b) AB

(c) A - B

(d) B - A

31. Suppose A is an idempotent matrix, and B = I - A, then (a) B is an idempotent

(b) AB = In

(c) BA¢ = In

(d) B2 = In

32. If A is an idempotent matrix and B = I - A, then (a) AB = O

(b) BA π O

(c) B2 = I

(d) AB = In

33. If A, B and A + B are idempotent matrices, then (a) AB = BA = O

(b) AB = BA = In

(c) AB = BA π O

(d) AB = BA π In

Paragraph for Question Nos. 34 to 38 Let A = (aij)n¥n be a matrix such that aij ∈ C trace of A, denoted by tr(A) as follows: n

tr(A) = Â aii i =1

Let A, B and C be three n ¥ n matrices such that C¢C = In and l, m ∈ C. 34. tr(lA + mB) equals

(a) l tr(A) + m tr(B)

(b) (l + m) tr(A + B)

(c) lm tr(A + B)

(d) (l + m) tr(AB)

35. tr(AB) equals (a) tr(A) tr(B)

(b) tr(BA)

(c) tr(A) + tr(B)

(d) none of these

36. If aij Œ R " i, j, then tr(AA¢) is (a) positive

(b) negative

(c) non-negative

(d) non-positive

37. tr(C¢ A C) equals (a) tr(A)

(b) tr(CA)

(c) tr(AC¢)

(d) tr(C)

38. tr(ABC) equals (a) tr(CBA)

(b) tr(BAC)

(c) tr(ACB)

(d) tr(ABC)

INTEGER-ANSWER TYPE QUESTIONS 39. If matrix A is such that A2 = A and (I + A)3 = I + kA, then k is È1 0 0 ˘ Í ˙ 40. If A = Í0 1 1 ˙ , I = ÍÎ0 - 2 4˙˚

È1 0 0 ˘ Í0 1 0 ˙ Í ˙ and ÍÎ0 0 1˙˚

1 (A2 + aA + bI), then b – 7 = 6 41. Suppose A π O, A2 = O and A(I + A)100 = kA, then k is 1 3˘ È1 Í5 2 6 ˙˙ . Then the smallest n ∈ N, 42. Let A = Í ÍÎ- 2 - 1 - 3˙˚ A-1 =

such that An = O is È7 1 ˘ 43. Let A be a square matrix and A3 = Í A|. ˙ Î1 4 ˚ 44. Let A be a non-singular matrix and X π O, Y π O be two column matrices such that AX = (1/100)X and A-1Y = lY, then one of the possible values of l/25 is 45. Let A = (aij)3¥3 be such that det(A) = 5, then det(A adj(A))/125 is

IIT JEE eBooks: www.crackjee.xyz 7.34 Comprehensive Mathematics—JEE Advanced

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS È 1. If A = Í Î1

6. Let w π 1 be a cube root of unity and S be the set of all non-singular matrices of the form

˘ È1 0˘ and B = Í ˙ ˙ , then 1˚ Î3 1˚

value of α for which A2 = B is (a) 1 (b) – 1 (c) 4 (d) no real values. [2003] Èa 2 ˘ 3 2. If A = Í ˙ and |A | = 125 then the value of α is Î2 a˚ (a) ±1 (c) ±3

(b) ±2 (d) ±5

[2004]

È1 0 0 ˘ Í ˙ 3. If A = Í0 1 1 ˙ , 6 A–1 = A2 + cA + dI, then ÍÎ0 –2 4˙˚ (c, d) is (a) (–11, 6) (b) (–6, 11) (c) (6, 11) (d) (11, 6) [2005] È 3/2 4. If P = Í ÎÍ –1 / 2

1/ 2 ˘ È1 1˘ ˙, A = Í ˙ 3 / 2˚˙ Î0 1˚

and Q = PAP¢ and X¢ = P¢ Q2005 P then X equals ˘ 1 1 È2 + 3 Í ˙ (a) 4 ÍÎ –1 – 2 – 3 ˙˚ (b)

1 4

È 2005 2 – 3 ˘ Í ˙ ÎÍ2 + 3 2005 ˚˙

[2005]

5. The number of 3 ¥ 3 matrices A whose entries are È x ˘ È1 ˘ Í ˙ Í ˙ either 0 or 1 and for which the system A Í y ˙ = Í0˙ ÎÍ z ˚˙ ÎÍ0˚˙ has exactly two distinct solutions, is (a) 0 (b) 29 – 1 (c) 168 (d) 2 [2010]

b˘ ˙ c˙. ˙ 1˚

where each of a, b and c is either ω or w2. Then the number of distinct matrices in the set S is (a) 2 (b) 6 (c) 4 (d) 8 [2011] T 7. If P is a 3 ¥ 3 matrix such that P = 2P + I, where PT is the transpose of P and I is the 3 ¥ 3 identity matrix, then there exists a column matrix X = È x ˘ È0 ˘ Í y ˙ π Í0 ˙ Í ˙ Í ˙ such that ÍÎ z ˙˚ ÍÎ0˙˚ È0 ˘ Í ˙ (a) PX = Í0˙ ÍÎ0˙˚

(b) PX = X

(c) PX = 2X

(d) PX = –X

[2012]

È 1 0 0˘ Í ˙ 8. Let P = Í 4 1 0˙ and I be the identity matrix of ÍÎ16 4 1˙˚ order 3. If Q = [qij] is a matrix such that P50 – Q = q +q I, then 31 32 equals q21 (a) 52 (c) 201

È4 + 2005 3 6015 ˘ ˙ (c) Í 4 – 2005 3 ˙˚ ÍÎ 2005 È1 2005˘ (d) Í 1 ˙˚ Î0

È1 a Í Íw 1 Í 2 Îw w

(b) 103 (d) 205

[2016]

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 1*. Let M and N be two 3 ¥ 3 non-singular skew-symmetric matrices such that MN = NM. If PT denotes the transpose of P, then M2N2(MT N)–1 (MN–1)T is equal to (b) – N2 (a) M2 (c) – M2 (d) MN [2011] 2. For 3 ¥ 3 matrices M and N, which of the following statement(s) is (are) NOT correct?

* Question is incorrect. There does not exist any 3 × 3 non-singular skew-symmetric matrix. In fact M¢ = – M fi | M¢| = |–M| or |M| = – | M |fi | M | = 0

IIT JEE eBooks: www.crackjee.xyz Matrices 7.35

3.

4.

5.

6.

(a) NTMN is symmetric or skew symmetric, according as M is symmetric or skew symmetric (b) MN – NM is skew symmetric for all symmetric matrices M and N (c) MN is symmetric for all symmetric matrices M and N (d) (adj M) (adj N) = adj(MN) for all invertible matrices M and N [2013] Let ω be a complex cube root of unity with w π 1 and P = [pij] be a n ¥ n matrix with pij = w i+j. Then P2 π O, when n = (a) 57 (b) 55 (c) 58 (d) 56 [2013] Let M be a 2 × 2 symmetric matrix with integer entries. Then M is invertible if M is the transpose of the second row of M (b) the second row of M is the transpose of the M (c) M is a diagonal matrix with nonzero entries in the main diagonal (d) the product of entries in the main diagonal of M is not the square of an integer [2014] Let M and N be two 3 ¥ 3 matrices such that MN = NM. Further, if M π N2 and M2 = N4, then (a) determinant of (M2 + MN2) is 0 (b) there is a 3 ¥ 3 non-zero matrix U such that (M2 + MN2) U is the zero matrix (c) determinant of (M2 + MN2) ≥ 1 (d) for a 3 ¥ 3 matrix U, if (M2 + MN2) U equals the zero matrix then U is the zero matrix [2014] Let X and Y be two arbitrary, 3 ¥ 3, non-zero, skew-symmetric matrices and Z be an arbitrary 3 ¥ 3, non-zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric? (a) Y3Z4 – Z4Y3 (b) X44 + Y44 (c) X4Z3 – Z3Y4 (d) X23 + Y23 [2015]

È3 -1 -2˘ Í ˙ 7. Let P = Í2 0 a ˙ , where a ∈ R ÍÎ3 -5 0 ˙˚ Suppose Q = [qij] is a matrix such that PQ = kI, where k ∈ R, k π 0 and I is identity matrix of order. k k2 If q23 = - , and det (Q) = , then 8 2 (a) a = 0, k = 8 (b) 4a – k + 8 = 0 9 (c) det (P adj (Q)) = 2 (d) det (Q adj (P)) = 213

8. Let a, l, m ∈ R. Consider the system of linear equations ax + 2y = l 3x – 2y = m Which of the following statements (s) is (are) correct? (a) If a tions for all values of l and m (b) If a π – 3, then the systems has a unique solution for all values of l and m (c) If l + m solutions for a = – 3 (d) If l + m π 0, then the systems has no solution for a =–3 [2016]

COMPREHESION-TYPE QUESTIONS Paragraph for Questions Nos. 1 to 3 È1 0 0 ˘ Í ˙ Let A = Í2 1 0˙ and X1, X2, X3 be three column ÍÎ3 2 1˙˚ matrices such that È2˘ È2˘ È1 ˘ Í ˙ Í ˙ Í ˙ = 0 , AX2 = Í3˙ and AX3 = Í3˙ AX1 Í ˙ ÍÎ0˙˚ ÍÎ1 ˙˚ ÎÍ0 ˚˙ and X is a 3 ¥ 3 matrix whose columns are X1, X2, X 3. 1. Value of det(X) is (a) – 2

(b) – 1

(c) 3

(d) 0

–1

2. Sum of the elements of X is (a) – 1 (b) 0 (c) 4

(d) 3/4

È3˘ Í ˙ 3. If [a] = [3 2 0] X Í2˙ , then a equals ÍÎ0˙˚ (a) 5

(b) 4

(c) 3/2

(d) 5/2 [2006]

Paragraph for Question Nos. 4 to 6 Let A be the set of all 3 ¥ 3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0. 4. The number of matrices in A is (a) 12 (b) 6 (c) 9 (d) 3 5. The number of matrices A in A for which the system of linear equations

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È x ˘ È1 ˘ A ÍÍ y ˙˙ = ÍÍ0˙˙ ÍÎ z ˙˚ ÍÎ0˙˚ has a unique solution, is (a) less than 4 (b) at least 4 but less than 7 (c) at least 7 but less than 10 (d) at least 10 6. The number of matrices A in A for which the system of linear equations È x ˘ È1 ˘ A ÍÍ y ˙˙ = ÍÍ0˙˙ ÎÍ z ˚˙ ÎÍ0˚˙ is inconsistent, is (a) 0 (c) 2

(a) – 2 (b) 2 (c) 3 (d) – 3 12. Let b = 6, with a and c satisfying (E). If α and β are the roots of the quadratic equation ax2 + bx + c = 0, Ê 1 1ˆ then  Á + ˜ b¯ n=0 Ë a •

(b) more than 2 (d) 1 [2009]

Paragraph for Questions Nos 7 to 9 Let p be an odd prime number and Tp be the following set of 2 ¥ 2 matrices: Ï ¸ Èa b ˘ : a, b, c Œ{0, 1, 2, ..., p – 1}˝ Tp = Ì A = Í ˙ Îc a˚ ÓÔ ˛Ô 7. The number of A in Tp such that A is either symmetric or skew-symmetric or both, and det (A) divisibly by p is (b) 2 (p – 1) (a) (p – 1)2 (d) 2p – 1 (c) (p – 1)2 + 1 8. The number of A in Tp such that the trace of A is not divisible by p but det (A) is divisible by p is [Note: The trace of a matrix is the sum of its diagonal entries.] (a) (p – 1) (p2 – p + 1) (b) p3 – (p – 1)2 (c) (p – 1)2 (d) (p – 1) (p2 – 2) 9. The number of A in Tp such that det (A) is not divisible by p is (b) p3 – 5p (a) 2p2 (d) p3 – p2 [2010] (c) p3 – 3p Paragraph for Question Nos. 10 to 12 Let a, b and c be three real numbers satisfying È1 9 7 ˘ [a b c] ÍÍ8 2 7 ˙˙ = [0 0 0] ÍÎ7 3 7 ˙˚

10. If the point P(a, b, c), with reference to (E), lies on the plane 2x + y + z = 1, then the value of 7a + b + c is (a) 0 (b) 12 (c) 7 (d) 6 3 11. Let w be a solution of x – 1 = 0 with Im(ω) > 0. If a = 2 with b and c satisfying (E), then the value of 3 1 3 + b + c is equal to a w w w

(E)

n

is

(a) 6

(b) 7

6 7

(d) •

(c)

[2011]

INTEGER-ANSWER TYPE QUESTIONS 1. Let M be a 3 ¥ 3 matrix satisfying È0˘ È-1˘ È1˘ È1˘ È1˘ È 0 ˘ Í ˙ Í ˙ Í ˙ Í ˙ M Í1˙ = Í 2 ˙ , M Í-1˙ = Í 1 ˙ , and M ÍÍ1˙˙ = ÍÍ 0 ˙˙ ÍÎ1˙˚ ÍÎ12˙˚ ÍÎ0˙˚ ÍÎ 3 ˙˚ ÍÎ 0 ˙˚ ÍÎ-1˙˚ [2011] Then sum of diagonal entries of M is -1 + 3i , where i = -1 , and r, s ∈ {1, 2, 2 È( - z )r z 2 s ˘ ˙ and I be the identity matrix 3}. Let P = Í 2 s ÍÎ z z r ˙˚ of order 2. Then the total number of ordered pairs (r, s) for which P2 = – I is [2016]

2. Let z =

SUBJECTIVE-TYPE QUESTIONS Èa b c ˘ Í ˙ 1. If matrix A = Íb c a ˙ where a, b, c are real numÍÎ c a b ˙˚ bers such that a + b + c > 0 and abc = 1 and A¢A = [2003] I a 3 + b 3 + c 3. 2. If M is a 3 ¥ 3 matrix, and det (M) = 1 and MM¢ = I, then prove that det (M – I) = 0

IIT JEE eBooks: www.crackjee.xyz Matrices 7.37

MATRIX-MATCH TYPE QUESTIONS

3. Let A, B, U, V and X follows: Èa 0 1˘ Í ˙ A = Í1 c b ˙ , B = ÍÎ1 d b ˙˚ Èa 2 ˘ Í ˙ V = Í 0 ˙, X = Í0˙ Î ˚

Èa Í0 Í ÍÎ f

1 d g

1˘ c ˙˙ , U = h ˙˚

Èf˘ Íg˙ Í ˙, ÍÎ h ˙˚

È x˘ Í y˙ Í ˙ ÍÎ z ˙˚

p

q

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46.

47.

If AX = U BX = V cannot have unique solution. Also prove that if afd π 0, then BX = V has n no solution. [2004]

Answers 48. LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17. 21. 25.

(a) (a) (a) (d) (d) (a) (b)

2. 6. 10. 14. 18. 22.

(b) (a) (a) (c) (b) (a)

3. 7. 11. 15. 19. 23.

(b) (b) (a) (d) (a) (a)

4. 8. 12. 16. 20. 24.

(c) (a) (c) (a) (d) (a)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 26. 28. 30. 32. 34. 36. 38. 40. 42. 44.

(a), (a), (b), (a), (a), (a), (a), (a), (a), (a),

(c) (b), (c), (b), (c) (c), (b) (c), (b), (b),

(c) (d) (c) (d) (d) (c) (c), (d)

27. 29. 31. 33. 35. 37. 39. 41. 43. 45.

(a), (a), (a), (a), (a), (a), (a), (a), (a), (a),

(c), (b), (b), (c), (c), (b), (b), (b), (d), (b),

(d) (d) (c), (d) (d) (d) (c) (c), (d) (c), (d) (d) (d)

49.

50.

ASSERTION-REASON TYPE QUESTIONS 51. (d) 55. (b)

52. (a)

53. (a)

54. (b)

COMPREHENSION-TYPE QUESTIONS 56. (c) 60. (b) 64. (a)

57. (a) 61. (b) 65. (d)

58. (c) 62. (b) 66. (c)

59. (d) 63. (b) 67. (a)

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INTEGER-ANSWER TYPE QUESTIONS 68. 72. 76. 80.

0 7 2 0

69. 4 73. 3 77. 3

70. 0 74. 1 78. 1

71. 1 75. 0 79. 4

INTEGER-ANSWER TYPE QUESTIONS 39. 7 43. 3

3. (d) 7. (a) 11. (b)

4. (a) 8. (b) 12. (c)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 13. 15. 17. 19.

(a), (a), (a), (a),

(b) (b), (c) (b), (d) (c)

14. 16. 18. 20.

(a) (b), (c) (a), (c), (d) (a), (c)

MATRIX-MATCH TYPE QUESTIONS p

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22.

24. (b) 28. (a)

30. (b) 34. (a) 38. (a)

2. (c) 6. (a)

3. (b) 7. (d)

4. (d) 8. (b)

MULTIPLE CORRECT ANSWERS TYPES QUESTIONS 1. 3. 5. 7.

Question is incorrect (b), (c) (d) (a), (b) (b), (c)

2. 4. 6. 8.

(c), (c), (c), (b),

(d) (d) (d) (c)

COMPREHESION-TYPE QUESTIONS 1. (c) 5. (b) 9. (d)

1. 9

2. (b) 6. (b) 10. (d)

3. (a) 7. (d) 11. (a)

4. (a) 8. (c) 12. (b)

2. 1

SUBJECTIVE-TYPE QUESTIONS 1. 4

Hints and Solutions LEVEL 1

Èa1 Èa b ˘ 1. Let A = Í B= Í ˙ Îc d ˚ Î c1

25. (c)

26. (b)

COMPREHENSION-TYPE QUESTIONS 29. (a) 33. (a) 37. (a)

1. (d) 5. (a)

INTEGER-ANSWER TYPE QUESTIONS

ASSERTION-REASON TYPE QUESTIONS 23. (a) 27. (a)

42. 3

SINGLE CORRECT ANSWERS TYPE QUESTIONS

SINGLE CORRECT ANSWER TYPE QUESTIONS 2. (d) 6. (a) 10. (c)

41. 1 45. 1

PAST YEARS IIT QUESTIONS

LEVEL 2

1. (a) 5. (c) 9. (b)

40. 4 44. 4

31. (a) 35. (b)

32. (a) 36. (c)

b1 ˘ d1 ˙˚

Then a = b = bc1 – b1c – a1d – ad1 2. Let X = (1 1 1), then XA = X fi XA2 = XA = X Continuing in this way we get XA2018 = X Sum of the entries in A2018 is 3. 3. Use A = - A¢ 4. Set |At| = 0

IIT JEE eBooks: www.crackjee.xyz Matrices 7.39

5. A2 = I fi a2 + b2 + c2 = 1 and bc + ca + ab = 0 Also, |A2| = 1 fi |A| = ± 1. But |A| = 3abc – (a3 + b3 + c3) = – (a + b + c) (a2 + b2 + c2 – bc – ca – ab) = – (a + b + c) As a + b + c > 0, we get |A| = – 1. \ a3 + b3 + c3 = 7 Ê 0 – b – cˆ 6. Let A = Á b 0 – a˜ , then tr(A2) = 0 Á ˜ 0 ¯ Ëc a – 2 (a2 + b2 + c2) = 0

25. (A– 1 + B– 1) [A – A (A + B)– 1 A] = B– 1 (B + A) A–1 A[I – (A + B)–1 A] = B– 1(A + B) (A + B)–1 [A + B – A] 26. 27. 28. 29. 30. 31. 32.

fi a = b = c = 0. \A=O 1 (B - A) 7. Use A + X = B – X fi X = 2 8. Calculate directly 9. A2 = O 10. |A| = 1 for each x. 11. Calculate directly Èa b ˘ 12. If A = O, tr(A) = 0. Suppose A π O and A = Í ˙, Îc d ˚ then |A| = 0 and A2 - (a + d)A + ad - bc = 0 fia+d=0 13. Use A2 - (a + d)A + ad - bc = 0 to obtain ad - bc = 0 1 È1 - 2 ˘ 14. Use A-1 = Í 3 Î0 3 ˙˚

33.

= B– 1 I B = I |A| = 0 See Example 48. See Theory See Theory See Theory See Theory |A| π 0 fi A-1 exists. Ècos a sin a 0˘ Ècos a Í ˙Í A = Ícos b sin b 0˙ Í sin a ÎÍ cos g sin g 0˚˙ ÎÍ 0

cos g ˘ sin g ˙˙ 0 ˚˙

n

2 34. Use  Ck = 2nCn k=0

35. Use |A(q)| = 1, A(p + q) = - A(q) and A(p - q) = A(q)-1 36. AD = DA for each 3 ¥ 3 matrix if and only if D is a scalar matrix, and D-1 if exists is a diagonal matrix. 37. See Theory. 38. Find A3 to see that A3 = I fi A2 = A-1. However, A5 = A2 π A. 39. Take D1 = diag (a1, a2, a3) and D2 = diag (b1, b2, b3) and verify. È 1 0˘ È 1 0˘ -n 40. A-1 = Í ˙ fi A = Í- n 1˙ 1 1 Î ˚ Î ˚

16. Calculate directly. 17. Use A = wB.

1 -n È1/n 0 ˘ A = Í ˙ n Î - 1 1/n ˚

19. Binomial theorem is applicable if and only if AB = BA. 20. Use A(q)3 = A(3q) 21. Use |AB| = |A| |B| 22. From the last two equations we get x = 1, y = 1. This gives a = 2. p+9 1 3 23. If p π - 2, x = , y= ,z= p+2 p+2 p+2

and

1 1 1 24. 0 π 2 1 - 1 = - k 3 2 k

cos b sin b 0

1 n2

È1/n 2 0 ˘ ˙ A-n = Í 2 ÎÍ- 1/n 1/n ˚˙

43. Use A is a skew symmetric matrix of odd order. 44. A2B = A(AB) = A(BA) = (AB)A = (BA)A = BA2 Similarly, A3B = BA3 In general AnB = BAn " n ≥ 1 (b) and (c) hold as AB = BA. Also, (An - Bn) (An + Bn) = An An - Bn An + An Bn + Bn Bn

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= A2n - B2n 45. AB2 = (AB)B = OB = O (A + B)2 = A2 + AB + BA + B2 =A+O+O+B=A+B Also, (A - B)2 = A + B π A - B 46. (a) A is skew symmetric matrix of size 3 ¥ 3. \ |A| = 0 fi A is singular (b) |A| = – (p + q + r) [p2 + q2 + r2 – qr – rp – pq] = – (p + q + r) [(p + q + r)2 – 3(qr + rp + pq)] = – (– 1) [(– 1)2] = 1 fi A is non-singular. (c) |A| = pqr = k π 0 fi A is non-singular. (d) Similar to (c) x = z, y = 3 – 2z Putting this in the last equation, we get (l – 5)z = m – 9. If l π 5, the system of equations has unique solution and hence consistent. When l = 5, m π 9, the system has no solution When l = 5, m of solutions and hence consistent. 48. Use |A – lI| = 0. 49. a, b, c are three roots of the quadratic equation (4f(– 1) – 3)x2 + (4f(1) – 3))x + f(2) = 0 fi 4f(– 1) = 3, 4f(1) = 3 and f(2) = 0. Let f(x) = (x – 2) (Ax + B) Now, 3 = 4f(– 1) = 4(– 3) (– A + B) fi A – B = 1/4 3 = 4f(– 1) = 4(– 1) (A + B) fi A + B = – 3/4 \ A = – 1/4, B = – 2/4 Thus, f(x) = 1 –

2

Ê 3 21 3 x2 ˆ ˘ (b) Area = Ú (4 – x2) dx = (2) Á 4 x – ˜ ˙ 8 3 ¯ ˙˚ 2 –2 4 Ë 0 =4 (c) Maximum value of f(x) is 1 (d) Length of intercept on the x-axis is 4. 50. Use A3 = I and A2 = A–1 51. Statement-2 is true since |A| π 0, implies A–1 exists. \ AX = B fi A–1 (AX) = A–1 B fi (A–1 A)X = A–1 B fi IX = A–1 B fi X = A–1 B That statement-1 is false can be seen by the following example. È0 0 1 ˘ È0 1 0 ˘ Í0 0 0 ˙ Í ˙ Let A = Í ˙ and B = Í0 0 0˙ then AB = ÍÎ0 0 0˙˚ ÍÎ0 0 0˙˚ O but neither A = O nor B = O. 52. Since B commutes with I, we can use binomial theorem to obtain Ê nˆ Ê nˆ 2 Ê nˆ n (I + B)n = I + Á ˜ B + Á ˜ B + ... + Á ˜ B Ë 1¯ Ë 1¯ Ë n¯ Since B2 = O, we get Br = O " r ≥ 2. Thus, (I + B)n = I + nB Now, È0 p ˘ A = I + B where B = Í ˙ Î0 0 ˚ Since B2 = O, we get

1 2 x. 4

È1 100p ˘ A100 = I + 100B = Í 1 ˙˚ Î0

1

-2

2 0

53. Since a0 An + a1 An – 1 + . . . + an – 1A + anI = 0, and an π 0, we get AB = I where B= –

Fig. 7.1

The graph of y = f(x) is given in Fig. 7.1. (a) x coordinates of points of intersection of y = f(x) when the x-axis are ± 2.



an – 1 a0 n – 1 a1 n – 2 A A I – – ...– an an an

B = A–1.

54. If |A| π 0, A is invertible and we can write AX = B as X = A–1 B. \ AX = B has a unique solution and hence is consistent. Subtracting (2) from (3) and (1) from (2),

IIT JEE eBooks: www.crackjee.xyz Matrices 7.41

we get the system of equation as 3x + 4y + 5z = a x+y+z =b–a x+y+z =c–b As a, b, c are in A.P. b – a = c – b \ the last two equations are identical. From (4) and (5) we obtain x = 4b – 5a + k y = 4a – 3b – 2k z =k

(4) (5) (6)

65. A¢A = In, (kA¢) (kA) = In fi k2 = 1 66. A¢ = - A fi A2 = - A¢A fi A¢A = In fi A is orthogonal. Now, A¢A = In fi det(A¢A) = 1 fi det(A¢) detA = 1 fi (-1)n det (A)2 = 1 1 ˆÊ 1 ˆ Ê 67. ÁË A¢ - I ˜¯ ÁË A - I ˜¯ = I and 2 2 1 ˆÊ 1 ˆ Ê ÁË A¢ + I ˜¯ ÁË A + I ˜¯ = I 2 2

where k is an arbitrary complex number. Thus, the system of equations in statement-1 is consistent. 55. Suppose A is symmetric, then

fi A + A¢ = 0 fi A¢ = - A fi A2 = n

Ê - 3ˆ fi ÁË ˜¯ = (det(A))2 fi n is even. 4

A¢ = A. Since X¢AY is a 1 ¥ 1, matrix, X¢AY = (X¢AY)¢ = Y¢A¢(X¢)¢ = Y¢AX Next, suppose X¢AY = Y¢AX for each pair of X and Y. È1 ˘ È0 ˘ Let E1 = Í ˙ , E2 = Í ˙ . Î0 ˚ Î1 ˚ Taking X = E1 and Y = E2, then E1¢AE2 = E2¢AE1 fi a12 = a21 Thus, A is symmetric. 56. det (AB) = det (A) det (B) 57. det (-A) = (-1)n det (A) 58. adj (APB) = (adj B) (adj P) (adj A) = B-1 (adj P) A-1 59. det (A¢A) = det (In) = 1 fi (det (A))2 = 1 fi det A = ± 1 60. If Ar = 0, Bs = O and AB = BA, then (AB)p = 0 where p = min (r, s). 61. We can use binomial theorem if AB = BA. r+s

(A + B)r+s = Â

k=0

r+s

68. |A| = 1, |B| = 1 and A = B -1 \ x = B31 = - y fi x + y = 0 69. A¢ = A, thus, A¢A = 41/3I fi A2 = 41/3I fi |A2| = 4 fi |A| = 2 as |A| > 0 Thus, a3 + b3 + c3 = 4 70. Let B = A - I fi B¢ = A¢ - I = A¢ - A¢A = A¢(I - A) = - A¢B

71. 72. 73. 74. 75.

Now, |B| = |- A¢| |B| = (-1)3 |A| |B| = - |B| fi |B| = |A - I| = 0 x = 2 8, y = 2 8 - 1 Use B-1 AB = A2 Use Ar = A " r ≥ 1 to show k = 1023. x = - 1/13, y = 7/13 A3 = O fi |A|3 = 0 fi |A| = 0 fi ad - bc = 0

Next, show that A2 - (a + d)A = O and a + d = 0 to conclude that A2 = O. 76. t = - 1/2

Ck Ar+s-k Bk = O

62. (I + A) (I - A + A2 - ... + (-1)r-1 Ar-1) = I 63. (A - I) (Ar-1 + Ar-2 + ... + A + I) = - I 64. (AB)¢ (AB) = (B¢A¢) (AB) = B¢(A¢A)B = B¢ In B = BB¢ = In

3 I 4

to obtain a cubic equation. 78. |A - lI| = 0 fi l = 2, - 1. 79. Show k = 4 80. Show det(AB) = 0

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8 Determinants 8.1 EVALUATION OF DETERMINANTS

A determinant of order two is written as a11 a21

a12 a22

Minor

(ai j ΠC " i, j)

and is equal to a11 a22 – a12 a21. A determinant of order three is written as a11 a21 a31

a12 a22 a32

a13 a23 a33

a22 a32

a a23 - a12 21 a31 a33

Let A = (aij)n × n the minor Mij of the element aij

n. Then A is the n – 1) A.

obtained by deleting ith row and j Illustration 1

(ai j ΠC " i, j)

and is equal to a11

8.2 MINORS AND COFACTORS

a a23 + a13 21 a31 a33

a22 a32

= a11 (a22 a33 – a23 a32) – a12 (a21 a33 – a23 a31) + a13(a21 a32 – a22 a31) = a11 a22 a33 + a12 a23 a31 + a13 a32 a21 – a13 a31 a22 – a32 a23 a11 – a12 a21 a33 A determinant of order 3 can also be evaluated by using the following diagram, due to Sarrus:

Minor of element a23 in the determinant a11 a12 a13 a21 a22 a23 a31 a32 a33 is a a M23 = 11 12 a31 a32

Cofactor Let A = (aij)n ¥ n n. Then the A is denoted by cofactor of the element aij Cij and is equal to (- 1)i + j Mij where Mij is the minor of A. the element aij Note that a11 a12 a13 a21 a22 a23 = a11 M11 – a12 M12 + a13 M13 a31

The product of the three terms on each of the three product of the three terms on each of the three double

If then

Remark This method does not work for higher order determinants.

a32

a33

= a11 C11 + a12C12 + a13C13

a11 a12 a13 D = a21 a22 a23 a31 a32 a33 D = ai1 Ci1 + ai2 Ci2 + ai3 Ci3 = a1j C1j + a2j C2j + a3j C3j

i = 1, 2, 3 j = 1, 2, 3

Note and

ai1 Cj1 + ai2 Cj2 + ai3 Cj3 = 0 a1i C1j + a2i C2j + a3i C3j = 0

iπj iπj

IIT JEE eBooks: www.crackjee.xyz 8.2 Comprehensive Mathematics—JEE Advanced

Remark

Illustration 2

The above results remain true for determinants of every order.

To evaluate

8.3 PROPERTIES OF DETERMINANTS 1.

Reflection Property

write

The determinant remains unaltered if its rows are changed into columns and the columns into rows. In other words, if A is a square matrix, then |A| = |A¢| where A¢ is transpose of A. 2.

= (1 + b) (1 + c) – bc = 1 + b + c

Switching Property

The interchange of any two rows (columns) of the determinant changes its sign. 5.

Scalar Multiple Property

If all the elements of a row (column) of a determinant are multiplied by a non-zero constant, then the determinant gets multiplied by the same constant. 6.

1 b c c D2 = 1 1 + b 1 b 1+ c

b1 b2 b3

c1 a1 + a b1 + b c1 c2 = a2 + a b2 + b c2 c3 a3 + a b3 + b c3

b1 b2 b3

Sum Property

c1 c2 c3

d1 a1 d2 = a2 d3 a3

c1 c2 c3

8.

d1 b1 d2 + b2 d3 b3

Factor Property

If a determinant D becomes zero when we put x = a, then (x – a) is a factor of D. Triangle Property

If all the elements of a determinant above or below the main diagonal consists of zeros, then the determinant is equal to the product of diagonal elements. That is,

c1 c2 c3

That is, a determinant remains unaltered under an operation of the form Ci Æ Ci + a Cj + b Ck, where j, k π i, or an operation of the form Ri Æ Ri + a Rj + b Rk, where j, k π i. a1 + b1 a2 + b2 a3 + b3

Using C2 Æ C2 – bC1 and C3 Æ C3 – cC1, we get 1 0 0 D2 = 1 1 0 = 1 1 0 1 \ D= 1+a+b+c

9.

Property of Invariance

a1 a2 a3

7.

and

Proportionality [Repetition] Property

If the elements of a row (column) are proportional or Two words the elements of the some other row (column), then the determinant is zero. 4.

1 b c D1 = 0 1 + b c 0 b 1+ c

All-zero Property

If all the elements of a row (column) are zero, then the determinant is zero. 3.

1+ a b c a 1+ b c D= a b 1+ c D = D1 + aD2 where

c1 c2 c3

d1 d2 d3

Remark It is one of the most under used property. But evaluation of some of the determinants became very easy when we use it.

a1 0 0 10.

a2 b2 0

a3 a1 b3 = a2 c3 a3

0 b2 b3

0 0 = a1 b2 c3 c3

Product of Two Determinants

a1 a2 a3

b1 b2 b3

c1 c2 c3

a1 a2 a3

a1a1 + b1b1 + c1g 1 = a2a1 + b2 b1 + c2g 1 a3a1 + b3 b1 + c3g 1

b1 b2 b3

g1 g2 g3

a1a 2 + b1b2 + c1g 2 a2a 2 + b2 b2 + c2g 2 a3a 2 + b3 b2 + c3g 2 a1a 3 + b1b3 + c1g 3 a2a 3 + b2 b3 + c2g 3 a3a 3 + b3 b3 + c3g 3

IIT JEE eBooks: www.crackjee.xyz Determinants 8.3

Here we have multiplied rows by rows. We can also multiply rows by columns, or columns by rows, or columns by columns. 11.

where a21, a22, a23, a31, a32 and a33 are constants, then a11 ¢ ( x ) a12 ¢ ( x ) a13 ¢ ( x) and D ¢(x) = a21 a22 a23 a31 a32 a33

Conjugate of a Determinant

If ai, bi, ci ΠC (i = 1, 2, 3), and Z= 12.

a1 a2 a3

b1 b2 b3

c1 c2 c3

a1 then Z = a2 a3

b1 b2 b3

c1 c2 c3

Ú D( x)dx

then

D(x) = D¢(x) =

a1 ( x ) a2 ( x ) a3 ( x ) a4 ( x )

If we write D(x) = [C1, C2], where Ci denotes ith column, then D¢(x) = [C 1¢ , C2] + [C1, C 2¢ ] where C i¢ denotes the column which contains the derivative of all the functions in the ith column Ci. Similarly, if ÈR ˘ D(x) = Í 1 ˙ Î R2 ˚

then

=

then

È R¢ ˘ È R ˘ D¢(x) = Í 1 ˙ + Í 1 ˙ Î R2 ˚ Î R2¢ ˚

a11 ( x ) a12 ( x ) a13 ( x ) D(x) = a21 ( x ) a22 ( x ) a23 ( x ) a31 ( x ) a32 ( x ) a33 ( x ) D¢(x) a11 ( x ) a12 a11 ¢ ( x ) a13 ( x ) ¢ ( x ) a12 ( x ) a13 ( x ) a21 ¢ ( x ) a23 ( x ) ¢ ( x ) a22 ( x ) a23 ( x ) + a21 ( x ) a22 a31 ( x ) a32 a31 ¢ ( x ) a33 ( x ) ¢ ( x ) a32 ( x ) a33 ( x )

a11 ( x ) a12 ( x ) a13 ¢ ( x) + a21 ( x ) a22 ( x ) a23 ¢ ( x) a31 ( x ) a32 ( x ) a33 ¢ ( x) = [C 1¢ , C2, C3] + [C1, C 2¢ , C3] + [C1, C2, C ¢3]

If

a11 ( x ) a12 ( x ) a13 ( x ) a22 a23 D(x) = a21 a31 a32 a33

a23 a33

Determinant of Cofactor Matrix

a11 If D = a21 a31

a12 a22 a32

a13 a23 a33

C11 C12 then D1 = C21 C22 C31 C32

C13 C23 = D2 C33

where Ci j denotes the co-factor of the element ai j in D. 8.4 SOME TIPS FOR QUICK EVALUATION OF DETERMINANTS

1. If D is a skew symmetric determinant of odd order, then D = 0 Illustration 3

Let

0 a b D = -a 0 c -b -c 0

0 - a -b D = a 0 -c b c 0 Taking –1 common from R1, R2 and R3 we get 0 a b D = (–1) - a 0 c = –D -b -c 0 3

Similarly, if È R1 ˘ È R1¢ ˘ È R1 ˘ È R1 ˘ Í ˙ D(x) = R2 then D¢(x) = Í R2 ˙ + Í R2¢ ˙ + Í R2 ˙ Í ˙ Í ˙ Í ˙ Í ˙ ÍÎ R3 ˙˚ ÍÎ R3 ˙˚ ÍÎ R3 ˙˚ ÍÎ R3¢ ˙˚ Corollary (Differentiation and Integration of Determinant)

a22 a32

m m m a11 ( x ) a12 ( x ) a13 ( x) Dm(x) = a21 a22 a23 a31 a32 a33

13.

a1¢ ( x ) a2 ( x ) a ( x ) a2¢ ( x ) + 1 a3¢ ( x ) a4 ( x ) a3 ( x ) a4¢ ( x )

a21 a31

=

In general, for any positive integer m

Differentiation of a Determinant

If each ai(x) is differentiable function and

Ú a11 ( x)d x Ú a12 ( x)d x Ú a13 ( x)d x



2D = 0



D=0

2. If a1, a2, a3 are in A.P.; b1, b2, b3 are in A.P. and c1, c2, c3 are also in A.P. Then a1 a2 a3 D = b1 b2 b3 = 0 c1 c2 c3 Use C1 Æ C1 + C3 – 2C2

IIT JEE eBooks: www.crackjee.xyz 8.4 Comprehensive Mathematics—JEE Advanced

3. If a1, a2, a3 are in G.P.; b1, b2, b3 are also in G.P., with the same common ratio, then a1 a2 a3 D = b1 b2 b3 = 0 c1 c2 c3

a1 a2 a3

a1 a2 a3

D=

If

1 a

1 b

a2

b2

1 1 1 1 c = a b c c 2 bc ca ab

1 a

1 b

= (a – b) (b – c) (c – a) 1 c = (a – b) (b – c) (c – a) (a + b + c)

a3

b3

c3

is given by x =

1

1

1

where

3. a

2

b

2

c

2

a

3

b

3

c

3

1.

2.

=0

b1 b2 b3

c1 c2 c3

π0

then the solution of the system of linear equations a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d2 a3 x + b3 y + c3 z = d3

= (a – b) (b – c) (c – a) (bc + ca + ab)

a b c 4. b c a = 3abc – a3 – b3 – c3 c a b

D D1 D ,y= 2 ,z= 3 D D D

d1 d2 d3

D1 =

b1 b2 b3

a1 D 3 = a2 a3

= (a + b + c) (bc + ca + ab – a2 – b2 – c2) 1 = - (a + b + c)[(b - c)2 + (c - a)2 + (a - b)2 ] 2

c1 a1 c2 , D 2 = a2 c3 a3 b1 b2 b3

d1 d2 d3

c1 c2 and c3

d1 d2 d3

SOLVED EXAMPLES

SINGLE CORRECT ANSWER TYPE QUESTIONS

8.5 LINEAR EQUATIONS

The system of linear homogeneous equations a1 x + b1 y + c1 z = 0

c1 c2 c3

8.6 CRAMER’S RULE

[c1, c2, c3 Some Frequently used Determinants

b1 b2 b3

Suppose a, b > 0 and let

Example 1

a2 x + b2 y + c2 z = 0 x a b D(x) = a x b , x ΠR a b x

a3 x + b3 y + c3 z = 0 has a non-trivial solution (i.e., at least one of the x, y, z is different from zero) if and only if D = 0, where D=

a1 a2 a3

b1 b2 b3

c1 c2 c3

If D π 0, then the only solution of the above system of equations is x = 0, y = 0 and z = 0. Corollary If at least one of x, y, z is non-zero and x, y and z are connected by the three given equations, then the elimination of x, y and z leads to the relation

then

D ¢ (0) is equal to D (0) (a)

1 a+b

(c) 0 Ans. (d)

Ê 1 1ˆ (b) 1 - Á + ˜ Ë a b¯ (d)

1 1 1 - a+b a b

IIT JEE eBooks: www.crackjee.xyz Determinants 8.5

Solution:

( x - a) + a a b a x b = (x – a) D1 + aD2 D(x) = a b x

such that tan A tan B tan B tan C tan C tan A + + =0 tan B tan A tan C tan B tan A tan C

where 1 a b D1 = 0 x b = x2 – b2 0 b x a 1 a b 1 D2 = 1 x b = 0 x - a 1 b x 0 b-a

and

b 0 x-b

[use R2 Æ R2 – R1, R3 Æ R3 – R1] = (x – a) (x – b) \

then (a) DABC is an equilateral triangle (b) DABC is an isosceles triangle (c) one of the angles of DABC is p/3 (d) one of the angle of DABC must be p/4 Ans. (b) Solution: Let x = tan A, y = tan B, z = tan C, then x y y z z x - + - + - =0 y x z y x z

D(x) = (x – a) (x2 – b2) + a(x – a) (x – b) = (x – a) (x – b) (x + b + a)



1 1 1 D ¢ ( x) + + = D ( x) x-a x-b x+a+b

\

D ¢ (0) 1 1 1 - = D (0) a+b a b

Example 2

then D is equal to (b) [ln(60)]3 (a) [ln(30)]3 (c) [ln(30)]2 (d) [ln(60)]2 Ans. (a) Solution: Let a = ln(2), b = ln(3), c = ln(5), then a-b-c 2a 2a 2b b-c-a 2b D= 2c 2c c-a-b Applying R1 Æ R1 + R2 + R3, we get 1 1 1 2b D = (a + b + c) 2b b - c - a 2c 2c c-a-b Using C2 Æ C2 – C1, C3 Æ C3 – C1, we get 1 0 0 0 D = (a + b + c) 2b -(a + b + c) 2c 0 -(a + b + c) 3

= (a + b + c) = [ln(30)]



x2 - y 2 y 2 - z 2 z 2 - x2 + + =0 xz xy yz



(x2 – y2)z + (y2 – z2)x + (z2 – x2)y = 0



ln(2/15) ln(4) ln(4) ln(3/10) ln(9) Let D = ln(9) ln(25) ln(25) ln(5/ 6)

3

Let DABC be an acute angled triangle

Example 3

fi fi fi

x

y

z

x2 1

y2 1

z2 = 0 1

(x – y) (y – z) (z – x) = 0 x = y or y = z or z = x A = B or B = C or C = A Thus, DABC is an isosceles triangle.

Example 4

Let a, b, c be distinct positive real numbers

such that S= Â

1 1 and T = , then a (a - b)(a - c) abc

(a) S = T (b) S + T = 0 (c) S = (a + b + c)T (d) S = (bc + ca + ab)T Ans. (a) 1 1 + Solution: S = a (a - b)(a - c) b(b - a )(b - c) 1 + c(c - a )(c - b) =

S1 abc(a - b)(b - c)(c - a)

where S1 = – bc(b – c) – ca(c – a) – ab(a – b) bc ca ab = - a b c 1 1 1

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1 1 1 = a b c bc ca ab

D2 =

y2

y + z2

z + x2

y3

y 2 + z3

z 2 + x3

y4

y3 + z 4

z3 + x4

= (a – b) (b – c) (c – a) \S=

(a - b)(b - c)(c - a) abc(a - b)(b - c)(c - a)

1 = =T abc

D2 =

Let x, y, z be three real distinct real Example 5 numbers, and let x+ y

y+z

2

2

z+x

2

2 3 S= x +y

y 2 + z3

z 2 + x3

x3 + y 4

y3 + z 4

z3 + x4

and T =

1 x

1 y

1 z

x2

y2

z2

\

D = xyz D1

(b)

D1 1 = 2 2 2 D 1+ x y z

(c)

D1 1 = D 1 + xyz

(d)

D1 = xyz + x2y2z2 D

Ans. (d) Solution: Write S = D1 + D2, where y + z2

z + x2

2 D1 = x

y 2 + z3

z 2 + x3

x3

y3 + z 4

z3 + x4

Using C3 Æ C3 – xC1 we get y + z2

z

2 D1 = x

y 2 + z3

z2

x3

y3 + z 4

z3

Using C2 Æ C2 – zC3 we get

and

x

y

z

D1 = x

2

y

2

z

2

x

3

y

3

z

3

z2

z + x2

y3

z3

z 2 + x3

y4

z4

z3 + x4 1 C , we get z 3

y2

z2

x2

y3

z3

x3

y4

z4

x4

= x 2y 2z 2T

(a)

x

y2

Using C3 Æ C3 –

D2 =

then

x

1 C , we get y 1

Using C2 Æ C2 –

= xyz T

S = xyz + x2y2z2 T Let A = (aij)3 ¥ 3 be such that det(A) = 5.

Example 6

Suppose bij = 2 aij (1 £ i, j £ 3) and let B = (bij)3 ¥ 3, then det(B) is equal to (b) 215 (a) 216 14 (c) 2 (d) 212 Ans. (c) i+j

22 a11

23 a12

24 a13

3 Solution: det (B) = 2 a21

24 a22

25 a23

24 a31

25 a32

26 a33

a11 2a21

a12 2a22

a13 2a23

22 a31

22 a32

22 a33

2

3

= 2 2 2

4

= 29 ◊ 2 ◊ 22 det (A) = 212 ◊ 22 = 214 Suppose a, b, c are distinct non-zero real

Example 7 numbers and a2

2 2 D1 = b + c bc

1 a a3 3 and D2 = 1 b b

1 c

c3

b2

c2

c2 + a2 ca

a 2 + b2 ab

IIT JEE eBooks: www.crackjee.xyz Determinants 8.7

then

D 2 - D1 is equal to (a - b)(b - c)(c - a)

2

+ b2 + c2

)

abc

a3 b3 c3 a b c abc abc abc 3

3

a = (a + b + c ) a 1 2

2

b b 1

2

1 -1- w2 w2

D2 = 0 b - c 1

c

c c 1

1

D= 1

w

w

1 w2

w 1

= 0

w

w2

0 w2

w

2

Using R2 Æ R2 – R1, we get a 2 + ab + b 2

0 c 2 - a 2 + b (c - a )

= (a – b) (b – c) (c – a) (a + b + c) Now, D2 – D1 = (1 + a2 + b2 + c2) D2 D 2 - D1 fi = (1 + a2 + b2 + c2) (a + b + c) (a - b)(b - c)(c - a)

Example 8 the determinant

Let w = -

w

w2

1+ w2 + w w2

w

= 3 (w2 – w4) = 3w (w – 1)

If n ∈ N and

then lim

(a)

1 b 2 + bc + c 2

1

1

2

n! (n + 1)! (n + 2)! ( n + 1 )! ( n + 2)! (n + 3)! Δn = (n + 2)! (n + 3)! (n + 4)!

c3

D2 = (a – b) (b – c)

1

= 1+ w + w

1

Example 9

b3 - c 3

= (a – b) (b – c)

3

2

3

nƕ

1 a + ab + b

is

4

[using C1 Æ C1 + C2 + C3]

3

2

w

1

3

Using R1 Æ R1 – R2, R2 Æ R2 – R3, we get 0 a-b a -b

w

2

1

= – (a2 + b2 + c2) D2

3

1

(a) 3w (b) 3w (w – 1) 2 (c) 3w (d) 3w (1 – w) Ans. (b) Solution: As w is a cube root of unity, 1 + w + w2 = 0 and w3 = 1. Thus, we can write D as

a 2 b2 c2 D1 = (a2 + b2 + c2) 1 1 1 bc ca ab

(a

D=

1

1

(a) a + b + c (b) a2 + b2 + c2 (c) (a + b + c) (1 + a2 + b2 + c2) (d) 4(abc)2 Ans. (c) Solution: Using R2 Æ R2 + R1, we get

=

1

1 i 3 + . Then the value of 2 2

(3n3 - 5) D n Dn + 1

3 2

(c) -

equals

(b) 5 2

5 2

(d) 3

Ans. (d) Solution: Taking n! common from R1, (n + 1)! from R2 and (n + 2)! from R3, we obtain Δn = n! (n + 1)! (n + 2)! Δ 1 n + 1 (n + 1) (n + 2) where Δ = 1 n + 2 (n + 2) (n + 3) 1 n + 3 (n + 3) (n + 4) Using R3 Æ R3 – R2 and R2 Æ R2 – R1, we get Δ=

=

1 n + 1 (n + 1) (n + 2) 0 1 2 (n + 2) 0 1 2 (n + 3) 1 2n + 4 =2 1 2n + 6

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Thus, Δn = 2n! (n + 1)! (n + 2)! fi

Δn

fi lim

3

(3n

)

- 5 Dn

Dn + 1 3

nƕ

y + z - y -z Δ = 2 z + x 0 -z x + y -y 0

= 2(n + 1)! (n + 2)! (n + 3)!

+1

(3n

Now,

Using C2 Æ C2 – C1, C3 Æ C3 – C1, we get

)

- 5 Dn

Dn + 1

Example 10

=

(3n

3

-5

)

(n + 1) (n + 2) (n + 3)

5ˆ Ê ÁË 3 - 3 ˜¯ n = lim nÆ• Ê 1ˆ Ê 2ˆ Ê ÁË1 + ˜¯ ÁË1 + ˜¯ ÁË1 + n n

3ˆ ˜ n¯

0 - y -z = 2 x 0 -z x -y 0

=3

If

1 x x +1 2x x ( x - 1) ( x + 1) x f(x) = 3 x ( x - 1) x ( x - 1) ( x - 2) ( x + 1) x ( x - 1) then f(2018) is equal to (a) 0 (b) 1 (c) 2018 (d) – 2018 Ans. (a) Solution: Taking x common from R2 and x(x – 1) common from R3, we get

2 1 1 = 2 xyz 2 0 1 2 1 0 [using C1 Æ C1 + C2 + C3] Taking 2 common and applying C2 Æ C2 – C1, C3 Æ C3 – C1, we get 1 0 0 Δ = 4 xyz 1 -1 0 = 4 xyz 1 0 -1

1 x x +1 2 f(x) = x (x – 1) 2 x - 1 x + 1 3 x - 2 x +1

Now, 4xyz = 12 fi xyz = 3

Applying C3 Æ C3 – C2, we get 1 x 1 2 f(x) = x (x – 1) 2 x - 1 2 3 x-2 3

The number 3 can be assigned to any of x, y, z. Therefore, the number of positive integral solutions of Δ = 12 is 3. =0

Example 12

a1 Δ = a4 a7

The number of positive integral solutions

of the equation Δ = 12, where Δ=

y+z z y

z z+x x

(b) 15

(c) 4

Solution: Using C1 Æ C1 + C2 + C3, we get y+z Δ = 2 z+x x+ y

z z+x x

y x x+ y

a2 a5 a8

a3 a6 a9

equals

y x x+ y

(a)

1 20

(b)

50 21

Let ai =

1 for 1 £ i £ 9. a + id

(c)

3 20

Ans. (b)

is (a) 3 Ans. (a)

If a1, a2, a3, ............., a9 are in H.P. and

a4 = 5, a5 = 4, then value of the determinant

Thus, f(2018) = 0. Example 11

[using C1 Æ C1 + C2 + C3]

0 1 1 = 2 xyz 1 0 1 1 1 0

(d) 2332

Solution:

We are given 5 =

1 1 . and 4 = a + 4d a + 5d



1 1 , a + 5d = 5 4

a + 4d =

(d)

7 90

IIT JEE eBooks: www.crackjee.xyz Determinants 8.9



a = 0, d =

= z [(z2 – (w4 + w2 – 2w3) – 3]

1 . 20

= z (z2) = z3

20 Thus, ak = for 1 £ k £ 9, and hence k 1 3

Δ = 20

1 4 1 7

1 2 1 5 1 8

1 3 1 3 = 20 6 1 9

[Using C2 Æ C2 –

0 3 40 3 56

0 1 12 4 63

1 1 C1 and C3 Æ C3 – C1] 2 3

1 50 = . = 20 ¥ 15 ¥ 224 21 w

z satisfying z +1

w

w2

w

z +w2

1

w2

1

z +w

Ans. (a) Solution: Using C1 Æ C1 + C2 + C3 and 1 + w + w2 = 0, we get w2

2 D = z 1 z +w 1 1

1 z +w

Applying R2 Æ R2 – R1 and R3 Æ R3 – R1, we get 1

w

2 D = z 0 z +w -w

0

1- w

c +a

ac

bc

then value of k is (a) 1

(b) 2

ac 2

bc

= ka2 b2 c2,

a 2 + b2 (c) – 2

(d) 4

Ans. (d) - a2

ab

ac

ab

-b

bc

ac

bc

2

- c2

-a a a = abc b - b b c c -c

[taking a, b, c common from C1, C2, C3 respectively] 2

2

=a b c

2

-1 1 1 0 2 1 2 2 2 1 -1 1 = a b c 0 0 1 2 0 -1 1 1 -1

= 4a2 b2 c2 Thus, k = 4 Example 15

w

ab

2

[using C1 Æ C1 + C2 and C2 Æ C2 + C3] =0

is

1

ab

D=

D=

Ê 2p ˆ Ê 2p ˆ cos Á ˜ + i sin Á ˜ . Then the number of distinct Ë 3¯ Ë 3¯

D=

b2 + c2

Solution: Using a2 + b2 + c2 = 0, we can write D as

3

Let

If a, b, c

Example 14

such that a2 + b2 + c2 = 0 and 1 1 4 1 7

4 1 3˘ 3 È 3 - ¥ = 20 Í ¥ Î 40 63 12 56 ˙˚

Example 13

Thus, z3 = 0 fi z = 0.

w2 1- w2 z +w -w2

= z [(z + w2 – w) (z + w – w2) – (1 – w) (1 – w2)] = z [(z2 – (w2 – w)2 – (1 – w – w2 + 1)

The determinant

a2 a 1 D = cos(nx) cos (n + 1) x cos (n + 2) x sin (nx) sin (n + 1) x sin (n + 2) x is independent of (a) n (c) x Ans. (a)

(b) a (d) none of these

Solution: Applying C1 Æ C1 + C3 – 2 cos x C2, we get a 2 - 2a cos x + 1 a 1 0 cos (n + 1) x cos (n + 2) x D= 0 sin (n + 1) x sin (n + 2) x [

cos nx + cos (n + 2)x = 2cos (n + 1)x cos x and sin nx + sin (n + 2)x = 2 sin (n + 1) x cos x] C1, we get

IIT JEE eBooks: www.crackjee.xyz 8.10 Comprehensive Mathematics—JEE Advanced

D = (a2 – 2a cos x + 1)[cos (n + 1) x sin (n + 2)x – sin (n + 1) x cos (n + 2)x] 2 = (a – 2a cos x + 1) sin [(n + 2)x – (n + 1)x] = (a2 – 2a cos x + 1) sin x which is independent of n.

n

n

Cr -1

n

D(n, r) =

Cr

Cr

n

Cr +1

n

Cr +1

n

Cr + 2

then D(n, r) is equal to (a) 0 (c) n+4Cr+3 Ans. (a) Solution: Using

( (r + 2) ( (r + 3) ( (r + 1)

n+2 n+2 n+2

Cr +1

) ) )

Cr + 2 Cr + 3

(b) n! – r! (d) n+4Cr+4

and nCr–1 + nCr = n+1Cr, we get

D(n, r) =

n +1 n +1

Cr

Cr +1

n

Cr + 2

n

=0 Example 17

(n + 2)

n

Cr

Cr +1

Cr + 2

[

( (n + 2) ( (n + 2)

(

n +1

n +1 n +1

m

Cr = 2m and  1 = m + 1. Therefore, r =0

m 2 - 1 2m

m

 Dr = m - 1 2 2

a

Cr

)

Cr +1

Cr + 2

) )

C1 and C3 are proptional]

Let m be a positive integer, a, b, c be

m -1 a 2

Δr =

m

Cr

1

m

m +1 c

2 b

(0 £ r £ m)

If a, b and g are real numbers, then

Example 18

1 cos (b - a ) cos (g - a ) 1 cos (g - b ) D = cos (a - b ) cos (a - g ) cos (b - g ) 1

Solution: We can write D as a product of two determinants as follows: cos a cos b cos g

sin a sin b sin g

(b) m2 – 1 d) 2m (a + b + c)

Solution: Using the sum property we get m

m

r =0

r =0

 (2r - 1)  mCr

r =0 m

m2 - 1 a

But  (2r – 1) = r =0

cos a cos b cos g

sin a sin b sin g

0 0 =0 0

If a + b + c > 0, and

Example 19

(a) D < 0

r =0

m

0 0 0

2m b

(b) D £ 0

(c) D > 0 (d) D = 0

Ans. (b) Solution: Applying C1 Æ C1 + C2 + C3, we get

m

 Dr =

m +1 = 0 c

a b c D = b c a , then c a b

Then the value of  Dr is given by (a) 0 (c) 2m abc Ans. (a)

b

m +1

[repetition property]

three real numbers and 2r - 1

m

is equal to (a) –1 (b) cos a cos b cos g (c) cos a + cos b + cos g (d) 0 Ans. (d)

(r + k) (n+2Cr+k) = (n + 2) (n+1Cr+k–1) n +1

r =0

m

r =0

Suppose n Œ N and 1 £ r £ n – 3 and

Example 16

m

Â

D=

a+b+c b c 1 b c a + b + c c a = (a + b + c) 1 c a a+b+c a b 1 a b

Applying R2 Æ R2 – R1 and R3 Æ R3 – R1, we get

r =0

Â1

1 b c D = (a + b + c) 0 c - b a - c 0 a-b b-c

m +1 c

C1, we get D = (a + b + c) [– (b – c)2 – (a – b) (a – c)]

m

1 (m + 1) (2m – 1 – 1) = m2 – 1, 2

= – (a + b + c) (b2 + c2 – 2bc + a2 – ab – ac + bc)

IIT JEE eBooks: www.crackjee.xyz Determinants 8.11

= –

Putting x = 0, we get

1 (a + b + c) [(b2 + c2 – 2bc) 2 2

2

2

2

+ (c + a – 2ca) + (a + b – 2ab)] 1 (a + b + c) [(b – c)2 + (c – a)2 2 + (a – b)2] £ 0 [ a + b + c > 0]

= –

If

Example 20

pl4 + ql3 + rl2 + sl + t =

l 2 + 3l

l -1

l +3

l2 +1

2 + 5l

l -3

l -3

l+4

3l

2

then p is equal to (a) 5

(b) 8

(c) 3

(d) 2

Example 22 If l21 + m21 + n21 = 1, etc., and l1 l2 + m1 m2 + n1 n2 = 0, etc. and l1 D = l2 l3 D

2

2

3

Ê 1ˆ Ê 1ˆ Ê 1ˆ Ê 1ˆ p + q Á ˜ +rÁ ˜ +sÁ ˜ +tÁ ˜ Ë l¯ Ë l¯ Ë l¯ Ë l¯ 1+ 3 l 1+1 l

2

1- 3 l

2

1-1 l

1+ 3 l

1+ 4 l

l1 D = D ◊ D = l2 l3 l12 + m12 + n12

= l1l2 + m1m2 + n1n2 l1l3 + m1m3 + n1n3

2 l + 5 1- 3 l 3

=

1 1 1 1 1 1 1 5 1 = 0 4 0 1 1 3 0 0 2

x -1

D

D=0

5x

If x - 1 x - 1 8 = ax3 + bx2 + cx + d 2x 3x 0

(b) 12

m1 m2 m3

n1 n2 n3

l1l2 + m1m2 + n1n2

l1l3 + m1m3 + n1n3

l22 + m22 + n22

l2l3 + m2 m3 + n2 n3

l2l3 + m2 m3 + n2 n3

l32 + m32 + n32

2 3+i -3 0 -1 + i Z = 3-i - 3 -1 - i 4 is equal to (a) 3 – 4i Ans. (d)

(b) 5 + 4i

(c) – 5i

(d) – 23

Solution: We have (c) 15

(d) 17

Solution: Differentiating both the sides of (1), we get x - 1 5x 7 x - 1 5x 7 2 1 0 + x2 - 1 x - 1 8 x - 1 x - 1 8 + 2x 2x 3x 0 2 x 3x 0 2 3 0

2 Z =

3-i -3

0

= 3ax2 + 2bx + c

n1 l1 n2 ◊ l2 n3 l3

1 0 0 0 1 0 =1fiD=±1fi D 0 0 1

7

2

(a) – 1 Ans. (d)

m1 m2 m3

Example 23

=8

then value of c is given by

5

D

2

[using R2 Æ R2 – R1, R3 Æ R3 – R1]

1

n1 n2 then n3

4

Taking limit as l Æ •, we get

Example 21

m1 m2 m3

Solution: We have

Solution: We divide LHS by l and C1 by l , C2 by l and C3 by l on the RHS to obtain, 4

p=

fi c = 0 + 0 + 17 = 17

Ans. (c)

Ans. (b)

=

-1 0 7 -1 0 7 1 5 0 c = -1 -1 8 + 0 1 0 + -1 -1 8 2 3 0 0 0 0 0 0 0

=

3+i

-3

-3 2 3-i -1 + i = 3 + i -1 - i 0 - 3 -1 + i 4 4 -1 - i 0

2 3+i 3-i 0 - 3 -1 - i

\ Z is purely real.

-3 -1+ i = Z 4

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Ans. (a)

If a1, b2, c3 π 0,

Example 24

a12 + b1 + c1 a1a2 + b2 + c2 b1b2 + c1

D1 =

b22 + c2

b2b3 + c3

c3c2

c32

c3c1

and

a1 D2 = a2 a3 a1 a1 a3 a1

(a) (b) (c) (d) Ans. (a)

b2 a2 b2 b1

b1 b2 b3

a1 a3 + b3 + c3

c1 D c2 , then 1 is equal to D2 c3

Solution: We can write the determinant D1 as product of two determinants as follows:

D1 =

b22 c2

b1b2 c1

D1 = b2 c3

a1 a2 = a1 b2 c3 b1 b2 c1 c2

Example 25

a1a3 b3 c3

D1 = 1 2 y

y 2 1 b b2 c2

(1 + bx) 2

(1 + cx)2

2 = (1 + ay )

(1 + by )2

(1 + cy )2 = D2

(1 + az )2

(1 + bz )2

(1 + cz)2

Let

f(x) =

cos x

sec2 x + cot x cosec x

cos 2 x cos 2 x

cosec 2 x

cos 2 x

cosec 2 x

1 p /2

f(x)dx is (b) p/48

(a) 0 (c) -

p p 2 15 2

1 p 3 2

(d)

Ans. (d) Solution: Applying R2 Æ R2 – R3, we get sec x 2 f(x) = - sin x

(b - x) 2

(c - x ) 2

2 D1 = ( a - y )

(b - y ) 2

(c - y ) 2 and

(a - z )2

(b - z ) 2

(c - z ) 2

(1 + ax)

2

D2 = (1 + ay )

2

(1 + bx)

2

(1 + cx)

(1 + by )

2

(1 + cy )2 then

(1 + bz )2

(b) D1 + D2 = 0 (d) D1 = – 2D2

sec 2 x + cot x cosec x

0

0 cosec 2 x

cos x 2

= – (– sin x)

cos x

sec2 x + cot x cosec x

cos 2 x

cosec 2 x

= sin2 x [cos x cosec2 x – cos2 x (sec2 x + cot x cosec x)] 2 = cos x – sin x – cos3 x

2

(1 + cz) 2

cos x 2

1

(a - x) 2

(a) D1 = D2 (c) D1 = 2D2

z2 1 c

(1 + ax) 2

then value of Ú0

a3 b3 = a1b2c3 D2 c3

c2

x2 1 a a2

sec x

For a, b, c, x, y, z ∈ R, if

(1 + az )2

- 2z 1 1 c

Example 26

D1 = a 1b 2c 3 D2



z2

a1a3 + b3 b2b3 c3

a1a2 b2 c2

- 2 y 1 1 b b2

1 2z

Taking b2 common from R2 and applying R1 Æ R1 – R2 we get a12 b1 c1

y2

1 2x

Solution: Taking c3 common from R3 and applying R2 Æ R2 – R3 and R1 Æ R1 – R3. We obtain

D1 = c3

- 2x 1 1 a a2

We interchange C1 and C3 absorb the resulting negative sign in C2 to get

c3 a3 c1 c1 + a2 b2 c2 + a3 b3 c3

a12 + b1 a1a2 + b2

x2

Thus, p /2

p

2

Ú0 f ( x ) dx = 1 - 4 - 3 =

1 p 3 4

IIT JEE eBooks: www.crackjee.xyz Determinants 8.13

n, let

Example 27

D=

(n - 1)! (n + 1)! (n + 3)! / n ( n + 1) (n + 1) ! (n + 3)! (n + 5)! / ( n + 2) ( n + 3) (n + 3))! (n + 5)! (n + 7)! / ( n + 4) ( n + 5)

Applying R2 Æ R2 – we get

D=

D then is equal to (n - 1)!(n + 1)!(n + 3)! (a) – 8 Ans. (d)

(b) – 16

(c) – 32

(d) – 64

where D1 =

6 0

3

0

2

2

=

=

If D =

Example 28

8 8

(a) 0 Ans. (d)

D=

6

-3 2

(b) –1

(c) 1

(d) 2

a b- y c- z y 0 =0 D = -x z -x 0

(c – z)

3+ 6

2i

-x y a b- y +z =0 -x 0 -x y

12

3 + 8i

3 2 + 6i



(c – z) (xy) + z(ay + bx – xy) = 0

18

2 + 12i

27 + 2i



cxy – xyz + ayz + bxz – xyz = 0



ayz + bzx + cxy = 2xyz



a b c + + =2 x y z

2i

2 iC1]

Solution: Applying R2 Æ R2 – R1, R3 Æ R3 – R1, we get

cos x Example 30

6 common from C1, we get 1

2

C3, we get

a natural number a negative integer an irrational number an imaginary number

Solution: Taking

-2 3

a b- y c- z b c - z = 0, D = a-x a- x b- y c

= – 64

then D is (a) (b) (c) (d) Ans. (b)

3

a b c + + is x y z

4n + 6 = 4n + 14

6

6

If x, y, z are different from zero and

D = – 64 (n - 1)!(n + 1)!(n + 3)!

Thus,

2i - 3 2

=

)

Example 29

[C2 Æ C2 – C1] 4n + 6 8 8 0

6i - 2 3

(

1 (n + 1) n (n + 3) (n + 2) 0 4n + 6 4n + 14 0 4n + 14 4n + 22 4n + 6 4n + 14 4n + 14 4n + 22

2i - 3 2

6 -3 6 + 2 6 = – 6, which is an integer.

=

Applying R3 Æ R3 – R2 and R2 Æ R2 – R1, we get D1 =

6i - 2 3

[using C2 Æ C2 –

Solution: Taking (n – 1)! common from R 1, (n + 1)! from R2 and (n + 3)! from R3, we get D = (n – 1)! (n + 1)! (n + 3)! D1 (n + 1) n (n + 3) (n + 2) 1 1 (n + 3) (n + 2) (n + 5) (n + 4) 1 (n + 5) (n + 4) (n + 7) (n + 6)

2i

3

3 R 1,

3+ 6

1

6

=

2 R1, and R3 Æ R3 –

3+ 6

2

3 + 2 2i 3 2 + 6i

3

2 + 2 3i

3 3 + 2i

then lim

xÆ0

Let f(x) = 2 sin x tan x

x 2

x x

1 2x , 2

f ( x) is equal to x2

(a) – 1

(b) 0

(c) 2

(d) 3

IIT JEE eBooks: www.crackjee.xyz 8.14 Comprehensive Mathematics—JEE Advanced

Ans. (a)



Solution: For x π 0, we have cos x 1 1 2 sin x f ( x) 1 2 = x 2 x tan x 1 2 x

1 1 x D = p +1 p +1 p + x = 0 3 x +1 x + 2

1 0 0 = 2 -1 0 = – 1 1 0 1

[using C2 Æ C2 – C1 and C3 Æ C3 – C1] Example 31 equations

If p π a, q π b, r π c and the system of px + ay + az = 0 bx + qy + bz = 0

(c) 1

(d) 2

Ans. (d) Solution: As the given system of equations has a non-trivial solution D=

p a a b q b =0 c c r

Applying C2 Æ C2 – C1 and C3 Æ C3 – C2, we get D=

p a- p a- p b q-b 0 =0 c 0 r-c C3, we get

b q-b p a- p + (r – c) =0 (a – p) c 0 b q-b fi (a – p) (– c) (q – b) + (r – c) {p(q – b) – b(a – p)} = 0 fi (p – a) (q – b) c + p (r – c) (q – b) + b (r – c) (p – a) = 0 Dividing by (p – a) (q – b) (r – c) we get c p b + + =0 r -c p-a q-b

(a) {1, 2} (c) {1, p, 2}

(b) {2, 3} (d) {1, 2, – p}

Ans. (a) Solution: Applying C2 Æ C2 – C1, we get x p + x = – (x – 2) x+2

1 0 p +1 0 3 x-2

1 p

x p

1 x p +1 p + x

= (x – 2) p (1 – x) = 0

fi Thus D = 0 x = 1, 2.

p q r + + is p-a q-b r -c (b) 0

is

= (x – 2)

cx + cy + rz = 0 has a non-trivial solution, then the value of

(a) – 1

If p π 0, solution set of the equation

Example 32

[divide each of R2 and C2 by x] 1 1 1 f ( x) = 2 (1) 1 2 \ lim x Æ 0 x2 1 1 2

p q r q-b r -c + + + = =2 p-a q-b r -c q-b r -c

Example 33 Let a, b, c ∈ R be such that a + b + c π 0. If the system of equations ax + by + cz = 0 bx + cy + az = 0 cx + ay + bz = 0 has a non-trivial solution, then (a) a + c – b = 0 (b) b + c – a = 0 (c) a + b – c = 0 (d) a = b = c Ans. (d) Solution: As the system of equations has a non-trivial solution, Δ=

a b c b c a =0 c a b

Using C1 Æ C1 + C2 + C3, we get 1 b c Δ = (a + b + c) 1 c a 1 a b b c 1 = (a + b + c) 0 c - b a - c 0 a-c b-a

IIT JEE eBooks: www.crackjee.xyz Determinants 8.15

[Use C3 Æ C3 – C2 C2 Æ C2 – C1] = (a + b + c) [– (b – c) (b – a) – (a – c)2] 2

2

z z1 bz2 + cz3 b+c

2

= – (a + b + c) [a + b + c – bc – ca – ab] = -

1 (a + b + c ) ÈÎ(b - c )2 + (c - a )2 + (a - b)2 ˘˚ 2

(b – c)2 + (c – a)2 + (a – b)2 = 0

As a, b, c ∈ R, we get (b – c)2 = (c – a)2 = (a – b)2 = 0 fi b – c = 0, c – a = 0, a – b = 0 fi a =b=c Example 34

If the system of linear equations x + y + z

= 6, x + 2y + 3z = 14 and 2x + 5y + l z = m, (l, m ∈ R) has a unique solution, then (a) l π 8 (c) l = 8, m = 36 Ans. (a)

Example 36

1 a a3

2 S= 1 b b

3 S1 = 1 b b

1 c

z 1 z z1 1 + c z1 z2 1 z3

z 1 z1 1 = 0 z3 1

c2

1 a2 1 c2

c3

1 c

a3

2 S2 = 1 b

a a2

a3

b3 and S3 = b b 2

b3

c3

c3

c2

c

Then

of a triangle. Let a = BC, b = CA and c = AB, then

represents (a) (b) (c) (d)

1 a a2

(a)

S1 =a+b+c S

(b)

(c)

S3 = abc S

(d) S1S2S3 = S3

Let A(z1), B(z2) and C(Z3) be the vertices

z b z1 z2

Suppose a, b, c are distinct real numbers

and let

i.e., if l – 8 π 0 or l π 8. Example 35

1

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

(b) l = 8, m π 36 (d) none of these

Solution: The given system of linear equations has a unique solution if 1 1 1 1 2 3 π0 2 5 l

1 1 =0

Ê bz + cz3 ˆ It represents line joining A(z1) and D Á 2 , Ë b + c ˜¯ which is bisector of –A.

Now, Δ = 0, a + b + c π 0 fi

z z1 bz2 + cz3 b+c

S2 = ab + bc + ca S

Ans. (a), (b), (c) Solution: Consider the system of equations A + Ba + Ca2 = – a3 A + Bb + Cb2 = – b3

median of DABC through A angle bisector of –A altitude of DABC through A perpendicular bisector of side BC

A + Bc + Cc2 = – c3 1 a a2 Let

2 D= 1 b b =S

Ans. (b) Solution: Write the given equation as z z1 bz2 + cz3

z z1 bz2 + cz3

1 1 =0 b+c

1 c

c2

- a3

a a2

3 D1 = - b

- c3

or as D2 =

a a2

b b2 = - b b2 c2

a3 b 3 = – S3

c

c2

c

1 - a3

a2

1 a2

1 -b3

b2 = 1 b2

b 3 = S2

1 - c3

c2

c3

1 c2

c3 a3

IIT JEE eBooks: www.crackjee.xyz 8.16 Comprehensive Mathematics—JEE Advanced

and 1 a - a3 D3 = 1 b 1 c

- b 3 = - 1 b b 3 = – S1 - c3

1 c

c3 \

By the Cramer’s rule A=

D D1 D ,B= 2 ,C= 3 D D D

But a, b, c are roots of x3 + Cx2 + Bx + A = 0 \

C = – (a + b + c), B = bc + ca + ab

and

A = – abc

Thus, – abc = -

– (a + b + c) = -

S1 S fi 1 =a+b+c S S

Example 37 Suppose a, b, c, x, y, z > 0 and satisfy the system of linear equations cy + bz = a cx + az = b

0 a b D2 = c b a = b(a2 + c2 – b2) b c 0

z=

a 2 + b2 - b2 2ab b 2 + c 2 - a 2 + 2bc 2bc

(b + c) 2 - a 2 = 2bc =

(a + b + c)(b + c - a) 2bc

As x + 1 > 0, we get b + c > a Similarly, a + c > b, a + b > c Thus a, b, c are sides of a triangle. \ x = cos A, y = cos B, z = cos C As x, y, z > 0, triangle must be an acute angled triangle Also, each of x, y, z < 1. Example 38

For q Π[0, 2p], let

cos q D(q) = cos 2q cos 3q

then

a c b D1 = b 0 a = a(c2 + b2 – a2) c a 0

b2 + c2 - a 2 a 2 + c2 - b2 ,y= 2bc 2ac

x+1=

bx + ay = c (a) a, b, c are sides of an acute angled triangle (b) x, y, z are cosines of the angles of an acute angled triangle (c) Each of x, y, z < 1 (d) x, y, z are sines of the angles of an acute angled triangle. Ans. (a), (b), (c) 0 c b Solution: Let D = c 0 a = 2abc b a 0

x=

We have

S S3 fi 3 = abc S S

S bc + ca + ab = 2 S and

0 c a D3 = c 0 b = c(a2 + b2 – c2) b a c

1 a a3

cos 2q cos 3q cos 4q

cos 3q cos 4q cos 5q

then Ê 5p ˆ (a) D Á ˜ = 0 Ë 12 ¯

Ê 7p ˆ (b) D Á ˜ = –1 Ë 12 ¯

Ê 2p ˆ (c) D Á ˜ = 0 Ë 3¯

Ê 3p ˆ (d) D Á ˜ = Ë 4¯

3

Ans. (a), (c) Solution: Put z = cos q + isin q, then cos(nq) =

1 n (z + z n ) 2

Thus, 1 (z + z ) 2 1 2 (z + z 2 ) D(q) = 2 1 3 (z + z 3 ) 3

1 2 (z + z 2 ) 2 1 3 (z + z 3 ) 2 1 4 (z + z 4 ) 2

1 3 (z + z 3 ) 2 1 4 (z + z 4 ) 2 1 5 (z + z 5 ) 2

IIT JEE eBooks: www.crackjee.xyz Determinants 8.17

=

1 (D + D2) 8 1

If

Example 40

where z2 + z 2

z3 + z 3

2 D1 = z

z3 + z 3

z4 + z 4

z3

z4 + z 4

z5 + z 5

z

and D2 = D1 Using C2 Æ C2 – zC1, C3 Æ C3 – z2C1, so that

D(x) =

2x - 5

3

3x + x + 4

6x + 1

9

2

7 x - 6 x + 9 14 x - 6 21 2

= ax3 + bx2 + cx + d, then (a) a = 0 (c) c = 0

(b) b = 0 (d) d = 141

Ans. (a), (b), (c), (d)

z

z2

z3

Solution: Differentiate both the sides with respect to x

2 D1 = z

z3

z4

z3

z4

z5

x2 - 5x + 3 2 3 2x - 5 2x - 5 3 2 D¢(x) = 6 x + 1 6 x + 1 9 + 3x + x + 4 6 9 14 x - 6 14 x - 6 21 7 x 2 - 6 x + 9 14 21

Using C3 Æ C3 – z C2, we get z

z2

0

2 D1 = z

z3

0 =0

z3

z4

0

= 0+0=0 fi D(x) is a constant. Thus, a = 0, b = 0 and c = 0. For value of d, put x = 0 Example 41

Also, D2 = D1 = 0 \

x2 - 5x + 3

D(q) = 0 " q Π[0, 2p ]

Example 39

D=

The determinant

a b aa + b b c ba + c D= aa + b ba + c 0 is equal to zero if (a) a, b, c are in A.P. (b) a, b, c are in G.P. (c) a, b, c are in H.P. (d) a is a root of ax2 + 2bx + c = 0. Ans. (b), (d) Solution: Applying R3 Æ R3 – a R1 – R2, we get a b D = b c

aa + b ba + c

0 0 - (aa 2 + 2ba + c) = (b2 – ac) (aa2 + 2ba + c) R3. Now D = 0 if either b2 – ac = 0, or 2 aa + 2ba + c = 0, that is, if either a, b and c are in G.P., or a is a root of ax2 + 2bx + c = 0.

If

1/ z

1/ z

- ( x + y )/ z 2

- ( y + z )/ x 2

1/ x

1/ x

- y ( y + z )/ x 2 z ( x + 2 y + z )/ xz - ( x + y ) / xz 2 then (a) D is independent of x (b) D is independent of y (c) D depends only on z (d) D = 0 Ans. (a), (b), (d) Solution: Multiplying C1 by x, C2 by y and C3 by z, we obtain x z 1 y+z D= xyz x y( y + z) xz

y z y x y( x + 2 y + z) xz

x+ y z z x y( x + y) xz -

Applying C1 Æ C1 + C2 + C3 we get 0 1 0 D= xyz 0

- ( x + y )/ z y/ z y/ x z/x =0 y ( x + 2 y + z )/ zx - y ( x + y )/ xz

This shows that D is independent of x, y and z.

IIT JEE eBooks: www.crackjee.xyz 8.18 Comprehensive Mathematics—JEE Advanced

sin q cos f sin q sin f D = cos q cos f cos q sin f - sin q sin f sin q cos f

where D1 =

cos q - sin q 0

(a) D is independent of q (b) D is independent of f (c) D is a constant

=

Ans. (b), (d) Solution: Applying C1 Æ C1 – (cot f) C2, we get 0 sin q sin f 0 cos q sin f - sin q sin f sin q cos f

= -

Hence, D =

cos q - sin q 0

sin q [– sin f sin2 q – cos2 q sin f] sin f

= sin q which is independent of f.

D=

(a) (b) (c) (d)

1

a2 + x

ab

ac

ab

b +x

bc

bc

c +x

1 cos q sin q

sin 2 q cos f sin f

sin 2 q sin f - cos f

cos 2 q

sin q cos 2 f

sin q sin 2 f

(b) x2 (d) a2 + b2 + c2 + x

a3 + ax

ab

2

b +x

bc

bc

c +x

a b 2

then

ac

2

2

Applying C1 Æ C1 + bC2 + cC3 and taking a2 + b2 + c2 + x common, we get a ab ac 1 2 2 bc D= (a + b2 + c2 + x) b b + x a 2 c bc c +x

0

Applying C2 Æ C2 – bC1 and C3 Æ C3 – cC1, we get a 0 0 1 2 2 2 D= (a + b + c + x) b x 0 a c 0 x

Solution: Taking 1/sin q cos f, 1/sin q sin f, 1/cos q common from C1, C2, C3 respectively, we get sin 2 q cos q sin f cos f

is divisible by

2

(a) x (c) x3

a c

Ans. (a), (b)

1

2

ac

1 D= a 1 sin q sin f - cos q

2

The determinant

D=

Let 1 sin q cos f - cos q

.

sin q cos q sin 2 f cos 2 f 3

Solution: We can write D as

D is dependent on q D is dependent on f D is a constant none of these

D=

1 tan q 0

Ans. (a), (b), (d)

dD dD˘ = cos q fi = cos (p / 2) = 0 ˙ dq dq ˚q = p / 2

Example 43

1 - cot q - cot f

1 sin f cos f sin q cos q

Example 44

C 1]

Also,

1 tan q 0

using C1 Æ C1 – C2, we get 1 [tan q + cot q] Thus, D1 = sin f cos f

dD˘ =0 ˙ dq ˚q = p / 2

D=

1 - cot q - cot f

0 0 D1 = 1/sin f cos f

then

(d)

1 - cot q tan f

Let

Example 42

=

1 2 (a + b2 + c2 + x) (ax2) a

= x2 (a2 + b2 + c2 + x)

D1 Thus

x

D and x2

D.

IIT JEE eBooks: www.crackjee.xyz Determinants 8.19

Let

Example 45

pn x

sin p x

Ê np ˆ (-1) n n ! - sin Á ˜ Ë 2¯

f (x) =

dn

then value of

d xn

Ê np ˆ - cos Á ˜ Ë 2¯

2

(b) –1 (d) independent of x

2

Ans. (a), (d) Solution: We know that

(-1) n! p n ; d n Èp n ˘ Í ˙ = xn + 1 dx n Î x ˚ n

(a) (b) (c) (d)

dx

n

xn +1

f n(x) =

(-1)

n

q

C2

q

C3

C1

r

C2

r

C3

D is an even integer D is divisible by 12 D is an rational number. none of these

1 q+ p-3 1 r+q-3

= pqr (q – p) (r – q) (r – p) np ˆ Ê p n sin Á p x + ˜ Ë 2¯

np ˆ Ê p n cos Á p x + ˜ Ë 2¯

Ê np ˆ - sin Á ˜ Ë 2¯

Ê np ˆ - cos Á ˜ Ë 2¯

1

3 2

n!

2 np ˆ Ê p n sin Á p + ˜ Ë 2¯

np ˆ Ê p n cos Á p + ˜ Ë 2¯

(-1)n n!

Ê np ˆ - sin Á ˜ Ë 2¯

Ê np ˆ - cos Á ˜ Ë 2¯

-1

1

3 2

2

[

R1 and R3 are proportional]

This also show that f n(1) is independent of x. Example 46

C1

= pqr (q – p) (r – q)

(-1)n n!p n

=0

q

1 p -1 ( p - 1) ( p - 2) D = pqr 0 q - p (q - p ) (q + p - 3) 0 r - q (r - q ) (r + q - 3)

np ˆ Ê n = p cos Á p x + ˜ Ë 2¯

-1

fi f n(1) =

C3

Using R3 Æ R3 – R2 and R2 Æ R2 – R1, we get

Thus,

(-1) n!p n

p

Solution: We have 1 1 p C1 = p, pC2 = p(p – 1) and pC3 = p(p – 1) (p – 2). 2 6 Thus, 1 p - 1 ( p - 1) ( p - 2) D = pqr 1 q - 1 (q - 1) (q - 2) 1 r - 1 (r - 1) (r - 2)

np ˆ Ê n sin p x ] = p sin Á p x + ˜ , and n [ Ë 2¯ dx cos p x ] n [

C2

then

dn

dn

p

Ans. (a), (c)

[ f (x)] at x = 1 is

1

C1

r

3 2

(a) 0 (c)

cos p x

1

-1

D = 12

p

If p, q and r are positive integers, and

Example 47

If sec2 x

1

2

1

2

2 f (x) = cos x cos x cosec x then

cos 2 x

1 p /4

(a)

cot 2 x

1

Ú f ( x) dx = 32 (3p + 8) - p /4

(b) f ¢ (p/2) = 0 f(x) is 1 (d) Minimum value of f(x) is 0. Ans. (a), (b), (c), (d) Solution: Applying C2 Æ C2 – cos2 x C1, we get sec2 x

0

1

2 2 4 2 f (x) = cos x cos x - cos x cosec x

1

0

cot 2 x

IIT JEE eBooks: www.crackjee.xyz 8.20 Comprehensive Mathematics—JEE Advanced

x + (cos a) y + (sin a) z = 0 – x + (sin a) y – (cos a) z = 0

= (cos2 x – cos4 x) (sec2 x cot2 x – 1) C 2] 2 2 2 = cos x sin x (cosec x – 1) = cos2 x sin2 x cot2 x = cos4 x =

has a non-trivial solution then S contains

1 (3 + 4cos 2x + cos 4x) 8

Ú

(d) (– 2, 2)

Ans. (a), (b), (c) f (x) dx

Solution: The given system of linear equations will have a non-trivial solution if

-p 4

=

2]

(c) [1,

p /4

Thus

(b) [ - 2 , – 1]

(a) (– 1, 1)

1Ê 1 ˆ ÁË 3x + 2 sin 2 x + sin 4 x˜¯ 8 4

p /4

= 0

1 (3 p + 8) 32

l sin a 1 cos a - 1 sin a

It can be easily checked that each of (b), (c) and (d) also holds. Let

Example 48

C1 we get l (– cos a – sin a) – (– sin a cos a – sin a cos a) – (sin2 a – cos2 a) = 0 2

3 D(x) = 3 x 3 x 2 + 2a 2

3x

3x 2 + 2a 2

3 x 2 + 2a 2

3 x3 + 6 a 2 x

3 x3 + 6 a 2 x 3 x 4 + 12 a 2 x 2 + 2 a 4

cos a sin a = 0 - cos a

fi fi

2

– l + sin 2a + cos 2a = 0 l = sin 2a + cos 2a = 2 sin (p/4 + 2a)

fi–

2 £l£

2.

then Example 50

(a) D¢(x) = 0 (b) D(x) is independent of x

in x, y and z: (sin 3q) x – y + z = 0 (cos 2q) x + 4y + 3z = 0 2x + 7y + 7z = 0

1 0

(c) Ú D(x) dx = 4a6 (d) y = D(x) is a straight line Ans. (a), (b), (c), (d) Solution: Applying C3 Æ C3 – xC2, C2 Æ C2 – xC1 we obtain D(x) =

3

0

2a 2

3x

2a 2

4a 2 x

3 x 2 + 2a 2

4a 2 x 6a 2 x 2 + 2a 4

Applying C3 Æ C3 – xC2, we get D(x) = 4a

0 1

1 x

3 x 2 + 2a 2

2x

x2 + a2

Apply C1 Æ C1 – 3C3 to get D(x) = 4a

4

The values of q for which the system of equations has a non-trivial solution are (a) {np : n∈I} (b) {mp + (–1)m p/6 : m ∈ I} (c) {np + (–1)n p/3 : n ∈ I} (d) none of these Ans. (a), (b)

3 3x

4

Consider the system of linear equations

0 0

0 1

1 x

- a2

2x

x2 + a2

= 4a6

Example 49 Let l and a be real. Let S denote the set of all values of l for which the system of linear equations lx + (sin a) y + (cos a) z = 0

Solution: The given system of equations will have a non-trivial solution if D=

sin 3q cos 2 q 2

-1 1 4 3 =0 7 7

Applying R2 Æ R2 + 4R1 and R3 Æ R3 + 7R1, we get D=

sin 3q cos 2 q + 4sin 3q 2 + 7 sin 3q C2, we get

-1 1 0 7 =0 0 14

D = 7[2 (cos 2q + 4 sin 3q) – (2 + 7 sin 3q)] = 0

IIT JEE eBooks: www.crackjee.xyz Determinants 8.21

fi fi

2 cos 2 q + sin 3q – 2 = 0 sin q (2 sin q – 1) (2 sin q + 3) = 0

As 2 sin q + 3 can never be zero, we get sin q = 0 or sin q = 1/2 = sin (p/6) fiq = np or q = m p + (–1)m p/6 (n, m Œ I)

D1 3 1 Êpˆ = = = cos Á ˜ Ë 3¯ D2 2 3 2

(c)



(d) Â

D 2n

3n n =1 2

MATRIX-MATCH TYPE QUESTIONS For q ΠR, n ΠN, let

Example 51 2 Dn(q) = 2 2

=

n +1 n +1

cos q

cos(q + p / 3) sin(q + p / 3)

Match the entries in Column 1 with appropriate entries in Column 2 Column 1 Column 2 (p) is a rational number (a) Dn •

2n (b) Â ( D n /2 )

(q) is an irrational number

(c) D1/D2

(r) sin(p/3)

n =1

3a 2 + 2ab 3a 2 + 6ab + b 2

a2

Match the entries in Column 1 with entries in Column 2 Column 1 Column 2 (a) D(a, b) (p) 1 (b) D(a, b) + D(b, c) + D(c, a) (q) – 1 (c) lim D(sin x, cos x)

(r) (b – a)3 + (c – b)3 + (a – c)3

(d) lim D(sin x, cos x)

(s) (b – a)3 (t) 3(b – a) (c – b) (a – c)

(t) cos (p/3) r

s

t

a

p

q

r

s

t

b

p

q

r

s

t

c

p

q

r

s

t

d

p

q

r

s

t

n +1

0

0

cos q

sin q

cos(q + p / 3) sin(q + p / 3)

n

n–1

3 = 2 sin (p/3) = 2 \ Dn is an irrational number for each n Œ N •

2n (b) Â ( D n /2 ) = Â

n =1 2

q

r

s

t

a

p

q

r

s

t

b

p

q

r

s

t

c

p

q

r

s

t

d

p

q

r

s

t

Solution: (a) Using R2 Æ R2 – aR1 and R3 Æ R3 – a2R1, we get

Solution: (a) Using R1 Æ R1 + R3 – R2 we get n +1 Dn = 2

p

Ans.

q

2n

10 6a + 4b

(s) 3/4

p

n =1

5 4a + b

xÆp / 2

n =1



1 a

D(a, b) =



2

23n

xÆ0

2 3n (d) Â ( D n / 2 )

Ans.

n =1

For a, b ΠR, let

Example 52 sin q

3(22 n - 2 )

3 • 1 3 Â n = 4 n=1 2 4

cos(q - p / 3) sin(q - p / 3)

n



= Â

3 n +1

=

3/4 3 p = = sin ÊÁ ˆ˜ Ë 3¯ 1 - 1/ 2 2

1 D(a, b) = 0

5 b-a

0 2ab - 2a 2 = (b – a)2

10 4(b - a) b 2 + 6ab - 7a 2

1 4 2a b + 7 a

= (b – a)3 (b) D(a, b) + D(b, c) + D(c, a) = (b – a)3 + (c – b)3 + (a – c)3 = 3(b – a) (c – b) (a – c) [ x+y+z=0

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Using R2 Æ R2 – R1, R3 Æ R3 – R1, we get

fi x3 + y3 + z3 = 3xyz] (c) lim D(sin x, cos x)

1 x D =4 0 y 0 0

xÆ0

= lim (cos x – sin x)3 = 1 xÆ0

(d)

lim

xÆp / 2

D(sin x, cos x) = – 1

Example 53 Let x and y be real numbers. Match the determinants in Column 1 with their values in Column 2. Column 1 x y y x+ y (a) x+ y x

(c) Denote the determinant by D. Using C1 Æ C1 + C2 + C3, we get D = (x + y + 2)D1, where 1 x y y D1 = 1 1 + x x 1 1+ y

Column 2 x+ y x y

y 0 = 4xy x

(p) 4xy

1 x y = 0 1 0 =1 0 0 1

[using R2 Æ R2 – R1, R3 Æ R3 – R1] D = (x + y + 2)

\

1 2x 2y 2y (b) 1 2 x + 2 y 1 2x 2x + 2y

(q) x + y + 2

(d) Denote the determinant by D. Using R1 Æ R1 + R2 + R3, we get D = 2D1

2 x y y (c) 1 1 + x x 1 1+ y

(r) – 2(x3 + y3)

1+ x 1 where D1 = x

1+ x 1 (d) x

(s) – 4(x + y)

y y x+ y 1 x 1+ y

x + y 1+ y x+ y 1 x 1+ y

Applying R2 Æ R2 – R1, R3 Æ R3 – R1, we get 1+ x D1 = - x -1

Ans.

x + y 1+ y 0 x 1 0 - y = -x 0 - y -y 0 -1 - y 0

= 2xy [using R1 Æ R1 + R2 + R3] \ Solution: (a) Denote the determinant by D. Using C1 Æ C1 + C2 + C3, we can write D = 2(x + y)D1 1 y + 1 x y where D1 = 1 x

x+ y x y

Applying R2 Æ R2 – R1, R3 Æ R3 – R1, we get 1 y x+ y -y x D1 = 0 0 x- y -x = – x2 + y (x – y) = – (x2 – xy – y2) \

D = – 2(x3 + y3)

(b) Denote the determinant by D.

D = 4xy

Example 54

p(x) =

Let

3

3x

3 x2 + 2

3x

3 x2 + 2

3 x3 + 6 x

3 x 2 + 2 3 x3 + 6 x 3 x 4 + 12 x 2 + 2

q(x) = 2

1 2+ x

1 3+ x

1 4+ x

(2 + x )2 (3 + x )2 (4 + x )2 cos ( x + p 4)

r(x) =

sin ( x + p 4)

cos ( x + p 2) sin ( x + p 2) cos ( x + 3 p 4) sin ( x + 3 p 4)

1 2 and 1

IIT JEE eBooks: www.crackjee.xyz Determinants 8.23

s(x) =

1

2x + 1 2 x ( 2 x + 1)

2x 2 x ( 2 x - 1)

1 4x

(

0

=

)

6 x ( 2 x - 1) 4 x ( 2 x - 1) ( x - 1) 2 x 4 x 2 - 1

Column 1 with their properties/ values in Column 2. Column 1 (a) p(x) (b) q(x) (c) r(x) (d) s(x)

(p) (q) (r) (s)

Column 2 is independent of x equals 4 equals 0 equals x2 – x + 1

2 1

2

1

1

0

2 =0

2

2

[using R3 Æ R3 + R1] (d) Take 2x common from R2, 2x(2x – 1) from R3 and 2x + 1 from C3 to obtain s(x) = 4x2(2x – 1) (2x + 1) s1(x), where 1 2x 1 1 1 1 s1(x) = 2 2 x - 1 1 = (2x + 1) 2 1 1 = 0 3 2x - 2 1 3 1 1

Ans. \

[using C2 Æ C2 + C1] s(x) = 0 Let

Example 55

p(q) = Solution: (a) Using C3 Æ C3 – xC2, C2 Æ C2 – xC1, we get 3 3x

p(x) =

0 2

2 4x

3 = 0

0 2

2 2x

[using R3 Æ R3 – xR2 and R2 Æ R2 – xR1] 2

r(q) =

2

= 3(4x + 4 – 4x ) + 2(– 4) = 4 (b) Using C3 Æ C3 – C2 and C2 Æ C2 – C1, we get 1 2+ x

q(x) = 2

0 1

(2 + x )2 - sin ( x + p 4)

0 1

sin ( x + p 4)

cos ( x + p 4)

+

cos ( x + p 4)

cos ( x + p 2) cos ( x + p 2) cos ( x + 3 p 4) cos ( x + 3 p 4)

=0+0=0 fi r(x) is a constant function. 1 \r(x) = r(0) =

2

1

0 -1

2 1

2 1

1 2

2

1

cos q - sin q - cos q

cos q sin q , - cos q 1 1 2 3 3 5

sin q cos q - sin q

cos q sin q cos q

sec2 q

1

1

cos q

cos q

cos ec2 q

1

cos 2 q

cot 2 q

=4

2

2

and

Match the functions in Column 1 with their range in Column 2.

5+ x 7+ x

- sin ( x + p 2) sin ( x + p 2) - sin ( x + 3 p 4) sin ( x + 3 p 4)

(c) r'(x) =

s(q) =

sin q cos q sin q

sin 2 q 2 cos 2 q cos 2 q

q(q) =

2 2 x 2 x2 + 2

3 x2 + 2 4 x 6 x2 + 2

- 2 1 -1

1

Column 1 (a) p(q)

Column 2 (p) [0, 1]

2 1

(b) q(q)

(q) [0, 2 2 ]

(c) r(q)

(r) [– 2, 2]

1

(d) s(q)

(s) [–

2 1

Ans.

5 – 2, –

5 + 2]

IIT JEE eBooks: www.crackjee.xyz 8.24 Comprehensive Mathematics—JEE Advanced

Solution

C1 to obtain

p(q) =

(- 2 ) (– 1) + (– 1) (– 2 sin q cos q) + (– 1) (sin2q – cos2q)

=

2 + sin 2q + cos 2q

=

2 +

Ans. (b) Solution: Let ak = ark – 1 " k ≥ 1, then D n = a na n

2 sin (2q + p/4)

Dn =

2 (sin 2q - cos 2q )

Thus, range of q(q) is [–2, 2]. (c) Using C1 Æ C1 + C3, we get cos q sin q cos q

= 2 cos q

(d) Taking sec q common from R1, we get

1

Example 57

s(q) = sec q cos q 2

an an + 1

3d 3d

3d 3d

an + 2

3d

3d

=0

C2 and C3 are identical]

0 cos x 0 Let f(x) = sin x cos x sin x

2

sin x cos x . 0

Ans. (d)

cos ec2 q

Solution: Multiplication of two determinants leads us

cos 2 q

cot 2 q

y y 1

cos 2 q

cos 2 q

cos q

cos ec q

0

cot 2 q - cos 2 q

2

2

= sec2q(cot2q – cos2q) (cos2q – cos4q)

1 -y -y 1 y y

y = sin x cos x

where

Using C1 Æ C1 – C2, C2 Æ C2 + C3, we get f(x) = (1 + y)

= (cot q – cos q) sin q 2

2

2

1 0 y 1 0 y -1 1 y = (1 + y ) 2 0 1 2 y 0 1 1 0 1 1

= (1 + y)2 (1 – 2y)

= cos2q – cos2q sin2q = cos4q \ range of s(q) is [0, 1]

When sin 2x = 1, y = 1/2 and f(x) = 0

ASSERTION-REASON TYPE QUESTIONS an Example 56

r

cos 2 q

0

2

r

=0

cos 2 q

f(x) =

2

2

cos 2 q

R3 Æ R3 – R1, we get 1

2

Statement-1: If sin 2x = 1, then f(x) = 2/3 Statement-2: f(x) = 0 if sin x = cos x

2

2 s(q) = sec2q cos q

2

[

\ range of r(q) is [– 2, 2] 1

+ 6

1 r

ak = b + (k – 1)d, then using C2 Æ C3 – C1 and C2 Æ C2 – C1, we get

C1, we get

1 sin q r(q) = 2 cos q 0 cos q 0 - sin q

an

1 r

r

\ range of p(q) is [0, 2 2 ] q(θ) =

+ 3

1 r

an + 3

an + 6

Let Dn = an + 1

an + 4

an + 7

an + 2

an + 5

an + 8

Statement-1: If ak > 0 " k ≥ 1 and a1, a2, a3, . . . are in G.P. then Dn = 0 " n ≥ 1. Statement-2: If a1, a2, a3 . . . are in A.P. then Dn = 0 " n ≥ 1.

When sin x = cos x, 1 – 2y = 1 – 2 sin2x = cos 2x = 0 \ f(x) = 0. Example 58

Suppose x > 0, y > 0, z > 0 and x log 2 3 15 + log (a x )

D(a, b, c) =

y log 3 5 25 + log (b y ) z log 5 7 35 + log (c z )

Statement-1: D(8, 27, 125) = 0 Ê 1 1 1ˆ Statement-2: D Á , , ˜ = 0 Ë 2 3 5¯ Ans. (b)

IIT JEE eBooks: www.crackjee.xyz Determinants 8.25

Solution: Using log (bc) = c log b and applying C3 Æ C3 – 5C2 we get x log 2 3 x log a D(a, b, c) = y log 3 5 y log b z log 5 7 z log c D(8, 27, 125) = D(23, 33, 53) = 0 as in this case C1 and C3 are proportional. Similarly, D(1/2, 1/3, 1/5) = D(2–1, 3–1, 5–1) = 0 Example 59

Let a π 0, p π 0 and

a b c D = 0 p q p q 0 Statement-1: If the equations ax2 + bx + c = 0 and px + q = 0 have a common root, then D = 0. Statement-2: If D = 0, then the of equations ax2 + bx + c = 0 and px + q = 0

Èp Ê p Êp ˆ ˆ˘ cot Á + x˜ = cot Í - Á - x˜ ˙ Ë4 Ë ¯ ¯˚ 4 Î2 pˆ Êp ˆ Ê = tan Á - x˜ = - tan Á x - ˜ , Ë4 ¯ Ë 4¯ and

log (x/y) = – log (y/x) D is a skew symmetric determinant of odd order.

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 61 to 65 Let 2 x3 - 3 x 2 5 x + 7 2 4 x3 - 7 x

D(x) =

7 x - 8x 3

Ans. (b)

Eliminating l, we obtained D = 0. D along C1 we obtain – aq2 + p(bq – cp) = 0

3

To evaluate ai we differentiate D(x) i times w.r.t. x and put x = 0 or divide D(x) by x4 put 1/x = t, differentiate (4 – i) time w.r.t. t and put t = 0. a0 equals

(a) 0

Solution: If l is a common root of ax2 + bx + c = 0 and px + q = 0, then al2 + bl + c = 0, pl + q = 0 and pl2 + ql = 0

x -1

= a0 + a1 x + ... + a4 x4

Example 61

have a common root.

3x + 2 1

2

(b) 1

(c) 2

(d) 3

(c) 161

(d) 191

(c) – 41

(d) 43

a1 equals

Example 62 (a) 0

(b) 61 a4 equals

Example 63 (a) 41

(b) – 43 a3 equals

Example 64

2

or

Ê qˆ Ê qˆ aÁ - ˜ + bÁ - ˜ + c = 0 Ë p¯ Ë p¯

(a) – 70

(b) – 73

(c) – 74

(d) 0

(c) 41

(d) 0

a2 equals

Example 65

2

Thus, ax + bx + c = 0 and px + q = 0 have a common root. Example 60

Statement-1:

sin p cos( x + p / 4) tan( x - p / 4) log( x / y ) =0 D = sin( x - p / 4) - cos(p / 2) cot(p / 4 + x) log( y / x) tan p Statement-2: A skew symmetric determinant of odd order equals 0. Ans. (a) Solution: For statement-2, see theory on page 8.3. Now, using Èp Ê p pˆ ˆ˘ Ê cos Á x + ˜ = cos Í - Á - x˜ ˙ ¯˚ Ë 4¯ Î2 Ë 4 pˆ Êp ˆ Ê = sin Á - x˜ = - sin Á x - ˜ ; Ë4 ¯ Ë 4¯

(a) 46

(b) – 43

Ans. 61 (a), 62 (c), 63 (b), 64 (b) 65 (a) Solution: 61. a0 = D(0) = 0 62. D¢(x) =

6 x2 - 6 x

5x + 7 2

12 x - 7

3x + 2 1

2

21x 2 - 16 x

x -1

2 x3 - 3 x 2

5 2

+ 4x - 7x

3 1

7 x - 8x

1 3

3

3

a1 = D(0) = 161

2

3

IIT JEE eBooks: www.crackjee.xyz 8.26 Comprehensive Mathematics—JEE Advanced

63.

D( x) x4

2 - 3/ x 2 = 4 - 7/ x 7 - 8/ x

(a) abc (c) abc/4

5 + 7/ x 2 3 + 2/ x 1 1 - 1/ x 3

4 - 7t2 7 - 8t

a = BC is a rational number, then D must be (a) (b) (c) (d)

1 3

D1 = 7

2

2 1 -1 3

a3 = D¢1(0) = –73

65. As a0 = 0 we get a 1 + a 2x + a 3x 2 + a 4x 3 2 x2 - 3 x 5 x + 7 2 4 x2 - 7

=

7 x2 - 8 x

If zk = xk + iyk for k = 1, 2, 3 and

2 x2 - 3x 5 2

4x - 3 5x + 7 2 8x 3x + 2 1 + 4 x2 - 7 3 1 = 14 x - 8 x - 1 3 7 x2 - 8 x 1 3

z1 1 z2 1 z3 1 1 D1 4i 1 D1 (d) 2i (b)

If P(x, y) is such that

Example 70 x x1 x2

3

Differentiating we get a2 + 2a3x + 3a4x2

z1 z2 z3

then area of DABC is 1 D1 (a) 4 1 D1 (c) 2i

3x + 2 1 x -1

a natural number a integer number an irrational number an integer

Example 69

2 - 3t - 3 5 + 7t 2 - 14 t 3 + 2t 1 + 4 - 7 t 2 -8 1- t 3 7 - 8t

D¢1(t) = fi

3 + 2t 1- t

If DABC is an equilateral triangle and

Example 68

Taking limit as x Æ •, we get 2 5 2 a4 = 4 3 1 = – 43 7 1 3 64. Put 1/x = t in D(x)/x4 and write 2 - 3t 5 + 7 t 2 D1(t) =

(b) s(s – a) (s – b) (s – c) (d) 4s(s – a) (s – b) (s – c)

y 1 x y1 1 + x1 y2 1 x3

then P lies on (a) median of D ABC (b) bisector of – A

y 1 y1 1 = 0 y3 1 A

Paragraph for Question Nos. 66 to 70

A (d) perpendicular bisector of the side BC Ans. 66. (a), 67. (d), 68. (c), 69. (a), 70. (a) Solution: 66. Points A, B, C are collinear if and only if area of DABC is 0. 67. Use Heron’s formula.

Let A(x1, y1), B(x2, y2) and C(x3, y3) be three points. Area of triangle with vertices A, B and C is given by

68. Use

\

a2 = 46

1 D 2 where D =

x1 x2 x3

Example 66

y1 1 y2 1 y3 1 Points A, B, C are collinear if and only if

(a) D = 0 Example 67

(b) D > 0

If a = BC, b = CA, c = AB and 2s = a + b

+ c, then D equals 2

(c) D < 0 (d) D £ 0

3 2 a = D and a is rational. 4 69. Use if z = x + iy, then 2x = z + z and 2 iy = z – z 70. Write the sum of two determinants as x x1 x2 + x3

y y1 y2 + y3

1 1 =0 2

which is an equation of straight line through A and the mid point of BC.

IIT JEE eBooks: www.crackjee.xyz Determinants 8.27

INTEGER-ANSWER TYPE QUESTIONS Let w

Example 71 cos

2k - 1 2 k

2 k

2 k

1

-2k

-2 k

2k

-1

det A =

2p 2p + i sin 3 3

Using R2 Æ R2 + R3, we get z +1

w

w2

w

z + w2

1

w

1

z +w

numbers z satisfying

2

2k - 1 2 k 0 1 + 2k det A =

= 0 is equal to

-2 k

2 k -(2k + 1) -1

2k

Using C3 Æ C3 + C2, we get

Ans. 1

2k - 1 2 k 0 1 + 2k det A =

Solution: Denote the given determinant by D. Using C1 Æ C1 + C2 + C3 and 1 + w + w2 = 0, we get w

1

w

2 D = z 1 z +w 1 1

1 z +w

w

2 D = z 0 z +w -w

1- w

0

= (2k + 1)3 Also, as B is a skew symmetric, det B = 0. \ det (adj A) = (det A)2 = (2k + 1)6 and det (adj B) = (det B)2 = 0 Now, det (adj A) + det (adj B) = 106

w2 1- w2 z + w - w2

= z [(z + w2 – w) (z + w – w2) – (1 – w) (1 – w2)] = z [(z2 – (w2 – w)2 – (1 – w – w2 + 1)]



(2k + 1)6 + 0 = 106



2k + 1 = 10



= z(z2) = z3 Thus, z3 = 0 fi z = 0

È 2k - 1 2 k Í 1 A= Í 2 k Í ÍÎ-2 k 2k

0 -2 k

k ˘ ˙ 2 k˙. ˙ 0 ˙˚

If det (adj A) + det (adj B) = 106, then [k] is equal to (where [k] denote the greatest integer less than or equal to k) Ans. 4 Solution: If A n ¥ n, then det (adj A) = (det A)n–1. is 0. We have

e-2iA

eiC

eiB

eiC

e-2iB

eiA

eiB

eiA

e-2iC

D=

2k - 1

k > 0]

If A, B and C are angles of a triangle, and

Let k be a positive real number and let È 0 2 k˘ Í ˙ -2k ˙ and B = Í1 - 2k Í ˙ -1 ˙˚ ÍÎ - k

[

fi [k] = 4

k = 4.5

Example 73

= z [z2 – (w4 + w2 – 2w3) – 3]

Example 72

2k - 1

2k

= (1 + 2k) [(2k – 1)2 + 8k]

Applying R2 Æ R2 – R1 and R3 Æ R3 – R1, we get 1

-2 k

2

4 k 0

D Ans. 4 Solution: Let a = eiA, b = eiB and g = eiC then abg = ei(A + B + C) = eip = cos p + i sin p = –1, g

g

1/ b

b

a

D=

and

=

1/a 2

1 a 2 b 2g 2

1 b g 2

a 1/ g 2

a 2g

a 2b

1

2

g b g 2a 2

b 2

1 b a = - b /a - g /a 1

-a /b 1 -g /b

- a /g - b /g 1

IIT JEE eBooks: www.crackjee.xyz 8.28 Comprehensive Mathematics—JEE Advanced

a 1 b = a bg -g

-a b -g

x – 3 sin 2x is 5, we get

-a -b g

f(x) is 5. Let

Example 76

2 cos 2 x sin (2 x) - sin x

[multiplying C1{C2} [C3] by a {b} [g]

Then

= – 2 (1 + 1) = – 4 fi D cos x Example 74

then lim

xÆ0

-x

2 Let f (x) = 2 sin x - x tan x - x

cos x 0

1 p /2 Ú [ f (x) + f ¢(x)]dx is equal to p 0

Ans. 1 Solution: Applying C1 Æ C1 – 2sinx C3 and C2 Æ C2 + 2 cosx C3, we obtain

1 2x 1

2 0 - sin x 0 2 cos x f (x) = 0 sin x - cos x

f ¢ ( x) is x

= 2 sin2 x + 2 cos2 x = 2 fi f ¢(x) = 0

Ans. 2

p /2

\

Solution: Differentiating w.r.t. x, we get - sin x

x

– f ¢ (x) = 2 cos x

x

2

sec2 x

x

1

cos x 1 1 2 x + 2 sin x 2 x 2 x tan x 1 1 1 cos x + 2 sin x tan x

x x2 x

[

0

1 3 cos x 1 1 3 cos x – 5, If f(x) = sin x sin x 1 1

Ans. 5 Solution: We have f(x) = 9 cos2 x + sin2 x – 6 sin x cos x – 5 = 5 + 4 cos 2x – 3 sin 2x – 5

2

2 Êpˆ

cos 2 x

4 sin 2 x

1 + cos x

4 sin 2 x

2

sin x

f (x) =

2

1 + 4 sin 2 x

2

sin x

2 0

0 1 1 1 1 0 f ¢ ( x) =- 2 0 0 - 0 0 2 =2 fi lim xÆ0 x 1 1 1 0 1 0

p /2

Let 1 + sin 2 x

the second determinant is zero ]

f(x) is

1

Ú [ f (x) + f ¢(x)]dx = p Ú 2dx = p ÁË 2 ˜¯ = 1 0 0

Example 77

- sin x 1 1 cos x 1 0 f ¢ ( x) fi – = 2 cos x x 2 x + 2 sin x x 2 x tan x 1 0 sec2 x 1 1

Example 75

2 sin 2 x - cos x

f (x) = sin (2 x) sin x

1 -1 -1 0 -1 -1 a bg -1 1 -1 = 0 1 -1 = a bg -1 -1 1 - 2 -1 1

cos x f (x)

Ans. 6. Solution: Applying C1 Æ C1 + C2, we get cos 2 x

2

2 f (x) = 2 1 + cos x 2

cos x

1

4 sin 2 x 4 sin 2 x 1 + 4 sin 2 x

Applying R2 Æ R2 – R1 and R3 Æ R3 – R1, we get 2 cos 2 x 4 sin 2 x 1 0 f (x) = 0 = 2 + 4 sin 2x 0 1 -1 x value of f (x) is 6. Example 78

Let f1(x) = x + a1, f2(x) = x2 + b1 x + b2,

x1 = 2, x2 = 3 and x3 = 5 and 1 1 1 D = f1 ( x1 ) f1 ( x2 ) f1 ( x3 ) f2 ( x1 ) f2 ( x2 ) f2 ( x3 )

IIT JEE eBooks: www.crackjee.xyz Determinants 8.29

then D is equal to

Let f (n), g(n) and h(n) be polynomials

Example 80 in n, and let

Ans. 6 Solution: Applying C3 Æ C3 – C2 and C2 Æ C2 – C1, we obtain 1 f ( D = 1 x1 )

0 x2 - x1

2r + 1 Dr =

0 x3 - x2

1 x2 + x1 + b1

Solution: We have

r =1

= (x1 – x2) (x2 – x3) (x3 – x1) = 6 n

If D(x) = x - 1

( x + 1)

x

r =1

2

( x + 2)

Ans. 2

n

n

r =1

r =1

But  (2r + 1) = 2  r + n = 2 ◊ n

1 ( x + 1) 2

= n (n + 2),

 2r – 1 = 2n – 1,

r =1 n

n

n

r =1

r =1

r =1

x -1

+

( x + 2) 3

Thus, x3

n

( x + 1)

2x

x - 2 ( x - 1) 2 x -1

x

x

( x + 1)

2

r =1

f ( n)

2 -1

6(2 - 1)

g ( n) = 0

[

3 ( x + 2) 2

a1 = – 8 – 12 + 18 = – 2.

n

1 n(n + 1) (n + 2) n(n + 1) (n + 2) 6

2

1 1 0 -2 - 2 0 -2 1 0 fi D¢(0) = 1 0 1 + - 1 0 1 + - 1 0 3 1 1 8 0 2 8 0 1 12 fi

6 n ( n + 2)

n

3 x2 3 ( x + 1)

2

n ( n + 2)

 Dr =

3

2( x + 1) ( x + 2)3

x

+

(n + 1) n (n + 1) n(n + 1) (2n + 1) 2 6 1 = n(n + 1) (n + 2) 6 =

3

x - 2 2( x - 1)

(n + 1) +n 2

n

( x + 1)3

x2

D¢(x) = 1

x

g ( n)

)

r =1

Now, 1 ( x - 1)

(

3 2n + 1 - 6

and  r(n – r + 1) = (n + 1)  r –  r2

fi a1 = D¢(0). 2

f ( n)

n

+ a6 x6, then

+ 6a6 x5

6n(n + 2)

 r (n - r + 1) n(n + 1) (n + 2) h(n)

Solution: Note that D(x) is a polynomial of degree at most 6 in x. D¢(x) = a1 + 2 a2x +

r =1

3

x in D(x) is

If D(x) = a0 + a1 x + a2 x2 +

n

 2r - 1

 Dr =

3

( x + 1)3 ,

x2

h( n)

Ans. 0 n

x

g ( n)

r =1

 (2r + 1)

x - 2 ( x - 1)

)- 6

n

1 x3 + x2 + b1

2

f ( n)

then  D r is equal to

= (x2 – x1) (x3 – x2) (x3 – x1)

Example 79

(

3 2

n +1

r (n - r + 1) n(n + 1) (n + 2)

f2 ( x1 ) x22 - x12 + b1 ( x2 - x1 ) x32 - x22 + b1 ( x3 - x2 ) = (x2 – x1) (x3 – x2)

2

6 n ( n + 2)

r -1

Example 81 x= y= z= has a nontrivial Ans. 3

h( n)

C1 and C2 are proportional]

If 0 £ q £ p, and the system of equations (sin q) y + (cos q) z z + (cos q) x (sin q) x + y solution, then 4q/p is equal to

Solution: We rewrite the system of equations as follows: – x + (sin q) y + (cos q) z = 0

IIT JEE eBooks: www.crackjee.xyz 8.30 Comprehensive Mathematics—JEE Advanced

(cos q) x – y + z = 0 (sin q) x + y – z = 0 As this system of equations has a non-trivial solution -1 cos q sin q

sin q -1 1

cos q 1 -1

trivial solution if a + 2t b c b c + 2t a D= c a b + 2t

=0

Clearly D is a cubic polynomial in t and has 3 roots.

fi – 1 + sin2 q + cos2 q + 2 sin q cos q + 1 = 0 fi sin 2q = – 1 fi 2q = 3p/2 fi 4q/p = 3 If the system of equations ax + hy + g = 0 hx + by + f = 0 and ax2 + 2hxy + by2 + 2gx + 2f y + c + t = 0 has a unique solution, and

Example 84

h 2 - ab

(1) (2) (3)

= 8, then t is equal to

Ans. 3 Solution: Using C1 Æ C1 + C2 – C3, we obtain 0 0 4 - 4x 0 2- x 0 D(x) = 1+ x 0 3 - 5x

Ans. 8 Solution: Since we should be able to solve (1) and (2) for x and y, we must have ab – h2 π 0. We can rewrite (3) as x(ax + hy + g) + y (hx + by + f ) + gx + f y + c + t = 0 (4)

= (1 + x) (2 – x) (4x – 4) D(x) = 0 has three real roots.

\

Example 85

Since any solution of (1) and (2) must be a solution of (3), the given system of equations can be rewritten as ax + hy + g = 0, hx + by + f = 0 and gx + fy + c + t = 0 Since the above system of equations is consistent, we must have a h g h b f =0 g f c+t a h g

h b f

fi [abc + 2 fgh – af fi

t=

g a f + h c g 2

h b f

– bg2 – ch2] + t (ab – h2) = 0 h 2 - ab

D(q) =

If 0 £ q £ p/2, the number of roots of cos 2 2 q sin 2q 1

cos 2 4 q sin 4 q 1

cos 2 6 q sin 6 q is 1

Ans. 7 Solution: Using cos2 x = 1 – sin2 x and 1 a

1 b

1 c

2

2

2

a

0 0 =0 t

abc + 2 fgh - af 2 - bg 2 - ch 2

The number of real roots of the equation

6 - 5x x - 2 4 - 4x 0 =0 D(x) = x - 2 2 - x 4 - 4x 0 3 - 5x

Example 82

abc + 2 fgh - af 2 - bg 2 - ch 2

=0

b

c

= (b – c) (c – a) (a – b),

we get D(q) = (sin 4q – sin 6q) (sin 6q – sin 2q) ¥ (sin 2q – sin 4q) = 26 cos (5q) cos (4q) cos (3q) sin2 q sin 2q fi q = 0, p/2, p/6, p/8, 3p/8, p/10, 3p/10

=8

Example 83 The number of values of t for which the system of equations (a + 2t)x + by + cz = 0 bx + (c + 2t) + az = 0 cx + ay + (b + 2t) z = 0 has non-trivial solutions is Ans. 3 Solution: The given system of equations will have a non-

EXERCISE LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. Suppose x, y, z are three distinct non-zero real num-

IIT JEE eBooks: www.crackjee.xyz Determinants 8.31

xn

xn+2

x n +3

(b) g(a) g(b) g(g)

n bers and D = y

y n+2

y n +3 .

(c) f(a) g(a) + f(b) g(b) + f(g) g(g)

zn

z n+2

z n +3

If

(d) 0

D 1 1 1 = + + then ( x - y )( y - z )( z - x) x y z

(a) n = –1 (b) n = 0 (c) n = 1 (d) n = 2 2. Let a, b, c, d be four real numbers and let 2 D= a+b+c+d ab + cd

a+b+c+d ab + cd 2(a + b)(c + d ) ab(c + d ) + cd (a + b) ab(c + d ) + cd (a + b) 2abcd

then D is equal to (a) 2abcd (a + b) (c + d) (b) 2abcd (a + b + c + d) (c) (a + b) (c + d) (d) 0 3. If q ∈ R 1 1 1 1 is D = 1 1 + sin q 1 1 1 + cos q (a) 1/2

3 /2

(b)

x+a a 4. If D = a

(c)

2

1 ca

5. The determinant D = tan q 12 2

2

sec q 10

1 -1 equals 2

(a) 2 sin2q (b) 12 sec2 q – 10 tan2 q (c) 12 sec2 q – 10 tan2 q + 5 (d) 0 6. Let f and g R such that f(x + y) = f(x) g(y) + g(x) f(y) for each x, y ∈ R. Let f (q + a ) Δ = f (q + b ) f (q + g )

f (a ) g (a ) f (b ) g (b ) f (g ) g (g )

then Δ equals (a) f(a) f(b) f(g)

1 1 + a b 1 1 + is b c 1 1 + c a

(a) 0 (c) a–1 + b–1 + c–1

(d) 3 2 /4

tan 2 q

then Δ equals (b) 2(a3 + b3) (a) a3 + b3i (d) –2(a3 – b3i) (c) a3 – b3i 8. If a, b, c are non-zero real numbers, then

D = 1 bc

(b) – (a + b + c) (d) 0, – (a + b + c) sec2 q

a a + bi bi bi a Δ = a + bi bi a a + bi

1 ab

b c x+b c = 0, then x equals b x+c

(a) a + b + c (c) 0, a + b + c

7. If a, b ∈ R and

(b) bc + ca + ab (d) none of these

log a ( abc ) log a b log a c logb c is 1 9. If a, b, c > 1, D = logb ( abc ) log c ( abc ) log c b 1 (a) 0 (b) loga b + logb c + logc a (c) logabc (a + b + c) (d) 3 10. If a, b, c are positive and are the pth, qth and rth terms respectively of a G.P., then log a D = log b log c

p 1 q 1 is r 1

(a) 0 (b) log (abc) (c) – (p + q + r) (d) 3 11. If pth, qth and rth terms of an H.P. be a, b, c bc ca ab respectively, then D = p q r is 1 1 1 (a) – 1 (b) abcpqr (c) pa + qb + rc (d) 0

IIT JEE eBooks: www.crackjee.xyz 8.32 Comprehensive Mathematics—JEE Advanced

b 2 - ab b - c 2 12. If D = ab - a

bc - ac

bc - ac

a - b b 2 - ab then D equals c - a ab - a 2

(a) (b – c) (c – a) (a – b) (b) abc (b – c) (c – a) (a – b) (c) (a + b + c) (b – c) (c – a) (a – b) (d) 0 13. Let p, q, r be three distinct prime numbers, and - pi - qi

1

r3 pqr

-1 = x + iy, then 2i

qi 2q

(a) x = –p, y = q (b) x = r, y = 0 (c) x = 0, y = –qr (d) x = 0, y = 0 14. If a, b, c are three integers in A.P., lying between 1 and 9 and a31, b41 and c51 are three digit numbers, then the determinant 3 4 5 a 31 b 41 c 51 equals a b c (a) a + b + c (c) – 1 1 15. If

m m

C1 C2

(b) – 1 (d) 2

sin 2 x (1 + 2 cos x ) sin 2 x sin 3x 3 + 4 sin x 3 4 sin x 18. If f(x) = 1 + sin x sin x 1 p 2

then the value of Ú0

1

1 m+6

C1 C2

m+6

then a + b + g is equal (a) 3 (c) 7 16. Let a, b, c and let 0 -b - c z = b 0 -a c a 0

C1

a

b

g

=2 3 5,

C2 (b) 5m (d) m2

then z equals (a) 0 (b) purely imaginary (c) abc (d) 2abc 17. If A, B and C are the angles of a triangle, then the determinant -1 cos C cos B -1 cos A cos C -1 cos B cos A

f ( x) dx is

(a) 3 (b) 2/3 (c) 1/3 (d) 0 19. If a, b, g are the roots of x3 + ax2 + b = 0, then the determinant D, where a b g D = b g a equals g a b (b) a3 – 3b (c) a2 – 3b (d) a3 (a) – a3 20. If a, b, g are the roots of x3 + ax2 + b = 0, b π 0; then the determinant D, where 1 a 1 D= b 1 g

(b) 3b (d) 0

m +3 m +3

is equal to (a) 0 (c) 1

1 b 1 g 1 a

1 g 1 equals a 1 b

(a) – b3 (b) b3 – 3c (c) b2 – 3c (d) 0 21. The number of distinct real roots of tan x cot x cot x D = cot x tan x cot x = 0 cot x cot x tan x in the interval – p/4 £ x £ p/4 is (a) 0 (b) 2 (c) 1 22. If ω π 1

i

-w

x + iy = -i

1

w2

-w 2

1

w

(d) 3

then (a) x = – 1, y = 0 (b) x = 1, y = – 1 (c) x = 1, y = 1 (d) x = 0, y = 0 23. If eix = cos x + i sin x and 1 -p i / 4 x + iy = e

e -p i / 3

ep i / 4 1 e-2p i / 3

ep i / 3 e2p i / 3 , then e-2p i

IIT JEE eBooks: www.crackjee.xyz Determinants 8.33

(b) x = 1, y = - 2

2

(a) x = – 1, y = (c) x = - 2 , y =

2

(d) none of these

24. If a, b, c ∈ R, the number of real roots of the equation x c -b a = 0, is D = -c x b -a x (a) 0 (b) 1 (c) 2 (d) 3 2 2 25. If a, b, c ∈ R and a + b – ab – a – b + 1 £ 0 and a + b + g = 0, then cos g a cos b

1 cos g cos a

D= (a) ab

cos b cos a b

(b) 1

sin a 26. If D = sin b sin g

cos a cos a cos a

equals

(c) 0

(d) – 1

sin a + cos b sin b + cos b then D is equal to sin g + cos b

(a) sin a sin b sin g (b) cos a sec b tan g (c) sin a sin (a + b) + cos a cos (g + b) (d) 0 27. Let f(x) =

cos ec x

sin x

cos ec2 x + tan x sec x

sin 2 x

sin 2 x

sec2 x

1

sin 2 x

sin 2 x

p /2

then Ú0

f(x)dx equals

Êp 8 ˆ (a) - Á + ˜ Ë 4 15 ¯ (c)

(b)

p 1 + 4 5

p 4

(d) p

28. The values of q lying between q = 0 and q = p/2 and satisfying the equation 1 + sin 2 q

cos 2 q

4 sin 6q

sin 2 q

1 + cos 2 q

4 sin 6q

sin q

cos q

1 + 4 sin 6q

2

are given by (a) p/36, 5p/36 (c) 5p/36, 7p/36

2

29. If pqr π 0 and the system of equations (p + a)x + by + cz = 0 ax + (q + b)y + cz = 0 ax + by + (r + c)z = 0 has a non-trivial solution, then value of

is (a) – 1 (b) 0 (c) 1 (d) 2 A B C 30. Let a = 2 , b = 2 , C = 2 the system of equations ax + by + (aa + b)z = 0 bx + cy + (ba + c)z = 0 (aa + b)x + (ba + c)y = 0 has a non-zero solutions if A, B, C are in (a) A.P. (b) G.P. (c) H.P. (d) none of these 31. If the system of equations ax + ay – z =0 bx – y + bz =0 – x + cy + cz = 0 (where a, b, c π – 1) has a non-trivial solution, then value of 1 1 1 + + is 1+ a 1+ b 1+ c (a) 2 (b) – 1 (c) – 2 (d) 0 32. All the values of l for which the system of equations (l + 5)x + (l – 4)y + z = 0 (l – 2)x + (l + 3)y + z = 0 lx + ly + z = 0 has a non-trivial solution lie in the set (a) {– 1, 2} (b) {0, – 1} (c) {0} (d) R 33. Given the system of equations (b + c) (y + z) – ax = b – c (c + a) (z + x) – by = c – a (a + b) (x + y) – cz = a – b (where a + b + c π 0); then x : y : z is given by (a) b – c : c – a : a – b (b) b + c : c + a : a + b (c) a : b : c a b c : : b c a 34. Number of real values of l for which the system of equations (l + 3)x + (l + 2)y + z = 0 3x + (l + 3)y + z = 0 2x + 3y + z = 0 has a non-trivial solutions is (d)

=0

(b) 7p/36, 11p/36 (d) 11p/36, p/36

a b c + + p q r

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35. The value of l for which the system of equations 2x – y – 2z = 2, x – 2y + z = – 4 x + y + lz = 4 has no solution is (a) 3 (b) - 3 (c) 2 (d) – 2

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS sin a 36. The determinant sin b sin g (a) – 4 sin

cos a 1 cos b 1 is equal to cos g 1

a-b b -g g -a sin sin 2 2 2

(b) sin a + sin b + sin g (c) sin (a – b) + sin (b – g) + sin (g – a) (d) none of these. 37. If a, b, c > 0 and x, y, z ∈ R, then the determinant (a x + a - x )2

(a x - a - x )2 1

(b y + b - y ) 2

(b y - b - y ) 2 1

(c z + c - z ) 2

(c z - c - z ) 2

cos 2 a

2 40. The determinant sin b

sin b cos b

cos 2 b is

sin 2 g

sin g cos g

cos 2 g

(a) – sin (b – g) sin (g – a) sin (a – b) (b) sin (a – b) cos a cos b + sin (b – g) cos b cos g + sin (g – a) cos g cos a 1 [sin (2a – 2b) + sin (2b – 2g) + 4 sin (2g – 2a)] (d) none of these. (c)

41. If a2 + b2 + c2 = 1, then a 2 + (b 2 + c 2 ) cos f

ab(1 - cos f )

ba (1 - cos f ) ca (1 - cos f )

b + (c 2 + a 2 ) cos f cb (1 - cos f )

2 a + ab + b 3 a+b 39. If ai, bi, ci ∈ R (i = 1, 2, 3) and x ∈ R and 2

(d)

a1 + b1 x 2

a1 x 2 + b1

c1

2 D = a2 + b2 x

a2 x 2 + b2

c2

a3 + b3 x 2

a3 x 2 + b3

c3

ac(1 - cos f ) bc (1 - cos f )

2

42. If D =

(d) x = 0

cos a cos b cos g

sin (a + d ) sin (b + d ) sin (g + d ) (c) g

then D is

(d) d.

43. If f ∈ R then D(f) = sin f cos f sin 2 f sin (f + 2 p 3) cos (f + 2 p 3) sin (2 f + 4 p 3) sin (f - 2 p 3) cos (f - 2 p 3) sin (2 f - 4 p 3) (a) is independent of f (b) is equal to a constant

=0 (d) has no local minimum 44. If abc π 0, then D(x) =

=0

sin a sin b sin g

(d) f.

(c) c

independent of (a) a (b) b

(b) x = – 1 c1 c2 c3

2

is independent of (a) a (b) b

1

(b) x, y, z (d) none of these. x a a b b x 38. If a π b, the equation a x a + b x b = 0 is a a x x b b x equals (a) 0 (b) a – b

then (a) x = 1 a1 b1 (c) a2 b2 a3 b3

sin a cos a

c 2 + (a 2 + b 2 ) cos f

is independent of (a) a, b, c (c) a, b, c, x, y, z

(c) a + b

sin 2 a

a 2 (1 + x)

ab

ac

ab

b (1 + x)

bc

ac

bc

c 2 (1 + x)

is divisible by

2

IIT JEE eBooks: www.crackjee.xyz Determinants 8.35

(a) 3 + x (c) x2

(b) x (d) (1 + x)2

cos x sin x cos x cos sin 2x 2 x 2 cos 2 x . 50. Let f (x) = cos 3x sin 3x 3 cos 3x

2 cos x 1 0 1 2 cos x 1 45. Let f(x) = , then 0 1 2 cos x (a)

Êpˆ f Á ˜ =–1 Ë 3¯

Êpˆ (b) f ¢ Á ˜ = Ë 3¯

(d) Ú- p f(x) dx = 0

sin a cos b cos a cos b - sin b

then D is independent of (a) a (c) g

sin a sin b cos a sin b cos b 52.

(b) b (d) none of these

a2 b sin A c sin A 1 cos A is independent of D = b sin A 1 c sin A cos A (a) a (c) c

53.

(b) b (d) A, B, C

48. If A, B and C are the angles of a triangle and 1 1 + sin B

1 1 + sin C

54. =0

sin A + sin 2 A sin B + sin 2 B sin C + sin 2 C then the triangle must be (a) isosceles (b) equilateral (c) right angled (d) none of these 49. If ai, bi ∈ N for i the determinant

x in

(1 + x) a1 b1

(1 + x) a1b2

(1 + x)

(1 + x)

a2b1

(1 + x) a3b1 (a) (b) (c) (d)

a ax

-1 0 a - 1 then f (2x) – f (x)

a x2

ax

51. If f (x) =

47. If a, b and c are the sides of a triangle and A, B and C are the angles opposite to a, b and c respectively, then

1 1 + sin A

(b) f ¢(p/2) = 4 (d) f ¢(p) = 4

p

p

(c) Ú0 f(x) dx = 0 cos a 46. If D = - sin a 0

3

Then (a) f ¢(0) = 0 (c) f ¢(p) = 0

a2b2

(1 + x)a3b2

(1 + x)a1b3 (1 + x)

a2b3

(1 + x)a3b3

0 a1 + a2 + a3 independent of ai,s bi,s b1 + b2 + b3

55.

a

is divisible by (a) x (b) a (c) 2a + 3x (d) x2 If sin b π cos b and x, y and z satisfy the equations x cosa – y sin a + z = cos b + 1 x sina + y cos a + z = – sin b + 1 x cos(a + b ) – y sin (a + b ) + z = 2 then x2 + y2 + z2 equals (a) 1 (b) 2 (c) 3 (d) 5 If w π 1 is a cube root of unity and x + y + z = a, x + wy + w2z = b, x + w2y + wz = c, then (a) x = (a + b + c)/3 (b) y = (a + bw2 + cw)/3 (c) z = (a + bw + cw2)/3 (d) x3 + y3 + z3 = a3/3 If the system of equations x = cy + bz, y = az + cx and z = bx + ay has a non-trival solution, then value of a2 + b2 + c2 + 2abc is (a) 1 (b) 0 (c) – 1 (d) independent of a, b, c Factors of D=

x3 yz 1

(a) y – z (c) x – y

y3 zx 1

z3 x y are 1 (b) z – x (d) x2 + y2 + z2 + yz + zx + xy

MATRIX-MATCH TYPE QUESTIONS 56. Suppose f(x) is a function satisfying the following conditions.

IIT JEE eBooks: www.crackjee.xyz 8.36 Comprehensive Mathematics—JEE Advanced

(a) f(0) = 2, f(1) =1 (b) f(x (c) for all x ∈ R

x = 5/2, and (c)

2 ax 2 ax - 1 2 ax + b + 1 b b +1 -1 f ¢(x) = 2(a x + b) 2 ax + 2b + 1 2 ax + b where a, b are constants, then Column 1 Column 2 (a) a (p) 1/2 (b) b (q) – 5/4 (c) f(2) (r) 1/4 (d) f'(0) (s) 1/8 x 1 1 57. If 0 £ x £ 1 and f(x) = - 1 x 1 , then f(x) has -1 -1 x Column 1 (a) least value (b) greatest value

Column 2 (p) 4 (q) 1

(d) a minimum at

x

x2

x3

x4

x5 + a + b + c

6

7

x

(s) –1

2 cot x - 1 0 1 cot x -1 0 1 2 cot x

20 5

1 2x 5x

p /3 p /6

(a) Ú

-

5

 f ( j ) equals

and f(0) = 0, then

j = -5

(b) If sin 2x = 1, then

(q) – 16/3

(c) f ¢(p/4) (d) f(p/2)

(r) 8 (s) 0

a2

b

x2

b2

b +a+b+c

x3

c2

c

x3 + a 3

x2 + a2

x+a

x +b

x +b

2

x+b

x2 + c2

x+c

x3 + c3

equals

sec x x sin x

x2 - 1 cos x

x tan x

x +1

(r) 2

2

Ú f ( x) dx equals

1 + sin 2 x 2

sin x 2

sin x

cos 2 x

sin 2 x

1 + cos x

sin 2 x

2

2

cos x

1 + sin 2 x

ASSERTION-REASON TYPE QUESTIONS 61. Statement-1: If

x

3

2

- p /3

59. If a, b, c are distinct, match the right side with degree of the polynomial in x in Column 1 with number in Column 2 Column 1 Column 2

3

(q) 0

p /3

then

(p) – 16

(b) f(p/4)

Column 2

sin ( x + a ) sin ( x + b ) sin ( x + g ) (a) If f(x) = cos ( x + a ) cos ( x + b ) cos ( x + g ) (p) – 1 sin (b + g ) sin (g + a ) sin (a + b )

sin x (c) If f(x) = cosec x

Column 2

f(x)dx

(s) 3

2

priate value in Column 2 Column 1

tan x Column 1

(b)

x

1 4 (d) 1 - 2

then

(a)

x

(r) 2

8

0 cos x - sin x 0 cos x f(x) = sin x 0 cos x sin x

58. Let f (x) =

1

2

(p) 0 D(x) = (q) 1

sin x - cos x

pˆ Ê sin Á x - ˜ Ë 3¯

pˆ Ê cos Á x - ˜ Ë 3¯

Êp ˆ sin Á - x˜ Ë3 ¯

pˆ Ê tan Á x - ˜ Ë 4¯

pˆ Ê sec Á x - ˜ Ë 4¯

Ê 2p ˆ + x˜ cos Á Ë 3 ¯

Ê 2p ˆ + x˜ sec Á Ë 3 ¯

pˆ Ê cot Á x + ˜ Ë 4¯

Êpˆ then D Á ˜ = 0 Ë 4¯

IIT JEE eBooks: www.crackjee.xyz Determinants 8.37

Statement-2: A 62. Statement-1: Let p < 0 and a1, a2, . . ., a9 be the nine roots of x9 = p, then a1 a 2 a 3 D = a 4 a5 a6 = 0 a 7 a8 a9 Statement-2: If two rows of a determinant are identical, then determinant equals zero. 63. Statement-1: w =

3 + 6i

2i

6 12

3 + 2 2i 3 2 + 6i

18

2 + 2 3i

is a purely

27 + 2i

imaginary number. z

Statement-2: z

z.

64. Statement-1: If ai, bi ∈ N, for i = 1, 2, 3 and (1 + x) a1 b1

(1 + x) a1 b2

(1 + x) a1 b3

a b D(x) = (1 + x) 2 1

(1 + x) a2 b2

(1 + x) a2 b3 ,

(1 + x) a3 b1

(1 + x) a3 b2

(1 + x) a3 b3

x D(x) is 0. n Statement-2: If P(x) = (1 + x) , n Œ N x P(x) is P¢(0). 65. Statement-1: If a + b + c = 0 and a2 + b2 + c2 π bc + ca + ab, then the system of homogenous equations ax + by + cz = 0 bx + cy + az = 0 cx + ay + bz = 0

then (a) x = y = z (b) x = y = z = 0 (c) x + y + z = 0 (d) x + y + z π 0 2 2 67. If (1, a, a ), (1, b, b ) and (1, c, c2) are linearly independent, then (a) a + b + c π 0 (b) (b – a) (c – b) π 0 (c) (b – c) (c – a) (a – b) π 0 (d) none of these 68. If a, b, c are distinct and (a, a2, a3 + 1), (b, b2, b3 + 1), (c, c2, c3 + 1) are linearly dependent, then value of abc is (a) – 1 (b) – 2 (c) 2 (d) 1 69. If (a1, a2, a3), (b1, b2, b3) and (c1, c2, c3) are linearly independent and x, y, z ∈ R, then a, b, g ∈ R such that (x, y, z) + a(a1, a2, a3) + b(b1, b2, b3) + g(c1, c2, c3) = (0, 0, 0) (b) if x(a1, a2, a3) + y(b1, b2, b3) + z(c1, c2, c3) = (0, 0, 0) fi x + y + z π 0. (c) a(a1, a2, a3) + b(b1, b2, b3) + g(c1, c2, c3) = (x, y, z) cannot hold for any values of a, b, g. (d) none of these. 70. If ABC is a triangle, then the vectors (– 1, cos C, cos B), (cos C, – 1, cos A) and (cos B, cos C, – 1) are (a) linearly independent for all triangles (b) linearly dependent for all triangles (c) linearly independent for all isosceles triangles

A

Statement-2:

AX = B has

(d) none of these Paragraph for Question Nos. 71 to 75

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 66 to 70 A set of vector {(a1, a2, a3), (b1, b2, b3), (c1, c2, c3)} is said to be linearly independent if and only if a1 b1 c1

a2 b2 c2

a3 b3 π 0, c3

otherwise the set is said to be linearly dependent. A similar result holds for {(a1, a2), (b1, b2)}. 66. If (a1, a2, a3), (b1, b2, b3) and (c1, c2, c3) are linearly independent and x(a1, a2, a3) + y(b1, b2, b3) + z(c1, c2, c3) = 0,

For a, b, g, q ∈ R. Let cos (a + q ) sin (a + q ) 1 Aq (a, b, g) = cos (b + q ) sin (b + q ) 1 cos (g + q ) sin (g + q ) 1 71. If a = Ap/2 (a, b, g), b = Ap/3 (a, b, g). Which of the following is true? (a) a = b (b) a < b (c) a > b (d) 2a = b 72. dAq/dq when q = p/6 equals (a) – 1 (c) 1

(b) 0 (d) p/6

IIT JEE eBooks: www.crackjee.xyz 8.38 Comprehensive Mathematics—JEE Advanced

73. If a = b + 2p/3, then Aq (a) a + p/3 (c) a + 2p/3 2

g equals (b) a – p/3 (d) none of these

81. If – p < q < p, the number of solutions of D=

2

2

74. Aq + Af – 2(Aq+f) equals (b) Aq + Af (a) – 2Aq Af (c) 2Aq – Af (d) 0

f(x, y) = (b) a straight line through the origin

INTEGER-ANSWER TYPE QUESTIONS D(q) =

76. If xyz = – 2019 and a+x b c a b+ y c = 0, a b c+z

cos 2 q 77. Let f(q) = cos q sin q sin q

cos q sin q

- sin q

sin 2 q - cos q

cos q 0

D=

cos A sin A sin ( A + D + E ) D = cos B sin B sin ( B + D + E ) cos C sin C sin (C + D + E ) then D(p/3) + D¢(p/6) is 79. Let Ak(x) denote a polynomial of degree k where k ≥ 1, and let A3¢ ( x)

( x A2 ( x)) ¢

( x 2 A1 ( x)) ¢

A3¢¢( x)

( x A2 ( x)) ¢¢

( x 2 A1 ( x)) ¢¢

A3¢¢¢ ( x) ( x A2 ( x)) ¢¢¢ ( x 2 A1 ( x)) ¢¢¢ then D¢(100!) is

D=

D= then L – 25 is

x y -1

x- y

2

x y -1

y - x2

x y -1

y - x2

x - y2

1 - sin q -1

sin q 1 - sin q

1 sin q 1

1

2a

a2 3

1 4a 2 2a 1

1

2

- 2b

2 ab

2 ab

1- a + b

2b

- 2a

2

2

a1 a2 a3

b1 b2 b3

c1 c2 c3

then an integer m by which D is necessarily divisible is x -1 2 86. Let D(x) = 2 x + 5

x3 - 1

2 x2 - 5

x3 - 1

2x + 2

x3 + 3

x +1

3x2 - 2

and ax + b

be the remainder when D(x) is divided by x2 – 1, then 4a + b is 87. Let

80. Suppose ab = 1, and L be the least possible value of 1+ a - b

x - y2

then D – 27 is 85. Suppose a1 b1 c1, a2 b2 c2 and a3 b3 c3 are three digit even positive integers and

then f(p/3) is 78. Let A, B, C, D, E pentagon and

2

y - x2

=0

If l is the length of interval of q for which [D(q)] = 2 (where [x] denotes the greatest integer £ x). Then 4l/p is 84. If a

then value of ayz + bzx + cxy + 2018 is

D(x) =

cos q sin q cos q

then f(2, 2) – 20 is 83. Let – p £ q £ p and

(d) none of these

D=

sin q cos q - sin q

is 82. Let

y = Ax (a, b, g) represents

75. If a, b, g

cos q - sin q - cos q

D1 =

1 cos a cos b

cos a 1 cos a

cos b cos g 1

and D2 =

0 cos a cos b

cos a 0 cos g

cos b cos g 0

2a 1 - a 2 - b2

IIT JEE eBooks: www.crackjee.xyz Determinants 8.39

If Δ1 = Δ2, then sin2a + sin2b + sin2g. 88. If 0 < q < p and a denotes the value of q for which the system of equations (sin q) x + y + z = 0 x + (cos q)y + z = 0 (sin q)x + (cos q)y + z = 0 has a non trivial solution, then 2 sin a + 3 cos a is 89. The sum of all the values of l for which the system of equations lx + y + z = 1, x + ly + z = l, x + y + lz =l2 90. If a π b and the system of equations ax + by + bz = 0, bx + ay + bz = 0, bx + by + az = 0 has a non-trivial solution, then the value of a + 2b is LEVEL 2

SINGLE CORRECT ANSWER TYPE QUESTIONS x +1 x + 2 x + a 1. If D = x + 2 x + 3 x + b = 0, then x+3 x+4 x+c the family of lines ax + by + c = 0 passes through (a) (1, –1) (b) (1, –2) (c) (2, –3) (d) (0, 0) 2. If R is the circum-radius of DABC, and D1 =

1 sin A

1 sin B

1 sin C

2

2

2

(c) 4

If (D + 2) is divisible by 10, then the digit at the unit’s place of x is (a) 1 5. Let

(b) 2 sin 2 35∞

tan (55∞ + 80∞)

2

tan (79∞ + 66∞)

sin 2 79∞

cos 2 18∞

cos 2 72∞

cot (72∞ + 63∞)

then D equals (a) – 1

(b) 0

(c) 2

(d)

(

)

5 +1 8

6. Let a, b, c be three real numbers such that 0 < a, b, c < 1. Let

D=

a

b

c

1 - a2 bc

1 - b2 ca

1 - c2 ab

then D equals

(d) none of these.

(d) 8

for 1 £ k £ 3 and F(x) =

sin 2 55∞

D = sin 11∞

7. Let an = 5n + 7n, and D=

x

f 2¢( x) f 2¢¢ ( x) f 2¢¢¢ ( x)

(d) 3

2 2 2 (c) a 1 - a + b 1 - b + c 1 - c

3. Let fk(x) = Ú0 (akt2 + bkt + ck)dt f1¢( x) f1¢¢( x) f1¢¢¢( x)

(c) 8

(b) abc

R3 then D = D (b - c) (c - a) (a - b) 1 (b) 2

D=

(a) 0

sin A sin B sin C

equals (a) 0

x 4 1 y 0 1 z 1 0

f3¢( x) f3¢¢ ( x) f3¢¢¢( x)

The curve y = F(x) is (a) a straight line (b) a parabola (c) an ellipse (d) a hyperbola 4. Let x, y z be three natural numbers such that y has 1 at its unit’s place and z has 0 at its unit’s place. Let

an + 1

an + 2

an + 3

an + 4

an + 5

an + 6

an + 7

an + 8

an + 9

then D equals (a) an + 9 – an

(b) 0

2

(c) an – an+1 an+9

(d) none of these

8. If a2 + b2 + c2 = – 2, and 1 + a2 x 2 f(x) = (1 + a ) x

(1 + b 2 ) x (1 + c 2 ) x 1 + b2 x

(1 + a 2 ) x (1 + b 2 ) x then f(x) is a polynomial of degree (a) 0 (b) 1 (c) 2

(1 + c 2 ) x , 1 + c2 x (d) 3

IIT JEE eBooks: www.crackjee.xyz 8.40 Comprehensive Mathematics—JEE Advanced

9. If a, b, c ∈ R, d = a + b + c, and a Δ = d d

d b d

cos ( x + a ) - sin ( x + a ) cos 2 a sin x cos x sin a f(a)= - cos x sin x cos a

d d c

then Δ equals (a) a3b3c3 (c) d 3 10. If x π a, y π b, z π c and a y z x b z x y c then value of

14. Given that

3

3

= A + B cos 2a then (a) A = 0, B = 1 (b) A = 1, B = 1 (c) A = 1, B = 0 (d) A = 1, B = – 1 15. If p and q are distinct primes, and

3

(b) a + b + c (d) none of these.

= 0,

pq D=

x+a y+b z+c + + is x-a y-b z-c

(a) 0 (b) 1 (c) – 1 (d) 2 2 2 11. Let x = bc – a , y = ca – b , z = ab – c2, r = a2 + b2 + c2, s = bc + ca + ab, and x y z

D1 =

y z x

z x , y

D2 =

then (a) D1 = D2

,

ac

ab

c2 + a2

bc

ac

bc

a 2 + b2

D1 =

D2 =

and

- a2

ab

ac

ab

-b

bc

ac then (a) D1 + D2 = 0 (c) D12 = D2 a2

2

-c

bc

c2

13. If (a + l )

2

(b + l )

2

(a - l )

2

(b - l )

2

then D is (a) an integer (c) non-zero rational

q q +q qi

(b) irrational (d) imaginary

16. If x + y + z = p and

(c) 0

(d) 4

(d) 1

18. If ABC is a triangle and

2

2

a (c + l ) = k l a 1 (c - l ) 2 2

l π 0, then k is equal to: (a) 4 l abc (c) 4 l2

p + p qi

a b c D = b c a = 0, c a b then value of cos2 A + cos2 B + cos2 C equals (a) 1/4 (b) 1/2 (c) 3/4 a2 b sin A c sin A 1 cos A D = b sin A 1 c sin A cos A

(b) D1 = D2 (d) D1 = D22

b2

q p

p qi

17. If for a triangle ABC,

12. Let ab

p + p pi q p +

then D equals (a) 3 (b) – 1

(d) D1 + D2 = 0

b2 + c2

p q

pqi

sin 3 x sin 3 y sin 3 z D = sin x sin y sin z cos x cos y cos z

(b) D12 = D2

D 22

(c) D1 =

r s s s r s s s r

q+

pi

(b) –4 l abc (d) –4 l2

2

b b 1

2

c c 1

then Δ equals to (a) 0 (b) – 1

(c) 1

(d) abc

19. Let 0 £ x < 4, – 2 £ y < 3 and – 1 £ z < 5. If [a] denotes the greatest integer £ a possible value of [ x + 2] [ y ] [ z] [ x] [ y + 1] [ z ] D= [ x] [ y ] [ z + 1]

IIT JEE eBooks: www.crackjee.xyz Determinants 8.41

is (a) 13 (b) 15 (c) 17 (d) 19 20. If DABC is not a right triangle, then value of tan A 1 1 1 tan B 1 D= 1 1 tan C is (a) – 1

(b) 2

(c) 3

(d) 0

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 21. Let 0 < q < p/2 and tan q -x 1

x D(x, q) = - tan q cot q

cot q 1 x

then (a) D(0, q) = 0 (b) D(x, p/4) = – x3 + x (c)

D(1, q) = 0

Min

0 0 38. Suppose a, b, c > 0 and let 2 x2 + a2

x 2 + bx + ay

ay + xy + cx

2 2 2 2 D(x, y) = x + bx + ay b + x + y

y 2 + by + cx

ay + xy + cx

y 2 + by + cx

c2 + 2 y 2

IIT JEE eBooks: www.crackjee.xyz Determinants 8.43

Statement 1: D(–1, –2) ≥ 0 Statement 2: D(x, y) ≥ 0 ∀ x, y Œ R

COMPREHESION-TYPE QUESTIONS Paragraph for Question Nos. 39 to 42 a11 If D = a21 a31

a12 a22 a32

and cij = (–1)i+j (determinant obtained by deleting ith row and jth column), then

39. If

1

x

x2

x

2

1

1

x

x

x

2

Sometimes it is easier for us to evaluate a determinant. For instance, if Sk = ak + bk + ck, then S0 S1 S2

a13 a23 a33

c11 c12 c21 c22 c31 c32

Paragraph for Question Nos. 43 to 46

c13 c23 c33

= D 2.

0

x - x4

x - x4

x3 - 1

D=

B C A b c a

x - x4 x3 - 1 , then 0

C A = 49, B c a equals b

(a) – 7 (b) 2401 (c) – 2401 (d) 7 3 3 41. If a + b + c3 – 3abc = – 5 and A, B, C are as in Question 40, value of aA + bB + cC is (a) – 5 (b) 5 (c) 25 (d) – 25 42. If A, B, C are as in Question 40, 3ABC – A3 – B3 – C3 = p2 and b c a D= c a b a b c 3

3

1 b

1 c

43. If a1 > a2 > a3, b1 > b2 > b3 and ai bj π 1 for 1 £ i, j £ 3, then the determinant

(a) D = 7 (b) D = 343 (c) D = – 49 (d) D = 49 40. Suppose a, b, c ∈ R, a + b + c > 0, A = bc – a2, B = ca – b2 and C = ab – c2, A B and C a then b c

1 a

a 2 b2 c2 which is quite easy to evaluate.

D= 0

S2 S3 can be written as D2 S4

where D =

= 7 and

x3 - 1

S1 S2 S3

then sin (D) + cos (D) equals (a) 1 (b) – 1 (c) 0 (d) none of these

1 - a13b13 1 - a1b1

1 - a13b23 1 - a1b2

1 - a13b33 1 - a1b3

1 - a23 b13 1 - a2b1

1 - a23 b23 1 - a2b2

1 - a23 b33 1 - a2 b3

1 - a33b13 1 - a3b1

1 - a33 b23 1 - a3b2

1 - a33 b33 1 - a3 b3

is (a) positive (b) non-negative (c) negative (d) non-positive 44. If A, B, C, P, Q, R, ∈ R, then cos ( A + P) cos ( A + Q) cos ( A + R) D = cos ( B + P) cos ( B + Q) cos ( B + R) cos (C + P) cos (C + Q) cos (C + R) equals (a) cos (A + B + C) + cos (P + Q + R) (b) cos (A + P) cos (B + Q) cos (C + R) (c) cos A cos P + cos B cos Q + cos C cos R (d) 0 a a+b a+b+c 45. D = 3a 4a + 3b 5a + 4b + 3c then D is equal to 6a 9a + 6b 11a + 9b + 6c (a) abc (b) a2b2c2 (c) a3 + b3 + c3 (d) a3 46. If ak, bk, ck ∈ R for k = 1, 2, 3 and a1 D = a2 a3

b1 b2 b3

c1 c2 c3

< 0,

then sum of the roots of the equation

IIT JEE eBooks: www.crackjee.xyz 8.44 Comprehensive Mathematics—JEE Advanced

a1 + ib1 x ia1 x + b1 a f (x) = 2 + ib2 x ia2 x + b2 a3 + ib3 x ia3 x + b3 is (a) a1 + a2 + a3 (c) a1b1 + a2b2 + a3b3

c1 c2 c3

51. Let a1 π 0 and A1, B1, C1 ... be respectively the cofactors of the elements a1, b1, c1 ... of

=0

D=

(b) b1 + b2 + b3 (d) 0

x2

1 + x3

2x 4x2

1 + 8 x3

2

1 + 27 x

x

3x 9 x

= 10 is

3

1 p p 48. Suppose x = - (1 + 7i ) and y = cos + i sin . 3 4 4 1 x Let D = 1 x + y 1 x

x y x+ y

D

2

equals

49. If ab = 25, then the point where x a a D(x) = b x a b b x

1 B2 a1 B3

D=

2 5

15 + 26

10

5

3 + 65

5

15

a2 – a3

53. The value of l for which the system of equations (1 + l)x1 + x2 + x3 = 1 x1 + (1 + l)x2 + x3 = 1 x1 + x2 + (1 + l)x3 = 1 54. If p, q, r, b, c are non zero real numbers such that p + q + r = 16, b + c = 12, then value of x + y + z if x, y, z satisfy the system of equations: x y z + + =1 p p-b p-c

)

5

C2 C3

1 a3 a2 2 a2 - x is a3 f(x) = 1 a2 1 2 a3 - x

50. Value of D 5 3 - 3 2 where 13 + 3

c1 c2 c3

52. If a2, a3 ∈ R value of

has a local minimum is

(

b1 b2 b3

If D

INTEGER-ANSWER TYPE QUESTIONS 47. The total number of distinct x ΠR for which

a1 a2 a3

x y z + + =1 q q-b q-c x y z + + =1 r r-b r-c

is

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS

2. If ω (π 1) is a cube root of unity, then 1

1. The equations x + 2y + 2z = 1 and 2x + 4y + 4z = 9 have (a) only one solution (b) only two solutions (d) none of these

[1979]

1+ i + w2

w2

-1 w2 -1 = Δ = 1- i -i -i + w 1 -1 (a) 0

(b) 1

(c) i

(d) w.

[1995] 3. Let a, b, c be the real numbers. Then following

IIT JEE eBooks: www.crackjee.xyz Determinants 8.45

system of equations in x, y and z x

2

a

2

+

y

2

b

2

-

z

2

c

2

= 1,

x

2

a

2



x2 a2

y

2

b

2

+

+ y2

b2

z

2

= 1,

c2 +

z2 c2

= 1 has

(a) no solution (b) unique solution

sin x cos x cos x Δ = cos x sin x cos x = 0 in the interval cos x cos x sin x

4. The determinant xp + y Δ = yp + z 0

x y xp + y

–p/4 £ x £ p/4 is (a) 0 (b) 2

y z = 0, if yp + z

(a) x, y, z are in A.P. (b) x, y, z are in G.P. (c) x, y, z are in H.P. (d) xy, yz, zx are in A.P. [1997] 5. The parameter, on which the value of the determinant 1 a a2 Δ = cos( p - d ) x cos px cos( p + d ) x sin( p - d ) x sin px sin( p + d ) x does not depend upon is (a) a (b) p (c) d

p2

1

Δ = 1 -1 - w w

2

1 2

w2 w

(b) x = 1, y = 3 (d) x = 0, y = 0 [1998]

1 x x +1 2x x( x - 1) ( x + 1) x 8. If f(x) = 3x( x - 1) x( x - 1)( x - 2) x( x + 1) ( x - 1)

(d) – 100

is

4

(c) 3w2 (d) 3w(1 – w) [2002] 13. If the system of homogeneous equation x + ay = 0, az + y = 0 and ax + z solutions, then the value of a is (a) – 1 (b) 1 (c) 0 (d) no real values [2003] 14. The value of λ for which the system of equations x + y + λz = 4, x – 2y + z + 4 = 0, 2x – y – z = 2 has no solution is (a) – 3 (b) – 2 (c) 0 (d) 3 [2004] 15. Let P = [aij] be a 3 ¥ 3 matrix and let Q = [bij], where bij = 2i+jaij for 1 £ i, j £ 3. If the determinant of P is 2, then the determinant of the matrix Q is (b) 211 (a) 210 (c) 212 (d) 213 [2012] (a) zero (b) 3w(w – 1)

p3

(c) 100

kx + (k + 3)y = 3k – 1

1

6i - 3i 1 3i -1 = x + iy, then 7. If 4 20 3 i

then f(100) equals (a) 0 (b) 1

[2001] 11. The number of values of k for which the system of equations (k + 1)x + 8y = 4k

1

(d) x

where p is a constant. Then f ¢¢¢(0) equals (a) p (b) p + p2 (c) p + p3 (d) independent of p

(a) x = 3, y = 1 (c) x = 0, y = 3

(d) 3

[2002]

sin x cos x -1 0

p

(c) 1

1 12. Let w = (- 1 + 3i ), then value of 2

[1997] x3 6. Let f(x) = 6

[1999] 9. If the system of equations x – ky – z = 0, kx – y – z = 0, x + y – z = 0 has a non-zero solutions, then possible value of k are (a) – 1, 2 (b) 1, 2 (c) 0, 1 (d) – 1, 1 [2000] 10. The number of distinct real roots of

IIT JEE eBooks: www.crackjee.xyz 8.46 Comprehensive Mathematics—JEE Advanced

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

Statement-2: The determinant

a b aa +b b c b a + c is 1. The determinant Δ = 0 aa + b ba + c equal to 0, if (a) a, b, c are in A.P. (b) a, b, c are in G.P. (c) a, b, c are in H.P. (d) a is a root of the equation ax2 + bx + c = 0 (e) (x – a) is a factor of ax2 + 2bx + c [1986] 2. The value of q lying between q = 0 and θ = p/2 and satisfying the equation

Δ=

1 + sin 2 q

cos 2 q

4 sin 4q

sin q

1 + cos q

4 sin 4q

sin q

cos q

1 + 4 sin 4q

2

2

2

(a)

7p 24

2

(b)

5p 24

(c)

11p 24

= 0 are

(d)

p 24

È1 4 Í P is, Í2 1 ÍÎ1 1

4˘ 7 ˙˙ , 3˙˚

then the possible value(s) of the determinant of P is(are) (a) –2 (b) –1 (c) 1 (d) 2 [2012] 4. Which of the following values of α satisfy the equation

INTEGER-ANSWER TYPE QUESTIONS 2p 2p + i sin . 3 3 z

1. Let w z +1

w

w2

w

z + w2

1

w

1

z +w

satisfying

2

2 k˘ ˙ -2k ˙ and ˙ -1 ˙˚

È 2k - 1 2 k Í A= Í 2 k 1 Í ÍÎ -2 k 2k È 0 Í B = Í 1 - 2k Í ÍÎ - k

2k - 1 0 -2 k

x

x2

1 + x3

(1 + 3a )2

2x 4x2

1 + 8 x3

(2 + a )2

(2 + 2a ) 2

(2 + 3a )2 = 648α ?

3x 9 x

2

1 + 27 x

(3 + a )2

(3 + 2a )2

(3 + 3a ) 2

FILL [2015]

ASSERTION-REASON TYPE QUESTIONS 1. Consider the system of equations x – 2y + 3z = – 1 – x + y – 2z = k x – 3y + 4z = 1 Statement-1: The system of equations has no solution for k π 3.

k ˘ ˙ 2 k ˙. ˙ 0 ˙˚

If det (adj A) + det (adj B) = 106, then [k] is equal to [Note : adj M M and [k] denotes the largest integer less than or equal to k]. [2010] 3. The total number of distinct x ∈ R for which

(1 + 2a ) 2

(b) 9 (d) 4

= 0 is equal to [2010]

(1 + a )2

(a) –4 (c) –9

[2008]

2. Let k be a positive real number and let

[1988] 3. If the adjoint of a 3 ¥

1 3 -1 -1 -2 k π 0, for k π 3. 1 4 1

IN THE

= 10 is

[2016]

3

BLANKS TYPE QUESTIONS

1. Let l 2 + 3l pλ4 + qλ3 + rλ2 + sλ + t = l + 1 l -3

l -1 l + 3 -2l l - 4 l + 4 3l

be an identity in l where p, q, r, s, t are constants. Then the value of t is ______ [1981] 2. The solution set of the equation

IIT JEE eBooks: www.crackjee.xyz Determinants 8.47

1 4 Δ = 1 -2

SUBJECTIVE-TYPE QUESTIONS

20 5 = 0 is ______

1 2 x 5x

[1981]

2

x 3 7 3. Given that x = – 9 is a root of 2 x 2 = 0 the 7 6 x other two roots are ______ and ______

[1983]

4. The system of equations lx + y + z = 0, – x + ly + z = 0, – x – y + lz = 0 will have a non-zero solution if real value of λ are given by ______ [1984] sec x

sec 2 x + cot x cosec x

cos x

2 2 5. Let f(x) = cos x cos x

1

cos 2 x

cos ec2 x cos 2 x

p /2

Then Ú f ( x)dx = ______

[1987]

0

1 a a 2 - bc Δ = 1 b b - ac is ______ 2

[1988]

1 c c 2 - ab 7*. For positive numbers x, y and z, the numerical value of the determinant 1 log x y log x z 1 log y z is ______ Δ = log y x

[1993]

a b c Δ = b c a is negative c a b

2 Δ = 2 x + 3x - 1

x + 2x + 3

1

1 a a2 1 a bc 2 Δ = 1 b ca and D¢ = 1 b b 1 c ab 1 c c2 tically equal.

are not iden-

[1983] b1 1 b2 1 b3 1

then the two triangles with vertices (x1, y1), (x2, y2), (x3, y3), and (a1, b1), (a2, b2), (a3, b3) must be congruent. [1985] * x, y, z should be different from 1.

[1981]

show that

2

1. The determinants

a1 y1 1 y2 1 = a2 a3 y3 1

[1979] 3. For what values of m does the system of equations 3x + my = m and 2x – 5y = 20 has a solution satisfying the conditions x > 0, y > 0. [1980] 4. Find the solution set of the system x + 2y + z = 1 2x – 3y – w = 2 x ≥ 0, y ≥ 0, z ≥ 0, w ≥ 0 [1980]

x2 + x

TRUE/FALSE TYPE QUESTIONS

x1 2. If x2 x3

For that value of k

5. Let a, b, c be positive and not all equal show that the value of the determinant

6. The value of the determinant

log z x log z y

1. Given x = cy + bz, y = az + cx, z = bx + ay where x, y, z are not all zero, prove that a2 + b2 + c2 + 2abc = 1. [1978] 2. For what value of k does the following system of equations posses a non-trivial solution over the set of rationals. x + ky + 3z = 0 3x + ky – 2z = 0 2x + 3y – 4z = 0

x +1 3x

x-2 3x - 3 = xA + B

2x - 1 2x - 1

where A and B are determinants of order 3 not involving x. [1982] 7. Show that the system of equations 3x – y + 4z = 3 x + 2y – 3z = –2 6x + 5y + λz = –3 has at least one solution for any real number l π – 5. Find the set of solutions if l = – 5 [1983] 8. Let α be a repeated root of a quadratic equation f(x) = 0 and A(x), B(x), C(x) be polynomials of degree 3, 4 and 5 respectively. Then show that

F(x) =

A( x) B( x) C ( x) A(a ) B(a ) C (a ) is divisible by f(x) A¢ (a ) B ¢ (a ) C ¢ (a )

IIT JEE eBooks: www.crackjee.xyz 8.48 Comprehensive Mathematics—JEE Advanced

where prime denotes the derivatives

9. Show that Δ =

x

Cr

x

y

Cr

z

Cr

=

x

Cr

y

Cr

z

Cr

[1984]

Cr + 1

Cr + 2

y

Cr + 1

y

Cr + 2

x + (cosa)y + (sina)z = 0

z

Cr + 1

z

Cr + 2

– x + (sina)y – (cosa)z = 0

x +1 y +1 z +1

Cr + 1

x+2 y +2

Cr + 1 Cr + 1

z+2

Cr + 2 Cr + 2 Cr + 2 [1985]

10. Consider the system of equations in x, y, z: (sin 3q) x – y + z = 0 (cos 2q)x + 4y + 3z = 0 2x + 7y + 7z = 0

11. Let Δa =

n

(a - 1)

2

(a - 1)

3

2n

4n - 2

3n

3

3n - 3n

[1993]

16. For all A, B, C, P, Q, R show that cos( A - P ) cos( A - Q) cos( A - R) Δ = cos( B - P ) cos( B - Q) cos( B - R) = 0 cos(C - P) cos(C - Q) cos(C - R) 17. For a > 0, d

6 2

has a non-trivial solution. For l values of a.

[1994]

Find the values of q for which the system has a non-trivial solution. [1986] a -1

15. Let l and a be real. Find the set of all values of l for which the system of linear equations lx + (sina)y + (cosa)z = 0

x

. Show that

2

1 a 1 Δ = a+d 1 a + 2d

1 a(a + d ) 1 (a + d )(a + 2d ) 1 (a + 2d )(a + 3d )

1 (a + d )(a + 2d ) 1 (a + 2d )(a + 3d ) 1 (a + 3d )(a + 4d )

n

 D a = c, a constant.

a =1

12. Let the three digit numbers A28, 3B9 and 62C, where A, B and C are integers between 0 and 9, be k. Show that the determinant A 3 6 8 9 C 2 B 2 is divisible by k. 13. If a π p, b π q, c π r and

18. Find the value of the determinant bc ca ab p q r 1 1 1 where a, b and c are respectively the pth, qth and rth terms of a harmonic progression. [1997] 19. Prove that for all values of q

[1990]

p b c Δ= a q c =0 a b r E=

[1996]

[1989]

p q r + + p-a q-b r-c [1991] n, if

cos q

sin q

2p ˆ Ê sin Á q + ˜ Ë 3¯

2p ˆ Ê cos Á q + ˜ Ë 3¯

4p ˆ Ê sin Á q + ˜ Ë 3¯

2p ˆ Ê sin Á q ˜ Ë 3¯

2p ˆ Ê cos Á q ˜ Ë 3¯

4p ˆ Ê sin Á q ˜ Ë 3¯

=0

[2000] 20. Let a, b, c be real numbers with a + b + c2 = 1. Show that the equation 2

2

ax - by - c bx + ay cx + a bx + ay - ax + by - c cy + b Δ= = 0. cx + a cy + b - ax - by + c

n! (n + 1)! (n + 2)! D = (n + 1)! (n + 2)! (n + 3)! (n + 2)! (n + 3)! (n + 4)! Ê D ˆ - 4˜ is divisible by n show that Á 3 Ë (n !) ¯

sin q

represents a straight line. [1992]

[2001]

IIT JEE eBooks: www.crackjee.xyz Determinants 8.49

58.

Answers LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17. 21. 25. 29. 33.

(a) (d) (a) (d) (a) (b) (c) (a) (a)

2. 6. 10. 14. 18. 22. 26. 30. 34.

(d) (d) (a) (d) (c) (a) (d) (a) (a)

3. 7. 11. 15. 19. 23. 27. 31. 35.

(a) (d) (d) (a) (a) (d) (a) (a) (b)

4. 8. 12. 16. 20. 24. 28. 32.

59. (d) (a) (d) (a) (d) (b) (b) (d)

60.

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 36. 38. 40. 42. 44. 46. 48. 51. 53. 55.

(a), (d) (a), (a), (a), (a), (a) (a), (a), (a),

(c) (b), (b), (b), (b), (b), (b), (b),

37. 39. (c) 41. (c), (d) 43. (c) 45. (c) 47. 49. (a), (c) 50. (c) 52. (c), (d) 54. (c), (d)

(a), (b), (c) (a), (b), (c) (a), (b), (c) (a), (b), (c), (d) (a), (b), (c), (d) (a), (b), (c), (d) (a), (b), (c) (b) (a), (d)

MATRIX-MATCH TYPE QUESTIONS 56.

ASSERTION-REASON TYPE QUESTIONS 61. (a) 65. (c)

62. (d)

63. (b)

64. (a)

COMPREHENSION-TYPE QUESTIONS 66. (b) 70. (b) 74. (d)

67. (c) 71. (a) 75. (a)

68. (a) 72. (b)

69. (a) 73. (d)

INTEGER-ANSWER TYPE QUESTIONS 76. 80. 84. 88.

1 2 3 2

77. 81. 85. 89.

1 2 2 1

78. 82. 86. 90.

0 5 3 0

79. 0 83. 2 87. 2

LEVEL 2

57.

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 4. 7. 10.

(b) (d) (b) (c)

2. 5. 8. 11.

(d) (b) (c) (a)

3. 6. 9. 12.

(a) (d) (d) (b)

IIT JEE eBooks: www.crackjee.xyz 8.50 Comprehensive Mathematics—JEE Advanced

13. (c) 16. (c) 19. (c)

14. (b) 17. (c) 20. (b)

INTEGER-ANSWER TYPE QUESTIONS

15. (d) 18. (a)

47. 2 50. 5 53. 0

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 21. 23. 25. 27. 29.

(a), (a), (a), (a), (a),

(b), (b), (b), (c), (b)

(c) (c), (d) (c) (d)

22. 24. 26. 28. 30.

(a), (a), (a), (a), (a),

(b), (c) (b) (b), (c), (d) (b), (c) (c)

MATRIX-MATCH TYPE QUESTIONS p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

31.

32.

33.

34.

36. (d)

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWERS TYPE QUESTIONS 1. 5. 9. 13.

(d) (b) (d) (a)

1. (b), (e)

2. 6. 10. 14.

(a) (d) (c) (b)

3. 7. 11. 15.

(d) (d) (b) (d)

4. (b) 8. (a) 12. (b)

2. (a), (c)

3. (a), (d)

4. (b), (c)

ASSERTION-REASON TYPE QUESTIONS 1. (a)

INTEGER ANSWER TYPE QUESTIONS 1. 0

FILL

2. 4 IN THE

3. 2

BLANKS TYPE QUESTIONS

1. 0 2. {–1, 2} 5. –(8 + p)/15

3. 2, 7 6. 0

4. 0 7. 0

TRUE FALSE TYPE QUESTIONS 1. False

2. False

SUBJECTIVE TYPE QUESTIONS

37. (c)

40. (a) 43. (a) 46. (d)

2. 3. 4. 7. 10. 13. 15.

38. (a)

COMPREHENSION-TYPE QUESTIONS 39. (d) 42. (b) 45. (d)

49. 5 52. 9

MULTIPLE CORRECT ANSWERS TYPES QUESTIONS

ASSERTION-REASON TYPE QUESTIONS 35. (b)

48. 5 51. 3 54. 4

41. (a) 44. (d)

17.

k = 33/2, x = 5l/2, y = –l/3, z = l, l Œ Q m > 30 or m < –15/2 x = 1, y = 0, z = 0, w = 0. x = (4 – 5k)/7, y = (13k – 9)/7, z = k, k Œ R q = np or q = mp + (–1)m p/6, n, m Œ I. 2 l = cos 2a + sin 2a; a = np or a = mp + p/4, n, m ŒI 4d 4 a(a + d ) 2 (a + 2d )3 (a + 3d ) 2 (a + 4d )

18. 0

IIT JEE eBooks: www.crackjee.xyz Determinants 8.51

Hints and Solutions LEVEL 1

10. Use a = ARp-1 fi log a = log A + (p - 1) log R etc. and apply C1 Æ C1 - log RC2 - log AC3.

12. 1 x

2

x

1. D = x y z 1 y

2

y3

1 z2

z3

n n n

3

= xn yn zn (x – y) (y – z) (z – x) (xy + yz + zx) fi

D 1 1ˆ n +1 n +1 n +1 Ê 1 =x y z Á + + ˜ Ë x y z¯ ( x - y )( y - z )( z - x)

\ n + 1 = 0 or n = –1 2 Write D = D1 D2 where 1 1 D1 = a + b c + d ab cd 1 and D2 = c + d cd

0 0 0

1 0 a+b 0 ab 0

\D=0 3. Apply C2 Æ C2 - C1, C3 Æ C3 - C1 to obtain 1 0 0 1 0 sin q D= = sin q cos q 1 0 cos q = (sin 2q)/2 4. Use C1 Æ C1 + C2 + C3. 5. Use C1 Æ C1 - C2 6. Use C1 Æ C1 – g(q) C2 – f(q) C3 7. Use C1 Æ C1 + C2 + C3 8. Use C2 Æ

1 C abc 2

1 1 C1 and use = A + (p - 1) D etc. abc a Take b - a common from C1 and C3 and then use C2 Æ C 2 - C 1 + C 3. Take – qi comman from C2. Use a31 = 100a + 30 + 1 etc. and R2 Æ R2 - 100R3 - 10R2 and then C1 Æ C1 - 2C2 + C3. Apply C3 Æ C3 - C2 and C2 Æ C2 - C1 and simplify to obtain value of determinant as 33.

11. Apply C1 Æ

13. 14. 15.

16. z is a skew symmetric determinant. 17. Use laws of projection from trigonometry. 18. Use C1 Æ C1 - C2 - C3 to obtain f(x) = sin 3x. 19. D = (a + b + g) [(a + b + g)2 - 3(bg + ga + ab)] 20. Use the fact

21. D = (tan x + 2 cot x) (tan x - cot x)2 22. Simplify to obtain x + iy = - 1 23. Take the conjugate of the determinant to obtain x + iy = x – iy fi y = 0 24. D = x3 + (a2 + b2 + c2)x 25. First show that a = b = 1, then put a = B - C, b = C - A, g = A - B and write D = D1 D1 cos A sin A 0 where D1 = cos B sin B 0 = 0 cos C sin C 0 C2 and C3 26. Apply C3 Æ C3 - C1 are proportional. 27. Apply C2 Æ C2 - sin2 x C1 and simplify to obtain f(x) = - sin5 x - cos2 x 28. By applying the operation R1 Æ R1 - R3 and R2 Æ R2 - R3, we get 1 0

followed by C3 Æ C3 + C2 9. Use C1 Æ C1 – C2 – C3 D=

1 log a b log a c logb a 1 logb c log c a log c b 1

Now use change of base.

1 1 1 + + = 0. a b g

sin q 2



1 0 sin 2 q

0 1

-1 -1

=0

cos q 1 + 4 sin 6q 2

0 1

0 -1

cos 2 q 1 + 4 sin 6q + sin 2 q

=0

IIT JEE eBooks: www.crackjee.xyz 8.52 Comprehensive Mathematics—JEE Advanced

[use C3 Æ C3 + C1] fi 1 + 4 sin 6q + sin2 q + cos2 q = 0 -1 fi 4 sin 4q = - 2 fi sin 6q = 2 29. D =

p+a b c a q+b c =0 a b r+c

Applying R2 Æ R2 - R1 and R3 Æ R3 - R1, we get p+a b c -p q 0 -p 0 r

For l = - 3, sum of the last two equation gives 2x - y - 2z C3, the given determinant equals sin b cos b sin a cos a (1) - (1) + sin g cos g sin g cos g (1)

Divide by pqr to obtain

a b c + + = - 1. p q r

a b aa + b b c ba + c = 0 aa + b ba + c 0 Applying R3 Æ R3 - aR1 - R2 and simplify to show that a, b, c are in G.P. fi A, B, C are in A.P.

31.

2 -1 - 2 35. D = 1 - 2 1 = - 3(3 + l) l 1 1

=0

fi pqc + [q(p + a) + bp]r = 0

30.

fi 2l2 + l + 2 = 0

a a -1 b -1 b = 0 -1 c c

= 2 sin

A+ B A- B C C cos + 2 sin cos 2 2 2 2

A- B C C Ê Cˆ = 2 sin ÁË - ˜¯ cos + 2 sin cos 2 2 2 2

a(b + 1) (c + 1) – b(a + 1) (c + 1) - (a + 1) (b + 1) =0 Divide by (a + 1) (b + 1) (c + 1). l +5 l -4 1 32. Let D = l - 2 l + 3 1 1 l l Apply R1 Æ R1 - R3, R2 Æ R2 - R3 to get

[A + B = - C] = - 2 sin

A-B A + Bˆ CÊ - cos Á cos ˜ 2Ë 2 2 ¯

= - 2 sin

CÊ A Bˆ Á 2 sin sin ˜¯ 2Ë 2 2

= - 4 sin

A B C sin sin 2 2 2

= - 4 sin

b -g g -a a-b sin sin 2 2 2

37. Use C1 Æ C1 - C2 - 4C3 =7π0

33. Add the three equations.

34.

cos a cos b

= (sin b cos g - sin g cos b) - (sin a cos g - sin g cos a) + (sin a cos b - sin b cos a) = sin (b - g) + sin (g - a) + sin (a - b), proving the answer (c). But (a) is also correct because, by putting A = b - g, B = g - a and C = a - b, we have A + B + C = 0, so that sin (b - g) + sin (g - a) + sin (a. - b) = sin A + sin B + sin C

Apply C2 Æ C2 - C1 and C3 Æ C3 - C1

5 -4 0 -2 3 0 l l 1

sin a sin b

l +3 l +2 1 3 l +3 1 =0 2 3 2

39. D = D1 D2 where

D1 =

a1 a2 a3

b1 b2 b3

c1 c2 c3

IIT JEE eBooks: www.crackjee.xyz Determinants 8.53

D2 =

and

1

x2

0

x2 0

1 0

0 1

40. Multiplying C2 by 2 and doing C3 Æ C3 + C1, the given determinant becomes sin a 1 sin 2 b 2 sin 2 g 2

sin 2a 1

sin (a - b) cos a cos b + sin (b - g) cos b cos g + sin (g - a) cos g cos a =

1 [sin (2b - 2g) + sin (2g - 2a) + sin (2a - 2b)], 4

proving the answer (b). 41. Multiplying C1 by a, C2 by b and C3 by c, the given determinant becomes

sin 2b 1 , sin 2g

We can derive the result

1

and after R2 Æ R2 - R1 and R3 Æ R3 - R1, it is

a3 + a (b 2 + c 2 ) cos f

ab 2 (1 - cos f )

ba 2 (1 - cos f )

b3 + b (c 2 + a 2 ) cos f

ca 2 (1 - cos f )

cb 2 (1 - cos f )

1 abc

1 sin a sin 2a 1 sin ( b - a ) sin ( b + a ) 2 cos ( b + a ) sin ( b - a ) 0 2 sin (g - a ) sin (g + a ) 2 cos (g + a ) sin (g - a ) 0 2

C3 to give sin (b + a ) cos (b + a ) 1 (2) sin (b - a) sin(g - a) sin (g + a ) cos (g + a ) 2 = sin (b - a) sin (g - a) [sin (b + a) cos (g + a) - sin (g + a) cos (b + a)]

ac 2 (1 - cos f ) bc 2 (1 - cos f ) c3 + c (a 2 + b 2 ) cos f Now take a, b and c common from R1, R2 and R3, respectively, to give a 2 + (b 2 + c 2 ) cos f

b 2 (1 - cos f )

a 2 (1 - cos f )

b 2 + (c 2 + a 2 ) cos f

a 2 (1 - cos f )

b 2 (1 - cos f )

abc abc

= sin (b - a) sin (g - a) sin (b + a - g - a) = - sin (b - g) sin (g - a) sin (a - b),

c 2 (1 - cos f )

proving the answer (a). Now, as shown in Problem 36, if A + B + C = 0,

c 2 (1 - cos f )

sin A + sin B + sin C = - 4 sin

A B C sin sin . 2 2 2

c 2 + (a 2 + b 2 ) cos f After C1 Æ C1 + C2 + C3, this becomes 1

In this case, therefore, we have - sin (b - g) sin (g - a) sin (a - b) =

1 [sin (2b - 2g) + sin (2g - 2a) + sin (2a - 2b)], 4

which shows that answer (c) is correct. Finally, noting that sin (a - b) cos a cos b =

1 sin (a - b) (2 cos a cos b) 2

1 = sin (a - b) [cos (a + b) + cos (a - b)] 2 1 = [2 sin (a - b) cos (a + b) + 4 2 sin(a - b) cos (a - b)] 1 [sin 2a - sin 2b + sin (2a - 2b)], = 4

b 2 (1 - cos f )

c 2 (1 - cos f )

2 2 2 = 1 b + (c + a ) cos f

1

b 2 (1 - cos f )

c 2 (1 - cos f ) c 2 + (a 2 + b 2 ) cos f

since a2 + b2 + c2 = 1. R2 Æ R2 - R1 and R3 Æ R3 - R1 now gives 1

b 2 (1 - cos f )

c 2 (1 - cos f )

0 (a 2 + b 2 + c 2 ) cos f 0

0 (a + b + c 2 ) cos f 2

0

2

C1, we get cos f 0 0 cos f

= cos2 f,

i.e., the given determinant depends on f only. 42. Use R3 Æ R3 – cosd R1 – sind R2.

IIT JEE eBooks: www.crackjee.xyz 8.54 Comprehensive Mathematics—JEE Advanced

43. Use R2 Æ R2 + R3 and sum to product identities to obtain sin f cos f 2 f 2 p 3 2 f sin cos ( / ) cos cos (2p /3) D= sin (f - 2p /3) cos (f - 2p /3) sin 2 f 2 sin (2f ) cos (4p /3) sin (2f - 4p /3) But cos (2p/3) = cos (p - p/3) = - cos (p/3) = - 1/2. and cos (4p/3) = cos (p + p/3) = - cos (p/3) = - 1/2. 44. First take a, b, c common from C1, C2, C3 and then from R1, R2, R3 D(x) = 3a2b2c2x2 (3 + x)

f(x) = 2 cosx

C1, we get 2 cos x 1 1 0 – 1 2 cos x 1 2 cos x

= (2 cos x) (4 cos2 x - 1) - 2 cos x = 8 cos3 x - 4 cos x 46. Multiplying C3 by sin b, the given determinant is cos a 1 - sin a D= sin b 0

sin a cos b sin a sin b cos a cos b cos a sin 2 b sin b cos b - sin b 2

Then C3 Æ C3 + (cos b)C2 gives D = D1/sin b, where cos a sin a cos b D1 = - sin a cos a cos b 0 - sin b sin a (cos 2 b + sin 2 b ) cos a (cos 2 b + sin 2 b ) 0 cos a 1 - sin a D= sin b 0

sin a cos b cos a cos b - sin b

sin a cos a . 0

R3, the determinant is cos a 1 (sin b ) D= - sin a sin b

sin a cos a

= cos2 a + sin2 a = 1. 47. By the law of sines we can write sin A = ak, sin B = bk and sin C = ck, so that the given determinant equals

a2 bak cak

bak cak 1 2 1 cos A = a bk ck cos A 1

=a

2

bk ck 1 cos A cos A 1

1 sin B sin C sin B 1 cos A . 1 sin C cos A

After C2 Æ C2 - (sin B) C1 and C3 Æ C3 - (sin C) C1, this becomes

a

2

1

0

0

sin B

1 - sin B

cos A - sin B sin C

2

sin C cos A - sin B sin C

1 - sin 2 C

R1 to give a [cos B cos C - (cos A - sin B sin C)2]. 2

2

2

(1) But since A, B and C are the angles of a triangle, we have cos A = cos [p - (B + C)] = - cos (B + C) = - (cos B cos C - sin B sin C) fi cos A - sin B sin C = - cos B cos C. Putting this back in (1), we see that the value of the determinant is 0. 48. Doing C2 Æ C2 - C1 and C3 Æ C3 – C1 to the given determinant, we get 1 1 + sin A

0 sin B - sin A

0 sin C - sin A

,

sin A + sin A sin B - sin A sin C - sin A 2

2

2

2

2

which, after taking sin B - sin A common from C2 and sin C - sin A common from C3, becomes (sin B - sin A) (sin C - sin A) 1 1 + sin A

0 1

0 1

sin A + sin 2 A sin B + sin A sin C + sin A = (sin B - sin A) (sin C - sin A) (sin C - sin B). Since the determinant is zero, we must have sin B = sin A or sin C = sin A or sin C = sin B, i.e., B = A or C = A or C = B. In all three cases we will have an isosceles triangle. x)m = g(x) (say). The x in each of these terms is then m

IIT JEE eBooks: www.crackjee.xyz Determinants 8.55

= g¢ determinant D(x) is D¢(0). Now, a1 b1 (1 + x)

a1 b1 - 1

(1 + x)

x in the given of equations is a1b2

(1 + x)

a1b3

D=

a b -1 (1 + x) a2b2 (1 + x) a2b3 D¢(x) = a2 b1 (1 + x) 2 1

a3 b1 (1 + x) a3 b1 - 1 (1 + x) a1b1

(1 + x) a3b2 (1 + x) a3b3

a1 b2 (1 + x) a1 b2 - 1

(1 + x) a1b3

ab + (1 + x) 2 1

a2 b2 (1 + x) a2 b2 - 1 (1 + x) a2b3

(1 + x) a3b1

a3 b2 (1 + x) a3 b2 - 1 (1 + x) a3b3

(1 + x) a1b1

(1 + x) a1b2

ab + (1 + x) 2 1

(1 + x)a3b1

a1 b3 (1 + x) a1 b3 - 1

which, after R3 Æ R3 - (cos b)R1 + (sin b) R2, becomes cos a - sin a 1 1 D = sin a cos a 0 0 1 + sin b - cos b

(1 + x)a2b2 a2 b3 (1 + x) a2 b3 - 1 (1 + x)a3b2

= (1 + sin b - cos b)

a3 b3 (1 + x) a3 b3 - 1

1 a1b2 1 a1b1 1 1 fi D¢ (0) = a2b1 1 1 + 1 a2b2 1 1 a3b2 1 a3b1 1 1 1 1 a1b3 + 1 1 a2b3 . 1 1 a2b3 Since each of the determinant on the RHS is zero, D¢ x in D(x) is zero. 50. Differentiating with respect to x, we get sin x cos x - sin x f¢ (x) = - 2 sin 2 x sin 2 x 2 cos 2 x - 3 sin 3x sin 3x 3 cos 3x cos x cos x cos x + cos 2 x 2 cos 2 x 2 cos 2 x cos 3x 3 cos 3x 3 cos 3x cos x sin x - sin x + cos 2 x sin 2 x - 4 sin 2 x cos 3x sin 3x - 9 sin 3x Now, put x = 0, p/4, p/2 and p. 51. Doing R3 Æ R3 - xR2 and R2 Æ R2 - xR1, we get 0 a -1 f(x) = 0 a + x - 1 = a(a + x)2, 0 0 a+x so that f(2x) - f(x) = a[(a + 2x)2 - (a + x)2] = a(a + 2x - a - x) (a + 2x + a + x) = ax(2a + 3x).

cos a - sin a 1 sin a cos a 1 cos(a + b ) - sin(a + b ) 1

cos a sin a

- sin a cos a

= (1 + sin b - cos b) (cos2 a + sin2 a) = 1 + sin b - cos b. To solve the equations by Cramer’s rule, we also need the determinants D1, D2 and D3 calculated below. cos b + 1 - sin a 1 cos a 1 D1 = - sin b + 1 - sin(a + b ) 1 2

=

cos b - sin b 1

- sin a 1 cos a 1 , - sin(a + b ) 1

after C1 Æ C1 - C3. Now R3 Æ R3 - (cos b) R1 + (sin b)R2 R3 gives D1 =

=

cos b - sin b

- sin a cos a

1 1

1 - cos 2 b - sin 2 b

0

1 + sin b - cos b

cos b - sin b 0

1 - sin a 1 cos a 0 1 + sin b - cos b

= (1 + sin b - cos b) (cos a cos b - sin a sin b) = (1 + sin b - cos b) cos(a + b). Similarly, D2 = - (1 + sin b - cos b) sin (a + b), and - sin a cos a cos b + 1 - sin b + 1 sin a cos a D3 = 2 cos (a + b ) - sin (a + b ) which, after R3 Æ R3 - (cos b)R1 + (sin b)R2, becomes

IIT JEE eBooks: www.crackjee.xyz 8.56 Comprehensive Mathematics—JEE Advanced

cos a sin a 0

D3 =

- sin a cos a 0

0 a + bw 2 + cw

cos b + 1 - sin b + 1

2

= 1 + sin b - cos b. Cramer’s rule now gives x=

D1 D = cos (a + b), y = 2 = - sin (a + b) and D D

z=

D3 = 1, D

fi x2 + y2 + z2 = cos2 (a + b) + sin2 (a + b) + 1 = 2. 53. The following determinants can be evaluated for the given set of equations. 1

1

D= 1

w

w

1 w

w

2

=

3

1

1

0

w

w

0 w

w

2

where we have used C1 Æ C1 + C2 + C3 and 1 + w + w2 = 0. a

1

1

b

w

w2

c w2

w

a+b+c

0

0

b

w

w2

c

w2

w

=

= (a + b + c)

w

w2

w2

w

1

-1 a c a c -1 -c +b =0 a -1 b -1 b a

x3 yz 1

a + bw 2 + cw 1 + w 4 + w 2 b c

w2 w

y 3 - x3 z 3 - y 3 z ( x - y) x ( y - z) 0 0

= (y - x) (z - y) D1 where D1 =

D2 = 1 b w 1 c w

1 1

R1 to given

fi - (1 - a2) - c(- c - ab) + b(ac + b) = 0 fi a2 + b2 + c2 + 2abc = 1. 55. Applying C3 Æ C2 - C2, C2 Æ C2 - C1 to obtain

-

2

=

-1 c b c - 1 a = 0. b a -1

D =

where R1 Æ R1 + R2 + R3

1+ w2 + w

54. Rewriting the given system of equations in the form - x + cy + bz = 0 cx - y + az = 0 bx + ay - z = 0,

(- 1)

= (a + b + c) (w2 - w),

1 a

1 (a + bw + cw2). 3

we see that they will have a non-trivial solution if and only if

2

= 3(w2 - w4) = 3(w2 - w),

D1 =

where we have done R1 Æ R1 + w2R2 + wR3 and R1. Similarly, the last determinant required is D3 = (a + bw + cw2) (w2 - w). Now we can use Cramer’s rule to get 1 1 x = (a + b + c), y = (a + bw2 + cw), 3 3 z=

1

1 w2 1 w

= - (a + bw2 + cw) (w - w2),

= (1 + sin b - cos b) (cos a + sin a) 2

w2 w

= - (a + bw2 + cw)

2 - cos 2 b - cos b - sin 2 b + sin b

2

b c

= 1 1

0

=

y 2 + x 2 + yx -z

z 2 + y 2 + zy -x

y 2 + x 2 + yx ( z - x)( x + y + z ) -z z-x

= (z - x) (x2 + y2 + z2 + yz + zx + xy) 56. Applying R3 Æ R3 - R1 – 2R2 R3, we get

IIT JEE eBooks: www.crackjee.xyz Determinants 8.57

f¢(x) =

Using R2 Æ R2 – R1 and R3 Æ R3 – R1. \ Maximum value of f(x) is 3.

2 ax 2 ax - 1 2 ax - 1 = b b +1 b 1

fi f¢(x) = 2ax + b Integrating, we get f (x) = ax2 + bx + c where c is an arbitrary constant. Since f has a maximum at x = 5/2, f ¢(5/2) = 0 fi 5a + b = 0 (1) f (0) = 2 fi c = 2 f (1) = 1 fi a + b + c = 1

Also and

\ a + b = - 1. Solving (1) and (2) for a, b we get a = 1/4, b = - 5/4

61. Let A be a skew symmetric matrix of order n, then |A¢| = (– 1)n |A| fi |A| = (– 1)n |A| If n is odd, then |A| = – |A| fi 2|A| = 0 fi |A| = 0. We have 0 sin(p /12) cos(p /12) Êpˆ D Á ˜ = - sin(p /12) 0 sec(p /12) Ë 4¯ - cos(p /12) - sec(p /12) 0 =0

(2)

Now, f¢ (0) = b = - 5/4 57. Use f(x) = x3 + 3x 58. Use R1 Æ R1 + R3 followed by C1 Æ C1 - C3, to get f (x) = 4 cot x cosec2 x

[using statement-2]

62. Statement-2 is true, See theory. The roots of x9 = p are p1/9 wr where 2p 2p + i sin . 9 9

w = cos

value of determinant D depends an a1, a2, . . . a9. If we put ak = p1/9 wk, then 1 w8 w7

fi f (p/4) = 8

D = (p1/9)3 1

w

w2

1

w

w

f ¢(x) = - 4 cosec4x - 8 cot2x cosec2x fi f ¢(p/4) = - 16 p /3 p /6

Finally, Ú

f(x) dx = - 2 cot2 x]p/3p/6 = - 16/3

However, if a1= p1/9, a2 = p1/9 w8, a3 = p1/9 w7, a4 = p1/9 w, a5 = p1/9 w5 a6 = p1/9 w4, a7 = p1/9 w2, a8 = p1/9 w3,

and f (p/2) = 0 59. (a) Expand along C1. (b) Use R3 Æ R3 – R2, R2 Æ R2 – R1 and the expand along R1. (c) Use R2 Æ R2 – x3R1. (d) Expand along R3. 60. (a) Show f ¢(x) = 0. 1 -1 / 2 1 / 2 1 1/ 2 = 0 (b) f(x) = - 1 / 2 1/ 2 1/ 2 1

a9 = p1/9 w6, then

cos 2 x

2 D1 = 1 1 + cos x

1

cos 2 x

1

w8 w7

D = (p1/9)3 w

w5 w4

w2 w3 w6 = p1/3 (2w2 – w7 – w6) π 0. 63. Statement-2 is true. Using R2 Æ R2 – get 6

(c) Show f is an odd function. (d) Use C1 Æ C1 + C2 + C3 to obtain f(x) = (2 + sin 2x) D1 where 1

w =

sin 2 x =

sin 2 x 1 + sin 2 x

2 R1 and R3 Æ R3 – 2i

0

3 ( 6 - 2 3 )i

0

2

(2 - 3 2 )i

6 ÈÎ2 3 - 3 6 - 2 3 + 2 6 ˘˚ i = – 6i

Ê nˆ 2 Ê nˆ P(x) = 1 + Á ˜ x + Á ˜ x + Ë 2¯ Ë1 ¯



3 R1, we

3 + 6i

64. We have

2

1 cos x sin 2 x 1 0 = 0 =1 0 0 1

=0

2

Ê nˆ Ê nˆ P¢(x) = Á ˜ + 2 Á ˜ x + Ë 2¯ Ë1 ¯

Ê nˆ + Á ˜ xn Ë n¯

Ê nˆ + n Á ˜ xn - 1 Ë n¯

IIT JEE eBooks: www.crackjee.xyz 8.58 Comprehensive Mathematics—JEE Advanced

Ê nˆ P¢(0) = Á ˜ Ë1 ¯



1 a a2

x in the ex-

2 0 π 1 b b

pansion of P(x)

1 c

\ Statement-2 is true. Note that D(x) consists of 6 terms of the form (1 + x)n. x in D(x) = D¢(0) a1b1 But D¢(0) = 1 1 1 + a2b1 1

1 a2b2 1

a1b2 1 1

a1b3 1 1

1 1 a2b3 + 1 1 a3b1

= (b - c) (c - a) (a - b) 68. Use a a2

a3 + 1

2 0= b b

b3 + 1

c2

c3 + 1

c 1 1 a3b2

1 1 =0 a3b3

65. Statement-2 is not always true. For instance, the system of equations x + 2y + 3z = 1 2x + 3y + 4z = 2 3x + 4y + 5z = 4

c2

= (b - c) (c - a) (a - b) (abc + 1) 69. Use Cramer’s to solve the system of equations a1a + b1b + c1g = - x a2a + b2b + c2g = - y a3a + b3b + c3g = - z 70. Use law of projections. 71. From Fig. 8.1, it is clear that Aq(a, b, g) is independent of q.

has no solution but |A| = 0 For statement-1, let a b c b c a D = c a b Using C1 Æ C1 + C2 + C3, we obtain 1 b c D = (a + b + c) 1 c a = 0 1 a b Also, note that x = y = z equations. Thus, the system of equations has 66. Rewrite the given relation as a 1x + b 1y + c 1z = 0 a 2x + b 2y + c 2z = 0 a 3x + b 3y + c 3z = 0 As

a1 a2 a3

b1 b2 b3

c1 c2 c3

π 0,

x = 0, y = 0, z = 0 by the Cramer’s rule 67. Use

Fig. 8.1

72. As Aq is independent of q, dAq/dq = 0. DABC is maximum 73. Fixing A and B when C is furthest from AB. 74. Expression equals zero in view of 61. 75. As Ax(a, b, g) is a constant is y = Ax(a, b, g) is a straight line parallel to the x-axis. 76. Take x, y, z common from C1, C2, C3 respectively. 77. Write f(q) as cos 2 sin q 1 cos q sin 2 q sin q sin 2 q

cos q sin q

- sin q

sin q

cos q

- cos q

0

2

Apply C1 Æ C1 - cos q C2 and simplify to get f(q) =1 78. Let D + E = q and apply C3 Æ C3 - sin q C1 - cos q C2 to get

IIT JEE eBooks: www.crackjee.xyz Determinants 8.59

D=0 79. Show D¢(x) = 0 80. C1 Æ C1 - bC3, C2 Æ C2 + aC3 to obtain D = (1 + a2 + b2)2 D1 1 where D1 = 0

0 1

- 2b 2a

84. Apply R3 Æ R3 - R1 and simplify to get D = 2(8a3 - 1) = 30 85. 2 divides each of c1, c2, c3. 86. Let D(x) = (x2 - 1) P(x) + (ax + b) where P(x) is a polynomial in x. Putting x = 1 and x = - 1, we obtain 21 = a + b, 33 = - a + b. This gives b = 27, a = - 6.

b - a 1 - a 2 - b2 Apply C3 Æ C3 + 2bC1 - 2aC2 to get D 1 = 1 + a 2 + b 2. Now use A.M. ≥ G.M. 81. Apply C1 Æ C1 + C2 to obtain D = 2 cos q

sin q 1 88. D = sin q

Apply R3 Æ R3 - R1 to obtain D = (1 - cosq) (sinq - 1) = 0. fi a = p/2

82. f(x, y) = D2 where x y 1 D = y 1 x = 3xy - 1 - x3 - y3 1 x y

89.

since f(x, y) is a determinant obtained by replacing (i, j)th element in D by its cofactor. 83. Applying C1 Æ C1 + C3 and simplifying we obtain D(q) = 2(1 + sin2 q) Now, [D(q)] = 2 ¤ 0 £ 2 sin2 q < 1 fi -1/ 2 < sin q < 1/ 2 ¤ - p/4 < q < p/4

1 1 cos q 1 = 0 cos q 1

l 1 1

1 l 1

1 1 l

= 0 fi (l + 2) (l - 1)2 = 0

fi l = - 2 or l = 1. For l = - 2, the system has no solution. 90.

a b b b a b = 0 fi (a + 2b) (a - b)2 = 0 b b a fi a + 2b = 0.

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9 Probability 9.1 DEFINITIONS

1. A random experiment is an experiment in which (i) all the outcomes of the experiment are known in advance, and (ii) the exact outcome of any specific performance of the experiment is unpredictable, i.e. not known in advance. 2. A sample space, denoted by S, associated with a random experiment is a set of points such that (i) each element of S denotes an outcome of the experiment, and (ii) any performance of the experiment results in an outcome that corresponds to exactly one element of S. In other words, sample space consists of all possible outcomes associated with the random experiment. The element of the sample space associated with a random experiment are called the elementary or simple or indecomposable events of that experiment. 3. An event (associated with a random experiment) is a subset of the sample space S (associated with that experiment). We say that an event E (Õ S) has occurred provided the outcome w of the experiment is an element of E. If the outcome w of the experiment is such that w œ E, we say that the event E has not occurred. The empty set Ø, which is a subset of every set, also represents an event and is said to be an impossible event. The set S is itself a subset of S and it represents the sure event. 4. A set of events is said to be mutually exclusive if the occurrence of one of them precludes the occurrence of any of the remaining events. In set-theoretic notation, events E1, E2, , Em are mutually exclusive if Ei « Ej = Ø for i π j and 1 £ i, j £ m. See Fig. 9.1 E4 E1

Fig. 9.1

6. A set of events is said to be exhaustive if the performance of the experiment always results in the occurrence of at least one of them. 9.2 CLASSICAL DEFINITION OF PROBABILITY

If a random experiment results in N mutually exclusive, equally likely and exhaustive outcomes, out of which n are favourable to the occurrence of an event A, then the probability of occurrence of A, usually denoted by P(A), is given by n P(A) = N Since the number of favourable outcomes can never exceed the total number of outcomes, we have 0 £ P(A) £ 1. Note that P(Ø) = 0 and P(S ) = 1. 9.3 NOTATIONS

Let A and B be two events. Then (i) A¢ or A or Ac stands for the non-occurrence or negation of A. (ii) A » B stands for the occurrence of at least one of A and B. (iii) A « B stands for the simultaneous occurrence of A and B. (iv) A¢ « B¢ stands for the non-occurrence of both A and B. (v) A Õ B stands for “the occurrence of A implies occurrence of B”. 9.4 A FEW THEOREMS ON PROBABILITY

E3 E2

5. A set of events is said to be equally likely if, taking into consideration all the relevant factors, there is no reason to expect one of them in preference to the others.

1. If A and B are two mutually exclusive events, then P(A » B) = P(A) + P(B). 2. If A is any event, then P(A¢) = 1 – P(A) 3. If A and B are two events, then P(A « B¢) = P(A) – P(A « B)

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(ii) P(A1 « A2 « .... « An) > 1 – P(A¢1) – P (A¢2) – .... – P(A¢n) 10. If A1, A2, , An are n events, then P(A1 « A2 « .... « An) ≥ P(A1) + P(A2) + + P(An) – (n – 1) 11. If A and B are two events such that A Õ B, then P(A) £ P(B). Odds in Favour and against an Event If x is the number of cases favourable to the occurrence of an event A and y that for the event A¢, then the odds in favour of A are x : y and the odds against A are y : x. In this case

4. If A and B are two events, then P(A » B) = P(A) + P(B) – P(A « B) 5. If A and B are two events, then P(exactly one of A, B occurs) = P[(A « B¢) » (A¢ « B)] [Fig. 9.2] A

B

Fig. 9.2

= P(A) – P(A « B) + P(B) – P(A « B) = P(A) + P(B) – 2P(A « B) = P(A » B) – P(A « B) Also, P(exactly one of A, B occurs) = P(A « B¢) + P (A¢ « B) = P(B¢) – P(A¢ « B¢) + P(A¢) – P(A¢ « B¢) = P(A¢) + P(B¢) – 2P(A¢ « B¢) = P(A¢ » B¢) – P(A¢« B¢) 6. If A and B are two events P(A¢ » B¢) = 1 – P(A « B) and P(A¢ « B¢) = 1 – P(A » B) 7. If A1, A2, , An are n events, then P(A1 » A2 » » An) n

=

Â

i =1

P(Ai) –

Â

P(A) =

Â

y x+y

If the event B has already occurred, then the sample space reduces to B. Now the outcomes favourable to the occurrence of A A (given that B has already occurred) are those that B are common to both A A«B and B, that is, those which belong to A « B. Thus, Fig. 9.3 +

1£ i < j < k £ n

P(A1 « A2 « A n) (–1) 8. If A, B and C are three events, then (i) P (A » B » C) = P(A) + P(B) + P(C) – P(B « C) – P(C « A) – P(A « B) + P(A « B « C) (ii) P (at least two of A, B, C occur) = P(B « C) + P(C « A) + P(A « B) – 2P (A « B « C) (iii) P (exactly two of A, B, C occur) = P(B « C) + P(C « A) + P(A « B) – 3P(A « B « C) (iv) P(exactly one of A, B, C occurs) = P(A) + P(B) + P(C) – 2P(B « C) – 2P(C « A) – 2P(A « B) + 3P(A « B « C) 9. If A1, A2, , An are n events, then (i) P(A1 » A2 » » An) < P(A1) + P(A2) + + P(An) n–1

P(A¢) =

The probability of occurrence of an event A, given that B has already occurred is called the conditional probability of occurrence of A. It is denoted by P(A | B).

P(Ai « Aj) P(Ai « Aj « Ak) –

and

9.5 CONDITIONAL PROBABILITY

1£ i < j £ n

+

x x+y

P(A | B) =

N A« B

NB where NA« B is the number of elements in A « B and NB π 0 is the number of elements in B and N the total number of elements in S. fi P(A | B) =

N A« B / N

Hence

NB / N

=

P ( A « B) P ( B)

Ï P( B) P( A | B) if P( B) π 0 P(A « B) = Ì Ó P( A) P( B | A) if P( A) π 0 Important Note The function P ( |A) is a probability function and hence all theorems on probability apply to this function. For instance, P (B » C | A) = P(B | A) + P(C | A) – P(B « C | A).

IIT JEE eBooks: www.crackjee.xyz Probability 9.3

9.6 MULTIPLICATION THEOREM

P(A) > 0. Then for j = 1, 2,

Let A1 A2, , An, An+1 be (n + 1) events such that « An) > 0. P(A1 « A2 « Then Ê n +1 ˆ PÁ A j ˜ = P(A1) P(A2 | A1) P(A3 | A1 « A2) Ë j =1 ¯ P(An+1 | A1 «

« A n)

9.7 TOTAL PROBABILITY THEOREM

Let {Hi}, i = 1, 2, for i π j and

n

, n, be n events such that Hi « Hj = Ø

Hi = S. Suppose that P(Hi) > 0 for 1 £ i £ n.

i =1

Then for any event A n

P(A) =

Â

i =1

P(Hi) P(A | Hi)

H1

Hn

H2

H3

H4

Fig. 9.4

9.8 BAYES’ RULE

Let S be a sample space and H1,

, Hn be n mutually

n

Hj = S and P(Hj) > 0 for

exclusive events such that j =1

j = 1, 2, , n. We can think of the Hj’s as the ‘causes’ (or the hypotheses) that lead to the outcome of an experiment. The probabilities P(Hj), j = 1, 2, , n are called prior probabilities. Suppose the experiment results in an outcome of event A, where P(A) > 0. What is the probability that the observed event was due to cause Hj? In other words, we seek the conditional probability P(Hj | A). These probabilities are frequently called posterior probabilities. The information that A has occurred makes us reassess the probability P(Hj) assigned to Hj. 9.9 THEOREM (BAYES’ RULE)

Let {Hj} be mutually exclusive events such that P(Hj) > 0 n

for j = 1, 2,

, n and S =

Hj. Let A be an event with j =1

P(Hj | A) =

,n P( H j ) P( A | H j )

n

 P ( H k ) P( A | H k )

k =1

9.10 INDEPENDENT EVENTS

We say that two events A and B are independent if the occurrence or non-occurrence of A(B) does not affect the probability of occurrence or non-occurrence of B(A), that is, if P(B | A) = P(B) provided P(A) π 0 P( B « A) = P(B) fi P(A « B) = P(A) P(B) P( A) Thus, two events A and B are independent if and only if P(A « B) = P(A) P(B). If two events are not independent, they are said to be dependent. We note that if P(A) = 0 then for any event B, 0 £ P (A « B) £ P(A) = 0, so that P(A « B) = P(A) P(B) and A is independent of any other event. Difference between Mutually Exclusiveness and Indepen dence fi

Important Note We advise the reader to distinguish between independent and mutually exclusive events. Independence is a property of probability, whereas mutual exclusion is a set-theoretic property. If A and B are mutually exclusive events with P(A) > 0 and P(B) > 0, then P (A « B) = 0 π P(A) P(B), so that A and B cannot be independent. In fact, P(A | B) = 0 = P(B | A), so that if A occurs, B cannot occur and vice-versa. Thus, A and B are strongly dependent when A « B = Ø. We also note the following fact; two events A and B are independent if and only if (i) A and B¢ are independent, or (ii) A¢ and B are independent, or (iii) A¢ and B¢ are independent. Three events A, B and C are said to be pairwise independent if P(A « B) = P(A) P(B), P(B « C) = P(B) P(C) and P(C « A) = P(C)P(A) Three events A, B and C are said to be mutually indepen dent if P(A « B) = P(A) P(B), P(B « C) = P(B) P(C), P(C « A) = P(C) P(A) and P(A « B « C) = P(A) P(B) P(C)

IIT JEE eBooks: www.crackjee.xyz 9.4 Comprehensive Mathematics—JEE Advanced 9.11 RANDOM VARIABLES

Let S be a sample space. A random variable X is a function from the set S to R, the set of real numbers. If X is a S and r is a real number, then {X = r} = {wŒS | X (w) = r } is an event. If the random variable X takes n distinct values x1, x2, , xn, then {X = x1}, {X = x2}, , {X = xn} are mutually exclusive and exhaustive events (Fig. 9.5). S X = x1

X = x2

Fig. 9.5

Now, since (X = xi) is an event, we can talk of P(X = xi). If P(X = xi) = pi (1 £ i £ n), then the system of numbers xn ˆ pn ˜¯

x2 p2

is said to be the probability distribution of the random variable X. Mean and Variance of a Random Variable

The expectation (mean) of the random variable X is n

Â

pi xi

i =1

and the variance of X n

var (X) =

Â

i =1

n

pi (xi – E(X))2 =

Â

pi xi2 – (E (X ))2

i =1

9.12 BINOMIAL DISTRIBUTIONS

Consider a random experiment with sample space S and an event E (a subset of S) associated with it. Then the event “not E ” may be denoted by E¢, the complement (in S) of the subset E. Let P(E) = p and P(E¢) = q, so that p, q > 0 and p + q = 1. If the experiment results in the event E, we say a success, denoted by S, has occurred. If on the other hand, the event E does not occur (i.e., E¢ occurs) the experiment is said to have resulted in a failure, denoted by F. The probability of a success is equal to p and that of a failure is q = 1 – p.

FSF

There are 2 such outcomes, and these constitute the sample space of the combined experiment. Since the successive experiments are independent, the probability of the outcome above is q p q = pr qn–r

pqqppqq

X = xn

E (X ) =

SFFSSFF n

X = x3

X = x4

Ê x1 ÁË p 1

Suppose the above experiment is carried out n times under identical conditions. If the success (occurrence of E) and failure (non-occurrence of E) are recorded successively as the experiment is repeated, we will get a result of the type

where r is the number of successes in the outcome. If X denotes the random variable “number of successes”, then the possible values of X are 0, 1, 2, , n. We shall now calculate the probability of the event (X = r). Out of the 2n outcomes, r successes and (n – r) failures can occur in nCr ways. Also, the probability of r successes and (n – r) failures is pr qn–r. Thus, P(X = r) = nCr pr qn–r. The probability distribution of the random variable X is given by r

0

1

P(x = r) nC 0 qn nC1 pq n–1

r

2 n

C 2 p 2 q n–2

n

n

Cr p rq n–r

n

Cn p n

Note that the probabilities that the random variable takes the values 0, 1, 2, , n are given by the terms in the binomial expansion of (q + p)n. Because of this, the probability distribution of the random variable X is said to be a binomial distribution and X is said to be a binomial random variable. A binomial distribution B(n, p) is the probability distribution of a random variable X which takes values 0, 1, 2, , n with probabilities n

C0 p0q n, nC1pqn–1,

, nCr p rqn–r,

, nC n p nq 0

where p, q > 0 and p + q = 1. The numbers n and p are said to be parameters of the binomial distribution B(n, p). We usually write, X ~ B (n, p). Mode of Binomial Distribution

For the value of r so that P(X = r) is maximum, we have two cases. Case 1: (n + 1)p is not an integer then, P(X = r) is maximum when r = m, where m is the greatest integer £ (n + 1)p, that is, m = [(n + 1)p]. Case 2: (n + 1)p is an integer then, P(X = r) is maximum when r = m – 1 or m, where

IIT JEE eBooks: www.crackjee.xyz Probability 9.5

(n + 1) p = m and m is an integer.

For a = 3, 3n – 2, b can be chosen in (n + 1) ways. … For a = n – 1, 2n + 2, b can be chosen in (2n – 3) ways. For n £ a £ 2n + 1, b can be chosen in (2n – 2) ways. Thus, number of favourable ways of choosing a, b is 2[(n – 1) + n + + (2n – 3)] + (n + 2)(2n – 2)

If X ~ B (n, p), then E(X) = np and var (X) = npq. SOLVED EXAMPLES

SINGLE CORRECT ANSWER TYPE QUESTIONS

Ê n - 1ˆ (n – 1 + 2n – 3) + 2(n + 2)(n – 1) = 2Á Ë 2 ˜¯

A fair coin is tossed 2019 times. Example 1 Probability of getting at least 1010 heads is (a) (c)

1 2

(b) 1

(d)

1009

2

1 4

= 1

1010

2

Ans. (a) Solution: Let n = 1009, then 2n + 1 = 2019 Let X = number of heads obtained when a fair coin is tossed (2n + 1) times. Then X ~ B(2n + 1, p) where p = 1/2. 1 P(X ≥ n + 1) = 2 n +1 ( S ) where 2 S = 2n + 1Cn + 1 + 2n + 1Cn + 2 + + 2n + 1C2n + 1 Using 2n + 1Cr = 2n + 1C2n + 1 – r, we get S = 2n + 1Cn + 2n + 1Cn – 1 + + 2n + 1C0 Adding (1) and (2), we get 2S = 2n + 1C0 + 2n + 1C1 + = 22n + 1 S = 22n P(X ≥ n + 1) =

22 n 22 n+1

=

= (n – 1)(3n – 4 + 2n +4) = 5n(n – 1) \ probability of required event

(2)

1 2

From the first 3n(n ≥ 2) natural numbers Example 2 two numbers are picked up at random without replacement. Let P denote the probability that the absolute value of the difference between the numbers is less than n, then P is equal to 2(n - 1) 3n - 1

(b)

5(n - 1) 3(3n - 1)

(c)

n -1 3n - 1

(d)

n +1 3(3n - 1)

A box conatins 2n cards numbered 20, 21, Example 3 22, …, 22n – 1. Two cards are taken out at random from the box, at random without replacement. Probability that sum of the numbers on the card is divisible by 3 is (a)

n 2n - 1

(b)

n -1 2n - 1

(c)

n +1 2n

(d)

1 2

(1)

+ 2n + 1C2n + 1

(a)

5n(n - 1) 5(n - 1) = 3n(3n - 1) 3(3n - 1)

Ans. (b) Solution: Two numbers a, b n natural numbers can be chosen in 3n(3n – 1) ways. For a = 1, 3n, b can be chosen in (n – 1) ways. For a = 2, 3n – 1, b can be chosen in n ways.

Ans. (b) Solution: Let the smaller card bear number a and the larger one bear number b. b = 2, 23, …, 22n – 1 When a = 20, When a = 2, b = 22, 24, …, 22n When a = 22, b = 23, 25, …, 22n – 1 3 When a = 2 , b = 24, 26, …, 22n When a = 22n – 3, b = 22n – 2 When a = 22n – 2, b = 22n – 1 Thus number of favourable cases = 2[n + (n – 1) + + 2 + 1] = n(n – 1) Total number of ways = 2nC2 = n(2n – 1) Thus, required probability =

n(n - 1) n -1 = n(2n - 1) 2n - 1

An unbiased cubical die is thrown n times. Example 4 If xi = outcome on the ith throw and X = gcd(x1, x2, …, xn), then E(X) is equal to (a) (b)

1 6

n

1 6n

(6n + 3n + 2n) (6n + 3n + 2n + 1)

IIT JEE eBooks: www.crackjee.xyz 9.6 Comprehensive Mathematics—JEE Advanced

(c) (d)

1 6

n

1 6n

Solution: The following cases arise:

(6n + 3n + 2n + 1 + 8) (6n + 3n + 2n + 2 + 8)

Ans. (c) Solution: Sample space of the random experiment is S = {(x1, x2, …, xn) | 1 £ xi £ 6} \ n(S) = 6n Note that X can take values 1, 2, 3, 4, 5, 6. P(X = 6) = P(X = 5) = P(X = 4) =

1

[all xi’s are equal to 6]

6n 1

[all xi’s are equal to 5]

6n 1

[all xi’s are equal to 4]

6n

Now, X = 3 if at least one of xi is 3 and rest of them 3 or 6 \

P(X = 3) =

2n - 1 6

n

Next, X = 2 if xi’s are 2, 4, 6 but not all of them 4 or not all of them 6. fi

P(X = 2) =

3n - 1 - 1

Thus, P(X = 1) = =

6n

=

3n - 2 6n

1 n [6 – (1 + 1 + 1 + 2n – 1 + 3n – 2)] 6n

2nd draw

3rd draw

4th draw

Blue

Blue

Red

Red

Case 2

Blue

Red

Blue

Red

Case 3

Red

Blue

Blue

Red

\ required probability = P(case 1) + P(case 2) + P(case 3) 2

2

2 1 8 2 9 1 2 Ê 9ˆ 1 Ê 8ˆ = Á ˜ ¥ ¥ + ¥ ¥ ¥ + ¥Á ˜ ¥ Ë 10 ¯ 10 10 10 10 10 10 10 Ë 10 ¯ 10

= 0.0434 There are m persons sitting in a row. Two Example 6 of them are selected at random. The probability that the two selected persons are not together is (a) 2/m (b) 1/m (c)

m (m - 1) (m + 1) (m + 2)

(d) 1 -

2 . m

Ans. (d) Solution: The total number of ways of selecting two persons out of m is m (m - 1) m C2 = 2 The number of ways in which the two selected persons are together is (m – 1). Therefore, the number of ways in which the two selected persons are not together is (m - 1) (m - 2) m C2 – (m – 1) = 2 Thus, the probability of the required event is

1 n (6 – 2n – 3n) 6n

Now, E(X) = 1P(X = 1) + 2P(X = 2) +

1st draw Case 1

(m - 1) (m - 2)/2 m - 2 2 = =1m (m - 1)/2 m m + 6P(X = 6)

1 = n [(6n – 2n – 3n) + 2(3n – 2) + 3(2n – 1) + 4(1) + 6 5(1) + 6(1)] 1 = n (6n + 3n + 2n + 1 + 8) 6 There are 8 blue and 2 red balls in a bag. Example 5 Each time one ball is drawn and replaced by a blue one. The probability of drawing the last red ball on the fourth draw is (a) 0.0434 (b) 0.0438 (c) 0.0444 (d) 0.0453 Ans. (a)

Three six-faced fair dice are thrown Example 7 together. The probability that the sum of the numbers appearing on the dice is k (3 £ k £ 8) is (a)

(k - 1) (k - 2) 432

(b)

k (k - 1) 432

(c)

k2 432

(d) k2/216

Ans. (a) Solution: The total number of cases is 6 ¥ 6 ¥ 6 = 63 = 216. The number of favourable ways xk in (x + x2 + + x 6) 3 xk – 3 in (1 – x6)3 (1 – x)–3

IIT JEE eBooks: www.crackjee.xyz Probability 9.7

= k – 3 + 2C k – 3

xk – 3 in (1 – x)–3 [ 0 £ k – 3 £ 5] k–3 3 in (1 + C1x + 4C2 x2 + 5C3 x3 + x ) (k - 1) (k - 2) = k – 1C 2 = 2

Thus, the probability of the required event is (k – 1) (k – 2)/432. Two squares are chosen at random on Example 8 a chessboard. The probability that they have a side in common is (a) 1/9 (b) 2/7 (c) 1/18 (d) 2/9 Ans. (c) Solution is 64, and that for the second square is 63. Therefore, the

=

P(W ) 1 - P( B)

2

=

a ( a + b) a + 2ab 2

Also P(B wins the game) = 1 –

=

a+b a + 2b

a+b b = a + 2b a + 2b

According to the given condition, fi

a+b b = 3◊ a + 2b a + 2b

fi a = 2b fi a : b = 2 : 1

Example 10 A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The probability that the determinant chosen is nonzero is (a) 3/16 (b) 3/8 (c) 1/4 (d) 1/8. Ans. (b) Solution: A determinant of order 2 is of the form

is 64 ¥

D=

four squares in the corner, the second square can be chosen (non-corner) squares on either side of the chessboard, the happens to be any of the 36 remaining squares, the second square can be chosen in 4 ways. Therefore, the number of favourable ways is (4) (2) + (24) (3) + (36) (4) = 224. Thus, the probability of the required event is 224/4032 = 1/18. A bag contains a white and b black balls. Example 9 Two players, A and B alternately draw a ball from the bag, replacing the ball each time after the draw till one of them draws a white ball and wins the game. A begins the game. If the probability of A winning the game is three times that of B, the ratio a : b is (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 1 : 3 Ans. (c) Solution: Let W denote the event of drawing a white ball at any draw and B that for a black ball. Then P(W) =

a b and P(B) = a+b a+b

P(A wins the game) = P(W or BBW or BBBBW or = P(W) + P(BBW) + P(BBBBW) + = P(W) + P(B) P(B) P(W) + P(B) P(B) P(B) P(B) P(W) + = P(W) + P(W) ◊ P(B)2 + P(W) ◊ P(B)4 +

It is equal to ad – bc. The total number of ways of choosing a, b, c and d is 2 ¥ 2 ¥ 2 ¥ 2 = 16. Now D π 0 if and only if either ad = 1, bc = 0 or ad = 0, bc = 1. But ad = 1, bc = 0 if and only if a = d = 1 and at least one of b, c is zero. Thus, ad = 1, bc = 0 in three cases. Similarly, ad = 0, bc = 1 in three cases. Thus, the probability of the required event is 6/16 = 3/8. One hundred identical coins, each with Example 11 probability p of showing heads are tossed once. If 0 < p < 1 and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, the value of p is (a) 1/2 (b) 51/101 (c) 49/101 (d) 3/101. Ans. (b) Solution: Let X be the number of coins showing heads. Then X follows a binomial distribution with parameters n = 100 and p. Since P(X = 50) = P(X = 51), we get 100

fi )

a b c d

C50 p50 (1 – p)50 =

100

C51 p51 (1 – p)49

100! 51!49! p 51 p = ◊ = fi 50!50! 100! 1 - p 50 1 - p

fi 51 – 51p = 50p fi p = 51/101 Example 12 Suppose X follows a binomial distribution with parameters n and p, where 0 < p < 1.

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If P(X = r)/P(X = n – r) is independent of n for every value of r, then (a) p = 1/2 (b) p = 1/3 (c) p = 1/4 (d) p = 1/5. Ans. (a)

Similarly, P(B) + P(C) – 2P(B « C) = p and P(C) + P(A) – 2P(C « A) = p Adding (1), (2) and (3) we get

Solution: We have



n-r

n - 2r

C p (1 - p) P( X = r ) (1 - p) = n r n-r = r P( X = n - r ) Cn - r p (1 - p ) p n - 2r n

r

Ê1 ˆ = Á - 1˜ Ëp ¯

2[P(A) + P(B) + P(C) – P(A « B) – P(B « C) – P(C « A)] = 3p P(A) + P(B) + P(C) – P(A « B) – P(B « C) – P(C « A) = 3p/2

n - 2r

Example 13 The minimum number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.8 is (b) 6

(c) 5

(d) 3.

Solution: Suppose the coin is tossed n times. Let X be the number of heads obtained. Then X follows a binomial distribution with parameters n and p = 1/2. We have P(X ≥ 1) ≥ 0.8 fi 1 – P(X = 0) ≥ 0.8 Ê 1ˆ fi 1 – nC0 p0 (1 – p)n ≥ 0.8 fi Á ˜ Ë 2¯

n

£ 0.2 =

1 5

We are also given that P(A « B « C) = p (5) Now, P(at least one of A, B and C) = P(A) + P(B) + P(C) – P(A « B) – P(B « C) – P(C « A) + P(A « B « C)

For the three events A, B and C,

P (exactly one of the events A or B occurs) = P(exactly one of the events B or C occurs) = P(exactly one of the events C or A occurs) = p and P(all the three events occur simultaneously) = p2, where 0 < p < 1/2. Then the probability of at least one of the three events A, B and C occurring is

(c)

3 p + 2 p2 2

(b)

p + 3 p2 2

(d)

p + 3 p2 4 3 p + 2 p2 4

(a)

52 81

(b)

55 81

(c)

61 81

(d)

62 81

When When When When When

a = 1, 2, a = 3, 4, a = 5, 6, a = 7, 8, a=9

b∈ [1, 5] b∈ [1, 6] b∈ [1, 7] b∈ [1, 8] b∈ [1, 9]

Thus, the inequality a – 2b ¥5+ 2 ¥ 6 + 2 ¥ 7 + 2 ¥ 8 + 9 = 61 ordered pairs (a, b). Total number of ways of choosing (a, b) is 9 ¥ 9 = 81 61 81

Example 16 An unbiased die with faces marked 1, 2, 3, 4, 5 and 6 is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5 is

Solution: We know that P(exactly one of A or B occurs) = P(A) + P(B) – 2P(A « B) P(A) + P(B) – 2P(A « B) = p

[using (4) and (5)]

Example 15 Nine identical balls are numbers 1, 2, ...9. are put in a bag. A draws a ball and gets the number a. The ball is put back the bag. Next B draws a ball gets the number b. The probability that a and b inequality a – 2b + 10 > 0 is

\ required probability is

Ans. (a)

Therefore,

3p 3 p + 2 p2 + p2 = 2 2

Solution: a – 2b + 10 > 0 fi 1 £ b < (a + 10) /2 Thus, b ∈ [1, (a + 10)/2)

This show that the least value of n is 3.

(a)

=

Ans. (c)

fi 2n ≥ 5 Example 14

(4)

2

Note that (1/p) – 1 > 0. Therefore the ratio will be independent of n for each r if (1/p) – 1 = 1, or p = 1/2.

(a) 7 Ans. (d)

(2) (3)

(1)

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(a) 16/81

(b) 1/81

(c) 80/81 (d) 65/81

Ans. (a) Solution: Let p = probability of getting face value not less than 2 and not more than 5 in a single throw of die = 4/6 = 2/3. n = number of times die is rolled X = number of times we get a number not less than 2 and not more than 5.

We have P(A « (B « C)) = P(A « B « C) = P(A) P(B) P(C) = P(A) P(B « C). fi A and B « C are independent. Therefore, S2 is true. Also P[(A « (B » C)] = P[(A « B) » (A « C)] = P(A « B) + P(A « C) – P[(A « B) « (A « C)] = P(A « B) + P(A « C) – P(A « B « C)

Then X ~ B (n, p) Required probability = P(X = 4) = 4C4 p4 = (2/3)4 = 16/81. Example 17 If the papers of 4 students can be checked by any one of the 7 teachers, then the probability that all the 4 papers are checked by exactly 2 teachers is (a) 2/7 (b) 6/49 (c) 32/343 (d) 1/7 Ans. (b)

\ the number of favourable ways = (7C2) (14) = (21) (14). the required probability =

(21) (14) 74

6 = 49

Example 18 Let A, B, C, be three mutually independent events. Consider the two statements S1 and S2. S1 : A and B » C are independent S2 : A and B « C are independent Then (a) (b) (c) (d) Ans.

= P(A) [P (B) + P(C) – P(B) P(C)] = P(A) [P(B) + P(C) – P(B « C)] = P(A) P(B » C) \

A and B » C are independent. If m is a natural such that m £ 5, then the

Example 19

probability that the quadratic equation x2 + mx +

1 m + =0 2 2

has real roots is

Solution: The total number of ways in which papers of 4 students can be checked by seven teachers is 74. The number of ways of choosing two teachers out of 7 is 7C2. The number of ways in which they can check four papers is 24. But this includes two ways in which all the papers will be checked by a single teacher. Therefore, the number of ways in which 4 papers can be checked by exactly two teachers is 24 – 2 = 14.

Thus,

= P(A) P(B) + P(A) P(C) – P(A) P(B) P(C)

Both S1 and S2 are true Only S1 is true Only S2 is true Neither S1 nor S2 is true.

(a)

Solution: We are given that P(A « B) = P(A) P(B), P(B « C) = P(B) P(C), P(C « A) = P(C) P(A), and P(A « B « C) = P(A) P(B) P(C)

(a) 1/5

(b) 2/3

(c) 3/5

(d) 1/5

Ans. (c) Solution: Discriminant D of the quadratic equation x2 + mx +

1 m + =0 2 2

is given by Ê 1 mˆ D = m2 – 4 Á + ˜ = m2 – 2m – 2 Ë2 2¯ = (m – 1)2 – 3 Now, D ≥0 ¤ (m – 1)2 ≥ 3 This is possible for m = 3, 4 and 5. Also, the total number of ways of choosing m is 5. \ Probability of the required event = 3/5. Example 20 A letter is taken at random from the letters of the word ‘STATISTICS’ and another letter is taken at random from the letters of the word ‘ASSISTANT’. The probability that they are the same letter is (a) 1/45 Ans. (c)

(b) 13/90

Solution: Letters of the word

(c) 19/90 (d) 5/18 STATISTICS are

AIICSSSTTT Letters of the word ASSISTANT are AAINSSSTT Common letters are A, I, S and T

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1 2 2 ¥ = 10 9 90

Probability of choosing A is Probability of choosing I is

2 1 2 ¥ = 10 9 90

Thus, by the Bayes’ rule

3 2 6 ¥ = 10 9 90

\ Probability of required event =

2 2 9 6 19 + + + = . 90 90 90 90 90

Example 21 A computer producing factory has only two plants T1 and T2. Plant T1 produces 20% and plant T2 produces 80% of the total computers produced. 7% of computers produced in the factory turn out of be defective. It is known that P(computer turns out to be defective given that it is produced in plant T1) = 10 P (computer turns out to be defective given that it is produced in plant T2), where P(E) denotes the probability of an event E. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant T2 is (a)

36 73

(b)

47 79

(c)

78 93

(d)

75 83

Ans. (c) Solution: Let T1, T2, D denote the following events. T1: computer is produced by plant T1 T2: computer is produced by plant T2 D: computer is defective P(T1) = 0.2, P(T2) = 0.8 P(D) = 0.07, P(D|T1) = 10 P(D|T2) By the total probability rule, P(D) = P(T1) P(D|T1) + P(T2) P(D|T2) fi

0.07 = [0.2(10) + (0.8)] P(D|T2)



P(D/T2) =

1 10 fi P(D|T1) = 40 40

\

P(D¢/T2) =

39 30 . P(D¢/T1) = 40 40

P (T1 ) P ( D¢ | T1 ) + P (T2 ) P ( D¢ | T2 )

Ê 8 ˆ Ê 39 ˆ ÁË ˜¯ ÁË ˜¯ 78 9 40 = = Ê 2 ˆ Ê 30 ˆ Ê 8 ˆ Ê 39 ˆ 93 ÁË ˜¯ ÁË ˜¯ + ÁË ˜¯ ÁË ˜¯ 10 40 40 40

3 3 9 ¥ = Probability of choosing S is 10 9 90 Probability of choosing T is

P (T2 ) P ( D¢ | T2 )

P(T2/D¢) =

Example 22 Seven distinct white balls and three distinct black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals (a) 1/2

(b) 7/15

(c) 2/15

(d) 1/3

Ans. (b) Solution: The number of ways of placing 3 black balls at 10 places is 10C3. The number of ways in which two black balls are not together is equal to the number of ways of choosing 3 places marked with X out of eight places XWXWXWXWXWXWXWX This can be done in 8C3 ways. Thus, probability of the required event is 8

C3

10

C3

=

8¥7¥6 7 = . 10 ¥ 9 ¥ 8 15

Example 23 If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that two white and one black ball will be drawn is (a)

13 32

(b)

1 4

(c)

1 32

(d)

3 16

Ans. (a) Solution: P (2 white and 1 black) = P(W1 W2 B3 or W1 B2 W3 or B1 W2 W3) = P(W1) P(W2) P(B3) + P(W1) P(B2) P(W3) + P(B1) P(W2)P(W3) Ê 3ˆ = Á ˜ Ë 4¯

Ê 2ˆ ÁË ˜¯ 4

Ê 3ˆ Ê 3ˆ ÁË ˜¯ + ÁË ˜¯ 4 4

Ê 2ˆ ÁË ˜¯ 4

Ê 1ˆ Ê 1ˆ ÁË ˜¯ + ÁË ˜¯ 4 4

Ê 2ˆ ÁË ˜¯ 4

13 Ê 1ˆ ÁË ˜¯ = 4 32

Example 24 There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, in a random order till both the faulty machines needed is

IIT JEE eBooks: www.crackjee.xyz Probability 9.11

(a)

1 3

(b)

1 6

(c)

1 2

(d)

1 4

Ans. (a) Solution: Let A(B two tests results in faulity (no-faulty) machines. Exactly two test will be needed to identify the faulty machines if and only A » B occures. Now, P(A » B) = P(A) + P(B) Ê 2ˆ Ê 1ˆ Ê 2ˆ Ê 1ˆ 1 = Á ˜Á ˜ +Á ˜Á ˜ = . Ë 4 ¯ Ë 3¯ Ë 4 ¯ Ë 3¯ 3

1 (9 C3 ) 11 10 C3 14 P ( E9 ) P ( A | E9 ) = = . P (E9|A) = 1 P ( A) 55 4 A group of 2n boys and 2n girls is Example 26 randomly divided into two equal groups. The probability that each group contains the same number of boys and girls is (a) 1/2 (b) 1/n (c)

Example 25 A bag contains some white and some black balls, all combinations of balls being equally likely. The total number of balls in the bag is 10. If three balls are drawn at random without replacement and all of them are found to be black, the probability that the bag contains 1 white and 9 black balls is 14 (a) 55

12 (b) 55

2 (c) 11

8 (d) 55

Ans. (a) Solution: Let Ei denote the event that the bag contains i black and (10 – i) white balls (i = 0, 1, 2, …, 10). Let A denote the event that the three balls drawn at random from the bag are black. We have 1 P(Ei) = (i = 0, 1, 2, …, 10) 11 P(A|Ei) = 0 for i = 0, 1, 2 i

and P(A|Ei ) =

C3

10

C3

for i ≥ 3

Now, by the total probability rule 10

P(A)= Â P( Ei ) P( A | Ei )

2n

Cn

4n

Cn

(d)

(

2n

Cn

4n

)

2

Cn

Ans. (d) Solution: Total number of ways of choosing a group is 4n C2n.The number of ways in which each group contains equal number of boys and girls is (2nCn) (2nCn) \ Required probability =

( 2 n Cn ) 2 4n

C2 n

.

Example 27 An unbiased die with faces marked 1, 2, 3, 4, 5 and 6 is rolled four times out of the four faces obtained, the probability that the minimum value is exactly than 2 and the maximum value is exactly 5 is (a) 4/27

(b) 1/81

(c) 80/81 (d) 65/81

Ans. (a) Solution: Total number of outcomes of 4 dice is 64. We can choose two dice (one for 2 and one for 5) in 4C2 ways. We can arrange 2 and 5 at these places in 2 ways. For the remaining two dice, the number of possible outcomes is (4) (4) = 16. Thus, the number of favourable outcomes is (4C2) (2) (16). Thus, probability of required event = (6) (2) (16) 4 = . 27 64

i =0

1 1 ¥ 10 = [ 3C 3 + 4C 3 + … + 11 C3 But 3C3 + 4C3 + 5C3 + … + 10C3 = 4C4 + 4C3 + 5C3 + … + 10C3 = 5C4 + 5C3 + 6C3 + … + 10C3 = 6C4 + 6C3 + … + 10C3 = … = 11

Thus, P(A) =

C4

11 ¥ 10 C3

By the Bayes’ rule

=

1 4

10

C 3]

Example 28 Four fair dice D1, D2, D3 and D4, each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The probability that D4 shows a number appearing on one of D1, D2, and D3 is (a)

11

C4

91 216

(b)

108 216

(c)

125 216

(d)

127 216

Ans. (a) Solution: Probability of the required event = 1 – P (none of D1, D2, D3 show the number appearing on D4)

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( C ) (5)(5)(5) = 1 = 16

1

64

91 25 = 216 216

= (100C1 +

Example 29 A pair of unbiased dice is rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is (a) 2/5 (b) 3/5 (c) 4/5 (d) 1/5

But Also,

Ans. (a)



Solution: Let A denote the event that a sum of 5 occurs, B the event that a sum of 7 occurs and C the event that neither a sum of 5 nor a sum of 7 occurs. We have 4 1 6 1 26 13 = , P(B) = = and P(C) = = P(A) = . 36 9 36 6 36 18 Thus, P(A occurs before B) = P [A or (C « A) or (C « C « A) or

C1 + 100 100 C99 = C1, 100

C97 =

100

100

C3 +

C3 + 100

C 3,

100

+

100

+

100

C51 =

Ê 1ˆ C49) Á ˜ Ë 2¯

100

C99 = 299

C49

Thus, 2(100C1 +

100

100

C3 +

C1 +

100

C3 +

+ +

100

100

\ Probability of required event

C49) = 299

C49 = 298

298 100

2

=

1 . 4

Example 31 Each of two persons A and B toss three fair coins. The probability that both get the same number of heads is (a)

]

3 8

(b)

1 9

(c)

5 16

(d)

7 16

= P(A) + P(C « A) + P(C « C « A) +

Ans. (c)

= P(A) + P(C) P(A) + P(C)2 P(A) +

Solution: Let X be the number of heads obtained by A and Y be the number of heads obtained by B. Note that both X and Y are binomial variate with parameters n = 3 and p = 1/2. Probability that both A and B obtain the same number of heads is

2

=

1 Ê 13 ˆ 1 Ê 13 ˆ 1 +Á ˜ +Á ˜ +º 9 Ë 18 ¯ 9 Ë 18 ¯ 9

=

1/9 2 = 1 - 13/18 5

\

Example 30 A fair coin is tossed 100 times. The probability of getting tails 1, 3, , 49 times is (a) 1/2 (b) 1/4 (c) 1/8 (d) 1/16 Ans. (b) Solution

100

100

Let p = probability of getting a tail in a single trial = 1/2. n = number of trials = 100 X = number of tails in 100 trials.

and We have P(X = r) =

100

Cr pr qn

–r

r

Ê 1ˆ Ê 1ˆ = 100 Cr Á ˜ Á ˜ Ë 2¯ Ë 2¯

100 - r

Ê 1ˆ = 100 Cr Á ˜ Ë 2¯

100

Now, P(X = 1) + P(X = 3) + =

100

Ê 1ˆ C1 Á ˜ Ë 2¯

+

100

Ê 1ˆ C3 Á ˜ Ë 2¯

100

100

+

P(X = 2) P(Y = 2) + P(X = 3) P(Y = 3) 2

2

2

3 3 3 3 È Ê 1ˆ ˘ È Ê 1ˆ ˘ È Ê 1ˆ ˘ È Ê 1ˆ ˘ = Í 3 C0 Á ˜ ˙ + Í 3 C1 Á ˜ ˙ + Í 3 C2 Á ˜ ˙ + Í 3 C3 Á ˜ ˙ Ë 2¯ ˙ Í Ë 2¯ ˙ Í Ë 2¯ ˙ Í Ë 2¯ ˙ ÎÍ ˚ Î ˚ Î ˚ Î ˚

Ê 1ˆ = Á ˜ Ë 2¯

6

[1 + 9 + 9 + 1] =

2

20 5 = . 64 16

Example 32 Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is 1 1 2 3 (a) (b) (c) (d) 2 3 3 4

Ans. (a)

+ P(X = 49)

100

P(X = 0) P(Y = 0) + P(X = 1) P(Y = 1) +

+

Ê 1ˆ C49 Á ˜ Ë 2¯

100

Solution: Let E denote the event that the number of boys ahead of every girl is at least one more than the number of girls ahead of her. Note that E¢ consists of the following arrangements.

IIT JEE eBooks: www.crackjee.xyz Probability 9.13

(2C1) (4!) = 48 ways (ii) both girls are at the second and third place. This can happen in (3!)(2!) = 12 ways. \ n(E¢) = 48 + 12 = 60 Also, total number of ways of arranging 3 boys and 2 girls is 5! = 120 ways. 60 1 = \ P(E¢) = 120 2 fi

P(E) = 1 – P(E¢) =

that

( x - 20) ( x - 40) < 0 is x - 30 (a) 1/50

(b) 3/50

(c) 3/25

(d) 7/25

Ans. (d) Solution Let E =

1 2

( x - 20) ( x - 40) ( x - 20) ( x - 30) ( x - 40) = x - 30 ( x - 30) 2

Sign of E is same as that of sign of (x – 20) (x – 30) (x – 40) = F(say).

Six distinct numbers are selected from

Example 33

A natural number x is chosen at random

Example 35

Note that F < 0 if and only if numbers are divisible both by 3 and 5 is (a) 1/33 (b) 1/35 (c) 1/49 (d) none of these Ans. (d) Solution: Total number of ways of selecting 6 numbers out of 1, 2, , 150 is 150C6. A number is divisible by 3 and 5 both if and only if it is divisible by 15. Thee are 10 numbers viz. 15, 30, , 150 that are divisible by 15. Thus, probability of required event 10

=

C6

150

C6

.

Example 34 There are two balls in an urn whose colours are not known (each ball can be either white or black). A white ball is put into the urn. A ball is drawn from the urn. The probability that it is white is (a) 1/4 (b) 1/3 (c) 2/3 (d) 1/6 Ans. (c) Solution: Let Ei (0 £ i £ 2) denote the event that urn contains i white and (2 – i) black balls. Let A denote the event that a white ball is drawn from the urn. We have P(Ei) = 1/3 for i = 0, 1, 2. and P(A | E1) = 1/3, P(A | E2) = 2/3, P(A | E3) = 1. By the total probability rule, P(A) = P(E1) P(A | E1) + P(E2) P(A | E2) + P(E3) P(A | E3) =

1 È1 2 ˘ 2 + +1 = 3 ÍÎ 3 3 ˙˚ 3

0 < x < 20 or 30 < x < 40. \ E < 0 in (0, 20) » (30, 40) Thus E is negative for x = 1, 2, , 19, 31, 32, is E, < 0 for 28 natural numbers. 28 7 = \ Required probability = . 100 25

, 39, that

Example 36 An experiment has 10 equally likely outcomes. Let A and B be two non-empty events of the experiment. If A consists of 4 outcomes, the number of outcomes that B must have so that A and B are independent, is (a) 2, 4 or 8 (b) 3, 6 or 9 (c) 4 or 8 (d) 5 or 10. Ans. (d) Solution: Let nB and nA«B denote the number outcomes favourable to B and A « B respectively. As A and B are independent. P(A « B) = P(A) P(B) fi fi fi

nA « B

=

10

n 4 ¥ B 10 10

5 nA«B = 2nB fi 5|nB nB = 5 or 10.

Example 37 P(B) =

[

1 £ nB £ 10.]

If A, B and C are three events such that

3 1 1 , P(A « B « C¢) = and P(A¢ « B « C¢) = , 4 3 3

then P(B « C) is equal to (a) Ans. (a) Solution:

1 12

(b)

We have

1 6

(c)

1 15

(d)

1 9

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P(B « C¢) = P[(A » A¢) « (B « C¢)] = P(A « B « C¢) + P(A¢ « B « C¢) =

1 1 2 + = 3 3 3

A letter is known to have come from either

Example 40

consecutive letters, TA, are visible. The probability that the letter has come from CALCUTTA is

Now, P(B « C) = P(B) – P(B « C¢) 3 2 1 = = 4 3 12

(a) 4/11

(b) 1/3

(c) 5/12

(d) 1/7

Ans. (a)

B « C¢

B

Solution: Let E1 denote the event that the letter came E2 the event that the letter came from CALCUTTA. Let A denote the event that the two consecutive alphabets visible on the envelope are TA. We have P(E1) = 1/2, P(E2) = 1/2, P(A |E1) = 2/8, P(A|E2) = 1/7. Therefore, by Bayes' theorem we have

C

B«C Fig. 9.6

Example 38 A fair coin is tossed n times. If the probability that head occurs 6 times is equal to the probability that head occurs 8 times, then value of n is (a) 24 (b) 48 (c) 14 (d) 16 Ans. (c) Solution: Let X be the number of times head occurs. Then X follows a binomial distribution with parameter n and p = 1/2. We are given that P(X = 6) = P(X = 8). 6

Ê 1ˆ Ê 1ˆ n fi C6 Á ˜ Á ˜ Ë 2¯ Ë 2¯ n



n

n-6

Ê 1ˆ Ê 1ˆ C6 Á ˜ = n C8 Á ˜ Ë 2¯ Ë 2¯



n

8

Ê 1ˆ Ê 1ˆ = n C8 Á ˜ Á ˜ Ë 2¯ Ë 2¯

n -8

P( E2 ) P( A | E2 ) 4 = . P ( E1 ) P( A | E1 ) + P( E2 ) P( A | E2 ) 11

P(E2|A) =

Example 41 A group of 6 boys and 6 girls is randomly divided into two equal groups. The probability that each group contains 3 boys and 3 girls is (a) 10/231 (b) 5/231 (c) 90/231 (d) 100/231 Ans. (d) Solution: The number of ways of choosing 6 persons out of 12 for a group is 12C6. The number of ways in which this group can contain 3 boys and 3 girls is (6C3) (6C3). Therefore required probability

n

C6 = nC8 = nCn–8 fi 6 = n–8 or n = 14

Example 39 A person writes 4 letters and 4 addresses on 4 envelopes. If the letters are placed in the envelopes at random, the probability that not all letters are placed in correct envelopes is (a) 1/24

(b) 11/24

(c) 5/8

(d) 23/24

\ Required probability = 1 -

1 23 = . 24 24

6

3

12

3

C6

=

100 . 231

Example 42 In a hurdle race, a runner has probability p the runner succeeded 3 times, the conditional probability

Ans. (d) Solution: Required probability = 1 – P(all the letters are put in correct envelops) The number of ways of putting the letters in the envelops = 4P4 = 4! The number of ways of putting letters in correct envelops = 1

( C )( C ) 6

=

(a)

3 5

(b)

2 5

(c)

1 5

(d)

4 5

Ans. (a) Solution:

Let A denote the event that the runner succeeds B denote the event that the

P(B|A) =

P ( B « A) P ( A)

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But P(B « A) = P exactly once in two other trials) = p(4C2 p2 (1 – p)2) = 6p3 (1 – p)2 and P(A) = 5C3 p3 (1 – p)2 = 10p3 (1 – p)2 P(B|A) =

Thus,

6 p3 (1 - p) 2 10 p (1 - p) 3

2

=

3 . 5

Example 43 Let F denote the set of all onto functions from A = {a1, a2, a3, a4} to B = {x, y, z}. A function f is chosen at random from F. The probability that f–1{x} consists of exactly one element is (a) 2/3 (b) 1/3 (c) 1/6 (d) 0 Ans. (a) Solution F. 4 Total number of functions from A to B is 3 = 81. The number of functions which do not contain x(y) [z] in its range is 24. \ the number of functions which contain exactly two elements in the range is 3 ◊ 24 = 48. The number of functions which contain exactly one elements in its range is 3. Thus, the number of onto functions from A to B is 81 – 48 + 3 = 36 [using principle of inclusion exclusion] \ n(F) = 36 Let f∈ F. We now count the number of ways in which f–1(x) consists of single element. We can choose preimage of x in 4 ways. The remaining 3 elements can be mapped onto {y, z} is 23 – 2 = 6 ways. \ f –1 (x) will consists of exactly one element in 4 ¥ 6 = 24 ways. Thus,the probability of the required event is 24/36 = 2/3. Example 44

Three integers are chosen at random

probability that their product is even is (a) 7/19 (b) 2/19 (c) 17/19

20

C3 C3

=

10 ¥ 9 ¥ 8 2 = 20 ¥ 19 ¥ 18 19

17 19

Example 45 A box contains tickets numbered 1 to N. n tickets are drawn from the box with replacement. The probability that the largest number on the tickets is k is Êkˆ (a) Á ˜ Ë N¯

n

Ê k - 1ˆ (b) Á Ë N ˜¯

(c) 0

(d)

n

k n - ( k - 1)

n

Nn

Ans. (d) Solution: Let X denote the largest number on the n tickets drawn. We have Êkˆ P(X £ k) = Á ˜ Ë N¯

n

Ê k - 1ˆ and P(X £ k – 1) = Á Ë N ˜¯ n

Êkˆ Ê k - 1ˆ Thus P(X = k) = Á ˜ - Á Ë N¯ Ë N ˜¯

n

n

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS Example 46

A biased cubical die is thrown twice. Let

P = probability of getting the same number both the times. Q and an even number on the second throw. Then 1 1 2 (b) P ≥ - Q (a) Q £ 4 3 3

(d) 10/19

Solution: The product of any number of integers will be even if and only if at least one of the integers involved is even. Let A denote the event that the product of the three integers is even. Then A¢ denotes the event that the product of the three integers is odd. The total number of ways of choosing three integers out of 1, 2, , 20 is 20C3. The number of ways favourable to A¢ is 10C3. Therefore, 10

P(A) = 1 – P(A¢) =

(c) Q ≥

Ans. (c)

P(A¢) =



1 3

(d) P + 2Q £ 1

Ans. (a), (b), (d) Solution: Let pi = probability of getting number i on the die in a single throw (1 £ i £ 6) We have 2 2 P = p1 + p2 +

+ p62

Q = p1(p2 + p4 + p6) + p3(p2 + p4 + p6) + p5(p2 + p4 + p6) = (p1 + p3 + p5)(p2 + p4 + p6) As

ab £

1 (a + b)2, we get 4

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1 (p + p2 + 4 1

+ p 6) 2 =

1 4

2A =

k =0

We have P + 2Q =

p12

+

p22

+ ... +

A = 22n – 1 +

p62

+ 2(p2 + p4 + p6)(p1 + p3 + P5) £ (p1 + p2 + + p 6) 2 = 1

Thus,

Also, 3P + 2Q – 1 2 2 = 3( p1 + p2 +

+ p62 ) + 2(p1 + p3 + p5)(p2 + p4

+ p6) – (p1 + p2 + … + p6)2

1 2n ( C n) 2

1 1 2n 1 + ( C n) fi P > 2 22 n+1 2 (2n)! Also, 2nCn = n !n ! (1)(2)(3)(4)...(2n - 1)(2n) = n !n !

P=

2n (n!)[(1)(3)(5)...(2 n - 1)] n !n !

=

22 n -1 Ê 3 ˆ Ê 5 ˆ Ê 7 ˆ Ê 2n - 1 ˆ Á ˜ Á ˜ Á ˜ ... Á ˜ n Ë 2 ¯ Ë 4 ¯ Ë 6 ¯ Ë 2n - 2 ¯

2 2 2 2 + ( p2 – 2p2p4 + p4 ) + ( p2 – 2p2p6 + p6 )



22 n -1 n

2 2 2 2 + ( p3 – 2p3p5 + p5 ) + ( p4 – 2p4p6 + p6 )

From (3), we get

+ p62 ) – 2p1(p3 + p5) – 2p2 (p4 + p6) – 2p3p5 – 2p4p6

2 2 2 2 = ( p1 – 2p1p3 + p3 ) + ( p1 – 2p1p5 + p5 )

2



+ 2nCn = 22n + 2nCn

=

2 2 = 2( p1 + p2 +



2n

 ( 2 n Ck )

2

2

= (p1 – p3) + (p1 – p5) + (p2 – p4) + (p2 – p6)2 + (p3 – p5)2 + (p4 – p6)2 ≥ 0 3P + 2Q ≥ 1 1 2 P≥ - Q 3 3

P≥ For

Example 47 An unbiased cubical die is thrown 2n times, and let P denote the probability of getting even numbers at least n times, then 1 1 1 (a) P > (b) P ≥ + 2 2 4n 163 319 (c) P = if n = 4 (d) P = if n = 5 256 512 Ans. (a), (b), (c), (d) Solution: Let p = probability of getting an even number on a single throw of the die and let X = number of times an even number is obtained, then X ~ B(2n. p) Now, P = (X ≥ n) =

2n

2n

k =n

k =n

 P( X = k ) =  ÊÁË 2n Ck 22n ˆ˜¯

A = 2n 2 where A = 2nCn + 2nCn + 1 + A = 2nCn + 2nCn – 1 + Adding (1) and (2) we get

1

+ 2nCn + 2nC0

(1) (2)

and for

1 1 + 2 4n

1 1 8 + ( C4 ) 2 29 1 70 163 = + 9 = 2 2 256 1 1 10 319 n = 5, P = + 11 ( C5 ) = 2 2 512 n = 4, P =

Example 48 If A and B are two events, the probability that exactly one of them occurs is given by (a) P(A) + P(B) – 2P(A « B) (b) P (A « B¢) + P (A¢ « B) (c) P(A » B) – P(A « B) (d) P(A¢) + P(B¢) – 2P (A¢«B¢) Ans. (a), (b), (c), (d) Solution: We have P(exactly one of A, B occurs) = P[(A « B¢ ) » (A¢ « B)] = P(A « B¢) + P(A¢ « B) = P(A) – P(A « B) + P(B) – P(A « B) = P(A) + P(B) – 2P(A « B) = P(A » B) – P(A « B) Also P(exactly one of A, B occurs) = [1 – P(A¢ « B¢)] – [1 – P (A¢ » B¢] = P(A¢ » B¢) – P (A¢ « B¢) = P(A¢) + P(B¢) – 2P (A¢ « B¢). Example 49 If A and B are two events such that P(A) = 1/2 and P(B) = 2/3, then

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(a) P (A » B) ≥ 2/3 (b) P(A « B¢) £ 1/3 (c) 1/6 £ P (A « B) £ 1/2 (d) 1/6 £ P (A¢ « B) £ 1/2. Ans. (a), (b), (c), (d)

(a) (b) (c) (d)

A A P P

and B are independent and B are mutually exclusive (A¢ | B) = 3/4 (B¢ | A¢) = 1/2.

Ans. (a), (c), (d)

Solution: We have P(A » B) ≥ max {P(A), P(B)} = 2/3 Next P(A « B) = P(A) + P(B) – P(A » B) ≥ P(A) + P(B) – 1 = 1/6 and P (A « B) £ min {P(A), P(B)} = 1/2 fi 1/6 £ P(A « B) £ 1/2 Also P(A « B¢) = P(A) – P(A « B) £ 1/2 – 1/6 = 1/3 Lastly P (A¢ « B) = P(B) – P(A « B) Hence 2/3 – 1/2 £ P(A¢ « B) £ 2/3 –1/6 fi 1/6 £ P(A¢ « B) £ 1/2

Solution: We have

Example 50 If A, B and C are three events, then (a) P (exactly two of A, B, C occur) £ P(A « B) + P(B « C) + P(C « A) (b) P(A » B » C) £ P(A) + P(B) + P(C) (c) P(exactly one of A, B, C occur) £ P(A) + P(B) + P(C) – P(B « C) – P(C « A) – P(A « B) (d) P (A and at least one of B, C occurs) £ P(A « B) + P(A « C). Ans. (a), (b), (c), (d)

Since A and B are independent,

Solution: We have P(exactly two of A, B, C occur) = P(A « B) + P(B « C) + P(C « A) – 3P (A « B « C) £ P(A « B) + P(B « C) + P(C « A) Also P(A » B » C) £ P(A » B) + P(C) £ P(A) + P(B) + P(C) Next P(exactly one of A, B, C occurs) = P(A) + P(B) + P(C) – 2P(A « B) – 2P(B « C) – 2P(C « A) + 3P(A « B « C) = P(A) + P(B) + P(C) – P(A « B) – P(B « C) – P(C « A) – [P (A « B) + P(B « C) + P(C « A) – 3P(A « B « C)] = P(A) + P(B) + P(C) – P(A « B) – P(B « C) – P(C « A) – P (exactly two of A, B, C occur) £ P(A) + P(B) + P(C) – P(A « B) – P(B « C) – P(C « A) Lastly P(A and at least one of B, C occur) = P[A « (B » C)] = P [(A « B) » (A « C)] = P(A « B) + P(A « C) – P[(A « B) « (A « C)] = P(A « B) + P(A « C) – P(A « B « C) £ P(A « B) + P(A « C). Example 51 For two events A and B, if P(A) = P(A | B) = 1/4 and P(B | A) = 1/2, then

P(A) = P(A|B) =

P ( A « B) fi P(A « B) = P(A) P(B) P ( B)

Therefore, A and B are independent. Also P(A « B) = P(A) P(B | A) = (1 4) (1 2) = 1 8 π 0 \ A and B cannot be mutually exclusive. As A and B are independent P(A¢ | B) = P(A¢) = 1 – P(A) = 1 – 1 4 = 3 4 P(B) = P(B | A) = 1 2 fi

P(B¢ | A¢) = P(B¢) = 1 – P(B) = 1 2

Example 52

If A and B are two independent events

such that P(A¢ « B) = 2 15 and P(A « B¢) = 1/6, then P(B) is (b) 1 6 (c) 4 5 (d) 5 6 . (a) 1 5 Ans. (b), (c) Solution:

Since A and B are independent

2 15 = P(A¢ « B) = P(A¢) P(B) = [1 – P(A)] P(B) (1) and 1 6 = P(A « B¢) = P(A) P(B¢) = P(A) [1 – P(B)] (2) Subtracting (2) from (1) we get P(A) – P(B) = 1 30 or P(A) = P(B) + 1 30 Putting this value in (2), we get

[ P( B) + 1 30]

[1 – P(B)] = 1 6



30P(B)2 – 29P(B) + 4 = 0



P(B) = 1 6 , 4 5 .

Example 53 A student appears for tests I, II, and III. The student is successful if he passes either in tests I and II or tests I and III. The probabilities of the student passing in tests I, II, III are p, q and 1 2 , respectively. If the probability that the student is successful is 1 2 , then (a) p = 1, q = 0 (b) p = 2 3 , q = 1 2 (c) p = 3 5 , q = 2 3 p and q.

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Ans. (a), (b), (c), (d) Solution: Let A, B and C be the events that the student is successful in tests I, II and III, respectively. Then P(the student is successful) = P [(A « B « C¢) » (A « B¢ « C) » (A « B « C)] = P (A « B « C¢) + P(A « B¢ « C) + P(A « B « C) = P (A) P(B) P(C¢) + P(A) P(B¢) P(C) + P(A) P(B) P(C) [ A, B and C are independent] = pq (1 - 1 2) + p (1 - q ) (1 2) + ( pq ) (1 2) = \

1 1 [pq + p(1 – q) + pq] = p(1 + q) 2 2

1 1 = p(1 + q) 2 2



p (1 + q) = 1

(a) P(E) = 1/3, P(F) = 1/4 (b) P(E) = 1/2, P(F) = 1/6 (c) P(E) = 1/6, P(F) = 1/2 (d) P(E) = 1/4, P(F) = 1/3 Ans. (a), (d) Solution: We are given that P(E « F) = 1/12 and P(E¢ « F¢) = 1/2 As E and F are independent, we get P(E) P(F) = 1/12 and P(E¢) P(F¢) = 1/2 But P(A¢) = 1 – P(A). Therefore (1 – P(E)) (1 – P(F)) = 1/2 fi 1 – P (E) – P(F) + P(E) P(F) = 1/2 fi P(E) + P(F) = 7/12 The quadratic equation whose roots are P(E) and P(F) is x2 – (P(E) + P(F)) x + P(E) P(F) = 0 7 1 x+ =0 12 12

values of p and q. For instance, when p = n/(n + 1) and q = 1/n, where n is any positive integer.



x2 –

Let X be a set containing n elements. If two

fi fi

12x2 – 7x + 1 = 0 fi (3x – 1) (4x – 1) = 0 x = 1/3, 1/4.

Example 54

subsets A and B of X are picked at random, the probability that A and B have the same number of elements is 2n

(a) (c)

2

Cn 2n

1

(b)

1◊ 3 ◊ 5

(2n - 1)

(d)

2 ◊ (n !) n

2n

3n 4

n

Cn .

Ans. (a), (c) Solution: We know that the number of subsets of a set containing n elements is 2n. Therefore, the number of ways of choosing A and B is 2n ◊ 2n = 22n. We also know that the number of subsets (of X) which contain exactly r elements is nCr. Therefore, the number of ways of choosing A and B so that they have the same number of elements is + ( nC n) 2 ( nC 0) 2 + ( nC 1) 2 + ( nC 2) 2 + 1◊ 2 ◊ 3 (2n - 1) (2n) = 2nCn = n !n ! =

[1◊ 3 ◊ 5

(2n - 1)] [2 ◊ 4 ◊ 6 n !n !

(2n)]

2n (n !) [1◊ 3 ◊ 5 (2n - 1)] 2n [1◊ 3 ◊ 5 (2n - 1)] = n !n ! n! Thus, the probability of the required event is =

2n

2

Cn 2n

=

1◊ 3 ◊ 5

(2n - 1) n

2 (n !)

Example 56 A random variable X follows binomial distribution with mean a and variance b. Then a (a) a > b > 0 (b) >1 b

.

Example 55 Let E and F be two independent events. The probability that both E and F happen is 1/12 and the probability that neither E nor F happens is 1/2. Then,

(c)

a2 a2 is an integer (d) is an integer a-b a+b

Ans. (a), (b), (c) Solution: Suppose X ~ B(n, p). Then a = np, b = npq As 0 < q < 1, we get 0 < npq < np a >1 fi a>b>0fi b Also,

a2 n2 p 2 np = = = n, a-b 1- q np - npq

which is an integer. Example 57

If A1, A2, …, An are n independent events

1 , i = 1, 2, …, n. The probability that i +1 none of A1, A2, … An occurs is

such that P(Ai) =

(a)

n n +1

(c) less than

(b) 1 n

1 n +1

(d) none of these

Ans. (b), (c) Solution: P(A¢ « A¢ 2 … « A¢n) = P(A¢1) P(A¢2 ) … P(A¢n)

IIT JEE eBooks: www.crackjee.xyz Probability 9.19

[ Ê 1ˆ = Á1 - ˜ Ë 2¯ =

k +1

A1, A2, …, An are independent]

Pn =

1 ˆ Ê 1ˆ Ê ÁË1 - ˜¯ º ÁË1 ˜ 3 n + 1¯

1 2 3 n -1 n 1 1 ¥ ¥ ¥º¥ ¥ = < . 2 3 4 n n +1 n +1 n

Example 58

Let

=

(n + 1)(n - 5) 16(n)(n - 3)

n = 2k, k ≥ 4 k

Ï 9 ¥ 10 ¥ 19 ¸ =2Ì ˝ + 100 = 670 6 Ó ˛ \ Probability of required event =

670 = 0.067 . 10000

Note that 0.067 > 0.065 and 0.067 < 0.068. Example 59 A bag contains n (≥7) balls number 1, 2, … n. Four balls are taken out at random without replacement. Let pn denote the probability that all the four balls bear odd numbers, then (a) pn =

(n + 1)(n - 5) if n is odd 16n(n - 3)

(b) pn =

(n - 4)(n - 6) a if n is even 16(n - 1)(n - 3)

(c) p20


7 162

Ans. (a), (b), (c), (d) Solution: Let n = 2k + 1, k ≥ 3

C4

=

k (k - 1)(k - 2)(k - 3) 2k (2k - 1)(2k - 2)(2k - 3)

(k - 2)(k - 3) (2k - 4)(2k - 6) = 4(2k - 1)(2k - 3) 16(2k - 1)(2k - 3)

=

(n - 4)(n - 6) 16(n - 1)(n - 3)

p20 = < Also,

C4

2k

=

Ans. (a), (b), (d)

Thus, the number of favourable ways = 1 × 1 + 2 × 2 + … + 10 × 10 +9×9+8×8+…+1×1

C4

(k + 1)k (k - 1)(k - 2) (2k + 1)(2k )(2k - 1)(2k - 2)

(k + 1)(k - 2) (2k + 1 + 1)(2k + 1 - 5) = 16(2k + 1)(2k + 1 - 3) 4(2k + 1)(2k - 1)

pn =

Solution: The total number of ways of choosing the ticket is 1000. Let the four digit number on the ticket be x1, x2, x3, x4. Note that 0 £ x1 + x2 £ 18 and 0 £ x3 + x4 £ 18. Also, the number of non-negative integral solutions of x + y = m (with 0 £ x, y £ 9) is m + 1 if 0 £ m £ 9 and is 19 – m if 10 £ m £ 18.

=

=

A four digit number (numbered from

digits is equal to the sum of its last two digits. If a four digit number is picked up at random, then the probability that it is lucky is (a) > 0.065 (b) < 0.068 (c) 0.066 (d) 0.067

C4

2 k +1

(20 - 4)(20 - 6) 14 = 16(20 - 1)(20 - 3) (19)(17) 14 1 = 322 23

p20 >

Example 60

14 7 = 324 162 A ship is fitted with three engines E1, E2

and E3. The engines function independently of each other 1 1 1 with respective probabilities , and . For the ship to 2 4 4 be operational at least two of its engines must function. Let X denote the event that the ship is operational and let X1, X2 and X3 denote respectively the events that the engines E1, E2 and E3 are functioning. Which of the following is (are) true? 3 c (a) P ÈÎ X1 | X ˘˚ = 16 (b) P [Exactly two engines of the ship are func7 tioning |X] = 8 (c) P [X |X2] =

5 16

(d) P [X |X1] =

7 16

Ans. (b), (d) Solution: We are given 1 1 1 P(X1) = , P (X2) = , P (X3) = 2 4 4

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We have P (X) = P (X1 X2 X3) + P (Xc1 X2 X3) + P (X1 Xc2 X3) + P (X1 X2 Xc3) Ê 1ˆ Ê 1ˆ Ê 1ˆ Ê 1ˆ Ê 1ˆ Ê 1ˆ = Á ˜Á ˜Á ˜ + Á ˜Á ˜Á ˜ Ë 2¯ Ë 4¯ Ë 4¯ Ë 2¯ Ë 4¯ Ë 4¯ 1 Ê 1ˆ Ê 3ˆ Ê 1ˆ Ê 1ˆ Ê 1ˆ Ê 3ˆ + Á ˜Á ˜Á ˜ +Á ˜Á ˜Á ˜ = Ë 2¯ Ë 4¯ Ë 4¯ Ë 2¯ Ë 4¯ Ë 4¯ 4 c

Now, P (X 1|X) =

(

P X1c X 2 X 3 P(X )

)

P (Exactly two engines of the ship are functioning |X) = 1 – P (X1 X2 X3 |X)

(1/2)(1/4)(1/4)

=1– Next, P [X |X2] =

= = =

= =

P( X 2 )

and

P(Y|X) =

As

P (X

1 P(Y X ) 1 fi = P(X ) 3 3 1 1 1 , we get P (X) = , P (Y) = 6 2 3

Y) =

Now, P (X

Y) = P(X) + P (Y) – P (X

P (X

) (

)

1/4 5 8 X1 )

(

) (

P ( X1 X 2 X 3 ) + P X1 X 2c X 3 + P X1 X 2 X 3c

)

P( X1 )

(1/2)(1/4)(1/4) + (1/2)(3/4)(1/4) + (1/2)(1/4)(3/4) 1/2 7 16

following is(are) correct?

(b) P(E2) = 4 455

36 455

(d) P(E1 | E2) =

1 9

Ans. (a), (b), (c), (d) Solution P(E1) = P(choosing 5 and two tickets from 6 to 15) 10

1 1 1 , P (Y |X) = and P (X « Y) = . Which of the 2 3 6

9 91

(c) P(E1 « E2) =

= Let X and Y be two events such that P (X |Y)

1 1 1 1 - = π 3 6 6 3

Example 62 Three numbers are chosen at random without replacement from {1, 2, 15}. Let E1 be the event that minimum of the chosen numbers is 5 and E2 their maximum is 10 then (a) P(E1) =

P( X1 )

1 1 1 2 + – = , 2 3 6 3

fi X and Y are independent. Also, P(Xc Y) = P(Y) – P(X Y) =

(1/2)(1/4)(1/4) + (1/2)(1/4)(1/4) + (1/2)(1/4)(3/4)

Y)

1 Ê 1 ˆ Ê 1ˆ = Á ˜ Á ˜ = P(X) P(Y) Ë 2 ¯ Ë 3¯ 6

Y) =

P( X 2 )

P(X

P(X Y ) 1 1 fi = P (Y ) 2 2

=

X2 )

(

Example 61 =

P(X

Ans. (a) (b)

7 8

P ( X1 X 2 X 3 ) + P X1c X 2 X 3 + P X1 X 2 X 3c

P [X |X1] =

=

1/4

=

(b) X and Y are independent (c) X and Y are not independent 1 (d) P (Xc « Y) = 3

Solution: P (X|Y) =

Ê 1ˆ Ê 1ˆ Ê 1ˆ ÁË ˜¯ ÁË ˜¯ ÁË ˜¯ 1 2 4 4 = = 1 /4 8

2 3

(a) P(X « Y) =

15

C2 C3

=

10 ¥ 9 3¥ 2 9 ¥ = 2 15 ¥ 14 ¥ 13 91

P(E2) = P(choosing 10 and two tickets from 1 to 9) 9

=

C2

15

C3

=

9¥8 3¥ 2 36 ¥ = 2 15 ¥ 14 ¥ 13 455

IIT JEE eBooks: www.crackjee.xyz Probability 9.21

P(E1 « E2) = P(choosing 5 and 10 and one ticket from 6 to 9) 4

=

15

P(E1 | E2) =

C1 C3

=4¥

3¥ 2 4 = 15 ¥ 14 ¥ 13 455

P(A|E1) =

By Bayes’ rule P (E1|A) =

1 9

Example 63 One die has three faces marked with 1; two faces marked 2, and one face marked 3; another has one face marked 1, two marked 2 and three marked 3, then (a) The most probable throw with two dice is 4 (b) The probability of most probable throw is

1 4

(c) The probability of most probable throw is 7/18 (d) None of these. Ans. (a), (c) Solution: The minimum sum that can occur is 2 and maximum that can occur is 6. The number of ways in x 2, x 3, 4 5 6 2 3 2 x , x , x in the expansion of (3x + 2x + x ) (x + 2x + 3x3) = 3x2 + 8x3 + 14x4 + 8x5 + 3x6. This shows that the sum that occurs most often is 4. Also, the number of ways in which different sums can occur is (3 + 2 + 1) (1 + 2 + 3) = 36. \ The probability of 4 is 14/36 = 7/18. Example 64 A box contain N coins, m of which are fair and rest are biased. The probability of getting a head when a fair coin is tossed is 1/2, while it is 2/3 when a biased coin is tossed. A coin is drawn from the box at random and time it shows tail. The probability that the coin drawn is fair is (a)

9m 8N + m

(b)

m 8N + m

(c)

N 8n + m

(d)

1 m

1 1 1, 2 1 2 ¥ = P(A|E2) = ¥ = . 3 3 9 2 2 4

P( E1 ) P( A | E1 ) 9m = . P( E1 ) P ( A | E1 ) + P( E2 ) P( A | E2 ) 8 N + m

A, B and C are events such that Example 65 P(A) = P(B) = P(C) = 1/5, P(A « B) = P(B « C) = 0 and P(A « C) = 1/10. The probability that at least one of the events A, B or C occurs is. (a) 1/2 (b) 2/3 (c) 2/5

(d) 3/5

Ans. (a) Solution: We have A « B « C Õ A « B fi P(A « B « C) £ P(A « B) = 0 Since P(A « B « C) ≥ 0, we get P(A « B « C) = 0. Now, P(at least one of A, B, C) = P(A » B » C) = P(A) + P(B) + P(C) – P(B « C) – P(C « A) – P(A « B) + P(A « B « C) =

1 1 1 1 1 + + -0-0+0= . 5 5 5 10 2

MATRIX-MATCH TYPE QUESTIONS Example 66

Let P = {1, 2, 3, …., n}.

A nonempty subset s of P is said to be non-isolated if x ∈ S fi x – 1 or x + 1 ∈ S. Let a subset A of P containing exactly k elements be selected at random, and let pk denote the probability that A is a non-isolated subset of P. Match the entries in column 1 with entries in column 2 Column 1 Column 2 (p) 3/190 (a) n = 50, p2 (q) 8/663 (b) n = 25, p3 (r) 1/25 (c) n = 20, p4 (p) 1/100 (d) n = 20, p5 Ans. p q r s

Ans. (a)

a

p

q

r

s

Solution: Let E1, E2 and A denote the following events: E1 : coin selected is fair E2 : coin selected is biased A second toss results in a tail. m N -m , P(E2) = , P(E1) = N N

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: Number of ways of choosing a subset A of P containing exactly two elements is nC2.

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Set A will be isolated if both k, k + 1 ∈ A for 1 £ k £ n – 1. \ number of non-isolated set is (n – 1) 2 ( n - 1) 2 = Required probability = n ( n - 1) n 2 1 = When n = 50, required probability p2 = 50 25 Number of ways of choosing a subset A of P containing exactly 3 elements is nC3. Number of ways of choosing non-isolated set is (n – 2) n-2 6 = \ Required probability = n n n ( - 1) C3

Example 67 Consider the system of equations ax + by = 0 cx + dy = 0 where a, b, c, d, ∈ {0, 1}. The probability that system has (a) unique solution (p) 1 (b) no solution (q) 3/8 (c) exactly two solutions (r) 0 (d) a solutions (s) 5/8 Ans.

6 1 = When n = 25, p3 = 25 ( 24) 100 6

3 = When n = 20, p3 = ( 20)(19) 190 Number of ways of choosing subset A of P containing exactly 5 elements is nC5. Now, we choose number of favourable ways. We split into two cases: Case 1 For 3 £ k £ n – 2, we choose k, k + 1 and k + 2 and choose two non-isolated numbers from {1, 2, … , k – 1} This can be done in (k – 2) ways. \ number of ways n-2

n-4

k =3

k =1

= Â ( k - 2) = Â k =

1 ( n - 4)( n - 3) 2

Case 2 For 4 £ k £ n – 1 we choose k, k + 1 and 3 nonisolated numbers from {1, 2, …, k – 1}. This can be done in k – 3 ways. \ number of way's n -1

n-4

k =4

k =1

= Â ( k - 3) = Â k 1 = ( n - 4)( n - 3) 2 Thus, number of ways of choosing a non-isolated subset containing exactly 5 elements 1 2 = ( n - 4)( n - 3) - ( n - 4) = ( n - 4) 2 For n = 20, required probability ps =

( 20 - 4) 20

2

C5

162 (5)( 4)(3)( 2) = 20 ¥ 19 ¥ 18 ¥ 17 ¥ 16 8 = 663

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: a, b, c, d can be chosen in 24 ways. The system of equations ax + by = 0 cx + dy = 0 will have a unique solution if and only if Δ=

a b c d

= ad – bc π 0

As a, b, c, d ∈ {0, 1}. Now, ad – bc π 0 if and only if ad = 1 and bc = 0 or ad = 0 and bc = 1. But ad = 1 and bc = 0 ¤ a = d = 1 and (b = c = 0 or b = 1, c = 0 or b = 0, c = 1) that is, ad = 1, bc = 0 in 3 cases. Similarly, ad = 0, bc = 1 in 3 cases. Therefore, the probability the system has a unique solution is 6/16 = 3/8. Since x = 0, y = 0 satisfy the system of equations irrespective of the values of a, b, c, d, the probability that the system has a solution is 1. \ probability the system has no solution is 0. Also the probability the system has exactly two solutions is 0. Example 68

A is a set containing n elements. A subset

P of A is chosen at random. The set A is reconstructed by replacing the elements of the subset P. A subset Q of A is again chosen at random. The probability that (a) P « Q = f (p) n(3n-1)/4n (b) P « Q is a singleton (q) (3/4)n (c) P » Q = A (r) 2nCn/4n (d) |P| = |Q| (s) 9n(n-1)/2(4n)

IIT JEE eBooks: www.crackjee.xyz Probability 9.23

p1 = P(X = 5) + P(X = 6) =

where |X| = number of elements in X. Ans.

=

11

C5 (pq)5 =

11

11

C5 p5 q6 +

11

C 6 p 6q 5

C5 (0.24)5

p3 = P(X = 4) + P(X = 7) =

11

C4 p4 q7 +

11

C7 p7 q4

=

11

C4 (pq)4 (1 - 3pq)

=

11

C4 (0.24)4 (0.28)

p10 = 0 and p11 = P(X = 0) + P(X = 11) = (0.4)11 + (0.6)11 Solution: Let A = {a1, a2, ..., an}. For each ai ∈ A (1 £ i £ n) we have the following four cases: (i) ai ∈ P and ai ∈ Q (ii) ai ∉ P and ai ∈ Q (iii) ai ∈ P and ai ∉ Q (iv) ai ∉ P and ai ∉ Q

Example 70

set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} with replacement. The probability that the roots of x2 + px + q = 0 are (a) (b) (c) (d)

Thus, the total number of ways of choosing P and Q is 4n. Out of these four choices (i) is not favourable for P « Q = f. Thus, probability that P « Q is (3/4)n. Next, P « Q can contain exactly one element in nC1(3n-1) = n(3n-1) ways.

If p and q are chosen randomly from the

imaginary equal real negative

(p) 0.38 (q) 0.03 (r) 0.62 (s) 1

Ans.

\ Probability that P « Q is singleton = n(3n–1)/4n Also, P » Q = A in 3n ways. \ Probability P » Q = A is (3/4)n . Lastly, |P| = |Q| in the following ways (nC0)2 + (nC1)2 + ... + (nCn)2 =

2n

C n.

Example 69 A man takes a step forward with probability 0.4 and backward with probability 0.6. Suppose the man takes 11 steps and pr denotes the probability that the man is r steps away from his initial position, then value of (a) p1

(p) 0

(b) p3

(q)

(c) p10

(r) (0.4)11+(0.6)11

(d) p11

(s)

We enumerate the possible values of p and q for which this can happen in Table 9.1. Table 9.1

11

11

Solution: The roots of x2 + px + q = 0 will be imaginary if and only if p2 - 4 q < 0, i.e., p2 < 4q.

C5 (0.24)5

C4 (0.24)4 (0.28)

Ans.

Solution: Let X = the number of steps taken in the forward direction, then X ~ B (n, p) with n = 11, p = 0.4.

q

1 2 3 4 5 6 7 8 9 10 Total

p

1 1, 2 1, 2, 3 1, 2, 3 1, 2, 3, 4 1, 2, 3, 4 1, 2, 3, 4, 5 1, 2, 3, 4, 5 1, 2, 3, 4, 5 1, 2, 3, 4, 5, 6

Number of pairs of p, q

1 2 3 3 4 4 5 5 5 6 38

Thus, the number of possible pairs = 38. Also, the total number of possible pairs is 10 ¥ 10 = 100. \ P (roots are imaginary) = 0.38 P(roots are real) = 0.62

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The roots of x2 + px + q = 0 will be equal if and only if p2 - 4q = 0. The possible pairs of (p, q) are (2, 1), (4, 4), (6, 9). Thus, P(roots are equal) = 0.03. As p > 0, q > 0, roots of x2 + px +q = 0 are negative if these are real.

answer

Cards are dealt one by one from well-

Example 71

r (1 £ r £ 4) aces are obtained. If pr denotes the probability of drawing r nth draw (with n ≥ 4), and Pr = (52C4) pr, then (p) 52 - nC3 (a) P1 (b) P2 (q) (n-1) (52 – nC2) (r) (n-1C2) (52 – nC1) (c) P3 (s) n - 1C3 (d) P4 Ans.

(b) p2

(q)

(c) pn-2

(r)

(d) pr

(s)

( n - 4)( n - 5)( n - 6) n ( n - 1)( n - 2) 6 n ( n - 1) ( n - 2)

3 ( n - r )( n - r - 1) n ( n - 1)( n - 2)

Ans.

Solution: As (n - 3) people do not know the answer, n-3 n

C4

C4

=

( n - 4)( n - 5)( n - 6) n ( n - 1)( n - 2)

Next,

Solution: We must draw (r - 1) aces in the (n - 1) draws. The total number of ways of drawing (n - 1) cards out of 52 is 52Cn-1. The number of ways of drawing (r - 1) aces and n - r other cards is (48Cn-r) (4Cr-1). The probability of getting an ace at the rth draw is (5 - r)/(53 - n). Thus,

( p =

48

Cn - r

r

52

)(

4

Cr - 1

Cn - 1



3 ( n - 3) Ê n - 3ˆ Ê 3 ˆ = p2 = Á Ë n ¯˜ ÁË n - 1˜¯ n ( n - 1) r r – 1) persons are chosen from (n – 3) persons who do not know the answer and the rth person is chosen from 3 persons who know the answer. The probability of this event is n-3

5-r 53 - n

n

48!4!(52 - n )!( n - 1) = = ( n - r )! ( 48 + r - n)!( r - 1)!( 4 - r )!52! 1 =

52

C4

=

fi 52C4 pr = (n-1Cr-1) (52-nC4 – r). Example 72 Suppose n (> 6) people are asked a question successively in a random order and exactly 3 out of n people know the answer. Let pr denote the probability that r 3 ( n - 3) n ( n - 1) do not know the

\

Cr - 1



3 n - (r - 1)

(n - 3)! (r - 1)! (n - r + 1)! 3 ◊ ◊ (r - 1)! (n - r - 2)! n! n - r +1 =

[(n-1Cr-1) (52-nC48+r-n)]

Cr - 1

3 (n - r ) (n - r - 1) n(n - 1) (n - 2)

pn-2 =

Example 73

1 n

C3

=

6 n ( n - 1)( n - 2)

A coin has probability p of showing

head when tossed. It is tossed n times. Let pn denote the probability that no two (or more) consecutive heads occur, then (p) 1 - 2p2 + p3 (a) p1 (b) p2 (q) 1 - p2

IIT JEE eBooks: www.crackjee.xyz Probability 9.25

(c) p3 (d) pn

(r) 1 (s) (1 - p)pn-1 + p(1 - p) pn-2

Ans.

Solution: The number of ways in which the students can mark the answer is (3) (2) (1) = 6. All the three answer can be correct in just one way, exactly two correct is impossible, exactly one can be correct in exactly 3 ways and no correct in 2 ways. Example 75

Solution: When n = 1, then two possible outcomes viz. H and T satisfy the condition that two (or more consecutive heads do not occur. Thus, p1 = 1. When n = 2, the possible outcomes are HH, HT, TH and TT HH does not satisfy the condition that no two (or more) consecutive heads occur. Thus, p2 = 1 - P(HH) = 1 - pp = 1 - p2. For n ≥ 3, if the last outcome is T, then the probability n - 1) tosses do not contain two (or more consecutive heads is pn - 1 and if the last outcome is H, then (n - 1)th outcome must be T n - 2) tosses do not contain two (or more) consecutive heads is pn-2. Hence, pn = pn-1 ¥ P (nth toss results in a tail) + pn-2 ¥ P (nth toss results in a head and (n - 1)th toss results in a tail) = (1 - p) pn-1 + p(1 - p)pn - 2 For p3, put n = 3 in the above expression and simplify. Example 74

A student has to match three historical

events—Dandi March, Quit India Movement and Mahatma 1942. The student has no knowledge of the correct answers and decides to match the events and years randomly. If X denotes the number of correct answer obtained by the student, then (a) P(X = 3) (p) 1/2 (b) P(X = 2) (q) 0 (c) P(X = 1) (r) 1/6 (d) P(X = 0) (s) 1/3 Ans.

Letters of the word INDIANOIL are

arranged at random. Probability that the word formed 1 (a) contains the word (p) 9 C5 INDIAN 1 (b) contain the word (q) 5 7 C2 C2 (9!) OIL 1 (c) begins with I and (r) 9 C4 ends with L 1 (d) has vowels at the (s) 7 C3 9 C2 odd places

( )( ) ( )( )

Ans. p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: Total number of ways of arranging letters of the 9! word INDIANOIL is 3! 2! (a) Treating INDIAN as a single object we can permute INDIAN, O, I and L in 4! ways. \ probability of the required event =

4!3! 2! = 9!

1

( C )( C ) 7

9

3

2

(b) We can permute OIL I, N, D, I, A, N in

7! ways. 2! 2!

\ probability of the required event =

2! 2!3! 2! = 7!9!

1

( C ) ( C ) (9!) 5

7

2

2

(c) Fixing an I L at the last place, we can permute the remaining letters viz. A, D, I, I, N, N, O in

7! ways. 2! 2!

\ probability of the required event is

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=

2! 2!3! 2! = 7!9!

Ans. (a) Solution:

1

( C ) ( C ) (9!) 5

7

2

2

(d) Vowels can be arranged at odd places viz 1st, 3rd, 5! 5th, 7th and 9th in ways. 3! The remaining letters can be arranged at 4 even places in 4! ways. 2! \ probability of the required event =

5! 4! 3! 2! ¥ = 3! 2! 9!

1 9

C4

=

Statement-1: Let A and B be two events

such that P(A » B) = P(A « B) then P(A « B') = P(A' « B) = 0 Statement-2: Let A and B be two events such that P(A » B) = P(A « B) then P(A) + P(B) = 1. Ans. (c) Solution:

P(A » B) = P(A « B)

¤ P(A) + P(B) – P(A « B) = P(A « B) ¤ [P(A) – P(A « B)] + [P(B) – P(A « B)] = 0.

and

1 1 (2 + 2p) + (2 + p) £ 1 fi 6 + 6p + 10 + 5p £ 15 5 3

Example 78

But P(A), P(B) ≥ P(A « B). Thus, (1) is possible if and only if P(A) – P(A « B) = 0 and P(B) – P(A « B) = 0 ¤ P(A) = P(B) = P(A « B) \ P(A) + P(B) = 1 need not hold. Next, P(A « B') = P(A) – P(A « B) = 0

(2)

Statement-1: A natural number is

chosen at random. The probability that sum of the digits of its square is 93 is 0. Statement-2: A number is divisible by 31 if and only if sum of its digits is divisible by 31. Ans. (c) Solution: Let n ∈ N, be such that sum of digits of its square is 93. Then 3|n2 fi 3|n fi 9|n2. A contradiction. Example 79

(1)

(1)

1 1 1 1 (1 + 5p) + (1 + 2p) + (1 – p) + (1 – 3p) £ 1 5 3 3 5

fi 11p £ – 1 fi p £ – 1/11. From (1) and (2) we get – 1/5 £ p £ – 1/11 fi p.

C5

ASSERTION-REASON TYPE QUESTIONS Example 76

1 1 1 1 (1 + 5p), (1 + 2p), (1 – p) (1 – 3p) ≥ 0 5 3 3 5

fi p ≥ – 1/5, p ≥ – 1/2, p £ 1, p £ 1/3 fi – 1/5 £ p £ 1/3



1 9

Now,

Statement-2 is true. See theory.

Let m ∈ N, and suppose three numbers

are chosen at random from the numbers 1, 2, 3, ....., m. Statement-1: If m = 2n for some n ∈ N, then the chosen 3 numbers are in A.P. with probability . 2 ( 2n - 1) Statement-2: If m = 2n + 1, then the chosen numbers are 3n in A.P. with probability 2 4n – 1

1 1 (1 – p) and (1 – 3p) are probabilities of four mutually 3 5 exclusive events, then p Statement-2: If A, B, C and D are four mutually exclusive events, then

Ans. (b) Solution: We can choose three numbers out of m in mC3 ways. Let the numbers be x1, x2, x3. Now, x1, x2, x3 are in A.P. if and only if x1 + x3 = 2x2, that is, if and only if either both x1, x3 are odd or both x1, x3 are even. If m = 2n, x1 and x3 can be chosen in nC2 + nC2 = n(n – 1) ways. n ( n - 1) In this case probability of the required event is 2 n C3

P(A), P(B), P(C), P(D) ≥ 0

=

Similarly P(A' « B) = P(B) – P(A « B) = 0 1 1 Example 77 Statement-1: If (1 + 5p), (1 + 2p), 5 3

P(A) + P(B) + P(C) + P(D) £ 1.

6 n ( n - 1) 3 = 2 n ( 2 n - 1) ( 2 n - 2) 2 ( 2 n - 1)

IIT JEE eBooks: www.crackjee.xyz Probability 9.27

If m = 2n + 1, x1 and x3 can be chosen in n + 1C2 + nC2 = n2 ways. n2 In this case probability of the required event is 2 n + 1 C3 =

3n 4 n2 - 1

. Statement-1: Let A, B, C be three

Example 80

mutually independent events. Statement-1: A and B » C are independent. Statement-2: A and B « C are independent. Ans. (b) Solution: As A, B, C are mutually independent P(B « C) = P(B) P(C), P(C « A) = P(C) P(A), P(A « B) = P(A) P(B) and P(A « B « C) = P(A) P(B) P(C) Now, P[A « (B » C)] = P[(A « B) » (A « C)] = P(A « B) + P(A « C) – P(A « B « C) = P(A) P(B) + P(A) P(C) – P(A) P(B) P(C) = P(A) [P(B) + P(C) – P(B) P(C)] = P(A) [P(B) + P(C) – P(B « C)] = P(A) P(B » C) and P(A « B « C) = P(A) P(B) P(C) = P(A) P(B « C) Example 81 Let H1, H2, ...., Hn be mutually exclusive events with P(Hi) > 0, i = 1, 2 , ...., n. Let E be any other event with 0 < P(E) < 1. Statement-1: P(Hi|Ei) > P(E|Hi) P(Hi) for i =1, 2, ...., n. n

Statement-2: Â P ( H i ) = 1 i =1

Ans. (d) Solution: Statement-1 is not always true. For instance, if P(Ei « H) = 0 for some i, then P ( Hi « E ) P(Hi|E) = =0 P (E) and P(E|Hi) . P(Hi) =

P ( E « Hi ) P (Hi )

P(Hi) = 0

P(Hi|E) = P(E|Hi) P(Hi) = 0.

that is,

However, if 0 < P(Hi « E) < 1, for i = 1, 2, ...., n, then P(Hi|E) =

=

P ( Hi « E ) P (E)

=

P ( Hi « E ) P ( Hi )

P ( E | Hi ) . P ( Hi ) P (E)

P (Hi )

P (E)

> P(E|Hi) . P(Ei)

[

0 < P(E) < 1]

Statement-2 is true. Example 82 Statement-1: Let n ≥ 3 and A1, A2, ... An 1 be n independent events such that P(Ak) = for 1 £ k k +1 1 £ n, then P(A¢1 « A¢2 « ... « A¢n) = . n +1 Statement-2: n events A1, A2, ..., An (n ≥ 3) are independent if any only if P(A1 « A2 « ... « An) = P(A1) P(A2) ... P(An) Ans. (c) Solution: See theory on page 10.3 to see that the statement-2 is false. We have P(A¢1 « A¢2 « ... « A¢n) = P(A¢1) P(A¢2) ... P(A¢n) Ê 1ˆ Ê 2ˆ Ê 3ˆ Ê n ˆ = ÁË ˜¯ Á ˜ ÁË ˜¯ ◊ ◊ ◊ Á 2 Ë 3¯ 4 Ë n + 1˜¯ = Example 83

1 n +1 Statement-1: If A and B are two events

such that P(B) = 1, then A and B are independent. Statement-2: A and B are independent if and only if P(A « B) = P(A) P(B) Ans. (a) Solution: For truth of statement-2, see theory on page 10.3. Now, P(B) = 1 fi P(B¢) = 1 – P(B) = 0 We have A « B¢ £ B¢ fi 0 £ P(A « B¢) £ P(B¢) = 0 fi P(A « B¢) = 0 = P(A) P(B¢) fi A and B¢ are independent. [See page 10.3] fi A and B are independent. Example 84

Statement-1: The probability of drawing

either an ace or a king from a pack of 52 playing cards in 1 a single draw is . 13 Statement-2: If A and B are two events, then P(A » B) = P(A) + P(B) – P(A « B) Ans. (d) Solution: See theory on page 10.1 for truth of the statement-2.

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COMPREHESION-TYPE QUESTIONS

Let A denote the event of drawing an ace and B denote the event of drawing a king, then P(A) =

4 4 , P(B) = and P(A « B) = 0 52 52 4+4 2 1 π = 52 13 13

\ P(A » B) =

Statement-1: A natural number x is

Example 85

Paragraph for Question Nos. 87 to 91 A class consists of n students. For 0 £ k £ n, let Ek denote the event that exactly k students out of n pass in the examination. Let P(Ek) = pk and let A denote the event that a student X selected at random pass in the examination. Example 87 equals

probability that ( x - 10) ( x - 50) < 0 is 0.69 x - 30 Statement-2: If A is an event, then 0 < P(A) < 1 Ans. (c) Solution:

Statement-2 is false since 0 £ P(A) £ 1.

( x - 10) ( x - 50) < 0 ¤ 10 < x < 30 or x > 50 x - 30 Thus, x can take 69 values. \ probability of the desired event is 0.69 Example 86 In a T teams. Each team plays one match against every other team. Each team has 50% chance of winning any game it plays. No match ends in a tie. Statement-1 The probability that there is an undefeated team in the tournament is 5/16 Statement-2 The probability that there is a winless team in the tourament is 3/16. Ans. (c)

(a) 1/2

(b) 2/3

(c) 1/6

(d) 1/(n + 1)

Example 88

If P(Ek) = C for 0 £ k £ n, then

the probability that X is the only student to pass the examination is (a) 3/4n (b) 2/(n + 1) (c) 2/n(n + 1) Example 89

(d) 3/n(n + 1)

If P(Ek) μ k for 0 £ k £ n, then P(A)

equals

¤ x = 11, ..., 29 or x = 51, ... 100

(a) 3n/(4n + 1)

(b) (2n + 1)/3n

(c) 1/(n + 1)

(d) 1/n2

Example 90

If P(Ek) μ k for 0 £ k £ n, then

the probability that X is the only student to pass the examination is 3 6 (b) (a) n ( n + 1) n ( n + 1)( 2n + 1) (c)

1 n

Example 91

1 n ( 2n + 1)

(d)

If P(Ek) μ k for 0 £ k £ n, then

n

Solution: The total number of matches played in the tournament is 5C2 = 10. Note that, there can be at most one undefeated team. The probability that a particular team, say team A, wins all its Ê 1ˆ 4 matches = Á ˜ Ë 2¯

4

\ probability there is an undefeated time in the tournament

( )

If P(Ek) = C for 0 £ k £ n, then P(A)

4

5 Ê 1ˆ 5 = C1 Á ˜ = Ë 2¯ 16 Similarly, the probability that there is a winless team = 5/16.

lim Â

nƕ k = 0

P(Ek|A) equals

(a) 0 (c) 1/6

(b) 1/2 (d) 1

Ans. 87. (a), 88. (c), 89. (b), 90. (b), 91. (d) Solution: 87. P(Ek) = C for 0 £ k £ n and n

 P(Ek) = 1 fi C = 1/(n + 1).

k =0

n -1

Also

P(A|Ek) =

By the total probability rule

n

Ck - 1 Ck

=

k n

IIT JEE eBooks: www.crackjee.xyz Probability 9.29

then value of a is (a) 9 (c) 3

n

P(A) = Â P(Ek) P(A | Ek) k=0

= 88.

By the Bayes' rule P(E1 | A) =

n ( n + 1) 1 1 = . n ( n + 1) 2 2 P ( E1 ) P ( A | E1 )

Example 94

n

k=0

Now,

n

2

k=0

n 2 (n + 1)

P(A) = Â P(Ek)P(A|Ek)= =

90. P(E1 | A) = =

n

 k2

k=0

n ( n + 1)( 2n + 1) 2n + 1 = . 6 3n n ( n + 1) 2

2

P ( E1 ) P ( A | E1 )

6 = . n ( n + 1)( 2n + 1) 91.

P(Ek|A) =

P ( Ek ) P ( A | Ek )

=

1 (1 - 2p), then p belongs to 4 (b) (- 1/9, 1/4) (d) (- 1, 0)

Example 95 P(none of E1, E2, E3) equals (a) 0 (b) p1 + p2 + p3 (c) (1 - p1) (1 - p2) (1 - p3) (d) none of these Example 96

P(E1 « E2¢) + P(E2 « E3¢) + P(E3 « E1¢)

equals (a) (b) (c) (d)

P ( A)

2 1 3n ¥ n ( n + 1) n 2n + 1

1 1 (1 + 3p), p2 = (1 - p) and 3 4

(a) (- 1/3, 1) (c) (- 2/3, 2/3)

89. Let P(Ek) = k a where a is the constant of proportionality. Since  P(Ek) = 1, we get a = 2/n(n + 1)

If p1 = p3 =

2 = . n ( n + 1)

P ( A)

(b) 6 (d) [0, 1/5]

p1(1 - p2) + p2(1 - p3) + p3(1 - p1) p 1p 2 + p 2p 3 + p 3 p 1 p1 + p2 + p3 p1 – p2 + p3

Ans. 92. (d), 93.(a), 94. (b), 95. (d), 96. (c) Solution: 92.

p1 ≥ 0, p2 ≥ 0, p3 ≥ 0 p1 + p2 + p3 £ 1

and

P ( A)



p £ 1, p ≥ - 1 2 , p ≥ - 2 3

2k k 3n ¥ n ( n + 1) n 2n + 1

and

p £ - 7 23

6k 2 = n ( n + 1)( 2n + 1)

\

– 1/2 £ p £ 7/23 1 93. Use (p1 + p2 + p3) = (p1 p2 p3)1/3 to obtain 3 fi

n

fi  P(Ek | A) = 1. k=0

p1 = p2 = p3 = 1/3

94. Similar to example 92.

Paragraph for Question Nos. 92 to 96

95. P(none of E1, E2, E3) = 1 - P(E1 » E2 » E3)

Suppose E1, E2, E3 be three mutually exclusive events such that P(Ei) = pi for i = 1, 2, 3. 1 1 (1 – p), p2 = (1 + 2p) and p3 Example 92 If p1 = 2 3

96. Use E1 « E2¢ = E1 etc.

=

1 (2 + 3p), then p belongs to 5 (a) (- 1, 2) (c) (- 1, 1/47)

Example 93

(b) (- 1/3, 2/3) (d) ( - 1 2, - 7 23 )

If p1, p2, p3 are the roots of 27x3 – 27x2 + ax – 1 = 0

Paragraph for Question Nos. 97 to 101 A chess match between two grandmasters X and Y is won X's chances of winning, drawing or losing any particular game are a, b, c respectively. The games are independent and a + b + c = 1. Example 97

The probability that X wins the match

after (n + 1) games (n ≥ 1) is

IIT JEE eBooks: www.crackjee.xyz 9.30 Comprehensive Mathematics—JEE Advanced

(a) na2 bn-1 2

(b) a (nb

n-1

2

(c) na bc

+n(n - 1)b

99. n-2

X will win the match with probability •





n =1

n =1

n =1

c)

 pn = a 2  nbn –1 + a 2 c  n(n – 1)bn – 2

n-1

(d) none of these =

The probability that Y wins the match

Example 98

after the 4th game is 2

(a) 3bc (b + 2a)

(b) bc (3b + a)

(c) 2ac2 (b + c)

(d) abc (2a + 3b)

(c)

a3 + a 2 c

( a + c )3 a

(b)

a3 + 3a 2 c

( a + c )3

3

(a + c)

3

(d) none of these

b +b c 3

(a)

(c)

2

(b + c )

3

c3

(b + c )3

Example 101

c + 3c a 3

(b)

2

( a + c )3

(d) none of these

The probability that there is no winner is

(a) (1 - a) (1 - c)

97. X can win after the (n + 1)th game in the following two mutually exclusive ways (i) X n games draws (n – 1) games and wins the (n + 1)th game (ii) X

n games, wins exactn games and draws (n - 2) games and wins the (n + 1)th game.

For (i) the probability is (nP1 abn-1)a and for (ii) the probability is (nP2) (ac)bn - 2 a \ The probability X wins after the (n + 1)th game pn = na2 bn-1 + n(n - 1) a2 bn-2 c = a2(nbn-1 + n(n - 1) bn-2 c). 98. In Example 97, put n = 3, interchange a and c to obtain c2[3b2 + (3) (2) ba] = 3bc2 (b + 2a)

a 2 ( a + c ) + 2a 2 c

a3 + 3a 2 c

=

( a + c )3

( a + c )3

Since

a3 + 3a 2 c

( a + c )3

+

c3 + 3c 2 a

( a + c )3

.

= 1.

Paragraph for Question Nos. 102 to 105 A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required. The probability that X = 3 equals

25 216 5 (c) 36

25 36 125 (d) 216

(a)

Example 103

(b)

The probability that X ≥ 3 equals

(a)

125 216

(b)

25 36

(c)

5 36

(d)

25 216

Example 104

Solution:

(1 - b)2 (1 - b)3

101

(b) (1 - a) b(1 - c)

(c) b (d) 0 Ans. 97. (b), 98. (a), 99. (b), 100. (b), 101. (d)

2a 2 c

Interchange a by c in Example 97.

Example 102 The probability that Y wins the match is

Example 100

+

100.

The probability that X wins the match is

Example 99 (a)

=

2

a2

The conditional probability that X ≥ 6

given X > 3 equals 125 216 5 (c) 36

25 216 25 (d) 36

(a)

Example 105

(b)

Expectation of X equals

(a) 6 (c) 8

(b) 12 (d) 10

Ans. 102. (a), 103. (b), 104. (d), 105. (a) Solution: 102. Let p = 1/6, q = 5/6 We have P(X = r) = pqr – 1, r ≥ 1

IIT JEE eBooks: www.crackjee.xyz Probability 9.31

2

1 Ê 5ˆ 25 ÁË ˜¯ = 6 6 216

Now, P(X = 3) = 103.

104.

P(X ≥ 3) = 1 – [P(X = 1) + P(X = 2)]

Ans. 106. (b), 107. (c) Solution: 106. P(X > Y) = P(T1 win both the matches) + P(T1 wins one match and draws the other) 2

È1 1 Ê 5ˆ ˘ = 1– Í + Á ˜˙ Î 6 6 Ë 6¯ ˚

=

11 25 = =1– 36 36

=

P[( X ≥ 6) « ( X > 3)] P( X > 3)

P(X ≥ 6 | X > 3) =

=

P( X ≥ 6) P( X > 3)

=

=

and P(X > 3) = P(X ≥ 4) = q3 Thus, P(X ≥ 6 | X > 3) =

105.

q5 q

3

= q2 =





r =1

r =1

E(X) = Â rP( X = r ) = Â r pq =

p (1 – q)

2

=

p p

2

=

25 36

Example 107 11 (a) 36

1 =6 p

P(X > Y) is 1 (c) 2

7 (d) 12

1 1 13 + = 3 36 36

Paragraph for Question Nos. 108 and 109

If 1 ball is drawn from each of the boxes

(a)

82 648

(b)

90 648

(c)

558 648

(d)

566 648

Example 109 If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B2 is (a)

116 181

(b)

126 181

(c)

65 181

(d)

55 181

Ans. 108. (a), 109. (d) Solution: 108. P(balls of the same colour) = P(W1 W2 W3) + P(R1 R2 R3) + P(B1 B2 B3)

P(X = Y) is 1 (b) 3

Ê 1 ˆ Ê 1ˆ Ê 1ˆ Ê 1ˆ C1 Á ˜ Á ˜ + 2C2 Á ˜ Á ˜ Ë 2 ¯ Ë 3¯ Ë 6¯ Ë 6¯

B1, B2 and B3, the probability that all 3 drawn balls are of the same colour is

respectively. Each team gets 3 point for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T1 and T2, respectively, after two games.

5 (b) 12

2

Example 108 r –1

Football teams T1 and T2 have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T1 winning, 1 1 1 drawing and losing a game against T2 are , and , 2 6 3

1 (a) 4

1 1 5 + = 4 6 12

A box B1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2 contains 2 white balls, 3 red balls and 4 black balls. A third box B3 contains 3 white balls, 4 red balls and 5 black balls.

Paragraph for Question Nos. 106 to 107

Example 106

=

5

pq pq = = q5 p 1– q

Ê 1ˆ Ê 1ˆ Ê 1ˆ C2 Á ˜ + 2 C1 Á ˜ Á ˜ Ë 2¯ Ë 2¯ Ë 6¯

107. P(X = Y) = P(T1 and T2 win one match each) + P(T1 and T2 draw both matches)

But P(X ≥ 6) = pq5 + pq6 + pq7 + ... 5

2

13 (c) 36

1 (d) 2

Ê 1 ˆ Ê 2 ˆ Ê 3 ˆ Ê 3ˆ Ê 3ˆ Ê 4 ˆ Ê 2ˆ Ê 4ˆ Ê 5 ˆ = Á ˜Á ˜Á ˜ +Á ˜Á ˜Á ˜ +Á ˜Á ˜Á ˜ Ë 6 ¯ Ë 9 ¯ Ë 12 ¯ Ë 6 ¯ Ë 9 ¯ Ë 12 ¯ Ë 6 ¯ Ë 9 ¯ Ë 12 ¯

IIT JEE eBooks: www.crackjee.xyz 9.32 Comprehensive Mathematics—JEE Advanced

1 82 6 + 36 + 40] = [ 648 648

=

109. Let Ei denote the event that two balls are drawn from the Bi. Let A denote the event that balls drawn from the box are white and red. We have 1 P(E1) = P(E2) = P(E3) = 3 and P(A|E1) = P(A|E2) =

P(A|E3) =

(1)(3) = 1 6

( 2)(3) 9

5

C2

C2

(3)( 4) 12

C2

=

1 6

=

2 11

By the Bayes’ rule, P(E2|A) =

P ( E2 ) P ( A | E2 ) P( E1 ) P( A | E1 ) + P( E2 ) P( A | E2 ) + P( E3 ) P( A | E3 )

=

(1/3)(1/6) (1/3)(1/5) + (1/3)(1/6) + (1/3)( 2/11)

=

1 330 55 ¥ = 6 181 181

Paragraph for Question Nos. 110 and 111 Box 1 contains three cards bearing numbers 1, 2, 3; box box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xi be the number on the card drawn from the ith box, i = 1, 2, 3. Example 110 29 105 57 (c) 105 (a)

The probability that x1 + x2 + x3 is odd, is 53 105 1 (d) 2 (b)

Example 111 The probability that x1, x2, x3 are in an arithmetic progression, is (a)

9 105

(b)

10 105

11 7 (d) 105 105 * Examples 112 and 113 are multiple answer equestions. (c)

Ans. 110. (b), 111. (c) Solution: 110. Total number of ways of choosing x1, x2, x3 is (3) (5)(7) = 105 x1 + x2 + x3 is odd if exactly one of x1, x2, x3 is odd or all three of them are odd. Case 1 Exactly one of x1, x2, x3 is odd. (i) x1 is odd, x2, x3 are even. Number of ways is (2)(2) (3) = 12 (ii) x2 is odd, x1, x3 are even. Number of ways is (1)(3) (3) = 9 (iii) x3 is odd, x1, x2 are even. Number of ways is (1)(2) (4) = 8 Thus, in this case number of ways is 12 + 9 + 8 = 29. Case 2: All of x1, x2, x3 are odd. In this case number of ways is = (2)(3)(4) = 24 ∴ number of favourable ways = 29 + 24 = 53. 53 Thus, probability of the required event = 105 111. x1, x2, x3 are in A.P. ¤ 2x2 = x1 + x3. When x2 = 1, (x1, x3) = (1, 1) x2 = 2, (x1, x3) = (1, 3), (2, 2), (3, 1) x2 = 3, (x1, x3) = (1, 5), (2, 4), (3, 3) x2 = 4, (x1, x3) = (1, 7), (2, 6), (3, 5) x2 = 5, (x1, x3) = (3, 7) Thus, there are 11 ways in which x1, x2, x3 are in A.P. 11 . Therefore, probability of the required event is 105 Paragraph for Question Nos. 112* and 113* Let n1 and n2 be the number of red and black balls, respectively, in box I. Let n3 and n4 be the number of red and black balls, respectively, in box II. Example 112 One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability 1 that this red ball was drawn from box II is , then the 3 correct option(s) with the possible values of n1, n2, n3 and n4 is(are) (a) n1 = 3, n2 = 3, n3 = 5, n4 = 15 (b) n1 = 3, n2 = 6, n3 = 10, n4 = 50 (c) n1 = 8, n2 = 6, n3 = 5, n4 = 20 (d) n1 = 6, n2 = 12, n3 = 5, n4 = 20 Example 113

A ball is drawn at random from box I

and transferred to box II. If the probability of drawing a

IIT JEE eBooks: www.crackjee.xyz Probability 9.33

1 then the correct 3 option(s) with the possible values of n1 and n2 is(are) (b) n1 = 2, n2 = 3 (a) n1 = 4, n2 = 6 red ball from box I, after this transfer, is

(d) n1 = 3, n2 = 6 (c) n1 = 10, n2 = 20 Ans. 112. (a),(b), 113. (c), (d) Solution: 112. Let E1, E2 and A denote the following events. E1 : box I is selected E2 : box II is selected A : red ball is drawn. We have 1 1 P(E1) = , P(E2) = 2 2 P ( A | E1 ) =

P(E1) =

P ( A | E1 ) =

=

= ∴

n3 n1 , , P ( A | E2 ) = n1 + n2 n3 + n4



n1 n1 + n2

n1 1 = fi n1 + n2 = 3n1 n1 + n2 3

Example 114

Then

...(1)

Of the three independent events E1, E2

Let E1, E2 and A denote the following events. E1 : red ball is transferred from box I to box II. E2 : black ball is transferred from box I to box II

Probability of occurrence of E1 = Probability of occurrence of E3

Ans. 6 Solution: Let P (E1) = x, P(E2) = y, P(E3) = z. As E1, E2, E3 are independent, a = x(1 – y) (1 – z) b = (1 – x) y (1 – z) g = (1 – x) (1 – y) z and p = (1 – x) (1 – y) (1 – z) thus,

x b y g z a , = , = = p 1- x p 1- y p 1- z

As (a –2b) p = ab, and (b – 3g) p = 2bg a b Êaˆ Ê bˆ b g b g - 2 = Á ˜ Á ˜ and - 3 = 2 . p p Ë p¯ Ë p¯ p p p p

In the other two cases (1) does not hold.

A : red ball is drawn.

=

and E3, the probability that only E1 occurs is a, only E2 occurs is b and only E3 occurs is g. Let the probability p that none of events E1, E2 or E3 occurs satisfy the equations (a – 2b)p = ab and (b – 3g)p = 2bg. All the given probabilities are assumed to lie in the interval (0, 1).

When n1 = 3, n2 = 3, n3 = 5, n4 = 15 5 1 20 = is true 3 5 3 + 6 20 When n1 = 3, n2 = 6, n3 = 10, n4 = 50 10 1 60 = is true. 3 10 3 + 9 60 113.

n2 n1 ( n1 + n2 ) ( n1 + n2 – 1)

INTEGER-ANSWER TYPE QUESTIONS

P ( E1 ) P ( A | E1 ) + P ( E2 ) P ( A | E2 )

n3 n1 + n1 + n2 n3 + n4

( n1 + n2 ) ( n1 + n2 – 1) n1 ( n1 + n2 – 1) ( n1 + n2 ) ( n1 + n2 – 1)

+

n1 = 10, n2 = 20 and n1 = 3, n2 = 6.

1 n3 2 n3 + n4 = 1 n1 1 n3 + 2 n1 + n2 2 n3 + n4 1 = 3

n1 ( n1 – 1)

fi n2 = 2n1

P ( E2 ) P ( A | E2 )

n3 n3 + n4

n1 n1 – 1 , P ( A | E2 ) = n1 + n2 – 1 n1 + n2 – 1

By the total probability rule P(A) = P ( E1 ) P ( A | E1 ) + P ( E2 ) P ( A | E2 )

By the Bayes’ Rule P ( E2 |A ) =

n1 n2 , P(E2) = n1 + n2 n1 + n2



ˆb a Êa bÊ gˆ g = Á + 2˜ and Á1 - 2 ˜ = 3 ¯ p p Ëp pË p¯ p

Eliminating b/p we get a p 3g p = a p + 2 1 - 2g p

IIT JEE eBooks: www.crackjee.xyz 9.34 Comprehensive Mathematics—JEE Advanced



x (1 - x ) z (1 - z ) =3 x (1 - x ) + 2 1 - 2 z (1 - z )



x z =3 2- x 1 - 3z 2 1Ê1 ˆ 2 1 - 1 = Á - 3˜ fi = ¯ x 3Ë z x 3z



x =6 z

n

Ê 1ˆ Ê 1ˆ 1 – Á ˜ - n Á ˜ ≥ 0.96 Ë 2¯ Ë 2¯ n +1



n

£ 0.04 =

1 25

2 fi 2 ≥ 25(n + 1) ≥ 50 ...(1) fi n ≥ 6. Note that n = 6, 7 do not satisfy (1). When n Thus, minimum value of n is 8. n

Example 116

One ticket is selected at random from

100 tickets numbered 00, 01, 02, , 98, 99. If X and Y denote the sum and the product of the digits on the tickets, then the value of 57P(X = 9 | Y = 0) is Ans. 6 Solution: (Y = 0) is {00, 01,

xk – 3 in (1 – 3x6) (1 + 3C1x + 4 C 2 x 2 + 5C 3x 3 + ) [note that 6 £ k – 3 £ 11] k–3 + 2 Ck – 3 – (3) k – 3 + 2 – 6Ck – 3 – 6 = = P (k ) =

Now \

P[( X = 9) « (Y = 0)] 2 = 19 P(Y = 0)

Example 117 Three six-faced fair dice are thrown together let P(k) denote the probability that the sum of the numbers appearing on the dice is k (9 £ k £ 14), then 14

54S – 30 where S = Â P ( k ) is k =9

C2 – (3)

k–7

C2

21k - k 2 - 83 216 35 fi 54S – 30 = 5 54

Example 118 Let A and B be two events such that P(A) = 0.3 and P(A » B) = 0.8. If A and B are independent events, then 7P(B) is Ans. 5 Solution: As A and B are independent 0.8 = P(A » B) = P(A) + P(B) – P(A) P(B) = 0.3 + P(B) – (0.3) P(B) fi 0.5 = (0.7) P(B) fi 7P(B) = 5. Example 119

A pair of fair dice is rolled together

till a sum of either 5 or 7 is obtained. If p denotes the probability that 7 comes before 5, then 5p is equal Ans. 3 Solution: Let A denote the event that a sum of 7 occurs, B the event that a sum of 5 occurs and C the event that neither a sum of 5 nor a sum of 7 occurs. We have P(A) = 6 1 4 1 26 13 = , P(B) = = , P(C) = = . Thus, 36 6 36 9 36 18 p = P (A or (C « A) or (C « C « A) or = P (A) + P(C) P(A) + P(C)2 P(A) +

, 09, 10, 20,

and P[(X = 9) « (Y = 0)] = 2 100 .

k–1

S=

, 90}. Also, (X = 9) « (Y = 0) = {09, 90}. We have P(Y = 0) = 19 100

P(X = 9 | Y = 0) =

+ x 6) 3

x k –3 in (1 – x6)3 (1 – x)–3



n

Solution: The total number of cases is 6 ¥ 6 ¥ 6 = 63 = 216. The number of favourable ways xk in (x + x2 +

Example 115 The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two head is at least 0.96 is Ans. 8 Solution: Let n be the number of required tosses. Let X = number of head obtained We have X ~ B (n, 1/2) As P(X ≥ 2) ≥ 0.96 fi 1 – P (X = 0) – P (X = 1) ≥ 0.96 fi

Ans. 5

=

)

16 3 P( A) = = fi 5p =3 1 - P(C ) 1 - 13 18 5

Example 120 Two numbers are selected at random from the numbers 1, 2, , n. Let p denote the probability than m (where 0 < m < n). If n = 25 and m = 10 then 5p is equal Ans. 1. Solution Let A

x and the second be y.

IIT JEE eBooks: www.crackjee.xyz Probability 9.35

and second numbers is at least m. Let Ex denote the event x. We must have x – y ≥ m or y £ x – m. Therefore x > m and y < n – m. Thus, P(Ex) = 0 for 0 < x £ m and P(Ex) = 1/n for m < x £ n. Also, P(A | Ex) = (x – m)/(n – 1). Therefore, n

P(A) = Â P(Ex) P(A | Ex) x =1

n

=

x = m +1

Put

P(Ex) P(A | Ex) =

1 [1 + 2 + n(n - 1)

=

1 x-m ◊ x = m +1 n n - 1 n

Â

Â

+ (n - m)] =

(n - m) (n - m + 1) 2n(n - 1)

n = 25 and m = 10

Example 121

Let 0.8 the probability that a man aged

90 years will die in a year. If p is the probability that out of 4 men A1, A2, A3 and A4 each aged 90, A1 will die in a year p – 2490 is Ans. 6 Solution: Let Ei denote the event that Ai dies in a year. Then P(Ei) = 0.8 and P(E¢i) = 0.2 for i = 1, 2, 3, 4. Now, P(none of A1, A2, A3, A4 dies in a year) = P(E¢1 « E¢2 « E¢3 « E¢4) = (0.2)4 Let E denote the event that at least one of A1, A2, A3, A4 dies in a year. Then P(E) = 1 – P(E¢1 « E¢2 « E¢3 « E¢4) = 1- (0.2)4 = 0.9984. If F denote the event that A1 Then P (F | E) = 1/4. Also, 1 p = (0.9984) = 0.2496 4 fi

10000p – 2490 = 6

Example 122

There are n different objects, 1, 2, 3,

, n, distributed at random in n places marked 1, 2, 3, , n. If p is the probability that at least three of the objects occupy places corresponding to their number, then 6p is equal to Ans. 1.

p = nC 3 ◊

(n - 3)! n! (n - 3)! 1 = ◊ = n! n! 6 3! (n - 3)!

Example 123 Fifteen A persons, among whom are B1 A and B, sit down at random B2 at a round table. It p is the probability that there are exactly 4 persons between A and B then 14 p is equal to Fig. 9.1 Ans. 2 Solution: Let A occupy any seat at the round table. Then there are 14 seats available for B. If there are to be four persons between A and B, then B has only two ways to sit, as shown in Fig. 9.1. Thus, the probability of the required event is 2 14 = 1 7 . Example 124

A locker can be opened by dialing a

who does not know the code, tries to open the locker by dialing three digits at random. If p is the probability that the stranger succeeds at the kth trial, then 1000 p is equal to. (Assume that the stranger does not repeat unsuccessful combinations.) Ans. 1. Solution: Let A denote the event that the stranger succeeds at the kth trial. Then 999 998 1000 - k + 1 1 ¥ ¥ ¥ ¥ P(A) = 1000 999 1000 - k + 2 1000 - k + 1 fi

p=

1 1000

Example 125

fi 1000p = 1. If n positive integers are taken at random

and multiplied together, and pn is the probability that the last digit of the product is 2, 4, 6 or 8 then 125p3 – 50 is equal to. Ans. 6. Solution: The last digit of the product will be 1, 2, 3, 4, 6, 7, 8 or 9 if and only if each of the n positive integers ends

Solution: Let Ei denote the event that the ith object goes to the ith place. We have (n - 3)! for i < j < k P(Ei « Ej « Ek) = n!

in any of these digits. Now, the probability of an integer

Since we can choose 3 places out of n in nC3 ways, the probability of the required event is

the product will be 1, 3, 7 or 9 if and only if each of the n positive integers ends in 1, 3, 7 or 9. The probability for an integer to end in 1, 3, 7 or 9 is 4 10 = 2 5 . Therefore,

ending in 1, 2, 3, 4, 6, 7, 8 or 9 is 8 10 = 4 5 . Therefore, the probability that the last digit of the product of n integers is 1, 2, 3, 4, 6, 7, 8 or 9 is ( 4 5) . Next, the last digit of n

IIT JEE eBooks: www.crackjee.xyz 9.36 Comprehensive Mathematics—JEE Advanced

the probability for the product of n positive integers to end in 1, 3, 7 or 9 is ( 2 5) . Hence, the probability of the required event is n

n

n

4n - 2n Ê 4ˆ Ê 2ˆ = ÁË ˜¯ ÁË ˜¯ 5 5 5n

20 ¥ 18 ¥ 16 ¥ 14 224 = 20 ¥ 19 ¥ 18 ¥ 17 323 Hence, the probability of getting at least one pair is

When n = 3 we get 125p – 50 = 6 Example 126 Two integers r and s are drawn one at a time without replacement from the set 1, 2, , n. If pk = P(r £ k | s £ k), then 4p7 is if n = 25. Ans. 1. Solution: We have P[(r £ k ) and ( s £ k )] P(r £ k | s £ k) = P( s £ k ) k P(s £ k) = and n

Also,

k

P[(r £ k) and (s £ k)] =

n

C2 C2

=

k (k - 1) n(n - 1)

1 – 224 323 = 99 323 . Example 129

Put n = 25 and k = 7 Example 127

Ans. 1. Solution: fi fi fi fi

Ans. 1. Solution: Let X denote the number of tosses required. Then P(X = r) = (1 – p)r–1p for r = 1, 2, 3, Let E denote the event that the number of tosses required is even. Then P(E) = P[(X = 2) » (X = 4) » (X = 6) » ] = P(X = 2) + P(X = 4) + P(X = 6) + = (1 – p)p + (1 – p)3 p + (1 – p)5 p + =

p(1 - p) 1 - (1 - p)

2

=

p(1 - p) 2p - p

2

=

1- p 2- p

As we are given that P(E) = 2/5, we get 5(1 – p) = 2(2 – p) fi 1 – 3p = 0 or p = 1 3 Example 128

There are 10 pairs of shoes in a

cupboard, from which 4 shoes are picked at random. If p is the probability that there is at least one pair, then 323p – 97 is equal to

X

follows

a

binomial

9P(X = 4) = P(X = 2) 9 ◊ 6C 4 p 4 q 2 = 6C 2 p 2q 4 9p2 = q2 3p = q = 1 – p 4p = 1

[

[q = 1 – p] 6 C 4 = 6C 2] [ q > 0]

EXERCISE LEVEL 1

A biased coin with probability p(0 < p

time. If the probability that the number of tosses required is even is 2/5, then 3p is equal to

Suppose

distribution with parameters n = 6 and p. If 9P(X = 4) = P(X = 2), then 4p is equal to

pk = (k - 1) (n - 1)

Thus,

Ans. 2.

Solution: That total number of ways of choosing 4 shoes (in order) out of 10 pairs (or 20 shoes) is 20 ¥ 19 ¥ 18 ¥ 17. The number of ways in which no pair is selected is 20 ¥ 18 ¥ 16 ¥ 14. Thus, the probability of not getting a pair is

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. A group consists 2n students including you and your friend. If the group is split into two different sections A and B, each containing n students, then the probability that you and your friend are in the different sections is n n -1 (b) (a) 2n - 1 2n - 1 n +1 2 (c) (d) 2n - 1 2n - 1 2. Two persons A and B think of two numbers at random from the numbers 1, 2, …, n. Probability A think of a number smaller than thought by B is n -1 2n - 1 (b) 2n 2n n -1 n (c) (d) 2n 2n - 1 3. A box contains 20 cards of these 10 have letter J printed on them and the remaining 10 have E print(a)

IIT JEE eBooks: www.crackjee.xyz Probability 9.37

ed on them. 3 cards are drawn from the box, the probability that we can write JEE with these cards is (a) 9/80 (b) 1/8 (c) 4/27 (d) 15/38 4. A and B toss a coin alternatively till one of them gets a head and wins the game. If A begins the game, the probability B wins the game is (a) 1/2 (b) 1/3 (c) 1/4 (d) 2/3 5. A fair coin is tossed repeatedly. If head and tail apbility that head appears on the eighth toss is (a) 1/2 (b) 1/128 (c) 1/256 (d) 7/256 6. If P(A) > 0, then the event A is independent of itself if and only if P(A) is (a) 1/3 (b) 1/2 (c) 1

(d)

(

)

5 -1 2

7. Two numbers are selected at random from 1, 2, 3,…, 100 and are multiplied, then the probability (correct to two places of decimals) that the product thus obtained is divisible by 3 is (a) 0.22 (b) 0.33 (c) 0.44 (d) 0.55 8. The probability that an event A occurrs in a single trial of an experiment is 0.6. Three independent trials of the experiment are performed. The probability that the event A occurs at least twice is (a) 0.636 (b) 0.632 (c) 0.648 (d) 0.946 9. Suppose X ~ B(n, p) and P(X = 3) = P(X = 5). If p > 1/2, then (a) n £ 7 (b) n > 8 (c) n ≥ 9 (d) n ≥ 10 10. In three throw of a pair dice, the probability throwing doublets not more than twice is (a) 1/6 (b) 5/72 (c) 215/216 (d) 7/128 11. A bag contains 4 brown and 5 white socks. A man pulls two socks at random without replacement. The probability that the man gets both the socks of the same colour is (a) 5/108 (b) 1/6 (c) 5/18 (d) 4/9 12. A is a set containing n elements. Two subsets P and Q of A are chosen at random. (P and Q may have elements in common). The probability that P » Q π A is (b) 1/4n (a) (3/4)n (c) nC2/2n

(d) 1 - (3/4)n

13. Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is not equilateral is (a) 1/2 (b) 1/5 (c) 9/10 (d) 1/20 14. A positive integer is chosen at random. The probability that the sum of the digits of its square is 39 is (a) 1/39 (b) 2/39 (c) 1/11 (d) 0 2 15. A lottery sells n tickets and declares n prizes. If a man purchases n tickets, the probability of his winning at least one prize is (b) 1/2n (a) (n2 – n)!/(n2)! (d) none of these (c) (n – 1)!2/(n2 )! 16. Let x be a non-zero real number. A determinant is chosen from the set of all determinants of order 2 with entries x or – x only. The probability that the value of the determinant is non-zero is (a) 3/16 (b) 1/4 (c) 1/2

(d) 1/8

17. A die is rolled three times. The probability of getting a number larger than the previous number each time is (a) 5/72 (b) 5/54 (c) 13/216 (d) 1/18 18. In a game called “odd man out man out,” m (m > 2) persons toss a coin to determiane who will buy refreshments for the entire group. A person who gets an outcome different from that of the rest of the members of the group is called the odd man out. The probability that there is a loser in any game is (a) 1/2m (b) m/2m-1 (c) 2/m (d) 1/2m – 1 19. Two non-negative integers are chosen at random. The probability that the sum of the square is divisible by 11 is (a) 9/16 (b) 1/121 (c) 9/17 (d) 2/121 20. A natural number x is choosen at random from the

x+ is (a) 1/10

100 > 50 x

(b) 11/50

(c) 11/20 (d) 3/20

21. A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides, whereas the remaining (n + 1) coins are fair. A coin is picked up at random from the bag and tossed. If the

IIT JEE eBooks: www.crackjee.xyz 9.38 Comprehensive Mathematics—JEE Advanced

probability that the toss results in a head is 31/42, then n is equal to (a) 10 (b) 11 (c) 12 (d) 13 22. If X and Y are independent binomial variate B (5, 1/2) and B (7, 1/2), then P (X + Y = 3) is (a) 55/1024 (b) 55/4098 (c) 55/2048

(d) 55/512

23. Suppose n (≥ 3) persons are sitting in a row. Two of them are selected at random. The probability that they are not together is (a) 1 –

2 n

(b)

2 n -1

(c) 1 –

1 n

(d)

2 n

24. A bag contains four tickets marked with numbers 112, 121, 211, 222. One ticket is drawn at random from the bag. Let Ei (i = 1, 2, 3) denote the event that ith digit on the ticket is 2. Then which of the following is not true (a) E1 and E2 are independent (b) E2 and E3 are independent (c) E3 and E1 are independent (d) E1, E2, E3 are independent 25. If the letters of the word PROBABILITY are written down at random in a row, the probability that two B¢s are together is (a) 2/11 (b) 10/11 (c) 3/11 (d) 6/11 26. If four positive integers are taken at random and multiplied together, then the probability that the last digit is 1, 3, 7 or 9 is (a) 1/8 (b) 2/7 (c) 1/625 (d) 16/625 27. Two contestants play a game as follows: each is asked to select a digit from 1 to 9. If the two digits match they both win a prize. The probability that they will win a prize in a single trial is (a) 1/81 (b) 7/81 (c) 1/9 (d) 3/11 28. A fair coin is tossed 2n times. Probability of getting more heads than tails is (a) 1/2 Ê 1ˆ (c) 1 – 2nCn Á ˜ Ë 2¯

(b) 2n

(d)

2n

Ê 1ˆ Cn Á ˜ Ë 2¯

1 2

2n

(b) P(A « B) £ P(A) (c) P(A¢ « B¢ ) £ P(A¢ ) + P(B¢ ) – 1 (d) P(A « B) = P(A) P(B) 30. A sum of money is rounded off to the nearest rupee. The probability that the round off error is at most 10 paise is (a) 1/10 (b) 11/100 (c) 3/25 (d) 21/100 31. The probability that an electric bulb will last 150 days or more is 0.7 and that it will last at most 160 days is 0.8. The probability that the bulb will last 150 to 160 days is (a) 0.5 (b) 1/3 (c) 0.56 (d) 0.59 32. A bag contains three tickets numbered 1, 2 and 3. A ticket is drawn at random and put back in the bag, and this is done four times. The probability that the sum of the numbers drawn is even is (a) 40/81 (b) 41/81 (c) 14/27 (d) 13/81 33. Two numbers x and y are selected at random from the set {1, 2, 3,…, 3N}. The probability that x2 – y2 is divisible by 3 is (a) (3N – 1)/3N (b) (N – 1)/N (c) (5N – 3)/(9N – 3) (d) (N – 1)/2N 34. Three persons A, B and C are to speak at a function along with 7 other persons. The probability that A speaks before B and B speaks before C is (a) 3/70 (b) 1/6 (c) 3/7 (d) 1/9 35. n men and n women are seated at round table in random order. The probalility that they can be divided into n non-interrecting pairs so that each pair consists of a man and a woman is (a) 1/2n (b) 2(2n -1)/2ncn (c) 2n/2nCn (d) 1/(nCn)2 36. If a is an integer lying in [– 5, 30], then the probability that the graph of y = x2 + 2 (a + 4) x –5a + 64 is strictly above the x – axis is (b) 7 36 (a) 1 6

2n

Ê 1ˆ Cn Á ˜ Ë 2¯

(c) 2 9 2n +1

29. If A and B are two events, then which one of the following is not always true (a) P(A « B) £ P(A) + P(B) – 1

(d) 3 5 .

37. Two numbers X and Y are chosen at random (without replacement) from the set {1, 2, 5N}. The probability that X4 – Y4 is divisible by 5 is (a)

N -1 5N -1

(b)

4 (4 N - 1) 5(5 N - 1)

(c)

17 N - 5 5(5 N - 1)

(d)

3N - 5 . 5 (5 N - 1)

IIT JEE eBooks: www.crackjee.xyz Probability 9.39

arranged on a shelf in random order. The probability that these volumes stand in increasing order from left to right (the volumes are not necessarily kept side-by-side) is (a) 1 5

50 (b) 1 5

5 (c) 1 50

(d) 1 120 .

39. A speaks the truth in 70 percent cases and B in 80 percent cases. The probability that they will contradict each other in describing a single event is (a) 0.36 (b) 0.38 (c) 0.4 (d) 0.42. 40. 2n boys are randomly divided into two subgroups containing n boys each. The probability that the two tallest boys are in different groups is (a)

(c)

n 2n - 1 2n -1 4n

2

(b)

(d)

n -1 2n - 1 n -1 2n 2

.

(c)

1 n!

(b)

(n - k )! n!

(d)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 46. A and B are two events. The probability that at most one of A, B occurs is (a) 1 – P(A « B) (b) P(A¢) + P(B¢) – P(A¢ « B¢) (c) P(A¢) + P(B¢) + P(A » B) – 1 (d) P(A « B¢) + P (A¢ « B) + P(A¢ « B¢). 47. The probability of the simultaneous occurrence of two events A and B is p. If the probability that exactly one of A, B occurs is q, then (a) P(A¢) + P(B¢) = 2 + 2q – p (b) P(A¢) + P(B¢) = 2 – 2p – q (c) P(A « B | A » B) =

41. The numbers 1, 2, 3, , n are arranged in random order. The probability that the digits 1, 2, , k (k < n) appears as neighbours in that order is (a)

Then P(E' « F' | G) equals (a) P(E') + P(F') (b) P(E') – P(F') (c) P(E') – P(F) (d) P(E) – P(F')

p p+q

(d) P(A¢ « B¢) = 1 – p – q. 48. Suppose that P(A) = 3/5 and P(B) = 2 3 . Then (a) P(A » B) ≥ 2 3

k! n!

(n - k + 1)! n!

(b) 4 15 £ P(A « B) £ 3 5 .

42. A random variable X has the probability distribution: 1 2 3 4 5 6 7 8 x P(X=x) 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05

For the events E = {x|x is prime} and F = {x|x < 4}, the probability P(E » F) is (a) 0.35 (b) 0.77 (c) 0.87 (d) 0.50 43. Two boys and three girls stand in a queue. The probability the number of boys ahead of every girl is at least one more than the number of girls ahead of her (a) 1/2 (b) 1/3 (c) 2/3 (d) 0 44. A bag contains some white and some black balls, all combinations of balls being equally likely. The total number of balls in the bag is 10. If three balls are drawn at random without replacement and all of them are found to be black, the probability that the bag contains 1 white and 9 black balls is (a) 14/55 (b) 12/55 (c) 2/11 (d) 8/55 45. Let E, F, G be pairwise independent events with P (G) > 0 and P(E « F « G) = 0.

(c) 2 5 £ P(A | B) £ 9 10 (d) P(A « B¢) £ 1 3 . 49. If E and F are independent events such that 0 < P(E) < 1 and 0 < P(F) < 1, then (a) E and F are mutually exclusive (b) E and F¢ are independent (c) E¢ and F¢ are independent (d) P(E|F) + P(E¢ | F) = 1. 50. If A and B are two events, then (a) P(A « B) £ min {P(A), P(B)} (b) P(A « B) « max {0, 1 – P(A¢) – P(B¢)} (c) P(A « B) £ P(A » B) (d) P(A « B) = P(A) P(B) if A and B are independent. 51. Twenty children are standing in a line outside a these children have a one-rupee coin each and the remaining 10 have a two rupee coin each. The entry ticket is priced at Re. 1. If all the arrange-ments of the 20 children are equally likely, the probability that

IIT JEE eBooks: www.crackjee.xyz 9.40 Comprehensive Mathematics—JEE Advanced

(assume that cashier has no change to begin with.) (a) 0 (b) < 1/2 (c) > 1/2 (d) 1/2 52. Eight tickets numbered 000, 010, 011, 011, 100, 101, 101 and 110 are placed in a bag. One ticket is drawn from the bag at random. Let A, B and C denote the following events: A 0”, B—“the second digit is 0” and C— “the third digit is 0”. Then (a) A and B are independent (b) B and C are independent (c) C and A are independent (d) A, B, C are independent 53. Four natural numbers are taken at random and are multiplied. Let p1 denote the probability of the event that last digit of the product is 1, 3, 7 or 9, and p2 denote the probability that the last digit of the product is 5, then (a) p1 > p2 (b) p1 = p2 (c) p1 < p2 (d) p1 + p2 < 1/2 54. I wrote a letter to my friend X and gave it to my son to post it. The probability that my son will forget to post the letter is 1/10 and the letter will be lost in the post is 1/100. If my friend X did not receive the letter, then the probability that (a) my son forgot to post the letter is 891/1000 (b) the letter was lost in the post 891/991 (c) my son forgot to post the letter is 100/991 (d) none of these. 55. A number x (a) x2 - 25x £ 150 is 0.3 (b) x2 - 17x + 30 ≥ 0 is 0.88 (c) 30x - x2 is a perfect square of a natural number is 0.07. (d) 30x - x2 < 0 is 0.7 56. Two fair dice are thrown. The probability that difference between the number is (a) two is 2/9 (b) three is 1/6 (c) at least two is 1/2 (d) at most one is 4/9 57. I post a letter to my friend and do not receive a reply. It is known that one letter out of m letters do not reach its destination. If it is certain that my friend will reply if he receives the letter. If A denotes the event that my friend receives the letter and B that I get a reply, then

(a) P(B) = (1 - 1/m)2 (b) P(A « B) = (1 - 1/m)2 (c) P(A | B¢) = (m - 1)/(2m - 1) (d) P(A » B) = (m - 1)/m 58. An electric component manufactured by ‘KIAANElectronics’ is tested for its defectiveness by a sophisticated testing device. Let A denote the event “the device is defective” and B the event “the testing device reveals the component to be defective”. Suppose P(A) = a and P(B | A) = P(B¢ | A¢) = 1 - a, where 0 < a < 1, then (a) P(B) = 2a (1 - a) (b) P(A¢|B) = 1/2 (c) P(B¢) = (1 - a)2

(d) P(A¢|B¢) = [a/(1 -a)]2

59. Suppose X is a random variable which takes values 0, 1, 2, 3, ... and P(X = r) = pqr where 0 < p < 1, q = 1 – p and r = 0, 1, 2, .... Then (a) P(X ≥ n) = qn (b) P(X ≥ m + n | X ≥ m) = P(X ≥ n) (c) P(X = m + n | X ≥ m) = P(X = n) (d) P(X > n) = qn + 1 60. Let 0 < P(A) < 1, 0 < P(B) < 1 and P(A » B) = P(A) + P(B) – P(A) P(B). Then (a) P(B | A) = P(B) (b) P(A¢ » B¢) = 1 – P(A) P(B) (c) P((A » B)¢) = P(A¢) P(B¢) (d) P(A | B) = P(A) 61. A die is rolled three times. Let E1 denote the event of getting a number larger than the previous number each time and E2 denote the event that the numbers (in order) from an increasing A.P. then (a) P(E2) £ P(E1)

(b) P(E2 « E1) = 1/36

3 10 (d) P(E1) = P(E2) 10 3 62. Seven digits from the numbers 1, 2, 3, ..., 9 are written in random order to obtain a seven digit number N. The probability that N is divisible by (b) 3 is 35/9P7 (a) 9 is 41/9P7 (c) P(E2|E1) =

(c) 5 is 1/3 (d) 5 is 1/4 63. Eight children are standing in a queue outside a these children have a one-rupee coin each and the remaining 4 children have a two-rupee coin each. The entry ticket is priced at Re. 1. If all the arrangements of the eight children are equally likely, let pi be probability that the ith child will be the window has no change to begin with, then

IIT JEE eBooks: www.crackjee.xyz Probability 9.41

p1 = 1/2 p2 = 0 p3 = 1/7 probability that no child has to wait for the change is 1/5 64. If X follows a binomial distribution with parameter n = 101 and p = 1/3, then P(X = r) is maximum if r equals (a) 34 (b) 33 (c) 32 (d) 35 65. When a fair die is thrown twice, let (a, b) denote a and the second shows b. Further, let A, B and C be the following events: A = {(a, b) | a is odd}, B = {(a, b) | b is odd} and C = {(a, b) | a + b is odd}. Then (a) P(A « B) = 1/4 (b) P(B « C) = 1/4 (c) P(A « C) = 1/4 (d) P(A « B « C) = 0 (a) (b) (c) (d)

MATRIX-MATCH TYPE QUESTIONS 66. Three players A, B and C alternatively throw a die in the winner. A's die is fair whereas B and C throw dice with probabilities p1 and p2 respectively of throwing a 6. Column 1 Column 2 1 (p) (a) If p1 = 1/5, p2 = 1/4, 2 the probability that A wins the game. 1 (q) (b) If p1 = 1/5, p2 = 1/4, 3 the probability that C wins the game 1 (r) (c) Value of p1 so that 4 P(A wins) = P(B wins) 1 (d) Value of p2 so that the (s) 5 game is equiprobable to all the three players. 67. Consider a town with N people. A person spreads a rumor to a second, who in turn repeats it to a third, and so on. Suppose that at each stage, the recipient of the rumor is chosen at random from the remaining (N – 1) people. Suppose rumor is repeated n (≥ 3) times, then probability that it will not be repeated to

Column 1

(a) originator

Column 2 Ê 1 ˆ (p) Á1 N - 1˜¯ Ë Ê 1 ˆ ÁË1 - N - 1˜¯

(c) second recipient (d) any of the earlier recipient

Ê 1 ˆ (r) Á1 N + 1˜¯ Ë (s)

N -1

n -1

n-2

n-2

Pn ( N - 1)n

68. A prisoner escapes from a jail and is equally likely to choose one of the four roads I, II, III or IV to reach away from the hands of law. If he choose I road, he is successful with probability 1/6 and for II, III and IV this is 1/8, 1/10 and 1/12. If the prisoner is successful, the probability that he chose road Column 1

Column 2

(a) I

(p) 12/57

(b) II

(q) 15/57

(c) III

(r) 20/57

(d) IV

(s) 10/57

69. Sixteen players S1, S2, ....., S16 play in a tournament. They are divided into eight pairs at random. From each pair a winner is decided on the basis of a game played between the two players. Assume that all the players are of equal strength. The probability that Column 1 Column 2 1 (p) (a) S1 reaches the 2 second round. (b) S2 reaches the second round

(q)

1 3

(c) Exactly one of S1 and S2 reach the second round.

(r)

8 5

(d) Both S1 and S2 are among the winners

(s)

7 30

70. Three distinct numbers a, b, c are chosen at random from the numbers 1, 2, ......, 100. The probability that Column 1 Column 2 (a) a, b, c are in A.P.

(p)

53 161700

IIT JEE eBooks: www.crackjee.xyz 9.42 Comprehensive Mathematics—JEE Advanced

1 66

(b) a, b, c (c)

1 1 1 , , a b c

(d) a + b + c is

1 22 (s)

divisible by 2

1 2

ASSERTION-REASON TYPE QUESTIONS 71. Let A, B and C be three events such that P(C) = 0. Statement-1: P(A « B « C) = 0 Statement-2: P(A » B » C) = P(A » B) 72. Let A, B and C be three events such that A is independent of both B and C. Statement-1: A is independent of B » C. Statemtne-2: A is independent of B « C. 73. A set P contains n elements. Two boys Rakshit and Kiaan independently pick up two subsets Q and R of P . Ê 3ˆ Statement-1: Probability Q « R = f is Á ˜ Ë 4¯

n

n

Ê 1ˆ Statement-2: Probability Q » R = P is Á ˜ . Ë 2¯ 74. A fair die is thrown twice. Let (a, b) denote the a and the second shows b. Let A and B be the following two events. A = {(a, b)|a is even}, B = {(a, b)|b is even} Statement-1: If C = {(a, b)|a + b is odd}, then P(A « B « C) = 1/8 Statement-2: If D = {(a, b)|a + b is even}, then P(A « B « D|A « B) = 1/3 75. A man P speaks truth with probability p and another man Q speaks truth with probability 2p. Statement-1: If P and Q contradict each other with probability 1/2, then there are two values of p. Statement-2: has two real roots.

COMPREHENSION-TYPE QUESTIONS

76. p(0) equals (a) w/b (c) b/w

(b) w/(b + w) (d) 0

77. p(1) equals (a) w/b (c) w/(b + w)

(b) (w - 1)/(b + w - 1) (d) 0

78. p(2) equals (a) w/(b + w) (b) (w - 2)/(b + w - 2) (c) w(w - 2)/(b + w) (b + w - 1) (d) 0 79. If l > 2, p(l) equals (a) w/(b + w) w

(c)

Pl

w+b

(b) (w - 1)/b + w - 1) (d) none of these

Pl

lim p(l) equals

80. If b

wƕ

(a) 1 (c) 1/2

(b) 0 (d) l/w

Paragraph for Question Nos. 81 to 85 Let n = 10k + r where k, r ∈ N, 0 £ r £ 9. A number a is chosen at random from the set {1, 2, ..., n} and let pn denote the probability that a2 - 1 is divisible by 10. 81. If r = 0, pn equals (a) 2k/n (b) (k + 1)/n (c) (2k + 1)/n (d) k/n 82. If r = 9, pn equals (a) 2k/n (b) 2(k + 1)/n (c) (2k + 1)/n (d) k/n 83. If 1 £ r £ 8, pn equals (a) (2k – 1)/n (b) 2k/n (c) (2k + 1)/n (d) k/n 84. lim pn equals nÆ•

(a) 1/10 (b) 2/5 (c) 1/5 (d) 3/5 85. If qn denote the probability that a2 + 1 is divisible by 10, then lim qn equals nƕ

(a) 1/5 (c) 3/5

(b) 2/5 (d) 0

Paragraph for Question Nos. 76 to 80

Paragraph for Question Nos. 86 to 90

A box contains w white, b black balls. Out of the box l balls are lost. A ball is drawn at random from the box, and let p(l) denote the probability of the event A that the ball drawn is white.

A player throws a fair cubical die and scores the number appearing on the die. If he throws a 1, he gets a further throw. Let pr denote the probability of getting a total score of exactly r.

IIT JEE eBooks: www.crackjee.xyz Probability 9.43

86. p1 equals (a) 1/6 (c) 0 87. p2 equal (a) 1/6 (c) 0 88. If 2 £ r £ 6, pr equals Ê 1ˆ (a) 1 - Á ˜ Ë 6¯

(b) 1 (d) 1/5 (b) 1 (d) 1/5

r -1

(b)

(c) 5r/6r

1 5

È Ê 1ˆ r - 1˘ Í1 - Á ˜ ˙ ÍÎ Ë 6 ¯ ˙˚

(d) 1/5

89. If r > 6, pr equals Ê 1ˆ (a) 1 - Á ˜ Ë 6¯ (c)

1 5

r -1

(b)

1 5

È Ê 1ˆ r - 1˘ Í1 - Á ˜ ˙ ÍÎ Ë 6 ¯ ˙˚

ÈÊ 1 ˆ r - 6 Ê 1 ˆ r - 1 ˘ -Á ˜ ÍÁ ˜ ˙ (d) 1/5 Ë 6¯ ÍÎË 6 ¯ ˙˚ •

90. Sum of the series S = Â

r =1

(a) 1 (c) 1/5

pr is

96. Two squares are chosen at random on a chessboard. If p denotes the probability that they have exactly one side in common, then 144p is native answers, of which one or more than one are correct. A candidate will get marks on the question if he ticks all the correct answers. If he decides to tick answers at random, then the least number of choices should he be allowed so that the probability of his getting marks on the question exceeds 1/8 is 98. Two persons each make a single throw with a pair of dice. If p denotes the probability that their throws are unequal, then 648 p – 70 is 99. In a sequence of independent trials, the probability of success is 1/4. If p denotes the probability that the second success occurs on the fourth trial or later trial, then 32 p – 20 is 100. If X follows a binomial distribution with parameters n = 100 and p = 1/3, then r – 30 for which P(X = r) is maximum is LEVEL 2

(b) 1/6 (d) 2/3

INTEGER-ANSWER TYPE QUESTIONS 91. Let P(A) = 0.4 and P(A » B) = P(A « B), then 5P(B) is 92. A pair of fair dice is rolled together till a sum of 7 or 11 is obtained. Let p denote the probability that 7 comes before 11, then the value of 4 p is 93. In a test an examine either gusses or copies or knows the answer to a multiple choice question with m choices out of which exactly one is correct. The probability that he makes a guess is 1/3 and the probability that he copies the answer is 1/6. The probability that his answer is correct given that he copied it, is 1/8. If the probability that he knew the answer to the question given that he correctly answered it is 120/141, then m is 1 1 1 (1 + 4p), (1 - p) and (1 - 2p) are the 94. If 4 4 2 probabilities of three mutually exclusive events, then 2p is 95. The digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 are written in random order to form a nine digit number. Let p be the probability that this number is divisible by 36, then 9p is

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. Cards are drawn one by one with replacement from ing of a spade. The probability that the second success occurs at the (n + 2)th draw is (a) (c)

n +1 4n + 2 n(3n ) 4n + 2

(b) (d)

(n + 1) 3n 4n + 2 (n + 2)3n 4n + 2

2. A key bunch contain m keys exactly one of which opens the door. A man tries to open the door by choosing a key at random discarding the wrong key after each trial. The probability that the door opens at the rth trial (1 £ r £ m), is (a) 1/m (b) 1/r (c) r/m (d) (r - 1)/m 3. An urn contains 6 black balls and an unknown (£ 6) of white balls. Three balls are drawn one-byone without replacement. If all these three are found to be white. The probability that the black will be drawn in the next draw is (a) 7/6! (b) 677/999 (c) 677/4620 (d) 1/303

IIT JEE eBooks: www.crackjee.xyz 9.44 Comprehensive Mathematics—JEE Advanced

4. A natural number is chosen at random form. The probability that sum of the digits of its square is 51 is (a) 1/39 (b) 1/13! (c) 2/13! (d) 0 5. From a heap containing 15 pairs of shoes, 10 shoes are selected at random. The probability that there is no complete pair in the selected shoes is 30

(a)

(

C10 -

30 30

(c)

15

C10 2

C10

C10 - 215 30

C10

)

10

15

(b)

(d)

( ) 10

C10 2 30

C10

15

C10

30

C10

6. Fifteen coupons are marked 1, 2, ..., 15. Seven coupons are selected at random without replacement. The probability that the largest number on the selected coupon is 15 is 15

(a)

30

C7 C7

-

14 15

(c) (14/15)7

(b)

15

C7

30

C7

(d) 7/15

7. Let a = 33 where n ∈ N. If the exponent n is chosen at random, the probability that the unit’s digit of a is 9 is (a) 1/2 (b) 1/3 (c) 1/4 (d) 1/9 n

8. 10 different pens and two different books are distributed randomly to 3 students giving 4 things to each. The probability that books go to different students is (a) 5/11 (b) 6/11 (c) 7/11 (d) 8/11 9. A fair die is rolled (2n + 1) times. The probability that the faces with even number show odd number of times is (a)

3n + 1 4n + 2

(b)

3 4

(c)

2 3

(d)

1 2

10. Three numbers are chosen at random from the set {0, 1, 2, ..., 29} without replacement. The probability that sum of the numbers is 30 is 5 812 5 (c) 406

(a)

15 812 3 (d) 406 (b)

11. Let Ei, i = 1, 2, ... n be n independent events such 1 that P(Ei) = (1 £ i £ n), the probability at least i +1 one of E1, E2, ... En occurs is 1 1 (b) (a) n +1 n! (c)

n n +1

(d)

(n - 1) n!

12. Two natural numbers a and b are selected at random. The probability that a2 + b2 is divisible by 7 is (a) 3/8 (b) 1/7 (c) 3/49 (d) 1/49 13. A boy has 20% chance of hitting at a target. Let p denote the probability of hitting the target for the nth trial. If p 2 25p - 15p + 2 £ 0, then value of n is (a) 4 (b) 3 (c) 2 (d) 1 14. Shashi tosses a coin. If he gets a head he throws a fair cubical die and the number on it is noted. If he gets a tail he is asked to throw 5 coins and number of heads obtained is noted. If the number noted is 4, the probability that he threw a die is (a) 16/31 (b) 15/31 (c) 31/192 (d) 2/191 15. Let 0 < P(A), P(B) < 1 and P(B¢) = (P(A » B))2, then (a)

1 2

(

)

5 - 1 < P(A » B) £ 1

(b) P(A | B)
pe (d) pe > p0 24. Let xi (1 £ i £ 5) denote the probability of an event Ei. Let 5

5

i =1

i =1

y = 5 Â x i2 - 2 Â x i + 5 Then

(a) max y = 4 (b) min y = 4/5 (c) max y = 1 (c) min y = 0 25. Let pn denote the probability of getting n heads when a fair coin is tossed m times. If p4, p5, p6 are in A.P., then value of m can be (a) 5 (b) 7 (c) 10 (d) 14 26. A bag cantains tickets bearing numbers 1, 2, 3, . . ., 50. Five tickets are drown at random from the bag without replacement. Let pm denote the probability that m then (a) 50C5 p3 = 47C2 (b) 50C5 p24 = (23C2) (25C2) (c) 50C5 p48 = 47C2 (d) pr = p51 – r for 3 £ r £ 48 27. m distinct odd natural numbers and n distinct even natural numbers are written at random to form a number of (m + n) digits. If n > m, the probability two odd numbers are not together is (a)

n !(n + 1)! (m + n)! (n - m + 1)! n +1

(b)

(c)

(d)

(

m+n

Cm Cm )

m !( n + 1)! (m + n)! (n - m + 1)!

(m!)

(

2 m+n

1 Cn

)(

n - m +1

Cm

)

28. n (≥ 5) persons are sitting in a row. Three of these are selected at random, the probability that no two of the selected persons are together is n-3

(a)

(c)

n

P2

P2

(n - 3) (n - 4) n (n - 1)

n-3

(b)

n

C2

n-3

(d)

n

C2 C2

P2

29. Two players A and B of equal skill agree to play a table tennis match. The game will be won by the player who is 2 games ahead. Let pn denote the probability that A wins after the nth match, then (b) p3 = 0 (a) p2 = 1/4 (d) p7 = 0 (c) p4 = 1/8 30. A pack contains n distinct red cards and n distinct white cards. This pack of 2n cards is divided into two equal parts at random. A card is drawn from

IIT JEE eBooks: www.crackjee.xyz 9.46 Comprehensive Mathematics—JEE Advanced

each of the two parts, the probability that both the cards are of the same colour is n -1 (a) 2n (c)

1 n2

(a) 4

(p) 1

(b) 5

(q) 20/81

(c) 45

(r) 17/81

n (d) 2 ( 2n - 1)

(d) 12

(s) 2/81

31. A determinant D is chosen at random from the set of all determinant of order two with elements 0 and 1 only. Column 1 Column 2 Value of D Probability (a) 1 (p) 5/8 (b) 0 (q) 3/16 (c) 2 (r) 3/8 (d) nonzero (s) 0 32. Suppose A has (n + 1) and B has n fair coins which pk denote the probability that A obtains k (0 £ k £ n + 1) heads more than B. Value of Column 1 Column 2 2n + 1

(a) p0

(p)

(b) p2

(q) 1/2

(c) pn

(r)

(d) Â pk

(s)

Cn

2 2n + 1 2n + 1

Cn - 1

2 2n + 1 2n +1 2

ASSERTION-REASON TYPE QUESTIONS 35. From an urn containing a white and b black balls, k(< a, b) are drawn and laid aside, their colour unnoted. Then another ball, that is, (k + 1)th ball is drawn. Statement-1: Probability that (k + 1)th ball drawn is white a is . a+b Statement-2: Probability that (k + 1)th ball drawn is black b is . a+b 36. Let A and B be two events such that P(A) > 0. Statement-1: If P(A) + P(B) > 1, then P(B|A) ≥ 1 – P(B')/P(A) Statement-2: If P(A|B') ≥ P(A), then P(A) ≥ P(A|B). 37. Let A, B and C be three events. Statement-1: If A « B Õ C, then P(C) ≥ P(A) + P(B) – 1.

2n +1

33. Two numbers a and b are chosen at random from the set {1, 2, 3, ..., 9} with replacement. The probability that the equation x2 +

Column 2

n -1 (b) 2n - 1

MATRIX-MATCH TYPE QUESTIONS

k ≥1

Column 1

2 (a - b)x + b2 = 0

has

Statement-2: P[(A « B) » (B « C) » (C « A)] £ min{P(A » B), P(B » C), P(C » A)}

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 38 to 42 Let x and y be two numbers chosen at random from the set {1, 2, 3, ..., n } with replacement. Let Qn(p) denote the probability that (xp-1 - yp-1) is divisible by p. 38. Q25(3) equals (a) 0.5648

(b) 0.5678

(c) 0.5698

(d) 0.5628

Column 1 (a) real roots

Column 2 (p) 7/8

(b) equal roots

(q) 0

(a) 0.52

(b) 0.64

(c) imaginary roots

(r) 11/81

(c) 0.68

(d) 0.75

(d) positive roots

(s) 5/81

34. A ten digit number N is formed by using the digits 0 to 9 exactly once. The probability that N is divisible by

39. Q50(5) equals

40. If [x] denotes the greatest integer £ x then Qn(p) equals Èn˘ 2 Èn˘ (a) 1 + 2 Í ˙ - 2 Í ˙ Î p˚ n Î p˚

2

IIT JEE eBooks: www.crackjee.xyz Probability 9.47

Èn˘ 1 Èn˘ (b) 1 – 3 Í ˙ - 2 Í ˙ Î p˚ n Î p˚ Èn˘ 1 Èn˘ (c) 1 - Í ˙ + 2 Í ˙ Î p˚ n Î p˚

46. If pn denotes the probability that there is at least one cycle, then pn equals

2

n

(a) 1 - Â ( - 1)

2

k=2 n

(c) 1 - Â ( - 1)

2 n 2 n 2 (d) 1 – ÈÍ ˘˙ + 2 ÈÍ ˘˙ n Î p˚ n Î p˚

k =1

42.

(c) 1 -

2 p2

(b) 1 -

2 p

(d) 2/p

lim Qn(p) equals

p , n Æ•

(a) 1/2 (c) 1/p

(b) Â ( - 1)

1 k!

(d) 0

k =1

k

1 k!

nƕ

nƕ

2 2 + p p2

k

n

1 k!

47. lim pn equals

41. lim Qn(p) equals (a) 1 -

k

(b) 1 (d) 0

Paragraph for Question Nos. 43 to 47 A permutation of n objects is a one-to-one function from A = {1, 2, ..., n} onto itself. If a1, a2, ... ak are n distinct elements of A such that a function f sends a1 to a2, a2 to a3, ..., ak-1 Æ ak and ak Æ a1, then a1, a2, ..., ak are said to form a cycle of length k. A permutation is chosen at random from the set of all permutations of the set A = {1, 2, ..., n}. 43. The probability that chosen permutation is a cycle such that 1 Æ 2 Æ 3 Æ ... Æ (n - 1) Æ 1, is (a) 1/n! (b) 1/(n - 1)! (c) 2/n! (d) 0 44. Let a1, a2, ..., ak, b1, b2, ..., bk ∈ A be such that ai’s are distinct and bj’s are distinct. The probability that the chosen permutation sends ai Æ bi for i = 1, 2, ..., k, is (a) k!/n! (b) k! (n - k)!/n! (c) 0 (d) (n - k)!/n! 45. The probability that the element i forms a unit cycle i.e. i Æ i, is (a) 1/n! (b) 1/(n - 1)! (c) 1/n (d) 0

(a) 1/e (c) 0

(b) 1 - 1/e (d) 0

INTEGER-ANSWER TYPE QUESTIONS 48. Two persons each make a throw with a pair of cubical die and let p denote the probability that their throws are equal. Then value of 648p – 70 is 49. A bag contains n + 1 coins. It is known that one of these coins shows heads on both sides, whereas the other coins are fair. One coin is selected at random and tossed. If the probability that the toss results in heads is 7/12, then n is 50. Two squares are chosen at random on a chess board and let p denote the probability that they have exactly one side in common. Then 36p is 51. Two integers r and s are chosen one at a without replacement from the numbers 1, 2, 3, ... 100. Let p be the probability that r £ 25 given that s £ 25. then the value of 33p is 52. A child arranges 3 distinct black balls, four green p is the probability that two red balls are not together, then 99p is 53. Two numbers x and y are chosen at random from the numbers from the set {1, 2, 3, ..., 300}. Let p be the probability that x3 + y3 is divisible by 3, then 9p is 54. Two non-negative integers are chosen at random. If p denotes the probability that sum of their squares is divisible by 5, then 25p is 55. Out of 31 numbers 1, 2, 3, ..., 31, three numbers are chosen at random, without replacement. Let p denote the probability that the selected numbers are in A.P. then 899p – 40 is

IIT JEE eBooks: www.crackjee.xyz 9.48 Comprehensive Mathematics—JEE Advanced

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. Two fair dice are rolled. Let X be the event that Y be the event that the second die shows an odd number. The two events X and Y are (a) mutually exclusive (b) mutually exclusive and independent (c) dependent (d) none of these [1979] 2. Two events A and B have probabilities of 0.25 and 0.50 respectively. The probability that both A and B occurs simultaneously is 0.14. Then the probability that neither A nor B occurs is (a) 0.39 (b) 0.25 (c) 0.11 (d) none of these. [1980] 3. The probability that an event A happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that the event A happens at least once is (a) 0.936 (b) 0.784 (c) 0.904 (d) none of these [1980] 4. If A and B are two independent events such that P(A) > 0 and P(B) π 1 then P(A¢|B¢) is equal to (a) 1 – P(A|B) (b) 1 – P(A|B¢) (c)

1 - P( A » B) P( B)

(d)

P ( A¢ ) P( B ¢)

Ê 3ˆ (c) Á ˜ Ë 5¯

6

Ê 8ˆ (b) Á ˜ Ë 15 ¯

8.

9.

10.

[1980]

5. Fifteen coupons are numbered 1, 2, 3, ..., 15 respectively. Seven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on a selected coupon is 9 is Ê 9ˆ (a) Á ˜ Ë 16 ¯

7. A student appear for tests, I, II and III. The student is successful if he passes either in tests I and II or tests I and III. The probabilities of the student passing in 1 tests I, II and III are p, q and , respectively. If the 2 1 probability that the student is successful is , then 2

11.

6

7

(d) none of these.

[1983] 6. Three identical dice are rolled. The probability that the same number will appear on each of them is (a) 1/6 (b) 1/36 (c) 1/18 (d) 3/28 [1984]

12.

(a) p = q = 1 (b) p = q = 1/2 (c) p = 1, q = 0 (d) p = 1, q = 1/2 (e) none of these [1986] The probability that at least one of the events A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then P(A¢) + P(B¢) is (a) 0.4 (b) 0.8 (c) 1.2 (d) 1.4 (e) none of these [1987] One hundred identical coins, each with probability, p, of showing up heads are tossed once. If 0 < p < 1 and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, then the value of p is 1 49 50 51 (a) (b) (c) (d) 2 101 101 101 [1988] India plays two matches each with the West Indies and Australia. In any match, the probabilities of India getting 0, 1 and 2 points are 0.45, 0.05 and 0.50, respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is (a) 0.8750 (b) 0.0875 (c) 0.0625(d) 0.0250 [1992] An unbiased die with faces marked 1, 2, 3, 4, 5 and 6 is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5 is, (a) 16/81 (b) 1/81 (c) 80/81 (d) 65/81 [1993] A box contains 24 identical balls of which 12 are white and 12 are black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the 4th time on the 7th draw is (a) 5/64 (b) 27/32 (c) 5/32 (d) 1/2 [1994]

IIT JEE eBooks: www.crackjee.xyz Probability 9.49

13. An unbiased die is tossed until a number greater than 4 appears. The probability that an even number of tosses is needed is (a) 1/2 (b) 2/5 (c) 1/5 (d) 2/3 [1994] 14. Let A, B, C be three mutually independent events. Consider the two statements S1 and S2. S1 : A and B » C are independent S2 : A and B « C are independent Then (a) Both S1 and S2 are true (b) Only S1 is true (c) Only S2 is true (d) Neither S1 nor S2 is true. [1994] 15. Three of the six vertices of regular hexagon are chosen at random. The probability that the triangle with these three vertices is equilateral, equals (a) 1/2 (b) 1/5 (c) 1/10 (d) 1/20 [1995] 16. The probability of India winning a test match against West Indies is 1/2. Assuming independence from match to match the probability that in a 5 match series India’s second win occurs at the third test is (a) 1/8 (b) 1/4 (c) 1/2 (d) 2/3 [1995] 17. For the three events A, B and C, P (exactly one of the events A or B occurs) = P(exactly one of the events B or C occurs) = P(exactly one of the events C or A occurs) = p and P (all the three events occur simultaneously) = p2, where 0 < p < 1/2. Then the probability of at least one of the three events A, B and C occurring is (a)

3 p + 2 p2 2

(b)

p + 3 p2 4

(c)

p + 3 p2 2

(d)

3 p + 2 p2 4

[1996]

18. Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals (a) 1/2 (b) 7/15 (c) 2/15 (d) 1/3 [1998] 19. If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn is (a) 13/32 (b) 1/4 (c) 1/32 (d) 3/16 [1998]

20. There are four machines and it is known that exactly two of them are faulty. These are tested, one by one, in a random order till both the faulty only two tests are needed is (a) 1/3 (b) 1/6 (c) 1/2

(d) 1/4

[1998] 21. A fair coin is tossed repeatedly. If the tail appears

(a) 1/2 22.

23.

24.

25.

(b) 1/32

(c) 31/32 (d) 1/5

[1998] If E and F are events with P(E) £ P(F) and P(E « F) > 0, then (a) occurrence of E fi occurrence of F (b) occurrence of F fi occurrence of E (c) non-occurrence of E fi non-occurrence of F (d) none of the above implications hold [1998] If the integers m and n are chosen at random between 1 and 100, the probability that a number of the form 7m + 7n is divisible by 5 equals (a) 1/4 (b) 1/7 (c) 1/8 (d) 1/49 [1999] If A, B and C are three events such that P(B) = 3/4. P(A « B « C¢) = 1/3 and P(A¢ « B « C¢) = 1/3, then P(B « C) is equal to (a) 1/12 (b) 1/6 (c) 1/15 (d) 1/9 [2003] Two numbers are selected randomly from the set A = {1, 2, 3, 4, 5, 6} without replacement one by one. The probability that minimum of the two numbers is less than 4 is (a) 1/15 (b) 14/15 (c) 1/5 (d) 4/5 [2003]

chosen without replacement. The probability that all these numbers are divisible both by 2 and 3 is (a) 4/11 (b) 4/55 (c) 4/33 (d) 4/1155 [2004] 27. An unbiased cubical die is rolled until 1 appears. The probability that an even number of tosses is needed is (a) 1/6 (b) 5/11 (c) 6/11 (d) 5/6 [2005] 28. One Indian and four American men and their wives are to be seated randomly around a circular table. Then the conditional probability that the Indian man is seated adjacent to his wife given that each of the American man is seated adjacent to his wife is

IIT JEE eBooks: www.crackjee.xyz 9.50 Comprehensive Mathematics—JEE Advanced

(a) 1/2

29.

30.

31.

32.

33.

(d) 1/5 [2007] Let E¢ denote the complement of an event E. Let E, F, G be pairwise independent events such that P(G) > 0 and P(E « F « G) = 0. Then P(E¢ « F¢|G) equals (a) P(E¢) + P(F¢) (b) P(E¢) – P(F¢) (c) P(E¢) – P(F) (d) P(E) – P(F¢) [2007] An experiment has 10 equally likely outcomes. Let A and B be two non-empty events of the experiment. If A consists of 4 outcomes, the number of outcomes that B must have so that A and B are independent, is (a) 2, 4 or 8 (b) 3, 6 or 9 (c) 4 or 8 (d) 5 or 10. [2008] Let w be a complex cube root of unity with w π 1. A fair die is thrown three times. If r1, r2 and r3 are the numbers obtained on the die, then the probability that wr1 + wr2 + wr3 = 0 is 1 1 (a) (b) 18 9 2 1 (c) (d) [2010] 9 36 A signal which can be green or red with probabil4 1 and respectively, is received by station A ity 5 5 and then transmitted to station B. The probability 3 of each station receiving the signal correctly is . 4 If the signal received at station B is green, then the probability that the original signal was green is 3 6 (a) (b) 5 7 20 9 (c) (d) [2010] 23 20 Four fair dice D1, D2, D3 and D4, each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The probability that D4 shows a number appearing on one of D1, D2, and D3 is 91 216 125 (c) 216

(a)

(b) 1/3

(c) 2/5

108 216 127 (d) 216 (b)

[2012]

34. Four persons independently solve a certain problem 1 3 1 1 correctly with probabilities , , , . Then the 2 4 4 8 probability that the problem is solved correctly by at least one of them is

(a)

235 256

(b)

21 256

(c)

3 256

(d)

253 256

[2013]

35. Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is 1 1 (a) (b) 2 3 (c)

2 3

(d)

3 4

[2014]

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 1. If M and N are any two events, the probability that exactly one of them occurs is (a) P(M) + P(N) – 2P(M « N) (b) P(M) + P(N) – P(M « N) (c) P(M¢) + P(N¢) – 2P(M¢ « N¢) (d) P(M « N¢) + P(M¢ « N) [1984] 2. For two given events A and B, P(A « B) is (a) not less than P(A) + P(B) – 1 (b) not greater than P(A) + P(B) (c) equal to P(A) + P(B) – P(A » B) (d) equal to P(A) + P(B) + P(A » B) [1988] 3. If E and F are independent events such that 0 < P(E) < 1 and 0 < P(F) < 1, then (a) E and F are mutually exclusive (b) E and F¢ (the complement of the event F) are independent (c) E¢ and F¢ are independent (d) P(E|F) + P(E|F¢) = 1 [1989] 4. For any two events A and B in a sample space P( A) + P( B) - 1 (a) P(A|B) ≥ , P(B) π 0, is always P( B) true (b) P(A « B¢) = P(A) – P(A « B) does not hold (c) P(A » B) = 1 – P(A¢) P(B¢), if A and B are independent (d) P(A » B) = 1 – P(A¢) P(B¢) if A and B are disjoint. [1991] 5. Let E and F be two independent events. The probability that both E and F happen is 1/12 and the probability that neither E nor F happens is 1/2. Then, (a) P(E) = 1/3, P(F) = 1/4 (b) P(E) = 1/2, P(F) = 1/6

IIT JEE eBooks: www.crackjee.xyz Probability 9.51

6.

7.

8.

9.

(c) P(E) = 1/6, P(F) = 1/2 (d) P(E) = 1/4, P(F) = 1/3 [1993] Let 0 < P(A) < 1, 0 < P(B) < 1 and P(A » B) = P(A) + P(B) – P(A)P(B). Then (a) P(B – A) = P(B) – P(A) (b) P(A¢ » B¢) = P(A¢) + P(B¢) (c) P((A » B)¢) = P(A¢) P(B¢) (d) P(A|B) = P(A) [1995] If E¢ and F¢ are the complementary events of E and F respectively and if 0 < P(E), P(F) < 1, then (a) P(E|F) + P(E¢|F) = 1 (b) P(E|F) + P(E|F¢) = 1 (c) P(E¢|F) + P(E|F¢) = 1 (d) P(E|F¢) + P(E¢|F¢) = 1 [1998] The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c, respectively. Of these subjects, the student has a 75% chance of passing in at least one, a 50% chance of passing in at least two, and a 40% chance of passing in exactly two. Which of the following relations are true? (a) p + m + c = 19/20 (b) p + m + c = 27/20 (c) pmc = 1/10 (d) pmc = 1/4 [1999] Let E and F be two independent events. The proba11 bility that exactly one of them occurs is and the 25 2 . If P(T) probability of none of them occuring is 25 denotes the probability of occurrence of the event T, then 4 3 (a) P(E) = , P(F) = 5 5 2 2 (b) P(E) = , P( F ) = 5 5 2 1 (c) P(E) = , P( F ) = 5 5 (d) P(E) =

3 4 , P( F ) = 5 5

[2011]

E1, E2 and E3. The engines function independently of each other 1 1 1 with respective probabilities , and . For the 2 4 4 ship to be operational at least two of its engines must function. Let X denote the event that the ship is operational and let X1, X2 and X3 denote respec-

tively the events that the engines E1, E2 and E3 are functioning. Which of the following is (are) true? c (a) P ÈÎ X1 | X ˘˚ =

3 16

(b) P [Exactly two engines of the ship are func7 tioning |X] = 8 (c) P [X |X2] =

5 16

(d) P [X |X1] =

7 16

[2012]

11. Let X and Y be two events such that P (X |Y) =

1 , 2

1 1 and P (X « Y) = . Which of the 3 6 following is(are) correct? 2 (a) P(X « Y) = 3 (b) X and Y are independent (c) X and Y are not independent 1 (d) P (X ¢ « Y) = [2012] 3 P (Y |X) =

ASSERTION-REASON TYPE QUESTIONS 1. Let H1, H2, ...., Hn be mutually exclusive and exhaustive events with P(Hi) > 0, i = 1, 2, ...., n. Let E be any other event with 0 < P(E) < 1. Statement-1: P(Hi|E) > P(E|Hi) P(Hi) for i = 1, 2, ...., n. n

Statement-2: Â P (Hi) = 1.

[2007]

i =1

2. Consider the system of equations ax + by = 0, cx + dy = 0, where a, b, c d ∈ {0, 1}. Statement-1: The probability that the system of equations has a unique solution is 3/8. Statement-2: The probability that the system of equations has a solution is 1. [2008]

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 1 to 3. There are n urns numbered 1 to n. The ith urn contains i white and (n + 1 – i) black balls. Let Ei denote the event of selecting ith urn at random and let W denote the event that the ball drawn from the selected urn is white.

IIT JEE eBooks: www.crackjee.xyz 9.52 Comprehensive Mathematics—JEE Advanced

1. If P(Ei) μ i for i = 1, 2, ..., n, then lim P(W) is nÆ•

(a) 2/3 (b) 1/3 (c) 3/4 (d) 1/4 2. If P(Ei) = c, a constant ∀ i, then P(En|W) is (a)

2 n +1

(b)

1 n +1

(c)

n 1 (d) n +1 2

3. If n is even and P(Ei) = 1/n ∀ n and E denotes the event of choosing even numbered urn, then P(W|E) is 1 n+4 (a) (b) 2(n + 1) 2n (c)

n+2 2(n + 1)

(d)

n +1 2n

[2006]

Paragraph for Question Nos. 4 to 6. A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required. 4. The probability that X = 3 equals 25 (a) 216

25 (b) 36

5 (c) 36

125 (d) 216

5. The probability that X ≥ 3 equals 125 25 (b) 216 36 5 25 (c) (d) 36 216 6. The conditional probability that X ≥ 6 is given X > 3 equals 125 25 (b) (a) 216 216 5 25 (c) (d) [2009] 36 36 (a)

Paragraph for Question Nos. 7 to 8. Let U1 and U2 be two urns such that U1 contains 3 white and 2 red balls, and U2 contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U1 and put into U2. However, if tail appears then 2 balls are drawn at random from U1 and put into U2. Now I ball is drawn at random from U2. 7. The probability of the drawn ball from U2 being white is 13 23 (b) (a) 30 30 19 11 (c) (d) 30 30

U2 is white, the probability that head appeared on the coin is (a)

17 23

(b)

11 23

(c)

15 23

(d)

12 23

[2011]

Paragraph for Questions Nos. 9 and 10 A box B1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2 contains 2 white balls, 3 red balls and 4 black balls. A third box B3 contains 3 white balls, 4 red balls and 5 black balls. 9. If 1 ball is drawn from each of the boxes B1, B2 and B3, the probability that all 3 drawn balls are of the same colour is (a)

82 648

(b)

90 648

(c)

558 648

(d)

566 648

10. If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B2 is (a)

116 181

(b)

126 181

(c)

65 181

(d)

55 181

[2013]

Paragraph for Question Nos. 11 and 12 Box 1 contains three cards bearing numbers 1, 2, 3; box box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xi be the number on the card drawn from the ith box, i = 1, 2, 3. 11. The probability that x1 + x2 + x3 is odd, is 29 53 (b) (a) 105 105 (c)

57 105

(d)

1 2

12. The probability that x1, x2, x3 are in an arithmetic progression, is (a)

9 105

(b)

10 105

(c)

11 105

(d)

7 105

[2014]

IIT JEE eBooks: www.crackjee.xyz Probability 9.53

is b and only E3 occurs is g. Let the probability p that none of events E1, E2 or E3 occurs satisfy the equations (a – 2b)p = ab and (b – 3g)p = 2bg. All the given probabilities are assumed to lie in the interval (0, 1).

Paragraph for Question Nos. 13 and 14 Let n1 and n2 be the number of red and black balls, respectively, in box I. Let n3 and n4 be the number of red and black balls, respectively, in box II. 13. One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability 1 that this red ball was drawn from box II is , then 3 the correct option(s) with the possible values of n1, n2, n3 and n4 is(are) (a) n1 = 3, n2 = 3, n3 = 5, n4 = 15 (b) n1 = 3, n2 = 6, n3 = 10, n4 = 50 (c) n1 = 8, n2 = 6, n3 = 5, n4 = 20 (d) n1 = 6, n2 = 12, n3 = 5, n4 = 20 14. A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball 1 from box I, after this transfer, is then the correct 3 option(s) with the possible values of n1 and n2 is(are) (a) n1 = 4, n2 = 6

(b) n1 = 2, n2 = 3

(c) n1 = 10, n2 = 20

(d) n1 = 3, n2 = 6 [2015]

Paragraph for Question Nos. 15 and 16

respectively. Each team gets 3 point for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T1 and T2, respectively, after two games. 15. P(X > Y) is 1 4

(b)

5 12

(c)

1 2

(d)

7 12

FILL

11 36

(b)

1 3

(c)

13 36

(d)

1 2

Probability

2.

3.

5.

6.

[2016]

1. Of the three independent events E1, E2 and E3, the probability that only E1 occurs is a, only E2 occurs

IN THE

Face

7.

INTEGER-ANSWER TYPE QUESTIONS

[2013]

BLANKS TYPE QUESTIONS

1. For a biased die the probabilities for different faces to turn up are given below:

16. P(X = Y) is (a)

Probability of occurrence of E1 = Probability of occurrence of E3

2. The minimum number of time a fair coin needs to be tossed, so that the probability of getting at least two head is at least 0.96 is [2015]

4.

Football teams T1 and T2 have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T1 winning, 1 1 1 drawing and losing a game against T2 are , and , 2 6 3

(a)

Then

8.

1

2

3

4

5

6

0.1

0.32

0.21

0.15

0.05

0.17

The die is rolled and you are told that face 1 or face 2 has turned up. Then the probability that it is face 1 __________. [1981] A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The probability that the value of the determinant chosen is positive is __________. [1982] P(A » B) = P(A « B) if and only if the relation between P(A) and P(B) is __________. [1985] A box contains 100 tickets numbered 1, 2,... 100. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than 10. The minimum number on them is 5 with probability __________. [1985] 1+ 3p 1- p 1- 2p If , and are the probabilities of 3 2 4 the three mutually exclusive events, then the set of all values of p is __________. [1986] Urn A contains 6 red and 4 black balls and urn B contains 4 red and 6 black balls. One ball is drawn at random from urn A and placed in urn B. Then one ball is drawn at random from urn B and placed in urn A. If one ball is now drawn at random from urn A, the probability that it is found to be red is __________. [1988] A pair of fair dice is rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is __________. [1989] Let A and B be two event such that P(A) = 0.3 and P(A » B) = 0.8. If A and B are independent events, then P(B) = __________. [1990]

IIT JEE eBooks: www.crackjee.xyz 9.54 Comprehensive Mathematics—JEE Advanced

9. If the mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than one is equal to __________. [1991] 10. Three faces of a fair die are yellow, two faces red and one blue. The die is tossed three times. The probability that the colours yellow, red and blue respectively, is __________. [1992] 11. If two events A and B such that P(A¢) = 0.3, P(B) = 0.4 and P(A « B¢) = 0.5, then P(B|A » B¢) = __________. [1994] 12. Three numbers are chosen at random without replacement from {1, 2, ..., 10}. The probability that the minimum of the chosen numbers is 3, or their maximum is 7, is __________. [1997]

TRUE/FALSE TYPE QUESTIONS 1. If letters of the word “ASSASSIN” are written down at random in a row, the probability that no two S’s occur together is 1/35. [1983] 2. If the probability for A to fail in an examination is 0.2 and that for B is 0.3, then the probability that either A or B fails is 0.5. [1989]

SUBJECTIVE-TYPE QUESTIONS 1. A box contains 2 black, 4 white and 3 red balls. One ball is drawn at random from the box and kept aside. From the remaining balls in the box, another ball process is repeated till all the balls are drawn from the box. Find the probability that the balls drawn in the sequence 2 black, 4 white and 3 red. [1978] 2. Six boys and six girls sit in a row randomly. Find the probability that (i) the six girls sit together; (ii) the boys and girls alternate. [1979] 3. An anti-aircraft gun can take a maximum of four shots at an enemy plane moving away from it. second, third and fourth shot are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that the gun hits the plane? [1981] 4. A and B are two candidates seeking admission in IIT. The probabilities that A is selected is 0.5 and the probability that both A and B are selected is at most 0.3. Is it possible that the probability of B getting selected is 0.9? [1982]

5. Cards are drawn one-by-one at random from wellN is the number of cards required to be drawn, then show that P(N = n) =

(n - 1)(52 - n)(51 - n) 50 ¥ 49 ¥ 17 ¥ 13

where 2 £ n £ 50. [1983] 6. If A, B, C are events such that P(A) = 0.3, P(B) = 0.4, P(C) = 0.8 P(A « B) = 0.08, P(A « C) = 0.28, P(A « B « C) = 0.09 If P(A » B » C) ≥ 0.75, then show that P(B « C) lies in the interval 0.23 £ x £ 0.48. [1983] 7. In a certain city only two news papers A and B are published. It is known that 25% of the city population reads A and 20% reads B while 8% reads both A and B. It is known that 30% of those who read A but not B look into the advertisements and 40% of those who read B but not A look into the advertisements while 50% of those who read both A and B look into the advertisements. What is the percentage of population who reads an advertisements? [1984] 8. A and B are two independent events. The probability that both A and B occur is 1/6 and the probability that neither of them occurs is 1/3. Find the probability of occurrence of A. [1984] 9. In a multiple choice question there are four alternative answers, of which one or more is correct. A candidate will get marks in the question only if he ticks all the correct answers. The candidate decides to tick answers at random. If is he is allowed upto probability that we will get marks in the question. [1985] 10. A lot contains 20 articles. The probability that the lot contains exactly two defective articles is 0.4 and the probability that the lot contains exactly 3 defective articles is 0.6. Articles are drawn from the lot at random one by one, without replacement and are tested till all the defective articles are found. What is the probability that the testing procedure ends at the twelfth testing? [1986] 11. An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white, it is not replaced into the urn. Otherwise it is replaced along with another ball of the same colour. The process is repeated. Find the probability that the third ball drawn is black. [1987]

IIT JEE eBooks: www.crackjee.xyz Probability 9.55

12. A man takes a step forward with probability 0.4 and backwards with probability 0.6. Find the probability that at the end of eleven steps he is one step away from the starting point. [1987] 20. N (≥ 2) of ten box at random. Find the probability that the total 21. 14.

15.

16.

17.

18.

19.

paise. [1988] Suppose that probability for A to win game against B is 0.4. If A has an option of playing either “best of 3 games” or a “best of 5 games” match against B, which option should A choose so that his probability of winning is higher? (No game ends in a draw.) [1989] A is a set containing n elements. A subset P of A is chosen at random. The set A is reconstructed by replacing the elements of P. A subset Q of A is again chosen at random. Find the probability that P and Q have no common elements. [1990] In a test, an examine guesses, copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is 1/3 and the probability that he copies the answer is 1/6. The probability that his answer is correct given that he copied it is 1/8. Find the probability that he knew the answer to the question, given that he answered it correctly. [1991] A lot contains 50 defective and 50 non-defective bulbs. Two bulbs are drawn at random, one at a time, with replacement. The events A, B and C as A B = {the second bulb is non-defective} C = {the two bulbs are both defective or both nondefective} Determine whether (i) A, B and C are pairwise independent and (ii) A, B and C are independent. [1992] Numbers are selected at random, from the two digit numbers 00, 01, 02, ..., 99 with replacement. An event E occurs if and only if the product of the two digits of a selected number is 18. If four numbers are E occurs at least 3 times. [1993] An unbiased coin is tossed. If the result is head, a pair of unbiased dice is rolled and the number obtained by adding the number on the two faces is noted. If

22.

23.

of eleven cards numbered 2, 3, 4, ..., 12 is picked and the number on the card is noted. What is the probability that the noted number is either 7 or 8? [1994] If p and q are chosen randomly from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} with replacement, determine the probability that the roots of the equation x2 + px + q = 0 are real. [1997] Sixteen players S1, S2, ..., S16 play in a tournament. They are divided into eight pairs at random. From each pair a winner is decided on the basis of the game played between the two players. Assume that all the players are of equal strength. (a) Find the probability that the player S1 is among the eight winners. (b) Find the probability that exactly one of the two players S1 and S2 is among the eight winners. [1997] Three players A, B and C toss a coin cyclically in that order (that is, A, B, C, A, B, C, A, B ...) till a head shows up. Let p be the probability that coin shows a head. Let a, b and g be respectively that A, B and C b = (1 – p)a. Determine a, b and g (in terms of p). [1998] Eight players P1, P2, ..., P8 play a knock-out tournament. It is known that whenever the players Pi and Pj play, the player Pi will win if i < j. Assuming that the players are paired at random in each round, what is the probability that the player P4 reaches the

24. A coin has probability p of showing head when tossed. It is tossed n times. Let pn denote the probability that no two (or more) consecutive heads occur. Prove that p1 = 1, p2 = 1 – p2 and pn = (1 – p)pn – 1 + p(1 – p)pn – 2 for all n ≥ 3. [2000] 25. An urn contains m white and n black balls. A ball is drawn at random and is put into the urn along with k additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. What is the probability that the ball drawn now is white? [2001] 26. An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6 is thrown n times and the list of n numbers showing up is noted. What is the probability that among the numbers 1, 2, 3, 4, 5, 6, only three numbers appear in this list? [2001] 27. A box contains N coins, m of which are fair and rest are biased. The probability of getting a head when a fair coin is tossed is 1/2, while it is 2/3 when a

IIT JEE eBooks: www.crackjee.xyz 9.56 Comprehensive Mathematics—JEE Advanced

biased coin is tossed. A coin is drawn from the

28.

29.

30.

31.

shows head and the second time it shows tail. What is the probability that the coin drawn is fair? [2002] A is targetting B, B and C are targeting A. Probability of hitting the target by A, B and C are respectively 2/3, 1/2 and 1/3. If A probability B but not C hits A. [2003] For a student to quality a competitive exam., he must pass at least two out of the three exams. The probability that he will pass the 1st exam. is p. If he fails in one of the exams. then the probability of his passing in the next exam. is p/2 otherwise it remains the same. Find the probability that the student will qualify the competitive exam. [2003] Let A and B be two independent events and C denote the event that exactly one A, B takes place. Show that P(A » B) ◊ P(A¢ « B¢) £ P(C) [2004] A bag contains 6 white and 12 red balls. Six balls are drawn at random without replacement. If at least

41. (d) 45. (c)

46. 48. 50. 52. 54. 56. 58. 60. 62. 64.

66.

67. are 2/9, 4/9, 1/9 and 1/9 respectively. If the person

Answers LEVEL 1

68.

SINGLE CORRECT ANSWER TYPE QUESTIONS 2. 6. 10. 14. 18. 22. 26. 30. 34. 38.

(a) (c) (c) (d) (b) (a) (d) (d) (b) (d)

3. 7. 11. 15. 19. 23. 27. 31. 35. 39.

(d) (d) (d) (d) (b) (a) (c) (a) (b) (b)

4. 8. 12. 16. 20. 24. 28. 32. 36. 40.

(b) (c) (d) (c) (c) (d) (d) (b) (c) (a)

44. (a)

(a), (a), (a), (d) (a), (a), (a), (a), (a), (a),

(b), (c), (d) (b), (c), (d) (b), (c), (d) (b) (b), (b), (c), (b), (b)

(c), (d) (c), (d) (d) (c)

47. 49. 51. 53. 55. 57. 59. 61. 63. 65.

(b), (b), (a), (c), (a), (a), (a), (a), (a), (a),

(c), (c), (b) (d) (b), (b), (b), (b), (b), (b),

(d) (d)

(c), (c), (c) (c), (d) (c),

(d) (d) (d) (d)

MATRIX-MATCH TYPE QUESTIONS

scooter or by bus or by train. The probabilities of his using car, scooter, bus and train are respectively 1/7, 2/7, 3/7 and 1/7. The probabilities of his reaching

(a) (a) (a) (c) (b) (a) (a) (d) (c) (c)

43. (d)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

the next two draws exactly one white ball is drawn. (Do not simplify the answer.) [2004]

1. 5. 9. 13. 17. 21. 25. 29. 33. 37.

42. (b)

69.

IIT JEE eBooks: www.crackjee.xyz Probability 9.57

MATRIX-MATCH TYPE QUESTIONS

70. 31.

ASSERTION-REASON TYPE QUESTIONS 71. (b) 75. (c)

72. (a)

73. (c)

74. (d)

32.

COMPREHENSION-TYPE QUESTIONS 76. 80. 84. 88.

(b) (a) (c) (b)

77. 81. 85. 89.

(c) (a) (a) (c)

78. 82. 86. 90.

(a) (b) (c) (c)

79. (a) 83. (c) 87. (a)

INTEGER-ANSWER TYPE QUESTIONS 91. 2 95. 2 99. 7

92. 3 96. 7 100. 3

93. 5 97. 4

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

33.

94. 1 98. 73 34.

LEVEL 2

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17.

(b) (b) (d) (d) (b)

2. 6. 10. 14. 18.

(a) (d) (b) (a) (d)

3. 7. 11. 15. 19.

(c) (c) (c) (a) (a)

4. 8. 12. 16. 20.

(d) (d) (d) (b) (d)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 21. 23. 25. 27. 29.

(a), (a), (a), (a), (a),

(c) (b), (d) (d) (b) (b), (c), (d)

22. 24. 26. 28. 30.

(a), (a), (a), (a), (b)

(b), (c), (d) (b) (b), (c), (d) (b), (c)

ASSERTION-REASON TYPE QUESTIONS 35. (b)

36. (b)

37. (b)

COMPREHENSION-TYPE QUESTIONS 38. (a) 42. (b) 46. (a)

39. (c) 43. (a) 47. (b)

40. (d) 44. (d)

41. (a) 45. (c)

INTEGER-ANSWER TYPE QUESTIONS 48. 3 52. 7

49. 5 53. 3

50. 2 54. 9

51. 8 55. 5

IIT JEE eBooks: www.crackjee.xyz 9.58 Comprehensive Mathematics—JEE Advanced

PAST YEARS’ IIT QUESTIONS

SUBJECTIVE TYPE QUESTIONS

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17. 21. 25. 29. 33.

(d) (d) (d) (b) (a) (a) (d) (c) (a)

2. 6. 10. 14. 18. 22. 26. 30. 34.

(a) (b) (b) (a) (b) (d) (d) (d) (a)

3. 7. 11. 15. 19. 23. 27. 31. 35.

(b) (c) (a) (c) (a) (a) (b) (c) (a)

4. 8. 12. 16. 20. 24. 28. 32.

1.

(a), (b), (a), (a), (a), (a),

(c), (d) (c) (d) (d) (d) (b)

2. 4. 6. 8. 10.

(a), (a), (c), (b), (b),

(b) (c) (c) (b) (a) (a) (c) (c)

g=

(b), (c) (c) (d) (c) (d)

26.

(a) (b) (a) (a), (b)

31.

n

FILL

3. 7. 11. 14.

(c) (b) (b) (c), (d)

1. 4. 6. 10.

5/21 1/9 23/35 1/36

66 [(6 C4 ) (12 C2 ) + (6 C5 ) (12 C1 ) + 1]

Hints and Solutions LEVEL 1

1. Number of ways of forming the sections ( 2n ) ! = (2nCn) (nCn) = n !n ! Number of ways in which both you and your friend are in different sections = (2C1) (2n – 2Cn – 1)

BLANKS TYPE QUESTIONS 2. 5. 7. 11.

3/16 3. P(A) = P(B) 1/3 £ p £ 1/2 2/5 8. 5/7 9. 11/16 1/4 12. 11/40

TRUE FALSE TYPE QUESTIONS 1. False

2. False

9m m + 8N

20 (6 C4 ) (12 C4 ) + 11(6 C5 ) (12 C1 )

4. (a) 8. (d) 12. (c)

2. 8 IN THE

27.

32. 1/7

INTEGER ANSWER TYPE QUESTIONS 1. 6

1 - (1 - p)3

6 28. 1/2 29. 2p2 – p3

2. (b)

2. (a) 6. (d) 10. (d)

3. 0.6976

p(1 - p)2

20[3n - 3(2n ) + 3]

COMPREHENSION TYPE QUESTIONS 1. 5. 9. 13.

1 1 (ii) 132 462

23. 4/35 25. m/(m + n)

ASSERTION-REASON TYPE QUESTIONS 1. (d)

2. (i)

4. No 7. 13.9% 8. 1/2, 1/3 9. 1/5 10. 99/1900 11. 23/30 12. 462(0.24)5 10( N + 2) 13. 1 – N + 7 C2 14. Best of 3 games 15. (3/4)n 16. 24/29 17. (i) Yes (ii) No 4 18. 97/25 19. 193/792 20. 0.62 21. (i) 1/2 (ii) 8/15 p p(1 - p) , b= 22. a = 3 1 - (1 - p ) 1 - (1 - p)3

MULTIPLE CORRECT ANSWERS TYPES QUESTIONS 1. 3. 5. 7. 9. 11.

1 1260

= \

2 ( 2n - 2 ) ! ( n - 1)!( n - 1)!

n 2n - 1 2. A and B can think of two numbers in n2 ways Two different numbers can be chosen in nC2 ways and assign smaller to A and larger to B. Probability of required event =

IIT JEE eBooks: www.crackjee.xyz Probability 9.59

\

probability of required event n

= 10

3.

C2

n2

P(A) =

n -1 = 2n

C3

4. P(B wins) 3

= 5. 6. 7. 8.

5

1 Ê 1ˆ Ê 1ˆ Ê 1ˆ Ê 1ˆ Ê 1ˆ 1 ÁË ˜¯ + ÁË ˜¯ ÁË ˜¯ + ÁË ˜¯ ÁË ˜¯ + ◊ ◊ ◊ = 2 2 2 2 2 2 3

Events are independent. P(A) P(A) = P(A « A) = P(A) Find the probability of the complement event. Use binomial distribution P(X ≥ 2) = P(X = 2) + P(X = 3)

9. nC3 p3(1 - p)n-3 = nC5 p5(1 - p)n-5 fi

( n - 3)( n - 4) 20

=

(1 - p )

2

p2

525 ¤ x = 1, 2 or x ≥ 48 21. E1: biased coin is chosen, E2: fair coin is choice and A: toss results in a head. P(E1) = n/(2n+1), P(E2) = (n+1)/(2n+1) P(A|E1) = 1 and P(A|E2) = 1/2 By the total probability rule n 31 (n + 1) 3n + 1 31 = = (1) + fi 2n + 1 21 2n + 1 2 (2n + 1) 42 fi n = 10 22. Use X + Y ~ B (12, 1/2) 23. 1 - P(selected person are together) =1-

12. Let A = {x1, x2, ..., xn}. Each xi ∈ A has four choices (ii) xi ∈ P, xi ∉ Q (i) xi ∈ P, xi ∈ Q (iv) xi ∉ P, xi ∉ Q. (iii) xi ∉ P, xi ∈ Q For P » Q = A, (iv) is not favourable. 13. Find probability of the complement event. 14. 3|x2 fi 3|x fi 9|x2. Contradiction.

16.

2

m

19. Use x = 11m + r where 0 £ r £ 10. 20. x2 - 50x + 100 > 0 ¤ (x - 25)2 > 525

C1 ¥ 10C2 20

2m

n

Cn

a b a b = 0 ¤ = = 1, - 1. c d c d Now, a = c = x fi a = c = x and b = d = x or b = d =-x Similarly, 6 other choices.

17. 6C3/63. 18. Let A denote the event that there is an odd man out in a game. The total number of possible cases is 2m. A person is odd man out if he is alone in getting a head or a tail. The number of ways in which there is exactly one tail (head) and the rest are heads (tails) is mC1 = m. Thus, the number of favourable ways is m + m = 2m. Therefore,

( n - 1) n

C2

24. P(E1) = P(E2) = P(E3) = 1/2, P(Ei « Ej) = 1/4 = P(Ei) P(Ej) " i π j and P(E1 « E2 « E3) = 1/4 π P(E1) P(E2) P(E3) 10! 25. Two B’s can be together in ways and the total 2! 11! number of ways = . 2!2! 26. Unit’s digit of the product will be 1, 3, 7 or 9 if and only if each of the four numbers end in 1, 3, 7 or 9. Thus, required probability is (4/10)4. 27. Total number of ways is 81 out of which 9 are favourable. 28. Probability of getting number of heads not equal to tails = 1 –

2n

Ê 1ˆ Cn Á ˜ Ë 2¯

2n

. Multiply by

1 so that 2

number of heads is > number of tails 29. (d) is true if and only if A and B are independent. 30. Sample space is {- 0.50, - 0.49, ..., - 0.0, 0, 0.01, ..., 0.49} and the event is {- 0.10, - 0.09, ..., 0.10}. 31. P(X ≥ 150) = 0.7, P(X £ 160) = 0.8, P(150 £ X £ 160) = 0.7 + 0.8 - 1 = 0.5. 32. Let X = number of odd numbers chosen. P(sum even) = P(X = 4) + P(X = 2) + P(X = 0)

IIT JEE eBooks: www.crackjee.xyz 9.60 Comprehensive Mathematics—JEE Advanced 4

2

2

Ê 2ˆ Ê 2ˆ Ê 1ˆ Ê 1ˆ 4 = Á ˜ + C2 Á ˜ Á ˜ + Á ˜ Ë 3¯ Ë 3 ¯ Ë 3¯ Ë 3¯

i

4

= 41/81 33. Use either both x and y are divisible by 3 or none of them is divisible by 3. Thus, required probability

=

2n

Cn

=

2n + 1 - 2 2n

Cn

.

36. (a + 4)2 - (- 5a + 64) < 0 ¤ a2 + 13a - 48 < 0 ¤ (a + 16) (a - 3) < 0 ¤ - 16 < a < 3 ¤ - 5 £ a £ 2 37. x - y is divisible by 5 if and only if both x and y are divisible by 5 or none of them is divisible by 5. Thus, favourable ways is NC2 + 4NC2. 38. (50C5) (45!)/50! = 1/120. 39. P(A « B¢) + P(A¢ « B) = (0.7) (0.2) + (0.3) (0.8) = 0.38. 40. Two tallest boy can be in different groups in (2C1) (2n-2Cn-1) ways. 41. 1, 2, 3, ... k appear together in that order in (n - k + 1)! ways. 42. P(E) = 0.62, P(F) = 0.5, P(E « F) = 0.35. 43. Impossible event. 4

4

44. Let Ei denote the event that the bag contains i black and (10 - i) white balls (i = 0, 1, 2, ..., 10). Let A denote the event that the three balls drawn at random from the bag are black. We have 1 P(Ei) = (i = 0, 1, 2, ..., 10) 11 P(A|Ei) = 0 for i = 0, 1, 2

C3

for i ≥ 3

But 3C3 + 4C3 + 5C3 + ... + 10C3 = 4C4 + 4C3 + 5C3 + ... + 10C3 = 5C4 + 5C3 + 6C3 + ... + 10C3 = ... = 11C4

10

34. ( C3) (7!)/10! = 1/6. 35. The women can be seated in 2nCn ways. Let A denote the event that pairs occupy the seats (1, 2), (3, 4), ..., (2n - 1, 2n) and B denote the event that pairs occupy the seats (2, 3), (4, 5), (6, 7), ..., (2n - 2, 2n - 1), (2n, 1). The number of cases favourable to A (B) is 2n. (For each man in the pair there are two choices.) However, there are just two cases common to A and B. One is the case {(M, W), (M, W), ......, (M, W)} of A(B) and {(W, M), (W, M), ..., (W, M)} of B(A). Therefore, P(A » B) = P(A) + P(B) - P(A » B)

C3

10

Now, by the total probability rule 1 1 ¥ 10 P(A) = [3C3 + 4C3 + ... + 10C3] 11 C3

= (NC2 + 2NC2)/3NC2.

2n + 2n - 2

P(A|Ei) =

and

P(E9|A) =

Use

45. P(E¢ « F¢ |G ) =

(

P ( E9 ) P A E9 P ( A)

)

P( E ¢ « F ¢ « G ) P(G )

=

P(G ) – P{( E « G ) » ( F « G )} P(G )

=

P(G ) – P( E « G ) – P ( F « G ) P(G )[1 – P( E ) – P( F )] = P(G ) P(G )

= P(E¢) – P(F)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 46—50. See Theory. 51. The child standing at the even numbered place can52. P(A) = P(B) = P(C) = 1/2 P(A « B) = P(B « C) = P(C « A) = 1/8 and P(A « B « C) = 1/8. 4

4

Ê 2ˆ Ê 1ˆ 53. p1 = Á ˜ , p2 = Á ˜ . Ë 5¯ Ë 2¯ 54. Let E1(E2) denote the event that my son posted (did not post) the letter and A be the event that my friend did not receive the letter. P(E1) = 9/10, P(E2) = 1/10 P(A|E1) = 99/100, P(A|E2) = 1 P(E1|A) and P(E2|A). 2 55. x - 25x - 150 £ 0 ¤ (x - 30) (x + 5) £ 0 ¤ x £ 30 x2 - 17x + 30 ≥ 0 ¤ (x - 2) (x - 15) ≥ 0 ¤ x £ 2 or x ≥ 15 2 30x - x is a perfect square of a natural number if x = 3, 6, 12, 15, 18, 21 or 24. Lastly 30x - x2 < 0 ¤ x(30 - x) < 0 or x > 30.

IIT JEE eBooks: www.crackjee.xyz Probability 9.61

56. Difference is two if the outcomes are (1, 3) (2, 4), (3, 5), (4, 6), (3, 1), (4, 2), (5, 3), (6, 4). Similarly for others m -1 1 57. P(A) = , P(A¢) = m m P(B | A) =

m -1 , P(B | A¢) = 0 m

Now P(B | A) =

P ( A « B) m -1 m -1 fi = P ( A) m m

( m - 1)2

fi P(A « B) = Also

7

P(B) = P(A) P(B | A) + P(A¢) P(B | A¢)

\

P(B¢) =1 - P(B) = 1 -

( m - 1)

2

=

2

2m - 1

m2 m2 P( A « B¢) P( A) - P( A « B) P(A | B¢) = = P( B¢) P( B¢) =

63. p1 =

C4

=

1 , p2 = 0. 2

For (d), list all the possible 14 combinations. 64. See Theory. 65. P(A « B) = P(both show odd numbers)= (3/6)2 etc. A « B « C means a, b and a + b are odd. Impossible.

MATRIX-MATCH TYPE QUESTIONS

2

P(A wins) =

= P(A) P(B | A) + P(A¢) [1 - P(B¢ | A¢)] = a(1 - a) + (1 - a) [1 - (1 - a)] = 2a(1 - a) P( A¢ « B) P( B) - P( A « B) = P(A¢|B) = P( B) P( B) =

P( B) - P( A) P( B | A) P( B)

=

2 a (1 - a ) - a (1 - a ) 1 = 2 a (1 - a ) 2 P( A¢ « B¢ ) 1 - P( A » B) = P( B¢) 1 - P( B)

and P(A » B) = P(A) + P(B) - P(A « B) = a2 •

pq n 59. P(X ≥ n) = Â P(X = k) = = qn 1 q k=n P ( X ≥ m + n) qm + n = = qn m P ( X ≥ m) q

1 Ê 5ˆ Ê 1ˆ Ê 5ˆ Ê 1ˆ + Á ˜ q 1 q 2 Á ˜ + Á ˜ q 12 q 22 Á ˜ Ë 6¯ Ë 6¯ Ë 6¯ 6 Ë 6¯ + .....

58. P(B) = P(A) P(B | A) + P(A¢) P(B | A¢)

P(X ≥ m + n | X ≥ m) =

C3

2

2

P(A¢ | B¢) =

8

66. Let q1 = 1 - p1 and q2 = 1 - p2. Now, A can win the game at the 1st, 4th, 7th, ..., trials.

m -1 2m - 1

m - 1 Ê m - 1ˆ Ê m - 1ˆ +Á -Á P(A » B) = . ˜ Ë m ¯ Ë m ˜¯ m

and

P ( X = m + n) = pqn. P ( X ≥ m)

60. A and B are independent events. 61. P(E1) = 6C3/63 = 5/54, P(E2) = 6/63 = 1/36, and P(E1 « E2) = 1/36. 62. For divisibility with 9, leave one of (1, 8), (2, 7), (3, 6) or (4, 5). For divisibility with 3, leave one of (1, 2), (1, 5), (1, 8), (2, 4), (2, 7), (3, 6), (3, 9), (4, 5), (4, 8), (5, 7), (6, 9), (7, 8). For divisibility with 5 unit’s digit must be 5.

m2

Ê m - 1ˆ Ê m - 1ˆ Ê 1 ˆ Ê m - 1ˆ + Á ˜ ( 0) = Á = Á ˜ Á ˜ Ë m ¯ Ë m ¯ Ë m¯ Ë m ˜¯ fi

P(X = m + n | X ≥ m) =

=

1/ 6 6 = 1 - (5q1 q2 / 36) 36 - 5q1 q2

Similarly P(B wins) =

5 Ê5 ˆÊ5 ˆ p1 + Á q1 q2 ˜ Á q1 ˜ + Ë ¯ Ë6 ¯ 6 6 2

Ê5 ˆ Ê5 ˆ ÁË q1 q2 ˜¯ ÁË p1 ˜¯ + ... 6 6 = and

5 p1 / 6 30 p1 = 1 - (5q1 q2 / 36) 36 - 5q1 q2 P(C wins) =

30 q1 p2 36 - 5q1 q2

The game will be equiprobable for all three players if P(A wins) = P(B wins) = P(C wins) fi

5 p1 5 q1 p2 1 = = 36 - 5q1 q2 36 - 5q1 q2 36 - 5q1 q2

IIT JEE eBooks: www.crackjee.xyz 9.62 Comprehensive Mathematics—JEE Advanced



p1 = 1/5, p2 = 1/4

When p1 = 1/5, p2 = 1/4 each player has the same probability of winning. 67. (a) In this case the originator has (N - 1) choices and each of the remaining (n - 1) narrators has (N - 2) choices. Thus, the number of favourable ways is (N - 1) (N - 2)n-1. Hence, the probability of the required event is ( N - 1) ( N - 2)n - 1 ( N - 1) n

Ê N - 2ˆ =Á Ë N - 1 ˜¯

n -1

Ê 1 ˆ = Á1 N - 1˜¯ Ë

n -1

(b) and (c) are similar to (a). (d) Since at each stage the recipient is chosen at random from the remaining (N - 1) people, the total number of ways is (N - 1)n number of favourable ways. The originator has (N - 1) choices, the second person has (N - 2) cipient), the third person has (N - 3) choices, ..., and the nth person has (N - n) choices. Thus, the number of favourable ways is (N - 1) (N - 2) ... (N - n). Hence, the probability of the required event is ( N - 1)( N - 2)...( N - n) ( N - 1)

n

N -1

=

Pn

( N - 1) n

.

68. P(E1) = P(E2) = P(E3) = P(E4) = 1/4 P(S|E1) = 1/6, P(S|E2) = 1/8, P(S|E3) = 1/10 and P(S|E4) = 1/12 Now use Bayes’ rule. 69. Since all the players are of equal strength, with whom S1 is paired will have no bearing on the probability of his winning. The number of ways of choosing 8 winners out of 16 is 16C8. The number of ways of choosing S1 and 7 other winners out of 15 is 15C7. \ Probability that S1 will win 15

=

16

15! 8!8! 1 C7 = = . 7!8! 16! 2 C8

Let E1(E2) denote the event that S1 and S2 are paired (not paired) together and the A denote the event that one of two players S1 and S2 is among the winners. Since S1 can be paired with any of the remaining 15 players, We have P(E1) = 1/15 and P(E2) = 14/15.

In case E1 occurs it is certain that one of S1 and S2 will be among the winners. In case E2 occurs the probability that exactly one of S1 and S2 is among the winners is P [(S1 « S2¢) » (S1¢ « S2)] = P(S1 « S2¢) + P(S1¢« S2) = P(S1) P(S2¢) + P(S¢1) P(S2) 1ˆ Ê 1ˆ 1 1 Ê 1ˆ Ê = Á ˜ Á1 - ˜ + Á1 - ˜ = . Ë 2¯ Ë 2¯ Ë 2¯ 2 2 That is, P(A|E1) = 1 and P(A|E2) = 1/2. Now, use the total probability rule. Both S1 and S2 will reach stage 2 if S1 and S2 are not paired against each other and each of them wins his game. 14 1 1 7 ¥ ¥ = This is possible with probability . 15 2 2 30 70. a, b, c can be chosen in 100C3 ways. (a) a, b, c are in A.P. fi 2b = a + c fi a and c are both even or both odd. Thus, number of favourable ways = 50C2 + 50C2 = 50 ¥ 49 50 ¥ 49 ¥ 3 ¥ 2 1 \ probability = = 100 ¥ 99 ¥ 98 66 (b) Taking r = 2, 3, ....., 10 we can show a, b, c can Thus, probability of required event 53 ¥ 3 ¥ 2 53 = = 100 ¥ 99 ¥ 98 161700 (c)

1 1 1 , , a b c

¤ a, b, c (d) P(a + b + c is even) = 1 – P(a + b + c is odd) 50

=1– =1–

C3 +

(

)(

50

C1

100

C3

50

C2

)

50 ¥ 49 ¥ 48 + 50 ¥ 50 ¥ 49 ¥ 3 1 = 100 ¥ 99 ¥ 98 2

ASSERTION-REASON TYPE QUESTIONS 71. A « B « C Õ C fi P(A « B « C) £ P(C) = 0 \ P(A « B « C) = 0 Similarly, P(A « C) = 0, P(B « C) = 0

IIT JEE eBooks: www.crackjee.xyz Probability 9.63

COMPREHENSION-TYPE QUESTIONS

Next, P(A » B » C) = P(A » B) + P(C) – P[(A » B) « C)] But P[(A » B) « C] = P[(A « C) » (B « C)] = P(A « C) + P(B « C) – P(A « B « C) = 0

77. Let E denote the event that the ball lost is white.

Thus, P(A » B » C) = P(A » B)

P(E) =

72. P[A « (B » C)] = P[(A « B) » (A « C)] = P(A « B) + P(A « C) – P(A « B « C)

w b P(E¢) = w+b w+b

P(A | E) =

= P(A) P(B) + P(A) P(C) – P(A « B « C) A and B » C are independent ¤ P[A « (B » C)] = P(A) P(B » C) ¤ P(A) P(B) + P(A) P(C) – P(A « B « C) = P(A) [P(B) + P(C) – P(B « C)]

Now use the total probability rule. 78. Similar to 77. 79. Let Er (0 £ r £ l) denote the event that r white balls are lost. w

¤ P(A « B « C) = P(A) P(B « C)

P(Er) =

¤ A and B « C are independent 73. Let P = {a1, a2, ...., an}. For ai ∈ P (1 £ i £ n), we have four choices: (i) ai ∈ Q and ai ∈ R (ii) ai ∈ Q and ai ∉ R (iii) ai ∉ Q and ai ∈ R (iv) ai ∉ Q and ai ∉ R Thus, total number of choices for picking up Q and R is 4n. For ai not to belong Q « R, we have 3 choices. Therefore, Q « R = f in 3n ways. Thus, probability Q « R = f is (3/4)n. Next, ai ∈ Q » R in 3 ways. Therefore, Q » R = P in 3n ways. Thus, probability Q » R = P is (3/4)n. 74. If a and b are both even, then a + b is even, therefore P(A « B « C) = 0. We have P(A) =

1 1 1 P(B) = , P(A « B) = . 2 2 4

Now, P[(A « B « D)|A « B] = =

P ÈÎ( A « B « D ) « ( A » B )˘˚ P ( A » B)

P ( A « B) = P ( A » B)

14 1 = 1 2 +1 2 -1 4 3

75. Statement-2 is false. P(P and Q contradict each other) = p(1 – 2p) + 2p(1 – p) = 1/2 fi 8p2 – 6p + 1 = 0. fi (2p – 1) (4p – 1) = 0 fi p = 1/2, 1/4.

w -1 w , P(A|E¢) = w + b -1 w + b -1

Cr b Cl - r w+ b

Cl

and P(A | Er) =

w-r w+b-l

Now use the total probability rule and the identity l

 (w - r) (wCr) (bCl - r) = w (w + b - 1Cl).

r=0

80.

1 1 = . wƕ 1 + b / w 1+ 0

lim p(l) = lim

wƕ

81. a2 - 1 is divisible by 10 if and only if last digit of a is 1 or 9. If r = 0, there are 2k ways to choose a. \ pn = 2k/n. 82. If r = 9, then there are 2(k + 1) ways to choose a and pn = 2(k+1)/n. 83. If 1 £ r £ 8, a can be chosen in (2k + 1) ways. \ pn = (2k + 1)/n. 84. When r = 0, pn = 1/5, 2(1 + 1 / k ) 1 When r = 9, pn = Æ 10 + r / k 5 as n and hence k Æ •. Similarly, for 1 £ r £ 8. 85. Similar to above. Divide r as 0 £ r £ 2; 3 £ r £ 6; and 7 £ r £ 9. 86. Least score is 3 as the player gets a further throw if he gets 1. throw results in a 2. 88. The player scores 6 or less the following ways: the player scores r r - 1 (if r - 1 > 1) at the second r - 2 (if

IIT JEE eBooks: www.crackjee.xyz 9.64 Comprehensive Mathematics—JEE Advanced

r - 2 > 1) at the third trial; and so on. This leads to the probability 1 Ê 1ˆ Ê 1ˆ Ê 1ˆ Ê 1ˆ Ê 1ˆ Ê 1ˆ + Á ˜ Á ˜ + Á ˜ Á ˜ Á ˜ + ... + Á ˜ Ë 6¯ 6 Ë 6¯ Ë 6¯ Ë 6¯ Ë 6¯ Ë 6¯

r-2

Ê 1ˆ ÁË ˜¯ 6

È Ê 1ˆ r - 1˘ Í1 - Á ˜ ˙ if 2 £ r £ 6. ÍÎ Ë 6 ¯ ˙˚ 89. We now enumerate the ways in which the player can score a value r > 6. He can score 1 at each of the r - 2 trials and 2 at the (r - 1)th trial; 1 at each r - 3 trials and 3 at the (r - 2)th trial; 1 r - 4 trials and 4 at the (r - 3)th trial; and so on. This gives the probability 1 = 5

Ê 1ˆ ÁË ˜¯ 6 =

r-2

1 5

Ê 1ˆ Ê 1ˆ ÁË ˜¯ + ÁË ˜¯ 6 6

Ê 1ˆ Ê 1ˆ ÁË ˜¯ + ... + ÁË ˜¯ 6 6

r-6

Ê 1ˆ ÁË ˜¯ 6

ÈÊ 1 ˆ r - 6 Ê 1 ˆ r - 1 ˘ -Á ˜ ÍÁ ˜ ˙ for r > 6. Ë 6¯ ÍÎË 6 ¯ ˙˚ S=

90.

r-3

=

• 1È Ê 1ˆ Í1 + Â Á ˜ 5 ÎÍ r = 7 Ë 6 ¯

r-6

• Ê 1ˆ - ÂÁ ˜ r = 2 Ë 6¯

r -1˘

˙ ˚˙

1 . 5

INTEGER-ANSWER TYPE QUESTIONS 91. P(A « B) £ P(A), P(B) £ P(A » B) = P(A « B) fi P(A) = P(B) = P(A « B). 92. Let A, B, C denote the events a sum of 7, a sum of 11, neither 7 nor 11 occurs. Then P(A) = 1/6, P(B) = 1/18 and P(C) = 7/9. p = P(A) + P(C « A) + P(C « C « A) + ... =

P ( A) 1/ 6 3 = = . 1 - P (C ) 1 - 7 / 9 4

93. Let G denote the event the examinee guesses, C the event that the examinee copies and K the event that the examinee knows. Let R denote the event that the answer is right. We have 1 1 P(G) = , P(C) = 3 6 and

P(K) = 1 -

1 1 1 - = . 3 6 2

Also, P(R|G) =

1 1 , P(R|C) = , P(R|K) = 1. m 8

By Bayes’ rule P(K|R) = 24m/(16 + 25m) = 120/141. 1+ 4p 1- p 1- 2p 94. 0 £ £ 1, 0 £ £ 1, 0 £ £1 4 2 2 and fi and fi

0 £

1- 4p 1- p 1- 2p + + £ 1 4 2 2

1 3 1 1 £ p £ , - 1 £ p £ 1, - £ p £ 4 4 2 2 1 5 £ p£ 2 2

1 1 £ p £ fi 2p = 1. 2 2

95. Since 1 + 2 + ... + 9 = 36, a number consisting all these digits will be divisible by 9. Thus, the number will be divisible by 36 if and only if it is divisible by 4. There are 9P9 = 9! ways of arranging the nine digits to form a nine-digit number. If this number is to be divisible by 4, its last digit must be even, and the number formed by its last two digits must be divisible by 4. This limits the possible values of the last pair to the following: 12, 32, 52, 72, 92, 24, 64, 84, 16, 36, 56, 76, 96, 28, 48 68. That is, there are 16 ways of choosing the last two digits. The remaining digits can be arranged in 7P7 = 7! ways. Therefore the number of favourable arrangements is (7!) (16) out of a total of 9!, so that the required probability is (16) (7!) 16 2 = = . 9! 9¥8 9 which the second can be chosen in 63 ways. Now, if square is one of the 24 non-corner squares along the sides of the chessboard, the second square can any of the 36 remaining squares, the second square can be chosen in four ways. Therefore the number of favourable choices is (4) (1) + (24) (2) + (36) (4) = 196, so that the required probability is 196 7 = . 64 ¥ 63 144

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97. The total number of ways of ticking one or more 5

C1 + 5C2 + 5C3 + 5C4 + 5C5 = 31

Each of these is equally likely to be chosen, so the probability of ticking the alternatives correctly at Ê 30 ˆ Ê 1 ˆ ; ÁË ˜¯ ÁË ˜¯ = 31 30 31 and that at the third trial is also Ê 30 ˆ Ê 29 ˆ Ê 1 ˆ 1 . ÁË ˜¯ ÁË ˜¯ ÁË ˜¯ = 31 30 29 31 Continuing this way, we see that the probability for r trials is r/31. Now, r 1 31 > fir> fi r ≥ 4. 31 8 8 Hence 4 is the least number of trials required. that is, their throws are equal. There are 36 outcomes for one roll of a pair of dice, so the total number of outcomes for the two persons making one throw each is 36 x 36 = 1296. Now let ai, with 2 £ i £ 12, be the number of ways to get a sum of i showing on the pair of dice when they are rolled. Then a2 = a12 = 1, a3 = a11 = 2, a4 = a10 = 3, a6 = a8 = 5, a7 = 6. a5 = a9 = 4, Each player can throw an i in ai ways, so both of them will throw an i in ai2 ways. Summing over

all values of i, we see that the number of ways the throws of the two persons will be equal is a22 + a32 + ... + a122 = 2(12 + 22 + 32 + 42 + 52) + 62. 2 = (5) (6) (11) + 62 = 146 6 \

1-p=

146 73 = 1296 648

648p = 73. 99. Let the second success occur at the nth trial. This n - 1 trials, so that the probability of getting the second success at the nth trial is pn = (n-1C1pqn-1-1) p = (n - 1)p2 qn-2. Therefore the probability of the required event is p4 + p5 + p6 + ... = p2q2 (3 + 4q + 5q2 + 6q3 + ...) = p2q2 [3(1 + q + q2 + q3 + ...) + q(1 + 2q + 3q2 + ...)] 2 2 -1 -2 = p q [3(1 - q) + q(1 - q) ] 1 Ê3 q ˆ 27 = p 2q 2 Á + 2 ˜ = . 32 Ëp p ¯ 2 is not an integer. Therefore, 3 P(X = r) is maximum when r = [(n + 1)p] = 33.

100. (n + 1)p = 33

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10 Trigonometry 10.1 INTRODUCTION

An angle is the amount of revolution which a line OP revolving about the point O has undergone in passing from its initial position OA OB.

10.2 DOMAIN AND RANGE OF TRIGONOMETRICAL FUNCTIONS sin A cos A tan A cosec A sec A cot A

Domain R R R – {(2n + 1) p/2, n ∈ I} R – {np, n ∈ I} R – {(2n + 1) p/2, n ∈ I} R – {np, n ∈ I}

A £ Fig. 10.1

If the rotation is in the clockwise sense, the angle measured is negative and it is positive if the rotation is in the anti-clockwise sense. The two commonly used systems of measuring an angle are 1. Sexagesimal system in which 1 right angle = 90 degrees (90°) 1 degree = 60 minutes (60¢) 1 minute = 60 seconds (60¢) 2. Circular systems in which the unit of measurement is the angle subtended at the centre of a circle by an arc whose length is equal to the radius and is called a radian.

Range [– 1, 1] [– 1, 1] R = (– •, •) (– •, – 1] » [1, •) (– •, – 1] » [1, •) R = (– •, •)

A £1

sec A ≥ 1 or sec A £ – 1 and cosec A ≥ 1 or cosec A £ – 1 10.3 SOME BASIC FORMULAE

1. sin2 A + cos2 A = 1 or cos2 A = 1 – sin2 A or sin2 A = 1 – cos2 A. 2. 1 + tan2 A = sec2 A or sec2 A – tan2 A = 1. 3. 1 + cot2 A = cosec2 A or cosec2 A – cot2 A = 1. 4. sin A cosec A = tan A cot A = cos A sec A = 1. A system of rectangular coordinate axes divides a plane into four quadrants. An angle q lies in one and only one of these quadrants. The values of the trigonometric ratios in the various quadrants are shown in Fig. 11.1.

Relation between degree and radian p radian = 180 degrees 180 degrees p = 57° 17¢ 45≤ (approximately)

1 radian =

The trigonometrical ratios of an angle are numerical quantities. Each one of them represents the ratio of the length of one side to another of a right angled triangle.

Fig. 10.2

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becomes cotangent, secant becomes cosecant and vice versa.

10.4 ALLIED OR RELATED ANGLES

The angles np/2 + q and np/2 – q, where n is any integer, are known as allied or related angles. The trigonometric functions of these angles can be expressed as trigonometric functions of q, with either a plus or a minus sign. The following working rules can be used in determining these functions : 1. Assuming that 0 < q < 90°, note the quadrant in which the given angle lies. The result has a plus or minus sign according as the given function is positive or negative in that quadrant. 2. If n is even, the result contains the same trigonometric function as the given function. But if n is odd, the result contains the corresponding co-function, i.e., sine becomes cosine, tangent

Illustration 1 (a) To determine sin (540° – q), we note that 540° – q = 6 ¥ 90° – q is a second quadrant angle if 0° < q < 90°. In this quadrant, sine is positive and, since the given angle contains an even multiple of p 2 , the sine function is retained. Hence, sin (540° – q) = sin q. (b) To determine cos (630° – q), we note 630° – q = 7 ¥ 90° – q is a third quadrant angle if 0° < q < 90°. In this quadrant, cosine is negative and, since the given angle contains an odd multiple of p 2 , cosine is replaced by sine. Hence, cos (630° – q) = – sin q.

Sine, cosine and tangent of some angles less than 90°



15°

18°

30°

sin

0

6- 2 4

5 -1 4

1 2

cos

1

6+ 2 4

10 + 2 5 4

3 2

tan

0

2- 3

25 - 10 5 5

1

36°

45°

60°

90°

sin

10 - 2 5 4

1

3 2

1

cos

5 +1 4

1

1 2

0

tan

5-2 5

1

10.5 COMPOUND ANGLES

An angle made up of the algebraic sum of two or more angles is called a compound angle. Some formulae and results regarding compound angles: 1. sin (A + B) = sin A cos B + cos A sin B. 2. sin (A – B) = sin A cos B – cos A sin B. 3. cos (A + B) = cos A cos B – sin A sin B. 4. cos (A – B) = cos A cos B + sin A sin B. 5. tan (A + B) =

tan A + tan B . 1 - tan A tan B

2 2

3

3

6. tan (A – B) =

tan A - tan B . 1 + tan A tan B

7. sin (A + B) sin (A – B) = sin2 A – sin2 B = cos2 B – cos2 A. 8. cos (A + B) cos (A – B) = cos2 A – sin2 B = cos2 B – sin2 A. 9. sin 2A = 2 sin A cos A =

2 tan A 1 + tan 2 A

.

10. cos 2A = cos2 A – sin2 A = 1 – 2 sin2 A

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= 2 cos2 A – 1 =

1 - tan 2 A 1 + tan A 2

.

sin (a + (n – 1)b) =

11. 1 + cos 2A = 2 cos2 A, 1 – cos 2A = 2 sin2 A. 12. tan 2A = 13. (a) (c)

2 tan A 1 - tan 2 A

cos (a + (n – 1)b) = (b)

1 + cos A A = cot sin A 2

1 - cos A A = tan 2 . 1 + cos A 2

24. tan (A1 + A2 +

3tan A - tan 3 A 1 - 3 tan A 2

(b) sin C – sin D = 2 cos

C+D C-D sin 2 2

(c) cos C + cos D = 2 cos

C+D C-D cos 2 2

(d) cos C – cos D = 2 sin

C+D D-C sin . 2 2

S3 = tan A1 tan A2 tan A3 + tan A2 tan A3 tan A4 + (sum of the tangents taken three at a time, there are nC3 such terms and so on.

1. sin q sin (60° – q) sin (60° + q) = ( 1 4 ) sin 3q. 2. cos q cos (60° – q) cos (60° + q) = ( 1 4 ) cos 3q. 3. tan q tan (60° – q) tan (60° + q) = tan 3q Ê 1ˆ 4. sin 9° = Á ˜ ÈÍ 3 + 5 - 5 - 5 ˘˙ = cos 81° Ë 4¯ Î ˚

sin ( A + B) cos A cos B

Ê 1ˆ 5. cos 9° = Á ˜ ÈÍ 3 + 5 + 5 - 5 ˘˙ = sin 81° Ë 4¯ Î ˚ 6. cos 36° – cos 72° = 1/2 7. cos 36° cos 72° = 1/4

sin ( A - B) . cos A cos B

2 sin A cos B = sin (A + B) + sin (A – B) 2 cos A sin B = sin (A + B) – sin (A – B) 2 cos A cos B = cos (A + B) + cos (A – B) 2 sin A sin B = cos (A – B) – cos (A + B) (Note) n n n 3 19. sin nA = cos A ( C1 tan A – C3 tan A +

18. (a) (b) (c) (d)

)

20. cos nA = cosn A (1 – nC2 tan2 A + nC4 tan4 A

)

C1 tan A - n C3 tan 3 A + n C5 tan 5 A º

n

1 - n C2 tan 2 A + n C4 tan 4 A º

22. sin a + sin (a + b) + sin (a + 2b) +

+ tan An

10.6 SOME IMPORTANT RESULTS

(Note)

21. tan nA =

S1 - S3 + S5 - S7 + º 1 - S 2 + S 4 - S6 + º

(sum of the tangents taken two at a time, there are nC2 such terms)

.

C+D C-D cos 2 2

C5 tan5 A –

sin (nb/2)

S2 = tan A1 tan A2 + tan A2 tan A3 +

16. (a) sin C + sin D = 2 sin

n

+ An) =

sin ( b 2)

+

(sum of the tangents taken one at a time)

15. tan (A + B + C) tan A + tan B + tan C - tan A tan B tan C . = 1 - tan B tan C - tan C tan A - tan A tan B

(b) tan A – tan B =

sin (nb/2)

cos (a + ( n - 1) b 2)

where S1 = tan A1 + tan A2 +

14. (a) sin 3A = 3 sin A – 4 sin3 A (b) cos 3A = 4 cos3 A – 3 cos A

17. (a) tan A + tan B =

sin ( b 2)

23. cos a + cos (a + b) + cos (a + 2b) +

.

1 - cos A A = tan sin A 2

(c) tan 3A =

sin (a + ( n - 1) b 2)

+

8. sin 22

1 ° Ê 1ˆ È = Á ˜ Í 2 - 2 ˘˙ Ë 2¯ Î ˚ 2

9. cos 22

1° = 2

10. tan 22

1° = 2

2 –1

11. cot 22

1° = 2

2 +1

12.

Ê 1ˆ È ˘ ÁË ˜¯ ÍÎ 2 + 2 ˙˚ 2

- a 2 + b 2 £ a sin x + b cos x £ ∈ R.

a 2 + b 2 for all x

10.7 IDENTITIES

A trigonometric equation is an identity if it is true for all values of the angle or angles involved. A given identity

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may be established by (i) reducing either side to the other one, or (ii) reducing each side to the same expression, or in (i) and (ii). 10.8 CONDITIONAL IDENTITIES

When the angles A, B and C satisfy a given relation, many interesting identities can be established connecting the trigonometric functions of these angles. In proving these identities, we require the properties of complementary and supplementary angles. For example, if A + B + C = p, then 1. sin (B + C) = sin A, cos B = – cos (C + A) 2. cos (A + B) = – cos C, sin C = sin (A + B) 3. tan (C + A) = – tan B, cot A = – cot (B + C). 4. cos

A+ B C = sin , 2 2

5. sin

C+A B A B+C = cos , sin = cos . 2 2 2 2

cos

C A+ B = sin . 2 2

(a) 3 (b) 2 (c) 1 (d) 0 Ans. (d) Solution: Let 4 sin q – 3 cos q = a then (3 sin q + 4 cos q)2 + (4 sin q – 3 cos q)2 = 25 + a2 fi 25 = 25 + a2 fi a2 = 0 fi 4 sin q – 3 cos q = 0. Example 2

sin2 12° + sin2 21° + sin2 39° + sin2 48°

– sin2 9° – sin2 18° is equal to (a) – 1 (b) 0 (c) 1/2 (d) 1 Ans. (d) Solution: we can write the given expression as sin2 12° + sin2 21° + (sin2 39° – sin2 9°) + (sin2 48° – sin2 18°) = sin2 12° + sin2 21° + sin(48°) sin(30°) + sin(66°) sin(30°) 1 1 = sin2 12° + sin2 21° + sin 48° + sin 66° 2 2

A B B+C C+A 6. tan = cot , tan = cot . 2 2 2 2

=

1 1 1 (1 – cos 24°) + (1 – cos 42°) + cos 42° 2 2 2

Some Important Identities If A + B + C = p, then 1. tan A + tan B + tan C = tan A tan B tan C. 2. cot B cot C + cot C cot A + cot A cot B = 1.

+

1 cos 24° 2

B C C A A B 3. tan tan + tan tan + tan tan = 1. 2 2 2 2 2 2 4. cot

A B C A B C + cot + cot = cot cot cot . 2 2 2 2 2 2

5. sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C. 6. cos 2A + cos 2B + cos 2C = – 1 – 4 cos A cos B cos C. 7. cos2 A + cos2 B + cos2 C = 1 – 2 cos A cos B cos C. A B C 8. sin A + sin B + sin C = 4 cos cos cos . 2 2 2 9. cos A + cos B + cos C = 1 + 4 sin

A B C sin sin . 2 2 2

SOLVED EXAMPLES

SINGLE CORRECT ANSWER TYPE QUESTIONS If 3 sin q + 4 cos q = 5, then the value of Example 1 4 sin q – 3 cos q is equal to

= 1. Example 3

cos 2a – 2 sin2b – 4 cos (a +b) sina sinb

– cos 2 (a +b) is independent of (a) a (b) b (c) both a and b (d) none of these Ans. (c) Solution: Given expression can be writtenes cos 2 a – (1 – cos 2 b) – 4 cos (a +b) sina sinb – cos 2 (a + b) = (cos 2a + cos 2b) – [1 + cos 2 (a + b)] – 4 cos (a + b) sina sin b = 2 cos (a + b) cos (a – b) – 2 cos2 (a + b) – 4 cos (a + b) sina sinb = 2 cos (a + b) [cos (a – b) – cos (a + b) – 2 sina sinb] = 2 cos (a + b) [2 sina sinb – 2 sina sinb] = 0 which is independent of both a and b Example 4

If

sin q sin 3q sin 9q + + cos 3q cos 9q cos 27q

1 (tan x – tanq) then the value of x is 2 (a) 3q (b) 6q (c) 9q Ans. (d)

=

(d) 27q

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sin q 1 2sin q cos q 1 sin 2q = = cos 3q 2 cos 3q cos q 2 cos 3q cos q

Solution:

= Similarly and so

1 sin (3q - q ) 1 = ( tan 3q - tan q ) 2 cos 3q cos q 2

sin 3q 1 = (tan 9q – tan 3q) cos 9q 2

sin 9q 1 = (tan 27q – tan 9q) cos 27q 2

1 1 (tan x – tanq) = (tan 27q – tanq) 2 2

fi x = 27q If 2 sina cosb sin g = sinb sin (a + g) then Example 5 tana, tanb, tan g are in (a) Arithmetic progression (b) Geometric progression (c) Harmonic progression (d) none of these Ans. (c) Solution: 2 sina cosb sin g = sinb (sina cos g + cosa sin g) sina cosb sin g = sina sinb cos g + cosa sinb sin g fi 2 cot b = cot g + cot a

If a cos3 a + 3a cos a sin2 a = m Example 7 and a sin3 a + 3a cos2 a sin a = n, then (m + n)2/3 + (m – n)2/3 is equal to (b) 2a1/3 (c) 2a2/3 (d) 2a3. (a) 2a2 Ans. (c). Solution: From the given relations, we get m + n = a cos3 a + 3a cos a sin2 a + 3a cos2 a sin a + a sin3 a = a(cos a + sin a)3 Similarly m – n = a(cos a – sin a)3 \ (m + n)2/3 + (m – n)2/3 = a2/3 [(cos a + sin a)2 + (cos a – sin 2 a) ] = a2/3 [2 (cos2 a + sin2 a)] = 2a2/3 Example 8

If

2 sin a = y, then 1 + cos a + sin a 1 - cos a + sin a is equal to 1 + sin a

(a) 1 y Ans. (b) Solution:

(b) y

(c) 1 – y (d) 1 + y.

1 - cos a + sin a 1 + sin a



2 1 1 = + tan b tan a tan g

=

1 - cos a + sin a 1 + cos a + sin a ◊ 1 + sin a 1 + cos a + sin a



tana, tanb, tan g are in H.P.

=

(1 + sin a ) 2 - cos 2 a (1 + sin a ) (1 + cos a + sin a )

=

1 + 2sin a + sin 2 a - 1 + sin 2 a (1 + sin a ) (1 + cos a + sin a )

=

2sin a (1 + sin a ) (1 + sin a ) (1 + cos a + sin a )

=

2 sin a = y. 1 + cos a + sin a

Example 6

If cot a + tan a = m and

1 – cos a cos a

= n, then (a) m (mn2)1/3 – n(nm2)1/3 = 1 (b) m(m2n)1/3 – n(mn2)1/3 = 1 (c) n(mn2)1/3 – m(nm2)1/3 = 1 (d) n(m2n)1/3 – m(mn2)1/3 = 1. Ans. (a) Solution: Clearly a π 0. cot a + tan a = m fi 1 + tan2 a = m tan a (1) fi sec2 a = m tan a 1 and – cos a = n fi sec2 a – 1 = n sec a cos a fi fi

tan a = n sec a fi tan a = n sec a [by (1)] tan4 a = n2m tan a 2

4

2

2

fi tan3 a = n2m fi tan a = (n2m)1/3 2 2 1/3 and sec a = m(n m) (by (1)] 2 2 2 1/3 2 Nowsec a – tan a = 1 fi m(n m) – (n m)2/3 = 1 fi m(mn2)1/3 – n(nm2)1/3 = 1.

Example 9 is equal to (a) –2 Ans. (b) Solution:

Minimum value of 4x2 – 4x (b) –1

(c) –1/2

q

2

q

(d) 0

4x2 ± 4x sinq – (1 – sin2q)

= 4x2 ± 4x sinq + sin2q – 1 = (2x ± sinq)2 – 1 ≥ –1 Hence the required value is –1. Example 10 If sin q and cos q are the roots of the equation ax2 – bx + c = 0, then a, b and c satisfy the relation

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(a) a2 + b2 + 2ac = 0 (b) a2 – b2 + 2ac = 0 (c) a2 + c2 + 2ab = 0 (d) a2 – b2 – 2ac = 0. Ans. (b). Solution: Since sin q and cos q are roots of the given quadratic equation, we have sin q + cos q = b/a and sin q cos q = c/a. 2 2 fi (sin q + cos q)2 = b a 2 2 sin2 q + cos2 q + 2 sin q cos q = b a



c b2 fi 1+2 = 2 a a Example 11

fi a2 + 2ac – b2 = 0.

If sin x + sin2 x = 1, then the value of

If q

q

= 8 17 , then the value of cos (30° + q) + cos (45° – q) + cos (120° – q) is Ê 3 - 1 1 ˆ 23 + (a) Á ˜ 2 ¯ 17 Ë 2

Ê 3 + 1 1 ˆ 23 + (b) Á ˜ 2 ¯ 17 Ë 2

Ê 3 - 1 1 ˆ 23 (c) Á ˜ 2 ¯ 17 Ë 2

Ê 3 + 1 1 ˆ 23 (d) Á ˜ 2 ¯ 17 Ë 2

Ans. (a). Solution:

cos q =

8 fi sin q = 17

(17) 2 - 82 15 = 17 17

Now the given expression is equal to cos 30° cos q – sin 30° sin q + cos 45° cos q + sin 45° sin q + cos 120° cos q + sin 120° sin q = cos q (cos 30° + cos 45° + cos 120°) – sin q (sin 30° – sin 45° – sin 120°) 8 Ê 3 1 1 ˆ 15 Ê 1 1 3ˆ + - ˜= Á Á 17 Ë 2 2 ˜¯ 2 2 ¯ 17 Ë 2 2 Ê 3 - 1 1 ˆ 23 + = Á . ˜ 2 ¯ 17 Ë 2

Let -

2sec q ± 4sec2 q - 4 2 = sec q ± tan q p p As - < q < and a1 > b1 6 12 we get a1 = sec q – tan q

Solution:

cos12 x + 3 cos10 x + 3 cos8 x + cos6 x – 1 is equal to (a) 0 (b) 1 (c) –1 (d) 2. Ans. (a). Solution: From sin x + sin2 x = 1, we get sin x = cos2 x. Now, the given expression is equal to cos6 x (cos6 x + 3cos4 x + 3 cos2 x + 1) – 1 = cos6 x (cos2 x + 1)3 – 1 = sin3 x (sin x + 1)3 – 1 = (sin2 x + sin x)3 – 1 = 1 – 1 = 0. Example 12

p p < q < – - . Suppose a1 and 6 12 b1 are the roots of the equation x2 – 2x secq + 1 = 0 and a2, b2 are the roots of the equation x2 + 2x tanq – 1 = 0. If a1>b1 and a2>b2, then a1 + b2 equals (a) 2 (sec q – tan q) (b) 2 sec q (c) – 2 tan q (d) 0 Ans. (c) Example 13

a 1, b 1 =

-2 tan q ± 4 tan 2 q + 4 2 = – tan q ± sec q p p - b2 6 12 We get b2 = – tan q – sec q So a1 + b2 = sec q – tan q – tan q – sec q = – 2 tan q Next, a2, b2 =

Example 14

The value of the determinant

a a2 1 cos (n - 1) x cos nx cos (n + 1) x sin (n - 1) x sin nx sin (n + 1) x is zero if (a) sin x = 0 (c) a = 0

(a π 1)

(b) cos x = 0 (d) cos x =

1 + a2 2a

Ans. (a). Solution: Applying C1 Æ C1 + C3 – 2cos x C2, the given determinant is equal to a a2 1 + a 2 - 2a cos x 0 cos nx cos (n + 1) x 0 sin nx sin (n + 1) x = (1 + a2 – 2a cos x) [cos n x sin(n + 1)x – sin nx cos (n + 1) x] = (1 + a2 – 2a cos x) sin (n + 1 – n) x = (1 + a2 – 2a cos x) sin x which is zero if sin x = 0 or cos x = (1 + a2)/2a. As a π 1, (1 + a 2 ) 2a > 1 Therefore, cos x = (1 + a2)/2a is not possible.

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Example 15

Ê 3p ˆ Êpˆ The value of cos4 Á ˜ + cos4 Á ˜ + Ë 8¯ Ë 8¯

Ê 5p ˆ Ê 7p ˆ cos4 Á ˜ + cos4 Á ˜ is equal to Ë 8¯ Ë 8 ¯ (a) 1/2 (b) 3/4 (c) 3/2 (d) 1 Ans. (c) Solution: We can write the given expression as 3p ˆ pˆ Ê 3p ˆ Ê Êpˆ Ê cos4 Á ˜ + cos4 Á ˜ + cos4 Á p - ˜ + cos4 Á p - ˜ Ë 8¯ Ë 8¯ Ë ¯ Ë 8 8¯ È 4Êpˆ 4 Ê 3p ˆ ˘ = 2 Ícos Á ˜ + cos Á ˜ ˙ Ë ¯ Ë 8 ¯˚ 8 Î

An angle a is divided into two parts so

that the ratio of the tangents of these parts is l. If the difference between these parts is x then sinx/sina is equal to (a) l/(l + 1) (b) (l – 1)/l l -1 2l (c) (d) l +1 l +1 Ans. (c) tan q1 tan q1 - tan q 2 l - 1 = =lfi tan q 2 tan q1 + tan q 2 l + 1 sin(q1 - q 2 ) l -1 = sin(q1 + q 2 ) l +1

x2 + x

(a) 2 tan a (c) 1 Ans. (a).

Given q ∈ (0, p/4) and t1 = (tanq)tanq,

q ∈ (0, p/4)

t1 = (tanq)tanq fi logt1 = tanq log (tanq) 1 fi log t1 = tanq log = tanq [log1 – log(cotq)] cotq = – tanq log(cotq) Similarly log t2 = – cotq log (cotq) log t3 = tanq log(cotq), log t4 = cotq log (cotq) As cotq > tanq, we have log t4 > log t3 > log t1 > log t2 fi t 4 > t 3 > t 1 > t 2. Ê pˆ Alternately q Œ Á 0, ˜ fi 0 < tan q < 1 fi cot q > 1 Ë 4¯ Take tan q = 1 fi cot q = 2, and we get t4 > t3 > t1 > t2 2 Example 19

If x = sin a, y = sin b, z = sin (a + b) then

cos(a + b) = (a)

x2 + y 2 + z 2 2 xy

(b)

x2 + y 2 - z 2 xy

(c)

z 2 - x2 - y 2 2 xy

(d)

z 2 - x2 - y 2 xy

Solution:

z2 – x2 – y2 = sin2(a + b) – sin2a – sin2b = sin (a + b + a) sin(a + b – a) – sin2b = sinb [sin(2a + b) – sinb]

If a ∈ (0, p/2), then the expression

tan 2 x

a

Ans. (d)

sin x l -1 fi = sin a l +1

x2 + x +

x2 + x

t2 = (tanq)cotq, t3 = (cotq)tanq and t4 = (cotq)cotq then (a) t1 > t2 > t3 > t4 (b) t4 > t3 > t1 > t2 (c) t3 > t1 > t2 > t4 (d) t2 > t3 > t1 > t4 Ans. (b)

Let q1 + q2 = a and q1 – q2 = x

Example 17

tan 2 x

Now

2 ÈÊ p p ˆ p p ˘ = 2 ÍÁ cos 2 ÁÊ ˜ˆ + sin 2 ÁÊ ˜ˆ ˜ - 2 cos 2 ÁÊ ˜ˆ sin 2 ÁÊ ˜ˆ ˙ Ë 8¯ Ë 8 ¯¯ Ë 8¯ Ë 8 ¯ ˙˚ ÎÍË 3 È 1 2 Êp ˆ˘ È 1 1˘ = 2 Í1 - sin Á ˜ ˙ = 2 Í1 - ¥ ˙ = Ë ¯ 2 2 2 2 Î ˚ 4 Î ˚



Example 18

x2 + x ¥

0 < tanq < 1 and cotq > 1 fi log cotq > 0

È 4Êpˆ 4p˘ = 2 Ícos Á ˜ + sin ˙ Ë 8¯ 8˚ Î

Solution:

Since A.M ≥ G.M, we get

tan 2 a ˘ 1È 2 Í x +x+ ˙≥ 2 ÍÎ x 2 + x ˙˚

Solution:

È 4Êpˆ 3p ˆ ˘ 4Êp = 2 Ícos Á ˜ + sin Á - ˜ ˙ Ë ¯ Ë 8 2 8 ¯˚ Î

Example 16

Solution:

is always greater than or equal to (b) 2 (d) sec2a

(2a + b + b ) (2a + b - b ) ˘ È sin = sinb Í2 cos ˙˚ Î 2 2 = 2sina sinb cos(a + b) fi cos(a + b) =

z 2 - x2 - y 2 . 2 xy

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Example 20 The radius of the circle 2x2 + 2y2 – 4xcosq + 4ysinq – 1 – 4cosq – cos2q = 0 is (a) 1 – cosq (b) 1 + cosq (c) 1 – sinq (d) none of these Ans. (b). Solution:

Equation of the circle can be written as

x2 + y2 – 2x cosq + 2y sinq = fi (x – cosq)2 + (y + sinq)2 =

fi r2 =

1 + 4 cos q + cos 2q 2

1 + 4 cos q + cos 2q + sin2q 2 + cos2q = r2

= (1 + cosq)2 fi The radius = r = 1 + cosq. If tan x + tan ( x + p 3) + tan ( x + 2p 3) = 3, then (a) tan x = 1 (c) tan 3x = 1 Ans. (c).

(b) tan 2x = 1 (d) none of these.

Solution: The given equation can be written as tan x + tan (p 3) tan x + tan (2p 3) + =3 tan x + 1 - tan x tan (p 3) 1 - tan x tan (2p 3) fi

tan x +

tan x + 3 1 - 3 tan x

+

tan x - 3 1 + 3 tan x



2 sin2 x – a sin x + 2a – 8 = 0

fi sin x =

a ± a 2 - 8(2a - 8) a ± a 2 - 16a + 64 = 4 4

a ± (a - 8) 4 Hence, sin x = (a – 4)/2 (the other value is not possible x £ 1). This value is possible only when a-4 –1£ £ 1 or – 2 £ a – 4 £ 2 2 =

1 + 4 cosq + cos 2q + 2 3 + 4 cos q + 2 cos 2 q - 1 = 2 2

Example 21

Solution: The given equation can be written as 1 – 2 sin2 x + a sin x = 2a – 7

fi 2 £ a £ 6. Example 23 sin 47° + sin 61° – sin 11° – sin 25° is equal to (a) sin 36° (b) cos 36° (c) sin 7° (d) cos 7°. Ans. (d). Solution: The given expression is equal to (sin 47° + sin 61°) – (sin 11° + sin 25°) = 2 sin 54° cos 7° – = 2 cos 7°(sin 54° – È 5 +1 = 2 cos 7∞ Í Î 4

(tan x + 3) (1 + 3 tan x) + fi fi fi fi fi

tan x + tan x +

1 - 3tan 2 x 8 tan x 1 - 3 tan 2 x 1 - 3 tan 2 x

3(3 tan x - tan 3 x) 1 - 3 tan 2 x

=3

(a) p 4 – a

(b) 3p 4 – a

(c) p 8 - a 2

(d) 3p 8 - a 2 .

Ans. (a). Solution:

=3

tan x (1 - 3tan 2 x) + 8 tan x

If tan a = 1/7 and sin b = 1/ 10

Example 24

where 0 < a , b < p 2 , then 2b is equal to

=3

(1 - 3 tan x) (tan x - 3)

2 sin 18° cos 7° sin 18°) 5 - 1˘ ˙ = cos 7°. 4 ˚

sin b =

1 10

fi cos b =

1-

1 = 10

3 10

fi tan b = 1/3 =3

\

tan 2b =

and = 3 fi 3 tan 3x = 3

tan 3x = 1.

Example 22 The equation cos 2x + a sin x = 2a – 7 possesses a solution if (a) a < 2 (b) 2 £ a £ 6 (c) a > 6 (d) a is any integer. Ans (b).

2 tan b 1 - tan b 2

=

2 ◊1 3 3 = 1-1 9 4

tan (a + 2b) =

tan a + tan 2b 17+3 4 25 = = =1 1 - tan a tan 2b 1 - (1 7) (3 4) 25

Since 0 < b < p 2 and tan 2b = 3 4 > 0, we get 0 < 2b < p 2 . Also, 0 < a < p 4 . Hence, 0 < a + 2b < p and tan (a + 2b) = 1, so that a + 2b = p 4 fi 2b = p 4 – a.

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2 cot a +

Ans. (b).

If 3p/4 < a < p, then

Example 25

Solution: The given expression can be written as

1 sin 2 a

(a) 1 + cot a (c) 1 – cot a Ans. (b). Solution: 2 cot a +

(cos 6 x + cos 4 x) + 5(cos 4 x + cos 2 x) + 10(cos 2 x + 1) cos 5 x + 5cos 3x + 10 cos x

is equal to (b) – 1 – cot a (d) – 1 + cot a.

1 sin 2 a

= 2 cot a + cosec2 a

2 cot a + 1 + cot 2 a =

=

2 cos 5 x cos x + 5 ◊ 2 cos 3x cos x + 10 ◊ 2 cos 2 x cos 5 x + 5cos 3x + 10 cos x

=

2 cos x (cos 5 x + 5 cos 3x + 10 cos x) = 2 cos x . cos 5 x + 5cos 3x + 10 cos x

(1 + cot a ) 2

a

cos (q – p/4) is equal to (a) ±

Since cot a < – 1 when 3p 4 < a < p, we have 1 + cot a

a.

(0 < q – a, q – b < p/2), then cos (a – b) + 2ab sin (a – b) is equal to 2

2

2

2

(a) 4a b (c) a2 + b2 Ans. (c).

= sin [(q – b) – (q – a)] = sin (q – b) cos (q – a) – cos (q – b) sin (q – a) = ba –

1- b

2

1- a

1 - b2 + b 1 - a 2

Substituting these values in the given expression, we get cos2 (a – b) + 2ab sin (a – b)

(

)

2

+ 2ab ÈÍab - (1 - a 2 ) (1 - b 2 ) ˘˙ Î ˚

= a2 (1 – b2) + b2 (1 – a2) + 2ab 2

2

+ 2a b – 2ab Example 27

(1 - a 2 ) (1 - b 2 )

(1 - a ) (1 - b ) = a + b . 2

2

Example 29

2

cos 6 x + 6 cos 4 x + 15 cos 2 x + 10 is equal to cos 5 x + 5 cos 3x + 10 cos x (b) 2 cos x (d) 1 + cos x.

p 1 - p sin q fi cos q + sin q = ± 2 2

If tan q1, tan q2, tan q3 and tan q4 are the

roots of the equation x4 – x3 sin 2b + x2 cos 2b – x cos b – sin b = 0 then tan (q1 + q2 + q3 + q4) is equal to (a) sin b Ans. (d)

(b) cos b

(c) tan b (d) cot b

Solution: S1 = S2 = S3 = and S4 =

From the given equation we get tan q1 + tan q2 + tan q3 + tan q4 = sin 2 b. S tan q1 tan q2 = cos 2 b S tan q1 tan q2 tan q3 = cos b tan q1 tan q2 tan q3 tan q4 = – sin b S1 - S3 Now tan (q1 + q2 + q3 + q4) = . 1 - S2 + S4 =

sin 2b - cos b cos b (2 sin b - 1) = = cot b. 1 - cos 2b - sin b sin b (2sin b - 1)

Example 30

The expression

(a) cos 2x (c) cos2 x

2

(d) ± 2 2

pˆ 1 1 pˆ Ê Ê 2 cos Á q - ˜ = ± fi cos Á q - ˜ = ± . Ë Ë 4¯ 2 4¯ 2 2

2

and cos (a – b) = cos [(q – b) – (q – a)] = cos (q – b) cos (q – a) + sin (q – b) sin (q – a)

2

Ê p ˆ = cot (p sin q) = tan Á ± - p sin q ˜ Ë 2 ¯



sin (a – b) = sin (a – q + q – b)

1

tan (p cos q)

fi p cos q = ±

Solution: We have

= a 1 - b2 + b 1 - a 2

(b) ±

2 2

Ans. (a). Solution

2

(b) a – b (d) –a2 b2.

=a

1

(c) ± 2

If cos (q – a) = a and sin (q – b) = b

Example 26

If tan (p cos q) = cot (p sin q) then

Example 28

The expression cos2 f + cos2 (a + f) –

2 cos a cos f cos (a + f) is independent of (a) f (b) a (c) both a and f (d) none of a and f Ans. (a)

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Solution: The given expression is equal to cos2 f + cos2 (a + f) – [cos (a + f) + cos (a – f)] cos (a + f) = cos2 f – cos (a + f) cos (a – f) = cos2 f – (cos2 f – sin2 a) = sin2 a which is independent of f. If the value of

Example 31

p 3p 5p 7p 9p 11 p 13 p sin sin sin sin sin sin 14 14 14 14 14 14 14 2 is equal to k , then k is equal to (a) –1/8 (b) 1/8 (c) 1/64 (d) 1 Ans. (a) sin

Solution: The given expression can be written as sin

p 3p 5p 5p ˆ Êpˆ Ê sin sin sin Á ˜ sin Á p ˜ Ë ¯ Ë 14 14 14 2 14 ¯ 3p ˆ pˆ Ê Ê sin Á p - ˜ = k2 sin Á p ˜ Ë Ë 14 ¯ 14 ¯

where k = sin

p 3p 5p sin sin 14 14 14

Êp p ˆ Ê p 3p ˆ Ê p 5p ˆ cos Á - ˜ = cos Á - ˜ cos Á Ë 2 14 ¯ Ë 2 14 ˜¯ Ë 2 14 ¯ = cos

p 3p 2p cos cos 7 7 7

1

=

2 sin =

p 7

1 p 4 sin 7

¥ 2 sin

sin

p p 2p 4p cos cos cos 7 7 7 7

4p 4p cos 7 7

Solution: From the given relations, we have sin 2B = (3/2) sin 2A and 3 sin2 A = 1 – 2 sin2 B = cos 2B so that cos (A + 2B) = cos A cos 2B – sin A sin 2B = cos A ◊ 3 sin2 A – (3/2) sin A sin 2A = 3 cos A sin2 A – 3 sin2 A cos A = 0 fi A + 2B = p/2. Example 34

Ê ˆ sin 2 b ˆ Ê sin 2 g ˆ Ê sin b sin g - sin b sin g ˜ fi Á1 - 2 ˜ Á1 - 2 ˜ = Á 2 sin a ¯ Ë sin a ¯ Ë sin a ¯ Ë

\

(d) 1/18

Solution: k = sin p/18 sin 5p/18 sin 7p/18 = sin 10° sin 50° sin 70° 1 = [cos 40° – cos 60°] sin 70° 2

(sin2 a – sin2 b) (sin2 a – sin2 g) = sin2 b sin2 g (1 – sin2 a)2 sin4 a (1 – sin2 b sin2 g) – sin2 a (sin2 b + sin2 g – 2 sin2 b sin2 g) = 0 sin2 a =

1 1 [sin 110° + sin 30°] – sin 70° 4 4

sin 2 b + sin 2 g - 2 sin 2 b sin 2 g 1 - sin 2 b sin 2 g [

and cos2 a = fi tan2 a =

=

1 1 = cos 40° sin 70° – sin 70° 2 4 =

If a, b, g are acute angles and cos q =

sin b/sin a, cos f = sin g /sin a and cos (q – f) = sin b sin g, then tan2 a – tan2 b – tan2 g is equal to (a) – 1 (b) 0 (c) 1 (d) none of these Ans. (b) Solution: From the third relation we get cos q cos f + sin q sin f = sin b sin g fi sin2 q sin2 f = (cos q cos f – sin b sin g )2

If k = sin p/18 sin 5p/18 sin 7p/18, then

the numerical value of k is equal to (a) 1/2 (b) 1/4 (c) 1/8 Ans. (c)

If A and B are acute positive angles

satisfying the equations 3 sin2 A + 2 sin2 B = 1 and 3 sin 2A – 2 sin 2B = 0, then A + 2B is equal to (a) p/4 (b) p/2 (c) 3p/4 (d) 2p/3 Ans. (b)



8p 1 ¥ sin = =– . p 7 8 8 sin 7

1 1 1 1 1 sin (180° – 70°) + ¥ - sin 70° = 4 4 2 4 8

Example 33



1

Example 32

=

1 - sin 2 b - sin 2 g + sin 2 b sin 2 g 1 - sin 2 b sin 2 g sin 2 b - sin 2 b sin 2 g + sin 2 g - sin 2 b sin 2 g cos 2 b - sin 2 g (1 - sin 2 b )

sin 2 b cos 2 g + cos 2 b sin 2 g cos 2 b cos 2 g

2 2 = tan b + tan g



sin a π 0]

tan2 a – tan2 b – tan2 g = 0.

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Example 35

È cos 2 a If A = Í ÎÍcos a sin a

cos a sin a ˘ ˙ sin 2 a ˚˙

È cos b cos b sin b ˘ and B = Í ˙ sin 2 b ˙˚ ÍÎcos b sin b are two matrices such that AB is the null matrix, then (a) a = b (b) cos (a – b) = 0 (c) sin (a – b) = 0 (d) none of these Ans. (b) 2

Solution: AB = 0 È cos 2 a cos a sin a ˘ È cos 2 b Í ˙Í fi sin 2 a ˚˙ ÎÍcos b sin b ÎÍcos a sin a

cos b sin b ˘ ˙ sin 2 b ˚˙

Ècos a cos b cos(a – b ) cos a sin b cos(a – b ) ˘ Í cos b sin a cos(a – b ) sin a sin b cos(a – b ) ˙ Î ˚ È0 0 ˘ = Í ˙ Î0 0 ˚



cos (a – b) = 0

Example 36

cos q sin q a b + = , then sec 2 q cos ec 2 q a b

is equal to (a) a (b) b (c) a/b (d) a + b Ans. (a) Solution: Let (cos q)/a = (sin q)/b = k, so that cos q = ak and sin q = bk. Then a cos 2q + b sin 2q = a(1 – 2 sin2 q) + 2b sin q cos q = a – 2ab2 k2 + 2b ◊ bk ◊ ak = a – 2ab2 k2 + 2ab2 k2 = a. If sin a + sin b + sin g = 0 and cos a +

Example 38

cos b + cos g = 0, then value of cos (a - b) + cos (b - g) + cos (g - a) is (a) –3/2 (c) –1/2

È0 0 ˘ = Í ˙ Î0 0 ˚ fi

If

Example 37

n sin a cos a

If tan b =

1 - n sin 2 a

is equal to (a) n tan a (c) (1 + n) tan a Ans. (b) Solution: tan b = =

, then tan (a – b)

n tan a n tan a

tan a + (1 - n) tan a - n tan a 1 + (1 - n) tan 2 a + n tan 2 a (1 - n) tan a (1 + tan a ) 2

1 + tan 2 a

= (1 – n) tan a .

13

1 is equal to Ê p ( k - 1) p ˆ k =1 Ê p kp ˆ sin Á + sin + ÁË ˜ 6 ˜¯ 4 6¯ Ë4

(c) 2

(

( ) 2 (2 + 3 )

(b) 2 3 - 3

3

)

3 -1

(d)

Ans. (c)

1 + (1 - n) tan 2 a

3

=

The value of

Â

2

n tan a tan a 1 + (1 - n) tan 2 a fi tan (a – b) = tan a ◊ n tan a 1+ 1 + (1 - n) tan 2 a =

= sin2 g + cos2 g fi 1 + 1 + 2 cos (a - b) = 1 fi cos (a - b) = -1/2. Similarly, cos (b - g) = -1/2 and cos(g - a ) = -1/2. Thus value of given expression is –3/2.

(a) 3 –

1 + tan a - n tan a 2

Ans.(a) Solution: We have (sin a + sin b)2 + (cos a + cos b)2

Example 39

(b) (1 – n) tan a (d) none of these

(b) –1 (d) 0

Let qk =

Solution:

p kp + , 0 £ k £ 13 4 6

Note that qk – qk – 1 = 13

Let S = Â

p 6

1

k =1 sin q k -1

sin q k

p 6 = 2Â k =1 sin q k -1 sin q k 13

13

= 2Â

sin

sin (q k - q k -1 )

k =1 sin q k -1

sin q k

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= 2 Â ( cot q k -1 - cot q k )

f ¢(q) =

k =1

= 2 (cot q0 – cot q13)

- sin (q + a ) sin (q + a ) 1 - sin (q + b ) sin (q + b ) 1 - sin (q + g ) sin (q + g ) 1 cos (q + a ) cos (q + a ) 1 + cos (q + b ) cos (q + b ) 1 cos (q + g ) cos (q + g ) 1

È Ê p 13p ˆ ˘ = 2 Í1 - cot Á + ˜ Ë4 6 ¯ ˙˚ Î

È p p ˆ˘ Ê = 2 Í1 - cot Á 2p + + ˜ ˙ Ë 4 6¯˚ Î

=0+0=0 where f(q) denotes the given determinant fi f(q) is independent of q. Alternately expanding the determinant along last column we get sin (q + g – q – b) – sin (q + g – q – a) + sin (q + b – q – a) = sin (g – b) + sin (b – a) + sin (a – g) which is independent of q.

p p È ˘ Í cot 4 cos 6 - 1 ˙ = 2 Í1 p p ˙ Í cot + cot ˙ 4 6 ˚ Î È 3 - 1˘ = 2 Í1 ˙ 3 + 1˚ Î =

4 3 +1

Example 40

=2

(

Example 42

)

3 -1

If k1= tan 27 q – tan q sin q sin 3 q sin 9 q + + cos 3 q cos 9 q cos 27 q

k2 =

and then,

(a) k1 = k2 (c) k1 + k2 = 2 Ans. (b)

(b) k1 = 2k2 (d) k2 = 2k1

= 1/2 and cos f = 1/3, then q + f lies in (a) ] p/3, p/2 [ (b) ] p/2, 2p/3 [ (c) ] 2p/3, 5p/3 [ (d) ] 5p/6, p [ Ans. (b) Solution: sin q = 1/2 fi q = p/6 and cos f = 1/3 fi p/3 < f < p/2 so that

p 2p < (q + f) < . 2 3

Example 43

Solution: We can write k1= tan 27q – tan 9q + tan 9q – tan 3q + tan 3q – tan q But sin 3q cos q - cos 3q sin q tan 3q – tan q = cos 3q cos q

È sin 9q sin 3q sin q ˘ + + k1 = 2 Í ˙ = 2k2. Î cos 27q cos 9q cos 3q ˚

Example 41

cos (q + a ) sin (q + a ) 1 cos (q + b ) sin (q + b ) 1 cos (q + g ) sin (q + g ) 1

is independent of (a) a Ans. (d) Solution: we get

(b) b

(c) g

(d) q

Differentiating the given determinant w.r.t. q

The value of tan 3a cot a cannot lie in

(a) ] 0, 2/3 [ (c) ] 4/3, 4 [ Ans. (b) Solution:

(b) ] 1/3, 3 [ (d) ] 2, 10/3 [

tan 3a cot a =

sin 2q 2 sin q = = cos 3q cos q cos 3q \

If q and f are acute angles such that sin q

= fi

tan2 a =

3 tan a - tan 3 a tan a (1 - 3 tan 2 a ) 3 - tan 2 a 1 - 3 tan 2 a

= x (say)

x-3 (3 x - 1) ( x - 3) = 3x -1 (3 x - 1)2

Since tan2 a is non-negative, either x < 1/3 or x ≥ 3, so x cannot lie between 1/3 and 3. Example 44

For a given pair of values x and y satisfying

x = sin a, y = sin b, there are four different values of z = sin (a + b) whose product is equal to (b) x2 + y2 (a) x2 – y2 (c) (x2 + y2)2 (d) (x2 – y2)2 Ans. (d)

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Solution:

z = sin (a + b) = sin a cos b + cos a sin b

(

) (

2 2 = ± x 1- y ± y 1- x

(a) 2 (a – b)

)

(c)

There are four values of z, given by ± ÈÍ x 1 - y 2 + y 1 - x 2 ˘˙ and ± ÈÍ x 1 - y 2 - y 1 - x 2 ˘˙ Î ˚ Î ˚ and their product is equal to [x2 (1 – y2) – y2(1 – x2)]2 = (x2 – y2)2. Example 45

2 2 b a

2 cos 2 x ( 1 + sin 2 x)

È ˘ 1 1 =2 Í ˙ Î1 + sin 2 x 1 + cos 2 x ˚

(b) 0 (d) sin A + cos A

È1 1˘ =2 Í - ˙ a b Î ˚

sin 2 A cos 2 A + cos A (sin A - cos A) sin A(cos A - sin A) – sec A cosec A sin 3 A - cos3 A – sec A cosec A sin A cos A(sin A - cos A)

sin 2 A + cos 2 A + sin A cos A = – sec A cosec A sin A cos A = sec A cosec A + 1 – sec A cosec A = 1. If a2 – 2a cos x + 1 = 674 and tan (x/2) = 7

then the integral value of a is (a) 25 (b) 49 Ans. (a) Solution

value of

1 + cos 290∞ (a)

1 3 sin 250∞

3 4

is equal to

(b) 4

3

(c) 2

3 (d)

3 2

Ans. (b) Solution: The given expression is equal to 1 1 + = cos (270∞ + 20∞) 3 sin (270∞ - 20∞) =

1 1 + sin 20∞ 3 (- cos 20∞)

= a2 – 2a ¥

1 - 49 +1 1 + 49

= a2 + 2a ¥

48 +1 50

=

(c) 67

1 - tan 2 ( x 2) 1 + tan 2 ( x 2)

(d) 74

+1

25a2 + 48a – 673 ¥ 25 = 0 (a – 25) (25a + 673) = 0 a = 25 (Taking the integral value of a).

Example 47

The value of the expression

Example 48

3 1 cos 20∞ - sin 20∞ 2 = 2 = 3 sin 20∞ cos 20∞ 3 sin 40∞ 4 sin 60∞ cos 20∞ - cos 60∞ sin 20∞ = 3 sin 40∞ 4

674 = a2 – 2a

fi fi fi

2 (cos 2 x - sin 2 x) (1 + cos 2 x) (1 + sin 2 x)

=

Solution: The given expression is equal to

Example 46

(d) 2 (a + b)

Solution: The given expression can be written as 2 (cos 2 x - sin 2 x)

tan A cot A + – sec A cosec A is equal to 1 - cot A 1 - tan A

=

2 2 a b

Ans. (b)

If 0 < A < p/2 the value of the expression

(a) – 1 (c) 1 Ans. (c)

(b)

If sin 2x = a – 1 and cos 2x = b – 1, then the

sec2 x [(cos 2 x - sin 2 x) - 2sin x cos x] is equal to 1 + sin 2 x

3 cos 20∞ - sin 20∞

sin 40∞ 3 sin 40∞ 4

Example 49

=

4 3

.

Let f(q) = sin q (sin q + sin 3q), then f(q)

(a) ≥ 0 only when q ≥ 0

(b) £ 0 for all real q

(c) ≥ 0 for all real q (d) £ 0 only when q £ 0 Ans. (c) Solution: f(q) = (1/2) (1 – cos 2q) + (1/2) (cos 2q – cos 4q)

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= (1/2) (1 – cos 4q) = sin2 2q ≥ 0 for all real q.



Example 50 The maximum value of (cos a1) (cos a2) (cos an) under the restriction 0 £ a1, a2, , an £ p/2 and (cot a1) (cot a2) (cot an) = 1 is (a) Ans. (a) Solution:

1 2

n2

1

(b)

2

(c)

n

1 2n

(d) 1

n

i =1

i =1

i =1

1 2

n

as max. value of sin 2ai is 1

n

’ (cos ai) £

i =1

1 2

n2

.

1 (1 + xn ) , then 2

(c) x0

(b) 1

Let x 0 = cos q, then x 1 =

Solution:

Example 51

1 2

n2

.

q q q cos 2 º cos n º infinite 2 2 2 q q 2 sin cos 2 2 = q q q cos cos 2 º cos n º infinite 2 2 2 q q 22 sin 2 cos 2 2 2 = q q q cos 2 cos 3 º cos n º infinite 2 2 2 q 2n sin n 2 = lim q nÆ• cos n + 1 2 cos

equals (b) tan b + tan g (d) 2 tan b + tan g

Solution: a + b = p/2 fi a = p/2 – b fi tan a = cot b fi tan a tan b = 1 Next, b + g= a a – b= g fi

tan a - tan b = tan g 1 + tan a tan b

fi tan a – tan b = 2 tan g fi tan a = tan b + 2 tan g. Example 52

The number of integral values of k for

q ˆ Ê sin n Á 1 2 ˜ = lim q Á =q q ˜ q nÆ• Á ˜ cos Ë 2n ¯ 2n + 1

which the equation 7 cos x + 5 sin x = 2k + 1 has a solution is (a) 4 (b) 8 (c) 10 (d) 12 Ans. (b) Solution: The given equation can be written as r cos (x – a) = 2k + 1 where r cos a = 7, r sin a = 5 fi cos (x – a) =

2 k +1 74

so that as r2 = 72 + 52 = 74

1 (1 + cos q ) 2 and so on.

sin q

=

If a + b = p/2 and b + g = a, then tan a

(a) 2 (tan b + tan g) (c) tan b + 2 tan g Ans. (c)

(d) 1/x0

È ˘ Í ˙ Í x1 x2 x3 º to infinite ˙ Î ˚

so that

So the maximum value of the given expression is



If xn + 1 =

= cos q/2, x2 = cos (q/22), x3 = cos (q/23),

for all i. fi

74

– 8 £ 2k + 1 £ 8 (For integral values of k) – 4 £k £ 3 k = – 4, – 3, – 2, – 1, 0, 1, 2, 3

(a) – 1 Ans. (c)

0 £ ai £ p/2 fi 0 £ 2ai £ p

n

£1

È ˘ 1 - x02 Í ˙ (– 1 < x0 < 1) is equal to cos Í x1 x2 x3 º to infinite ˙ Î ˚

n n n Ê sin 2 a ˆ i fi ’ (cos2 ai) = ’ (cos ai sin ai) = ’ Á 2 ˜¯ i =1 Ë i =1 i =1

\ ’ (cos2 ai) £

fi fi fi

74

74 £ 2k + 1 £



Example 53

’ (cos ai) = ’ (sin ai)

Since



2 k +1

which gives 8 integral values of k.

From the given relations we have

n

–1£

È 1 - x2 ˘ 0 ˙ = cos q = x0. cos Í Í x1 x2 º inf . ˙ Î ˚

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The simplest value of

Example 54 2

2 + 2 + 2 + 2 cos 4x (a) sec(x/2) (c) cosecx Ans. (a) Solution:

(b) secx (d) 1

=

2 + 2 + 2 ¥ 2 cos 2 2x 2

=

2 + 2 cos x

2 2 + 2 + 2 cos 2x

2

= sec (x/2)

4 cos 2 x / 2

If a, b are positive acute angles and cos2a

Example 55 =

Example 57

Given expression is equal to 2

=



is

(c) k = 1 Ans. (b)

(b) k =

2

(d) k =

3

m 3 - 4 sin 2 q = n 1 - 4 sin 2 q

(b) cos 2q =

(c)

m-n = 2 sec 2q m+n

(d) sin 2q =

Ans. (a), (b)

=

=

1 - cos 2a (3 - cos 2b ) - (3cos 2b - 1) = (3 - cos 2b ) + (3cos 2b - 1) 1 + cos 2a

Solution:

3(1 - cos 2b ) + (1 - cos 2b ) fi tan a = 3(1 + cos 2b ) - (1 + cos 2b )

If m tan (q – 30°) = n tan (q + 120°), then

(a)

Solution:

3cos 2b - 1 , then tana = k tanb such that 3 - cos 2b (a) k = - 2

(tan2a + cot2a) (tan3a + cot3a) = tan5a + cot5a + cot a + tan a tan5 a + cot5 a = (p2 – 2) (p3 – 3p) – p = p5 – 5p3 + 5 (tan a – cot a)2= (tan a + cot a)2 – 4 tan a cot a = p2 – 4

=

m tan (q + 120∞) - cot (30∞ + q ) = = n tan (q - 30∞) tan (q - 30∞)

cos (30∞ + q ) cos (30∞ - q ) sin (30∞ + q ) sin (30∞ - q ) cos 2 30∞ - sin 2 q sin 2 30∞ - sin 2 q

=

3/4 - sin 2 q 1/4 - sin 2 q

3 - 4 sin 2 q 1 - 4 sin 2 q

2

= fi tan2a =

4(1 - cos 2b ) = 2tan2b 2(1 + cos 2b ) (Rejecting the minus sign as a, b are positive acute angle)

2 tanb

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS If tan a + cot a = p, then

Example 56

(a) tan a + cot a = (b) tan3a + cot3a = (c) tan5a + cot5a = (d) (tan a – cot a)2 Ans. (a), (b), (c), (d) 2

2

2

p –2 p(p2 – 3) p5 – 5p3 + 5 = p2 – 4

Solution: tan2a + cot2a = (tan a + cot a)2 – 2 tan a cot a 2

=p –2 tan a + cot a = (tan a + cos a)3 – 3 tan a◊ cot a (tan a + cot a) 3 = p – 3p 3

3



m + n 4 - 8 sin 2 q = = 2 cos 2q m-n 2



cos 2q =



m-n 1 = sec 2q m+n 2

Example 58

m+n 2 ( m - n)

If

tan 3 A = k, then tan A

(a) tan2A =

k -3 3k - 1

(b) cot2A =

k -3 3k - 1

(c)

sin 3 A 2k = sin A k - 1

(d)

cos 3 A 2 = cos A k - 1

Ans. (a), (c), (d)

m+n 2 ( m - n) m-n 2 ( m + n)

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Solution: fi

tan 3 A 3tan A - tan 3 A 1 = ¥ =k 2 tan A tan A 1 - 3tan A

3 - tan 2 A 1 - 3tan 2 A

a +b = fi tan 2

=k

Êa + bˆ 676 1 - tan 2 Á Ë 2 ˜¯ 1 - 49 -627 = = fi cos (a + b) = 725 Ê a + b ˆ 1 + 676 1 + tan 2 Á Ë 2 ˜¯ 49

k -3 fi tan2A = 3k - 1 sin 3 A Next = 3 – 4 sin2A sin A =3–

a +b 2 sin (a + b) = 2 Êa + bˆ 1 - tan Á Ë 2 ˜¯ 2 tan

4 1 + cot 2 A

4 3k - 1 1+ k -3 3k - 3 - k + 3 2k = = k -1 k -1 =3–

=

4 1 + tan 2 A

a +b 2 tan (a + b) = + bˆ a Ê 1 - tan 2 Á Ë 2 ˜¯

–3

=

4 –3 = k -3 1+ 3k - 1 =

4 (3k - 1) –3 4 ( k - 1)

3k - 1 - 3 ( k - 1) k -1 2 = k -1

=

a b , tan are the roots of the 2 2 2 equation 8x – 26x + 15 = 0, then a + b -26 = (a) tan 2 7 Example 59

2 ( -26) -364 = 725 Ê 676 ˆ 7 Á1 + ˜¯ Ë 49 2 tan

cos 3 A and = 4 cos2A – 3 cos A =

13 4 = -26 15 7 18

If tan

-627 (b) cos (a + b) = 725 -364 (c) sin (a + b) = 725 364 (d) tan (a + b) = 627 Ans. (a), (b), (c), (d) a b 13 a b 15 Solution: tan + tan = , tan tan = 2 2 4 2 2 8

2 ( -26) 364 = Ê 676 ˆ 627 7 Á1 ˜ Ë 49 ¯

(m + 2) sin q + (2m – 1)cos q = 2m + 1, if

Example 60 (a) tan q (b) tan q (c) tan q (d) tan q Ans. (b) and (c)

= = = =

3/4 4/3 2m/(m2 –1) 2m/(m2 + 1)

Solution: The given relation can be written as (m + 2) tan q + (2m – 1) = (2m + 1) sec q fi

(m + 2)2 tan2 q + 2(m + 2) (2m – 1) tan q + (2m – 1)2 = (2m + 1)2 (1 + tan2 q)



[(m + 2)2 – (2m + 1)2] tan2 q + 2(m + 2) (2m – 1) tan q + (2m – 1)2 – (2m + 1)2 = 0

fi 3 (1 – m2) tan2 q + (4m2 + 6m – 4) tan q – 8m = 0 fi

(3 tan q – 4) [(1 – m2) tan q + 2m] = 0

which is true if tan q = 4/3 or tan q = 2m/(m2 – 1). If x = Example 61 y = cosec f + cot f, then

sec

f



tan

f

and

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(a) x =

y +1 y -1

(b) x =

1+ x (c) y = 1- x

y -1 y +1

1 - sin f 1 + cos f ,y= cos f sin f

xy =

(1 - sin f ) (1 + cos f ) cos f sin f

1 - sin f + cos f = cos f sin f (1 - sin f ) sin f - cos f (1 + cos f ) cos f sin f

sin f - sin 2 f - cos f - cos 2 f = cos f sin f sin f - cos f - 1 = - ( xy + 1) cos f sin f

Thus, xy + x – y + 1 = 0. fi

x=

Example 62

sin 2 q

1 + cos 2 q

4 sin 4q

sin q

cos q

1 + 4sin 4q

2

(a) 7p 24 (b) 5p 24 Ans. (a) and (c).

=0

y -1 1+ x and y = . y +1 1- x If tan x = 2b/(a – c) (a π c),

y = a cos2 x + 2b sin x cos x + c sin2 x and z = a sin2 x – 2b sin x cos x + c cos2 x, then (a) y = z (b) y + z = a + c (c) y – z = a – c (d) y – z = (a – c)2 + 4b2. Ans. (b) and (c). Solution: Adding the expression for y and z, we get y + z = a(cos2 x + sin2 x) + c(sin2 x + cos2 x) =a+c and subtracting them y – z = a (cos2 x – sin2 x) + 4b sin x cos x – c(cos2 x – sin2 x) = a cos 2x + 2b sin 2x – c cos 2x = (a – c) cos 2 x + 2b sin 2x = (a – c) [cos 2x + tan x sin 2x] = (a – c) [cos2x – sin2x + 2 sin2x] = (a – c) As a π c, we get y π z. Example 63 The values of q lying between 0 and p/2 and satisfying the equation

(c) 11p 24 (d) p 24 .

Solution: Applying R1 Æ R1 – R3 and R2 Æ R2 – R3 on the LHS, the given equation can be written as 1 0

1 - sin f + cos f - sin f cos f + sin f cos f fi xy + 1 = cos f sin f

=

4 sin 4q

are

Solution: We have x =

and x – y =

cos 2 q

2

(d) x y + x – y + 1 = 0.

Ans. (b), (c) and (d)

Multiplying, we get

1 + sin 2 q

-1 -1

0 1

=0

cos q 1 + 4sin 4q

sin q 2

2

Expanding the LHS along R1, we get 1 + 4 sin 4q + cos2 q + sin2 q = 0 \

4 sin 4q = – 2 fi sin 4q = – 1 2

\

4q = 7p 6 or 11p 6 [

\

0 0, we get fi

1+ b

and sin f =

2

Therefore, we have from the given relations

[where y = tan2 x/2]



1- b2

Similarly cos f =

= =

1- a2 1+ a2 2a 1+ a2

and

z = Â cos2n f sin2n f n=0

=

1 1 - cos f sin f 2

2

=

1 xy = 1 - 1 xy xy - 1

fi xyz – z = xy fi xyz = xy + z Also x y =

1 sin 2 f cos 2 f

=

sin 2 f + cos 2 f sin 2 f cos 2 f

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=

1 cos f 2

+

1 sin 2 f

(because a, b ∈ [0, p 2 ]), and from sin (a – b) = 1 2 , we

So that we can write xyz = x + y + z. Example 69

get a – b = p 6 . Therefore, a = p 3 and b = p 6 , so that a + 2b = 2p 3 and 2a + b = 5p 6 .

If

a 2 cos 2 a + b 2 sin 2 a + a 2 sin 2 a + b 2 cos 2 a

x =

From sin (a + b) = 1, we get a + b = p 2

Solution:

= x + y.

tan (a + 2b) = tan 2p 3 = tan (p - p 3)



= – tan p 3 = –

2 2 2 then x2 = a2 + b2 + 2 P(a + b ) - P where P is equal to

= – tan p 6 = – 1 Example 71

a 2 cos 2 a + b 2 sin 2 a + a 2 sin 2 a + b 2 cos 2 a

x=

(a 2 cos 2 a + b 2 sin 2 a )

fi x2 = a2 + b2 + 2

(a 2 sin 2 a + b 2 cos 2 a )

a +b a –b Ê a +b ˆ 3 cos - Á 2 cos 2 - 1˜ = Ë ¯ 2 2 2 2

[(a 2 + b 2 ) - (a 2 sin 2 a + b 2 cos 2 a )]

where P = a2 sin2 a + b2 cos2 a. or P = (a2/2) (1 – cos 2a) + (b2/2) (1 + cos 2a) = (1/2) [a2 + b2 – (a2 – b2) cos 2a] Also x2 = a2 + b2 + 2

(a cos a + b sin a ) [(a + b ) 2

2

2

2

2

- (a cos a + b sin a )] 2

2

2

where P = a2 cos2 a + b2 sin2 a or P = (1/2) [a2 + b2 + (a2 – b2) cos 2a] If sin (a + b) = 1 and sin (a – b) = 1/2

where a, b ∈ [0, p/2] then 3

(a) tan (a + 2b) = – (b) tan (2a + b) = – 1 (c) tan (a + 2b) = (d) tan (2a + b) = 1 Ans. (a) and (b)

3

3 3

4 cos2



a +b a - b˘ È 2 a -b =0 Í2 cos 2 - cos 2 ˙ + sin 2 Î ˚



sin

a –b a+b a-b = 0 and 2cos = cos 2 2 2

(1)

Also, since 0 < a, b < p, we have a = b. Therefore, from (1) we get cos a = 1 2 , so that a = b = p 3 .

2

2 2 2 = a2 + b2 + 2 ( a + b ) P - P

Example 70

a+b a –b a+b - 4 cos cos +1= 0 2 2 2



2

2 2 2 \ x = a2 + b2 + 2 ( a + b ) P - P

2

(b) b = p/3 (d) a + b = p/3

Solution: The given equation can be written as 2 cos

¥ (a 2 sin 2 a + b 2 cos 2 a )

3.

If 0 < a, b < p and cos a + cos b –

cos (a + b) = 3/2 then (a) a = p/3 (c) a = b Ans. (a), (b) and (c)

= a2 + b2 + k, where k=2

5p = tan (p - p 6) 6

tan (2a + b) = tan

and

(a) a2 cos2 a + b2 sin2 a (b) a2 sin2 a + b2 cos2 a. (c) (1/2) [a2 + b2 + (a2 – b2) cos 2a] (d) (1/2) [a2 + b2 – (a2 – b2) cos 2a] Ans. (a), (b), (c) and (d) Solution:

3

Example 72

If A and B are acute angles such that

2

sin A = sin B, 2 cos2 A = 3 cos2 B; then (a) A = p/6 (b) A = p/2 (c) B = p/4 (d) B = p/3 Ans. (a) and (c) Solution: From the given conditions 2(1 – sin2 A) = 3(1 – sin2 B) = 3 (1 – sin A) fi fi

2 sin2 A – 3 sin A + 1 = 0 (2 sin A – 1) (sin A – 1) = 0



sin A = 1 or sin A = 1 2



A = p 2 or p 6 But since A is acute, we have A = p 6 .



sin2 B = sin ( p 6 ) = 1 2

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sin B = 1

Example 73

2

fi B= p 4

If A and B are acute angles such that

A + B and A – B satisfy the equation tan q – 4 tan q + 1 = 0, then (a) A = p/4 (b) A = p/6 (c) B = p/4 (d) B = p/6 Ans. (a) and (d) 2

(1) (2)

and from (1) we get 1 + tan B 1 - tan B (1 + tan B) 2 + (1 - tan B) 2 + =4 fi =4 1 - tan B 1 + tan B 1 - tan 2 B

If cos x – sin a cot b sin x = cos a, then

the value of tan (x/2) is. (a) – tan (a/2) cot (b/2) (b) tan (a/2) tan (b/2) (c) – cot (a/2) tan (b/2) (d) cot (a/2) cot (b/2) Ans. (a) and (b)

1 + tan ( x /2) 2



fi fi

– sin a cot b

fi tan2

2

(b)

2

(c) 1/2

(d) – 1/2

Solution: Since cos (q – a), cos q and cos (q + a) are in H.P., we have

fi cos2 q =

sin 2 a 4 sin 2 (a /2) cos 2 (a /2) = 1 - cos a 2 sin 2 (a /2)

If cos (b – g) + cos (g – a) + cos (a – b)

= – 3/2 then (a) S cos a = 0 (b) S sin a = 0 (c) S cos a sin a = 0 (d) S (cos a + sin a) = 0 Ans. (a), (b) and (d) 2 [cos b cos g + cos g cos a + cos a cos b] + 2 [sin b sin g + sin g sin a + sin a sin b] + (sin2 a + cos2 a) + (sin2 b + cos2 b) + (sin2 g + cos2 g) = 0

2 tan ( x /2) 1 + tan 2 ( x /2)

= cos a

x x tan2 (1 + cos a) + sin a cot b ◊ 2 tan 2 2 – (1 – cos a) = 0 x 2 sin a cot b x 1 cos a tan =0 tan2 + 2 1 + cos a 2 1 + cos a tan2

(a) –

Solution: The given expression can be written as

Solution: The given equation can be written as 1 - tan ( x /2)

Example 75 If cos (q – a), cos q and cos (q + a) are in harmonic progression, then cos q sec (a/2) is equal to

Example 76

cos 2B = 1 2 fi 2B = p 3 fi B = p 6 .

2

tan (x/2) = – tan (a/2) cot (b/2) tan (x/2) = tan (a/2) tan (b/2).

fi cos2 q = 2 cos2 ( a 2 ) fi cos q sec (μ/2) = ± 2 .

1 - tan 2 B

1 =4 fi = 2 2 1 - tan B 1 + tan B 2

Example 74

fi or

fi cos2 q cos a = cos2 q – sin2 a fi (1 – cos a) cos2 q = sin2 a

fi 2A = p 2 fi A = p 4



x b aˆ Ê ÁË tan - tan tan ˜¯ = 0 2 2 2

2 cos (q - a ) cos (q + a ) 2(cos 2 q - sin 2 a ) = cos q = cos (q - a ) + cos (q + a ) 2 cos q cos a

tan [A + B + A – B] = tan p 2



x b aˆ Ê ÁË tan + cot tan ˜¯ 2 2 2

Ans. (a) and (b)

Solution: From the given equation, we have tan (A + B) + tan (A – B) = 4 tan (A + B) tan (A – B) = 1 From (1) and (2) we get

2(1 + tan 2 B)



x a x a + 2 tan cot b tan - tan 2 = 0 2 2 2 2

x a 1Ê b bˆ x a + 2 tan ◊ Á cot - tan ˜ tan - tan 2 = 0 2 2 2Ë 2 2¯ 2 2



(cos a + cos b + cos g)2 + (sin a + sin b +sin g)2 = 0



S cos a = 0 and S sin a = 0



S (cos a + sin a) = 0.

Example 77

If tan 1° tan 2° ... tan 89° = x2 – 8, then

the value of x can be (a) – 1 (b) 1 (c) – 3 (d) 3 Ans. (c), (d) Solution: x2 – 8 = (tan 1° tan 89°) (tan 2° tan 88°) ... (tan 44° tan 46°) tan 45° = (tan 1° cot 1°) (tan 2° cot 2°) ... (tan 44° cot 44°) tan 45° = 1 fi x2 = 9 fi x = ± 3.

IIT JEE eBooks: www.crackjee.xyz Trigonometry 10.21

Which of the following statements are

Example 78 correct

(a) sin 1 > sin 1° (b) tan 2 < 0 (c) tan 1 > tan 2 (d) tan 2 < tan 1 < 0 Ans. (a), (b), (c) Solution: Since 1 radian lies between 57° and 58° and sin 57° > sin 1°, so sin 1 > sin 1°. Again 1 radian is an acute angle and 2 radian is an obtuse angle, tan 1 > 0, tan 2 < 0, so that tan 1 > tan 2.

tan (a/2) tan (b/2) = n(π 1), then (a) cos (a – b) =

( xy ) q

q – f = a π 0 then (a) ab > 4

2

=

l = sin a + sin b + 2 sin a sin b and

2

2

fi 2 cos (a – b) = l2 + m2 – 2

(by adding) 2

fi 2 cos (a + b) cos (a – b) + 2 cos (a + b) = m – l 2

ficos (a + b) =

m -l m +l

2

Next,

Example 80 ( x2 + y 2 ) p ( xy ) q

(a)

2

(b) .

1+ n cos[(a – b ) / 2] = = 1– n cos[(a + b ) / 2] =

and

2

(c)

1 + cos (a – b ) 1 + cos (a + b )

[1 + (a / b)]2 ab (ab – 4) ( a + b) 2

sin 4 a a2 sin 4 a b2 sin 8 a a3

= = +

(d) sin4 a =

l 2 + m2 . 2m If x = a cos q sin q, y = a sin q cos q 3

2

3

2

cos 4 a b2 cos 4 a a2

cos8 a b3

=

1 ( a + b )3

a2 ( a + b) 2

Ans. (a), (c), (d) Solution: We are given that Ê sin 4 a cos 4 a ˆ + (a + b) Á =1 b ˜¯ Ë a

(p, q ∈ N) is independent of q, then

(a) p = 4 (b) p = 5 (c) q = 4 (d) q = 5 Ans. (b), (c) Solution: x2 + y2 = a2 sin4 q cos4 q xy = a2 sin5 q cos5 q

2

as tan2 a > 0, ab > 4. sin 4 a cos 4 a 1 + = , then Example 82 If a b a+b

and cos 2a + cos 2b + 2 cos (a + b) = m – l (by subtracting) 2

2

( a – b) 2

a 2 – 4(a / b)

2

2

ab (ab + 4)

2

=

m = cos a + cos b + 2 cos a cos b 2

(d) cot2 a =

È tan q – tan f ˘ tan a = tan (q – f) = Í ˙ Î1 + tan q tan f ˚ 2

(d) a + b = p/2 if l = m Ans. (a), (b), (c), (d) Solution:

( a + b)

2

a = tan q tan f b

Solution:

m2 + l 2

2

(b) ab = 4 ab (ab – 4)

(c) tan2 a =

l 2 + m2 1+ n = 1– n 2n

2

a 2 q (sin q cos q )5q

If tan q + tan f = a, cot q + cot f = b,

Example 81

m2 – l 2

(b) cos (a + b) =

a 2 p (sin q cos q ) 4 p

=

Ans. (a), (c)

l +m –2 2 2

( x2 + y 2 ) p

which is independent of q if 4p = 5q i.e. p = 5, q = 4.

If sin a + sin b = l, cos a + cos b = m and

Example 79

(c)

\

fi sin4 a + cos4 a +

b a sin4 a + cos4 a = 1 a b

= (sin2 a + cos2 a)2 fi

b a sin4 a – 2 sin2 a cos2 a + cos4 a = 0 a b

IIT JEE eBooks: www.crackjee.xyz 10.22 Comprehensive Mathematics—JEE Advanced

Ê b sin 2 a – fi Á Ë a fi fi

Example 84 The equation sin4 x + cos4 x = a has a real solution for (a) all values of a (b) a = 1/2 (c) a = 7/10 (d) a = 1 Ans. (b), (c), (d) Solution: We have sin4 x + cos4 x £ sin2 x + cos2 x, as x £ x £1 fi a£1 (1) Next, sin4 x + cos4 x = a fi (sin2 x + cos2 x)2 – 2 sin2 x cos2 x = a

2

=0

b a sin4 a = cos4 a a b sin 4 a a2

Since

=

cos 4 a b2

= k say

sin 4 a cos 4 a 1 + = , we get a b a+b

ak + bk = \

ˆ a cos 2 a ˜ b ¯

sin 8 a a3

1 a+b +

fi k=

cos8 a b3

=

1

(a 2 k )2 a3



( a + b) 2 +

fi 1 – a £ 1/2 [ sin2 2x £ 1] fi 1/2 £ a (2) From (1) and (2) we get 1/2 £ a £ 1. Note that a = 1/2 for x = p/4 and a = 1 for x = p/2.

(b 2 k ) 2 b3

= ak2 + bk2 = (a + b)k2 = (a + b) ◊

1 ( a + b)

4

=

1 ( a + b )3

sin2 2x = 1 – a

.

Example 83 If Pn = cosn q + sinn q, then (a) 2P6 – 3P4 = – 1 (b) 2P6 – 3P4 = 1 (c) 6P10 – 15P8 + 10P6 = 0 (d) 6P10 – 15P8 + 10P6 = 1 Ans. (a), (d) Solution: 2P6 – 3P4 + 1 = 2(cos6 q + sin6 q) – 3(cos4 q + sin4 q) + 1 = 2[(cos2 q + sin2 q)3 – 3 sin2 q cos2 q (cos2 q + sin2 q)] 2 2 2 – 3[(cos q + sin q) – 2 sin2 q cos2 q]+ 1 = 2(1 – 3 sin2 q cos2 q) – 3(1 – 2 sin2 q cos2 q) + 1 = 0 Again for n ≥ 4, we have Pn – Pn–2 = cosn q + sinn q – (cosn–2 q + sinn–2 q) = cosn–2 q (cos2 q – 1) + sinn–2 q (sin2 q – 1) 2 n–2 2 = – sin q cos q – cos q sinn–2 q = – sin2 q cos2 q (cosn–4 q + sinn–4 q) = – sin2 q cos2 q Pn–4 \ 6P10 – 15P8 + 10P6 – 1 = 6(P10 – P8) – 9(P8 – P6) + (P6 – P4) + P4 – P2 = – sin2 q cos2 q (6P6 – 9P4 + P2 + P0) = – 3 sin2 q cos2 q (2P6 – 3P4) – sin2 q cos2 q (1 + 2) [\ P2 = 1, P0 = 2] = – 3 sin2 q cos2 q (– 1) – 3 sin2 qcos2 q = 0. [\ 2P6 – 3P4 + 1 = 0 (as proved)]

Example 85

For a positive integer n, let

fn(q) = tan (q/2) (1 + sec q) (1 + sec 2q) ... (1 + sec 2nq) then (b) f3(p/32) = 1 (a) f2(p/16) = 1 (d) f5(p/128) = 1 (c) f4(p/64) = 1 Ans. (a), (b), (c), (d) sin (q / 2) 1 + cos q ¥ Solution: fn(q) = cos q cos (q / 2) (1 + sec 2q) ... (1 + sec 2n q)

sin (q /2) ¥ 2 cos 2 (q /2) = (1 + sec 2q) cos (q /2) cosq



... (1 + sec 2n q) = tan q (1 + sec 2q) (1 + sec 4q) ... (1 + sec 2n q) n = tan 2 q f2(p/16) = f3(p/32) = f4(p/64) = f5(p/128) = tan p/4.

MATRIX-MATCH TYPE QUESTIONS Example 86 Column 1

Column 2

(a) sin 15°

(p) 2 + 3

(b) cos 15°

(q)

6- 2 4

(c) tan 15°

(r)

6+ 2 4

(d) cot 15°

(s) 2 - 3

IIT JEE eBooks: www.crackjee.xyz Trigonometry 10.23

Ans.

a

Solution:

p

q

r

s

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

tan 15° =

6+ 2

(p)

(b) sin 2A

(q)

(c) cos 2A

(r)

(d) sec2A

(s) r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

2 = 1-

( m - 1)

( m + 1)2 - ( m - 1)2 ( m + 1)2 + ( m - 1)2

m2 + 1

m2 + 1 2 m + 1) ( +1=

2m m2 + 1

(

m2 + 1

)

2 m2 + 1

( m + 1)

2

Column 2

Column 2 2m m2 + 1

9

(d) Â sin

m2 - 1

r =1

m +1 2

(

)

2 m +1 2

( m + 1)

2

rp =0 18

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

m2 - 1 2m

2

Solution: (a) tan2q = sin q cos q fi sin q = cos3q Given expression is equal to 1 – sin2q + 1 – 3 sin2q + 3 sin4q – sin6q = cos2q + (1 – sin2q)3 = cos2q + cos6q

(p) 5

(s) 3

(1)

(from 1) = cos2q + sin2q = 1

)

2 m2 - 1 4m

2m

(b) 4 cos 20° – 3 cot 20° = (q) – 1 (c) If A, B, C are the angles of a triangle such that sin A, sin B, A C (r) 1 sin C are in A.P. then cot cot 2 2

1 - tan 2 A

(

=

2m

(a) If tan q is the G.M. between sin q and cos q, then 2– 4 sin2q + 3 sin4q – sin6q =

2 tan A

m +1 = ( m - 1)2

( m + 1)2

fi sec A =

m -1 m +1

(a) tan 2A

tan 2A =

fi 2 cos A =

=

m2 + 1

Example 88

Column 1

q

1 + tan A 2

2

= 2- 3

If tan A =

p

1 - tan 2 A

m2 - 1

Column 1

cot 15° = 2 + 3 Example 87

=

1 + tan 2 A

2

6+ 2 4

6- 2

2 tan A

2 cos2A – 1 =

1 Ê 3 1ˆ 6- 2 Á 2 - 2˜ = 4 2Ë ¯

Similarly cos 15° =

Solution:

cos 2A =

sin 15° = sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30° =

Ans.

sin 2A =

m2 - 1 = 2m

(b) 4 cos 20° – =

3 cot 20°

2sin 40∞ - 3 cos 20∞ 2sin ( 60∞ - 20∞) - 3 cos 20∞ = sin 20∞ sin 20∞

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3 cos 20∞ - sin 20∞ - 3 cos 20∞ = -1 sin 20∞

=

(c) 2 sin B = sin A + sin C fi 2 sin (A + C) = sin A + sin C A+C A+C A+C A-C cos cos fi 2 sin = 2 sin 2 2 2 2 A-C 2 =2 A+C cos 2 cos fi

Solution: Let A = 20° fi 3A = 60° fi 2 cos 3A = 1 fi 2 (4 cos3A – 3 cos A) – 1 = 0 fi cos 20° is a root of the equation 8x3 – 6x – 1 = 0 (b) Let A = 10° 1 fi sin 3A = sin 30° = 2 fi fi

A C cos 2 2 = 3 (using componedo and dividendo) A C sin sin 2 2

2(3 sin A – 4 sin3A) – 1 = 0 sin 10° is a root of the equations 8x3 – 6x + 1 = 0 (c) Let A = 15° fi tan 3A = tan 45° = 1

cos fi

fi cot 2 (d) sin

+ sin 2

9p 18

8p ˆ Ê 2p 7p ˆ Ê 2p + sin 2 ˜ + Á sin 2 + sin 2 = Á sin ˜ Ë 18 18 ¯ Ë 18 18 ¯ +

p 5p ˆ Ê 2 4p + sin 2 ˜ + sin 2 . ÁË sin ¯ 18 18 2

pˆ Ê 2p + cos 2 ˜ + = Á sin Ë 18 18 ¯

p + sin 2 2

=5 Example 89 Match the trigonometric ratio with the equations whose one of the roots is given Column 1 Column 2 (a) cos 20° (p) x3 – 3x2 – 3x + 1 =0 (b) sin 10° (q) 32x5 – 40x3 + 10x –1=0 3 (c) tan 15° (r) 8x – 6x – 1 = 0 (d) sin 6° (s) 8x3 – 6x + 1 = 0 p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

=1 1 - 3tan 2 A tan 15° is a root of the equation x3 – 3x2 – 3x + 1 = 0 1 (d) Let A = 6° fi sin 5A = 2



A C cot =3 2 2

p 2p + sin 2 + 18 18

3tan A - tan 3 A





2 (16 sin5 A – 20 sin3A + 5 sin A) = 1 So sin 6° is a root of 32x5 – 40x3 + 10x – 1 = 0

Example 90 Column 1 (a) (b) (c) (d)

Column 2

sin 70° sin 10° cos 70° cos 10° cos 20° cos 80° sin 20° sin 80° p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

(p) (q) (r) (s)

sin2 40° – sin2 30° cos2 30° – cos2 40° cos2 40° – sin2 30° cos2 30° – sin2 40°

Solution: sin 70° sin 10° = sin (40° + 30°) sin (40° – 30°) = sin2 40° – sin2 30° = cos2 30° – cos2 40° cos 20° cos 80° = sin 70° sin 10° cos 70° cos 10° = cos (40° + 30°) cos (40° – 30°) = cos2 40° – sin2 30° = cos2 30° – sin2 40° sin 20° sin 80° = cos 70° cos 10°

IIT JEE eBooks: www.crackjee.xyz Trigonometry 10.25

If A and B are acute angles.

Example 91

Column 1

Column 2

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

Column 1 (a) cos (a + b) (b) sin (a + b) (c) cos (a – b)

(p) A = p/4

(a) sin2A = sinB, 3 cos2 A = 2 cos2B (b) A + B and A – B satisfy the equation tan2 q – 4 tanq + 1 = 0 (c) cos A + cos B – cos (A + B) = 3/2 (d) sin (A + B) = 1, sin (A – B) = 1/2

cos a + cos b = a, sin a + sin b = b

Example 93

(q) B = p/6

Column 2 (p) 2ab/(a2 + b2) (q) b/a (r) (a2 – b2)/(a2 + b2)

a+b 2 p q r

s

(s) (a2 + b2 – 2)/2

(d) tan (r) A = p/3

Ans.

(s) B = p/3

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: cos

Solution: Substitute the values of A and B given on R.H.S. to verify the equations on L.H.S.

a = 2 cos

a+b a+b a –b cos , b = 2 sin 2 2 2

a+b a –b b fi tan = 2 2 a



cos (a + b) =

1 – (b 2 /a 2 ) 1 + (b 2 /a 2 )

Example 92 Column 1 (a) (b) (c) (d) Ans.

sin 85° + sin 55° sin 85° – sin 55° cos 85° + cos 55° cos 85° – cos 55° p q r s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution:

and sin (a + b) =

Column 2 (p) (q) (r) (s)

2 2 2 2

cos 70° sin 15° cos 20° cos 15° cos 70° sin 75° sin 70° cos 105°

sin 85° + sin 55° 85∞ + 55∞ 85∞ – 55∞ = 2 sin cos 2 2

= 2 sin 70° cos 15° = 2 cos 20° cos 15° sin 85° – sin 55° = 2 cos 70° sin 15° cos 85° + cos 55° = 2 cos 70° cos 15° = 2 cos 70° sin 75° cos 85° – cos 55° = – 2 sin 70° sin 15° = 2 sin 70° cos 105°

2 (b /a) 1 + (b /a ) 2

2

=

=

a 2 – b2 a 2 + b2 2ab a + b2 2

a2 + b2 = sin2 a + sin2 b + cos2 a + cos2 b + 2 cos a cos b + 2 sin a sin b a2 + b2 – 2 = cos (a – b)



Example 94 Column 1 (a) sin q = 1/2 (b) cos q = 1/2

Column 2 (p) q = 13p/6 (q) q = 4p/3

(c) tan q =

(r) q = 5p/6

3

(d) sin q = 1/2 and

Ans.

cos q = 3 / 2 p q r s a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution

sin q =

(s) q = 7p/3, 5p/3

p 1 5p = sin = sin 6 2 6

IIT JEE eBooks: www.crackjee.xyz 10.26 Comprehensive Mathematics—JEE Advanced

cos q = tan q =



p pˆ 1 Ê = cos = cos Á 2p ± ˜ Ë 3 3¯ 2 3 = tan

fi cos x + sin x =

p pˆ Ê = tan Á p + ˜ Ë 3 3¯



p 1 3 sin q = and cos q = fiq= 2 6 2 or 2p +

sin 2x = 3/4 fi (cos x + sin x)2 = 1 + 3/4 = 7/4

p . 6

(b) cos 3A = sin 7A (c) tan A = cot 3A (d) cot A = tan 2A p q r s Ans. a p q r s b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

1∞ (p) A = 22 2 (q) A = 30° (r) A = 9° (s) A = 18°

7 2= 7 4.

1 ak + ck 1 + bk q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: a2 + b2 =

1

=

(p) 3/4

(b) sin 2x

(q)

(

)

7 +1 4

(c) cos 2x

(r)

7 4

(d) cos x p q r

(s)

7 2

2

k 1

Column 1 (a) sin (1/2) (A – B) (b) sin 2A + sin 2B (c) cos (A + B) (d) tan ((A + B)/2) p q r s

p

q

r

s

b

p

q

r

s

a

p

q

r

s

c

p

q

r

s

b

p

q

r

s

d

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans

2

If sin A = sin B and cos A = cos B, then

Example 98

s

cos x – sin x = 1/2 fi 1 – 2 sin x cos x = 1/4

1

cos x (1 + cos x) + sin 2 x 1 1 = = . sin x (1 + cos x) sin x ak

a

Solution:

(sin2 x + cos2 x) =

k2 1 sec x 1 = 2 4 a2 + b2 + c2 = 2 + 2 tan2 x = k k k2 b k 1 bc = 2 sin x = a/k k 1 ak sin x + = cot x + 1 + cos x ck 1 + bk

Column 2

(a) cos x + sin x

)

7 +1 4

(s) a/k

p

Ans.

1∞ , 2A + A = 90° fi A = 30°. 2

Column 1

(

Given

If cos x – sin x = 1/2

Example 96

Ans.

cos 2x = 1/2 ¥

(d)

Solution: 2A + 3A = 90° fi A = 18°, 3A + 7A = 90° fi A = 9° A + 3A = 90° fi A = 22

7 2 fi cos x fi

sin x cos x tan x = = =k a b c Column 1 Column 2 2 2 (p) 1/b2k4 (a) a + b (b) a2 + b2 + c2 (q) 1/k2 (c) bc (r) 1/ak

Column 2

(a) cos 2A = sin 3A

2 cos x = 1/2 +

Example 97

Example 95 Column 1

7 2

(p) (q) (r) (s)

Column 2 0 cos 2A 1 2 sin (A + B)

IIT JEE eBooks: www.crackjee.xyz Trigonometry 10.27

Solution:

sin A – sin B = cos B – cos A = 0

A+ B A+ B A–B A–B fi 2 cos sin = 2 sin sin 2 2 2 2 fi sin

A–B 2

A+ B A + B˘ È Ícos 2 – sin 2 ˙ = 0 Î ˚

A+ B A–B fi Either sin = 0 or tan =1 2 2 sin 2A + sin 2B = 2 sin A cos A + 2 sin B cos B = 2 sin A cos B + 2 cos A sin B = 2 sin (A + B) cos (A + B) = cos A cos B – sin A sin B = cos2 A – sin2 A = cos 2A. sin 3 a is Example 99 cos 2 a Column 1

so

sin 3 a can be positive or negative. cos 2 a

Ê 18p 23p ˆ , , sin 3a is – ve If a ∈ Á Ë 48 48 ˜¯ and cos 2a is – ve so

(b) negative

Ê 14p 18p ˆ , (q) a ∈ Á Ë 48 48 ˜¯

(c) positive or negative

Ê 18p 23p ˆ , (r) a ∈ Á Ë 48 48 ˜¯

sin 3a is positive cos 2a

and if a = 23p/4 then cos 2a = 0 so

Column 1 (a) sin 16° + sin 44° +

a = 23p/4

q

r

s

a

p

q

r

s

b

p

q

r

s

a

p

q

r

s

c

p

q

r

s

b

p

q

r

s

d

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution:

Ans.

Ê 13p 14p ˆ Ê 13p 14p ˆ , , a∈ Á fi 3a ∈ Á ˜ Ë 48 48 ¯ Ë 16 16 ˜¯

= (p – 3p/16, p – 2p/16) i.e. 3a ∈ II quadrant fi sin 3a > 0 11p 10p ˆ Ê 13p 14p ˆ Ê , ,p – 2a ∈ Á = Áp – ˜ Ë 24 24 ˜¯ Ë 24 24 ¯ i.e. 2a ∈ II quadrant fi cos 2a < 0

Ê 14p 18p ˆ , fi 3a ∈ Similarly a ∈ Á Ë 48 48 ˜¯

Solution: since 16° + 44° + 120° = 180° = p Use conditional identities. See text.

ASSERTION-REASON TYPE QUESTIONS Example 101

2p 2p ˆ Ê ,p + ÁË p – ˜ 16 16 ¯

fi sin 3a can be positive or negative.

Statement 1: If f(x) = 0 be an equation of

p , then the sum of all the roots of f¢(x) = 0 is zero. 18 p Statement 2: 2 cos x6 – 6x4 + 18 9x2 – 3 = 0 Ans. (a) p p 1 fi 6q = fi cos 6q = Solution: Let q = 18 3 2

is 2 cos

sin 3 a is negative. cos 2 a

Column 2

sin 120° (p) 3 tan 16° tan 44° (b) tan 16° + tan 44° + tan 120° (q) 1 + cos 16° cos 44° 2 2 (c) cos 16° + cos 44° (r) 2 cos 8° cos 22° + cos2 120° (d) cos 16° + cos 44° + cos 120° (s) 1 + 2 3 sin 8° sin 22° p q r s

p

Ans.

sin 3a is not cos 2a

Example 100

Column 2 Ê 13p 14p ˆ , (p) a ∈ Á Ë 48 48 ˜¯

(a) positive

\

10p 61p ˆ Ê ,p – 2a ∈ Á p – ˜ cos 2a is negative. Ë 24 24 ¯

IIT JEE eBooks: www.crackjee.xyz 10.28 Comprehensive Mathematics—JEE Advanced

1 2 2 3 fi 8 (2 cos q – 1) – 6 (2 cos2q – 1) = 1 Let 2 cos q = x fi 4 cos3 2q – 3 cos 2q =

3

So

È Ê x2 ˘ x2 ˆ 8 Í2 ¥ - 1˙ - 6 Á 2 ¥ - 1˜ = 1 4 4 Ë ¯ Î ˚

x xˆ Ê Solution: f(x) = 3 + sin2x + Á sin + cos ˜ Ë 2 2¯ now 3 £ sin2x + 3 £ 4 x x - 2 £ sin + cos £ 2 and 2 2 3 - 2 tan 36°

Statement-2: cos 36° > sin 36° Ans. (b) Solution: Statement-2 is true as cos q > sin q for 0 £ q < p/4 Statement-1 is true if cos2 36° > sin 36° or if 1 + cos 72° > 2 sin 36° = 2 sin (30° + 6°) or if 1 + sin 18° > 2 (sin 30° cos 6° + cos 30° sin 6°) or if 1 + 2 sin 9° cos 9° > cos 6° + 2 cos 30° sin 6° which is true because 1 > cos 6°, sin 9° > sin 6°, cos 9° > cos 30° so statement-1 is also true but does not follow from statement-2 Example 107

the expression sin a + sin b + sin g where a, b, g are real numbers such that a + b + g = p is negative. Statement-2: If a + b + g = p, then a, b, g are the angles of a triangle. Ans. (c)

so statement–1 is correct. Also

Statement-1: The minimum value of

Statement-1: The numbers sin 18° and

– sin 54° are the roots of a quadratic equation with integer Statement-2: If x = 18°, cos 3x = sin 2x and if y = – 54° sin 2y = cos 3y.

Solution: The minimum value of the sum can be – 3 provided sin a = sin b = sin g = – 1 fi a = (4l – 1) p/2, b = (4m – 1) p/2, g = (4n – 1) p/2 Now a + b + g = p fi [4 (l + m + n) – 3] p/2 = p fi 4 (l + m + n) = 5 which is not possible as l, m, n are integers. 1. minimum value can not be – 3. But for a = 3p/2, b = 3p/2, g = – 2p; a + b + g = p and sin a + sin b + sin g = – 2 So sin a + sin b + sin g can have negative values and thus the minimum value of the sum is negative proving that statement-1 is correct. But the statement-2 is false as a + b + g = p for a = b = 3p/2, g = – 2p which are not the angles of a triangle. Example 109 +

Statement-1: If 2 sin (q/2) = 1 + sinq

1 - sinq then q/2 lies between 2np + p/4 and 2np +

3p/4. Statement-2: If

p q 3p £q£ then sin > 0. 4 2 4

Ans. (b) Solution: We have 2 sin (q/2) =

(cos(q / 2) + sin(q / 2)) 2 + q/2) + sin (q

(cos(q / 2) - sin(q / 2)) 2 q/2) – sin (q

Ans. (a)

fi cos (q/2) + sin (q/2) > 0 and cos (q/2) – sin (q/2) < 0

Solution Statement-2: is correct, using it we have

fi sin (q/2 + p/4) > 0 and cos (q/2 + p/4) < 0

cos 3x = sin 2x fi

3

4 cos x – 3 cos x = 2 sin x cos x

Similarly So fi

3

4 cos y – 3 cos y = 2 sin y cos y 2

4(1 – sin x) – 3 = 2 sin x 4 sin2 x + 2 sin x – 1 = 0 and 4 sin2 y + 2 sin y – 1 = 0

Hence sin x = sin 18° and sin y = sin (– 54°) = – sin 54° are

fi 2np + p/2 < q/2 + p/4 < 2np + p fi 2np + p/4 < q/2 p/2 fi B + C < p/2 tan B + tan C fi tan (B + C) > 0 fi >0 1 – tan B tan C

= 23 = 8.

If

sin 20∞

1

=

+

3 cos 20∞ + sin 20∞ 3 sin 20∞ cos 20∞ 3 1 cos 20∞ + sin 20∞ 2 2 3 sin 40∞ 4

=

2p 4p 6p ˘ È + cos + cos ˙ If x + 270 Ícos 7 7 7 ˚ Î

= 139 then the value of x is Ans. 4 2p 4p 6p + cos + cos Solution: cos 7 7 7 1 = ¥ 2p 2sin 7 6p 2p 8p 4p ˘ È 4p Ísin 7 + sin 7 – sin 7 + sin 7 – sin 7 ˙ Î ˚

1 – b2

=

fi tan B tan C < 1 as tan B > 0, tan C > 0 fi [x] = 6 – 1 = 5. Example 145

2a

= 2k cos 40°, 3 cos 20∞ then 18k4 + 162k2 – 1370 is equal to Ans. 6 1 1 + Solution: 2k cos 40° = sin 20∞ 3 cos 20∞ Example 147

= 1 – (625 + 961) sin q + (775 sin q) 2

135 = 4.

Example 146 If a + b = g and tan g = 4, a is the arithmetic and b is the geometric mean respectively a3 between tan a and tan b, then the value of is (1 – b 2 )3 equal to Ans. 8 tan a + tan b Solution: tan g = tan (a + b) = 1 – tan a tan b fi

fi (a + 25 cos q)2 = 1 – (25 sin q)2

6p 8p ˘ È Í\ sin 7 = – sin 7 ˙ Î ˚

1

sin 60∞ cos 20∞ + cos 60∞ sin 20∞

= (4

(

)

3/4 sin 40∞

3 ) 2 cos 40∞

fi 3k2 = 16 so 18k4 + 162k2 – 1370 = 6. Example 148

If [16 cos x + 12 sin x] = 2k + 10, then

the maximum value of k is Ans. 5 Solution:

16 cos x + 12 sin x =

Ê 3ˆ a = tan–1 Á ˜ . Ë 4¯

162 + 122 cos (x – a),

IIT JEE eBooks: www.crackjee.xyz Trigonometry 10.37

fi fi

£ k £ 5.

k

x–a £1

Example 149 Value of 6736 cos2 18° + 421 tan2 36° – 6312 is Ans. 3 Solution: 421 (16 cos2 18° + tan2 36°) = 421 ÈÎ10 + 2 5 + 5 – 2 5 ˘˚ = 421 ¥ 15 = 6315. Example 150

If A + B + C = 180°,

A B C sin 2 A + sin 2 B + sin 2C = k sin sin sin sin A + sin B + sin C 2 2 2 then the value of k3 – 8k2 + 1 is equal to Ans. 1 Solution: From conditional identities we have sin 2 A + sin 2 B + sin 2C sin A + sin B + sin C =

4sin A sin B sin C 4 cos( A/2) cos( B /2) cos(C /2)

= 8 sin (A/2) sin (B/2) sin (C/2) k=8



and k3 – 8k2 + 1 = 1

3. If xi > 0 for 1 £ i £ n and x1 + x2 + … + xn = p then the greatest value of the sum sin x1 + sin x2 + … + sin xn is equal to (a) n (b) 0 Êpˆ (c) n sin Á ˜ Ë n¯

(c) a + b = p/3 9.

1. The value of

(b)

(d)

(b) y = 0 3p 4

2. If sin A, cos A and tan A are in G.P, then Cot6A – cot2A is equal to (a) – 1 (b) a (c) 1 (d) none of these

11.

3 (1 – a)x + a = 0

3 x2 – (1 – a)x + a 3 = 0

(c) x2 +

Êp ˆ Êp ˆ + sin y cos Á - y˜ + cos x sin Á - 4˜ is zero if Ë2 ¯ Ë2 ¯

(d) x – y =

2 + 2 + 2 cos 4q is equal to

(a) x2 –

Êp ˆ Êp ˆ cos y cos Á - x˜ – cos Á - y˜ cos x Ë2 ¯ Ë2 ¯

(c) x = y

(d) none of these.

(a) cos q (b) 2 cos q (c) cos 2q (d) 2 cos 2q. 10. If tan x tan y = a and x + y = p/6, then tan x and tan y satisfy the equation

SINGLE CORRECT ANSWER TYPE QUESTIONS

(a) x = 0

n 2

4. If sin x + sin2x + sin3x = 1, then cos6x – 4 cos4x + 8 cos2x is equal to (a) 0 (b) 2 (c) 4 (d) 8 a a 5. If sin a = p p £ 1, then tan , cot are the roots 2 2 of the equation (b) px2 – x + p = 0 (a) px2 – 2x + p = 0 (d) px2 – 2x – p = 0 (c) px2 + 2x + p = 0 6. 6(sin6q + cos6q) – 9 (sin4q + cos4q) is equal to (a) – 1 (b) 1 (c) – 3 (d) 3 2 7. (1 + tan a tan b) + (tan a – tan b)2 is equal to (b) cos2a cos2b (a) tan2a + tan2b (d) tan2a tan2b (c) sec2a sec2b 5 1 tanb = then 8. If tana = 6 11 (a) a + b = p/6 (b) a + b = p/4

EXERCISE LEVEL 1

(d)

3 (1 + a)x – a = 0

3 x2 + (1 + a)x – a 3 = 0

sin 7 x + 6sin 5 x + 17 sin 3x + 12 sin x is equal to sin 6 x + 5sin 4 x + 12 sin 2 x (a) cos x

(b) 2 cos x (c) sin x (d) 2 sin x. n sin a cos a 12. If tan b = then, tan (a + b) is equal to 1 - n cos 2 a (a) (n – 1) tan a (c)

1 tan a n +1

(b) (n + 1) tan a (d)

–1 tan a n -1

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p a 7 , 0 < a < , then tan is 2 6 2

13. If sin a + cos a = equal to 7 –2

(a)

(c) 2 –

(b) (1/3) ( 7 – 2)

7

7)

(d) (1/3) (2 –

336 and 450° < a < 540°, then 625 sin (a/4) is equal to

14. If sin a =

(a)

1

7 25

(b)

5 2

(c)

4 5

(d)

3 . 5

15. If sin (q + a) = a and sin (q + b) = b, ((0 < a, b, q < p/2) then cos 2 (a – b) – 4 ab cos (a – b) is equal to (a) 1 – a2 – b2 2

(c) 2 + a + b

(b) 1 – 2a2 – 2b2

2

2

2

(d) 2 – a – b .

16. (sin a + cosec a) + (cos a + sec a)2 – (tan2 a + cot2 a) for all values of a is equal to 2

(a) 0

(b) 2

(c) 4

(d) 7

17. The expression È ˘ Ê 3p ˆ - a ˜ + sin 4 (3p + a ) ˙ 3 Ísin 4 Á Ë 2 ¯ Î ˚ È ˘ Êp ˆ - 2 Ísin 6 Á + a ˜ + sin 6 (5p - a ) ˙ Ë ¯ 2 Î ˚ is equal to (a) 0 (b) –1 (c) 1 (d) 3. 18. If sin (x + 3a) = 3 sin (a – x), then (a) tan x = tan a (b) tan x = tan2 a 3 (d) tan x = 3 tan a. (c) tan x = tan a 19. If

cos (q - a ) m +1 = , then m is equal to sin (q + a ) m -1

Êp ˆ Êp ˆ (a) tan Á - q ˜ tan Á - a ˜ Ë4 ¯ Ë4 ¯ Êp ˆ Êp ˆ (b) tan Á - q ˜ tan Á + a ˜ Ë4 ¯ Ë4 ¯ Êp ˆ Êp ˆ (c) tan Á + q ˜ tan Á + a ˜ Ë4 ¯ Ë4 ¯ Êp ˆ Êp ˆ (d) tan Á + q ˜ tan Á - a ˜ Ë4 ¯ Ë4 ¯

20. The value of tan 130° tan 140° is equal to (a) – 1 (b) 1 (d) 1 + 3 (c) 1/ 3 21. If cos 1° cos 2° cos 3° cos 179° = x + 1, then x is equal to (a) – 1 (b) 0 (c) 1 (d) none of these 22. If – p/4 £ x < p/4 and 1 + tan x = 1 + sin 2x, then tan x is equal to 1 - tan x (b) –1

(a) – 1 3. If tan a =

3

(c) 1

(d) 2

1 m and tan b = , then m -1 2m -1

(a) a + b = p/4 (c) a + b = p/6

(b) a – b = p/4 (d) a – b = p/3

24. If m tan (q – 30°) = n tan (q + 120°), then

m-n is m+n

equal to (a) 2 cos 2q (c) 1/(2 cos 2q)

(b) 2 sin2 q (d) 1/(2 sin 2q)

25. If cos x + sin x = 2 cos x, then tan2 x + 2 tan x is equal to (a) 0 (b) 1 (c) 2 (d) 3 26. If A = sin8 q + cos14 q, then for all values of q (a) A > 1 (b) A ≥ 1 (c) A < 1 (d) A £ 1 27. If p/2 < a < p, then the expression 1 - sin a 1 + sin a + 1 + sin a 1 - sin a

is equal to

(a) 2/cos a (b) – 2/cos a (c) 2/sin a (d) 2 tan a 28. If 0 < q < p/2 and sin q + cos q + tan q + cot q + sec q + cosec q is equal to 7, then sin 2q is a root of the equation (a) x2 + 44x + 36 = 0 (b) x2 – 44x – 36 = 0 (c) x2 – 44x + 36 = 0 (d) x2 + 44x – 36 = 0 cos ( q1 - q 2 ) cos ( q3 + q 4 ) + 29. If = 0, cos (q1 + q 2 ) cos (q3 - q 4 ) then tan q1 tan q2 tan q3 tan q4 is equal to (a) – 1 (b) 1 (c) 2 (d) 4 30. If cos 2b = tan g are in (a) A.P. (c) H.P.

cos (a + g ) , then tan a, tan b and cos (a - g ) (b) G.P. (d) none of these

IIT JEE eBooks: www.crackjee.xyz Trigonometry 10.39

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

37. If tan q =

31. If x cos a + y sin a = x cos b + y sin b = 2a (0 < a, b < p/2) then (a) cos a + cos b =

x2 + y 2

(b) sin q =

(c) cos q =

4ay x2 + y 2

(d) cos q =

4a 2 - x 2 x2 + y 2

1 - sin 4 A + 1 1 + sin 4 A - 1

Êp ˆ (d) – cot Á + A˜ Ë4 ¯

33. For 0 < q < p/2, tan q + tan 2q + tan 3q = 0 if (a) tan q = 0 (b) tan 2q = 0 (c) tan 3q = 0 (d) tan q tan 2q = 2 34. If sin q (1 + sin q) + cos q (1 + cos q) = x and sin q (1 – sin q) + cos q(1 – cos q) = y, then (b) y2 + 2y = sin 2q (a) x2 – 2x = sin 2q (c) xy = sin 2q (d) x – y = 2 p , then q

1 (p cosec 2q – q sec 2q) = (b) 2

p +q

(c) p sin 6q – q cos 6q = 0 2 pq p2 + q2

2

Ê 1 + sin q - cos q ˆ ÁË 1 + sin q + cos q ˜¯ is equal to (a)

1 - cos q 1 + cos q

(c) tan2

q 2

(b)

2 ( x 4 + 1) x2 - 1 2 ( x 4 + 1) x2 + 1 2 ( x 4 + 1) x2 - 1 2 ( x 4 + 1)

1 - sin q 1 + sin q

(d) cot2

q 2

(a) sin q cos q =

1 x

39. If cos a =

.

(b) sin q tan q = y

3 5 and cos b = , then 5 13

(a) cos (a + b) = (c) sin2

2

2

33 65

a-b 1 = 2 65

(b) sin (a + b) =

56 65

(d) cos (a – b) =

63 65

40. The equation sin6 x + cos6 x = a2 has real solutions if 1˘ È (a) a ∈ (–1, 1) (b) a ∈ Í-1, - ˙ 2˚ Î Ê 1 1ˆ (c) a ∈ Á - , ˜ Ë 2 2¯

p2 + q2

(a) p sin 6q +q cos 6q =

36.

x2 + 1

(c) (x2y)2/3 – (xy2)2/3 = 1 (d) (x2y)1/3 + (xy2)1/3 = 1.

(b) cot A

(d) p cos 6q + q sin 6q =

, then

38. If cot q + tan q = x and sec q – cos q = y, then

, then one of the values of y is

(a) – tan A Êp ˆ (c) tan Á + A˜ Ë4 ¯

35. If tan 6q =

(a) sin q =

4ax

x2 + y 2

(c) sin a + sin b =

32. If y =

x2 - 1

4a 2 - y 2

(b) cos a cos b =

(d) sin a sin b =

x2 + 1

È1 ˘ (d) a ∈ Í , 1˙ . Î2 ˚

41. If tan a and tan b are the roots of the equation x2 + px + q = 0 ( p π 0), then (a) sin2 (a + b ) + p sin (a + b) cos (a + b) + q cos2 (a + b) = q p (b) tan (a + b ) = q -1 (c) cos (a + b) = 1 – q (d) sin (a + b) = – p. 42. If sin q + sin f = a and cos q + cos f = b, then q -f 1 =± a 2 + b2 2 2 q -f 1 =± a 2 - b2 (b) cos 2 2 (a) cos

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(c) tan

q -f =± 2

49. If a £ 5 cos q + 3 cos (q + p/3) + 5 £ b, then (a) a = – 2 (b) a = 2 (c) b = 12 (d) b = 7 2 50. The value of tan a – tan2 b – (1/2) sin (a – b) sec2 a sec2 b is zero if (a) sin (a + b) = 0 (b) sin (a + b) = 1/2 (c) sin (a – b) = 0 (d) sin (a - b) = 1/2

4 - a 2 - b2 a 2 + b2

a 2 + b2 - 2 . 2 sin a - cos a , then 43. If tan q = sin a + cos a (d) cos (q – f) =

MATRIX-MATCH TYPE QUESTIONS

(a) sin a – cos a = ± 2 sin q (b) sin a + cos a = ± 2 cos q (c) cos 2q = sin 2a (d) sin 2q + cos 2a = 0. 44. Which of the following statements are possible; a, b, m and n being non-zero real numbers? (a) 4 sin2 q = 5 (b) (a2 + b2) cos q = 2ab (c) (m2 + n2) cosec q = m2 – n2 (d) sin q = 2.375. 45. If A lies between 270° and 360° and sin A = –7/25, then 336 A 2 = (a) sin 2A = (b) cos 2 5 625 (c) tan 46. If

A 1 =2 7

(d) sin

A 2 =. 2 10

x cos A = where A π B, then y cos B

A + B x tan A + y tan B = (a) tan 2 x+ y A - B x tan A - y tan B = (b) tan 2 x+ y sin ( A + B) y sin A + x sin B = (c) sin ( A - B) y sin A - x sin B (d) x cos A + y cos B = 0. 47. If a, b and g are connected by the relation 2 tan2 a tan2 b tan2 g + tan2 a tan2 b + tan2 b tan2 g + tan2 g tan2 a = 1 then (a) sin2 a + sin2 b + sin2 g = 1 (b) cos2 a + cos2 b + cos2 g = 2 (c) cos 2a + cos 2b + cos 2g = 1 (d) cos (a + b) cos (a – b) = cos2 g. 48. If sin (p cos q) = cos (p sin q), then sin 2q is equal to (a) – 1/4 (b) – 3/4 (c) 1/4 (d) 3/4

51. Column 1

Column 2

(a) cos 36° – cos 72° (b) cos 36° cos 72° (c) tan 36° tan 18°

(p) 5 4 (q) 1/2 (r) 1/4

(

(d) sin 36° cos 18° (s) 5 – 2 5 52. cos q + sin q = x, cos q – sin q = y Column 1

5

Column 2

(a) cos 2q (b) sin 2q

(p) 1 – y2 (q) xy

(c) cot q

(r)

x2 – 1 xy

(d) tan 2q

(s)

x+ y x– y

53. sin 2q = k Column 1 (a) cosec 2q + cot 2q cos 2q Ê 1 + tan q ˆ (b) Á Ë 1 – tan q ˜¯

)

Column 2 (p) (1 + k)/(1 – k)

2

(q) k2 + k – 1

(c) sin 2q – (1/2) (1 + cos 4q) (r) 3k – 4k3 (s) (2 – k2)/k

(d) sin 6q 54. Column 1

Column 2 1 – tan 2 ( x / 2)

(a)

1 – sin x cos x

(p)

(b)

1 + sin x cos x

(q) tan2 (x/2)

(c)

1 – cos x 1 + cos x

(r)

1 – tan ( x / 2) 1 + tan ( x / 2)

(s)

1 + tan ( x / 2) 1 – tan ( x / 2)

(d) cos x

1 + tan 2 ( x / 2)

IIT JEE eBooks: www.crackjee.xyz Trigonometry 10.41

55. Column 1 (a) sin q + sin 2q =

Column 2 3 +1 2 3 +1 2

(b) cos q + cos 2q = (c) tan q + tan 2q = (d) cot q + cot 2q =

4 3 4 3

(p) q = p/6

62. sin (a + b + g + d) is equal to (a) sin (3b + d) (b) sin (a + 3g) (c) sin (a + 3d) (d) sin (3b + g) 63. cos b is equal to (a) cos (a + g – d) (b) cos (a + d – g)

(q) q = 13p/6

(c) cos (2g – d)

Êa + g ˆ (d) cos Á Ë 2 ˜¯

(r) q = 25p/6

Paragraph for Question Nos. 64 to 67

(s) q = 37p/6

x = cosec2 q, y = sec2 q, z =

ASSERTION REASON TYPE QUESTIONS 56. Statement-1: A, B, C are the angles of a triangle such that angle A is obtuse then tan B tan C > 1 Statement-2: In a triangle ABC tan B + tan C tan A = tan B tan C - 1 sin ( A + B ) + sin ( A - B ) 57. Statement-1: = tan A cos ( A + B ) + cos ( A - B ) Statement-2: sin (A + B) + sin (A – B) = 2 sin A cos (A + B) + cos (A – B) = 2 cos A 58. Statement-1: If x and y are real numbers such that x2 + y2 = 27, then the maximum possible value of x – y is 3 6 Statement-2: – 1 £ cos (q + p/4) £ 1 p 59. a, b, g > 0 and a + b + g = 2 Statement-1: If p = 7 tana tanb, q = 5 + tanb tan g and r = 3 + tan g tana, then the maximum value of p + q + r is 4. Statement-2: If tana tanb + tan b tan g + tan g tan a = 1 60. Statement-1: If x sin3q + y cos3q = sinq cosq and x sinq = y cosq then x2 + y2 = 1 Statement-2: 4 sin3q = 3 sinq – sin 3q and 4 cos3q = 3 cosq + cos 3q

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 61 to 63 If a, b, g, d are in arithmetic progression. 61. (a) tan (a + d) = tan (b + g) (b) tan (a + g) = tan (b + d) (c) tan (a + b) = tan (g + d) (d) none of these

64.

1 x

2

+

1 y2

1 1 – sin q cos 2 q 2

is equal to

2+ z z z (c) 2+ z

(a)

65. xyz is equal to (a) x + y + z (c) x + y – z

2–z z z (d) . 2–z (b)

(b) xy + z (d)

x+ y z

66. tan2 2q is equal to (a)

(c) 67.

4( z + 1) y 2 z (2 – y ) 2 4( z – 1) x 2 z (2 – x) 2

(b)

(d)

4( z – 1) y 2 z (2 – y ) 2 4 x 2 ( y – 1) y 2 ( x – 1)

tan 3q is equal to tan q (a)

(3x – 4) y (4 – 3 y ) x

(b)

(4 – 3 y ) x (3x – 4) y

(c)

(4 – 3x) y (4 – 3 y ) x

(d) (y/x)3/2

Paragraph for Question Nos. 68 to 60 AB is a vertical line and BC is horizontal. D and E are two points on BC. –ACB= q –ADB = 2q, –AEB = 3q. DL and EM are perpendiculars on BC meeting AC at L and M respectively. DL = x, EM = y, BA = z. cot 2q – cot 3q is equal to 68. cot q z–y x z–x (c) y

(a)

(b)

y–x z

(d) none of these

IIT JEE eBooks: www.crackjee.xyz 10.42 Comprehensive Mathematics—JEE Advanced

69.

cot q – cot 2q is equal to cot q (a)

x z

(b)

70. AD is equal to (a) x cot q (c) z cot q

y z

(c)

y x

(d)

x y

(b) y cot q (d) none of these

INTEGER-ANSWER TYPE QUESTIONS 71. If cos A =

2p 4p 6p + cos + cos is equal to 7 7 7 (a) – 1/2 (b) 0 (c) 1/2 (d) 3/2 6. cosec2 A cot2 A – sec2 A tan2 A – (cot2 A – tan2 A) (sec2 A + cosec2 A – 1) is equal to (a) 0 (b) 1 2 (d) cosec2 A (c) sec A 7. If sin A = x, cos B = y and A + B + C = 0, then x2 + 2xy sin C + y2 is equal to (b) cos2 C (a) sin2 C (d) 1 + cos2 C (c) 1 + sin2 C 2 2 2 8. If x tan 60° – 4x cos 45° = 4 sin 30° cos 60° tan 45° and x is not an integer then x is equal to (a) 2/3 (b) –1/3 (c) 1/3 (d) 2/3 5. 1 + cos

3 , then the value of 16 cos2 (A/2) 4

– 32 sin (A/2) sin (5A/2) is equal to 72. If 0 < a, b < p and cos a + cos b – cos (a + b) =

3 2

then 3 sin a + cos a is equal to 73. If sin a equals the slope of the bisector of the b equals the integral solution of the inequality 4x2 – 16x + 15 < 0, then 10 sin (a + b) sin (a – b) is equal to p p˘ È 6p - 33 tan 4 + 27 tan 2 ˙ = 21 then the 74. If k Í tan 9 9 9˚ Î value of k is 75. If sin 32° = k and cos x = 1 – 2k2; a, b are the values of x between 0° and 360° with a y (b) x ≥ y (c) x < y (d) x £ y 10. If sinq, cosq are the roots of the equation 2x2 – 2 2 x + 1 = 0, then q is equal to (a) 15° (b) 30° (c) 45° (d) 60° 11. If x =

sin 3 p cos 2 p

,y=

cos3 p sin 2 p

; and

sin p + cos p = 1/2 then x + y is equal to (a) 75/18 (b) 44/9 (c) 79/18 (d) 48/9 12. The values of x between 0 and 2p which satisfy the equation sinx

8 cos 2 x = 1 are in A.P. The

common difference of the A.P. is (a) p/8 (b) p/4 (c) 3p/8 (d) 5p/8 13. If A.M. of y and 1/y is cos q and that of x and 1/x is cos a then A.M. of x/y + y/x is (a) cos (q + a) (b) cos (q – a) (c) 2 cos (q – a) (d) cos ((q – a)/2) 14. Least value of the expression 1 , x ∈ [– 1, 0], b ∈ [2, 3] is 2 2bx – ( x + b 2 + sin 2 x ) (a) – (c)

1 4

(b) 1

1 4

(d) –

1 9

8 + sin 1 15. If sin a, sin b, sin g are in A.P. and cos a, cos b, cos g are in G.P., then 2

IIT JEE eBooks: www.crackjee.xyz Trigonometry 10.43

cos 2 a + cos 2 g - 4 cos a cos g is equal to 1 - sin a sin g (a) – 2 (b) – 1 (c) 0 (d) 2 16. The greatest value of f(x) = 2 sin x + sin 2x on [0, 3p/2] is (a) 9/2 (b) 5/2 (c) 3 3 2 (d) 3/2 2 17. If sin A = x, then sin A sin 2A sin 3A sin 4A is a polynomial in x (a) 0 (b) 40 (c) 168 (d) 336 18. If a = cos a cos b + sin a sin b cos g b = cos a sin b – sin a cos b cos g and c = sin a sin g, then a2 + b2 + c2 is equal to (a) – 1 (b) 0 (c) 1 (d) none of these 19. If cos a + cos b = a, sin a + sin b = b and cos 3q is equal to a – b = 2q then cos q (b) a2 + b2 – 3 (a) a2 + b2 – 2 (d) (a2 + b2)/4 (c) 3 – a2 – b2 20. If tan q, 2 tan q + 2, 3 tan q + 3 are in G.P., then the 7 - 5 cot q is value of 9 - 4 sec2 q - 1 (a) 12/5 (b) – 33/28 (c) 33/100 (d) 12/13 21. If sin q + cos q = a, cos q – sin q = b, then sin q (sin q – cos q) + sin2 q (sin2 q – cos2 q) + sin3q (sin3 q – cos3 q) + is equal to (a)

(c)

1- ab 1+ ab 1 - a b 1 - a2 + 1 + a b 3 - a2

(b)

(d)

1 - a2

sin n q cos x1 cos xn

(b)

(c) sin nq cos x1 cos xn(d)

(

5+ 5 8

(c)

1 + a b a2 - 1 + 1 - a b 3 - a2

sin (n - 1) q cos x1 cos xn cos (n - 1) q cos x1 cos xn

)

(b) 1 + 5 4

(a) 0

x 24. If x > 0 and - sin q cos q

(d)

5 +1 4

sin q x 1

cos q 1 x

= 0, then

(a) x


2

(d) none of these

2

25. Let n n Êpˆ Êpˆ sin Á ˜ + cos Á ˜ = , then Ë 2n ¯ Ë 2n ¯ 2 (a) n < 4 (c) n = 6

(b) n > 8 (d) none of these

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 26. If

cos a sin a = a, = b, then cos b sin b a2 - 1

(a) sin2 b =

a 2 - b2

(c) cos2 a = 27. If tan q =

(b 2 - 1) a 2 b2 - a 2

(b) cos2 b =

(d) sin2 a =

a2 - 1 a 2 - b2 (b 2 - 1) a 2 b2 - a 2

p2 - q2 , then 2 pq

3 - a2

22. If x1, x2, x3, xn are in A.P. whose common difference is q, then the value of + sin q (sec x1 sec x2 + sec x2 sec x3 + sec xn – 1 sec xn) is (a)

23. The greatest value of cos q for which cos 5q = 0 is

(a) tan (q/2) =

p-q p+q (b) tan (q/2) = – p +q p -q

(c) cot (q/2) =

p-q p+q (d) cot (q/2) = p +q p -q

28. If x and y are acute angles such that x + y and x – y satisfy the equation tan2 q – 4 tan q + 1 = 0 then (a) x = p/4 (b) x = p/6 (c) y = p/6 (d) y = p/4 29. A root of the equation 2x3 – 3x2 cos (A – B) – 2x cos2 (A + B) + sin 2A sin 2B cos (A – B) = 0 is (a) 2 cos A cos B (b) 2 sin A sin B (c) 2 sin A cos B (d) – (1/2) cos (A – B)

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30. Let x = (1 + sin A) (1 + sin B) (1 + sin C) y = (1 – sin A) (1 – sin B) (1 – sin C) and if x = y, then (a) x = cos A cos B cos C (b) y = sin A sin B sin C (c) y = cos A cos B cos C (d) x = – sin A sin B sin C 31. If sin q (1 + sin q) + cos q (1 + cos q) = x and sin q (1 – sin q) + cos q (1 – cos q) = y, then (a) x2 – 2x – sin 2q = 0 (b) y2 + 2y – sin 2q = 0 (c) xy = sin 2q (d) xy = sin q + cos q 32. If f(q) = 5 cos q + 3 cos (q + p/3) + 7 and a £ f(q) £ b, then (a) a = 0 (b) a = – 7 (c) b = 14 (d) b = 7 33. If 3 cos 2q + 4 sin 2q = 5 has a and b as its solutions, then (a) tan a + tan b = 1 (b) tan a + tan b = 4 (c) tan a tan b = 4 (d) tan a tan b = 1/4 34. The value of the determinant 1 cos (b – a ) cos (g – a ) cos (a – b ) 1 cos (g – b ) cos (a – g ) cos (b – g ) 1

sin 2 2 x + 4sin 4 x – 4sin 2 x cos 2 x 4 – sin 2 2 x – 4sin 2 x

( a + b )2

( a - b )2 + a 2 b 2 ( a + b )2

(d) sec a = 2

39. If x = tan (A + B) + tan (A – B) y = tan (A + B) tan (A – B) then 2sin 2 A (a) x = cos 2 A + sin 2 B (b) x =

sin 2 A cos A - sin 2 B 2

(c) y =

cos 2 B - cos 2 A cos 2 B + cos 2 A

(d) y =

cos 2 A - cos 2 B cos 2 A + cos 2 B

40. (a tan a + y cot a) (x cot a + y tan a) – 4xy cot2 2a (a) is equal to (x + y)2 (b) is independent of a (c) is equal to x cos a + y sina (d) is independent of x and y

MATRIX-MATCH TYPE QUESTIONS 41.

is (a) independent of a (b) independent of ‘b’ (c) constant (d) cos a + cos b + cos g 35. If

ab ( ab - 4)

(c) tan2a =

=

1 and 9

0 < x < p, then the value of x is (a) p/3 (b) p/6 (c) 2p/3 (d) 5p/6 2 2 36. If sin a, 3 cos a are the roots of the equation 4x2 – 3px + k = 0, then (a) p = 4 – (8/3) sin2 a (b) p = 4/3 (d) k = 3 sin 2a (c) k = 3 sin2 2a 37. If the angles of a triangle are in arithmetic progression such that sin (2A + B) = 1/2, then (a) A = 45° (b) C = 75° (c) sin (B + 2C) = – 1/2 (d) cos 2B = – 1/2 38. If tan q + tan f = a, cot q + cot f = b, q – f = a π 0, then (a) ab > 4 (b) ab < 4

Column 1 (a) (sin q + sec q)2 + (cosq + cosecq)2 2sin q cos q - cos q (b) 1 - sin q + sin 2 q - cos q

Column 2 (p) 2 secq + 5 sinq (q) cot q

(c) cosq (tanq + 2)(2 tanq + 1) (r) (1 + secq cosecq)2 Ê 1 + sin q ˆ (d) Á (secq – tanq) Ë cos q ˜¯ + tanq – 1 3 5 and cosf = 5 13 Column 1 a -b (a) sin2 2 a -b (b) cos2 2

(s) tan q

42. If cosa =

(c) tan2

a -b 2

(d) sin (a + b)

Column 2 64 (p) 65 56 (q) 65 (r)

1 65

(s)

1 64

IIT JEE eBooks: www.crackjee.xyz Trigonometry 10.45

43. If A lies in the second quadrant and 5 sin A – 4 = 0 Column 1

Column 2 31 25

(a) 3 tan A + 5 cos A + 10 sin A

(p) -

(b) sin 2A + cos 2A

(q)

44 125

(c) sin 3A

(r)

24 7

(d) tan 2A

(s) 1

44. Column 1

Column 2

(a) sin 36°

(p)

5-2 5

(b) tan 36°

(q)

6+ 2 4

(c) cos 15°

(r)

10 - 2 5 4

(s)

25 - 10 5 5

(d) tan 18° 45.

Column 1 Column 2 (a) sin 68° + sin 94° + sin 198° (p) cot 22° cot 25° cot 43° (b) tan 47° + tan 34° (q) 1 + 4 sin 22°sin 25° + tan 99° sin 43° (c) cos 44° + cos 50° (r) 4 sin 47° sin 34° + cos 86° sin 99° (d) cot 22° + cot 25° (s) tan 47° tan 34° + cot 43° tan 90° 46. Column 1 Column 2 (a) sin 3x = 4 sin x cos2x 5p – sin x (p) if x = 6 (b) sin 2x – cos 2x

( tan x + 1) - 1 (q) ( tan x + 1)2 - 2 tan x 2

=

(c) 2 sin3x + cos 2x 1 = 2

if x =

p 4

47.

(d) sin x + sin 2x + sin 3x + sin 4x 1+ 2 + 3 + 6 25p = (s) if x = 12 2 2

Column 1 (a) f(x) = sin2x + sin2 pˆ pˆ Ê Ê ÁË x + ˜¯ + cos x cos ÁË x + ˜¯ 3 3

p 12

(p) 3

q (1 + sec q) 2 (1 + sec2q) … (1 + sec2n q) (c) If cot (q – a), 3 cot q, cot (q + a) are in A.P. and q is p not an integral multiple of , 2 then sinq coseca is equal to (d) Number of ordered pairs (a, x) satisfying the equation sec2 (a + 2)x + a2 – 1 = 0 – p t3 > t4 (b) t4 > t3 > t1 > t2 (c) t3 > t1 > t2 > t4 (d) t2 > t3 > t1 > t4 [2006] 22. Let P = {q : sin q – cos q = {q : sin q + cos q =

2 cos q} and Q =

2 sin q} be two sets. Then

P ⊂ Q and Q – P π Δ QÀP PÀQ P=Q [2011] p p 23. Let - < q < - . Suppose a1 and b1 are the roots 6 12 of the equation x2 – 2x secq + 1 = 0 and a2, b2 are the roots of the equation x2 + 2x tanq – 1 = 0. If a1 > b1 and a2 > b2then a1 + b2 equals (a) 2(secq – tanq) (b) 2 secq (c) – 2 tanq (d) 0 [2016] (a) (b) (c) (d)

13

1 is Ê p ( k - 1) p ˆ k =1 Ê p kp ˆ sin Á + sin Á + Ë 4 6 ˜¯ 6 ˜¯ Ë4

24. The value of  equal to (a) 3 - 3 (c) 2

(

( ) 2 (2 + 3 )

(b) 2 3 - 3

)

3 -1

(d)

[2016]

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 1. If f(x) = cos [p2]x + cos [–p2]x where [x] stands for the greatest integer function then (a) f(p/2) = –1 (b) f(p) = 1 (c) f(– p) = 0 (d) f(p/4) = 1 [1991] 2. For a positive integer n, let fn (q) = (tan (q/2)) (1 + sec q) (1 + sec 2q) (1 + sec 4q) ... (1 + sec 2nq). Then, (a) f2(p/16) = 1 (b) f3(p/32) = 1 (d) f5(p/128) = 1 (c) f4(p/64) = 1 [1998] 4 4 sin x cos x 1 + = , then 3. If 2 3 5 2 2 (a) tan x = 3 sin 8 x cos8 x 1 + = 8 27 125 1 2 (c) tan x = 3

(b)

sin 8 x cos8 x 2 + = [2009] 8 27 125 4. Let f : (–1, 1) Æ R be such that f(cos 4q) = (d)

Ê pˆ for q ∈ Á 0, ˜ Ë 4¯ 2 - sec q 2

2

Êp pˆ ÁË , ˜¯ .Then the 4 2

Ê 1ˆ value(s) of f Á ˜ is (are) Ë 3¯ (a) 1 –

3 2

(b) 1 +

3 2

(c) 1 –

2 3

(d) 1 +

2 3 [2012]

INTEGER-ANSWER TYPE QUESTIONS 1. The maximum value of the expression 1 is 2 sin q + 3sin q cos q + 5cos 2 q

[2010] 2. The positive integer value of n > 3 satisfying the equation 1 1 1 = + Êpˆ Ê 2p ˆ Ê 3p ˆ sin Á ˜ sin Á ˜ sin Á ˜ Ë n¯ Ë n¯ Ë n¯ is [2011]

IIT JEE eBooks: www.crackjee.xyz Trigonometry 10.49

FILL

IN THE

BLANKS TYPE QUESTIONS

1. Suppose sin3x sin 3x =

n

 Cm cos (mx) is an

m=0

identity in x, where C0, C1, ..., Cn are constants and Cn π 0. Then the value of n is _______. [1981] 2. The value of sin

p 3p 5p 7p 9p 11p sin sin sin sin sin 14 14 14 14 14 14

13 p 14 is equal to _______.

[1983]

sin

[1991]

3. If k = sin (p/18) sin (5p/18) sin (7p/18), then the numerical value of k is _______. [1993] 4. If A > 0, B > 0 and A + B = p/3, then the maximum value of tan A tan B is _______. [1993] 5. For each natural number k, let Ck denote the circle with radius k centimeter and centre at the origin. On the circle Ck, a particle moves k centimeters in the counter clockwise direction. After completing its motion on Ck, the particle moves to Ck + 1, in the radial direction. The motion of the particle continues in this manner. The particle starts at (1, 0). If the particle crosses the positive direction of the x-axis Cn then n = _______. [1997]

TRUE/FALSE TYPE QUESTIONS 1 - cos B , sin B then tan 2A = tan B.

4. Prove that 5 cos q + 3 cos (q + p/3) + 3 lies between – 4 and 10. [1979] 5. Given a + b – g = p, prove that sin2 a + sin2 b – sin2 g = 2 sin a sin b cos g [1980] 6. Without using tables, prove that sin 12° sin 48° sin 54° = 1/8 [1982] 7. Show that Ê 2p ˆ Ê 4p ˆ Ê 8p ˆ Ê 16 p ˆ 16 cos Á cos Á cos Á ˜ cos Á =1 ˜ ˜ Ë 15 ¯ Ë 15 ¯ Ë 15 ¯ Ë 15 ˜¯ 8. Prove that tan a + 2 tan 2a + 4 tan 4a + 8 cot 8a = cot a [1988]

Answers LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17. 21. 25. 29.

[1983]

1. Prove that a = sin x sin y sin (x – y) + sin y sin z sin (y – z) + sin z sin x sin (z – x) + sin (x – y) sin (y – z) sin (z – x) = 0 [1978] 2. If tan a =

1 m and tan b = show that m +1 2m +1

a + b = p/4 [1978] 3. If cos (a + b) = 4/5, sin (a – b) = 5/13, and a, b lie between 0 and p a. [1978]

2. 6. 10. 14. 18. 22. 26. 30.

(c) (c) (b) (c) (c) (a) (d) (b)

3. 7. 11. 15. 19. 23. 27.

(c) (c) (b) (b) (c) (b) (b)

4. 8. 12. 16. 20. 24. 28.

(c) (b) (d) (d) (b) (c) (c)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

1. If tan A =

SUBJECTIVE-TYPE QUESTIONS

(d) (a) (b) (b) (c) (a) (b) (a)

31. 33. 35. 37. 39. 41. 43. 45. 47. 49.

(a), (c), (a), (a), (b), (a), (a), (a), (a), (a),

(b), (d) (b), (d) (c), (b) (b), (c) (b), (c)

(c), (d) (d) (d) (c), (d) (c)

32. 34. 36. 38. 40. 42. 44. 46. 48. 50.

(a), (a), (a), (a), (b), (a), (b) (a), (b), (b),

(b), (b), (c) (b), (d) (c),

(c), (d) (c), (d) (c) (d)

(b), (c) (d) (c)

IIT JEE eBooks: www.crackjee.xyz 10.50 Comprehensive Mathematics—JEE Advanced

MATRIX-MATCH TYPE QUESTIONS p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

51.

52.

53.

54.

55.

71. 3 75. 5

57. (c)

62. (b) 66. (a) 70. (d)

73. 8

74. 7

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17. 21. 25.

(a) (c) (b) (b) (a) (c) (c)

2. 6. 10. 14. 18. 22.

(b) (a) (c) (a) (c) (b)

3. 7. 11. 15. 19. 23.

(c) (b) (c) (a) (b) (c)

4. 8. 12. 16. 20. 24.

(a) (b) (b) (c) (c) (d)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 26. 28. 30. 32. 34. 36. 38. 40.

(a), (a), (a), (a), (a), (a), (a), (a),

(c) (c) (c) (c) (b), (c) (c) (c), (d) (b)

27. 29. 31. 33. 35. 37. 39.

(a), (a), (a), (a), (b), (a), (a),

(b), (b), (b), (d) (d) (b), (b),

(d) (d) (c)

(c), (d) (c)

MATRIX-MATCH TYPE QUESTIONS p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

41.

58. (a)

59. (d)

COMPREHENSION-TYPE QUESTIONS 61. (a) 65. (b) 69. (b)

72. 2

LEVEL 2

ASSERTION-REASON TYPE QUESTIONS 56. (d) 60. (b)

INTEGER-ANSWER TYPE QUESTIONS

63. (b) 67. (c)

64. (c) 68. (c)

42.

IIT JEE eBooks: www.crackjee.xyz Trigonometry 10.51

p

q

r

s

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

43. a

44.

45.

46.

47.

INTEGER-ANSWER TYPE QUESTIONS 61. 5 65. 5

62. 1

63. 3

64. 6

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17. 21.

(b) (b) (a) (b) (c) (b)

2. 6. 10. 14. 18. 22.

(a) (d) (c) (c) (a) (d)

3. 7. 11. 15. 19. 23.

(b) (c) (b) (c) (a) (c)

4. 8. 12. 16. 20. 24.

(c) (b) (d) (a) (b) (c)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 1. (a), (c) 3. (a), (b)

2. (a), (b), (c), (d) 4. (a), (b)

INTEGER ANSWER TYPE QUESTIONS 1. 2

FILL

2. 7 IN THE

1. 6 5. 7

BLANKS TYPE QUESTIONS 2. 1/64

3. 1/8

4. 1/3

TRUE FALSE TYPE QUESTIONS 1. True

SUBJECTIVE TYPE QUESTIONS 3. 56/33

Hints and Solutions ASSERTION-REASON TYPE QUESTIONS 48. (a)

49. (d)

COMPREHENSION-TYPE QUESTIONS 51. 53. 55. 57. 59.

(a) (b), (c), (d) (a), (b) (a) (a)

LEVEL 1

50. (b)

52. 54. 56. 58. 60.

(a), (b) (b) (b), (c) (b) (a)

1. Given expression is equal to cos y sin x – sin y cos x +sin y sin x + cos x cos y = sin (x – y) + cos (x – y) = if

Èp ˘ 2 sin Í + ( x - y )˙ = 0 Î4 ˚ p + (x – y) = np for some nŒI 4

IIT JEE eBooks: www.crackjee.xyz 10.52 Comprehensive Mathematics—JEE Advanced

3p for n = 1 4 We have cos2A = sin A tan A fi cos3A = sin2A fi cot2A = sec A fi cot4A = 1 + tan2A fi cot6A – cot2A = 1 Value of the sum is greatest when x1 = x2 … = xn = p p and the required value = n sin . n n Given relation can be written as sin x(1 + sin2x) = 1 – sin2x = cos2x fi sin x(2 – cos2x) = cos2x fi sin2x(2 – cos2x)2 = cos4x fi (1 – cos2x) (4 – 4 cos2x + cos4x) = cos4x fi cos6x – 4 cos4x + 8 cos2x = 4 a a sin 2 + cos 2 a a 2 2 = 2 = 2 tan + cot = a a 2 2 sin a p sin cos 2 2 a a tan cot = 1 2 2 So the required equation is 2 x2 – x + 1 = 0 p

n tan a 1 – n + tan 2 a n tan 2 a 1– 1 – n + tan 2 a

tan a +

fix–y=

2.

3.

4.

5.

or px2 – 2x + p = 0 6. 6[(sin2 q + cos2 q)2 – 3 sin2 q cos2 q] – 9 [(sin2 q + cos2 q)2 – 2 sin2 q cos2 q] = 6 – 9 = – 3. 7. 1 + tan2 a tan2 b + tan2 a + tan2 b = (1 + tan2 a) (1 + tan2 b) = sec2 a sec2 b 5/6 + 1/11 8. tan (a + b) = = 1 fi a + b = p/4 1 – (5/6) (1/11) 2 + 2.2 cos 2 2q =

9.

2 + 2 cos 2q = 2 cos q.

10. tan (x + y) = tan (p/6) = 1

(

fi tan x + tan y = 1

3

)

3 (1 – a) and tan x tan y = a

The required equation is 1 (1 – a)x + a = 0 x2 – 3 =0 or 3 x2 – (1 – a)x + a sin 7 x + sin 5 x + 5sin 5 x + 5sin 3x + 12sin 3x + 12sin x 11. sin 6 x + 5sin 4 x + 12sin 2 x =

2 cos x[sin 6 x + 5sin 4 x + 12sin 2 x] = 2 cos x sin 6 x + 5sin 4 x + 12 sin 2 x

12. tan (a + b) =

tan a (1 + tan 2 a )

=

13.

2 tan (a /2) 1 + tan (a /2) 2

fi 0

(

+

(1 – n) (1 + tan a ) 2

1 – tan 2 (a /2) 1 + tan (a /2) 2

)

=

2

(

4±2 7+2

fi tan (a/2) = (1 3)

(

tan a . n –1

7 2

7 + 2 tan2 (a/2) – 4 tan (a/2) +

fi tan (a/2) =

= –

(

)

7–2 =

) )

7 – 2 as 0 < a < p/6.

14. Let sin (a/4) = x fi cos a = 1 – 8x2 + 8x4 = Ê 336 ˆ – 1– Á Ë 625 ˜¯

2

=–

527 625 (

a lies in the II quadrant)

1 (1 ± 7/25) 2 fi x2 = 16/25, 9/25 fi sin (a/4) = 4/5 or 3/5 sin 135° < sin (a/4) < sin 112.5° as 112.5° < a/4 < 135° fi x2 =

fi sin (a/4) > 1

2 fi sin (a/4) > 0.7. Hence

sin (a/4) = 4/5 = 0.8. 15. cos (a – b) = [cos (a + q)– – (b + q)] = 1 – a 2 1 – b 2 + ab so 2 cos2 (a – b) – 1 – 4ab cos (a – b) = 2 [(1 – a2) (1 – b2) – a2b2] – 1 = 1 – 2a2 – 2b2 16. sin2 a + cosec2 a + 2 + cos2 a + sec2 a + 2 – tan2 a – cot2 a = 4 + sin2 a + cos2 a + sec2 a – tan2 a + cosec2 a – cot2 a = 7 17. 3 [cos4 a + sin4 a] – 2 [cos6 a + sin6 a] = 3 [1 – 2 sin2 a cos2 a] – 2 [1 – 3 sin2 a cos2 a] =1 18. sin x cos 3a + cos x sin 3a = 3 (sin a cos x – cos a sin x) fi sin x (cos 3a + 3 cos a) + cos x (sin 3a – 3 sin a) = 0 fi sin x (4 cos3a) + cos x (– 4 sin3 a) = 0 fi tan x = tan3 a.

IIT JEE eBooks: www.crackjee.xyz Trigonometry 10.53

19. m =

cos (q – a ) + sin (q + a ) = cos (q – a ) – sin (q + a )

29.

(cos q + sin q ) (cos a + sin a ) (cos q – sin q ) (cos a – sin a ) =

1 + tan q 1 + tan a ◊ = tan (p/4 + q) tan (p/4 + a). 1 – tan q 1 – tan a

20. tan 270° =

tan130∞ + tan140∞ 1 – tan130∞ tan140∞

fi tan 130° tan 140° = 1. 21. cos 1° cos 2° ... cos 90° cos 91° ... cos 179° = 0 as cos 90° = 0 È (1 + tan x) 2 ˘ 22. 1 + tan x = (1 – tan x) Í ˙ 2 Î 1 + tan x ˚ fi 2 tan2 x [1 + tan x] = 0 fi tan x = – 1 as –p/4 £ x < p/4 23. tan (a – b) =

2m 2 – 2m + 1 2m 2 – 2m + 1

= 1 fi a – b = p/4.

24. m [sin (2q + 90°) + sin (– 150°)] = n [sin (2q + 90°) + sin 150°] fi

m cos 2q + 1 / 2 m–n 1 = fi = . cos 2q – 1 / 2 2 cos 2q n m+n

25. (cos x + sin x)2 = 2 cos2 x fi 1 + tan2 x + 2 tan x = 2 fi tan2 x + 2 tan x = 1. 26. sin8 q £ sin2 q, cos14 q £ cos2 q fi sin8 q + cos14 q £ sin2 q + cos2 q = 1

30.

1 + tan q1 tan q 2 1 – tan q3 tan q 4 + =0 1 – tan q1 tan q 2 1 + tan q3 tan q 4 fi 2 + 2 tan q1, tan q2 tan q3 tan q4 = 0 1 – tan 2 b 1 + tan b 2

27.

cos(a / 2) – sin(a / 2) cos(a / 2) + sin(a / 2) + cos(a / 2) + sin(a / 2) cos(a / 2) – sin(a / 2) =

(

2 cos 2 (a / 2) + sin 2 (a / 2) cos 2 (a / 2) – sin 2 (a / 2)

)

2 –2 = = cosa cosa

cos a + cos b =

4ax

sin a + sin b =

4ay

sin q + cos q +

1+ x ,

1 + sin q + cos q =7 sin q cos q

1 + x = 7x – 2

fi (x + 2)

fi x – 44x + 36x = 0 fi x2 – 44x + 36 = 0 as x π 0. 3

2

and cos a cos b =

4a 2 - y 2

x2 + y 2 x2 + y 2 similarly writing the given relation as a quadratic in sin q, we have

32. y =

x2 + y 2

and sin a sin b =

4a 2 - x 2 x2 + y 2

(cos 2 A - sin 2 A)2 + 1 (cos 2 A + sin 2 A)2 - 1

fiy=

± ( cos 2 A - sin 2 A) + 1 ± ( cos 2 A + sin 2 A) - 1

Let y1 = = =

cos 2 A - sin 2 A + 1 cos 2 A + sin 2 A - 1 2 cos 2 A - 2sin A cos A -2sin 2 A + 2sin A cos A 2 cos A ( cos A - sin A) = cot A 2sin A ( cos A - sin A)

33. Clearly tan q π 0, tan 2q π for 0 < q
0 as A = 0 if sin q = cos q = 0 which is not possible.

=

IIT JEE eBooks: www.crackjee.xyz 10.54 Comprehensive Mathematics—JEE Advanced

35.

sin 6q cos 6q = = p q

1

x2y =

p2 + q2 p2 + q2

p sin 6q + q cos 6q =

p2 + q2

=

p2 + q2

1 (p cosec 2q – q sec 2q) 2 =

1 È p cos 2q - q sin 2q ˘ 2 ÍÎ sin 2q cos 2q ˙˚

=

1 È cos 2q sin 6q - sin 2q cos 6q ˘ p2 + q2 Í ˙ 2 sin 2q cos 2q Î ˚

=

sin 4q = sin 4q

p2 + q2

p2 + q2

p2 - q2

p sin 6q – q cos 6q =

=

pq

p2 + q2

=

fi tan2

sin 2 (q /2) cos (q /2) 2

– (cos 2 (q /2) ˘ ˙ – sin 2 (q /2)) ˙ ˙ + (cos 2 (q /2) ˙ – sin 2 (q /2)) ˙˚ =

2

1 – cos q 1 + cos q

2

È x 2 + 1˘ 2 ( x 4 + 1) 37. sec q = 1 + Í 2 ˙ = 2 ( x – 1)2 Î x – 1˚ 2

x2 – 1 2( x + 1) 4

x2 + 1 2( x 4 + 1)

38. cot q + tan q = x fi =

p 2 + p 2 (q – 1) + q (q – 1)2 (q – 1)2 + p 2

= q.

42. a2 + b2 = 2 + 2 cos (q – f) = 4 cos2 ((q – f)/2).

È (cos(q /2) + sin(q /2)) 2 Í Í 36. Í 2 Í (cos(q /2) + sin(q /2)) Í Î

sin q =

1 + tan 2 (a + b )

p2 + q2

p2 + q2

fi cos q =

tan 2 (a + b ) + p tan(a + b ) + q

=

2 pq

=

sin 2 q = sec3 q, xy2 = tan3 q cos q

39. cos a = 3/5 fi sin a = 4/5; cos b = 5/13 fi sin b = 12/13 sin (a + b) = 56/65, cos (a – b) = 63/65, sin2 [(a – b)/2] = 1/65. 40. a2 = 1 – 3 sin2 x cos2 x = 1 – 3 sin2 x + 3 sin4 x fi 3 sin4 x – 3 sin2 x + 1 – a2 = 0, for real solution 9 – 12 (1 – a2) ≥ 0 fi a2 ≥ 1/4 and 1 – a2 = 3 sin2 x cos2 x ≥ 0 fi a2 £ 1 – 1 £ a £ – 1/2, 1/2 £ a £ 1 41. tan a + tan b = – p, tan a tan b = q fi tan (a + b) = p/(q – 1) sin2 (a + b) + p sin (a + b) cos (a + b) + q cos2 (a + b)

qp

+

p2 + q2

sin 2 q cos 2 q

¥

fi (x2y)2/3 – (xy2)2/3 = 1

and p cos 6q + q sin 6q =

1

.

43.

q –f 4 – a 2 – b2 . = 2 a 2 + b2

sin a – cos a = tan q sin a + cos a fi cos2 q = sin2 q =

sin 2 q = sin q tan q cos q

1 + tan 2 q

=

(sin a + cos a ) 2 2

(sin a – cos a ) 2 2

cos 2q = cos2 q – sin2 q = sin 2a, sin 2q = – cos 2a. 44. |sin q| < 1 so sin2 q π 5/4, cos q =

2ab a + b2

cosec q = 1 1 – cos 2 q = x, y = sin q cos q cos q

1

2

£ 1 fi (a – b)2 ≥ 0 which is true.

m2 – n2 m2 + n2

< 1 which is not possible and

sin q = 2.375 > 1 again not possible. 45. sin A = – 7/25, cos A = 24/25 fi sin 2A = –

336 625

cos2 (A/2) = 49/50, sin2(A/2) = 1/50, fi sin (A/2) = +

2 /10, tan (A/2) = – 1/7, as 135° < A/2 < 180°.

IIT JEE eBooks: www.crackjee.xyz Trigonometry 10.55

46.

cos A cos B = y x then

5 +1 5 –1 1 – = 4 4 2

51. cos 36° – cos 72° =

A+ B x tan A + y tan B sin A + sin B = = tan x+ y cos A + cos B 2

cos 36° cos 72° = 1/4 tan 36° tan 18° =

x tan A – y tan B sin A – sin B A–B = = tan x+ y cos A + cos B 2

sin 36° cos 18° =

y sin A + x sin B sin A cos B + cos A sin B = y sin A – x sin B sin A cos B – cos A sin B =

sin( A + B) sin( A – B)

47. S tan2 b tan2 g = 1 – 2 tan2 a tan2 b tan2 g. sin2 a + sin2 b + sin2 g =

tan 2 a 1 + tan 2 a

+

tan 2 b 1 + tan 2 b

53. cosec 2q + cot 2q cos 2q =

=

=

=

1 + S tan 2 a + S tan 2 b tan 2 g + tan 2 a tan 2 b tan 2 g 2

2

2

2 + S tan 2 a – tan 2 a tan 2 b tan 2 g

(

)

49. a – 5 £ (5 + 3/2) cos q + 3 3 / 2 sin q £ b – 5. Ê 3 3ˆ Ê 13 ˆ a–5= – Á ˜ +Á ˜ Ë 2¯ Ë 2 ¯ 2

and b – 5 =

(

Ê 1ˆ sin 2q – ÁË ˜¯ (1 + cos 4q) = sin 2q – cos2 2q 2 = sin 2q – 1 + sin2 2 q = k2 + k – 1 sin 6q = 3 sin 2q – 4 sin3 2q = 3k – 4k3.

=1

fi cos2 a + cos2 b + cos2 g = 2 cos 2a + cos 2b + cos 2 g = 1 and cos (a + b) cos (a – b) = – cos2 g fi cos2 a – sin2 b = – cos2 g fi cos2 a + cos2 b + cos2 g = 1 which is not correct. 48. sin (p cos q) = cos (p sin q) = sin (p/2 ± p sin q) fi p (cos q ± sin q) = p/2 fi 1 ± sin 2q = 1/4 fi sin 2q = ± 3/4.

2 –k 2 k

2

2

1 + S tan 2 a + 1 - tan 2 a tan 2 b tan 2 g 2 + S tan 2 a – tan 2 a tan 2 b tan 2 g

1 + cos 2 2q 2 – sin 2 2q = sin 2q sin 2q

Ê 1 + tan q ˆ 1 + sin 2q 1+ k ÁË 1 – tan q ˜¯ = 1 – sin 2q = 1 – k .

S tan a + 2 S tan b tan g + 3tan a tan b tan g 2

5

x2 – 1 = tan 2q. xy

tan 2 a (1 + tan 2 b + tan 2 g + tan 2 b tan 2 g ) 2

5–2 5

10 – 2 5 10 + 2 5 5 ¥ = . 4 4 4

1 + tan 2 g =Â

5

=

52. cos 2q = cos2 q – sin2 q = xy sin 2q = 1 – (cos q – sin q)2 = 1 – y2 = x2 – 1 2 cos q x+ y cot q = = 2sin q x– y

+

tan 2 g

5–2 5

5–2 5 ¥

54.

1 – 2sin( x /2) cos( x /2) cos 2 ( x /2) – sin 2 ( x /2)

=

(cos x /2 – sin x /2) 2 cos 2 ( x /2) – sin 2 ( x /2) =

=

1 – tan ( x /2) 1 + tan( x /2)

(cos x /2 – sin x /2) cos( x /2) + sin( x /2)

1 + sin x 1 + tan ( x /2) = 1 – tan ( x /2) cos x

2

(13 / 2)2 + 3 3 / 2

1 – cos x 2sin 2 ( x /2) = = tan2 (x/2) 1 + cos x 2 cos 2 ( x /2)

)

2

fi a = – 2, b = 12. 50. sec2 a – sec2 b – (1/2) sin (a – b) sec2 a sec2 b = 0 fi cos2 b – cos2 a – (1/2) sin (a – b) = 0 fi sin (a + b) sin (a – b) – (1/2) sin (a – b) = 0 fi Either sin (a – b) = 0 or sin (a + b) = 1/2.

cos x =

cos 2 ( x /2) – sin 2 ( x /2) cos 2 ( x /2) + sin 2 ( x /2)

=

1 – tan 2 ( x /2) 1 + tan 2 ( x /2)

.

55. 13p/6 = 2p + p/6, 25p/6 = 4p + p/6, 37p/6 = 6p + p/6 So for all values q in p, q, r and s sin q = sin p/6 = 1/2, sin 2q = sin p/3 =

3 2

IIT JEE eBooks: www.crackjee.xyz 10.56 Comprehensive Mathematics—JEE Advanced

cos q = tan 2q =

3 2 , cos 2q = 1/2, tan q = 1 3 cot q = 3 , cot 2q = 1 3

3,

tan B + tan C tan B + tan C = 1 - tan A tan B tan B tan C - 1

( )

2

sin q cos q 2

=

1 sin q 2

+

1 cos 2 q

= x + y.

1ˆ 4( z – 1) ˜¯ = z z

cos 2q = 2 cos2 q – 1 =

fi Statement 2 is true Since A is obtuse tan A < 0 and tan B, tan C > 0 fi tan B tan C < 1 fi Statement-1 is false. 57. sin (A + B) + sin (A – B) = 2 sin A cos B cos (A + B) + cos (A – B) = 2 cos A cos B Thus statement-1 is true and statement-2 is false 58. x2 + y2 = 27 = 3 3 y = 3 3 sin q

1 2

Ê 66. sin2 2q = 4 Á1 – Ë

56. tan A = – tan (B + C) in any triangle ABC =–

as xy =

2– y y = 1 – 2 sin2 q =

tan2 2q =

67. tan 3q =

fi x = 3 3 cos q,

pˆ Ê and x – y = 3 3 (cos q – sin q) = 3 3 ¥ 2 cos Á q + ˜ Ë 4¯ Applying statement-2 which is true

4( z – 1) y 2 z (2 – y ) 2

or

4( z – 1) x 2 z ( x – 2)2

x–2 x

.

È 3 – 4sin 2 q ˘ = tan q Í ˙ 2 4 cos3 q – 3cos q Î 4 cos q – 3 ˚ 3sin q – 4sin 3 q

4 3– tan 3q x = (3x – 4) y fi = 4 (4 – 3 y ) x tan q –3 y 68.

x – y £ 3 6 and statement-1 is also true 59.

(

p+ q+ r

)

2

=p+q+r+2

(

pq + qr + rp

)

Statement-2 is true fi p + q + r = 16 So

(

p + r +r

)

2

> 16 and statement-1 is False.

60. x sin q sin2 q + y cos q cos2q = sin q cos q fi x sin q = sin q cos q fi x = cos q and y = sin q fi x2 + y2 = 1, so statement 1 is true Statement-2 is also true (see text) but does not imply statement-1 61. Since a, b, g, d are in A.P., a + d = b + g. 62. Let b = a + r, g = a + 2r, d = a + 3r fi a + b + g + d = 4a + 6r 3b + d = 3a + 3r + a + 3r = 4a + 6r a + 3g = a + 3a + 6r = 4a + 6r a + 3d = 4a + 9r, 3b + g = 4a + 5r. a +g . 63. a + g – d π b, a + d – g = b = 2g – d = 2 64.

1 x

2

+

1 y2

= sin4 q + cos4 q = 1 – 2 sin2 q cos2 q

1ˆ 2–z Ê . = 1 – 2 Á1 – ˜ = Ë z¯ z 65. z =

xy fi xyz = xy + z = x + y + z xy – 1

Fig. 10.7

z cot 2q – z cot 3q BD – BE DE = = z cot q BC BC = 69.

CE – CD y cot q – x cot q y–x = = BC z cot q z

cot q – cot 2q z cot q – z cot 2q BC – BD = = = cot q z cot q BC

CD x = . BC z 70. AD = CD = x cot q. 71. cos A = 3/4 fi cos2 (A/2) = 7/8, cos 2A = 1/8, cos 3A = –9/16 6 [cos2 (A/2) – (cos 2A – cos 3A)] 3 È7 1 9 ˘ = 16 Í – – ˙ = 16 ¥ = 3. 16 Î 8 8 16 ˚ a+b ˆ 3 a+b a –b 72. 2 cos – ÊÁ 2 cos 2 – 1˜ = cos Ë ¯ 2 2 2 2

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fi 4 cos2 fi

a+b a+b a –b – 4 cos cos +1=0 2 2 2

2 È2 cos a + b – cos a – b ˘ + sin2 a – b = 0 ÍÎ 2 2 ˙˚ 2

a+b a –b a –b = 0 and 2 cos = cos 2 2 2 fi a = b and 2 cos a = 1 fi a = p /3 (0 < a, b < p ) fi

74.

3 = tan

sin

so sin a = and

73. sin a = tan (p /4) = 1 and cot b = 2 as 4x2 — 16x + 15 < 0 fi (2x – 4)2 < 1 fi x = 2. So 4 = 8. 10 [sin2 a – sin2 b ] = 10 ¥ 5

3 2 , cos a = 1/2

3 sin a + cos a = 2.



p 3 tan(p / 9) – tan 3 (p / 9) = 3 1 – 3tan 2 (p / 9)

3 (1 – 3 tan2 (p/9))2 = tan2 (p /9) [(3 – tan2 (p /9))2] 6 4 fi tan (p /9) – 33 tan (p /9) + 27 tan2 (p /9) = 3 fi 3k = 21 fi k = 37. 75. cos x = cos 64° = cos 296° fi a = 64°, b = 296° a b + =5 so 64 74

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11 Trigonometric Equations 11.1 GENERAL SOLUTIONS OF TRIGONOMETRICAL EQUATIONS

The following formulae are used in solving trigonometrical equations: 1. 2. 3. 4.

If sin q = sin a, then q = np + (– 1)n a (n ∈ I) If cos q = cos a, then q = 2np ± a (n ∈ I) If tan q = tan a, then q = np + a (n ∈ I) If sin q = sin a, cos q = cos a, then q = 2np + a (n ∈ I)

Illustration 1

Illustration 3 If tan q = – 1, q lies in 2nd or 4th quadrant. For 2nd quadrant we will select anticlockwise and for 4th quadrant we will select clockwise direction. So we get two values 3p/4 and – p/4, of which – p/4 is numerically least angle. Hence principal value is – p/4. Similarly for sin q = – 1/2, the principal value of q is –p/6 and for cos q = 1/2, the principal value of q is p/3. [Note cos q is positive in 1st and 4th quadrant, so we get two values – p/3 and p/3 of q satisfying the relation. In such case, we select the angle with positive sign as principal value].

If sin q = 1/2, then sin q = sin (p/6) so that q = np + (– 1)n p/6 (n ∈ I) In solving trigonometrical equations, the general values of the angle should be given, unless the solution is required

11.2 PRINCIPAL VALUE

Since all the trigonometrical ratios are periodic functions, an equation of the form sin q = k, cos q = k or tan q = k etc. of all such values is called general value of q. Numerically least value in this set is called Principal Value.

SOLVED EXAMPLES

SINGLE CORRECT ANSWER TYPE QUESTIONS The complete solution of the equation Example 1 2 7 cos x + sin x cos x – 3 = 0 is given by (a) np + p 2 (n ∈ I) (b) np – p 4 (n ∈ I) (c) np + tan–1 ( 4 3) (n ∈ I)

(d) np + 3p/4, kp + tan–1 ( 4 3) (n, k ∈ I) Ans. (d)

Illustration 2 If sin q = 1/ 2 , the general value of q is np + (– 1)n p/4 and the principal value is p/4. Method for Finding Principal Value

(i) First note the quadrant in which the angle lies. (ii) For 1st and 2nd quadrants consider anticlockwise direction and for 3rd and 4th clockwise direction. (iii) Find the angles in the Ist rotation. (iv) Select the numerically least angle among these two values. The angle thus found will be the principal value.

Solution: The given equation is equivalent to 7 cos2 x + sin x cos x – 3 (sin2 x + cos2 x) = 0 or 3 sin2 x – sin x cos x – 4 cos2 x = 0 Since cos x = 0 does not satisfy the given equation we divide throughout by cos2 x and get 3 tan2 x – tan x – 4 = 0 fi (3 tan x – 4) (tan x + 1) = 0 fi tan x = 4 3 or tan x = –1. That is, x = kp + tan–1 ( 4 3 ) or x = np + 3p 4 (n, k ∈ I). It should be noted that the solutions given by (b) or (c) are not complete; the general solution is given by (d) only.

IIT JEE eBooks: www.crackjee.xyz 11.2 Comprehensive Mathematics—JEE Advanced

Example 2

The number of solutions of the equation sin 5x cos 3x = sin 6x cos 2x

in the interval [0, p] are (a) 3 (b) 4 (c) 5 (d) 6. Ans. (c) Solution: The given equation can be written as fi (1/2) (sin 8x + sin 2x) = (1/2) (sin 8x + sin 4x) fi sin 2x – sin 4x = 0 fi – 2 sin x cos 3x = 0 fi sin x = 0 or cos 3x = 0. That is, x = np (n ∈ I), or 3x = kp + p 2 (k ∈ I). Therefore, since x ∈ [0, p], the given x = 0, p, p 6, p 2 or 5p 6 . Number of solutions of the equation tan x + sec x = 2 cos x lying in the interval [0, 2p] is (a) 0 (b) 1 (c) 2 (d) 3 Ans. (c) Solution: The given equation can be written as Example 3

1 + sin x = 2 cos x cos x fi 1 + sin x 2 fi 2 sin x + sin x – 1 fi (1 + sin x)(2 sin x – 1) fi sin x

= 2 cos2 x = 2(1 – sin2 x) =0 =0 = – 1 or 1 2

\ The required number of solutions is 2. The value of the determinant

1 a a2 cos (n - 1) x cos n x cos (n + 1) x sin (n - 1) x sin n x sin (n + 1) x is zero if (a) x = np

(b) x = np/2

(c) x = (2n + 1) p/2

1 + a2 n∈I (d) x = 2a

Ans. (a) Solution: Applying C1 Æ C1 – 2 cos x C2 + C3 to the given determinant, we get 1 - 2a cos x + a 0 0

2

if

2

a a cos n x cos (n + 1) x sin n x sin (n + 1) x

x = np, n ∈ I.

i.e. if

Example 5

sin 3a < 0 if a lies in cos 2a

(a) (13p/48, 14p/48) (b) (14p/48, 18p/48) (c) (18p/48, 23p/48) (d) any of these intervals Ans. (a) Solution:

sin 3a 0 and cos2a < 0 or sin3a < 0 and cos2a > 0

i.e. i.e. i.e.

Now sin x = –1 fi x = 3p 2 for which the given equation is not meaningful. and sin x = 1/2 fi x = p 6 or 5p 6 . Example 4

= (1 – 2 a cos x + a2) sin x = 0 sin x = 0 or cos x = (1 + a2)/2a

if

3a ∈ (0, p) and 2a ∈ (p/2, 3p/2)

or

3a ∈ (p, 2p) and 2a ∈ (–p/2, p/2)

if

a ∈ (0, p/3) and a ∈ (p/4, 3p/4)

or

a ∈ (p/3, 2p/3) and a ∈(–p/4, p/4)

if

a ∈ (p/4, p/3)

since (13p/48, 14p/48) Ã (p/4, p/3), (a) is correct sin 2 q Example 6

cos 2 q

x

x

sin 2 q

sin 2 q

cos 2 q

2 If f(x) = cos q

x

q ∈ (0, p/2) then roots of f(x) = 0 are (a) 1/2, – 1

(b) 1/2, – 1, 0

(c) – 1/2, 1, 0

(d) – 1/2, – 1, 0

Ans. (a) 1 cos 2 q Solution: f(x) = (sin2 q + cos2 q + x) 1

x

1 sin 2 q 1 = (x + 1) 0

cos 2 q

x

x - cos 2 q

sin 2 q - x

0 sin 2 q - cos 2 q

x sin 2 q cos 2 q

cos 2 q - x

= (x + 1) [(x – cos2 q) (cos2 q – x) – (sin2 q – cos2 q) (sin2 q – x)] = (x + 1) [– x2 – cos4 q + 2x cos2 q – x cos2 q + x sin2 q – sin4 q + sin2 q cos2 q] = (– 1/2) (x + 1) [(x – sin2 q)2 + (x – cos2 q)2 + (sin2 q – cos2 q)2]

IIT JEE eBooks: www.crackjee.xyz Trigonometric Equations 11.3

So f(x) = 0 if x = – 1 or x = sin2 q = cos2 q sin2 q = cos2 q fi q = p/4 fi x = 1/2 Hence x = – 1, 1/2. Example 7

So the required values of x are p/8, 3p/8, 5p/8, 7p/8 which form an A.P. with common difference p/4.

The arithmetic mean of the roots of the

equation 4 cos3 x – 4 cos2 x – cos (p + x) – 1 = 0 in the interval (0, 315) is equal to (a) 49 p (b) 50 p (c) 51 p (d) 100 p

The equation (cos p – 1)x2 + (cos p)x

Example 10

Ans. (c)

+ sin p = 0 where x is a variable, has real roots if p lies in the interval (a) (0, 2p) (b) (– p, 0) (c) (– p/2, p/2) (d) (0, p)

Solution: The given equation can be written as

Ans. (d)

(4 cos2 x + 1) (cos x – 1) = 0 fi cos x = 1 fi x = 2p, 4p, 100 p as 100 p < 315 < 101 p So the required arithmetic mean 2 (p + 2p + + 50p ) = 50

Solution: The equation has real roots if cos2 p – 4(cos p – 1) sin p ≥ 0 or cos2 p – 4 cos p sin p + 4 sin p ≥ 0 or (cos p – 2 sin p)2 – 4 sin2 p + 4 sin p ≥ 0 or (cos p – 2 sin p)2 + 4 sin p (1 – sin p) ≥ 0 since (cos p – 2 sin p)2 ≥ 0, 1 – sin p ≥ 0 for all values of p, and for p ∈ (0, p), sin p ≥ 0 so that the discriminant is non-negative.

= Example 8

2 ¥ 50 ¥ 51 p = 51p. 2 ¥ 50

(

)

2 If 2 sin2 (p /2) cos x = 1 – cos (p sin 2x),

x π (2n + 1) p/2, n ∈ I, then cos 2x is equal to (a) 1/5

(b) 3/5

(c) 4/5

(d) 1

Example 11 The solution of (secq + 1) = (2 + 3 )tanq (0 < q < 2p) are (a) p/6, p (b) p/3, p/4 (c) p/6, 2p/3 (d) none of these Ans. (a) Solution:

Ans. (b)



Solution: The given equation is equivalent to



2 sin2 ((p/2) cos2 x) = 2 sin2 ((p/2) sin 2x) fi fi fi fi

cos2 x = sin 2x cos x (cos x – 2 sin x) = 0 1 – 2 tan x = 0 as cos x π 0, x π (2n + 1)p/2 tan x = 1/2 1 - tan 2 x

3 = . cos 2x = 2 5 1 + tan x



Example 9

The values of x between 0 and 2p which

2 satisfy the equation sin x 8cos x = 1 are in A.P. The common difference of the A.P. is (a) p/8 (b) p/4 (c) 3p/8 (d) 5p/8

Ans. (b) Solution:

From the given equation we have

2 sin x |cos x| = 1/ 2 fi sin 2x = 1/ 2 if cos x > 0 and sin2 x = –1/ 2 if cos x < 0 \ When cos x > 0, sin 2x = 1/ 2 fi x = p/8, 3p/8 when cos x < 0, sin 2x = – 1/ 2 fi x = 5p/8, 7p/8.

fi fi or fi

(secq + 1) = (2 +

3 )tanq

(1 + cosq) = (2 +

3 )sinq

2

2cos (q/2) = (2 +

3 ) ¥ 2sin(q/2)cos(q/2)

cos(q/2)[cos(q/2) – (2 + 3 )sin(q/2)] = 0 Either cos(q/2) = 0 fi q = (2n + 1)p, n ∈ I …(i) cosq – (2 + 3 )sin(q/2) = 0 tan(q/2) =

1 2+ 3

=2–

Êp ˆ 3 = tan Á ˜ Ë 12 ¯

q p = mp + fi q = 2mp + p/6, 2 12 m∈I …(ii) Since 0 < q < 2p, From (i) q = p and From (ii) q = p/6



Example 12 The smallest positive number p for which the equation cos(p sinx) = sin(p cosx) has a solution x ∈ [0, 2p] (a) 2 (c) p/2 Ans. (b)

(b) p 2 /4 (d) none of these

Êp ˆ Solution: We have cos(p sinx) = cos Á - p cos x˜ Ë2 ¯

IIT JEE eBooks: www.crackjee.xyz 11.4 Comprehensive Mathematics—JEE Advanced

p – p cos x = 2np ± p sin x, n ∈ I 2



sin x 0 fi

p – 2np 2 p cos x ± sin x = (1 – 4n) 2p

sin x >0 + sin x |

fi p(cos x ± sin x) =





So the given equation becomes sin x 2 1– > 1+ | sin x | 3

Since



2 £ cosx ± sinx £



2 £ (1 – 4n)



p £ 2p

| p | > |(1 – 4n)|

The smallest positive value of p is

p 2 2

2 2

p 2 2 =

p 2 4

Example 13 Number of solutions of pair x, y of the equation sin x sin y = min{–1, a2 – 4a + 5}, a ∈ R (where 0 < x < p, –p < y < 0) is Ans. (a) Solution a2 – 4a + 5 = (a – 2)2 + 1 > 1 So min {–1, a2 – 4a + 5} = – 1 and sin x sin y = – 1 fi sin x = 1, siny = – 1 or sinx = –1, siny = 1 fi x = p/2, y = – p/2 or x = – p/2, y = p/2 So there is only one pair – p/2 & p/2 satisfying the given equation. Example 14 2

1 | sin x | 1 ≥ fi | sin x | £ 3 | + | sin x | 2

fi -



1 1 È 1 1˘ £ sinx £ fi sinx ∈ Í - , ˙ 2 2 Î 2 2˚

Number of solutions of the equation

Example 16

cos(p x ) ◊ cos (p x - 4) = 1 are (a) 1 (b) 2 (c) 3 Ans. (a) Solution:

(

cos p x – 4

(

)

)

(d) 0 x>4

(

)

cos p x cos p x – 4 = 1 fi fi

(

)

(

)

cos p x = 1, cos p x – 4 = 1 p x = 2mp, p x – 4 = 2np (m, n ∈ I)

fi x = 4m2, x – 4 = 4n2 2 fi m – n2 = 1 fi m = 1, n = 0 fi x = 4 So x = 4 is the only solution of the equation

The number of solutions of the equation

10

tan x – sec x + 1 = 0 in (0, 10) is (a) 3 (b) 6 (c) 10 (d) 0 Ans. (a) Solution: sec2x – sec10x = 0 fi sec2x (1 – sec8x) = 0 fi 1 – sec8x = 0 as sec2x > 1 fi sec8x = 1 fi cos8x = 1 fi cosx = ±1 fi x = np, n ∈ I fi x = {p, 2p, 3p} ∈ (0, 10) So the required number is 3. Example 15

1–

If 1 –

sin x 2 > , then sin x lies in | + | sin x 3

Ê - 1˘ È 1 ˆ Ê 1 1ˆ (a) Á – •, » , •˜ (b) Á - , ˜ Ë 2 2¯ Ë 2 ˙˚ ÍÎ 2 ¯ È 1 1˘ (c) Í - , ˙ Î 2 2˚ Ans. (c)

(d) none of these

Note

(

)

(

)

If we take cos p x = –1, cos p x – 4 = –1 2

2

We will get (2m + 1) – (2n + 1) = 4 fi (m + n + 1)(m – n) = 1 which does not give any integral solution of m & n. Let a b ∈[– p p] be such that Example 17 cos (a – b) = 1 and cos(a + b) = 1/e. The number of pairs of a, b satisfying the above system of equations is (a) 0 (b) 1 (c) 2 (d) 4 Ans. (d) Solution: – p £ a, b £ p fi – 2p £ a – b £ 2p So cos(a – b) = 1 fi a – b = – 2p, 0. 2p fi b = a – 2p, a, a + 2p As cos(a + b) = 1/e, cos2a = 1/e As – 2p £ 2a £ 2p or – p £ a £ p And cos2a = 1/e, there are four values a and hence at least four pairs (a, b).

IIT JEE eBooks: www.crackjee.xyz Trigonometric Equations 11.5

Example 18 The number of solutions of the pair of equations 2sin2q – cos2q = 0 and 2cos2q – 3sinq = 0 in the interval [0, 2p] is (a) 0 (b) 1 (c) 2 (d) 4 Ans. (c) Solution: 2sin2q – cos2q = 0 fi 2sin2q – (1 – 2sin2q) = 0 fi 4sin2q = 1 fi sinq = ±1/2 and 2cos2q – 3sinq = 0 fi 2(1 – sin2q) – 3sinq = 0 fi 2(1 – sin2q) – 3sinq = 0 fi 2sin2q + 3sinq – 2= 0 fi fi So sinq

-3±5 sinq = - 3 ± 9 + 16 = 4 4 sinq = 1/2 as –1 £ sinq £ 1



q=

p , 6

as q Π[0, 2p]

Example 19 For x Π(0, p), the equation sin x + 2 sin 2x Рsin 3x = 3 has



1 ˆ Ê 4 cos2 x – 4 cos x + 1 = 3 Á1 Ë sin x ˜¯



1 ˆ Ê (2 cos x – 1)2 = 3 Á1 Ë sin x ˜¯

As

0 < x < p, x π p/2, 0 < sin x < 1

1 1 (c) a = 1/2 (d) a is any real number Ans. (a), (b) fi

sin x = ±

Solution: fi

a2 1 - tan 2 x

=

(sin 2 x + a 2 - 2)(1 + tan 2 x) 1 – tan 2 x

a2cos2x = sin2x + a2 – 2

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2 = sin2x (1 + a2)



sin2x =



2





1+a 2

2 1+a2

£1

a2 > 1 fi a £ – 1 or a > 1 The equation 4sin(x + p/3)cos(x – p/6)

Example 27 = a2 +

3 sin2x – cos2x has a solution if the value of

(a) – 2 (b) 0 (c) 2 (d) a,aŒ]–2, 2[ Ans. (a), (b), (c), (d) Solution: 4[sin x cos p/3 + cos x sin p/3] ¥ [cos x cos p/6 + sin x sin p/6] = a2 + 3 sin 2x – cos 2x È1 ˘ È 3 3 cos x ˙ Í cos x fi 4 Í sin x + 2 Î2 ˚ Î 2 = a2 +

1 sin 2

˘ x˙ ˚

3 sin 2x – cos 2x

3 sin 2x + 3cos2x + sin2x = a2 +

3 sin 2x – cos 2x

fi cos 2x + 2 = a – cos 2x 2

fi cos 2x =

a2 – 2 2

a2 – 2 £1 2 fi 0 £ a2 £ 4 fi –2 £ a £ 2. all values of a given in (a), (b), (c), (d) satisfy this relation. The equation

Example 28

b + sin x (cos x - 3 sin 2 x) tan x 2

(a) b
0 as q/ quadrant and sine in both the quadrants is positive. Example 42

Let q Π(p/4, p/2), then statement-1:

(cosq)sinq < (cosq)cosq < (sinq)cosq Statement-2 The equation esinq – e–sinq = 4 has a unique solution. Ans. (c) 1 < sinq < 1 Solution: For (p/4, p/2), 0 < cosq < 2 So (cosq)cosq < (sinq)cosq and (cosq)sinq < (cosq)cosq fi (cosq)sinq < (cosq)cosq < (sinq)cosq Showing that statement-1 is true. In statement-2, let esinq = t Then t2 – 4t – 1 = 0 fi

t=

4 ± 16 + 4 =2± 5 2



esinq = 2 ± 5



sinq = log 2 ± 5

(

)

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since 2 –

5 < 0, sinq = log(2 +

5 ) > loge

fi sinq > 1 which is not possible So the given equation has no solution and the statement-2 is false. Statement-1: The values of q, 0 < q 6 È ( m – 1) p p/2, satisfying the equation  cosec Íq + 4 m =1 Î Example 43

È mp Ê (m - 1) p ˆ ˘ sin Íq + - Áq + ˜¯ ˙ 6 Ë 4 4 Î ˚  ( m 1) p m p p È ˘ È ˘ m =1 sin Íq + sin Íq + sin ˙ ˙ 4 4 ˚ 4 Î ˚ Î

3 3+2 6

Example 45 (a)

(b)

(given) 3

...(i) 1 4± 3 3 = 3 ±1 = 1 2 3 1 3 =2±

3

So from (i) tanq = tan(45° ± 30°) p 5p or 12 12 Showing that statement-1 and 2 both are true and statement-2 leads to statement-1. fi q = 15° or 75° or q =

COMPREHENSION-TYPE QUESTIONS

3 3 +8 6

(d) none of these

sin a + sin b + sin g can be equal to

14 + 3 2

(b) 5/6

6 2 3+4 2 6

(d)

1+ 2 2

sin (a – b) is equal to (b) 0

1- 2 6 (d) 6 Ans. 44. (a), (b) 45. (a), (c) 46. (c) (c)

fi cotq + tanq = 4 fi tan2q – 4tanq + 1 = 0

1

6 2

(a) 1

2 [cotq + tanq] = 4 2

We have tan(45° ± tan 30°) =

(c)

cos a + cos b + cos g can be equal to

3 6 +2 2 +6

Example 46

È pˆ pˆ 6p ˆ ˘ Ê Ê Ê 2 Í cot q – cot q Áq + ˜ + cot Áq + ˜ .....cot Áq + ˜ Ë Ë Ë 4¯ 4¯ 4 ¯ ˙˚ Î



(a)

(c)

6 Ê È ( m - 1) p ˘ mp ˘ˆ È 2  Á cot Íq ˙ – cot Íq + 4 4 ˙˚˜¯ Î m = 1Ë Î ˚

fi tanq = 2 ±

Example 44

< ˘ ˙ ˚

mp ˘ È cosec Íq + = 4 2 are p/12 and 5p/12 4 ˙˚ Î Statement-2 If tanq + cotq = 4, 0 < q < p/2 then q = p/12 or 5p/12. Ans. (a) Solution. In statement-1, L.H.S. is equal to

È 3p ˆ ˘ Ê 2 Ícot q – cot Á q + ˜ ˙ = Ë 2 ¯˚ Î

b is a root of the equation 3 cos2 x – 10 cos x + 3 = 0 and g is a root of the equation 1 – sin 2x = cos x – sin x. 0 £ a, b, g £ p/2

3-2 2 6

Solution: (2 sin x – cos x) (1 + cos x) = sin2 x fi (1 + cos x) [2 sin x – cos x – 1 + cos x] = 0 fi fi so

(1 + cos x) (2 sin x – 1) = 0 cos x = – 1 or sin x = 1/2 sin a = 1/2 [as 0 £ a £ p/2]

fi Next, fi fi

cos a = 3 2 3 cos2 x – 10 cos x + 3 = 0 (3 cos x – 1) (cos x – 3) = 0 cos x = 1/3 as cos x π 3

So

cos b = 1/3, sin b =

(1)

and fi fi

2 2 3 1 – sin 2x = cos x – sin x sin2 x + cos2 x – 2 sin x cos x = cos x – sin x (cos x – sin x) (cos x – sin x – 1) = 0



Either sin x = cos x fi sin g = cos g = 1

or cos x – sin x = 1 fi cos x = 1, sin x = 0 fi cos g = 1, sin g = 0 So that cos a + cos b + cos g can be equal to

Paragraph for Question Nos. 44 to 46

3 1 1 3 1 + + + +1 or 2 3 2 3 2

a is a root of the equation (2 sin x – cos x) (1 + cos x) = sin2 x i.e.

3 6 +2 2 +6 6 2

or

3 3 +8 6

2

(2)

(3) (4)

IIT JEE eBooks: www.crackjee.xyz Trigonometric Equations 11.13

sin a + sin b + sin g can be equal to 1 2 2 1 1 2 2 +0 + + or + 2 3 2 3 2 3 2 + 14

i.e. and

6 2

or

3+4 2 6

=

1 1 fi tan2 q = 3 2

1 + tan q So from 32 tan8 q = 2 cos2 a – 3 cos, a we get 2 = 2 cos2 a – 3 cos a fi cos a = –1/2 as cos a π 2. a = 2p/3, a root of the given equation.

sin (a – b) is equal to sin a cos b – cos a sin b

2

INTEGER-ANSWER TYPE QUESTIONS

1 1 3 2 2 1- 2 6 ¥ ¥ = . = 2 3 2 3 6 Paragraph for Question Nos. 47 to 49 cos q cos 2q cos 3q = 1/4 (0 £ q £ p) Example 47 (a) p

Sum of the roots of this equation is (b) 2p (c) 3p (d) 4p

Example 48

If a is a root of this equation, 2 cos a is

a root of the equation (a) x2 – 1 = 0 (c) x4 – 4x2 + 2 = 0

1 - tan 2 q



(b) x2 + 1 = 0 (d) x4 – 4x2 + 3 = 0

Example 49 One of the roots a the system of equations 32 tan8 q = 2 cos2 a – 3 cos a and 3 cos 2 q = 1; true or false. Ans. 47. (c), 48. (a), (c), 49. true Solution: We have 2 cos 2q (cos 4q + cos 2q) = 1 fi 2 cos 2q cos 4q + 2 cos2 2q – 1 = 0 fi cos 4q (2 cos 2q + 1) = 0 fi Either cos 4q = 0 or cos 2q = – 1/2 cos 4q = 0 fi q = (2n + 1) p/8 n Œ I. p 3p 5p 7p fi q = , , , as 0 £ q £ p and 8 8 8 8 cos 2q = – 1/2 fi q = p/3 and 2p/3 again as 0 £ q £ p So the sum of the roots = 3p Next, Let x = 2 cos a where a is a root of the equation i.e. either cos 4a = 0 or cos 2a = – 1/2. so x2 = 4 cos2 a = 2 ( 1 + cos 2a ) = 1 if cos 2a = – 1/2 fi 2 cos a is a root of the equation x2 – 1 = 0 or x2 – 2 = 2 cos 2a. fi (x2 – 2)2 = 4 cos2 2a = 2 (1 + cos 4a) = 2 if cos 4a = 0 4 2 fi x – 4x + 2 = 0 fi 2 cos a is a root of the equation x4 – 4x2 + 2 = 0. Now 3 cos 2q = 1 fi cos 2q = 1/3

If 3 sin x + 4 cos x = 5, then the value of

Example 50 2

90 tan (x/2) – 60 tan (x/2) + 15 is equal to Ans. 5 Solution: 3 fi fi fi

From the given equation we have 2 tan ( x / 2) 1 + tan 2 ( x / 2)

+4

1 - tan 2 ( x / 2) 1 + tan 2 ( x / 2)

=5

6 tan (x/2) + 4 – 4 tan2 (x/2) = 5 + 5 tan2 (x/2) 9 tan2 (x/2) – 6 tan (x/2) + 1 = 0 90 tan2 (x/2) – 60 tan (x/2) = – 10. If tan (p cos q) = cot (p sin q) then

Example 51

72 cos (q – p/4) is equal to 2

Ans. 9 Solution: tan (p cos q) = tan (p/2 – p sin q) fi p cos q = np + p/2 – p sin q (n ∈ I) fi p (sin q + cos q) = (2n + 1) p/2 fi

sin q + cos q =



cos (p/4 – q) =

2n + 1 2 2n +1 2 2

Since – 1 £ cos (p/4 – q) £ 1 fi

–1£



2n +1 2 2

£1

n = 0 or – 1 as n is an integer

(



cos (p/4 – q) = ± 1 2 2



8 cos (p/4 – q) = 1

)

2



72 cos2 (p/4 – q) = 9 If x and y are the solutions of the Example 52

equation 12 sin x + 5 cos x = 2y2 – 8y + 21, the value of 12 cot (xy/2) is Ans. 5 Solution: We can write r sin (x + a) = 2y2 – 8y + 21 where r =

144 + 25 = 13

and tan a = 5/12 L.H.S £ 13 and R.H.S = 2(y – 2)2 + 13 ≥ 13

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Roots of the equation exist if L.H.S. = R.H.S = 13 fi

–1

y = 2 and sin (x + tan (5/12)) = 1



x=

p 5 5 – tan–1 fi cot x = 2 12 12

5 Ê xy ˆ \ 12 cot Á ˜ = 12 × =5 Ë 2¯ 12 Example 53 The number of all possible values of q, where 0 < q < p, for which the system of equations (y + z) cos 3q = (xyz) sin 3q 2 cos 3q 2 sin 3q + y z (xyz) sin 3q = (y + 2z) cos 3q + y sin 3q Have a solution (x0, y0, z0) with y0z0 π 0 Ans. 3. Solution: The three equations can be written as (y + z) cos 3q = xyz sin 3q 2y sin 3q + 2z cos 3q = xyz sin 3q (y + 2z) cos 3q + y sin 3q = xyz sin 3q Thus, (y + z) cos 3q = 2y sin 3q + 2z cos 3q = (y + 2z) cos 3q + y sin 3q fi y (2 sin 3q – cos 3q) + z cos 3q = 0 (1) and y (cos 3q – sin 3q) = 0 As y0 z0 π 0, we take y = y0 π 0, so that cos 3q = sin 3q. p fi tan 3q = 1 fi 3q = np + 4 p 1 fi q = np + 3 12 As 0 < q < p, we get, three value of q. x sin 3q =

Example 54

The number of value of q in the interval

np Ê p pˆ for n = 0, ±1, ±2 and tan q = ÁË - , ˜¯ such that q π 2 2 5 cot 5q as well as sin 2q = cos 4q is. Ans. 3 Solution: tan q = cot 5q fi fi fi fi As -

sin q = cos q 2 cos 5q cos 6q + 2 cos 6q

cos 5q sin 5q cos q = 2 sin 5q sin q cos 4q = cos 4q – cos 6q = 0 fi cos 6q = 0

p p < q < we get –3p < 6q < 3p 2 2

Now, cos 6q = 0 fi 6q = ± q= ±



5p 3p p ,± ,± 2 2 2

5p p p ,± ,± 12 4 12

5p p p ,and do not satisfy the relation 12 12 4 sin 2q = cos 4q. Note that 5p p p , - and q= 12 12 4 Satisfy the relation sin 2q = cos 4q. But q = -

Example 55 The number of distinct solutions of the equation 5 cos2 2x + cos4x + sin4x + cos6x + sin6x = 2 in the interval 4 [0, 2p] is Ans. 8 Solution: cos4x + sin4x = (cos2x + sin2x)2 – 2 cos2x sin2x 1 2 = 1 - sin 2 x 2 6 and cos x + sin6x = (cos2x + sin2x)3 – 3cos2x sin2x (cos2x + sin2x) 3 2 = 1 - sin 2 x 4 Thus, the given equation becomes 3 1 5 cos 2 2 x + 1 - sin 2 2 x + 1 - sin 2 2 x = 2 4 2 4 fi fi fi

sin2 2x = cos2 2x tan (2x) = ±1 p 3p 5p 7p 9p 11p 13p 15p , , , , , , , 2x = 4 4 4 4 4 4 4 4

1 (2k - 1)p where k = 1, 2, 3, 4, 5, 6, 7, 8. 8 Thus, number of distinct solution is 8.



x=

EXERCISE LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. cos 2x + a sin x = 2a – 7 possesses a solution for (a) all value of a (b) a > 6 (c) a < 2 (d) 2 £ a £ 6

IIT JEE eBooks: www.crackjee.xyz Trigonometric Equations 11.15

2. A solution (x, y) of the system of equations x – y = 1/3 and cos2 (px) – sin2 (py) = 1/2 is given by (a) (2/3, 1/3) (b) (7/6, 1/6) (c) (13/6, 11/6) (d) (1/6, 5/6) 3. If sin x = cos y, 6 sin y = tan z and 2 sin z = 3 cos x; u, v, w denote respectively sin2 x, sin2 y, sin2 z, then the value of the triplet (u, v, w) is (a) (1, 0, 0) (b) (0, 1, 0) (c) (1/2, 1/2, 3/4) (d) (1/2, 3/4, 1/2) 4. The least difference between the roots of the equation 4 cos x (2 – 3 sin2 x) + (cos 2x + 1) = 0 (0 £ x £ p/2) is (a) p/6 (b) p/4 (c) p/3 (d) p/2 5. The sum of the roots of the equation 4 cos3 (p + x) – 4 cos2 (p – x) + cos (p + x) – 1 = 0 in the interval [0, 320] is pp where p is equal to (a) 2500 (b) 2601 (c) 2600 (d) 2651

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 6. cos x – 2 cos x = 4 sin x – sin 2x if (a) tan (x/2) = – 1/2 (b) tan x = – 1/2 (c) tan (x/2) = 2 + 5 (d) tan (x/2) = 2 – 7. sin x + cos x = 1 + sin x cos x, if

(r) cos x =

+ 2 sin2 x = 0 (d) 2 sin 2x + 1 = 2 (sin x + cos x)

(s) cos x = 1/2

10. Column 1 (a) cos2 x – 2 cos x = 4 sin x – sin 2x (b) 1 - 2 tan x –

Column 2 (p) cot x – tan x = 2 (q) cot x – tan x = 2

5

(c)

5 - 2sin x

= 6 sin x – 1 (d) tan2 x + 4 tan x – 1 =0

(r) cot x – tan x = – 3/2 (s) cot x – tan x = 4

ASSERTION-REASON TYPE QUESTIONS 11. Statement-1: If exp{(sin2x + sin4x + sin6x + inf) loge x2 – 9x + 8 = 0, then the cos x 3 –1 value of is . (0 < x < p/2) cos x + sin x 2

2

(b) sin (x – p/4) = 1

2

(c) cos (x + p/4) = 1

2

Paragraph for Question Nos. 13 to 15

(d) cos (x – p/4) = 1

2

Consider the equations 5sin2 x + 3 sin x cos x – 3 cos2 x = 2 sin2 x – cos 2x = 2 – sin 2x

MATRIX-MATCH TYPE QUESTIONS 9. Column 1 Column 2 2 (a) 2 cos x + 4 cos x (p) sin x = 1/2 = 3 sin2 x (b) tan x + sec x = 2 cos x (q) cos x = – 3/5

inf

12. Statement-1: cos7x + sin4x = 1 has only two nonzero solutions in the interval – p < x < p Statement-2: cos5x + cos2x – 2 = 0 is possible only when cos x = 1

(a) sin (x + p/4) = 1

8. If x + y + z = p, tan x tan y = 2, tan x + tan y + tan z = 6 then (a) x = mp + p/4 (b) y = np + tan–1 2 (c) z = lp + tan–1 3 (d) all are correct (l, m, n Œ I)

3

1 + 2 cot x = 2

Statement-2: sin2x + sin4x + sin6x + = sec2x

2

-2 + 19 5

(c) 6 cos 2x + 2 cos2 (x/2)

COMPREHENSION-TYPE QUESTIONS

(1) (2)

13. If a is a root of (1) and b is a root of (2) then tan a + tan b can be equal to (a) 1 +

69 6

(b) 1 –

69 6

-3 + 69 -3 - 69 (d) 6 3 14. If tan a, tan b satisfy (1) and cos g, cos d satisfy (2) then tan a tan b + cos g + cos d can be equal to 5 2 (a) –1 (b) - + 3 13 5 2 5 2 (c) (d) - 3 3 13 13 (c)

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15. The number of solutions common to (1) and (2) is (a) 0 (b) 1

INTEGER-ANSWER TYPE QUESTIONS 16. cos 5q = 0 if 64 cos q – 80 cos q + 25 is equal to 4

2

2

17. If 3 tan 2x – 4 tan 3x = tan 3x tan 2x then tan 3x + 3 tan x + 3 is equal to

MATRIX-MATCH TYPE QUESTIONS Column 1

Column 2 (p) tan x = -1

9. (a) sin2 x + 2 sin x cos x – 3 cos x = 0 (b) 5 sin2 x +

3 sin x

(q) tan x = 4/3

2

cos x + 6 cos x = 5 (c) sin x + 7 cos x = 5

(r) tan x = –3

2

(d) 4 sin x + 15 sin x cos x

LEVEL 2

3

2

(s) tan x = –3/4

2

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. The number of solutions of the equation sin4 2x + cos4 2x + 3 sin2 x cos2 x = 0 is (a) 0 (b) 1 (c) 2 (d) 3 x 0

2. The equation Ú (t2 – 8t + 13) dt = x sin (a/x) has a solution if sin (a/6) is equal to (a) 0 (b) 1 (c) 1/2 (d) 2/3 3. A solution (x, y) of x2 + 2x sin xy + 1 = 0 is (a) (1, 0) (b) (1, 7p/2) (c) (– 1, 7p/2) (d) (– 1, 0) 4. The number of pairs (x, y) satisfying the equation sin x + sin y = sin (x + y) and |x| + |y| = 1 is 5. The least positive value of x satisfying sin 2 2 x + 4sin 4 x - 4sin 2 x cos 2 x 4 - sin 2 x - 4sin x 2

(a) p/3

(b) p/6

2

=

1 is 9

(c) 2p/3 (d) 5p/6

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 6. sin 2x = 2 cos x if (a) x = np + p/2 (b) x = 2np + p/4 (c) x = 2np – p/4 (d) x = np – p/2 (n Œ I) 7. A value of q in the interval (– p/2, p/2) satisfying the equation (1 – tan q) (1 + tan q) sec2 q + 2 tan2 q = 0 is (a) – p/6 (b) – p/3 (c) p/6 (d) p/3 8. In a right angled triangle the hypotenuse is four times the length of the perpendicular drawn from the opposite vertex on the hypotenuse, then one of the other angle is 1 1 (a) 15° (b) 75° (c) 22 ° (d) 67 ° 2 2

+ 9 cos x 10. If 0 < x < 2p Column 1

Column 2

(a) sin x + sin 2x + sin 3x

(p) 3

= cos x + cos 2x + cos 3x (b) sin x + sin 2x + sin 3x = 0 2

(q) 6

(c) sin x – cos 2x = 2 – sin 2x

(r) 4

(d) 3 sin 2x + 4 sinx – 3 cosx = 2

(s) 5

ASSERTION-REASON TYPE QUESTIONS 11. Statement-1: If 2 sin2x + 3 sin x – 2 > 0 and x2 – x – 2 < 0, then – 1 < x < 2 Statement-2: x2 – x – 2 < 0 if – 1 < x < 2 Ê px ˆ = x2 – 2 3 x + 4 has only 12. Statement-1: sin Á Ë 2 3 ˜¯ one solution Statement-2: The smallest positive value of x in degrees, for which tan (x + 100°) = tan (x + 50°) tan x tan (x – 50°) is 30°.

COMPREHESION-TYPE QUESTIONS Paragraph for Question Nos. 13 to 15 Consider cosn x – sinn x = 1, where n is a natural number and – p < x £ p 13. When n = 1, the sum of the values of x satisfying the equation is (a) – p/2 (b) 0

(c) p/4

(d) p/2

14. When n is an even natural number then the value of x is (a) 0

(b) p/2

(c) 2p/3

(d) p

15. When n is an odd natural number other than 1, then the value of x is (a) – p/2

(b) 0

(c) p

(d) 3p

IIT JEE eBooks: www.crackjee.xyz Trigonometric Equations 11.17

INTEGER-ANSWER TYPE QUESTIONS 16. If cos 2q + 9 sin 2q – 6 sin q + 54 cos q = 1 then the value of 100 tan2 q + 9 tan q is equal to 909 k, then k is equal to

17. Four times the sum of the roots of the equation sin 2x + 5 sin x + 5 cos x + 1 = 0 in the interval. [0, 50p] is pp where p is equal to 1010 a, where a is equal to

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. The general solution of the trigonometric equation sin x + cos x = 1 is given by (a) x = 2np ; n = 0, ± 1, ± 2, .... (b) x = 2np + p/2; n = 0, ± 1, ± 2, .... (c) x = np + (–1)n p/4; n = 0, ± 1, ± 2, .... (d) none of these [1981] 2. The smallest positive root of the equation tan x – x = 0 lies in (a) (0, p/2) (b) (p/2, p) (c) (p, 3p/2) (d) (3p/2, 2p) [1987] 3. The general solution of sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x is (a) np + p/8 (b) np/2 + p/8 (d) 2np + cos–1 (2/3) (c) (–1)n np/2 + p/8 where n ∈ I. [1989] 4. In a triangle ABC, angle A is greater than angle B. If the measures of angles A and B satisfy the equation 3 sin x – 4 sin3 x – k = 0, 0 < k < 1, then the measure of angle C is (a) p/3 (b) p/2 (c) 2p/3 (d) 5p/6 [1990] 5. Number of solutions of the equation tan x + sec x = 2 cos x lying in the interval [0, 2p] is: (a) 0 (b) 1 (c) 2 (d) 3 [1993] 6. Let n be a positive integer such that Êp ˆ Êp ˆ sin Á ˜ + cos Á ˜ = 2 n Ë ¯ Ë 2n¯

n . Then 2

(a) 6 £ n £ 8 (b) 4 < n £ 8 (c) 4 £ n £ 8 (d) 4 < n < 8 [1994] 7. Let 2 sin2 x + 3 sin x – 2 > 0 and x2 – x – 2 < 0 (x is measured in radians). Then x lies in the interval (a) (p/6, 5p/6) (b) (–1, 5p/6) (c) (–1, 2) (d) (p/6, 2) [1994] 8. The number of values of x in the interval [0, 5p] satisfying the equation 3 sin2 x – 7 sin x + 2 = 0 is (a) 0 (b) 5 (c) 6 (d) 10 [1998]

n

9. Let n be an odd integer. If sin (nq) = Â br sinr q, for r=0

10.

11.

12.

13.

14.

each value of q, then: (a) b0 = 1, b1 = 3 (b) b0 = 0, b1 = n (c) b0 = –1, b1 = n [1998] (d) b0 = 0, b1 = n2 – 3n + 3 The number of values of k for which the equation 7 cos x + 5 sin x = 2k + 1 has a solution is (a) 4 (b) 8 (c) 10 (d) 12 [2002] Let a, b ∈ [– p, p] be such that cos (a – b) = 1 and cos (a + b) = 1/e. The number of pairs of a, b satisfying the above system of equations is (a) 0 (b) 1 (c) 2 (d) 4 [2005] 2 If 0 £ q £ 2p, 2 sin q – 5 sin q + 2 > 0, then the range of q is (a) (0, p/6) » (5p/6, 2p) (b) (0, 5p/6) » (p, 2p) (c) (0, p/6) » (p, 2p) (d) none of these [2006] The number of solutions of the pair of equations 2 sin2 q – cos 2q = 0 and 2 cos2 q – 3 sin q = 0 in the interval [0, 2p] is (a) 0 (b) 1 (c) 2 (d) 4 [2007] For x ∈ (0, p), the equation sin x + 2 sin 2x – sin 3x = 3 has (b) three solutions (c) one solution (d) no solution

[2014]

p¸ Ï 15. Let S = Ì x Œ (-p , p ) : x π 0, ± ˝ , the sum of all dis2˛ Ó tinct solutions of the equation 3 sec x + cosec x + 2 (tan x – cot x) = 0 in the set S is equal to 7p 2p (a) (b) 9 9 (c) 0

(d) -

5p 9

[2016]

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MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

Column I (a) The minimum value

p 1. For 0 < q < , the solution(s) of 2

of

Ê ( m – 1) p ˆ cosec Êq + mp ˆ = 4 2  cosec ÁËq + ˜¯ ÁË ˜ 4 4 ¯ m =1 6

is(are) p (a) 4 p (c) 12

p 6 5p (d) 12

(b)

[2009]

2. Let q, f ∈ [ 0, 2p] be such that qˆ Ê q 2 cos q (1 – sin f) = sin2q Á tan + cot ˜ cos f – 1, Ë 2 2¯ tan (2p – q) > 0 and – 1 < sin q < – Then f cannot satisfy p (a) 0 < f < 2

p 4p 0. Then for all natural numbers n, f ¢(x) vanishes at 1ˆ Ê (a) A unique point in the interval Á n, n + ˜ Ë 2¯ (c)

4p 3p 0 fi 2tan2x > 1] 3

2

12. cos x + (1 – cos x) = 1 fi cos7 x + cos4 x – 2 cos2 x = 0 fi cos2 x (cos5 x + cos2 x – 2) = 0 fi cos2 x = 0 as statement-2 is true. fi x = ± p/2 fi

Statement 1 is also true.

13. 5 sin2 x + 3 sin x cos x – 3 cos2 x = 2 (sin2 x + cos2 x) fi 3 tan2 x + 3 tan x – 5 = 0 -3 ± 69 fi tan x = 6

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and sin2 x – cos 2x = 2 – sin 2x

15. (1) and (2) have no solution common.

fi 3 sin x + 2 sin x cos x = 3 (sin x + cos x) 2

2

2

fi cos x (2 sin x – 3 cos x) = 0 3 2 fi Either cos x = 0 or tan x = fi cos x = ± 2 13 Taking tan a =

-3 ± 69 3 , tan b = 6 2

we get tan a + tan b = 1 ± 69 6 -3 + 69 -3 - 69 14. Taking tan a = , tan b = 6 6 cos g = 0, cos d = ±

2 13

5 2 we get tan a tan b + cos g + cos d = - ± 3 13

16. cos 5q = cos 4q cos q – sin 4q sin q = (2 cos2 2q – 1) cos q – 2 sin 2q cos 2q sin q = cos q [2 cos2 2q – 1 – 4 sin2 q cos 2 q] = cos q [4 cos2 2q – 2 cos 2q – 1] = 0 if cos q = 0 or cos 2q =

1± 5 5± 5 fi cos2 q = 4 8

(8 cos2 q – 5)2 = 5 fi 64 cos4 q – 80 cos2 q + 20 = 0 so that 64 cos4 q – 80 cos2 q + 25 = 5 17. 3 tan 2x – 3 tan 3x = tan 3x + tan2 3x tan 2x tan 2 x - tan 3x = tan 3x 1 + tan 3x tan 2 x fi – 3 tan x = tan 3x Hence tan 3x + 3 tan x + 3 = 3 fi 3

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12 Solution of Triangles and Applications of Trigonometry fi

12.1 IMPORTANT FORMULAE

In a triangle ABC, the angles are denoted by the capital letters A, B, C and the lengths of the sides opposite these angles are denoted by a, b, c, respectively (Fig. 12.1).

Some important formulae regarding the sides and angles of a triangle are given below. 1. The law of sines In any triangle ABC a b c = = = 2R sin A sin B sin C where R is the radius of the circumcircle of the triangle ABC. R is also known as the circumradius of the triangle. 2. The law of cosines In any triangle ABC b2 + c 2 - a 2 2bc 2 2 or a = b + c2 – 2bc cos A Similar formulae for cos B, cos C exist.

sin

A = 2

( s - b) ( s - c ) , bc

cos

A = 2

s( s - a) , bc

tan

A = 2

( s - b)( s - c) s( s - a)

similar formulae for sin (B/2), cos (B/2), tan (B/2) and sin (C/2), cos (C/2), tan (C/2) exist

D B tan ÊÁ ˆ˜ = Ë 2¯ s ( s - b)

Illustration 1

D C tan ÊÁ ˆ˜ = , Ë 2¯ s ( s - c) where D is the area of the DABC. 4. Projection rule In any triangle ABC a = b cos C + c cos B, b = c cos A + a cos C and c = a cos B + b cos A

and

Find the angles of the triangle whose sides are 2 3 , and 3 + 3 .

6

Solution: Let the sides of the triangle ABC be a = 2 3 , b = 6 and c = 3 + 3 a 2 + b2 - c2 12 + 6 - (12 + 6 3 ) = 2ab 2¥2 3¥ 6 6-6 3

b sin C 6 ( 3 + 1) 1 = = c 2 3+ 3 2 2

D Ê Aˆ , Note that tan Á ˜ = Ë 2¯ s( s - a)

cos A =

=

fi sin B =

fi B = 30° fi A = 180° – (105° + 30°) = 45° Hence the required angles are 45°, 30°, 105°. 3. Trigonometric ratios of half-angles in terms of the sides Let 2s = a + b + c, so that s is the semi-perimeter of D ABC. Then

Fig. 12.1

cos c =

C = 105° b c = Now sin B sin C

=

1

12 2 2 2 = cos (60° + 45°)

-

3 2 2

Illustration 2 C B + c cos 2 in terms of the semi2 2 perimeter s of a triangle ABC 2 Find the value of b cos

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2 Solution: b cos

C B + c cos 2 2 2

=

b c (1 + cos C) + (1 + cos B3) 2 2

=

b+c 1 + (b cos C + c cos B) 2 2

=

b+c 1 + a 2 2

of these six elements (except the three angles) of a triangle are given, the triangle is known completely; that is, the other three elements can be expressed in terms of the given elements and hence evaluated. This process is called the solution of the triangle. Let ABC be a right-angled triangle, right-angled at C (Fig. 12.2).

[Using Projection Rule]

a+b+c =s 2 5. Napier’s analogy In any triangle ABC =

tan tan

B-C b-c A cot , = b+c 2 2

Fig. 12.2

C-A c-a B cot = etc. c+a 2 2

The triangle can be solved as follows: 1. If the hypotenuse c and one side, say b, are given, then sin B = b/c gives the value of B, A = 90° – B and a =

Illustration 3

2. If two sides a and b are given, then tan B = b/a

p , determine the angle C. 3 p Solution: a = 2b fi A > B fi A – B = . 3 So using Napier’s analogy

In a triangle ABC, a = 2b, |A – B| =

A- B a-b C cot = 2 2 a+b C p 2b - b cot tan = 6 2b + b 2 1 C p cot = 3¥ = 3 = cot 2 6 3 tan

fi fi

p . 3 6. Area of a triangle. If D represents the area of a triangle ABC, then D = (1/2) bc sin A = (1/2) ca sin B fi

C=

= (1/2) ab sin C = =

c2 - b2 .

abc = rs 4R

s ( s - a ) ( s - b) ( s - c )

where 2s = a + b + c, R is the radius of the circumcircle of triangle ABC and r is the radius of the circle inscribed in triangle ABC. 12.2 SOLUTION OF TRIANGLES

The three sides a, b, c and the three angles A, B, C are called the elements of the triangle ABC. When any three

2 2 gives the value of B, A = 90° – B and c = a + b . 3. If an angle B and a side a are given, then A = 90° – B, side b is obtained from the relation tan B = b/a and side c from the relation cos B = a/c. 4. If an angle B and the hypotenuse c are given, then A = 90° – B, a = c cos B and b = c sin B. If the triangle is not right-angled, the different cases to be considered are as follows: 1. If the three sides a, b and c are given, angle A can be found from the formulae

tan

A ( s - b) ( s - c ) = s( s - a) 2

or cos A =

b2 + c2 - a 2 2bc

Angles B and C can be found using the corresponding formulae. 2. If two sides b and c and the included angle A are given, and if b > c then tan

B-C b-c A = cot 2 b+c 2

gives the value of (B – C)/2, while B+C A = 90° – 2 2 so that B and C can be evaluated. The third side a is given by sin A or a2 = b2 + c2 – 2bc cos A a=b sin B 3. If two sides b and c and the angle B opposite one of them are given, then angle C is given by sin C = (c/b)

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sin B, A = 180° – (B + C) and the third side a is given by a = b (sin A/sin B). If b < c and B is an acute angle, then there are two values of C and hence two values of A, so that there are two triangles having the given elements. This is called an ambiguous case in the solution of triangles. 4. If one side a and two angles B and C are given, then A = 180° – (B + C) and the sides b, c are obtained from b=

a sin B sin A

and

c=

a sin C . sin A

5. If the three angles A, B and C are given, we can only a, b and c, using a b c = = . sin A sin B sin C

3. R =

a b c abc = = = 2 sin A 2 sin B 2 sin C 4D

D . s 5. r1 + r2 + r3 = 4R + r. 4. r =

6. r1 r2 + r2 r3 + r3 r1 = s2. Illustration 4 Prove that in a right angled triangle r + 2R = s. Solution: Let A be a rightangle of the triangle ABC. Circumcentre O lies on the hypotenuse BC = a. C

12.3 GEOMETRICAL PROPERTIES AND FORMULAE FOR A TRIANGLE

(i) In a tiangle the internal bisector of an angle divdes the opposite side in the ratio of the sides containing the angle. (ii) In an isosceles triangle the median is perpendicular to the base. (iii) In similar triangles the corresponding sides are proportional. (iv) The exterior angle of a triangle is equal to the sum of interior opppsite angles. 12.4 TRIANGLES, QUADRILATERALS, POLYGONS

Escribed circles The circle touching BC and the two sides AB and AC produced of a triangle ABC is called the escribed circle opposite angle A. Its radius is denoted by r1. Similarly, r2 and r3 denote the radii of the escribed circles opposite angles B and C, respectively. r1, r2, r3, are known as the ex-radii of triangle ABC. Some important formulae connecting the sides, angles and radii of a triangle ABC are given as follows. A B C = (s – b) tan = (s – c) tan 1. r = (s – a) tan 2 2 2 A B C = 4R sin sin sin 2 2 2 B C a cos cos A D 2 2 2. r1 = = s tan = A s-a 2 cos 2 = 4R sin

A B C cos cos 2 2 2

Similar formulae for r2 and r3 exist.

O

A

B Fig. 12.3

So,

R=

a 2

A 2 = (s – a) tan 45° = s – a r + 2R = s.

Also r = (s – a) tan fi

Illustration 5 If the ex-radii r1, r2, r3 of a triangle ABC are in H.P. Show that its sides a, b, c are in A.P. Solution: fi fi fi

2 1 1 = + r2 r1 r2 2( s - b) s-a s-c = + D D D 2b = a + c a, b, c are in A.P.

Area of a cyclic quadrilateral. Let ABCD be a cyclic quadrilateral, whose sides AB, BC, CD and DA are a, b, c, and d, respectively (Fig. 12.4). The area of this quadrilateral is ( s - a ) ( s - b) ( s - c) ( s - d ) , where 2s = a + b + c + d.

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Illustration 6 A regular pentagon and a regular decagon have the same perimeter. Find the ratio of their areas. Solution: Let the perimeter be 10x. Each side of the pentagon is 2x, so its area is 1 p ¥ 5 ¥ (2 x)2 cot = 5x2 cot 36° D1 = 4 5

Fig. 12.4

Also,

cos B =

a 2 + b2 - c2 - d 2 2 (ab + cd )

There are corresponding formulae for the cosines of the other angles. Radii of the inscribed and circumscribed circles of a regular polygon. Let AB, BC and CD be three successive sides of a regular polygon of n sides, each of length a (Fig. 12.5). Let the bisectors of angles ABC and BCD meet at O and draw OL perpendicular to BC. O is the centre of both the incircle and the circumcircle of the polygon. Also, BL = LC = a/2, OB = OC = R (the radius of the circumcircle) and OL = r (the radius of the incircle).

Each side of the decagon is x, so its area is 1 p 5 2 ¥ 10( x)2 cot D2 = = x cot 18∞ 4 10 2 D1 2 cot 36∞ 2 cos 36∞ sin 18∞ = = D2 cot 18∞ 2 sin 18∞ cos18∞.cos18∞

So, = =

cos 36∞ cos 18∞ 2

=

2 cos 36∞ 1 + cos 36∞

2( 5 + 1) 2( 5 + 1) 2 = = . È ˘ 5+ 5 5 5 +1 4 Í1 + ˙ 4 ˚ Î

12.5 ANGLES OF ELEVATION AND DEPRESSION

Let OP be a horizontal line in the same vertical plane as an object R, and let OR be joined.

Fig. 12.6A

Fig. 12.5

Further, –BOC = 2p /n and –BOL = –LOC = p/n, so that from DCOL, we get a a Êpˆ Êpˆ cosec Á ˜ and r = cot Á ˜ R= Ë n¯ Ë n¯ 2 2 and the area of the polygon = n ¥ area of DOBC = n ¥ =n ¥ =

Fig. 12.6B

1 BC ¥ OL 2

a a Êpˆ ¥ cot Á ˜ Ë n¯ 2 2

1 2 Êpˆ na cot Á ˜ Ë n¯ 4

1 Êpˆ Ê 2p ˆ = nr2 tan Á ˜ = nR2 sin Á ˜ Ë n¯ Ë n¯ 2

In Fig. 12.6, where the object R is above the horizontal line OP, the angle POR is called the angle of elevation of the object R as seen from the point O. In Fig. 12.6, where the object R is below the horizontal line OP, the angle POR is called the angle of depression of the object R as seen from the point O. Remark Unless stated to the contrary, it is assumed that the height of the observer is neglected, and that the angles of elevation are measured from the ground.

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2B = A + C Also, as A + B + C = 180°, we get 3B = 180° or B = 60°

Illustration 7 Find the height of the tower standing in the centre of a an angle of 75° at each corner. Solution: Let h be the height of the tower OP standing at O ABCD. P A

75°

C 5m

O A

Fig. 12.8

B

10 m

By the law of sines a b c = = sin A sin B sin C Thus, a c sin 2C + sin 2A c a sin A sin C (2 sin C cos C) + (2 sin A cos A) = sin C sin A = 2 (sin A cos C + cos A sin C) = 2 sin (A + C) = 2 sin (π – B)

Fig. 12.7

1 1 5 5 (10) 2 + (5)2 = AC = . 2 2 2 Now POC is a right angled triangle, right angle at O and PCO = 75°. OC =

OP 1 + tan 30∞ = tan 75° = = 1 - tan 30∞ OC fi

3 +1 3 -1

h = OP = (2 + 3 )OC = (2 + 3 )

=

= 2+ 3

5 5 2

5(2 5 + 15 ) m 2

Example 2

SOLVED EXAMPLES

SINGLE CORRECT ANSWER TYPE QUESTIONS If the angles A, B and C of a triangle are Example 1 in an arithmetic progression and if a, b and c denote the lengths of the sides opposite to A, B and C respectively, a c then the value of the expression sin 2C + sin 2A is c a 1 2

(b)

3 2

(c) 1

(d)

3

(a)

Ê 3ˆ = 2 sin B = 2 sin 60° = 2 Á ˜ = Ë 2 ¯

Ans. (d) Solution: As the angles A, B, C are in arithmetic progressions,

3

Let PQR be a triangle of area D with a = 2,

7 5 and c = , where a, b and c are the lengths of the 2 2 sides of the triangle opposite to the angles at P, Q and R 2sin P - sin 2 P equals respectively. Then 2sin P + sin 2 P b=

(a)

3 4D

(b)

2 Ê 3 ˆ (c) Á ˜ Ë 4D ¯

45 4D

2 Ê 45 ˆ (d) Á ˜ Ë 4D ¯

Ans. (c) Solution: =

2sin P - sin 2 P 2sin P (1 - cos P ) = 2sin P (1 + cos P ) 2sin P + sin 2 P 2 2sin 2 ( P/2) Ê tan P ˆ = Á ˜ Ë 2¯ 2 cos 2 ( P/2)

Ê ( s - b )( s - c ) ˆ = Á ˜¯ Ë D

2

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where s =

1 (a + b + c) 2

2 Ê 3 ˆ = Á ˜ Ë 4D ¯

Example 3

– 1 < ( l - 2) 2 < 1 In a triangle the sum of two sides is x and

3y (c) 4 x( x + c) Ans. (d) Solution: Suppose a + b= and x2 – c2 = fi (a + b)2 – fi a2 +

3y (d) 4c( x + c)

a 2 + b2 - c2 1 =cos C = 2ab 2

R=

c = 2 sin C

Ê a + b + c - cˆ tan Ê C ˆ = Á ˜¯ ÁË ˜¯ Ë 2 2

Now,

1 ( x - c) 3 2

r ( x - c) 3 3( x - c) = = R 2c 2c / 3 =

(b)

3( x 2 - c 2 ) 3y = . 2c( x + c) 2c( x + c)

In a triangle ABC, (a + b + c) (b + c – a) = Example 4 l bc if (a) l < 0 (b) l > 6 (c) 0 < l < 4 (d) l > 4 Ans. (c) Solution: We are given that (a + b + c) (b + c – a) = l bc, or (b + c)2 – a2 = l bc. That is, b2 + c2 + 2bc – a2 = l bc, or b2 + c2 – a2 = (l – 2)bc.

3 /2

(c) 1/ 3 (d) 2 3 Ans. (b) Solution: The roots of the given equation are given by

=

2 6 ± 24 - 16 8 2 6±2 2 6± 2 = 8 4

Now, if a represents the third side, then 2

2

Ê 6 + 2ˆ Ê 6 - 2ˆ 2 +Á ÁË ˜ ˜¯ - a 4 4 ¯ Ë 2¥

Ê Cˆ r = (s – c) tan Á ˜ Ë 2¯

=

3

(a)

cos 60° = c c = . p 2 Ê ˆ 3 2 sin Ë 3¯

0 < l < 4.

is 60°, then the third side is

x= x, ab = y y c2 = ab b2 – c2 = – ab



If two sides of a triangle are the roots of Example 5 2 the equation 4x – ( 2 6 )x + 1 = 0 and the included angle

2p fi C= 3 We know that

Also,

b2 + c2 - a 2 l - 2 = 2 bc 2

As A is the angle of a triangle, – 1 < cos A < 1. Therefore

the product of the same two sides is y. If x2 – c2 = y, where c is the third side of the triangle, then the ratio of the inradius to the circum-radius of the triangle is 3y 3y (b) (a) 2 x( x + c) 2c( x + c)



cos A =

Therefore,

6+ 2 6- 2 ¥ 4 4



1 2 ( 6 + 2) - 16a 2 = 2 2 ( 6 - 2)



a2 =

3 3 fia= . 2 4

In a triangle, if the sum of two sides is x Example 6 and their product is y such that (x + z) (x – z) = y where z is the third side of the triangle then the triangle is (a) equilateral (c) obtuse angled Ans. (c) Solution:

(b) right angled (d) none of these

Let ABC be the triangle with b + c = x and

bc = y then a = z, and from the given relations we have (b + c + a) (b + c – a) = bc fi fi fi fi

b2 + c2 – a2 = – bc b2 + c2 - a 2 1 = 2bc 2 1 cos A = - = cos 120° 2 A =120° and the triangle is obtuse angled.

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If D is the mid-point of side BC of a Example 7 triangle ABC and AD is perpendicular to AC, then (b) 3a2 = b2 – 3c2 (a) 3b2 = a2 – c2 2 2 2 (c) b = a – c (d) a2 + b2 = 5c2 Ans. (a)

If in a triangle ABC, sin A, sin B, sin C are Example 9 in A.P., then (a) the altitudes are in A.P. (b) the altitudes are in H.P. (c) the altitudes are in G.P. (d) none of these. Ans. (b) Solution: If p1, p2, p3, are altitudes from A, B, C respectively, then

Fig. 12.9

Solution: From the right-angled triangle CAD (Fig. 12.9), we have cos C = fi

b 2b a 2 + b 2 - c 2 fi = a /2 a 2 ab

a2 + b2 – c2 = 4b2 fi a2 – c2 = 3b2.

Example 8

In a triangle with one angle 2 p 3 , the

lengths of the sides form an A.P. If the length of the greatest side is 7 cm, the radius of the circumcircle of the triangle is (a) 7 3 3 cm

(b) 5 3 3 cm

(d) 3 cm (c) 2 3 3 cm Ans. (a) Solution: Let the sides of the triangle be 7, 7 – d, 7 – 2d. Since the given angle is the greatest (being obtuse) angle of the triangle, it is opposite to the greatest side of the triangle and we have, 72 = (7 – d)2 + (7 – 2d)2 – 2 (7 – d) (7 – 2d) cos 2 p 3 .



D= p1 =

1 1 1 ap1 = bp2 = cp3 2 2 2 2D 2D 2D , p2 = , p3 = a b c

By the law of sines a b c = = = k (say) sin A sin B sin C \

p1 =

2D 2D 2D , p2 = , p3 = k sin A k sin B k sin C

Now, sin A, sin B, sin C, are in A.P. fi

p1, p2, p3 are in H.P.

Example 10

If ABC is a triangle in which B = 45°,

C = 120° and a = 40, the length of the perpendicular from A on BC produced is (a) 20 (3 –

3)

(c) 20 ( 3 + 1)

(b) 20 (3 +

(d) 20 ( 3 – 1)

Ans. (b) Solution:

72 = 2 ¥ 72 – 42 d + 5d2 – 2 (72 – 21d + 2d2) ( - 1 2)

fi fi fi fi

d2 – 9d + 14 = 0 (d – 7) (d – 2) = 0 d= 2

(d = 7 is not possible)

Therefore, the sides of the triangle are 7 cm, 5 cm, 3 cm. Area of the triangle D = 2

1 2p 15 3 ¥ 5 ¥ 3 ¥ sin = 4 2 3

cm and the radius of the circumcircle R = =

7¥5¥3 15 3

=

7 3 cm. 3

7¥5¥3 4(15 3/4)

Fig. 12.10

Referring to Fig.12.10, we have from D ABC, AC/sin 45° = 40/sin 15°. 40 sin 45∞ fi AC = sin 15∞

3)

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=

40 2 sin (45∞ - 30∞)

40 ◊ 2 2

=

2 ( 3 - 1)

so that AC = 40 ( 3 + 1) . Therefore, from right-angled triangle ADC, we get the length of the perpendicular AD = AC sin 60° = 40

(

)

3 +1 ◊

3 = 20 (3 + 2

In a triangle ABC, if

Example 11

3 ).

1 1 + = a+c b+c

3 then C is equal to a+b+c (a) 30°

(b) 60°

Fig. 12.11

(a) 113° Ans. (a)

(c) 75°

(d) 90°



We have AD = b sin C abc = b sin C 2 b - c2

The given relation can be written as



a + b + 2c 3 = (a + c) (b + c) a+b+c

sin C a = 2 c b - c2



sin A a = 2 a b - c2



sin A =

fi (a + b + 2c) (a + b + c) = 3(a + c) (b + c) fi (a + b)2 + 3c(a + b) + 2c2 = 3(ab + ac + bc + c2) fi a2 + b2 – c2 = ab a 2 + b2 - c2 ab 1 = = cos C = 2ab 2ab 2 C = 60°

\ fi

In a triangle ABC, if C = 90°, then

Example 12

a 2 + b2 sin (A – B) = a 2 - b2 (a) sin (A + B) (c) cos (A – B) Ans. (a) Solution: Since C = 90°

(b) cos (A + B) (d) sin A + sin B fi A + B = 90°, we have

a +b sin (A – B) a 2 - b2 2

=

In a triangle ABC, AD is the altitude from abc A (Fig. 12.11). Given b > c, – C = 23° and AD = 2 b - c2 then – B. is equal to Example 13

È sin C = sin A ˘ ÍÎ c a ˙˚

=

sin 2 A sin 2 B - sin 2 C

=

sin 2 A sin ( B + C ) sin ( B - C )

Example 14 In a triangle ABC, a : b : c = 4 : 5 : 6. The ratio of the radius of the circumcircle to that of the incircle is (a) 15/4 (b) 11/5 (c) 16/7 (d) 16/3 Ans. (c) Solution:

sin 2 A + sin 2 B sin ( A - B) sin 2 A - sin 2 B

sin 2 A + cos 2 A 1 = = 1 = sin (A + B). sin ( A + B) sin 90∞

a2 b2 - c2

fi sin (B – C) = 1 [ sin (B + C) = sin A] fi B – C = 90° fi B = 90° + C = 90° + 23° = 113°.

2

sin 2 A + sin 2 (90∞ - A) = sin (A – B) sin ( A + B) sin ( A - B) =

(c) 147° (d) 157°

Solution:

Ans. (b) Solution:

(b) 123°

R= fi

Let abc 4D

a b c = = =k 4 5 6 and

r=

D s

R abc s s abc ¥ = = 4 D D 4 D2 r =

s abc 4 s ( s - a ) ( s - b) ( s - c )

=

4k ¥ 5k ¥ 6k 15 15 15 4 Ê k - 4k ˆ Ê k - 5k ˆ Ê k - 6 k ˆ Ë 2 ¯Ë 2 ¯Ë 2 ¯

=

5 ¥ 6 ¥ 8 16 = . 7¥5¥3 7

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Example 15

The sides of a triangle are sin a, cos a

and 1 + sin a cos a for some 0 < a < p/2; the greatest angle of the triangle is (a) 120° (b) 90° (c) 60° (d) 150°

So, that cos B = Example 18

(a)

Let a = sin a, b = cos a and c = 1 + sin a cos a for a triangle ABC. a +b -c 2 ab 2

cos C =

then fi

lim

2

=

- sin a cos a 1 = 2sin a cos a 2

C = 120°. which is the greatest angle.

Example 16 AÆ C

2

1 1 2R r

fi fi

|a – c| =

fi fi 2 cos fi

2 sin

so that lim

AÆ C

A-C = 2

sin ( A/2) 4 R sin ( B /2) sin (C /2)

=

sin 2 ( A/2) 4 R sin ( A/2)sin ( B /2)sin (C /2)

sin 2 ( A/2) r r r r 1 so that 1 + 2 + 3 = [(sin2 (A/2) + sin2 (B/2) + bc ca ab r sin2 (C/2)]

b 2 - ac

A+C A-C sin = 2 2

=

=

1 c2 + a 2 - b2 = 2 ca 2

|sin A – sin C| =

1 1 r 2R

r1 4 R sin ( A/2) cos ( B /2) cos (C /2) = bc 2 R sin B ◊ 2 R sin C

Solution:

(d) 4

a2 + c2 = b2 + ac (a – c)2 = b2 – ac



(d)

Ans. (d)

3 - 4 sin A sin C is | A-C|

fi cos B = cos 60° =

(b) 2R – r

(c) r – 2R

In a D ABC, angles A, B, C are in A.P. then

(a) 1 (b) 2 (c) 3 Ans. (a) Solution: A, B, C are in A.P. fi B = 60°

In a triangle ABC, r r1 r + 2 + 3 is equal to bc ca ab

Ans. (a) Solution:

c2 + a 2 - b2 1 + 1 - (1/4) 7 = = . 2 ¥1¥1 2 ca 8

=

1 (1 – cos A + 1 – cos B + 1 – cos C) 2r

=

1 [3 – (cos A + cos B + cos C)] 2r

=

1 [3 – (1 + 4 sin (A/2) sin (B/2) sin (C/2))] 2r

=

1 2r

sin 2 B - sin A sin C 3 - sin A sin C 4 3 - 4 sin A sin C

3 - 4 sin A sin C = lim AÆ C | A - C|

A-C 2 = 1. |A - C |

r˘ 1 1 È ÍÎ2 - R ˙˚ = r - 2 R .

2sin

Example 17 In a triangle ABC, 2a2 + 4b2 + c2 = 4ab + 2ac, then numerical value of cos B is equal to (a) 0 (b) 1/8 (c) 3/8 (d) 7/8

Example 19

triangle ABC then

1 1 1 + 2 + 2 is equal to 2 a b g

(a)

cot A + cot B + cot C D

(b)

D cot A + cot B + cot C

Ans. (d) Solution: 2a2 + 4b2 + c2 = 4ab + 2ac fi (a – 2b)2 + (a – c)2 = 0 fi a = 2b and a = c a b c = fi a = 2b = c or = 1 1/2 1

If a, b, g are lengths of the altitudes of a

(c) D (cot A + cot B + cot C) (d) none of these Ans. (a)

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Solution: fi

D=

1 1 1 aa = bb = cg 2 2 2

a 2 + b2 + c2 1 1 1 + + = 4D 2 a2 b2 g 2

and the height of D AOC is one-third that of D ABC. Now, in D AOC, AO = (2/3) AD = 10/3. Therefore, applying the sine rule to D AOC, we get OC AO 10 sin (p /8) = fi OC = ◊ sin (p /8) sin (p /4) 3 sin (p /4)

and cot A + cot B + cot C 2R = [b2 + c2 – a2 + c2 + a2 – b2 + a2 + b2 – c2] 2 abc

Area of D AOC =

R (a 2 + b 2 + c 2 ) a 2 + b2 + c2 = = abc 4D 1 1 1 cot A + cot B + cot C Hence, 2 + 2 + 2 = . D a b g Example 20 in A.P. then

If in a triangle ABC the sides a, b, c are

(a) B > 60°

(b) B < 60°

(c) B £ 60°

(d) B = |A – C|

Ans. (c) Solution: since a, b, c are in A.P., 2b = a + c fi 2 sin B = sin A + sin C fi 4 sin (B/2) cos (B/2) = 2 sin ((A + C)/2) cos ((A – C)/2) = 2 cos (B/2) cos ((A – C)/2) 1 cos ((A – C)/2) £ 1/2 fi sin (B/2) = 2

=

1 10 10 sin (p /8) p p ◊ ◊ ◊ ◊ sin ÊÁ + ˆ˜ Ë 2 3 3 sin (p /4) 2 8¯

=

50 sin (p /8) cos (p /8) 50 25 ◊ = = 9 sin (p /4) 18 9 Area of D ABC = 3 ◊

fi Example 22

In a triangle ABC, medians AD and CE

are drawn. If AD = 5, DAC = p/8 and ACE = p/4, the area of the triangle is (a) 50/9 (b) 25/9 (c) 25/3 (d) 25/7 Ans. (c) Solution: Let O be the point of intersection of the medians of triangle ABC (Fig. 12.12). Then the area of D ABC is three times that of D AOC, O being the centroid of D ABC, divides the median through B in the ratio 2 : 1

25 25 = 9 3

In a triangle ABC, if tan (A/2) = 5/6 and

tan (b/2) = 20/37, the sides a, b, c of the triangle are in (a) A.P. (b) G.P. (c) H.P. (d) none of these Ans. (a) Solution: We have tan (C/2) = tan (90° - (A + B)/2) = cot

A+B cot ( A/2) cot ( B/2) - 1 = cot ( A/2) + cot ( B/2) 2

6 37 ◊ -1 222 - 100 122 2 = = = = 5 20 6 37 120 + 185 305 5 + 5 20

fi B £ 60° Example 21

1 ◊ AO◊ OC◊sin –AOC 2

Also tan

A C tan = 2 2

( s - b) ( s - c ) s( s - a)

( s - a ) ( s - b) s ( s - c)

5 2 s-b ◊ = fi 3(s – b) = s fi 2s = 3b 6 5 s



fi a + b + c = 3b fi a + c = 2 b which shows that a, b and c are in A.P. Example 23

If in a triangle ABC, a = 5, b = 4 and

cos (A - B) = 31/32, then the third side c is equal to (a) 3 (b) 6 (c) 7 (d) 9 Ans. (b) Solution:

fi Fig. 12.12

31 = cos (A – B) = 32

63 tan2

A- B =1 2

A- B 2 2 A- B 1 + tan 2 1 - tan 2

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(a) 2a sin (C/3) (c) 2c sin (B/3)



tan

A- B = 2

Now

tan

A- B a-b C cot = 2 a+b 2 1



63

fi Also

=

C = 2

tan cos C =

1 63

5-4 C cot 5+4 2

Ans. (c) Solution:

From DABP

AP AB = sin(2B /3) sin(p /2 + B /3) A

63 9

1 - tan 2 (C /2) 1 + tan 2 (C /2)

P

B/3

1 - 63 81 18 1 = = = 1 + 63 81 144 8 \

c2 = a2 + b2 – 2ab cos C = 25 + 16 – 2◊5◊4◊ ( 1 8 ) = 36

B

Example 24 A circle is inscribed in an equilateral triangle of side a. The area of any square inscribed in this circle is (a) a 2 (b) a2/4 (c) a 2/3 (d) a2/6 Ans. (d) Solution: Let a be the length of each side of the equilateral triangle ABC (Fig. 12.13). Then r, the radius of the in- circle = (1/3) AP (the altitude, median and the angle bisector of angle A) r=

a2 a 1 a2 = . 3 4 2 3

D

C

Fig. 12.14



AP =

Hence, c = 6.



(b) 2b sin (A/3) (d) 2c sin (C/3)

2c sin (B /3) cos(B /3) cos(B /3)

= 2c sin (B/3) Example 26

For any triangle ABC, the expression

(b + c – a) (c + a – b) (a + b – c) – abc is (a) positive (b) negative (c) non-positive (d) non-negative Ans. (b) Solution: We know that in any triangle b + c – a > 0, c + a – b > 0 and a + b – c > 0 Since A.M > G.M. we get (b + c – a ) + (c + a – b) > [(b + c – a) (c + a – b)]1/2 2 fi

c > [(b + c – a) (c + a – b)]1/2

Similarlyb > [(b + c – a) (a + b – c)]1/2 and a > [(c + a – b) (a + b – c)]1/2 Multiplying the above inequalities we get abc > (b + c – a) (c + a – b) (a + b – c) Note It a, b, c are any distinct positive numbers at most one of b + c – a, c + a – b, a + b – c can be non-positive and in that case also the given expression is negative).

Fig. 12.13

Area of the square PQRS inscribed in this circle = PQ2 = OP2 + OQ2 = 2r2 = 2 ¥

a2 a2 = . 4¥3 6

Example 25 If P is a point on the altitude AD of the triangle ABC such that –CBP = B/3, then AP is equal to

Example 27 The perimeter of a triangle is 6 times the arithmetic mean of the sines of its angles. If the side a is 1, then A is equal to (a) 30° (b) 60° (c) 90° (d) 120° Ans. (a)

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a+b+c= 6

fi fi 1+

C

(sinA + sinB + sinC ) 3 a + b + c = 2(sinA + sinB + sinC)

Solution

sinA + sinB + sinC = 2 sinA (sinA + sinB + sinC)

fi sin A =

1 2

Fig. 12.15

Since A.M. ≥ G.M. AD + DB ≥ 2

fi A = 30°

Example 28 If in a triangle ABC, cosA + 2 cosB + cos C = 2 then a, b, c are in (a) A.P. (b) G.P. (c) H.P. (d) none of these

B

E

a b c (using = = ) sinA sinB sinC fi

D

A

sinB sinC + = 2(sinA + sinB + sinC) sinA sinA

or

AD ◊ DB

c2 ≥ AD . DB 4

fiAD.DB £ c2/4 and the required value is c2/4.

Ans. (a)

Example 30

Solution:

a 3 + b3 + c 3 = 8, then the maximum value of a is sin 3 A + sin 3 B + sin 3 C

cosA + cosC = 2(1 – cosB) B A+C A–C cos = 2 ¥ 2 sin2 fi 2 cos 2 2 2 fi 2 sin

B B A–C A+C cos = 2 ¥ 2 sin cos 2 2 2 2 [\ A + B + C = p]

fi fi fi

cos A 2 A tan 2

cos

A–C A+C = 2 cos 2 2 C A C cos = 3 sin sin 2 2 2 C 1 tan = 2 3

In a triangle ABC, a ≥ b ≥ c. If

(a) 1/2

(b) 2

Solution: We have

If p denotes the perimeter of the triangle

Example 31 ABC, then b cos2



3 (s – b) = s fi 2s = 3b a+c a + b + c = 3b fi b = 2 a, b, c are in A.P.

(a) p

Example 29 In a triangle ABC, bisector of angle C meets the side AB at D and circumcentre at E. The maximum value of CD. DE is equal to (a) a2/4 (b) b2/4 2 (d) (a + b)2/4 (c) c /4 Ans. (c) Solution CD. DE = AD.DB AD + DB = c

(2 R)3 (sin 3 A + sin 3 B + sin 3C ) =8 sin 3 A + sin 3 B + sin 3C

fi R = 1, the radius of the circumcircle is 1. Greatest length of a side of a triangle inscribed in a circle can be equal to the diameter of the circle and hence the maximum value of the greatest side a is equal to 2.

(s – b)(s – c) (s – a )(s – b) 1 ¥ = s (s – a ) s (s – c) 3



(d) 64

Ans. (b)





(c) 8

C B + c cos2 is equal to 2 2 (b) 2p

(c) p/2

Ans. (c) Solution:

b cos2

C B + c cos2 2 2

= b¥

s ( s - c) s ( s - b) +c× ab ac

=

s ( s - c ) + s ( s - b) a

=

2s 2 - s (b + c) a

=

2s 2 - s (2 s - a) a

=

a+b+c p as =s= = ; 2 2 a

(d) p 2

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(

)

(

)

y = 15

3.

3 -1 x



x = 15 3 + 3

(

)



Example 32 In a triangle ABC sin3 A cos (B – C) + sin3 B cos (C – A) + sin3 C cos (A – B) is equal to



(a) 3 cos A cos B cos C (b) 3 sin A sin B sin C (c) 3 tan A tan B tan C (d) none of these Ans. (b) Solution: sin3 A cos (B – C) = sin2 A sin (B + C) sin (B – C) 1 = sin2 A [sin 2B + sin 2C] = sin2 A 2 [sin B cos B + sin C cos C ] So, the given expression is equal to sin A sin B (sin A cos B + cos A sin B) + sin B sin C (sin B cos C + cos B sin C) + sin C sin A (sin A cos C + cos A sin C) = sin A sin B sin (A + B) + sin B sin C sin (B + C) + sin C sin A sin (C + A)

So that PQ = BP + BQ = x + y / 3 = 15 3 + 3 + 15

3 =

30

30 = y

Similarly

3 - 1/ 3

(

= (60 + 15

)

3 ) m.

Example 34 In a cubical hall A B C D P Q R S with each side 10 m, G is the centre of the wall B C R Q and T is the mid point of the side AB. The angle of elevation of G at the point T is (a) sin–1 ( 1/ 3 ) (b) cos–1 ( 1/ 3 ) (c) tan–1 ( 1/ 3 )

(d) cot–1 ( 1/ 3 )

Ans. (a)

= 3 sin A sin B sin C Example 33 A and B are two points on the line joining their feet and between them. The angles of elevation of the tops of the A are 30° and 60° and as seen from B are 60° and 45°. If AB is 30 m, the distance between the (a) 30 + 15 3

(b) 45 + 15 3

(c) 60 – 15 3 Ans. (d)

(d) 60 + 15 3

Solution:

Let x and y

P

and Q respectively. Then AP = x cot 60° = x/ 3 , AQ = y cot 30° = y 3 BP = x cot 45° = x, BQ = y cot 60° = y/ 3 (Fig. 12.16) fi BP – AP= x – x / 3 = AB

Fig. 12.17

Solution:

Let H be the mid point of BC since

–TBH = 90°, (TH)2 = (BT)2 + (BH)2 = 52 + 52 = 50 Also since –THG = 90°, (TG)2 = (TH)2 + (GH)2 = 50 + 25 = 75 Let q be the required angle of elevation of G at T. (Fig. 12.17) then sin q =

GH 5 1 = = fi q = sin–1 ( 1/ 3 ) TG 5 3 3

Example 35 Each side of an equilateral triangle subtends an angle of 60° at the top of a tower h m high located at the centre of the triangle. If a is the length of each side of the triangle, then (a) 3a2 = 2h2 (b) 2a2 = 3h2 2 2 (c) a = 3h (d) 3a2 = h2 Ans. (b) Fig. 12.16

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Thus, 3 ( x 2 - 1) (2 x 2 + 2 x - 1) = 2 2( x 2 + x + 1) ( x 2 - 1)

Fig. 12.19



Fig. 12.18

Solution: Let O be the centre of the equilateral triangle ABC and OP the tower of height h. Then each of the triangles PAB, PBC and PCA are equilateral. Thus, PA = PB = PC = a. Therefore, from right-angled triangle POA, we have PA2 = PO2 + OA2. (Fig. 12.18) 2 a2 4 Êa ˆ 2 a = h + Á sec 30∞˜ = h + ◊ Ë2 ¯ 4 3



2

2

=

Let ABC be a triangle such that

p and let a, b and c denote the lengths of 6 the sides opposite to A, B and C respectively. The value(s) of x for which a = x2 + x + 1, b = x2 – 1 and c = 2x + 1 is are –ACB =

3)

(a) – (2 +

(b) 1 +

3

3 )x – (1 +

3)=0

-(2 - 3) ± (2 - 3)2 + 4(2 - 3)(1 + 3) 2(2 - 3)

-(2 - 3) ± 3 2(2 - 3)

= – (2 +

3 ),

3 +1

3

3 +1 In a triangle PQR, P is the largest angle and

Example 37

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

(c) 2 +

fix=

3 )x2 + (2 –

Thus, x =

2a2 = 3h2

Example 36

fi (2 –

But x > 1.

a2 2 2 = h + fi a2 = h2 3 3 fi

3 (x2 + x + 1) = 2x2 + 2x – 1

1 . Further the incircle of the triangle touches the 3 sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is are (a) 16 (b) 18 (c) 24 (d) 22 cos P =

Ans. (b), (d) Solution: Let PN = 2n – 2 QL = 2n RM = 2n + 2 P

(d) 4 3

M

Ans. (b)

N

2 2 2 Êp ˆ a + b - c Solution: By the law of cosines cos Ë ¯ = 2ab 6 But a2 + b2 – c2

= (x2 + x + 1)2 + (x2 – 1)2 – (2x + 1)2 = (x2 + 3x + 2) (x2 – x) + (x2 – 1)2 = (x + 1) (x + 2) x (x – 1) + (x2 – 1)2 2

2

2

= (x – 1) (x + 2x + x – 1) = (x2 – 1) (2x2 + 2x – 1)

Q

R

L

Fig. 12.20

We have PM = PN = 2n – 2 QN = QL = 2n RL = RM = 2n + 2

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PQ = PN + NQ = 4n – 2 QR = QL + LR = 4n + 2 RP = RM + MP = 4n

Now,

Also, PQ + PR - QR 2( PQ)( PR) 2

cos P =

2

2

1 (4n - 2) 2 + (4n) 2 - (4n + 2) 2 = 3 2(4n - 2)(4n) n-2 1 fi n=5 = 2n - 1 3



8 Ê8 ˆ p = p r2 = p Á k 2 ˜ Ë3 ¯ 3 fi

Thus, area of DXYZ = 6 6 . Next, x = s – 4k = 5k, y = 6k, z = 7k fi x = 5, y = 6, z = 7 sin

So, the sides of the triangle are 18, 20, 22. In a triangle XYZ, let x, y, z be the lengths

Example 38

of sides opposite to the angles X, Y, Z, respectively, and 2s s-x s- y s-z = = and area of incircle = x + y + z. If 4 3 2 8p of the triangle XYZ is , then 3 (a) area of the triangle XYZ is 6 6 (b) the radius of circumcircle of the triangle XYZ 35 6 is 6 (c) sin

Ans. (a), (c), (d) Solution:

s-x s- y s-z = = = k(say) 4 3 2



s – x = 4k, s – y = 3k, s – z = 2k Adding we get 3s – (x + y + z) = 9k fi s = 9k Also, area of DXYZ is given by D= =

s ( s - x)( s - y )( s - z )

r=

=

( s - x)( s - y )( s - z ) (4)(3)(2) = xyz (5)(6)(7)

=

4 35

s(s - z ) = xy

=

9(9 - 7) = (5)(6)

3 5

3 Ê X +Yˆ sin 2 Á = ˜ Ë 2 ¯ 5



Finally, R = =

xyz (5)(6)(7) = 4D 4( 6 6 ) 35 4 6

π

35 6 6

Example 39 In a triangle, the lengths of the two larger sides are 10 and 9, respectively. If the angles are in A.P., the length of the third side can be (a) 5 –

6

(b) 3 3

(c) 5 Ans. (a) and (d)

(d) 5 +

6

Solution: If the angles A, B and C of triangle ABC are in A.P. (A < B < C), then 2B = A + C. Also, A + B + C = 180°, so that B = 60°. Since the greater side has the greater angle opposite it, we have b = 9 and c = 10. Therefore, by the law of cosines

(9k )(4k )(3k )(2k ) = 6 6k 2

D 6 6k 2 2 2 k = = s 9k 3 We are given \

X Y Z ( s - y )( s - z )( s - z )( s - x)( s - x)( s - y ) sin sin = 2 2 2 ( yz )( zx)( xy )

Z Ê X +Yˆ Êp Zˆ sin Á = sin Á - ˜ = cos Ë 2 ˜¯ Ë 2 2¯ 2

X Y Z 4 sin sin = 2 2 2 35

3 2 Ê X +Yˆ (d) sin Á = ˜ Ë 2 ¯ 5

k2 = 1 fi k = 1

cos B = fi or fi

c2 + a 2 - b2 2ca

1 100 + a 2 - 81 = 2 2 ¥ 10 ¥ a a2 – 10 a + 19 = 0 a =5 ± 6.

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There exists a triangle ABC satisfying

Example 40

(a)

2 bc sin ( A/2) b+c

(b)

2 bc cos ( A/2) b+c

(c)

abc A cosec 2 R (b + c) 2

(d)

4D A cosec . b+c 2

(a) tan A + tan B + tan C = 0 sin A sin B sin C = = (b) 2 3 7 (c) (a + b)2 = c2 + ab and 3 (d) sin A + sin B =

2 (sin A + cos A) =

3 +1 3 , cos A cos B = 4 2

= sin A sin B. Fig. 12.21

Ans. (c) and (d) Solution: (a) In a triangle ABC, A + B + C = p, so tan A + tan B + tan C = tan A tan B tan C. Since none of tan A, tan B, tan C can be zero, a triangle given in (a) is not possible. (b) (sin A)/2 = (sin B)/3 = (sin C)/7, so that by the laws of sines a b c a+b 5 = = fi = 2 3 7 c 7 which is not possible, as the sum of two sides of a triangle is greater than the third. (c) We have (a + b)2 = c2 + ab, so that a 2 + b2 - c2 1 = 2ab 2 1 2

Ans. (b) and (c). Solution: From D ABD (Fig. 12.21), we get by the sine rule AD BD = (1) sin B sin ( A/2) Since the bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle, we have BD/DC = c/b. Also, BD + DC = BC = a. Therefore, c ◊a (2) BD = b+c From (1) and (2) we get ca sin B ◊ AD = b + c sin ( A/2)



cos C = -

and

2 (sin A + cos A) =

3

=

1

3 2

=2◊

fi fi

1 2

sin A +

2



cos A =

C = 120°

pˆ 3 Ê sin Á A + ˜ = Ë ¯ 2 4

p p = 4 3 That is, A = p /12 = 15° and B = 45°. Such a triangle is possible. (d) Since cos A cos B = sin A sin B = 3 /4, we get cos (A + B) = 0, i.e., A + B = p 2 . Also, from cos (A – B) = 3 /2, we get A – B = p 6 , so that A = p 3 , B = p 6 and fi

A+

sin A + sin B =

3 1 3 +1 + = 2 2 2

ca sin B ◊ 2 cos ( A/2) (b + c) sin ( A/2) ◊ 2 cos ( A/2)

=2 =

c a A ◊ ◊ sin B cos b + c sin A 2

c b A ◊ ◊ sin B cos 2 b + c sin B

2bc cos ( A/2) b+c

Thus, (b) is correct. 2bc cos ( A/2) sin ( A/2) ◊ Also, AD = b+c sin ( A/2) =

bc sin A (b + c) sin ( A/2)

=

abc A cosec 2 R (b + c) 2

Hence, such a triangle is also possible.

Therefore, (c) is also correct. Example 41 In a triangle ABC, the length of the bisector of angle A is

[by the law of sines]

Again AD =

abc ◊ 4D A cosec 2abc (b + c) 2

a ˘ È Í sin A = 2 R ˙ Î ˚

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=

2D A cosec b+c 2

abc ˘ È Í R = 4D ˙ Î ˚

Hence (d) is not correct. Example 42

Given an isosceles triangle with equal

sides of length b, base angle a < p 4 , R, r the radii and O, I the centres of the circumcircle and incircle, respectively. Then 1 (a) R = b cosec a 2

1 2 b sin 2a D b sin 2a 2 = = Also, r = s 1 2(1 + cos a ) (b + b + 2b cos a ) 2 so that (c) is correct. Further OI = | OD + DI | = |OD + r | because a < p 4 , A > p 2 and O lies on AD produced. Now, from right-angled triangle ODB, we get OD2 = OB2 – BD2 = R2 – (b cos a)2

(b) D = 2b2 sin 2a b sin 2a (c) r = 2(1 + cos a )

=

b 2 (1 - 4 sin 2 a cos 2 a ) = 4 sin 2 a

b cos (3a /2) (d) OI = . 2 sin a cos (a /2)

=

b 2 (cos 2 a - sin 2 a ) 2 4 sin 2 a

=

b 2 cos 2 2a (2sin a ) 2

Ans. (a), (c) and (d).

Fig. 12.22

Solution: Let ABC be the isosceles triangle with AB = AC = b and –B = –C = a (Fig. 12.22). Let AD be the perpendicular bisector of the side BC. Since DABC is isosceles, AD is also the bisector of angle A, so that O and I both lie on AD. We have OB = R and ID = r. Also, since O is the circumcentre, we get OA = OB = R. Therefore, from isosceles triangle OAB OB AB = sin (90∞ - a ) sin 2a cos 1 = b cosec a 2 sin a cos a 2 so that (a) is correct. 1 Again D = BD◊AD = b cos a◊b sin a = b2 sin 2a 2 fi

R=

so that (b) is not correct.

1 b2 - b 2 cos 2 a 4 sin 2 a



OD =

\

OI =

[from (a)]

b cos 2a 2 sin a b sin 2a b cos 2a + 2 (1 + cos a ) 2 sin a

=

b sin 2a b cos 2a + 2 4 cos (a /2) 4sin (a /2) cos (a /2)

=

b sin 2a sin (a /2) + cos 2a cos (a /2) ◊ 4 cos (a /2) sin (a /2) cos (a /2)

=

b cos (3a /2) 2 sin a cos (a /2)

Thus, (d) is also correct. Example 43 If H is the orthocentre of triangle ABC, then AH is equal to (a) 2R cos A (b) 2R sin A 2 abc (c) a cot A (d) cos A. D Ans. (a) and (c) Solution:

Referring to DAHB (Fig. 12.23), we have

c AH = sin (90∞ - A) sin ( A + B) fi

AH =

c cos A sin (180∞ - C )

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=

c c cos A È ˘ = 2 R˙ = 2 R cos A Í sin C Î sin C ˚

=

a 2 + b2 +1=1+1=2 c2 [

a 2 + b 2 = c 2]

Example 45 If in a triangle ABC, a, b, c are in A.P. and p1, p2, p3 are the altitudes from the vertices A, B, C respectively then (a) sin A, sin B, sin C are in A.P. (b) sin A, sin B, sin C are in H.P. 3R D 1 1 1 3R + + £ (d) p1 p2 p3 D (c) p1 + p2 + p3 £

Fig. 12.23

so that (a) is correct while (b) is not correct. Also, abc abc AH = 2R cos A = 2◊ cos A = cos A 4D 2D a ◊cos A = a cot A and AH = 2R cos A = sin A so that (c) is correct, while (d) is not correct. Example 44 If the tangents of the angles A and B of a triangle ABC satisfy the equation abx2 – c2x + ab = 0, then (a) tan A = a b (b) tan B = b a (c) cos C = 0 (d) sin2 A + sin2 B + sin2 C = 2. Ans. (a), (b), (c) and (d). Solution:

From the given equation, we get

tan A + tan B = c2/ab tan A tan B = 1. Since tan A + tan B tan (A + B) = 1 - tan A tan B

and

Solution: D = fi

p1 =

1 1 1 bp2 = ap1 = cp3 2 2 2

2D 2D 2D , p2 = , p3 = a b c

p1, p2, p3 are in H.P. 2D 2D 2D Now p1 = , p2 = , p3 = 2 R sin A 2 R sin B 2 R sin C fi fi fi \

1 R sin A 1 R sin B 1 R sin C , , = = = p1 D p2 D p3 D sin A, sin B, sin C are in A.P. sin A + sin C = 2 sin B 1 1 1 R + + = (sin A + sin B + sin C) p1 p2 p3 D =

Example 46

A

c

Ans. (a) and (d)

b

3R 3R sin B £ . D D

If a, b, c are the sides of the DABC and a2,

b2, c2 are the roots of x3 – px2 + qx – k = 0, then p cos A cos B cos C + + = (a) a b c 2 k (b) a cos A + b cos B + c cos C =

a

B

C

Fig. 12.24

we get A + B = p 2 and C = p 2 . Therefore, triangle ABC is right-angled at C (Fig. 12.24). Hence, tan A = a/b, tan B = b/a, cos C = 0, sin A = a/c, sin B = b/c and sin C = 1, so that a 2 b2 sin A + sin B + sin C = 2 + 2 + 1 c c 2

2

2

(c) a sin A + b sin B + c sin C = 8 D3 (d) sin A sin B sin C = k Ans. (a), (b), (c) and (d)

4q - p 2 2 k 2 pD k

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Again, from D ACE, we get

Solution: cos A cos B cos C + + (a) a b c

cos q =

=

b2 + c2 - a 2 c2 + a 2 - b2 a 2 + b2 - c2 + + 2 abc 2 abc 2 abc

=

p a 2 + b2 + c2 = 2 abc 2 k

73 25 + 16 9 = 9 73 2◊ ◊4 9

(b) a cos A + b cos B + c cos C

=

a 2 (b 2 + c 2 - a 2 ) + b 2 (c 2 + a 2 - b 2 ) + c 2 ( a 2 + b 2 - c 2 ) 2 abc

=

4 (a 2 b 2 + b 2 c 2 + c 2 a 2 ) - (a 2 + b 2 + c 2 ) = 2 abc =

2

AE 2 + AC 2 - CE 2 2 AE ◊ AC

\ tan q =

192 3 8 ◊ = 9 2 ◊ 4 ◊ 73 73 sec2 q - 1 = 3 8 .

Example 48 PQR is an isosceles triangle whose equal sides PQ and PR are at right angles. S and T are points on PQ such that QS = 6 SP and QT = 2TP. PRS = q, PRT = f.

4 q - p2

(a) sin 2f = 3/5

(b) 2f + q = 45°

2 k

(c) sin 2q = 3/5

(d) sin f + cos f =

(c) a sin A + b sin B + c sin C a 2 + b2 + c2 (a 2 + b 2 + c 2 ) 4 D 2 pD = = = . 2R 2 abc k (d) sin A sin B sin C = Example 47

abc ¥ 64 D 3 8 D3 abc = = . k 8(abc)3 8 R3

If in a triangle ABC, BC = 5, CA = 4, AB

= 3 and D, E are points on BC such that BD = DE = EC, CAE = q (b) AE2 = 73/9 (a) AE2 = 73/3 (c) tan q = 3/8 Ans. (b), (d)

Solution: Referring to Fig. 12.26, if SP = x, then QS = 6x. Also, PQ = PR = 7x, RS = 49 + 1 x and – PRQ = – PQR = 45°. –PRS = q, – SRQ = 45° – q, – PRT = f and – TRQ = 45° – f. Then from D SRQ, we get SR SQ = sin (45∞ - q ) sin 45∞ fi

sin (45° – q) =

1 2



6x 50 x

=

3 5

(d) cos q = 3 73

Fig. 12.25

Fig. 12.26

–CAE = q and CE = ( 5 3 ). Then (AE) = 4 + (5 3) - 2 ◊ 4 ◊ (5 3) cos C 2

2

2

25 40 Ê 42 + 52 - 32 ˆ = 16 + m 9 3 ÁË 2 ◊ 4 ◊ 5 ˜¯ =

73 9

[from D ABC]

10

Ans. (a), (b), (d)

In D ACE (Fig. 12.25),

Solution:

4

Also

Ê7 ˆ PT = Ë ¯ x 3

and

RT = =

7 (7 x) 2 + ÊÁ xˆ˜ Ë3 ¯ 7 10 x 3

2

(1)

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sin f =

PT = RT

cos f =

PR 3 = RT 10

1 10

Ans. (a), (d) ,

Solution:

[by the law of sine] 2 cos[(1 2) ( A + B)] sin [(1 2) ( A - B)] = 2sin [(1 2) C ] cos [(1 2) C ]

4

sin f + cos f =

10

sin 2f = 2 sin f cos f

and

=

= 3 5 = sin (45° – q) [from 1] 2f = 45° – q

fi Example 49

cos A + 2 cos C sin B = , then the triangle can be m cos A + 2 cos B sin C (b) isosceles (d) obtuse angled

Solution: The given relation can be written as cos A (sin B – sin C) + sin 2B – sin 2C = 0 or cos A (sin B – sin C) + 2 cos (B + C) sin (B – C) = 0 or cos A (sin B – sin C) + 2 cos (180° – A) sin (B – C) = 0 or cos A [sin B – sin C – 2 sin (B – C)] = 0 from which it follows that either cos A = 0, so that A = p /2 and D ABC is right-angled, or sin B – sin C – 2 sin (B – C) = 0 fi (b – c) – 2(b cos C – c cos B) = 0 [by the law of sines] fi

Ê a 2 + b2 - c2 c2 + a 2 - b2 ˆ (b – c) – 2 Á ˜¯ = 0 2a 2a Ë

fi fi fi fi

a (b – c) –2 (b2 – c2) = 0 (b – c) [a – 2(b + c)] = 0 b–c =0 triangle is isosceles

(c)

fi sin q =

(a + b) sin q 2 ab c sin q 2 ab

= cos

= cos

A- B 2

A+ B (d) = cos 2 2 ab c sin q

A+ B 2

=

sin A sin B sin [(1 2) ( A + B)] (a + b) sin q

fi =

sin 2 [(1 2) ( A + B)] - sin 2 [(1 2) ( A - B)] sin [(1 2) ( A + B)]

2 ab

(

sin A + sin B

sin A sin B sin[(1 2) ( A + B)] 2 sin A sin B

=

2sin[(1 2) ( A + B)]cos[(1 2) ( A - B)] 2 sin [(1 2) ( A + B)]

= cos[(1 2) ( A - B)] c sin q

and

2 ab =

=

sin C

sin A sin B sin [(1 2) ( A + B)] 2 sin A sin B

2 sin (C 2) cos (C 2) = cos ((A + B)/2) 2 sin [ ( A + B) / 2]

Example 51 If D is the area of a triangle with sides of lengths a, b, c, then 1 (a + b + c) a b c (a) D £ 4 (b) D =

1 4

(a + b + c) a b c if a = b = c

1 (a + b + c) a b c 4 1 (d) D = abc (a + b + c) . 4

b + c > a)

Example 50 If in a triangle ABC, q is the angle determined by cos q = (a - b)/c, then A- B (a + b) sin q (a) = cos 2 2 ab (b)

2 sin [(1 2) C ] sin [(1 2) ( A - B)] sin [(1 2) ( A - B)] = 2 sin [(1 2) C ] sin [(1 2) ( A + B)] sin [(1 2) ( A + B)]

If in a triangle ABC

(a) equilateral (c) right angled Ans. (b), (c)

a - b sin A - sin B = sin C c

cos q =

(c) D ≥

Ans. (a), (b) Solution: We have a2 ≥ a2 – (b – c)2 = (a + b – c) (a – b + c) fi

a2 ≥ (2s – 2c) (2s – 2b) = 4 (s – b) (s – c)

Similarly b2 ≥ 4(s – c) (s – a) c2 ≥ 4 (s – a) (s – b) So that a2 b2 c2 ≥ 64 [(s – a) (s – b) (s – c)]2 fi

abc ≥ 8 (s – a) (s – b) (s – c)

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(a + b + c) (abc) ≥ 8 × 2s (s – a) (s – b) (s – c) = 16 D





2

1 4

(a) x = 25 6 m

(a + b + c) a b c

and if a = b = c, the triangle is equilateral and its area is 1 4

2

3a . 4

3a ¥ a3 =

Example 52 A and B are two points 30 m apart in a line on the horizontal plane through the foot of a tower lying on opposite sides of the tower. The distances of the top of the tower from A and B are 20 m and 15 m respectively, if the angle of elevation of the top of the tower at A is q and the height of the tower is h, then (a) cos q = 29/36 (b) cos q = 43/48 (c) h = 5 5 / 3 Ans. (b) and (d)

building and the pole subtend equal angle q. If the horizontal distance of the observer from the pole is x, then (c) tan q = 2/ 3 (d) tan q = 2 6 Ans. (a) and (c) Solution: Let PQ be the pole on the building QR and O be the observer. Then PQ = 50, QR = 250 (Fig. 12.28)

(d) h = 5 455 /12

Solution: Let OP be the tower of height h, then AB = 30, AP = 20 and BP = 15 (Fig. 12.27)

Fig. 12.28

fi PR = 300 so the observer is at the same height as the top P of the pole. So OP = x. Then from right angled triangles OPQ and OPR, 50 300 and tan 2q = . x x

tan q = so that Fig. 12.27

so from D ABP cos q =

202 + 302 - 152 2 ¥ 20 ¥ 30

1075 43 = = 1200 48 fi

and

Ê 43 ˆ sin q = 1 - Á ˜ Ë 48 ¯

2

2 tan q 300 = 2 x 1 - tan q

50 2 x = 300 fi 3 Ï1 - Ê 50 ˆ ¸ = 1 fi Ì ÁË ˜¯ ˝ x x ˛Ô 50 2 ÓÔ 1- Ê ˆ Ë x¯ 2¥

2 2 50 3 Ê 50 ˆ = 25 6 fi Á ˜ = fix= Ë x¯ 3 2

and tan q =

=

2304 - 1849 48

=

455 48

h = 20 sin q = 20 ¥

(b) x = 25 3 m

455 5 455 = 48 12

Example 53 A pole 50 m high stands on a building 250 m high. To an observer at a height of 300 m, the

50 25 6

=

2 . 3

Example 54 A monument ABCD stands at a point A on a level ground. At a point P on the ground the portions AB, AC, AD subtend angles a, b, g respectively. If AB = a, AC = b, AD = c, AP = x and a + b + g = 180° then (a) x 2 = (a + b + c)/abc (b) x2 = abc/(a + b + c) (c) tan a + tan b + tan g = (a + b + c)3/2 (d) tan a tan b tan g = (a + b + c)3/2 Ans. (b), (c) and (d)

abc abc

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Solution: We have a = AB = AP tan a = x tan a (Fig. 12.28) b = AC = x tan b and c = AD = x tan g so that a + b + c = x (tan a + tan b + tan g ) = x tan a tan b tan g [ a + b + g = 180°] and abc = x3 tan a tan b tan g abc fi x2 = a+b+c so that tan a + tan b + tan g = tan a tan b tan g = (a + b + c)/x = (a + b + c)3/2/ abc .

Solution: PQ = RS = k, the length of the ladder. Then – PQO = a, – RSO = b, SQ = a and PR = b. From right-angled triangle ROS, we have OS = k cos b and OR = k sin b. From right-angled triangle POQ, we have OQ = k cos a and OP = k sin a. \ a = OS – OQ = k(cos b – cos a) and b = OP – OR = k(sin a – sin b) a cos b - cos a = = fi sin a - sin b b

a +b a -b sin a +b 2 2 a+b a-b 2 2 cos sin 2 2 2 sin

a2 + b2 = k2 [2 - 2 cos (a - b)] a 2 + b2 4k 2



Êa - bˆ = sin2 Á Ë 2 ˜¯

MATRIX-MATCH TYPE QUESTIONS Example 56 In a triangle ABC Column 1 (a) b cosC + c cos B Fig. 12.30

Example 55 A ladder of length k rests against a wall at an angle a to the horizontal. If foot is pulled away through a distance a, so that it slides a distance b down the wall, b with the horizontal. Then a Êa + bˆ (a) tan Á = Ë 2 ˜¯ b a Êa - bˆ (b) tan Á = Ë 2 ˜¯ b Êa - bˆ (c) cosec Á = Ë 2 ˜¯ (d) sin (a + b) = ab/k

2k a +b 2

(b) b2 + c2 - 2bc cos A (c) bc sin A

(q) (a + b + c)r (r) a2

(d) b sin A – a sin B

(s) 0

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

Solution: 2

b cos C + c cos B

= CD + DB = BC = a

2

cos A =

Ans. (a), (c)

Column 2 (p) a

b2 + c2 – a 2 fi b2 + c2 - 2bc cos A = a2 2bc

1 bc sin A = sr 2 \ bc sin A = 2sr = (a + b + c) r. D =

Also

sin A sin B = a b

Fig. 12.31

Fig. 12.32

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Example 57 If R, r denote respectively the radius of the circumcircle and incircle of a triangle ABC, then Column 1 Column 2 (a) a cos A + b cos B + c cos C (p) 2(R + r) (b) a cot A + b cot B + c cot C (q) r (c) 4R sin (A/2) sin (B/2) sin (C/2) a b c + + sin A sin B sin C p q r s Ans. a p q r s (d)

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(r) 4R sin A sin B sin C (s) 6 R

Fig. 12.33

fi BD2 = 49 fi BD = 7 fi D BCD is isosceles fi ED = 1/2 CD = 2 7 2 - 22 =

and BE =

45 , AE = 6

Example 59 In a triangle ABC, sin A cos B = 1/4, 3 tan A = tan B. Column 1 Column 2

Solution: a cos A + b cos B + c cos C = 2R (sin A cos A + sin B cos B + sin C cos C) = R (sin 2A + sin 2B + sin 2C) = 4R sin A sin B sin C Next, a cot A + b cot B + c cot C = 2R (cos A + cos B + cos C)

B A A A p

–A + B/2 +B + C/2 q r

s

(a) (b) (c) (d) Ans.

(p) (q) (r) (s)

= 2R [1 + 4 sin (A/2) sin (B/2) sin (C/2)]

a

p

q

r

s

= 2(R + r)

b

p

q

r

s

as 4R sin (A/2) sin (B/2) sin (C/2) = r (See the Text)

c

p

q

r

s

Example 58 In a triangle ABC, a = 7, b = 8, c = 9, BD is the median and BE the altitude from the vertex B, then Column 1 Column 2 (a) BD (p) 2 (b) BE (q) 7

d

p

q

r

s

(c) ED (d) AE p q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

(r) 45 (s) 6

Solution:

3 tan A = tan B fi 3 sin A cos B = cos A sin B

fi cos A sin B = 3/4 fi sin (A + B) = 1/4 + 3/4 = 1 fi C = 90° and tan A = tan (90° - B) = cot B fi tan2 A = 1/3 fi tan A =

cos A = =

42 + 92 - BD 2 2¥4¥9

82 + 9 2 - 7 2 (From D ABC) 2 ¥8¥ 7

1 3

fi A = 30° fi B = 60º

Example 60 The angles of a triangle are in the ratio 2:3:7 Column 1 Column 2 (a) smallest side = 2 (b) smallest side =

Solution: AD = 4 From D ABD

60° 90° 30° 75°

(c) s = 3 + 3 + 2 (d) D =

(

)

3 -1 / 4

2

(p) largest side =

3+ 1

(q) largest side =

6+ 2

(r) largest side = 1 (s) largest side = 2 + 3

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p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

Solution: D = 400

a b = = 1/2 1/ 2



a 2

b = 2

=

If a =

3 +1

= k (say)

2 k and largest side c =

(

)

3 +1 k

6+ 2 3 +1

2 + 3 , k = 2, c = 2 + 2 3

If s = 3 + If D =

)

c

2 , k = 1, c =

3 -1 1 , then from D = bc sin A we have 4 2

3 -1 1 = ¥ 2k ¥ 4 2 fi k2 =

2

(

3 –1

)

3 +1

(

=

)

3 + 1 k sin 30°

(

)

(

)

3 –1 4

D 400 3 = = 20 s-a 20 3

a + b - c = 40 (2 -

3 ) fi 2s - 2c = 40 (2 -

)

3)

abc = 40 4D

Example 62 In a triangle ABC, AD is perpendicular to BC and DE is perpendicular to AB. Column 1 Column 2 (a) Area of D ADB (p) (b2/4) sin 2C (b) Area of D ADC (q) (c2 /4) cos2 B sin 2B (c) Area of D ADE (r) (c2/4) sin 2B (d) Area of D BDE (s) (c2 /4) sin2 B sin 2B p q r s Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

2

fik=

3 –1 2

Solution:

ADE = B

From D ABD, AD = c sin B

and hence c = 1

BD = c cos B

The area of a triangle ABC is 400

Example 61

(

s-a = 20 3 fi r1 =

abc = 64000 3 fi R =

3 + 1 /2 2

2,c=

If a = 2, k =

D 400 3 = s 20 2 + 3

fi r3 = 20 (2 3 + 3)

c

(

The smallest side a =

3)fir=

= 20 3 2 - 3

Solution: Angles of the triangle are 30°, 45°, 105°. a b c then = = sin 30∞ sin 45∞ sin 105∞ fi

3 , s = 20 (2 +

Column 1

3.

Column 2 3)

(a) s = 20 (2 +

(p) R = 40

(b) s - a = 20 3

(q) r = 20 ( 2 3 - 3)

(c) a + b - c = 40 (2 (d) abc = 64000 3 p q r s Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

3)

(

(r) r3 = 20 2 + 3 (s) r1 = 20

)

Fig. 12.34

3

fi Area of D ABD c2 1 sin 2B = c2 sin B cos B = 4 2 From D ACD, CD = b cos C, AD = b sin C b2 sin 2c 4 AE = AD sin B fi AE = c sin B sin B, ED = c sin B cos B fi Area of D ADE = (1/4)c2 sin2 B sin 2B Area of D BDE = (1/4)c2 cos2 B sin 2B

fi Area of D ADC =

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Example 63 PQ and RS are two vertical towers of the same height h. O is a point on the horizontal line through the feet P and R of the towers. The angle of elevation at O, of the middle point T of RS and the top Q of PQ is a, and that of top S of RS is a + b. Then Column 1 Column 2 (a) OR (b) OT (c) OS (d) RP

(h/2) cos (a + b) cosec b (h/2) cos a cosec b h cot (a + b) (h/2) cot a s

p

(p) (q) (r) (s) q r

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

Solution:

p

Ans.

the top of the tower q r s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: Let OP be the tower of height h. PQ of height x and y be the distance of the point A from the foot O of the tower

From D OST

OT OS ST = = cos (a + b ) cos a sin b

Fig. 12.36

h cos (a + b ) h cos a OT = , OS = (ST = h/2) 2 sin b 2 sin b

PAQ = a where tan a = 1/9

then

PAO = b where tan b = 4/5

OR = h cot (a + b) (from D ORS) = (h/2) cota (from D OTR) = PR

h 4 = tan b = y 5 h+x = tan (a + b) y =

tan a + tan b 1 - tan a tan b

=

(1 / 9) + (4 / 5) =1 1 - (1 / 9) (4 / 5)

h h/2

h x x 4 1 + =1fi = 1- = y 5 5 y y

fi Fig. 12.35

Example 64 subtends an angle tan-1 (1/9) at a point on the horizontal line through the foot of the tower where the tower subtends an angle tan-1 (4/5), then Column 1 Column 2

So that if y = 100 m, x = 20 m, h = 80 m and AP = = 20 41 m Example 65 2

2

If a, b, c are the sides of the triangle ABC

2

and a , b , c are the roots of x3 – px 2 + qx – k = 0 Column 1

(b) Height of the tower. (c) Distance of the point from the foot of the tower (d) Distance of the point from

h2 + y 2

(q) 100 m (r) 20 m

(a)

(s) 20 41 m

= 1/2 (b) a cos A + b cos B

cos A cos B cos C + + a b c

Column 2 (p)

p = 2, q = 1, k = 4

(q) p = 1, q = 1/4, k = 1

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+ c cos C = 0 (c) sin 2A + sin 2B

c2 + a 2 - b2 2ac = 1/2 fi cos B = 1/2 fi B = 60°

(a) c 2 + a 2 – 2ac = b 2 – ac fi

(r) p = 3, q = 9/4, k = 9

Solution:

(s) p = 4, q = 4, k =2

and cos B + sin C = 3/2 fi sin C = 1 fi C = 90° and A = 30° A+C (b) A + B + C = 180°, B = fi B = 60°, C + A = 2 120°

+ sin 2C = 0 (d) a sin A + b sin B + c sin C = 2D p q r s Ans. a p q r s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

C = 3A fi A = 30°, C = 90°. (c) Length of the bisector of angle B is (see Ex. 34) 2ca 3ca 3 cos (B/2) = fi cos (B /2) = c+a 2 c+a = cos 30°

p cos A cos B cos C + + = a b c 2 k

Solution:



(d) We have (a + b + c) (a + 2b + c) = 3(a + b) (b + c) (See Ex. 40)

= 1/2 if p2 = k which is true in p, q, r, a cos A + b cos B + c cos C = kπ0

4q - p 2 2 k

2

= 0 if p = 4q,

which is true in p, q, r, s sin 2A + sin 2B + sin 2C = (1/R)(a cos A + b cos B + c cos C ) 2 pD = 2D if p 2 = k a sin A + b sin B + c sin C = k

Column 1

(a + c)2 + 3b (a + c) + 2b2 = 3(ab + bc + ca + b2)



a 2 + c2 - b2 1 = 1 fi cos B = fi B = 60° ac 2

Example 67 Column 1 Column (a) In a triangle ABC, (a + b + c) (b + c – a) = lbc if l = (p) (b) If the area of a triangle ABC is (q) given by D = a2 – (b – c)2 then cot A/2 is equal to (c) In a triangle ABC, B = 60° then (r) a2 + b2 + c2 – ac = lb2 if l = tan A tan B tan C = = π 0 for (d) If 1 2 3 a triangle ABC, then tan B + tan C – tan A = (s) p q r s

Column 2

(a) (c – a) 2 = b2 – ac and (p) A = 30° cos B + sin C = 3/2 (b) A, B, C are in A.P. (q) B = 60° and C = 3A (c) The length of the bisector (r) C = 90° of angle B = 3 ca / (c + a) and a = b 1 1 3 + = (s) A = B = C = 60° (d) a+b b+c a+b+c

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.



In a triangle ABC

Example 66

B = 60°, a = b fi A = B = 60° = C

Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

2 1 2

3

4

Solution: (a) we have (b + c)2 – a2 = lbc fi b2 + c2 – a2 = (l – 2) bc b2 + c2 - a 2 l-2 fi cos A = = , As A is angle of the 2bc 2 l-2 1, its largest angle is twice the smallest. The value of 3a3 + 20a2 + 35a is equal to 210k, k = Ans. 5 Solution: Let a be the smallest angle then 2a is largest. So from the triangle we have (Fig. 12.45) sin a sin 2a a +1 = fi = 2 cos a a +1 a -1 a -1 Also cos a = \

(a + 1)2 + a 2 - (a - 1)2 a 2 + 4a = 2(a + 1) a 2(a + 1) a

a +1 a+4 = fia=5 a +1 a -1

Hence 3a3 + 20a2 + 35a = 375 + 500 + 175 = 1050 = 210 ¥ 5 k= 5

Fig. 12.45 r

Example 92 C/2

Fig. 12.44

fi fi

BC = 2y – 1, CA = 2y + 1 and AB = 2y, s = 3y C B r r = and tan = Now tan y 2 2 y -1

In a triangle ABC, B = 30°, C = 150°,

b = 50 3 . a1, a2 are the two values of the third side a. x and y are respectively the arithmetic and geometric means 2 2 Ê x ˆ Ê y ˆ of a1 and a2 then Á ˜ - Á is equal to (k)k ˜ Ë 3¯ Ë 6¯ where k = Ans. 5

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Solution:

b2 = a2 + c2 - 2ac cos B

(

fi 50 3

)

2

= a2 + (150)2 - 2a.150.

(

3/2

)

fi a - 150 3 a + 15000 = 0 2

fi a = 50 3 or 100

19 ¥ 32 ¥ 19 ¥ 11 = (19 ¥ 3)2 = 3249 11

= 26 ¥ 53 – 1 fik=1

3

3 , a2 = 100

Let a1 = 50

fi x2 =

3

a1 + a2 = 75 3 , y = 50 6 2

then x =

2 2 Ê x ˆ Ê y ˆ fi Á ˜ -Á = (75)2 - (50)2 = (75 + 50) (75 - 50) Ë 3¯ Ë 6 ˜¯ = 125 ¥ 25 = 3125 = 55 fi k = 5

In a triangle ABC, if r1 = 2r2 = 3r3 then

Example 93

[100 - (4a/5b)] is equal to 9800 + k, k = Ans. 1 Solution: We are given s tan (A/2) = 2 s tan (B/2) = 3s tan (C/2) 2

Fig. 12.46

Example 95 A vertical tower PQ subtends the same angle 30° at each of two places A and B, 60 m. apart on the ground, AB subtends an angle 120° at the foot of the tower. If h is the height of the tower then 9h2 + h + 1 is equal to 1207k, k = Ans. 3 Solution: AP = PB = h cot 30° = 3 h – 1/2 = cos 120° =

3h 2 + 3h 2 - 602 2 . 3h 2

tan ( A / 2) tan ( B / 2) tan (C / 2) = = = k (say) 6 3 2 Since A/2 + B/2 + C/2 = 90°

fi 9h2 = 602 fi h = 20 fi 9h2 + h + 1 = 3621

tan (A/2) tan (B/2) + tan (B/2) tan (C/2) + tan (C/2)

= 1207 ¥ 3



tan (A/2) = 1

fik=3

fi (6.3 + 3.2 + 2.6)k = 1 fi k = 1/6 2 tan ( A / 2) =1 so that sin A = 1 + tan 2 ( A / 2) 2

Similarly sin B = 4/5 a sin A 5 4a = = fi =1 and sin B b 4 5b \ [100 - (4a/5b)]2 = (100 - 1)2 = 992 = 9801 = 9800 + 1 fi k = 1 ABC is a triangle, B = 60°, C = 30° BC is 8 produced to D so that ADB = 15°, then given 3 = 1 11 and BC ¥ CD = 23 ¥ 32 ¥ 19 ¥ 11, the square of the altitude from A to BC is 26 ¥ 53 – k, k = Example 94

Fig. 12.47

Example 96 ABCD is a trapezium such that AB and CD are parallel and CB is perpendicular to then. If ADB = q, BC = 45, CD = 30 and tan q = 1/5, then 51 AB is equal to 65k where k = Ans. 9 Solution:

Ans. 1 BC = x (cot 60° + cot 30°) = x

(

) (

)

3 +1/ 3 = 4 / 3 x

CD = x (cot 15° - cot 30°)

(

)

= x 2 + 3 - 3 = 2x

(

)

BC ¥ CD = 4 / 3 . 2x2 = 23 ¥ 32 ¥ 19 ¥ 11 (given)

Fig. 12.48

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AB BD = sin q sin (p - (q + a )) BD sin q BD sin q = sin (q + a ) sin q cos a + cos q sin a

fi AB =

=

BD tan q sin a + cos a tan q

=

( BD )2 tan q

=

fi 51AB = [(30)2 + (45)2] ¥

1 = 585 = 65 ¥ 9 5

fik=9 Example 97 In a triangle ABC, the sides b and c are the roots of the equation x2 - 61x + 820 = 0 and A = tan-1 (4/3), then a2 is equal to 1100 – k, k =

1 1 a1c sin 30∞ - a2 c sin 30∞ 2 2

1 Ê 1ˆ (4) Á ˜ |a1 - a2 | = 4 2 Ë 2¯

Example 99

È sin a = 45 , cos a = 30 ˘ ÍÎ BD BD ˙˚

1 5

45 + 30 ¥

|D1 – D2| =

Now,

Consider a triangle ABC and let a, b and

c denote the lengths of the sides opposite to the vertices A, B and C respectively. Suppose a = 6, b = 10 and the area of the triangle is 15 3 . If ABC is obtuse and if r denote the radius of the incircle of the triangle, then r2 is equal to Ans. 3 Solution: We know that 1 Area of DABC = ab sin C 2 fi 15 3 =

1 (6) (10) sin C fi sin C = 2

3/2 A

Ans. 3 Solution:

x2 - 61x + 820 = 0 fi x = 41, 20

fi b = 41, c = 20, tan A = 4/3 fi cos A = 3/5

b = 10

so a = b + c - 2bc cos A 2

2

2

= 1681 + 400 - 2 ¥ 41 ¥ 20 ¥ (3/5) = 1097 = 1100 – 3 fik=3 Example 98 Let ABC and ABC¢ be two non-congruent triangles with sides AB = 4, AC = AC¢ = 2 2 and angle B = 30°. The absolute value of the difference between the areas of these triangles is Ans. 4 Solution: By the law of cosines cos B =

a 2 + c2 - b2 2ac a 2 + 16 - 8 2(4)(a )



cos 30° =



a2 – 4 3a + 8 = 0

C

As C is an obtused angle, we get C = 2p/3. We have c2 = a2 + b2 – 2ab cos (2p/3) = 100 + 36 + (2) (6) (10) (1/2) = 196 fi c = 14. Thus, 2s = a + b + c = 6 + 10 + 14 = 30 fi s = 15 We know that

(1) r=

C 2÷2

fi a 30°

A

4

B

Fig 12.49

Let a1, a2 be the roots of (1), then |a1 – a2|2 = (a1 + a2)2 – 4a1a2 = (16) (3) – 32 fi |a1 – a2| = 4

B

Fig 12.50

C¢ 2÷2

a=6

D 15 3 = = s 15

3

r2 = 3

Example 100 If the sides of a triangle are in G.P. and the largest angle is twice the smallest angle, then if r is the inradius of the triangle, r2 – r is equal to Ans. 1 Solution: Let the sides of the triangle be a, b = ar, c = ar2, where r > 1 and C = 2A fi B = p – 3A a b c = = So sin A sin B sin C

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1 r r2 = = sin B sin C sin A



1 r r2 = = sin 3 A sin 2 A sin A



r2 =

and

r=



r = r4 – 1 fi r4 – r = 1

sin 2A = 2 cos A sin A sin 3A = 3 – 4 sin2 A = 4 cos2 A – 1 sin A

(a) A.P. (b) G.P. (c) H.P. (d) none of these 6. If the area of a triangle ABC is given by D = a2 - (b - c)2, then tan A is equal to (a) 1/4 (b) 8/15 (c) 4/15 (d) 3/4 7. If a, b, c, d are the sides of a quadrilateral, then a 2 + b2 + c2 can be d2 (a)

EXERCISE

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. The expression (a + b + c) (b + c - a) (c + a - b) (a + b - c) 4b 2 c 2 is equal to (b) sin2 A (a) cos2 A (d) sin A sin B sin C

2. If the median AD of a triangle ABC makes an angle a with AB, then sin (A - a) is equal to b sin a c b (c) c sin a (a)

c sin a b c (d) b sin a (b)

3. If the bisector of angle A of triangle ABC makes an angle q with BC, then sin q is equal to B-C B-C (b) sin (a) cos 2 2 Aˆ Ê (c) sin Á B - ˜ Ë 2¯

(b)

1 3

(c)

1 4

(d)

1 6

8. If B = 45°, C = 60° and a = 2 ( 3 + 1) cm. The area of triangle ABC in sq. cm. is

LEVEL 1

(c) cos A cos B cos C

1 2

Aˆ Ê (d) sin Á C - ˜ Ë 2¯

1 1 4. If the base angle of a triangle are 22 ° and 112 °, 2 2 then height of the triangle is equal to (a) half the base (b) the base (c) twice the base (d) four times the base 5. If in a triangle ABC, a cos2 (C/2) + c cos2 (A/2) = 3b/2 then the sides of the triangle are in

(a)

3 +1

(b) 4

(c) 6 + 2 3 (d) None of these 9. If r1, r2, r3 are in A.P. for the triangle ABC, then cot (A/2), cot (B/2), cot (C/2) are in (a) A.P. (b) G.P. (c) H.P. (d) none of these 10. If in a triangle ABC r1 = 2r2 = 3r3, then the perimeter of the triangle is (a) 3a (b) 3b (c) 3c (d) none of these sin A sin B sin C = = , the 11. If in a triangle ABC, 4 5 6 value of cos A + cos B + cos C is equal to (a) 4/15 (b) 1/3 (c) 2/5 (d) 23/16 12. In a triangle ABC, (a2 - b2 - c2) tan A + (a2 - b2 + c2) tan B is equal to (a) – 1 (b) 0 (c) 1 (d) 2 13. If two sides of a triangle are 12 and 8 , and the angle opposite the shorter side is 45°, the maximum value of the third side can be (a)

2

(b)

6

(c) 6 - 2 (d) 6 + 2 14. In triangle ABC, a cos A + b cos B + c cos C is equal to a+b+c (a) r + R (c) r/R

(b) R/r (d) rR

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15. If in a triangle ABC, the angles A, B, C are in A.P. a+c then is equal to 2 a - ac + c 2 A-C (a) 2 cos 2 (c) 2 cos

A+C 2

A-C (b) 2 sin 2 (d) 2 sin

A+C 2

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 16. In a triangle ABC BC, the vertex A moves such that cos B + cos C = 4 sin2 (A/2) If a, b, c denote the lengths of the sides of the triangle opposite to the angles A, B and C, respectively, then (a) b + c = 4a (b) b + c = 2a (c) locus of point A is an ellipse (d) locus of point A is a pair of straight lines 17. In a triangle ABC, points D and E are taken on side BC such that BD = DE = EC. If ADE = AED = q, then (a) tan q = 3 tan B (b) tan q = 3 tan C 6 tan q (d) 9 cot2 (A/2) = tan2q (c) tan A = tan 2 q - 9 18. If the angles A, B, C of a triangle ABC are in A.P. and the sides a, b, c opposite these angles are in G.P., then a2, b2, c2 are in (a) A.P. (b) G.P. (c) H.P. (d) none of these 19. For a triangle ABC, which of the following is true? cos A cos B cos C (a) = = a b c (b) (c) (d)

cos A cos B cos C a 2 + b2 + c2 + + = 2 abc a b c sin A sin B sin C 3 + + = a b c 2R sin 2 A sin 2 B sin 2C = = a2 b2 c2

20. If in a right angled triangle ABC, 4 sin A cos B – 1 = 0 and tan A is real, then (a) angles are in A.P. (b) the triangle is isosceles

(c) sin A + sin B + sin C =

3+ 3 2

a b c = = 1 3 2 21. If one angle of a triangle is 30° and the lengths of the sides adjacent to it are 40 and 40 3 , the triangle is (a) right angled (b) isosceles (c) obtuse angled (d) none of these 22. In a triangle ABC, 2R is equal to (d)

(a)

b2 - c2 a sin ( B - C )

(b)

c2 - a2 b sin (C - A)

(c)

a 2 - b2 c sin ( A - B)

(d)

a sin A

23. In a triangle ABC, a cos A + b cos B + c cos C is equal to (a) 2a sin B sin C (b) 2b sin C sin A (c) 2c sin A sin B (d) none of these 24. A triangle ABC is right angled if rˆÊ rˆ Ê (a) Á1 - 1 ˜ Á1 - 1 ˜ = 2 Ë r2 ¯ Ë r3 ¯ (b) 8R2 = a2 + b2 + c2 (c) cos2 A + cos2 B + cos2 C =1 (d) sin2 A + sin2 B + sin2 C = 2 25. Internal bisector of –A of a triangle ABC meets the side BC at D. A line drawn through D perpendicular to AD intersects the side AC at E and the side AB at F. If a, b, c represent the sides of D ABC, then (a) AE is the H.M. of b and c 2bc A cos (b) AD = b+c 2 (c) EF =

4bc A sin b+c 2

(d) the triangle AEF is isosceles

MATRIX-MATCH TYPE QUESTIONS 26. In a triangle ABC (a) tan A + tan B + tan C (b) sin A + sin B + sin C (c) cos A + cos B + cos C (d) cot B cot C + cot C cot A + cot A cot B

(p) 1 + 4 sin (A/2) sin (B/2) sin (C/2) (q) tan A tan B tan C (r) 4 cos (A/2) cos (B/2) cos (C/2) (s) 1

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27. In a triangle ABC (a) b cos C + c cos B (p) a 2 2 (q) (a + b + c)r (b) b + c - 2bc cos A (c) bc sin A (r) a2 A B–C b–c (d) tan tan (s) 2 2 b+c 28. In a triangle ABC, D and E are points in BC such that BAD = DAE = EAC = A/3 and AD = 1. (a) BD (b) DE

(p) sin (B + (A/3))/sin(C + (A/3)) (q) sin (A/3) sin (B + (A/3))/sin C sin (C + (A/3)) (c) EC (r) sin (A/3)/sin (C + (A/3)) (d) AE (s) sin (A/3)/sin B 29. In a triangle ABC, if A, B, C are in A.P. and b : c = 3 : 2 then (a) cos (A - C) (p) sin (B/4) (b) sin (A - C) (q) sin (B/2) (c) sin (A + C) (r) sin B (d) sin (2C - A) (s) sin 2B 30. A tower AB stands on a horizontal plane. AC, AD are its shadows at noon and 6 P.M. When the altitude of the sun are tan-1 (45/28) and 45° respectively. It is given that AD is 17 m longer than AC, then (a) AB (p) 45 m (b) BC (q) 28 m (c) CA (r) 53 m (d) BD (r) 45 2 m 31. In a DABC a b (a) (p) sin A sin B (b)

c sin C

(q) 2R

(c)

abc 2D

(r) rs

(d) D (s) (1/2) ca sin B 32. In a DABC if (a) a = 5, b = 7, c = 3 (p) A is acute (b) a = 7, b = 4, c = 5 (q) A is obtuse (r) C is acute (c) a2 + b2 > c2 (s) B is obtuse (d) c2 + a2 < b2

(b) The perimeter of triangle is numerically equal to the area of the triangle, it is given that a < b. Statement-1: The number of ordered pairs (a, b) is equal to 2. Statement-2: Maximum possible perimeter of the triangle is 30 34. Statement-1: If lengths of two sides of a triangle are the roots of the equation x2 – 12x + 35 = 0 and the angle opposite to third side is obtuse, then square of the length of the third side is greater than 74. Statement-2: In a triangle ABC a 2 + b2 – c2 2ab 35. Statement-1: In any DABC, if A is obtuse, then tan B tan C < 1 Statement-2: If any triangle ABC cot A + cot B + cot C = cot A cot B cot C cos C =

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 36 to 38 AL, BM and CN are diameter of the circumcircle of a triangle ABC. D1, D2, D3 and D are the areas of the triangles BLC, CMA, ANB and ABC respectively. 36. D1 is equal to (a) 2R2 sin A cos B cos C (b) 2R2 sin A sin B cos C (c) 2R2 cos A cos B sin C (d) 2R2 sin A sin B sin C 37. D1 + D2 + D3 is equal to (a) 2D (b) 3D (c) D (d) none of these 2 2 2 38. If BL + CM + AN = x and CL2 + AM2 + BN2 = y then (a) x + y = 0 (b) x - y = 0 (c) x + y = 2(a2 + b2 + c2) (d) none of these Paragraph for Question Nos. 39 and 40 A tower OP of height b

PQ on its top. The A, distant a from the foot of the tower on the ground.

ASSERTION-REASON TYPE QUESTIONS 33. Let a and b be length of the legs of a right triangle with following properties (a) All three sides of the triangle are integer.

(a)

a (a 2 + b 2 ) a 2 - b2

(b)

b (a 2 + b 2 ) a 2 - b2

(c)

b (a 2 - b 2 ) a 2 + b2

(d)

a (a 2 - b 2 ) a 2 + b2

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40. The angle subtended by the upper half of the tower at the point A is ab 2a 2 + b 2 (b) cos-1 (a) tan-1 2 ab 2a + b 2 (c) tan-1

ab 2 2a + b 2

(d) tan-1

2b a 2 a 2 + b2

INTEGER-ANSWER TYPE QUESTIONS 41. If the radius of the circumcircle of a triangle is 10 and that of the incircle is 5, then the square of the sum of radii of the escribed circles is (5k)2, k = 42. If two sides of a triangle are 50 and 40 and sine of the included angle is 15 / 4 , then the square of the third side is equal to 3107 – k, k = 43. If the area of a triangle is 91, circumradius is 7 then the product of the lengths of the sides is equal to 50 ¥ 51 – k, k = 44. In a triangle ABC, a = 7, b = 5, c = 3. p1, p2, p3 are the attitudes from A, B, C. Then 4(ap1 + bp2 + cp3)2 is equal to 1215k, k = 45. The sides of a cyclic quadrilateral are in A.P. the shortest is 6 and the difference of the longest and the shortest is also 6. The square of the area of the quadrilateral is 160k2, k = LEVEL 2

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. In a triangle ABC, if a is the arithmetic mean and b, c (b π c) are two geometric means between, any sin 3 B + sin 3 C two positive real numbers then sin A sin B sin C is equal to (a) 0 (b) 1 (c) 2 (d) 4 2. In a triangle ABC, if cot A = (x3 + x2 + x)1/2, cot B = (x + x–1 + 1)1/2 and cot C = (x–3 + x–2 + x–1)–1/2, then the triangle is (a) isosceles (b) obtuse angled (c) right angled (d) equilateral

2 abc D

(b)

a+b cD

(c)

abc 2D

(d)

cos A cos B cos C = = and a a b c

= 1 6 then the area of the triangle in square units is equal to (a) 1/24 (b) 1/8 (c) 1 8 3 (d) 1 24 3 5. In a triangle ABC, if 5 cos C + 6 cos B = 4 and 6 cos A + 4 cos C = 5, then tan (A/2) tan (B/2) is equal to (a) 2/3 (b) 3/2 (c) 1/5 (d) 5 a ( a + c - b) is equal to 6. In a triangle ABC, b (b + c - a) (a)

1 - cos A 1 - cos B

(b)

1 + cos A 1 + cos B

(c)

cos 2 ( A/2) sin 2 ( B /2)

(d)

sin 2 A sin 2 B

7. If r, R are respectively the radii of the inscribed and circumscribed circles of a regular polygon of n R = 5 - 1 , then n is equal to sides such that r (a) 5 (b) 6 (c) 10 (d) none of these 8. In a triangle ABC, if cos A + 2 cos B + cos C = 1, then a, b, c are in (a) A.P. (b) G.P. (c) H.P. (d) none of these a b c 9. In a triangle ABC, if b c a = 0, then c a b sin A sin B + sin B sin C + sin C sin A is equal to (a) 0 (b) 9/4 (c) 1 (d) none of these 10. In a triangle ABC, if A = 18°, b – a = 2, ab = 4, then the triangle is (a) acute angled

(b) right angled

(c) obtuse angled (d) isosceles 11. If for a triangle ABC, a, b, A are given, then which of the following gives us two such triangles (a) a < b sin A

(b) a = b sin A

(c) a > b sin A and a < b (d) a > b sin A and

r +r 3. In a triangle ABC, 1 2 is equal to 1 + cos C (a)

4. In a triangle ABC, if

abc D2

a>b 12. The distance of the incentre of the triangle ABC from A is (a) 4R sin (A/2)

(b) 4R sin (B/2) sin (C/2)

(c) 4R cos (A/2)

(d) 4R cos (B/2) cos (C/2)

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13. If d1, d2, d3 are the diameters of the three escribed circles of a triangle ABC, then d1 d2 + d2 d3 + d3d1 is equal to a b c + + (a) ab + bc + ca (b) b c a (c) (a + b + c)2

(b) (c)

s-a s-b s-c = = , then tan2 (A/2) is equal to 11 12 13 (a) 143/432 (b) 13/33 (c) 11/39 (d) 12/37 15. A point P is inside an equilateral triangle. If the distances of P from the sides are 8, 10, 12 units, the length of a side of the triangle is (a) 10

(b) 10 3

(c) 20 3

(d) 30 3

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 16. If ABC is not a right angled triangle, then which of the following is (are) not possible. (a) sin 2A + sin 2B + sin 2C = 4 cos A cos B sin C (b) cos 2A + cos 2B + cos 2C = – 3/2 (c) sin A + sin B + sin C = 2 2 cos (A/2) cos (B/2) (d) cos A + cos B + cos C = 1 17. In a triangle ABC, if the median AD through A is perpendicular to AB; then AD is equal to (a) c tan B (b) – c tan A (c) 2c tan A (d) – (c/2) tan A 18. If the angles of a triangle ABC satisfy the equation 2

81sin x + 81cos x = 30, then the triangle can be (a) equilateral (b) isosceles (c) obtuse angled (d) right angled 19. In a triangle ABC, cos A cos B + sin A sin B sin C = 1, then the triangle is (a) equilateral (b) isosceles (c) right angled (d) obtuse angled 20. If a, b, g are the distances of the vertices of a triangle ABC from the corresponding points of contacts with the incircle, then a bg (a) s = a + b + g (b) r2 = a +b +g (c) D2 = (a + b + g)abg

(a)

(d) a2 + b2 + c2

14. In a triangle ABC, if

2

21. If A, A1, A2, A3 are the areas of the inscribed and escribed circles of a triangle ABC, then

(d) 2s = a + b + g

(d)

A1 +

A2 +

A3 =

1

+

1

+

1

+

1

+

1

A1 1 A1

A1 +

A2 A2

A2 +

A3 A3

p (r1 + r2 + r3)

=

1 A s2

=

p r1 r2 r3 p (4R + r)

A3 =

22. If I is the incentre of the triangle ABC; P1, P2 and P3 are the radii of the circumcircle of the triangles IBC, ICA and IAB respectively, then (b) 2P2 = b sec (B/2) (a) 2P1 = a sec (A/2) (c) 2P3 = c sec (C/2) (d) P1 P2 P3 = 2R2 r 23. The angle of elevation of the top of a pillar of height h at a point A on the ground distant x from the pillar is 15°. If on walking 100 m towards the pillar, the angle becomes 30°, then (a) h = 50 m

(b) h = 50 3 m

(c) x = 100 + h 3 (d) x = 50 (2 + 3 ) 24. If a tower subtends angles q, 2q and 3q at three points A, B and C respectively, lying on the same side of a horizontal line through the foot of the tower, AB/BC is equal to sin 2q sin 3q (b) (a) sin q sin 2q (c)

sin 3q sin q

(d)

cot q - cot 2q cot 2q - cot 3q

25. A person stands at a point A due south of a tower of height h and observes that its elevation is 60°. He then walks westwards towards B, where the elevation is 45°. At a point C on AB to be 30°. (b) (AC)2 = 8h2/3 (a) (AB)2 = 2h2/3 (c) (BC)2 = 2h2/3

(D) OA, OB, OC are in G.P.

MATRIX-MATCH TYPE QUESTIONS 26. If in a triangle ABC, b + c = 3a Column 1 cos (( B - C ) / 2) (a) cos (( B + C ) / 2)

Column 2 (p) 2

(b) 1 - tan (A/2) tan (B/2)

(q) 3

(c) cot (B/2) cot (C/2)

(r) c/2a

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sin B + sin C (s) 1 3sin A 27. The angles A, B, C of a triangle ABC are in decreasing A.P. The largest is twice the smallest and the median of length 2 3 cm to the largest side divides the angle at the vertex in the ratio 2 : 3. (d)

Column 1

Column 2

(a) AB (b) BC

(p) 4 sin 72° (q) 4 sin 48° sin 72° sin 32°

(c) CA (r) 8 sin 48° (d) CD (s) 4 cos 42° 28. The vertical angle A of a triangle is divided into two parts, a and b such that tan a = 3 tan b and a - b = 30°. Column 1 Column 2 (a) 4 sin a cos b (p) 1 (b) 4 cos a sin b (q) 2 (c) 2 sin A (r) 3 (d) 4 tan (A/2) (s) 4 29. From the top of a lift of height h, the angles of depression of the top and bottom of a tower of height x are observed 45° and 60°. Column 1 Column 2 (p) x = 100 ( 3 - 1) /3 m (q) x = 100 ( 3 - 1) m

(a) h = 100 3 m (b) h = 100

3 m

(c) h = 100 ( 3 + 1) m

(r) x = 200 / 3 m

(d) h = 300 (s) x = 100 (3 - 3 ) m 30. ABCD AB = a, BC = b. A lamp-post of height h at A subtends an angle a at P, the middle point of CD and another lamp-post of equal height at D subtends an angle b at Q, the middle point of BC. If PQ subtends an angle q at A, then Column 1 Column 2 2 (p) (4a2 + b2)/4h2 (a) cot a 4 (a 2 + b 2 )2 (b) cot2 b (q) (a 2 + 4b 2 ) (4a 2 + b 2 ) (c) cos q

a 2 + 4b 2 (r) 4h 2

(d) cot2 a + cot b

(s) 5(a2 + b2)/4h2

2

ASSERTION-REASON TYPE QUESTIONS 31. Statement-1: If two angles of a triangle are 45° and 60°, then the ratio of the smallest and the greatest sides are ( 3 – 1) : 1. Statement-2: The greatest side of a triangle is opposite to its greatest angle. 32. Statement-1: In a DABC, (a + b + c) (tan (A/2) + tan (B/2)) = 2c cot (C/2) Statement-2: In a DABC a = b cos C + c cos B 33. Statement-1: In a DABC, r1 + r2 + r3 – r = 4R Statement-2: r1 r2 + r2 r3 + r3 r1 = D2 34. Statement-1: If the sines of the angles of a triangle are in A.P., then the altitudes of the triangles are also in A.P. Statement-2: Twice the area of a triangle is equal to the product of the lengths of a side and the altitude on it. 35. Statement-1: In a DABC, if 2a2 + 4b2 + c2 = 4ab + 2ac then cos A = 1/4 Statement-2: In a DABC, if cos A = 1/4 then (a + b + c) (b + c – a) = (5/2) bc

COMPREHESION-TYPE QUESTIONS Paragraph for Question Nos. 36 to 38 ABC is a triangle in which the line joining I, the incentre and O the circumcentre of the triangle is parallel to BC, b > c. OE and ID are perpendicular on BC. 36. cos B + cos C is equal to (a) r/R (b) R/r (c) 1 (d) 1 + r/R 37. Distance between the incentre and circumcentre is (a) r (tan A - cot (B/2)) (b) r (cot A - tan (B/2)) (c) r (tan (B/2) – sin A) (d) none of these 38. Area of the triangle BIC is equal to (a) Area of D BOC (b) 2 Area of DBID (c) (1/2) Area of D ABC (d) Area of DAIB Paragraph for Question Nos. 39 to 43 In a triangle the sum of two sides is x and their product is y such that (x + z) (x - z) = y where z is the third side of the triangle.

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39. Greatest angle of the triangle is (a) 90°

(b) 110°

(c) 120° (d) 135° 40. Square of the length of the third side of the triangle is (b) x 2 - y (a) x 2 + y (c) x 2 + y2 41. Area of the triangle is

(d) x - y2

)

(

2 È ˘ 2 (a) ÍÎ y + (1 / 8) x + x - 4 y ˙˚ (b) (1/4) y

3 4

(c) (1/4) x (d) x - z 42. In radius of the triangle is x+z y (a) (b) y 2 ( x + z) x y

(d)

(b) 24/25 (d) - 24/25 (b) (1/2) AC (d) (1/2) AD (b) (1/2) D (d) (1/8) D

INTEGER-ANSWER TYPE QUESTIONS 2

(c)

44. sin q is equal to (a) 7/25 (c) - 7/25 45. DL is equal to (a) (1/2) AB (c) (1/2) BC 46. D¢ is equal to (a) (1/4) D (c) (1/3) D

47. If in a triangle ABC, a is the arithmetic mean and b, c are the geometric means between any two positive real numbers, b + c È sin B sin C ˘ + - 1˙ , then 90k3 + 87k2 + k = Í a Î sin C sin B ˚

z y

43. Circum radius of the triangle is (a) x (b) y (c) z (d) none of these Paragraph for Question Nos. 44 to 46 In a triangle ABC, r1 = 2r2 = 3r3; D is the mid-point of BC, ADC = q. DL is perpendicular to AB, Area of the triangle ALD is D¢.

30k + 29 is equal to 1150 + a, a = 48. If in a triangle ABC, Rr (sin A + sin B + sin C) = 96 then square of the area of the triangle is (32)2 a, a = s-a s-b s-c 49. If in a triangle ABC, = = . Then 11 12 13 33 tan2 (A/2) – 10 = 50. The angle of elevation of a tower of height h standand 70 m, then h2 + h + 1 is equal to 711a – 6, a =

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. If the lengths of the sides of triangle are 3, 5, 7, then the largest angle of the triangle is (a) π/2 (b) 5π/6 (c) 2π/3 (d) 3π/4 [1994] 2. In a triangle ABC, –B = π/3 and –C = π/4. Let D divide BC internally in the ratio 1 : 3 then sin –BAD equals sin –CAD 1 1 (a) (b) 3 6 (c)

1 3

(d)

2 3

[1995]

3. If in a triangle PQR; sin P, sin Q, sin R are in A.P. then (a) the altitudes are in A.P. (b) the altitudes are in H.P. (c) the medians are in G.P. (d) the medians are in A.P. [1998] A B + C Ê ˆ 4. In a triangle ABC, 2ac sin Á ˜¯ Ë 2 (a) a2 + b2 – c2 (b) c2 + a2 – b2 (c) b2 – c2 – a2 (d) c2 – a2 – b2 [2001] 5. In a triangle ABC, let –C = π/2. If r is the inradius and R is the circumradius of the triangle then 2(r + R) is equal to (a) a + b (b) b + c (c) c + a (d) a + b + c [2000]

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6. Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at a point X on the circumference of the circle, then 2r equals PQ + RS (b) (a) PQ ◊ RS 2 (c)

2PQ ◊ RS PQ + RS

(d)

PQ 2 + RS 2 2

[2001] 7. Which of the following pieces of data does NOT uniquely determine an acute angled triangle ABC, R being the radius of the circumcircle? (a) a, sin A, sin B (b) a, b, c (c) a, sin B, R (d) a, sin A, R [2002] 8. If the angles of a triangle are in the ratio 4 : 1 :1, then the ratio of the longest side to the perimeter is (a)

3 : (2 +

3)

(b) 1 :

3

(b) (1/4) (12 + 7 3 ) cm2 (c) (1/4) (48 + 7 3 ) cm2 [2005]

11. If a, b, c denote the lengths of the sides opposite to angles A, B, C of a triangle ABC, then the correct relation among a, b, c, A, B and C is given by (a) (b + c) sin ((B + C )/2) = a cos (A/2) (b) (b – c) cos(A/2) = a sin((B – C)/2) (c) (b – c) cos(A/2) = 2a sin((B – C )/2) (d) (b – c) sin((B – C )/2) = a cos (A/2) [2005] 12. Given an isosceles triangle, whose one angle is 2π/3 and radius of its incircle = 3 . Then the area of the triangle in square units is (a) 7 + 12 3

(b) 12 – 7 3

(c) 12 + 7 3

(d) 4π

a c sin 2C + sin 2A is c a 1 2

(b)

3 2

(c) 1

(d)

3

(a)

[2006]

[2010]

14. Let PQR be a triangle of area D with a = 2, 7 5 b= and c = , where a, b and c are the lengths 2 2 of the sides of the triangle opposite to the angles 2sin P - sin 2 P at P, Q and R respectively. Then 2sin P + sin 2 P equals (a)

(c) 1 : (2 + 3 ) (d) 2 : 3 [2003] 9. If a, b, c are the sides of a triangle such that a : b : c = 1 : 3 : 2, then ratio A : B : C is equal to (a) 3 : 2 : 1 (b) 3 : 1 : 2 (c) 1 : 2 : 3 (d) 1 : 3 : 2 [2004] 10. Three circular coins each of radii 1 cm are kept in an equilateral triangle so that all the three coins touch each other and also the sides of the triangle. Area of the triangle is (a) (4 + 2 3 ) cm2

(d) (6 + 4 3 ) cm2

13. If the angles A, B and C of a triangle are in an arithmetic progression and if a, b and c denote the lengths of the sides opposite to A, B and C respectively, then the value of the expression

3 4D

(b)

2 Ê 3 ˆ (c) Á ˜ Ë 4D ¯

45 4D

2 Ê 45 ˆ (d) Á ˜ Ë 4D ¯

[2012]

15. In a triangle the sum of two sides is x and the product of the same two sides is y. If x2 – c2 = y, where c is the third side of the triangle, then the ratio of the inradius to the circum-radius of the triangle is 3y 3y (a) (b) 2c( x + c) 2 x( x + c) (c)

3y 4 x( x + c)

(d)

3y 4c( x + c)

[2014]

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 1. There exists a triangle ABC satisfying the conditions: (a) b sin A = a, A < π /2 (b) b sin A > a, A > π/ 2 (c) b sin A > a, A < π /2 (d) b sin A < a, A < π/2, b > a (e) b sin A < a, A > π /2, b = a [1986] 2. In a triangle, the lengths of the two larger sides are 10 and 9 respectively. If the angles are in A.P. then the length of the third side can be (a) 5 –

6

(c) 5 (e) none of these

(b) 3 3 (d) 5 +

6 [1987]

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3. Internal bisector AD of angle –A of triangle ABC meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at E and the side AB at F. If a, b, c represent the sides of D ABC then (a) AE is H.M. of b 2bc cos (b) AD = b+c

and c Ê Aˆ ÁË ˜¯ 2

(c) sin

2bc (c) EF = sin A b+c (d) The triangle AEF is isosceles. [2006] 4. In a triangle ABC BC, the vertex A moves such that A cos B + cos C = 4sin 2 . 2 If a, b and c denote the lengths of the sides of the triangle opposite to the angles A, B and C, respectively, then (a) (b) (c) (d)

b + c = 4a b + c = 2a locus of point A is an ellipse locus of point A is a pair of straight lines [2009]

p 6 and let a, b and c denote the lengths of the sides opposite to A, B and C respectively. The value(s) of x for which a = x2 + x + 1, b = x2 – 1 and c = 2x + 1 is (are)

5. Let ABC be a triangle such that –ACB =

(a) – (2 +

3)

(b) 1 +

(a) area of the triangle XYZ is 6 6 (b) the radius of the circum circle of the triangle 35 6 XYZ is 6

3

(c) 2 + 3 (d) 4 3 [2010] 6. In a triangle PQR, P is the largest angle and 1 cos P = . Further the incircle of the triangle 3 touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are) (a) 16 (b) 18 (c) 24 (d) 22 [2013] 7. In a triangle XYZ, let x, y, z be the lengths of the sides opposite to the angles X, Y, Z respectively and s-x s- y s-z = = and area of 2s = x + y + z. If 4 3 2 8p the triangle XYZ is , then 3

X Y Z 4 sin sin = 2 2 2 35

3 2 ÈX +Y ˘ (d) sin Í = ˙ 5 Î 2 ˚

[2016]

INTEGER-ANSWER TYPE QUESTIONS 1. Let ABC and ABC¢ be two non-congruent triangles with sides AB = 4, AC = AC¢ = 2 2 and angle B = 30°. The absolute value of the difference between the areas of these triangles is [2009] 2. The centres of two circles C1 and C2 each of unit radius are at a distance of 6 units from each other. Let P be the mid point of the line segment joining the centres of C1 and C2 and C be a circle touching circles C1 and C2 externally. If a common tangent to C1 and C passing through P is also a common tangent to C2 and C, then the radius of the circle C is [2009] 3. Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B and C respectively. Suppose a = 6, b = 10 and the area of the triangle is 15 3 . If –ABC is obtuse and if r denotes the radius of the incircle of the triangle, then [2010] r2 is equal to

FILL

IN THE

BLANKS TYPE QUESTIONS

1. In a triangle ABC, if cot A, cot B, cot C are in A.P., then a2, b2, c2 are in progression. [1985] 2. The set of all real numbers a such that a2 + 2a, 2a + 3 and a2 + 3a + 8 are the sides of a triangle [1985] is 3. The sides of a triangle inscribed in a given circle subtend angles α, β and γ at the centre. The minimum value of the arithmetic mean of cos (α + π/2), cos (β + π/2) and cos (γ + π/2) is equal . to 4. If the angles of a triangle are 30° and 45° and included side is ( 3 + 1) cm, then the area of the triangle [1988] is 5. ABC is an isosceles triangle inscribed in a circle of radius r. If AB = AC and h is the altitude from A to

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BC, then the triangle ABC has perimeter 2 P = 2 ( 2hr - h + 2hr ) and area A = A also lim 3 = [1989] hÆ 0 P 6. If in a triangle ABC 2 cos A cos B 2 cos C a b + + = + then the a b c bc ca value of the angle A is degrees. [1993] 7. In a triangle ABC, AD is the altitude from A. abc Given b > c, –C = 23° and AD = 2 b + c2 then –B = [1994] 8. A circle is inscribed in an equilateral triangle of side a. The area of any square inscribed in this circle [1994]

9. In a triangle ABC, a : b : c = 4 : 5 : 6. The ratio of the radius of circumcircle to that of incircle is [1996]

SUBJECTIVE-TYPE QUESTIONS 1. If p1, p2, p3 are the altitudes of a triangle from the vertices A, B, C respectively and D, the area of the triangle, prove that C 1 1 1 2ab + – = cos2 p1 p2 p3 2 (a + b + c) D

[1978]

2. A quadrilateral ABCD is inscribed in a circle S and A, B, C, D are the points of contacts with S of another quadrilateral which is circumscribed about S. If this quadrilateral is also cyclic, prove that [1978] (AB)2 + (CD)2 = (BC)2 + (AD)2. 3. If two sides of a triangle and the included angle are given by a = (1 + 3 ) cm, b = 2 cm, C the other two angles and the third side. [1979] 4. If a circle is inscribed in a right angled triangle ABC with right angle at B, show that the diameter of the circle is equal to AB + BC – AC. [1979] 5. ABC is a triangle. D is the middle point of BC. If AD is perpendicular to AC, then prove that 2(c 2 - a 2 ) . [1980] 3ac 6. Let the angles A, B, C of a triangle ABC be in A.P. and let b : c = 3 : 2 . Find the angle A. [1981] 7. The ex-radii r1, r2, r3 of DABC are in H.P. Show that the sides a, b, c are in A.P. [1983] cos A cos C =

8. For a triangle ABC it is given that cos A + cos B + cos C = 3/2, prove that the triangle is equilateral. [1984] 9. With usual notation, if in a triangle ABC b+c c+a a+b = = , then prove that 11 12 13 cos A cos B cos C = = [1984] 7 19 25 10. In a triangle ABC, the median to the side BC is of 1 length and it divides the angle A into 11 - 6 3 angles of 30°and 45°. Find the length of the side BC. [1985] 11. In a triangle ABC, cos A cos B + sin A sin B sin C [1986] = 1. Show that a : b : c = 1 : 1 : 2 12. ABC is a triangle such that sin (2A + B) = sin (C – A) = – sin (B + 2C ) = 1/2 If A, B and C are in arithmetic progression, determine the values of A, B, and C. [1990] 13. The sides of a triangles are three consecutive natural numbers and its largest angle is twice the smallest one. Determine the sides the triangle. [1991] 14. In a triangle of base a, the ratio of the other two sides is r (< 1) show that the altitude of the triangle ar . [1991] is less than or equal to 1 - r2 15. Let A1, A2, … An be the vertices of an n-sided regular polygon such that 1 1 = + A1 A2 A1 A3 A1 A4

n. [1994]

16. Consider the following statements concerning a triangle ABC. (i) the sides a, b, c and area D are rational (ii) a, tan (B/2), tan (C/2) are rational (iii) a, sin A, sin B, sin C are rational. Prove that (i) fi (ii) fi (iii) fi (i) [1994] 17. Let A, B, C be three angles such that A = π/4 and tan B tan C = p. Find all possible values of p such that A, B, C are the angles of a triangle. [1997] 18. Prove that a triangle ABC is equilateral if and only [1998] if tan A + tan B + tan C = 3 3 19. Let ABC be a triangle having O and I as its circumcentre and incentre respectively. If R and r are

IIT JEE eBooks: www.crackjee.xyz Solution of Triangles and Applications of Trigonometry 12.45

the circumradius and inradius, respectively, then prove that (IO)2 = R2 – 2Rr. Further show that the triangle BIO is a right angled triangle if and only if b is the arithmetic mean of a and c. [1999] 20. In any triangle ABC, prove that cot (A/2) + cot (B/2) + cot (C/2) = cot (A/2) cot (B/2) cot (C/2) [2000] 21. Let ABC be a triangle with incentre I and inradius r. Let D, E, F be the feet of the perpendiculars from I to the sides BC, CA and AB respectively. If r1, r2 and r3 are the radii of circles inscribed in the quadrilaterals AFIE, BDIF and CEID respectively, prove that r r1 r2 + + 3 r - r1 r - r2 r - r3 r1r2 r3 = (r - r1 )(r - r2 )(r - r3 )

[2000]

MATRIX-MATCH TYPE QUESTIONS

In =

On È 2I Í1 + 1 - ÊÁ n ˆ˜ Ë n ¯ 2 Í Î

˘ ˙ ˙˚

[2003]

SINGLE CORRECT ANSWER TYPE QUESTIONS 2. 6. 10. 14.

(b) (b) (b) (c)

3. 7. 11. 15.

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p p

q q

r r

s s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

29.

LEVEL 1

(b) (a) (c) (d)

r

28.

Answers

1. 5. 9. 13.

q

27.

22. If In is the area of n sided regular polygon inscribed in a circle of unit radius and On be the area of the polygon circumscribing the given circle, show that 2

p

26.

(a) (a) (d) (a)

30.

4. (a) 8. (c) 12. (b) 31.

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 16. 18. 20. 22. 24.

(b), (a), (a), (a), (a),

(c) (b), (c), (b), (b),

(c) (d) (c), (d) (c), (d)

17. 19. 21. 23. 25.

(a), (b), (b), (a), (a),

(b), (c), (d) (c) (c) (b), (c) (b), (c), (d)

32.

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ASSERTION-REASON TYPE QUESTIONS 33. (b)

34. (a)

35. (c)

COMPREHENSION-TYPE QUESTIONS 36. (a) 40. (c)

37. (c)

38. (b)

39. (b)

INTEGER-ANSWER TYPE QUESTIONS 41. 9 45. 6

42. 7

43. 2

LEVEL 2

1. 5. 9. 13.

(c) (c) (b) (c)

2. 6. 10. 14.

(c) (a) (c) (b)

3. 7. 11. 15.

(c) (a) (c) (c)

16. 18. 20. 22. 24.

(a), (a), (a), (a), (c),

(c), (b), (b), (b), (d)

(d) (c) (c) (c), (d)

17. 19. 21. 23. 25.

(a), (b), (a), (a), (a),

(d) (c) (b), (c), (d) (c), (d) (b), (c), (d)

MATRIX-MATCH TYPE QUESTIONS p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

26.

27.

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

30.

4. (c) 8. (a) 12. (b)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

q

29.

44. 5

SINGLE CORRECT ANSWER TYPE QUESTIONS

p

28.

ASSERTION-REASON TYPE QUESTIONS 31. (a) 35. (b)

32. (b)

33. (c)

34. (d)

COMPREHESION-TYPE QUESTIONS 36. (c) 40. (b) 44. (b)

37. (a) 41. (b) 45. (b)

38. (a) 42. (b) 46. (a)

39. (c) 43. (c)

INTEGER-ANSWER TYPE QUESTIONS 47. 7

48. 9

49. 3

50. 7

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13.

(c) (a) (c) (d)

2. 6. 10. 14.

(a) (a) (d) (c)

3. 7. 11. 15.

(b) (d) (b) (b)

4. (b) 8. (a) 12. (c)

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MULTIPLE CORRECT ANSWERS TYPES QUESTIONS 1. 3. 5. 7.

(a), (d) (a), (b), (c) (b) (a), (c), (d)

a/2 b = sin ( A – a ) sin ( A + C - a ) a/2 c = sin a sin ( A + C - a )

2. (a), (d) 4. (b), (c) 6. (b), (d)

\ b sin (A - a) = c sin a

INTEGER-ANSWER TYPE QUESTIONS 1. 4

FILL

2. 8 IN THE

3. 3

BLANKS TYPE QUESTIONS

Ê c ˆ Ê b ˆ a, DC = Á a, 3. BD = Á Ë b + c ˜¯ Ë b + c ˜¯ sin q sin ( A / 2) = ba b b+c

2. (5, •)

1. arithmetic 3. –

3 /2.

4.

1 ( 3 + 1)cm2 2

2 5. h 2hr - h cm2; 1/128r 6. 90° 7. 113° 2 2 9. 16/7 8. (a /6) (unit)

Fig. 12.52

A b+c sin 2 a A B-C sin B + sin C = . sin = cos sin A 2 2 a b c 4. = 1 = sin 45∞ sin 222∞ sin112 12∞

SUBJECTIVE-TYPE QUESTIONS 3. 6. 12. 13.

Fig. 12.51

fi sin q =

A = 105°, B = 15°, c = 6 cm A = 75° 10. 2 units A = 45°, B = 60°, C = 75° 4, 5, 6 units 15. 7

17. (– •, 3 – 2 2 ] » [3 + 2, •)

Hints and Solutions LEVEL 1

È(b + c )2 - a 2 ˘ Èa 2 - (b - c )2 ˘ Î ˚Î ˚ 1. 2 2 4b c

(

)

Fig. 12.53

(

)

ÈÎ2bc + b 2 + c 2 - a 2 ˘˚ ÈÎ2bc - b 2 + c 2 - a 2 ˘˚ = 4b 2 c 2 2bc (1 + cos A) 2bc (1 - cos A) = = 1 - cos2 A 4b 2 c 2 = sin2 A

a =

c sin 45∞ c sin 45∞ 1 = sin112 2∞ cos 22 12∞ 1

= 2c sin 22 2 = 2 AD 5. a (1 + cos C) + c (1 + cos A) = 3b fi a + c + a cos C + c cos A = 3b fi a + c + b = 3b fi a + c = 2b fi a, b, c are in A.P. 6. D = a2 - b2 - c2 + 2bc = 2bc - 2bc cos A = 2bc ◊ 2 sin2 (A/2) fi (1/2) bc sin A = 4bc sin2 (A/2) fi tan (A/2) = 1/4 fi tan A = 8/15

2. Fig. 12.51

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7. Let AB = a, BC = b, CD = c, AD = d We know (a – b)2 + (b – c)2 + (c – a)2 ≥ 0 fi 2(a2 + b2 + c2) ≥ 2(ab + bc + ca) fi 3(a2 + b2 + c2) ≥ (a + b + c)2 As the sum of the three sides of a quadrilateral is greater than the third. 3(a2 + b2 + c2) ≥ (a + b + c)2 > d2 fi

a 2 + b2 + c2 d2

2 R(sin A cos A + sin B cos B + sin C cos C ) 2 R (sin A + sin B + sin C ) =

(sin 2 A + sin 2 B + sin 2 C ) 2 (sin A + sin B + sin C ) =

)

2 3 +1 2 2 b a 8. = fi b= =4 sin 45∞ sin 75∞ 2 3 +1

(

)

1 Area of the triangle = ab sin C 2

=6+ 2 3 9. s tan (A/2), s tan (B/2), s tan (C/2) are in A.P. fi cot (A/2), cot (B/2), cot (C/2) are in H.P. D 2D 3D 10. = = s-a s-b s-c

11.

s-a 1 s-c 3 = , = s-b 2 s-b 2



s-a+s-c =2 s-b



a + b + c = 3b

a b c sin A sin B sin C = = = = fi 4 5 6 4 5 6 cos A =

3 9 1 b2 + c2 - a 2 = ; cos B = , cos C = 2b c 4 16 8

23 fi cos A + cos B + cos C = 16 12.

(a 2 - b 2 - c 2 ) sin A ◊ 2 b c b2 + c2 - a 2 +

(a 2 - b 2 + c 2 ) sin B ◊ 2 c a c2 + a 2 - b2

= - 2bc sin A + 2ca sin B = 0 13. cos 45° =

x 2 + 12 - 8 2 x 12

\

a+c sin A + sin C = = sin B b 2sin

A+C A-C cos A+C 2 2 = 2 cos sin B 2

16. cos B + cos C = 4 sin2 (A/2) B+C B–C A fi 2 cos cos = 4 sin 2 2 2 2 p A B-C fi 2 cos cos 2 2 2 p B+C A = 4 sin sin – 2 2 2 A B–C fi 2 sin cos 2 2 A B+C = 4 sin cos 2 2 B C B C fi cos cos + sin sin 2 2 2 2 B C B C = 2 cos cos – sin sin 2 2 2 2

(

()

)

() (

)

()

(

)

fi 3 sin (B/2) sin (C/2) = cos (B/2) cos (C/2) B C 1 fi tan tan = 2 2 3 D D 1 ◊ = fi s ( s – b) s ( s – c) 3 fi 3s(s – a) (s – b) (s – c) = s 2 (s – b) (s – c) fi 3(s – a) = s fi 2s = 3a fi a + b + c = 3a fi b + c = 2a A

fi x2 - 2 6 x + 4 = 0

2 6 ± 24 - 16 fix= = 2

r R

15. B = 60°, A + C = 120°, b2 = c2 + a2 - 2ac cos B = c2 + a2 - ac

1 3 = ¥ 2 ( 3 + 1) ¥ 4 ¥ 2 2



4 sin A sin B sin C 8 cos ( A / 2) cos ( B / 2) cos (C / 2)

= 4 sin (A/2) sin (B/2) sin (C/2) =

1 1 1 > and > 3 2 3

(

14.

c

b

6± 2 B

a

Fig. 12.54

C

IIT JEE eBooks: www.crackjee.xyz Solution of Triangles and Applications of Trigonometry 12.49

fi AB + AC = 2a

21. a = 40, b = 40 3 , C = 30°

Also, AB + AC > BC = a

2 fi c2 = 40 2 + (40 3) - 2 (40) (40 3) cos 30°

Thus, locus of A is an ellipse with focii at B and C and length of major axis as 2a.

= 1600 fi c = 40 = a fi D is isosceles. fi A = C = 30° fi B = 120° fi D is obtuse angled.

17. D ADE is isosceles. 22.

(

2 R sin 2 B - sin 2 C b2 - c2 = sin A sin ( B - C ) a sin ( B - C ) =

)

2 R sin ( B + C ) sin ( B - C ) = 2R similarly for (b) sin ( B + C ) sin ( B - C )

and (c). 23. a cos A + b cos B + c cos C = R (sin 2A + sin 2B + sin 2C)

Fig. 12.55

= R.4 sin A sin B sin C

D ABC is isosceles, B = C AL (a / 2) tan B tan q = = = 3 tan B = 3 tan C (a / 6) DL tan A = tan (180° - 2B) = - tan 2B =

2 tan B tan 2 B - 1 6 tan q = tan 2 q - 9

cot (A/2) = tan B fi 9 cot2 (A/2) = tan2 q 18. B = 60°, b2 = ac fi sin2 B = sin A sin C fi 3/4 = sin A sin C cos (A + C) = - cos B = - 1/2 fi cos A cos C - sin A sin C = - 1/2 fi cos A cos C = 1/4 fi cos (A - C) = 1/4 + 3/4 = 1 fi A = C = 60° and the triangle ABC is equilateral fia=b=c 19.

cos A b2 + c2 – a 2 = , (a) will hold if a2 = b2 = c 2 a 2 abc

= 2a sin B sin C = 2b sin C sin A = 2c sin A sin B s - bˆ Ê 24. Á1 Ë s - a ˜¯

s - cˆ b-a c-a Ê ÁË1 - s - a ˜¯ = 2 fi s - a ◊ s - a = 2

fi 2(s - a)2 = (b - a) (c - a) fi (2s - 2a)2 = 2(bc - ac - ab + a2) fi (b + c -a)2 = 2(bc - ac - ab + a2) fi b2 + c2 = a2 fi triangle is right angled. 8R2 = 4R2 (sin2 A + sin2 B + sin2 C) fi sin2 A + sin2 B + sin2 C = 2 fi cos2 A + cos2 B + cos2 C = 1 fi cos 2A + cos 2B + cos 2C = - 1 fi - 1 - 4 cos A cos B cos C = - 1 fi cos A cos B cos C = 0 fi one of the three angles is p/2 25. D ABC = DABD + D ACD 1 1 1 fi bc sin A = c AD sin (A/2) + b AD sin (A/2) 2 2 2

cos A cos B cos C a 2 + b2 + c2 + + = 2 abc a b c sin A sin B cos C 1 = = = 2R a b c sin 2 A sin 2 B cos 2 C = = a2 b2 c2 cos A cos B cos C = = a b c 20. Since 4 sin A cos B = 1, A or B can not be p /2 [as if B = p/2, cos B = 0 and if A = p/2, tan A is not C = p /2 fi B = p /2 – A and 4 sin A cos B = 1 fi sin2A = 1/4 fi sinA = 1/2 fi A = 30°, B = 60°, C = 90°. fi

Fig. 12.56

fi AD =

A 2bc cos 2 b+c

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A 2bc = 2 b+c fi AE is H.M. between b and c A A 4bc EF = ED + DF = 2 ¥ AD tan = sin 2 2 b+c As AD is perp. to EF, DE = DF, D AEF is isosceles. 26. See the Text. Conditional Identities. 27. b cos C + c cos B = CD + DB = BC = a

and A = 75° fi cos (A - C) = cos 30° = sin 60° = sin B sin (A - C) = sin (B/2), sin (A + C) = sin 2B and sin (2C - A) = sin 15° = sin (B/4)

AE = AD sec

30. AD = h, tan a = 45/28 AC = h - 17 = h cot a =

28 h 45

fi 45 (h - 17) = 28 h fi h = 45 = AB AC = 45 - 17 = 28

bc sin A = 2 D = 2 sr = (a + b + c)r b2 + c2 = (AD)2 + (CD)2 + (BD)2 + (AD)2

BC =

(45) 2 + (28) 2 = 53.

BD = 45 2

= (BD + CD)2 - 2BD ◊ CD + 2(AD)2 = a2 - 2[bc cos B cos C - bc sin B sin C] = a2 - 2bc cos (B + C) = a2 + 2bc cos A fi b2 + c2 - 2bc cos A = a2

Fig. 12.57

Fig. 12.59

BD sin ( A / 3) 28. = (From D ADE) sin B AD sin ( A / 3) AD DE = sin (C + ( A / 3))

31.

a b c abc ca sin B = = = 2R, D = = rs = sin A sin B sin C 4R 2

32. A is acute if b2 + c2 > a2 and obtuse if b2 + c2 < a2. 33. a + b +

sin ( B + ( A / 3)) AD AE = sin (C + ( A / 3)) AE sin ( A / 3) EC = sin C sin ( B + ( A / 3)) sin ( A / 3) AD = sin (C + ( A / 3)) sin C

a 2 + b 2 = (1/2) ab

fi (2a + 2b – ab)2 = 4(a2 + b2) fi ab + 8 – 4a – 4b = 0 fi (a – 4) (b – 4) = 8 fi a – 4 = 1, b – 4 = 8, or a – 4 = 2, b – 4 = 4 ordered pairs are (5, 12) or (6, 8) So Statement 1 is True Perimeter is maximum when (a, b) = (5, 12) and is equal to 5 + 12 +

52 + 122 = 30

so statement 2 is also true. But statement-1 does not follow from statement-2. 34. a = 5, b = 7, cos C =

a 2 + b2 – c2 (true) < 0 2ab

fi 25 + 49 – c 2 < 0 or c 2 > 74 35. Statement-2 is False Fig. 12.58

c 29. B = 60°, sin C = sin B = b

2

3 1 ¥ = 2 3 2 fi C = 45°

as tan A + tan B + tan C = tan A tan B tan C tan B + tan C fi tan A = < 0 [ A is obtuse] tan B tan C – 1 fi tan B tan C < 1 fi Statement-1 is True.

IIT JEE eBooks: www.crackjee.xyz Solution of Triangles and Applications of Trigonometry 12.51

36.

b b+h , tan 2q = a a

39. tan q = fi

2 tan q h+b = 2 a 1 - tan q

fih=

a ¥ 2 (b / a) b (a 2 + b 2 ) b = 1 - (b 2 / a 2 ) a 2 - b2

40. tan (q - a) = fi Fig. 12.60

ALC = ABC = B, ALB = ACB = C Circumcircle of triangle BLC is same as that of ABC. \ D1 = Area of triangle BLC a ◊ c cot C ◊ b cot B BC ¥ BL ¥ CL = 4R 4R

=

(2 R) ◊ sin A sin B sin C cos B cos C 4 R sin B sin C

= 2R2 sin A cos B cos C abc [cot B cot C + cot C cot A 4R + cot A cot B] abc = =D 4R

37. D1 + D2 + D3 =

38. x = BL + CM + AN = 4R - c + 4R - a + 4R - b 2

2

2

2

2

2

fi tan a =

ab b + 2a 2 2

41. r1 + r2 + r3 = 4R + r fi (r1 + r2 + r3)2 = (4 ¥ 10 + 5)2 = 2025 = (5 ¥ 9)2 fik=9 42. b = 50, c = 40, sin A =

3

2

b (b / a) - tan a = 2 a 1 + (b / a) tan a

È b2 ˘ b b fi tan a Í 2 + 1˙ = a 2a Î 2a ˚

ACL = ABL = 90° (angles in a semicircle)

=

tan q - tan a 1 + tan q tan a

2

2

15 1 , cos A = 4 4

a2 = b2 + c2 - 2bc cos A = (50)2 + (40)2 - 2 ¥ 50 ¥ 40 ¥ (1/4) = 3100 = 3107 – 7 fi k = 7 abc fi abc = 91 ¥ 4 ¥ 7 = 2548 = 50 ¥ 51 – 2 43. D = 4R fik=2 44. D =

s ( s - a ) ( s - b) ( s - c ) =

15 1 5 9 ¥ ¥ ¥ 2 2 2 2

y = CL2 + AM2 + BN2 = 4R2 - b2 + 4R2 - c2 + 4R2 – a 2

=

x - y = 0, x + y = 24R2 - 2 (a2 + b2 + c2) Also D =

15 3 4

1 1 1 ap1 = bp2 = cp3 2 2 2

\ (ap1 + bp2 + cp3)2 = (2D + 2D + 2D)2 225 ¥ 3 = 36 ¥ 16

Fig. 12.61

fi 4(ap1 + bp2 + cp3)2 = 6075 = 1215 ¥ 5 fi k = 5 45. The sides are 6, 8, 10, 12, 2s = 36, s = 18 (Area)2 = (s – a) (s – b) (s – c) (s – d) = 12 ¥ 10 ¥ 8 ¥ 6 = 5760 = 160 ¥ 62 fi k = 6

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13 Inverse Trigonometric Functions INVERSE TRIGONOMETRIC FUNCTIONS AND FORMULAE Function

y y y y y

= = = = =

Domain

sin–1 x cos–1 x tan–1 x cot–1 x sec–1 x

–1 £ x £ 1 –1 £ x £ 1 –• 0, and x2 + y2 >1 2 2 = p – sin–1 ÎÈ x 1 - y - y 1 - x ˘˚

if 0 < x £ 1, –1 £ y < 0, x2 + y2 >1

2 2 = – sin–1 ÎÈ x 1 - y - y 1 - x ˘˚ – p

if –1 £ x < 0, 0 < y £ 1, x2 + y2 >1

(iii) cos–1 x + cos–1 y

(iv)

2 2˘ È = cos–1 Î xy - 1 - x 1 - y ˚

if –1 £ x, y £ 1, x + y ≥ 0

2 2 = 2p – cos–1 ÈÎ xy - 1 - x 1 - y ˘˚

if –1 £ x, y < 0, x + y £ 0

cos–1 x – cos–1 y

(

= cos–1 xy + 1 - x 2 1 - y 2

(

)

= – cos–1 xy + 1 - x 2 1 - y 2

if –1 £ x, y £ 1, x £ y

)

if –1 £ y < 0, 0 < x £ 1, x ≥ y

Ê x+ yˆ if x y < 1 (v) tan–1 x + tan–1 y = tan–1 Á Ë 1 - xy ˜¯ Ê x+ yˆ = p + tan–1 Á Ë 1 - xy ˜¯

if x y > 1, x > 0, y > 0

Ê x+ yˆ –p = tan–1 Á Ë 1 - xy ˜¯

if xy > 1, x < 0, y < 0

Illustration 3 Find the value of tan–1 (1/2) – tan–1 (– 3) – tan – 1 (– 7) Solution: –3 < 0, – 7 < 0, (–3)(–7) > 1 So tan–1 (1/2) – tan–1 (–3) – tan–1 (–7) -3-7 ˘ +p=p = tan–1 (1/2) – Ètan -1 ÍÎ 1 - (-3) (-7) ˙˚

Ê x- yˆ (vi) tan–1 x – tan–1 y = tan–1 Á Ë 1 + xy ˜¯

if xy > –1

IIT JEE eBooks: www.crackjee.xyz Inverse Trigonometric Functions 13.3

Ê x- yˆ = tan–1 Á –p Ë 1 + xy ˜¯

if x < 0, y > 0, xy < –1

Ê x- yˆ = p + tan–1 Á Ë 1 + xy ˜¯

if x > 0, y < 0, xy < –1

Ê1 + xˆ p = + tan -1 x (vii) tan–1 Á Ë 1 - x ˜¯ 4 = tan–1 x –

if x £ 1

3p 4

if x > 1

Ê1 - xˆ p = - tan -1 x (viii) tan–1 Á Ë 1 + x ˜¯ 4 = tan–1 x – (f)(i) 2sin–1 x = sin–1 (2x

p 4

1 - x2 )

= p – sin–1 (2x = sin–1 (2x (ii) 2 cos

–1

x = cos

–1

if x ≥ –1 if x < –1 if – 1 - x2 )

1 - x2 ) – p

2

(2x – 1)

= p – cos–1 (2x2 – 1)

if

1 2 1 2

1

£x£

2

1

IIT JEE eBooks: www.crackjee.xyz 13.4 Comprehensive Mathematics—JEE Advanced

Ê 2x ˆ = tan–1 Á –p Ë 1 - x 2 ˜¯

if x < –1

Ê 2x ˆ (iv) 2 tan–1 x = sin–1 Á Ë 1 + x 2 ˜¯

if –1 £ x £ 1

Ê 2x ˆ if x > 1 = p + sin–1 Á Ë 1 + x 2 ˜¯ Ê 2x ˆ – pif x < –1 = sin–1 Á Ë 1 + x 2 ˜¯ Ê 1 - x2 ˆ (v) 2 tan–1 x = cos–1 Á Ë 1 + x 2 ˜¯

if 0 £ x < •

Ê 1 - x2 ˆ = - cos - 1 Á Ë 1 + x 2 ˜¯ (g)

if - • < x £ 0

(i) 3 sin–1 x = sin–1 (3x – 4x3)

if –

(ii) 3 cos–1 x = cos–1 (4x3 – 3x)

if

Ê 3 x - x3 ˆ (iii) 3 tan–1 x = tan–1 Á Ë 1 - 3x 2 ˜¯

if –

Illustration 5 Find the values of 1 2 1 sin Ê 3 sin -1 Ê ˆ ˆ , cos Ê 3 cos-1 Ê ˆ ˆ , tan Ê 3 tan -1 Ê ˆ ˆ Ë Ë 2¯¯ Ë Ë 3¯ ¯ Ë Ë 3¯ ¯ Solution:

1 1 1 23 sin Ê 3 sin -1 Ê ˆ ˆ = sin Ê sin -1 Ê 3 ¥ - 4 ¥ ˆ ˆ = Ë Ë 3¯ ¯ Ë Ë 3 27 ¯ ¯ 27 2 8 cos Ê 3 cos-1 Ê ˆ ˆ = cos Ê cos-1 Ê 4 ¥ -3¥ Ë Ë 3¯ ¯ Ë Ë 27 Ê 22 ˆ = cos cos-1 Ë - ¯ 27 22 = cos Ê p - cos-1 Ê ˆ ˆ Ë Ë 27 ¯ ¯ = -

22 . 27

Ê Ê 3 ¥ 1 - 1 ˆˆ 1ˆˆ Á Ê Ê 1 1 2 8 ˜˜ tan 3 tan = tan tan Á Á Á Ë Ë 2¯¯ 1 ˜˜ ÁË ÁË 1 - 3 ¥ ˜¯ ˜¯ 4 11 11 = tan Ê tan -1 Ê ˆ ˆ = Ë Ë 2 ¯¯ 2

If an expression contains

2ˆˆ 3¯ ¯

1 1 £x£ 2 2

1 £x£1 2 1 3

£x£

1 3

IIT JEE eBooks: www.crackjee.xyz Inverse Trigonometric Functions 13.5

a2 - x2 , a2 + x2 , x2 - a2 simplify them by applying the following substitution. Expression

Substitution

a2 - x2

x = a sin q

Ê- p 1/3 For x = 1, l = 1, y = 2 For x = 2, l = 5, y = 7 For x > 3, l < 0. Hence the required solutions are (x, y) = (1, 2), (2, 7)

Let cos–1 (1 8) = q, where 0 < q < p. Then

1 1 1 1ˆ 1 Ê1 cos -1 = q fi cos Á cos -1 ˜ = cos q Ë2 2 8 2 8¯ 2 Now cos–1

q 1 1 1 = q fi cos q = fi 2 cos 2 - 1 = 8 8 2 8

q q 3 9 = fi cos = 2 16 2 4 q p q 3˘ È Í 0 < 2 < 2 , so cos 2 π - 4 ˙ Î ˚

Example 9

The inequality sin–1 (sin 5) > x2 – 4x

holds if (a) x = 2 –

9 - 2p

(b) x = 2 +

9 - 2p

(c) x Œ (2 –

9 - 2p , 2 +

(d) x > 2 +

9 - 2p

9 - 2p )

IIT JEE eBooks: www.crackjee.xyz Inverse Trigonometric Functions 13.7

Ans. (c) Since 3p 2 < 5 < 2p, we have sin 5 < 0, so

Solution: –1

sin (sin 5) = 5 – 2p. Therefore, the given inequality can be written as 5 – 2p > x2 – 4x or x2 – 4x + (2p – 5) < 0 È 4 - 16 - 4(2p - 5) ˘ È 4 + 16 - 4(2p - 5) ˘ fi Íx ˙ Íx ˙ 2 2 ÍÎ ˙˚ ÍÎ ˙˚ 0, y > 0 and x > y, then tan–1 (x/y) + tan–1 [(x + y)/(x – y)] is equal to (a) – p /4 (b) p /4 (c) 3p /4 (d) none of these Ans. (c) Solution: Since

and fi fi

bp 2

a sin–1 x – b cos–1 x = c (given) b p (a + b) sin–1 x = +c 2 (b p ) 2 + c sin–1 x = a+b

x x+ y . > 1 The given expression is y x- y

È x x+ y ˘ Í y + x- y ˙ -1 ˙ p + tan Í Í1 - x ¥ x + y ˙ ÍÎ y x - y ˙˚ = p + tan 1 Example 14

x2 + y 2 -( x + y ) 2

2

= p + tan -1 ( -1) = 3p /4.

The equation 2 cos–1 x =

sin–1 (2 x 1 - x 2 ) is valid for all values of x satisfying (a) – 1 £ x £ 1

Ans. (d) Solution: We have b sin–1 x + b cos–1 x =

3 sin 2q 6 tan q 2 tan f = = = sin 2 f 5 + 4 cos 2q 9 + tan 2 q 1 + tan 2 f

where tan q = 3 tan f

2

p a b + c (b - a) a+b

(b) 3 tan q (d) 3 cot q

Ans. (c)

If a sin–1 x – b cos–1 x = c, then a sin–1 x +

(b)

1 È 3 sin 2q ˘ -1 sin–1 Í ˙ = tan x, then 2 Î 5 + 4 cos 2q ˚

(a) tan 3q (c) (1/3) tan q

b cos–1 x is equal to (a) 0

If

x=

fi x = 0, x = ±1 2 x
0 (b) sin b < 0 (c) cos (a + b) > 0 (d) cos a < 0 Ans. (b), (c), (d) 4 1 Solution: As 0 < < , we get 9 2 p Ê 1ˆ Ê 4ˆ p = cos -1 Á ˜ < cos -1 Á ˜ < Ë 2¯ Ë 9¯ 2 3 fi \ Also

Ê 4ˆ 3 p < b = 3 cos -1 Á ˜ < p Ë 9¯ 2 cos b < 0 and sin b < 0. 1 6 1 < < 2 11 2



p Ê 6ˆ p < sin -1 Á ˜ < Ë 11¯ 4 6



p Ê 6 ˆ 3p < a = 3 sin -1 Á ˜ < Ë 11¯ 4 2

Let cos–1 (4/5) = a, that is, cos a = 4/5, so that 2

tan a =

and

so

Example 26

(b) a – b = 11 (d) 2a = 3b

Ans. (a), (b) and (c)

= tan–1 cot b = tan–1 (tan (p/2 – b)) = p/2 – b

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

If the numerical value of tan (cos–1 (4/5)

Example 27

3 Ê 5ˆ ÁË ˜¯ - 1 = 4 4

(

0 < a < p and cos a > 0)

4 2ˆ Ê tan Á cos - 1 + tan - 1 ˜ = Ë 5 3¯

2 3 2 1 - tan a ◊ 3 tan a +

3 2 + 17 a = 4 3 = (given) = 2 3 6 b 1- ◊ 3 4 a = 17, b = 6, a + b = 23, a – b = 11 and 3b = a + 1 Ê 2ˆ Ê 1ˆ + cos–1 Á - ˜ If a = sin–1 Á ˜ Ë 2¯ Ë 2 ¯

Example 28

Ê and b = tan–1 - 3 – cot–1 Á Ë (a) a – b = 17p/12 (c) a + b = –7p/12 Ans. (a) and (c)

(

)

1 ˆ ˜ , then 3¯ (b) a + b = 17p/12 (d) a – b = p/12

Solution we have a = – p/4 + 2p/3 b = – p/3 – 2p/3 p p -7p = 3 4 12 p 2p p 2p 17 p a – b= - + = + + 4 3 3 3 12

so that a + b = –

Example 29

If a, b are the roots of the equation (tan–1

(x/5))2 + ( 3 – 1) tan–1 (x/5) – 3 = 0, |a| > |b| then (a) a + b = – 5p/12 (b) |a – b| = 35 p/12 2 (d) 3a + 4b = 0 (c) a b = – 25p /12 Ans. (a), (b), (c), (d) Solution: (tan–1 (x/5) + 3 ) (tan–1 (x/5) – 1) = 0 x p 5p =– fix=– fi tan–1 (x/5) = – 3 fi 5 3 3

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and

tan–1 (x/5) = 1 fi

Let a = – fi

x p 5p = fix= 5 4 4

p 2 , so that x = 1/12 is an extraneous root of the given equation. Hence x = –1/12 is the only root. Example 32 If A = tan–1 (1 7 ) and B = tan–1 (1 3) , then (a) cos 2A = 24/25 (b) cos 2B = 4/5 (c) cos 2A = sin 4B (d) tan 2B = 3/4

5p 5p ,b= 3 4

a + b = – 5p/12, | a – b | = 35p/12 a b = –25p2/12 and 3a + 4b = 0.

Example 30 q = tan–1 (2 tan2 q) – tan–1((1/3) tan q ) if tan q is equal to (a) – 2 (b) 1 (c) 2/3 (d) 2

Ans. (a), (b), (c), (d) Solution: We have A = tan–1 (1 7 ) , or tan A = 1 7 .

Ans. (a), (b) Solution: q = tan fi tan q = fi tan q

-1

(2 tan q) - tan 2

-1

2 tan 2 q - (1/3) tan q

2 tan q - (1/3)

Example 31 sin–1 6x + sin–1 6 3 x = – p 2 if x is equal to (a) –1/12 (b) 1/6 (c) 1/12 (d) – 1/6 Ans. (a) Solution: The given equation can be written as sin–1 6x = – p/2 – sin–1 6 3 x

(



6x = – cos (sin–1 6 3 x)



6x = –

) (1)

x = ± 1 12

Substituting x = –1/12 in the given equation, we get LHS = Ê 3ˆ p p p Ê 1ˆ sin–1 Á - ˜ + sin -1 Á - ˜ = - - = - = RHS Ë 2¯ 6 3 2 Ë 2 ¯ Therefore, x = –1/12 is a root of the given equation. Again, substituting x = 1/12 in the given equation, we get LHS =

1 + tan B 2

=

2(1/3) 2 9 3 = ◊ = 1 + (1/9) 3 10 5

1 - 1/9 8 4 = = 1 + 1/9 10 5

\

3 4 24 sin 4B = 2 sin 2B cos 2B = 2 ◊ ◊ = 5 5 25 cos–1 x is equal to

Example 33

(a) 2 sin–1

1- x 2

(b) 2 cos–1

1- x 2

(c) 2 cos–1

1+ x 2

(d) 2 sin–1

1+ x 2

Ans. (a), (c) Solution: Let cos–1 x = y, so that x = cos y. Then



Squaring both sides, we get 36x2 = 1 – 108x2 fi 144x2 = 1 fi

1 - 1/49 48 24 = = 1 + 1/49 50 25

cos 2B =



1 - 108 x 2

=

and

(tan q - 1)2 (tan q + 2) = 0 which holds if tan q = 1 or tan q = - 2.

sin (sin–1 6x) = sin - p 2 - sin -1 6 3 x

2 tan B

sin 2B =

1 + (2/3) tan 3 q



1 + tan A 2

Also, from B = tan–1 (1 3) , or tan B = 1 3 , we get

((1/3) tan q)

–1=0 1 + (2/3) tan 3 q which is true if tan q = 0 2 tan q - (1/3) or = 1 fi tan3 q - 3 tan q + 2 = 0 1 + ( 2/3) tan 2 q fi

1 - tan 2 A

cos 2A =

Also

1- x 1 - cos y = = 2 2 sin–1

y 2 sin 2 ( y /2) = sin 2 2

1- x y 1- x = fi 2 sin -1 =y 2 2 2

1+ x 1 + cos y = = 2 2

1+ x y 1+ x = fi 2 cos -1 =y 2 2 2



cos–1

\

y = cos–1 x = 2 sin–1

Example 34

y 2 cos 2 ( y /2) = cos 2 2

1- x 1+ x = 2 cos -1 2 2

If sin [2 cos–1 {cot (2 tan–1 x)}] = 0, x >

0, then (a) x = 1 (c) x = Ans. (a), (b), (c)

(b) x = 2 –1

(d) x = 3

2 +1

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Solution: The given equation can be written as 2 cos–1 [cot (2 tan–1 x)] = np fi

=

cos–1 [cot (2 tan–1 x)] = np 2 (n Œ I)

\ fi

Note that n can take only three values 0, 1, 2. fi

cot (2 tan–1 x) = cos

if n = 1 np Ï 0 =Ì 2 Ó ± 1 if n = 0, 2

Now, from cot (2 tan–1 x) = 0, we get 2 tan–1 x = kp +

p/2 + 2a = tan–1 (–4/3) 2 tan–1 (– 3) = –p/2 + tan–1(–4/3)

MATRIX-MATCH TYPE QUESTIONS p , 2

Example 36 Column 1

where k = – 1 or 0. fi

tan–1 x = p 4 or - p /4 fi x = ± 1

and

cot (2 tan 2x



1- x

2

–1

x) = ± 1 fi tan (2 tan

1/ 2

–1

x) = ± 1

È 1 È cos(tan -1 y ) + y sin(tan -1 y ) ˘ ˘ 4 (a) Í 2 Í ˙+ y ˙ -1 -1 ÍÎ y Î cot(sin y ) + tan(sin y ) ˚ ˙˚ takes value

x2 ± 2 x – 1 = 0



( x ± 1)

2

= 2 fi x = ± 2 ±1

–1

Example 35 2 tan (– 3) is equal to (a) – cos–1 (– 4/5) (b) – p + cos–1 (4/5) –1 (c) – p/2 + tan (– 4/3) (d) cot–1 (4/3)

Êp ˆ sec x = cos x sin 2x sec x + cos Á + x˜ Ë4 ¯

cos (– 2a) = cos 2a =



– 2a = cos–1

1 - tan a 1 + tan a 2

Ê 4ˆ ÁË - ˜¯ 5

fi 2 tan–1 (– 3) = – cos–1 (– 4/5) Again – p < 2a < 0 fi 0 < 2a + p < p 4 So cos (p + 2a) = – cos 2a = 5 4 5 4 (– 3) = – p + cos–1 5



p + 2a = cos–1



–1

2 tan

Finally fi

cos 2x then possible value of secx is

Let tan–1 (– 3) = a fi tan a = – 3 p – 0, 1 – 3 ¥ 4 < 0]

tan–1 3 + tan–1 4 = p + tan–1

Êp ˆ (c) cos Á - x˜ cos (2x) + sin x sin (2x) sec x Ë4 ¯

fi sin (2x) sec x [sin x – cos x] = cos (2x) [cos (p/4 + x) – cos (p/4 – x)] fi 2 sin x (sin x – cos x) = cos(2x) [–2 sin (p/4)sin x]

1 + 6 x2

Example 37 Column 1 Column 2 (a) tan–1 3 + tan–1 4 (p) p/2 (b) tan–1 (1/3) + tan–1 (1/4) (q) p – tan–1 (7/11) (c) sin–1 (1/3) + cos–1 (1/3) (r) tan–1 (7/11) (s) tan–1 (13) (d) tan–1 (3) + cot–1 (4) p q r s Ans. a

ˆ Êp = cos x sin (2x) sec x + cos Á + x˜ cos(2 x) ¯ Ë4

x 6

1 + 6 x2

fi 12x2 = 5 fi x = ±

=1

=

x 6

=

1 - x2 As x π 0, we get 1 + 6x2 = 6 – 6x2

1/ 2

{

x

Thus,

2 ˘ È Ê 2 2 1 y 1+ y 1- y ˆ 4˙ Í +y = Í y 2 ÁÁ 1 - y 2 + y 2 ˜˜ ˙ ¯ ÍÎ Ë ˙˚ 1/ 2

1 - x2

tan q

RHS = sin q =

2 ˘ È Ê 2 2 2 1 1/ y + 1 + y / y + 1 ˆ 4˙ Í = ˜ +y ˙ Í y 2 ÁÁ 2 2 ˜ ÍÎ Ë 1 - y /y + y / 1 - y ¯ ˙˚

È1 ˘ = Í 2 y 2 (1 - y 4 ) + y 4 ˙ Îy ˚

x

and for RHS, put x 6 = tan q to obtain

1/ 2

2

cos q = sin q

[

= p – tan–1 (7/11) tan–1 (1/3) + tan–1 (1/4) = tan–1

1 3 +1 4 1 - (1 3) (1 4)

= tan–1 (7/11), and sin–1 (1/3) + cos–1 (1/3) = p/2 = tan–1 (3) + cot–1 (4) = tan–1 3 + tan–1 (1/4) = tan– 1 13

)

2 - sin x - cos x = 0

fi sin x = 0 or tan x = 1 or sin x + cos x = 2 fi x = np or x = np + p/4 or x = 2np + p/4 fi sec x = ± 1 or sec x = ± 2 Thus, the possible values of sec x are 1 and 2 . (d) For LHS, put 1 - x 2 = sin q or x = cos q to obtain

Example 38 Column 1 –1

Column 2 2

(a) (tan (x/3)) – 4 tan–1 (x/3) – 5 = 0

(p) x = – 3 tan 1

(b) (tan–1 (3x + 2))2 + 2 tan–1 (3x + 2) = 0

(q) x =

(c) 3 (tan–1 x)2 – 4p tan–1 x + p2 = 0

(r) x = – 2/3

(d) (tan–1 (x/3))2 –

(s) x = 3 tan5

3

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(5 + sin–1 (1/2)) tan–1 (x/2) + 5 sin–1 (1/2) p q r s

Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

so x = – 3 tan 1, 3 tan 5 so x = – 2/3

fi tan–1 x = p or p/3

so x =

(d) fi tan–1 (x/3) = 5 or p/6

(b) sin

–1

(c) tan

–1

Ans.

–1

(–1/2) + cos ( 3) + cot

(d) sin (1 p q r

3

Column 2

(1/2) + cos

–1

3

so x = 3 tan 5,

Example 39 Column 1 (a) sin

(–1/2)

–1

–1

2) + cos s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(p) 7p/6

(1/2)

(q) 5p/6

( - 3)

(r) p/6

–1

(–1

2) (s) p

sin–1 (1/2) + cos–1 (–1/2) = p/6 + 2p/3 = 5p/6 sin–1 (–1/2) + cos–1 (1/2) = – p/6 + p/3 = p/6 tan–1 ( 3) + cot–1 ( - 3) = p/3 + p – p/6 = 7p/6 2) = p/4 + 3p/4 = p

2) + cos–1 (–1

Number of real solutions of

Example 40

Column 1

Column 2 –1

(a) tan (p/4 + (1/2) cos x) + tan (p/4 – (1/2) cos–1 x) = 1 1 1 + tan–1 2x + 1 4x + 1 = tan–1 (2/x2)

(b) tan–1

(c) tan–1 ( x + ( 2 x ) ) – tan–1 (4/x) – tan (d) tan

–1

–1

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

( x - ( 2 x))

(1 – x) + tan

(p) 0 (q) 2

(r) 3

=0 –1

(1 + x) = tan–1 2x (s)1

2 (1 + tan 2 a )

2 2 =1 fi =1 cos 2a x 1 - tan a fi x =2, But for x = 2, cos–1 x has no solution. 1 1 + 2 6x + 2 2 x + x +1 2 1 4 = 2 fi = 2 Next, 2 1 x 8x + 6 x x 1(2 x + 1) (4 x + 1) fi Either x = 0 or 6x2 – 14x – 12 = 0 fi x = 0, 3, – 2/3 So (b) has 3 solutions. (c) can be written as 4 tan–1 ( x + ( 2 x ) ) – tan–1 ( x - ( 2 x ) ) = tan–1 x x + ( 2 x) - ( x - ( 2 x)) 4 fi = 2 2 x 1+ x - 4 x fi



Solution:

sin–1 (1

r

1 + tan a 1 - tan a + =1 1 - tan a 1 + tan a

(a) fi tan–1 (x/3) = 5 or – 1 (b) fi tan–1 (3x + 2) = – 2 or 0

–1

q

Solution: Taking (1/2) cos–1 x = a in (a)

Solution:

(c)

p

Ans.

2

=1fi

( ( ))

(

1 + x2 – 4 x 2

)

= 1 fi x4 = 4 fi x = ± 2

(c) has only 2 real solutions. (d) fi tan–1 (2/x2) = tan–1 (2x) fi x = 1 only one solution Let (x, y) be such that sin–1 (ax) + cos–1 (y) + cos–1 (bxy) = p/2 Column 1 Column 2 (a) If a = 1, b = 0, then (x, y) (p) lies on the circle x2 + y2 = 1 (b) If a = 1, b = 1, then (x, y) (q) lies on (x2 – 1) (y2 – 1) = 0 (c) If a = 1, b = 2, then (x, y) (r) lies on y = x (d) If a = 2, b = 2, then (x, y) (s) lies on (4x2 – 1) (y2 – 1) = 0 p q r s Ans. Example 41

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

IIT JEE eBooks: www.crackjee.xyz Inverse Trigonometric Functions 13.15

Solution: (a) sin–1 x + cos-1 (y) + cos-1 (0) = p/2 fi sin-1 x + cos-1 y = 0 fi sin-1 x = - sin-1

1 - y2

x = - 1 - y 2 fi x2 = 1 - y2 = x2 + y2 = 1



(b) sin-1 x + cos-1 y + cos-1 (xy) = p/2 fi cos-1 (xy) = p/2 - sin-1 x – cos-1 y = cos-1 x – cos-1 y

( xy +

1 - x2 1 - y 2

fi xy = xy +

1 - x2 1 - y 2

= cos-1

(

fi cos-1 (2xy) = cos-1 xy + 1 - x 2 1 - y 2 fi xy =

y2 + y - 56 = 0 fi (y + 8) (y - 7) = 0 fi y = - 8 or 7 fi x = 5 or x = 2 (c) cos-1 x = tan-1 (tan q) where y = tan q 1 1 = fi x = cos q = 2 1+ 3 1+ y

)

fi (1 – x2) (1 - y2) = 0 (c) sin-1 x + cos-1 y + cos-1 (2xy) = p/2

fi x = ± 1/2

)

(d) sin-1 (1) - sin-1

fi x2 y2 = 1 - x2 - y2 + x2 y2

fi sin (p/3) =

fi x2 + y2 = 1 (d) sin-1 (2x) + cos-1 (y) + cos-1 (2xy) = p/2 fi cos-1 (2xy) = sin-1 y - sin-1 (2x)



1- y

2

1 - 4x

2

( 1- y

2

1 - 4 x 2 + 2 xy

=0

2

Example 42 Column 1

Column 2

(a) If x = cosec2 (cot-1 3) - sec2 (tan-1 2), then (b) If tan-1 x + tan-1 (1/y) = tan-1 3 and y2 + y - 56 = 0 (c) If cos-1 x = tan-1 y and y2 = 3 (d) If sin-1 (tan p/4) - sin Ans.

-1

3/ y - p/6 = 0

and x2 = y p q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(

)

- p/6 = 0

3/ y

) 3 = 2

3 y

fi y = 4 fi x = ± 2.

)

ASSERTION-REASON TYPE QUESTIONS Example 43

Statement-1: If 1/2 £ x £ 1 then cos–1

(

˘ ˙ is equal to – p/3 ˙ ˚

Statement-2: sin–1 2 x 1 - x 2

(p) x = 2 (q) x = 5

3/ y

3/ y fi

Èx 3 - 3x 2 x – sin–1 Í + 2 Í2 Î

fi (4x - 1) (y - 1) = 0 2

(

p p = sin-1 2 6



1 - x2 1 - y 2

fi cos-1 (2xy) = cos-1

Solution: (a) x = 1 + cot2 (cot-1 3) - 1 - tan2 (tan-1 2) =9-4=5 x + 1/ y = tan-1 3 if 1 - x/y > 0 (b) tan-1 1 - x/ y 3y - 1 fi x + 1/y = 3(1 - x/y) fi x = 3+ y

) = 2 sin

–1

x

if x Œ ÈÎ - 1

2, 1

2, ˘˚ .

Ans. (b) Solution: In statement-1 put x = cos q then 0 £ q £ p/3

(r) x = 1/2 (s) x = - 1/2

È1 ˘ 3 L.H.S. = [cos–1 (cos q) – sin–1 Í cos q + sin q ˙ 2 Î2 ˚ = q – sin–1 (sin (q + p/3)) = q – q – p/3 – q = – p/3 so statement-1 is true. In statement-2, put x = sin q then – p/4 £ q £ p/4

(

so sin–1 2 x 1 - x 2

) = sin

–1

(2 sin q cos q) = sin–1 (sin 2q)

= 2q = 2 sin–1 x so statement-2 is also true but statement-2 does not lead to statement-1. Example 44

Statement-1: If a2 + b2 = c2, c π 0, ab π 0

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then the non zero Solution of the equation ax bx sin–1 + sin–1 = sin–1 x is ± 1 c c Statement-2: sin–1 x + sin–1 y = sin–1 (x + y) Ans. (c) Solution: Statement-2 is not true (see the theory) For statement-1, since c2 = a2 + b2; Put a = c cos a, b = c sin a. ax bx then sin–1 + sin–1 = sin–1 x c c fi sin–1 (x cos a) + sin–1 (x sin a) = sin–1 x fi sin

–1

fi cos a

Example 47

È x 4 x6 + – p/2 – cos– 1 Í x 2 – 2 4 Î unique solution.

Ans. (c)

cos2 a (1 – x2 sin2 a) + sin2 a (1 – x2 cos2 a)

1 - x 2 sin 2 a

1 - x 2 cos 2 a = x2 sin a cos a

Statement-1: cosec

x 2 x3 + + 2 4

(3/2) + cos

(2/3)

x x2 = 1 + ( x 2) 1 + x 2 2

(

fi x = 1 as 0 < |x|
p/3 Satement-2: If 3 sin–1 x = p/6, then 6x – 8x3 – 1 = 0 Ans. (b) Ê 1ˆ Ê 3ˆ Solution: 3 sin–1 Á ˜ + sin–1 Á ˜ Ë 3¯ Ë 5¯ 1 Ê ˆ Ê 3ˆ = sin–1 Á 3 ¥ - 4 (1 / 3)3 ˜ + sin–1 Á ˜ Ë ¯ Ë 5¯ 3

= – sin (cos–1x) = - 1 - cos 2 (cos -1 x)

4ˆ Ê = sin–1 Á1 - ˜ Ë 27 ¯

= - 1 - x2 fi x2 + y2 = 1 So statement-2 is also true but does not justify statement-1 Example 49 determinant

The

Statement-1:

tan -1 x

value

the

p /2

-1

sin (4 / 5)

of

cos -1 (3 / 5) cos -1 (4 / 5)

1

is equal to zero for all values of x. Statement-2: 2 cos–1 x = cos–1 (2x2 – 1) if –1 £ x £ 1. Ans. (c) Solution: Statement-2 is true if 0 £ x £ 1 For –1 £ x < 0, 2 cos–1 x = p – cos–1 (2x2 – 1) So statement-2 is false. If D denotes the given determinant Then D tan -1 x + cot -1 x =

-1

sin (4 / 5) + sin (3 / 5)

cot -1 x -1

sin (3 / 5)

cos -1 (3 / 5) + cos -1 (4 / 5) cos -1 (4 / 5)

p /2 sin -1 1 1

(Applying C1 Æ C1 + C2) cot -1 x

p /2 =

p /2

È4 9 3 16 ˘ + sin -1 Í 1 1- ˙ sin -1 (3 / 5) sin -1 1 25 5 25 ˚ Î5 È3 4 9 16 ˘ -1 cos -1 Í ¥ - 1 11 ˙ cos (4 / 5) 25 25 ˚ Î5 5 p /2 = sin -1 1

cot -1 x

p /2

sin -1 (3 / 5)

sin -1 1

cos -1 0 cos -1 (4 / 5) = 0 ( cos–1 0 = 1) (C1 and C2 are identical) So statement-1 is true.

1

Ê 3ˆ + sin–1 Á ˜ Ë 5¯

23 3 + sin–1 27 5 3 3 + sin–1 2 2

È 23 ˘ 3 3 = 0.85, = 0.6 and = 0.86 ˙ Í 5 2 Î 27 ˚

sin -1 1

sin (3 / 5)

-1

= sin–1 < sin–1

cot -1 x

-1

Ê 1ˆ Ê 3ˆ Statement-1: 3 sin–1 Á ˜ + sin–1 Á ˜ < Ë 3¯ Ë 5¯

Example 50

fi x2y2 – x2 + 2y2 = 1

2

=

p p 2p + = 3 3 3

and tan–1 (2 2 – 1) > tan–1 3 = [

p 3

2 2 – 1 = 1.8,

3 = 1.7]

So statement-1 is true In statement-2, 3 sin–1 x = fi

sin (3 sin–1 x) =

p 6

1 2

fi 3x – 4x3 = 1/2 fi 6x – 8x3 – 1 = 0 Showing that statement-2 is also true but does not lead to statement-1.

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 51 to 55 a = cos–1 (4/5), b = tan–1 (2/3), 0 < a, b < p/2 a + b is equal to

Example 51 (a) tan

–1

(17/6)

(c) sin–1 (3/5) Example 52

(b) sin–1 (17/ 5 13 ) (d) cos–1 (3/ 13 )

a – b is equal to

(a) cos–1 (18/ 5 13 )

(b) sin–1 (2/ 13 )

(c) tan–1 (1/18)

(d) cos–1 (1/ 5 13 )

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= sin cot–1 tan b = sin cot–1 cot (p/2 – b) = sin (p/2 – b) = cos b

Example 53 cos–1 (44/125) is equal to (a) 2a (b) 3a (c) p – 3a (d) p – 2a

Paragraph for Question Nos. 56 to 58

–1

sin (117/125) is equal to

Example 54

(a) 2a (c) p/2 – 2a

f(x) = sin {cot–1 (x + 1)} – cos (tan–1 x)

(b) 3a (d) p – 3a –1

a = cos tan–1 sin cot–1 x b = cos (2 cos–1 x + sin–1 x)

–1

sin cot tan cos ( 3 13 ) is equal to

Example 55 (a) sin b (c) tan b

(b) cos b (d) none of these

Ans. 51. (a), (b) 52. (a), (c) 53. (c) 54. (b) 55. (b) Solution:

cos a = 4/5, sin a = 3/5, tan a = 3/4

tan b = 2/3, sin b = 2

13 , cos b = 3 13

cos (a + b) = cos a cos b – sin a sin b =

17 5 13 6 5 13

18

1 cos (a – b) = , tan (a – b) = , 18 5 13 sin (a – b) =

1 5 13

cos 2a = 2 cos2 a – 1 = 7/25, cos 3a = 4 cos3 a – 3 cos a = 4 ×

Example 57 (a) 1/2 (c) 5/9

If f(x) = 0 then a2 is equal to (b) 2/3 (d) 9/5

Example 58 If a2 = 26/51, then b2 is equal to (a) 1/25 (b) 24/25 (c) 25/26 (d) 50/51

3 2 + tan a + tan b 17 tan (a + b) = = 4 3 = 3 2 1 - tan a tan b 6 1- ¥ 4 3 sin (a + b) = sin a cos b + cos a sin b =

Example 56 The value of x for which f(x) = 0 is (a) – 1/2 (b) 0 (c) 1/2 (d) 1

64 4 -3¥ 125 5

44 = – 125 44 cos (p – 3a) = – cos 3a = 125 44 fi cos–1 = p – 3a 125 3 4 24 ¥ = 5 5 25 3 27 sin 3a = 3 sin a – 4 sin3 a = 3 ¥ - 4 ¥ 5 125 225 - 108 117 = = 125 125 117 fi sin–1 = 3a 125 sin 2a = 2 sin a cos a = 2 ×

sin cot–1 tan cos–1 (3 13) = sin cot–1 tan cos–1 (cos b)

Ans. 56. (a) 57. (c) 58. (b) Solution: f(x) = 0 fi sin {cot–1 (x + 1)} = cos (tan–1 x) fi sin sin–1

1 1 + ( x + 1)

1



1 + ( x + 1)

2

=

2

= cos cos–1

1 1+ x

2

1 1 + x2

fi 1 + x2 = 2 + x2 + 2x

fi x = – 1/2, so f(x) = 0 for x = – 1/2 a = cos tan–1 sin cot–1 x = cos tan–1 sin a where x = cot a. 1 1 = cos b where tan b = = cos tan–1 2 1+ x 1 + x2 1

= 1+

1 1 + x2

=

x2 + 1 x +2 2

fi a2 =

x2 + 1 x +2 2

=

5 9

for x = – 1/2 Now a = 26/51 fi x = ± 1/5 and b = cos (2 cos–1 x + sin–1 x) = cos (cos–1 x + p/2) 2

fi b = – sin (cos–1 x) = - 1 - x 2 fi b2 = 1 – x2 = 24/25 for x = ±1/5 Paragraph for Question Nos. 59 and 60 a = tan–1 (3 tan3 q), b =

1 cos–1 2

Example 59 b = f, where (a) tan f = 3 tan q (c) sin f = 3 sin q

Ê 5 cos 2q + 4 ˆ ÁË 5 + 4 cos 2q ˜¯

(b) tan f = (1/3) tan q (d) cos f = (1/3) cos q

IIT JEE eBooks: www.crackjee.xyz Inverse Trigonometric Functions 13.19

q = a – b if

Example 60

Example 63

(a) tan q = 0

(b) tan2 q =

9 - 33 6 Ans. 59. (b), 60. (b), (c).

(d) tan q = 1

(c) tan2 q =

9 + 33 6

Solution:

= cos (cos x). The number of points x Œ [0, 4p] satisfying 10 - x the equation f(x) = is 10 Ans. 3 Solution: We have f(x) = cos–1 (cos x)

5 cos 2q + 4 5 (cos 2 q - sin 2 q ) + 4 (cos 2 q + sin 2 q ) 5 + 4 cos 2q 5 + 4 (2 cos 2 q - 1) =

9 cos 2 q - sin 2 q 1 + 8 cos 2 q =

1 - tan 2 f 1 + tan 2 f

=



9 + tan 2 q

where tanf = (1/3) tanq

tan q =

fi fi

tan q [9 tan2q – 1– 3– 3 tan4q] = 0 tan q = 0 or 3tan4q – 9tan2q + 4 = 0 tan2 q =

Ï x Ô2p - x Ô = Ì Ô x - 2p ÔÓ4p - x

Graph of y = f(x) and y =

1 + tan 4 q

1 4 ˘ È - sin -1 If x = tan Í cos -1 ˙ then 5 2 17 ˚ Î

p

\

1

4

= tan–1 7, sin–1

5 2 7-4 3 = x= 1+ 7 ¥ 4 29

17

= tan–1 4

p

O

Solution: y = sin sin–1

y

3p £ x £ 4p

0£ x£p p £ x £ 2p 2p £ x £ 3p 3p £ x £ 4p x 10 - x = 1= g(x) are shown 10 10

y = f (x)

1 1 + x2

2p

3p

4p

f(x) =

10 - x has x

3 solutions.

y = tan–1

If x = sec2 (tan–1 2) + cosec2 (cot–1 3) and

1 + tan 2 8 - 1 then 15x2 + 4y2 is equal to 3434 tan 8

Solution

=

1 1 + x2

= 1 + x2 = 1 + (99)2 = 9802

x = 1 + tan2 (tan–1 2) + 1 + cot2 (cot–1 3) = 1 + 4 + 1 + 9 = 15 1 - cos8 y = tan–1 = tan–1 tan 4 = 4 sin 8

Ans. 1

2

if

+ l where l = Ans. 5

Example 62 If y = sin (cot–1 x) and x = 99, then 9802 y2 is equal to

1

p £ x £ 2p 2p £ x £ 3p

y = g (x )

Example 64

Ans. 3 cos–1

if if

9 ± 33 6

29x is equal to Solution:

0£ x£p

y

INTEGER-ANSWER TYPE QUESTIONS Example 61

if if if if

if

3 tan 3 q - (1 / 3) tan q





Ï cos -1 (cos x) Ô -1 Ôcos (cos( 2p - x)) = Ì -1 Ôcos (cos( x - 2p )) Ô -1 Ócos (cos(4p - x))

9 - tan 2 q

= cos 2f 1 b= cos–1 (cos 2f) = f 2 q = a – b = tan–1 (3 tan3q) – f = tan–1 (3 tan3 q) – tan–1 [(1/3 tan q]

Let f : [0, 4p] Æ [0, p] be defined by f(x)

–1

\ 15x2 + 4y2 = (15)3 + (4)3 = 3439 = 3434 + l fi l = 5 If sin–1 x + sin–1 y + sin–1 z = 3p/2 then 816 is equal to the value of 3000 (x + y + z) – 2 x + y2 + z2 8730 – a where a = Example 65

IIT JEE eBooks: www.crackjee.xyz 13.20 Comprehensive Mathematics—JEE Advanced

Ans. 2 Solution:

Since |sin x | £ p/2, we have –1

sin–1 x = sin–1 y = sin–1 z = p/2 fi x = y = z = 1 816 and 3000 (x + y + z) – 2 = 9000 – 272 = 8728 x + y2 + z2 = 8730 – a fi a =2

8190 (cot S – 1) = 2r 2 + r2 + r4

=

(1 + r + r 2 ) - (1 - r + r 2 ) 1 + (1 + r + r 2 ) (1 - r + r 2 )

90

S = Â [tan–1 (1 + r + r2) – tan–1 (1 – r + r2)] r =1

= Â [tan–1 (1 + r + r2) – tan–1 (1 + (r – 1) + r =1

2

(r – 1) )] = tan–1 (1 + 90 + 902) – tan–1 1

2 tan–1 (1/5) = tan–1 5/12 x = tan (tan

–1

5 -1 –7 (5/12) – p/4) = 12 = 5 17 1+ 12

Example 70

Maximum value of 2997 sin x + 3996 cos x

is equal to 999 k, where k = Ans. 5 2997 sin x + 3996 cos x = 999 (3 sin x + 4 cos x) = 999r sin (x + a) where r sin a = 4, r cos a = 3 and r = 5 £ 999 ¥ 5 = 999k fi k = 5

EXERCISE

2

LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS

If q = cot–1 7 + cot–1 8 + cot–1 18, then

cot8 q is equal to 9x, where x = Ans. 4 7 ¥ 8 ¥ 18 - (7 + 8 + 18) 975 = =3 56 + 144 + 126 - 1 325 fi cot8 q = 38 = (9)4 = (9)x fi x = 4 cot q =

Example 68

Solution:

90 + 902

90 + 90 + 2 8190 fi tan S = 8192 fi 8190 cot S = 8192 fi 8190 (cots S – 1) = 2.

Solution:

Ans. 1

Solution:

90

Example 67

1 p - , then 5 4

fi 17x + 7 = 0 fi 289x2 + 238x + 49 = 0 So the required value = 50 – 49 = 1

Ans. 2 Solution: Since

tan–1 x = 2 tan–1

289x2 + 238x + 50 is equal to

90 2r Ê ˆ , then If S = Â tan -1 Á 2 4˜ Ë ¯ 2 r r + + r =1

Example 66

= tan–1

Example 69

2ˆ 2ˆ 4 Ê Ê If tan–1 Á x + ˜ – tan–1 Á x - ˜ = tan–1 Ë Ë x¯ x¯ x

then the value of 250x4 + 320x2 – 1635 is equal to Ans. 5 Solution: We have 2 Ê 2ˆ -Áx - ˜ x Ë x¯ = 4ˆ Ê 1 + Á x2 - 2 ˜ Ë x ¯ 2 fi x = 2 x+

fi x2 -

4 x2

=0

So250x4 + 320x2 + 137 = 250 ¥ 4 + 320 ¥ 2 – 1635 = 5

1. Number of solutions of the equation 5p 2 is (tan–1x)2 + (cot–1x)2 = 8 (a) 0 (b) 1 (c) 2 (d) 3 2a 2b + sin–1 = 2 tan–1 x, then x = 2. If sin–1 2 2 1+ a 1+ b a-b a+b (a) (b) 1 + ab 1 - ab a-b a+b (c) (d) 1 - ab 1 + ab 2x 3. If y = 2 tan–1 x + sin–1 then 1 + x2 (a) – p/2 < y < p/2 (b) – 3p/2 < y < 3p/2 (c) – p < y < p –1

(d) – p/4 < y < p/4

4. If 0 < sin x < 1 and 1 + sin–1 x + (sin–1 x)2 + ........ x is equal to (a) p/6 (b) p/4 (c) p/3 (d) none of these

IIT JEE eBooks: www.crackjee.xyz Inverse Trigonometric Functions 13.21

p q p 2 2 pq q2 5. If cos–1 + cos–1 = a, then 2 cos a + 2 a b ab a b is equal to (a) sin2 a (b) cos2 a (c) tan2 a (d) cot2 a –1

–1

6. The inequality sin x > cos x holds for (b) x Π(0, 1

(c) x Π(1

(d) no value of x

2)

7. 2 tan–1 (cosec tan–1 x – tan cot–1 x) is equal to (a) sin–1 x

(b) cos–1 x

(c) tan–1 x

(d) cot–1 x

8. sin2 (cos–1 (1/2)) + cos2 (sin–1 (1/3) is equal to (a)

13 36

(b)

59 36

5 (c) (d) none of these 36 9. tan–1 1 + tan–1 2 + tan–1 3 is equal to (a) p/2 (b) p (c) 2p (d) 26p/24 10. If cos–1 x + cos–1 y + cos–1 z = p, then x2 + y2 + z2 + 2xyz is equal to (a) – 1 (b) 0 (c) 1 (d) 5

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 11. A solution of sin–1 (1) – sin–1 ( 3 /x2) – p/6 = 0 is (a) x = - 2

(b) x = 1

(c) x =

(d) x = 1

2

2

È x + 1˘ È x - 1˘ 12. If y = sec–1 Í + sin–1 Í ˙ ˙ Î x - 1˚ Î x + 1˚ È 2x + 3 ˘ –1 z = cosec–1 Í ˙ + cos + 3 x 2 Î ˚

È 3x + 2 ˘ Í 2 x + 3 ˙ then Î ˚

(a) y = p/2 (b) z = p/2 (c) y + z = p (d) y + z = p/2 13. If a and b are two real values of x which satisfy the equation sin–1 x + sin–1 (1 – x) = cos–1 x, then (a) a + b = 1/2 (b) ab = 0 (c) a + b =1 (d) ab = 1/2 14. If q = cos–1 (4/5) + tan–1 (2/3) then (a) sin q =

17

5 13 (c) tan q = 17/6

(b) cos q =

MATRIX-MATCH TYPE QUESTIONS Column 1

(a) all values of x 2 , 1)

15. If A = sin–1 (sin 10), B = cos–1 (cos 10) then (a) A = 3p – 10 (b) A = 3p + 10 (c) A > B (d) A < B

6

5 13 (d) cot q = 17/6

Column 2

16. (a) cos (3 cos–1 x)

(p) 2x

1 - x2

(b) sin (2 sin–1 x)

(q) 4x3 – 3x

(c) tan (3 tan–1 x)

(r)

(d) sin (2 cos–1 x) Column 1 17. (a) cos (2 sin–1 x) = 1/9 (b) cos–1x + sin–1 (x/2) = p/6 (c) tan–11 + cos–1 ( -1

3 x - x3

1 - 3x 2 (s) 2x2 – 1 Column 2 (p) x = 1 (q) x = – 1

2) = cos–1x (r) x = –2/3

(d) tan–1x + sin–1x + cot–1 x = p

(s) x = p/2

ASSERTION-REASON TYPE QUESTIONS 18. Statement-1: If x2 – px + q = 0 where p is twice the tangent of the arithmetic mean of sin–1 x and cos–1 x; q is the geometric mean of tan–1 x and cot–1 x then x = 1 Statement-2: tan (sin–1 x + cos–1 x) = 1 19. Statement-1: If a, b are the roots of the equation 18(tan–1 x)2 – 9p tan–1 x + p2 = 0 then a + b = 4 3 . Statement-2: sec2 (cos–1(1/4)) + cosec2 (sin–1(1/5) = 41 20. Statement-1: sin–1 tan (tan–1 x + tan–1(1 – x)) = p/2 has no non-zero integral solution. Statement-2: The greatest and least value of (sin–1 x)3 + (cos–1 x)3 are respectively 7p3/8 and p3/32.

COMPREHENSION-TYPE QUESTIONS Paragraph for Qustions Nos. 21 to 25 a = tan–1 (1/2) + tan–1 (1/3), b = cos–1 (2/3) + cos–1 ( 5 3) 1 cos–1 (cos (2p/3)) 2 21. cos (a + b + g) is equal to (a) cos (5p/12) (b) cos (7p/12) (c) – cos (5p/12) (d) – cos (7p/12)

and g = sin–1 (sin (2p/3)) +

IIT JEE eBooks: www.crackjee.xyz 13.22 Comprehensive Mathematics—JEE Advanced

22. tan a – tan (b/2) + 3 tan (g/4) is equal to (a) 4 (b) 3 (c) 2 (d) 1 23. sin cot–1 tan cos–1 (sin g) is equal to (a) sin g (b) sin (g/2) (c) (1/2) sin g (d) cos g 24. cos a + cos b + cos g is equal to (a)

2 -1 2

(b)

2 +1 2

(c)

2+ 3 2

(d)

3- 2 2

25. 4 (sin2 a + sin2 b + sin2 g) = a2 where a is equal to (a) – 1 (b) 2 (c) – 3 (d) 3

INTEGER-ANSWER TYPE QUESTIONS 26. If x = cos2 (tan–1 (sin (cot–1 3))) then 1331x3 – 3630x2 + 3300x – 993 is equal to 27. If tan–1 4 = 4 tan–1 x then x5 – 7x3 + 5x2 + 2x + 7 is equal to 28. If a = 2 sin–1 (2/3) and b = 2 tan–1 9 then 80 cosec2 a + 81 cosec2 b is equal to 881l, l = n p 29. If cot -1 > , n Œ N. then the maximum value of p 6 n is 30. If tan–1

2x 1 - x2

+ cot–1

1 - x2 = p/3, x > 0 then 2x

Ê 4 1ˆ ÁË x + 4 ˜¯ – 190 is equal to x

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. A non zero integral solution of the equation is

(a) 1

(c)

x+ y 1 - xy

1 1 2 + tan–1 = tan–1 2 is 2x + 1 4x + 1 x (b) 2

(c) 3

(d) 4

2. tan (3 tan–1 3) + cos (3 cos–1 (1/3)) + 1 is equal to (a) 11 (b) 9/13 (c) 4/27 (d) 295/351 3. sec2 (sin–1 (3 10) + cosec2 (cos–1 (4 17) ) is equal to (a) 7 (b) 27 (c) 10 (d) 17

(d)

x- y 1 + xy

5. If sin–1 (tan 17 p/4) – sin–1 3/ x – p/6 = 0, then x is a root of the equation (a) x2 – x – 6 = 0 (b) x2 + x – 6 = 0 (d) x2 + x – 12 = 0 (c) x2 – x – 12 = 0 6. x = np – tan–1 3 is a solution of the equation 12 tan 2x +

10 + 1 = 0 if cos x

(a) n is any integer (b) n is an even integer (c) n is a positive integer (d) n is an odd integer 7. If sin–1 (x/a) + cos–1 (y/b) = a. Then x2/a2 + y2/b2 is equal to (a) (2xy/ab) cos a + sin2 a (b) (2xy/ab) sin a + cos2 a (c) (2xy/ab) cos2 a + sin a (d) (2xy/ab) sin2 a + cos a

(

)

8. If a = 2 tan–1 2 2 - 1 and b = 3 sin–1 (1/3) + –1

sin (3/5) then (a) a < b (c) a > b

(b) a = b (d) none of these

a ( a + b + c) + tan–1 bc

9. tan–1

LEVEL 2

tan–1

2 È1 2x 1 –1 1 – y ˘ 4. tan Í sin –1 + cos ˙ is equal to 1 + x2 2 1 + y2 ˚ Î2 (a) x + y (b) x – y

b ( a + b + c) + ca

c ( a + b + c) is equal to ab (a) p/4 (b) p/2 (c) 0 tan–1

(d) p –1

10. Two angles of a triangle are sin

(

)

(1 5 )

and

sin–1 1 10 , then third angle is (a) p/6

(b) p/4

(c) p/3

(d) 5p/6

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 11. If tan–1 (x + 1) + cot–1 (x – 1) = sin–1 (4/5) + cot–1 (3/5) then (a) x = 3/4 (b) x = 8/5 (d) x = -4 3/7 (c) x = 4 3/7

IIT JEE eBooks: www.crackjee.xyz Inverse Trigonometric Functions 13.23

12. If sin–1 x + sin–1 y = 2p/3 cos

–1

x – cos

–1

(b) tan–1

y = p/3, then

(a) x = 1/2

(b) x = 1

(c) y = 1/2

(d) y = 1 2

(c) x = 1 –

2

(d) sin–1

(b) x = 1 + (d) x = –1

2

2 –1

14. The equation sin (3x/5) + sin (4x/5) = sin–1 x has (a) 3 roots (b) the roots whose sum is zero (c) roots which form an A.P. (d) no root a˘ Èp 1 15. If x = tan Í + cos -1 ˙ b˚ Î4 2

–1

a˘ Èp 1 and y = tan Í - cos -1 ˙ , then b˚ Î4 2 (a) x + y = 2b/a (b) x + y = 2a/b 2 b -a a 2

(c) x – y =

(q)

Ê sin x ˆ (c) 2 tan–1 Á Ë 1 + cos x ˜¯

13. sin [2 cos–1 cot (2 tan–1 x)] = 0 if (a) x = 1 –

cos x ± sin x cos x ± sin x

2

2 a -b b 2

(d) x – y =

p –x 4

(r) x/2

cos 2 x

(s) p/4 + x

2 ( cos x ± sin x )

ASSERTION-REASON TYPE QUESTIONS 19. Statement-1:

( ) ˘˙ + cot ( ) ˙˚

È log e x 2 cot –1 Í Í log ex 2 Î

( )

È log ex 4

˘ ˙ Í log e2 x 2 ˙ Î ˚

–1 Í

(

)

= p – tan–1 3 Statement-2 È1 + log x 2 ˘ tan–1 Í = tan–1 1 + tan–1(log x2) 2˙ Î1 – log x ˚

2

MATRIX-MATCH TYPE QUESTIONS 16. Column 1 (a) sin–1 (2/3) + cos–1 (2/3) (b) sin–1 (sin (7p/6))

Column 2 (p) sin–1 sin (p/6) (q) sin–1 sin (11p/6)

(c) cos–1 (cos (p/6)) (d) tan–1 3 + tan–1 (1/3)

(r) tan–1 2 + cot–1 2 (s) p/2

Ê 3 x – x3 ˆ 20. Statement-1: Let f(x) = cot–1 Á ˜ , g(x) = Ë 1 – 3x 2 ¯ Ê 1 – x2 ˆ cos–1 Á ˜ Ë 1 + x2 ¯ then f(x) = g(x) if x = Statement-2: sin 18° =

25 – 10 5 . 5 5 –1 . 4

COMPREHESION-TYPE QUESTIONS Paragraph for Question Nos. 21 to 23

17. Column 1

Column 2

(a) sin x + sin 2x = p/3 –1

(b) tan–1

–1

x -1 2x - 1 + tan–1 x +1 2x + 1

= tan–1

(p) 4/3, (q)

1 3 2 7

23 36

(c) tan–1 cos x = sin (tan–1 2)

(r)

(d) sin tan–1 x = – 3

(s) – 3/8

73

5 3

18. Column 1 (a) tan–1

Column 2 1 - cos x 1 + cos x

(p) x

Ê ˆ y I. cot–1 Á ˜ = 2 tan–1 ÁË 1 - x 2 - y 2 ¯˜

3 - 4 x2 4 x2 – tan–1

3 - 4x 2 x2

ax bx + sin–1 = sin–1 x, (c2 = a2 + b2) c c 21. Expression of I as a rational integral equation in x and y is (a) 27x2 = y2(9 – 8y2) (b) 27y2 = x2(9 – 8x2) (c) 27x4 – 9x2 + 8y2 = 0 (d) 27y4 – 9y2 + 8x2 = 0 II. sin–1

IIT JEE eBooks: www.crackjee.xyz 13.24 Comprehensive Mathematics—JEE Advanced

22. Sum of the roots of the equation II is (a) –1 (b) 0 (c) 1 (d) 3 23. For non zero the values of x satisfying II, the values of y in I are 1 1 (b) (a) 2 2 3 3 -1

(c)

(d) imaginary 3 3 Paragraph for Question Nos. 24 and 25 cos–1 24. (a) (b) (c)

x2 a

2

x2 a

2

x

2

a

2

+ +

x2

y2 b

2

y2 b

2

y2 b

2

2 xy cos a = sin2 a ab

+

2 xy cos a = sin2 a ab

-

2 xy cos a = sin2 a ab

(a) a = 0 (c) a = p/2

(b) a = p/4 (d) a = p

INTEGER-ANSWER TYPE QUESTIONS 26. If a = sin–1 (4/5), b = cos–1 (5/13) and x = sin (2a + b) then the value of 325x is equal to a2 where a = 1 + cos x and p = 22/7, then the value 1 - cos x

27. If y = tan–1

x y + cos–1 =a a b

+

25. Given equation represents an ellipse if

of (7x + 14y)3 – 73 is equal to 10300 + k, k = 28. x = sin [sin–1 (3/5) + sin–1 (8/17)] y = cos [cos–1 (4/5) + cos–1 (12/13)] then the value of

2541 is equal to 221l2, l = xy

(

29. If 2 tan–1 x = cos–1 3/ 13 4

2

)

then the value of

60x – 540x + 360x – 51 is equal to 30. If x = sin (2 tan–1 (1/3)) + cos (tan–1 (2 2) ) then

y2

2 xy (d) 2 - 2 cos a = sin2 a ab a b

1 (1 - x)

3

+

1 (1 - x)

2

+

1 is equal to 1205k, k = 1- x

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS Ï Ê 4ˆ Ê 2ˆ ¸ 1. The value of a = tan Ìcos –1 Á ˜ + tan –1 Á ˜ ˝ is Ë ¯ Ë 3¯ ˛ 5 Ó (a) 6/17 (c) 16/7

(b) 7/16 (d) none of these [1983] 2p ˘ –1 È 2. The principal value of sin Ísin is 3 ˙˚ Î (a) – 2p/3 (c) 4p/3 (e) none of these.

(a)

29 3

(b) 29/3

(c)

3 29

(d) 3/29

4. The number of real solutions of tan–1 x ( x + 1) + sin–1 (a) 0

x 2 + x + 1 = π/2 is (b) one

Ê ˆ x 2 x3 + ........˜ + 5. If sin–1 Á x 2 4 Ë ¯ Ê ˆ p x 4 x6 + ........˜ = cos–1 Á x 2 2 2 4 Ë ¯

(b) 2p/3 (d) 5p/3 [1986]

3. If we consider only the principal values of the inverse trigonometric functions, the value of 1 4 ˘ È - sin –1 tan Í cos –1 ˙ is 5 2 17 ˚ Î

[1994]

for 0 < |x| < (a) 1/2 (c) – 1/2 f (x) =

2 , then x equals (b) 1 (d) – 1

sin -1 ( 2 x ) + p / 6 is

[2001]

IIT JEE eBooks: www.crackjee.xyz Inverse Trigonometric Functions 13.25

(a) [– 1/4, 1/2] (c) [– 1/2, 1/2]

(b) [– 1/2, 1/9] (d) [– 1/4, 1/4]

(a)

23 25

(b)

25 23

(c)

23 24

(d)

24 23

[2003] 7. Value of x for which sin {cot–1 (x + 1)} = cos (tan–1 x) is (a) – 1/2 (b) 0 (c) 1/2 (d) 1 8. If 0 < x < 1, then

x

(c) x 1 + x

2

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

[2004]

1 + x 2 [{x cos (cot–1 x) + sin

Ê 6ˆ Ê 4ˆ 1. If a = 3sin –1 Á ˜ and b = 3cos –1 Á ˜ , where the Ë 11¯ Ë 9¯

(cot–1 x)}2 – 1]1/2 equals (a) x

(b) x 1+ x

(d)

[2013]

2

inverse trigonometric functions take only the principal values, then the correct option(s) is(are) (a) cosβ > 0 (b) sinβ < 0 (c) cos(α + β) > 0 (d) cosα < 0

[2008]

n Ê 23 Ê ˆˆ 9. The value of cot Á  cot -1 Á1 +  2k ˜ ˜ is Ë k =1 ¯ ¯ Ë n =1

[2015]

MATRIX-MATCH TYPE QUESTIONS 1. Let (x, y) be such that sin–1 (ax) + cos–1 (y) + cos–1 (bxy) =

(a) (b) (c) (d)

p . 2

Column I If a = 1 and b = 0, then (x, y) If a = 1 and b = 1, then (x, y) If a = 1 and b = 2, then (x, y) If a = 2 and b = 2, then (x, y)

(p) (q) (r) (s)

Column II lies on the circle x2 + y2 = 1 lies on (x2 – 1) (y2 – 1) = 0 lies on y = x lies on (4x2 – 1) (y2 – 1) = 0

[2007]

2. Match List I with List II and select the correct answer using the code given below the lists: List I P.

(

)

(

Ê Ê cos tan -1 y + y sin tan -1 y Á 1 Á Á y 2 Á cot sin -1 y + tan sin -1 y ÁË Ë

(

(

)

)

) ˆ˜

ˆ

2

˜¯

+y



˜ ˜¯

List II

1/2

takes value

Q. If cos x + cos y + cos z = 0 = sin x + sin y + sin z then possible value of cos

x- y is 2

Êp ˆ Êp ˆ R. If cos Á - x˜ cos 2x + sin x sin 2x sec x = cos x sin 2x sec x + cos Á + x˜ Ë4 ¯ Ë4 ¯

1.

1 5 2 3

2.

2

3.

1 2

cos 2x then possible value of sec x is

(

) (

( ))

S. If cot sin -1 1 - x 2 = sin tan -1 x 6 . x π 0, then possible value of x is Codes : P (a) 4 (b) 4 (c) 3 (d) 3

Q 3 3 4 4

R 1 2 2 1

S 2 1 1 2

4. 1

[2013]

IIT JEE eBooks: www.crackjee.xyz 13.26 Comprehensive Mathematics—JEE Advanced

INTEGER-ANSWER TYPE QUESTIONS 1. Let f : [0, 4π] Æ [0, π f(x) = –1 cos (cos x). The number of points x Œ [0, 4π] satisfying the equation 10 – x is f(x) = 10 [2014]

FILL

IN THE

13. (a), (b) 15. (a), (d)

MATRIX-MATCH TYPE QUESTIONS

c (a + b + c) then tan θ equals ________ tan ab [1981] 2. The numerical value of –1

Ï Ê 1ˆ p ¸ α = tan Ì2 tan -1 Á ˜ - ˝ is equal to _______ Ë 5¯ 4 ˛ Ó [1984] 3. The greater of two angles

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

17.

ASSERTION-REASON TYPE QUESTIONS 18. (c)

A = 2 tan–1 (2 2 – 1) and Ê 1ˆ Ê 3ˆ B = 3 sin–1 Á ˜ +sin–1 Á ˜ is ______ Ë 3¯ Ë 5¯

[1989]

1. Find the value of α = cos (2 cos–1 x + sin–1 x) at x = 1/5 where 0 ≤ cos–1 x ≤ π and – π/2 ≤ sin–1 x ≤ π/2. 2. Prove that

[1981]

24. (a)

26. 7 30. 4

27. 8

28. 2

29. 5

[2002]

x2 + 2

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. (c) 5. (c) 9. (d)

SINGLE CORRECT ANSWER TYPE QUESTIONS 4. (a) 8. (b)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 11. (a), (c)

23. (b)

LEVEL 2

x2 + 1

3. (b) 7. (c)

22. (d)

INTEGER-ANSWER TYPE QUESTIONS

LEVEL 1

2. (b) 6. (c) 10. (c)

20. (d)

COMPREHENSION-TYPE QUESTIONS

Answers

1. (b) 5. (a) 9. (b)

19. (b)

21. (b), (c) 25. (c), (d)

SUBJECTIVE-TYPE QUESTIONS

y = cos tan–1 sin cot–1 x =

p

16.

BLANKS TYPE QUESTIONS

1. Let a, b, c be three positive real numbers and a (a + b + c) b (a + b + c) + tan–1 + θ = tan–1 bc ca

14. (a), (b), (c)

12. (a), (b), (c)

2. (d) 6. (d) 10. (b)

3. (b) 7. (b)

4. (c) 8. (c)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 11. (c), (d) 13. (a), (b), (c), (d) 14. (a), (b), (c)

12. (a), (d) 15. (a), (c)

IIT JEE eBooks: www.crackjee.xyz Inverse Trigonometric Functions 13.27

MATRIX-MATCH TYPE QUESTIONS

MATRIX-MATCH TYPE QUESTIONS

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

FILL

p

q

r

s

1. 0

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

16.

17.

18.

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

1.

2. (b)

INTEGER-ANSWER TYPE QUESTIONS 1. 3

BLANKS TYPE QUESTIONS

IN THE

2. – 7/17

SUBJECTIVE-TYPE QUESTIONS 1. – 2 6 5

Hints and Solutions

ASSERTION-REASON TYPE QUESTIONS 19. (a)

22. (b)

23. (b), (c)

24. (c)

INTEGER-ANSWER TYPE QUESTIONS 26. 6 30. 3

LEVEL 1

20. (a)

COMPREHENSION-TYPE QUESTIONS 21. (b) 25. (c)

3. A

27. 5

28. 5

29. 9

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. (d) 5. (b) 9. (b)

2. (e) 6. (a)

3. (d) 7. (a)

4. (c) 8. (c)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 1. (b), (c), (d)

1. We have 2

5p 2 Êpˆ -1 Ê p -1 ˆ ÁË ˜¯ - 2 tan x ÁË - tan x˜¯ = 2 2 8 fi

p2 5p 2 – p tan–1x + 2(tan–1x)2 = 4 8

fi 2(tan–1x)2 – p tan–1x –

3p 2 =0 8

p 3p , 4 4 p p p fi tan–1x = - as - < tan -1 x < 4 2 2 fi x = –1, only one solution

fi tan–1x = -

2. sin–1 (sin 2a) + sin–1 (sin 2b) = 2 tan–1 x where a = tan a, b = tan b fi 2a + 2b = 2 tan–1 x fi tan (a + b) = x a+b fi x= 1 - ab 2x 3. – p/2 < tan–1 x < p/2, – p/2, £ sin–1 £ p/2 1 + x2

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fi – 3p/2 < 2 tan–1 x + sin–1 4.

2x

< 3p/2

1 + x2

1

Ê p q p2 ˆ Ê q2 ˆ . - Á1 - 2 ˜ Á1 - 2 ˜ a b a ¯Ë b ¯ Ë

a2

-

2 pq q2 cos a + 2 = 1 – cos2 a = sin2 a ab b

£ x £ 1. If –1 £ x < 0, sin x < 0 and cos x > 0 so there is no solution. When 0 £ x £ 1, both functions have values in the interval [0, p/2]. Since the sine function increases monotonically in this interval. –1

–1

sin (sin–1 x) > sin (cos–1 x) fi x > fi 2x2 > 1 or x2 > 1/2 fi x Œ (1 1

i.e.

2

1 - x2 2 ,1]

< x £ 1.

(x = tan q) secq - 1 = 2 tan–1 tan q

Ê qˆ –1 ÁË tan ˜¯ = q = tan x. 2

8. sin2 (cos–1 (1/2)) + cos2 (sin–1(1/3)) = 1 – 1/4 + 1 – 1/9 = 59/36 9. p + tan–1

1+ 2 + tan–1 3 = p – tan–1 3 + tan–1 3 = p 1- 2

10. cos x + cos y + cos z = p, –1

–1

–1

fi cos–1 x + cos–1 y = p – cos–1 z fi cos (cos x + cos y) = – z –1

fi xy –

–1

1 - x2 1 - y 2 = – z

fi (xy + z)2 = (1 – x2) (1 – y2) fi x2 + y2 + z2 + 2xyz = 1 11. sin–1 1 – sin–1 fi

(

13. x = sin q = cos (p/2 – q) p sin–1(1– x) = - 2q fi 1 – x = cos 2q = 2 cos2 q – 1 2 fi 1 – x = 1 – 2x2 fi x (2x – 1) = 0 fi x = 0 or 1/2 \ if a = 0, b = 1/2

)

3 x 2 – p/6 = 0

p p - = sin–1 2 6

(

3 x2

)

a + b = 1/2, ab = 0

14. cos (4/5) = a fi cos a = 4/5, sin a = 3/5, –1

tan a = 3/4 tan–1 (2/3) = b fi tan b = 2/3, sin b = cos b =

3 13

2 13

,

.

q = a + b fi tan q = cos q =

Ê 1 + x2 1 ˆ 1 + x2 - 1 7. 2 tan–1 Á - ˜ = 2 tan–1 ÁË x x ˜¯ x

= 2 tan–1

3 3 = 2 fi x2 = 2 2 x

x -1 x -1 + sin–1 = p/2 x +1 x +1 similarly z = p/2 and y + z = p

2

p2

x



12. y = cos–1

p2 q2 p2q2 Ê pq ˆ - cos a ˜ = 1 - 2 - 2 + 2 2 fi Á Ë ab ¯ a b a b fi

3 2

fi x= ± 2

= 2 fi 1 – sin–1 x = 1/2 1 - sin -1 x fi sin–1 x = 1/2 fi x = p/6

5. cos a =

fi sin (p/3) =

6 5 13

17 17 , sin q = 6 5 13

.

p so let 10 = 3p + a, 0 < a < p/2 2 the sin–1 sin (10) = sin–1 (– sin a) = – a = 3p – 10 and cos–1 cos (10) = cos–1 (cos a) = a = 10 – 3p

15. 3 p < 10 < 3p +

A < 0 and B > 0 so B > A. 16. cos (3 cos x) = cos 3q = 4 cos3 q – 3 cos q = 4x3 – 3x [x = cos q] sin (2 sin–1 x) = sin 2q = 2 sin q cos q = –1

2x

1 - x2

tan (3 tan–1 x) = tan 3q =

[x = sin q] 3x - x

3

1 - 3x 2

(x = tan q)

sin (2 cos–1 x) = 2 x 1 – x 2 17. x = sin q fi cos 2q = 1/9 fi 1 – 2 sin2 q = 1/9 fi sin q = ± 2/3 x x tan–1 x + sin–1 x + cot–1 x = p p p fi + – cos–1x = p fi cos–1 x = 0 2 2 fix=1

IIT JEE eBooks: www.crackjee.xyz Inverse Trigonometric Functions 13.29

18. Statement-2 is false as tan (sin–1 x + cos–1 x) = tan (p/2)

cos (a + b + g) = cos (p/4 + p/2 + 2p/3) = cos

In statement-1, p = 2 tan q=

sin –1 x + cos –1 x = 2, 2

5p ˆ 5p Ê = cos Á p + = – cos = cos ˜ Ë 12 ¯ 12

tan –1 x cot –1 x = 1 and we get x2 – 2x + 1 = 0 = cos

fi x = 1, so it is True. Êp p ˆ x)2 – Á + ˜ tan–1 x + Ë 6 3¯

19. In statement-1, (tan–1 p p ¥ =1 6 3

17p 12 5p ˆ Ê ÁË p ˜ 12 ¯

7p 12

22. tan a – tan (b/2) + 3 tan (g/4) 1 =1 =1–1+ 3¥ 3 23. sin cot–1 tan cos–1 (sin g)

fi tan–1 x = p/6 or tan–1 x = p/3 fi x = 1

(

fia+b= 1

)

3 + 3 = 4

3.

3 fi statement-1 is

True

2p ˆ Ê = sin cot–1 tan cos–1 Á sin ˜ Ë 3 ¯ = sin cot–1 tan cos–1 [sin (p – 2p/3)] = sin cot–1 tan cos–1 (cos (p/2 – p/3)) = sin cot–1 tan (p/2 – p/3)

L.H.S of Statement-2 = sec2 (sec–1 4) + cosec2 (cosec–1 5)

= sin cot–1 cot (p/3) = sin

= 16 + 25 = 41 and the statement-2 is True.

= sin

20. sin–1 x tan (tan–1 x + tan–1 (1 – x)) = p/2 x +1– x fi = sin p/2 = 1 fi 1 – x(1 – x) = 1 1 – x (1 – x ) fi x = 1 is a non zero solution and thus the statement is false. Next, (sin–1 x)3 + (cos–1 x)3 = (sin–1 x + cos–1 x)3 – 3 sin–1 x cos–1 x (sin–1 x + cos–1 x) = (p/2)3 – (3p/2) sin–1 x(p/2 – sin–1 x) = (p3/8) – (3p2/4) sin–1 x + (3p/2) (sin–1 x)2 = (p3/8) + (3p/2) [sin–1 x – (p/4)]2 – (3p3/32) = (p3/32) + (3p/2) [sin–1 x – (p/4)]2 Greatest and least value of sin–1 x – p/4 is (3p/4) and 0 respectively. 1 2 +1 3 21. a = tan–1 (1/2) + tan–1 (1/3) = tan–1 1 - (1 2) (1 3)

g 2

p 3

1

24. cos a + cos b + cos g =

2

(

)

5 3 = cos–1 (2/3) + sin–1

1 2p ˆ Ê 2p ˆ Ê + cos–1 Á cos ˜ g = sin–1 Á sin ˜ Ë Ë 3 ¯ 3 ¯ 2 = sin–1 sin (p – p/3) + p 1 2p 2p = = + ¥ 3 2 3 3

1 2p cos–1 cos 2 3

2 -1 2

1 = 2

3˘ È1 25. 4 (sin2 a + sin2 b + sin2 g) = 4 Í + 1 + ˙ = 9 = a2 4˚ Î2 2 –1 –1 26. If x = cos (tan (sin (cot 3))) then x=

1 + 32 2+3

2

=

10 11

1331x3 – 3630x2 + 3300x – 993 is equal = (11x – 10)3 + 7 = 7 as 11x – 10 = 0 27. 4 = tan 4q where x = tan q

tan 2q =

2x 1 - x2

fi tan 4q =

Ê 2x ˆ 2Á Ë 1 - x 2 ˜¯ 1-

= tan–1 1 = p/4 b = cos–1 (2/3) + cos–1 (2/3) = p/2

+0-

=

4 x2 (1 - x 2 ) 2

4 x - 4 x3 1 - 6 x2 + x4

fi x4 + x3 – 6x2 – x + 1 = 0 Now x5 – 7x3 + 5x2 + 2x + 7 = (x – 1) (x4 + x3 – 6x2 – x + 1) + 8 = 8 È 2 4˘ 28. a = 2 sin–1 (2/3) = sin–1 Í 2 ¥ 1- ˙ 3 9˚ Î

= 4 (given)

IIT JEE eBooks: www.crackjee.xyz 13.30 Comprehensive Mathematics—JEE Advanced

= sin–1

4 5 9

fi 81 sin2 a = 80 fi 80 cosec2 a = 81 2¥9 9 = sin–1 b = 2 tan–1 9 = sin–1 1 + 81 41 9 fi 81 cosec2 b = 1681 41 80 cosec2 a + 81 cosec2 b = 1681 + 81 = 1762 = 881 l fil=2 n p 29. cot -1 > p 6 fi sin b =

nˆ p Ê fi cot Á cot -1 ˜ < cot Ë ¯ p 6 fi

n < 3 p

fi n < 3p = 5.5 So the maximum value of n is 5 2x 2x + tan–1 = p/3 30. tan–1 2 1- x 1 - x2 2x fi 2 tan–1 = p/3 1 - x2 2x 1 = tan (p/6) = fi 2 1- x 3 fi x4 – 14x2 + 1 = 0 fi x4 + 1 = 14x2 1 fi x8 + 1 = 194x4 fi x4 + 4 = 194 x 1ˆ Ê fi Á x 4 + 4 ˜ – 190 = 4 Ë x ¯

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14 Cartesian System of Coordinates and Locus 14.1 RESULTS REGARDING POINTS IN A PLANE Distance formula

The distance between two points P(x1, y1) and Q(x2, y2) is given by PQ =

( x1 - x2 ) 2 + ( y1 - y2 ) 2 . The distance from the origin

O (0, 0) to the point P(x1, y1) is OP =

x12 + y12 .

Illustration 2 The line x – y + k = 0 passes through the point which divides the segment joining the points (2, 3) and (4, 5) in the ratio 2 : 3. Find the value of k. Solution: Using the section formula, we get the coordinates of the point of division as Ê 2 ¥ 4 + 3 ¥ 2 2 ¥ 5 + 3 ¥ 3 ˆ Ê 14 19 ˆ , Ë ¯ =Ë 5 , 5¯ 5 5

Since it lies on the line x – y + k = 0 Illustration 1 Show that the triangle with vertices (3, 0), (–1, –1) and (2, 4) is isosceles and right angled.h Solution: Let the vertices of the triangle be A (3, 0), B (–1, –1) and C (2, 4) then using distance formula, we have AB =

(3 + 1)2 + (0 + 1)2 = 17

BC =

(-1 - 2 )2 + (-1 - 4 )2 = 9 + 25 = 34

and CA =

(2 - 3)2 + (4 - 0 )2 = 17 .

Since AB = AC = 17 , the triangle ABC is isosceles. Also (AB)2 + (AC)2 = (BC)2 shows the triangle is right angled. Section formula

If R(x, y) divides the join of P(x1, y1) and Q(x2, y2) in the ratio m : n (m, n > 0, m π n) then my2 ± ny1 mx2 ± nx1 x = and y = m±n m±n The positive sign is taken for internal division and the negative sign for external division. The mid-point of P(x1, y1) and Q(x2, y2) is Ê x1 + x2 y1 + y2 ˆ , ÁË ˜ which corresponds to internal division 2 2 ¯ when m = n. Note that for external division m π n.

14 19 +k=0fik=1 5 5

Centroid of a triangle

If G(x, y) is the centroid of the triangle with vertices A(x1, y1), B(x2, y2) and C(x3, y3), then x =

x1 + x2 + x3 3

and

y=

y1 + y2 + y3 3

Illustration 3 Show that the centroid of the triangle with vertices (5 cos q, 4 sin q), (4 cos q, 5 sin q) and (0, 0) lies on the circle x2 + y2 = 9. Solution: Coordinates of the centroid of the given triangle are, Ê 5 cosq + 4 cosq + 0 4 sin q + 5 sin q + 0 ˆ , Ë ¯ 3 3

= (3cos q, 3sin q ) which lies on the circle x2 + y2 = 9 as (3 cos q )2 + (3 sin q)2 = 9.

Note When A, B, C are taken as vertices of a triangle, it is assumed that they are not collinear.

IIT JEE eBooks: www.crackjee.xyz 14.2 Comprehensive Mathematics—JEE Advanced

Incentre of a triangle

Illustration 5

If I(x, y) is the incentre of the triangle with vertices A(x1, y1), B(x2, y2) and C(x3, y3), then x =

a x1 + bx2 + c x3 a+b+c

and

a y1 + b y2 + c y3 y = a+b+c a, b and c being the lengths of the sides BC, CA and AB, respectively of the triangle ABC.

If the area of the triangle with vertices at the points whose coordinates are (2, 5), (0, 3) and (4, k) is 4 units, then find the value of k. Solution: We have 2 5 1 1 | 0 3 1 |= 4 2 4 k 1

fi 2(3 – k) + 4(5 – 3) = ± 8 fi 7 – k = ± 4 fi k = 3 or 11

Illustration 4 Condition of collinearity

Find the coordinates of the in center of the triangle with vertices A (–1, 12), B (–1, 0) and C (4, 0). Solution: We have

Three points A(x1, y1), B(x2, y2) and C(x3, y3) are collinear if and only if x1 x2 x3

A(–1, 12)

b

c

B(–1, 0)

a

y1 1 y2 1 = 0 y3 1

Circumcentre of a Triangle

The circumcentre of a triangle is the centre of the circle passing through the vertices of the triangle, so it is equidistant from the vertices of the triangle. For example– The circumcentre (x, y) of the triangle with vertices (0, 0), (3, 4) and (4, 3) is given by x2 + y2 = (x – 3)2 + (y – 4)2 = (x – 4)2 + (y – 3)2 Solving, we get (x, y) = (25/14, 25/14) as the required circumcentre of the triangle.

C(4, 0)

Fig. 14.1

a = BC =

(-1 - 4 )2 + 0 = 5

b = CA =

(4 + 1)2 + (0 - 12 )2 =13

c = AB =

(-1 + 1)2 + (12 - 0 )2 = 12

14.2 SOME USEFUL POINTS

So using the formula for the coordinates of the in-center we get the required coordinates

To show that A, B, C, D are the vertices of a

as ÊÁ 5(-1) + 13(- 1) + 12(4) , 5(12) + 13(0) + 12(0) ˆ˜

Parallelogram

Ë

5 + 13 + 12

5 + 13 + 12

¯

Show that the diagonals AC and BD bisect each other.

= (1, 2) Area of triangle

ABC with vertices A(x1, y1), B(x2, y2) and C(x3, y3) is x1 y1 1 1 | x2 y2 1 | 2 x3 y3 1 1 = | x1 ( y2 – y3) + x2( y3 – y1) + x3( y1 – y2)| 2 and is generally denoted by D. Note that if one of the vertex 1 (x3, y3) is at O(0, 0), then D = | x1 y2 – x2 y1 |. 2

Illustration 6 If P(4, 7), Q(7, 2) R(a, b) and S(3, 8) are the vertices of a parallelogram then find the value of a and b. Solution: Since the diagonals of a parallelogram bisect each other. Mid points of PR and QS are same. 4 + a 7 + bˆ = ÊË 7 + 3 , 8 + 2 ˆ¯ , so ÊË 2 2 ¯ 2 2

fi a = 6, b = 3

Rhombus

Show that the diagonals AC and BD bisect each other and a pair of adjacent sides, say, AB and BC are equal.

IIT JEE eBooks: www.crackjee.xyz Cartesian System of Coordinates and Locus 14.3

Square

Translation of Axes

Show that the diagonals AC and BD are equal and bisect each other, a pair of adjacent sides say AB and BC are equal.

The shifting of origin of axes without rotation of axes is called Translation of axes. If the origin (0, 0) is shifted to the point (h, k) without rotation of the axes then the coordinates (X, Y ) of a point P (x, y) with respect to the new system of coordinates are given by X = x – h, Y = y – k.

Rectangle

Show that the diagonals AC and BD are equal and bisect each other.

Illustration 8

14.3 LOCUS OF A POINT

If origin is shifted to the point (2, 3) and the axes are rotated through an angle p /4 in the anticlockwise direction, then find the coordinates of the point (7, 11) in the new system of coordinates. Solution: When origin is shifted to the point (2, 3), the coordinates of P(7, 11) are (7 – 2, 11 – 3) = (5, 8). Now when the axes are rotated through an angle p /4 in the anticlockwise direction, the coordinates of P are (5 cos p/4 + 8 sin p/4, 8 cos p/4 – 5 sin p/4.) =

To obtain the equation of a set of points satisfying some given condition(s) called locus, proceed as follows. 1. Let P (h, k) be any point on the locus. 2. Write the given condition involving h and k and 3. Eliminate the unknowns, if any. 4. Replace h by x and k by y and obtain an equation in terms of (x, y) and the known quantities. This is the required locus.

Ê 13 3 ˆ , ÁË ˜ 2 2¯ which are the required coordinates.

Illustration 7 Find the locus of the mid point of the portion between the axes of the line x cos a + y sin a = sin a tan a. Solution: The given line meets x-axis at the point A (tan2 a, 0) and y-axis at the point B (0, tan a) Let P(h, k) be the mid point of AB, then Y

B(0, tana)

A(tan2a,0)

O

Fig. 14.2 2 h = tan a , k = tana .

2

2

Eliminating a, we get 2k2 = h So the required locus is 2y2 – x = 0.

SOLVED EXAMPLES

SINGLE CORRECT ANSWER TYPE QUESTIONS If P(1, 2), Q(4, 6), R(5, 7) and S(a, b) are Example 1 the vertices of a parallelogram PQRS, then (a) a = 2, b = 4 (b) a = 3, b = 4 (c) a = 2, b = 3 (d) a = 3, b = 5 Ans. (c) Solution: PQRS represents a parallelogram so mid points of the diagonals PR and QS are same 1+ 5 4+a 2+7 6+b = , = 2 2 2 2 a = 2, b = 3.

fi fi

Example 2 14.4 CHANGE OF AXES Rotation of Axes

If the axes are rotated through an angle q in the anticlockwise X, Y ) of a point P (x, y) with respect to the new system of coordinate are given by X = x cos q + y sin q and Y = y cos q – x sin q.

The incentre of the triangle with vertices

(1, 3 ) , (0, 0) and (2, 0) is Ê 3ˆ (a) Á1, 2 ˜¯ Ë

Ê2 1 ˆ (b) Á , ˜ Ë 3 3¯

Ê 2 3ˆ (c) Á , ˜ Ë3 2 ¯

Ê 1 ˆ (d) Á1, ˜ Ë 3¯

Ans. (d) Solution:

AB =

1+ 3 = 2

IIT JEE eBooks: www.crackjee.xyz 14.4 Comprehensive Mathematics—JEE Advanced

BC = 2, CA =

A (2,20)

(2 - 1) 2 + ( 3 )2 = 2 A (1, ÷3) I

So the triangle is equilateral hence the incentre I of the triangle is at the centroid. Ê 1 + 0 + 2 3 + 0 + 0ˆ Ê 1 ˆ , = Á1, ˜ Á ˜ Ë 3 3 3¯ Ë ¯

C

Fig. 14.4

C (2,0) Fig. 14.3

H

D(x,y)

B B (0,0)

G

From Geometry Circumcentre, centroid and ortho centre of a triangle lie on a line and the centroid divides the line joining the circumcentre and the ortho centre in the ratio 1:2. Ê 1 - 2 4 + 4ˆ Ê 1 8ˆ , So coordinates of G are Á ˜¯ = ÁË - , ˜¯ Ë 3 3 3 3 Also G divides AD in the ratio 2 : 1

Example 3

The area of a triangle is 5, two of its vertices

are (2, 1) and (3, –2). If the third vertex lies on y = x + 3, coordinates of the third vertex lying in the I quadrant are Ê 2 11ˆ Ê 7 13 ˆ (a) Á , ˜ (b) Á , ˜ Ë3 3¯ Ë2 2¯ Ê 2 23 ˆ (c) Á , ˜ Ë7 7 ¯



3 x = - , y = –6 2

Example 5 (d) (2, 5)

Ans. (b) Solution: x x+3 1 2 1 2 3 -2

Let the third vertex be (x, x + 3) then 1 1 =±5 1 7 fi 3x + x + 3 – 7 = ±10 fi x = or x = –3 2 Ê 7 13 ˆ So the coordinates of the third vertex are Á , ˜ Ë2 2¯ Example 4

Ê 2 x + 2 2 y + 20 ˆ Ê 1 8ˆ , ÁË ˜¯ = ÁË - , ˜¯ 3 3 2 3



the equation of the locus of the point P(x, y) such that | AP – BP| = 6 is (a) x2/7 – y2/9 = 1 (c) x2/9 – y2/7 = 1 Ans. (b) Solution: and

ˆ Ê3 (c) Á , -6˜ ¯ Ë2

Ê 1 8ˆ (d) Á , ˜ Ë 3 3¯

Ans. (c) Solution: Let D(x, y) be the mid point of BC. G, the centroid of the DABC

(b) y2/9 – x2/7 = 1 (d) y2/7 – x2/9 = 1

x 2 + ( y – 4) – x 2 + ( y + 4) 2

2

=6

È x 2 + ( y – 4 )2 ˘ – È x 2 + ( y + 4 )2 ˘ = –16 y Î ˚ Î ˚



x 2 + ( y – 4) + x 2 + ( y + 4)



2 x 2 + ( y – 4)

If in a triangle ABC, vertex is A(2, 20)

circumcentre I is (–1, 2) and the ortho centre H is (1, 4), the coordinates of mid point of side BC are Ê 1ˆ (a) Á 0, ˜ (b) (1, 2) Ë 2¯

Given the points A(0, 4) and B(0, – 4),

2

2

2

–8 y 3 8y = 6– 3 =



4yˆ 2 Ê x2 + (y – 4)2 = Á 3 – ˜ Ë 3¯



x2 + y2 + 16 = 9 +

⇒ Example 6

16 y 2 9

y2 x 2 – = 1. 9 7 Coordinates (x, y) of a point P satisfy the

relations 3x + 4y = 9, y = mx + 1. The number of integral values of m for which the x-coordinate of P is also an integer is (a) 0 (b) 1 (c) 2 (d) 4

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Ans. (c) Solution

Solving the given relations we have 5 , which is an integer if x = 3 + 4m

3 + 4m = ± 1 or 3 + 4m = ± 5. ⇒ m = –1 or m = –2, two integral values of m. Example 7

The points A(2, 3), B(3, 5), C(7, 7) and

D(4, 5) are such that (a) ABCD is a parallelogram (b) A, B, C and D are collinear (c) D lies inside the triangle ABC (d) D lies on the boundary of the triangle ABC. Ans. (c) 2+3+7 3+5+7 = 4, = 5 D(4, 5) is the Solution Since 3 3 centroid of the triangle ABC and hence lies inside it. Example 8

(a) PQ (c) RS Ans. (b)

a+b a+b + y sin = 0. 2 2

Example 10 OPQR is a square and M, N are the middle points of the sides PQ and QR respectively then the ratio of the areas of the square and the triangle OMN is (a) 4 : 1 (b) 2 : 1 (c) 8 : 3 (d) 4 : 3 Ans. (c) Solution

Taking the coordinates of vertices O, P, Q, R

as (0, 0), (a, 0), (a, a), (0, a) respectively we get the coordinates of M as (a, a/2) and of N as (a/2, a) (Fig. 14.5)

the ratio 3 : 5 and hence is the middle point of QR. L P (a, y) Q

Fig. 14.5

\

(b) QR (d) ST

Ê 5a + 3b 5 x + 3 y ˆ , Solution The point L Á ˜ divides PT in Ë 8 8 ¯

R

S

T (b, y)

The line joining A (b cos a, b sin a) and B Example 9 (a cos b, a sin b ) is produced to the point M (x, y) so that a +b a +b + y sin = AM : MB = b : a, then x cos 2 2 (a) – 1 (b) 0 (c) 1 (d) a2 + b2 Ans. (b) Solution As M divides AB externally in the ratio b : a b ( a cos b ) - a (b cos a ) and x= b-a



x cos

Q, R and S are the points on the line joining

the points P(a, x) and T(b, y) such that PQ = QR = RS = ST, Ê 5a + 3b 5 x + 3 y ˆ , then Á ˜ is the mid point of the segment Ë 8 8 ¯

y=



b ( a sin b ) - a ( b sin a ) b-a x cos b - cos a = = y sin b - sin a

a +b a -b sin 2 2 a +b b -a 2 cos sin 2 2

\

1 Area of the DOMN = | 2

0 0 1 3a 2 a a /2 1 |= 8 a /2 a 1

Area of the square = a2 the required ratio is 8 : 3.

If p, x1, x2 xi, and q, y1, y2 , yi, Example 11 are in A.P., with common difference a and b respectively, then locus of the centre of mean position of the points Ai (xi, yi), i = 1, 2 n is (a) ax – by = aq – bp (b) bx – ay = ap – bq (c) bx – ay = bp – aq (d) ax – by = bq – ap È Ê Sxi Syi ˆ ˘ , Ans. (c) ÍNote. Centre of Mean Position is Á Ë n n ˜¯ ˙˚ Î Solution Let the coordinates of the centre of mean position of the points Ai, i = 1, 2, n be (x, y), then x= fi

2 sin



x1 + x2 + + xn y + y2 + + yn ,y= 1 n n n p + a (1 + 2 + + n) x= , n n q + b (1 + 2 + + n) y= n n ( n + 1) n ( n + 1) x=p+ a,y=q+ b 2n 2n

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x=p+



2

Ans. (d) Solution

n +1 n +1 a, y = q + b 2 2

( x - p) ( y - q) =2 fi bx – ay = bp – aq. a b

Example 12 The number of points (p, q) such that p, q Œ {1, 2, 3, 4} and the equation px2 + qx + 1 = 0 has real roots is (a) 7 (b) 8 (c) 9 (d) none of these Ans. (a) Solution px2 + qx + 1 = 0 has real roots if q2 – 4p ≥ 0 or q2 ≥ 4p Since p, q Œ {1, 2, 3, 4} The required points are (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4), (4, 4) So the required number is 7. Example 13 If G is the centroid and I the incentre of the triangle with vertices A(–36, 7), B(20, 7) and C(0, –8), then GI is equal to. (a) 173 / 3 (b) 397 / 3 (c) 205 / 3 Ans. (c) Solution We have a = BC = b = CA =

(d) None of these (20)2 + (7 + 8)2 = 25,

(36)2 + (8 + 7)2 = 39

and

2 c = AB = (56) = 56 Therefore, the coordinates of I are

Ê 25(- 36) + 39 (20) + 56 (0) 25 (7) + 39 (7) + 56 (- 8) ˆ , ÁË ˜¯ 25 + 39 + 56 25 + 39 + 56 = (–1, 0) and the coordinates of G are Ê - 36 + 20 + 0 7 + 7 - 8 ˆ Ê 16 ˆ , ÁË ˜¯ = ÁË - , 2˜¯ 3 3 3 \ GI =

16 ˆ 2 1 Ê 2 205 ÁË - 1 + ˜¯ + (0 - 2) = 3 3

Example 14 Consider the points P = (– sin (b – a), – cos b), Q = (cos (b – a), sin b) and R = (cos (b – a + q), sin (b – q)), where 0 < a, b < p/4 then (a) P lies on the line segment RQ (b) Q lies on the line segment PR (c) R lies on the line segment QP (d) P, Q, R are non-collinear.

Put β – α = φ and consider the determinant – sin f – cos b 1 cos f sin b 1 Δ = cos (f + q ) sin ( b – q ) 1

Using R3 → R3 – cos θ R2 – sin θ R1

Δ =

– sin f cos f

– cos b sin b

0

0

1 1 1 – cos q – sin q

= (1 – cos θ – sin θ) cos (φ + β) = (1 – cos θ – sin θ) cos (2β – α) = [1 –

2 sin (θ + π /4)]

cos (2β – α ) As 0 < θ < π/4 ⇒ π /4 < θ + π/4 < π/2 1 < sin (θ + π/4) < 1 ⇒ 2 and 0 < α, β < π /4 ⇒ – π /4 < 2 β – α < π/2 ⇒ cos (2β – α) ≠ 0 Thus Δ ≠ 0 and the points P, Q, R are non-collinear. Example 15 The x-coordinates of the vertices of a square of unit area are the roots of the equation x2 – 3|x| + 2 = 0 and the y-coordinates of the vertices are the roots of the equation y2 – 3y + 2 = 0. The number of such square is (a) 1 (b) 2 (c) 3 (d) 4 Ans. (b) Solution: x2 – 3|x| + 2 = 0 fi |x| = 1, 2 fi x = ±1, ±2 and y2 – 3y + 2 = 0 fi y = 1, 2 Y

x = -2 x = -1 (-2, 2)

(-1, 2)

(-2, 1)

(-1, 1)

x=1

O

x=2

(1, 2)

(2, 2)

(1, 1)

(2, 1)

y=2 y=1 X

Fig. 14.6

So the vertices can be (–2, 1), (–1, 1), (–1, 2), (–2, 2) or (1, 1), (2, 1), (2, 2), (1, 2) Both are of unit area. Hence there are 2 such squares.

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MULTIPLE CORRECT ANSWERS TYPE QUESTIONS Example 16

A, B, C are three points on the axis of x

whose distances from the origin O are in A. P and OA + OB + OC = 12. If the distance of P(0, 6) on the y-axis from the point C is 6 2 units, then the coordinates of A, B, C are respectively. (a) (–6, 0), (4, 0), (14, 0) (b) (14, 0), (4, 0), (–6, 0) (c) (2, 0), (4, 0), (6, 0) (d) (6, 0), (4, 0), (2, 0) Ans. (b), (c) Solution: Set OA = a – d, OB = a Y P(0, 6)

O

A (a, - d)

B C (a, 0) (a, + d, 0)

X

Fig. 14.7

And OC = a + d So OA + OB + OC = 12 fi a – d + a + a + d = 12 fi 3a = 12 fi a = 4 So the coordinates of B are (4, 0) 2 Next, (PC)2 = (a + d)2 + 62 = (6 2 ) fi (4 + d)2 = 36 fi d = 2 or –10 When d = 2, Coordinates of A are (2, 0) and of C are (6, 0) When d = –10, coordinates of A are (14, 0) and of C are (–6, 0).

Example 17 2

If points A(3, 4), B(7, 12) and P(x, x) are

such that (PA) > (PB)2 > (AB)2 then integral value of x can be. (a) 7 (b) 12 (c) 16 (d) 20 Ans. (c), (d) Solution: (PA)2 > (PB)2 fi (x – 3)2 + (x – 4)2 > (x – 7)2 + (x – 12)2 fi x>7 (PB)2 > (AB)2 fi (x – 7)2 + (x – 12)2 > (3 – 7)2 + (4 – 12)2 fi 2x2 – 38x + 113 > 0 x = 16 and x = 20.

Example 18 If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b), then (a) ax = by (b) bx = ay (c) x2 – y2 = 2(ax + by) (d) P can be (a, b). Ans. (b) and (d) Solution We have PA = PB, i.e., (PA)2 = (PB)2. fi [x – (a + b)]2 + [y – (b – a)]2 = [x – (a – b)]2 + [y – (a + b)]2 2 fi [(x – a) – b] + [( y – b) + a]2 = [(x – a) + b]2 + [( y – b) – a]2 2 fi [(x – a) + b] – [(x – a) – b]2 = [( y – b) + a]2 – [( y – b) – a]2 fi 4b (x – a) = 4 a ( y – b) fi b x = a y (1) Therefore, (b) is correct. Also, P(a, b condition (1), so that P can be (a, b) and hence (d) is also correct. Example 19 The coordinates of three points O, A and B are (0, 0), (0, 4) and (6, 0), respectively. If a point P moves so that the area of D POA is always twice the area of D POB, then locus of P is (a) x – 3y = 0 (b) x + 3y = 0 (c) 3x + 4y = 0 (d) 3x – 4y = 0. Ans. (a) and (b) Solution:

Let P(x, y) be the moving point. Then the area 1 1 of D POA is | x◊4 – y◊0 | = | 2x |, and that of D POB is 2 2 | x◊0 – y◊6 | = | –3y |, because one vertex of the triangle is at the origin. Therefore, from the given condition, we have | 2 x | = 2| – 3y | or x = ± (3y) fi x – 3y = 0 or x + 3y = 0 Example 20

The points (k, 2 – 2k), (–k + 1, 2k) and

(– 4 – k, 6 – 2k) are collinear for (a) all values of k (b) k = –1 (c) k = 1/2 (d) no value of k. Ans. (b) and (c) Solution The given points are collinear if k 2 - 2k 1 k 2 - 2k 1 2k 1 = 0 fi - 2k + 1 4k - 2 0 = 0 fi -k +1 - 4 - k 6 - 2k 1 - 4 - 2k 4 0 fi fi fi

[R2 Æ R2 – R1, R3 Æ R3 – R1] 4(–2k + 1) – (– 4 – 2k) (4k – 2) = 0 (1 – 2k ) (4 – 8 – 4k) = 0 (1 – 2k) (k + 1) = 0 fi k = – 1 or k = 1/2

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ABC is an isosceles triangle. If the Example 21 coordinates of the base are B(1, 3) and C(–2, 7), the coordinates of vertex A can be (a) (1, 6) (b) (– 1/2, 5) (c) (5/6, 6) (d) (–7, 1/8). Ans. (c) and (d) Solution The vertex A(x, y) is equidistant from B and C. \ (x – 1)2 + ( y – 3)2 = (x + 2)2 + ( y – 7)2

(a) (– 3, – 2) (b) (3, – 2) (c) (3, 2) (d) (– 3, 2) Ans. (a), (b), (c), (d) Solution Centre of the rectangle is (3, 2), coordinates of the vertices in the new system of coordinates are (– 3, –2), (6 – 3, – 2), (6 – 3, 4 – 2) and (– 3, 4 – 2) i.e. (– 3, – 2), (3, – 2), (3, 2) and (– 3, 2).

fi fi

is 10 sq. units. Coordinates of O are (0, 0) and of A are (4, 0). It is given that B is not a right angle, then centroid of the triangle OAB is Ê 4 5ˆ Ê 8 5ˆ (a) Á , ˜ (b) Á , ˜ Ë 3 3¯ Ë 3 3¯

–2x – 6y + 10 = 4x – 14y + 53 6x – 8y + 43 = 0 (1) Any point satisfying this can be the vertex A, except the mid-point (– 1/2, 5) of BC. Since the coordinates given in (c) and (d) satisfy Equation (1), these are therefore the correct answers. Example 22

Coordinates of a point P are (a, b) where a

is a root of the equation x2 – x – 42 = 0 and b is an integral root of the equation x2 + ax + a2 – 37 = 0. The coordinates of P can be (a) (6, 4) (b) (– 7, 4) (c) (– 7, – 3) (d) (6, – 3) Ans. (b), (c) Solution a2 – a – 42 = 0 ⇒ a = – 7 or a = 6 Since b is a root of x2 + ax + a2 – 37 = 0 For a = –7, we have x2 – 7x + 49 – 37 = 0 ⇒ x2 – 7x + 12 = 0 ⇒ x = 4, – 3 i.e. b = 4 or – 3 So the coordinates of P can be (– 7, 4) or (– 7, – 3), for a = 6 we have x2 + 6x – 1 = 0 which does not give an integral value, so a ≠ 6. Example 23 If origin is shifted to the point (2, 3) without rotation of axes then the coordinates of the point P which divides the join of A(4, 8) and B(7, 14) in the ratio 1 : 2 or 2 : 1 with respect to the new system of coordinates can be (a) (2, 5) (b) (3, 7) (c) (4, 9) (d) (5, 11) Ans. (b), (c) Solution Coordinates of P can be Ê 8 + 7 16 + 14 ˆ Ê 14 + 4 28 + 8 ˆ , , ÁË ˜¯ or ÁË ˜ 3 3 3 3 ¯ = (5, 10) or (6, 12) in old system of coordinates. = (5 – 2, 10 – 3) or (6 – 2, 12 – 3) in new system of coordinates = (3, 7) or (4, 9). Example 24 If the origin is shifted to the centre of the rectangle with vertices (0, 0), (6, 0), (6, 4) and (0, 4) without rotation of axes then coordinates of a vertex of the rectangle with respect to the new system of coordinates is

Example 25

OAB is a right angled triangle whose area

Ê 4 5ˆ (c) Á , - ˜ Ë 3 3¯

Ê 8 5ˆ (d) Á , - ˜ Ë 3 3¯

Ans. (a), (b), (c), (d) Solution: If –AOB is a right angle then let the coordinates of B be (0, ± b) B (0, 5)

B (4, 5)

O (0, 0)

A (4, 0)

B (0, -5)

B (4, -5) Fig. 14.8

Ê 1ˆ Area of the triangle = Á ˜ (OA) (OB) = 10 Ë 2¯ fi 4b = 20 fi b = 5 and the coordinates of B are (0, ±5). So the coordinates of centroid of the triangle AOB are Ê 0 + 4 + 0 0 + 0 ± 5ˆ Ê 4 ±5 ˆ , ÁË ˜¯ = ÁË , ˜¯ 3 3 3 3 Next, if –OAB is a right angle, the coordinates of B are (4, ±5) and the coordinates of the centroid of the triangle OAB are Ê 0 + 4 + 4 0 + 0 ± 5ˆ Ê 8 ±5 ˆ , ÁË ˜ = ÁË , ˜¯ 3 3 ¯ 3 3

MATRIX-MATCH TYPE QUESTIONS Example 26

A(2, 8), B(1, 1), C(3, 3) are three given

points. M is the mid point of BC. AM is produced to meet x-axis at D, O is the origin.

IIT JEE eBooks: www.crackjee.xyz Cartesian System of Coordinates and Locus 14.9

Column 1

Column 2

(a) Coordinates of D (p) (1, 1) (b) Coordinates of the circumcentre of the triangle OMD (q) (2, 0) (c) Coordinates of the orthocenter Ê 7 13 ˆ of the triangle OMD (r) Á , ˜ Ë3 3¯ (d) Coordinates of the centroid Ê 7 5ˆ (s) Á , ˜ Ë 3 3¯

of the triangle CMD

ABCD is a rectangle in the clockwise

Example 27

direction. The coordinates of A are (1, 3) and of C are (5, 1) coordinates of vertices B and D satisfy y = 2x + c, the coordinates of the Column 1 Column 2 (a) middle point of BD (p) (2, 0) (b) middle point of AB (q) (3, 2) (c) middle point of BC (r) (5/2, 7/2) (d) the point D (s) (9/2, 5/2) p q r s Ans.

p

q

r

s

a

p

q

r

s

a

p

q

r

s

b

p

q

r

s

b

p

q

r

s

c

p

q

r

s

c

p

q

r

s

d

p

q

r

s

d

p

q

r

s

Ans.

Solution: Let the coordinates of D be (x, 0), then A, M, D are collinear. A(2, 8)

M (2, 2)

B( 1

,1

)

C (3, 3)

D(2, 0)

O

Fig. 14.9

Ê 3 + 1 3 + 1ˆ , Coordinates of M are Á ˜ = (2, 2) Ë 2 2 ¯ x 0 1 2 8 1 =0 2 2 1 fi x (8 – 2) + (4 – 16) = 0 fi x=2 So the coordinates of D are (2, 0) (Note. (OD)2 = 4, (MD)2 = 4, (OM)2 = 4 + 4 = 8 = (OD)2 + (MD)2 And, thus the triangle OMD is rightangel D being the right angle and OM, the hypotenuse circumcentre of the triangle OMD is (1, 1), the mid point of the hypotenuse and orthocenter of this triangle is D(2, 0). Centroid of the Ê 3 + 2 + 2 3 + 2 + 0ˆ Ê 7 5ˆ , = Á , ˜ triangle CMD is Á ˜ Ë Ë 3 3¯ 3 3 ¯

Solution: (a) Let M be the point of intersection of the diagonals. M bisects each of the diagonals. (Fig. 14.4) \ The coordinates of the middle point M of Ê 5 + 1 1 + 3ˆ , BD = Á ˜ = (3, 2) Ë 2 2 ¯ (d) Next, as M y = 2x + c fi 2 = 6 + c fi c = - 4. Let the coordinates of D y = 2x - 4 be (x, 2x - 4). Then from (AD)2 + (DC)2 = (AC)2 we get (x - 1)2 + (2x - 4 - 3)2 + (x - 5)2 + (2x - 4 - 1)2 = (5 - 1)2 + (1 - 3)2 fi x2 - 6x + 8 = 0 fi x = 2 or x = 4 when x = 2, y ABCD is in the clockwise direction. We get coordinates at D as (2, 0). (b) When x = 4, y = 4 so the coordinates of B are (4, 4) and the coordinates of the middle point of AB are Ê 4 + 1 4 + 3ˆ Ê 5 7 ˆ , ÁË ˜ = Á , ˜. 2 2 ¯ Ë 2 2¯ (c) and the coordinates of the middle point of BC are Ê 5 + 4 1 + 4ˆ , ÁË ˜ = (9/2, 5/2). 2 2 ¯

Fig. 14.10

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If P(x, y) is a point in the coordinate plane

Example 28 such that Column 1

Column 2 locus of P is (p) 3x2 + 3y2 - 2xy =0

(a) P is equidistant from (a + b, a - b) and (a - b), (a + b) (b) P is at a distance a + b, (q) from (a, b) (c) distance of P from x-axis (r) is twice its distance from y-axis (d) distance of P from the origin (s) is the mean of the its distances from the coordinate axes. p q r s Ans. q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

x=y

49 ˆ 2 Ê 39 87 ˆ 2 Ê – ˜ (ST)2 = Á 5 – ˜ + Á Ë Ë 5 5¯ 5¯ 2 2 576 Ê 24 ˆ Ê 48 ˆ = Á ˜ +Á ˜ = Ë 5¯ Ë 5¯ 5 2 ⇒ 5(ST) = 576 (PQ)2 = 29, (QR)2 = 205, (PR)2 = 320

116 . 25 261 . (QS)2 = [(3/5) (PQ)]2 = 25 369 . (QT)2 = [(3/5) (QR)]2 = 5 (PS)2 = [(2/5) (PQ)]2 =

P (3.7) 2

p

x2 + y2 - 2ax -2by - 2ab = 0

Coordinates of S are 16 Ê + 9 18 + 21ˆ Ê 39 ˆ , ÁË ˜ = Á 5, ˜ 5 5 ¯ Ë 5¯ Coordinates of T are Ê 33 + 16 69 + 18 ˆ Ê 49 87 ˆ , ÁË ˜ = ÁË , ˜¯ 5 5 ¯ 5 5

Solution (a) [x - (a + b)]2 + [y - (a - b)]2 = [x - (a - b)]2 + [y - (a + b)]2 fix=y (b) (x - a)2 + (y - b)2 = (a + b)2 fi x2 + y2 - 2ax - 2by - 2ab = 0 (c) y = 2x 1 (x + y) fi 4(x2 + y2) = (x + y)2 (d) x + y = 2 fi 3x2 + 3y2 - 2xy = 0 2

Example 29

2

P(3, 7), Q(8, 9) and R(11, 23) are three

vertices of the triangle PQR. S divides PQ in the ratio 2 : 3 and T divides QR in the ratio 3 : 2. Column 1 Column 2 (p) 576 (a) 5(ST)2 (q) 464 (b) (PR)2 + (QR)2 + 51 2 (r) 2 (PR)2 – 64 (c) 100 (PS) 2 2 (d) (PQ) + 2(QR) + 25 (s) 50 (QS)2 – 2(PQ)2 (t) 5 (QT)2 + (QR)2 + 2 p q s t r Ans. a

p

q

r

s

t

b

p

q

r

s

t

c

p

q

r

s

t

d

p

q

r

s

t

3

S

a

y - 2x = 0

Solution

Q (8.9)

T

R (11.23)

3: 2 Fig. 14.11

So that (PR)2 + (QR)2 + 51 = 320 + 205 + 51 = 576 2 100 (PS) = 4 ¥ 116 = 464 and (PQ)2 + 2(QR)2 + 25 = 29 + 410 + 25 = 464 Next, 2(PR)2 – 64 = 2 ¥ 320 – 64 = 576 50 (QS)2 – 2(PQ)2 = 2 ¥ 261 – 2 ¥ 29 = 464 and 5 (QT)2 + (QR)2 + 2 = 369 + 205 + 2 = 576 Combining these results we get the required answer.

(r) (s) (t)

Example 30 A is a point on the +ve x-axis and B is a point on the +ve y-axis such that the product of their distances from the origin O is 36 units. P divides OA on it the ratio 1 : 2 and Q divides it in the ratio 2 : 1. R and S divide OB in the ratio 1 : 2 and 2 : 1, respectively. Column 1 Column 2 (a) Area of the triangle OQR (p) 2 ¥ Area of the triangle OPR (b) Area of the triangle OPS (q) 8 sq. units

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(c) Area of the OQS (r) 4 sq. units (d) Area of the triangle PQR (s) Area of the triangle PRS p q r s Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d p q r s Solution: Set the coordinates of A be (a, 0) and of B be (0, b) such that ab = 36. Coordinates of P, Q, R, S are respectively

R ÊR ˆ ÁË [cos a + cos b + cos g ], [sin a + sin b + sin g ]˜¯ 3 3 Using Statement - 2 R 1¥ h + 2 ¥ 0 (cos a + cos b + cos g ) = 3 3 R 1¥ k + 2 ¥ 0 (sin a + sin b + sin g ) = 3 3 fi h = R (cos a + cos b + cos g) k = R (sin a + sin b + sin g) k sin a + sin b + sin g = fi h cos a + cos b + cos g Êg ˆ Ê bˆ Êaˆ 4 cos Á ˜ sin Á ˜ cos Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ = Ê bˆ Êaˆ Êg ˆ 1 + 4 sin Á ˜ sin Á ˜ sin Á ˜ Ë 2¯ Ë 2¯ Ë 2¯

Ê 2b ˆ Ê a ˆ Ê 2a ˆ Ê b ˆ ÁË , 0˜¯ , ÁË , 0˜¯ , ÁË 0, ˜¯ and ÁË 0, ˜¯ 3 3 3 3 Area of the triangle OQR =

1 2a b ab 36 ¥ ¥ = = =4 2 3 3 9 9

1 a 2b =4 Area of the triangle OPS = ¥ ¥ 2 3 3 1 2a 2b ¥ =8 Area of the OQS = ¥ 2 3 3 1 1 a b Area of the PQR = ¥ PQ ¥ OR = ¥ ¥ = 2 2 2 3 3 1 a b Area of the triangle OPR = ¥ ¥ = 2 2 3 3 Area of the triangle PRS =

1 1 b a SR ¥ OP = ¥ ¥ = 2 2 2 3 3

ASSERTION-REASON TYPE QUESTIONS Example 31

Statement-1: The circumcentre of a

triangle with vertices (a, a tan a), (b, b tan b) and C(c, tan p and a + b + g = p. If g) lies at the origin, where 0 < a, b, 2 the coordinates of the orthocenter are (h, k), then a b g a b g cos cos .h - 4 sin sin sin .k = k 2 2 2 2 2 2 Statement-2: The centroid G of a triangle divides the join of its orthocenter H and circumcentre O in the ratio 2 : 1. Ans. (a) Solution: From geometry, statement 2 is true. In Statement-1 Coordinates of the circumcentre O are (0, 0) If R is the radius of the circumcircle. Then OA = OB = OC = R (OA)2 = R2 fi a2 + a2 tan2a = R2 fi a = R cos a Coordinates of the centroid G are 4 cos

[As a + b + g = p, using conditional identities from trigonometry] fi

Êaˆ Ê bˆ Êg ˆ k = 4 cos Á ˜ sin Á ˜ cos Á ˜ . h Ë 2¯ Ë 2¯ Ë 2¯ Êaˆ Ê bˆ Êg ˆ - 4 sin Á ˜ sin Á ˜ sin Á ˜ k Ë 2¯ Ë 2¯ Ë 2¯

Example 32

Statement-1: If (x, y) and (X, Y) be

the coordinates of the same point reffered to two sets of rectangular axes with same origin and ax + by becomes AX + BY then a2 + b2 = A2 + B2 Statement-2: If axes of coordinates are rotated through an angle q, without changing the origin, then the coordinates (X, Y) of a point with respect to the new axes and coordinates (x, y) with respect to old axes are related as x + iy = (X + iY) eiq = (X + iY) (cos q + isinq) Ans. (a) Solution: In statement-2 we have. fi X = x cosq + y sinq, Y = y cosq – x sinq (See text) fi x = X cosq – Y sinq, y = X sinq + Y cos q fi x + iy = (X cosq – Y sinq) + i (X sinq + Y cosq) = (cosq + i sinq) X + iY (cosq + i sinq) = (X + iY) eiq = (cosq + i sinq) (X + iY) = (X + iY) eiq So the statement-2 is true In statement-1: If the axes are rotated through an angle q, then using statement-2 ax + by = a (X cosq – Y sinq) + b (X sinq + Y cosq) = (a cosq + b sinq) X + (b cos q – a sinq) Y = AX + BY fi A2 + B2 = (a cosq + b sinq)2+ (b cosq – a sinq)2 = a2 + b2

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Example 33

Statement-1: The quadrilateral whose

vertices (in order) are A(1, 0), B(0, 3), C(– 2, 0) and D(0, 2) can not be convex. Statement-2: A quadrilateral ABCD (in order) is convex if and only if when any diagonal is taken then the remaining vertices must be on the opposite sides of it. Ans. (a) Solution Statement-2 is correct because if one angle say D of the quadrilateral ABCD which is not covex is more than 180° then B and D lie on the same side of the diagonal AC of the quadrilateral. Statement-1 is true, using statement-2, the quadrilateral ABCD is not convex.

mid-points of the sides BC, CA and AB respectively then area of ΔAEF. =

1 1 1 ¥ BC × l 2 2 2

(l being the altitude from A)

1 1 1 ¥ BC × l = area of the triangle ABC. 4 2 4 similarly for triangles BDF and CDE. 1 area of ΔABC So that area of ΔDEF = 4 =

2 2 1 1 1 19 = ¥ 6 –1 1 = . 4 2 8 7 3 1 and thus statement-1 is also true. A

F

E

Fig. 14.12

Example 34

y2) and C(x3, y3) are collinear if x1 + x2 + x3 = y1 + y2 + y3. Statement-2: The points A(x, – x), B(7, – 5) C(– 5, 3) are collinear if x = 1. Solution

Statement-1 x1 y1 collinearity is x2 y2 x3 y3

is False as the condition for 1 1 = 0, using in statement-2 we 1

x –x 1 have 7 –5 1 = 0 –5 3 1 4x – 4 = 0 ⇒ x = 1 and statement-2 is True.

Example 35

D

C

Fig. 14.13

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 36 to 38

Ans. (d)



B

Statement-1: Three points A(x1, y1) B(x2,

Statement-1: A(2, 2), B(6, – 1) and

C(7, 3) are the vertices of a triangle ABC. D, E, F are the mid-points of the sides BC, CA and AB respectively. Area of the DDEF is equal to 19/8 square units. Statement-2: Lines joining the mid-points of the sides of a triangle divide the triangle into four triangles of equal areas. Ans. (a) Solution Statement-2 is true. Because if ABC is a given triangle and D, E, F are the

A1, A2, A3, ... An are n points in a plane whose coordinates are (x1, y1), (x2, y2), (x3, y3) ..., (xn, yn) respectively. Example 36

A1 A2 is bisected at the point G1, G1 A3 is

divided in the ratio 1 : 2 at G2, G2 A4 is divided in the ratio 1 : 3 at G3, G3A5 is divided in the ratio 1:4 at G4, and so on until all n point G so obtained are (a) (n (x1 + x2 + ... + xn ), n(y1 + y2 + ... yn)) Ê x + x2 + ... + xn y1 + y2 + ... + yn ˆ , (b) Á 1 ˜¯ Ë n n Ê x + 2 x2 + 3 x3 + ... + nxn y1 + 2 y2 + 3 y3 + ... + nyn ˆ , (c) Á 1 ˜¯ Ë n n (d) none of these Example 37 If x1 = a, y1 = b; x1, x2, ... xn and y1, y2, ... yn form an ascending arithmetic progression with common difference 2 and 4 respectively, then the coordinates of G in Ex. 36 are (a) (a + n - 1, b + 2(n - 1)) (b) (a + 2n - 2, b + n - 1)

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(c) (a + n - 1, b + n + 1) (d) (a + n + 1, b + n + 2) Example 38 If P1 is the centroid of the triangle A1, A2, A3, P2 is the centroid of the triangle A2 A3 A4, P3 is that of the triangle A3 A4 A5 and so on Pn-2 is the centroid of the triangle An-2 An-1 An. Then the centre of Mean position of the points P1, P2, ... Pn-2 is Ê x + 2 x2 + 3 x3 + ... + nxn y1 + 2 y2 + 3 y3 + ... + nyn ˆ , (a) Á 1 ˜¯ Ë n-2 n-2 Ê x1 + 2 x2 + 3 x3 + ... + 3 xn - 2 + 2 xn - 1 + xn ˆ , ˜ Á n-2 ˜ (b) Á y1 + 2 y2 + 3 y3 + ... + 3 yn - 2 + 2 yn - 1 + yn ˜ Á Á ˜ Ë ¯ n-2 Ê 3( x1 + x2 + ... + xn ) 3( y1 + y2 + ... + yn ) ˆ , (c) Á ˜¯ Ë n-2 n-2 (d) at the centre of mean position of the points A1, A2, ... An.

= (a + n - 1, b + 2(n - 1)) In Ex. 38. Ê x + x2 + x3 y1 + y2 + y3 ˆ , P1 Á 1 ˜¯ Ë 3 3 Ê x + x3 + x4 y2 + y3 + y4 ˆ , P2 Á 2 ˜¯ Ë 3 3 Ê x + x4 + x5 y3 + y4 + y5 ˆ , P3 Á 3 ˜¯ Ë 3 3 . . . . + xn -1 + xn yn - 2 + yn -1 + yn ˆ Êx , Pn-2 Á n - 2 ˜¯ Ë 3 3 So the coordinates of the centre of mean position of P1, P2 ... Pn-2 is ( x , y ) where x =

x1 + 2 x2 + 3 x3 + ... + 3 xn - 2 + 2 xn -1 + xn n-2

y =

y1 + 2 y2 + 3 y3 + ... + 3 yn - 2 + 2 yn -1 + yn n-2

Ans. 36. (b), 37. (a), 38. (b) Solution The coordinates of G1 are Ê x1 + x2 y1 + y2 ˆ , ÁË ˜ 2 2 ¯ Now, G2 divides G1A3 in the ratio 1 : 2. Therefore, the coordinates of G2 are Ê 1 Ê 2( x1 + x2 ) ˆ 1 Ê 2( y + y2 ) ˆˆ + x3 ˜ , Á 1 + y3 ˜ ˜ ÁË ÁË ¯ 3Ë ¯¯ 3 2 2 or

Ê x1 + x2 + x3 y1 + y2 + y3 ˆ , ÁË ˜¯ 3 3

Again, G3 divided G2A4 in the ratio 1 : 3. Therefore, the coordinates of G3 are Ê 1 Ê 3( x1 + x2 + x3 ) ˆ 1 Ê 3( y + y2 + y3 ) ˆˆ + x4 ˜ , Á 1 + y4 ˜ ˜ ÁË ÁË ¯ 4Ë ¯¯ 4 3 3 or

Ê x1 + x2 + x3 + x4 y1 + y2 + y3 + y4 ˆ , ÁË ˜¯ 4 4

Proceeding in this manner, we can show that the coordinates G so obtained will be 1 Ê1 ˆ ÁË ( x1 + x2 + ... + xn ), ( y1 + y2 + ... + yn )˜¯ n n Now if x1 = a, y1 = b (given in Ex. 37) then x2 = a + 2, x3 = a + 4 ... xn = a + (n - 1)2 y2 = b + 4, y3 = b + 8, ... yn = b + (n - 1)4 and the coordinates of G are 1 n Ê1 n ˆ ÁË ¥ [ a + a + (n - 1)2 ] , ¥ [ b + b + (n - 1)4 ]˜¯ n 2 n 2

Paragraph for Question Nos. 39 to 42 Given two points A(– 2, 0) and B (0, 4), M is a point with coordinates (x, x), x ≥ 0. P divides the joint A and B in the ratio 2 : 1. C and D are the mid-points of BM and AM respectively. Example 39 Area of the DAMB is minimum, if the coordinates of M are (a) (1, 1) (b) (0, 0) (c) (2, 2) (d) (3, 3) Example 40 Ratio of the areas of the triangles APM and BPM is (a) 2 : 1 (b) 1 : 2 (c) 2 : 3 (d) 1 : 3 Example 41

Perimeter of the quadrilateral ABCD is

(a) 2 5

(b) 3 + 3 5

(c) 2 + 5

(d) none of these

Example 42 Area of the quadrilateral ABCD in units is (a) 1 (b) 2 (c) 3 (d) 4 Ans. 39. (b), 40. (a), 41. (b), 42. (c)

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x x 1 1 –2 0 1 Solution Area of the ΔAMB = 2 0 4 1 1 ( –4 x + 2 x – 8) = | – (x + 4) | 2 which is minimum for x = 0 and thus the coordinates of M are (0, 0). In example 40, as P divides AB in the ratio 2:1. The base of the triangles APM and BPM are in the ratio 2:1 and the length of the perpendicular from the vertex M on the base is same. So, the ratio of the areas of the triangles APM and BPM is also 2 : 1. =

B (0.4)

P C A

(0.0)

M

D

(–2.0)

Example 45

yˆ Ê (–2b, 0) and Á c, - ˜ is denoted by D, then rD is equal to Ë z¯ (a) 6b (b) 3b (c) a + c (d) a – c Ans. 43. (d) 44. (b) 45. (b) Solution: (a, x), (b, y), (c, z) are collinear if a b c

Ê Ê xˆ Ê Á a - 2b + c ÁË y ˜¯ + 0 - ÁË Á , 3 3 Á Á Ë

In example 41, ABCD is a trapezium with AD = 1, BC = 2 1 1 22 + 42 = 5 AB = 2 2

So, the required perimeter is 1+2+

5+2 5 =3+ 3 5

In example 42, Area of the trapezium ABCD = Area of the ΔAMB – Area of the ΔCMD. 1 (8 – 2) = 3 = 2 Paragraph for Questions Nos. 43 to 45 a, b, c are in A.P with common difference d (non-zero) x, y, z are in G.P with common ratio r. Example 43 are collinear if (a) x2 = y Example 44

The points with vertices (a, x), (b, y), (c, z) (b) y2 = z

Centroid of the triangle with vertices

Ê xˆ ÁË a, y ˜¯ (–2b, 0) and (a) (a, 0)

(c) z2 = x (d) x = y = z

Ê ÁË c, -

yˆ ˜ has the coordinates z¯

(b) (0, 0)

(c) (c, 0) (d) (0, b)

x 1 y 1 =0 z 1

fi x(b – c) + y(c – a) + z(a – b) = 0 As d is the common difference of the A.P. So we get. x (–d) + y (2d) + z (–d) = 0 fi d (x + z – 2y) = 0 fi x + z = 2y as d π 0 Also as x, y, z are in G.P y2 = xz fi (x – z)2 = (x + z)2 – 4xz = 0 fi x=z=y Next, centroid of the triangle with the given vertices is

Fig. 14.14

DC =

Ê xˆ If area of the triangle with vertices Á a, ˜ , Ë y¯

yˆ ˆ ˜ z¯˜ ˜ ˜ ˜ ¯

(0, 0) Because a + c = 2b and xz = y2 x 1 a y 1 1 È Ê x yˆ y x˘ -2b 0 1 = Í2b Á + ˜ + a + c ˙ Now D = 2 2Î Ë y z¯ z y˚ y 1 c z =

1È 2 1˘ 3b 2b ¥ + 2b ¥ ˙ = Í 2Î r r˚ r

INTEGER-ANSWER TYPE QUESTIONS Example 46 Let the sides of a triangle ABC are all integers with A as the origin. If (2, – 1) and (3, 6) are points on the line AB and AC respectively (lines AB and AC may be extended to contain these points), and lengths of any two sides are primes that differ by 50. If a is least possible length of the third side and S is the least possible perimeter of the triangle then aS is equal to 495 k2 where k = Ans. 4

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Solution:

Example 49

(2, – 1)

C(a3 + 1, a3 – 1) are given points D(11, 9) is the mid point of AB and E (41, 39) is the mid point of BC. If F is the mid point of AC then (BF)2 is equal to 8k2 where k = Ans. 648 (3, 6)

Fig. 14.15

Since (AB)2 + (AC)2 = 50 = (BC)2 the lines AB and AC are at right angles. It is given that all sides are integers and since the triangle is right angled, they form a pythagorean triplet, so one leg must be even. As two sides are primes, both cannot be the legs of the right angled triangle. So, let AC = p, a prime, BC = p + 50 and AB = a (even) then a2 = (p + 50)2 – p2 = 100 (p + 25) fi a = 10 p + 25 . Since a is least possible integer and p is a prime we get a = 60 when p = 11. So smallest value of the third side is 60. S = 60 + 11 + 61 = 132 aS = 60 ¥ 132 = 7920 = 495 ¥ 42 Example 47 If O is the origin and the coordinates of A and B are (51, 65) and (75, 81) respectively. Then OA ¥ OB cos AOB is equal to 1010k, k = Ans. 9 Solution

(OA)2 = (51)2 + (65)2, (OB)2 = (75)2 + (81)2 (AB)2 = (75 - 51)2 + (81 - 65)2 and from triangle OAB (AB)2 = (OA)2 + (OB)2 - 2OA ¥ OB cos AOB . So

Solution We have

fi OA ¥ OB cos AOB = 3825 + 5265 = 9090 = 1010k fik=9 If O is the origin and An is the point with

coordinates (n, n + 1) then (OA1)2 + (OA2)2 + … + (OA7)2 is equal to 114k + 1 where k = Ans. 3 Solution (OA1)2 + (OA2)2 + … + (OA7)2 = 12 + 22 + 22 + 32 + … + 72 + 82 = 2(12 + 22 + … + 72) – 1 + 82 2 ¥ 7 ¥ 8 ¥ 15 – 1 + 82 = 343 = 114 ¥ 3 + 1 = 6 n ( n + 1) ( 2n + 1) ˆ Ê 2 2 2 Ë using 1 + 2 + ... + n = ¯. 6

a –1 + a 2 –1 a + 1 + a2 + 1 = 11, =9 2 2

⇒ a + a2 – 20 = 0 ⇒ a = – 5 or 4 a 2 – 1 + a3 – 1 a 2 + 1 + a3 + 1 = 41, = 39 2 2 ⇒ a2 + a3 = 80 which holds for a = 4. So, the given points are A(5, 3), B(17, 15), C(65, 63) and Ê 65 + 5 63 + 3 ˆ , coordinates of F are Á ˜ = (35, 33) and Ë 2 2 ¯ and

(BF)2 = (35 – 17)2 + (33 – 15)2 = (18)2 + (18)2 = 648 = 8(9)2 Example 50

When the axes of coordinates are rotated

through an angle p/4 without shifting the origin, the equation 2x2 + 2y2 + 3xy = 4 is transformed to the equation 7x2 + y2 = k where the value of k is Ans. 8 Solution: Replace x by x cosq – y sinq and y by x sinq + y cosq, where q = p/4 So 2x2 + 2y2 + 3xy = 4 becomes 2

2

ÈÊ x + y ˆ ˘ ÈÊ x - y ˆ ˘ Èx + y x - y˘ 2 ÍÁ ¥ ˜¯ ˙ + 3 Í ˜¯ ˙ + 2 ÍÁË ˙ =4 Ë 2 ˚ 2 ˚ 2 ˚ Î 2 Î Î fi 2(2x2 + 2y2) + 3 (x2 – y2) = 8 fi 7x2 + y2 = 8 And the required value of k is 8.

2OA ¥ OB cos AOB = 2(75 ¥ 51 + 81 ¥ 65)

Example 48

A(a + 1, a – 1), B(a2 + 1, a2 – 1) and

EXERCISE LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. If A(1, a), B(a, a2), C(a2, a2) are the vertices of a triangle which are equidistant from the origin, then the centroid of the triangle ABC is at the point (a) (1, –1) (b) (–1, 1) (c) (1, 1) (d) (1/3, 1/3) 2. If A(2, 0) and B(0, 2) are given points and P is a point such that PA : PB = 2 : 3 then the locus of P passes through the point (a, a) for

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(a) ΔA + ΔB = Δ + ΔC (c) ΔA + ΔB = 2ΔC

(a) a > 2 (b) a = 1 (c) no real value of a a 3. O(0, 0), P(– 2, – 2) and Q (1, – 2) are the vertices of a triangle, R is a point on PQ such that PR : RQ = 2 2 : 5, , then OR is (a) a median of the triangle (b) an altitude of the triangle (c) bisector of the angle at O (d) none of then 4. If three vertices of a rectangle are (0, 0), (a, 0) and (0, b), length of each diagonal is 5 and the perimeter is 14, then the area of the rectangle is (a) 35 (c) 12

(b) 24 (d) 7

5. If the line joining the points A (a2, 1) and B(b2, 1) is divided in the ratio b : a at the point P whose x-coordinate is 7, their (a) ab = 7 (c) a3 + b3 = 14

(b) a2 + b2 = 7 (d) a2 – ab + b2 = 7

6. a is a root of the equation x2 – 5x + 6 = 0 and b is a root of the equation x2 – x – 30 = 0, then coordinates (a, b) of the point P farthest from the origin are (a) (2, 6) (c) (6, – 5)

(b) (2, 3) (d) (3, 6)

7. P (sin θ, cos θ ) and Q (cos θ, sin θ) are two points whose mid point is at the origin. R (sin 2θ, cos θ ) is a point on the plane whose distance from the origin is (a) 2/3

(b) 3/2 1 (d) (c) 3 / 2 2 8. Locus of the point P (2t2 + 2, 4t + 3), where t is a variable is 2

(a) y = 8x (c) (x – 3)2 = 8(y – 2)

2

(b) (y – 3) = 8(x – 2) (d) y – 3 = 2(x – 2)

9. If the coordinates of An are (n, n2) and the ordinate of the centre of mean position of the points A1, A2, … An is 46, then n is equal to (a) 5 (c) 7

(b) 6 (d) 11

10. Area of the triangle with vertices A(3, 7) B(– 5, 2) and C(2, 5) is denoted by D. If ΔA, ΔB, ΔC denote the areas of the triangles with vertices OBC, AOC and ABO respectively, O being the origin, then

(b) ΔA + ΔB = ΔC – Δ (d) ΔA + ΔB + ΔC = 2Δ

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 11. The points A(–4, –1), B(–2, –4), C(4, 0) and D(2, 3) are the vertices of a (a) parallelogram (b) rectangle (c) rhombus (d) square. 12. The area of a triangle is 5. Two of its vertices are y = x + 3. The coordinates of the third vertex can be Ê 3 3ˆ (a) Á - , ˜ Ë 2 2¯

Ê 3 3ˆ (b) Á , - ˜ Ë 2 2¯

Ê 7 13 ˆ (c) Á , ˜ Ë2 2¯

Ê 1 11ˆ (d) Á - , ˜ . Ë 4 4¯

13. Two points (a, 3) and (5, b) are the opposite vertices of a rectangle. If the coordinates (x, y) of the other two vertices satisfy the relation y = 2x + c where c2 + 2a – b = 0, then the value c can be (a) 2 2 + 1

(b) 2 2 – 1

(c) 1 – 2 2

(d) –1 – 2 2

14. OPQR is a square, M and N are the middle points of the sides PQ and QR respectively. (a) ratio of the areas of quadrilateral OMNR and the square is 5:8 (b) ratio of the areas of the DMQN and OMN is 1:3 (c) ratio of the areas of the DOPM and ORN is 1:1 (d) ratio of the areas of quadrilateral OMQN and the square is 1:2 15. If a, b, c form a G.P. with common ratio r .P and Q are two points whose coordinates (x, y) satisfy the relations ax + by + c = 0 and x2 – y2 – 4, then (a) sum of the ordinates of P and Q is (b) sum of the ordinates of P and Q is

2r 3 1 – r2 2r 3 r2 – 1

(c) product of the ordinates of P and Q is

(d) product of the ordinates of P and Q is

r4 – 4 r2 – 1 4 – r4 r2 – 1

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16. If α, β, γ are the real roots of the equation x3 – 3px2 + 3qx – 1 = 0, (p, q) = (± 1, ± 1) then the centroid of the triangle with vertices (α 2, 1/α2), (β 2, 1/β2) and (γ 2, 1/γ 2) can be (a) (1, 1) (b) (5, 5) (c) (1, 5) (d) (5, 1) 17. a, b, c are in A.P. and the points A(a, 1), B(b, 2) and C(c, 3) are such that (OA)2, (OB)2 and (OC)2 are also in A.P. 0 being the origin, then (b) ac = b2 + 1 (a) a2 + c2 = 2b2 – 2 (d) a + b + c = 3b (c) (a + c)2 = 4b2 18. If the coordinates (x, y) of the centroid of the triangle with vertices (0, 0), (cos t, sin t) and (sin t, – cos t) satisfy y = 2x or x = 2y then t can be (b) tan–1 (1/3) (a) tan–1 (–1/3) –1 (d) tan–1 (3) (c) tan (–3) 19. If area of the triangle with vertices (m2, 2m), ( mm ¢, m + m ¢ ) and m ¢ 2 , 2m ¢ is 32, then

(

)

(a) m – m¢ = 4

(b) m¢ – m = 4

(c) m + m¢ = 4

(d) m + m¢ = 2

20. P(x1, y1) and Q(x2, y2) are points in the plane such that PQ subtends a right angle at the origin O, then (a) P(1, 3), Q(– 3, 1) (b) P(3, 1), Q(1, – 3) (c) P(2, 5), Q(5, – 2) (d) P(1, 1), Q(– 1, 1)

MATRIX-MATCH TYPE QUESTIONS 21. If A(2a, 4a) and B(2a, 6a) are two vertices of a triangle ABC and the vertex C is given by Column 1 Column 2 (a) (4a, 5a)

(p) equilateral

(b) ((2 + 3 )a, 5a)

(q) right angled

(c) (6a, 4a) (r) obtuse angled (d) (a, 3a) (s) isosceles 22. Coordinates of A are (1, 1) and of An are (n, 2n + 1), O is the origin Column 1 Column 2 2 (p) 5n2 – 2n + 1 (a) (OAn) (q) 5n2 – 10n + 5 (b) (AAn)2 (r) 5n2 + 4n + 1 (c) (A1An)2 (s) 5 (d) (An–1 An )2, n > 1

ASSERTION-REASON TYPE QUESTIONS 23. Statement-1: ABCD is a square with vertices (0, 0), (1, 0), (1, 1) and (0, 1). P, Q, R and S are the points which divide AB, BC, CD, DA respectively in the

ratio 2 : 1. If the origin is shifted to the centre of the square ABCD without rotation of axes, area of the square PQRS in the new system of coordinates is 5/9 square units. Statement-2: If the origin is shifted to the point P(a, b) without rotation of axes then the distance between any two given points remains unchanged in the new system of coordinates. 24. Statement-1: Vertices of a triangle are A(a, b), B(a + 1, b – 2), C(a – 2, b + 1). If the origin is shifted to the centroid of the triangle without rotation of axes, coordinates of A in the new system of coordinates are (1/3, 1/3). Statement-2: Centroid of an equilateral triangle is equidistant from its vertices. 25. Statement-1: a, b are the roots of the equation x2 – 7x + 2 = 0, r is the arithmetic mean of a, b. The locus of the point P which is equidistant from the points (a, b) and (a + r, b + r) is 2x + 2y – 2l = 0. Statement-2: Locus of a point which moves such constant always passes through the mid-point of the

COMPREHENSION-TYPE QUESTIONS Paragraph for Question No 26 to 28 Vertices of a triangle ABC are such that the x-coordinate is a root of the equation x2 + 2x – 3 = 0 and y-coordinate is a root of the equation and y 2 + y 2 – 2 = 0; x π y. A is farthest from the origin and C is nearest to the origin. 26. Coordinates of the vertex A are (a) (2, –3) (b) (–3, –2) (c) (–3, 1) (d) (3, –2) 27. Circumcentre of the triangle ABC has the coordinates -1ˆ Ê (a) (–1, –2) (b) Á -3, ˜ Ë 2¯ -1ˆ Ê (c) Á -1, ˜ Ë 2¯

(d) none of these

28. Distance of the centroid of the triangle ABC from the vertex B is (a)

2 13 3

(b)

5 3

(c)

73 (d) 3

34 3

Paragraph for Question Nos. 29 and 30 a and b are real numbers between 0 and 1 A(a, 1), B(1, b) and C(0, 0) are the vertices of triangle.

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29. If the triangle ABC is equilateral, its area is equal to (a) (c)

7 3 + 12 4 3 4

(b)

7 3 – 12 4

(d) none of these

30. If the triangle ABC is isosceles with AC = BC and 5(AB)2 = 2(AC)2 then (a) a = 1/2 = b

(b) a = 1/4

(c) b = 1

(d) b = 1/4

INTEGER-ANSWER TYPE QUESTIONS 31. (p, q) is a point such that p and q are integers, p ≥ 50 and the equation px2 + qx + 1 = 0 has real roots. Square of the least distance of the point from the origin is 109 k2 where k = 32. The coordinates of a point An is (n, n n ) where n is 12

a natural number. If O is the origin, then

 (OAi )2

i =1

= 6730 + k, where k = 33. The point P(a, b) is such that b – 25a = 4 and the arithmetic mean of a an b is 28. Q(x, y) is the point such that x and y are two geometric means between a and b, if O is the origin then OP2 + OQ2 is equal to 205k2 when k = 34. When origin is shifted to the point (4, 5) without changing the direction of the coordinate axes, the equation x2 + y2 – 10y + 5 = 0 is transformed to the equation x2 + y2 = a2. Value of |a| is 35. Number of integral values of m for which the x-coordinate of the point of intersection of 2x + 3y = 40 and y = mx is also an integer are. LEVEL 2

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. If the points (a, 1), (1, b) and (a – 1, b – 1) are collinear, a, b are respectively the arithmetic and geometric means of a and b then 4a – b2 is equal to (a) – 1 (b) 0 (c) 3 (d) 2 2. If P (1, 0), Q (– 1, 0) and R (2, 0) are three given points. The point S SQ2 + SR2 = 2 2SP . The locus of S meets PQ at the point (a) (0, 0) (b) (2/3, 0) (c) (– 3/2, 0) (d) (0, – 2/3)

3. A1, A2, … An are points whose coordinates (x, y) satisfy y = x and lie in the positive quadrant such that O An = n O An–1, O being the origin. If OA1 = 1 and the coordinates of An are (2520 2 , 2520 2 ) then n = (a) 5 (b) 6 (c) 7 (d) 8 4. The coordinates of the point An are (na, a n). If a denotes the arithmetic mean of x coordinates of the points A1, A2 … An and b denotes the geometric mean of the ordinates of these points then if a is a variable, locus of the point P(a, b) is (a) xn = nn y2 Ê n + 1ˆ (b) xn + 1 = Á Ë 2 ˜¯

n +1

y2

(c) xn = nn y (d) none of these 5. Coordinates of the point An are (n2, 2n) and of Bn are (n2, – 2n). Area of the quadrilateral with vertices A1 B1 Bn An is (a) (n + 1)2 (b) (n – 1)2 2 (c) 2(n + 1) (d) 2(n – 1)(n + 1)2

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 6. ABC is an equilateral triangle. If the coordinates of two of its vertices are (1, 3) and (– 2, 7), the coordinates of the third vertex can be Ê 1 3 3ˆ (a) Á - + 2 3, 5 + 2 ˜¯ Ë 2 Ê 1 3 3ˆ (b) Á - + 2 3, 5 2 ˜¯ Ë 2 Ê1 3 3ˆ (c) Á - 2 3, - 5 + 2 ˜¯ Ë2 Ê1 3 3ˆ (d) Á + 2 3, - 5 2 2 ˜¯ Ë 7. The coordinates of a point at a distance 5 from the point (3, 4) and having the same abcissa and ordinate are (a) (2, 2) (b) (– 2, – 2) (c) (5, 5)

(d) (– 5, – 5)

8. If two vertices of a triangle of area a2 are (0, 0) and (a, 0) then the coordinates of the third vertex can be (a) (a/2, 2a) (b) (3a/2, – 2a) (c) (a/2, a/2) (d) (a/2, – 2a)

IIT JEE eBooks: www.crackjee.xyz Cartesian System of Coordinates and Locus 14.19

9. If a > 0 and P(– a, 0), Q(a, 0) and R (1, 1) are three points such that |(PR)2 – (QR)2| = 12 then (b) (QR)2 = 5 (a) (PR)2 = 17 (d) (QR)2 = 17 (c) (PR)2 = 5 10. A(0, b), B(0, 0) and C(a, 0) are the vertices of a triangle ABC. D, E, F are the mid-points of the sides BC, CA and AB respectively. If a2 + b2 = 20, then (a) (AD)2 = 9 (b) (BE)2 = 4 (c) (AD)2 + (CF)2 = 25 (d) (AD)2 + (CF)2 = (BE)2

Statement-2: If r is a root of the equation. x2 + x – 6 = 0 and s is a root of the equation. x2 + x – 12 = 0, r π s, the centroid of the triangle with vertices (a, b), (ar, bs) and (ar2, bs2) lies at the point (7a/3, 13b/3).

COMPREHESION-TYPE QUESTIONS Paragraph for Questions Nos. 16 to 18 Vertices of a parallelogram are A(0, 0), B(– 2/3, 7/3), C(a, b) and D(5/3, – 4/3) 16. Length of the larger diagonal of ABCD is (a)

MATRIX-MATCH TYPE QUESTIONS 11. Points with vertices Column 1 Column 2 (a) (0, 4), (2, 10), (p) form a right angled (7, 25) triangle (b) (0, 4), (4, 0), (q) form an isosceles (4, 4) triangle (c) (4, 0), (– 2, 2), (r) are collinear (– 17, 7) (d) (– 2, 2), (2, 2), (s) form a triangle of area (2, – 2) 8 sq. units 12. Two points A (– 1, 0) and B(1, 0) are given, locus of P Column 1 Column 2 (a) PA = 3PB (p) 4x2 + 4y2 + 10x + 4 = 0 (b) 3PA = PB (q) x2 + y2 – 11 = 0 (r) x = 6 (c) (PA)2 + (PB)2 = 24 (d) (PA)2 – (PB)2 = 24(s) 4x2 + 4y2 – 10x + 4 = 0

ASSERTION-REASON TYPE QUESTIONS 13. Statement-1: Points P(– sin (b – a), – cos b), Q(cos (b – a), sin b) and R(cos (b – a + q), sin (b – q)), where b = p/4 + a/2 are non-collinear. Statement-2: Three given points are non-collinear if they form a triangle of non-zero area. 14. Statement-1: Let a, b be the roots of the equation x2 – ax + b = 0. If the coordinates of An are (a n/2, bn/2) then (OAn + 1)2 – a(OAn)2 + b(OAn–1)2 is equal to zero. Statement-2: If a, b are the roots of the equation x2 – ax + b = 0, then an + b n = an – nb. 15. Statement-1: Centroid of the triangle with vertices 3, - 3 lies at the (1, tan a), (– tan a, 1) and point (1/3, 1/3) if a = p/3.

(

)

2

(b)

58 / 3

(c) 41 / 3 (d) 170 / 3 17. If P and Q are two points on the diagonal DB such that DP = PQ = QB then area of the DCPQ in square units is (a) 1/3 (b) 1/2 (c) 1/4 (d) 1 18. Sum of the areas of the triangles ABP, APD, BQC and CQD is (a) 3 (c) 1

(b) 3/2 (d) 1/2

Paragraph for Question Nos. 19 and 20 ABC is a triangle with vertices A(x, x + 3), B(2, 1) and C(3, – 2). Area of the triangle is 5. 19. Length of the largest side is (a)

10

(b)

65 / 2

(c)

145 / 2

(d) none of these

20. Coordinates of the centroid of the triangle ABC are (a) (7/2, 13/2) (c) (5/2, 3/2)

(b) (17/6, 11/6) (d) (19/6, 21/6)

INTEGER-ANSWER TYPE QUESTIONS 21. If the coordinate of a point An are (2n, 2n + 1) then (A1A5)2 – 4498 = 22. A line is such that its segment between the axes is bisected at the point (23, 27) the difference of the intercepts made by the line on the coordinate axes is 23. If the circumcentre of the triangle whose vertices are (0, 2), (3, 5) and (5, 8) is (h, k) then h + k is equal to

IIT JEE eBooks: www.crackjee.xyz 14.20 Comprehensive Mathematics—JEE Advanced

24. If D denotes the area of the triangle with vertices (0, 0), (5, 0) and (5/6, 25/6) then 12D/25 is equal to

m ˆ Ê 1 , , 25. Vertices of a parallelogram are (0, 0), Á Ë m - n m - n ˜¯ m ˆ Ê -1 ÁË m - n , m - n ˜¯ and (0, 1) where m, n are the roots of the equation 441x2 + 42x – 8 = 0. If area of the parallelogram is A then 2A is equal to

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS

(c) Q lies on the line segment QP (d) P, Q, R are non-collinear

1. The points (– a, – b), (0, 0), (a, b) and (a2, ab) are (a) collinear (b) vertices of a parallelogram (c) vertices of a rectangle (d) none of these [1979] 2. The points (0, 8/3), (1, 3) and (82, 30) are the vertices of (a) an obtuse angled triangle (b) an acute angled triangle (c) right angled triangle (d) an isosceles triangle (e) none of these [1986] 3. If P(1, 2), Q(4, 6), R(5, 7) and S(a, b) are the vertices of a parallelogram PQRS, then (a) a = 2, b = 4 (b) a = 3, b = 4 (c) a = 2, b = 3 (d) a = 3, b = 5 [1998] 4. The incentre of the triangle with vertices (1, (0, 0) and (2, 0) is (a) (1,

3 /2)

3 ),

(b) (2/3, 1/ 3 )

(c) (2/3, 3 /2) (d) (1, 1/ 3 ) [2000] 5. The number of integer values of m, for which the x-coordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1 is also an integer is (a) 2 (b) 0 (c) 4 (d) 1 [2001] 6. Consider three points P = (– sin (b – a), – cos b), Q = (cos b – a), sin b) and R = (cos (b – a + q ), sin (b – q)) where 0 < a, b, q < p/4. Then (a) P lies on the line segment RQ (b) Q lies on the line segment PR

[2008]

TRUE/FALSE TYPE QUESTIONS x1 1. If x2 x3

y1 1 a1 y2 1 = a2 a3 y3 1

b1 1 b2 1 then the two triangles b3 1

with vertices (x1, y1), (x2, y2), (x3, y3) and (a1, b1), (a2, b2), (a3, b3) must be congruent (True/False) [1985]

SUBJECTIVE-TYPE QUESTIONS 1. The area of a triangle is 5, two of its vertices are (2, 1) and (3, –2). The third vertex lies on y = x + 3. Find the third vertex. [1978] 2. Given the points A(0, 4) and B(0, –4). Find the equation of the Locus of the point P(x, y) such that |AP – BP| = 6 [1983] 3. The coordinates of A, B, C are (6, 3), (–3, 5) and (4, –2) respectively, and P is any point (x, y). Show that the ratio of the areas of the triangles PBC and ABC x+ y-2 . [1985] is 7

Answers LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. (d) 5. (d) 9. (d)

2. (c) 6. (d) 10. (b)

3. (c) 7. (c)

4. (c) 8. (b)

IIT JEE eBooks: www.crackjee.xyz Cartesian System of Coordinates and Locus 14.21

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 11. 13. 15. 17. 19.

(a), (a), (a), (a), (a),

(b) (c) (c) (b), (c), (d) (b)

12. 14. 16. 18. 20.

(a), (a), (a), (c), (a),

(c) (b), (c), (d) (b), (c), (d) (d) (b), (c), (d)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 6. (a), (b) 8. (a), (b), (d)

7. (a), (c) 9. (a), (b)

10. (c), (d)

MATRIX-MATCH TYPE QUESTIONS p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

11.

MATRIX-MATCH TYPE QUESTIONS p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

ASSERTION-REASON TYPE QUESTIONS

r

s

13. (d)

21.

22.

12.

ASSERTION-REASON TYPE QUESTIONS 23. (a)

24. (b)

25. (c)

COMPREHENSION-TYPE QUESTIONS 26. (b) 30. (a)

27. (c)

28. (a)

29. (b)

INTEGER-ANSWER TYPE QUESTIONS 31. 5 35. 3

32. 4

33. 4

34. 6

16. (d) 20. (b)

3. (c)

17. (b)

18. (a)

19. (c)

INTEGER-ANSWER TYPE QUESTIONS 21. 2 25. 7

22. 8

23. 5

24. 5

PAST YEARS' IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. (a) 5. (a)

SINGLE CORRECT ANSWER TYPE QUESTIONS 2. (c)

15. (b)

COMPREHENSION-TYPE QUESTIONS

LEVEL 2

1. (c) 5. (d)

14. (c)

2. (e) 6. (d)

3. (c)

4. (d)

TRUE-FALSE TYPE QUESTIONS 4. (b)

1. False

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SUBJECTIVE-TYPE QUESTIONS Ê 7 13 ˆ Ê 3 3 ˆ 1. Á , ˜ or Á - , ˜ Ë 2 2 ¯ Ë 2 2¯

x2 y 2 +1 = 0 2. 7 9

Hints and Solutions LEVEL 1

1. (OA)2 = (OB)2 = (OC)2 fi 1 + a2 = a2 + a4 = a4 + a4 fi a2 = 1, but a π +1 So a = –1 and the required centroid is Ê 1 - 1 + 1 -1 + 1 + 1ˆ Ê 1 1ˆ , ÁË ˜¯ = ÁË , ˜¯ 3 3 3 3 2. PA : PB = 2 : 3 fi 9(PA)2 = 4(PB)2 fi 9[(x – 2)2 + y2] = 4[x2 + (y – 2)2] fi 5x2 + 5y2 – 36x + 16y + 20 = 0 Which passes through (a, a) if 10a2 – 20a + 20 = 0 or if (a – 1)2 + 1 = 0 which is not possible for any real value of a. 3. OP = 2 2 , OQ = 5 so that OR divides the PQ in the ratio of the sides containing O and hence is the bisector of the angle POQ . 4. a2 + b2 = 25, 2(a + b) = 14 fi ab = 12 5.

a 3 + b3 = 7 fi a2 – ab + b2 = 7 a+b

6. a = 2, 3 and b = 6, – 5, P is farthest from the origin if P = (3, 6) 7. sin q + cos q = 0 fi sin (q + p/4) = 0 fi q = – p/4. So the coordinate of R are (1, – 1/ 2 ), OR = 3 / 2 . 8. x = 2t2 + 2, y = 4t + 3 fi (y – 3)2 = 8(x – 2) 12 + 22 + + x 2 x( x + 1)(2 x + 1) 9. = 46 fi = 46 n 6n fi n = 11 10. D = 11/2, DA = 29/2, DB = 1/2, DC = 41/2. 1ˆ Ê 11. Mid point of AC is Á 0, - ˜ Ë 2¯ 1ˆ Ê Mid point of BD is Á 0, - ˜ Ë 2¯ fi diagonals AC and BD bisect each other

fi ABCD is a parallelogram Next, AC = BD = 65 fi diagonals are equal fi ABCD is a rectangle As AB = 13 , BC = 52 adjacent sides are not equal fi ABCD is not a square or rhombus x x+3 1 1 1 2 1 1 = | 4x - 4 | = 5 12. 2 2 3 -2 1 3 7 fi 2x – 2 = ±5 fi x = - or 2 2 Ê 3 3ˆ Ê 7 13 ˆ Coordinate of third vertex are Á - , ˜ or Á , ˜ Ë 2 2¯ Ë2 2¯ 13. Mid-point of the other vertices is also the mid point relation. b+3 Ê a + 5ˆ =2 Á +c so Ë 2 ˜¯ 2 fi 2a + 2c – b + 7 = 0 Also c2 + 2a – b = 0 fi c2 – 2c – 7 = 0 fi c = 1 ± 2 2 14. Let O (0, 0), P(a, 0), Q(a, a), R(0, a) be the vertices of the square OPQR of area = a2, M(a, a/2), N(a/2, a). Area of D OMN = 3a2/8 Area of D MQN = a2/8, Area of D OPM = Area of D ORN = a2/4 Area of the quadrilateral O MQN = a2 – a2/2 = a2/2 Area of the quadrilateral O MNR = 3a2/8 + a2/4 = 5a2/8 15. ax + ary + ar2 = 0 fi x + ry + r2 = 0 ordinates of P and Q are the roots of (r2 – 1) y2 + 2r3 y + r4 – 4 = 0 Ê a 2 + b 2 + g 2 a 2 b 2 + b 2 g 2 + g 2a 2 ˆ , 16. Á ˜¯ 3 Ë 3a 2 b 2g 2 Ê (3 p)2 - 2 ¥ 3q (3q)2 - 2 ¥ 3 p ˆ , = Á ˜¯ 3 3 Ë = (3p2 – 2q, 3q2 – 2p) 17. a + c = 2b, a2 + 1 + c2 + 9 = 2(b2 + 4) cos t + sin t 2(sin t - cos t ) sin t - cos t = or 18. 3 3 3 2(cos t + sin t ) = 3 3 cos t = sin t or sin t = 3 cos t

IIT JEE eBooks: www.crackjee.xyz Cartesian System of Coordinates and Locus 14.23

m2 1 mm ¢ 19. Area of the triangle 2 ( m ¢ )2 m2 1 (m ¢ - m) = 2 m ¢2 - m2

2m m¢ - m

Statement-1 is true.

2m 1 m + m¢ 1 2m ¢

(1, 0) whose mid point is (0, 0).

1

2 2 2 2 Locus is ( x + 1) + y + ( x - 1) + y = c (constant) which does not pass through (0, 0) for all verlues of c. So statement-2 is False. 26. x2 + 2x – 3 = 0 fi x = 1, –3 y2 + y – 2 = 0 fi y = 1, –2 So coordinates of the vertices of the triangle are (1, –2), (–3, 1), (–3, –2) as (x π y) Distances of these points from the origin are respec-

1 0

2(m ¢ - m) 0

1 | (m¢ – m)3 | 2 = 32 fi (m¢ – m)3 = ± 64 20. (OP)2 + (OQ)2 = (PQ)2 fi x1x2 + y1 y2 = 0 =

21. Length of AB = 2a

tively 5 , 10 , 13 So nearest is (1, –2) farthest is (–3, –2) and vertices of the triangle are A(–3, –2), B(–3, 1), C(1, –2) 27. (AB)2 = 9, (AC)2 = 16, (BC)2 = 25 fi (AB)2 + (AC)2 = (BC)2 DABC is right angled Circumcentre of the triangle is the mid point

(a) BC = 5a , CA = 5a fi BC = CA π AB DABC is isosceles (b) BC = 2a = CA = AB DABC is equilateral (c) BC = 2 2a , CA = 2a = AB fi (BC)2 = (AB)2 + (AC)2 DABC is right angled isosceles (d) BC = cos A =

10a , CA =

−1⎞ ⎛ ⎜⎝ −1, ⎟⎠ of BC, the hypotenuse. 2 ⎛ 5 ⎞ 28. Centroid of the DABC is ⎜ − , − 1⎟ whose dis⎝ ⎠ 3 tance from B(–3, 1) is

2a

( AB)2 + ( AC ) 2 - ( BC ) 2 2 AB ¥ AC =

2

4a 2 + 2a 2 - 10a 2 2 ¥ 2a ¥ 2a

b > c > 0, the distance between (1, 1) and the point of intersection of the lines ax + by + c = 0 and bx + ay + c = 0 is less than 2 2 . Then (a) a + b – c > 0 (b) a – b + c < 0 (c) a – b + c > 0 (d) a + b – c < 0 Ans. (a) Solution: The lines ax + by + c = 0 and bx + ay + c -c ˆ Ê -c . = 0 intersect in Á , Ë a + b a + b ˜¯ -c ˆ Ê -c Distance between (1, 1) and Á , Ë a + b a + b ˜¯

We know from geometry that the circumcentre,

centroid and orthocentre of a triangle lie on a line. So the orthocentre of the triangle lies on the line joining the circumcentre (0, 0) and the centroid

i.e.,

Eliminating l, m, n, we get abc = – 3(a + b + c) + bc + ca + ab fi bc + ca + ab – abc = 3(a + b + c)

( a + 1)2 ( a - 1)2 y = x 2 2 (a – 1)2x – (a + 1)2 y = 0.

Example 12

If a, b, c are unequal and different from

Ê a3 a 2 - 3 ˆ Ê b3 b 2 - 3 ˆ 1 such that the points Á , , ˜,Á ˜ , and Ë a -1 a -1 ¯ Ë b -1 b -1 ¯ Ê c3 c 2 - 3 ˆ Á c - 1 , c - 1 ˜ are collinear, then Ë ¯

= As

2

2 1+

c = a+b

2

a+b+c . a+b

a+b+c < 2 2 , we get a+b

a + b + c < 2(a + b) fi

a + b – c > 0.

Example 14 The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated p and q are the intercepts of the line L on the new axes, then 1 1 1 1 - 2 + 2 - 2 is equal to 2 a p b q (a) – 1 (c) 1 Ans. (b)

(b) 0 (d) none of these

(a) bc + ca + ab + abc = 0 (b) a + b + c = abc

Equation of the line L in the two coordinate x y X Y systems is + = 1, + =1 a b p q

(c) bc + ca + ab = abc (d) bc + ca + ab – abc = 3 (a + b + c)

where (X, Y) are the new coordinates of a point (x, y) when

Ans. (d) Solution: Suppose the given points lie on the line lx + my + n = 0 then a, b, c are the roots of the equation. lt3 + m(t2 – 3) + n(t – 1) = 0 or

lt3 + mt2 + nt – (3m + n) = 0



a + b + c = – m/l bc + ca + ab = n/l abc = (3m + n)/l

Solution:

origin has not changed. 1 1 = 1 1 1 1 + + a2 b2 p2 q2 or



1 1 1 1 + 2 = 2+ 2 2 a b p q

1 1 1 1 - 2 + 2 - 2 = 0. 2 a p b q

Example 15 If P is a point (x, y) on the line. y = – 3x such that P and the point (3, 4) are on the opposite sides of the line 3x – 4y = 8, then

IIT JEE eBooks: www.crackjee.xyz Straight Lines 15.9

(a) x > 8/15, y < – 8/5

Ans. (a)

(b) x > 8/5, y < – 8/15

Solution: Since xy > 0, P or in the third quadrant. The inequality x + y < 1 represents all points below the line x + y = 1. So that xy > 0 and x + y < 1 imply that either P lies inside the triangle OAB or in the third quadrant.

(c) x = 8/15, y = – 8/5 (d) none of these Ans. (a) Solution: Let k = 3x – 4y – 8 then the value of k at (3, 4) = 3 × 3 – 4 × 4 – 8 = – 15 < 0 \ For the point P (x, y) we should have k > 0 fi 3x – 4y – 8 > 0 fi 3x – 4(– 3x) – 8 > 0 [∵ P (x, y) lies on y = – 3x] fi x > 8/15 and – y – 4y – 8 > 0 fi y < – 8/5. Example 16 The area enclosed by 2|x| + 3|y| £ 6 is (a) 3 sq units (b) 4 sq units (c) 12 sq units

(d) 24 sq units

Ans. (c) Solution: The given inequality is equivalent to the following system of inequalities. (Fig. 15.5) 2x + 3y £ 6, when x ≥ 0, y ≥ 0 2x – 3y £ 6, when x ≥ 0, y £ 0 – 2x + 3y £ 6, when x £ 0, y ≥ 0 – 2x – 3y £ 6, when x £ 0, y £ 0 which represents a rhombus with sides 2x + 3y = 6 and 2x + 3y = – 6 Length of the diagonals is 6 and 4 units along x-axis and y-axis. 1 \ The required area = × 6 × 4 = 12 sq units. 2

Example 18 If a line joining two points A (2, 0) and B (3, 1) is rotated about A in anticlockwise direction through an angle 15º, then equation of the line in the new position is (a)

3x +y= 2 3

(c) x +

3y = 2 3

(b)

3x –y= 2 3

(d) x –

3y = 2 3

Ans. (b) Solution: fi

1- 0 =1 3- 2 –BAX = 45º (Fig. 15.6) Slope of AB =

If AC is the new position of the line AB then –CAX = 45° + 15º = 60º. and thus its equation is y = tan 60º (x – 2) fi

y=

3 (x – 2)



3x - y = 2 3

Fig. 15.6

Example 19 An equation of a line through the point (1, 2) whose distance from the point (3, 1) has the greatest value is

Fig. 15.5

Example 17 Let O be the origin, A (1, 0) and B (0, 1) and P (x, y) are points such that xy > 0 and x + y < 1, then (a) P lies either inside the triangle OAB or in the third quadrant (b) P can not lie inside the triangle OAB (c) P lies inside the triangle OAB (d) P

(a) y = 2x

(b) y = x + 1

(c) x + 2y = 5

(d) y = 3x – 1

Ans. (a) Solution: Let the equation of the line through (1, 2) be y – 2 = m(x – 1) If p denotes the length of the perpendicular from (3, 1) on this line, then p =

2m + 1 m2 + 1

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p2 =

4 m2 + 4 m + 1 m2 + 1

=4+

4m - 3 m2 + 1

= s (say),

then p2 is greatest if and only if s is greatest Now

ds ( m2 + 1)(4) - 2m(4m - 3) = dm ( m2 + 1)2 =

- 2(2m + 1) ( m - 2) ( m2 + 1)2

ds ds = 0 fi m = -1/2. Also < 0 if m < – 1/2 dm dm > 0 if -1/2 < m < 2 < 0 if m >2. So, s is greatest for m = 2. and thus the equation of the required line is y = 2x. Note p will be least when (3, 1) lies on the line through (1, 2) and the equation of the line in that case is x + 2y = 5.

So that –POQ = a and thus Q is obtained from P by clockwise rotation through an angle a around the origin. Note In the anticlockwise rotation, OQ makes (2p – a + q ) with the + ve direction of x-axis. Example 21 On the portion of the straight line x + y = 2 which is intercepted between the axes, a square is constructed, away from the origin, with this portion as one of its side. If p denotes the perpendicular distance of a side of this square from the origin, then the maximum value of p is (b) 3 2 (a) 2 3 (c) 2 / 3 (d) 3 / 2 Ans. (b) Solution: Clearly p = perpendicular distance from (0, 0) to AD + side of the square (Fig. 15.8) =

2 +2 2 =3 2 2

Example 20 Let 0 < a < p If P = (cos q, sin q ) and Q = (cos (a – q ), sin (a – q )) then Q is obtained from P by (a) clockwise rotation around the origin through an angle a (b) anticlockwise rotation around the origin through an angle a tan a tan (a/2) Ans. (a) Fig. 15.8

Example 22 The line x + y = 1 meets x-axis at A and y-axis at B. P is the mid-point of AB (Fig. 15.9). P1 is the foot of the perpendicular from P to OA; M1 is that from P1 to OP; P2 is that from M1 to OA; M2 is that from P2 to OP; P3 is that from M2 to OA and so on. If Pn denotes the n th foot of the perpendicular on OA from Mn – 1, then OPn = (a) 1/2 (b) 1/2n (c) 1/2n/2 (d) 1/ 2 Ans. (b) Fig. 15.7

Solution: OP makes an angle q with the positive direction of x-axis and OQ makes an angle (a – q ) with the positive direction of x-axis.(Fig.15.7)

Solution: x + y = 1 meets x-axis at A(1, 0) and y-axis at B(0, 1) (Fig. 15.9).

IIT JEE eBooks: www.crackjee.xyz Straight Lines 15.11

x – y=

or

1- l a. 1+ l

Ê l -1 ˆ So, the coordinates of M are Á 0, a Ë l + 1 ˜¯ Therefore, area of the triangle AMN is =

The coordinates of P are (1/2, 1/2) and PP1 is perpendicular to OA. fi OP1 = P1P = 1/2 Equation of line OP is y = x. We have (OMn – 1)2 = (OPn)2 + (Pn Mn – 1)2 =2 (O Pn)2 = 2pn2 (say) (OPn – 1)2 = (OMn – 1)2 + (Pn – 1 Mn – 1)2 = fi p2n= \

1 2 pn–1 4

OPn = pn =



2p2n

l a2

3 1 = ◊ a2 8 2 (1 + l ) 2

fi fi

3 l2 – 10 l + 3 = 0 l = 3 or l = 1/3.

For l = 1/3, M lies outside the segment OB and hence the required value of l is 3.

1 2 + p n–1 2

pn =

È Ê -a ˆ 1- l 2˘ l a2 + = a a Í Á ˙ ˜ 2 2 Î Ë l + 1¯ (1 + l ) ˚ (1 + l )

Also area of the triangle OAB = a2/2. So that according to the given condition.

Fig. 15.9

Also,

1 2

1 pn – 1 2

1 1 pn – 1 = 2 pn – 2 =  2 2 1 2 2n The line x + y = a, meets the axis of x and =

1

n -1

p1 =

Example 23 y at A and B respectively. A triangle AMN is inscribed in the triangle OAB, O being the origin, with right angle at N. M and N lie respectively on OB and AB. If the area of the triangle AMN is 3/8 of the area of the triangle OAB, then AN/BN is equal to (a) 3 (b) 1/3 (c) 2 (d) 1/2 Ans. (a) Solution:

Let

AN = l . Then the coordinates of N are BN Ê a la ˆ ÁË 1 + l , 1 + l ˜¯

where (a, 0) and (0, a) are the coordinates of A and B respectively. Now equation of MN perpendicular to AB is y–

la a =x– . 1+ l 1+ l

Fig. 15.10

Example 24 The point (4, 1) undergoes the following transformation successively. y = x. (ii) translation through a distance 2 units along the positive direction of x-axis. (iii) rotation through an angle p /4 about the origin in the anticlockwise direction. x=0 sition of the given point is (a) (1/ 2 , 7/2)

(

(c) 1/ 2, 7/ 2

(b) (1/2, 7 2 )

)

(d) (1/2, 7/2)

Ans. (c) Solution: Let B, C, D, E be the positions of the given point A (4, 1) after the transformations (i), (ii), (iii) and (iv) successively (Fig. 15.11).

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fi x2 + y2 – 7x + 5y = 0 which is the locus of R(x, y). Q (0, b) R (x , y )

Fig. 15.11

A (7, 0)

P (a, 0)

O (0, 0) B (0, -5)

The coordinates of B are (1, 4) and that of C are ( 1 + 2, 4 + 0), i.e., (3,4). Now if OC makes an angle q with x-axis, OD makes and an angle q + p/4 with x-axis. If (h, k) denote the coordinates of D, then h = OD cos (q + 45°), k = sin (q + 45°) and OD = OC = 5, sin q = 4/5, cos q = 3/5 fi h = (5/ 2 ) (cos q - sin q) = - 1/ 2

Fig. 15.12

Example 26 Equation of the line which bisects the obtuse angle between the lines x – 2y + 4 = 0 and 4x – 3y + 2 = 0 is (a) (4 + 5 ) x - (3 + 2 5 ) y + 2 + 4 5 = 0 (b) (4 - 5 ) x - (3 + 2 5 ) y + 2 - 4 5 = 0

k = (5/ 2 ) (cos q + sin q) = 7/ 2 coordinates of D are (- 1/ 2 , 7/ 2

(c) (4 - 5 ) x - (3 - 2 5 ) y + 2 - 4 5 = 0

about x = 0 in (1/ 2 , 7/ 2 ).

(d) (4 + 5 ) x - (3 - 2 5 ) y + 2 + 4 5 = 0

Example 25 A line cuts the x-axis at A(7, 0) and the y-axis at B(0, – 5). A variable line PQ is drawn perpendicular to AB. Cutting the x-axis at P and the y-axis at Q. If AQ and BP intersect at R, the locus of R is (a) (b) (c) (d)

x2 + y2 + 7x – 5y = 0 x2 + y2 – 7x + 5y = 0 5x – 7y = 35 none of these

Ans. (a) Solution: Let us form a triangle with sides x – 2y + 4 = 0, 4x – 3y + 2 = 0 and x = 0. Coordinates of the vertices of this triangle are A(8/5, 14/5), B(0, 2/3) and C(0, 2). Let

x1 = 8/5, y1 = 14/5, y2 = 2/3 and y3 = 2. y

Ans. (b) Solution:

Slope of PQ = – fi

A(8/5, 14/5 )

Let P(a, 0) and Q(0, b). (0, 2 )

–b . a

b 5 =– 1 ¥ a 7

(0, 2 /3 )

a 5 = . b 7 x y + =1 7 b



b=

7y 7–x

Equation of BP is

x y – =1 a 5



a=

5x 5+ y

7–x a 5 5x = = ¥ b 7 5+ y 7y

fi x (7 – x) = y (5 + y)

B

Fig. 15.3

Equation of AQ is

so that

C

We have AB2 + AC2 – BC2 = x12 + (y2 – y1)2 + x12 + (y3 – y1)2 – (y2 – y3)2 = 2x12 + 2y12 – 2y1 (y2 + y3) + 2y2y3 = 2x12 + 2 (y1 – y2) (y1 – y3) > 0 Thus, –BAC is an acute angle. Equation of angle bisectors of two lines are given

IIT JEE eBooks: www.crackjee.xyz Straight Lines 15.13

x - 2y + 4

4x - 3y + 2 5



5 (x – 2y + 4) = ± (4x – 3y + 2)



(4 - 5 ) x - (3 - 2 5 ) y + 2 - 4 5 = 0

(1)

Example 28 If the lines joining the origin to the intersection of the line y = mx + 2 and the curve x2 + y2 = 1 are at right angles, then (b) m2 = 3 (a) m2 = 1 (d) 2m2 = 1 (c) m2 = 7

or

(4 + 5 ) x - (3 + 2 5 ) y + 2 + 4 5 = 0

(2)

Ans. (c)

5

= ±

Line (1) meets x = 0 in E (0, y) where y= =

4 5-2 2 5 -3

=

(4 5 - 2)(2 5 + 3) 20 - 9

34 + 8 5 >2. 11

Thus, E lies outside BC.

Solution: Joint equation of the lines joining the origin and the point of intersection of the line y = mx + 2 and the curve x2 + y2 = 1 is 2 Ê y - mx ˆ x2 + y2 = Á Ë 2 ˜¯ fi

x2 (4 – m2) + 2mxy + 3y2 = 0

Since these lines are at right angles 4 – m2 + 3 = 0 fi m2 = 7.

\ (1) is the required obtuse angle bisector. Example 27 If the pairs of lines x2 + 2xy + ay2 = 0 and ax2 + 2xy + y2 = 0 have exactly one line in common then the joint equation of the other two lines is given by (a) 3x2 + 8xy – 3y2 = 0

Example 29 Let PQR be a right angled isosceles triangle right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is (a) 3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0

(b) 3x2 + 10xy + 3y2 = 0

(b) 3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0

(c) y2 + 2xy – 3x2 = 0

(c) 3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0

(d) x2 + 2xy – 3y2 = 0

(d) 3x2 – 3y2 – 8xy – 10x – 15y – 20 = 0

Ans. (b) Solution:

Let y = mx be a line common to the given pairs

Solution:

of lines, then 2

2

am + 2m + 1 = 0 and m + 2m + a = 0 fi

m2 m 1 = 2 = 2 (1 - a ) a - 1 2 (1 - a )



m2 = 1 and m = -



a = 1 or –3

Ans. (b) Let the slopes of PQ and PR be m and

– 1/m respectively. Since PQR is an isosceles triangle PQR = PRQ

a +1 fi (a + 1)2 = 4 2

2x

But for a = 1, the two pairs have both the lines common, so a = –3 and the slope m of the line common to both the pairs is 1. Now x2 + 2xy + ay2 = x2 + 2xy – 3y2 = (x – y) (x + 3y) and

ax2 + 2xy + y2 = – 3x2 + 2xy + y2 = – (x – y) (3x + y)

So, the equation of the required lines is (x + 3y) (3x + y) = 0 fi 3x2 + 10xy + 3y2 = 0.

+

y=

3

Fig. 15.14



m+2 = 1 - 2m

1 +2 m 2 1+ m

-

[∵ slope of QR = – 2] fi

m + 2 = ± (1 – 2m) fi m = 3 or – 1/3

So the equations of PQ and PR are (y – 1)= 3(x – 2) and y – 1 = (– 1/3) (x – 2)

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Thus, joint equation representing PQ and PR is [3(x – 2) – (y – 1)] [(x – 2) + 3(y – 1)] = 0 fi

3(x – 2)2 – 3(y – 1)2 + 8(x – 2) (y – 1) = 0



2

Solution:

Writing the given equation as a quadratic in x 2

we have 6x + (5y + 7)x – (4y2 – 13y + 3) = 0 x=

- (5 y + 7) ±

(5 y + 7)2 + 24 ( 4 y 2 - 13 y + 3) 12

=

- (5 y + 7) ± 121y 2 - 242 y + 121 12

=

- (5 y + 7) ± 11( y - 1) 6 y - 18 -16 y + 4 = , 12 12 12

fi 2x – y + 3 = 0 and 3x + 4y – 1 = 0 which are the two lines represented by the given equation and the point of intersection is (– 1, 1), obtained by solving these equations. [The point of intersection of the given lives is also obtained by solving 12x + 5y + 7 = 0 and 5x – 8y + 13 = 0 (see Text, 15.5(4)] Also tan q =

2 h2 - ab where a = 6, b = – 4, h = 5/2 a+b 2 (5 2) - 6 ( -4 ) 2

=

6-4

=

121 11 = 4 2

So the equation of the required line is y – 1=

11 (x + 1) 2



11x – 2y + 13 = 0

Example 31 If one of the lines given by the equation 2x2 + axy + 3y2 = 0 coincide with one of those given by 2x2 + bxy – 3y2 = 0 and the other lines represented by them be perpendicular, then (a) a = – 5, b = 1 (b) a = 5, b = – 1 (c) a = 5, b = 1 (d) none of these Ans. (c)

2 2 a x + x y + y2 = (y – mx)(y – m¢x) 3 3

2 2 b 1 ˆ Ê x + x y + y2 = Á y + x˜ (y – m¢x) Ë m ¯ -3 -3

then

a 2 m + m¢ = - , mm¢ = 3 3

(i)

-b 1 m¢ 2 – m¢ = , = 3 m m 3

(ii)

3x – 3y + 8xy – 20x – 10y + 25 = 0.

Ans. (b)

Let

and

2

Example 30 If q is an angle between the lines given by the equation 6x2 + 5xy – 4y2 + 7x + 13y – 3 = 0, then equation of the line passing through the point of intersection of these lines and making an angle q with the positive x-axis is (a) 2x + 11y + 13 = 0 (b) 11x – 2y + 13 = 0 (c) 2x – 11y + 2 = 0 (d) 11x + 2y – 11 = 0



Solution:

m2 = 1 fi m = ± 1



2 fi a = – 5, b = – 1 3 2 If m = – 1, m¢ = - fi a = 5, b = 1. 3 If m = 1, m¢ =

Example 32 The equation x – y = 4 and x2 + 4xy + y2 = 0 represent the sides of (a) an equilateral triangle (b) a right angled triangle (c) an isosceles triangle (d) none of these Ans. (a) Solution:

Acute angle between the lines x2 + 4xy + y2 = 0 is tan–1

2 4 -1 = tan–1 1+1

3 = p/3

Angle bisectors of x2 + 4xy + y2 = 0 are given by x2 - y2 xy = fi x2 – y2 = 0 1-1 2 fi x= ± y As x + y = 0 is perpendicular to x – y = 4, the given triangle is isosceles with vertical angle equal to p/3 and hence it is equilateral. Example 33 If the equation of the pair of straight lines passing through the point (1, 1), one making an angle q with the positive direction of x-axis and the other making the same angle with the positive direction of y-axis is x2 – (a + 2) xy + y2 + a (x + y – 1) = 0, a π –2, then the value of sin 2q is (a) a – 2 (b) a + 2 (c) 2/(a + 2) (d) 2/a Ans. (c) Solution:

Equations of the given lines are

y – 1 = tan q (x – 1) and y – 1 = cot q (x – 1) so their joint equation is [(y – 1) – tan q (x – 1)] [(y – 1) – cot q (x – 1)] = 0

IIT JEE eBooks: www.crackjee.xyz Straight Lines 15.15

fi (y – 1)2 – (tan q + cot q) (x – 1) (y –1) + (x – 1)2 = 0 fi

= ah(x4 + y4) + 2(h2 – a 2 ) (x3y – xy3) – 6ahx2y2

x2 – (tan q + cot q) xy + y2 + (tan q + cot q – 2) ah = 1 and c = – 6ah = – 6 (x + y – 1) = 0

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

Comparing with the given equation we get tan q + cot q = a + 2 1 =a+2 fi sin q cos q



sin 2q =

2 a+2

Example 34 If q1 and q2 be the angles which the lines (x2 +y2) (cos2 q sin2 a + sin2 q ) = (x tan a – y sin q)2 make

Example 36 The coordinates of a point at unit distance from the lines 3x – 4y + 1 = 0 and 8x + 6y + 1 = 0 are

with the axis of x, then if q = p/6, tan q1 + tan q2 is equal to (a) (– 8/3) sin2 a (b) (– 8/3) cosec 2a (c) – 8 3 cosec 2a

(d) – 4 cosec 2a

Solution:

The given equation can be written as

(x +y ) (cos q sin a + sin q) = x tan a – 2xy tan a sinq + y2 sin2 q or (cos2 q sin2 a + sin2 q – tan2 a) x2 + 2 (tan a sin q) xy + cos2 q sin2ay2 = 0. Since the slope of these lines are given as tan q1 and tan q2. -2 tan a sin q and sum of the slopes = cos2 q sin 2 a fi

2

Ê 3ˆ (b) Á 0, ˜ Ë 2¯

Ê -2 -13 ˆ (c) Á , Ë 5 10 ˜¯

Ê -8 3 ˆ (d) Á , ˜ Ë 5 10 ¯

Ans. (a), (b), (c), (d)

Ans. (b) 2

Ê 6 -1ˆ (a) Á , ˜ Ë 5 10 ¯

2

2

2

tan q1 + tan q2 =

2

2

-2 tan a

(∵ q = p/6)

2 ¥ (3 4) ¥ sin 2 a

= – (8/3) cosec 2a. Example 35

If two of the lines represented by x4 + x3 y + cx2 y2 – xy3 + y4 = 0

bisect the angle between the other two, then the value of c is (a) 0

(b) – 1

(c) 1

(d) – 6

Solution: given by

3x - 4 y + 1 32 + 42 fi

xy x2 - y2 The equation of its bisectors is = . 2a h By hypothesis x4 + x3y + cx2y2 – xy3 + y4 = (ax2 + 2hxy – ay2) (hx2 – 2axy – hy2)

= ±1 and

8x + 6 y + 1 82 + 6 2

= ±1

3x - 4 y - 4 = 0 ¸ (1) 8 x + 6 y - 9 = 0 ¸ (3) and ˝ ˝ 3 x - 4 y + 6 = 0 ˛ ( 2) 8 x + 6 y + 11 = 0 ˛ (4)

Ê 6 -1ˆ Solving (1) and (3) we get (x, y) = Á , ˜ Ë 5 10 ¯ Ê -2 -13 ˆ (1) and (4) we get (x, y) = Á , Ë 5 10 ˜¯ Ê 3ˆ (2) and (3) we get (x, y) = Á 0, ˜ Ë 2¯ Ê -8 3 ˆ and (2) and (4) we get (x, y) = Á , ˜ Ë 5 10 ¯ Example 37

Ans. (d) Solution: Since the product of the slopes of the four lines represented by the given equation is 1 and a pair of lines represent the bisectors of the angles between the other two, the product of the slopes of each pair is –1. So let the equation of one pair be ax2 + 2hxy – ay2 = 0.

Coordinates (x, y) of the required point are

and

Equations

(b – c)x + (c – a)y + (a – b) = 0 (b3 – c3)x + (c3 – a3)y + a3 – b3 = 0

will represent the same line if (a) b = c

(b) c = a

(c) a = b

(d) a + b + c = 0.

Ans. (a), (b), (c) and (d) Solution: The two lines will be identical if there exists some real number k, such that b3 – c3 = k(b – c), c3 – a3 = k(c – a) and a3 – b3 = k (a – b). fi b – c = 0 or b2 + c2 + bc = k

IIT JEE eBooks: www.crackjee.xyz 15.16 Comprehensive Mathematics—JEE Advanced

c – a = 0 or c2 + a2 + ca = k and a – b = 0 or a2 + b2 + ab = k That is, b = c or c = a or a = b. Next b2 + c2 + bc = c2 + a2 + ca fi b2 – a2 = c(a – b) Hence, a = b or a + b + c = 0. Example 38 A rectangle ABCD has its side AB parallel to the line y = 2x and vertices A, B and D on lines y = 1, x = 1 and x = –1 respectively. The coordinates of C can be (a) (3, 8) (c) (–3, –1)

Let the equation of AB be y = 2x + l

coordinates of C be (h, k) B(1, k1) and D(–1, k2) Since the diagonals of a rectangle bisect each other and the x-coordinates of the midpoint of BD lies on the y-axis, the mid-point of AC also lies on the y-axis, so the coordinates of A are (–h, 1) y

Solution: The required lines are parallel to the angle bisectors of the given lines. The angle bisectors of the given lines are 3x - 4 y - 7 9 + 16

Ê 12 x - 5 y + 6 ˆ =±Á ˜ Ë 144 + 25 ¯

Equations of the lines passing through (4, 5) and parallel to these lines are 21 (x – 4) + 27 (y – 5) = 0 and 99 (x – 4) – 77 (y – 5) = 0 i.e., the required lines are 7x + 9y – 73 = 0 and 9x – 7y – 1 = 0 Example 40 For all values of q, the lines represented by the equation

C(h, k)

B(1, k1)

D (- 1, k2)

(b) 9x + 7y = 71 (d) 7x – 9y + 17 = 0.

fi 13 (3x – 4y – 7) = ± 5 (12x – 5y + 6) fi 21x + 27y + 121 = 0 or 99x – 77y – 61 = 0

(b) (–3, 8) (d) (3, –1)

Ans. (a), (c) Solution:

(a) 9x – 7y = 1 (c) 7x + 9y = 73 Ans. (a), (c)

(2 cos q + 3 sin q) x + (3 cos q - 5 sin q) y - (5 cos q - 2 sin q) = 0

y = 2x A(- h, 1) O

(b) pass through the point (1, 1)

x

Fig. 15.15

the line

Now B lies on AB fi k1 = 2 + l fi l = k1 – 2

x+y=

Also A lies on AB fi 1 = –2h + l fi 2h = k1 – 3



2 - 1)

(d) pass through the origin Ans. (a), (b), (c)

Next, AB is perpendicular to BC k1 - 1 k - k1 ¥ = –1 1+ h h -1

so

2 is ( 2 - 1,

(k1 – 1) (k – k1) + (h – 1) (h + 1) = 0

fi (2h + 3 – 1) (k – 2h – 3) + (h – 1) (h + 1) = 0 fi

(h + 1) (2k – 4h – 6 + h – 1) = 0



3h – 2k + 7 = 0

Solution:

The given equation can be written as

(2x + 3y - 5) cos q + (3x - 5y + 2) sin q = 0 (2x + 3y - 5) + tan q (3x - 5y + 2) = 0

or

This passes through the point of intersection of the lines 2x + 3y - 5 = 0 and 3x - 5y + 2 = 0 for all values of q. The coordinates of the point P of intersection are (1, 1). Let Q(h, k P(1, 1) in the line x + y=

coordinates in (a) and (c) satisfy this relation. Note h π –1 as the slope of CD is 2. Example 39 Equation of a straight line passing through the point (4, 5) and equally inclined to the lines 3x = 4y + 7 and 5y = 12x + 6 is

2

(1)

Then PQ is perpendicular to (1) and the mid-point of PQ lies on (1). \

k -1 =1fik=h h -1

and

h +1 k +1 + = 2 2

2 fih=k=

2 -1

IIT JEE eBooks: www.crackjee.xyz Straight Lines 15.17

pˆ Ê y = tan Á a + ˜ x Ë 4¯

Equation of a bisector of the angle Example 41 between the lines 2m y-b= (x - a) and 1 - m2 y-b=

2 m¢

or

y(cos a – sin a) = x(sin a + cos a)

(x - a) is

1 - m¢ 2

(a) (y - b) (m + m¢) + (x - a) (1 - mm¢) = 0 (b) (y - b) (1 - mm¢) + (x - a) (m + m¢) = 0 (c) (x - a) (m + m¢) - (y - b) (1 - mm¢) = 0

p

(d) (x - a) (m + m¢) - (y - b) (1 - mm¢) = 0 Ans. (a), (d) Solution:

2m ( x - a) - ( y - b) 1 - m2 2

Ê 2m ˆ Á 1 - m2 ˜ + 1 Ë ¯ fi

Fig. 15.16

Equations of the bisectors are given by



2m ¢ ( x - a) - ( y - b) 1 - m¢ 2 2

Ê 2m ¢ ˆ Á 1 - m¢ 2 ˜ + 1 Ë ¯

2m( x - a) - (1 - m2 )( y - b) 1 + m2 =±

2m¢( x - a) - (1 - m¢ 2 )( y - b)

S i n ce AC is perpendicular to OB, its slope is –cot (a + p /4), and as it passes through A (a cos a, a sin a), its equation is y – a sin a = – cot (p /4 + a) (x – a cos a) or y (sin a + cos a) + x (cos a – sin a) = a Example 43 The coordinates of the feet of the perpendiculars from the vertices of a triangle on the opposite sides are (20, 25), (8, 16) and (8, 9). The coordinates of a vertex of the triangle are

1 + m¢ 2

Taking the positive sign, we get ( y – b) [(1 – m¢2) (1 + m2) – (1 + m¢2) (1 – m2)] + (x – a) [2m (1 + m¢2) – 2m¢ (1 + m2)] = 0 or ( y – b) (m + m¢) + (x – a) (1 – mm¢) = 0 Similarly, by taking the negative sign, we get ( y – b) (1 – mm¢) – (x – a)(m + m¢) = 0

(a) (5, 10)

(b) (50, -5)

(c) (15, 30)

(d) (10,15)

Ans. (a), (b), (c) Solution: We use the fact that the orthocentre O of the triangle ABC is the incentre of the pedal triangle DEF. Let (h, k) be the coordinates of O. Now ED =

Example 42 A square with each side equal to a lies above the x-axis and has one vertex at the origin. One of the sides passing through the origin makes an angle a (0 < a < p/4) with the positive direction of the x-axis. Equation of a diagonal of the square is

(20 - 8)2 + (25 - 16)2

= 15

(a) y (cos a – sin a) = x (sin a + cos a) (b) y (sin a + cos a) + x (cos a – sin a) = a (c) x (cos a - sin a) = y (cos a + sin a) (d) x (cos a - sin a) - y (cos a + sin a) = a Ans: (a), (b) Solution: Let the side OA make an angle a with the x-axis (see Fig. 15.17). Then the coordinates of A are (a cos a, a sin a). Also, the diagonal OB makes an angle a + p /4 with the x-axis, so that its equation is

Fig. 15.17

FD = 20 and EF = 7 7 ¥ 20 + 20 ¥ 8 + 15 ¥ 8 so that h = = 10 7 + 20 + 15 and

k=

7 ¥ 25 + 20 ¥ 16 + 15 ¥ 9 = 15 7 + 20 + 15

IIT JEE eBooks: www.crackjee.xyz 15.18 Comprehensive Mathematics—JEE Advanced

thus, the coordinates of 0 are (10, 15) Since AC is perpendicular to OE, equation of AC is 10 - 8 (x – 8) y – 16 = 15 - 16 fi y – 2x = 0 Similarly equation of AB is

(a) a = 1

(b) b = 2

(c) a = 2

(d) b = 4

Ans. (c), (d) y=

(1)

20 - 10 (x – 20) y – 25 = 25 - 15 fi y + x – 45 = 0 (3) Solving (1) and (2) we get A (5, 10) From (2) and (3) we get B (50, – 5) and from (3) and (1) we get C (15, 30) Example 44 If the area of the triangle formed by the lines y = x, x + y = 2 and the line through P(h, k) and parallel to x-axis is 4h2, the locus of P can be (a) 2x – y + 1 = 0 (b) 2x + y – 1 = 0 (c) x – 2y + 1 = 0 (d) x + 2y – 1 = 0 Ans. (a), (b) Solution: Coordinates of A are (1, 1) which is the point of intersection of the given lines. y = k is the line through P parallel to x-axis which meets the given lines at B and C. So coordinates of B are (k, k) and C are (2 – k, k). 1 1 k 1 = 4h2 k 1

fi (k – 1)2 = 4h2 fi k – 1 = ± 2h Locus of P(h, k) is y – 1 = ± 2x fi 2x – y + 1 = 0 or 2x + y – 1 = 0 y

o

1 2

X

B

c

b

C

1 y=– 2

Fig. 15.19

Solution:

Let the squares of unit area be bounded by the 1 1 line x = ± , y = ± 2 2 A(x1, 1/2), B(1/2, y1), C(x2, – 1/2), D(– 1/2, y2) a2 = (x1 – 1/2)2 + (1/2 – y1)2 = x12 + y21 – x1 – y1 + 1/2 b2 = x22 + y12 – x2 – y1 + 1/2 c2 = x22 + y22 – x2 – y2 + 1/2 d2 = x12 + y22 – x1 – y2 + 1/2 fi a2 + b2 + c2 + d2 = 2 (x12 + x22 + y12 + y22) + 2 As 0 £ x12, x22, y12, y22 £ 1/4 0 £ x12 + x22 + y12 + y22 £ 1 Thus, 2 £ a2 + b2 + c2 + d2 £ 4 so a = 2 and b = 4 Example 46 Three lines px + qy + r = 0, qx + ry + p = 0 and rx + py + q = 0 are concurrent if (b) p2 + q2 + r2 = pq + qr + rp (c) p3 + q3 + r3 = 3pqr (d) none of these Ans. (a), (b), (c)

A B

x=

(a) p + q + r = 0

y=x

(0, 2) C

y=k

P O

1 x=– 2

(2)

2–k

a

D

fi 3y + x – 35 = 0 and equation of BC is

1 k

A

d

10 - 8 y–9 = (x – 8) 15 - 9

1 Area of the triangle ABC = 2

Y

1 2

(2, 0)

X

Solution:

p Lines are concurrent if q r

q r p

r p =0 q

or if (p + q + r) (p2 + q2 + r2 – pq – qr – rp) = 0 Fig. 15.18

Example 45 Let S be a square with unit area. Consider any quadrilateral which has one vertex on each side of S. If a, b, c, d denote the lengths of the sides of the quadrilateral, then a £ a2 + b2 + c2 + d2 £ b where

or if p3 + q3 + r3 – 3pqr = 0 Example 47 An equation of a straight line passing through the point (2, 3) and having an intercept of length 2 units between the straight lines 2x + y = 3, 2x + y = 5 is

IIT JEE eBooks: www.crackjee.xyz Straight Lines 15.19

(a) x – 2 = 0 (c) 3x + 4y – 18 = 0

(b) y – 3 = 0 (d) 4x – 3y + 1 = 0

Since P and Q lie on the line y = m(x – a) y1 = m(x1 – a)

Ans. (a), (c)

y2 = m(x2 – a) (2, 3) A

fi L

B

È a (2am 2 + 4a) ˘ = m 2 Í 2a 2 ˙ m2 Î ˚

2x + y = 5

M

2x + y = 3

y1y2 = m2[x1x2 – a(x1 + x2) + a2]

= –4a2 and

x1x2 – y1y2 = 5a2

Note Fig. 15.20

If we consider that the point P and Q lie on the curve y2 = 4ax, we get

Solution: Any line through (2, 3) is

(y1y2)2 = (4a)2 (x1x2) = (4a2)2

y – 3 = m(x – 2), which meets the line 2x + y = 5 at



Ê 2m 6 - m ˆ AÁ , Ë m + 2 m + 2 ˜¯

But y1y2 = 4a2 is not possible as the point (a, 0) lies on x-axis and the curve is symmetrical about x-axis. So y1 and y2 are of opposite signs and thus their product is negative.

and the line 2x + y = 3 at

Ê 2m + 2 m + 6 ˆ , BÁ Ë m + 2 m + 2 ˜¯

x (tan q + cos q) – 2x y tan q + y2 sin2 q = 0 make angles 2



Ê 2 ˆ Ê 2m ˆ ÁË ˜ + ÁË ˜ m + 2¯ m + 2¯

2

=4

-3 4 So, when the line is not parallel to y-axis its equation is 3x + 4y – 18 = 0.



1 + m2 = m2 + 4m + 4 fi m =

If the line is parallel to y-axis its equation is x = 2, which meets the line 2x + y = 5 at L(2, 1) and the line 2x + y = 3 at M(2, –1) and LM = 2. Example 48 A line through the point (a, 0) meets the curve y2 = 4ax at P(x1, y1) and Q(x2, y2), then 2

(a) x1x2 = a (c) y1y2 = –4a2 Ans. (a), (c), (d)

(b) y1y2 = 4a (d) x1x2 – y1y2 = 5a2

2

(b) tan a tan b = sec2 q + tan2 q (c) tan a – tan b = 2 (d)

tan a 2 + sin 2q = . tan b 2 - sin 2q

Ans. (a), (c) and (d) Solution: be Then

Let the lines represented by the given equation y = x tan a and y = x tan b

tan a + tan b =

x 1x 2 =

a 2 m2 m

2

= a2

2 tan q sin q 2

=

2 sin q cos q

= 4 cosec 2q tan a tan b =

tan 2 q + cos2 q sin 2 q

= sec2 q + cot2 q

m2x2 – (2am2 + 4a)x + a2m2 = 0

which gives x-coordinates x1, x2 of P and Q as the roots of this equation. So

2

a, b with the x-axis, then (a) tan a + tan b = 4 cosec 2q

2

Solution: Let the equation of the line through (a, 0) be y = m(x – a), which meets the curve y2 = 4ax at points for which m2(x – a)2 = 4ax fi

If the two lines represented by

Example 49

AB = 2 (given) 2

y1y2 = ±4a2

tan a – tan b = ±

= ±2

4 - 4 (sec2 q + cot 2 q ) sin 2 q cos2 q 1 - (sin 2 q + cos 4 q ) sin 2 q cos2 q

IIT JEE eBooks: www.crackjee.xyz 15.20 Comprehensive Mathematics—JEE Advanced

= ±2

cos2 q - cos 4 q sin q cos q 2

2

Example 52 Equation of a line which is parallel to the line common to the pair of lines given by 6x2 – xy – 12y2 = 0 and 15x2 + 14xy – 8y2 = 0 and at a distance 7 from it is

= ± 2.

tan a 4 cosec 2q + 2 2 + sin 2q = = tan b 4 cosec 2q - 2 2 - sin 2q

and

3

2

Example 50 If two of the lines given by 3x + 3x y – 3xy2 + dy3 = 0 are at right angles then the slope of one of them is (a) – 1 (b) 1 (c) 3 (d) – 3 Ans. (a) and (b) Solution: Let the lines represented by the given equations be y = m1 x, y = m2 x and y = m3 x. Then m1 m2 m3 = – 3 d . If two of the lines are perpendicular, Let m1 m2 = – 1 fi m3 = 3 d . fi y = ( 3 d )x fi

d (3 d ) - 3 (3 d ) + 3 (3 d ) + 3 = 0



d = –3

3

The given equation then becomes x3 + x2 y – xy2 – y3 = 0 (x + y) (x2 – y2) = 0

Showing that the slope of the other two are 1 and – 1. Example 51 9x2 + 2hxy + 4y2 + 6x + 2f y – 3 = 0 represent two parallel lines if (a) h = 6, f = 2

(b) h = – 6, f = – 2

(c) h = – 6, f = 2

(d) h = 6, f = – 2

Ans. (a) and (b) Solution: Since the given parallel lines, we have h2 = 9 ¥ 4 9 h h 4 and 3 f fi fi fi

equation represents a pair of fi h = ± 6. 3 f = 0. -3

9 (–12 – f 2 ) – h(–3h – 3f ) + 3(h f – 12) = 0 3h2 + 6h f – 9f 2 – 144 = 0 (∵ h = ± 6) 108 ± 36 f – 9 f 2 – 144 = 0

fi 9f ∓ 36 f + 36 = 0

if h = ± 6



if h = 6

2

and

f=2 f=–2

(b) 5x – 2y = 7

(c) 3x + 4y = – 35

(d) 2x – 3y = 7

Ans. (a) and (c) Solution: 6x2 – xy – 12y2 = 0 fi (2x – 3y) (3x + 4y) = 0 and 15x2 + 14xy – 8y2 = 0 fi (5x – 2y) (3x + 4y) = 0 Equation of the line common to (i) and (ii) is 3x + 4y = 0 Equation of any line parallel to (ii) is

if h = – 6

(i) (ii) (iii)

3x + 4y = k Since its distance from (iii) is 7 k

2

and the slope of one of the lines is m3 = – 1

or

(a) 3x + 4y = 35

3 + 42 2

=7 fi

k = ± (7 ¥ 5) = ± 35

Example 53 The lines joining the origin to the point of intersection of 3x2 + lxy – 4x + 1 = 0 and 2x + y – 1 = 0 are at right angles for (a) l = – 4

(b) l = 4

(c) l = 7

(d) no value of l

Ans. (a), (b) and (c) Solution: Equation of the lines joining the origin to the points of intersection of the given lines is 3x2 + lxy – 4x(2x + y) + 1◊(2x + y)2 = 0 (Making the equation of the pair of lines homogeneous with the help of the equation of the line) fi x2 – lxy – y2 = 0 which are perpendicular for all values of l. Example 54 If x 2 + 2hxy + y2 = 0 represents the equations of the straight lines through the origin which make an angle a with the straight line y + x = 0, then (a) sec 2 a = h (c) 2 sin a =

1+ h h

(b) cos a =

1+ h 2h

(d) cot a =

h +1 h -1

Ans. (a), (b) and (d) Solution: Let equation of the lines given by x2 + 2hxy + y2 = 0 be y = m1x and y = m2x. Since these make an angle a with y + x = 0 whose slope is –1,

IIT JEE eBooks: www.crackjee.xyz Straight Lines 15.21

MATRIX-MATCH TYPE QUESTIONS

m1 + 1 -1 - m2 = tan a = 1 - m1 1 - m2 fi fi

m1 + m2 =

( tan a - 1)2 + ( tan a + 1)2 tan a - 1 2

= fi

sec2 a = h



cos2 a =



1 tan a + 1 and m2 = 1 tan a - 1

tan m1 = tan

cos a =

2sec a ¥ cos2 a = – 2 sec 2a = – 2h. cos 2a 2

1 h



2cos2 a – 1 =

1+ h and cot a = 2h

1 h

h +1 h -1

Example 55 If y = mx bisects the angle between the lines x2 (tan2 q + cos2 q ) + 2xy tan q – y2 sin2 q = 0 when q = p/3 the value of m is (a)

-2 - 7

7 -2

(b)

3

(c) 2 7

3

(d) 2 3

Ans. (a) and (b) Solution: Equation of the bisectors of the angles between the given lines is xy x2 - y2 = 2 2 2 tan q tan q + cos q + sin q x -y 2

1 + tan q 2

=

xy tan q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

3 m2 + 4m –

m=

-2 ± 7 3

3 = 0.

(s)

5 7 0)

length of the line segment AB intercepted between the coordinates axes 2 2 c a2 + b2 Ê cˆ Ê cˆ = + ÁË ˜¯ ÁË ˜¯ ab a b area of the triangle formed by the line and the coordinates axes = c 2/2ab.

=

Example 59 The vertices of triangle ABC are A(1, - 2), B(-7, 6) and C(11/5, 2/5). Column 1 Column 2 (a) Equation of the perpend- (p) 26x + 17y + 8 = 0 icular bisector of AB

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(b) Equation of the median through A (c) Equation of altitude through C (d) Equation of BC

(q) 14x + 23y - 40 = 0 (r) x - y + 5 = 0 (s) 5x - 5y - 9 = 0

p

q

r

s

a

p

q

r

s

a

p

q

r

s

b

p

q

r

s

b

p

q

r

s

c

p

q

r

s

c

p

q

r

s

d

p

q

r

s

d

p

q

r

s

Ans.

Solution:

Ê 1 - 7 -2 + 6 ˆ , Middle point of AB is Á ˜ = (- 3, 2) and Ë 2 2 ¯ 6+2 slope of AB is =-1 -7 - 1 (a) Equation of the perpendicular bisector of AB is y - 2 = 1 ◊ (x + 3) fi x - y + 5 = 0 Ê -7 + (11 / 5) (2 / 5) + 6 ˆ Middle point of BC is Á ˜¯ Ë 2 2 Ê 12 16 ˆ = Á- , ˜ Ë 5 5¯ y+2 x -1 y+2 x -1 = fi = (16 / 5) + 2 ( -12 / 5) - 1 26 -17 26x + 17y + 8 = 0. Slope of AB = - 1

Solution: Solving the equation 2m3 - 3m2 - 3m + 2 = 0 we get 2(m3 + 1) - 3m(m + 1) = 0 fi (m + 1) (2m2 - 5m + 2) = 0 fi (m + 1) (2m - 1) (m - 2) = 0 fi m = - 1, 1/2 or 2. Equation of the given lines can be written as m2ix - miy = - 1 (a) algebraic sum of the intercepts made by the lines on x-axis 21 È 1 ˘ 1 = - Í1 + + 4 ˙ = = -Â 4 mi Î 4 ˚ (b) algebraic sum of the intercepts made by the line on y-axis = Â

(b) Equation of the median through A is



lines from the origin. (d) sum of the lengths of the lines (s) (5 2 + 9 5) / 10 intercepted between the coordinate axes. p q r s Ans.

(c) Let pi denote the perpendicular distance of the line from the origin. - 1 / mi

then pi =

1+

(from (a))

Equation of the altitude through C is y - 2/5 = 1(x - 11/5) fi 5x - 5y - 9 = 0 (d) Equation of BC is y -2/5 x - 11 / 5 5y - 2 5 x - 11 = fi = 6-2/5 -7 - 11 / 5 - 46 28

=

(c)

1 (i = 1, 2, 3) represent mi three straight lines whose slopes are the roots of the equation 2m3 - 3m2 - 3m + 2 = 0, then Column 1 Column 2 (a) algebraic sum of the intercepts (p) (4 2 + 9 5) / 4 made by the lines on x-axis. (b) algebraic sum of the intercepts (q) 3/2 made by the lines on y-axis (c) sum of the distances of the (r) -21/4 If y = mi x +

mi2

fi  pi =

1 2 1/ 2 + + 1+1 1 + (1 / 4) 1+ 4

1 4 1 1 = + + (5 2 + 9 5) 10 2 5 2 5

(d) li = length of the line intercepted between the coordinates axes 2

=

fi 14x + 23y - 40 = 0 Example 60

1 = - 1 + 2 + 1/2 = 3/2. mi



Ê 1 ˆ Ê 1ˆ Á 2 ˜ + ÁË m ˜¯ Ëm i¯ i

2

 li = 1 + 1 + 16 + 4 + 1 / 16 + 1 / 4 =

2 + 2 5 + 5 / 4 = ( 4 2 + 9 5 )/4 x + y + 1 in the line

Example 61 (a) (b) (c) (d)

Column 1 2x + y + 1 x - 2y + 1 x + 2y - 1 2x + y - 1

= = = =

0 0 0 0

(p) (q) (r) (s)

Column 2 x + 7y - 11 = 0 7x + y + 1 = 0 7x + y - 11 = 0 7x + y + 7 = 0

IIT JEE eBooks: www.crackjee.xyz 15.24 Comprehensive Mathematics—JEE Advanced

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

(d) incentre of the triangle OAB (s) 2x + y = 4 is (1, 1) p q r s Ans.

x+y+1 in the line lx + my + n = 0 be x + y + 1 + l (lx + my + n) = 0 Then (ii) equally inclined to (i) and (iii), l ( l l + 1) l - + -1+ lm + 1 m m = fi ll + 1 l l 1+ 1+ ¥ m m lm + 1 fi

(i) (ii) (iii)

m-l l-m -2(l + m) = fil= 2 2 2 l+m l (l + m ) + (l + m) l + m2

x + y + 1 = 0 in the line lx + my + n = 0 is (x + y + 1) (l2 + m2) - 2(l + m) (lx + my + n) = 0. (a) Taking l = 2, m = n = 1 we get the equation of the required line as (x + y + 1) (4 + 1) - 2(2 + 1) (2x + y + 1) = 0 or 7x + y + 1 = 0 (b) Taking l = 1; m = - 2, n = 1; the equation of the required line is (x + y + 1) (1 + 4) - 2(1 - 2) (x - 2y + 1) = 0 fi 7x + y + 7 = 0 (c) Taking l = 1, m = 2, n = - 1, the equation of the required line is (x + y + 1) (1 + 4) - 2(1 + 2) (x + 2y - 1) = 0 fi x + 7y - 11 = 0. (d) Taking l = 2, m = 1, n = - 1, the equation of the required line is (x + y + 1) (4 + 1) - 2(2 + 1) (2x + y - 1) = 0 fi 7x + y - 11 = 0 Example 62

A line meets X-axis at A and Y-axis at B,

O is the origin such that Column 1

Column 2 Equation of AB (p) 2x + y = 2

(a) centroid of the triangle OAB is (1, 2) (b) circumcentre of the triangle (q) 3x + 4y = 12 OAB is (1, 2) (c) distance of the orthocentre of (r) 2x + y = 6 the triangle OAB from A and B is 1 and 2 respectively

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: Let OA = a and OB = b then coordinate of A are (a, 0) and of B are (0, b) (a) Centroid of the triangle OAB Ê 0 + a + 0 0 + 0 + bˆ , = Á ˜¯ = (1, 2) fi a = 3, b = 6 Ë 3 3 x y + = 1 fi 2x + y = 6. 3 6 (b) Circumcentre of the triangle OAB is the middle point of the hypotenuse AB

so that equation of AB is



Ê a bˆ ÁË , ˜¯ = (1, 2) fi a = 2, b = 4 2 2



Ê x yˆ Equation of AB is Á + ˜ = 1 fi 2x + y = 4. Ë 2 4¯

(c) Orthocentre of the triangle OAB is O so that OA = 1 and OB = 2 fi a = 1 and b = 2 Ê x yˆ fi Equation of AB is Á + ˜ = 1 fi 2x + y = 2. Ë 1 2¯ (d) Distance of (1, 1) from OA or OB is 1 Since (1, 1) is the incentre of the triangle OAB distance of (1, 1) from AB is also 1. The equation of AB conditions as 3 ¥ 1 + 4 ¥ 1 - 12 (3)2 + (4)2

=1

Example 63 Column 1 (a) An integer value of m for which the x-coordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx +1is also an integer is (b) The line 2x - 3y - 6 = 0 meets the coordinate axes at (a, 0) and (0, b). If the line y = mx + m2 passes through (a, b), then

Column 2 (p) m = -1

(q) m = -2

IIT JEE eBooks: www.crackjee.xyz Straight Lines 15.25

(r) m = 4

(c) If the point (3, 4) lies on the locus of the point of intersection of the lines x cos a + y sin a = m and x sin a - y cos a = n such that 3m - 4n = 0, then (d) If the line y = mx - 8/m meets the curve y2 = 4x at a point where the curve meets the line y = 2, then p q r s Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(s) m = - 4

5 3 + 4m Since x is an integer 3 + 4m = ± 1 or ± 5 fi m = - 1 or m = - 2 (b) We get a = 3, b = – 2, y = mx + m2 passes through (3, - 2) fi - 2 = 3m + m2 fi m2 + 3m + 2 = 0 fi m = - 1 or - 2 (c) Squaring and adding we get the locus as x2 + y2 = m2 + n2, the point (3, 4) lies on it 9 so 9 + 16 = m2 + m2 (as 3m = 4n) 16 fi (d) y = fi fi

(a) (b) (c) (d) Ans.

x=

25 ¥ 16 = 25m2 fi m = ± 4 mx - 8/m meets the curve y2 = 4x at (1, 2) 2 = m - 8/m fi m2 - 2m - 8 = 0 m = 4 or m = - 2 The lines y = ± x meets

Example 64

Column 1 2x + 3y - 15 = 0 at x - 5y + 12 = 0 at x2 = 4y - 4 at x2 + y2 - x - 5y = 0 at p q r s a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

= x, 5x = 15 fi x = 3, y = 3 = - x, y = 15, x = - 15 = x, - 4y + 12 = 0 fi y = 3, x = 3 = - x, 6x + 12 = 0 fi x = - 2, y = 2 = x, x2 - 4x + 4 = 0 fi x = 2, y = 2 = - x, x2 + 4x + 4 = 0 fi x = - 2, y = 2 = x, 2x2 - 6x = 0 fi x = 0 or x = 3, y = 3 = - x, 2x2 + 4x = 0 fi x = 0 or x = - 2, y = 2

Example 65

Solution: (a) 3x + 4 (mx + 1) = 9 fi

Solution (a) If y If y (b) If y If y (c) If y If y (d) If y If y

Column 2 (p) (3, 3) (q) (-15, 15) (r) (-2, 2) (s) (2, 2)

Consider the lines given by

L1 : x + 3y – 5 = 0, L2 : 3x – ky – 1 = 0 L3 : 5x + 2y – 12 = 0. Column 1 Column 2 (p) k = – 9 (a) L1, L2, L3 are concurrent if (q) k = – 6/5 (b) one of L1, L2, L3 is parallel to at least one of the other two, if (r) k = 5/6 (c) L1, L2, L3 form a triangle, if (d) L1, L2, L3 do not form a triangle if (s) k = 5 Ans.

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: L1 and L3 intersect at (2, 1). L1, L2, L3 are concurrent if L2 passes through (2, 1) that is if k = 5. Note that L1 is not parallel L2 So L2 is parallel to L1 if 3/K = – 3 fi K = – 9 and L2 is parallel to L3 if 3/K = – 5/2 fi k = – 6/5. The lines L1, L2, L3 will form a triangle if no two of them are parallel and they are not concurrent. Thus k = 5/6. Lines L1, L2, L3 will not form a triangle if two of them are parallel or all of them are concurrent. Example 66

Distance between the lines

Column 1 (a) 4x2 + 12xy + 9y2 – 16x – 24y + 7 = 0 (b) 9x2 + 24xy + 16y2 + 12x + 16y – 5 = 0 (c) x2 – 2xy + y2 – 4x + 4y – 21 = 0 (d) 16x 2 – 24xy + 9y2 + 16x – 12y – 5 = 0

Column 2 (p) 6/5 (q) 6

13

(r) 5 2 (s) 3/5

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p

q

r

s

a

p

q

r

s

s

b

p

q

r

s

r

s

c

p

q

r

s

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

c

p

q

d

p

q

Ans.

Ans.

Solution (a) 2

64 - 28 = 4 (4 + 9)

6 13

Ê g 2 - ac ˆ Á Using 2 ˜ ÁË a ( a + b) ˜¯

Solution: (a)

2

(5 2)2 - (6)( -4) 6-4

36 + 45 6 (b) 2 = 9(9 + 16) 5 (c) 2 (d) 2

(b)

64 + 80 6 = 5 16(9 + 16)

Example 67 Equation of the pair of lines joining the origin to the points of intersection of the line and the conic. Column 1 Column 2 2 2 (p) 4x2 – 9xy + 4y2 = 0 (a) x + y = 5, 2x + 3y – 1 = 0 (q) 19x2 + 60xy + 44y2 (b) x 2 + y2 – 2x + 2y + 1 = 0, x + y + 1 =0 (c) xy = 4, x – y = 1 (r) xy + 2y2 = 0 (s) 4x 2 – y 2 = 0 (d) y2 = 4ax, x = a p q r s Ans.

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(c)

(a)

(b)

(d)

Tangent of the angle between the lines

Column 1 (a) 6x2 + 5xy – 4y2 = 0

Column 2 (p) 11/2

(b) 2x2 + 5xy + 3y2 = 0

(q)

(c) x2 + 4xy + y2 = 0

(r) 1/5

(d) 4x2 + 5xy – 6y2 = 0

(s) 11/5

3

2

(5 2)2 - (2)(3) 2+3

2 4 -1 = 1+1 those in (a)

=

1 5

3 , lines in (d) are perpendicular to

Example 69

(c)

Solution (a) x 2 + y2 = 5 (2x + 3y)2 fi 19x2 + 60xy + 44y2 = 0 (b) x 2 + y2 – 2 (x – y) (x + y) + (x + y)2 = 0 fi xy + 2y2 = 0 (c) xy = 4 (x – y)2 fi 4x2 – 9xy + 4y2 = 0 (d) y2 = 4 a ¥ x/a fi 4x2 – y2 = 0 Example 68

11 2 Ê 2 h2 - ab ˆ Á Using tan q = ˜ ÁË a + b ˜¯

4 + 21 = 5 2 1.(1 + 1)

a

=

Ans.

Column 1 If the slope of one of the lines represented by ax2 - 6xy + y2 = 0 is square of the other then If 9x2 + 2(a - 2) xy + 4y2 + 6x + 4y - 3 represent two parallel lines, then If a denotes the slope of a line given by 108x2 + 31xy + y2 = 0, then If a2x2 + 15xy - 16y2 represent a pair of two perpendicular lines then p q r s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution (a) m + m2 = 6, m3 = a fi and a = 8 or - 27.

Column 2 (p) a = -4

(q) a = 8

(r) a = -27

(s) a = 4

m = 2 or -3

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(b) The lines are parallel if 9 ¥ 4 = (a _ 2)2 fi a - 2 = ± 6 fi a = 8 or a = - 4 But the equation represents a pair of lines if 9 a-2 3 a-2 4 2 = 0 or if a = ± 8 3

(b) The lines are neither parallel nor perpendicular meet x-axis at 30x2 - 35x + 5 = 0 fi x = 1 or x = 1/6 so the intercept on x-axis = 1 – 1/6 = 5/6◊ Similarly intercept on y-axis = 1 – 5/6 = 1/6◊ (c) As in (b). (d) The lines are parallel as 6 ¥ 6 = (6)2 (h2 = ab) meet x-axis at 6x2 - 7x + 1 = 0 fi x = 1 or x = 1/6 so intercept on x-axis = 1 – 1/6 = 5/6◊ intercept on y-axis is also 5/6◊

-3

2

So we get a = 8 (c) We have (27x + y) (4x + y) = 0 Slopes are - 27 or - 4 i.e. a = - 27 or a=-4 (d) Lines are perpendicular if a2 - 16 = 0 fi a=±4

ASSERTION-REASON TYPE QUESTIONS Example 71 Statement-1: If the circumcentre of a triangle lies at the origin and centroid is the middle point of the line joining the points (2, 3) and (4, 7), then its orthocentre lies on the line 5x – 3y = 0. Statement-2: The circumcentre, centroid and the orthocentre of a triangle lie on the same line.

Example 70

(a) (b) (c)

(d)

The lines given by Column 1 6x2 + 5xy - 6y2 - x + 5y - 1 = 0 30x2 + 36xy + 6y2 - 35x - 11y + 5 = 0 30x2 + 41xy + 6y2 – 35x cept - 11y + 5 = 0 6x2 + 12xy + 6y2 - 7x - 7y + 1 = 0

Column 2 (p) are parallel

Ans. (a) (q) are perpendicular

Solution:

(r) make an

joining (0, 0) and (3, 5) i.e. 5x – 3y = 0

inter-

5/6 on x-axis (s) make an intercept 1/6 on y-axis.

Statement-2 is correct from Geometry and

Example 72 Statement-1: L1 : 3x + 4y = 16, L2 : 3x + 4y = 12, L3 : 3x + 4y + 4 = 0 are three parallel straight lines. A line making an angle q with them meets them PQ 1 at P, Q, R respectively, then = . QR 4

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

Statement-2: If a line meets three parallel lines at A, B AB is independent of the slope of the line. and C, then BC Ans. (a)

d

p

q

r

s

Solution:

Ans.

Solution

A

L1 2

2

y =0 a fi line are perpendicular Meet x-axis at 6x2 - x - 1 = 0 fi (2x - 1) (3x + 1) = 0 fi x = 1/2 or x = - 1/3 fi intercept on x-axis =

Statement-2 is true from geometry

1 1 5 + = 2 3 6

Meet y-axis at - 6y2 + 5y - 1 = 0 fi (2y - 1) (3y - 1) = 0 fi y = 1/2 or y = 1/3 fi intercept on y-axis = 1/2 – 1/3 = 1/6◊

B

L2 L3

L

C M Fig. 15.23

Let a line meet three parallel lines L1, L2, L3 at A, B, C, respectively. AB AL Draw ALM perpendicular, then = . BC LM =

Distance betwwen L1 and L2 , which is constant Distance between L2 and L3

Using this result in statement-1.

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Distance betwwen L1 and L2 PQ = Distance between L2 and L3 QR =

| 16 - 12 | 1 = | 12 + 4 | 4

And, the statement-1 also true. Example 73 The line x = a meets the curve y2 = 4ax at P and Q. If PQ subtends an angle q at the origin O, then tan q =

-4 . 3

Statement-2: The curve y2 = 4ax is symmetrical about y-axis. Ans. (c) P(a, 2a)

Solution:

Statement-1 is false. Since

x – 3 + 2x – 7 + 10 – 3x = 0, but the lines x – 3 = 0, 2x – 7 = 0, 10 – 3x = 0 are parallel. Statement-2 is correct. Example 75 Lines L1: y – x = 0 and L2: 2x + y = 0 intersect the line L3: y + 2 = 0 at P and Q respectively. The bisector of the acute angle between L1 and L2 intersect at R. Statement-1: The ratio PR : RQ equals 2 2 : 5 Statement-2: In any triangle bisector of an angle of divides the triangle into two similar triangles. Ans. (c) Solution: The vertices of the triangle formed by the lines L1, L2, L 3 are O (0, 0), P(– 2, – 2) and Q(1, – 2). Since the bisector of an angle of a triangle divides the opposite side in the ratio of the side containing the angle, we have PR : RQ = OP : OQ = 2 2 : 5 .

O

S(a, 0)

X

Q (a, -2a)

So statement-1 is correct. But the Statement-2 is false. For example if A = 50°, B = 60°, C = 70°. Showing that bisector of angle B does not divided the triangle into two similar triangles.

Fig. 15.24

Sol.

Coordinates of P are (a, 2a) and of Q are (a, –2a)

So OPQ is an isosceles triangle with OP = OQ and S(a, 0) is the mid-point of PQ POQ = q fi POS =

q 2

2a Êqˆ tan Á ˜ = Slope of OP = =2 Ë 2¯ a Êqˆ 2 tan Á ˜ Ë 2¯ 4 tan q = = - , so statement-1 true. 3 Êqˆ 1 - tan 2 Á ˜ Ë 2¯ But the Statement-2 is false, the curve is symmetrical about x-axis. Statement-1: If (a1 x + b1 y + c1) Example 74 + (a2 x + b2 y + c2 ) + (a3 x + b3 y + c3) = 0, then the lines a1 x + b1 y + c1 = 0, a2 x + b2 y + c2 = 0 and a 3 x + b3 y + c3 = 0 can not be parallel. Statement-2: If L1 = 0, L2 = 0, L3 = 0 be three lines and there exist l, m, n; not all zero, such than lL1 + mL2 + n L3 = 0 then the lines L1 = 0, L2 = 0 and L3 = 0 are either parallel or concurrent. Ans. (d)

Fig. 15.22

Statement-1: If the perpendicular Example 76 bisector of the line segment joining P(1, 4) and Q(K, 3) has y-intercept – 4, then K2 – 16 = 0 Statement-2: Locus of a point equidistant from two given points is the perpendicular bisector of the line joining the given points. Ans. (a) Solution: fi

Statement-2 is True, using in statement-1 (x – 1)2 + (y – 4)2 = (x – K)2 + (y – 3)2 2(K + 1)x – 2y = K 2 – 8

K2 – 8 =–4 2 showing that statement-1 is True. y-intercept = –



K 2 – 16 = 0

Statement-1: Circumcentre of the Example 77 triangle represented by the lines x + y = 0, x – y = 0 and x – 7 = 0 is (7, 0)

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Statement-2: Circumcentre of a triangle lies inside the triangle. Ans. (c) Solution: Statement-2 is false as the circumcentre of an obtuse-angled triangle lies outside the triangle and that of a right angled triangle is the mid-point of the hypotenuse. For statement-1, the triangle is right angled, the vertices of the hypotenuse x – 7 = 0 are (7, 7) and (7, – 7) and the circumcentre is (7, 0). So it is true. Example 78

Statement-1: Slope of the line 3 –1

(cos q + sin q)x + sin 2q y = 1 is

if q = p /6 3 Statement-2: 3x + y = 30 represents the line making an angle p /6 with the + ve direction of x-axis at a distance 15 units from the origin. Ans. (d) Solution:

=

Slope of line in statement-1 is Ê 3 1ˆ –Á + ˜ 2¯ Ë 2

=



(

3 1 ¥ 2¥ 2 2 So statement-1 is False. In statement-2, we can write

)

3 +1 3

– ( cos q + sin q ) 2sin q cos q

for q = p /6

Example 80 PQR is a right angled isosceles triangle, right angled at P(2, 1). Equation of the line QR is 2x + y = 3. Statement-1: Equation representing the pair of lines PQ and PR is 3(x – 2)2 – 8(x – 2) (y – 1) – 3(y – 1) = 0. Statement-2: If the origin is shifted to the point P(2, 1), then equation representing PQ and PR in the new coordinate system is xy = 0. Ans. (c) Solution: PQ, PR subtend equal angle p /4 with the hypotenuse QR. Let the slope of PQ be m. then as the slope of QR is – 2, m+2 = ± 1 fi 3m2 – 8m – 3 = 0 1 – 2m fi 3(y – 1)2 – 8(y – 1) (x – 2) – 3(x – 2)2 = 0, as the equation of PQ is y – 1 = m(x – 2) which shows that the statement-1 is true. If the origin is shifted to the point P(2, 1). Equation of the lines PQ and PR in the new system is 3y2 – 8xy – 3x2 = 0 so the statement-2 is false.

COMPREHENSION-TYPE QUESTIONS 3 1 x + y = 15 2 2

Paragraph for Question Nos. 81 to 83 A(x1, y1), B(x2, y2), C(x3, y3) are the vertices of a triangle ABC. lx + my + n = 0 is an equation of the line L.

or x cos p /6 + y sin p/6 = 15 i.e. x cos a + y sin a = p where a = p /6, p = 15 So statement-2 True. 2

Example 79 Statement-1: A chord of the curve 3x – y2 – 2x + 4y = 0 passing through the point (1, – 2) subtend a right angle at the origin. Statement-2: Lines represented by the equation (3c + 2m) x2 – 2 (1 + 2m) xy + (4 – c)y2 = 0 are perpendicular if c + m + 2 = 0 Ans. (a) Solution: of x2 and y2 = 3c + 2m + 4 – c = 0 fi c + m + 2 = 0 so the lines are perpendicular if c + m + 2 = 0. In statement-1, let the equation of the chord by y = mx + c then equation of the pair of lines joining the origin to the points of intersection of the chord and the curve is Ê y – mx ˆ Ê y – mx ˆ + 4y Á =0 3x2 – y2 – 2x Á Ë c ˜¯ Ë c ˜¯ fi

which are at right angles if c + m + 2 = 0 (using statement2) and since the line y = mx + c passes through (1, – 2) c + m + 2 = 0. So statement-1 is also true.

(3c + 2m)x2 – 2(1 + 2m)xy + (4 – c)y2 = 0

Example 81 If L intersects the sides BC, CA and AB of the triangle ABC at P, Q, R respectively then BP CQ AR ¥ ¥ is equal to PC QA RB (a) - 1 (c) 1/2

(b) - 1/2 (d) 1

If the centroid of the triangle ABC Example 82 is at the origin and algebraic sum of the lengths of the perpendiculars from the vertices of the triangle ABC on the line L is equal to 1 then sum of the squares of the intercepts made by L on the coordinate axes is equal to (a) 0 (b) 4 (c) 9 (d) 16 Example 83 If P divides BC in the ratio 2 : 1 and Q divides CA in the ratio 1 : 3 then R divides AB in the ratio (P,Q,R are the points as in Ex. 87)

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(a) 2 : 3 internally (c) 3 : 2 internally

(b) 2 : 3 enternally (d) 3 : 2 enternally

Ans. 81. (a), 82. (c), 83. (d) Solution: If P dives BC in the ratio l : 1, then the coordinates of P are Ê l x3 + x2 l y3 + y2 ˆ ÁË l + 1 , l + 1 ˜¯

(c) (0, 3y1/(3 - x1))

(d) (0, 6y1/(6 - x1))

(c) (1, 0)

l x + my2 + n BP - 2 =l= PC l x3 + my3 + n

(d) (1, 1)

Ans. 84. (b), 85 (a), (b)

Ê l x + x2 ˆ Ê l y + y2 ˆ l Á 3 +m Á 3 +n=0 ˜ l + 1 Ë ¯ Ë l + 1 ˜¯

Solution:

Equation of AU is 0 - y1 (x – x1) 3 - x1

y – y1 =

(1)

(0, 3 y1 (3 - x1 ) ) Similarly, the coordinates of D are ( 0, 6 y1 (6 - x1 ) )

so that the coordinates of C are

Similarly, we obtain l x + my3 + n CQ = - 3 l x1 + my1 + n QA and

Ê 6 x1 6 y1 ˆ , (b) Á Ë 6 + x1 6 + x1 ˜¯

Example 85 For all positions of U in the plane, CV passes through

Also, as P lies on L, we have



Ê 6 x1 6 y1 ˆ , (a) Á Ë 6 + x1 6 + y1 ˜¯

(2)

l x + my1 + n AR = - 1 RB l x2 + my2 + n

Now, equation of AD is x y(6 - x1 ) + =1 3 6 y1

(3)

(1)

Multiplying (1), (2) and (3) we get BP CQ AR ¥ ¥ =-1 PC QA RB x + x2 + x3 y + y2 + y3 In Ex. 57. 1 = 0, 1 =0 (4) 3 3 l x + my1 + n l x2 + my2 + n l x3 + my3 + n and 1 =1 + + l 2 + m2 l 2 + m2 l 2 + m2 fi l(x1 + x2 + x3) + m(y1 + y2 + y3) + 3n = l 2 + m2 fi

l2 + m2 = 9n2 (from (4))

Fig. 15.21

sum of the squares of the intercept made by L on the 2

2

Ê l ˆ Ê mˆ l +m = 9. coordinate axis is Á ˜ + Á ˜ = Ë - n¯ Ë - n¯ n2 2

2

and equation of OU is y1x = x 1y

Solving (1) and (2), we get x1 y y(6 - x1 ) + = 1 fi y (2 x1 + 6 – x1) = 6y1 3 y1 6 y1

BP CQ 1 = = 2, In Ex. 58. QA 3 PC fi

AR 3 = RB 2



R divides AB externally in the ratio 3 : 2

Paragraph for Question Nos. 84 and 85 A(3, 0) and B U (x1, y1) is a variable point of the plane. AU and BU meet the Y-axis at C and D respectively. AD meets OU at V. Example 84

The coordinates of V are

(2)



y=

6 y1 6 + x1

Hence the coordinates of V are Therefore, equation of CV is



x=

6 x1 6 + x1

Ê 6 x1 6 y1 ˆ ÁË 6 + x , 6 + x ˜¯ 1 1

6 y1 3 y1 3 y1 6 + x1 3 – x1 y– = ( x - 0) 6 x1 3 - x1 -0 6 + x1

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fi y=

xˆ Ê ÁË1 - ˜¯ 2

3 y1 9 x1 y1 3 y1 x fi y= 3 - x1 6 x1 (3 - x1 ) 3 - x1

or

40 = 3k – 3h

Locus of P(h, k) is 3x – 3y + 40 = 0

which passes through the point (2, 0) for all values of (x1, y1).

87. If (OP)2 = (OA) ¥ (OB)

Paragraph for Question Nos. 86 to 88



r2 =

L1 : x – y + 10 = 0 L2 : x – y + 20 = 0



(r sin q – r cos q)2 = 200

L is a variable line through the origin O meeting L1 at A and L2 at B. P is a point on L



(k – h)2 = 200

2 1 1 = , then the locus of P is + Example 86 If OP OA OB (a) 3x + 3y = 40 (b) 3x + 3y + 40 = 0 (c) 3x – 3y = 40 (d) 3x – 3y + 40 = 0 Example 87

If (OP) = (OA) ¥ (OB), then locus of P is 2

(a) (x – y)2 = 100 (c) (x – y)2 = 200 Example 88 of P is

If

1 (OP)

2

(b) (x + y)2 = 50 (d) (x + y)2 = 100 =

1 (OA)

2

+

1 (OB)2

, then locus

Locus of P(h, k) is (x – y)2 = 200. 88. If fi

1 (OP) 1 r2

2

=

1

=

(OA)

2

(a) (x – y) = 80 (c) (x – y)2 = 64

2

(b) (x – y) = 100 (d) (x – y)2 = 400

Ans. 86. (d) 87. (c) 88. (a)

(OB)2

(sin q - cos q )2 (sin q - cos q )2 + 100 400

400 = 5(r sin q – r cos q)2



400 = 5 (k – h)2

Locus of P(h, k) is (x – y)2 = 80.

L1 : 3x + 2y + l = 0, L2 : lx – y + 1 = 0 Example 89 intersect at

Let equation of L be

Substituting in L1, r (cos q – sin q) + 10 = 0 10 = OA sin q - cos q



r=



1 sin q - cos q = 10 OA

If L1 and L2 are perpendicular, they

Ê 8 -23 ˆ (a) Á - , Ë 13 39 ˜¯

Ê 8 -23 ˆ (b) Á , Ë 13 39 ˜¯

Ê 8 23 ˆ (c) Á - , ˜ Ë 13 39 ¯

Ê 8 -23 ˆ (d) Á - , Ë 13 13 ˜¯

Example 90 If L1 and L2 intersect on the line 2x + 3y = 0, coordinates of the point of intersection are

Substituting in L2, we get 1 sin q - cos q = 20 OB

Ê -3 2 ˆ (a) Á , ˜ Ë 5 5¯

2ˆ Ê (b) Á -1, ˜ Ë 3¯

Ê -2 ˆ (c) Á1, ˜ Ë 3¯

Ê 3 -2 ˆ (d) Á , ˜ Ë5 5 ¯

Let the coordinates of P be (r cos q, r sin q) = (h, k)

Ans. 89. (a) 90. (a), (c)



Solution:

OP = r

2 = OP 2 fi = r fi 40 =

1



x y = = r or x cosq sinq = r cos q, y = r sin q are the parametric coordinates of any point on L at a distance r from O.

86.

+

Paragraph for Question Nos. 89 and 90. 2

Solution.

10 20 ¥ sin q - cos q sin q - cos q

1 1 + OA OB sin q - cos q sin q - cos q + 10 20 3r sin q – 3r cos q



L1 and L2 are perpendicular

2 Ê 3ˆ ÁË - ˜¯ l = –1 fi l = 2 3

Equation of L1 is 9x + 6y + 2 = 0 and of L2 is 2x – 3y + 3 = 0 Ê 8 -23 ˆ Solving we get their point of intersection as Á - , Ë 13 39 ˜¯

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Ê 3 - l2 ˆ Ê 2+l ˆ 2Á + 3Á ˜ ˜ =1 Ë 3 + 2l ¯ Ë 3 + 2l ¯

Solution: L 1: 6x2 + xy – y2 = 0 fi (3x – y) (2x + y) = 0. If 2x + y coincides with L 2: 3x 2 – a xy + y 2 = 0 then 3 + 2a + 4 = 0 fi a = – 7/2, which is not possible as a > 0. So if y = 3x coincides with L2 then 3 – 3a + 9 = 0 fi a = 4. L 2: 3x2 – 4xy + y2 = 0 fi (3x – y) (x – y) = 0 and L3: (2x + y) (x – y) = 0



3l2 + 2l – 5 = 0

p1 = ±



l = 1, -

L1 and L2 intersect on 2x + 3y = 0 fi The point of intersection of L1 and L2 lies on 2x + 3y = 0

5 3



Ê -3 2 ˆ So the coordinates of the point are Á , ˜ for l = 1 Ë 5 5¯ Ê -2 ˆ 5 and ÁË1, ˜¯ for l = - . 3 3 Paragraph for Question Nos. 91 to 95 L1: 6x 2 + xy – y 2 = 0,

L2: 3x 2 – axy – y2 = 0, a > 0.

Example 91 If one of the lines given by L1 coincides with one of the lines given by L2 , then the value of a is equal to (a) – 7/2 (b) 1/2 (c) 1 (d) 4 Example 92 If L 3 denotes the equation of the pair of other two lines of L1 and L2; p1, p2 denote the lengths of the perpendiculars from the point (8, 4) on the lines given by L3, then |p12 – p22| is equal to (a) 8 (b) 32 (c) 40 (d) none of these Example 93 The centroid of the triangle formed by the pair of lines given by L 3 and x = 1 is (a) (2/3, – 2/3) (b) (2/3, – 1) (c) (1/3, – 1/3) (d) (2/3, – 1/3) Example 94 Coordinates of the points of intersection of the lines given by L 1, L2 and L3 with x 2 + y2 = 5 are (a) (± 1, (c)



5 2,

2)

(b) 52

)

(d)

( ±1 2 , ± 3 2 ) (± 5 2, ± 5 2 )

Example 95 Equation of the pair of straight lines passing through (1, –2) perpendicular to the lines L2 is (a) x 2 + 4xy + 3y2 + 6x + 8y + 5 = 0 (b) x 2 – 4xy + 3y2 + 6x + 8y + 5 = 0 (c) 3x 2 + 4xy + y2 + 6x + 8y + 5 = 0 (d) 3x 2 – 4xy + y2 – 14x + 8y + 15 = 0 Ans.

91. (d) 92. (d) 93. (d) 94. (a), (b), (d) 95. (a)

16 + 4 8-4 = ±4 5 , p 2 = ± = ±2 2 5 2

|p12 – p22| = 80 – 8 = 72

Vertices of the triangle formed by the lines L3 and x = 1 are (0, 0), (1, – 2), (1, 1), so the coordinates of the centroid of the triangle are (2/3, – 1/3). The circle x 2 + y2 = 5 meets the line 3x – y = 0 at ± 1 2 , ± 3 2 the line 2x + y = 0 at ( ±1, 2) and the

(

) line x – y = 0 at ( ±

)

5 2, ± 5 2 .

Let the equations of the lines perpendicular to the lines in L 2 be x + 3y + C1 = 0, x + y + C2 = 0. Since they pass through (1, –2), C1 = 5, C2 = 1. Joint equation of these lines is (x + 3y + 5) (x + y + 1) = 0 or x2 + 4xy + 3y2 + 6x + 8y + 5 = 0

INTEGER-ANSWER TYPE QUESTIONS Example 96 The straight line L ∫ x + y + 1 = 0 and L1 ∫ x + 2y + 3 = 0 are intersecting. m is the slope of the straight line L2 such that L is the bisector of the angle between L1 and L2. The value of m2 + 1 is equal to Ans. 5 Solution: Let the equation of L2 be L1 + lL = 0 fi (1 + l) x + (2 + l) y + 3 + l = 0 Slopes of L2, L and L1 are 1+ l , - 1, - 1/2 2+l Since L is the bisector of the angle between L1 and L2 1+ l +1 2+l -1+1/ 2 = 1+ l 1+1/ 2 1+ 2+l

\



l=-3

So the equation of L2 is y + 2x = 0 fi m = -2 and m2 +1 = 5 Example 97 If the coordinates of the orthocentre of the triangle formed by the lines y = 0, 37x - 36y + 37 ¥ 36 = 0 and 64x - 63y + 64 ¥ 63 = 0 is (a, b) then a + b is equal to Ans. 0 Solution: Let the equation of BC be y = 0, then the coordinates of B are (- 36, 0) and of C are (- 63, 0). Where

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equation of AB is 37x - 36y + 37 ¥ 36 = 0 and of AC is 64x - 63y + 64 ¥ 63 = 0. Equation of the altitudes through B and C are - 63 - 36 (x + 36), y = (x + 63) 64 37 Solving these equations we get the coordinates of the orthocentre of the triangle ABC so a = 36 ¥ 63, b = - 36 ¥ 63 fia+b =0 y=

Example 98

One diagonal of a square is the portion

x y + = 1 intercepted between the axes. p1, 97 79 p2 are the lengths of the perpendiculars from the vertices of the other diagonal on the axis of y then p1 + p 2 – 90 is equal to Ans. 7 x y + Solution: Let AB be the line whose equation is 97 79 = 1 where A is (97, 0) and B is (0, 79). Equation of the other Ê 97 79 ˆ diagonal passing through Á , ˜ , the mid-point of AB Ë 2 2¯

Example 99 A straight line passing through the point (87, 33) cuts the positive direction of the coordinate axes at the points P and Q. If O is the origin and the minimum A area of the triangle OPQ is A, then 957 Ans. 6 Solution: Equation of a line through (87, 33) is y - 33 = m(x - 87) which meets y-axis at P(0, 33 - 87m) and x-axis 33 ˆ Ê at Q Á 87 - , 0˜ Ë m ¯ A = Area of the triangle OPQ

of the line

and perpendicular to AB is 79 97 Ê 97 ˆ = (1) Á x - ˜¯ 79 Ë 2 2 Let the coordinates of the vertex C of the other diagonal be (x1, y1) then (x1, y1) lies on (1) and BC2 = AB2/2 y-

fi 2[x12 + (y1 - 79)2] = (97)2 + (79)2 2 È 97 ˆ 79 ¸ ˘ Ï 97 Ê 2 fi 2 Í x1 + Ì Á x1 - ˜ - ˝ ˙ = (97)2 + (79)2 2 ¯ 2 ˛ ˙˚ ÍÎ Ó 79 Ë

ÈÊ ˘ 972 ˆ 2 97 Ê 972 + 792 ˆ x x1 + ¥ 1 2 ÍÁ ˙ Á ˜ 2˜ 1 79 Ë 2 ¥ 79 ¯ 79 ¯ ÍË ˙ fi2 Í ˙ 2 Ê (97)2 + ( 79)2 ˆ ˙ Í +Á Í ˜ ˙ 2 ( 79) ÍÎ Ë ¯ ˙˚ = (97)2 + (79)2 fi 4x12 - 4 ¥ 97x1 + (97)2 + (79)2 = 2 ¥ (79)2 fi (2x1 - 97)2 = (79)2

97 ± 79 97 ± 79 y1 = 2 2 \ The coordinates of the vertices of the other diagonal are (88, 88) and (9, 9) fi 2x1 = 97 ± 79 fi x1 =

fi p1 = 88, p2 = 9 fi p1 + p2 – 90 = 7

= fi

A= -

1 (33 - 87m)(87m - 33) 2m

1 1 m(87)2 + 33 ¥ 87 . (33)2 2 2m

dA 1 1 (33)2 2 2 = - (87)2 + (33) = 0 fi m = dm 2 2m2 (87)2 33 fi m= ± 87 d2 A 1 33 = - 3 (33)2 > 0 if m = 2 87 dm m So the minimum value of A is when m = - 33/87 1 È 33 87 ˘ ¥ 872 + ¥ 332 ˙ 2 ÍÎ 87 33 ˚ A = 2 ¥ 33 ¥ 87 = 5742 fi =6 957 A1, A2, ... An are points on the line y = x Example 100 A = 33 ¥ 87 +

lying in the positive quadrant such that OAn = n O An-1, O being the origin. If OA1 = 1 then the coordinates of A8 are 8! (3a 2 , 3a ) where a is equal to , K = K Ans. 6 Solution: OA8 = 8 ¥ 7 ¥ 6 ¥ 5 ¥ 4 ¥ 3 ¥ 2 ¥ OA1 = Coordinate of A8 (OA8 cos 45°, OA8 sin 45°) 1 \ 3a 2 = 8 ¥ 7 ¥ 6 ¥ 5 ¥ 4 ¥ 3 ¥ 2 ¥ 2 (as OA1 = 1) 8! fi a = 8 ¥ 7 ¥ 6 ¥ 5 ¥ 4 = 6720 = 6 A line passes through the point P(2, 3) Example 101 and makes an angle of 30° with the positive direction of x-axis. It meets the lines represented by x 2 – 2xy – y2 = 0 at the points A and B such that PA ◊ PB = a, the value of a( 3 - 1) is equal to 17

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Ans. 2 Solution:

Equation of the line through P(2, 3) is

x-2 y-3 = (q = 30°) cos q sin q If PA = r1 , PB = r2 then r1, r2 are the roots of the equation.

fi (2 × 69 × 25)2 (x 2 + y 2) – 4 × 69 × (25) 2 x (69x + 25y) – 4 ¥ (69)2 × 25y (69x + 25y) + [(69)2 + (25) 2 – c2 )] [(69) 2 x2 + (25)2 y2 + 2 × 69 × 25xy] = 0 which is the equation of the required lines. Since they are at right angles. x2

(2 + r cos q)2 – 2 (2 + r cos q) (3 + r sin q) – (3 + r sin q)2 = 0

y2 = 0



r 2 (cos 2q – sin 2q ) – 2r (cos q + 5 sin q) – 17 = 0



[(69) 2 + (25) 2 – c2] [(69)2 + (25)2] = 0

So

a = PA ◊ PB = r1 r2 =



c2 = (69)2 + 54



a( 3 - 1) =2 17

-17 = 17( 3 + 1) cos 2q - sin 2q

If the slope of one of the lines given by Example 102 2 36x + 2hxy + 72y2 = 0 is four times the other, value of h2 is equal to 450 Ans. 9 Solution: Let the lines represented by the given equation be y = mx and y = 4mx, then 5m = – 2h/72 and 4m2 = 36/72. fi

h2 =

25 ¥ (36)2 = 4050 fi 8

h2 =9 450

If ax2 – 2hxy – ay2 = 0 represents a pair Example 105 of straight lines forming with 2x + 3y = 6 an isosceles right angled triangle at the origin then h – a is equal to Ans. 7 Solution: Let y = mx be one of the lines through the origin. Since the triangle is isosceles right angled, this line makes an angle of ± p/4 with 2x + 3y = 6 whose slope is – 2/3. So

tan (± p/4) =

fi fi

m + ( 2 3)

1 + m ( -2 3)

=

3m + 2 3 - 2m

(3 – 2m) 2 = (3m + 2) 2 [3 – 2 (y/x)] 2 = [3 ( y/x) + 2] 2[∵ m = y/x]



(3x – 2y)2 = (3y + 2x)2

If the equation 12x2 – 10xy + 2y2 + 11x Example 103 – 5y + l = 0 represents a pair of straight lines, then the value of l =

which is same as ax2 – 2hxy – ay2 = 0

Ans. 2

fi a = 5 and h = 12 fi h – a = 7

Solution: a = 12, h = – 5, g = 11/2, b = 2, f = – 5/2, c = l Since the equation represents a pair of straight lines. 12 -5 11 2 -5 2 -5 2 = 0 11 2 -5 2 l fi

–l–

25 33 = 0 fi l =2 + 4 4

If the straight lines joining the origin to Example 104 the points of intersection of the straight line 69x + 25y = 3450 and the circle (x – 25) 2 + (y – 69)2 = c2 are at right angle then c2 – (69)2 = r4, where r = Ans. 5 Solution: Making the equation of the circle homogeneous with the help of the equation of the line we get (69 x + 25 y ) x 2 + y2 – 2 (25x + 69y) 2 ¥ 69 ¥ 25 2 È 69 x + 25 y ˘ 2 + [(69) + (25) 2 – c2] Í =0 Î 2 ¥ 69 ¥ 25 ˙˚



5x 2 – 24xy – 5y 2 = 0

EXERCISE LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. If the lines 2(sin a + sin b)x – 2(sin a – b)y = 3 and 2(cos a + cos b)x + 2cos(a – b)y = 5 are perpendicular, then sin 2a + sin 2b is equal to (a) sin (a – b) – 2sin(a + b) (b) sin 2(a – b) – 2sin(a + b) (c) 2sin (a – b) – sin(a + b) (d) sin 2(a – b) – sin(a + b) 2. A(0, 0), B(2, 0), C(2, 2), D(0, 2) are the vertices of a square ABCD. The square is rotated through an angle of 30° in the anticlockwise direction so that AB makes an angle of 30° with the positive direction of x-axis. Equation of the diagonal BD in the new position is

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(a) ( 3 + 1) x + ( 2 - 1) y = 3 (b) (2 - 3 ) x + y = 2( 3 - 1) (c) ( 3 - 1) x + y = 2 - 3 (d) ( 2 - 1) x + ( 3 + 1) y = 3 3. If the lines x + 2ay + a = 0, x + 3by + b = 0 and x + 4cy + c = 0 are concurrent, then a, b, c are in (a) A.P. (b) G.P. (c) H.P. (d) none of these 4. If the lines 2 (sin a + sin b) x - 2 sin (a - b) y = 3 and 2 (cos a + cos b) x + 2 cos (a - b)y = 5 are perpendicular, then sin 2a + sin 2b is equal to (a) sin (a - b) - 2 sin (a + b) (b) sin 2 (a - b) - 2 sin (a + b) (c) 2 sin (a - b) - sin (a + b) (d) sin 2 (a - b) - sin (a + b) 5. If p1, p2 denote the lengths of the perpendiculars from the origin on the lines x sec a + y cosec a = 2a and x cos a + y sin a = a cos 2a respectively, then 2 Ê p1 p2 ˆ + is equal to ÁË p P1 ˜¯ 2

6.

7.

8.

9.

(a) 4 sin2 4a (b) 4 cos2 4a 2 (d) 4 sec2 4a (c) 4 cosec 4a The locus of the point of intersection of the lines x sin q + (1 - cos q) y = a sin q and x sin q - (1 + cos q) y + a sin q = 0 is (b) x2 + y2 = a2 (a) x2 - y2 = a2 2 (c) y = ax (d) none of these The straight lines 4x - 3y - 5 = 0, x - 2y - 10 = 0, 7x + y - 40 = 0 and x + 3y + 10 = 0 form the sides of a (a) quadrilateral (b) cyclic quadrilateral (c) rectangle (d) parallelogram The lengths of the perpendicular from the points (m2, 2m), (mm¢, m + m¢) and (m¢2, 2m¢) to the line x + y + 1 = 0 form (a) an A.P. (b) a G.P. (c) a H.P. (d) none of these The sine of the angle between the pair of lines represented by the equation x2 – 7xy + 12y2 = 0 is (a) 1/12 (b) 1/13 (c) 1/ 170

(d) none of these

10. The square of the differences of the slopes of the lines represented by the equation x2(sec2 q – sin2 q) – 2xy tan q + y2 sin2 q = 0 is (a) 1 (b) 2 (c) 4 (d) 8

11. The lines joining the origin to the points of intersection of x2 + y2 + 2gx + c = 0 and x2 + y2 + 2fy – c = 0 are at right angles, if (a) g2 + f2 = c (b) g2 – f2 = c (c) g2 – f2 = 2c (d) g2 + f2 = c2 12. Two of the lines represented by x3 – 6x2y + 3xy2 + dy3 = 0 are perpendicular for (a) all real values of d (b) two real values of d (c) three real values of d (d) no real value of d 13. If pairs of lines 3x2 – 2pxy – 3y2 = 0 and 5x2 – 2qxy – 5y2 = 0 are such that each pair bisects the angle between the other pair, then pq is equal to (a) – 1 (b) – 3 (c) – 5 (d) – 15 14. The line x + y = 1 meets the lines represented by the equation y3 – xy2 – 14x2y + 24x3 = 0 at the points A, B, C. If O is the origin, then OA2 + OB2 + OC2 is equal to (a) 22/9 (b) 85/72 (c) 181/72 (d) 221/72 15. If the area of the triangle formed by the pair of lines 8x2 – 6xy + y2 = 0 and the line 2x + 3y = a is 7, then a is equal to (a) 14 (c) 28

(b) 14 2 (d) none of these

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 16. The lines ax + (b + c)y = p, bx + (c + a)y = p and cx + (a + b)y = p intersect at the point (3/4, 3/4) if (a) p = 3, b + c = 4 (b) p = 3, a + b + c = 4 (c) p = 6, a = b + c = 4 (d) p = 2, a + b + c = 3 17. A line passes through the point (3, 4) and the point x y x y of intersection of the lines + = 1 and + 3 4 4 3 = 1 and makes intercepts of lengths a and b on the coordinate axes then (a) a + b = 25/12 (b) |a - b| = 7/12 (c) ab = 1 (d) ab = 9 18. If p is the length of the perpendicular from the x y origin on the line + = 1 and a2, p2, b2 are in A.P. a b then ab is equal to (a) p2 (b) 2 p2 (c) - 2 p2 (d) 2p2

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19. Given four lines with equations x + 2y – 3 = 0, 3x + 4y – 7 = 0, 2x + 3y – 4 = 0, 4x + 5y – 6 = 0. (a) They are all concurrent (b) Three of them are concurrent (c) They do not enclose a quadrilateral (d) Two of these pass through (1, 1) 20. One side of a rectangle lies along the line 4x + 7y + 5 = 0. Two of its vertices are (– 3, 1) and (1, 1). Equation of the other side can be. (a) 7x – 4y + 25 = 0 (b) 4x + 7y – 11 = 0 (c) 4x – 7y + 5 = 0 (d) 7x – 4y – 3 = 0 21. Straight line 3x + 4y = 5 and 4x – 3y = 15 intersect at the point A. Points B and C are chosen on these two lines such that AB = AC. Possible equation of BC passing through (1, 2) can be (a) 5x + 2y – 9 = 0 (b) 7x + y – 9 = 0 (c) x – 7y + 13 = 0 (d) 7x + 3y – 13 = 0 22. Two vertices of an equilateral triangle are (– 1, 0), (1, 0), third vertex can be

23.

24.

25.

26.

27.

28.

(a) ( 3 , 0) (b) (0, 3 ) (d) (0, 2 3 ) (c) (0, – 3 ) Two equal sides of an isosceles triangle are given by the equations 7x + y + 3 = 0 and x + y – 3 = 0 and its third side passes through the point (1, – 10), equation of the third side can be (a) 3x + y + 7 = 0 (b) x – 3y – 31 = 0 (c) 3x – y – 13 = 0 (d) x + 3y + 29 = 0 Equation of a line passing through (1, – 1) and perpendicular to a line given by x2 – 5xy + 4y2 = 0 is (a) 4x + y = 3 (b) x – 4y = 5 (c) x + y = 0 (d) x – y = 2 If two of the lines given by the equation ax3 – 9yx2 – y2x + 4y3 = 0 are perpendicular then a is equal to (a) – 4 (b) 4 (c) – 5 (d) 5 2 2 If lx – 10xy + 12y + 5x – 16y – 3 = 0 represents a pair of straight lines, then the equation of one of them is (a) x – 2y + 3 = 0 (b) x – 2y – 1 = 0 (c) 2x – 6y + 3 = 0 (d) 2x – 6y – 1 = 0 Equation of a line passing through the point of intersection of the lines given by the equation x2 – 5xy + 4y2 + x + 2y – 2 = 0 is (a) 2x + 3y = 7 (b) x = 2y (c) 3x – 7y + 1 = 0 (d) all of these If the pair of lines represented by the equation 6x2 + 17xy + 12y2 + 22x + 31y + 20 = 0 be 2x + 3y + p = 0 and 3x + 4y + q = 0, then (a) p + q = 9 (b) p2 + q2 = 41 (c) 3p + 2q = 22 (d) 4p + 3q = 31

29. A line passing through the point (3, 4) and making an angle q with the positive direction of x-axis, in1 tersects the line x + y = 8 at a distance 6 units 3 from the point, then q is equal to p 12 5p (c) 12

p 12 7p (d) 12 a, b) in the line ax + by = p + q

(a) -

(b)

is (p, q), then (a) a2 + b2 + pa + qb = 2(p + q) (b) a2 + b2 – pa – qb = 0 (c) pb – qa = 0 (d) pa – qb = 0

MATRIX-MATCH TYPE QUESTIONS 31.

y = mx + c Column 1 (a) y = (b)

2x + 5

2 x + 2y + 6 = 0

(c) 2x +

2y + 8 = 0

(d) 15x – 3y + 3 2 = 0

m, c Column 2 (p) -1/ 2 , - 3 (q)

2, 5

(r) - 2 , - 4 2 (s) 5,

2

32. Sum of the squares of the intercepts on the coordinate axes Column 1 Column 2 (a) y = 2x (p) 0 (b) 2x + 3y = 6 (q) 1666/225 (c) 3x + 5y = 7 (r) 13 (d) 3x + 2y = 6 (s) 34 33. Let L1: x + 3y – 5 = 0, L2: 3x – Ky – 1 = 0 L3: 5x + 2y – 12 = 0. Column 1 (a) L1, L2, L3 are concurrent, if (b) One of L1, L2, L3 is parallel to at least one of the other two, if (c) L1, L2, L3 form a triangle, if (d) L1, L2, L3 do not form a triangle, if

Column 2 (p) K = – 9 (q) K = – 6/5

(r) K = 5/6 (s) K = 5

34. Point at a distance 14/5 from the line Column 1 Column 2 (a) 3x + 4y + 7 = 0 (p) (1, 2) (b) 4x - 3y + 3 = 0 (q) (1, 1) (c) 3x - 4y + 19 = 0 (r) (2, - 1) (d) 4x + 3y + 9 = 0 (s) (- 3, - 3)

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35. Point of intersection of Column 1 Column 2 (p) (1, – 1) (a) 10x2 + xy – 21y2 – 21x + 41y – 10 = 0 (q) (1, 1) (b) 72x2 + 17xy – 72y2 – 127x – 161y – 17 = 0 (r) (– 1, 1) (c) 3x2 – 22xy + 24y2 + 28x – 70y + 49 = 0 (s) (0, 0) (d) x2 – y2 – 2x – 2y = 0 36. Area of the triangle Column 1 Column 2 (p) 5/2 (a) 6x2 + xy – y2 = 0 x=1 (q) 2 (b) x2 + xy – 2y2 = 0 y=1 (c) xy = 0, x + y = 2 (r) 3/2 (d) xy – x – y + 1 = 0 (s) 9/2 x+y–4=0 37. Slope of a line represented by Column 1 Column 2 (p) 1 (a) 2x2 – xy – 3y2 (q) – 1 (b) 6x2 – xy – 12y2 (r) 2/3 (c) 2x2 – 5xy + 3y2 (s) – 3/4 (d) 3x2 + xy – 4y2 38. In a triangle formed by the lines xy = 0 and 2x + 3y = k. Column 1 Column 2 (a) centroid is at (3, 2) (p) if k = 12 (b) orthocentre is at the origin (q) if k = 48 (c) circumcentre is at (3, 2) (r) if k = 18 (d) Area is 192 sq. units (s) if k = – 48 39. A line has intercepts a and b on the axis of x and of y respectively. When the axes are rotated through an angle a in the anticlockwise direction keeping the coordinate axes. Column 1 (a) a = 1, b = 0 (b) a = 0, b = 1 (c)

a 1+ 3 = b 1- 3

Column 2 p (p) a = 3 (q) a =

p 4

(c) a = -

p 4

b 3 +1 p = (d) a = a 6 3 -1 40. Locus of the mid-points of the intercepts between the coordinate axes by the lines passing through a a, 0) Column 1 Column 2 (a) does not intersect (p) x-axis (d)

(b) intersect (q) y-axis (c) is neither parallel (r) y = x nor perpendicular to (d) is parallel to (s) x = a

ASSERTION-REASON TYPE QUESTIONS 41. Statement-1: The straight line (sin q + 3 cos q) x + ( 3 sin q – cos q) y + (5 sin q – 7 cos q) = 0 passes through the point of intersection of the lines x + 3 y + 5 = 0 and x – y – 7 = 0 for all values of q except q = np/2, n is an integer. Statement-2: L1 + l L2 = 0 represents the equation of a line through the points of intersection of the lines L1 = 0 and L2 of l. 42. Statement-1: One side of a rectangle lies along the line 4x + 7y + 5 = 0. Two of its vertices are (– 3, 1) and (1, 1). Equation of the side farthest from the origin is 7x – 4y + 25 = 0 Statement-2: If a and b are constants (a, b π 0) and c is a variable, then from the lines ax + by + c = 0 and bx – ay – c = 0 the line farthest from the origin is the one for which |c| is least. 43. Statement-1: If the area of the triangle formed by the lines y = x, x + y = 4 and the lines through P(h, k) parallel to y-axis is k2, then P lies on x2 + y2 – 4x + 4 = 0 Statement-2: Area of the triangle formed by the lines y = x, x + y = 4 and the x-axis is equal to half the area of the triangle formed by the line x + y = 4 and the coordinate axes. 44. Statement-1: x + y = 0 is (– 1, – 6). Statement-2: P(a, b) in the line a + a ¢ b + b¢ˆ Ê , ax + by + c = 0 is Q(a ¢, b¢ ) if Á ˜ lies Ë 2 2 ¯ on the line. 45. Statement-1: Angle between the lines represented by 3x2 + 10xy + 3y2 – 2x + 2y – 1 = 0 and 3x2 + 10xy + 3y2 – 11x + 7y – 20 = 0 is same. Statement-2: Equations of the bisectors of the angles between the lines 3x2 + 10xy + 3y2 = 0 is x2 – y2 = 0

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 46 to 48 L1 : 3x + 4y + 8 = 0 L2 : 2x + 7y - 1 = 0

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46. If L1, L2 represent the sides AB and AC of the isosceles triangle ABC with AB = AC = 2 then the coordinates of (a) B are (28/5, - 11/5) (b) B are (28/5, 1/5) Ê 14 + 4 53 4 + 5 ˆ (c) C are Á , ˜ 53 5 ¯ Ë Ê 14 + 4 53 4 - 5 ˆ (d) C are Á , ˜ 53 5 ¯ Ë 47. Equation of the line through B parallel to AC is (a) 2x + 7y + 21 = 0 (b) 10x + 35y = 63 (c) 10x + 35y + 21 = 0 (d) 2x + 7y = 63 48. If D is the mid-point of BC and E is the mid-point of CA then DE is equal to (a) 1/4 (b) 1/2 (c) 1 (d) 2 Paragraph for Questions Nos 49 to 51 P: x2 – y2 + 2y – 1 = 0 L: x + y = 3 49. Equation of the angle bisectors of the pairs of lines P is (a) xy – y = 0 (b) xy – x = 0 (c) xy = 0 (d) xy + y = 0 50. Area of the triangle formed by the angle bisectors of the pair of lines P and the line L (in square units) is (a) 1 (b) 2 (c) 3 (d) 4 51. If L¢ represents the line perpendicular to L passing through the point of intersection of the pair of lines P, then equation of the pair of lines representing L and L¢ is (a) x2 – y2 – 4x + 2y + 3 = 0 (b) x2 – y2 – 2x + 4y – 3 = 0 (c) x2 – y2 + 2x – 4y – 3 = 0 (d) x2 – y2 + 4x – 2y + 3 = 0

INTEGER-ANSWER TYPE QUESTIONS 52. If A denotes the area enclosed by 3| x| + 4| y| £ 12 and A + 1 = a2, then a = 53. The locus of the mid-point of the portion intercepted between the axes by the line x cos a + y sin a = p passes through the point (p + 1, p - 1), then p4 - 5p2 + 5 is equal to 54. If the point (3, 4) lies on the locus of the point of intersection of the lines x cos a + y sin a = a and x sin a - y cos a = b, (a is a variable), the point (a, b) lies on the line 3x - 4y = 0 then a2 – b2 is equal to

55. If the slopes of the lines given by the equation 24x3 – ax2y + 26xy2 – 3y3 = 0 are in G.P., p is the greatest perpendicular distance of the point (1, 1) from these lines then p 37 is equal to LEVEL 2

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. Triangle ABC has vertices (0, 0), (11, 60) and (91, 0). If the line y = kx cuts the triangle into two triangles of equal area, then k is equal to (a) 30/51 (b) 4/7 (c) 7/4 (d) 30/91 2. The distance of the point (1, 2) from the line x + y + 5 = 0 measured along the line parallel to 3x – y = 7 is equal to (a) 4/ 10

(b) 40

(c) 40 (d) 10 2 3. A ray of light travels along the line 2x – 3y + 5 = 0 and strikes a plane mirror lying along the line x + y = 2. The equation of the straight line containing (a) 2x – 3y + 3 = 0 (b) 3x – 2y + 3 = 0 (c) 21x – 7y + 1 = 0 (d) 21x + 7y – 1 = 0 4. Given two points A = (– 2, 0) and B = (0, 4), the coordinates of a point M lying on the line x = y so that the perimeters of the DAMB is least; is (a) (1, 1) (b) (0, 0) (c) (2, 2) (d) (3, 3) 5. If the point (a, a2) lies inside the triangle formed by the lines 2x + 3y – 1 = 0, x + 2y – 3 = 0 and 5x – 6y – 1 = 0, then (a) – 1/2 < a < 1/2 (b) – 3/2 < a < –1 (c) – 3/2 < a < 3/2 (d) – 3/2 < a < – 1/2 6. If u = a1x + b1y + c1 = 0 and v = a2x + b2y + c2 = 0 and

a1 b1 c1 = = , then u + kv = 0 represents a2 b2 c2

(a) u = 0 (b) a family of concurrent lines (c) a family of parallel lines (d) none of these 7. The point (4, 1) undergoes the following successive transformations: y=x (ii) translation through a distance 2 units along the positive x-axis.

IIT JEE eBooks: www.crackjee.xyz Straight Lines 15.39

(a) (4, 3)

(b) (3, 4)

(c) (1, 4) (d) (4, 4) 8. A line has intercepts a and b on the coordinate axes. When the axes are rotated through an angle a intercepts on the coordinate axes, then tan a = a+b a-b (b) a-b a+b 2 2 (c) a – b (d) none of these If h denotes the arithmetic mean and k denotes the geometric mean of the intercepts made on the coordinate axes by the lines passing through the point (1, 1), then the point (h, k) lies on (a) a circle (b) a parabola (c) a straight line (d) a hyperbola From the vertex of the isosceles triangle with vertices (0, 0), (p, 0) and (p/2, q) an arc is described with a radius equal to half the base, meeting the base at a point which divides it into two parts whose lengths are the roots of the equation. (a) x2 + px + q2 = 0 (b) x2 – px + q2 = 0 2 2 (d) x2 – px – q2 = 0 (c) x + px – q = 0 A line which is parallel to x-axis and crosses the curve y = x at angle of 45º is (a) x = 1/ 4 (b) y = 1/4 (c) y = 1/2 (d) y = 1 If t1 + t2 + t3 = – t1 t2 t3, then the orthocentre of the triangle formed by the points (at1 t2, a (t1 + t2)) (at2 t3, a (t2 + t3)) and (at3 t1, a (t3+ t1)) lies on (a) x-axis (b) y-axis (c) y = x (d) x = a (a)

9.

10.

11.

12.

13. Area of the parallelogram formed by the lines y = mx, y = mx + 1, y = nx and y = nx + 1 equals. (a) (c)

m+n

( m - n )2 1 m+n

(b)

2 m+n

(d)

1 m-n

14. If the line 5 x = y meets the lines x = 1, x = 2  x = n, at points A1, A2,  An respectively then (OA1)2 + (OA2)2  + (OAn)2 is equal to (0 is the origin) (a) 3n2 + 3n (b) 2n3 + 3n2 + n 3 2 (d) (3/2) (n4 + 2n3 + n2) (c) 3n + 3n + 2 15. Joint equation of a pair of lines passing through the point of intersection of the lines x2 + xy – 2y2 – 4x + 7y – 5 = 0 and perpendicular to these lines is

(a) 2x2 – xy + y2 – 4x + 7y – 5 = 0 (b) 4x2 – 7xy – 2y2 + 6x + 15y – 18 = 0 (c) 2x2 + xy – y2 + 2x + y + 5 = 0 (d) none of these 16. If two of the lines represented by the equation ax3 + bx2y + cxy2 + dy3 = 0 are at right angles then a2 + d2 + ac + bd is equal to (a) – 1 (b) 0 (c) 1 (d) ab + cd 17. If the centroid of the triangle formed by the lines 2y2 + 5xy – 3x2 = 0 and x + y = k is (1/18, 11/18), then the value of k is (a) – 1 (b) 0 (c) 1 (d) none of these 18. The angle between the lines represented by (1 – cos q tan a)y2 – (2 cos q + sin2 q tan a) xy + cos q (cos q + tan a)x 2 = 0 is (a) a (b) q (c) q + a (d) q – a 19. Equation of the bisectors of the angles between the lines through the origin, the sum and product of whose slopes are respectively the arithmetic and the geometric means of 9 and 16 is (a) 24x2 – 25xy + 2y2 = 0 (b) 25x2 + 44xy – 25y2 = 0 (c) 11x2 – 25xy – 11y2 = 0 (d) none of these 20. If the pairs of lines x2 + 2xy + ay2 = 0 and ax2 + 2xy + y2 = 0 have exactly one line in common then the joint equation of the other two lines is given by (a) 3x2 + 8xy – 3y2 = 0 (b) 3x2 + 10xy + 3y2 = 0 (c) y2 + 2xy – 3x2 = 0 (d) x2 + 2xy – 3y2 = 0

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 21. Coordinates of a point at unit distance from the lines 3x – 4y + 1 = 0 and 8x + 6y + 1 = 0 are (a) (6/5, – 1/10) (b) (– 2/5, – 13/10) (c) (0, 3/2) (d) (– 8/5, 3/10) 22. The base of a triangle lie along the line x = a and is of length a. If the area of the triangle is a2, the vertex lies of the line. (a) x = – a (b) x = 0 (c) x = a/2 (d) x = 3a

IIT JEE eBooks: www.crackjee.xyz 15.40 Comprehensive Mathematics—JEE Advanced

23. If the medians AD and BE of the triangle with vertices A(0, b), B (0, 0) and C (a, 0) are perpendicular, then an equation of the line through (a, b) perpendicular to AC is (a) y = 24.

25.

26.

27.

28.

29.

30.

2x + b

(b) y = – 2x – b

(c) y = 2x – b (d) x = 2 y – a If the vertices P, Q, R of a triangle PQR are rational points, which of the following points of the triangle PQR is/are always rational point(s)? (Note: A rational point has rational numbers as its coordinates) (a) centroid (b) incentre (c) circumcentre (d) orthocentre All points lying inside the triangle formed by the points (1, 3), (5, 0) and (– 1, 2) satisfy: (a) 3x + 2y ≥ 0 (b) 2x + y – 13 ≥ 0 (c) 2x – 3y – 12 £ 0 (d) – 2x + y ≥ 0 Two sides of a rhombus ABCD are parallel to the lines y = x + 2 and y = 7x + 3. If the diagonals of the rhombus intersect at the point (1, 2) and the vertex A is on the y-axis coordinates of A can be (a) (0, 3/2) (b) (0, 0) (c) (0, 5/2) (d) (0, 2/5) A line PQ makes intercepts of length 2 units between the lines y + 2x = 3 and y + 2x = 5. If the coordinates of P are (2, 3), coordinates of Q can be (a) (6, 0) (b) (2, 3) (c) (0, 9/2) (d) (3, 2) If p and q are the roots of the equation x2 + 2x – 3 = 0 and x2 – 7x + 10 = 0 respectively then the orthocentre of the triangle formed by the lines (1 + p) x – py + p (1 + p) = 0 (1 + q) x – qy + q (1 + q) = 0 and y = 0, can be (a) (5, – 5) (b) (– 2, 7) (c) (6, – 6) (d) (– 15, 15) Line x = a meets the curve y2 = 4ax at the points A and B. If the area of the triangle OAB, O being the origin, is 8 square units, a is root of the equation (a) x2 – 2x – 8 = 0 (b) x2 + 6x + 8 = 0 (c) x2 + 2x – 8 = 0 (d) x2 – 6x + 8 = 0 A and B are two points on the line 2x + 3y – 1 = 0 whose coordinates satisfy the relation y = x2. If APB is an isosceles triangle the coordinates of P can be

(a) (0, 19/18) (b) (– 19/27, 0) (c) (– 19/9, – 19/9) (d) (– 2/3, 1/18) 31. One of the lines given by the equation ax2 + 2hxy + by2 = 0 will bisect the angle between the coordinate axes if (a) a + b = 2h (b) a + b = – 2h (c) a + b = 0 (d) a – b = 0 32. Slope of a bisector of the angle between the lines 4x2 – 16xy – 7y2 = 0 is (a) (c)

11 + 377 16 -3+2 3

(b) (d)

11 - 377 16 -3-2 3

7 7 2 2 33. The straight lines 7x + 6xy + 4y = 0 have the same pair of bisectors as those of the lines given by (a) 49x2 + 66xy + 16y2 = 0 (b) 10x2 + 6xy + 7y2 = 0 (c) 5x2 + 6xy + 2y2 = 0 (d) 4x2 – 6xy + 7y2 = 0 34. The equation 3x2 – 8xy – 3y2 = 0 and x + 2y = 3 represent the sides of (a) an equilateral triangle (b) isosceles triangle (c) right angled triangle (d) obtuse angled triangle 35. The equation x3 + x2y – xy2 – y3 – x2 + y2 = 0 represents three lines. (a) Two of which are parallel (b) Two of which are perpendicular (c) all of which are concurrent (d) which do not form a triangle

MATRIX-MATCH TYPE QUESTIONS 36. A line has intercepts a and b on the coordinate axes. When the axes are rotated through an angle a, keepon the coordinate axes. Column 1 (a) a = 1, b = 0 (b) a = 0, b = 1 (c)

1+ 3 a = b 1- 3

Column 2 (p) a = p /3 (q) a = p /4 (r) a = - p /4

(d) a = 2, b = 0 (s) a = p/6 37. Locus of the mid-points of the intercepts between the coordinate axes by the lines passing through a a, 0)

IIT JEE eBooks: www.crackjee.xyz Straight Lines 15.41

38.

39.

40.

41.

42.

43.

Column 1 Column 2 (a) does not intersect (p) x-axis (b) intersects (q) y-axis (c) is neither parallel nor (r) y = x perpendicular to (d) is parallel to (s) x = a For all values of a, b, c Œ R, the lines ax + by + c are concurrent at Column 1 Column 2 (a) (3/4, 1/2) (p) 5a - 4b + 3c = 0 (b) (- 2, - 3) (q) 3a + 2b + 4c = 0 (c) (5/3, - 4/3) (r) 2a + 3b - c = 0 (d) (0, 0) (s) c = 0 The lines ax + by + c = 0 and a¢x + b¢y + c¢ = 0 Column 1 Column 2 (a) parallel (p) aa¢ + bb¢ = 0 (b) perpendicular (q) ab¢c¢ = a¢b¢c = a¢c¢b (c) identical (r) ab¢ - a¢b = 0 (d) intersect at (1, 1) (s) a + b + c = a¢ + b¢ + c¢ = 0 Equation of a line passing through the intersection of Column 1 Column 2 (a) 2x + 3y - 4 = 0, (p) x + y + 2 = 0 3x + 4y - 2 = 0 (b) x + y + 2 = 0, (q) x - 2y = 0 7x - 5y + 6 = 0 (c) x - 2y = 0, (r) 2x + 3y - 4 = 0 3x + y - 4 = 0 (d) x + y + 1 = 0, (s) y + 1 = 0 x–y–1=0 One of the lines represented by the pair of lines is Column 1 Column 2 (a) 6x2 – xy + 4cy2 = 0 (p) c = 0 3x + 4y = 0 (b) x2 – 2cxy – 9y2 = 0 (q) c = 15 x + 3y = 0 (c) cx2 + 16xy – 15y2 = 0 (r) c = –3 5x – 3y = 0 (d) x2 + 7xy – cy2 = 0 (s) c = 1 x + 7y = 0 Slope of a line Column 1 Column 2 (a) 2x2 + 5xy + 3y2 = 0 (p) 3/2 (b) 3x2 – 5xy + 2y2 = 0 (q) – 1 2 2 (c) 4x – 5xy + y = 0 (r) 4 (d) 3x2 + xy – 2y 2 = 0 (s) 1 Lines represented by (a) 2x2 + 3xy – 5y2 = 0 (p) x – 4y = 0 (b) 2x2 – 3xy – 20y2 = 0 (q) x – y = 0

(c) 3x2 – 5xy + 2y2 = 0 (d) 6x2 + 11xy – 10y2 = 0

(r) 3x – 2y = 0 (s) 2x + 5y = 0

ASSERTION-REASON TYPE QUESTIONS 44. Statement-1: Two vertices of a triangle are (– 2, 3) and (– 4, – 7). If the orthocentre of the triangle is the origin, the coordinates of the third vertex are (5, – 1). Statement-2: Orthocentre of a triangle is equidistant from the sides of the triangle. 45. Statement-1: If x + ky = 1 and x = a are the equations of the hypotenuse and a side of a right angled isosceles triangle then k = ± a. Statement-2: Each side of a right angled isosceles triangle makes an angle ± p /4 with the hypotenuse. 46. Statement-1: Locus of the centroid of the triangle whose vertices are (a cos t, a sin t), (b sin t, – b cos t) and (1, 0), where t is a parameter is (3x – 1)2 + (3y)2 = a2 + b2 Statement-2: The centroid of a triangle is equidistance from the vertices of the triangle. 47. Statement-1: If non-zero numbers a, b, c are in H.P x y 1 then the straight line + + = 0 always passes a b c Statement-2: If a, b, c are in H.P (a, b, c π 0) then 2 1 1 = + . b a c 48. Statement-1: If (a, a2) falls inside the angle made by the lines 2y = x, x > 0 and y = 3x, x > 0 then a Œ (1/2, 3). Statement-2 Sum of the squares of the length of the perpendiculars from the point (a, a) on the lines 2y = x and y = 3x is 3a2/5. 49. Statement-1: Lines represented by the equations xy – 8x – 6y + 48 = 0 and xy + 4x – 18y – 72 = 0 enclose a square of area 144 square units. Statement-2: For all values of a and h, the lines represented by the equation ax2 + 2hxy – ay2 = 0 are perpendicular. 50. Statement-1: Pair of straight lines represented by the equation (2l + 7)x 2 + 2lxy – l2y2 = 0 are not perpendicular for any real value of l. Statement-2: If for a non-zero value of l, one of the lines represented by the equation (2l + 1)x2 + lxy – (l2 + 1)y2 = 0 is x + y = 0, the other line is 3x – 2y = 0.

IIT JEE eBooks: www.crackjee.xyz 15.42 Comprehensive Mathematics—JEE Advanced

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 51 and 52 L1 : 2x + y = 50 and L2 : y = mx + 1 are two lines. 51. The greatest of integer values of m for which the point of intersection of L1 and L2 has an integer as its x-coordinate is (a) 5 (b) – 9 (c) 47 (d) – 51 52. Square of the distance between the points of intersection of L1 and L2 for the greatest and the least integer values of m is. (a) 5 (b) 20 (c) 25 (d) 30 Paragraph for Question Nos. 53 to 55 Two adjacent sides of a parallelogram are 4x + 5y = 0 and 7x + 2y = 0 one of its diagonal is 11x + 7y – 9 = 0 53. Equation of the other diagonals is (a) x + y = 0 (b) 7x – 11y = 0 (c) x – y = 0 (d) 11x + 7y = 0 54. Length of the given diagonal is 2

(a)

(b)

170 /3

(c) 13/3 (d) 41 /3 55. Area of the parallelogram is (a) 3 (b) 3/2 (c) 6

(d) 2 85 /3

Paragraph for Questions Nos. 56 to 58 P: 2x2 – axy – 6y2 = 0 Q: 3x2 – 8xy + 4y2 = 0 56. If one of the lines represented by P coincides with one of the lines represented by Q and the other two are perpendicular, then the value of a is (a) 1 (b) 2 (c) – 1 (d) none of these 57. If the bisectors of the angles between the lines represented by P and Q are same, the value of a is

(a) 8 (b) 16/3 (c) – 64 (d) – 16 58. The angle between the lines given by P is same as the angle between the lines given by Q for (a) a = 1 (b) a = – 1 (c) a = 8/7 (d) no real value of a

INTEGER-ANSWER TYPE QUESTIONS 59. The coordinates of the feet of the perpendiculars from the vertices of a triangle on the opposite sides are (20, 25), (8, 16) and (8, 9). If the orthocentre of the triangle is (h, k) then k – h is equal to [Hint: orthocentre of the triangle is the incentre of the pedal triangle.] 60. Through the point P(3, - 5), a line is drawn inclined at 45 ° with the positive direction of x-axis. It meets the line x + y - 6 = 0 at the point Q, O being the origin, then if 9(PQ)2 + 14(OP)2 + 10(OQ)2 = 79k2, k = 61. If P(1, 2), Q(a, b), R(5, 7) and S(2, 3) are the vertices of a parallelogram, then sum of the squares of the lengths of its diagonals is bl where l = 62. Medians AD and BE of the triangle ABC with vertices A(0, b), B(0, 0) and C(a, 0) are perpendicular. If b is the geometric mean between 7 and 17 then a2 = 61k – 6, where k = 63. If 9x2 + 2hxy + 4y2 + 2gx + 2fy – 3 = 0 represents a pair of parallel lines, g is the slope of the line represented by the equation 3x2 + 5xy – 2y2 = 0 and making an acute angle with the positive direction of x-axis, then (g2 + f2 + h2)2 + g2 f 2 h2 = 3700 – k where k = 64. If y = mx bisects the angle between the lines x2 (tan2 q + cos2 q) + 2xy tan q – y2 sin2 q = 0 when 4

2

1ˆ 1ˆ Ê Ê q = p/3 then 81 Á m - ˜ + 27 Á m - ˜ =306k Ë ¯ Ë m m¯ where k = 65. If the distance between the parallel lines given by x 2 + 2xy + y2 – 8ax – 8ay – 9a2 = 0 is 90 2 then a3 + 8a2 + 1 = 337 r2 where r =

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. Given four lines with the equations x + 2y – 3 = 0, 3x + 4y – 7 = 0, 2x + 3y – 4 = 0, 4x + 5y – 6 = 0

(a) they are all concurrent (b) they are sides of a quadrilateral (c) none of these [1980] 2. The point (4, 1) undergoes the following three transformations successively y=x

IIT JEE eBooks: www.crackjee.xyz Straight Lines 15.43

(ii) transformation through a distance 2 units along the positive direction of x-axis. p (iii) rotation through an angle about the origin 4 in the counter clockwise direction. Then the ordinates Ê 1 7 ˆ (a) Á , Ë 2 2 ˜¯

(b) (– 2 , 7 2 )

Ê 1 7 ˆ (c) Á , ˜ Ë 2 2¯

(d) ( 2 , 7 2 )

[1980]

3. The straight lines x + y = 0, 3x + y – 4 = 0, x + 3y – 4 = 0 form a triangle which is (a) isosceles (b) equilateral (c) right angled (d) none of these [1983] 4. If P = (1, 0), Q = (– 1, 0) and R = (2, 0) are three given points, then locus of the point S satisfying the relation SQ2 + SR2 = 2SP2 is (a) a straight line parallel to x-axis (b) a circle passing through the origin (c) a circle with the centre at the origin (d) a straight line parallel to y-axis. [1988] 5. Line L has intercepts a and b on the coordinates axes. When the axes are rotated through a given L has intercepts p and q, then (a) a2 + b2 = p2 + q2 (b)

2

2

(c) a + p = b + q

(a) 3/4

(c)

1 |m + n|

1 1 1 1 [1990] + = + a2 p2 b2 q2 6. If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is (a) square (b) circle (c) straight line (d) two intersecting lines [1992] 7. The orthocentre of the triangle formed by the lines x = 0, y = 0 and x + y = 1 is (d)

Ê 1 1ˆ (a) Á , ˜ Ë 2 2¯

Ê 1 1ˆ (b) Á , ˜ Ë 3 3¯

(c) (0, 0)

Ê 1 1ˆ (d) Á , ˜ Ë 4 4¯

[1995]

8. The diagonals of a parallelogram PQRS are along the lines x + 3y = 4 and 6x – 2y = 7. Then PQRS must be a (a) rectangle (b) square (c) cyclic quadrilateral (d) rhombus [1998]

(d)

1 |m - n|

[2001]

13. Let P = (–1, 0), Q = (0, 0) and R = (3, 3 3 ) be three points. Then the equation of the bisector of the angle PQR is (b) x +

3y = 0

(d) x + ( 3 /2) y = 0 [2002] 14. A straight line through the origin O meets the parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at points P and Q respectively. Then the point O divides the segment PQ in the ratio (a) 1 : 2 (b) 3 : 4 (c) 2 : 1 (d) 4 : 3 [2002] 15. Let 0 < α < π P = (cos θ, sin θ) and Q = (cos (α – θ ), sin (α – θ)) then Q is obtained from P by (a) clockwise rotation around origin through an angle α. (b) anti clock wise rotation around origin through an angle α (c)

2

(b) 3 3

(c) 3 (d) 3 3 / 2 [1998] 10. If x1, x2, x3 as well as y1, y2, y3 are in G.P., with the same common ratio, then the points (x1, y1), (x1, y2), (x3, y3) (a) lie on a straight line (b) lie on an ellipse (c) lie on a circle (d) are vertices of a triangle [1999] 11. Let PS be the median of the triangle with vertices P (2, 2), Q(6, – 1) and R (7, 3). The equation of the line passing through (1, – 1) and parallel to PS is (a) 2x – 9y – 7 = 0 (b) 2x – 9y – 11 = 0 (c) 2x + 9y – 11 = 0 (d) 2x + 9y + 7 = 0 [2000] 12. Area of the parallelogram formed by the lines y = mx, y = mx + 1, y = nx and y = nx + 1 equals (m + n) 2 (b) (a) 2 |m + n| ( m – n)

(a) ( 3 /2) x + y = 0

1 1 1 1 + 2 = 2 + 2 2 a b p q 2

9. Let A0 A1 A2 A3 A4 A5 be a regular hexagon inscribed in a circle of unit radius. Then, the product of the length of the line segments A0 A1, A0 A2 and A0 A4 is

3x + y = 0

tan α tan (α/2) [2002] 16. Orthocentre of the triangle with vertices (0, 0), (3, 4) and (4, 0) is

IIT JEE eBooks: www.crackjee.xyz 15.44 Comprehensive Mathematics—JEE Advanced

(a) (3, 5/4)

(b) (3, 12)

(c) (3, 3/4)

(d) (3, 9)

[2003]

17. The locus of the orthocentre of the triangle formed by the lines (1 + p)x – py + p(1 + p) = 0, (1 + q)x – qy + q(1 + q) = 0, and y = 0, where p

q, is

(a) a hyperbola

(b) a parabola

(c) an ellipse

(d) a straight line

[2009] 18. A straight line L through the point (3, –2) is inclined at an angle 60º to the line 3x + y = 1. If L also intersects the x-axis, then the equation of L is (a) y + 3x + 2 - 3 3 = 0 3y - x + 2 3 = 0 3y + x - 3 + 2 3 = 0

[2011]

19. For a > b > c > 0, the distance between (1, 1) and the point of intersection of the lines ax + by + c = 0 and bx + ay + c = 0 is less then 2 2 . Then (a) a + b – c > 0 (b) a – b + c < 0 (c) a – b + c > 0 (d) a + b – c < 0 [2013]

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 1. Three lines px + qy + r = 0, qx + ry + p = 0 and rx + py + q = 0 are concurrent if (a) p + q + r = 0 (b) p2 + q2 + r2 = pq + qr + rp (c) p3 + q3 + r3 = 3pqr (d) none of these [1985] 2. All points lying inside the triangle formed by the points (1, 3), (5, 0) and (– 1, 2) satisfy: (a) 3x + 2y 0 (b) 2x + y – 13 0 (c) 2x – 3y – 12 0 (d) – 2x + y 0 (e) none of these [1986]

MATRIX-MATCH TYPE QUESTIONS Consider the lines given by L1 : x + 3y – 5 = 0, L2 : 3x – ky – 1 = 0, L3 : 5x + 2y – 12 = 0 Column I (a) L1, L2, L3 are concurrent, if (b) One of L1, L2, L3 is parallel to at least of the other two, if

(d) L1, L2, L3 do not form a triangle, if

Column II (p) k = – 9 (q) k = –

6 5

5 6 (s) k = 5 [2008] (r) k =

ASSERTION-REASON TYPE QUESTIONS 1. Lines L1: y – x = 0 and L2: 2x + y = 0 intersect the line L3: y + 2 = 0 at P and Q respectively. The bisector of the acute angle between L1 and L2 intersect L3 at R. Statement-1: The ratio PR : RQ equals 2 2 : 5 Statement-2: In any triangle bisector of an angle divides the triangle into two similar triangles. [2007]

FILL

(b) y - 3x + 2 + 3 3 = 0 (c) (d)

(c) L1, L2, L3 form a triangle, if

IN THE

BLANKS TYPE QUESTIONS

1. The area enclosed within the curve |x| + |y| = 1 [1983] is 2. If a, b, c are in A.P. then the straight line ax + by +c [1984] coordinates are 3. The orthocentre of the triangle formed by the lines x + y = 1, 2x + 3y = 6, 4x – y + 4 = 0 lies in the . [1985] quadrant no. 4. Let the algebraic sum of the perpendicular distance from the points (2, 0), (0, 2) and (1, 1) to a variable straight line be zero; then the line passes through a . [1991] 5. The vertices of a triangle are A (– 1, – 7), B (5, 1) and C (1, 4). The equation of the bisector of the angle . [1993] ABC is

TRUE/FALSE TYPE QUESTIONS 1. The straight line 5x + 4y = 0 passes through the point of intersection of the straight lines x + 2y – 10 = 0 and 2x + y + 5 = 0 [1983]

SUBJECTIVE-TYPE QUESTIONS 1. One side of a rectangle lies along the line 4x + 7y + 5 = 0. Two of its vertices are (– 3, 1) and (1, 1). Find the equation of the other three sides. [1978] 2. A straight line segment of length l moves with its ends on two mutually perpendicular lines. Find the locus of the point which divides the line in the ratio 1 : 2. [1978] 3. Two vertices of a triangle are (5, – 1) and (– 2, 3). If coordinates of the third vertex.

[1979]

IIT JEE eBooks: www.crackjee.xyz Straight Lines 15.45

4. Find the equation of the line which bisects the obtuse angle between the lines x – 2y + 4 = 0 and 4x – 3y + 2 = 0 [1979] 5. A straight-line L is perpendicular to the line 5x – y = 1. The area of the triangle formed by the line L and the coordinate axes is 5. Find the equation of the line L. [1980] 6. The ends A and B of a straight line segment of length c OX and OY respectively. If the rectangle OABP be completed, then show that the locus of the perpendicular drawn from P to AB is x2/3 + y2/3 = c2/3. [1983] 7. The vertices of a triangle are (a t1 t2, a (t1 + t2 )), (a t2 t3, a (t2 + t3 )) (a t3 t1, a (t3 + t1)). Find the orthocentre of the triangle. [1983] 8. Two equal sides of an isosceles triangle are given by the equations 7x – y + 3 = 0 and x + y – 3 = 0 and its third side passes through the point (1, – 10). Determine the equation of the third side. [1984] 9. Two sides of a rhombus ABCD are parallel to the lines y = x + 2 and y = 7x + 3. If the diagonals of the rhombus intersect at the point (1, 2) and the vertex A is on the y A. [1985] 10. The equations of the perpendicular bisectors of the sides AB and AC of a triangle ABC are x – y + 5 = 0 and x + 2y = 0, respectively. If the point A is (1, – 2), BC. [1986] 11. Lines L1 ax + by + c = 0 and L2 lx + my + n = 0 intersect at the point P and make an angle θ with each other. Find the equation of a line L different from L2 which passes through P and makes the same angle θ with L1. [1988] 12. Let ABC be a triangle with AB = AC. If D is the midpoint of BC, E the foot of the perpendicular drawn from D to AC and F the mid-point of DE. Prove that AF is perpendicular to BE. [1989] 13. Straight lines 3x + 4y = 5 and 4x – 3y = 15 intersect at the point A. Points B and C are chosen on these two lines such that AB = AC. Determine the possible equations of the line BC passing through the point (1, 2) [1990] 14. A line cuts the x-axis at A (7, 0) and the y-axis at B (0, –5). A variable line PQ is drawn perpendicular to AB cutting the x-axis at P and the y-axis in Q. If AQ and BP intersect at R R. [1990] 15. Find the equation of the line passing through the point (2, 3) and making intercepts of length 2 units between the lines y + 2x = 3 and y + 2x = 5. [1991]

16. Determine all values of α for which the point (α, α 2) lies inside the triangle formed by the lines 2x + 3y – 1 = 0, x + 2y – 3 = 0 and 5x – 6y – 1 = 0. [1992] 17. A line through A(– 5, –4) meets the lines x + 3y + 2 = 0, 2x + y + 4 = 0 and x – y – 5 = 0 at the points B, C and D respectively. If (15/AB)2 + (10/AC)2 = (6/ AD)2 18. Let (h, k h > 0, k > 0. A straight line passing through this point cuts the positive direction of coordinate axes at the points P and Q. Find the minimum area of the triangle OPQ, O being the origin. [1995] 19. A rectangle PQRS has its side PQ parallel to the line y = mx and vertices P, Q, and S on the lines y = a, x = b and x = – b, respectively. Find the locus of the vertex R. [1996] 20. Using coordinate geometry prove that the three altitudes of any triangle are concurrent [1998] 21. For points P = (x1, y1) and Q = (x2, y2) of the co-ordinate plane, a new distance d (P, Q by d (P, Q) = |x1 – x2| + |y1 – y2|. Let O = (0, 0) and A quadrant which are equidistant (with respect to the new distance) from O and A consists of the union Sketch this set in a labelled diagram.

[2000]

22. Let ABC and PQR be any two triangles in the same plane. Assume that the perpendiculars from the points A, B, C to the sides QR, RP, PQ respectively are concurrent. Using vector methods or otherwise, prove that perpendiculars from P, Q, R to BC, CA, AB respectively are also concurrent. [2000] 23. A straight line L through the origin meets the lines x + y = 1 and x + y = 3 at P and Q respectively. Through P and Q two straight lines L1 and L2 are drawn parallel to 2x – y = 5, 3x + y = 5 respectively. Lines L1 and L2 intersect at R. Show that the locus of R, as L varies, is a straight line. [2002] 24. A straight line L with negative slope passes through the point (8, 2) and meets the positive coordinates axes at points P and Q. Find the absolute minimum value of OP + OQ as L varies, where O is the origin. [2002] 25. Find the locus of point P(h, k) if area of the triangle formed by the lines y = x, x + y = 2 and line through [2005] P (h, k) and parallel to the x-axis is 4h2.

IIT JEE eBooks: www.crackjee.xyz 15.46 Comprehensive Mathematics—JEE Advanced

LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13.

(b) (c) (c) (d)

2. 6. 10. 14.

(b) (b) (c) (d)

3. 7. 11. 15.

(c) (b) (c) (c)

(b), (b), (a), (b), (a), (a), (a), (a),

(c) (c) (b), (d) (c) (c) (d) (b), (c), (d) (c)

17. 19. 21. 23. 25. 27. 29.

q

r

s

a

p

q

r

s

b

p

q

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c

p

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d

p

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p

q

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s

(a), (b), (b), (a), (a), (a), (b),

a

p

q

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b

p

q

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c

p

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s

d

p

q

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p

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a

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b

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c

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d

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p

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s

35.

4. (b) 8. (b) 12. (b)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 16. 18. 20. 22. 24. 26. 28. 30.

p

34.

Answers

(b), (c) (c), (d) (c) (b) (d) (b), (c), (d) (c)

36.

37.

MATRIX-MATCH TYPE QUESTIONS 31. 38.

p

q

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a

p

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b

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a

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d

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d

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32.

33.

39.

40.

IIT JEE eBooks: www.crackjee.xyz Straight Lines 15.47

ASSERTION-REASON TYPE QUESTIONS

p

q

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a

p

q

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b

p

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c

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d

p

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p

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a

p

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b

p

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LEVEL 2

c

p

q

r

s

SINGLE CORRECT ANSWER TYPE QUESTIONS

d

p

q

r

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p

q

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a

p

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b

p

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c

p

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d

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a

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41. (a)

42. (c)

43. (b)

44. (c)

38.

45. (b)

COMPREHENSION-TYPE QUESTIONS 46. (a) 51. (b)

47. (c)

48. (c)

49. (b)

50. (b)

INTEGER-ANSWER TYPE QUESTIONS 52. 5

1. 6. 11. 16.

53. 3

(a) (a) (c) (b)

2. 7. 12. 17.

54. 7

(c) (b) (a) (c)

3. 8. 13. 18.

55. 5

(b) (b) (d) (a)

4. 9. 14. 19.

(b) (b) (b) (b)

5. 10. 15. 20.

40. (b) (b) (d) (b)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 21. 23. 25. 27. 29. 31. 33. 35.

(a), (b), (a), (a), (a), (a), (a), (a),

(b), (c) (c) (c) (b), (b) (b), (b),

(c), (d)

22. 24. 26. 28. 30. 32. 34.

(c), (d) (c), (d) (d)

39.

(a), (a) (b), (a), (a), (a), (b),

41.

(d) (c) (d) (b), (c), (d) (b) (c)

42.

MATRIX-MATCH TYPE QUESTIONS p

q

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a

p

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b

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c

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d

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36.

37.

43.

ASSERTION-REASON TYPE QUESTIONS 44. (c) 48. (b)

45. (d) 49. (b)

46. (c) 50. (b)

47. (a)

IIT JEE eBooks: www.crackjee.xyz 15.48 Comprehensive Mathematics—JEE Advanced

COMPREHENSION-TYPE QUESTIONS 51. (c) 55. (a)

52. (b) 56. (a)

53. (c) 57. (c)

54. (b) 58. (d)

INTEGER-ANSWER TYPE QUESTIONS 59. 5 63. 3

60. 4 64. 8

61. 9 65. 5

62. 4

SINGLE CORRECT ANSWER TYPE QUESTIONS (c) (b) (c) (c) (d)

2. 6. 10. 14. 18.

(c) (a) (a) (b) (b)

3. 7. 11. 15. 19.

(a) (c) (d) (d) (a)

4. 8. 12. 16.

(d) (d) (d) (c)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 1. (a), (b), (c)

5. 7. 8. 9. 10.

x + 5y ± 5 2 = 0 (–a, a (t1 + t2 + t3 + t1t2t3)) 3x + y + 7 = 0, x + 3y – 31 = 0 (0, 0), (0, 5/2) 14x + 23y – 40 = 0 a 2 + b2 (lx + my + n) = 0 1 + al + bm x – 7y + 13 = 0 or 7x + y – 9 = 0 x2 + y2 – 7x + 5y = 0 x – 2 = 0; 3x + 4y – 18 = 0 –3/2 < a < 1 or 1/2 < a < 1 2x + 3y + 22 = 0 18. 2hk 2 2 (m – 1)x + my + b (m + 1) + am = 0 {(x, y)|x = 1/2, y ≥ 2} » {(x, y)| x + y = 5/2, 0 < x < 3, 0 < y < 2} 18 2x – y + 1 = 0; 2x + y – 1 = 0

11. ax + by + c –

PAST YEARS’ IIT QUESTIONS

1. 5. 9. 13. 17.

4. (4 - 5 ) x - (3 - 2 5 ) y + 2 - 4 5 = 0

13. 14. 15. 16. 17. 19. 21. 24. 25.

Hints and Solutions

2. (a), (c)

MATRIX-MATCH TYPE QUESTIONS

LEVEL 1

p

q

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s

a

p

q

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1. -

b

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c

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1.

P(2cos 30°, 2sin 30°) = P( 3,1) R(2cos (90° + 30°), 2sin (90° + 30°)) = R(-1, 3 )

1. (c) IN THE

sin 2a + sin 2b = sin 2(a – b) – 2 sin (a + b)

2. Let APQR be the new position of the square.

ASSERTION-REASON TYPE QUESTIONS FILL

2(sin a + sin b) 2(cos a + cos b) ¥ = –1 2 sin(a - b) 2 cos(a - b)

Q

BLANKS TYPE QUESTIONS

1. 2(units)2 2. (1, –2) 5. 13x – 16y = 49.

3. First

4. (1, 1)

R

P

TRUE/FALSE TYPE QUESTIONS 1. True.

SUBJECTIVE-TYPE QUESTIONS 1. 4x + 7y – 11 = 0; 7x – 4y+ 25 = 0, 7x – 4y – 3 = 0 3. (–4, –7) 2. 9x2 + 36y2 = 4 l2

C (2, 2)

D (0, 2)

30° A (0, 0)

B (2, 0)

Equation of PR is y–1=

3 -1 -1 - 3

( x - 3)

IIT JEE eBooks: www.crackjee.xyz Straight Lines 15.49

fi 3.

(2 - 3 ) x + y = 2( 3 - 1)

1 2a a 1 3 b b = 0 fi - bc + 2ac - ab = 0 1 4c c 1 1 2 fi + = fi a, b, c are in H.P. a c b

4. -

2 (sin a + sin b) 2 (cos a + cos b) ¥ =-1 2 sin ( a - b) 2 cos ( a - b)

fi 2(sin a + sin b) (cos a + cos b) = 2 sin (a - b) cos (a - b) fi sin 2a + sin 2b = sin2 (a - b) - 2 sin (a + b) 5. p1 =

2a

, p2 =

sec2 a + cos ec 2 a

a cos 2a sin 2 a + cos2 a

11. Equation of common chord is fy – gx = c, equation of the line joining the origin to the points of ( fy - gx ) intersection of the circles is x2 + y2 + 2g x c ( f y - g x )2 +c = 0 which are at right angle if c2 c + c – 2g2 + f 2 + g 2 = 0 fi g2 – f 2 = 2c 1 12. Product of the slopes is given by m1 m2 m3 = - . d If m1 m2 = – 1, then m3 = 1/d and m3 equation d m 3 + 3m 2 – 6m + 1 = 0 fi d2 – 6d + 4 = 0 which gives two real values of d. 13. Equation of bisectors of angle between 3x2 – 2pxy x2 - y2 xy = or – p (x 2 – y 2) – 6xy = 0 – 3y 2 = 0 is -p 3+3 Comparing with 5 (x2 – y2) – 2qxy = 0, p = – 5, q = 3

2

Ê p1 p2 ˆ 2 2 ÁË p + p ˜¯ = (tan 2a + cot 2a) = 4 cosec 4a 2 1

14. x-coordinates of the points are given by

6. The point of intersection is x = a cos q, y = a sin q fi x2 + y2 = a2 7. Slopes of the lines are 4/3, 1/2, - 7 and - 1/3, respectively. If a b is the angle between second and fourth. tan a = -1, tan b = 1 fi a = 135°, b = 45° fi a+ b = 180° So the quadrilateral in cyclic. Since no two sides are parallel, it can not be a parallelogram or a rectangle. 8. p1 =

m2 + 2 m + 1 2

, p2 =

m m¢ + m + m¢ + 1 2 p3 =



p1 p3 =

9. tan q =

2

p22

,

( 7 2)

- 12

1 + 12

=

1 13



m¢ 2 + 2m¢ + 1

sin q =

2

1 170

10. x 2 (1 – sin2 q cos2 q) – 2xy sin q cos q + y2 sin2 q cos2 q = 0 m1 + m2 = m1 m2 = \

2sin q cos q sin q cos q 2

2

=

2 , sin q cos q

1 - cos2 q sin 2 q cos2 q sin 2 q

(m1 – m 2 ) 2 = (m1 + m2)2 – 4m1m2 = 4



36x3 – 9x2 – 4x + 1 = 0



(3x – 1) (3x + 1) (4x – 1) = 0



x = 1/3, –1/3, 1/4



A (1/3, 2/3), B (– 1/3, 4/3) and C (1/4, 3/4)

OA2 + OB2 + OC2 =

1 4 1 16 1 9 221 = . + + + + + 9 9 9 9 16 16 72

Ê a aˆ 15. Vertices of the triangle are (0, 0), Á , ˜ Ë 8 4¯ Area of the triangle =

fi p1, p2, p3 are in G.P. 2

24x3 + 14x2 (x – 1) – x (x – 1)2 – (x – 1)3 = 0

Ê a 2a ˆ ,Á , ˜. Ë 14 7 ¯

1 È1 2 1 1 ˘ 2 ¥ - ¥ a =7 2 ÍÎ 8 7 4 14 ˙˚

fi a2 = 7 × 2 × 56 fi a = 28 16. The lines intersect at the point Ê ˆ p p ÁË a + b + c , a + b + c ¯˜ 17. Equation of the line is y - 4 = fi

18. p =

(12/7) - 4 (x - 3) (12/7) - 3

16x - 9y = 12 fi a = 3/4, b = 4/3 or a = 4/3, b = 3/4 ab a2 + b2

fi p2 =

a2 b2 a2 + b2

, Also p2 =

a2 + b2 2

fi a2b2 = 2p4 19. All lines except 3x + 4y – 7 = 0 pass through (– 1, 2) x + 2y – 3 = 0 & 3x + 4y – 7 = 0 pass through (1, 1).

IIT JEE eBooks: www.crackjee.xyz 15.50 Comprehensive Mathematics—JEE Advanced

20. (– 3, 1) lies on the given line Equation of the line through this vertex perpendicular to the given line is 7x – 4y + 25 = 0 others two sides pass through (1, 1) and are parallel to these two sides. 21. Given lines are at right angles. So ABC is an isosceles right angled triangle BC makes an angle of p/4 with both the lines. 22. Third vertex lies on y-axis, and is at a distance (2)2 – (1)2 from the origin. 23. Third side makes equal angles with the given lines. m +1 m–7 = – . If its slope is m, then 1 + 7m 1– m (m – 7) (1 – m) + (1 + m) (1 + 7m) = 0 6m2 + 16m – 6 = 0 fi m = – 3 or 1/3 24. x – 5xy + 4y = 0 fi (x – y) (x – 4y) = 0 Lines perpendicular to them passing through (1, – 1) are x + y = 0 and 4x + y = 4 – 1 = 3 2

2

25. m1 m2 m3 = – a/4, m1 m2 = – 1 fi m3 = a/4 3



26.

2

a a a – 2 –9× + a = 0 fi a = 5, – 4. 3 4 4 4



27. The point of intersection is given by h f - bg -5 2 - 2 = =2 x1 = 2 4 - 25 4 ab - h y1 =

gh - a f ab - h

2

=

-5 4 - 1 =1 4 - 25 4

28. (2x + 3y + p) (3x + 4y + q) = 6x2 + 17xy + 12y2 + 22x + 31y + 20 fi 2q + 3p = 22, 3q + 4p = 31, pq = 20 fi

p + q = 9 and p2 + q2 = 81 – 2 × 20 = 41

So fi



6 6 cos q + 4 + sin q = 8 3 3

cosq + sinq =

3 6

=

3 2

q=

a (a + p ) b (b + q) =p+q + 2 2 a2 + b2 + pa + qb = 2(p + q)

and

q-b b = p -a a



qa – pb = 0

31. If y = mx + c then (a) m = c = -3. similarly for (c) and (d)

2 , c = 5, (b) m = -

1 , 2

x y 32. Write the equation in the form + = 1 and the a b required sum is a2 + b2 33. (a) The lines are concurrent if 1 3 5

3 –K 2

–5 –1 = 0 fi K = 5 –12

(b) Slope of L2 = and

3 5 = – if K = – 6/5 K 2

3 1 = – if K = – 9 K 3

(c), (d) The lines do not form a triangle if they are concurrent or any two of them are parallel. 34. (a) 3.1 + 4.1 + 7 = 14 3(–3) + 4(–3)+ 7 = – 14 Similarly for other parts. 35. Match with the point of intersection of the lines 36. Area of the triangle formed by the lines in (a)

x-3 y-4 1 = = 6 cosq sinq 3 3+

pˆ Ê cos Á q - ˜ = cos p Ë 4¯ 6 p p q- = ± 4 6

p 5p or 12 12 30. Line joining (a, b) and (p, q) is perpendicular to ax + by = p + q and is bisected by it fi

l -5 5 2 -5 12 -8 = 0 fi l = 2 5 2 -8 -3 2x2 – 10xy + 12y2 + 5x – 16y – 3 = (2x – 6y) (x – 2y) + 3 (2x – 6y) – (x – 2y) – 3 = 0 fi (2x – 6y – 1) (x – 2y + 3) = 0

29.



37. (a) (b) (c) (d)

0 0 1 1 5 1 -2 1 = 2 2 1 3 1 Similarly for other parts. (x + y) (2x – 3y) (2x – 3y) (3x + 4y) (x – y) (2x – 3y) (3x + 4y) (x – y)

IIT JEE eBooks: www.crackjee.xyz Straight Lines 15.51

38. Vertices are (0, 0), (k/2, 0), (0, k/3). It is a right angled triangle, right angle at the origin. So orthocentre is at the origin for all values of k Ê k kˆ Circum-centre is Á , ˜ = (3, 2) if k = 12 Ë 4 6¯ kˆ ˜ = (3, 2) if k = 18 9¯ 1 k k Area of the triangle = ¥ ¥ = 192 if k2 = (48)2 2 2 3

43. In statement 1, let the third side be x = h, then the vertices are (2, 2) (h, h) and (h, 4 – h) fi k2 =

Êk Centroid is Á , Ë6

39. Equation of the line in new system of coordinates x cos a - y sin a x sin a + y cos a =1 + a b As it makes equal intercepts on the axes. b cos a + a sin a = a cos a – b sin a fi

tan a =

a-b a+b

p 4 -p (b) a = 0, b = 1 fi tan a = –1 fi a = 4 (a) a = 1, b = 0 fi tan a = 1 fi a =

(c)

a 1+ 3 = fi tan a = b 1- 3

(d)

b = a

3 +1 3 -1

fi tan a =

3 fia= 1

fia=

p 3 p 6

3 a 40. Required locus is x = , which 2 (a) does not intersect y-axis (b) intersects x-axis and y = x (c) is neither parallel nor perpendicular to y = x (d) is parallel to y-axis and x = a. 41. Statement-2 is true (See Text) Equation in statement-1 is x + 3 y + 5 + cot q ( 3 x – y – 7) = 0 where cot q As q π np /2 Using statement-2, Statement-1 is True. 42. Statement-2 is false as the distance of any such line from the origin is ±c a + b2 2

which is maximum when |c | is maximum

In statement-1, the other sides of the rectangle are 4x + 7y = 4 + 7 = 11, 7x – 4y = 7 – 4 = 3 and 7x – 4y = – 21 – 4 = – 25. So the farthest from the origin is 7x – 4y + 25 = 0.

2 2 1 1 h h 1 2 h 4–h 1 2

2

1

2

2k = h – 2 h – 2 0 h 2–h 1 fi K2 + (h – z)2 = 0 So P(h, k) lies on x2 + y2 – 4x + 4 = 0 Statement: 1 is True. of the second is 2. So it is also true. 44. Statement-2 is false, PQ should be perpendicular to the line and the mid-point of PQ lies on the line then Q P. Using the correct statement, statement-1 is True. 45. Statement-2: is true as the required equation is x2 – y2 xy = 5 3–3 fi x2 – y2 = 0 Statement-1 is true as the angle between the lines represented by ax2 + 2hxy + by2 + 2gx + 2dy + c = 0 is same as the angle between the lines represented by ax2 + 2hxy + by2 = 0, but it does not follow from the statement-2. 46. L1 and L2 intersect at A(4, -1) Equation of a line through A is x-4 y +1 = (1) cos q sin q coordinates of any point on this at a distance 2 from A is (2 cos q + 4, 2 sin q - 1) (2) If (1) represents L1 = AB then tan q = - 3/4 3 4 , cos q = ∓ 5 5 Taking cos q = 4/5, sin q = - 3/5 in (2) we get the coordinates of B (28/5, - 11/5) Similarly for the coordinates of C 47. Let the equation be 2x + 7y = k. Since it passes through B, k = - 21/5. fi sin q = ±

1 AB = 1 2 49. P: x2 – (y – 1)2 = 0 48. DE =

IIT JEE eBooks: www.crackjee.xyz 15.52 Comprehensive Mathematics—JEE Advanced

50. 51. 52.

53.

fi x + y – 1 = 0 and x – y + 1 = 0 Equation of the angle bisectors is x + y –1 x – y +1 =± 2 2 fi x = 0 or y – 1 = 0 Joint equation is xy – x = 0 Vertices of the triangle are (0, 1), (0, 3) and (2, 1). Point of intersection of P is (0, 1). Required equation is (x + y – 3) (x – y + 1) = 0 Given equations represent a rhombus with sides 3x ± 4y = 12 and 3x ± 4y = - 12, Length of the diag1 ¥ 6 ¥ 8 = 24 onals are 6 and 8 units. A = 2 2 2 A + 1 = 25 = (5) p p h= ,k= 2 cos a 2 sin a fi

1 1 4 1 1 4 + 2 = 2 fi locus of (h, k) 2 + 2 = 2 2 h k p x y p

it passes through (p + 1, p - 1) 1 1 4 = 2 fi p4 - 5p2 + 2 = 0 + 2 2 ( p + 1) ( p - 1) p 54. Locus is x2 + y2 = a2 + b2, since it passes through (3, 4), a2 + b2 = 25. Also 3a - 4b = 0 fi a2 = 16, b2 = 9 fi a2 – b2 = 7 m 55. Let the slopes be , m and mr, r 24 =8fim=2 product of slopes = m3 = 3 \ 24 – am + 26m2 – 3m3 = 0 fi 24 – 2a + 26 × 4 – 3 × 8 = 0 fi a = 52 24x3 – 52 x 2 y + 26xy2 – 3y3 = 0 fi (2x – y) (12x2 – 20xy + 3y2) = 0 fi (2x – y) (6x – y) (2x – 3y) = 0 The lines are 2x – y = 0, 6x – y = 0 and 2x – 3y = 0 Perpendicular distances from (1, 1) are 1 5 1 5 , so p = , , 5 37 13 37

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16 Circles and Systems of Circles 16.1

DEFINITION OF A CIRCLE

A circle is the locus of a point which moves in a plane, so that its distance from a in the plane is always . of the circle and the constant distance is called its .

2

+ y2 + 2 + 2 + c = 0 where g, f and c are constants. (i) Centre of this circle is (– g, – ). (ii) Its radius is

Illustration 1 A circle of area 9p square units has two of its diameters along the lines + y = 5 and – y = 1. Find the equation of the circle. Solution: The centre of the circle is the point of intersection of the diameters + y = 5 and – y = 1, which is (3, 2). If is the radius of the circle, then p 2 = 9p fi = 3 and the equation of the circle is ( – 3)2 + (y – 2)2 = 32 fi 2 + y2 – 6 – 4y + 4 = 0 2. An equation of a circle with centre (0, 0) and radius is 2 + y2 = 2 3. An equation of the circle on the line segment joining ( 1, y1) and ( 2, y2) as diameter is ( – 1) ( – 2) + ( y – y 1) ( y – y 2) = 0 Illustration 2 The line 3 + 4y A and B. Find the equation of the circle drawn on AB as diameter. Solution: 3 + 4y – 12 = 0 meets A(4, 0) and y B(0, 3). Equation of the circle on the line joining A and B as diameter is ( – 0) ( – 4) + (y – 0) (y – 3) = 0 fi 2 + y2 – 4 – 3y = 0 4. General equation of a circle is

≥ c).

g 2 - c if g2 – c ≥ 0, and that on the

EQUATIONS OF A CIRCLE

1. An equation of a circle with centre (h, k) and radius is ( – h)2 + ( y – k)2 = 2

2

(iii) Length of the intercept made by this circle on the -

16.2

g 2 + f 2 - c , ( g2 +

y

is 2

f 2 - c if f 2 – c ≥ 0.

Illustration 3 Show that + y – 1 = 0 is the equation of a diameter of the circle 2 + y2 – 6 + 4y + c = 0 for all values of c. Ê 1 1ˆ Solution: Centre of the circle is Ë- , - ¯ 2 2 i.e (3, – 2) which lies on the line + y – 1 = 0 and any line passing through the centre of the circle is a diameter of the circle. Hence + y – 1 = 0 is a diameter of the circle. 5. General equation of second degree 2 +2 + by2 + 2 + 2 + c = 0 in and y, represents a circle if and only if 2 y2, i.e., = b π 0. is zero, i.e., h = 0. 2 2 (iii) g + – ≥ 0 Illustration 4 Find the centre and radius of the circle 3 2 + ( + 1)y2 + 6 – 9y + + 4 = 0 Solution: Since the given equation represents a circle, 2 y2 fi 3 = + 1 fi = 2 and the equation of the circle becomes 3 2 + 3y2 + 6 – 9y + 6 = 0 or 2 + y2 + 2 – 3y + 2 = 0 whose centre is (– 1, 3/2) and radius =

1 + (3 / 2 ) - 2 = 2

5 . 2

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SOME RESULTS REGARDING CIRCLES

1. y1) lies outside, on or inside a circle ∫ + c = 0, according as S1 ∫ 21 + y21 + 2 = or < 0.

2

Point P( 1, + y2 + 2 + 2 1+2 1 + c >,

Find the values of for which the point ( , ), > 0, lies outside the circle 2 + y2 – 2 + 6y – 6 = 0 Solution: The point ( , a) lies outside the given circle if 2 + 2–2 +6 –6>0 fi 2 ( 2 + 2 – 3) > 0 fi ( + 3) ( – 1) > 0 fi > 1 as > 0 2. of any point on the circle ( – h)2 + ( y – k)2 = 2 are given by (h + cos q, k + sin q), with 0 £ q < 2p. In particular, parametric coordinates of any point on the circle 2 + y2 = 2 are ( cos q , sin q ) with 0 £ q < 2p. 3. An equation of the to the circle 2 + y2 + 2 + 2 + c = 0 at the point ( 1, y1) on the circle is 1 + 1 + ( + 1) + ( y + y 1) + c = 0 4. An equation of the to the circle 2 + y2 + 2 + 2 + c = 0 at point ( 1, y1) on the circle is y - y1 x - x1 = y1 + f x1 + g 5. Equations of the and to the circle 2 + y2 = at the point ( 1, y1) on the circle are, respectively, x y 2 and = 1 + 1 = x1 y1

2

+ y2 = 25 at – 3y = 0

meets the circle. Solution: The diameter meets the given circle at the point Equation of the tangent to the circle at this point is (3) + y(4) = 25 fi 3 + 4y = 25 6. The line = + c is a tangent to the circle 2 + y2 = 2 2 2 if and only if c = (1 + 2). 2



2

=4fi

=2

and S1 ∫ 12 + y12 + 2 1 + 2fy1 + c In particular, an equation of the chord of the circle 2 + y2 = 2, whose mid-point is ( 1, y1), is 1 + yy1 = 12 + y12. 9. An equation of the of the tangents drawn from a point ( 1, y1) outside the circle S = 0, is T = 0. (S and T drawn from a point ( 1, y1)

10.

outside the circle S = 0, to the circle, is

S1 . (S and S1 are

as Illustration 8 Find the length of a tangent drawn from the point (3, 4) to the circle 2 + y2 – 4 + 6y – 3 = 0. T 4 P(3, 4)

C(2, -3)

2

Illustration 6

7. The lines y =

(4)2 = 2 (1 + ( 3 )2) so the required radius is 2.

8. An equation of the of the circle S = 2 + y2 + 2 + 2fy + c = 0, whose mid-point is ( 1, y1), is T = S1, where T ∫ 1 + yy1 + ( + 1) + ( + y1) + c

Illustration 5

Find the equation of the tangent to the circle

Solution: Let the radius of the circle be , so its equation is 2 + y2 = 2. As the line y = 3 x + 4 touches this circle,

Fig. 16.1

Solution: Centre of the circle is (2, – 3) and radius = 22 + (-3)2 + 3 = 4. So if PT is the tangent from P(3, 4) to the circle, then (PT)2 = (PC)2 – (CT)2 = (3 – 2)2 + (4 + 3)2 – 42 = 34. The required length of the tangent is 34 . Note If S ∫ 2 + y2 – 4 + 6y – 3 then length of the tangent from P(3, 4) to this circle is S=

(3)2 + (4)2 - 4(3) + 6(4) - 3 = 34 .

± r 1 + m 2 are tangents to the circle

+ y2 = 2 tangents are ± = 0.

. If

Illustration 7 The line y = 3 x + 4 touches a circle with centre at the origin. Find the radius of the circle.

11. Two circles with centres C1( 1, y1) and C2( 2, y2), and radii 1, 2 respectively, (i) if |C1 C2| = 1 + 2. The point of contact is Ê r1 x2 + r2 x1 r1 y2 + r2 y1 ˆ , ÁË r + r r1 + r2 ˜¯ 1 2

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Illustration 9 Show that the circles 2 + y2 – 2 = 0 and 2 + y2 – 8 + 12 = 0 touch each other. Find the point of contact. Solution: 1. Centre of the second circle is (4, 0) and its radius in 2. Since the distance between the centres = 3, which is equal to the sum of the radii. The two circles touch each other divides the join of (1, 0) and (4, 0) in the ratio 1 : 2. Y

(1, 0)

(4, 0)

X

O

(i) Equation of the polar of P (x1, y1) with respect to the circle S ∫ x2 + y2 + 2gx + 2fy + c = 0, is T = 0, where T is as defined in (8) above. (ii) If the polar of P with respect to a circle passes through Q, then the polar of Q with respect to the same circle passes through P. Two such points P and Q are called conjugate points of the circle. 15. If lengths of the tangents drawn from a point P to the two circles S1 ∫ 2 + y2 + 2g1 + 2f1y + c1 = 0 and S2 ∫ 2 + y2 + 2g2 + 2f2y + c2 = 0 are equal, then the locus of P is called the of the two circles S1 = 0 and S2 = 0, and its equation is S1 – S2 = 0, i.e., 2(g1 – g2) + 2(f1 – f2)y + c1 – c2 = 0 line joining their centres.

Fig. 16.2

(ii) π 2. The point of contact is

if | C1 C2 | = |

1



Ê r1 x2 - r2 x1 r1 y2 - r2 y1 ˆ , ÁË r - r r1 - r2 ˜¯ 1 2 12. An equation of the points ( 1, y1) and ( 2, y2) is

16.4

passing through the

x ( – 1) ( – 2) + (y – y1) (y – y2) + l x1 x2

y 1 y1 1 = 0 y2 1

13. An equation of the family of circles which touch the line y – y1 = ( – 1) at ( 1, y1 , is 2 2 ( – 1) + (y – y1) + l[(y – y1) – ( – 1)] = 0 If – 1)2 + (y – y1)2 + l( – 1) = 0. 14. Let QR be a chord of a circle passing through the point Q and R of P( 1, y1 this chord intersect at the point L(h, k) (Fig. 16.3). Then locus of L is called as of P with respect to the circle, and P is called the of its polar. R L (h, k)

of the three circles, if the centres of these circles are noncollinear.

2 |, 1

SPECIAL FORMS OF EQUATION OF A CIRCLE

1. An equation of a circle with centre ( , ), radius ( – )2 + (y – )2 = 2 the points ( , 0) and (0, ). Illustration 10

equation. Solution: of the circle can be (± 3, ± 3) and hence there are four equations are 2 + y2 ± 6 ± 6y + 9 = 0. 2. An equation of a circle with centre ( 1, ), radius is ( – 1)2 + (y – )2 = 2. This touches the 1, 0). is 3. An equation of a circle with centre ( , y1), radius y1). ( – )2 + ( – y1)2 = 2. This touches the y 4. An equation of a circle with centre ( /2, b/2) and radius

(a2 + b2 )/ 4 is

Q P (x1, y1) Fig. 16.3

is

2

+ y2 – – by = 0 This circle passes through the origin (0, 0), and has intercepts and b on the and y

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SYSTEMS OF CIRCLES

2

Let S ∫ + y + 2 + 2fy + c, S¢ ∫ + y + 2g¢ + 2f ¢y + c¢ and L ∫ + by + k¢. 1. If two circles S = 0 and S¢ = 0 intersect at real and distinct points, then S + lS¢ = 0 (l π –1) represents a passing through these points (l being a parameter), and S – S¢ = 0 (for l = –1) represents the of the circles. 2. If two circles S = 0 and S¢ = 0 touch each other, then to S – S¢ = 0 represents equation of the the two circles at their point of contact. 3. If two circles S = 0 and S¢ = 0 intersect each other (the tangents at the point of intersection of the two circles are at right angles), then 2gg¢ + 2ff¢ = c + c¢. 4. If the circle S = 0 intersects the line L = 0 at two real and distinct points, then + lL = 0 represents a family of circles passing through these points. 2

2

2

2

Illustration 11 Find the equation of the circle described on the chord 3 + y + 5 = 0 of the circle 2 + y2 = 16 as diameter. Solution: Equation of any circle passing through the points of intersection of the chord and the circle is 2 + y2 – 16 + l(3 + y + 5) = 0 The chord 3 + y + 5 = 0 is a diameter of this circle if -3 l - l , of the circle lies on the chord. the centre 2 2

(

( )

)

-3l l - +5=0 2 2 fi l = 1 and the required equation of the circle is 2 + y2 + 3 + y – 11 = 0.



3

5. If L = 0 is a tangent to the circle S = 0 at P, then S + lL = 0 represents a family of circles touching S = 0 at P, and having L = 0 as the common tangent at P. 6. if every pair of circles of the system have the same radical

circles is

2

+ y2 + 2

+ c = 0, where g is a variable and c

y members of the system which are of zero radius. Thus the

+ y2 + 2

(

)

+ c = 0 are ± c, 0 if c ≥ 0.

The equation S + lS¢ = 0 (l π –1) represents a family of S = 0 and S¢ = 0. Two system of circles such that every circle of one system cuts every circle of the other system orthogonally are said to be conjugate system of circles. For instance, 2 + y2 + 2 + c = 0 and 2 + y2 + 2 – c = 0, where g and f are variables and c is constant, represent two

16.6

COMMON TANGENTS TO TWO CIRCLES

If ( – g1)2 + (y – f1)2 = 12 and ( – g2)2 + (y – f2)2 = 22 are two circles with centres C1 (g1, f1) and C2 (g2, f2) and radii 1 and 2 respectively, then we have the following results regarding their common tangents. 1. When C1 C2 > a1 + a2 i.e., distance between the centres is greater than the sum of their radii, the two circles do not intersect with each other and four common tangents can be drawn to two circles. Two of them are direct common tangents and the other two are transverse common tangents. The points T1, T2 of intersection of direct common tangents and transverse common tangents respectively, always lie on the line joining the centres of the two circles and divide it externally and internally respectively in the ratio of their radii. Direct Transverse common tangent

comm on tan gent

C1 T2

C2

T1

gent on tan m m o c Direct Fig. 16.4

C1T1 a1 = T1C2 a2

C1T2 a1 = internally T2C2 a2

2. When C1 C2 = a1 + a2 i.e., the distance between the centres is equal to the sum of the radii, the two circles touch each other externally, two direct common tangents are real and distinct and the transverse common tangents coincide.

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Illustration 12

C1

T 2 C2

T1

Fig. 16.5

3. When C1 C2 < a1 + a2 i.e., the distance between the centres is less than the sum of the radii, the circles intersect at two real and distinct points, the two direct common tangents are real and distinct while the transverse common tangents are imaginary.

Find the number of common tangents to the circles 2 + y2 = 16 and 2 + y2 – 2y = 0. Solution: Centre of the circle 2 + y2 = 16 is (0, 0) and its radius is 4. Centre of the circle 2 + y2 – 2y = 0 is (0, 1) and its radius is 1. Distance between the centre = 1 which is less than the difference between the radii. common tangent to the given circles. (0, 4)

O C1

T1

C2

Fig. 16.9 Fig. 16.6

SOLVED EXAMPLES

SINGLE CORRECT ANSWER TYPE QUESTIONS The circle passing through (–1, 0) and

Example 1 touching the y Fig. 16.7

C1 C2

Fig. 16.8

4. When C1 C2 = |a1 – a2| (a1 π a2) i.e., the distance between the centres is equal to the difference of their radii, the circles touch each other internally, two direct common tangents are real and coincident while the transverse common tangents are imaginary. (Fig 16.7) 5. When C1 C2 < |a1 – a2| a1 π a2 i.e., the distance between the centres is less than the difference of the radii, one circle with smaller radius lies inside the other and the four common tangents are all imaginary. (Fig 16.8)

Ê 3 ˆ (a) Á - , 0˜ Ë 2 ¯

Ê 5 ˆ (b) Á - , 0˜ Ë 2 ¯

Ê 3 5ˆ (c) Á - , ˜ Ë 2 2¯

(d) (–4, 0)

(d) Centre of the circle lies on the line y = 2. Let the coordinates of the centre by (h, 2) where h < 0. Radius of the circle is –h. fi

–h =

(h + 1)2 + (2 - 0)2 as it passes through (–1, 0)



h2 = h2 + 2h + 5 fi h = -

5 2

So the equation of the circle is 5ˆ Ê ÁË + ˜¯ 2

2

Ê 5ˆ + (y – 2)2 = Á ˜ Ë 2¯

which passes through (–4, 0)

2

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Locus of the mid-points of the chord of Example 2 contact of tangents drawn from the point lying on the straight line 4 – 5y = 20 to the circle 2 + y2 = 9 is (a) (b) (c) (d)

20( 20( 36( 36(

2 2 2 2

+ + + +

y 2) y 2) y 2) y 2)

– + – +

36 36 20 20

+ – + –

45y 45y 45y 45y

= = = =

0 0 0 0

\ (AB)2= (OA)2 + (OB)2 = 4 + 4 = 8 fi AM = (1 2) AB =

2

fi (OM)2= (OA)2 – (AM)2 = 4 – 2 = 2 fi h2 + k2 = 2 Therefore, the locus of (h, k) is

2

+ y2 = 2.

(a) Coordinates of any point P on the line Ê 4a - 20 ˆ 4 – 5y = 20 are Á a , ˜ Ë 5 ¯ Equation of the chord of contact of P with respect to the 1 circle 2 + y2 = 9 is a + (4a – 20)y = 9 5 fi

4 ˆ Ê a Á x + y˜ + (-4 y - 9) = 0 Ë 5 ¯

This chord passes through the point of intersection of the lines 4 x + y = 0 and 4y + 9 = 0 5 Ê 9 9ˆ that is Á , - ˜ Ë 5 4¯ Let (h, k) be the mid point of this chord, then its equation is + ky – 9 = h2 + k2 – 9 Ê 9 9ˆ Since it passes through Á , - ˜ Ë 5 4¯ 9 9 h - k = h2 + k2 5 4 fi

20(h2 + k2) – 36h + 45k = 0

Fig. 16.10

A circle C touches the Example 4 2 circle + (y – 1)2 of the circle is given by (a) {( , y) : 2 = 4y} » {(0, y) : y £ 0} (b) {( , y) : y = 2} » {(0, ) : £ 0} (c) {( , y) : 2 + (y – 1)2 = 4} » {(0, y) : y £ 0} (d) {( , y) : 2 + 4y = 0} » {(0, y) : y £ 0} (a)

y

All the circles having centre on the negative y the origin touch the circle 2 + (y – 1)2 = 1 and the Let A1(0, 1) be the centre of the given circle and A2 ( , y) be the centre of C. If 1, 2 be the radii of these circles then 1 + 2 = A1 A2 fi

1+|y|=



1 + y2 + 2|y| =

A1 r1

2

+ y2 – 2y + 1

2

= 21y1 + 2y



2

= 4y If y ≥ 0

Locus of the mid-points of the chords of Example 3 2 2 the circle + y = 4 which subtend a right angle at the centre is (a) + y = 2 (b) 2 + y2 = 1 (d) – y = 0. (c) 2 + y2 = 2

Thus the locus of the centre is

(c) Let O be the centre of the circle 2 + y2 = 4, and let AB be a chord of this circle, so that –AOB = p 2. Let M(h, k) be the mid-point of AB. Then OM is perpendicular to AB (Fig. 16.10).

Example 5

x

x 2 + ( y – 1)2



2

C A2

Fig. 16.11

So the Locus of mid-points of the chord of contacts is 20( 2 + y2) – 36 + 45y = 0

{( , y) :

r2

= 4y} » {(0, y) : y £ 0} Three circles with radii 3 cm, 4 cm and

A is the point of intersection of tangents to these circles at their points of contact, then the distance of A from the points of contact is (a) (c)

3

(b) 2

(c)

5

(d)

6

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R 3

4

3

A

5

Let the coordinates of A be ( , 0) and that of B be (0, b) (Fig. 16.13). Since –AOB = p/2, the line AB is a diameter of the circle circumscribing the triangle OAB, its centre is the mid-point of AB, i.e., ( /2, b/2), and its radius is

Q 4

(1 2 ) AB = (1 2) a2 + b2 . Therefore, equation of the – by = 0, and circle through O, A and B is 2 + y2 – the equation of the tangent at the origin to this circle is + by = 0. If AL and BM are the perpendiculars from A and B to this tangent, then

5 P

AL =

a2

=

a2 + b2

Fig. 16.12

From the Fig. 16.12 it is clear that A is the centre of the incircle of the triangle PQR formed by the lines joining the centres of the given circles. We have PQ = 5 + 4 = 9, QR = 4 + 3 = 7, RP = 3 + 5 = 8, distance of A from the points of contact is the radius of the incircle of triangle From trigonometry = D/ where D is the area and 2 is the perimeter of the triangle PQR D=

12(12 – 7)(12 – 8)(12 – 9) =

12 ¥ 5 ¥ 4 ¥ 3

= 12 5 = (1/2) (9 + 7 + 8) = 12, Hence

=

5

A and Example 6 B. A circle is circumscribed about the triangle OAB. If and are the distances of the tangent to the circle at the origin from the points A and B respectively, the diameter of the circle is (a) ( + ) (b) + (c) ( + ) (d) (1 2) ( + ). (b)

Hence

+

and BM =

=

=

b2 a2 + b2

a2 + b2 is the diameter of the circle.

If a circle passes through the point ( , b) Example 7 and cuts the circle 2 + y2 = k2 orthogonally, equation of the locus of its centre is (a) 2 + 2by = 2 + b2 + k2 (b) + by = 2 + b2 + k2 2 (c) + y2 + 2 + 2by + k2 = 0 (d) 2 + y2 – 2 – 2by + 2 + b2 – k2 = 0. (a) Let equation of the circle through ( , b) be 2 + y2 + 2 + 2fy + c = 0 (1) 2 2 Then + b + 2 + 2fb + c = 0 (2) Since (1) cuts the circle

2

+ y2 = k2 orthogonally, we have

2g ¥ 0 + 2f ¥ 0 = c – k2



c = k2

so that from (2), we get 2 + b2 + 2 + 2fb + k2 = 0, and the locus of the centre (–g, –f) of (1) is 2

+ 2by – (

2

+ b2 + k2) = 0.

Equation of the circle which passes Example 8 through the origin, has its centre on the line + y = 4 and cuts the circle 2 + y2 – 4 + 2y + 4 = 0 orthogonally, is (a) 2 + y2 – 2 – 6y = 0 (b) 2 + y2 – 6 – 3y = 0 (c) 2 + y2 – 4 – 4y = 0 (d) none of these. .

(c)

Since the centre of the required circle lies on + y = 4, let (g, 4 – g) be this centre. Since the circle passes through the origin, let its equation be Fig. 16.13

2

+ y2 – 2

– 2(4 – g)y = 0

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As this circle cuts the given circle orthogonally, we have 2 ¥ g ¥ 2 – 2(4 – g) = 4 fi 6g = 12 fi g = 2 so that equation of the required circle is

2

+ y2 – 4 – 4y = 0.

If O is the origin and OP, OQ are distinct Example 9 tangents to the circle 2 + y2 + 2 + 2fy + c = 0, the circumcentre of the triangle OPQ is (a) (– , – ) (b) (g, f) (d) none of these. (c) (–f, –g) .

which passes through (1, 1) for all values of , while the points given by (b), (c) and (d) do not satisfy this equation. If OA and OB are the tangents from the Example 11 origin to the circle 2 + y2 + 2 + 2fy + c = 0, and C is the centre of the circle, the area of the quadrilateral OACB is (a)

(d)

Since PQ is the chord of contact of the tangents from the origin O to the circle 2

Finally, from (6), we get c = – 2 2 + 2 ( + 1) = 2 , so that the equation of the circle (1) can be written as 2 + y2 – ( + 1) – ( + 1) + 2 = 0

+ y2 + 2

+ 2fy + c = 0,

(1)

equation of PQ is + fy + c = 0

(2)

An equation of a circle through the intersection of (1) and (2) is given by 2 + y2 + 2 + 2fy + c + l (g + fy + c) = 0 (3) If the circle (3) passes through O, the origin, then c + l c = 0, i.e., l = –1, and the equation of the circle (3) 2 + y2 + + fy = 0 becomes

1 c( g 2 + f 2 - c ) 2

(c) c g 2 + f 2 - c

(b) (d)

c( g 2 + f 2 - c ) g2 + f 2 - c c

.

(b) Since OA = OB and CA = CB, the diagonal OC divides the quadrilateral OACB in two equal right-angled triangles, OAC and OBC (Fig. 16.14). Therefore, the area of the quadrilateral OACB is 2(area of triangle OAC) = 2 ¥ (1 2) OA ¥ AC =

0 + 0 + 2g ¥ 0 + 2 f ¥ 0 + c

g2 + f 2 - c

= c( g 2 + f 2 - c )

Centre of this circle is (–g/2, –f/2), and hence it is the circumcentre of the triangle OPQ. Example 10 The circle passing through the distinct points (1, ), ( , 1) and ( , ) for all values of , passes through the point (a) (1, 1) (b) (–1, –1) (c) (1, –1) (d) (–1, 1). (a) Let equation of the circle passing through the given points be 2

Then and

+ y2 + 2 + 2 1 + 2 + 2g + 2 2 + 1 + 2 + 2f 2 + 2+2 +2

From (2) and (3), we have 2g (1 – ) + 2f( – 1) = 0

+ + + +

c= c= c= c=



0 0, 0 0

(1) (2) (3) (4)

2( – 1) (f – g) = 0

Fig. 16.14

Example 12 The angle between a pair of tangents drawn from a point P to the circle 2 + y2 + 4 – 6y + 9 sin2 a + 13 cos2 a = 0 is 2a. The equation of the locus of the point P is (a) 2 + y2 + 4 – 6y + 4 = 0 (b) 2 + y2 + 4 – 6y – 9 = 0 (c) 2 + y2 + 4 – 6y – 4 = 0 (d) 2 + y2 + 4 – 6y + 9 = 0 (d)

That is, f = g, because π 1, otherwise the three given points would be the same. Again, from (3) and (4), we get 2 + 1 + 2 + 2g + c = 0 (5) (6) and 2 2 + 2 + 2 + c= 0 Subtracting these, we get 2 – 1 + 2g( – 1) = 0



2g = – ( + 1) Fig. 16.15

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Let PA and PB be the tangents drawn from the point P(h, k) to the given circle with centre C(–2, 3). So that –APB = 2a and –APC = –CPB = a (Fig. 16.15) –PAC = –PBC = 90° From triangle PCA, CA and CA = 4 + 9 - (9sin 2 a + 13cos2 a ) CP = 2 sin a fi CP = 2. 4 = h2 + k2 + 4h – 6k + 13.

fi sin a = fi

The locus of P(h, k) is therefore 2 + y2 + 4 – 6y + 9 = 0. Example 13

Equation of a circle through the origin

points are (1, 2), (4, 3) is (a) 2 + y2 – 2 + 4y = 0 (b) 2 + y2 – 8 – 6y = 0 (c) 2 2 + 2y2 – – 7y = 0 (d) 2 + y2 – 6 – 10y = 0 (c) Since the limiting points of a system of comembers of the system are ( – 1)2 + (y – 2)2 = 0 fi and ( – 4)2 + (y – 3)2 = 0 fi

Fig. 16.16

Since AM is parallel to OX, a = + and b = fi

+ y – 2 – 4y + 5 + l ( 2

2

and y = b

2

As A( , y) lies on the circle + y = we have (a – )2 + b 2 = 2 fi a2 – 2 a + b2 = 0 fi locus of M (a, b ) is 2 + y2 = 2 . Example 15 If common chord of the circle C with centre at (2, 1) and radius and the circle 2 + y2 – 2 – 6y + 6 = 0 is a diameter of the second circle, then value of is (a) 3 (b) 2 (c) 3/2 (d) 1 (a)

2

2

2

+ y – 2 – 4y + 5 = 0 + y2 – 8 – 6y + 25 = 0. fi

2

=a–

2

2

2

+ y – 8 – 6y + 25) = 0

Equation of the circle C is ( – 2)2 + (y – 1)2 = 2 2 + y2 – 4 – 2y + 5 – 2 = 0

Equation of the common chord is 2

+ y2 – 4 – 2y + 5 –

2

2

+ y2 – 2 – 6y + 6) = 0

It passes through the origin if 5 + 25l = 0 or l = – (1 5) ,

(

which gives the equation of the required circle as 5( 2 + y2 – 2 – 4y + 5) – ( 2 + y2 – 8 – 6y + 25) = 0 fi 4 2 + 4y2 – 2 – 14y = 0 fi 2 2 + 2y2 – – 7y = 0.

If it is a diameter of the second circle, it passes through the centre (1, 3) of the circle

Example 14 If a line segment AM = moves in the plane XOY remaining parallel to OX so that the left end point A slides along the circle 2 + y2 = 2, the locus of M is (a) 2 + y2 = 4 2 (b) 2 + y2 = 2 (c) 2 + y2 = 2 (d) 2 + y2 – 2 – 2 = 0 (b). Let the coordinates of A be ( , y) and M be (a, b ) (Fig. 16.16).

fi 2 – 4y +

2

So 2 – 4 ¥ 3 +

)–(

+1=0

2

+1=0fi

2

=9fi =3

Example 16 Tangents drawn from the point P(1, 8) to the circle 2 + y2 – 6 – 4y – 11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle in PAB is (a) 2 + y2 + 4 – 6y + 19 = 0 (b) 2 + y2 – 4 – 10y + 19 = 0 (c) 2 + y2 – 2 + 6y – 29 = 0 (d) 2 + y2 – 6 – 4y + 19 = 0 (b) Equation of AB the Chord of contact of 1◊ + 8y – 3( + 1) – 2(y + 8) – 11 = 0 fi

– 2 + 6y – 30 = 0

is

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– 3y + 15 = 0

Example 18 circle 2 + y2=

A P B Fig. 16.17

Equation of any circle through. AB is 2 + y2 – 6 – 4y – 11 + l( – 3y + 15) = 0 It will pass through P(1, 8) if 1 + 64 – 6 – 32 – 11 + l (1 – 24 + 15) = 0 fi

16 – 8l = 0



l=2

Thus, equation of the required circle is fi

2

+ y2 – 6 – 4y – 11 + 2( – 3y + 15) = 0

2

+ y2 – 4 – 10y + 19 = 0

be a quadrilateral with area Example 17 Let 18, with side AB parallel to CD and AB = 2CD. Let AD be perpendicular to AB and CD. If a circle is drawn inside the quadrilateral ABCD touching all the sides, then its radius is (a) 3 (b) 2 (c) 3/2 (d) 1 (b) Y Let A(0, C (a, 2r) (0, 2 r ) D 0), (2 , 0) C( , 2 ), D(0, 2 ). Since the circle touches the parallel sides AB and B (2a, 0) A (0, 0) CD its diameter is 2 so that the radius of the circle is . Since it touches AD, distance of the centre from AB and AD is and the coordinates of the centre are ( , ). 2r ( –2 ) Equation of BC is y = –a or 2 + – 4 = 0

As the circle touches BC, distance of the centre from BC is also . so fi fi

2r 2 + ar – 4ar 4r 2 + a2

=

(2 – 3 )2 = 4 12 = 8

2

Area of the quadrilateral fi 3 = 18 fi = 6 (2) From (1) and (2) we get

+

2

2

(1) 1 = (2 + ) ¥ 2 = 18 2

= 3, = 2

So the required radius is 2.

An equilateral triangle is inscribed in the 2

, 0). The equation of

(a) 2 – = 0 (b) + = 0 (c) 2 + = 0 (d) 3 – 2 = 0 (c) ( ABC inscribed in the circle 2 + y2 = 2 Let M be the middle point of the side BC, then MOA is perpendicular to BC and O being the centroid of the triangle OA = 2 (Fig. 16.18) So if (h, k) be the coordinates of M, 2h + a 2k + 0 then = 0 and =0 3 3 fi h = - ( /2) and k = 0 and hence the equation of BC is = – /2 or 2 + = 0.

Fig. 16.18

Example 19 The lines 2 – 3y = 5 and 3 – 4y = 7 are the diameters of a circle of area 154 square units. An equation of this circle is (p = 22/7) (a) (b) (c) (d)

2

+ y2 + 2 + y2 + 2 2 + y2 – 2 2 + y2 – 2 2

– 2y = 62 – 2y = 47 + 2y = 47 + 2y = 62

(c) The centre of the circle is the point of intersection of the given diameters 2 – 3y = 5 and 3 – 4y = 7, which is (1, – 1) and if is the radius of the circle, then 7 p 2 = 154 fi 2 = 154 × fi = 7. 22 Hence an equation of the required circle is ( –1)2 + (y + 1)2 = 72 fi 2 + y2 – 2 + 2y = 47. Example 20 The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3 is (b) 2 + y2 = 16 2 (a) 2 + y2 = 9 2 2 2 2 (c) +y =4 (d) 2 + y2 = 2 (c)

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As the centre of the circumcircle of an equilateral triangle is its centroid and the distance of the from ¥ 3 = 2 and hence the equation of the circumcircle is 2

+ y 2 = 4 2.

Example 21 A line is drawn through the point P(3, 11) to cut the circle 2 + y2 = 9 at A and B. Then PA ◊ PB is equal to (a) 9 (b) 121 (c) 205 (d) 139 (b) PA . = (PT)2 where PT is the length of the tangent from P to the circle. Hence PA . = (3)2 + (11)2 – 9 = 112 = 121.

Fig. 16.19

AL = (1/2) AB = /2

then

PM = (1/2) PQ = b/2 and CA = CP (radii of the same circle) fi

2

+

b2 a2 = h2 + 4 4

Alternately



Equation of any line through (3, 11) is

so that locus of (h, k) is 4(

x -3 y - 11 = = cos q sin q

(a) 0 < (c) 9


9 + 25 - p fi

> 25

(1)

5>

9 + 25 - p fi

>9

(2)

and the point (1, 4) lies inside the circle fi

1 + 16 – 6 – 10 ¥ 4 +

b) as shown in the Fig. 16.20, then AP is equal to (a)

a2 + p2 + b2 + p2

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(b)

a2 - p2 + b2 - p2

(c)

a2 - p2 - b2 - p2

(d)

a2 + p2 - b2 + p2

So, if (h, ) be any point on the locus, then 3(h2 + k2) – 9 + 8 2 = b2 So the required locus of (h, k) is 3( 2 + y2) – 9 + 8 2 – b2 = 0. are two equal chords of the Example 26 If OA and 2 2 circle + y – 2 + 4y = 0 perpendicular to each other and passing through the origin O, the slopes of OA and OB are the roots of the equation (b) 3 2 – 8 – 3 = 0 (a) 3 2 + 8 – 3 = 0 (d) 8 2 + 3 – 8 = 0 (c) 8 2 – 3 – 8 = 0

(c)

(b)

Fig. 16.20

The given circles are concentric with centre at (0, 0) and the length of the perpendicular from (0, 0) on the given line is . Let OL = then

AL =

(O A)2 - (O L )2

=

a2 - p2

and

PL =

( O P )2 - ( O L )2

=

b2 - p2



AP =

a2 - p2 -

b2 - p2

Example 25 Two points P and Q joining the points A (0, 0) and B (3 , 0) such that AP = PQ = QB. Circles are drawn on AP, PQ, and QB as diameters. The locus of the point S, the sum of the squares of the lengths of the tangents from which to the three circles is equal to b2, is (a) 2 + y2 – 3 + 2 2 – b2 = 0 (b) 3( 2 + y2) – 9 + 8 2 – b2 = 0 (c) 2 + y2 – 5 + 6 2 – b2 = 0 – b2 = 0 (d) 2 + y2 – (b)

Let the equations of OA and OB be y – = 0 and + = 0 since OA = OB, lengths of the perpendiculars from the center (1, – 2) of the circle on OA and OB are also equal. –2–m – 2m + 1 = fi ( + 2)2 = (2 – 1)2 fi 2 1+ m 1 + m2 fi

3

2

–8 –3=0

Example 27 An equation of the chord of the circle 2 + y2 = 2 passing through the point (2, 3) farthest from the centre is (a) 2 + 3y = 13 (b) 3 – y = 3 (c) – 2y + 4 = 0 (d) – y + 1 = 0 (a) Let P (2, 3) be the given point, M be the middle point of a chord of the circle 2 + y2 = 2 through P. Then the distance of the centre O of the circle from the chord is OM. and

(OM)2 = (OP)2 - (PM)2

PM is minimum, i.e. M coincides with P i.e. P is the middle point of the chord. Hence the equation of the chord is 2◊ + 3◊y–

2

= (2)2 + (3)2 –

2

Since AP = PQ = QB. The coordinates of P are ( , 0) and of Q are (2 , 0). Equations of the circles on AP, PQ and QB as diameters are respectively.

Fig. 16.21

( – 0) ( – ) + y2 = 0, ( – ) ( – 2 ) + y2 = 0 and ( – 2 ) ( – 3 ) + y2 = 0

(Fig. 16.22)

Fig. 16.22



2 + 3 = 13.

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The lengths of the intercepts made by any

Example 28

lies on the line (s) represented by (a) + 1=0 2 2 (c) – y = 0 (c)

+y=1 –y=1

(b) (d)

Example 30 Let PQ and RS emities of a diameter PR of a circle of radius . Such that PS and RQ intersect at a point X on the circumference of the circle, then diameter of the circle equals. PQ + RS (b) (a) PQ.RS 2

Let the equation of any circle be 2

+ y2 + 2

+2

+c=0

If

2 1,

+



1

2

g -c = 2

Since the lengths of these intercepts are equal f -c 2

and fi fi

2

2

=

RS = tan q PR PQ RS . =1 PR PR (PR)2 = PQ . PS

= (– g)2 = (– f)2

Therefore, centre lies on Example 29

PQ 2 + RS 2 2

PQ = tan (p /2 –q) = cot q. PR

2 f2 -c

Similarly length of the intercept of the y



(d)

From the Fig. 16.23 we have

=0

( x1 + x2 )2 - 4 x1 x2 = 2 g 2 - c

|=

2 PQ.RS PQ + RS

(a)

are roots of ( ), then length of the intercept on

2

|

+2

(i) y = 0 in (i)

For intercept made by the circle on fi

(c)

If

2

– y2 = 0

> 2b > 0 then the positive value of

for which y = – b 1 + m2 is a common tangent to 2 + y2 = b2 and ( – )2 + y2 = b2 is 2b

(a)

a 2 - 4b 2 2b a - 2b

(c)

a 2 - 4b 2 2b

(b)

(d)

b a - 2b

(a) = – b 1 + m2 is a tangent to the circle + y = b for all values of . If it also touches the circle ( – )2 + y2 = b2, then the length of the perpendicular from its centre ( , 0) on this line is equal to the radius b of the circle, which gives 2

2

2

ma - b 1 + m2 1 + m2

=±b = 0, so we

neglect it. = 2b 1 + m2 fi fi

2

(

2

– 4b2) = 4b2 =

2b a - 4b 2 2

.

Fig. 16.23

Example 31 A triangle PQR is inscribed in the circle 2 + y2 = 25. If Q and have coordinates (3, 4) and (– 4, 3) respectively, then –QPR is equal to (a) p /2 (b) p /3 (c) p /4 (d) p /6 (c) 1 –QPR = –QOR; O being the 2 2 centre (0, 0) of the given circle + y2 = 25. (Fig. 16.24) Let 1 = slope of OQ = 4/3 and 2 = slope of OR = – 3/4 As 1 2 = –1, –QOR = p/2 fi

–QPR = p/4

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Let OA, OB be the tangents from the origin to the given circle with centre C(- 3, 5) and radius 9 + 25 - c = 34 - c .

Fig. 16.24

Example 32 For each natural number k, let Ck denote the circle with radius k centimeters and centre at the origin on the circle Ck a particle moves k centimetres in the on Ck, the particle moves to Ck +1 in the radial direction. The motion of the particle continues in this manner. The particle starts at (1, 0). If the particle crosses the positive direction of C, then = (a) 4 (b) 5 (c) 6 (d) 7

Then area of the quadrilateral OACB = 2 ¥ area of the triangle OAC = 2 ¥ (1/2) ¥ AC OA = length of the tangent from the origin to the given circle = c and AC = radius of the circle = 34 - c so that c 34 - c = 8 (given) fi c (34 – c) = 64 fi 2 – 34c + 64 = 0 Example 34 If two distinct chords, drawn from the point ( , q) on the circle 2 + y2 = + qy (where π 0) are bisected by the (b) 2 = 8q2 (a) 2 = q2 2 2 (c) < 8q (d) 2 > 8q2 (d)

(d)

Let

( , q) and the coordinates of Q be ( , y). Since

through that on every circle the particle travels just one radian. The particle crosses the positive direction of C , where is the least positive integer such that ≥ 2p



= 7.

be a chord of the given circle passing

PQ is bisected by the the mid-point of PQ lies on the y = – q. 2

Q lies on the circle so fi

2

2

+q – 2



+ y2 –

– qy = 0

2

+q =0 + 2q2 = 0

(i)

Fig. 16.26

Fig. 16.25

Example 33 If the area of the quadrilateral formed by the tangent from the origin to the circle 2 + y2 + 6 – 10y + c = 0 and the pair of radii at the points of contact of these tangents to the circle is 8 square units, then c is a root of the equation (b) 2 – 34c + 64 = 0 (a) c2 – 32c + 64 = 0 (d) c2 + 34c – 64 = 0 (c) c2 + 2c – 64 = 0 (b)

which gives two values of and hence the coordinates of two points Q and R (say), so that the chords PQ and PR are bisected by PQ and PR are distinct, the roots of (i) are real distinct. fi

the discriminant

2

– 8q2 > 0



2

> 8q2

Example 35 Let A0 A1 A2 A3 A4 A5 be a regular inscribed in a unit circle with centre at the origin. Then the product of the lengths of the line segments A0 A1, A0 A2 and A0 A4 is

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(a) 3/4

(b) 3 3

(c) 3

(d) 3 3 /2

(c)

Fig. 16.28

Equation of any circle through (0, 0) and (1, 0)

Fig. 16.27

is

Let O be the centre of the circle of unit radius and the coordinates of A0 be (1, 0). 60º at the centre O. Coordinates of A1 are (cos 60º, sin 60º) = 1 2, 3 2

(

(

A2 are (cos 120º, sin 120º) = - 1 2, 3 2 A3 are (– 1, 0) 4

(

)

)

(

are - 1 2, - 3 2 and A5 are 1 2, - 3 2

If it represents C3, its radius = 1 fi

)

2

Ê 3ˆ 1 3 Ê 1ˆ + =1 ÁË1 - ˜¯ + Á ˜ = 2 4 4 Ë 2 ¯ 2

A 0 A 1=

)

x y 1 ( – 0) ( – 1) + (y – 0) (y – 0) + l 0 0 1 = 0 1 0 1 fi 2 + y2 – + ly = 0 1= (1/4) + (l2/4)

A0 A2 = So that

– y + k= 0

3

It will touch C1, if k =1fi 3 +1

Example 36 C1 and C2 are circles of unit radius with centres at (0, 0) and (1, 0) respectively. C3 is a circle

3y+2=0

(a)



(b)

3 –y+2=0

(c)

3 –y–2=0

(d)

+ 3y+2=0

(b)

l = - 3y = 0

Since C1 and C3 intersect and are of unit radius, their common tangents are parallel to the line joining their centres (0, 0) and (1/2, 3 /2). So, let the equation of a common tangent be

(A0 A1) (A0 A2) (A0 A4) = 3

of unit radius, passes through the centres of the circles C1 and C2 and have its centre above the common tangent to C1 and C3 which does not pass through C2 is

l= ± 3

As the centre of C3, lies above the - 3 and thus an equation of C3 is 2 + y2 –

2

2 Ê 3ˆ 9 3 Ê 1ˆ + = 3 = A0 A4 ÁË1 + ˜¯ + Á ˜ = 2 4 4 Ë 2 ¯



k=+2

y negative on the 3 x – y + 2 = 0. Example 37 A chord of the circle 2 + y2 – 4 – 6y = 0 passing through the origin subtends an angle tan–1 (7/4) at the point where the circle meets positive y of the chord is (a) 2 + 3y = 0 (c) – 2y = 0 (c)

(b) + 2y = 0 (d) 2 – 3y = 0

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The given circle passes through the origin O and meets the positive Y (0, 6). Let OP be the chord of the circle passing through the origin subtending an angle q at B, where tan q = 7/4 fi –OBP = q Equation of the tangent OT at O to the given circle is 2 + 3y = 0 fi slope of the tangent = – 2/3 So that, if –XOT = a, tan a = 2/3 From geometry, –POT = –OBP = q fi – =q–a

OL = OB + BL = 3 + 5 sin q = 3 + 5 × (4/5) = 7 CL = 5 cos q = 5 × (3/5) = 3 So the coordinate of C are (7, 3) and the equation of the circle having C ( – 7)2 + (y –3)2 = (CL)2 = 9 2 + y2 – 14 – 6y + 49 = 0. fi Example 39 A circle with centre at the origin and radius equal to at A and B. (a) and (b) are two points on this circle so that a – b = 2g, where g is a constant. The locus of the point of intersection of AP and BQ is (a) 2 – y2 – 2 tan g = 2 (b) 2 + y2 – 2 tan g = 2 (c) 2 + y2 + 2 tan g = 2 (d) 2 – y2 + 2 tan g = 2 (b)

Fig. 16.29

7 2 tan q - tan a 13 1 and tan (q – a) = = 4 3 = = 1 + tan q tan a 1 + 7 ¥ 2 26 2 4 3 Hence the equation of OP is y = tan (q – a)fi –2y = 0

( cos a, \

or

Coordinates of A are (– , 0) and of P are sin a)

Equation of AP is a sin a y= ( + ) a (cos a + 1) y = tan (a /2) ( + )

Example 38 On the line joining the points (0, 4) and B (3, 0), a square ABCD is constructed on the side of the line away from the origin. Equation of the circle having centre at C is 2 2 + y – 14 – 6y + 49 = 0 (a) (b) 2 + y2 – 14 – 6y + 9 = 0 (c) 2 + y2 – 6 – 14y + 49 = 0 (d) 2 + y2 – 6 – 14y + 9 = 0 (a) Let –ABO = q then –CBL= 90º –q, CL being perpendicular to C are (OL, LC)

Fig. 16.30

(i)

Fig. 16.31

Similarly equation of BQ is y=

a sin b ( x - a) a (cos b - 1)

or y = – cot (b /2) ( – ) (ii) We now eliminate a, b from (i) and (ii) y a-x , tan (b/2) = From (i) and (ii) tan (a /2) = y a+x a – b = 2g y a-x tan(a /2) - tan ( b / 2) a+x y = fi tan g = y a -x 1 + tan (a /2) tan( b /2) 1 + . a+x y

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MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

y 2 - ( a2 - x 2 ) x 2 + y 2 - a2 = (a + x) y + (a - x) y 2ay

tan g =

2 fi + y2 – 2 tan g = which is the required locus.

2

from the origin and having intercept of length 2 7 on y

A circle C1 of radius b touches the circle

Example 40 2 + y2 = 2

Circles touching

Example 41

(a) (b) (c) (d)

C2 of radius c touches the circle C1 < b < c, then the three circles have a common tangent if , b, are in (a) A.P. (b) G.P. (c) H.P. (d) none of these (b) The centre of C1 is ( + b, 0) and the centre of C2 is ( + 2b + c, 0)

2 2 2 2

+ + + +

y2 y2 y2 y2

– – – –

6 6 6 6

+ 8y + 9 = 0 + 7y + 9 = 0 –8y + 9 = 0 – 7y + 9 = 0

(a), (c) Centre of circle is (3, k) and radius is |k|. Equation of circle is ( – 3)2 + (y – k)2 = |k|2 It meets the y satisfy

y1), (0, y2) where y1, y2

(0 – 3)2 + y2 – 2ky = 0 fi

c a O (0, 0)

b C1 (a + b, 0)

C2

y1 + y2 = 2k, y1 y2 = 9 |y1 – y2|2 = (y1 + y2)2 – 4y1 y2

(a + 2b, + c, 0)



28 = 4k2 – 4(9)



k = ±4

Fig. 16.32

Let y = + k be a tangent common to the three circles. Since it touches 2 + y2 = 2, C1 and 2 k 1+ m

2

= ±a ,

m( a + b) + k 1+ m

m ( a + 2b + c ) + k

and

2

3

=±b

O

=±c

As the centre of the three circles lie on the same side of the line y = +k three relations we get,

Fig 16.33

Thus, required circles are

k =

=b–

( + b) = c –

( + 2b + c)

1+ m a + b a + 2b + c = (eliminating ) fi b-a c-a fi ( + b) (c – ) = (b – ) ( + 2b + c) 2

fi fi2



2

+ bc –

= 2b

2



fi , b, c are in G.P.

= =b

x

– 2

2

+ 2b2 – 2

and

+ bc –

2

+ y2 – 6 + 8y + 9 = 0

2

+ y2 – 6 – 8y + 9 = 0

Example 42 A circle S passes through the point (0, 1) and is orthogonal to the circles ( – 1)2 + y2 = 16 and 2 + y2 = 1. Then (a) (b) (c) (d)

radius radius centre centre

of of of of

S S S S

is is is is

8 7 (–7, 1) (–8, 1)

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. (b), (c) Solution: 2

Let equation of S be + y2 + 2

+ 2fy + c = 0

and

+ y2 – 1 = 0

2

+ y2 – 2 – 15 = 0,



\ Locus of E is 2 = 1 – y2 or 2 + y2 = 1

As S is orthogonal to 2

Êqˆ 1 - tan 2 Á ˜ Ë 2¯ = 2

1 ˆ Ê1 1 ˆ Ê1 It passes through Á , ˜ and Á , - ˜ Ë 3 3¯ Ë3 3¯

we get

y

(2g) (0) + (2f)(0) = c – 1 fi c = 1. and

E

(2g)(–1) + (2f)(0) = c – 15 fi g = 7

Q

Also, S passes through (0, 1), we get 1 + 2f + c = 0 fi f = –1.

P

\ Equation of S is 2

S (1, 0)

- (1, 0) R O

+ y2 + 14 – 2y + 1 = 0.

x

Its centre is (–7, 1) and radius is 7. Example 43 Let RS be the diameter of the circle 2 + y2 = 1, where S is the point (1, 0). Let P be a variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. Then the locus of E passes through the point Ê1 1 ˆ (a) Á , ˜ Ë 3 3¯

Ê 1 1ˆ (b) Á , ˜ Ë 4 2¯

1 ˆ Ê1 (c) Á , - ˜ Ë3 3¯

Ê 1 1ˆ (d) Á , - ˜ Ë 4 2¯

Fig 16.34

Example 44 The circle C1 : 2 + y2 = 3, with centre at O, intersects the parabola 2 = 2y at the point P in the C1 at P touches other two circles C2 and C3 at R2 and R3, respectively. Suppose C2 and C3 have equal radii 2 3 and centre Q2 and Q3 respectively. If Q2 and Q3 lie on the y (a) Q2 Q3 = 12 (b) R2R3 = 4 6 (c) area of the triangle OR2R3 is 6 2 (d) area of the triangle PQ2Q3 is 4 2

(a), (c)

(a), (b), (c), (d)

Let coordinate of P be (cosq, sinq). An equation of tangent at P is cos q + y sin q = 1.

ola

2

C1 and parab= 2y, we solve

= 1. Ê 1 - cos q ˆ Thus, coordinates of Q are Á1, ˜ Ë sin q ¯

+ y2 = 3 and simultaneously,

A equation of line parallel to RS and through Q is 1 - cos q Êqˆ = = tan Á ˜ Ë 2¯ sin q P is y = (tan q)

Thus, coordinates of P are ( 2 ,1) .

An equation of tangent at S is

For coordinate of E, set Êqˆ 2 tan Á ˜ Ë 2¯ Êqˆ tan Á ˜ = Ë 2¯ Êqˆ 1 - tan 2 Á ˜ Ë 2¯

2

2

= 2y

+ 2y = 3 fi y = 1 as y > 0.

2

An equation of tangent to C1 at P is 2x + y = 3 Let coordinates of Q2 be (0, a) We have | 2 (0) + a - 3 |



= 2 3 2 +1 |a – 3| = 6 fi a – 3 = ±6

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a = 3 ± 6 fi a = –3, 9

Thus, coordinates of Q2 are (0, –3) and that of Q3 are (0, 9) An equation of Q2R2 is – (–3) =

1 2

Therefore, the point (0, ) will lie on the circle (5) if 2 – 19 + 34 = 0, i.e., = 2 or 17.

y

P of a Example 46 triangle PQR intersects the side QR at the point S and the circumcircle of the triangle at the point T. If S is not the centre of the circumcircle then

Q2 R2

( – 0) (a)

1 1 2 + < PS ST QS ¥ SR

(b)

1 1 2 + > PS ST QS ¥ SR

(c)

1 1 4 + < PS ST QR

(d)

1 1 4 + > PS ST QR

2y - x + 3 2 = 0 Similarly, equation of Q3R3 is

O

x

ge

Q3

nt

R2R3 = distance between parallel lines (1) and (2) =

n Ta

2y - x - 9 2 = 0

R3

Fig 16.35

| 3 2 - (-9 2 ) | 1+1

(b), (d)

= 4 6 1 Area of DOR2R3 = (R2R3) (length of perpendicular 2 from O to R2R3) 1 = (4 6 ) 2

2 (0) + 0 - 3 2 +1

= 6 2 Area of DPQ2Q3 = =

1 Q2Q3 ( 2 ) 2

1 (12)( 2 ) = 6 2 2

Example 45 The points (2, 3), (0, 2), (4, 5) and (0, ) are concyclic if the value of is (a) 2 (b) 1 (c) 17 (d) 19.

Since S is not the centre of the circumcircle PS π ST, QS π SR Since A.M. > G.M. 1Ê 1 1 ˆ + Á ˜> 2 Ë PS ST ¯ fi

1 1 2 + > PS ST QS ¥ SR

Substituting the coordinates in (1), we get 4 + 9 + 4g + 6 f + c = 0 fi 4g + 6 f + c = – 13, (2) 0 + 4 + 0g + 4f + c = 0 fi 4f + c = – 4 (3) and 16 + 25 + 8g + 10f + c = 0 fi 8g + 10f + c = – 41 (4) Solving (2), (3) and (4), we get g = 5/2, f = – 19/2 and c = 34, so that equation (1) becomes 2 + y2 + 5 – 19y + 34 = 0 (5)

[∵ PS ¥ ST = QS ¥ SR]

+ = , a constant, = y = k/2. 1 (QR)2 Thus QS ¥ SR < 4

P

then

1 4 > fi QS ¥ SR (QR)2 1 1 2 4 + > > fi PS ST QS ¥ SR QR

(a) and (c) Let the equation of the circle passing through (2, 3), (0, 2) and (4, 5) be 2 + y2 + 2 + 2fy + c = 0 (1)

1 PS ¥ ST

Example 47

Q

S R T Fig. 16.36

The locus of the point of intersection of

2

+ y2 = 2 which touches the circle 2 + y2 – 2 = 0 passes through the point (a) ( /2, 0) (b) (0, /2) (c) (0, ) (d) ( , 0) (a) and (c) Let P(h, k) be the point of intersection of the AB of the circle 2 + y2 = 2. Since AB is the chord of contact of the tangents from P to this circle, its equation is + ky = 2. If this line touches the circle 2 + y2 – 2 = 0, then

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h ◊ a + k ◊ 0 - a2 h2 + k 2



fi (h – )2 = h2 + k2

Therefore, the locus of (h, k) is ( – )2 = 2 + y2, or y2 = ( – 2 ), which passes through the points given in (a) and (c) Example 48 Equation of the circle with centre (4, 3) and touching the circle 2 + y2 = 1 is (a) 2 + y2 – 8 – 6y + 9 = 0 (b) 2 + y2 + 8 + 6y – 11 = 0 (c) 2 + y2 – 8 – 6y – 11 = 0 (d) 2 + y2 + 8 + 6y – 9 = 0.

c and y = square is equal to the diameter of the circle , i.e., 2 98. Hence the length of each side of the square is 14, and the perpendicular distance of each side from the centre (1, –2) of the circle is 7. This gives = – 6, b = 8, c = – 9 and = 5, so the vertices of the square are (8, 5), (8, –9), (–6, 5) and (– 6, – 9). Example 51 the circle

If q is the angle subtended at P( 1, y1) by

S∫

+ 2fy + c = 0, then

2

+ y2 + 2

(a) cot q =

(a) and (c) Let equation of the required circle be ( – 4)2 + (y – 3)2 = 2

(b) cot (1)

If the circle (1) touches the circle 2 + y2 = 1, the distance between the centres (4, 3) and (0, 0) of these circles is equal to the sum or difference of their radii, and 1. fi

42 + 32 = 1 ±



(c) tan q =

g + f2 -c S1 g + f2 -c 2

2 g2 + f 2 - c S1

=

g2 + f 2 - c S1

Ê g2 + f 2 - c ˆ (d) q = 2 tan–1 Á ˜ ÁË ˜¯ S1

±1=5

fi = 4 or 6 so that the equations of the required circles from (1) are 2 + y2 – 8 – 6y + 9 = 0 and 2 + y2 – 8 – 6y – 11 = 0.

q = 2

S1 2

where S1 =

2 1

+ y 12 + 2

1

+ 2fy1 + c.

(b) and (d)

Example 49 The tangents drawn from the origin to the circle 2 + y2 – 2 – 2qy + q2 = 0 are perpendicular if (a) = q (b) 2 = q2 (c) q = – (d) 2 + q2 = 1. (a), (b) and (c) Equation of the given circle can be written as ( – )2 + (y – q)2 = 2, so that the centre of the circle is ( , q) and its radius is . This shows that = 0 is a tangent to the circle from the origin. Since tangents from the origin are perpendicular, equation of the other tangent must be y = 0, which is possible if q = ± or 2 = q2. Example 50 A square is inscribed in the circle 2 + y2 – 2 + 4y of coordinates. The coordinates of the vertices are (a) (8, 5) (b) (8, –9) (c) (–6, 5) (d) (–6, –9) (a), (b), (c) and (d) The centre of the given circle is (1, –2) and its radius is 1 + 4 + 93 = 98 . Since the sides of the square inscribed in the circle are parallel to the coordinate = , = b, y =

Fig. 16.37

Let PA and PB be the tangents from P( 1, y1) to the given circle with centre C(– – ), such that –APB = q. Then –APC = q/2 (Fig. 16.37). Therefore, from DPAC, we get cot

q PA = = 2 AC



tan

q = 2



Ê g2 + f 2 - c ˆ q = 2 tan–1 Á ˜ ÁË ˜¯ S1

S1 g2 + f 2 - c

g2 + f 2 - c S1

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Example 52 The centre of the circle(s) passing through the points (0, 0), (1, 0) and touching the circle 2 + y2 = 9, is (are) (a)

(3 2, 1 2)

(b) (1 2 3 2)

(

(d) 1 2, – 21/ 2

(c) 1 2, 21/ 2

)

(

)

4 – 3y = 0

(c) and (d) Since the required circle touches the given circle and passes through its centre (0, 0), its radius is half that of the given circle. Let the equation of the required circle be ( – h) + (y – k) = (3 2) 2

2

)

Example 53 y (a) 2 + (b) 2 + (c) 2 + (d) 2 +

(1 2, 2 ) 1/ 2

and

An equation of a circle which touches the y2 y2 y2 y2

+ + – –

4 5 5 5

– – – +

5y 4y 4y 4y

+ + + +

4 4 4 4

= = = =

16 + 9

= 3 (The radius of the circle whose centre is (–3, 1))

Example 55 If the circle 2 + y2 + 2 + 2fy + c = 0 cuts each of the circles 2 + y2 – 4 = 0, 2 + y2 – 6 – 8y + 10 = 0 and 2 + y2 + 2 – 4y diameter, then (a) c = – 4 (b) g + f = – 1 (d) gf = 6. (c) g2 + f 2 – c = 17

2

2

– 2a + , , such that 1 + 1 2 1 – 2 | = 3.

\

2)

+

2)

2

–4

1 2

=(

1



2

=9

25 5 fia=± 4 2 Hence the equation of the required circle is 2 + y2 ± 5 – 4y + 4 = 0. 4a 2 – 16 = 9 fi a 2 =

Example 54 The equation of a common tangent to the circles 2 + y2 – 2 – 6y + 9 = 0 and 2

-12 - 3

+ y2 + 6 – 2y + 1 = 0 is (a) y = 4 (c) = 0 (a), (b), (c) and (d)

2

+ y2 + 2

+ 2fy + c = 0

the common chords will pass through the centre of the respective circles, so that

This circle meets the 4 = 0, which gives two values of , say 2 = 2a and 1 2



and the distance of (–3, 1) from the same line is

Since the circle

0 0 0 0.

As the required circle touches the y (0, 2), let its equation be ( – a)2 + (y – 2)2 = a2 or 2 + y2 – 2a – 4y + 4 = 0

1

= 1 (The radius of the circle whose centre is (1, 3))

(a), (b), (c) and (d)

(b) and (c)

(

16 + 9

This shows that 4 – 3y = 0 is a common tangent to the given circles.

so that (1 2)2 + k2 = 9 4 fi k2 = 2 fi k = ±21/2

(

4 -9

(in (b)) is

2

Since it passes through (0, 0) and (1, 0), we have h2 + k2 = 9 4 and (h – 1)2 + k2 = 9 4 fi h2 – (h – 1)2 = 0 fi 2h – 1 = 0 fi h = 1 2 Therefore, the required centres are 1 2, – 21/ 2 .

The centres of the given circles are (1, 3) and (–3, 1) and radii are 1 and 3 respectively. Verify that the lengthof the perpendicular from the centre of each circle to the lines given by (a), (b), (c) and (d) is equal to its radius.

(b) 4 – 3y = 0 (d) 3 + 4y = 10



+ 2fy + c + 4 = 0 passes through (0, 0) c=–4 (1) + 2fy + c + 6 + 8y – 10 = 0 passess through

(3, 4) fi fi

(2g + 6)3 + (2f + 8)4 – 14 = 0 3g + 4f + 18 = 0

(ii)

and 2 + 2fy + c – 2 + 4y + 2 = 0 passes through (–1, 2) fi (2g – 2) (–1) + (2f + 4)2 – 2 = 0 fi g – 2f – 4 = 0 (iii) From (ii) and (iii) we get g = –2 and f = – 3. fi g + f = c – 1 = –5 gf = 6. g2 + 2 – c = 4 + 9 + 4 = 17. Example 56 If two vertices of an equilateral triangle are (– 1, 0) and (1, 0), equation of its circumcircle is (a)

3x 2 + 3 y 2 + 2 y - 3 = 0

(b) 2

2

(c)

3x 2 + 3 y 2 - 2 y - 3 = 0

(d) 2

2

+ 2y2 +

+ 2y2 –

3y – 2 = 0

3y – 2 = 0

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(a), (c)

2

Ê 5h + 2 ˆ Ê 5k - 2 ˆ ÁË ˜¯ + ÁË ˜ 6 6 ¯

2

=1

25(h2 + k2) + 20(h – k) – 28 = 0

or

Hence the locus of (h, k) is 25 (

2

+ y2) + 20( – y) – 28 = 0.

which meets the line y = 50

Fig. 16.38

Since two of the vertices A (–1, 0) and B(1, 0) C lies on y

lie on

fi 1+

=4fi

2

=3fi

We have a + b = –b/ , ab = c/ , = ± 3 fi OC = ± 3

Since triangle is equilateral, circumcentre of the triangle coincides with its centroid, i.e. with (0, ± 1/ 3 ) and circumradius is 2/ 3 . Hence possible equations of the circumcircle is 2

1 ˆ Ê + Áy± ˜ Ë 3¯ 2

or fi

2

Ê 2 ˆ = Á ˜ Ë 3¯

2

2 +y ± y–1=0 3 2

3x + 3 y ± 2 y - 3 = 0 2

2

Example 57 The locus of a point which divides the join of A(- 1, 1) and a variable point P on the circle 2 + y2 = 4 in the ratio 3 : 2 meets the line y = at the point (a) (14/25, 14/25) (b) (8/25, 8/25) (c) (- 14/25, - 14/25) (d) (- 8/25, - 8/25) . (a) (c) Let the coordinates of P be (2 cos q, 2 sin q) and those of the point which divides AP in the ratio 3 : 2 be (h, k). 6 cos q - 2 5 Eliminating q, we get

Then

h=

and

- 28 = 0 fi = ± 14/25

. (a), (d)

(CA)2 = (AB)2 = (BC)2 = 4. 2

at

Example 58 If a, b are the roots of the equation 2 + + c = 0 and a¢, b¢ those of ¢ 2 + b¢ + c¢ = 0 and the circle having A(a, a¢) and B(b, b¢) as diameter passes Ê b b¢ ˆ through the origin, and the point Á , ˜ then Ë a a¢ ¯ (a) ¢c + ¢ = 0 (b) ¢b + ¢ = 0 (c) b¢c + bc¢ = 0 (d) 2b¢2 + ¢2b2 = 0

Let the coordinate of C be (0, ) Then

2

k=

6 sin q + 2 5

a ¢ + b ¢ = –b¢/ ¢ and a ¢b ¢ = c¢/ ¢. Therefore, equation of the circle having AB as diameter is ( – a) ( – b ) + (y – a ¢) (y – b ¢) = 0 fi 2 – (a + b ) + ab + y2 – (a ¢ + b ¢)y + a ¢b ¢ = 0 b c b¢ c¢ =0 fi 2 + x+ + y+ y+ a a a¢ a¢ fi

¢(

2

+ y 2) + ¢

+

¢y + ¢c +

Since it passes through the origin, ¢c + (b/ , b¢/ ¢) Ê b 2 b¢ 2 ˆ ¢ Á 2 + 2˜ + ¢ a¢ ¯ Ëa

¥

b + a

¢ ¥

¢=0 ¢ = 0 and through

b¢ =0 a¢

fi ¢2 b2 + 2 b¢2 = 0 Hence equation of the required tangent is + y = ± 2 2 . Example 59 If is the length of the tangent to the circle 2 + y2 + 2g = 1, 2, 3 from any point and g1, g2, g3 are in A.P., A = (g - 2). Then (a) A1, A2, A3 are collinear (b) A2 is the mid-point of A1 and A3 (c) A1A2 is perpendicular to A2 A3 (d) A2 divides A1 A3 in the ratio 2 : 5 . (a), (b) 2 2 + y2 + 2g + 5 where ( , y) is any point. i = Since g1, g2, g3 are in A.P. 2g2 = g1 + g3 fi 2 22 = 12 + 32 fi 12, 22, 32 are in A.P. and A2 is the mid-point of A1 and A3. fi A1, A2, A3 are collinear.

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(b) radius of the circle = 3/2 (c) centre of the circle lies on 12 - 16y + 1 = 0 (d) centre of the circle lies on 12 - 16y + 31 = 0

Example 60 The circle 2 + y2 - 2 intercept AB on the line y = . The circle on AB as diameters

. (a), (c)

(a) 1 unit on (c) 1 unit on y

2 unit on

=

. (a), (c), (d) Equation of a circle passing through the intersection of the line y = and the circle 2 + y2 – 2 = 0 is 2 + y2 – 2 + l ( – ) = 0. Since y = is a diameter of Ê l + 2 lˆ , - ˜ lies on it this circle, its centre Á Ë 2 2¯ fi l = –1 and the required equation is 2 + y2 – – y = 0. intercept on X

Y

and intercept on y =

is = length of AB = 2

2. 2

Example 61 If a chord of the circle + y – 4 – 2y – c = 0 is trisected at the points (1 3,1 3) and (8 3,8 3) , then (a) (b) (c) (d)

Length of the chord = 7 2 c = 20 radius of the circle = 25 c = 25

The given lines being parallel tangents to a circle, the diameter of the circle is equal to the distance between these lines, so that the required radius is 7 4+ 1 1 15 1 3 2 ¥ = ¥ ¥ = 2 9 + 16 2 2 5 4 The centre of the circle lies on the line parallel to the given lines at a distance of 3/4 from each of them. So let the equation be 3 - 4y + k = 0 (i) then

k-4 9 + 16



3 fi k = 4 ± (15/4) fi k = 1/4 or 31/4 4

For k = 1/4, distance of (i) from the other line is also 3/4. Thus the centre lies on the line 12 - 16y + 1 = 0. Example 63 In a right angled triangle, the length of the sides are and b (0 < < b). A circle passes through the mid-point of the smaller side and touches the hypotenuse at its mid-point.

(a), (b), (c)

Ê 2a 2 - b 2 b ˆ (a) centre of the circle is Á , ˜ 4¯ Ë 4a

Equation of the line joining the given points is y = and the distance between them is i.e., 7 2 3 .

(7 3)2 + (7 3)2

Ê b2 - a2 b ˆ (b) centre of the circle is Á , ˜ 4¯ Ë 4a (c) radius of the circle is b a2 + b2 4a (d) radius of the circle is a a2 + b2 4b . (a), (c)

Fig. 16.39

If PQ is the chord of the given circle which is trisected at these points then equation of PQ is y = and the length 7 2 of this chord is 3 ¥ = 7 2 . Let C(2, 1) be the centre 3 of the given circle and CL be the perpendicular from C to PQ, then the radius of the given circle is PL2 + CL2 fi (2)2 + (1)2 +

(

= 7

2

) + (1 2 ) 2

2

Let the vertices of the triangle be O(0, 0), A ( , 0) and B(0, b) (Fig. 16.40). Let C(h, k) be the centre and the radius of the given circle. Since the circle touches AB at the point D( /2, b/2), the centre C(h, k) lies on the line y– \

= 25 fi c = 20.

Example 62 The lines 3 - 4y + 4 = 0 and 6 - 8y - 7 = 0 are tangents to the same circle. (a) radius of the circle = 3/4

Also

b aÊ aˆ = Áx - ˜ 2 bË 2¯ 2

–2

or =

2

– 2by =

– b2

2

Ê = CD2 = Á h Ë

2

Ê = Áh Ë

aˆ ˜ 2¯

2

2

aˆ bˆ Ê ˜¯ + ÁË k - ˜¯ 2 2

2

– b2 (1)

2

(2)

2

+ k21

(3)

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(because the circle passes through ( /2, 0), the mid-point of the smaller side). From (2) and (3), we get k = b/4, and from (1), we get h = (2 2 – b2)/4 . Putting these values in (3), we get

This has one end at the origin O and the other end P is given by solving (2) and (3). Êb ÁË a fi

y=

ˆ y˜ ¯

2

2a2 b a2 + b2

Fig. 16.40

b 4 + a2 b2



16a2

=

b2 ( a2 + b2 )

4a

Example 64 A and B are two points on the y the origin and having centre at A and B. (a) Equation of the common chord is - by = 0

Example 65 Equation of the straight line which meets the circle 2 + y2 = 2 at points which are at a distance from a point A(a, b) on the circle is (a) 2a + 2by = 2 2 - 2 (b) 2a – 2by = 2 2 + 2 (c) 2a + 2by = 2 2 + 2 (d) 2a + 2by + 2 2 = 2 . (a), (d)

(b) mid-point of the common chord is Ê ab2 a2 b ˆ , Á 2 ˜ Ë a + b2 a2 + b2 ¯ (c) AB bisects the common chord. (d) AB is perpendicular to the common chord. . (a), (b), (c), (d)

Let M and N be the points on the circle + y2 = 2 such that AM = AN = d (Fig. 16.42). The line joining A(a, b) and O(0, 0), the centre of the circle, meets the chord MN at B, the mid-point of MN of OA is b /a, so that the slope of MN is –a /b. An equation of MN is a + b y = k. Then, since (a, b) lies on the given circle, we have 2

Let OA = and OB = b (Fig. 16.41), so that the equation of the line AB is / + y/b = 1. Equations of the circles passing through the origin and having centres at A( , 0) and B(0, b) are, respectively, 2

and

2

2ab2 a2 + b2

which lies on AB

(4 a)2

b a2 + b2

=

=

Therefore, the mid-point of the common chord OP is Ê ab2 a2 b ˆ , Á 2 ˜ Ë a + b2 a2 + b2 ¯

2 Ê 2a 2 - b 2 a ˆ b4 b2 Ê bˆ - ˜ +Á ˜ = + = Á Ë 4¯ 2¯ 16a2 16 Ë 4a

=



Fig. 16.41

2

2

– 2by + y2 = 0

– 2 + y2 = 0 – 2by + y2 = 0

so that the equation of the common chord OP is – by = 0 which is perpendicular to AB

(1) (2) (3)

Fig. 16.42

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-k

OB =

a2 + b2

=



2

Ê ar 2 ˆ Ê br 2 ˆ = Á 2 - a˜ + Á 2 - b˜ 2 2 Ëa +b ¯ Ëa +b ¯

2

k , BM2 = OM2 – OB2 = a

2

k a2



AM2 = AB2 + BM2

and fi

±

PAB is PC ¥ AC, where (PC 2) = (h – )2 + (k – b)2

-k , = a

2

kˆ Ê = Áa ± ˜ Ë a¯

2

= 2

+

2



k a2



2

2

=2

± 2k

( r 2 - a 2 - b 2 )2 ( a 2 + b 2 ) (a + b ) 2

2 2

and (AC2) = (OA2) – (OC2) =

= 2

2

( r 2 - a 2 - b 2 )2 a2 + b2 –(

2

+ b 2) =

2



2

– b2

Area of D

2a 2 - d 2 =± 2

r 2 - a2 - b2

Example 66 AB is the chord of the circle 2 + y2 = 2 whose mid-point is C( , b). O is the origin and P is the point of intersection of the tangents to the circle at the AB, then (a) Area of D PAB = (b) Area of D PAB = (c) Area of D OAB = (d) Area of D OAB =

( r 2 - a2 - b2 )3/2 a +b 2

a2 + b2

=

2

)( + b )( r

) +b )

+ b2 r 2 - a2 - b2 2

2

+a

a2 + b2

r 2 - a2 - b2

Example 67 Equation of a common tangent to the circles 2 + y2 - 6 = 0 and 2 + y2 + 2 = 0 is

2

a2 + b2 2

a2 + b2

Area of D OAB = OC ¥ AC

( r 2 + a2 + b2 )3/2

(a (a

( r 2 - a2 - b2 )3/2

◊ r 2 - a2 - b2 =

2

(a) (b)

=1 =0

(c)

+

3y + 3 = 0

(d)

-

3y + 3 = 0

2

. (b), (c), (d) Equations of the given circles can be written as

. (a), (c) Equation of AB is

+ by =

2

+b

2

and

( – 3)2 + y2 = 32 ( + 1)2 + y2 = 12

(1) (2)

Equation of any tangent to circle (2) is ( + 1) cos q + y sin q = 1

(3)

This will be a tangent to circle (1) also if (3 + 1) cos q - 1 cos2 q + sin 2 q

= ±3



4 cos q – 1 = ± 3

1 . When cos q = 1, we 2 have sin q = 0, and the equation of the common tangent (3) becomes +1=1 or =0 (4) That is, cos q = 1 or cos q = –

Fig. 16.43

Let the tangents at A and B to the circle intersect at P(h, k). Then AB is the chord of contact of the tangents drawn from P(h, k) to the circle 2 + y2 = 2 and its equation is (2) + ky = 2

When cos q = –1/2 , we have sin q = ± 3/2 , and the equations of the common tangents are -

As (1) and (2) represent the same line, we have h k r2 ar 2 br 2 fi h = and k = = = 2 a b a + b2 a2 + b2 a2 + b2

and

1 3 y =1 ( x + 1) ± 2 2 +





3y + 3 = 0

3y + 3 = 0

(5) (6)

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P and Example 68 A line L1 intersects and y Q respectively. Another line L2, perpendicular to L1, cuts and y R and S respectively. The locus of the point of intersection of the lines PS and QR is a circle passing through the (a) origin (b) point P (c) point Q (d) point R

so that (h + 6)2 = (h – 4)2. That is, h = –1, and therefore, k = 3. Hence radius of the circle is

. (a), (b), (c)

Fig. 16.45

( -1 + 6)2 + (3 - 7)2

OA =

25 + 16 = 41

=

so that AC, the diameter of the circle, is 2 41 . =

Fig. 16.44

Let the equation of L1 be / + y/b = 1, so that the coordinates of P and Q are ( , 0) and (0, b), L1 is –b/ and therefore, that of L2 is /b. Let the equation of L2 be x y =1 bk ak Then the coordinates of R and S are (bk, 0) and (0, – ), respectively. Also, the equation of PS is / – y/ = 1, and that of RQ is /bk + y/b = 1. Therefore, eliminating k, we get the required locus as x-a b- y = y x fi ( – ) + y(y – b) = 0 2 + y2 – – by = 0 fi which is a circle passing through the origin, points P and Q.

( -6 - 4)2 + (7 - 7)2 = 10

fi = 164 - 100 = 8 Thus required area of the rectangle is AB ¥ BC = 10 ¥ 8 = 80 sq. units. Example 70

An equation of a circle through the origin,

10 on the line y = 2 + 5/ 2 , which subtends an angle of 45° at the origin is (a) 2 + y2 - 4 - 2y = 0 (b) 2 + y2 - 2 - 4y = 0 (c) 2 + y2 + 4 + 2y = 0 (d) 2 + y2 + 2 + 4y = 0 . (b), (d) Let an equation of the circle through the origin be 2

+ y2 + 2

+ 2fy = 0.

If is the radius of this circle, then 2 + 2 = 2.

Example 69 A rectangle ABCD is inscribed in a circle with a diameter lying along the line 3y = + 10. If A and B are the points ((a) centre of the circle is (- 1, 3) (b) centre of the circle (- 1, 0) (c) area of the rectangle is 100 sq. units (d) area of the rectangle is 80 sq. units . (a), (d) Let O(h, k) be the centre of the circle (Fig. 16.45). Then O(h, k) lies on the diameter 3y = + 10, so that 3k = h + 10. Also, OA = OB (radii of the same circle),

Fig. 16.46

IIT JEE eBooks: www.crackjee.xyz Circles and Systems of Circles 16.27

Let PQ be the intercept of length 10 made by the line y = 2 + 5 2 on this circle (Fig. 16.46). Since it is given that PQ subtends an angle of 45° at the origin (a point on the circle). It subtends a right angle at the centre (– g, – f ) of the circle. fi PQ2 = CP2 + CQ2 = 2 2 fi so that

10 = 2

2

fi g + f2 = 5

=

5.

2

(i)

Let CL be the perpendicular from C on PQ. CL = LP = LQ =

Then

- f + 2g - 5



2



1+ 4



2g – f = 0

10 2

5 5 + 2 2

2g – f = ±



10 2

2g – f = 5 2 .

or

2g – f = 5 2 gives imaginary values of g and f from (i). So 2g – f = 0 fi f = 2g and from (i) g = ± 1 and the required equations are 2

+ y - 2 - 4y = 0 or 2

2

2

+ y + 2 + 4y = 0

Then centre (2, - 1) of the circle lies on 3 + 4y - 2 = 0. = 2 and each diameter is perpendicular to a tangent to S Example 72 Column 1 (a) locus of point of intersection of the lines = 2, y = 2 (b) locus of the point of intersection of the perpendicular tangents to the circle 2 + y2 = 2 (c) locus of the point of intersection of the lines cos q = y cot q = (d) The locus of the mid points of the chords of the circle 2 + y2 - 2 = 0 passing through the origin. p q r s . a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(p)

Column 2 2 + y2 = 2

2

(q) y2 = 4

(r)

2

+ y2 =

(s)

2

- y2 =

2

MATRIX-MATCH TYPE QUESTIONS Example 71 If S ∫ ( - 2)2 + (y + 1)2 = 1 is a circle, then the equation of a (a) (b) (c) (d)

Column 1 tangent to S diameter of S line perpendicular to a tangent to S chord of with

Column 2 (p) 3 + 4y - 2 = 0 (q) = 2 (r) y = 0 (s) y + 2 = 0

.

Equation of the given circle can be written as S∫

2

+ y2 - 4 + 2y + 4 = 0

The centre of the circle is (2, - 1) and the radius is 1 so the , i.e. y = 0 and thus y = 0 is a tangent to S, y + 2 = 0 also touches the circle.

(a) Eliminating (b) Let y =

we get y2 = 4

2 2

=4 .

+ a 1 + m2 and

= (–1/ ) + 1 + 1/m2 be a pair of perpendicular tangents to the circle 2 + y2 = 2. Eliminating we get + )2 = 2 2 (1 + 2) fi 2 + y2 (y )2 + ( 2 =2 (c) cos q = y cot q = fi = sec q, y = tan q Eliminating q we get 2 - y2 = 2 (d) Equation of the chord of the circle 2 + y2 - 2 = 0 in terms of its middle point (a, b) is a + by - ( + a) = a2 + b2 - 2 a. Since it passes through the origin. a2 + b2 - a = 0. Locus of (a, b) is 2 + y2 = Example 73

For the circle

2

+ y2 + 4 + 6y - 19 = 0

Column 1 (a) Length of the tangent from (6, 4) (p) to the circle.

Column 2 72 226 113

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(b) Length of the chord of contact from (6, 4) to the circle

(q)

113

(c) Distance of (6, 4) from the centre of the circle (d) Shortest distance of (6, 4) from the circle

(r)

113 - 32

If S ∫

Example 74

Column 1

(s) 9

.

2

+ y2 + - y - 2 = 0, then Column 2

(a) (- 2, 1) lies

(p) on the circle S.

(b) (2, - 1) lies

(q) outside the circle S.

(c) (0, 1) lies

(r) on the tangent at (1, 0) to S (s) inside the circle S.

(d) (2, 3) lies .

(a) Required length of the tangent PT =

36 + 16 + 24 + 24 - 19 = 9.

(b) Equation of the chord of contact TT¢ is 6 + 4 + 2 ( + 6) + 3(y + 4) - 19 = 0 fi 8 + 7y + 5 = 0 perpendicular distance of TT¢ from the centre C (- 2, - 3) of the circle is - 16 - 21 + 5 32 = CS = 113 64 + 49 4 + 9 + 19 =

Column 1 Column 2 (p) if c = 1 (a) The circle 2 + y2 + 2 + c = 0 and 2 + y2 + 2y + c = 0 touch each other (b) The circles 2 + y2 + 2 + 3y + c2 (q) if c = – 1 = 0 and 2 + y2 - + 2y + c 2 = 0 intersect orthogonally 1 (c) The circle 2 + y2 = 9 contains the (r) if c = 2 circle 2 + y2 - 2 + 1 - c2 = 0 (s) If c > 8 (d) The circle 2 + y2 = 9 is contained in the circle 2 + y2 - 6 - 8y + 25 - c2 = 0 . p q r s

32

Fig. 16.47

ST)2 = (CT)2 - (CS)2 = (32) -

(32)2 32 ¥ 81 = 113 113 2 72 226 = 113 113

Length of TT¢ = 2ST = 2 ¥ 36 (c) required distance is CP =

(6 + 2)2 + (4 + 3)2 =

113

(d) required distance is PR where R is the point of intersection of CP with the circle. PR = CP - CR =

Tangent at (1, 0) is 3 - y - 3 = 0 which passes through (2, 3) but not through (2, - 1) Example 75

and the radius of the circle is CT =

Let S( , y) = 2 + y2 + - y - 2 then S(- 2, 1) = 4 + 1 - 2 - 1 - 2 = 0 fi (- 2, 1) lies on S. S(2, - 1) = 4 + 1 + 2 + 1 - 2 > 0 fi (2, - 1) lies outside S. S(0, 1) = 0 + 1 + 0 - 1 - 2 < 0 fi (0, 1) lies inside S. S(2, 3) = 4 + 9 + 2 - 3 - 2 > 0

113 -

32

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(a) Distance between the centres (- 1, 0) and (0, - 1) = 1 + 1 = 2

IIT JEE eBooks: www.crackjee.xyz Circles and Systems of Circles 16.29

Sum of the radii =

1- c + 1- c

The circles touch each other if 2 1 - c = 2 fi c = 1/2 (b) The circles intersect orthogonally if 2(1) (- 1/2) + 2(3/2) = c2 + c2 or if 2 = 1 (c) Centre of 2 + y2 = 9 is O(0, 0) and radius is 3 = OA, centre of 2 + y2 - 2 + 1 - c2 is O¢ (1,0) and radius is c. Since (1, 0) lies inside the circle 2 + y2 = 9, the circle with O¢(1, 0) lies inside the circle with centre O if the radius c < O¢A = OA - OO¢ = 2 (d) Centre of 2 + y2 = 9 is O(0, 0) and radius is 3 = OA and that of the circle 2 + y2 - 6 - 8y + 25 c2 = 0 are O¢(3, 4) and c. Since the point O¢(3, 4) lies outside the circle with centre O. The circle with centre O will be contained in the circle with centre O¢(3, 4) if the radius c > OA + OO¢ = 8

f1 - f 2 g1 - g2

Slope of the line joining the centres =

ing the centre (b) Common tangent to the intersecting circles of equal radii is at the same distance from the centres of the two circles and hence is parallel to the line joining the centres. (c) Since the line joining the centres of the circles to the mid-point of the common chord is perpendicular to the chord, it bisects the chord. (d) The line joining the centre of one to a point of intersection is tangent to the other circle. So by 2

Example 77

+ y2 - 14 - 10y

Column 1

Column 2

(a) intercept on Example 76

(b) intercept on y

Column 1

Column 2

circles (b) The common tangent to two intersecting circles of equal radii (c) The common chord of two intersecting circles (d) The line joining the centres of two circles intersecting orthogonally. p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

.

angle at a point of intersection. (q) is perpendicular to the line joining the centres. (r) is parallel to the line joining the centres. (s) is bisected by the line joining the centres.

(r) 8 3

(d) intercept on 7 + y – 4 = 0 p q r s

(s) 10

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

2

(a) intercept on - 14 + 24 fi ( - 12) ( - 2) = 0

Required intercept = 12 - 2 = 10 (b) intercept on Y -4=2 (c) intercept of y = is given by 2 2 - 24 + 24 = 0 fi

= y = 6± 2 6

Required intercept = (2 ¥ 2 6)2 + (2 ¥ 2 6)2

Let the equations of the circles be 2 + y2 + 2g1 + 2f1y + c1 = 0 and 2 + y2 + 2g2 + 2f2y + c2 = 0. 2(g1 - g2)

.

(c) intercept on y =

+ 2(f1 - f2)y + c1 - c2 = 0 -

g1 - g2 f1 - f 2

= 8 3 (d) intercept is given by

+ (4 - 7 )2 - 14 - 10(4 - 7 ) + 24 = 0 2 fi 50 = 0 fi = 0, y = 4. Showing that line meets the circles at one point and hence the required intercepts is zero. 2

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Example 78 Column 1

Column 2

(a) 3 - y + 10 = 0 (b)

(c) 3 - y + 3 10 = 0 (d) - y + 3 2 = 0

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

y=

the lines – 2y – 4 = 0 and – ly – 2 = 0, then, value of l is

= 1,

+ y2 = 9 is

– y ±3 2 = 0

= 3, 3 - y ± 3 10 = 0 Any tangent to 2 + y2 = 9 lies inside 2 + y2 = 10 which has the same centre but a larger radius. Similarly for the other circle 2 + y2 = 10. Example 79 Column 1

1 2

tangents to the circle C1: 2 + y2 = 9 and C 2: 2 + y 2 – 6 – 8y – 24 = 0 is (d) If circles 2 + y2 + 2 + 2ky + 6 = 0 and 2 + y2 + 2ky + k = 0 intersect orthogonally. Then value of k is

± 3 1 + m2 For

Column 2 (p) 0

can be drawn from points (–2, 3) to the circle passing through A(–1, 1), B(2, 1) and C(2, –3) is

2

Any tangent to

Column 1 (a) If a circle passes through the points of intersection

(p) is a tangent to 2 + y2 = 9 (q) is a tangent to 2 + y2 = 10 (r) lies inside 2 + y2 = 10 (s) lies outside 2 + y2 = 9

- 3y - 10 = 0

.

Example 80 Match the statements in Column I with values in Column II.

Column 2

(a) Two intersecting circles (p) have a common tangent circles normal (c) Two circles, one strictly (r) do not have cominside the other mon tangent (d) Two branches of a (s) do not have a hyperbola common normal q r p s . a p q r s

p

q

r

s

t

a

p

q

r

s

t

b

p

q

r

s

t

c

p

q

r

s

t

d

p

q

r

s

t

-3 2

(t)

(a) – 2y in A(4, 0) and C(0, –2), for l π 0, – ly – 2 = 0 meets Ê -2 ˆ B(2, 0) and D Á 0, ˜ . Ë l¯

b

p

q

r

s

Let the centre of circle through A, B, C be S. As S lies on the perpendicular bisector of AB, that is, = 3, we let the centre to be S(3, k).

c

p

q

r

s

As AS = SC, we get

d

p

q

r

s

(3 – 2)2 + k2 = 32 + (k + 2)2

The line joining the centres of the circles in (a), (b), or (c) is a common normal and in (d) the line joining the two focii is a common normal. For tangents, see the

(s) 2



k = –3

\ Centre of circle is (3, –3) As SA = SD,

y

-2

C

-2 l

D

2

4

A

B S

Fig. 16.48

x

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fi fi

Ê -2 ˆ (2 – 3)2 + (0 + 3)2 = (3 – 0)2 + Á + 3˜ Ë l ¯ 2 2ˆ Ê 1 = Á3 - ˜ Ë l¯

2 fi 3= ±1 l 1 fi l = 1, 2 For l = 0, we get the points of intersection as A(4, 0), B(2, A(-1, 1) 0) and C(0, –2), which clearly lie on a circle. (b) As DABC is a right triangle with right angle B, centre of the circle is the mid-point of the diameter AC, viz.

Statement-2: If a point P lies on 3 + 3y – 8 = 0, then either tangents cannot be drawn to S1 and S2 from P or the length of tangents from P to S1 and S2 are equal.

2

(a) If lengths of tangents from P(h, k) to S1 and S2 are equal, then 2

fi y

x

O D

C(2, -3) Fig. 16.49

Ê1 ˆ D Á , -1˜ and radius of cirË2 ¯ 5 cle is = DA = 2 2

, S(–2, 3) lies outside

the circle and hence two tangents can be drawn from S to the circle. (c) Centre of C1 is O(0, 0) and that of C2 is A(3, 4). Radii of C1 and C2 are 1 = 3 and 2 = 7. Since OA = 5 and |

1



2|

< OA
0 Ë 3¯ Ë 3¯ Ê 23 ˆ Ê 23 ˆ And (-5)2 + Á ˜ + 4(-5) + 2 Á ˜ - 20 > 0 fi This Ë 3¯ Ë 3¯ point lies outside S1 and S2 using statement-2, statement-1 is also true. Example 82 Statement-1: The circle + 16 = 0 touches the Statement-2: The circle ( – the 1, 0)

1)

2

2

+ y2 – 8 – 4y

+ (y – )2 =

2

touches

. (a)

2,

C1 and C2 intersect each other, thus C1 and C2 have



However, no tangent can be drawn to S1 and S2 from a point lying on 3 + 3y – 8 = 0 and lying inside S1 and S2. So statement-2 is true. 23 ˆ Ê 23 ˆ Ê As 3(–5) + 3 Á ˜ – 8 = 0, Á -5, ˜ Ë 3¯ Ë 3¯

2

2

1ˆ Ê As SD2 = Á -2 - ˜ + (3 + 1)2 > Ë 2¯

3h + 3k – 8 = 0

\ (h, k) lies on the line 3 + 3y – 8 = 0 B(2, 1)

-2

+ k2 – 2h – 4k – 4 = h2 + k2 + 4h + 2k – 20

Statement 2 is true as the centre of the circle ( 1, ) is at a distance , which is the radius of the circle, from the point of contact is ( 1, 0). Circle in statement-1 is ( – 4) 2 + ( y – 2) 2 = (2)2 using statement-2, statement-1 is true.

2

Example 83

ASSERTION-REASON TYPE QUESTIONS 2

2

+ y – 2 – 4y Example 81 Let S1 bet the circle 2 2 – 4 = 0 and S2 be the circle + y + 4 + 2y – 20 = 0 be two circles. 23 ˆ Ê Statement-1: Lengths of tangents from Á -5, ˜ to S1 Ë 3¯ and S2 are equal

Statement-1: Equation of a circle

of which the limiting points are (1, 1) and (3, 3) is 2 2 + 2y2 – 3 – 3y = 0 Statement–2: Equation of a circle passing through the points (1, 1) and (3, 3) is 2 + y2 – 2 – 6y + 6 = 0 . (b) Two members of the system of circles in statement-1 are the circles with centres at the limiting points and radius equal to zero . . ( – 1) 2 + ( y – 1) 2 = 0

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( – 3) 2 + ( y – 3) 2 = 0

and 2

or 2

and 2

2

+ y – 2 – 2y + 2 = 0 2

+ y – 6 – 6y + 18 = 0

+ y2 – 6 – 6y + 18 + l (

2

squaring we get (k –

) 2 = 169(1 +



2

(169 – h2)

+ y 2 – 2 – 2y + 2) = 0

Statement-2 is also true as the circle in it passes through (1, 1) and (3, 3) but does not lead to statement-1. Example 84 Statement-1: The line + 9y – 12 = 0 is a chord of contact of a point P with respect to the circle 2 2 + 2y2 – 3 + 5y – 7 = 0. Statement-2: The line joining the points of contacts of the tangents drawn from a point P outside a circle to the circle is the chord of contact of P with respect to the circle. . (d) P(h, k) be a point whose chord of contact is the line in statement-1, then 3 5 2 + 2 yk – ( + h) + (y + k) – 7 = 0 is same as 2 2 + 9y – 12 = 0 4h - 3 4k + 5 3h - 5k + 14 = = 1 9 12 h = 1, k = 1 so the point P is (1, 1) S(1, 1) = 2 + 2 – 3 + 5 – 7 < 0

fi But where

S= 2

2

)

+ (169 – k2) = 0

+2

giving us the slopes of the two tangents to the circle from the point (h, k) If these tangent are perpendicular then

which passes through the origin if l = – 9 and the equation of the required circle is 2 2 + 2y2 – 3 – 3y = 0. So that statement-1 is true.



2

2

+ 2 y – 3 + 5y – 7.

169 - k 2 169 - h

2

=–1fi

h 2 + k 2 = 338 2

and the locus of (h, k) is

+ y2 = 338

Statement-1 is true as the point (17, 7) lies on the circle. Example 86

Consider L1: 2 + 3y +

L2: 2 + 3y + + 3 = 0, where C: 2 + y2 + 6 – 10y + 30 = 0

–3=0

is a real number, and

Statement-1: If line L1 is a chord of circle C, then L2 is not always a diameter of circle C. Statement-2: If line L1 is a diameter of circle C, then L2 is not a chord of circle C. . (c) C: ( + 3)2 + (y – 5)2 = 4 centre of C is (– 3, 5) and radius is 2. If L1 is a chord of C then fi fi

– 6 + 15 + p – 3 4+9

0 If the locus of the circumcentre of the triangle passes through the point (2, -1) then 2 - 2 is equal to

fi = 0 or 3 Let the slope of QR be 0 and the coordinates of E be ( , y), x– 3 =0 then since IE is perpendicular to QR y –1

P

3 /2)

2

Êpˆ ± tan Á ˜ = Ë 3¯ 1 + m (– 3)

fi =

1 3 1 fiq= or 2 2 2

2

is the slope of QR m – (– 3)

3)=

3 ) + (q – 1)2 = 1

Example 105

3 )2 + (y – 1)2 = 1

fi ± 3 (1 –

3 (q – 1) 2

if q =

1 – 6 < 0 we get centre of the circle C to be I ( 3 , 1) and hence its equation, is ( –

3 =–

fi 4 (q – 1)2 = 1 fi q – 1 = ±

/2)

Ê3 ˆ 4 Á – y1 ˜ = 1 fi Ë2 ¯ y1 = 1, 2



So that A is ( , 0) and B is (0, q) (Fig. 16.55) and the line AB touches the given circle. Since –AOB is a right angle, AB is a diameter of the circumcircle of the triangle AOB. So the circumcentre P(h, k) of the triangle AOB is the mid point of AB. i.e. 2h = , 2k = q

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x y = 1, which touches the + p q

AB is given circle a( p + q) - pq \ = p2 + q2 fi

2

fi2

2

( + q)2 +

2

q2 – 2

– 2 ( + q) +

=0

( + q) =

2

(

2

+ q 2) fi

1 2 1 D = 1 7 1 = 75 16 7 1

Let the centre of the square be the origin O and the lines through O parallel to the sides of the square be the A(16, 16), B(- 16, 16), C(- 16, - 16) and D(16, - 16) The radii of the inscribed and circumscribed circles are respectively 16 and OA = 162 + 162 = 16 2 and their centre is at the origin. Let the coordinate of P be (16 cos q, 16 sin q) and that of Q be (16 2 cos f, 16 2 sin f). then S(PA)2 = 162 [(cos q - 1)2 + (sin q - 1)2 + (cos q + 1)2 + (sin q - 1)2 + (cos q + 1)2 + (sin q + 1)2 + (cos q - 1)2 + (sin q + 1)2] = 2 ¥ 162 [2 cos2 q + 2 + 2 sin2 q + 2] = 12 ¥ (16)2 similarly S(QA)2 = 16 ¥ (16)2 1 [S(QA)2 – S(PA)2] = 4 16

Let A be the centre of the circle 2 + y2 - 2 Example 107 - 4y -20 = 0. The tangents at the points B(1, 7) and C(4, - 2) on the circle meet at the point D. If D denotes the D area of the quadrilateral ABCD, then is equal to 25 .3 Tangents at B and C are respectively and

+ 7y - ( + 1) - 2(y + 7) - 20 = 0 4 - 2y - ( + 4) - 2(y - 2) - 20 = 0



y = 7 and 3 - 4y - 20 = 0 They intersect at D (16, 7). Coordinates of A are (1, 2)

D =3 25

If Four distinct points (2, 3), (0, 2), Example 108 (4, 5) and (0, ) are concyclic, then – 10 is equal to

-2 =2¥2¥1=4

Two circles are inscribed and Example 106 circumscribed about a square ABCD, length of each side of the square is 32. P and Q are two points respectively on 1 [S(QA)2 – S(PA)2] is equal to these circles, then 16 .4



= 2 Area of the triangle ABD fi

fi 2 2 – 2 (2h + 2k) + 2h ◊ 2k = 0. Hence the locus of P(h, k) is 2 – 2 ( + y) + 2 = 0. Since it passes through (2, - 1) 2

Area of the quadrilateral ABCD

.7 Let equation of the circle be 2 + y2 + 2 + 2fy + c = 0. Since it passes through the points (2, 3), (0, 2), (4, 5) we have 4g + 6f + c = - 13, 4f + c = - 4, 8g + 10f + c = - 41. Solving these equations we get g = 5/2, f = –19/2, c = 34 and the equation of the circle is 2 + y2 + 5 - 19y + 34 = 0. As this circle passes through the point (0, ), 2 - 19 + 34 = 0 fi = 2 or = 17. But = 2 corresponds to the point (0, 2) which is different from (0, t). Therefore = 17 and – 10 = 7. Equation of a tangent to the circle with Example 109 centre (2, - 1) is 3 + y = 0. The square of the length of the tangent to the circle from the point (3, –3) is .5 Length of the perpendicular from the centre 6 -1 5 (2, - 1) on the tangent 3 + y = 0 is = , which 9 +1 10 is the radius of the circle. Equation of the circle is ( - 2)2 + (y + 1)2 = 5/2 2[( - 2)2 + (y + 1)2] - 5 = 0

or

The required length = 2[(3 - 2)2 + (–3 + 1)2] - 5 = 5 A circle passes through the point (3, 4) Example 110 and cuts the circle 2 + y2 = 2 orthogonally. The locus of its centre is a straight line. If the distance of the straight line from the origin is 3, then 2 is equal to .5 Let the equation of the circle be 2 + y2 + 2 + 2fy + c = 0. Since it passes through (3, 4), 6g + 8f + c = - 25. As it cuts the circle 2g ¥ 0 + 2f ¥ 0 = c -

2

+ y2 = 2

fic=

2

orthogonally 2

fi 6g + 8f +

Locus of the centre (- g, - f) is 6 + 8y - ( Distance of the line from the origin is 3=

a2 + 25 36 + 64



2

+ 25 = 30 fi

2

=5

2

2

+ 25 = 0

+ 25) = 0

IIT JEE eBooks: www.crackjee.xyz Circles and Systems of Circles 16.39

The centres of two circles C1 and C2 Example 111 each of unit radius are at a distance of 6 units from each other. Let P be the mid point of the line segment joining the centres of C1 and C2 and C be a circle touching C1 and C1 and C passing C2 through P is also a common tangent to C2 and C1, then the radius of the circle C is

O

1/2

-1/3

- ÷6

÷6

.8 Let A1, A2 and M be the centres of the circles C1, C2 and C respectively. Let the common tangent through P to C1 and C touch C1 at B1, C at B2 and C2 also at B2. From right angled triangle A1 B1 P if A1 P B1 = a, sin a =

A1 B1 1 = A1 P 3

Two parallel chords of a circle of radiExample 113 us 2 are at a distance 3 + 1 apart. If the chords subtend p 2p at the centre, angles of and , where k > 0, then k k the value of k is

2 2 fi PB1 = 2 2 = PB2 3 From triangle MPB2 PB2 2 2 = MB2 r

1

2 2 = =fi r 2 2



Ê -3 ˆ Ê 3 ˆ Ê 1 -1ˆ Out of the given points Á 2, ˜ , Á , ˜ and Á1, ˜ lie Ë 2¯ Ë 4¯ Ë 4 4 ¯ inside the smaller part of S.

fi cos a =

tan a =

Fig. 16.57

3

=8 side of the centre.

C

N

C M a

B2

r a

A2

2

a a

C2

O 2a 2a

2

2

P

A1 C1

2

D

B1

A

a

B

M Fig. 16.58

Fig. 16.56

Let

The straight line 2 – 3y = 1 divides Example 112 the circular region 2 + y2 £ 6 into two parts. If ÏÊ 3 ˆ Ê 5 3 ˆ Ê 1 1 ˆ Ê 1 1 ˆ Ê -3 ˆ ¸ S = ÌÁ 2, ˜ , Á , ˜ , Á , - ˜ , Á , ˜ Á1, ˜ ˝ ÓË 4 ¯ Ë 2 4 ¯ Ë 4 4 ¯ Ë 8 4 ¯ Ë 2 ¯ ˛ then the number of point(s) in S lying inside the smaller part is 3 As O does not lie in the smaller part and 2(0) 2 – 3y – 1 > 0,

2

+ y2 £ 6.

p 2k = OA cos 2a

a=

= 2 cos 2a and

ON = 2 cos a

We are given + ON =

3 +1



2 cos 2a + 2 cos a =



2(2 cos2a – 1) + 2 cos a =



4 cos2 a + 2 cos a – (3 + 3 ) = 0



cos a =

3 +1

-2 ± 4 + 16(3 + 3 ) 8

3 +1

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(

1 -2 ± 2 (2 3 + 1) 2 8 As cos a > 0, we get =

)

C B

1 3 Êpˆ = cos Á ˜ cos a = (2 3 + 1 - 1) = Ë 6¯ 4 2 p p = 2k 6

fi fi

2

If and q be the longest and the Example 114 shortest distances respectively of the point (–7, 2) from 2

+ y2 – 10 – 14y – 51 and q, then

= 0. If is the geometric mean of is equal to

q D

O

Fig. 16.60

k=3

any point (a, b) on the circle

q r

A

= fi

+

Ê ˆ -Á ˜ Ë 2¯ 2 ¥

2

2

=

7 8

cos 2q = 2cos2 q – 1 =

In triangle AOC, 11

2

34 64

(AC)2 = (OA)2 + (OC)2 – 2(OA) (OC) cos 2q = In right angled triangle ACD,

Centre of the circle is C(5, 7). Let P(–7, 2) be the given point and the line joining P and C meet the circle at A and B fi A ( 7,2) P

B

49 16

(CD)2 = (AD)2 – (AC)2 =

15 16

2

2

4CD =7 r

C (5, 7)

EXERCISE LEVEL 1

Fig. 16.59

PC =

SINGLE CORRECT ANSWER TYPE QUESTIONS

(12) 2 + (5)2 = 13 radius of the circle is

52 + 7 2 + 51 = 5 5

= PB = 13 + 5 5 , q = PA = 13 - 5 5 G.M of , q = fi

pq =

(13) 2 - (5 5 ) 2 = 2 11 =

(a) 11

=2

Let S be a circle with diameter AD and Example 115 radius . B and C are points on the circumference such 4CD that chord AB = chord BC = then is equal to 2 r 7 Let O be the centre of the circle and

1. Let A and B are two points outside a circle S such that the chord of contact from A to S passes through B. If the length of tangent from A to S is 1 and length of tangent from B to S is 2, then length of AB is

AOB = BOC = q

then cos q =

(OA)2 + (OB)2 - ( AB)2 2(OA)(OB)

2 1

+

2 2

(b)

2 1

+

2 2

+

12

(c) 1 + 2 (d) | 1 – 2| 2. Angle of intersection of the circles S1 : 2 + y2 – 2 – 6y – 39 = 0 and S2 : 2 + y2 + 10 – 4y + 20 = 0 is p p (a) (b) 6 4 p p (c) (d) 3 2 3. The locus of the point of intersection of the tangents to the circle = cos q, y = sin q at points whose parametric angles differ by p /3 is (a) 2 + y2 = 4(2 – (b) 3( 2 + 2) = 2

2

3)

(c) 2 + y2 = (2 – 3 ) (d) 3( 2 + y2) = 4 2.

2

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4. The locus of a point which moves such that the tangents from it to the two circles 2 + y2 – 5 – 3 = 0 and 3 2 + 3y2 + 2 + 4y – 6 = 0 are equal, is (a) 2 2 + 2y2 + 7 + 4y – 3 = 0 (b) 17 + 4y + 3 = 0 (c) 4 2 + 4y2 – 3 + 4y – 9 = 0 (d) 13 – 4y + 15 = 0. + 2 = 0 and 5. If the two circles 2 + y2 + 2 2 + y 2 + 2g1 + 2f1 y = 0 touch each other, then (b) 1 = (a) f1 g = 1 1 2 2 2 2 (c) f + g = f1 + g1 (d) none of these 6. If two lines 1 + b1 y + c1 = 0 and 2 + b2 y + c2 = (b) 1 2 – b1 b2 = 0 (a) 1 2 + b1b2 = 0 (d) 1 b1 – 2 b2 = 0. (c) 1 b1 + 2 b2 = 0 7. The locus of a point which moves in a plane so that the sum of the squares of its distances from the lines + by + c = 0 and – + = 0 is 2, is a circle of radius. (a)

( a2 + b2 )

(b)

(c)

a -b

(d) 2

2

2

8. The locus of the center of the circle passing through the origin O and the points of intersection of any line through ( , b (a) circle (b) straight line (c) rectangular hyperbola (d) none of these 9. Four distinct points (1, 0), (0, 1), (0, 0) and ( , ) are concyclic for (a) = 0 (b) = 1 (c) = – 1 (d) none of these 10. If two circles which pass through the points (0, ) and (0, – ) cut each other orthogonally and touch the straight line y = + c, then (b) c2 = 2 |1 – 2| (a) c2 = 2(1 + 2) (d) c2 = 2 2 (1 + 2) (c) c2 = 2 (2 + 2) 11. The coordinates of two points P and Q are (2, 3) and (3, 2) respectively. Circles are described on OP and OQ O being the origin, then length of the common chord is 2 (b) 13 (a) (c) 5

2

(d) none of these

12. The circle 2 + y2 – 6 – 4y + 9 = 0 bisects the circumference of the circle 2 + y2 – (l + 4) – (l + 2) y + (5l + 3) = 0 if l is equal to (a) -1 (b) 1 (c) 2 (d) 4

13. The locus of the middle points of the chords of the circle of radius which subtend an angle p/4 at any point on the circumference of the circle is a concentric circle with radius equal to (a) /2 (b) 2 /3 (c) r

(d) r

2

3

14. The lengths of the tangents from two points A and B to a circle are and ¢ respectively. If the points are conjugate with respect to the circle, then (AB)2 is equal to (a) 2 + ¢2 (b) | 2 – ¢2| 2 (c) ( + ¢ ) (d) ¢ 15. If two circles, each of radius 5 units, touch each other at (1, 2) and the equation of their common tangent is 4 + 3y = 10, then equation of the circle, a portion of which lies in all the quadrants is (a) 2 + y2 – 10 – 10y + 25 = 0 (b) 2 + y2 + 6 + 2y – 15 = 0 (c) 2 + y2 + 2 + 6y – 15 = 0 (d) 2 + y2 + 10 + 10y + 25 = 0 16. A point moves such that the sum of the squares of its distances from the sides of a square of side unity is equal to 9. The locus of such a point is a circle (a) inscribed in the square (b) circumscribing the square (c) inside the square (d) containing the square 17. Radical centre of the three circles 2 + y2 = 9, 2 + y2 – 2 – 2y = 5, 2 + y2 + 4 + 6y = 19 lies on the line y = if is equal to (a) – 1 (b) – 2/3 (c) – 3/4 (d) 1 18. The coordinates of the point on the circle 2 + y2 – 2 – 4y – 11 = 0 farthest from the origin are (a) = (2 + 8

5,1+ 4

5)

(b) (1 + 4

5,2+ 8

5)

(c) (1 + 8

5,2+ 4

5)

(d) none of these 19. A circle passes through the origin O and cuts the A ( , 0) and B (0, b O in the line AB is the point Ê 2 ab2 2 a2 b ˆ , (a) Á 2 ˜ Ë a + b2 a2 + b2 ¯ Ê 2 a2 b 2 ab2 ˆ (b) Á 2 , 2 ˜ 2 Ë a + b a + b2 ¯

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26. If two circles 2 + y2 - 6 - 12y + 1 = 0 and 2 + y2 4 - 2y - 11 = 0 cut a third circle orthogonally then

Ê 2 ab 2 ab ˆ , 2 (c) Á 2 ˜ 2 Ë a + b a + b2 ¯ (d) ( , b) 20. The length of the longest ray drawn from the point (4, 3) to the circle 2 + y2 + 16 + 18y + 1 = 0 is equal to (a) the radius of the circle (b) the diameter of the circle (c) circumference of the circle (d) distance of the centre from the origin

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 21. Centre (s) of the circle passing through (0, 0), (1, 0) and touching the circle 2 + y2 = 9 is (are) Ê1 Ê1 ˆ ˆ (b) Á , - 2 ˜ (a) Á , 2 ˜ Ë2 ¯ Ë2 ¯ Ê1 ˆ (c) Á , 3 ˜ Ë2 ¯

Ê1 ˆ (d) Á , - 3 ˜ Ë2 ¯

22. If the length of the common chord of the circles 2 + y2 + 8y + 1 = 0 circles 2 + y2 + 2 – 1 = 0 is 2 6 then the value of is (a) 3 (b) – 3 (d) 2 3 (c) -2 3 23. If the length of the tangent drawn from any point on the circle 2 + y2 + 15 - 17y + c2 = 0 to the circle 2 + y2 + 15 - 17y + 21 = 0 is 5 units, then c is equal to (a) - 3 (b) 3 (c) - 4 (d) 4 24. The area of the quadrilateral formed by the tangent from the point (4, 5) to the circle 2 + y2 - 4 - 2y - c = 0 with a pair of radii joining the points of contacts of these tangents is 8 sq. units. The value of c is (a) - 11 (b) - 1 (c) 0 (d) 11 25. a, b, g are parametric angles of three points P, Q and R respectively, on the circle 2 + y2 = 1. A is the point (- 1, 0). If a, b, g are in A.P. and AP, AQ, AR are in G.P. then (a) cos (a/2), cos (b/2), cos (g/2) are in A.P. (b) cos (a/2), cos (b/2), cos (g/2) are in G.P. (c) sin (a/2), sin (b/2), sin (g/2) are in A.P. (d) sin (a/2), sin (b/2), sin (g/2) are in G.P.

(a) (1, 1) (b) (0, 6) (c) centre of the third circle (d) mid-point of the line joining the centres of the given circles. 27. A circle C passes through (2 , 0) and has the line 2 2 + y2 = 2, then (a) centre of the circle C is ( , 0) (b) centre of the circle C is (- , 0) (c) C passes through (0, 0) (d) C passes through ( , ) 28. C1: 2 + y2 + 2 - 3 = 0, C2: 2 + y2 - 2 - 3 = 0, C3: 2 + y2 - 2y - 3 = 0 are three circles (a) C1 passes through the centre of C2 (b) C2 passes through the centre of C1 (c) centres of the three circles form an equilateral triangle. (d) sum of the radii of C1 and C2 is twice the radius of C3. 29. If P is a point on the circle 2 + y2 = 9, Q is a point on the line 7 + y + 3 = 0, and the line - y + 1 = 0 is perpendicular bisector of PQ, then the coordinates of P are (a) (3, 0)

(b) (- 3, 0)

(c) (- 72/25, 21/25)

(d) (72/25, 21/25)

30. If the circle C1: 2 + y2 = 16 intersects another circle C2 of radius 5 in such a manner that the common coordinates of the centre of C2 are (a) (9/5, - 12/5) (c) (9/5, 12/5)

(b) (- 9/5, 12/5) (d) (- 9/5, - 12/5) 2

+ y2 31. If a chords of the circle intercepts of length can be (a) 2 (b) 3 (c) 4 (d) 5 32. Two equal chords of the circle 2 + y2 - 2 + 4y = 0, passing through the origin are perpendicular to each (a)

- 2y = 0

(b) 2 + y = 0

(c)

+ 3y = 0

(d) 3 - y = 0

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33. Equation of a circle of radius 3, touching the

(a)

2

+ y2 + 6 + 6y + 9 = 0

(b) (c)

2

+ y - 6 + 6y + 9 = 0 + y2 + 6 - 6y + 9 = 0

2

(a) y =

1+ a

+

(b) y = -

+ -

(c) y = (d) y = -

-

+ y2 =

2

is

2

1+ a 1+ a

2

2

1 + a2

MATRIX-MATCH TYPE QUESTIONS 36. Square of the radius (a) 2 + y2 - 2 + 2y + 1 = 0

(p) 6

(b)

2

+ y - 4 + 6y + 9 = 0

(q) 1

(c)

2

+ y - 6 + 6y + 12 = 0

(r) 4

(d)

2

+ y + 2 – 2y – 2 = 0

(s) 2

38.

2 2 2

+ y + 6 + 8y - 11 = 0 (chord of contact) (a) 3 - 4y + 10 = 0 (p) (1, 1) (b) 4 - 3y + 18 = 0 (q) (0, 0) 2

(c) 4 - 5y - 4 = 0

(r) (1, - 1)

(d) 2 - 5y + 12 = 0

(s) (- 1, - 1)

+ y2 + 2 - 4y + k = 0 passes through (a) k = 0 (p) (3, 4) (b) k = 1 (q) (1, 2) (c) k = - 15 (r) (0, 0) 2

(d) k = - 1 39.

2

2

(b) (c) (d)

2

(q) 15 (r) 9 (s) 5

+ y2 - 10 - 2ky + 25 = 0 (p) touches both

+ y2 - 2 - 20y + 100 = 0 (q) 2 + y2 + 14 + 14y + 49 = 0 (r) y 2 + y2 + 7 – 7y + 49/4 = 0 (s) none

ASSERTION-REASON TYPE QUESTIONS 2

35. The locus of a point, which is such that the tangents from its two concentric circles of radii and b are inversely proportional to their radii, is a circle C of area 16 p. (a) centre of C is at the centre of the given circles. (b) centre of C is at a distance 4 from the centre of the given circles (c) 2 + b2 = 16 (d) 2 + b2 = 4

37.

40. (a)

2

(d) 2 + y2 - 6 - 6y + 9 = 0 34. Equation of a tangent to the circle

2

(b) (5, 3) (c) (4, 3) (d) (0, 5)

(s) (0, 4)

+ y2 = 25, (square of the length of the tangent from) (a) (6, 2) (p) 0

41. Statement-1: The locus of the point of intersection of the tangents to the circle = cos q, y = sin q at points whose parametric angles differ by p /2 is 2 + y 2 = 2 2. Statement-2: diameter of a circle are parallel. 42. Statement-1: The centre of a circle passing through the points (0, 0), (1, 0) and touching the circle C : 2 + y2 = 9 lies inside the circle C. Statement-2: If a circle C1 passes through the centre of the circle C2 radius of the circle C2 is twice the radius of the circle C 1. 43. C1: ( – 3)2 + (y – 4)2 = 2 Statement-1: C1 if = 4 Statement-2: C1 touches the line y = if = 3. 44. Statement-1: The common chord of the circles 2 + y2 – 10 + 16 = 0 and 2 + y2 = 2 length if 2 = 34. 44. Statement-2: The common chord of two circles is of the circle with smaller radius. 45. Suppose , b, c are three distinct real numbers such that A( 2, 2 + 3), B(b2, 2b + 3) C(c2, 2c + 3) and D(0, 3) are concyclic. Statement-1: + b + c = 0 Statement-2: =1

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 46 to 48 L1 is a line intersecting and y P( , 0) and Q(0, b). L2 is a line perpendicular to L1 intersecting and y R and S respectively. 46. Locus of the point of intersection of the line PS and QR is a circle with (a) centre at ( /2, b/2) (b) + by = 0 as a tangent

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(c)

and b as the intercepts on respectively

y

(d) radius equal to (1/2) a2 + b2 47. Common chord of the circles on QS and PR as diameters passes through the point ( , b) if (a)

= 2b

(b) 2 = b

(c)

=b

(d)

55. The distance between the chords of contact of the tangents to the circle 2 + y2 + 32 + 24y - 1 = 0 from the origin and the point (16, 12) is k. The value of 40k – 400 is LEVEL 2

=-b

SINGLE CORRECT ANSWER TYPE QUESTIONS

48. If the area of the triangle ORS is 4 times the area of the triangle then equation of PS is (a)

+ 2y = 2b

(b) 2 + y = 2

(c)

- 2y + 2b = 0

(d) 2 - y - 2 = 0

Paragraph for Question Nos. 49 to 51 A(3, 7) and B(6, 5) are two points. C:

2

+ y2 - 4 - 6y - 3 = 0 is a circle.

49. The chords in which the circle C cuts the members of the family S of circles through A and B are concurrent at (a) (2, 3) (b) (2, 23/3) (c) (3, 23/2)

(d) (3, 2)

50. Equation of the member of the family S which bisects the circumference of C is

+ y2 = 3, which are form a rhombus, the length of its sides is

1. The tangents to the circle

(a) 2 3

2

(b) 3

(d) 4 (c) 3 3 2. A circle that can be drawn to touch the coordinate + 3y = 12 can not lie in (c) third quadrant (d) fourth quadrant 3. Two circles are such that one is inscribed in and the other is circumscribed about a square A1 A2 A3 A4. If the length of each side of the square is and P, Q are two points respectively on these circles, then 4

4

i =1

i =1

| Â ( PAi )2 - Â (QAi )2 |=

(a)

2

+ y2 - 5 - 1 = 0

(b)

2

+ y2 - 5 + 6y - 1 = 0

(c)

2

+ y2 - 5 - 6y - 1 = 0

(d)

2

+ y2 + 5 - 6y - 1 = 0

51. If O is the origin and P is the centre of C, then difference of the squares of the lengths of the tangents from A and B to the circle C is equal to (a) (AB)2

(b) (OP)2

(c) |(AP)2 - (BP)2|

(d) none of these

INTEGER-ANSWER TYPE QUESTIONS 52. Sum of the squares of the lengths of the tangents from the points (20, 30), (30, 40) and (40, 50) to the circle 2 + y2 = 16 is 7850 + k, k = 53. If , , denote the lengths of the intercepts made by the circle 2 + y2 - 8 + 10y + 16 = 0 on y y = - respectively, then 2 + 10 2 + 26 2 is equal to 2900 + k, k = 54. The point P(10, 7) lies outside the circle C: 2 + y2 - 4 - 2y - 20 = 0. , denote respectively the least and greatest distance of P from the points on the circle C. Then (9 + 2 )2 is equal to (25k)2, k =

(a) 2/4 (b) 2/2 (d) 2 2 (c) 2 4. Two tangents T1 and T2 are drawn from (– 2, 0) to the circle : 2 + y2 = 1. Equation of a circle touching C, having T1, T2 as a pair of tangents from (–2, 0) and radius greater than the radius of C is (a) 2 + y2 – 6 + 5 = 0 (b) 2 + y2 – 8 + 7 = 0 (c) 9 2 + 9y2 + 24 + 15 = 0 (d) none of these nates and the line cos a + y sin a = 2 is – 2 + 2gy + g2 = 0 where g = (a) 2 (cos a + sin a + 1)–1 (b) 2 (cos a – sin a + 1)–1 (c) 2 (cos a + sin a – 1)–1 (d) – 2 (cos a – sin a – 1)–1

2

+ y2

6. If q is the angle subtended by the circle ∫ 2 + y2 + 2 + 2fy + c = 0 at a point P( 1, y1) outside the circle and S1 ∫ 21 + y21 + 2 1 + 2fy1 + c, then cos q is equal to

IIT JEE eBooks: www.crackjee.xyz Circles and Systems of Circles 16.45

(a)

S1 + c - g 2 - f 2

(c)

S1 + c + g 2 - f 2

S1 - c + g 2 + f 2 S1 - c + g - f 2

2

(b)

S1 - c + g 2 + f 2

(d)

S1 - c + g 2 + f 2

S1 + c - g 2 + f 2 S1 + c + g - f

7. A square is inscribed in the circle

8.

9.

10.

11.

12.

2

2

2

+ y2–2 + 4 –93

gin are (a) (9, – 8) (b) (8, – 9) (c) (8, 5) (d) (– 6, 9) The point of intersection of the common tangents to the circle 2 + y2 + 14 – 4y + 28 = 0 and 2 + y2 – 14 + 4y – 28 = 0 is (a) (0, 0) (b) (7, – 7) (c) (– 343/14, 7) (d) none of these The locus of the point, the sum of the squares of A ( , y ), = 1, 2, …, is equal to k2 is a circle (a) passing through the origin (b) with centre at the origin (c) with centre at the point of mean position of the given points (d) none of these If the line 3 – 4y – k = 0, (k > 0) touches the circle 2 + y2 – 4 – 8y – 5 = 0 at ( , b), then k + + b is equal to (a) 20 (b) 22 (c) – 30 (d) – 28 2 2 The circles + y + 2g1 – 2 = 0 and 2 + y2 + 2g2 – 2 = 0 cut each other orthogonally. If 1, 2 are perpendiculars from (0, ) and (0, – ) on a common tangent of these circles then 1 2 (b) 2 (a) 2/2 2 (c) 2 (d) 2 + 2 The centre of the circle passing through the points (0, 0), (1, 0) and touching the circle 2 + y2 = 9 is (a) (3/2, 1/2) (b) (1/2, 3/2) (c) (1/2, 1/2) (d) (1/2, ± 2 )

parallel to the line y = , the locus of the points of contact is (b) 2 + y2 = 1 (a) 2 – y2 – 2 = 1 2 2 +y +2 =0 (d) 2 – y2 + 2 + 1 = 0 (c) 14. Equation of a circle which passes through the point (2, 0) and whose centre is the limit of the point of

intersection of the lines 3 + 5y = 1 and (2 + a ) + 5a2y = 1 as a tends to 1 is (a) 25( 2 + y2) + 20 – 2y – 140 = 0 (b) 25( 2 + y2) – 20 + 2y – 60 = 0 (c) 9( 2 + y2) – 20 + 2y + 4 = 0 (d) 9( 2 + y2) – 2 – 20y + 4 = 0 15. If the limiting points of the system of circles 2 + y2 + 2 + l ( 2 + y2 + 2fy + k) = 0, where l is a parameter, subtend a right angle at the origin, then k/ 2 = (a) – 1 (b) 1 (c) 2 (d) none of these 16. Three concentric circles of which the biggest is 2 + y2 = 1 have their radii in A.P with common difference (> 0). If the line y = + 1 cuts all the circles in real distinct points, then (a)

>

2- 2 4

(b)

>

2+ 2 4

1 2 (d) is any real number 17. If a circle having centre at (a, b), radius , completely lies with in two lines + y = 2 and + y = – 2, then min (| a + b + 2 |, | a + b – 2 |) is (c)

> 1+

(b) less than 2 (a) greater than 2 (c) greater than 2 (d) less than 2 18. Equation of a circle touching the line | – 2| + | y – 3 | = 4 will be (a) ( – 2)2 + ( y – 3)2 = 12 (b) ( – 2)2 + ( y – 3)2 = 4 (c) ( – 2)2 + ( y – 3)2 = 10 (d) ( – 2)2 + ( y – 3)2 = 8 19. The radius of circle C which envelopes the curve = 2 cos 2q + 1 – 3 cos q is (a) 4 (b) 5 (c) 6 (d) 8 20. If the tangent at the point P on the circle 2 + y2 + 6 + 6y = 2 meets the straight line 5 – 2y + 6 = 0 at a point Q on the y PQ is (a) 4

(b) 2 5

(c) 5

(d) 3 5

IIT JEE eBooks: www.crackjee.xyz 16.46 Comprehensive Mathematics—JEE Advanced

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

(a) c = – 5, g2 < 5 (c) c = – 9, – 3 < g < 3

21. The equation of a common tangent to the circles 2 + y2 + 14 – 4y + 28 = 0 and 2 + y2 – 14 + 4y – 28 = 0 is (a) – 7 = 0 (b) y – 7 = 0 (c) 28 + 45y + 371 = 0 (d) 7 – 2y + 14 = 0. 22. The circles 2 + y2 + 2 – 2y + 1 = 0 and 2 + y2 – 2 – 2y + 1 = 0 (b) touch each other internally (c) intersect on the y (d) touch each other at the point (0, 1). 23. The tangent at any point to the circle 2 + y2 = 2 A and B. If lines drawn A and B intersect at P, then the locus of P is (a) 2 + y2 = –2 (b) 2y2 – 2 ( 2 + y2) = 0 (c)

1 1 1 + 2 = 2 2 x y r

(d)

1 1 + 2 = r2 . 2 x y

24. Two circles 2 + y2 + each other if (a) + c = 0 (c) 2 = c2

= 0 and

2

+ y2 = c2 touch

(b) – c = 0 (d) none of these.

25. The circle on A(– 4, 3) and B ities of a diameter intercepts a length equal to (a) 16 on y – 3 = 0 (b) 16 on y + 1 = 0 (c) 4 on + 4 = 0 (d) 4 on – 12 = 0 26. The circle described on the chord sin a – = 0 of the circle 2 + y2 = passes through the origin if

2

-

cos a + y as diameter

(a)

=

2

(b)

=– 2

(c)

=

2

(d)

=–

2

27. The polar of the point (2, 2) with respect to the circle 2 + y2 = 2 touches ( – 1)2 + (y – 1)2 = b2 if (a)

2

+2 2 b=4

(c) b2 + 2 2 28. The circle (0, g) if

2

=4 + y2 + 2

(b)

2

–2 2 b=4

(d) b2 – 2 2

=4

+ c = 0 contains the point

(b) c = 5, g2 < 5 (d) c = 9, g2 < 9

29. The tangent at the origin to the circle with centre (a, b) and passing through the origin passes through the point (a) (b, a) (b) (b, – a) 2 (d) (– b, a) (c) (–a b, a ) 30. The line y = + 2c touches the circle – 4y + 4 = 0 for all valus of if

2

+ y2 + 4

1 + m2

(a) c = (1 + ) + (b) c = 1 + +

2

(c) c = (1 + ) – (d) c = 1 + –

2

1 + m2

31. The circle passing through the points of interesec+y+1 = 0 and – 2y + 3 = 0 passes through the point (a) (– 3, 0) (b) (0, 3/2) (c) (– 1/2, 0) (d) (0, – 1) 32. An isoscels triangle is inscribed in the circle 2 + y2 = 2 A( , 0) and the base angles B and C be each 75°, then the coordinates of B and C are Ê a 3 aˆ , ˜ (a) Á 2¯ Ë 2

Ê a 3 aˆ , ˜ (b) Á 2 2¯ Ë

Êa 3 aˆ , - ˜ (c) Á 2¯ Ë 2

Ê a 3 aˆ , - ˜ (d) Á 2 2¯ Ë

33. Equation of a circle which passes through the origin and cuts off intercepts 2/3 and 3/2 respectively from and y is 2 2 (a) 6 + 6y + 4 + 9y = 0 (b) 6 2 + 6y2 – 4 + 9y = 0 (c) 6 2 + 6y2 + 4 – 9y = 0 (d) 6 2 + 6y2 – 4 – 9y = 0 34. If the circles 2 + y2 + 2 + 2ky + 6 = 0 and 2 + y2 + 2ky + k = 0 intersect orthogonally then k is (a) 2 (b) – 2 (c) 3/2 (d) – 3/2 35. Equation of a tangent drawn from the origin to the circle 2 + y2 – 2 – 2hy + h2 = 0 is (a) = 0 (b) y = 0 =0 (c) (h2 – 2) = 2 (d) (h2 – 2) + 2 =0

IIT JEE eBooks: www.crackjee.xyz Circles and Systems of Circles 16.47

MATRIX-MATCH TYPE QUESTIONS Column 1 36. (a) 2 + y2 - 4 - 6y = 0 (b) 2 + y2 + 6 - 4y = 0

Column 2 (p) 2 (q) 50

(c) 2 + y2 - 7 - 2y = 0 (r) 81/2 (s) 0 (d) 2 + y2 - - y = 0 Squares of the lengths of intercepts on the line y = 37. The circle 2 + y2 the point A(1, 0), B(0, 1), C(- 1, 0) and D(0, - 1) diameter Column 1 (a) (b)

2

(c) (d)

2

38.

2

2

Column 2

+y + y2 +

-y=0 -y=0

(p) AD (q) DC

+ y2 + y2 +

+y=0 +y=0

(r) CB (s) BA

2

orthogonal Column 1

+ y2 + 2 + 4y + 10 = 0 2 2 (q) 2 + y2 + 4 (b) + y + 6y - 7 = 0 + 7 =0 2 2 (c) + y + 4 + 6y + 6 = 0 (r) 2 + y2 + 6y -5=0 2 2 (s) 2 + y2 – 2 (d) + y + 6 – 2y – 8 + 4y + 2 = 0 39. ( > 0) a portion of the circle lies in Column 1 Column 2 (p) I quadrant (a) 2 + y2 + 2 +2 - 2=0 (q) II quadrant (b) 2 + y2 - 2 + 2 =0 2 (r) III quadrant (c) 2 + y2 - 2 = 0 (a)

(d)

2

Column 2

2

+ y2 + 4 + 5 = 0

+ y2 - 2

=0

(p)

2

(s) IV quadrant

Column 1 Column 2 2 + y2 = 25 (p) 2 + y2 - 10 = 0 2 + y2 - 2 - 4y + 1 = 0 (q) 2 + y2 - 6 + 4y +9=0 (c) 2 + y2 - 6 + 4y + 6 = 0 (r) 2 + y2 + 6 - 2y +3=0 (d) 2 – 2 + y 2 + 8y – 8 = 0 (s) 2 + y2 = 4 Radii are equal

40. (a) (b)

ASSERTION-REASON TYPE QUESTIONS 41. Statement-1: The chords of contact of the pair of tangents drawn from each point on the line 2 + y = 4 to the circle 2 + y2 = 1 pass through the point (1/2, 1/4). Statement-2: The locus of the point of intersection circle C, passing through a point P is the polar of P with respect to the circle C and P is the pole of its polar. 42. C: 2 + y2 + 6 + 6y = 2 L: 5 – 2y + 6 = 0 Statement-1: If the tangent at the point P on the circle C meets the straight line L at a point Q on the y PQ = 5. Statement-2: Equation of the chord of contact of Q with respect to the circle C is 3 + 6y + 7 = 0, where Q is the point where the line L y. 2 2 2 2 43. C1: + y + 2 + 2ky + 6 = 0, C2: + y + 2ky + =0 Statement-1: C1 and C2 do not intersect orthogonally for any integer value of . Statement-2: Y of the circles C1 and C2 for an integer value of k. 44. C1: 2 + y2 = 2, C2: 2 + y2 = b2 > b Statement-1: Equation of a circle touching C1 2 internally and C2 + y2 – ( + b) + = 0. Statement-2: Length of the tangent from any point on the circle C1 to the circle C2 is – b. 45. Statement-1: From a point P outside a circle a line is drawn to meet the circle at Q and R. The product PQ. PR is independent of the slope of the line. Statement-2: Mid-points of the chords of the circle 4 2 + 4y2 – 12 + 4y + 1 = 0 that subtend an angle of 2p /3 at its centre lie on a circle concentric with the given circle and radius equal to half that of the given circle.

COMPREHENSION-TYPE QUESTIONS A circle C1 of radius 2 units rolls on the outerside of the circle C2: 2 + y2 + 4 46. The locus of the centre of this outer circle C1 is the circle C3: (a) 2 + y2 + 4y - 12 = 0 (b) 2 + y2 + 4 - 12 = 0

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(c) 2 + y2 + 4 + 4y - 4 = 0 (d) 2 + y2 - 4 = 0 47. If the line joining the centres of C1 and C2 angle of 60° with the tangent to them is (a) + 3 y - 2 = 0 (b) 3 - y + 4 + 2 3 = 0 (c)

3

INTEGER-ANSWER TYPE QUESTIONS 51.

-y-4+2 3 =0

(d) 3 - y - 4 - 2 3 = 0 48. Area of the quadrilateral formed by a pair of tangents from a point on C3 to the circle C2 with a pair of radii at the points of contact of the tangents is (b) 4 3 sq. units (a) 2 3 sq. units (d) 3 3 sq. units (c) 3 sq. units 49. If the line joining the centres of C1 and C2 is perpendicular to the contact of the tangents drawn from the centre of C2 to the circle C1 is (a) y - 2 = 0 (b) y + 2 = 0 (c) y - 3 = 0 (d) y + 3 = 0 50. Square of the length of the intercept made by the circle C3 on any tangent to C1 is (a) 12 sq. units (b) 24 sq. units (c) 48 sq. units

(d) 16 sq. units

and are the lengths of the tangents from the origin and the point (9, - 1) to the circle 2 + y2 + 18 - 2y + 32 = 0. If 2 + 2 – k2 = 301, then k =

52. A chord of the circle 2 + y2 = 5 passing through the point O(73, 23) meets the circle in points A and B. P is a point on this chord such that OP is the Geometric mean of OA and OB, the locus of P is a circle with centre at O, the square of whose radius is k8 k3, k = 53. A circle touches the sides AB and AD of a rectangle at P and Q respectively and passes C. If the distance of C from PQ is 97 units, then area of the rectangle in sq. units is (100 – k)2 where k = 54. From points on the line 4 - 3y = 6 tangents are drawn to the circle 2 + y2 - 6 - 4y + 4 = 0 which -1 (24/7) between them, denotes the sum of the squares of the distances of all such points from the origin then the value of s is equal to 19 55. Let A be the centre of the circle 2 + y2 - 2 - 4y - 20 = 0. The tangents at the points B(1, 7) and D (4, - 2) on the circle meet at the point C. If D is the area of the quadrilateral ABCD, then D = 3k2, where k =

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 2

1. A square is inscribed in the circle

(a) 2 + y2 – 6 + 4 = 0 (b) 2 + y2 – 3 + 1 = 0 (c) 2 + y2 – 4 + 2 = 0 (d) none of these

+ y2 – 2 + 4

is a diameter of a circle and C is any point on the circumference of the circle. Then (a) The area of triangle ABC isosceles. (b) The area of triangle ABC is minimum when it is isosceles. (c) The perimeter of triangle D ABC is minimum when it is isosceles. (d) none of these [1983] 4. The equation of the circle passing through (1, 1) and the points of intersection of 2 + y2 + 13x 3.

(a) (1 + 2, – 2) (b) (1 - 2, – 2) (c) (1, – 2 + 2) (d) none of these 2

2

[1980] 2

2

2. Two circles + y = 6, and + y – 6 + 8 = 0 are given, then the equation of the circle through their points of intersection and the point (1, 1) is

[1980]

IIT JEE eBooks: www.crackjee.xyz Circles and Systems of Circles 16.49

–3 0 and 2 2 + 2y2+ 4 – 7y – 25 = 0 is (a) 4 2 + 4y2 – 30 – 10y – 25 = 0 (b) 4 2 + 4y2 + 30 – 13y – 25 = 0 (c) 4 2 + 4y2 – 10 – 10y – 25 = 0 (d) none of these [1983] 5. The locus of the mid point of a chord of the circle 2 + y2 = 4, which subtends a right angle at the origin is (a) + 2 (b) 2 + y2 = 1 (d) + y = 1 [1984] (c) 2 + y2 = 2 6. If a circle passes through the point ( ) and cuts the circle 2 + y2 = k2 orthogonally, then the equation of the locus of its centre is (a) 2 + 2by – ( 2 + 2 + 2) = 0 (b) 2 + 2by – ( 2 – 2 + 2) = 0 (c) 2 + y2 – 3 + 4by – ( 2 + 2 + 2) = 0 (d) 2 + y2 – 2 – 3by + ( 2 – 2 + 2) = 0 [1988] 2 2 2 2 7. If two circles ( – 1) + ( – 3) = and + y2– 8 + 2 + 8 = 0 intersect in two distinct points, then (a) 2 < < 8 (b) < 2 (c) = 2 (d) > 2 [1989] 8. The lines 2 – 3y = 5 and 3 – 4y = 7 are diameters of a circle of area 154 sq. units. Then the equation of the circle is (a) 2 + y2 + 2 – 2y = 62 (b) 2 + y2 + 2 – 2y = 47 (c) 2 + y2 2 + 2y = 47 (d) 2 + y2 2 + 2y = 62 [1989] 9. The centre of a circle passing through the points (0, 0), (1, 0) and touching the circle 2 + 2 = 9 is (a) (3/2, 1/2) (b) (1/2, 3/2) (c) (1/2, 1/2)

(d) (1/2, –21/2)

[1992] 10. The locus of the centre of a circle which touches 2 + y2 – 6 – 6y +14 = 0 and also touches the y (a) 2 – 6 – 10y + 14 = 0 (b)

2

– 10 – 6y + 14 = 0

2

(a) – 6 – 10y + 14 = 0 2 – 10 – 6y + 14 = 0 [1993] (d) 2 2 2 2 + y – 10 + 16 = 0 and + y = 2 11. The circles intersect each other in two distinct points if (a) < 2 (b) >8 (c) 2
2b > 0 then the positive value of

for which

y= – b 1 + m is a common tangent to 2 = b and ( – )2+ y2 = 2 is 2

(a) 2b / a2 - 4b2 (c) 2b/( – 2b)

(b) (d)

2

+ y2

( a2 - 4b2 ) 2b /( – 2b) [2002]

21. The centre of circle inscribed in square formed by the lines 2 – 8 + 12 = 0 and y2–14 + 45 = 0 (a) (4, 7) (b) (7, 4) (c) (9, 4)

(d) (4, 9) 2

[2003] 2

22. If one of the diameters of the circle + y – 2 – 6y + 6 = 0 is a chord to the circle C with centre (2, 1) then the radius of the circle is (b) 2 (c) 3 (d) 2 [2004] (a) 3 2 2 23. A circle is given by + (y – 1) = 1 another circle locus of its centre is (a) {( , y) : 2 = 4y} » {( , ) :

0}

(b) {( , y) :

2

+ ( –1)2 = 4} » {( , ) :

(c) {( , y) :

2

= y} » {(0, ) :

0}

(d) {( , y) :

2

= 4y} » {(0, ) :

0}

0} [2005]

24. Let ABCD be quadrilateral with area 18, with side AB parallel to the side CD and = 2CD. Let AD be perpendicular to AB and CD. If a circle is drawn inside the quadrilateral touching all the sides, then its radius is (a) 3 (b) 2 (c) 3/2 (d) 1. [2007] 25. Tangents drawn from the point P(1, 8) to the circle 2 + y 2 – 6 – 4y – 11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is (a) 2 + y2 + 4 – 6y + 19 = 0 (b) 2 + y2 – 4 – 10y + 19 = 0 (c) 2 + y 2 – 2 + 6y – 29 = 0 [2009] (d) 2 + y2 – 6 – 4y + 19 = 0 26. The circle passing through the point (–1, 0) and touching the y point Ê 3 ˆ Ê 5 ˆ (b) Á - , 2˜ (a) Á - , 0˜ Ë 2 ¯ Ë 2 ¯

Ê 3 5ˆ (c) Á - , ˜ Ë 2 2¯

(d) (–4, 0)

[2011]

27. The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line 4 – 5y = 20 to the circle 2 + y2 = 9 is (a) 20( 2 + y2) – 36 + 45y = 0 (b) 20( 2 + y2) + 36 – 45y = 0 (c) 36( 2 + y2) – 20 + 45y = 0 [2012] (d) 36( 2 + y2) + 20 – 45y = 0

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 1. The equation of the tangents drawn from the origin to the circle 2 + y2 +2 – 2hy + h2 = 0 are (a) = 0 (b) = 0 =0 (c) (h2 – 2) –2 2 2 =0 [1988] (d) (h – ) +2 2. Circle(s) touching origin and having an intercept of length 2 7 on y (a)

2

+ y2 – 6 + 8y + 9 = 0

(b)

2

+ y2 – 6 + 7y + 9 = 0

(c)

2

+ y2 – 6 – 8y + 9 = 0

(d) 2 + y2 – 6 – 7y + 9 = 0 [2013] 3. A circle S passes through the point (0, 1) and is orthogonal to the circles ( – 1)2 + y2 = 16 and 2 + y2 = 1. Then (a) radius of S is 8 (b) radius of S is 7 (c) centre of S is (– 7, 1) (d) centre of S is (– 8, 1) [2014] 2 2 + y = 3, with centre at O, in4. The circle C1 : tersects the parabola 2 = 2y at the point P in the C1 at P touches other two circles C2 and C3 at R2 and R3, respectively. Suppose C2 and C3 have equal radii 2 3 and centres Q2 and Q3, respectively. If Q2 and Q3 lie on the y (a) Q2Q3 = 12 (b) R2R3 = 4 6 (c) area of the triangle OR2R3 is 6 2 (d) area of the triangle PQ2Q3 is 4 2 [2016] 2 + y2 = 1, 5. Let RS be the diameter of the circle where S is the point (1, 0). Let P be a variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal

IIT JEE eBooks: www.crackjee.xyz Circles and Systems of Circles 16.51

to the circle at P intersects a line drawn through Q parallel to RS at point E. Then the locus of E passes through the point(s) Ê1 1 ˆ (a) Á , ˜ Ë 3 3¯

Ê 1 1ˆ (b) Á , ˜ Ë 4 2¯

1 ˆ Ê1 (c) Á , - ˜ Ë3 3¯

Ê 1 1ˆ (d) Á , - ˜ Ë 4 2¯

[2016]

1. Tangents are drawn from the point (17,7) to the circle 2 + y2 = 169 Statement-1 The tangents are mutually perpendicular. Statement-2 The locus of the point from which mutually perpendicular tangents can be drawn to the given circle is 2 + 2 = 338. [2007] 1:2

+ 3y + :2 + 3y + 2

–3=0 +3=0

where is a real number, and C: 2 + y2 + 6 – 10 30 = 0 Statement-1 If L1 is a chord of circle C, then line L2 is not always a diameter of circle Statement-2 If L1 is a diameter of circle C, the [2008] line L2 is not a chord of circle

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 1 to 3 A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the side PQ, QR, RP are respectively. The line PQ is given by the equation 3x + y - 6 = 0 and the point D is (3 3 /2, 3/2). Further it is given that the origin and the centre C are on the same side PQ. 1. The equation of circle C is (a) ( x - 2 3)2 + ( y - 1)2 = 1 (b) ( x - 2 3) + ( y - 1/2) = 1 2

2

(c) ( x - 3)2 + ( y + 1)2 = 1 (d) ( x - 3)2 + ( y - 1)2 = 1 2. Points E and

are given by

(a) ( 3/2,3/2),( 3, 0) (b) ( 3/2,1/2),( 3, 0)

are

(a) y = (2/ 3) x + 1, y = ( -2 3) x - 1 (b) y = (1/ 3) x + 1, y = 0 (c) y = ( 3/2) x + 1, y = ( - 3/2) x - 1

ASSERTION-REASON TYPE QUESTIONS

2. Consider

3. Equations of the sides

(d) y = 3 x, y = 0

[2008]

Paragraph for Question Nos. 4 and 5 A tangent PT is drawn to the circle 2 + y2 = 4 at the point P 3, 1 . A straight line L, perpendicular to PT is a tangent to the circle ( – 3)2 + y2 = 1. 4. A possible equation of L is

(

)

(a)



3y = 1

(b)

+

3y = 1

(c) – 3 y = – 1 (d) + 3 y = 5 5. A common tangent of the two circles is (a) = 4 (b) y = 2 (c)

+

3y = 4

(d)

+ 2 2y = 6 [2012]

INTEGER-ANSWER TYPE QUESTIONS 1. The centres of two circles C1 and C2 each of unit radius are at a distance 6 units from each other. Let P be the mid-point of the line segment joining the centres of C1 and C2 and C be a circle touching cirC1 cles C1 and C2 and C passing through P is also a common tangent to C2 and C, then the radius of the circle C is[2009] 2. Two parallel chords of a circle of radius 2 are at a distance 3 + 1 apart. If the chords subtend at the p 2p center, angles of and , where k > 0, then the k k value of [k] is Note [k] denotes the largest integer less than or equal to k [2010] 3. The straight line 2 – 3y = 1 divides the circular region 2 + y2 6 into two parts. If ÏÊ 3 ˆ Ê 5 3 ˆ Ê 1 1 ˆ Ê 1 1 ˆ ¸ S = ÌÁ 2, ˜ , Á , ˜ , Á , - ˜ , Á , ˜ ˝ , ÓË 4 ¯ Ë 2 4 ¯ Ë 4 4 ¯ Ë 8 4 ¯ ˛ then the number of point(s) in S lying inside the smaller part is [2011]

FILL

IN THE

BLANKS TYPE QUESTIONS

PA =k PB (constant) for all P on a given circle, then the value of k cannot be equal to ______. [1982]

1. If A and B are point in the plane such that (c) ( 3/2,3/2),( 3/2,1/2) (d) (3/2, 3/2),( 3 / 2,1/2)

IIT JEE eBooks: www.crackjee.xyz 16.52 Comprehensive Mathematics—JEE Advanced

2. The points of intersection the line 4 – 3y –10 = 0 and the circle 2 + y2 – 2 + 4y – 20 = 0 are _____ and_____. [1983] 3. The lines 3 4 + 4 0, and 6 – 8y – 7 = 0 are tangents to the same circles. The radius of the circle is _____. [1984] 4. From the origin, chords are drawn to the circle ( – 1)2 + y2 = 1. The equation of the locus of the midpoints of these chords is______. [1985] 2 2 5. Let + y – 4 – 2y – 11 = 0 be a circle. A pair of tangents from the point (4, 5) with a pair of radii form a quadrilateral of area_______. [1985] 6. The equation of the line passing through the points of intersection of the circles 3 2 + 3y2 – 2 +12y – 9 = 0, and 2 + 2 + 6 + 2y – 15 = 0 is______. [1986] 7. From the point (0, 3) on the circle 2 + 4 + (y – 3)2 = 0 a chord AB point M such that AM = 2AB. The equation of the locus of M is_______. [1986] 8. The area of the triangle formed by the tangents from the points (4, 3) to the circle 2 + y2 = 9 and the line joining their points of contact is _______. [1987] 9. If the circle C1 : 2 + y2 = 16 intersects another circle C2 of radius 5 in such a manner that the common 3 , then the coordinates of the centre of C2 are 4 ______ . [1988]

to

10. The area of the triangle formed by the positive of the positive [1989] circle 2 + y2 = 4 at (1, 3 ) is ______. 11. If a circle passes through the points of intersection λ – y +1= 0 and – 2y + 3 = 0, then the value of λ is____. [1991] 12. The equation of the locus of the mid-point of chords of the circle of 4 2 + 4y2 – 12 + 4y + 1= 0 that subtend an angle of 2π /3 at its centre is ____. [1993] 13. Intercept on the line y = by the circle 2 + y2 –2 0 is AB. Equation of the circle with AB as diameter is ______. [1996] 14. The chords of contact of the pair of tangents drawn from each point on the line 2 + y = 4 to the circle 2 + y2 = 1 pass through the point ______. [1997] 15. Two vertices of an equilateral triangle (–1, 0) and tion of its circumcircle is ________. [1997]

TRUE/FALSE TYPE QUESTIONS to the circumcircle of the triangle with vertices (1, 3), (1, – 3), (3, – 3) [1985] 2. The line + 3y = 0 is a diameter of the +2 =0

2

+ y2–6 [1989]

SUBJECTIVE-TYPE QUESTIONS 1. Find the equation of the circle which passes through the point (2, 0) and whose centre is the limit of the point of intersection of the lines 3 + 5 = 1, (2 + c) + 5c2 = 1 as c tends to 1. [1979] 2. AB is a diameter of a circle. CD is a chord parallel to AB and 2CD = AB. The tangent at B meets the line AC produced at E. Prove that AE = 2AB. [1980] 3. Let A be the centre of the circle 2 + y2 – 2 – 4y – 20 = 0. Suppose that the tangents at the points (1, 7) and (4, –2) on the circle meet at the point . Find the area of the quadrilateral ABCD. [1981] 4. Find the equation of the circle passing through (– 4, 3) and touching the lines + y = 2 and – y = 2. [1982] h, k) secants are drawn to = 2 Show that the locus of the the circle 2 middle point of the secants intercepted by the circle [1983] is 2 + y2 – – ky = 0. 6. The abscissae of the two points A and B are the roots of the equation 2 + 2 – b2 = 0 and their ordinates are the roots of the equation 2 + 2 – q2 = 0. Find the equation of the circle on AB as diameter. [1984] 7. Lines 5 + 12y –10 = 0 and 5 – 12y – 40 = 0, touch a circle C1 of diameter 6. If the centre of C1 lies in C2 which is concentric with C1 and cuts intercepts of length 8 on these lines. [1986] 8. Let a given line L1 intersect the and y at and Q respectively. Let another line L2, perpendicular to L1 , cut the and y and respectively. Show that the locus the point of intersection of the lines PS and QR is a circle passing through the origin [1987] 2 2 9. The circle + y – 4 – 4y – 4 = 0 is inscribed in a triangle which has two of its sides along the the triangle is + y –

+k(

2

+ y2)1/2 = 0. Find k.

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[1987] 2 10. Let S + y2 – 2 – 2fy + c = 0 be a given circle. Find the locus of the foot of the perpendicular drawn from the origin upon any chord of S which subtends a right angle at the origin. [1988] 11. If ( , 1/ ), > 0, = 1, 2, 3, 4 are four distinct points on a circle, then show that 1 2 3 4 = 1. [1989] 12. A circle touches the line y = at a point P such that OP = 4 2 , where O is the origin. The circle contains the point (–10, 2) in its interior and the length of the its chord on the line + y = 0 is 6 2 . Determine the equation of the circle. [1990] 13. Two circles, each of radius 5 units, touch each other at (1, 2). If the equation of their common tangent is 4 +3y [1991] 14. Let a circle be given by 2 ( – ) + y(2y – b) = 0 ( 0, b 0). Find the condition on and b if two chords, each bisected by the the circle from ( , b/2). [1992] 15. Consider a family of circles passing through two A (3, 7) and (6, 5). Show that the chords in which the circle 2 + y2 – 4 – 6y – 3 = 0 cuts the members of the family are concurrent at a point. Find the coordinate of this point.[1993] 16. Find the coordinates of the point at which the circle 2 + y2 – 4 – 2y + 4 = 0 and 2 + y2 – 12 – 8y + 36 = 0 tangents touching the circles in distinct points. [1993] 17. Find the intervals of values of for which the line y + = 0 bisects two chords drawn from a point Ê 1 + 2a 1 - 2a ˆ Á 2 , 2 ˜¯ to the circle Ë 2

+ 2y2 – (1 +

2 ) y = 0. [1996] 2 2 18. Consider a curve +2 + by = 1 and a point P not on the curve. A line drawn from the point P intersects the curve at points Q and R. If the product × PR is independent of the slope of the line, then show that the curve is a circle. [1997] 2

2 )

20. C1 and C2 are two concentric circles, the radius of C2 being twice that of C1. From a point P on C2 tangents PA and are drawn to C1 Prove that the centroid of the triangle PAB lies on C1. [1998] 21. Let T1, T2 be two tangents drawn from (– 2,0) on to the circle C = 2 + y2 = 1. Determine the circles touching C and having T1, T2 as their pairs of

at a time. [1999] 2 2 22. Let 2 + y – 3 0 be the equation of pair of tangents drawn from the origin O to a circle of A is one OA. [2001] 23. Let C1 and C2 be two circle with C2 lying inside C1. A circle C lying inside C1 touches C1 internally and C2 C. [2001] 24. Find an equation of the circle C which touches the line 2 + 3y +1 = 0 at the point (1, –1) and is orthogonal to the circle C1 which has (0, –1) and (–2, 3) as the end points of a diameter. [2004]

Answers LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17.

(1 –

19. Let C be any circle with centre (0, 2 ). Prove that at the most two rational points can be there on C. (A is a point both of whose coordinates are rational numbers). [1997]

(a) (a) (b) (c) (d)

2. 6. 10. 14. 18.

(c) (b) (c) (a) (b)

3. 7. 11. 15. 19.

(d) (b) (c) (b) (a)

4. 8. 12. 16. 20.

(b) (d) (d) (d) (a)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 21. 23. 25. 27. 29. 31. 33. 35.

(a), (c), (b), (a), (a), (a), (c), (a),

(b) (d) (d) (c), (d) (c) (b) (d) (c)

22. 24. 26. 28. 30. 32. 34.

(a), (b), (a), (a), (a), (c), (a),

(b) (d) (c) (b), (d) (b) (d) (b), (c), (d)

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MATRIX-MATCH TYPE QUESTIONS p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

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p

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a

p

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b

p

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c

p

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a

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b

p

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s

c

p

q

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s

d

p

q

r

s

36.

37.

38.

39.

INTEGER-ANSWER TYPE QUESTIONS 52. 2

1. 5. 9. 13. 17.

(d) (b) (c) (a) (a)

2. 6. 10. 14. 18.

(c) (a) (a) (b) (d)

3. 7. 11. 15. 19.

(c) (b) (b) (c) (c)

4. 8. 12. 16. 20.

(b) (c) (d) (b) (c)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 21. 23. 25. 27. 29. 31. 33. 35.

(b), (b), (a), (a), (b), (a), (a), (a),

(c) (c) (b), (b) (c), (b), (b), (c)

22. 24. 26. 28. 30. 32. 34.

(c), (d) (d) (c), (d) (c), (d)

(a), (a), (a), (a), (a), (b), (a),

(d) (b), (c) (b) (c) (c) (d) (d)

MATRIX-MATCH TYPE QUESTIONS 36.

ASSERTION-REASON TYPE QUESTIONS 43. (c)

44. (a)

COMPREHENSION-TYPE QUESTIONS 46. (a), (b), (c), (d) 48. (b), (d) 50. (c)

55. 1

SINGLE CORRECT ANSWER TYPE QUESTIONS

p

q

r

s

a

p

q

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s

b

p

q

r

s

c

p

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d

p

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p

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a

p

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b

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c

p

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d

p

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p

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a

p

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b

p

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c

p

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d

p

q

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s

37.

42. (a)

54. 3 LEVEL 2

40.

41. (b) 45. (c)

53. 8

47. (c), (d) 49. (b) 51. (c)

38.

IIT JEE eBooks: www.crackjee.xyz Circles and Systems of Circles 16.55

ASSERTION-REASON TYPE QUESTIONS

p

q

r

s

a

p

q

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s

b

p

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c

p

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d

p

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p

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a

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b

p

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FILL

c

p

q

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1. 1

d

p

q

r

s

39.

40.

1. (a)

COMPREHENSION-TYPE QUESTIONS 1. (d)

42. (b)

1. 8

43. (d)

44. (c)

45. (b)

COMPREHENSION-TYPE QUESTIONS 46. (b) 48. (b) 50. (c)

47. (a), (b), (c) 49. (c), (d)

INTEGER-ANSWER TYPE QUESTIONS 51. 3 55. 5

52. 5

53. 3

54. 4

PAST YEAR’S IIT QUESTIONS

(d) (c) (d) (b) (b) (a) (b)

2. 6. 10. 14. 18. 22. 26.

4. (a)

5. (d)

(b) (a) (d) (d) (a) (c) (d)

3. 7. 11. 15. 19. 23. 27.

(a) (a) (c) (c) (c) (d) (a)

2. (a), (c) 4. (a), (b), (c), (d)

2. 3 IN THE

3. 2

BLANKS TYPE QUESTIONS

3/4 2 + y2 – = 0 8 sq. units 10 – 3y – 18 = 0 2 + (y – 3)2 + 8 = 0 (192/25) sq. unit Ê 9 12 ˆ 9. Á ± , ∓ ˜ Ë 5 5¯

10. 11. 12. 13. 14.

2 3 sq. unit 0, –1/3, 2 16( 2 + y2) – 48 + 16y + 31 = 0 2 + y2 – – y = 0 (1/2, 1/4) 2

+ ( y - 1/ 3 ) 2 = 4/3

TRUE/FALSE TYPE QUESTIONS 1. True

4. 8. 12. 16. 20. 24.

(b) (c) (d) (a) (a) (b)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 1. (a), (d) 3. (b), (c)

3. (d)

3. 4. 5. 6. 7. 8.

15.

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17. 21. 25.

2. (a)

INTEGER-ANSWER TYPE QUESTIONS

ASSERTION-REASON TYPE QUESTIONS 41. (a)

2. (c)

5. (a), (c)

2. True.

SUBJECTIVE-TYPE QUESTIONS 1. ( – 2/5)2 + (y + 1/25)2 = 1601/625 3. 75 sq. units. ( – k)2 + y2 = 9 + (4 + k)2 where k = -10 ± 54 2 + y2 – 2 – 2 – 2 – q2 = 0 2 + y2 – 10 – 4y – 4 = 0 1 2 + y2 + + fy + c/2 = 0 2 2 + y + 18 + 2y + 32 = 0 – 3)2 + (y + 1)2 = 25 ( – 5)2 + (y – 5)2 2 > 2b2 (2, 23/3) 24 Ê 2ˆ 16. y y= ÁË - ˜¯ 7 3

4. 6. 7. 9. 10. 12. 13. 14. 15.

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17. Œ (–•, – 2) » (2, •) 21. ( + 4/3)2 + y2 = (1/3)2 ( – 4)2 + y2 = 9

\

6. Equation of a curve through the intersection of the

5 Ê 4ˆ y= ± ÁË + ˜¯ 5 39 22. 9 + 3 10 23. Ellipse 24. 2( 2 + y2) – 10 – 5y + 1 = 0

7.

Hints and Solutions 8.

LEVEL 1

1. Let equation of the circle S be 2 + y2 = 2 and coordinates of A and B be ( 1, y1) and ( 2, y2) respectively. Chord of contact from A to S is 1 + yy1 = 2 . Since it passes through B 1 2 2 1

+ y 1y 2 = 2

=

1

and (AB)2 = ( =

2 1

+

2

+ y 12 – 1



2)

2 2

,

2 2

=

2 2

+ y 22 –

10.

+ (y1 – y2)2

2 2

If q is the required angle, then

= fi

r12 + r22 - (c1c2 ) 2 2r1r2

49 + 9 - (36 + 1) 1 = 2(7)(3) 2

q=

( 1 + b 1y + c 1) ( 2 + b 2y + c 2) + l = 0 if it represents a circle, co-eff of 2 and y2 are equal fi 1 2 = b 1b 2. The given lines are perpendicular. So shifting the origin to their point of intersection if (h, k) are the new coordinates of any point on the locus then h2 + k2 = 2 and the locus is 2 + y2 = 2 which is a circle of radius . x y a b = 1 and + = 1. Equation of the line be + l m l m Equation of the circle through (0, 0), ( , 0) and = 0, centre is ( /2, /2). (0, ) is 2 + y2 - a b whose locus is + = 1. 2x 2 y Equation of the circle through (1, 0), (0, 1), (0, 0) is 2 + y2 - - y = 0 which passes through ( , ) for = 1. For = 0, the point is (0, 0). Equation of a family of circles through (0, ) and (0, - ) is 2 + y2 + 2l - 2 = 0. If two members are for l = l1 and l = l2 then since they intersect orthogonally 2l1l2 2 = - 2 2 fi l1l2 = - 1 Since the two circles touch the line y = +c 2

È - l am + c ˘ = l2 Í 2 ˙ ÍÎ 1 + m ˙˚ fi

2 2



2

+

2

l - c2 +

l +2 (1 +

2

2

)-c =2

2 2

(1 +

2

)=0

fi c = (2 + 2

2

)

2

11. Equations of the circles are 2 + y2 - 2 - 3y = 0 and 2 + y2 - 3 - 2y = 0. Equation of the common chord is y = , for the points of intersection 2 2 - 5 = 0 fi

p 3

3. Tangents at q = 0 and q = p/3 are cos (p/3) + y sin (p/3) =

9.

2

2. Centre of S1 is C1 (1, 3) and of S2 is C2 (–5, 2) radius of S1 is 1 = 7, and of S2 is 2 = 3

cos q =

g f fi f1g = fg1 = g1 f1

=

and

fi = ,y= r 3, fi 3( 2 + y2) = 4 2 4. 2 + y2 - 5 - 3 = 2 + y2 + (2/3) + (4/3)y - 2 fi 17 + 4y + 3 = 0 5. Both the circles pass through the origin. If they touch each other then the tangents to the two circle at the origin is same. 2 + 2fy = 0 and 2g1 + 2f1y = 0 represent the same line.

= 0, 5/2, length of the chord = (5/2)2 + (5/2)2 = 5/ 2 . 12. Equation of the common chord is (l - 2) ( + y) + 6 - 5l = 0, which is a diameter of the 2nd circle if Èl + 4 l + 2˘ + + 6 - 5l = 0 (l - 2) Í Î 2 2 ˙˚ fi (l - 2) (l + 3) + 6 - 5l = 0 fi l2 - 4l = 0 fi l = 0 or l = 4. 13. Equation of the circle be 2 + y2 = 2. The chord which subtends an angle p/4 at the circumference will subtend a right angle at the centre. Chord joining ( , 0) and (0, ) subtends a right angle at the centre so (h, k) the mid-point of the chord is ( /2,

IIT JEE eBooks: www.crackjee.xyz Circles and Systems of Circles 16.57

/2) and locus of (h, ) is 2 + y2 = 2/2. 14. Equation of the circle be 2 + y2 = 2. A( 1, y1), B( 2, y 2) Since they are conjugate 1 2 + y1y2 = 2 2

=

2 1 2

+ y 12 -

(AB) = (

1

-

2 2)

, ¢2 =

2

2 2

+ y 22 -

+ (y1 - y2)2 =

2

A

O

2

1

+ ¢2

C

15. Centre of the circles are given by x -1 y - 2 3 4 = = 5 where tan q = , cos q = , cos q sin q 4 5 -4 -3 3 or cos q = , sin q = 5 5 5 fi Centres are (5, 5) and (- 3, - 1) and the circles are 2 + y2 - 10 - 10y + 25 = 0 and 2 + y2 + 6 + 2y - 15 = 0 sin q =

16. The sides of the square be = 0, = 1, y = 0, y = 1 (h, k) be any point on the locus, then

Fig. 16.61

So let the coordinates of the centre of circle be Ê1 ˆ ÁË , h˜¯ . 2 1 + h2 = 4 As it touches 2 + y2 = 9 internally, distance between the centres = difference of the radii. Radius of the circle is OC =



OC = 3 – = 3 – OC

(h) + (1 - h) + k + (1 - k) = 9



4(OC)2 = 9

fi h2 + k2 - h - k - 7/2 = 0



1 + 4h2 = 9 fi h = ± 2

2

2

2

2

Locus + y - - y - 7/2 = 0 is a circle with centre (1/2, 1/2), the centre of the square and radius equal to 4, greater than the diagonal of the square and hence contains the square. 2

2

17. The radical centre is the point of intersection of 2 + 2y = 4 and 4 + 6y = 10 i.e. (1, 1) which lies on y= if = 1.

1ˆ Ê and the required coordinates are Á ± 2 , ˜ Ë 2¯ 22. Equation of the common chord of the given circles is 2 – 8y – 2 = 0 or – 4y – 1 = 0 1 1 AB = ¥ 2 6 = 2 2

=

18. The point lies on the line joining the origin to the centre (1, 2) i.e. y = 2 which meets the given circle at the points. (1 ± 4 / 5, 2 ± 8 / 5) , the one which is farthest from the origin is (1 + 4 / 5, 2 + 8 / 5) .

A

C1 (0, - 4)

19. Equation of AB is / + y/b = 1. If P(h, k) is the reO in AB then (h/2, k/2) lies on AB and AB and OP are at right angles.

B

h k b k \ + = 2 and - ¥ = - 1 a b a h fih=

2ab2

k=

Fig. 16.62

2a2 b

1(0,

a +b a +b 20. Length of the longest say is equal to the length of the tangent from the point (4, 3) which is 2

2

2

2

1M

16 + 9 + 64 + 54 + 1 = 12 radius of the circle =

64 + 81 - 1 = 12

21. Centre of the required circle lies on the perpendicular bisector of the segment joining O(0, 0) and A(1, 1 0) i.e. on the line = 2

C2

M

–4), AC1 =

=

15

(0) - 4(-4) - 1 2

+ (4) 2

(C1M)2 = (AC1)2 – (AM)2 fi

(15)2 = (



2

2

+ (42)) (15 – 6)

=9fi

= ±3

6

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23. Required length is x 2 + y 2 + 15 x - 17 y + 21 2 2 where + y + 15 - 17y + c2 = 0 =

21 - c 2 =

5 fi c2 = 16 fi c = ± 4

24. Area of the quadrilateral =

42 + 52 - 4 ¥ 4 - 2 ¥ 5 - c ¥ 4 + 1 + c

fi 82 = (15 - c) (5 + c) fi c2 - 10c - 11 = 0 fi c = 11, - 1 25. P (cos a, sin a), Q (cos b, sin b), R (cos g, sin g) \ AP = =

(cos a + 1)2 + sin 2 a 2 + 2 cosa = 2 cos (a/2)

fi cos2 (b/2) - cos b = sin (a/2) sin (g/2) fi 1 - cos2 (b/2) = sin (a/2) sin (g/2) fi sin2 (b/2) = sin (a/2) sin (g/2) fi sin (a/2), sin (b/2), sin (g/2) are in G.P. + 5y - 6 = 0 which passes through (1, 1). Let the given circles intersect the circle 2 + y2 + 2 + 2fy + c = 0 orthogonally then 2g(- 3) + 2f(- 6) = c + 1 2g(- 2) + 2f(- 1) = c - 11 fi 2g(- 1) + 2f(- 5) = 12 fi - g - 5f - 6 = 0 fi - g, - f) of the third circle. Verify it does not pass through the mid-point of the line joining the centres. 27. Equation of the circle be 2 + y2 + 2 + 2fy + c = 0, it passes through (2 , 0) fi 4 2 + 4 + c = 0 + 2fy + c + 2 = 0 which is 2 - =0

fi 2+ = c fi c = 0, g = the circle is 2 + y2 - 2 = 0

1

+ y1 = h + k

h + x1 k + y1 + 1 = 0 fi 1 - y1 = - (h - k + 2) 2 2 1 = k - 1, y1 = h + 1 fi 7(k - 1) + h + 1 + 3 = 0 fi h = 3 - 7k fi (3 - 7k)2 + k2 = 9 fi k = 0, 21/25, h = 3, - 72/25

Since AP, AQ, AR are in G.P. cos (a/2), cos (b/2), cos (g/2) are in G.P. fi cos (a/2) cos (g/2) = cos2 (b/2) a +g =b Since a, b, g, are in A.P 2 fi cos (a/2) cos (g/2) - sin (a/2) sin (g/2) = cos b

=c+

y1 - k =-1fi x1 - h

fi 50k2 - 42k = 0

AQ = 2 cos (b/2), AR = 2 cos (g/2)

g f c + a2 fi f = 0, = = 1 0 -a

Centre is ( , 0), passes through (0, 0) and ( , ) 28. Centres C1(- 1, 0) C2(1, 0), C3(0, 1) Vertices of an isosceles triangle. C1 passes through (1, 0) and C2 passes through (- 1, 0) radii of each = 2. 29. P(h, k), Q( 1, y1) then h2 + k2 = 9, 7 1 + y1 + 3 = 0

2

and the equation of

diameter of the circle with smaller radius 4. So its equation is 3 - 4y = 0 as the slope is 3/4. Equation of the circle be 2 + y2 - 16 + l (3 - 4y) = 0 2

Ê - 3l ˆ (Radius)2 = Á + (2l)2 + 16 = 25 fi 25l2 = 36 Ë 2 ˜¯ fi l = ± 6/5 and the centre of the circle is (9/5, - 12/5) or (- 9/5, 12/5) 31. Equation of the chord be ± y = ± length of the perpendicular from the centre (0, 0) is less than the radius 8 . fi

±a 1+1


0 be two points, then for all values of x, PP¢ is a double ordi6. are concurrent, are called co-normal points of x1, y1), (x2, y2) and (x3, y3) are conormal points of the parabola y2 = 4ax, then y1 + y2 + y3 7. A line which bisects a system of parallel chords of a parabola is called a diameter standard results for the parabola y2 = 4ax parametric equations of the parabola or the coordinates of any point on it are x = at2, y = 2at P2. tangent to the parabola at (x¢, y¢ ) is yy¢ = 2a (x + x¢ ) and that at (at2, 2at) is ty = x + at2 P3. y = mx + c is a tangent to the parabola is c = a/m and the equation of any tangent to it (not parallel to the y y = m x + (a/m

Illustration 3 Find the equation of the tangent with slopes 5 to the parabola y2 = 20x Solution Equation of the parabola can be written as y2 = 4ax, a Equation of the tangent with slope 5 is 5 y = 5x + fi y = 5x 5 Parametric coordinates of any point on the parabola are (5t2, 10t) and the equation of the tangent at this point is ty = x + 5t2 1 Comparing with y = 5x + 1, we get t = 5 So the coordinates of the point of contact are Ê Ê 1ˆ Ê1 ˆ Ê 1ˆ ˆ ÁË 5 Ë ¯ , 10 Ë ¯ ˜¯ = Ë , 2¯ 5 5 5 2

P4.

P6.

pair of tangents from (x¢, y¢ ) to the parabola is T2 = SS¢ S = y2 – 4ax.

P8.

normal at (at2, 2at) to the parabola is y = – tx + 2at + at3 m is the slope of this normal, then its equation is y = mx – 2am – am3, which is the normal to the parabola at (am2, –2am

Illustration 4 A normal of slope 4 at a point P on the parabola y2 = 28x, Q

m

PQ

2

P1.

chord of contact (x¢, y¢ P5. polar parabola is yy¢ = 2a(x+ x¢

P7.

yy¢ = 2a (x + x¢ x¢, y¢

chord with mid-Point (x¢, y¢ ) of the parabola is T = S¢, where T = yy¢ – 2a (x + x¢ ) and S¢ = y¢ 2 – 4ax¢

Solution y = 28x = 4ax, a = 7 also m = 4, equation of the normal at P(am2 – 2am) is y = mx – 2am – am3 So, coordinates of P are (112, – 56) and the equation of the normal is y = 4x – 56 – 448 or y = 4x and

y = 0 at Q (126, 0) (PQ) = (126 –112)2 + (0 + 56)2 = 142 × 17 2



PQ = 14 17

P9. A diameter of the parabola is the locus of the middle points of a system of parallel chords of the parabola and the equation of a diameter is y = 2a/m where m is the slope of the parallel chords at21, 2at1) and

P10. (at22,

2at2) is y(t1 + t2) = 2x + 2at1t2

P11. t1 and t2 passes through the focus, then t1 t2 P12. (at2, 2at), then the coordinates of the other end are (a/t2, – 2a/t Illustration 5 PP¢ is a focal chord of the parabola y2 = 8x of P P¢ Solution y2 = 8x = 4ax, a So, coordinates of the focus are S P and S is 3x– 4y – 6 = 0 which meets the parabola at points for which 3y2– 32y – 48 = 0 fi y = 12, – 4/3 2 4 y = 12, represents P P¢ are Ê , - ˆ Ë9 3¯

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Alternately Coordinates of P are (at2, 2 at) = (18, 12) where a = 2, t 4ˆ Ê a - 2a ˆ Ê 2 of P¢ are Á 2 , ˜= , - ¯ Ët t ¯ Ë9 3 P13. For the end of the latus rectum, the values of the parameter t are ± P14. at21, 2at1) and 2 ( at2 ; 2at2) intersect at (at1 t2, a (t1 + t2 P15.

Illustration 6

is parallel to the system of chords bisected by the P25. harmonic mean between the segments of any focal P on the T and G

P26.

respectively, then (i) ST = SG = SP, S being the focus (ii) PSK = p/2, where K is the point where the tangent at P P of the parabola and the focal distance of P Illustration 7

PP¢ Solution Equations of the tangents at P and P¢ are 4 2 12y = 4(x + 18) and y Ê - ˆ = 4 Ê x + ˆ Ë 3¯ Ë 9¯ fi x – 3y + 18 = 0 and 3x + y + 2/3 = 0 16 ˆ Ê which are perpendiculars and intersect at Ë -2, ¯ which 3 l x P16. P17. on the parabola is twice the area of the triangle

Find the locus of the intersection of the perpendicular tangents to the parabola x2 = 4y Let x = my + c of the equation (my + c)2 = 4y are equal, as the line meets 1 m 1 So equation of any tangent to the parabola is x = my + m and equation of the perpendicular tangent is 1 x= - y–m m Solving for the point of intersection we get y+1=0 fi (2 mc – 4)2 = 4m2c2 fi c =

So the perpendicular tangents to the parabola intersect on the line y

P18. P19. parabola again at

(at22,

at21, 2at1) meets the 2at2) then t2 = – t1 – 2/t1 x1, y1)

P20.

meet the parabola are called feet of the normals or conormal points normals is zero and the sum of the ordinates of the t1 and t2 meet on the parabola then

P21. t1 t2 P22.

P23. P24. the tangent at the point where it meets the parabola

Illustration 8 P (36, 36) on the parabola y2 = 36x meets the parabola again at a point Q of Q Solution y2 = 4ax, a = 9 and P(36, 36) = (at2, 2at), t = 2 Equation of the normal at (at2, 2at) is y = –tx + 2at + at3 So equation of the normal at P is y = –2x + 36 + 72 or y = –2x + 108 Which meets the parabola at points for which (–2x + 108)2 = 36x fi x2 – 117x + (54)2 = 0 fi x

IIT JEE eBooks: www.crackjee.xyz Parabola 17.5

x = 36, gives the point P, so other end Q of the normal at P is (81, –54) y-coordinate of Q cannot be +ve Coordinates of Q are (at ¢2, 2at ¢) 2 Where t ¢ = – 3 = –t – t (using result P19)

SOLVED EXAMPLES

SINGLE CORRECT ANSWER TYPE QUESTIONS Length of the chord of the parabola Example 1 y = 4ax passing thr q (0 < q < p (a) 4a|cot q | cosec q (b) 2a |cot q | cosec q (c) a |cot q | cosec q (d) a cot3 q Ans Solution x = r cos q, y = r sin q in y2 = 4ax, we get r2 sin2 q = 4ar cos q 4a cos q fi r= = 4a cot q cosec q sin 2 q \ Length of the required chord = 4a |cot q | cosec q 2

Illustration 9 P (at12, 2at1) and Q (at22, 2at2) to the

2

parabola y = 4ax value of t1 t2 Solution Equation of normal at (at2, 2at) to the parabola is y = –tx + 2at + at3 h, k) then k = –th + 2at + at3 fi at3 + (2a – h)t – k = 0 (1) Which gives three values of t showing that the normal at three points with t = t1, t2, t3 pass through (h, k So if the normals at P and Q intersect on the parabola, then h = at32, k = 2at3

y

r q

2at3 k = From (7) we have t1 t2 t3 = a a fi t1 t2

O

Illustration 10

x

Fig. 17.6

P on the parabola T and G respectively, then show that the focus S of the parabola is the centre of the circle passing through T, P and G Solution Equations of the tangent and normal at P(at2, 2at) are respectively ty = x + at2 y = –tx + 2at + at3 2,

at P(

T

y2 = 4ax

S

)

2at

Example 2

r1 and r2 are length of two perpendicular

2 2 È - ˘ of 16a 2 Í(r1r22 ) 3 + (r12 r2 ) 3 ˙ is ˙˚ ÍÎ 1 (b) 1 (a) 4 2 (c) 1 (d) 2 Ans

Solution G

Fig. 17.5

So the coordinates of T are (–at2, 0) and of G are (2a + at2, 0) Focus S is (a, 0) SG = ST = SP = a(1 + t2) Which shows that S is the centre of the circle through T, P and G

p - q with the positive direction of x 2 4a cos q r1 = 4a cot q cosec q = sin 2 q p Replacing q by - q , we get 2 4a sin q r2 = cos 2 q

pˆ Ê q Á 0 < q < ˜ and Ë 2¯

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(r1, r22)

= ( 4a )

3

cos q sin 2 q

¥

fi x = 0 or x2 = c – 1/2 when x = 0, S = c, when x2 = c – 1/2, c ≥ 1/2 1 1 1 and S = c - + = c 2 4 4 Hence, the least distance is

sin 2 q cos 4 q

= (4a)3 (cos q)–3 2 2 -3 1 2

(r r )

fi fi

16a

2 r1r22 3

( )

2

Similarly 16a

2

= (4a)–2 cos2q = cos q

( )

= sin q 2

( )

2

-

2˘ 3˙

˙˚

A point P moves such that the difference x is always a constant c P is a (a) straight line having equal intercepts c (b) circle having its centre at (0, –c/2) and passing Example 3

(

through c 2 , -c / 2

(

through c 2 , c / 2

)

c/2) and passing

)

Ans Solution

P be (h, k)

h +k -k = c 2

2

2

2

(

passing through c 2 , c / 2

(d) Ans

)

c

which is least fi S is least dS 2 = 2x + 2(x2 – c) × 2x = 0 dx 2

(at - a ) + (2at ) 2

= SP =

2

2

L1

2

= a(t + 1)

PQ is a focal chord, then Q ( a t 2 , -2a t ) Ê1 ˆ a(1 + t 2 ) So that L2 = SQ = a Á 2 + 1˜ = Ët ¯ t2 1 1 1 + and = L1 L2 a

(c)

a (1 + t 2 )

(b) a 1 + t 2 (d) none of these

Ans

c - 1 / 4 if 1/2 £ c £ 5, c if 0 £ c £ 1/2

x 2 + ( y - c) 2 = x 2 + ( x 2 - c) 2

Solution Any point on the parabola is P(at2, 2at) where a S(a

c/2) and

Solution S is the distance of the point (x, y) on the parabola y = x2 from (0, c S=

L1 and L2 are the length of the segments 1 1 of any focal chord of the parabola y2 = x, then + is L1 L2 equal to (a) 2 (b) 3 (c) 4 (d) none of these Ans

(a) a(1 + t2)

Shortest distance of the point (0, c) from Example 4 the parabola y = x2 where 0 £ c £ 5 is (a) c if 0 £ c £ 1/2 (b) c if 3 £ c £ 5 (c)

1 1 if £c£5 4 2

Example 6 at the point (at2, 2at) of the parabola y2 = 4ax made by the circle which is described on the focal distance of the given point as diameter is

fi h + k = (k + c) fi h2 = 2c(k + c/2) Locus of (h, k) is x2 = 2c(y + c/2) 2

c-

Example 5

2 r12 r2 3

2 È 2 3 16a Í(r1r ) + r12 r2 ÍÎ



c if 0 £ C £ 1/2 and

2

…(1)

Solution

Ê a(1 + t 2 ) ˆ , at ˜ and the ÁË ¯ 2

a(1 + t 2 ) 1 =r (at 2 - a)2 + 4a 2t 2 = 2 2 Equation of the normal is y = –tx + 2at + at3 p is the length of the perperndicular from the centre on the normal Ê 2 ˆ -t Á t + 1˜ a + 2at + at 3 - at Ë 2 ¯ at 1 + t 2 = then p = 2 1 + t2 radius is

x is the required intercept

IIT JEE eBooks: www.crackjee.xyz Parabola 17.7

a 2 (1 + t 2 )2 a 2t 2 (1 + t 2 ) then x = r – p = 4 4 2

2

=

2

i

t1, t2, t3 are t22

point

A line bisecting the ordinate PN of a point Example 7 2 P (at , 2at), t > 0, on the parabola y2 = 4ax is drawn parallel Q NQ meets the tangent T, then the coordinates of T are (a) (0, (4/3)at) (b) (0, 2at) 2

(c) ((1/4)at , at)

(d) (0, at)

Ans. (a) Solution bisecting the ordinate PN of the point P (at2, 2at) is 2

y = at which meet the parabola y = 4ax at the point Q (1 / 4) a t 2 , a t

(

Let the coordinates of P, Q, R be (ati2, 2ati)

t 1t 3 = at P and R are t1 y = x + at21 and t3 y = x + at23, which intersect at the

a 2 (1 + t 2 ) 4

2 2x = a 1 + t



Solution

)

x + a t32 x + a t12 = t1 t3



x = at1t3 = at22

which is a line through Q parallel to Y A, B, C on the

Example 9 parabola y2 = 4x

P, Q

and R D, D¢ be the areas of the triangles ABC and PQR respectively, then (a) D = 2D¢ (b) D¢ = 2D (c) D = D¢ (d) none of these Ans. (a) Solution Let the coordinate of A, B, C be (t i2, 2t i ) i A and B are t1y = x + t12 and t2y = x + t22 which intersect at x = t1 t2, y = t1 + t2 So, the vertices are P (t1t2, t1+t2), Q (t2t3, t2+t3) and R (t1t3, t1+t3) D = |(t1 - t2) t1 t2 1 D¢ = t2 t3 2 t3 t1

(t1 - t3 ) t2 1 = (t2 - t1 ) t3 2 t3 t1

Fig. 17.7

Coordinates of N are (at2 Equation of NQ is y =

0 - at

a t - (1 / 4) a t 2 2

(x – at2 )

=

(t3 - t1)| 1 1 1

t1 - t3 0 t2 - t1 0 t3 + t1 1

1 (t1 - t3 ) (t2 - t1 ) (t2 - t3 ) 2



D = 2D¢

x = 0, at the point

which meets the y = ( 4 / 3) at

P, Q, R are three points on a parabola Example 8 2 y = 4ax whose ordinates are in geometrical progression, then the tangents at P and R meet on (a) the line through Q parallel to x (b) the line through Q parallel to y Q Q Ans. (b)

(t2 - t3) t1 + t2 t2 + t3 t3 + t1

Example 10 y2 = 4ax (a) x = – a (c) x = 0

(b) x = – a/2 (d) x = a/2

Ans. (c) y2 = 4ax is S (a, 0), let

Solution

P (at2, 2at) be any point on the parabola then co-ordinates of the mid-point of SP are given by x=

a (t 2 + 1) , 2

y=

2at + 0 2

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Eliminating t we get the locus of the mid-point (1) as y2 = 2ax – a2 or y2 = 2a (x – a/2) (2) which is a parabola of the form Y2 = 4AX Where Y = y, X = x – a/2 and A = a/2 X=–A x – a/2 = – a/2 fi x= 0

(a) l . p is constant (c) l2 p

(b) l

p2 is constant

Ans. (b) Solution Let P (at2, 2at) and Q(a/t2, –2a/t) be a focal chord of the parabola (as t1t2 = –1) Length of PQ = l = (at 2 - a / t 2 ) 2 + (2at + 2a / t ) 2 = a (t 2 - 1 / t 2 ) 2 + 4(t + 1 / t ) 2

Example 11 Equation of a common tangent to the curves y2 = 8x and xy = – 1 is (a) 3y = 9x + 2 (b) y = 2x + 1 (c) 2y = x + 8 (d) y = x + 2

2 = a (t + 1 / t ) (t - 1 / t ) + 4

= a(t + 1/t)2

Ans. (d) Solution Equation of a tangent at (at2, 2at) to y2 = 8x is ty = x + at2 where 4a = 8 i.e. a = 2 fi ty = x + 2t2 which intersects the curve xy = – 1 at 2 the points given by x( x + 2t ) = – 1 clearly t π 0 t or x2 + 2t2x + t = 0 and will be a tangent to the curve if the roots of this quadratic equation are equal, for which 4t4 – 4t = 0 fi t = 0 or t = 1 and an equation of a common tangent is y = x

P(x1, y1) to the Example 12 2 2 parabola y = 4ax meets the parabola y = 4a (x + b) at Q and R, the coordinates of the mid-point of QR are (b) (x1, y1) (a) (x1 – a, y1 + b) (d) (x1 – b, y1 – b) (c) (x1 + b, y1 + a)

line PQ whose equation is Ê 1ˆ y Á t - ˜ = 2(x – a) is given by Ë t¯ 2a 2a = p= (t - 1 / t ) 2 + 22 (t + 1 / t ) so that

l p2 =

4a 2 (t + 1 / t )

2

¥ a(t + 1 / t ) 2 = 4a3

which is c Example 14 point T P, Q are the points of contact, then perpendicular distance from P, T and Q the parabola are in

Ans. (b) Solution

Equation of the tangent at P (x1, y1) to the

parabola y2 = 4ax is yy1 = 2a (x + x1) (i) or 2ax – y1y + 2ax1 = 0 M (h, k) is the mid-point of QR, then equation of QR a chord of the parabola y2= 4a(x + b) in terms of its mid-point is ky – 2a (x + h) – 4ab = k2 – 4a (h+ b) (ii) (using T = S¢ ) or 2ax – ky + k2 – 2ah = 0 Since (i) and (ii) represent the same line, we have

fi fi fi

2ax1 2a y = 1 = 2 2a k - 2ah k k = y1 and k2 – 2ah = 2ax1 y21 – 2ah = 2ax1



4ax1 – 2ax1 = 2ah

h = x1

Example 13 Consider a parabola y2 = 4ax, the length of focal chord is l and the length of the perpendicular from p then

Ans. (b) Solution Let P(at12, 2at), Q(at22, 2at2), then the point of intersection of the tangents t1y = x + at12 and t2y = x + at22 is T(at1t2, a(t1 + t2 Distance of P, T and Q from y at12, at1t2, at22 Example 15

Chords of the parabola y2 + 4y =

(a) (7/3, –2) (b) (1/3, 0) (c) (4/3, 0) (d) (0, 4/3) Ans. (a) Solution Equation of the parabola is 4 (y + 2)2 = (x – 1) 3 1 or Y2 = 4aX where Y = y + 2, a = , X = x – 1 3

16 4 x– 3 3

IIT JEE eBooks: www.crackjee.xyz Parabola 17.9

Equation of the chord of this parabola Joining P(at12, 2at1) and Q(at22, 2at2) is Y(t1 + t2) = 2X + 2at1t2

Similarly they have a common tangent y = – (x + 1) at (1, –2)

PQ

y

(X, Y) = (0, 0) so fi

2at1 at12

¥

2at2 at22

= –1

y2 = 4an

(1, 2)

x

t1t2 = –4 So the equation of the chord PQ is

o

x2 + y2 – 6x + 1

Y(t1 + t2) = 2(X – 4a) (1, – 2)

which passes through X = 4a, Y = 0 fi fi

x –1 = 4 × 1/3, y + 2 = 0 x = 1 + 4/3 = 7/3, y

Example 16

Example 18

3 2 2 parabolas y = a x + a x – 2a is 3 2

105 64

(b) xy =

3 4

(c) xy =

35 16

(d) xy =

64 105

Solution

2

35a 105 3 ,y=– fi xy = x= 16 64 4a Example 17 C1 (a) (b) (c)

y = 4ax with focus F, then harmonic mean FA and FB is (a) a (b) 2a (c) 4a (d) 8a Ans Solution Ê a -2a ˆ A(at2, 2at), B Á 2 , ˜ and F(a, 0) Ët t ¯

Equation of the parabola is

a3 Ê 35a ˆ 3ˆ Ê ÁË y + ˜¯ = ÁË x + ˜¯ 3 16 4a

AB is a focal chord of the parabola

2

(a) xy =

Ans. (a)

Fig. 17.8

(FA)2 = (at2 – a)2 + (2at)2 = a2(t2 + 1)2

So

(FB)2 =

and

a 2 (t 2 + 1)2 t4

H is the harmonic mean of FA and FB then H =

Consider the two curves

y2 = 4x ; C2 x2 + y2 – 6x + 1, then C1 and C2 C1 and C2 C1 and C2

Equation of the tangent at (1, 2) to C1 is y(2) = 2(x + 1) fi y=x+1 and Equation of the tangent at (1, 2) to C2 is x y(2) – 3(x + 1) + 1 = 0 fi y = x + 1 Showing that C1 and C2 have a common tangent at the

1 1 + FA FB

1 1 1 + = fi H = 2a a FA FB

Now

Example 19 Length of the shortest normal chord of the parabola y2 = 4x is (b) 6 3

(a) 6 (d) C1 and C2 Ans. (b) Solution Solving the two equations, we get x2 + 4x – 6x + 1 = 0 fi x2 + 2x + 1 = 0 fi x = 1 and y = ±2

2

(c) 1 (d) 3 3 Ans Solution A(t2, 2t) is y = –tx + 2t + t3 Suppose it meets the parabola at b(t12, 2t1 t1 = –tt12 + 2t + t3 fi 2(t1 – t) = – t(t12 – t2) As t1 π t, we get (t1 + t)t = –2 fi Now,

t1 = - t -

2 t

AB2 = (t12 – t2)2 + (2t1 – 2t)2

y2 = 4x at

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= (t1 – t)2[(t1 + t)2 2

2ˆ Ê 4 Ê ˆ = Á -2t - ˜ Á 2 + 4˜ Ë ¯ t¯ Ët Ê 2 1 ˆ Ê1 ˆ = 4 ÁË t + 2 + 2˜¯ 4 ÁË 2 + 1˜¯ t t = 16 (t2 + 3t–2 + t–4 + 3) = s (say) ds = 16 (2t – 6t–3 – 4t–5) dt 32 = 5 (t6 – 3t2 – 2) t =

32

2

2

ˆ Ê y2 = y12 - 4ax1 Á 1 + 4˜ ¯ Ë a2

(

2

(t + 1) (t – 2)

t5



ds = 0 fi t = ± 2 dt

But

Solution A(at12, 2at1) and B(at22, 2at2) pass through (x1, y1 Equations of tangents at A and B are (1) t1y = x + at12 2 (2) and t2y = x + at2 Solving (1) and (2) we get the point of intersection of (1) and (2) as (at1t2, a(t1 + t2 x1 = at1t2, y1 = a (t1 + t2) Now, AB2 = a2 (t12 – t22)2 + 4a2 (t1 – t2)2 = a2 (t1 – t2)2 [(t1 + t2)2 = a2 [(t1 + t2)2 – 4t1t2 t1 + t2)2

AB =

t2, 2t) x t12, 2t1)

y = x2 is of the form

Equation of the tangent to y = x2 at (t, t2) is y – t2 = 2t (x – t) fi y = 2tx – t2 Which will be a tangent to the parabola y = –(x – 2)2 if it

> 0 if t > 2 S is least when t = 2 Hence, least value of AB2 is 108 and that of AB is 6 3 Let (x1, y1) be a point outside the parabola

2

y = 4ax from point (x1, y1) to y2 = 4ax is 1 y12 - 4ax1 y12 + 4a 2 (a) a 1 (b) y12 - 4ax1 x12 + a 2 a 1 (c) y12 - 4ax1 x12 + 4a 2 a

Ans

(

)(

( (

)( )(

)

)(

)

(y

2 1

)

dy dy = 2x fi = 2t dx dx (t ,t 2 )

< 0 if - 2 < t < 2

1 a

)(

- 4ax1 y12 + 4a 2

Example 21 the parabola y = x2 and y = –(x – 2)2 (a) y = 4(x – 1) (b) y = 0 (c) y = –4(x – 1) (d) y = –30x – 50

Also, ds > 0 if t < - 2 dt

(d)

2 1

Solution (t, t2)

Fig. 17.9

Example 20

(y

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

y

O

1 a

)

- 4ax1 y12 + a 2

)

x-coordinate of the point of intersection is given by 2tx – t2 = –(x – 2)2 fi x2 + 2(t – 2)x + (4 – t2) = 0 Which will have equal roots if (t – 2)2 – (4 – t2) = 0 fi t = 0 or 2 y = 0 and y = 4(x – 1) Example 22 2

parabola y = 4ax at a point P

)

PT and the normal PN to the T and N PTN

is a parabola whose a x=0 (c) latus rectum is 2a/3 (d) focus is (a, 0) Ans

IIT JEE eBooks: www.crackjee.xyz Parabola 17.11

Solution P(at2, 2at) to the parabola y2 = 4ax are ty = x + at2 and y = –tx + 2at + at3 y = 0 at T(–at2, 0) and N(2a + at2, 0) Let centroid of triangle PTN be (h, k) 1 1 then h = (at2 – at2 + 2a + at2) and k = (0 + 0 + 2at) 3 3 1 2at (2a + at2), k = 3 3 Locus of (h, k) the centroid is fi

h=

y2 =

4 Ê 2a ˆ aÁ x - ˜ 3 Ë 3¯

a/3, 0) Length of the latus rectum is 4a/3 Ê 2a a ˆ + , 0 = (a, 0) Focus is Á Ë 3 3 ˜¯ x=

2a a a - = 3 3 3

Example 23 y = mx bisects two distinct chords drawn from (4, 4) on y2 = 4x if (a) m = –1/2 (b) m = 0 (c) m = 1/2 (d) m = 1 Ans Solution y = mx is (t, mt) Equation of the chord of the parabola y2 = 4x having (t, mt) as the mid point is y(mt) – 2(x + t) = m2t2 – 4t fi m2t2 – 2(2m + 1)t + 8 = 0 For two distinct chords, (D > 0) 4(2m + 1)2 – 32m2 > 0 fi 4m2 – 4m – 1 < 0 fi (2m – 1)2 Example 24 All chords of the curve 3x2 – y2 – 2x + 4y = 0 which subtend a right angle at the origin pass through (a) (1, –2) (b) the point of intersection of the lines y + 2x = 0 and x=1 x2 – 2x – 4y – 7 = 0 (d) centre of the circle x2 + y2 + 2x – 4y – 4 = 0 Ans Solution y = mx + c Equation of the pair of lines through the origin and the points of intersection of the chords and the curve is

y - mx ˆ =0 3x2 – y2 – (2x – 4y) ÊÁ Ë c ˜¯ 3+

2m Ê 4ˆ + Á -1 + ˜ = 0 c Ë c¯

fi m So, the equation of the chord is y + (c + 2)x – c = 0 fi (y + 2x) + c(x Showing that the chord passes through the point of intersection of the lines y + 2x = 0 and x – 1 = 0 fi through the point (1, –2) and the equation of the parabola in (c) is (x – 1)2 = 4(y Note: d Example 25 Let P be the point on the parabola y2 = 4x which is at a shortest distance from the centre S of the circle x2 + y2 – 4x – 16y + 64 = 0 Let Q be the point on the circle dividing the line segment SP internally, then (a) SP = 2 5

(

)

5 +1 (b) SQ QP = (c) the x-intercept of the normal to the parabola at P (d) the slope of the tangent to the circle at Q is 1 2 Ans Solution S Let the coordinate of P be (t2, 2t), then (SP)2 = (t2 – 2)2 + (2t – 8)2 = t4 – 32t + 68 = f (t) (say) Now f ¢(t) = 4t3 – 32 = 0 fi t = 2 and f ≤(t) = 12t2, f ≤(2) > 0 fi SP is least when t = 2 fi Coordinate of P are (4, 4) (a) Least value of SP = 20 = 2 5 Suppose Q divides SP in the ratio k 4k + 2 4k + 8 ˆ Q are ÊÁ , Ë k + 1 k + 1 ˜¯ As Q lies on the circle x2 + y2 – 4x – 16y + 64 = 0 or (x – 2)2 + (y – 8)2 = 4, 2

2

4k + 2 ˆ Ê 4k + 8 ˆ = 4 We get ÊÁ - 2˜ + Á - 8˜ Ë k +1 ¯ Ë k +1 ¯ fi

2

Ê 2k ˆ Ê -4k ˆ ÁË ˜¯ + ÁË ˜ k +1 k + 1¯

2

=4

IIT JEE eBooks: www.crackjee.xyz 17.12 Comprehensive Mathematics—JEE Advanced

fi fi

y4 = (4a)3y fi y(y – 4a) (y2 + 4ay + 16a2) = 0 y = 0 or y = 4a a, 4a) Since 2bx + 3cy + 4d = 0 passes through these points, d = 0 and 2b + 3c = 0 (as a π 0)

fi 4k2 – 2k – 1 = 0 fi k = 2 ± 2 5 8 As k > 0, k =

5 +1 4

(b) SQ QP = 5 + 1 Equation of the normal to y2 = 4x at (4, 4) is y = –2x + 2(2) + 23 or y = –2x + 12 x-intercept of the normal is 6 (d) Equation of the tangent to the circle at Q is x

Example 28 Let P and Q be distinct points on the parabola y2 = 2x such that a circle with PQ as diameters O P lies in the OPQ in 3 2 , then which of the following is are the coordinates of P

4k + 2 4k + 8 Ê 4k + 2 ˆ Ê 4k + 8 ˆ +y - 2Á x + - 8Á y + + 64 = 0 ˜ k +1 k +1 k +1 ¯ Ë k + 1 ˜¯ Ë

2k k +1 1 ¥ Slope of the tangent = = k + 1 4k 2 Example 26 Suppose (xi, yi), i = 1, 2, 3, 4 be the point of intersections of the parabola y2 = 4ax and the circle x2 + y2 + 2gx + 2fy + c = 0, c π 0, then (a) y1 + y2 + y3 + y4 = 0 (b) x1 + x2 + x3 + x4 = 0 (c) y1y2y3y4 = 16a2c (d) x1x2x3x4 = c2 y2 in x2 + y2 + 2gx + 2fy + c = 0, we Solution x= 4a get y1, y2, y3, y4 are roots of y4

+ y2 + 2g

16a 2

y2 + 2 fy + c = 0 4a

Or y4 + 4a (4a + 2g)y2 + 32fa2y + 16a2c = 0 \ y1 + y2 + y3 + y4 = 0, y1y2y3y4 = 16a2c Note that x1 + x2 + x3 + x4 π 0 as xi ≥ 0 for i = 1, 2, 3, 4 Also, x1x2x3x4 =

( y1 y2 y3 y4 )2 256a

4

(b)

Ê1 1 ˆ (c) Á , Ë 4 2 ˜¯

(d)

(

)

P be t 2 , 2t , t > 0, and

Solution

(

the coordinates of Q be t12 , 2t1 Let m1 = slope of OP =

)

2 t

m2 = slope of OQ =

2 t1

p , m1m2 = –1 2 fi tt1 = –2 1 Area of OPQ = |x1y2 – x2y1| 2 1 = t 2 2t1 - t12 2t 2

As –POQ =

fi fi

=

2 (2) |t - t1 | 2

=

2 t+

3 2 =

[

tt1

Ê 2ˆ 2 Át + ˜ Ë t¯

2 = t

Ê 2ˆ 2 Át + ˜ Ë t¯

t2 – 3t + 2 = 0 fi t = 1, 2 P are 1, 2 or 4, 2 2

(

) (

)

Example 29 y2 = 36x at point A (t2, 6t) is 100, then value of t is (a) 1 (b) –1

2

Ê y2 ˆ Á 4a ˜ = 4ay Ë ¯

(9, 3 2 ) (1, 2 )

Ans

= c2

a π 0 and the line 2bx + 3cy + 4d = 0 Example 27 passes through the points of intersection of the parabolas y2 = 4ax and x2 = 4ay, then (a) 2b + 3c = 0 (b) d = 0 (c) 2b = a + c (d) d = a Ans y2 in x2 = 4ay, we get Solution x= 4a

(4, 2 2 )

(a)

(c) 2

(d) -

1 2

IIT JEE eBooks: www.crackjee.xyz Parabola 17.13

Ans Solution y2 = 36x point of the focal chord be B(t12, 6t1), then equation of AB is x - t2 t12

-t

2

y - 6t 6t1 - 6t

=

6(x – t2) = (t1 + t) (y – 6t)

or

6(9 – t2) = (t1 + t) (–6t) fi t1 = -

9 t

From (i) and (ii), we get t2 – t1 – 4(t1 – t2) = t1t2 (t1 – t2) + t3 (t12 – t22) [ t1 π t2 fi t1t2 + t2t3 + t3t1 = –5 From (i) t1 (t1t2 + t1t3) = t2 + t3 – 4t1 fi t1 [–5 – t2t3 t2 + t3 – 4t1 fi t 1 + t 2 + t 3 + t 1t 2t 3 = 0 1 1 1 = –1 fi + + t2t3 t3t1 t1t2 Also, (1 + t1) (1 + t2) (1 + t3) = 1 + t1 + t2 + t3 + (t1t2 + t2t3 + t 3t 1) + t 1t 2t 3 = 1 + 0 – 5 = –4

Length of AB is given by AB2 = (t12 – t2)2 + (6t1 – 6t)2 = (t1 – t)2 [(t1 + t)2

A (t12, 2t1)

2 2 ˘ Ê 9 ˆ ÈÊ 9 ˆ + t t + 36˙ ÍÁ = Á ˜ ˜ Ë t ¯ ÍË t¯ ˙˚ Î 2

Ê 9 ˆ Ê 9ˆ = Á + t˜ Á t + ˜ Ët ¯ Ë t¯ Ê 9ˆ AB = Á t + ˜ Ë t¯



t+

fi fi Example 30

2

Ê 9ˆ = Át + ˜ Ë t¯

E S (1, 0)

4

B (t22, 2t2)

Fig. 17.10

= 100

MATRIX MATCH TYPE QUESTIONS

9 = ±10 t t = 1, –1 Let A(t12, 2t1), B(t22, 2t2) C(t32, 2t3) be

three distinct points on the parabola y2 = 4x of DABC is the focus S(1, 0) of the parabola y2 = 4x, then (a) t1t2 + t2t3 + t3t1 = –5 (b) t1 + t2 + t3 + t1t2t3 = 0 (c)

C (t32, 2t3)

D

2

1 1 1 + + = –1 t1t2 t2t3 t3t1

(d) (1 + t1) (1 + t2) (1 + t3) = – 4

Example 31 y2 = 4ax be the equation of a parabola, then Column 1 Column 2 (p) Equation of the nor(a) yy1 = 2a (x + x1) mal at (x1, y1) (b) xy1 = 2a (y1 – y) + x1 y1 (q) Equation of the focal chord through (x1, y1) (r) Equation of the line (c) xy1 = y (x1 – a) + ay1 through the point of

Ans Solution

m1 = slope of BC =

2 t2 + t3

(d) (x + a) y1 = (x1 + a) y Ans

p

q

r

s

a

p

q

r

s

As AD ^ BC, m1m2 = –1

b

p

q

r

s

Ê 2 ˆ Ê 2t ˆ \ Á Á ˜ = –1 Ë t2 + t3 ˜¯ Ë t12 - 1¯

c

p

q

r

s

d

p

q

r

s

and m2 = slope of AD =

fi fi Similarly,

2

0 - 2t1 1 - t12

=

2t1 t12

-1

2

(s) Equation of the tangent at (x1, y1)

–4t1 = t1 t2 + t1 t3 – t2 – t3 t2 + t3 – 4t1 = t12t2 + t12t3

(i)

Solution y2 = 4ax Equation of the tangent at (x1, y1) is yy1 = 2a (x + x1)

t3 + t1 – 4t2 = t22t3 + t22t1

(ii)

Equation of the normal at (x1, y1) is y – y1 = -

y1 (x – x1) 2a

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fi xy1 = 2a (y1 – y) + x1y1 x1, y1) and (a, 0) is y1 - 0 (x – a) fi xy1 = y (x1 – a) + ay1 x1 - a a, 0), the point of y x + a = 0 with (x1, y1) is (x + a) y1 = (x1 + a) y

y=

circles described on AB and CD as diameter always passes through

P is the parabola y2 = 4a(x + a)

Example 32

L is the line x cos a + y sin a = p Column 1 (a) L touches the parabola P if (b) L of P if (c) L passes through the focus of P if (d) L passes through the point p

q

P if r s

t

a

p

q

r

s

t

b

p

q

r

s

t

c

p

q

r

s

t

d

p

q

r

s

t

Ans

that sum of reciprocals of two parts of the chord through P Coordinates of P (c) AB and CD are the chords of the parabola which intersect at a point E

Column 2 (p) p + a cos a = 0 p cos a + a = 0 (r) p + 2a cos a = 0 (s) p = 0

Solution x1, y1) to P is 2ax – yy1 + 2a (x1 + 2a) = 0 Comparing with L, we get sin a -p cosa 2a = - y1 = 2a( x1 + 2a) fi y1 = –2a tan a, x1 = – p sec a – 2a So, y12 = 4a (x1 + a) fi 4a2 tan2 a = –4a (p sec a + a) fi p cos a + a = 0 P is (–a, 0) fi –a cos a = p (c) Focus of P is (0, 0) fi p = 0 P is x = –2a y= 0 at (–2a, 0) So, –2a cos a = p Example 33 Consider the parabola y2 = 12x Column 1 Column 2

T and G the middle point of T and G (b) Variable chords of the parabola passing P

(d) circle described on the latus rectum of the parabola as a diameter meets p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans

Solution (a) Equation of the tangent at the end (3, 6) of the latus rectum is y = x + 3 and of the normal is y = –x + 9 So T(–3, 0), G(9, 0) fi Coordinates of the middle point of T and G are (3, 0) l1, l2 are the lengths of the segments of a focal chord of a parabola, then its latus rectum is 4l1l2 l1 + l2 1 1 + is Constant fi l1 l2 (c) Let A(t1), B(t2), C(t3), D(t4) AB and CD intersect at a point E t 1t 2 = t 3t 4 Equation of the circles on AB and CD as diameters are S ∫ (x – at12) (x – at22) + (y – 2at1) (y – 2at2) = 0 and S¢ ∫ (x – at32) (x – at42) + (y – 2at3) (y – 2at1) = 0 S – S¢ = 0 which passes through (0, 0) (d) Equation of the circle is (x – 3)2 + y2 = (6)2 y = 0 at (9, 0) and (–3, 0) Example 34 Locus of mid-point of chords of the parabola y2 = 12x which Column 1 Column 2 4ˆ Ê (a) pass through (3, 4) (p) y2 = 6 Á x - ˜ Ë 3¯ y2 = 6(x – 3)

IIT JEE eBooks: www.crackjee.xyz Parabola 17.15

(r) y2 = 6(x – 12) 8ˆ Ê (s) (y – 2)2 = 6 Á x - ˜ Ë 3¯ y2 = 6x

(c) pass through the focus (d) Subtend a right angle p

q

r

s

t

a

p

q

r

s

t

b

p

q

r

s

t

c

p

q

r

s

t

d

p

q

r

s

t

Ans

Column 1

2

DEFG is (c) y0 = (d) y1 = p q r s

– 12h Ans.

4k – 6(3 + h) = k2 – 12h 8ˆ Ê (k – 2)2 = 6 Á h - ˜ Ë 3¯

8ˆ Ê (y – 2)2 = 6 Á x - ˜ Ë 3¯ (b) (1) will pass through (0, 0) if 0 – 6(0 + h) = k2 – 12h fi k2 = 6h y2 = 6x (c) (1) will pass through the focus (3, 0) if 0 – 6(3 + h) = k2 –12h fi k2 = 6(h – 3)

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

m2 = slope of OB =

p 2 m1m2 = –1 fi t1t2 = –4 As (h, k) is the mid point of AB, 3 2 2 t1 + t2 h= 2 and k = 3 (t1 + t2)

6t2 3t22

y

E

=

)

k2 = 9(t12 + t22 + 2t1t2) 2h ˆ = 9 ÊÁ - 8˜ = 6(h – 12) Ë 3 ¯ y2 = 6 (x – 12)

(0, 3) F (x0, y0)

G x

0 y (mx + 3)

2 t2

Since, –AOB =

(

(r) 2 (s) 1

Solution: y = mx + 3 and y2 = 16x, we put y = mx + 3 in y2 = 16x, we get (mx + 3)2 = 16x fi m2x2 + 2(3m – 8) x + 9 = 0

y2 = 6(x – 3) (d) Suppose end points of a chord are A(3t12, 6t1) and B(3t22, 6t2) 6t 2 Let m1 = slope of OA = 12 = t 3t1 1 and

Column 2 1 (p) 2

(a) m =

Solution y2 = 4(3) x whose mid-point is (h, k) is ky – 6(x + h (a) (1) will pass through (3, 4) if



Example 35 A line L y = mx + 3 meets y E(0, 3) and the arc of the parabola y2 = 16x, 0 £ y £ 6 at the point F(x0, y0 G(0, y1 m of the F(x0, y0) intersects y line L is chosen such that the area of the triangle EFG has

Fig. 17.11

One of the roots of this equation is x0 to y2 = 16x at (x0, y0) is y y0 = 8(x + x0) y y1) where y1 = 8x0/y0 1 Now, D = Area of DEFG = (x0)(3 – y1) 2 2 8 x0 ˆ 1Ê = Á 3x0 y0 ˜¯ 2Ë =

8 x02 ˆ 1Ê x 3 0 y0 ˜¯ 2 ÁË

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(

)

1 3x0 - 2 x03/ 2 2 dD 1 fi = 3 - 3 x0 dx0 2 dD As, > 0 for 0 < x0 < 1 dx0 < 0 for x0 > 1 D x0 8 (1) when x0 = 1, y0 = 4, y1 = =2 4 Since y0 = mx0 + 3, we get m =

(

Example 37

)

= 0 fi x0

Let P be the parabola y2 = 4ax and

Example 36

A(at12, 2at1), B(at22, 2at2) be two points on P Column 1 Column 2 (a) AB is a focal chord of P (p) t1t2 = –4 p Slope of AB is 4

(b)

(q)

AB subtends a right angle (r) at the origin (s) AB is normal to P at A

(c) (d)

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans

Solution

p

q

r

s

a

p

q

r

s

b

p

q

r

s

t1t2 = –1

c

p

q

r

s

t12 + t1t2 + 2 = 0

d

p

q

r

s

t1 + t2 = 2

at22

-

at12

or

=

Solution

fi fi fi

y - 2at1 2at2 - 2at1

fi For For

2(x + at1t2) = (t1 + t2)y

(a, 0), that is, if a + at1t2 = 0 fi t1t2 = –1 (b) Slope of AB =

Ans

We get

AB is x - at12

Column 1 (a) Length of common chord of 2y2 = 3(x + 1) and x2 + y2 + 2x = 0 (b) Radius of the largest circle passing through the focus of y2 = 4x and lying entirely in the parabola (c) segments of a focal chord of the parabola y2 = 16x (d) Radius of circle passing through (3, 4) and (5, 2) and having centre on the line y = 2x

2 Êpˆ = tan Á ˜ Ë 4¯ t1 + t2

fi t1 + t2 = 2 (c) Let, m1 = slope of OA = 2/t1 and m2 = slope of OB = 2/t2 p As –AOB = , m1m2 = –1 2 fi t1t2 = –4 (d) Slope of normal at A = –t1 2 \ = –t1 t1 + t2 fi t 12 + t 1t 2 + 2 = 0



y2 =

Column 2 (p) 4 (q)

2 13

(r)

(s)

3 8

3 ( x + 1) in x2 + y2 + 2x = 0, 2

3 x 2 + ( x + 1) + 2 x = 0 2 2 2x + 3x + 3 + 4x = 0 2x2 + 6x + x + 3 = 0 (x + 3) (2x + 1) = 0 1 x= - , 3 2 2 x = –3, 2y = 3(–3 + 1) < 0 3 1 Ê 1 ˆ x = - , 2y2 = 3 Á - + 1˜ = Ë ¯ 2 2 2 y= ±

3 2

Ê 1 3ˆ Ê 1 3ˆ A Á - , - ˜ and B Á - , ˜ 2 ¯ Ë 2 2 ¯ Ë 2 \ Length of common chord = 3 (b) Let the centre of circle be (a, 0), then its radius is |a – 1|, and its equation (x – a)2 + y2 = (a – 1)2 Putting y2 = 4x, we get (x – a)2 + 4x = (a – 1)2 2 fi x + 2(2 – a)x + 2a – 1 = 0 (2 – a)2 – (2a – 1) = 0

IIT JEE eBooks: www.crackjee.xyz Parabola 17.17

a2 – 6a + 5 = 0 (a – 1) (a – 5) = 0 a = 1, 5

or fi fi

p

q

r

s

t

a

p

q

r

s

t

b

p

q

r

s

t

c

p

q

r

s

t

d

p

q

r

s

t

Ans

(1, 0)

Solution equation is Fig. 17.12

When a When a Let one end point of the focal chord of y2 = 16x be 4 8 A(4t2, 8t), then the other end point is B ÊÁ , - ˆ˜ Ë t2 t ¯ We have

FA2 = (4t2 – 4)2 + (8t – 0)2 = 42 (t2 + 1)2 fi FA = 4(t2 + 1) Ê1 ˆ Similarly, FB = 4 Á 2 + 1˜ Ët ¯

Harmonic mean H of FA and FB is given by t2 ˘ 1È 1 1 1 1 1 ˆ = ÊÁ = + + Í ˙ ˜ 2 Î 4(t 2 + 1) 4(t 2 + 1) ˚ H 2 Ë FA FB ¯ fi H=8 (d) Let the centre of circle be (a, 2a), then (a – 3)2 + (2a – 4)2 = (a – 5)2 + (2a – 2)2 fi a2 – 6a + 9 + 4a2 – 16a + 16 = a2 – 10a + 25 + 4a2 – 8a + 4 fi a = –1

Ê 1 1ˆ ÁË - , ˜¯ 2 2 Let the equation of dire (which is parallel to the x–y+a=0 Ê a aˆ AÁ - , ˜ Ë 2 2¯ 1 1 As ÊÁ - , ˆ˜ Ë 2 2¯ Ê a a ˆ and focus (0, 0) we get ÁË - , ˜¯ 2 2 fi

P is a parabola whose focus is (0, 0) and x–y+1=0

Column 1 (a) (b) (c) (d)

the parabola Equation of direcEquation of latus rectum Equation of tangent at the end point of latus rectum

= 4

0 - 0 +1 2

= 2 2

Equation of latus rectum is x = y points of latus rectum are 2 , 2 and - 2 , - 2

(

(p) (q)

x=y

(r)

x+y=0

(s)

( 2 + 1) x - ( 2 - 1) y

or

+2 2=0

and

( 2 - 1) x - ( 2 + 1) y

or

= 2 2

a= 2

Length of latus rectum = 4 [length of perpendicular from

Column 2 x–y+2=0

(t)

1 1Ê aˆ = Á0 - ˜ ’ 2 2Ë 2¯ x–y

(-1 - 3) 2 + (-2 - 4) 2 = 2 13 Example 38

(x – 0) + (y – 0) = 0 x+y=0

or

(

)

)

tangents at end points of latus rectum pass through the tangents at the end point of latus rectum are x +1 y -1 = 2 +1 2 -1 x +1 - 2 +1

(

(

=

) (

2 -1 x y -1

)

2 +1 y = 2 2

- 2 -1

) (

2 +1 x -

)

2 -1 y + 2 2 = 0

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ASSERTION-REASON TYPE QUESTIONS x2 + 2y2 = 5 and a

Example 39

parabola y2 = 4 5x Statement-1: An equation of a common tangent to these curve is y = x + 5

m4 – 3m2

common tangent, then m Ans

y2 = 4 5x is

Solution y = mx + x2 + y2 = m(0) - 0 +

fi fi fi Also m

5 , (m π 0) is their m

y = mx +

Statement-2:

5 m

m2 + 1

=

5 m

(m π 0)

5

a will pass through (–a, h) if m a h = - ma + m fi am2 + hm – a = 0 As h2 – 4a(–a) = h2 + 4a2 > 0, this equation has real and distinct roots, say m1 and m2 and product of the roots is m1m2 = –1

y = mx +

[

mŒR

Let P be the parabola y2 = 2x P pass through h £

P passes y2 = 2x is

x – 2y = 2 meets

Example 42 Statement-1: the parabola y2 + 2x

1 , (m π 0) is a tangent 2m 1 1ˆ Ê to the parabola y2 = – 2x at point Á - 2 , ˜ Ë 2m m ¯ Ans 1 Solution y = mx in y2 = – 2x, we get 2m y = mx -

Statement-2:

2

= – 2x

2

¤



¤ h

Let P be the parabola (y – 3)2 = 2(x a to P m

5 2

Ê 5 ˆ ÁË - ,1˜¯ to P 2

h – 1 > 0, that

from point mx +

x= -

1 ˆ Ê Ê 1 ˆ ¤ Á mx ˜¯ + 4(mx) ÁË ˜ =0 Ë 2m 2m ¯

m3 = 2m (h – 1) = 0 m [m2 – 2(h

Statement-1:

1 2

1 ˆ Ê ÁË mx ˜ 2m ¯

1 3 0 = mh - m - m 2

Example 41

x + 2= or

1 3 y = mx - m - m 2 h, 0) if

is, h that is, P

y2 = 4ax a, h

y – 3)2 = 2(x + 2) is

y = x + 5 and y = - x - 5

fi fi

Ans Solution x+a

5 , if 2

2 (m2 + 1)m2 = 2 (m2 + 2) (m2 – 1) = 0 m = ±1, m4 – 3m2 + 2 = 0

Example 40 Statement-1: (h, 0), then h Statement-2: through (h Ans Solution

Statement-2:

when

1 ˆ Ê ÁË mx + ˜ 2m ¯ x =-

2

=0

1

2m 2 1 x= - 2 2m 1 ˆ 1 Ê 1 y = mÁ - 2 ˜ = Ë 2m ¯ 2m m

IIT JEE eBooks: www.crackjee.xyz Parabola 17.19

1 y = mx and y2 = – 2x is 2m

1 1ˆ Ê ÁË - 2 , - ˜¯ m 2m \ We can write x – 2y

1 1 x2 Ê 1ˆ 2Á ˜ Ë 2¯ By Statement-2, it touches the parabola y2 = –2x at Ê ˆ Á ˜ Á - 1 , - 1 ˜ = (–2, –2) Á Ê 1ˆ 2 1 ˜ Á 2Á ˜ 2 ˜¯ Ë Ë 2¯ as

Example 43

y=

Statement-1:

1 points (8, 8) and ÊÁ , -2ˆ˜ to the parabola y2 = 8x are at Ë2 ¯ Statement-2: Ans Solution

y2 = 4ax at the

Ê a 2a ˆ end points A (at2, 2at) and B Á 2 , - ˜ are respectively Ët t ¯ y = tx + at2 y= –

(1)

1 a x+ 2 t t

(2)

Ê1 ˆ As (8, 8) and Á , -2˜ are end points of a focal chord of the Ë2 ¯ parabola y2 – 8x, the tangents and hence, normal at these Example 44

Statement-1:

y2 = 4ax,

Statement-2:

F(a, 0) is the focus of the parabola y2 = 4ax, then the 2at 2t slope of PF is m1 = 2 = 2 and slope of QF is at - a t -1 Ê 1ˆ aÁt - ˜ Ë t¯ t2 -1 = m2 = (- a) - a -2t As (m1) (m2) = –1 We get the portion intercepted between the point of contact An equation of circle with FP as diameter is (x – a) (x – at2) + (y – 0) (y – 2at) = 0 when x = 0, (1) becomes a2t2 + y (y – 2at) = 0 or (y – at)2 = 0 fi (1) touches the y at

(1)

x + 2y = 5 and 2x – y = x + y = 3, x–y Statement-2: Perpendicular tangents to the parabola Example 45

Statement-1:

Ans Solution

y2 = 4ax is

a (1) m 2 and an equation of tangent to the parabola y = 4ax which is perpendicular to (1) is 1 y = - x - am (2) m Note that (1) and (2) intersect on the line x + a = 0, that is, y = mx +

\ As x + 2y = 5 and 2x – y = 0 are perpendicular to each other,

parabola, its equation is (x – 1) – (y – 2) = 0 or x–y+1=0

COMPREHENSION-TYPE QUESTIONS Ans Solution P(at2, 2at) 2

to the to the parabola y = 4ax is ty = x + at2, (t π 0, ±1) Ê Ê 1ˆ ˆ x + a = 0 in Q Á - a, a Á t - ˜ ˜ Ë t¯¯ Ë

Paragraph for Question Nos. 46 and 47 Let a, r, s, t P(at2, 2at), 2 2 Q, R(ar , 2ar) and S(as , 2as) be distinct points on the PQ is the focal chord and parabola y2 = 4ax line QR and PK are parallel, where K is the point (2a Example 46 (a) -

1 t

r is 2 (b) t + 1 t

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y = – sx + 2as + as3 As st = 1, we can write (2) as

2 (d) t - 1 t

1 (c) t

st = 1, then the tangent at P and the Example 47 normal at S to the parabola meet at a point whose ordinate is (a) (c)

(t 2 + 1)2 2t 3 a (t 2 + 1)2

Ans

a (t 2 + 1)2

(b)

2t 3

1 2a a + 3 y = - x+ t t t a fi ty = - x + 2a + 2 t Adding (1) and (3), we get 2ty = 2a + at2 + a/t2 Ê 1ˆ = aÁt + ˜ Ë t¯

a(t 2 + 2) 2

(d)

t3

t3 fi

y

F

K (a, 0)

(2a, 0)

x

R Q

a (t 2 + 1)2

(a) 1 : 2 Example 49 triangle PRS is

Solution Q be (at12, 2at1) An equation of PQ is x - at 2 y - 2at = 2 2 2at1 - 2at at1 - at

(a) 5 Example 50 PQR is (a) 4

As it passes through (a, 0), we get a(1 - t 2 ) -2at = a(t12 - t 2 ) 2a(t1 - t )

(b) 3 3

(c) 3 2

(d) 2 3

(b) 3

(c) 8/3

(d) 2

Solution we solve

fi fi

1 – t2 = –t(t1 + t) t1 = –1/t 0 - 2at -2t Slope of PK = 2 = 2a - at 2 - t2

and

P and Q x2 + y2 = 9 y2 = 8x

ar 2 - at12 2 2t = = r + t1 rt - 1

(1) (2)

y

2ar - 2at1

As PK is parallel to RQ -2t 2t 2 = 2-t rt - 1 fi – rt + 1 = 2 – t2 fi rt = t2 – 1 fi r = (t2 – 1)/t P(at2, 2at) is 2 ty = x + at and an equation of normal at S(as2, 2as)

2

Example 48 PQS and PQR is

Fig. 17.13

Slope of RQ =

y=

(3)

2t 3 Paragraph for Question Nos. 48 to 50 Consider the circle x2 + y2 = 9 and the parabola y2 = 8x P and Q P and Q intersect the x R and tangents to the parabola at P and Q intersect the x S

P(at2, 2at)

O

(2)

2

2

x

+y

(1, 2 2)

=9

P

O S (-1, 0)

y2 = 8x

T (1, 0)

Q (1, − 2 2)

(1) Fig. 17.14

R (9, 0)

x

IIT JEE eBooks: www.crackjee.xyz Parabola 17.21

Simultaneously from (1) and (2), We get x2 + 8x – 9 = 0 fi (x + 9) (x – 1) = 0 fi x= 1 \ y2 = 8 fi

y = ±2 2

Paragraph for Questions Nos. 51 and 52 Let PQ be a focal of the parabola y2 = 4ax the parabola at P and Q meet at a point lying on the line y = 2x + a, a Length of chord PQ is

Example 51 (a) 7a

(

P are 1, 2 2

(1, -2 2 )

x

[

)

and that of Q are

2 2 y = 4(x + 1) and -2 2 y = 4(x + 1)

2 7 3

(b)

Ê a -2a ˆ coordinates of Q are Á 2 , ˜ Ët t ¯ y

1 Let D1 = area of DPQS = ( PQ)( ST ) 2

O

1 (2)(4 2 ) = 4 2 2 1 and D2 = area of DPQR = (PQ) (TR) 2 1 4 2 (8) = 16 2 = 2

(

)

2

2

Equation of tangents at P and Q are ty = x + at2 1 Ê 1ˆ - y = x + aÁ 2 ˜ Ët ¯ t

= 4 + 8 = 12

Ê Ê 1ˆ ˆ R Á - a, a Á t - ˜ ˜ Ë t¯¯ Ë

= 64 + 8 = 72

SR = 10 1 1 ( SR )( PT ) = ¥ 10 ¥ 2 2 = 10 2 2 2 ( PS ( PR)( SR) R= 4D

D = area of DPSR =

= 2 3 ¥ 6 2 ¥ 10 = 3 3 4 ¥ 10 2

Since R lies on y = 2x + a, we get Ê 1ˆ a Á t - ˜ = –2a + a Ë t¯ fi

t-

1 = –1 t



t+

1 = t

2

a 2a PQ2 = ÊÁ at 2 - ˆ˜ + ÊÁ 2at + ˆ˜ 2¯ Ë Ë t ¯ t

PQ = 4 2 D2 = area of DPQR = 16 2 D2 r= s =

16 2 1 6 2 +6 2 +4 2 2

(

5

We have

PR = RQ = 6 2 and

x

Fig. 17.15

D2 = 4 2 : 16 2

)

(a, 0) Q (a/t, 2a/t)

( )

PR2 = (9 – 1)2 + 2 2 - 0

P (at, 2at)

R

=

(

-2 -2 2 5 7 (c) 5 (d) 3 3 3 P are (at2, 2at), then

S

PS2 = (1 + 1)2 + 2 2 - 0

(d) 3a

y2 = 4ax, then tan q =

Ans Solution

On solving these, we obtain coordinates of R Equations of tangents to parabola at P and Q

D1

(c) 2a

PQ subtends an angle q at the

Example 45

(a)

Equations of tangents to the circle at P and Q x + 2 2 y = 9 and x - 2 2 y = 9

(b) 5a

)

=2



2 2 ˘ 1ˆ ÈÊ 1ˆ 2Ê a t + t Í = ÁË ˜¯ ÁË ˜¯ + 4˙ t ÍÎ t ˚˙ 2 = 25a PQ = 5a

2

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m1 = slope of OP =

2at

=

Let image of P(t2, 2t) in y = x + 1 be P(x¢, y¢), then PP¢ is perpendicular to T and mid point of PP¢ lies on T, that Ê y ¢ - 2t ˆ ÁË ˜ (1) = –1 x¢ - t 2 ¯

2 t

at 2 2a t = – 2t and m2 = slope of OQ = a t2 tan q =

Now,

m1 - m2 1 + m1m2

Ê1 ˆ 2 Á + t˜ Ët ¯ = 1- 4 2 5 = 3 y

O

1 È ˘ Í t + t = 5˙ Î ˚

P (at, 2at)

q

Ê 2t + y ¢ ˆ t 2 + x¢ +1 ÁË ˜¯ = 2 2 fi x¢ + y¢ = 2t + t2 and –x¢ + y¢ = –2t + t2 + 2 fi x¢ + 1 = 2t, y¢ – 1 = t2 Eliminating t, we get 1 y¢ – 1 = ( x ¢ + 1)2 4 x + 1)2 = 4(y – 1) 2 y = 4x at (1, 2) is a circle touching T and

x

(a, 0)

Q (a/t, 2a/t)

(x – 1)2 + (y – 2)2 + l[x – y As it passes through the focus (1, 0), we get 02 + 4 + l (2) = 0 or l = –2 \ Required equation is (x – 1)2 + (y – 2)2 – 2x + 2y – 2 = 0 x2 + y2 – 4x – 2y + 3 = 0

or

\ radius of this circle is

Fig. 17.16

2 ABOA) = area (DABF) – area (OBFO) 1 1 = (2)(2) - Ú0 2 xdx 2

Paragraph for Question Nos. 53 to 55 A tangent T is drawn to the parabola y2 = 4x at point A Example 53 (a) (x + 1)2 = 4(y – 1) (c) (x + 1)2 = 4(1 – y)

y2 = 4x in T is (b) (x – 1)2 = 4(y + 1) (d) (x – 1)2 = 4(1 – y)

1

4 3/ 2 ˘ = 2- x ˙ 3 ˚0 2 2 = (unit) 3

Example 54 Radius of circle touching the parabola y2 = 4x at point A and passing through the focus of y2 = 4x is (a) 1 (c)

(c) Ans Solution or

y2 = 4x

(d) 2

3

Example 55 parabola is 4 (a) 3

B(1, 2)

2

(b)

Area enclosed by T, the x A

(b) 1

2 3

(d) T is 2y = 2(x + 1) y= x + 1

1 3

(-1, 0)

O

F(1, 0)

Fig. 17.17

INTEGER-ANSWER TYPE QUESTIONS x + y = 6 is a Example 56 normal to the parabola y2 = 8x at point (a, b), b > 0 then |a – b| is equal to

IIT JEE eBooks: www.crackjee.xyz Parabola 17.23

Ans Solution a + b= 6 and b2 = 8a \ b2 = 8(6 – b) fi (b + 12) (b – 4) = 0 As b > 0, b = 4 \ a= 2 fi |a – b| = |2 – 4| = 2

Example 59 (y – 7)2 = r2 least value of [r Ans Solution

y = 2x – k, k π 0, meets the p parabola y = x – 4x at points A and B –AOB = , then 2 value of k is, (0 bring the origin)

x2 = 4y and the circle x2 + then x

£x

y-coordinate of the point of intersections, we solve 4y + (y – 7)2 = r2 or y2 – 10y + 49 – r2 = 0

Example 57

2

Ans Solution

OA and OB is

1 x - (4 x + y ) (2 x - y ) = 0 k fi (k – 8) x2 + y2 – 2xy = 0 p As –AOB = 2 k–8+1=0 or k= 7

100 – 4 (49 – r2) > 0 fi r2 > 24 fi r > 24 fi least value of [r

2

Example 60 the parabola

AB is a focal chord of the parabola y2 1 = 4ax, then minimum possible value of l ( AB) is [l (AB) a stands for length of AB Example 58

Ans Solution Ê a 2a ˆ , then ÁË 2 , - ˜¯ t t

as 15x – 8y L = length of latus rectum | 15(3) - 8(1) + 13 | 17 100 = 17 = 2

2

2 2 ˘ Ê 1ˆ ÈÊ 1ˆ = a 2 Á t + ˜ ÍÁ t - ˜ + 4˙ Ë t ¯ ÍË t ¯ ˚˙ Î 2 2 1ˆ Ê 1ˆ 2Ê = a Át + ˜ Át + ˜ Ë t¯ Ë t¯

1ˆ 2Ê = a Át + ˜ Ë t¯



17 289 {(x – 3)2 + (y – 1)2} = (15x – 8y + 13)2, then L is 25 equal to Ans Solution | 15 x - 8 y + 13 | ( x - 3) 2 + ( y - 1) 2 = 17

A are (at2, 2at) and that of B are

2a ˆ Ê 2 aˆ Ê l(AB)2 = Á at - 2 ˜ + Á 2at + ˜ Ë ¯ Ë t ¯ t



L is the length of the latus rectum of

4

4

Ê 1ˆ l ( AB)2 = Á t + ˜ ≥ 16 2 Ë t¯ a 1

1 l ( AB) ≥ 4 a 1 l ( AB) a

2



17 L =4 25

y2 = 4x Example 61 drawn at the end of its latus rectum are tangents to the circle (x – 3)2 + (y + 2)2 = r2, then value of r2 is Ans Solution For (1, 2), t = 1 and equation of normal to y2 = 4x is y = –x + 2 + 1 or x + y= 3 For (1, –2), t = –1 and equation of normal to y2 = 4x is y= x – 2 – 1 or x – y= 3 As x + y = 3, x – y = 3 are tangents to (x – 3)2 + (y + 2)2 = r2, we get | 3 - (-2) - 3 | r= = 2 1+1 r2 = 2

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k for which the line (1 – k) x Example 62 – (k + 1) y + 2(k + 1) = 0, (k π on the parabola x2 = 4(y – 1) is Ans Solution x2 = 4(y the line (1 – k) x – (k + 1) y + 2(k + 1) = 0 (1) or (x – y + 2) – k (x + y – 2) = 0 x2 = 4(y – 1) will be shortest, if (1) is latus rectum of x = 4(y – 1), that is, if (1) becomes y k

(x – 1) + (y – 2) = 0 or x+y=3 Foot of perpendicular from (1, 2) to x – y = 0 is the point Ê 3 3ˆ of intersection of x – y = 0 and x + y = 3, that is A Á , ˜ Ë 2 2¯ Similarly, foot of perpendicular from (1, 2) to x + y = 0 is Ê 1 1ˆ BÁ- , ˜ Ë 2 2¯ parabola is (an equation of AB)

2

x + by + c = 0, (bc π 0), touches Example 63 both the parabolas y2 = 4x and x2 = – 32y, then b + c is equal to Ans Solution x + by + c = 0 can be written as 1 c (1) y = - xb b Note that (1) will be a tangent to y2 = 4x, if and only if c - =–b b ¤

3 =0 2 Now l = Length of latus rectum

or

x - 2y +

1 - 2( 2) = 4 fi

3 2

1+ 4

=

6 5

5l = 6

2

b =c 1 y = - x-b b

Ê 1ˆ Putting y = Á - ˜ x - b in x2 = – 32y, we get Ë b¯ ˆ Ê1 x2 = 32 Á x + b˜ ¯ Ëb

ax + by + c = 0 is a common Example 65 2 tangent of the parabolas y = 4x and x2 + 32y = 0, then |a| + |b| + |c| is equal to Ans Solution y2 = 4x is y = mx + Putting y = mx +

bx2 – 32x – 32b2 = 0

or

fi fi \

3 3 y2 2 = 1 3 1 3 - 2 2 2 2 x-

(–32)2 – 4 b(–32b2) = 0 b =–8 b= – 2 b + c = b + b2 = – 2 + 4 = 2 3

Example 64 Let l be the length of the latus rectum of the parabola whose focus is (1, 2) and two tangents are x – y = 0 and x + y 5l is equal to Ans Solution TIP

An equation of line through (1, 2) and perpendicular to x – y = 0 is

1 m

1 in x2 + 32y = 0, we get m 1ˆ Ê x 2 + 32 Á mx + ˜ = 0 Ë m¯

is possible if Ê 32 ˆ 322 m 2 - 4 Á ˜ = 0 Ë m¯ 1 2 \ an equation of common tangent is fi 8m3 = 1 fi m =

1 x + 2 = 0 or x – 2y + 4 = 0 2 a = 1, b = –2, c = 4 |a| + |b| + |c| = 7 y=

fi \

IIT JEE eBooks: www.crackjee.xyz Parabola 17.25

Example 66

x + y = k is a normal to the parabola y2 =

36x, then |k – 19| is equal to Ans Solution x + y = k as y = – x + k –1 and k = –2am – am3 = –2(9) (–1) – (9) (–1)3 = 27 fi |k – 19| = 8

m=

q is acute angle between the tangents Example 67 drawn from point (1, 4) to the parabola y2 = 4x, then 4 cos q is equal to Ans Solution y2 = 4x is 1 y = mx + m 1 fi m2 – 4m + 1 = 0 m Let m1, m2 be roots of this equation, then 4= m+

tan q =

m1 - m2 = 1 + m1m2

= q=

(m1 + m2 ) - 4m1m2 2

16 - 4 = 1+1

1 + m1m2 3

\ \

3

2

2

Example 70 of a common chord AB of the circle x2 + y2 = 5 and the parabola y2 = 4x intersect at the point T A square ABCD is constructed on this chord lying inside the parabola then 1 (TC + TD) is equal to 10 Ans Solution parabola are A (1, 2), B (1, –2) Equation of common chord is x = 1, which is the latus \

AB intersect on the x

Coordinates of T are (–1, 0) Since length of AB = 4, sides of the square ABCD are of length 4, coordinates of C are (5, –2) and of D (TC)2 = (TD)2 = (5 + 1)2 fi TC + TD = 4 10 x = -1

x + y = 6 is a normal to the parabola y2 =

T

8x at point (a, b) with b < 0, then |b Ans Solution a + b = 6, b2 = 8a \ b2 – 8(6 – b) = 0 fi b2 + 8b – 48 = 0 fi (b + 12)(b – 4) = 0 b = –12 and b = 4 As b < 0, |b| – 7 = 5 x + y = k is a normal to the parabola Example 69 2 y = 12x, p is the length of the perpendicular from the focus of the parabola on this normal; then 3k3 + 2p2 – 2220 is equal to Ans Solution y2 = 12x, equation of a normal with slope –1, is [m = –1, a y = – x – 2(3) (–1) – (3) (–1)3 fi x + y = 9 so k = 9 Focus of the parabola is (3, 0)

fi 2p2 = 36

3k + 2p = 2223

p fi 4 cos q = 2 3

Example 68

3-9

p=

t

A(C1, 2)

O

B(1, -2)

D(5, 2)

x=5 C(5, -2)

Fig 17.18

EXERCISE LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS y = x and the 2 and that from its focus is 2 2 (a) (x + y)2 = (x – y – 2) (b) (x – y)2 = (x + y – 2)

IIT JEE eBooks: www.crackjee.xyz 17.26 Comprehensive Mathematics—JEE Advanced

(c) (x – y)2 = 4(x + y – 2) (d) (x – y)2 = 8(x + y –2) y = mx – 2am – am3 to the parabola 2 y = 4ax (a) m = 1

(b) m =

2 +1

(c) m = ± 2

(d) m =

2 -1

2

2

y = 4x and x = 4y is (a) x – y + 1 = 0 (c) x + y + 1 = 0

(b) x + y – 1 = 0 (d) y = 0

of the parabola (y – 1)2 = 2(x + 2), which does not lie on the line 2x + y + 3 = 0 are (a) (–2, 1) (b) (–3/2, 1) (c) (–3/2, 2) (d) (–3/2, 0) y2 = 9x which passes through the point (4, 10) and q that tan q > 2 is (a) (4/9, 2) (b) (36, 18) (c) (4, 6) (d) (1/4, 3/2) y2 = 36x, whose ordinate is three times its abscissa is (a) 2x + 3y + 44 = 0 (b) 2x – 3y + 44 = 0 (c) 2x + 3y – 44 = 0 (d) 2x – 3y = 0 a, b > 0, then the angle of intersection of two parabolas y2 = a3x and x2 = b3y at a point other than the origin is Ê 3ab ˆ (a) tan -1 Á ˜ Ë 2( a 2 + b 2 ) ¯

ab ˆ -1 Ê (b) tan Á 2 ˜ Ë (a + b 2 ) ¯

Ê a 2 + b2 ˆ (c) tan -1 Á ˜ Ë ab ¯

Ê 3(a 2 + b 2 ) ˆ (d) tan -1 Á ˜ Ë 2ab ¯

O

=

LL¢ is the latus rectum of the P be a point on the parabola and Q be OPQ p 2

(a) a (c) –a

PQ on the abola be a, then a – |LL¢ | equals (b) 2a (d) 0

x2 + 4x + 2y = 0 is 3 (a) y = 2 3 (c) y = 2

(b) y =

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS P(5, b) to the parabola y2 = 4x, then (a) there are three normal if –2 < b < 2 if b slope if b > 2 (d) one of the normal is the x

2 3

b=0 -

2

2

2

ola y = 4x and x + 4y = 8 is are (a) x + 2y + 4 = 0 (b) x – 2y + 4 = 0 (c) x – 2y – 4 = 0 (d) x + 2y – 4 = 0 a the parabola y2 = 4ax points of intersection of the parabola and the circle (a) x + y + a = 0 (c) x – y + a = 0

(b) x + y – a = 0 (d) x + 2y – a = 0

parabola y2 = 4(x – 3) are (a) (4, 2) (b) (4, –2) (c) (39, 12) (d) (39, –12) y2 = 4ax at a distance 2ab ac, then 1 (b) c ≥ 4 (a) 0 < b £ 2 (d) b2c ≥ 2 (c) b2c = 1

MATRIX-MATCH TYPE QUESTIONS PQ be a chord of the parabola y2 = 4ax such that the normals at P and Q intersect at point R Match the statements in Column 1 with appropriate Column 1 (a) Length of chord PQ (b) Distance of PQ from

2 3

(d) y = -

y = (x – 11) cos q – cos 3q is a normal to the parabola y2 = 16x for (a) only one value of q (b) two values of q (c) all values of q (d) no value of q

Column 2 (p) 20 4 3

IIT JEE eBooks: www.crackjee.xyz Parabola 17.27

(c) Area of DPQR

4

(r)

Column 1

Column 2

(a)

13

(d) Area of triangle bounded by the line PQ and

(p) (–•, 0) » (0, •) of a for which two distinct tangents can be drawn from (a + 3, a) to the parabola y2 = 4x

13 (b)

Column 1

Column 2 k 0, intersect at points A and C O (0, 0), A, B (a, 0) and C are concyclic, then a value of a is (a) 5 (b) 7 (c) 8 (d) 12

D1 be the area y2 = 8x of the triangle formed by the end points of its latus 1 rectum and the point P ÊÁ , 2ˆ˜ on the parabola, and Ë2 ¯ D2 be the area of the triangle formed by drawing tangents at P D1 is D2 S be the focus to the parabola y2 = 8x and let PQ be the common chord of the circle x2 + y2 – 2x – 4y PQS is

(a) 5 2

(b) 7 2

(c) 8 2 (d) 12 2 O, A, B and C are concyclic, then equation of tangent to y2 = –8a (x – a) at (–a, 4a) is (a) x + y = 36 (b) x – y = 24 (c) x + y = 30 (d) x – y = 36 Paragraph for Questions Nos. 29 and 30 Lines x = y and x + y = 0 are two tangents to a parabola

(a) (c)

9

(b)

41 5

(d)

41

(a) 4x – 5y + 9 = 0 (c) 3x + 2y – 22 = 0

41 1 41

(b) 4x + 5y – 41 = 0 (d) 2x – 3y + 2 = 0

y2 = 4x from the x = 2 in 1 2 2

y = x2 – x and x = y2 – y is x + y = k is a normal to the parabola y2 = 12x, then k is LEVEL 2

7

INTEGER-ANSWER TYPE QUESTIONS

points A and B, then

C be the mirror image of the parabola A and y2 = 4x with respect to the line x + y B are the points of intersection of C with the line y = –5, then the distance between A and B is y = 2x + 3 is a tangent to the parabola 1 y2 = 4a ÊÁ x - ˆ˜ , then 3(a – 5) is equal to Ë 3¯

AB is equal to

a, b) to the parabola y2 = 4x are such that slope of one of the tangents is double the other, then 9a is equal to

b2 l be length of the normal chord of the parabola y2 = 4x 1 the parabola, then l is equal to 2 3 S is focus of the parabola y2 = 8x ASB is focal chord of the parabola with SA = 6, then SB is equal to

SINGLE CORRECT ANSWER TYPE QUESTIONS parabola y2 = 8ax whose distance from the focus is 8a an angle q x, then q is equal to (a) p/6 (b) p/4 (c) p/3 (d) none of these y2 = 4ax (at2, 2at (b) 2t2 + 8t – 5 = 0 (a) 2t2 – 8t + 5 = 0 (d) 2t2 – 8t – 5 = 0 (c) 2t2 + 8t + 5 = 0 PN is an ordinate of the parabola y2 = 9x line is drawn through the mid-point M of PN parallel Q NQ T, then AT/NP = (a) 3/2 (c) 2/3

A, at a point (b) 4/3 (d) 3/4

IIT JEE eBooks: www.crackjee.xyz Parabola 17.29

P is a point on the parabola y2 = 4ax; Q and R are A PQR is an equilateral triangle lying within the parabola and –AQP = q, then cos q = (a)

2- 3 2 5

8 5

5-2

(c)

9

(b)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

(d) none of these

2 5 A and B of a parabola y2 = 4ax lies on the line y + a = 0, then the point of intersection of the tangents at A and B lie on (a) x + a = 0 (b) x2 + y2 = a2 2 2 2 (c) x – y = a (d) xy = a2 y2 = 4ax and x2 = 4by is (a)

1 a3 x 1

1 + b3 y

1 + (ab) 3

1

=0

2

(b) a 3 x + b 3 y + (ab) 3 = 0 1

1

(d)

1 + b3 y

2 - (ab) 3

3 2 (c) 4

(a)

=0

)

(a)

1 2 y1 - 4ax1 a

(c)

1 2 y1 - 4ax1 4a

)

3 2

(b) x = 0

3y + x + 3a = 0 (d) x + y = 0 P(at2, 2at) to the parabola 2 y = 4ax subtends a right angle at the focus of the parabola, then t is equal to (a)

2

(

(b)

1 2 y1 - 4ax1 2a

(d)

2 2 y1 - 4ax1 a

(

)

)

3 2

3 2

(b) 2

(c) –2 (d) - 2 P is a point on the parabola (y – 2)2 = 2(x P has the slope m such that m2 = 1, the coordinates of P can be Ê 5 ˆ Ê 5 ˆ (b) Á - , 3˜ (a) Á - ,1˜ Ë 2 ¯ Ë 2 ¯ Ê3 ˆ (c) Á , 5˜ Ë2 ¯

a point P(x1, y1) outside the parabola y2 = 4ax and their chord of contact is 3 2

3y = x + 3a

(c)

(b) 2 5 (d) 2

(

PQ be a double ordinate of the parabola R y2 = –4x, where P divides PQ in the ratio k k > 0, then locus of R is (a) 9y2 = –4x if k = 2 (b) y2 = –4x if k = 2 (c) y2 = –x if k = 3 (d) y2 = –2x if k = 3

(a)

25 ax , then m is equal to 4

(

3- 2 2

y2 = 4ax and the circle x2 + y2 + 2ax = 0 are

p parabola y2 = 4ax which are inclined at angle of 4 is 2 2 2 2 (a) y – 4ax = (x + a) (b) y – 2ax = x (c) y2 – 4ax = (x – a)2 (d) y2 – 4ax = x2 – a2 y2 = 4ax drawn at points m whose abscissae are in the ratio m2 is y2 =

a, a + 1) lies in the area common to y2 £ 4x and x2 + y2 £ 4, then a can be

1

(c) a 3 x + b 3 y - (ab) 3 = 0 1 a3 x

P(x1, y1) and Q(x2, y2) on the parabola y2 = 4x intersect at a point R on the y1y2 > 0, then y1y2 = (a) 2 (b) 6 (c) 8 (d) 4

Ê3 ˆ (d) Á , -1˜ Ë2 ¯

MATRIX-MATCH TYPE QUESTIONS y2 = 12x at point P is inclined at an angle a Column 1 p (a) a = 4 p (b) a = 3

Column 2 (p) P(3, 6)

(

(q) P 1, 2 3

)

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(c) a =

3p 4

(r) P(3, –6)

Statement-2: Perpendicular tangents to the parabola y2 = 4ax

(d) a =

2p 3

(s) P(1, -2 3 )

COMPREHENSION-TYPE QUESTIONS

Column 1 (a) P(1, 2) (b) P(1, –2) (c) P(9, 6) (d) P(9, –6) P y2 = 8(x – 2) and Q olas Column 1 (a) (0, 0) (p) (b) (4, 0) (q) (c) (3, –1) (r) (d) (4, 3) (s)

y2 = 4x. P is a point on P Column 2 (p) y – 2x + 12 = 0 (q) 3x + y + 33 = 0 (r) x + y – 3 = 0 (s) x – y – 3 = 0 x – 4)2 = 8y are two parabColumn 2 lies inside P outside Q lies inside both P and Q lies outside both P and Q lies on Q and inside P

ASSERTION-REASON TYPE QUESTIONS Statement-1: ax + by + c = 0, ab π 0 is a normal to the parabola y2 = 4x at a point A, then equation of the tangent at point A is bx – ay + bc = 0 Statement-2: An equation of a normal to the parabola y2 = 4x at a point (t2, 2t) is y = – tx + 2t + t3 and an equation of the tangent at that point is ty = x + t2 x Statement-1: – a(5x) – a + 5 £ 0, a Œ R, has at least one negative solution then aŒ Statement-2: y = x2 + 5 intersects the line y = a(x + 1) at a point (x1, y1) where 0 < x1 < 1, then a Œ (3, 5) Statement-1: x – 2y + 5a = 0 touches the parabola y2 = 4a (x + 1) where a is a positive con2

Statement-2: m (x + a) – my – a = 0 touches the parabola y2 = 4a (x + a) for all non zero values Statement-1: x – 1)2 + (5y + 2)2 = 2 (4x – 3y + 7) represents a parabola with the line 8x – 6y Statement-2:

Statement-1: x2 + y2 = 1 from which two perpendicular tangents can be drawn to the parabola y2 = 8x

Paragraph for Question Nos. 24 to 26 Let P be the parabola y2 = 4x and L L is (a)

2

(c) 4

(b) 2 (d) 4 2 C described with L as diameter touches the

line (a) x + 1 = 0 (c) x – 1 = 0

(b) x = 0 (d) none of these C (above) on

(a) 2 2

(b) 2 3

(c) 4

(d) 3 2

Paragraph for Question Nos. 27 and 28

y2 = 4ax which bisects the chords with slope m is a 2a (a) y = (b) y = m m 2a a (c) y = (d) y = m m which bisects the chords with slope m is m (b) m (a) 2 m (c) – m (d) 2 Paragraph for Question Nos. 29 and 30 P1 y2 – 4ax = 0, P2 y2 + 4ax = 0 P3 x2 – 4ay = 0, P4 x2 + 4ay P3 and P4 meet the parabolas P1 and P2 at points (other than the origin) which enclose a square of (a) 4a2 (c) 64a2

(b) 16a2 (d) none of these

(a) a2 (c) 4a2

(b) 2a2 (d) 16a2

IIT JEE eBooks: www.crackjee.xyz Parabola 17.31

INTEGER-ANSWER TYPE QUESTIONS PQ be a double ordinate of the parabola y2 = –4x where P R divides PQ R is ky2 = –4x, then k is equal to x2 = 8y to a point of the pakx2 = 2y, then the value of k is

y = kx – 7 intersects the parabola AOB = p , O being y = x – 4x at points A and B 2 the origin, then k is equal to 2

l be the length of the normal chord of the parabp ola y2 = 8x 4 1 of x, then l 4 2 y2 = 4ax and x2 = 4ay is

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS (0, 1) and touching the curve y = x2 at (2, 4) is Ê -16 53 ˆ Ê -16 27 ˆ , ˜ , (a) Á (b) ÁË Ë 5 10 ˜¯ 7 10 ¯ Ê -16 53 ˆ , (c) Á Ë 5 10 ˜¯ y2 = 4ax at right angles? (a) x2 + y2 = a2 (c) y = ax

Ê p ˆ (c) Á - , p˜ Ë 2 ¯

y2 + 4y + 4x + 2 = 0 is (a) x = – 1 (c) x = – 3/2

(b) x = 1 (d) x

(d) none of these circle (x – 3)2 + y2 = 9 and the parabola y2 = 4x above the x (b) y = e–x/2a (d) x2 = 4ay

the parabola y2 = 2px circle and parabola is Êp ˆ Êp ˆ (a) Á , p˜ or Á , - p˜ Ë2 ¯ Ë2 ¯

x y2 – kx + 8 = 0, then one of the values of k is (a) 1/8 (b) 8

(a)

3y = 3x + 1

(b)

3y = –(x + 3)

(c)

3y = x + 3

(d)

3y = –(3x + 1)

the focus to a moving point on the parabola y2 = 4ax Êp ˆ (b) Á , - p˜ Ë2 ¯ Ê p pˆ (d) Á - , - ˜ Ë 2 2¯ x = t2 + t + 1,

(a) x = –a (c) x = 0

(b) x = –a/2 (d) x = a 2 y = 16x is tangent to (x – 6)2 + 2 y = 2, then the possible values of the slope of this chord, are (a) {–1, 1} (b) {–2, 2}

2

y = t – t + 1 represents (a) a pair of straight lines (b) an ellipse (c) a parabola x + y = k is normal to y2 = 12x, then k is (a) 3 (b) 9

point (1, 4) to the parabola y2 = 4x is p p (b) (a) 6 4 (c)

p 3

(d)

p 2

IIT JEE eBooks: www.crackjee.xyz 17.32 Comprehensive Mathematics—JEE Advanced

y = x and the 2 and 2 2 is (a) (b) (c) (d)

(x (x (x (x

+ – – –

y)2 y)2 y)2 y)2

= = = =

x–y x+y 4(x + 8(x +

–2 –2 y – 2) y C1 y2 = 4x, C2 x2 + y2 – 6x

(a) C1 and C2 touch each other only at one point (b) C1 and C2 points (c) C1 and C2 (d) C1 and C2 neither intersect nor touch each x, y) be any point on the parabola y2 = 4x P be the point that divides the line segment from (0, 0) to (x, y P is 2 2 (b) y = 2x (a) x = y (c) y2 = x (d) x2 = 2y x2 + y2 = 2 and 2 the parabola y = 8x touch the circle at the points P, Q and the parabola at the points R, S of the quadrilateral PQRS is (a) 3 (b) 6

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

of radius r having AB as its diameter, then the slope A and B can be 1 1 (b) (a) r r 2 2 (c) (d) r r L be a normal to the parabola y2 = 4x L passes through the point (9, 6), then L is given by (a) y – x + 3 = 0 (b) y + 3x – 33 = 0 (c) y + x – 15 = 0 (d) y – 2x + 12 = 0 P and Q be distinct points on the parabola y = 2x such that a circle with PQ as diameter passes O P lies in the DOPQ is 3 2 , then which of the following is/are the coordinates of P? 2

(4, 2 2 )

(b)

Ê1 1 ˆ (c) Á , Ë 4 2 ˜¯

(d)

(a)

(9, 3 2 ) (1, 2 )

P be the point on the parabola y2 = 4x which is at the shortest distance from the center S of the circle Q be the point on the x2 + y2 – 4x – 16y circle dividing the line segment SP (a) SP = 2 5

(

)

(b) SQ QP = 5 +1 : 2 (c) the x-intercept of the normal to the parabola at P is 6 1 (d) the slope of the tangent to the circle at Q is 2

2

2

ola y = x and y = –(x – 2) is/are (a) y = 4(x – 1) (b) y = 0 (c) y = –4(x – 1) (d) y = –30x – 50 PT and the normal PN to the parabola T and y2 = 4ax at a point P N angle PTN is a parabola whose a/3, 0) x=0 (c) latus rectum is 2a/3 (d) focus is (a A and B be two distinct points on the parabola y2 = 4x

MATRIX-MATCH TYPE QUESTIONS Normal are drawn at points P, Q and R lying on the parabola y2 = 4x Column 1 (a) Area of DPQR

Column 2 (p) 2

(b) Radius of circumcircle of DPQR

(q) 5/2

(c) Centroid of DPQR

(r) (5/2, 0)

(d) Circumcentre of DPQR

(s) 2/3, 0

L y = mx + 3 meets y E(0, 3) and the arc of the parabola y2 = 16x, 0 £ y £ 6 at the point

IIT JEE eBooks: www.crackjee.xyz Parabola 17.33

F(x0, y0 F(x0, y0) m of the intersects the y G(0, y1 line L is chosen such that the area of the triangle EFG List I m=

List II D EFG is

y0 = y1 =

A S cuts the line M at T2 and T3 and AC at T1, then area of DT1T2T3 is

Paragraph for Question Nos. 4 to 6 Consider the circle x2 + y2 + 9 and the parabola y2 = 8x P and Q P and Q intersect the x R and tangents to the parabola at P and Q intersect the x S PQS and PQR is 2

(a) (b) (c) (d)

P 4 3 1 1

Q 1 4 3 3

R 2 1 2 4

S 3 2 4 2

PRS is (a) 5

(b) 3 3

(c) 3 2

(d) 2 3 PQR is

(a) 4

(b) 3

ASSERTION-REASON TYPE QUESTIONS 2 Statement-1: y = - x + x + 1 is sym2 metric with respect to the line x = 1 Statement-2: A parabola is symmetric about its

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 1 to 3 Let ABCD C2 is the circle through vertices A, B, C, D and C1 is the circle touching all the side of the square ABCD L is a line through A P is a point on C1 and Q is another point of C2, PA2 + PB 2 + PC 2 + PD 2 is equal to then QA2 + QB 2 + QC 2 + QD 2

L and the circle C1 nally such that both the circles are on the same side of the line, then the locus of centre of the circle is (a) ellipse (b) parabola (c) hyperbola (d) parts of straight line M is drawn parallel to BD S moves such that its distances from the line BD and

Paragraph for Questions 7 and 8 Let PQ be a focal chord of the parabola y2 = 4ax P and Q meet at a point lying on the line y = 2x + a, a PQ is (a) 7a (b) 5a (c) 2a (d) 3a PQ subtends an angle q y2 = 4ax, then tan q = 2 -2 (a) (b) 7 7 3 3 (c)

2 5 3

(d) -

2 5 3

Paragraph for Question Nos. 9 and 10 Let a, r, s, t P(at2, 2at), Q, R(ar2, 2ar) and S(as2, 2as) be distinct points on the PQ is the focal chord and parabola y2 = 4ax lines QR and PK are parallel, where K is the point (2a r is 2 1 (b) t + 1 (a) t t (c)

1 t

(d)

t2 -1 t

IIT JEE eBooks: www.crackjee.xyz 17.34 Comprehensive Mathematics—JEE Advanced

st = 1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is (a)

(t 2 + 1)2

(c)

a (t 2 + 1)2

(b)

2t 3

(d)

t3

O of the parabola y2 = 4x chords OP and OQ are drawn at right angles, show that for all positions of P, PQ

a (t 2 + 1)2 2t 3 a(t 2 + 2) 2 t3

of PQ

INTEGER-ANSWER TYPE QUESTIONS y2 = 8x D1 be the area of the triangle formed by the end points of its latus rectum and the point P ÊÁ 1 , 2ˆ˜ on the parabola, and Ë2 ¯ D2 be the area of the triangle formed by drawing tangents at P D1 D2 S be the focus to the parabola y2 = 8x and let PQ be the common chord of the circle x2 + y2 – 2x – 4y PQS C be the mirror image of the parabola A and y2 = 4x with respect to the line x + y B are the point of intersection of C with the line y = –5, then the distance between A and B y2 = 4x drawn at the end points of its latus are tangents to the circle (x – 3)2 + (y + 2)2 = r2, then the value of r2

FILL

IN THE

the other two normal are perpendicular to each oth-

BLANKS TYPE QUESTIONS

of the latus rectum of the parabola y2 = 4x is

SUBJECTIVE TYPE QUESTIONS

of slope 2 of the parabola y2 = 4x internally in the A, B, C lie on the parabola y2 = 4ax tangent to the parabola at A, B and C intersect at the points P, Q and R ratio of the areas of the DABC and DPQR A common tangents are drawn to the circle x2 + y2 = a2/2 and the parabola y2 = 4ax mon tangents drawn from A and the chord of contact C1 and C2 be, respectively, the parabolas x2 = y – 1 and y2 = x – 1, let P be any point on C1 and Q P1 and Q1 by any point on C2 of P and Q respectively, with respect to the line y = x P1 lies on C2, Q1 lies on C1 and PQ ≥ min[PP1, QQ1 points P0 and Q0 on the parabolas C1 and C2 respectively such that P0Q0 £ PQ for all pairs of points (P, Q) with P on C1 and Q on C2 m1, m2 and m3 are drawn from a point P y2 = 4x m1m2 = a, results in the locus of P being a 2 A on the parabola y – 2y – 4x + 5 = 0, a B the locus of point C which divides BA

Answers

2

points on the parabola y = 4x pass through the point (h, k h A is a point on the parabola y2 = 4ax A cuts the parabola again at B AB subtends a right AB x2 = 4y c, 0) to the c curve y2 = x One normal is always the x

c for which

LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS

IIT JEE eBooks: www.crackjee.xyz Parabola 17.35

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 11. (a), (b), (c), (d) 13. (a), (c) 15. (a), (c)

12. (c), (d) 14. (b), (c)

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

a

p

q

b

p

c d

17.

18.

26. (d) 30. (a)

27. (c)

31. 3 35. 2 39. 1

32. 2 36. 4 40. 9

33. 3 37. 4

1. (c) 5. (d) 9. (b)

2. (b) 6. (b) 10. (c)

11. (a), (b), (c), (d) 13. (a), (b), (c) 15. (a), (b)

4. (a) 8. (c)

12. (a), (c) 14. (b), (c)

MATRIX-MATCH TYPE QUESTIONS s

a

p

q

r

s

s

p

q

r

b

r

s

s

p

q

r

c

r

s

q

s

q

r

r

s

q

d

p

p

s

p

q

r

p

q

s

r

s

a

p

q

r

s

p

q

r

s

b

p

q

r

s

a

p

q

r

s

c

p

q

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s

b

p

q

r

s

d

p

q

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c

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p

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a

p

q

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s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

16.

17.

18.

ASSERTION-REASON TYPE QUESTIONS 22. (b)

3. (c) 7. (a)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

r

21. (a) 25. (b)

34. 3 38. 7

SINGLE CORRECT ANSWER TYPE QUESTIONS

q

20.

29. (a)

LEVEL 2

p

19.

28. (a)

INTEGER-ANSWER TYPE QUESTIONS

MATRIX-MATCH TYPE QUESTIONS 16.

COMPREHENSION-TYPE QUESTIONS

23. (b)

24. (a)

IIT JEE eBooks: www.crackjee.xyz 17.36 Comprehensive Mathematics—JEE Advanced

ASSERTION-REASON TYPE QUESTIONS

COMPREHENSION-TYPE QUESTIONS

INTEGER-ANSWER TYPE QUESTIONS FILL

IN THE

BLANKS TYPE QUESTIONS

SUBJECTIVE-TYPE QUESTIONS INTEGER-ANSWER TYPE QUESTIONS

± 2

x+y Ê 2 8ˆ ÁË , ˜¯ 9 9

y2 = 2x

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS

15a 2 4 Ê 1 5ˆ Ê 5 1ˆ P0 Á , ˜ , Q0 Á , ˜ Ë 2 4¯ Ë 4 2¯

3 4

x + 1) (y + 1)2 + 4 = 0

Hints and Solutions LEVEL 1

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

x+y x+ y

=

2 fi (x – y)2 = 8(x + y – 2)

MATRIX-MATCH TYPE QUESTIONS p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

ASSERTION-REASON TYPE QUESTIONS COMPREHENSION-TYPE QUESTIONS

( x - 2) 2 + ( y - 2) 2

parabola y2 = 4ax, to the points of intersection of the normal and the curve is (mx - y ) y2 = 4ax 2am + am3 or 4amx2 – 4axy – (2am + am3) y2 = 0 4am – 2am – am3 = 0 fi m = ± 2 y2 = 4x is y = mx + 1/m, tangent to x2 = 4y with slope m is x = (1/m) y + m or y = mx – m2 1 So = –m2 fi m3 = –1 fi m = –1 and the equation m of common tangent is x + y + 1 = 0 a where a x its equation is x = –3/2, the coordinates of the end points of the LR are (–3/2, 1 + 1) and (–3/2, 1 – 1)

IIT JEE eBooks: www.crackjee.xyz Parabola 17.37

= Length of the latus rectum = LL¢

2x + y y2 = 9x at (9t2/4, 9t/2) is ty = x + 9t2/4 which passes through (4, 10) if 40t = 16 + 9t2 fi 9t2 – 40t + 16 = 0 fit But the slope 1/t of the tangent is > 2 fi t < 2 so t = m2, –18m) (For a = 9) to the parabola y2 = 4 ¥ 9x is y = mx – 2 ¥ 9m – 9m3 where – 18m = 3 (9m2) (given) fi m = –2/3 or 0 For m = –2/3, the normal is 2x + 3y – 44 = 0 ab2, a2b) 2 2 slopes of tangents to two parabolas at (ab , a b) are m1 =

a3

=

2

a 2a and m2 = 2b b

2a b So if q is the required angle, then tan q =

3ab m1 - m2 = 2( a 2 + b 2 ) 1 + m1m2

P be (at2, 2at) 2at 2 Slope of OP = 2 = t at t 2 An equation of PQ is t y – 2at = - ( x - at 2 ) 2 fi Slope of PQ = -

y

P(at2, 2at) L

(x + 2)2 = –2(y – 2) fi X2 = – 2Y where x + 2 = X, y – 2 = Y, equation of the latus rectum is 1 1 Y = (-2) = 4 2 3 1 = 2 2 x – 11) cos q – cos 3q = (x – 11) cos q – (4 cos3 q – 3 cosq) = x cos q – 8 cos q – 4 cos3 q = (cos q)x – 2(4) cos q – 4 cos3 q = mx – 2am – am3 for m = cos q, a = 4 So represents a normal to y2 = 4ax for all value of q y2 = 4x is y = mx – 2m – m3 b) if 3 b = 5m – 2m – m fi m3 – 3m + b = 0 (1) 3 Let f(m) = m – 3m + b; m Œ R f¢(m) = 3m2 – 3 = 3 (m + 1) (m – 1) Now, f¢(m) = 0 fi m = –1, 1 Also, f¢(m) > 0 if m < –1 < 0 if –1 < m < 1 > 0 if m > 1 f(m) = f(–1) = –1 + 3 + b =b+2 and Min f(m) = f(1) = 1 = b – 2 fi y = 2-

f m)

O

M

Q

(-1 b + 2)

x



b

-1

m

(1 b - 2)

Fig. 17.18

x Q (at2 + 4a foot of perpendicular from P to the x ordinates of M are (at2 a PQ on the x 2 = MQ = |at + 4a – at2| = 4a

b 1

M be the -

Fig. 17.19

Equation (1) will have three real roots if b + 2 > 0 and b b 2, then f(m

fi 2

y = mx + c is a tangent to the parabola y = 4x if 1 c= and will be a tangent to the curve x2 + 4y2 = m 8 if the roots of the equation x2 + 4(mx + c)2 = 8 are fi (8mc)2 = 4(1 + 4m2) (4c2 – 8) fi c2 = 8m2 1 1 fi = 8m2 So c = m m2 fi 8m + 2m – 1 = 0 fi (4m2 – 1) (2m2 + 1) = 0 4

2

5t = –3 – 3 + t2 fi t2 – 5t – 6 = 0 fi t = –1, 6 So the required points are (4, –2), (39, 12) P(at2, 2at) are coordinates of one end point of the focal chord, then the other end point has coordinates Ê a 2a ˆ PQ is QÁ 2 ,- ˜ Ët t ¯ a - at

2

=

y - 2at 0 - 2at

or 2tx – (t2 – 1)y – 2at = 0 2a | t | 4t + (t 2 - 1)2 2

t +1

=b 2

2

a 2a Also, ÊÁ at 2 - ˆ˜ + ÊÁ 2at + ˆ˜ = a2c2 2 Ë Ë t ¯ t ¯ 2

2

1 1 fi ÊÁ t 2 - ˆ˜ + 4 ÊÁ t + ˆ˜ = c2 2¯ Ë Ë t¯ t 2 2 ˘ Ê 1ˆ ÈÊ 1ˆ fi Á t + ˜ ÍÁ t - ˜ + 4˙ = c2 Ë t ¯ ÍË t ¯ ˙˚ Î 2

1 fi ÊÁ t + ˆ˜ = c Ë t¯ 2

1 fim= ± 2 and the required tangents are x – 2y – 4 = 0 and x + 2y x – a)2 + y2 = (2a)2 Which meets the parabola y2 = 4ax at points for which (x – a)2 + 4ax = 4a2 fi (x + a)2 = (2a)2 fi x + a = 2a as x > 0 fi x = a and the points of intersection are (a, 2a) and (a, –2a y = x + a and –y = x + a y2 = 4(x – 3) 2 2 at (t + 3, 2t) is ty = x – 3 + t

x - at 2

|t | 2

= 2ab

1 As ÊÁ t + ˆ˜ ≥ 4, c ≥ 4 Ë t¯ Also

1 b2

=

(t 2 + 1)2 t

2

2

Ê 1ˆ = Át + ˜ = c Ë t¯

fib c=1 2

t2 +1 1 1 Also, = = |t |+ ≥ 2 | t | |t | b fi 0 0 fi | k |
5

(d) A tangent to y2 = 4x is y2 = 4x is

1 y = mx + m a + 3, a) if 1 a = m(a + 3) + m fi m2 (a + 3) – ma + 1 = 0 As m is real, a2 – 4(a + 3) > 0 fi (a – 2)2 > 42 fi a – 2 < –2 or a – 2 > 2 fi a < 0 or a > 4 fi a Œ(–•, 0) » (4, •) (b) An equation of tangent to y2 = 4ax is a y = mx + (1) m x2 + y2 – 2ax – 4ay – a2 =0 (2) 2a = ma +

1 + m2 2 m 1 + m2

= ±4 -

a m

¤ a π 0 and m = 1 (c) An equation of tangent with positive slope to y2 = 4x is 1 (1) y = mx + m where m > 0 An equation of tangent with slope m to the circle x2 + y2 – 2x + 4y – 11 = 0

y = mx +

1 m

(1)

Putting y = mx +

1 1 in x 2 + y 2 = 1, we get m 2

2

1ˆ Ê 2 x 2 + Á mx + ˜ = 2 Ë m¯ fi (2 + m2)x2 + 2x +

1 m2

–2=0

For (1) to be a tangent to x 2 +

(2)

1 2 y = 1, 2

Ê 1 ˆ ¤ 1 – (2 + m2) Á 2 - 2˜ = 0 Ëm ¯ ¤m= ±

p 2 - 1 = ± tan ÊÁ ˆ˜ Ë 8¯

Length of perpendicular from (–1, 3) to 3x – 4y + 5 | 3(-1) - 4(3) + 5 | =2 = 0 is 32 + 42 \ length of latus rectum of the parabola = 2(2) = 4 x4 –x2 + 2 = 0 2

7 1ˆ 1 Ê fi Á x2 - ˜ = - 2 = Ë 4 2¯ 4 which gives no real value of x and the statement is y = mx + c touches y2 = 16x then c 4 4 = and y = mx + touches the circle x2 + y2 = 8 m m if

IIT JEE eBooks: www.crackjee.xyz Parabola 17.41

x2 + y2 – 4x + 3 = 0 if

4 = ± 8(m 2 + 1) m

1 2m

2m - 0 +

or m4 + m2 – 2 = 0 fi m2 = 1 fi m = ±1 Statement-1 is true but does not follow from State-

1 + m2

=

22 - 3

2

y2 = 4ax is given by y = mx – 2am – am2

(1) a, 0), if

-

We can write the equation (5x – 15)2 + (5y + 10)2 = (3x – 4y + 2)2 | 3x - 4 y + 2 | 5 which represents a parabola with focus (3, –2) and x – 4y x – 3) + 3(y + 2) = 0 or 4x + 3y – 6 = 0 ( x - 3) 2 + ( y + 2) 2 =

2

t > 0, t +

1 1ˆ Ê = Á t - ˜ +2≥2 Ë t t¯

t < 0, t +

1 1 = -t - ≥ 2 t t

fi 4m2 + 2 +

1 4m 2

1

fi 3m2 +

0 = ma – 2am – am3 fi 0 = –am (1 + m2) = 0 fi m = 0 y

as

fi ÊÁ 2m + 1 ˆ˜ = 1 + m2 Ë 2m ¯

4m 2

= 1 + m2

+1=0

fi 12m4 + 4m2 + 1 = 0 Which has no solution Hence Statetement-1 is also true but does not follow from Statement-2 A and C, we put y2 = 4ax in 2 y = –8(x – a), so that 4x = –8 (x – a) 2a fix= 3 2a 2 4 Ê 2a ˆ when x = ,y = Á ˜ Ë 3¯ 3 fi y = ±k, where k =

8a 3 A (2a/3, k)

For statement - 1, Length of the focal chord of Ê 1ˆ y2 = 4ax is a Á t + ˜ Ë t¯

2

O

D

B (a, 0)

x

2

Ê 1ˆ So 4 Á t + ˜ = 9 Ë t¯ fi t+

C (2a/3, -k) Fig. 17.20

3 1 = b

X

18.3

DEFINITION 2

An ellipse is the locus of a point which moves in a plane in Fig. 18.1

foci (ii) Mid-point of the line joining the foci is called the centre (iii) The line segment through the foci of the ellipse is called the major axis (iv) The line segment through the centre and perpendicminor axis of the

e, where 0 < e focus directrix (iii) The constant ratio e is called the eccentricity of the Standard Equation of an Ellipse referred to its principle

ver-

x2

tices (vi) A line segment through a focus perpendicular to the Latus rectum principal axis

a2

18.2

STANDARD EQUATION OF THE ELLIPSE

Let F1 (– c, 0) and F2 (c O (0, 0), the mid-point of F1 F2 Let P (x, y PF1 + PF2 = 2a (constant) fi

( x + c)2 + y 2 + ( x - c) 2 + y 2 = 2a. (2a > 2c)

b2 = a2 (1 – e2)

+

y2 b2

= 1,

Y B (0, - b) L M¢ S¢(-ae, 0) A¢ (-a,0)

x = - a/e

X

A (a,0) L¢

Ellipse (e , - b) Fig. 18.2

M

S(ae, 0)

C(0, 0)

B¢(0, - b)

P

x = a/e

IIT JEE eBooks: www.crackjee.xyz 18.2 Comprehensive Mathematics—JEE Advanced x=

Let P (x, y S = (ae, 0) be a focus and x =

a e

fix=

ae

11 = ± 11 6



and length of each latus rectum =

Êa ˆ (x – ae)2 + y2 = e2 Á - x˜ Ëe ¯ fi

x2 a2

+

y2 a 2 (1 - e2 )

2

=1

=

2b 2 2 ¥ 25 = a 6

25 units. 3

Illustration 2

Since e < 1, 1 – e2 > 0 so let a2 (1 – e2) = b2 x2 a2

+

y2 b2

=1

9x2 + 16y2 – 18x + 32y – 119 = 0 Solution: 9(x – 1)2 + 16 (y + 1)2 = 144 fi

(i) A¢ A is the major axis of the ellipse along x length 2a; A(a, 0), A¢ (–a, 0) are the vertices of the ellipse; (ii) BB¢ y length 2b (iii) O (iv) e =

( x - 1) 2 ( y + 1)2 + =1 16 9

Which can be written as

where X = x – 1, Y = y + 1, a = 4, b centre is X = 0, Y ae, Y = 0 fi x – 1 =

foci are X =

a 2 - b2

a2 From symmetry we observe that if

fix=1

-a a and x = are the two directrices of the e e

(vii) x = ae (viii) Latus rectum x = ae meets the ellipse at point

eccentricity =

1-

7 b2 = 4 a2

fix–1=

4¥4 fix=1 7

Find the length and equations of the latera recta of the x2 y 2 =1 ellipse + 36 25 x2 y 2 Solution: Let the equation of the ellipse be 2 + 2 = 1, a b 25 11 = where a2 = 36, b2 = 25 fi e2 = 1 – 36 36

a e

16 7

Illustration 3

Ê b2 ˆ 2b 2 Á ae, ± a ˜ , so length of each latus rectum is a Ë ¯ Illustration 1

b2 = ± 7 a2

a=8 b

X=

(vi) x =

1-

4

7 ; y = –1

-a S ¢ = (– ae, 0) be taken as a focus and x = is taken e (v) S ¢ (– ae, 0) and S (ae, 0) are the two foci of the el-

X2 Y2 + =1 a 2 b2

1 ,a 2 x Solution: (x – 2)2 + (y – 3)2 =

1 (7 – x)2 4

fi 3x2 + 4y2 – 2x – 24y + 3 = 0 which can be written as 2

1ˆ Ê 1 100 3 ÁË x - ˜¯ + 4 (y – 3)2 = 3 ¥ +4¥9–3= 3 9 3

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.3 Solution: Let the coordinates of P be (7 cos q, 5 sin q) P is x y (1) cos q + sin q = 1 7 5

2

1ˆ Ê ÁË x - ˜¯ ( y - 3) 2 3 or =1 + 100 100 9 12 100 20 = 9 3

x 3 y =1 + 14 10 Comparing (1) and (2) we get

100 20 10 = = 12 2 3 3

dinates of any point on the ellipse are x = a cos q, y = b sin q q q is called q ” is xx ¢ yy ¢ x y cos q + sin q = 1 and at (x¢, y¢) is 2 + 2 a b a b Illustration 4 Find an equation of the tangent to the ellipse

x2 y 2 = + 81 49

1 at the point P whose eccentric angle is p coordinates of P. p pˆ Ê Solution: Coordinates of P are Á 9 cos , 7 sin ˜ Ë 6 6¯ Ê 9 3 7ˆ Á 2 , 2˜ Ë ¯

=

An equation of the tangent at P to the ellipse is x p y p cos + sin = 1 9 6 7 6 fi

y 3x = 1 fi 7 3x + 9y + 9¥2 7¥2

x ¢/ a

=

fi 14x – 10 3 y =

3

y = mx + c is a tangent to the ellipse is c2 = a2m2 + b2 tangent to the ellipse (not parallel to y written as y = mx a 2 m2 + b2 Point of intersection of the tangents at the point ‘a’ and ‘b’ is Ê Êa + bˆ ˆ Êa + bˆ Á cos ÁË 2 ˜¯ sin ÁË 2 ˜¯ ˜ ,b Áa ˜ Á cos Ê a - b ˆ cos Ê a - b ˆ ˜ ˜ ÁË ˜ ÁË ÁË 2 ¯ 2 ¯ ˜¯ Illustration 6 If y = mx + 5 is a tangent to the ellipse 4x2 + 25y2 = 100, m2

Then (5)2 = 25 m2 + 4 fi 25m2 = 21 fi 100 m2

2

chord of contact of the point (x¢, y¢) joining the points of contact of the tangents drawn from (x¢, y¢) to the ellipse is

Illustration 5 5 3x + 7y = 70 is a tangent to the ellipse the point P the normal at P

7x 5y =1 cos q sin q

If y = mx

y - y¢ y ¢/ b

P is

x2 y 2 =1 + 25 4

ax by = a2 – b2 cos q sin q and at (x¢, y¢) is 2

Ê 7 3 5ˆ So the coordinates of P are Á 2 , 2 ˜ Ë ¯

Solution:

P(q) is

x - x¢

3 1 , sin q = fi q = 30° 2 2

cos q =

SOME PROPERTIES AND STANDARD x2 y2 RESULTS FOR THE ELLIPSE a 2 + b 2 = 1

(2)

P

x2 y 2 = 1 at + 49 25

xx ¢ a

2

+

yy ¢ b2

=1

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Note x¢, y¢) to the P (x¢, y¢) lies on the ellipse the two points of contact coincide with P and the tangent at P and the chord of contact of P

then ACQ = q is called the eccentric angle of the point P and the coordinate of P are (a cos q, b sin q PQ = (a – b) sin q Q P C q (0, 0)

point is (x¢, y¢) is T = S¢, where T

xx ¢

+

yy ¢

– 1 and S¢

x¢2

y ¢2

b2 P(x¢, y¢) outside the ellipse to the ellipse is SS¢ = T2, x2 y 2 + S a 2 b2 a2

b2

a2

+

Illustration 7 Find the equation of the chord of contact of the point (3, 1) to the ellipse x2 + 9y2 = 9. Also find the mid-point of this chord of contact. Solution: Equation of the ellipse is x2 y 2 =1 + 9 1 Equation of the chord of contact of (3, 1) is x(3) y (1) = 1 fi x + 3y = 3 + 9 1 If (h, k hx ky h2 k 2 – 1 –1= + + 9 1 9 1 From (1) and (2) we get

(1)

Fig. 18.3

Illustration 8 Find the equation of the ellipse whose auxiliary circle is the x2 y 2 director circle of the ellipse = 1 and the length of + 36 13 a latus rectum is 2 units. Solution: Equation of the director circle of the ellipse is x2 + y2 = 36 + 13 = 49. x2 y 2 + = 1, a 2 b2 x2 + y2 = a2 = 49 fi a2 = 49 fi a Also length of a latus rectum =

2b 2 =2 a

fi b2 = a = 7 (2)

h 9k h 2 + 9k 2 = = 1 3 3 fih=

A (a, 0)

3 1 ,k= 2 2

Ê 3 1ˆ So the mid point of the chord of contact is Á , ˜ Ë 2 2¯

Director circle of the ellipse is the locus of the point of intersection of the tangents to the ellipse x2 + y2 = a2 + b2 So a pair of tangent draw from any point on the Auxiliary circle of the ellipse is the circle on the x2 + y2 2 =a P is a point on the ellipse and Q is a point on Q lies on the ordinate produced of the point P C is the centre of the ellipse and CA

x2 y 2 + 49 7

A diameter of an ellipse is the locus of the mid points of a system of parallel chords of the ellipse and its y= -

b2 a2m

x,

where m is the slope of the parallel chords of the ellipse are said to be conjugate when each bisects that y = mx and y = m¢x of the ellipse are conjugate if mm¢ b2 =– 2 a Illustration 9 x2 y 2 =1 + 36 16 bisecting the chords parallel to the line y = x – 1. Also find Find the equation of the diameters of the ellipse

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.5 the equation of the conjugate diameter bisecting the chords parallel of this diameter. Solution: Slope of the given line is m = 1. 16 x fi 4x + 9y 1 is y = 36 diameter is m¢ such that –

then (ii) is the chord of contact of (h, k) with respect to hx ky =1 + 6 3

(iv)

Since (ii) and (iv) represents the same line

16 16 m¢ = – fi m¢ = 1 and the 36 36

h/ 6 k /3 = =1 (cos q ) / 2 sinq fi

y=x

h = 3 cos q, k = 3 sin q

and the locus of (h, k) is x2 + y2 = 9 The normal at an end of a latus rectum Example 2 2 2 of the ellipse x /a + y2/b2 = 1 passes through an end of W, W¢ are the feet of the perpendiculars from the foci S and S¢ respectively on the tangent at any point P of an ellipse with centre C, then CW is parallel to S¢ P and CW¢ is parallel to SP P of the ellipse x2/a2 + y2/ 2 G and g reb spectively and CF is perpendicular from the centre C on the normal then

(a) e4 + e2 = 1 (c) e2 + e = 1 Ans. (a)

x - ae ae /a 2

SINGLE CORRECT ANSWER TYPE QUESTIONS

2

–a =

The locus of the points of intersection of

or

(

-b 2 1 + 1 - e 2 1 - e2

) or b

1 - e2

2

a2

=

1 + 1 - e2

1 - e2

If an ellipse sides between two perpenExample 3 dicular lines, then the locus of the centre is

We can write x2 + 4y2 = 4 as 2

x y + =1 4 1

b 1 - e 2 /b 2

1 - e2 + 1 – e2 = 1 or e4 + e2 = 1

or

(b) x2 + y2 = 6 (d) none of these

Ans. (c) 2

y - b 1 - e2

(1 – e2) È1 + 1 - e2 ˘ = ÎÍ ˚˙

x2 + 2y2 = 6 which touch the ellipse x2 + 4y2 = 4 is

Solution:

=

It will pass through the end (0, – b) if

SOLVED EXAMPLES

(a) x2 + y2 = 4 (c) x2 + y2 = 9

)

(

Let an end of a latus rectum be ae, b 1 - e2 ,

Solution:

PF PG = b2 and PF Pg = a2

Example 1

(b) e3 + e2 = 1 (d) e3 + e = 1

(i)

x cos q + y sin q = 1 (ii) 2 x2 + 2y2 = 6 can be written as x2 y 2 + =1 (iii) 6 3 Suppose (ii) meets the ellipse (iii) at P and Q and the tangents at P and Q to the ellipse (iii) intersect at (h, k),

(a) a parabola (c) a circle

(b) an ellipse (d) none of these

Ans. (c) Solution: Let 2a, 2b be the length of the major and minor If the ellipse slides between two perpendicular lines, the point of intersection P of these lines being the point of intersection of perpendicular tangents lies on

B C 90° P

A Fig. 18.4

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This means that the centre C of the ellipse is always at a constant distance of C

a +b 2

2

from P

So that foci of the ellipse are (± ae F1 and F2 P is any point on the

If the tangent at a point (a cosq, b sinq) Example 4 2 2 on the ellipse x /a + y2/b2 in two points, the chord joining them subtends a right angle at the centre; then the eccentricity of the ellipse is given by (a) (1 + cos2 q)–1/2 (c) (1 + sin2 q)–1/2

(b) 1 + sin2 q (d) 1 + cos2 q

ellipse PF1 + PF2 = 2a = 2 ¥ 5 = 10 x2 y 2 + + y2 = 4, parabola y2 = x – 4 and the ellipse 16 9 = 1 is (a) x = –4 (c) y = –3

Ans. (c) a cos q, b sin q) to

Solution: the ellipse x2/a2 + y2/b2 = 1 is fi

Ans. (b)

(I)

y Èx ˘ x2 + y2 = a 2 Í cos q + sin q ˙ a b Î ˚ Since these lines are at right angles



1-

2

2

2

O

4

x

6

Fig. 18.5

y =0

Ê cos a2 Á 2

x2 a2

Ê a sin 2 q Á1 - 2 ˜ + 1 = 0 Ë b ¯

and



sin2 q (b2 – a2) + b2 = 0



sin q [a (1 – e ) – a ] + a (1 – e ) = 0



(1 + sin2 q) a2e2 = a2 fi e = (1 + sin2 q)– 1/2

2

2

2

2

2

If F1 = (3, 0), F2 = (–3, 0) and P is Example 5 any point on the curve 16x2 + 25y2 = 400, then PF1 + PF2 (a) 8 (b) 6 (c) 10 (d) 12 Ans. (c) Solution:

AB is a variable chord of the ellipse

Example 7

qˆ 2 Ê sin q ˆ ˜ +1- a Á 2 ˜ = 0 Ë a ¯ Ë b ¯ 2



2

-4

2

2

2

2

y2 = x - 4

y

x + y = a to the origin, which is the centre of the circle, is 2



(b) x = 4 (d) y = 3

Solution: tangent to the three curves is x

x y cos q + sin q = 1 a b

x

A common tangent to the circle (x – 6)2

Example 6

+

y2 b2 1

OA

2

(a) (c)

AB subtends a right angle at the origin, +

1 OB 1

=

2

3 a2

then eccentricity of the ellipse is (b)

3 1 2

(d)

1 2 1 3

Ans. (b) Solution: Suppose OB makes an angle q with the positive direction of the x OA makes an angle of p q+ with the positive direction of the x 2 y

2

x y + =1 25 16 a2 = 25, b2 = 16 But

b2 = a2(1 – e2)



16 = 25 (1 – e2) fi e =

B

A q O

3 5

Fig. 18.6

x

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.7

Let OA = r1 and OB = r2 B lies on the ellipse r2sinq x2

+

a2 r22 cos 2 q 1



OB Similarly,

=

2



b œ (2, 3)

a

2

+

sin q 2

(i)

b2

r12

a2 sin 2 q

+

OA2

+

b2

Solution: (ii)

OB 2 a2

= =

1 a2

+

x2

b2

a2

sects the ellipse

x

2 2

+

2 3

(b)

1

(d)

2

1 3

If b < c < a, then the circle x2 + y2 = c2 meets x2 a2

+

y2 b2

b2 1 2 a + b 2 = a or b 3 1 2 a + b 2 > b , therefore, 3

But y2 a2

a2 + b 2 = a2 fi 3b2 = 2a2 3 1 3a2 (1 – e2) = 2a2 fi e = 3

= 1 interfi

+ y2

(a) (–1, 1) (c) (2, 3)

Example 10

(b) (1, 4) (d) (–1, 3)

x2 a

2

+

y2 b

2

Ans. (c) (a)

As the ellipse intersect in four distinct points, y

(c) a

-1

1 -1 -a Fig. 18.7

If y = mx + c is a normal to the ellipse

= 1, then c2 (a 2 - b 2 )2 a 2 m2 (a 2 - b 2 )2 a 2 + b2 m2

(b) (d)

(a 2 - b 2 )2 m 2 a 2 m2 + b2 (a 2 - b 2 )2 m 2 a 2 + b2 m2

Ans. (d)

1 -2

= 1, (a > b

y2

a 2 (1 - e2 )

4 a = b2 – 5b + 7, then b cannot lie in the set

Solution: a

b2

Therefore,

2 - e2

Suppose the ellipse x 2 +

Example 8

y2

3

the ellipse

1

3 – 3e2 = 2 – e2 fi 1 = 2e2 1 e= 2



a2

+

If c = a or b, then the circle x2 + y2 = c2 meets the ellipse

3



Ans. (a)

b2

1



pˆ Ê sin 2 Á q + ˜ Ë 2¯

cos 2 q

a2 From (i) and (ii) 1

+

x2

1

(c)

pˆ Ê cos 2 Á q + ˜ Ë 2¯

The circle 3x2 + 3y2 = a2 + 3b2 meets

The eccentricity of the ellipse is

1

=

OA2

=

(b – 2) (b – 3) > 0

the ellipse

cos q

=



=1

2

r22

b2 – 5b + 7 > 1

Example 9

b2

1

Thus,

(a)

1

=

=1

b2

r22 sin 2 q

+

a2

y2

OB are (r2cosq,

2

x

Solution: x

2

+

y

Suppose y = mx + c is a normal to the ellipse

2

= 1 at point (a cosq, b sinq a b2 normal at (a cosq, b sinq) is 2

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ax secq – by cosecq = (a2 – b2)

(i)

x2 mx – y = –c

(ii)

Comparing (i) and (ii) we get m -1 -c = = 2 a secq -bcosecq a - b2 fi

- ac

cosq =

m( a 2 - b 2 )

, sinq =

Now, 1 = cos2q + sin2q = fi

(a - b ) m 2

c2 =

2 2

Two tangents are drawn to the ellipse

Example 12 a

2

+

y2 b2

= 1 from a point P (h, k

of P is

x2

(a)

2

+

y2 2

b a (c) xy = ab

-bc a 2 - b2

c 2 (a 2 + b 2 m 2 ) m 2 (a 2 - b 2 )2

(b) x2 + y2 = a2 + b2

=1

(d) x2 – y2 = a2 – b2

Ans. (d) Solution: P (h, k) is Ê hx ky ˆ ÁË 2 + 2 - 1˜¯ a b

2

a 2 + b2 m2

Ê x2 y 2 ˆ Ê h2 k 2 ˆ = Á 2 + 2 - 1˜ Á 2 + 2 - 1˜ (i) b b Ëa ¯Ëa ¯

2

A1(x1, 0), a + b moves

Example 11

A2(x2, 0), B1(0, y1) and B2(0, y2 Putting y = 0 in (i), we get x1, x2 are roots of Ê hx ˆ ÁË 2 - 1˜¯ a

the line into two parts of length a and b is (a) a parabola (c) an ellipse

(b) a circle (d) none of these

or

Ans. (c) Solution: Let coordinates of P that divide AB in the ratio a : b be (h, k lb ma Then, h = ,k= a+b a+b fi

l=

h( a + b) k ( a + b) ,m= b a



1-

k2 b2

Ê h2 k 2 ˆ 2 Á a 2 + b2 ˜ b Ë ¯

h2 a2 Suppose A1, A2, B1, B2 lie on the circle x2 + y2 + 2gx + 2fy + c

P

Then x1, x2 are roots of x2 + 2gx + c = 0

a A(l, 0)

x



Ê h2 k 2 ˆ Now, (a + b)2 = l2 + m2 = Á 2 + 2 ˜ (a + b) 2 b ¯ Ëa

y1 y2 = c

Êh k2 ˆ 2 + Á a 2 b2 ˜ a Ë ¯ 2

Thus

k2

=1 a 2 b2 \ locus of point P

x1 x2 = c

Similarly,

Fig. 18.8

+

x 1x 2 =

Ê h2 k 2 ˆ 2 Á a 2 + b2 ˜ a Ë ¯

1-

B(0, m) b

h2

Ê x2 ˆ Ê h2 k 2 ˆ = Á 2 - 1˜ Á 2 + 2 - 1˜ b Ëa ¯Ëa ¯

x 2 Ê k 2 ˆ 2hx h 2 k 2 + + =0 Á1 - ˜ a 2 Ë b2 ¯ a 2 a 2 b2

Similarly, y1y2 =

y



2

k2 1- 2 b

=

Ê h2 k 2 ˆ 2 Á a 2 + b2 ˜ b Ë ¯ 1-

h2 b2



Ê k2 ˆ Ê h2 ˆ b 2 Á1 - 2 ˜ = a 2 Á1 - 2 ˜ Ë b ¯ Ë a ¯

fi fi

h2 – k2 = a2 – b2 Locus of P is x2 – y2 = a2 – b2

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.9

Example 13 From a point P two perpendicular tangents PQ and PR are drawn to the ellipse x2 + 4y2 Locus of circumcentre of D PQR is (a) x2 + y2 =

5 2 (x + 4 y2 ) 4

OA and OB is

5 2 (b) x + y = ( x + 4 y 2 )2 16 (c) x2 + 4y2 = 16 (d) x2 + 4y2 = (x2 + y2 – 4)2 2

(c) a circle with centre at (a, 0) (d) none of these Ans. (a) Solution: AB be x cos a + y sin a = p



2

(

5 cos q , 5 sin q

P

)

x y 5 cos q + 5 sin q = 1 4 1

(i)

b2

ˆ Êx y = Á cos a + sin a ˜ ¯ Ëp p

2

1 a

2

+

1 b

OL2 =

2

=

1 p

2

=

1 OL2

a 2b2 a 2 + b2

fiL Thus, locus of L

Let M(h, k) be the mid point of QR (T = S1

QR is

hx ky h2 k 2 + -1 + -1 = 4 1 4 1 hx h2 + k2 + ky = 4 4 Comparing (i) and (ii), we get or



fi fi

of chord of contact QR of point P is

5 cosq / 4 = h 4

y2

Ê 1 1 1 2 ˆ Ê 1 2 ˆ Á 2 - 2 cos a ˜ + Á 2 - 2 sin a ˜ = 0 Ëa p ¯ Ëb p ¯

Solution:

on the director circle is P

a2

+

As OA and OB are perpendicular to each other

Ans. (b) x2 y 2 + 4 1 2 2 2 2 tor circle is x + y = 4 + 1 or x + y

x2

Example 15 A of the ellipse x2 + 9y2 the point M at A, M and the origin O is

(ii) (a)

31 10

(b)

B

29 10

(c)

21 10

(d)

Ans. (d)

5 sinq 1 = 2 k h + k2 4

Solution: x2 y 2 + =1 9 1

Ê h2 ˆ h Ê h2 k 2ˆ , Á + k 2 ˜ sinq = Á 4 + k ˜ cosq = 5 Ë 4 5 Ë ¯ ¯

x2 + y2 = 9 Ê h2 2ˆ Á 4 +k ˜ Ë ¯

2

(i)

AB is

h2 k 2 + = 5 5

y x2 + y 2 = 9

M

x2 + y2 = Example 14

B (0, 1)

5 2 ( x + 4 y 2 )2 16 Let AB be a chord of the ellipse

27 10

x2 a2

+

y2

N

O

b2

If L is the foot of perpendicular from the origin to AB, then locus of L is (a) a circle with centre at the origin



y¢ Fig. 18.9

a+b

A (3, 0)

x

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±

x y =1 + 3 1 fi

PQRS = 4[Area of DOPQ]

x = 3 (1 – y)

Putting this in (i), we get

Ê 1ˆ Ê 9ˆ = 4 Á ˜ Á ˜ (3) Ë 2¯ Ë 2¯

9(1 – y)2 + y2 = 9 fi

10y2 – 18y = 0 fi y = 0,

which are tangents to the ellipse

1 (OA) (MN) = 1 (3) ÊÁ 9 ˆ˜ = 27 2 2 Ë 5 ¯ 10

x2 y 2 + =1 2 4

(a)

Example 16

ellipse

(c)

3x

2

x y + 9 5

(a)

1

(c)

the tangents at the end points of the latus recta of the 2

Locus of the mid points of the segments

Example 17

9 5

Thus y-coordinate of M is Area of DOAM =

9 5

2

+

1 4y

27 4

(b) 9

1 2x

2

+

1 4 y2

=1

27 2

(d) 27

Solution: x2/2 + y2 = 1 y B

Solution:

a = 3, b = 2

P (÷2 cos q sin q)

5 , so that

2

2

x

A

O

b = a (1 – e ) fi 5 = 9(1 – e ) fi

(d)

x2 y 2 + =1 4 2

Ans. (d)

Ans. (d)

2

(b)

=1

2

x2 y 2 + = 1 and which 2 1

9e2 = 4 fi e2 = 4/9 fi e = 2/3 Q

B

Fig. 18.11

A

A point on the ellipse is P R

O C

P S

D

x 2

25 Ê 4ˆ When x = 2, y2 = 5 Á1 - ˜ = Ë 9¯ 9 fi y = ±5/3 \ Coordinates of A Similarly, coordinates of B, C, D are respectively (–2, A, B, C, D are given 2 1 x y Ê 5ˆ ± (2) ± Á ˜ = 1 or ± x ± y = 1 Ë ¯ 9 5 3 9 3

2 cos q ,sin q

A Mid point of AB is

(

)

2 sec q , 0 and B(0, cosec q

1 ˆ Ê 1 (h, k) = Á sec q , cosecq ˜ ¯ Ë 2 2 fi

2 h = sec q, 2k = cosec q 1

+

1

2

+

1

2

=1 2h 4k 2 Thus, locus of point (h, k) is 1 2x

)

cos q + y sin q = 1

Fig. 18.10

Thus coordinates of foci are (±ae, 0) = (±2, 0)

(

4 y2

=1

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.11

Example 18 Consider the ellipse x2 +2y2 = 2. Let L be the end of the latus rectum in the first quadrant. The tangent at L to the ellipse meets the right side directrix at T. The normal at L meets the major axis at N. Area of the triangle LNT in sq. units is 3 4

(a) Ans

(b)

3 2 4

(c)

3 4 2 2

3 2

2

x y + = 1 if e is 2 1 the eccentricity of the ellipse, then coordinates of L are Ê b2 ˆ ae , where a2 = 2, b2 = 1, a2e2 = a2 – b2 = 1 ÁË a ˜¯

Solution

+ 4y2 = 16 meets the x Q M is the mid point of the line segment PQ, then the locus of M intersects the latus rectums of the given ellipse at points

4

(d)

1 ˆ Thus coordinates of L are ÊÁ1, ˜ Ë 2¯ L to the ellipse is

The normal at a point P on the ellipse x2

Example 19

Ê 3 5 2ˆ (a) Á ± ,± ˜ 7¯ Ë 2

Ê 3 5 19 ˆ (b) Á ± ,± 4 ˜¯ Ë 2

1ˆ Ê (c) Á ±2 3 , ± ˜ Ë 7¯

Ê 4 3ˆ (d) Á ±2 3, ± 7 ˜¯ Ë

Ans. (c) Solution: x2 y 2 + =1 16 4 Let coordinates of P be (4 cos q, 2 sin q)

q, 2 sin q) to the ellipse (1) is

x y + =1 2 2

(i)

x= a = 2 =2 e 1/ 2 which meets the tangent at L at T (2, 0)

(ii)

4x sec q – 2y cosec q = 42 – 22 or

2x sec q – y cosec q = 6 P (4 cos q, 2 sin q)

L is

M

x - 1 y - 1/ 2 = 1/2 1/ 2 1 fi 2x – 2 =

(1)

Q (3 cos q, 0)

2y -1

1 N ÊÁ , 0ˆ˜ Ë2 ¯

So Area of the triangle LNT is

Fig. 18.13

It meets the line y = 0 in Q(3 cos q, 0) Let (h, k) be mid point of PQ, then

⎛ 1 ⎞ L ⎜1, ⎟ ⎝ 2⎠

h=

4 cos q + 3 cos q 2 sinq + 0 ,k= 2 2 2h , sin q = k 7



cos q =

\

Ê 2h ˆ 2 ÁË ˜¯ + k = 1 7

2

N (1/ 2, 0)

T (2, 0)

Fig. 18.12

1 NT ¥ y-coordinate of L 2 =

1Ê 1ˆ 1 3 = Á 2 - ˜¯ 2Ë 2 2 4 2

Thus, locus of M is 4 2 x + y2 = 1 49 If e is eccentricity of (1), then 4 = 16(1 – e2) fi e = x = ±2 3

(2)

3/2

IIT JEE eBooks: www.crackjee.xyz 18.12 Comprehensive Mathematics—JEE Advanced

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

Putting this in (2), we get y2 = 1 fi

48 1 = 49 49

Example 21 P(x1, y1) and Q(x2, y2), y1 < 0, y2 < 0 be the end points of the latus rectum of the ellipse x2 + 4y2 rectum PQ are

y = ±1/7

(±2

\

3, ±1/ 7

)

x2 y 2 + = 1 is inscribed 9 4 in a rectangle R whose sides are parallel to the coordiE2 passing through the point (0, 4) circumscribes the rectangle R the ellipse E2 is The ellipse E1:

Example 20

2 2 1 (c) 2

3 2 3 (d) 4

(a)

(b) x2 – 2 3y = 3 +

3

(c) x2 + 2 3y = 3 –

3

(d) x2 – 2 3y = 3 –

3

Solution:

Eccentricity e of the ellipse is given by

b2 = a2(1 – e2) fi

1 = 4(1 – e2) fi e =

3/2

Focii of the ellipse are ( 3, 0) and (- 3, 0)

Sides of R are given by

x = ± 3, y = ± 2

3

Ans. (b), (c)

(b)

Ans. (c) Solution:

(a) x2 + 2 3y = 3 +

Y

y E2

A2

(0, 4)

A1

o

(-2, 0)

(3, 2)

(2, 0) P

Q

x

0

Fig. 18.15

Length of a latus rectum of the ellipse is E1

Fig. 18.14

E2 be x

2

+

y

2

=1 a b2 As it passes through (0, 4) and (3, 2) we get 2

16 b2 9

= 1 fi b = 16 2

+

4

= 1 = a2 = 12 a b2 Eccentricity e of E2 is given by and

2

a2 = b2(1 – e2) fi

12 = 16(1 – e2) fi e = 1/2

2

b2 =1 a

( Q(x , y ) = Q (

Thus, P(x1, y1) = P - 3, -1/ 2 and

2

2

3, -1/ 2

)

)

Length of the latus rectum PQ of the parabola is |x2 – x1| = 2 3 = 4p(say) As focus of a parabola is the mid point of the latus rectum, focus of the desired parabola is (0, –1/2) and hence its vertices are (0, –1/2 ± p) Ê Ê 1 3ˆ 1 3ˆ Á 0, - 2 - 2 ˜ and Á 0, - 2 + 2 ˜ Ë ¯ Ë ¯ Thus there are two parabolas having PQ as the latus

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.13

Ê 1 3ˆ x2 = 4p Á y + + = 2 3y + 3 + 3 2 2 ˜¯ Ë and

Ê 1 3ˆ x = – 4p Á y + = -2 3 y - 3 + 3 2 2 ˜¯ Ë 2

Example 22 In a DABC A moves such that

BC, the ver-



Also AB + AC > BC = a So locus of A is an ellipse with foci at B and C and the a E1: x2 + 2y2 – 6x – 12y + 23 = 0 and

Example 23

E2: 4x2 + 2y2 – 20x – 12y points of intersection of E1 and E2 lie on a circle with Ê8 ˆ (a) centre at Á , 3˜ Ë3 ¯

cosB + cosC = 4sin2(A/2) If a, b and c denote the sides of the triangle opposite to the angles A, B and C respectively, then (a) (a) (c) (d)

AB + AC = 2a

Ê 8 ˆ (b) centre at Á - , 3˜ Ë 3 ¯

b + c = 4a b + c = 2a locus of point A is an ellipse locus of point A

1 47 3 2 1 47 2 3

Ans. (b), (c) Solution:

cosB + cosC = 4sin2(A/2)

Ans. (a), (c)



2 cos

B+C B-C A cos = 4 sin 2 2 2 2

Solution: intersection of the given ellipse is



A Êp B + Cˆ B-C Ê p Aˆ 2 cos Á - ˜ cos = 4 sin sin Á ˜ Ë2 Ë 2 2¯ 2 2 2 ¯

4x2 + 2y2 – 20x – 12y + 35 + l(x2 + 2y2 – 6x – 12y + 23) = 0 which represents a circle is

A

4 + l = 2 + 2l fi l = 2 b

c

6x2 + 6y2 – 32x – 36y + 81 = 0 B

a

C

Fig. 18.16





B-C cos 2 =2 B+C cos 2 B C B C cos cos + sin sin 2 2 2 2 =2 B C B C cos cos - sin sin 2 2 2 2 B C 1 tan = 2 2 3



tan



D D 1 = s ( s - b) s ( s - c ) 3



3s(s – a) (s – b) (s – c) = s2(s – b) (s – c)



3(s – a) = s fi 2s = 3a



a + b + c = 3a fi b + c = 2a

81 Ê 16 ˆ =0 x2 + y2 – Á ˜ x - 6 y + Ë 3¯ 6



centre of the circle is (8/3, 3) 2

and the radius is

128 + 162 - 243 1 47 = 18 3 2

=

If the normal at any point P on the

Example 24 ellipse

81 Ê 8ˆ 2 ÁË ˜¯ + (3) 3 6

x2 a

2

+

y2 b2 G2, then

b 2 b cos 2 q + a 2 sin 2 q a a 2 (b) PG2 = a cos 2 q + b 2 sin 2 q b (a) PG1 =

G1 and the

IIT JEE eBooks: www.crackjee.xyz 18.14 Comprehensive Mathematics—JEE Advanced

(c) PG1 : PG2 = a : b (d) PG1 : PG2 = b2 : a2 Ans. (a), (d) Solution: Let the coordinates of P be (a cos q, b sin q) Equation of the normal at P is ax sec q – by cosec q = a2 – b2 = a2e2 It meets the major axis at G1 (ae2 cos q, 0) and minor Ê ˆ a 2 e2 sin q ˜ axis at G2 Á 0, b Ë ¯ (PG1)2 = (a cos q – ae2 cos q)2 + b2 sin2 q

So

= a cos q (1 – e ) + b sin q 2

2

2 2

2

2

2

Ê b2 ˆ = a cos q Á 2 ˜ + b 2 sin 2 q Ëa ¯ 2

= and

b2 a

2

2

(b2 cos2 q + a2 sin2 q)

Ê a 2 e2 ˆ (PG2) = (a cos q)2 + Á b + b ˜¯ Ë = (a cos q)2 + =

a2 b2

a

4

b

2

We can write x + y = 3 as x –1 y–2 3p ˘ È = = r Í slope = –1 = tan 1 1 Î 4 ˙˚ – 2 2 For coordinates of Q and R, 2 2 r= ± 3

( )

\

2

2

Fig. 18.17

sin2 q

y–2 x –1 2 =± 2 = 1 1 3 – 2 2

5 4 1 8 ,y= and x = , y = 3 3 3 3 Let equation of E1 be fix=

sin2 q

x2 y 2 + =1 a 2 b2 where a > b

(b2 cos2 q + a2 sin2 q)

PG1 : PG2 = b2 : a2 Example 25 Let E1 and E2 be two ellipses whose centres are at the origin. The major axes of E1 and E2 lie along the x-axis and the y-axis, respectively. Let S be the circle x2 + (y – 1)2 = 2. The straight line x + y = 3 touches the curves S, E1 and E2 at P, Q and R, respectively. Suppose 2 2 that PQ = PR = . If e1 and e2 are the eccentricities 3 of E1 and E2, respectively, then the correct expression(s) is(are) (a) e12 + e22 =

43 40

(b) e1 + e2 =

(c) e12 – e22 =

5 8

(d) e1e2 =

7 2 10

3 4

Ans. (a), (b) Solution: P is the point of intersection of the tangent x + y = 3 to S and normal to S at P, that is, of (x – 0) – (y – 1) = 0. Thus, coordinates of P are (1, 2)

As x + y = 3 is a tangent to E1 at Q

( ) 5 4 , 3 3

5 4 x+ 2 y = 1 2 3a 3b and x + y = 3 are the same line. \ a2 = 5, b2 = 4 b2 4 1 fi e12 = 1 – 2 = 1 – = a 5 5 Let equation of E2 be x2 y 2 + =1 c2 d 2

( )

1 8 As x + y = 3 is a tangent to E2 at R , 3 3 Thus, x + y = 3 and 1 8 x+ 2 y = 1 2 3c 3d are the same line. \ c2 = 1, d2 = 8

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.15

fi e22 = 1 –

Thus, e12 + e22 = e12 – e22 Example 26 x

2

a2

+

y



c2 7 2 = d 8

And t2 – 5t + 6 π 5 – 4t – t2 fi

7 43 , e 1e 2 = 2 10 40

Let F1 and F2 be the foci of the ellipse

BF1F2 e is the eccentricity of the ellipse and D is the area of the triangle BF1F2, then 1 1 (a) e = (b) e = 2 3 3 2 a 4

(d) D =

Coordinates of F1 are (ae, 0) and of F2 are

a2e2 + b2 = 4a2e2



a2e2 + a2(1 – e2) = 4a2e2 1 e= fi F 1F 2 = a 2

(

l ≥0

Ê bˆ 1- Á ˜ Ë a¯

2

(c) e(l) is independent of l (d) e(l) has no minimum value

x2

If

+

(d)

)

2 7

We must have t2 – 5t + 6 > 0, 5 – 4t – t2 > 0

t – 5t + 6 π 5 – 4t – t

Now, t2 – 5t + 6 > 0 t < 2 or t > 3 5 – 4t – t2 > 0

2

e¢(l) =

1-

b2 + l a2 + l

a 2 - b2

=

a2 + l

a 2 - b2 Ê 1 ˆ - 0

Ê bˆ 1- Á ˜ Ë a¯

2

y2

(

3 16

e(l) =

As e(l) decreases in the interval [0, •), e(l) has no

)

2

Solution:

l ≥0

Ans. (a), (b), (c), (d)



(a) max e(l ) =

max e(l ) = e(0) =

= 1 represents t 2 - 5t + 6 5 - 4t - t 2 an ellipse but not a circle, then possible values (s) of t is (are) - 5 +1 - 3 +1 (b) (a) 4 3 Example 27

and

y2

fi e(l) decreases in the interval [0, •)

3 2 and the area of the DBF1F2 = a 4

Solution:

+

= 1, where a > b, l ≥ a 2 + l b2 + l (a) e(l) decreases in the interval [0, •)





(c)

x2

Let e(l) be the eccentricity of the ellipse

Ans. (a), (b), (d)

Also BF1 = BF2 = F1F2



Example 28

3a 2

Ans. (a), (c) Solution: (–ae, 0)

tŒR

Thus, –5 < t

= 1 and (0, b) be an end point of the minor

(c) D =

2t2 – t + 1 π 0



27 5 = π 40 8

2

b2

–5 3a + 15 [∵ f is a strictly decreasing function]



a2 + 2a – 12 > 0



(a + 6) (a – 2) > 0



a < –6 or a

Thus a Œ (–•, –6) or a Œ (2, •)

Solution: Also,

PF1 + PF2 = 2a = 10

(b) a2 = 4, b2 = 3, a2 (1 – e2) = b2 3 4 1 fi e2 = 4 1 fi e= 2 Now, PM1 + PM2 = e(PF1 + PF2) fi

1 – e2 =

=

1 1 (2a) = (4) = 2 2 2

(c) 2[x2 – 2x + 1] + 3[y2 – 4y + 4] = 1

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.17



or

( x - 1) 2 ( y - 2) 2 =1 + 1 1 2 3



e2 = 1 -



3e2 = 1

(d) Putting

(2

b2 a2

= 1-

y = 2 + 4 tan q (d) x – 1 = 4 sec q, y – 2 = 3 tan q

2 1 = 3 3



x = 1 + 4 sec q, y = 2 + 3 tan q

2x = 2 3 - 3 y in 2x2 + 3y2 = 6, we get 3 - 3y

)

2

+ 3y2 = 6



12 + 3y2 – 12y + 3y2 = 6



6y2 – 12y + 6 = 0



(y – 1)2 = 0



y=1

Example 32

x = 1 + tan2 q

Let E be the ellipse

Example 33 (a > b), and d =

Column I

Column II (p) x = 1 + 3 cos q y = 2 + 4 sin q

(b) a = 2, b = 3, g = 3

x = 1 + 4 sec q y = 2 + 3 tan q

(c) a = 2, b = 3, g = 4

(c) (y – 2)2 = 16(x – 1)

(r) x = 1 + 4 cos q y = 2 + 4 sin q

(d) a = 4, b = 3, g = 5

(d) 9(x – 1)2 – 16(y – 2)2 = 144

(s) x = 1 + tan2 q y = 2 + 4 tan q

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

Solution:

y – 2 = 4 sin q

fi x = 1 + 4 cos q,

y = 2 + 4 sin q

( x - 1) ( y - 2) + 9 16 x = 1 + 3cos q,



y = 2 + 4 sin q (c) We can take y – 2 = 4 tan q and x – 1 = tan2 q

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

Solution:

(a) x – 1 = 4 cos q,

2

y2 b2

= 1,

a a 2 + b b2 , a > 0, b > 0, g g

(a) a = 1, b = 2, g = 4

(b) 16(x – 1)2 + 9(y – 2)2 = 144

a2

+

E, which are at distance d a, b, g in column I with eccentricity e of the ellipse as given

Match the curves in column I with their

Column I (a) (x – 1)2 + (y – 2)2 = 16

x2

(a)

d=b



d2 = b2

Column II 1 (p) e = 3 1 e= 2 1 (r) e = 2 (s) e =

fi a2 + 2b2 = 4b2

2



a2 = 2b2



a2 = 2a2 (1 – e2) 1 e= 2



=1

2a 2 + 3b 2 > b2, so d = a 3 fi 2a2 + 3b2 = 3a2 fi 3a2 (1 – e2) = a2

(b) d2 =



e=

2 3

2 3

IIT JEE eBooks: www.crackjee.xyz 18.18 Comprehensive Mathematics—JEE Advanced

d2 = a2 2a2 + 3b2 = 4a2

(c) fi

a2 OT 2

fi 3a2 (1 – e2) = 2a2 fi

1 = 3e2 1 e= 3



(d) As d > b, d fi d2 = a2 fi 4a2 + 3b2 = 5a2 fi 3a2 (1 – e2) = a2 fi

2 = 3e2



e=

a

Ê ˆ a 2 - b2 N1 Á 0, sin q ˜ b Ë ¯

y

Now, (ON) (ON1) =

0£q £

Area of DOTT1 =

min (ON )(ON1 )

(r) 1

(d)

min (Area of DOTT1 )

(s)

p 2

p 0£q £ 2

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

(a 2 - b 2 )2 2ab

Solution: P to E is x y cos q + sin q = 1 a b P to E is ax sec q – by cosec q = a2 – b2 x T (a sec q T1 (0, b cosec q) We have

(a 2 - b 2 )2 2ab

1 ab (OT )(OT1 ) = 2 sin(2q )

Min (Area of DOTT1) = ab

ab

p 2

p 2

p 2

Match the entries in column I with en-

Example 35

(p) 2ab

Ê a 2 - b2 ˆ NÁ cos q , 0˜ and Ë a ¯

(a 2 - b 2 )2 sin 2q 2ab

Max (ON )(ON1 ) =

Column II

min (OT )(OT1 )

0£q £

p 2

x

2 3

Column I a2 b2 + (a) OT 2 OT12

2ab sin(2q )

Max (OT )(OT1 ) = 2ab

0£q £

0£q £

(c)

= cos2q + sin2q = 1

OT12

(OT)(OT1) =

x2 y 2 pˆ Ê b sin q) Á 0 < q < ˜ to the ellipse E : 2 + 2 = 1 Ë 2¯ a b meets the x T and y T1, and normal at point P to the ellipse E meets the x N and y N1

0£q £

b2

Suppose the tangent at point P (a cos q,

Example 34

(b)

+

Column I (a) Ordinate of the Foci of the ellipse 4x2 + 9y2 + 8x – 18y + 4 = 0

Column II (p) 4

(b) Abcissa of the centre of the ellipse 3x2 + 4y2 – 6x – 16y = 0

(b) 2

(c) If e is the eccentricity of ellipse 3x2 + 4y2 = 1 the 2e

(r) 1

two points of the ellipse 5x2 + 9y2 = 20 is p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans. (i)

(ii)

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.19

Solution: 4(x2 + 2x + 1) + 9(y2 – 2y + 1) = 9 ( x + 1) 2

or

2

Ê 3ˆ ÁË ˜¯ 2

+

( y - 1) 2 =1 1

(i)

Shift the origin to (–1, 1), so that X = x + 1, Y = y – 1 and (i) becames X2 Ê 3ˆ ÁË ˜¯ 2

2

+

Y2 =1 1

+

y2

Statement-2: NT is independent of q Ans. (c)

(ii)

Ê 9ˆ Eccentricity e of (ii) is given by 1 = Á ˜ (1 - e2 ) Ë 4¯ fie=

x2

= 1 at point P (a cos q, b sin q), a 2 b2 p 0 0, T and normal at P meets y1 N Statement-1: PN is internal bisector of –F1 PF2

x2 + y2 = a2

2

(i)

y

2

(iii)

P (x1, y1)

[Make (ii) homogeneous using (i)] fi

Ê a2 ˆ Êa ˆ x 2 sin 2 a - Á sin 2a ˜ xy + y 2 Á1 - 2 sin 2 a ˜ = 0 Ëb ¯ Ë b ¯

F1 O

Ê h 2 - ab ˆ As –AOB = tan -1 Á ˜ , it is a function of a ÁË a + b ˜¯ Now, –AOB = fi fi

b2 = (a2 – b2) sin2a



1 – e2 = e2 sin2 a 1 e= 1 + sin 2 a



Let E be the ellipse

T

Fig. 18.20

xx1 a

2

+

yy1 b2

=1

Ê a2 ˆ Coordinates of T are Á , 0˜ Ë x1 ¯ fi

PT2 =

(a 2 - x12 ) 2 x12

+ y12

Clearly PT depends on position of P x2

y2

P is

+ = 1 and L a 2 b2 be the line px + qy + r = 0, where r2 = (p2 + q2) (a2 + b2 Example 38

F2

Solution:

p 2

Ê a2 ˆ (sin2a) + Á1 - 2 sin 2 a ˜ = 0 Ë b ¯

N

Statement-1: L from which two perpendicular tangents can be drawn to E

a 2 x b2 y = a2 – b2 (= a2e2) y1 x1 N (e2x1, 0)

We have

Èa ˘ PF1 = e Í + x1 ˙ = a + ex1 Îe ˚

Statement-2: L touches the circle x2 + y2 = a2 + b2

and

PF2 = a – ex1, NF1 = e2x1 + ae = e (PF1)

Ans. (a)

and

NF2 = ae – e2x1 = e (PF2)

As

PF1 NF1 = PF2 NF2

Solution: As r2 = (a2 + b2) (p2 + q2),

|r | p +q 2

2

= a 2 + b2

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.21

PN bisects the angle –F1 NF2

Example 42 If L represents the line joining the point P on C to its centre O M (above) to the ellipse E is

Remark In fact PN is internal bisector –F1PF2

(a) x + 3y = 3 5

(b) 4x + 3y =

5

(c) x + 3y + 3 5 = 0

(d) 4x + 3y +

5 =0

x2 y 2 + = 1 whose 36 25 foci are F1 and F2 P (x1, y1) x1 > 0, y1 > 0 be a point on E L1 and L2 be the feet of perpendiculars from F1 and F2 to the tangent at P to E

Example 43 E conjugate to the diameter represented by L is

Statement-1: F1L1 + F2L2 ≥ 10

Example 44 If R is the point of intersection of the line L with the line x = 1, then

Example 40

Let E be the ellipse

Statement-2: (F1L1) (F2L2

(a) 9x + 2y = 0 (c) 4x + 9y = 0

(a) (b) (c) (d)

Ans. (a) pˆ Ê Let x1 = 6 cosq, y1 = 5 sinq Á 0 < q < ˜ Ë 2¯ P is x y cos q + sin q = 1 6 5

Solution:

F1L 1 =

| e cos q - 1 | Ê 1 - e cosq ˆ = Á ˜¯ and Ë b b

F2L 2 =

| - e cos q - 1 | Ê 1 + e cosq ˆ = Á ˜¯ Ë b b

where b2 = and

1 2

[1 - e2 cos 2 q ]

b 1 È 36 - 25 ˘ cos 2 q ˙ = 25 = 2 Í1 36 b Î ˚

Also, F1 L1 + F2 L2 ≥ 2 ( F1 L1 )( F2 L2 ) = 10

Paragraph for Question Nos. 41 to 44 2

C: x2 + y2 = 9, E : Example 41

x y + = 1, L: y = 2x 9 4

(a)

1 3

1 (c) 2

MQ = 2 sin q so

MQ 2 = PQ 3

L : y = 2x meets the circle C: x2 + y2 = 9 at 3 points for which x2 + 4x2 = 9 fi x = ± 5

(

Coordinate of P are ±3

(

coordinates of M are ±3

E meets the MQ PQ

5 , ±4

5

)

5

)

M to the ellipse E is x(±3)

+

y (±4) 4 5

=1

(mm¢ = –b2/a2) 2 3

(d) none of these

P

y = mx be the diameter conjugate to the diameter L : y = 2x of the ellipse E then 2m = – 4/9

(b)

5 , ±6

x + 3y = ±3 5

P is a point on the circle C, the perpen-

dicular PQ ellipse at M

P be (3 cos q, 3 sin q) then the eccentric angle of M, the point where the ordinate PQ through P meets the ellipse is q and the coordinates of M are (3 cos q, 2 sin q), PQ = 3 sin q

9 5

2

inside both C and E outside both C and E on both C and E inside C but outside E

Solutions:



COMPREHENSION-TYPE QUESTIONS

lies lies lies lies

Ans.

cos 2 q sin 2 q 1 11 = sin 2 q + 36 25 25 900

(F1L1) (F2L2) =

R R R R

(b) 2x + 9y = 0 (d) 4x – 9y = 0

fi m of the conjugate diameter is y = (–2/9) x or 2x + 9y = 0

M O

Q

Fig. 18.21

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R are (1, 2) C(1, 2) = 1 + 22 – 9 < 0 fi R lies inside C 1 + 1 – 1 > 0 fi R lies outside E E(1, 2) = 9

B

Paragraph for Question Nos. 45 to 47

2

A 3

O

-3

Tangents are drawn from the point P(3, 4) to the ellipse x2 y 2 + = 1 touching the ellipse at points A and B 9 4 Example 45

The coordinates of A and B are

(a) (3, 0) and (0, 2) Ê 8 2 161 ˆ Ê 9 8ˆ and Á - , ˜ (b) Á - , ˜ Ë 5 5¯ Ë 5 15 ¯

Fig. 18.22



1 2 1Ê 2 x2 ˆ x + Á1 - x + ˜ = 1 9¯ 9 4Ë 3



5x2 – 6x – 27 = 0



(5x + 9)(x – 3) = 0 fi x = –9/5 or x

When x = 3, y = 0

Ê 8 2 161 ˆ (c) Á - , ˜ and (0, 2) Ë 5 15 ¯

and when x = –9/5, y = 8/5 Thus, coordinates of A and B are (3, 0) and (–9/5, 8/5)

Ê 9 8ˆ (d) (3, 0) and Á - , ˜ Ë 5 5¯ Example 46

B is y

The orthocentre of the triangle PAB is

Ê 8ˆ (a) Á 5, ˜ Ë 7¯

Ê 7 25 ˆ (b) Á , ˜ Ë5 8 ¯

Ê 11 8 ˆ (c) Á , ˜ Ë 5 5¯

Ê 8 7ˆ (d) Á , ˜ Ë 25 5 ¯

Example 47 whose distance from the point P and the line AB are (a) (a) (c) (d)

9x2 + y2 – 6xy – 54x – 62y + 241 = 0 x2 + 9y2 + 6xy – 54x + 62y – 241 = 0 9x2 + 9y2 – 6xy – 54x – 62y – 241 = 0 x2 + y2 – 2xy + 27x + 31y – 120 = 0

Ans. Solution: T=0 fi

3 4 x+ y = 1 9 4



x +y =1 3

x x2 y 2 + Putting y = 1 - in = 1, 9 4 3 We get

xˆ 1 2 1Ê x + Á1 - ˜ 9 4 Ë 3¯

1 3 fi Slope of altitude through P Slope of AB is -

\

P is y – 4 = 3(x – 3)

When

y = 8/5, x = 11/5

Thus, orthocentre of DPAB is (11/5, 8/5) TIP. Since the orthocentre lies on the line 8/5, only Q(x, y) lies on the locus, then PQ = length of perpendicular from Q to AB



( x - 3) 2 + ( y - 4) 2 =

(1/ 3) 2 + 1



10[x2 + y2 – 6x – 8y + 25] = (x + 3y – 3)2



9x2 + y2 – 6xy – 54x – 62y + 241 = 0

Paragraph for Question Nos. 48 and 49 Let F1 (x1, 0) and F2 (x2, 0) for x1 < 0 and x2 > 0 be the foci of the ellipse

2

=1

x + y -1 3

ellipse at point M

x2 y 2 + 9 8 F2 intersects the N in

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.23

The orthocentre of the triangle F1MN is

Example 48

Ê3 Ê3 ˆ ˆ Coordinates of M are Á , 6 ˜ and that N are Á , - 6 ˜ Ë2 Ë2 ¯ ¯

Ê 9 ˆ (a) Á - , 0˜ Ë 10 ¯

Ê2 ˆ (b) Á , 0˜ Ë3 ¯

Altitude through F1 is y = 0

Ê9 ˆ (c) Á , 0˜ Ë 10 ¯

Ê2 ˆ (d) Á , 6 ˜ Ë3 ¯

Slope of F1N is

Example 49 If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the x Q, then ratio of area of triangle MQR to MF1NF2 is (a) 3 : 4 (c) 5 : 8

(b) 4 : 5 (d) 2 : 3

\ y- 6 =

5 Ê 3ˆ ÁË x - ˜¯ 2 2 6

When y = 0, x = -

9 10

Ê 9 ˆ Thus, orthocentre of DF1MN is Á - , 0˜ Ë 10 ¯

Ans. Solution:

M and N are

a2 = 9, b2 = 8

respectively

We have b2 = a2(1 – e2)

(3/ 2) x 6y =1 + 9 8

1 8 = 9 (1 – e2) fi e = 3 Ê Ê 1ˆ ˆ Thus, foci are Á ±3 Á ˜ , 0˜ = (±1, 0) Ë Ë 3¯ ¯



\

- 6 -0 -2 6 = 3 5 +1 2 M is

(3/ 2) x (- 6 ) y =1 + 9 8 These tangents intersect at R(6, 0)

x1 = –1, x2 = 1

Focus of parabola is (1, 0) y2 = 4x

M is

2 ÷2 M (3/2, ÷ 6)

-3

F1 (-1, 0)

y- 6 = -

3

-2 ÷2

N (3/2, - ÷ 6)

6Ê 3ˆ ÁË x - ˜¯ 2 2

It meets the x

Ê7 ˆ Q Á , 0˜ Ë2 ¯

Area of the DMQR =

1 Ê 7ˆ 5 6 ¥ Á6 - ˜ ¥ 6 = 4 2 Ë 2¯ MF1 NF2

Fig. 18.23

= 2 area of the DMF1F2 x2 y 2 + =1 9 8 y2 = 4x

and

= 2¥

(1)

1 ¥2¥ 6 = 2 6 2 5 6 : 2 6 or 5 : 8 4

(2)

2

Putting y = 4x in (1), we get x2 4 + x =1 9 8 fi

2x2 + 9x – 18 = 0



(2x – 3) (x + 6) = 0



x=

3 as x > 0 2

INTEGER-ANSWER TYPE QUESTIONS Example 50

The points of intersection of the per-

pendicular tangents to the ellipse Ans. 6

x2 y 2 + = 0 lies on a 25 11

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Solution:

on the ellipse is

Any tangent to the ellipse is

y = mx + 25m 2 + 11 perpendicular to it is

then (CF)2 =

1 25 y = - x+ + 11 m m2

=

7 2 ¥ 52 52 cos 2 q + 7 2 sin 2 q 25 ¥ 49

25 cos 2 q + 49 sin 2 q P is 7x 5y = 72 – 52 cos q sin q Ê -24 sin q ˆ Coordinates of G are Á 0, ˜¯ Ë 5

Eliminating m, we get (y – mx)2 + (my + x)2 = (1 + m2) (25 + 11) fi

x y cos q + sin q = 1 7 5

x2 + y2 = 36 = (6)2

So radius of the circle is 6 Note

24 sin q ˆ Ê (PG)2 = (7 cos q)2 + Á 5 sin q + ˜ Ë 5 ¯

Perpendicular tangents intersect on the director circle x2 + y2 = 25 + 11 = 36 Example 51 If l is the length of the intercept made by a common tangent to the circle x2 + y2 = 16 and the ellipse x2/25 + y2 3l2 Ans. 6

=

2

49 (25 cos2 q + 49 sin2 q) 25

So

(CF ◊ PG)2 = (49)2



CF ◊ PG = (7)2



CF PG = 7

So Geometric mean of CF and PG Solution: mx +

y=

25m 2 + 4 , which also touches the circle if m(0) - 0 + 25m + 4

fi fi

=4

25m2 + 4 = 16 (1 + m2) fi m2 = 4/3

Ans. 3

m = ±2/ 3 2

4 y= ± x + 25 ¥ + 4 3 3 fi

3y ± 2x = 4 7 2

fi fi Example 52

x2 y 2 + y = m¢x is the diameter of the el20 9 lipse bisecting the chords parallel to L m represents the slope of L, then the value of 20(m + m¢ ellipse

2

1 + m2

Ê 4 7ˆ Ê 4 7ˆ l = Á +Á ˜ ˜ Ë 2 ¯ Ë 3¯

2

2

Solution:

x y2 + = 1 on the tangent at any point of the ellipse 49 25 P, and G is the point where the normal at P meets the CF and PG is

9h 3 = 20k 4

Locus of (h, k) is y = fi

m¢ = -



9 4 ¥ x = m¢x 20 3

3 5

m + m¢ =

Ans. 7 P(7 cos q, 5 sin q)

3 4 h, k) as the mid

-

If CF is perpendicular from the centre C

m=

point is hx ky h2 k 2 + + = 20 9 20 9 which is parallel to L if

3l2 = 196 2

Solution:

L : 3x – 4y = 0 is a diameter of the

Example 53

3 3 3 - = 4 5 20

20(m + m¢) = 3

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.25

Note

Example 56

Two diameters of an ellipse are said to be conjugate when each bisects the chords parallel to other and product of their slope is -

Using this result fi

b2 a

, where 2

x2 a

2

+

y2 b

2

3 9 4 ¥ = 5 20 3

1 ellipse is 8 and the eccentricity is D denotes the 2 area of the rectangle formed by joining the vertices of the D latera recta of the ellipse then the value of 12 Ans. 8 x2 y 2 + =1 Solution: a 2 b2 Ê 1ˆ where a2 = 82 and b2 = 82 Á1 - ˜ Ë 4¯ Ê b2 ˆ Vertices of the latera recta are Á ± ae, ± ˜ a¯ Ë 2b 2 2 So D = 2ae ¥ = 4b e a 3 1 = 4 ¥ 82 ¥ ¥ = 12 ¥ 8 4 2 D =8 fi 12

2

The line y = 3x + c touches the ellipse

2

3x + 4y – 6x + 16y – 7 = 0 if (c + 5)

y2

P be any point on the ellipse,

p +4 2

PF1 ◊ PF2 – p2 is Ans. 4 Solution:

Example 54

Example 55

p +2

+

= 1 is

9 mm¢ = 20 m¢ = -

x2 2

Let F1, F2 be two foci of the ellipse

Let a2 = p2 + 2, b2 = p2 + 4

Let e are F1(0, –be), F2(0, be b b and y = e e nates of P are (a cos q, b sin q)

Its directrices are y = -

b We have PF1 = e (distance of P from y = - ) e -b = e - b sinq = |b + be sin q | e and

PF2 = |b – be sin q |

Now, PF1 ◊ PF2 = b2 – b2e2 sin2 q £ b2 = p2 + 4 fi

PF1 ◊ PF2 – p2 £ 4

Example 57

Suppose (h, k) lies on the circle x2 + y2

= 4, and (2h + 1, 3k + 2) lies on an ellipse with eccentricity e, then 9e2 Ans. 5 Solution:

Let a = 2h + 1, b = 3k + 2,

(a - 1) Ê b - 2ˆ ,k= Á Ë 3 ˜¯ 2 2 2 As (h, k) lies on x + y = 4, we get fi

h=

(a - 1)2 (b - 2) 2 =4 + 4 9 fi

2

(a1b) lies on the ellipse ( x - 1) 2 ( y - 2) 2 =1 + 16 36

Ans. 9 Solution:

Now, 16 = 36(1 – e2)

( x - 1) 2 ( y + 2) 2 =1 + 4 3 which can be written as



X2 Y2 + = 1 where X = x – 1, Y = y + 2 4 3 Line y = 3x + c or Y = 3X + c + 5 touches the ellipse if (c + 5)2 = 4(3)2 + 3 = 39 fi

(c + 5)2 – 30 = 9

-

9e2 = 5

Example 58

The line joining the foci F and F¢ of an

ellipse subtend a right angle at the positive end B of the e is the eccentricity of the ellipse then the value of 2e2 Ans. 2 Solution:

x2 a2

+

y2 b2

=1

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B(0, b), F(–ae, 0), F¢(ae, 0) Slope of BF =

2 Ê h2 ˆ Ê 4 ˆ = 3 Á1 - ˜ ÁË - h˜¯ 4¯ h Ë

-b b , slope of BF¢ = ae ae

fi fi fi

(

b 2 = a 2e 2 1 a2 (1 – e2) = a2e2 fi e2 = 2 2e2

A vertical line passing through the point x2 y 2 + (h, 0) intersects the ellipse = 1 at the points P 4 3 and Q. Let the tangents to the ellipse at P and Q meet at the point R. If D(h) = area of the triangle PQR, D1 = min 1/2£ h £1 D(h),

then =

8 5

max 1/2£ h£1

D1 - 8D 2

D(h)

[2013]

Ans Solution: y-coordinates of P and Q are k and –k respectively, where 1 2 1 2 h + k =1 4 3

P (h, k)

A -2

h

3

2

R

Q (h, - k)

P and Q 1 1 hx + ky = 1 4 3 1 1 and hx - ky = 1 4 3 Ê4 ˆ These tangents intersect at R ÁË , 0˜¯ h



x

)

(

)(

)

)

(

D = area of DPQR 1 Ê4 ˆ = (PQ) (AR) = k ÁË - h˜¯ h 2 2 4 D2 = k 2 ÊÁ - hˆ˜ Ëh ¯

)

3 9 D2 = 1 2min £ h£1 D(h) = D(1) = 2 3 3 = 2 (1 ) and D1 = 1 max 2 £ h £1 D(h) = D 2

\

= fi

8 5

3 45 15 15 = 5 8 8

D1 - 5D 2 = 45 – 36 = 9 Suppose that the foci of the ellipse

x2 y 2 + = 1 are (f1, 0) and (f2, 0) where f1 > 0 and f2 < 0. 9 5 Let P1 and P2 be two parabolas with a common vertex at (0, 0) and with foci at (f1, 0) and (2f2, 0), respectively. Let T1 be a tangent to P1 which passes through (2f2, 0) and T2 be a tangent to P2 which passes through (f1, 0). m1 is the slope of T1 and m2 is the slope of T2, then the value Ê 1 ˆ of Á 2 + m22 ˜ is. Ëm ¯ 1

Fig. 18.24

Let

)

(

Example 60

y

3

)

(

Example 59

and D2 =

(

3 3 1 4 - h2 2 4h 3 1 1 £ h £1 Let f(h) = 2 4 - h 2 , 2 h 3 2 -2 1 f ¢(h) = 3 4 - h 2 + 2 3 4 - h 2 ( -2h ) h h 4 1 = - 3 4 - h 2 2 + h 2 < 0 for < h < 1. 2 h Thus, f(h

=

b b p –FBF¢ = fi - ¥ =–1 ae ae 2

Ans Solution: We have a2 = 9, b2 = 5 \

e2 = 1 –

b2 5 4 = 2 = 1– a 9 9

2 3 Thus, f1 = ae = 2 and f2 = – 2 fi

e =

2

P1 is y2 = f1x = 8x

P2 is y = –16x T1 be y = m1 x +

2 m1

As T1 passes through (2f2, 0) = (–4, 0)

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.27

0 = –4m1 +

c for which y = 5x + c is a

2 1 ⇒ m12 = m1 2

2

tangent to the ellipse

T2 is (a) 1

4 m2 As it passes through (f1, 0) = (2, 0) 4 0 = 2m2 – m2 y = m2 x –



1 + m22 = 2 + 2 = 4 m12

q1) (tanq2) =

SINGLE CORRECT ANSWER TYPE QUESTIONS

(c) 1/2

(d) 2/3

(x1, y1) and (x2, y2) to the ellipse x2/a2 + y2/b2 = 1 are at right angles then x1 x2/y1 y2 (b) –b2/a2 (a) a2/b2 4 4 (c) –a /b (d) –b4/a4 E1 be the ellipse ellipse

x2

+

y2

x2 a2 + 2

+

y2 b2

=1 and E2 be the

a b +1 which to perpendicular tangents can be drawn to each of E1 and E2 is … (a) 0 (b) 1 (c) 2 (d) 4 2

eccentricity (a) (b) (c) (d)

16x2 25x2 25x2 16x2

+ + + +

-a 2 b2

, then the chord joining two

2

3 and passing through (4, 0) is 5 25y2 = 256 16y2 = 400 16y2 = 256 25y2 = 400

x2 a2

+

y2 b2

=1

p x2 y 2 be two points on the ellipse 2 + 2 = 2 a b The locus of point of intersections of normals at P and Q is (a) ax + by = 0 (b) ax – by = 0 (c) x + y = 0 (d) x + y = a + b

q+f= -

3/2

(d) 6

will subtend a right angle at (a) focus (ae, 0) (b) focus (–ae, 0) a, 0) P(a cos q, b sin q) and Q(a cos f, b sin f) where

LEVEL 1

(b)

(c) 4

points P1(q1) and P2(q2) on the ellipse

EXERCISE

ellipse is (a) 3/4

(b) 2

x2 y2 + = 1 represents an 10 - 2a 4 - 2a ellipse, then a lies in the interval (a) (–•, 5) (b) (2, 5) (c) (–•, 2) (d) (5, •)

m22 = 2

Thus,

x + y 2 = 1 is 25

x2 + 2y2 = 6 which touches the ellipse x + 4y = 4 is (b) x2 + y2 = 2 (a) x2 + y2 = 6 (d) x2 + y2 = 12 (c) x2 + y2 = 9 2

P is a point on the ellipse

x2

2

+

y2

= with foci S a 2 b2 and S¢ and eccentricity e, then locus of the incentre of the triangle PSS¢ is an ellipse of eccentricity (a)

1- e 1+ e

(b)

e 1+ e

(c)

2e 1+ e

(d)

1 - 2e 1 + 2e

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS x2 y 2 + =1 25 16 (a) x + y – 41 = 0 (c) x + y – 9 = 0

(b) x – y + 41 = 0 (d) x – y + 9 = 0

IIT JEE eBooks: www.crackjee.xyz 18.28 Comprehensive Mathematics—JEE Advanced

x+ 4y (a)

x2 y2 + =1 100 5

(b)

x2 y 2 + =1 80 5/ 4

x2 y 2 + =1 (d) none of these (c) 20 5 a and b two natural numbers such that a + b = ab x2 + 4y2 = 4 passing through (a, b) are (a) x = 2 (b) 3x + 4y – 14 = 0 (c) y = 2 (d) 3x – 8y + 10 = 0 pˆ Ê =1, Á 0 < q < ˜ repË 2¯ sec q tan q resents an ellipse with eccentricity e l (a) e is independent of q (b) e decreases as q increases (c) el is independent of q (d) l2 – e2 independent of q x2 + 4y2 – 6x + 8y + k = 0 represents (a) a point if k = 7 (b) an ellipse if k < 7 (c) no locus if k > 7 (d) a hyperbola if 3 < k < 7 x2

2

+

y2

2

2

x y + 25 16 = 1 at point A(a, b) and is at a distance d from the x + y = p touches the ellipse E :

Column 1 (a) p2 p2 (b) 2 d (c) a2 + b2 (d) Number of values of p C be the curve

Column 2 881 (p) 41

(p) -

3 3 4 4

conjugate to the diameter through P (c) Slope of the normal through P (d) Slope of the line joining P to the end of the latus rectum in the

3 3 (r) (s)

(

3 2 3- 7 4

)

3 4

ASSERTION-REASON TYPE QUESTIONS P on the ellipse

(r) 2 (s) 41

Column 2 (p) Ellipse

a2

+

y2 b2

= 1,

which T Statement 1: If the normal at an end L of a latus rectum of the ellipse x2/a2 + y2/b2 = 1 meets the major G, O is the centre of the ellipse, then OG = ae3, e Statement 2: The normal at a point on the ellipse x2/a2 + y2/b2 Statement 1: Two perpendicular tangents can be

(2

7, 6 2

)

x2 y 2 + = 1 from the point 74 26

Statement 2: Perpendicular tangents to the ellipse x2 y 2 + = 1 intersect at a point on the circle x2 + 74 26 y2 Statement 1: Feet of the perpendicular from the foci x2 y 2 + = 1 on the line y = 2x + 13 lie 36 25 on the circle x2 + y2 of the ellipse

(s) Parabola

x2

T Statement 1: The circle on PT as diameter passes through the focus of the ellipse corresponding to the T Statement 2: PT subtends a right angle at the fo-

drawn to the ellipse

5x2 + ay2 – 20x + 4y + 20 = 0 (a) (b) (c) (d)

(a) Slope of the diameter through P

2

MATRIX-MATCH TYPE QUESTIONS

Column 1 a 0, a π a=5

3ˆ x2 y 2 Ê + = 1 is an ellipse and P Á 2 3, ˜ is a Ë 16 9 2¯ point on E Column 1 Column 2 E:

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.29

Statement 2: Locus of the feet of the perpendiculars from the foci on any tangent to an ellipse is the Statement 1: The coordinates of the mid-point of the chord of contact of the tangents drawn from the point (6, 9) to the ellipse

x2 y 2 Ê 24 36 ˆ + = 1 are Á , ˜ Ë 25 25 ¯ 9 36

Statement 2: 8x + 5y = 41 represents the chord of x2 y 2 + = 1 whose mid-point is (2, 5) the ellipse 9 36

COMPREHENSION-TYPE QUESTIONS C2 is a circle on the

P:

x2 y 2 + 25 16

(a) 5/4

(a) (9 + 5 )3

(b) (9 + 9 5 )5

(c) 81 + 9 5

(d) 81 - 9 5

INTEGER-ANSWER TYPE QUESTIONS x2

+

y2

abscissa of Q abscissa of R (b) 4/5 (c) 3/4

(d) 3/5

any point on the circle C1 to the circle C2 is (a) 12/5 (b) 24/5 (c) 28/5 (d) 32/5 Paragraph for Question Nos. 27 to 30 If the lines through P(a, 2) meet the ellipse at A and D PC, PD

B and C

x

a x1 the determinant x2 x3

2

y

b2

y1 y2 y3

2 9 5 or 9

= 1 are concurrent, then x1 y1 x2 y2 x3 y3

x2 + y 2 = 1 and the ellipse 2 x2 y 2 + = 1 where a2 = b2 – 8b + 13 intersect in 5 a2 four distinct points, then number of integral value of b is … x2 y 2 + =1 e be the eccentricity of the ellipse 16 b 2 where 0 < b < p1, p2 be the lengths of perpendicular from (0, 4e) and (0, –4e) to any tangent to

x2 y 2 + =1 9 4 PA, PB,

6 b and C be

E at the circle x2 + y2 = a2 P(a cos q, b sin q) and to C at Q(a cos q, a sin q) -p p f is angle between BC, then A x2 a2

+

y2 b2

= 1 whose foci are F1

and F2 and let P tion of F1 in the tangent at P be Q Statement 1: Q Statement 2: F2, P and Q Statement 1: A parallelogram inscribed in the elx2 y 2 lipse 2 + 2 a b Statement 2: Every chord passing through the cenE DPQR be a right triangle with right angle at P P lies on director circle of E Statement 1: PQ and PR are tangents to E Statement 2: E lies inside the triangle PQR E be the ellipse

x2 2

+

y2 2

3y

2

=1

(c)

1 2 1 2 x + y =1 2 3

(a)

1 2

ASSERTION-REASON TYPE QUESTIONS

E be the ellipse

• < t < •)

(c)

3

(b) 4x2 + 3y2 = 1 1 2 1 2 x + y =1 4 3 C is 1 (b) 3 (d)

(d) 1

Ê 1 ˆ Ê 1 ˆ (a) Á , 0˜ , Á , 0˜ Ë2 3 ¯ Ë 2 3 ¯ Ê 1 ˆ Ê 1 ˆ (b) Á ,0 , , 0˜ Ë 3 ˜¯ ÁË 3 ¯ 1 ˆ Ê 1 ˆ Ê (c) Á 0, , Á 0, ˜ ˜ Ë 2 3¯ Ë 2 3¯ (d) (0, 1) (0, – 1) Paragraph for Question Nos. 27 to 29 Let, E1 and E2 be two ellipse having the same centre and such that foci of E1 lie on E2 and foci of E2 lie on E1 and E2 E1 inclined at an angle of q E1 be e1 and eccentricity of E2 be e2 ellipse is (a) parallelogram

(b) a rhombus

q (a) e12 + e22 (b) e–21 + e–22 (c) e–21 + e–22 – e–21 e–22 (d) e–21 e–22 p 2 q = , then 2 2 (e21 + e22 4 e1 e2 (a) – 1 (b) 0 2

= 1 and P(x1, y1),

a b where x1, y1 > 0 be a point on E

a x= in Q e

-

gent at P C be a circle through P and Q Statement 1: If PQ is diameter of C, then C passes through (ae Statement 2: passing through P and Q.

(c) 1

(d)

2

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.33

INTEGER-ANSWER TYPE QUESTIONS C be the centre of the ellipse E is 3x2 + 4y2 – 12x + 8y + k = 0 where k < A and B be two points on the ellipse such that –ACB p = , then (28 – k) ((CA)–2 + (CB)–2 2 e the eccentricity of the ellipse passing through A(1, – 1) and having foci at F1(–2, 3) and F2(5, 2), then 4e2 e be the eccentricity of the ellipse represented by x = 5 (2 cos q + 3 sin q), y = 4(2 sin q – 3 cos q),

x2 y 2 + = 1 whose 25 7 distance from the centre of the ellipse is 2 7 is … x2 y 2 + =1 coordinates as integers) on the ellipse 49 25 is … x2

+

y2

= 1 at a 2 b2 which the normal to the ellipse passes through at least one of the foci of the ellipse is …

325e2 13

then

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS x2 y 2 + = 1 and C be the circle 9 4 P and Q be the points (1, 2) and (2,

E be the ellipse x2 + y2 (a) (a) (c) (d)

Q lies inside C but outside E Q lies outside both C and E P lies inside both C and E P lies inside C but outside E

[1994]

(–2, 0) is 2/3 times its distance from the line x = –9/2 is (a) an ellipse (b) a parabola (c) a hyperbola (d) none of these [1994]

ellipse

x2 y 2 + = 1 and having its centre at (0, 3) is 16 9

(a) 4

(b) 3

(d) 7/2 [1995] c such that the straight line y = 4x + c touches the curve x2/4 + y2 = 1 is (a) 0 (b) 1 (c)

P = (x, y), F1 = (3, 0), F2 = (– 3, 0) and 16x2 + 25y2 = 400, then PF1 + PF2 (a) 8 (b) 6 (c) 10 (d) 12 [1998] gents at the end points of latus recta of the ellipse x2/9 + y2/5 = 1, is

tangents to the ellipse x2/2 + y2 = 1 and which are (a) x2/2 + y2/4 = 1 (c) 1/3x2 + 1/4y2 = 1

(b) x2/4 + y2/2 = 1 (d) 1/2x2 + 1/4y2 = 1 [2004] x2/a2

+ y2/b2 triangle cannot be less than (a) (1/2) (a2 + b2 (a) (1/3) (a2 + ab + b2 (c) (1/2) (a + b)2 (d) ab

12

A of the maB ellipse x2 + 9y2 = 9

IIT JEE eBooks: www.crackjee.xyz 18.34 Comprehensive Mathematics—JEE Advanced

M area of the triangle with vertices at A, M and the origin O is (a)

31 10

(b)

21 10

(d)

29 10

27 [2009] 10 P on the ellipse x2 + 4y2 = 16 meets the x Q M is the mid point of the line segment PQ, then the locus of M intersects the latus rectums of the given ellipse at the points (c)

Ê 3 5 2ˆ (a) Á ± ,± ˜ 7¯ Ë 2

Ê 3 5 19 ˆ (b) Á ± ,± 4 ˜¯ Ë 2

1ˆ Ê (c) Á ±2 3 , ± ˜ Ë 7¯

Ê 4 3ˆ (d) Á ±2 3, ± 7 ˜¯ Ë [2009]

E 1:

x2 y 2 + = 1 is inscribed in a rectan9 4

gle R Another ellipse E2 passing through the point (0, 4) circumscribes the rectangle R the ellipse E2 is 2 2 1 (c) 2

3 2 3 (d) 4

(a)

(b)

[2012]

2

x + 9y = 1, the points at which the tangents are parallel to the line 8x = 9y are Ê 2 1ˆ (a) Á , ˜ Ë 5 5¯

Ê -2 1 ˆ (b) Á , ˜ Ë 5 5¯

Ê -2 -1ˆ (c) Á , ˜ Ë 5 5¯

Ê 2 -1ˆ (d) Á , ˜ Ë5 5 ¯

[1999]

P(x1, y1), Q(x2, y2), y1 < 0, y2 < 0, be the end points of the latus rectum of the ellipse x2 + 4y2 PQ are (a) x 2 + 2 3 y = 3 + 3 (b) x 2 - 2 3 y = 3 + 3 (c) x 2 + 2 3 y = 3 - 3

(a) e21 + e22 =

43 40

(c) |e21 – e22| =

5 8

(b) e1 + e2 = (d) e1e2 =

7 2 10

3 [2015] 4

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 1 to 3 Tangents are drawn from the points P(3, 4) to the ellipse x2 y 2 + = 1 touching the ellipse at points A and B 9 4 A and B are (a) (3, 0) and (0, 2) Ê 8 2 161 ˆ Ê 9 8ˆ (b) Á - , and Á - , ˜ ˜ Ë 5 5¯ Ë 5 15 ¯ Ê 8 2 161 ˆ (c) Á - , ˜ and (0, 2) Ë 5 15 ¯

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 2

(d) x 2 - 2 3 y = 3 - 3 [2008] E1 and E2 be two ellipses whose centres are at E1 and E2 lie along the x y S be the circle x2 + (y – 1)2 x+y=3 touches the curves S, E1 and E2 at P, Q and R, re2 2 e1 and PQ = PR = 3 e2 are the eccentricities of E1 and E2, respectively,

Ê 9 8ˆ (d) (3, 0) and Á - , ˜ Ë 5 5¯ PAB is Ê 8ˆ (a) Á 5, ˜ Ë 7¯

Ê 7 25 ˆ (b) Á , ˜ Ë5 8 ¯

Ê 11 8 ˆ (c) Á , ˜ Ë 5 5¯

Ê 8 7ˆ (d) Á , ˜ Ë 25 5 ¯

tances from the point P and the line AB (a) 9x2 + y2 – 6xy – 54x – 62y + 241 = 0 (a) x2 + 9y2 + 6xy – 54x + 62y – 241 = 0 (c) 9x2 + 9y2 – 6xy – 54x – 62y – 241 = 0 (d) x2 + y2 – 2xy + 27x + 31y – 120 = 0 [2010]

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.35

Paragraph for Question Nos. 4 and 5

PF1 F2

Let F1 (x1, 0) and F2 (x2, 0) for x1 < 0, be the foci of the x2 y 2 + ellipse 9 8 at the origin and Focus at F2 intersects the ellipse at a point M N in the

OB foci, and the angle FBF¢

Ê 9 ˆ (a) Á - , 0˜ Ë 10 ¯

F1 MN is Ê2 ˆ (b) Á , 0˜ Ë3 ¯

Ê9 ˆ (c) Á , 0˜ Ë 10 ¯

Ê2 ˆ (d) Á , 6 ˜ Ë3 ¯

INTEGER-ANSWER TYPE QUESTIONS h, 0) in2

x y + = 1 at the points P and 4 3 Q P and Q meet at the point R D(h) = area of the triangle PQR, D1 = max D(h) and D2 = min D(h), then

tersects the ellipse

1/ 2£ h £1

1/ 2£ h £1

=

8 5

D1 – 8D2

[2013]

x2 y 2 + = 1 are 9 5 (f1, 0) and (f2, 0) where f1 > 0 and f2 < P1 and P2 and with foci at (f1, 0) and (2f2 T1 be a tangent to P1 which passes through (2f2, 0) and T2 be a tangent to P2 which passes through (f1, m1 is the slope of T1 and m2 is the slope of Ê 1 ˆ T2, then the value of Á 2 + m22 ˜ Ë m1 ¯

FILL

IN THE

BLANKS TYPE QUESTIONS

P be the variable point on the ellipse 1, with foci F1 and F2

F, F¢ are its

SUBJECTIVE-TYPE QUESTIONS d be the perpendicular distance from the centre x2 y 2 of the ellipse 2 + 2 = 1 to the tangent drawn at a a b point P F1 and F2 are the two foci of the ellipse then show that

M and N meet at R and the normal to the parabola at M meets the x at Q, then the ratio of area of triangle MQR to area MF1 NF2 is (a) 3 : 4 (b) 4 : 5 (c) 5 : 8 (d) 2 : 3 [2016]

2

A

x2

+

y2

= a 2 b2 A is the area of the triangle

Ê b2 ˆ (PF1 – PF2)2 = 4a 2 Á1 - 2 ˜ Ë d ¯ x2 + 4y2 = 4 meets the ellipse x + 2y = 6 at P and Q P and Q of the ellipse x2 + 2y2 [1997] ABC circle x2 + y2 = a2 A, 2 2 x y + 2 = 1, (a B, C 2 a b > b) meet the ellipse respectively at P, Q, R so that P, Q, R A, B, C drawn at the points P, Q and R [2000] 2

2

P be a point on the ellipse

x2 a2

+

y2 b2

= 1, 0 < b
0, 4 – 2a > 0



a < 5, a < 2 fi a < 2

IIT JEE eBooks: www.crackjee.xyz 18.38 Comprehensive Mathematics—JEE Advanced

m1 = slope of OP1 = m2 = slope of OP2 =

y

b sin q1 b = tan q1 and a cos q1 a

b tan q2 a



P(a cos q, b sin q)

O

x

S

Êb ˆÊb ˆ And m1m2 = Á tan q1 ˜ Á tan q 2 ˜ = – 1 Ëa ¯Ëa ¯ fi

–P1 OP2 =

p 2

Fig. 18.26

We have PS¢ + PS + SS¢ = 2a(1 + e)

P and Q are

If (h, k) is the incentre of DPSS¢, Then

ax sec q – by cosec q = a2 – b2 ax sec f – by cosec f = a2 – b2

p Suppose these intersect at (h, k) and f = - q , we 2 get

h= =

(2ae)(a cos q ) + ae[a(1 + e cos q ) - a(1 - e cos q )] 2a(1 + e)

=

2a 2 e cos q (1 + e) = ae cos q 2a(1 + e)

ah sec q – bk cosec q = ah cosec q – bk sec q fi

(ah + bk) (sec q – cosec q) = 0



ah + bk = 0 ax + by x2 + 4y2 = 4 as

1 2 x + y2 = 1 4

k= (i)

( SS ¢ )(a cos q ) + ( PS ¢ )(ae) + ( PS )(- ae) 2a(1 + e)

(2ae)(b sin q ) be sinq = 1+ e 2a (1 + e)

Thus, (h, k) lies on 2

1 x cos q + y sin q = 1 2 Suppose (ii) meets x2 + 2y2 = 6 in P and Q

(1 + e)2 2 Ê xˆ y =1 + ÁË ˜¯ ae b2 e2

(ii)

Let e1 be the eccentricity of the ellipse then

Let the tangents at P and Q to x2 + 2y2 = 6 meet at R(h, k R is 1 1 xh + yk = 1 6 3



(cos q ) sinq 2 = =1 k h 6 3 h = 3 cos q, k = 3 sin q



h2 + k2 = 9 2

(iii)

(1 + e)

2



1 – e21 =



e 21 =

= a2e2(1 – e21) b2

1

(1 + e) a 2

2

2e 1+ e

P is (a cos q, b sin q Coordinates of S¢ are (–ae, 0) and that of S are (ae, 0) We have PS¢ = a(1 + e cos q), PS = a(1 – e cos q) SS¢ = 2ae

=

1 - e2 (1 + e)

2

=

1- e 1+ e

x ± y = ± c, then

c2 = 25(±1)2 + 16 = 41

x±y± x2 it passes through (4, – 1)

2

x +y =9

Also,

b2 e2

16 a

2

+

1 b

2

a2

+

41 = 0

y2 b2

= 1 or a2 + 16b2 = a2b2

Since x + 4y – 10 = 0 touches the ellipse 2

2

Ê 10 ˆ 2 Ê 1ˆ 2 2 2 ÁË ˜¯ = a ÁË - ˜¯ + b fi a + 16b = 100 4 4

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.39

fi a 2b 2

| p|

Also, d = a, b are natural numbers such that a + b = ab then a = 2, b = 2 x2 + y2 = 1 4 2

T =



d2

252 + 162 p

2

=

881 41

a = 0, the curve becomes 5(x2 – 4x + 4) + 4y = 0

2 Ê x2 ˆÊ4 ˆ Ê 2x ˆ + + y 2 - 1˜ Á + 4 - 1˜ = 2 1 y ÁË ˜¯ Á ¯ 4 Ë 4 ¯Ë4

Which represents the parabola (x – 2)2

(x + 4y – 2)2 = (x2 + 4y2 – 4) (4)



x2 + 16y2 + 4 + 8xy – 4x – 16y = 4x2 + 16y2 – 16



3x2 – 4x(2y – 1) + 16y – 20 = 0



x=

1 1 [2(2y – 1) ± 4(y – 2)] = 2, (8y – 10) 3 3 x = 2, 3x – 8y + 10 = 0 2

As a < If a > 0, a π 5, the curve becomes 2

2ˆ 4 Ê 5(x – 2)2 + a Á y - ˜ = 2 Ë a¯ a

2

e sec q = 1 fi e sec q = 1 2

2

2ˆ 4 Ê 5(x – 2)2 + a Á y - ˜ = 2 Ë a¯ a

q = sec q (1 – e ) 2

y

If a < 0, the curve becomes





a2 + b2 =

p2

Also, number of values of p

SS1)



2

fi2=

2

fi e = cos q which decreases as q increases (0 < q < p/2)

As a > 0, a π If a 3 -0 2 = 2 3-0

q=l Also, el

through P (a) Slope of the diameter conjugate to the diameter through P

3(x2 – 2x + 1) + 4(y2 + 2y + 1) = 7 – k or

3(x – 1)2 + 4(y + 1)2 = 7 – k

or k and ellipse if 7 – k > 0 or k
7

x + y = p is a tangent at P(5 cos q, 4 sin q) to E P to E is x y cos q + sin q = 1 5 4 As (i) and (ii) are the same line cosq sinq 5 = 4 = 1 1 p 1 fi

cos q =



1=

5 4 , sin q = p p

25 + 16 p2

fi p2 = 41

3 4

= -

(mm¢ = -

3 3 4 (c) Slope of the normal through P = -

= (ii)

9 4 ¥ 16 3

16 3/ 2 4 = ¥ 9 2 3 3 3

9ˆ Ê L Á - 7, - ˜ Ë 4¯

PL

3 9 + 3 = 2 4 = (2 3 - 7 ) 4 2 3+ 7 x y cos q + sin q a b b(e - cos q ) ˆ Ê x = a/e at T Á a /e, ˜ Ë e sin q ¯

S (ae

b2 a2

)

IIT JEE eBooks: www.crackjee.xyz 18.40 Comprehensive Mathematics—JEE Advanced

Slope of SP = Slope of ST =

b sin q a (cos q - e)

Statement-2 is false as the point (2, 5) lies outside the ellipse C1 is (5 cos q, 5 sin q) C 1 : x2 + y2

b(e - sin q ) a sin q (1 - e2 )

Product of the slopes = –

b2 a 2 (1 - e2 )

25 cos 2 q 25 sin 2 q 3 3 + - 1 = sinq £ 25 16 4 4

l=

=–1

fi statement-2 is true using which statement-1 is L(ae, b2/a) = (a cos q,

Q (4 cos q, 4 sin q

C2 : x2 + y2

R lies on the line y = 4 sin q coordinate of R are (5 cos q, 4 sin q)

ax by = a2e2 which meets the ma2 e 1- e jor y = 0 at x = ae3 ax by Normal at (a cos q, b sin q) is = a2 – b2 cos q sin q = a2 e2

abcissa of Q 5 cos q 5 = = abcissa of R 4 cos q 4

b sin q) is

which passes through (±ae, 0) if cos q = ± 1/e which is not possible as e < 1 fi

q

O

x2 y 2 + =1 74 26 x2 +

Fig. 18.27

SQ

y2 = 74 + 26 = (10)2

P(5 cos q, 5 sin q)

As the point (2 7 , 6 2 )

QR is 5(x cos q + y sin q) = 16

-

(25 cos q + 25 sin = 2

(PS)2 Using it, statement-1 is true if y = 4x + 13 is a tangent to the ellipse as x2 + y2 Now (13)2 = 36(4) + 25 = 169 fi

2

2

(QS)2 = 9 -

(c2 = a2m2 + b2)

4h k 4h 2 + k 2 = = 8 3 12 h=

24 36 ,k= 25 25

2

=

81 25

81 12 fi QS = 25 5 Q S R

hx ky h k + + = 9 36 9 36 or 4hx + ky = 4h + k Comparing (1) and (2), we get

2

)

(PQ)2 = 25 – 16 = 9

P

2

q - 16

25 cos q + 25 sin q

(1) 2

2

2

x2 y 2 + =1 9 36

6x 9 y is =1 + 9 36 or 8x + 3y = 12 Let (h, k



R

Q

C2 C1

Fig. 18.28

(2)

P(a, 2) be x = a + r cos q, y = 2 + r sin q This line meets the ellipse fi

x2 y 2 + = 1 at A and D 9 4

4(a + r cos q)2 + 9(2 + r sin q)2

This gives two values of r, r1 = PA, r2 = PD

IIT JEE eBooks: www.crackjee.xyz Ellipse 18.41

So

PA ◊ PD = r1 r2 =

So

PB ◊ PC = -

4a 2 4 cos 2 q + 9 sin 2 q B and C

a -2 2a ¥ = sin q cos q cos q sin q

Since PA, PB, PC, PD

4a

2

=

4 cos q + 9 sin q 2



2



2a sin q cos q

-1
6

So that

4a 2 + 25

1. of the hyperbola. of the hyperbola. e, is called the eccentricity of the hyperbola.

2

=

2

referred to its principal x2 y 2 - 2 = 1 where 2 a

(e2 y

b

x = a/e L B (0, b) M

(-ae, 0) (-a, 0) S¢

O (0, 0)



P A (a, 0)

B¢ (0, -b)



x = - a/e

In the figure A¢A is the

of the hyperbola along the and is of length 2 . A( A1(– the vertices of the hyperbola. ¢ is the of the hyperbola along the y and is of length 2 . O e=

a 2 + b2

is the eccentricity of the hyperbola. a2 From symmetry we observe that if S ¢(–ae, 0) be taken as the focus and x = -

hyperbola is described. So S¢(– S( bola. a a are the two directrices of the = - and = e e hyperbola. = and hyperbola.

Hyperbola (e >1)

Ê

points Á ae,±

Fig. 19.1

Ë

Let P( , y

a is taken as the directrix, same e

=–

are the two latera recta of the

=

meets the hyperbola at the

b2 ˆ so length of each latus rectum is a ˜¯

2

2 / .

a = e

S(

Illustration 1 2

( – fi

x2 a2

-

aˆ Ê + y2 = e2 Á x - ˜ Ë e¯

2

of the latera recta of the hyperbola

y2 a 2 (e2 - 1) 2

x2 y 2 = 1. 16 9

=1

Since e > 1, e2 – 1 > 0 so 2

Find the eccentricity, length of a latus rectum, equations

x y equation is 2 - 2 = 1. a b

2

(e2

2

Solution Let the equation of the hyperbola be and the required

= 1, where a2 = 16, b2 = 9.

x2 y 2 a 2 b2

IIT JEE eBooks: www.crackjee.xyz 19.2

fi e2 =

16 + 9 5 fie= . 16 4

Lengths of a latus rectum =

2b 2 2¥9 9 = = . a 2 4

Equations of the latera recta are x = ± ae fi x = ± 4¥

5 fi x = ± 5. 4

P3. An equation of the normal at the point ‘q’ is = a2 + b2 and at (x¢, y¢) is

Illustration 2 Find the centre, foci and the eccentricity of the hyperbola. 11x2 – 25y2 – 44x + 50y – 256 = 0. Solution: The equation of the hyperbola can be written as 11(x – 2)2 – 25(y – 1)2 = 275 fi

( x - 2) 2 ( y - 1) 2 =1 25 11

2

2

25 + 11 25

6 . 5

P4. The condition that the line y = mx + c is a tangent to the hyperbola is c2 = a2m2 – b2. So equation of any tangent to the hyperbola (not parallel to y-axis) can be written as y = mx ± a 2m2 - b2

x2 y 2 =1 25 4

Solution: Equation of the hyperbola is y = mx + 1 is a tangent to the hyperbola fi (1)2 = 25m2 – 4 fi 5m2 = 1 fi 25m4 + 5m2 + 1 = 3. Illustration 5

Find the equation of the normal to the hyperbola 4x2 – 9y2 = 144 at the point whose eccentric angle q is p /3.

So the coordinates of the foci are (X = ± ae, Y = 0) = (x – 2 = ± 6, y – 1 = 0) = (8, 1) and (– 4, 1).

Solution: Equation of the hyperbola is

x2 y 2 =1 36 16

Coordinates of the point whose eccentric angle is p/3 are (6 sec (p/3), 4 tan (p/3)) = (12, 4 3 )

Illustration 3 e1 is the eccentricity of the ellipse

x2 y 2 + = 1 and e2 25 b 2

is the eccentricity of the hyperbola

x2 y 2 = 1 such 16 b 2

that e1e2 = 1. Find the value of b. Solution: e1 =

25 25

2

and e2 =

16 + 16

2

e1e2 = 1 fi (25 – b2) (16 + b2) = 25 ¥ 16. fi 9b2 – b4 = 0 fi b2 = 9 fi b = ± 3. Some Properties and Standard Results for the 2

Hyperbola

x - x¢ y - y ¢ + =0 x¢ / a 2 y ¢ / b 2

If y = mx + 1 is a tangent to the hyperbola 4x2 – 25y2 = 100, find the value of 25m4 + 5m2 + 1.

where X = x – 2, Y = y – 1, a = 25, b = 11. So centre of the hyperbola is (X = 0, Y = 0) = (2, 1) eccentricity e =

ax by + secq tan q

Illustration 4

X2 Y2 which can be written as 2 - 2 = 1 b a

fie=

P1. The parametric equation of the hyperbola or the coordinates of any point on the hyperbola are x = a secq, y = b tan q. The point is denoted by ‘q’. q is called the eccentric angle of the point. P2. An equation of the tangent at the above point ‘q’ is x y x x ¢ y y¢ sec q - tan q = 1 and at (x¢, y¢) is 2 - 2 = 1. a b a b

2

x y - 2 =1 2 a b

6x

4y

+ The required equation of the normal is = 36 2 3 + 16

fi 3 3 x + 4y = 52 3 . P5. An equation of the of the point ( ¢ , y ¢ joining the points of contact of the tangents drawn xx ¢ yy¢ =1 from ( ¢, y ¢ a 2 b2 P6. An equation of the chord of the hyperbola whose mid-point is ( ¢, ¢ T = ¢, where ∫

xx ¢ a2

-

yy¢ b2

– 1 and S ¢ ∫

x¢2 y¢2 - 2 – 1. a2 b

IIT JEE eBooks: www.crackjee.xyz Hyperbola 19.3

Equation of the hyperbola can be written as

P7. An equation of pair of tangents drawn from a point P( ¢, ¢ ¢ = 2, where. S∫

2

x2 y2 =1 2a 2 2b 2

2

x y - 2 –1 2 a b

Since this represents the director circle of the given circle. Subtracting (2) from (1) we get

Illustration 6

1 ˆ 2 Ê 1 ÁË 2 + 2 ˜¯ y = 0 fi y = 0 2a 2b

Find the equation of the chord of contact of the point (5, 1) to the hyperbola, x2 – 4y2 = 16. Also find the mid-point of this chord. x2 y 2 =1 Solution: Equation of the hyperbola is 16 4 5 x y(1) Equation of the chord of contact of (5, 1) is = 16 4

(

)

So the required points are ± 2 , 0 . Illustration 8 If y = mx + 2 5 is a tangent to the hyperbola 4x2 – y2 = 16. Find the equation of a diameter of the hyperbola bisecting the chords parallel of the tangent. Solution: Equation of the hyperbola can be written as

fi 5x – 4y = 16. (1) If (a, b) is the mid-point of this chord then its equation is ax by a2 b2 –1= –1 (2) 16 4 16 4 From (1) and (2) we get a=

(2)

x2 y 2 =1 4 16

y = mx + 2 5 touches the hyperbola.

80 16 ,b= . 21 21

fi (2 5 )2 = 4m2 – 16 fi m2 = 9 fi m = ± 3

So that coordinate of the mid-point of the chord are

So the required equation of a diameter is

Ê 80 16 ˆ ÁË , ˜¯ . 21 21

y=

P8.

of a hyperbola is the locus of the point of intersection of the tangents to the hyperbola which intersect at right angles and its equation is 2 + y 2 = 2 – 2. P9. A of a hyperbola is the locus of mid-points of a system of parallel chords of the hyperbola and b2 where is the slope of the its equation is y = 2

P10.

16 x fi 4x ± 3y = 0. 4(±3)

If the length of the perpendicular let fall from a point on a hyperbola. to a straight line tends to zero as the point on the hyperbola moves to called an

of the hyperbola. y y = -ab x

a m

parallel chords of the hyperbola which are bisected

y = ab x

A¢(-a, y)

by it. Two diameters of a hyperbola are said to be when each bisects the chords parallel to the other. Thus two diameters y = and y = ¢ of the b2 hyperbola are conjugate if ¢= 2 .

A(a, 0)

a

Fig. 19.2

Illustration 7 2

Find the points on the hyperbola

2

x y = 2 from which a 2 b2

two perpendicular tangents can be drawn to the circle x2 + y2 = a2. Solution: Director circle of x2 + y2 = a2 is x2 + y2 = 2a2 or

x2 y2 + 2 =1 2 2a 2a

x

0

(1)

(i) Equation of the asymptotes is i.e.

x2 a2

-

y2 b2

=0

x y ± =0 a b

Ê bˆ (ii) Angle between two asymptotes is 2 tan–1 Á ˜ . Ë a¯ Asymptotes are at right angle if and only if a = b.

IIT JEE eBooks: www.crackjee.xyz 19.4

(iii) If angle between two asymptotes is 2q, then eccentricity of the hyperbola is sec q. P11. A hyperbola in which the equal is called a rectangular hyperbola. (i) Equation of a rectangular hyperbola is x2 – y2 = a2. (ii) Eccentricity of the rectangular hyperbola is 2 . (iii) Asymptotes of a rectangular hyperbola are at right angles. (iv) Rotation of the system of coordinates through an angle p/4 in the clockwise direction (i.e. the axes are taken along the asymptotes of the hyperbola) gives another form of the equation of the rectangular hyperbola i.e. xy = c2. (v) Some facts about rectangular hyperbola xy = c2 (a) Vertices (c, c) and (– c, – c) (b) Foci ( 2 , 2 ) and (- 2 , - 2 ) (c) Directrices x + y = ±

2c

(d) length of the latus rectum = 2 2 c (e) parametric equation: x = ct, y = c/t. (i) Equation of tangent at this point is x + y2 = 2ct (ii) Equation of normal at this point is x + t3 – yt + c – ct4 = 0 Illustration 9

(c) The foci of a hyperbola and its conjugate are concentric and form the vertices of a square. P13. Two hyperbolas are said to be similar if they have the same eccentricity. P14. two hyperbolas are equal if they are similar and have the same latus rectum. 2 2 / – y 2/ 2 P15. =1 (i) If PN be the ordinate of a point P on the hyperbola and the tangent at P meets the tranverse axis in T, then ON. OT = a2, O being the origin. (ii) If PM be drawn perpendiculars to the conjugate axis from a point P on the hyperbola and the tangent at P meets the conjugate axis in T, then OM. OT = –b2; O, being the origin. (iii) If the normal at P on the hyperbola meets the transverse axis in G, then SG = eSP; S being a foci and e the eccentricity of the hyperbola. (iv) The tangent and normal at any point of a hyperbola bisect the angle between the focal radii to that point. (v) The locus of the feet of the perpendiculars from the foci on a tangent to a hyperbola is the auxillary circle. (vi) The product of the perpendiculars from the foci on any tangent to a hyperbola is constant and equal to b2.

If the circle x2 + y2 = a2 intersects the hyperbola xy = 25 in four points, then find the product of the ordinates of these points. Solution: The ordinates of the points of intersection of x2 + y2 = a2 and the hyperbola xy = 25 are given by Ê 25 ˆ ÁË y ˜¯

2

+ y2 = a2 or y4 – a2y2 + (25)2 = 0

SOLVED EXAMPLES

SINGLE CORRECT ANSWER TYPE QUESTIONS

2

If y1, y2, y3, y4 are the roots of this equation then y1 y2 y3 y4 = (25)2 = 625. P12. Two hyperbolas such that the

2

: other are called conjugate hyperbolas of each other. x2 y 2 (a) Equation of the hyperbola conjugate to 2 - 2 a b x2 y 2 = 1 is 2 - 2 = 1 b a (b) If e1 and e2 are the eccentricities of the hyperbola 1 1 and its conjugate, then 2 + 2 = 1. e1 e2

The equation of the hyperbola whose

Example 1

– 144y2 – 36y2

2 2

– 25y2 = 900 – 25y2 = 900

Let the equation of the hyperbola be

= 1, then 2 = 5 and 2 =

2



(e2

x2 y 2 =1 fi 25 36 25 4

2

a2

-

y2 b2

= 13

25 169 - 2 fi = 4 4 So the required equation of the hyperbola is 2

x2

– 144y2 = 900

2

= 36

IIT JEE eBooks: www.crackjee.xyz Hyperbola 19.5

So the required equation of the hyperbola is

Example 2

( x - 1) 2 ( y - 4) 2 =1 25 75 4 4

hyperbola is x2 y 2 =1 25 144



( x - 5)2 y 2 =1 25 144 x ( y - 5) 25 144 2

12

2

2

are the slopes of the tangents to

x2 y2 = 1 which pass through the point 144 25

the hyperbola =1

( x - 5)2 ( y - 5) 2 =1 25 144

1,

If

Example 4

2

– 4y2 – 24 + 32y – 127 = 0

1

+

2

=

42 13

1

2

=

11 52

1

+

2

=

51 52

1

2

=

42 13

of the vertices (10, 0) h

C

(0, 0)

A

S (18, 0)

(5, 0)

Fig. 19.3

If e is the eccentricity, then 8 13 fi e = 1+ = 5 5 2

=

2



fi =5 = 18 – 10 = 8

Ê 169 ˆ 25 Á - 1˜ Ë 25 ¯

(e2

2 2 2 2

– – – –

4y2 4y2 4y2 4y2

+ – + –

24 24 24 24

– + + +

32y 32y 32y 32y

– 127 = 0 – 127 = 0 + 127 = 0 + 127 = 0

If 2 = 6–1=5fi and

=

51 52

25 3 x2 y 2 is a normal to the hyperbola = 1 is 16 9 3 -

3

2

-

3

3 2

4x 3y + = 16 + 9 = 25 sec q tan q secq - tanq -25 3 = = 4 3 3 ¥ 25 -4

fi sec q =

3

(e

2

5 5 = = e 2

25 75 25 = 4 4

, tan q =

3 fiq=

4 p q= , sec q = 2 = 3 3 Example 6

2

=

The equation of the hyperbola whose foci

Example 3

2

2

The value of , for which the line

Example 5 y = mx +

= 144 = 12 and the required equation of the hyperbola ( x - 5)2 y 2 is =1 25 144



y–6= ( or y= + (6 – 14 which touches the given hyperbola if (6 – 14 2 = 144 2 – 25 fi 52 2 – 168 + 51 = 0 168 42 = , 1 fi 1 + 2 = 52 13

tan f 2

2

Let P( sec q,

fi tan q

p 4p , 3 3 = -

2 3

Q( sec f,

q + f = p/2, be two points on the hyperbola 2

2

/ – y / = 1. If (h, k normals at P and Q, then k is equal to

IIT JEE eBooks: www.crackjee.xyz 19.6

a 2 + b2 a

È a 2 + b2 ˘ -Í ˙ Î a ˚

a 2 + b2 b

È a 2 + b2 ˘ -Í ˙ Î b ˚

SP = e (distance of P from = / 2

q + cosec q = ( + and the equation of the normal at Q ( sec f, sec f f + cosec f = ( 2 + 2 2

Similarly S¢P =

2 | x + (a / 2 )|

x2

È a 2 + b2 ˘ a 2 + b 2 secq - cosecq = = -Í ˙ b cosecq - secq Î b ˚ 2

2

1 < e < 2/ 3

– y = , C is its centre and S, S¢ are the two foci, then SP. S¢P = CS 2 SS¢ 2 CP 2

Let the coordinates of P be (a, b

:

fi fi

a = ± 3b

Also since (a, b a2

2

a

2

2

b2

2



Example 9 S (a 2, 0)

b2 b2

-

b2 b

2

=1 =1fi

3 a

2

-

1 b

2

=

1 b2

>0

1 . 3 a 1 2 4 e2 - 1 > fi e2 > fi e > . 3 3 3



Y

-

3b 2

fi =

a2 + b2

a2 + b2 = 4b2 fi a2 = 3b2

: Let the coordinates of P be ( , y The coordinates of the centre C

2

e > 2/ 3

3/2

e=

a2

The eccentricity of the hyperbola is 1 +

e = 2/ 3

Since OPQ is an equilateras triangle OP = PQ

2

C (0, 0)

2

y2

Then PQ = 2b and OP =

If P is a point on the rectangular hyperbola

S¢ (- a 2, 0)

– y2 =

= 1 such that OPQ is an equilateral triangle, O a 2 b2 being the centre of the hyperbola. Then the eccentricity e

a 2 + b 2 sec q - sec(p / 2 - q ) So that k = y = cosecq - cosec(p /2 - q ) b [ q + f = p/2]

Example 7

-

2

If PQ is a double ordinate of the hyperbola

Example 8

a 2 + b2 sec q - sec f ◊ cosecq - cosecf b

=

2 | x - (a / 2 )|

2 So that SP. S¢.P = 2 | 2 – ( 2 – 2 2 2 2 = + y = (CP ( P lies on the hyperbola

P is 2

=

>

Let and be non-zero real numbers.

Then the equation ( represents

2

2

+

+

2

–5

+ 6y2

= 0 and ,

are

of the same sign. x = - a / 2, 0)

x = a / 2, 0)

=

Fig. 19.4

and

and

So the coordinates of the foci are S (a 2 , 0) and

and

are of opposite sign

S ¢ (- a 2 , 0) . = and S = - / 2 .

are of the same sign.

same sign and

/ 2 : 2

and are of the is of sign opposite to that of .

Given equation represents + 2 + = 0, – 2y = 0,

– 3y = 0

IIT JEE eBooks: www.crackjee.xyz Hyperbola 19.7

which is of the form (shifting the origin to ( 2 , - 2 )

Note When = 0, straight line as If = Then

2

+ and

2

= 0 do not represent a pair of are of same sign.

X2 a2

-

Y2

= y+ 2,

and and 2 + 2+

are of same sign = 0 fi 2 + y2 = – / which does not c represent a circle as - < 0 a If and are of opposite signs, then c c c c If - > 0, - < 0 and - > 0, - < 0 a b b a +

2

If

hyperbola is (SS1 = T2 .1 – y ( 2 – y2 fi –8 2 + 8y2 + 72 = 2 – 18 + 81 fi 9 2 – 8y2 – 18 + 9 = 0.

2

Example 11 – 2y2 – 2 2 x - 4 2 y A. Let B be one of the end points of its latus rectum. If C is 2

the focus of the hyperbola nearest to the point A, the area of the triangle ABC is 1- 2 / 3

3 / 2 -1

1+ 2 / 3

3 / 2 +1

C (ae, 0)

b2 1 1 (ae - a) ¥ (AC BC 2 a 2 1Ê 3 ˆ - 1 ¥ 2 = 3/ 2 - 1 = Á 2 Ë 2 ˜¯

Area of DABC =

Example 12

The tangent at any point P( secq, tanq

of the hyperbola 2/ 2 – y2/ 2 = 1 makes an intercept of length p of the hyperbola. p1, p2 are the lengths of the perpendiculars drawn from the foci on the normal at P, then p is an arithmetic mean between p1 and p2 p is a geometric mean between p1 and p2 p is a harmonic mean between p1 and p2

: Let hyperbola.

(

(– P are

x sec q y tan q a x b2 y + = 1 and = a sec q b tan q a b Q( cosq and the 2

p2 P p

be written as



( x - 2 )2 ( y + 2 )2 =1 4 2

1

G Q F (ae, 0) 1

p

( x 2 - 2 2 x + 2) - 2( y 2 + 2 2 y + 2) = 6 + 2 – 4

3 . 2

Fig. 19.5

= 9 is the chord of contact of the

: Let (h, k w.r.t the hyperbola 2 – y2 = 9 is = 9 fi – ky = 9 and = 9 represent the same line fi h = 1, k = 0

=2

2 /a)

A (a, 0)

hyperbola 2 – y2 = 9, then the equation of the corresponding pair of tangents is 2 – 8y2 + 18 – 9 = 0 2 – 8y2 – 18 + 9 = 0 2 – 8y2 – 18 – 9 = 0 2 – 8y2 + 18 + 9 = 0

2

= 4,

B (ae, b

=– fi

Example 10

2

4+2 = 4

x2 y2 + =1 - c /a - c /b Which represents a hyperbola 2

- 2

= 1 where X =

b2

(– ae, 0) F2

Fig. 19.6

2 2

e

P

IIT JEE eBooks: www.crackjee.xyz 19.8

G(

2

P meets

secq

p QG e a sec q - a cos q = = p1 F1G e2 a sec q - ae 2

DGPQ,

e2 - cos 2 q

cosq e + cosq = 2 = = 1+ e e e - e cos q Similarly

: We have

D s ( s - b) =k D s ( s - c)



cosq p = 1p2 e

A

p p 1 1 2 + + =2fi = . p1 p2 p p1 p2



Example 13

Let

> 0 and A(–

c

:

b

B( B

P moves such that base angles of the triangle PAB be such then –PAB = 2–PBA, then P traces



tan( B / 2) =k tan(C / 2)

Suppose coordinates of P be (h, k PA = tan (2q k = tan (2q h+a

a

C

Fig. 19.8



s-c =k s-b



( s - c ) - ( s - b) k -1 = ( s - c ) + ( s - b) k +1



b-c k -1 = a k +1 Ê k - 1ˆ a – AB = Á Ë k + 1˜¯

P (h, k)

fi A traces a hyperbola. =

Example 15 2q

,

q

A (-a, 0)

B (a, 0) Fig. 19.7

Example 14

0£q £

Suppose base BC of a triangle is of A moves such that the ratio

tan( B / 2) is a constant k π 1, then A traces tan(C / 2)

sinq, y =

sinq –

cos q,

> 0, represents a rectangular hyperbola if 3p p p 0.

if

3p p < 2q < 2 2

or

3p p a 3 4 Thus, p( for Also, P¢

2

x y =1 16 9 fi

e12 = 1 +



e1 =

b2 a

= 1+

2

25 9 = 16 16

5 4

When = 3, y 64 – 33 + 3 – 6 = 0 2 (2 fi 6( 2 2 2 – fi ( fi

x2 y 2 + =1 9 16 e22 = 1 -

Now,

9 7 = 16 16

Let H be the hyperbola

Example 35

2

– y2 = 36. Let

P(h, k point P normals at which pass through point P. Column 1 Column 2

H

When = 1, y fi 64 – fi 6( 4 fi (2 fi

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Column 1

Ê 1ˆ The parametric form of 2 – y2 = 36 is = 3 Á + ˜ , Ë ¯ Ê 1ˆ y = 3 Á - ˜ , π 0. Ë ¯ dy x Slope of tangent at ( , y = y dx

Ê 1ˆ -3 Á t - ˜ Ë t¯ 1 Ê ˆ y - 3Á t - ˜ = Ë t¯ Ê 1ˆ 3Á t + ˜ Ë t¯

È Ê 1ˆ ˘ Í x - 3 ÁË t + t ˜¯ ˙ Î ˚

64 – ( + y 3 + ( – y = 0, y 4 6 – 6 = 0 fi = ±1 = 2, y 4 – 33 + (



–6=0

Column 2 y–3=0 +y+1=0 +4=0 –y+7=0

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

.

Ê Ê 1ˆ Ê 1ˆ ˆ ÁË 3 ÁË + ˜¯ , 3 ÁË - ˜¯ ˜¯ is

When

–6=0 (2 2 –

3 + 4y

:

or When

+

Suppose equation of a hyperbola is

Example 36 .

3

( y Its asymptotes are + 4 = 0, y – 3 = 0. TIP bisectors of asymptotes of a rectangular hyperbola.

+4 fi

= ±

y-3

1 1 + y + 1 = 0 or

–y+7=0

IIT JEE eBooks: www.crackjee.xyz Hyperbola 19.15

which represents the parabola 2 = 4 where = y – 2, X = – 3 and 4 = 8. The coordinates of the focus are (X = , – 3 = 2, y

=–y (– y y 2 +4=0 fi (y It has no Solution. Thus, + y –y

Putting

2

x 2 y2 =1 4 21

which is a hyperbola with eccentricity e 2 =

+ y + 1 = 0. Example 37 Column 1

– 4y 2 = 84 fi

fie=

Column 2 2 /25 + y2/21 = 1

4 + 21 25 = 4 4

5 and the coordinates of a focus are ( 2

(

=1

Length of latus rectum.

Example 38

2

y – 4y – 8 + 28 = 0 2

2, 2 p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

.

Column 2 2 /9 + y2 /16 = 1 2 /16 – y 2 /25 = 1 2 – y2 = 4 y2 = 8 (

– 4y = 84

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

.

: 2

Column 1

2

2

/25 + y /21 = 1, coordinates of a focus are ( , 2 = 25, 25 - 21 4 e2 = = 25 25

2

/9 + y 2 along y

= 1 is the equation of a rectangular hyperbola,

X= and

p /4 we get x+y

y

2 y-x

y

= –

2

So that X – = 1/2 ◊ =2 which is a rectangular hyperbola whose focus is X = , = 0. 2+2 where 2 = 2, e2 = =2 2 2

2

fi fi

X= x+y 2

2¥ 2 ,

= 2 and

y-x 2 = 1 are

y 2– 4y – 8 + 28 = 0 2 = 8 – 24 = 8 ( fi (y

=0 =0fiy= =

(

2, 2

)

2

32 = 4.5 4 2 /16 – y 2/25 = 1 is an hyperbola with

rectum = 2 ×

y

.

So length of the latus rectum = 2 × 2

25 = 12.5 4

– y2 = 4 is a rectangular hyperbola with

to 2 each. So the length of the latus rectum = 2 ×

22 =4 2

y2 = 4 ( can be written as 2 = 4X where = y, X = –4 So the length of the latus rectum = 4

IIT JEE eBooks: www.crackjee.xyz 19.16

ASSERTION-REASON TYPE QUESTIONS Statement-1: If a hyperbola having the

Example 39

1 unit is confocal with the ellipse 2 3

2

+ 4y2 = 12, then its equation is

x2 y 2 1 = 1 15 16

Statement-2: x2 y 2 1 x2 y 2 1 hyperbola = is = 15 1 1 15 16 16

x2 y 2 + 4 3

So length of the latus rectum = 2 ¥

=

9 2

16 Thus statement-1 is true coordinates of a point on this hyperbola is + 4 = 4 secq, y + 1 = 3 tan q fi = 4 secq – 4, y = 3 tanq – 1 and the statement-2 is also true. Let H be the hyperbola 3

Example 41

= 1

9

2

– 4y2 = 12 and

C be the circle 2 + y2 = 1. Statement-1: Two perpendicular tangents can be drawn to Ê 5 - 1 5 + 1ˆ , the hyperbola from the point Á ˜. Ë 2 3 2 3 ¯ Statement-2: Two tangents drawn from any point on the

3 1 1 eccentricity is e2 = 1 - = fie= . 4 4 2 Hyperbola has same foci, if e1 is the eccentricity of the hyperbola then 2

1

= 2 and 2 =

: Let (h, k h2 + k2 to 3 2 – 4y2 = 12 is given by 2 (3 – 4ky – (3 2 – 4y2

1 2

or

2

2

=

= 9h2 – 3(3h2 – 4k2 = 12k2 + 36 y2 = 16k2 + 4 (3h2 – 4k2 = 12h2 – 48 2 y2 = 12k2 + 36 + 12h2

(e12

x2 y 2 =1 1 15 16 16

– 48

x2 y 2 1 = 1 15 16

Statement 2 is true.

= 12(k2 + h2

2

Thus statement-1 is true Statement-2 is false as the equation of the hyperbola x2 y 2 x2 y 2 conjugate to = 16 is = –16. 15 1 15 1 Statement-1: Length of the latus rectum

Example 40

of the hyperbola 9 2 – 16y2 + 72 – 32y – 16 = 0 is 9/2 Statement-2: Parametric coordinates of a point on the hyperbola 9 2 – 16y2 + 72 – 32y – 16 = 0 are (4 sec q – 4, 3 tan q

Ê 5 - 1ˆ Ê 5 + 1ˆ 2 Also, Á ˜ +Á ˜ = 12 (5 + 1) = 1 Ë 2 3 ¯ Ë 2 3 ¯ Ê 5 - 1 5 + 1ˆ , fi Á ˜ lies on the circle C. Ë 2 3 2 3 ¯ By the Statement-2, two tangents drawn from point Ê 5 - 1 5 + 1ˆ , Á ˜ to the hyperbola H are at right angles. Ë 2 3 2 3 ¯

fi fi

2

+8 2

y2 + 2y 2 – 16 (y = 144

( x + 4) 2 ( y + 1) 2 =1 16 9

Statement-1: If y =

Example 42 the hyperbola

9( 9(

h2 – 4k2

2

1 15 (16 - 1) = 16 16 So the equation of the hyperbola is

fi e1 = 4 and

C, so that h, k

2

+ is a tangent to

1 . 2 is a tangent to the hyperbola

– 4y2 = 16, then | |>

Statement-2: If y = + x2 y 2 = 1, then 2 = 2 a 2 b2

2



2

and π 0.

IIT JEE eBooks: www.crackjee.xyz Hyperbola 19.17

:

If y =

2

2

+ is a tangent to the hyperbola, then b = – . See theory. If = 0, then = ± ; in this a b case the lines y = ± x are asymptotes to the hyperbola. a Thus, Statement-2 is true. x2 y 2 = 1 if Thus, y = + is tangent to the hyperbola 16 4 2

2

1 . 2 Thus, both the statement are true and statement 2 is a 2

= 16

2

– 4 and π 0. Therefore, 16

2

– 4 > 0. fi | |>

fi a=±1 Thus, P lies on the pair of lines parallel to y

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 44 to 46 A point P moves such that the sum of the slopes of the normals drawn from it to the hyperbola = 16 is equal to the sum of the ordinates of fect of normals. The locus of P is a curve C. The equation of the curve C is

Example 44 Let H be the rectangular hyperbola

Example 43

Ê cˆ ÁË ct , ˜¯ on H is t

Ê ÁË 4

( , fi Statement-2 is false. Let P(a, b 1 from point P to H is 2

–a fi If 1, 2, 3, 4 4

then  =

3

a

2

=

2

760 3

2

768 3

2

44. Any point on the hyperbola

4 = t

2

= 16 is

( –4

If it passes through P(h, k fi44–

3

+

k-

4 = 2(h – 4 t

–4=0

Ê 4ˆ Let Pi Á 4ti , ˜ , = 1, 2, 3 be the fect of the normal through ti ¯ Ë P, then

(a –

4

h and  1 2 = 0 4 i=1 k  1 2 3 = - and 1 2 3 4 = –1 4 1 1 1 1 k + + + = 4 4 1 2 3

 ti =

Slopes of four normals through P(a, b Now, Â 12 = (Â 1 2 – 2Â 1 2 fi

776 3

y-

and  1 2 = 0

1

772 3

2

4ˆ , ˜ ¯

H that pass through

2

Area of the equilateral triangle inscribed C is

: ,

c = t +b –

= 16y y2 = 8

Example 45 If the tangent to the curve C, cuts the asymptotes of the hyperbola in A and B, then locus of the mid-point of AB is 2 2 = 4y = 2y 2 2 + 2y + 4y = 0

viz. 1 and –1.

b-

2

= 4y = 12y

in C

3

(

2

Example 46

c y - = 2( – t It will pass through point ( , 1 1 - = 2 (1 – fi

2

= 2.

Statement-1: All point P such that sum of slope of normals 1 from point P to H is 2 lie on a pair of lines parallel to y Statement-2: to H.

= ±1. These lines are

a2 2

fia =±1 2

2 1

,

2 2

,

2 3

and

2 4



fi Sum of the ordinates of the feet of the normals is 4. Sum of the slopes of the normals through P is

IIT JEE eBooks: www.crackjee.xyz 19.18

2 1

2 2

+

2 3

+

+

2 4

= (Â

2

– 2Â 1 2

2

h2 = 16

2 2

According to the given condition

2

2

h =k 16 2 = 16y. Locus of P(h, k

: C:

= 16y at (4 ,

– + – +

12 12 24 24

+ 24 = 0 + 24 = 0 – 12 = 0 – 12 = 0

47. A tangent to the hyperbola x2 y 2 =1 9 4

2

2

y+

(4

2

y2 y2 y2 y2

+ + + +

2 is of the form y = mx + 9m - 4 where, > 2/3

y A

B

A

30° 0 Fig. 19.13

fi = 2(y + 2 which meets the asymptotes B(0, – 2 Let M(h, k

B

AB then

2

+ 2y = 0 A be (–4 ,

4

=

2

1 3

=

This line will be a tangent to the circle 2 + y2 – 8 = 0, if length of perpendicular from the centre to the line = radius of the circle.

fi 2k = – h2

2 Locus of (h, k



Fig. 19.14

= 0 and y = 0 at A(2

2

h= ,k= -

4



2

B be (4 ,

2

= 4 3

Area of the triangle OAB =

3 (32 3 )2 4

Paragraph for Questions Nos. 47 and 48 2

2

+ y2 – 8 = 0 and hyperbola H:

(4



16

2

+ 9 2

2

-4 =4

+1 2

+ 8

-4 9

) 2

2

= 16 (

2

-4 + 9

2

x y 9 4

4 2 or = as 5 5 Thus equation of required tangent is fi

2

=

Example 47 fi

positive slope to C and H is

Example 48

2



= 1 intersect at A and B.

– 5y – 4y

4 -0+ 9

– 4 = 16

8 9 2 - 4 = 20 – 9 2 fi Squaring both the sides we obtain 64 2 (9 2 fi 495 4 + 104 2 – 400 = 0 2 fi (99 2

2 = 768 3 (unit)

The circle C :



4

= 32 3

Now AB = 2(4

x

8

O

– 5y + 4 = 0 – 3y + 4 = 0 AB as diameter is

=

2 5

+

2

>

+ 81

2

+ 16

4

2 . 3

4 5

2 x - 5 y + 4 = 0. Let coordinates of A be (3 sec q, 2 tan q As A lies

on the circle 2 + y2 – 8 = 0, we get 9 sec2q + 4tan2q – 24 secq = 0 fi 13 sec2q – 24 secq – 4 = 0

IIT JEE eBooks: www.crackjee.xyz Hyperbola 19.19

fi fi

(13 secq secq = 2

q



tanq = ± 3

[

|secq| ≥ 1]

Thus, the two curves intersect in A(6, 2 3 ) and B(6, -2 3 ) . AB as diameter is ( y - 2 3 )( y + 2 3 ) = 0

( 2

or

+ y2 – 12 + 24 = 0.

INTEGER-ANSWER TYPE QUESTIONS If the equation of the asymptotes of the

Example 49

hyperbola + 2 + 4y + 6 = 0 is the value of is .8

+ 2 + 4y + = 0, then

y



p b 1 = tan = 6 a 3

1 2 = 3 3 If e¢ is the eccentricity of the conjugate hyperbola then 1 1 + 2 =1 2 e e¢ fi eccentricity e of the hyperbola is

fi fi

1 e¢

2

= 1-

1+

3 1 = 4 4

e¢ = 2.

Example 52

Tanents are drawn from any point on the

2

hyperbola /9 = 4 to the circle 2 + y2 = 9. If the locus of the mid-point of the chord of contact is ( 2 + y2 2 = 2 – 2 , then the value of – – 2 is equal to .5

y or + 2 + 4y + 8 = 0 and the required value of is 8 Alternately + 2 + 4y + = 0 represents a pair of straight lines if = 8. If e is the eccentricity of the hyperbola 25e 2 2 2 – (5y = (12 – 5y then is equal to 13

Example 50 (5 .5

13 12 x - 5 y + 1 13 5



13 25e fi =5 5 13 If the angle between the asymptotes of

Example 51 the hyperbola

x2 2

-

y2

= 1 is

2

a b the conjugate hyperbola is .2

p tan = 3 where tan a = fi

sec q = fi

( x - 2)2 - ( y + 3) 2 = fie=

P(3 sec q, 2tan q circle is (3 sec q + (2 tan q y Let M(h, k terms of the mid-point + ky = h2 + k2

2a =

b a

p , then the eccentricity of 3 x y ± =0 a b

b a = 2 tan a = tan 2a 1 - tan 2 a b2 1- 2 a 2

p p -1 b fi 2 tan = 3 a 3

9h 3h , tan q = 2 2 2 h + k2 h +k

(

2

)

locus of (h, k 9 x2

( x2 + y 2 )

2



81y 2

(

4 x2 + y 2

)

2

or 4( 2 + y2 2 = 36 2 – 81y2 which is same as ( 2 + y2 2 = 2 – fi = 4, = 36, = 81 fi – –2 =5

=1

2

Suppose the normal to the hyperbola 2ˆ Ê A Á 2a, ˜ and the Ë a¯ normal to the hyperbola = 4 at A meets the hyperbola at B(h, k h is equal to .2 : An equation of normal to the hyperbola Ê cˆ = 2 at Á ct , ˜ is Ë t¯ c y- = 2 ( – t Ê cˆ Suppose it meets the hyperbola at Á ct1 , ˜ so that, t1 ¯ Ë Example 53

IIT JEE eBooks: www.crackjee.xyz 19.20

c c - = t1 t fi

1

2

(

1

-



= – –3

Ê cˆ hyperbola at Á ct2 , ˜ , then t2 ¯ Ë = 2,

16 9

4 3 5. Suppose P is a variable point on the hyperbola x2 y 2 = 1 with eccentricity e. Let A(a, b a 2 b2 AP lies on e -

Ê cˆ If the normal at Á ct1 , ˜ to the hyperbola meets the t1 ¯ Ë

Here

20 9

= 1,

1

2

= – 1–3.

= –1,

2

= 1.

e -1 e e

Therefore, h = 2 2 = 2.

EXERCISE

2 LEVEL 1

perbola

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. Locus of mid-point of chord of the hyperbola = 4 whose length is 4 units is 2 + y2 = 4 ( = 2 – y2 + y 2 4 2y 2 = ( 2 + y2 2. If (12 tan f, 5 secf centricity is 15 12 17 5 13 17 5 12 3. Let P(3 secq, 2 tan q Q(3 sec f, 2 tan f p q + f = , be two distinct points on the hyperbola 2 x2 y 2 = 1. Then the ordinate of the point of in9 4 tersection of the normals at P and Q is 11 3 13 2

11 3 13 2 -

x2 y 2 = 1, meets 4 5 y A and B, respectively. Then (OA 2 – (OB 2, where O is origin, equals latus rectum of the hyperbola

-

= 36 is

6 2

12 2

7. Let A and B be two points on the hyperbola 31 2 – 29y2 = 1 such that the chord AB subtends a right angle at O, the centre of the hyperbola then 1 1 + is 2 (OA) (OB)2 8. Let S1 and S2 dii with centre C1 and C2. The centre C of circle S which touches S1 and S2

centricity

5 be 4. The difference between length

asymptotes is 6± 2 2± 3

p is 6 6± 3 5± 3

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS perbola passing through the origin, then 1 386 12

IIT JEE eBooks: www.crackjee.xyz Hyperbola 19.21

11 2

+ 3 – 5y – 15 = 0 x2 y 2 and 25 9

+ 19y – 253 =0 2 y– , where > 0 is +y= + + 2 +y= + + +y= + – 2 +y= + – 13. If a hyperbola passes through the foci of the ellipse x2 y 2 + 25 16

la ( –

latus rectum Column 1 41

– 4y

–y + 2

7y –y

(

5, - 2

)

Column 2 2

/9 + y2/16 = 1

2

/16 – y2/25 = 1

2

– y2 = 4

y2 = 8 (

18. Column 1

and product of the eccentricities is 1, then x2 y 2 =1 9 16

3 2 – 2y2 = 1 where the tangent is parallel to 6 – 4y + 7 = 0

x2 y 2 =1 9 25

(y

2

= –4

(5 3, 0) y2 = x2 y 2 = 1 passes through 25 16

12 and the hyperbola the point

2

+ 3y2 = 4 where the normal is parallel to 3 – y – y + 2 = 0 to the hyperbola – 2y2 = 2, then all the chord of contact pass through

ASSERTION-REASON TYPE QUESTIONS 19. Statement-1: If the foci of an hyperbola are at

Statement-2: Distance between the foci of a hyperbola is equal to the product of its eccentricity and

MATRIX-MATCH TYPE QUESTIONS 16. Column 1 + y2 = 9 and

Column 2 2

x2 y 2 = 1 and 16 9

= 1 and

1ˆ Ê ÁË1, - ˜¯ 2

2

( x - 5)2 ( y - 3) 2 = 1 are 25 9

2

Column 2

x2 y 2 25 16

– y2

= 12 2 x2 y 2 + 25 16

(4

)

2, 3

20. Statement-1: Locus of the mid-points of normal chords of the rectangular hyperbola 2 – y2 = 25 is (y2 – 2 3 = 100 2y2. Statement-2: h, k a normal chord to the rectangular hyperbola 2 – y2 = 2 at the point ( secq, tanq +y sinq = 0 21. Statement-1: If two diameters are conjugate with respect to a hyperbola then they are conjugate with respect to the conjugate hyperbola.

IIT JEE eBooks: www.crackjee.xyz 19.22

Statement-2: Two diameters y = and y = ¢ of a x2 y 2 b2 hyperbola 2 - 2 = 1 are conjugate if ¢= 2 . a b a

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 22 to 24 Let

P( ,

y

( x - 6)2 + ( y - 6) 2 - ( x - 2) 2 + ( y - 3) 2 = 3. Then P traces a hyperbola, say H.

41

INTEGER-ANSWER TYPE QUESTIONS

x2 y 2 =1 25 16

29. Number of points on the hyperbola

2

common to the cricle 2 + y2 = 36 are 30. Product of the lengths of the perpendiculars from any point on the rectangular hyperbola 2 – y2 = 16 to its asymptotes is

H is + 6y = 43 + 6y = 47

+ 3y + 3y 24. An angle q

k is

28. Locus of the poles of the tangents to the hyperbola 2 – y2 = 1 with respect to the parabola y2 = 4 is the ellipse 4 2 + y2 = 2 where is equal to

H is 5 2

5 3 3 2

26. If k = 3, then equation of H1 is + 3 – 3y – 18 = 0 – 3 + 3y – 18 = 0 – 3 + 3y + 18 = 0 – 3 – 3y – 18 = 0 27. If k > 0 and H1 3

LEVEL 2

H Ê 2ˆ tan -1 Á ˜ Ë 3¯

Ê 3ˆ tan -1 Á ˜ Ë 2¯

Ê 3ˆ tan -1 Á ˜ Ë 4¯

Ê 4ˆ tan -1 Á ˜ Ë 3¯

SINGLE CORRECT ANSWER TYPE QUESTIONS

Paragraph for Question Nos. 25 to 27 Let L be the line y = Let H1 Ê aˆ Ê0 Á b˜ = Á 1 Á ˜ Á Ë 1¯ Ë0

+ k and H be the hyperbola H in L. P(a, b P¢( , 1 -k ˆ Ê a ˆ 0 k ˜ Á b˜ ˜Á ˜ 0 1 ¯ Ë 1¯

Ê aˆ Ê 0 1 k ˆ Êaˆ Á b˜ = Á 1 0 - k ˜ Á b ˜ Á ˜ Á ˜Á ˜ Ë 1¯ Ë 0 0 1 ¯ Ë 1¯ Ê aˆ Ê 0 1 1 ˆ Êaˆ Á b˜ = Á 1 0 -1˜ Á b ˜ Á ˜ Á ˜Á ˜ Ë k¯ Ë0 0 1 ¯ Ë k¯ Ê aˆ Ê 0 -1 k ˆ Ê a ˆ Á b˜ = Á 1 0 - k ˜ Á b ˜ Á ˜ Á ˜Á ˜ Ë 1¯ Ë 0 0 1 ¯ Ë 1¯

= 9.

b x2 y 2 1. If parabola y2 = - x and the hyperbola 2 - 2 a a b = –1 do not intersect then 1 < 0, 0 < > 2 > 2 1 < 2, < , >0 2 2. If q is an angle between two asymptotes of

x2 a

2

-

y2 b2

Êqˆ = 1, then cos Á ˜ equals Ë 2¯ b a +b a 2

ab a + b2 a-b a+b

2

2

a 2 + b2 3. If + + = 0 touches the hyperbola 2 2 = , then 2 2 – 2 2 equals 2

2 2

2

n2 (a 2 + b 2 )

-

n2 (a 2 + b 2 )



2 2

y

IIT JEE eBooks: www.crackjee.xyz Hyperbola 19.23

b > 0. If the conics y2 = - x , and a

4. Suppose , y2 b2

-

x2 a2

= 1 intersect, then

0< £

1 2

0< £

1 2

2

2 < 2 < 2 > 1, then sata 2 - b 2 a 3 - b3

≥ + + 2 + 2 + + ≥( + 2 2 2 2 – – ≥( + 3 2 2 + ≥( + 3 6. The parametric equation = 2 + + 1, y = represents

2

– +1

10. The locus of the foot of the perpendicular from the x2 y 2 centre upon any normal to the hyperbola 16 9 =1 2 2 – y2 y2 – 9 2 – 2 2 2y 2 2 2 2 2 2 2 + y (16y – 9 + 2 2 2y 2 2 2 + y2 y2 – 9 2 + 2 + 2 2y 2 + y 2 (16y2 – 9 2

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 11. A straight line touches the rectangular hyperbola 9 2 – 9y2 = 8 and the parabola y2 = 32 . An equation of the line is + 3y – 3y + 8 = 0 + 3y – 3y – 8 = 0 5 4

and having eccentricity

7. If for the hyperbola

x2 2

-

y2 2

a b tance between foci = 9, then 27 2

= 1, e =

3 and dis-

2

– 2 equals 27 4 9 2 8. A right triangle PQR, with right angle at P has its vertices on the rectangular hyperbola = 2. An angle between the tangent at P and the sides QR is p p 6 4 p p 3 2 a 2 m 2 - b 2 touches the hyper-

9. If the line y = + x2 y 2 bola 2 - 2 = 1 at the point ( secq, tan q a b q is equal to Ê b ˆ Ê b ˆ sin -1 Á cos -1 Á Ë am ˜¯ Ë am ˜¯ Ê a ˆ sin -1 Á ˜ Ë bm ¯

Ê a ˆ cos -1 Á ˜ Ë bm ¯

x2 y 2 64 36

9 x2 - 4 y2 = 1 4

x2 y 2 16 9

x2 y 2 =1 9 16

13. The equation 16 2 – 3y2 – 32 – 12y – 44 = 0 represents a hyperbola with 2 3

19 14. Suppose eccentricity of the hyperbola is sec q, where 0 < q < of

x2 b

2

y2

-

a

2

p 4 2 q + cos2f

x2 b2

-

y2 a2

x2 a2

a2

-

y2 b2

-

y2 b2

=1

p , and suppose eccentricity 2

= 1 is cosec f, where 0 < f
0 H and equation of the hyperbola H differ by a constant when these have identical second degree terms. 20. The value of so that 2 + 4 + y2 – 2 + 2y + = 0 represents a hyperbola lies in the set • • R f 21. If = –6, then equation of asymptotes of the hyperbola above are +1+

by H on the line y

h a f

=0

+4

+ y2 – 2 + 2y – 6

= 0 is

Paragraph for Question Nos. 23 and 24 The combined equation of the pair of tangents drawn from x2 y 2 =1 a point P( 1, y1 a 2 b2 to the hyperbola is SS1 = T2 where S ∫ x12 a

2

-

y12 b

2

P

- 1 and T ∫

xx1 a

2

-

yy1 b2

x2 a2

-

-1 x2 - y 2 = 1 is 4

y2 b2

- 1 , S1 =

IIT JEE eBooks: www.crackjee.xyz Hyperbola 19.25 2 2 2 2

+ – – –

4 4 4 4

+ 6y2 – 4 – 6y2 – 4 + 6y2 – 4 + 6y2 – 4

– 4y + 6 = 0 – 4y + 6 = 0 – 4y + 6 = 0 – 4y – 6 = 0

x2 - y2 4 = 1 meet the hyperbola at Q and R then equation of QR is –y –y–1=0 +y –y–3=0

24. If the tangents from P

INTEGER-ANSWER TYPE QUESTIONS ¢ = 0 is a pair of conjugate x2 y 2 diameters of the hyperbola = 1, then ¢= 16 96

25. If y –

x2 y 2 =1 9 3 G, A and A¢ are the verAG. A¢ G tices of the hyperbola, then 4 is equal to e sec2 q - 1 q’ on the hyperbola

= 0 and y +

2 2

– 16y – 36 + 96y – 252 = 0 is 28. If a straight line with slope is a normal to the 2 parabola y = 4 and is a tangent to the rectangular hyperbola 2 – y2 = 2 then 6 + 4 4 + 3 2 + 4 is equal to x2 y 2 + = 1 and the hyper25 b 2

29. If the foci of the ellipse

x2 y2 1 = coincide, then the value of is bola 144 81 25 30. Area of the triangle formed by the lines – y = 0, + y = 0 and any tangent to 2 – y2 = 9 in sq. units is

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 2

1. The equation

4. Let P( secq,

2

[1981]

points ( 1, y1

plane. One of these four regions 2,

y2

Ê x1 + x2 y1 + y2 ˆ ÁË 2 , 2 ˜¯ is also in the region. The in+ 2y2 £ |, |y|} £ 1 2 2 2 –y £ y – £ 0 [1981] 2 2 3. The equation 2 + 3y – 8 – 18y + 35 = k represents k>0 k0 [1994] 2

tanf

q + f = p/2, be two points on the hyperbola

x y = 1, > 1 represents 1- r 1+ r

a region in the

Q( secf,

tanq

If (h, k P and Q, then k is equal to 2 + 2 2 + 2

2 2

+ +

x2 a2

-

y2 b2

= 1.

2 2

[1999] 5. If = 9 is the chord of contact of the hyperbola 2 – y2 = 9, then the equation of the corresponding pair of tangents is 2 – 8y2 + 18 – 9 = 0 2 – 8y2 – 18 + 9 = 0 2 – 8y2 – 18 – 9 = 0 2 – 8y2 + 18 + 9 = 0 [1999] x2

-

y2

= 1 which of the cos 2 a sin 2 a following remains constant with change if a.

IIT JEE eBooks: www.crackjee.xyz 19.26

(4, - 6 )

x2 y 2 =1 9 25

6 y = 2 and

7. The point of contact of the line 2 + the hyperbola 2 – 2y2 = 4 is ( 6 ,1)

(5 3, 0)

(1/ 2,1/ 6 ) 2 sin q, is confocal with the ellipse 3 then its equation is 2 cosec2 q – y2 sec2 q = 1 2 sec2 q – y2 cosec2 q = 1 2 sin2 q – y2 cos2 q = 1 2 cos2 q – y2 sin2 q = 1

2

2

+ 4y = 12,

2

1 - 2/3

3/ 2 - 1

1 + 2/3

3/ 2 + 1

2

+ 2y2 = 2

2

+ 2y2 = 4

(± 2 , 0)

a2 = 1. If the normal at the point P intersects the

-

[2009] x

2

2

y2 b2

3 2 3

y2

x2 y 2 =1 3 2 5 3 2

5 2

-

=1 a 2 b2 be reciprocal of that of the ellipse 2 + 4y2 = 4. If the hyperbola passes through a focus of the ellipse, then.

4. Let the eccentricity of the hyperbola

[2008] x2

10. Let P

3. An ellipse intersects the hyperbola 2 2 – 2y2 = 1 orthogonally. The eccentricity of the ellipse is re-

[2007] – 2y2 –

2 2x - 4 2 y A. let B be one of the end points of its latums rectum. If C is focus of the hyperbola nearest to the point A, then area of the triangle ABC is

[2006]

[2011]

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 1. If the circle 2 + y2 = 2 intersects the hyperbola = 2 in four point P( 1, y1 Q( 2, y2 R( 3, y3 S( 4, y4 1 + 2 + 3 + 4 = 0 y1 + y2 + y3 + y4 = 0 4 1 2 3 4 = y 1y 2y 3y 4 = 4 [1998] 2. A hyperbola passes through the focus of the ellipse x2 y 2 + 25 16 this hyperbola coincide with the major and minor ities of given ellipse and hyperbola is 1, then x2 y 2 =1 9 16

– 3y2 = 3 [2011]

x2 y 2 = 1, 9 4 parallel to the straight line 2 – y = 1. The points of contact of the tangents on the hyperbola are

5. Tangents are drawn to the hyperbola

9 1 ˆ Ê ,˜ ÁË 2 2 2¯

1 ˆ Ê 9 , ˜ ÁË 2 2 2¯

(3

3, -2 2

)

(-3

)

3, 2 2 [2012]

MATRIX-MATCH TYPE QUESTIONS Column 1

Column 2 gent

circles normal inside the other

-

IIT JEE eBooks: www.crackjee.xyz Hyperbola 19.27

hyperbola

Column 1

mon tangent

FILL

mon normal[2007]

1. An ellipse has eccentricity 1/2 and one focus at the point P gent, nearer to the point P, to the circle 2 + y2 = 1 and the hyperbola 2 – y2 = 1. The equation of the ellipse in the standard from is ______. [1996]

Column 2 h, k for which the line + ky = 1 touches the circle 2 + y2 =4 z satisfying |z + 2| – |z – 2| = ±3 parametric representation Ê1- 2 ˆ 2 = 3Á 2˜ , y = 1+ 2 Ë1+ ¯ lies in the interval 1 £

0.

IIT JEE eBooks: www.crackjee.xyz Hyperbola 19.33

sec q – y tan q If (h, k w.r.t y2 = 4 is 2 – ky = – 2h secq tanq 1 = = 2 -2h k 2 Locus of (h, k + y2 = 4 =

h, k

So

2



= 2.

So the circle meets the hyperbola in 4 points. 30. Let P(4 sec q, 4 tan q + y = 0 and – y = 0. Product of the length of the perpendicular from P on these is 4 sec q + 4 tan q 4 sec q - 4 tan q 2

hyperbola and the diameter of the circle is greater

=

16 =8 2

2

IIT JEE eBooks: www.crackjee.xyz

20 Three Dimensional Geometry 20.1

Z

COORDINATES

A

Distance Formula , y, z P (x

Distance between A(x1, y1, z1) and B (x2, y2, z2) is given by AB =

r

( x2 - x1 )2 + ( y2 - y1 )2 + ( z2 - z1 )2 .

z

O

Section Formulae A¢

The coordinates of the point P which divides the join of A (x1, y1, z1) and B(x2, y2, z2) in the ratio m : n are Ê mx2 + nx1 my2 + ny1 mz2 + nz1 ˆ , , ÁË ˜ m+n m+n m+n ¯ By taking m = n of AB as

)

M

Y

x

N

y

X Fig. 20.1

Ê x1 + x2 y1 + y2 z1 + z2 ˆ , , ÁË ˜ 2 2 2 ¯ Coordinates of any point on the join of two points Ê l x + x1 l y2 + y1 l z2 + z1 ˆ A(x1,y1,z1)andB(x2,y2,z2)are Á 2 , , Ë l +1 l +1 l + 1 ˜¯ where l π – 1. If l takes only positive real values, we get AB (except A and B). Volume of a Tetrahedron

If Ai(xi, yi, zi), i = 1, 2, 3, 4 are the vertices of a tetrahedron,

1 6

x1 x2 x3 x4

y1 y2 y3 y4

z1 z2 z3 z4

1 1 1 1

Direction-Cosines

If a, b, g with the positive directions X¢OX, Y¢OY, Z¢OZ of the coordinate axes, then cos a, cos b, cos g are called the direction-cosines of the line.

Let P(x, y, z) be any point on A¢OA of OP be r. Let PN P on the plane XOY, and NM N on OX. Then PN = z, NM = y, OM = x x = r cos a, y = r cos b, z = r cos g. and cos2 a + cos2 b + cos2 g = 1. Direction-Ratios

If l, m, n OP and a, b, c l, m, n respectively, then a, b, c are called the direction ratios of OP. If l, m, n be the k (π kl, km, kn OP. If a, b, c a, cos b, cos g relations cos a cos b cos g = = a b c = ±

cos2 a + cos2 b + cos2 g a +b +c 2

2

2

=

±1 a + b2 + c 2 2

IIT JEE eBooks: www.crackjee.xyz 20.2

If P be the point (a, b, c), and cos a, cos b, cos g are the direction cosines of the directed line OP, then a b , cos b = , cos a = 2 2 2 2 (a + b + c ) (a + b2 + c 2 )

tan q = ±

[(a1b2 - a2 b1 )2 + (b1c2 - b2 c1 )2 + (c1a2 - c2 a1 )2 ]1 / 2 a1a2 + b1b2 + c1c2

The lines are parallel to each other if and only if

c

cos g =

a1 a2

(a + b + c ) 2

2

2

b1 c1 = b2 c2

=

PO are a

-

(a 2 + b2 + c 2 )

b

,-

(a 2 + b2 + c 2 )

a1a2 + b1b2 + c1c2 = 0.

c

,-

(a 2 + b2 + c 2 )

Illustration 1 Find the direction cosines of the line whose direction ratios are 12, 4, – 18. Solution: Direction cosines of the given line are 12

4

,

(12) + (4) + (-18) 2

2

,

(12) + (4) + (-18)

2

2

2

(12) + (4)2 + (-18)2

or 12 , 4 , -18 22 22 22

(1) + (-1)2 + (1)2 12 + 0 + l 2

20.2

THE PLANE

Equation of a Plane

or 6 , 2 , -9 .

by ax + by + cz + d x, y, z represents a plane. yz x = 0. zx y = 0. xy z = 0. (d) If l, m, n a plane and p be the length of the perpendicular

11

Angle Between Two Lines

(a) If q cosines are (l1, m1, n1) and (l2, m2, n2) then cos q = l1l2 + m1m2 + n1n2. The expressions for sin q, tan q are given below: sin q =

1(1) + (-1)(0) + (1)(l ) 2

fi l = –1

-18

2

Direction ratios of a line L1, are 1, –1, 1 and of L2 are 1, 0, l. Find the value of l so that L1 and L2 are perpendicular to each other. Solution: The line L1 and L2 are perpendicular if the angle q fi cos q =

2

2

11 11

Illustration 2

m1 m2

n1 n2

2

n + 1 n2

l1 l2

2

l + 1 l2

m1 m2

2

tan q = ± [S (l1 m2 – l2 m1) ] /(l1l2 + m1m2+ n 1n 2) The lines are parallel to each other if and only if l1/l2 = m1/m2 = n1/n2.

plane is lx + my + nz = p. Illustration 3

2 1/2

and

l1l2 + m1m2 + n1n2 = 0. q are a1, b1, c1 and a2, b2, c2 is given by cos q = ±

4x – 3y + 5z the origin on this plane Solution: 4, –3, 5. So direction cosines are l=

4 + (-3) + 5 2

2

-3

,m=

4 + (-3)2 + 52 2

n=

a1a2 + b1b2 + c1c2 a12 + b12 + c12

a22 + b22 + c22

or l =

sin q = ±

4 2

[(a1b2 - a2 b1 )2 + (b1c2 - b2 c1 )2 + (c1a2 - c2 a1 )2 ]1 / 2 a12

+

b12

+

c12

a22

+

b22

+

c22

4 5 2

4 5 2

x–

,m=

3 5 2

y+

–3 5 2 5

5 2

,n=

z =

5 5 2

25 5 2

, 5

4 + (-3)2 + 52 2

IIT JEE eBooks: www.crackjee.xyz 20.3

or lx + my + nz = p where p =

25 5 2

=

5 2

a, b, c on the axes of x,

a1x + b1y + c1z + d1 = 0, a2x + b2y + c2z + d2 = 0, is given by a1a2 + b1b2 + c1c2

cos q = ±

(a12

y, z x y z + + = 1. a b c Systems of Planes

conditions. If we consider a plane satisfying just two given If we consider a plane satisfying one given condition, its

+ b12 + c12 ) (a22 + b22 + c22 )

Parallelism and Perpendicularity of Two Planes

The planes a1x + b1y + c1z + d1 = 0, a2x + b2y + c2z + d2 = 0 are parallel if and only if a1 a2

=

b1 c1 = ; b2 c2

and perpendicular if and only if a1a2 + b1b2 + c1c2= 0. Two Sides of a Plane

ax + by + cz + k = 0 represents a ax + by + cz +

Two points P(x1, y1, z1) and Q(x2, y2, z2 different sides of the plane ax + by + cz + d = 0 according as ax1 + by1 + cz1 + d and ax2 + by2 + cz2 + d are of the

d = 0, k ax + by + cz + k x y z of planes perpendicular to the line = = . a b c a1x + b1y + c1z + d1) + k (a2x + b2y + c2z + d2 through the intersection of the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0, k

Length of the Perpendicular from a Point to a Plane

The perpendicular distance of the point (x1, y1, z1 plane lx + my + nz = p is |p – lx1 – my1 – nz1|, where l, m, n p is The perpendicular distance of the point (x1, y1, z1 the plane ax + by + cz + d = 0 is

Illustration 4

ax1 + by1 + cz1 + d

intersection of the planes 3x + 4y + 7z + 5 = 0 and 2x – 3y + 4z the axis of x. Solution: intersection of the given planes is 3x + 4y + 7z + 5 + l (2x – 3y + 4z – 6) = 0 Intercept on x

6l – 5 =1fil=2 3 + 2l

7x – 2y + 15z – 7 = 0 A(x – x1) + B(y – y1) + C(z – z1) = 0. point (x1, y1, z1), the ratios of A, B, C being the two

(a 2 + b2 + c 2 )

Bisectors of the Angles between Two Planes

a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 d1 and d2 are both positive. Then a1 x + b1 y + c1z + d1

(a12 + b12 + c12 )

=

a2 x + b2 y + c2 z + d2

(a22 + b22 + c22 )

given planes which contains the origin, and a1 x + b1 y + c1z + d1

(

a12

+

b12

+

c12

)

=–

a2 x + b2 y + c2 z + d2

(a22 + b22 + c22 )

given planes which does not contain the origin. Illustration 5 Angle Between Two Planes

q between the planes

Find the angle between the planes x + y + z + 1 = 0 and x–y+z–

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bisecting the angle between these planes which contain the origin. Solution: If q is the angle between these planes, then (1) ¥ (1) + (1)(-1) + (1)(1)

fi cos q =

12 + 12 + 12 (1)2 + (-1)2 + (1)2

= 1

P and Q are two points on a line whose direction ratios are 3, 5, 1. If the coordinates of P are (1, – 4, 2) and (PQ)2 = 35. Find the coordinates of Q. Solution: Direction cosines of the line are

between the given planes is 1 +1 +1 2

2

2

-x + y - z +1

=

a line. Illustration 6

3

fi q = cos– 1 (1/3)

x + y + z +1

x = x1 + lr, y = y1 + mr, z = z1 + nr. distance form

3

(1)2 + (-1)2 + (1)2

fix+z=0 angles between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are a1 x + b1 y + c1z + d1

(

a12

+

b12

+

c12

)

= ±

(

+

b22

+

c22

35

)

3 / 35

,

=

2

5 35

,

y+4 5 / 35

3 + 5 +1 2

2

2

,

1 3 + 52 + 12 2

1 35

=

z-2 1 / 35

= r = 35 = PQ .

and the coordinates of Q are. (3 + 1, 5 – 4, 1 + 2 ) = (4, 1, 3) If l, m, n

The bisector of the acute angle between the given planes given planes. 20.3

3

x -1

a2 x + b2 y + c2 z + d2 a22

or

2

5

,

3 + 5 +1 2

a straight line, but in this case r not the distance of (x, y, z x1, y1, z1). x-a y-b z-c = , = l n m

THE STRAIGHT LINE

Equations of a Line

represent a straight line passing through the point (a, b, c) and having direction ratios l n. Equation of Line through two Given Points

of every point on the line of intersection of the planes represent that line. Therefore ax + by + cz + d = 0 and a¢x + b¢y + c¢z + d¢ = 0 together represent a straight line.

If A(x1, y1, z1), B(x2, y2, z2 of the line AB is x - x1 y - y1 z - z1 = = x2 - x1 y2 - y1 z2 - z1

The Equations to the Axes of Coordinates

x y z

The coordinates of a variable point on AB can be l

y = 0, z = 0. x = 0, z = 0. x = 0, y = 0.

x=

Symmetrical form of the Equation of a Line

If a straight line passes through a given point (x1, y1, z1) and has direction cosines l, m, n, then the coordinates of x - x1 y - y1 z - z1 = (= r) = m l n of any point (x, y, z x 1, y 1, z 1

l x2 + x1 l y2 + y1 l z2 + z1 ,y= ,z= , l +1 l +1 l +1

l x, y, z) are the coordinates of the point which divides the join of A and B in the ratio l : 1.

Changing unsymmetrical form to symmetrical form r

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ax + by + cz + d = 0,

Angle between a Line and a Plane

The angle q between the line

a¢x + b¢y + c¢z + d¢ = 0.

x -a y-b z-g = = , l m n and the plane ax + by + cz + d = 0, is given by al + bm + cn sin q = a 2 + b2 + c 2 l 2 + m 2 + n2

bd ¢ - b ¢d da ¢ - d ¢a xyz ab ¢ - a ¢b ab ¢ - a ¢b = = bc ¢ - b ¢c ca ¢ - c ¢a ab ¢ - a ¢b

(

Illustration 7 Find the direction ratios of the line x + 2y – z = 3, 2x –y+z Solution: Direction ratios of the line are 2 -1 1 -1 , , -1 1 2 1

1 2 2 -1

r + 1, –3r + 1, – 5r r + 1 + 2(–3r + 1) – (–5r) = 3 for all r and 2(r + 1) – (–3r + 1) + (–5r) = 1 for all r. showing that each point on the line

(1)

Coplanar Lines

The lines

x -a y-b z-g = = , l m n

x - a¢ y - b¢ z - g ¢ = = l¢ m¢ n¢ are coplanar if and only if a - a¢ b - b¢ g - g ¢ l m n l¢ m¢ n¢

=0

the plane containing the lines is x -a l l¢

(1)

Number of Constants in the Equation of a Line

x-a y-b z-c = = , m n l of a line can be written as x = (l/n) z + a – (lc/n), y = (m/n) z + b – (mc/n), x = Az + B, y = Cz + D Therefore the equation of a line contains four arbitrary constants. A Plane and a Straight Line

x -a y - b z -g = = , l m n represent a given plane and a straight line respectively. (i) The line is perpendicular to the plane if and only if a b c = = , l m n (ii) The line is parallel to the plane if and only if al + bm + cn = 0. (iii) The line lies in the plane if and only if al + bm + cn = 0 and aa + bb + cg + d = 0. ax + by + cz + d = 0,

)

and

or 1, –3, –5 For any point on the line, let x = 0, then x = y = 1 and x -1 y -1 z - 0 = = 1 -3 -5

) (

y-b m m¢

z-g n n¢

= 0.

Illustration 8 Find the value of l, so that the lines x -1 y - 2 z + l and = = 1 2 3 x +1 y -1 z - 3 are coplanar = = 2 3 1

Solution: 1 + 1 2 - 1 -l - 3 1 2 3 =0 fil=6 2 3 1

lines is x -1 y - 2 z + 6 1 2 3 =0 2 3 1

or 7x –5y + z General Equation of a Plane Containing a Line

x -a y-b z-g = , = l m n

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A(x – a) + B(y – b) + C(z – g) = 0, Al + Bm + Cn = 0.

is where

Equation of a Line through the Intersection of Given Lines

Length of the Perpendicular from a Point to a Line

The length p x -a y-b z-g = = l m n 2 (l,m,n are direction cosines of the line), is given by p = (x1 – a)2 + (y1 – b)2 + (z1 – g)2 – [l(x1 – a) + m(y1 – b) + n(z1 – g)]2. p2 is given by P(x1, y1, z1) to a given line

2

p =

x1 - a l

y1 - b m

z1 - g n

+

2

y -b + 1 m

z1 - g n

2

2

x1 - a l

If L1 = 0 = L2 and L3 = 0 = L4, where LK = akx + bky + ckz + d k, k for all values of l and m L1 + lL2 = 0 = L3 + mL4 represents a line through their intersection. 20.4

VECTORIAL EQUATIONS

a point with position vector a and parallel to the vector b is r = a + tb where t two points with position vectors a and b is r = (1 – t)a + tb where t

Illustration 9 through the point with position vector a and which is parallel to the vectors b and is r = a + tb + p where, t and p

Find the perpendicular distance of the point x -1 y - 3 z - 4 = = 3 1 -5

Solution: Let P(2, 1, 3) be the given point A(1, 3, 4) is a point on the given line p=PM, be the perpendicular P on the line. P(2,1,3)

through two given points with position vectors a and b and parallel to the vectors is r = a + t(b – a) + p = (1 – t) a + tb + p where t and p with position vectors a, b, is r ◊ (b ¥ + ¥ a + a ¥ b) = [a b ].

q A(1, 3,4)

6. Normal form of the vector equation of a plane r◊n=p n p is the

M Fig. 20.2

Let PAM = q then

plane.

p2 = (PM)2 = (AP)2 – (AM)2 = (AP)2 – (AP cos q )2 fi p2 = (AP)2 sin2 q Direction ratios of AP are 2 –1, 1 – 3, 3 –4 or 1, –2, –1 and of the given line are 3, 1, –5 so cosq =

Illustration 10

(1)(3) + (-2)(1) + (-1)(-5) (1)2 + ( -2)2 + ( -1)2 (3)2 + (1)2 + ( -5)2

= (AP) = (2 –1) + (1 – 3) + (3 – 4) = 6 2

7. (r – a) ◊ N to the vector N and passing through a point with position vector a.

2

2

Hence p2 = 6 ¥ 29 fi p = 35

2

174 35

6 6 35

=

6 35

through the point (2, 5, –3) and perpendicular to the line with direction ratios 3, 7, 5. Solution: Let a = 2i + 5j – 3k and (r – a) ◊ N = 0

N = 3i + 7j + 5k

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b ¥ (b ¥ d) = 3i + 3j

or [r – (2i + 5j – 3k)] (3i + 7j + 5k) = 0 8. Angle between two planes r ◊ N1 = p1 and r ◊ N2 = p2 is cos- 1

d ¥ (b ¥ d) = 5i + j + 2k, ◊ [d ¥ (b ¥ d)] = 4

N1 ◊ N 2 N1 N 2

r ◊ (3i + 3j) = 0, r ◊ (5i + j + 2k) = 4 r = (i – j So the line of shortest distance is r = (i – j) + l (i – j – 2k)

Angle between a line and a plane whose vectorial r = a + bt and r ◊ N = q respectively is sin - 1

N ◊b . N b

Or x - 1 = y + 1 = z

10. The perpendicular distance of a point with position

r = a + tb and r = + pd is (r – a) ◊ b ¥ d =0 fi r ◊ b ¥ d = a ◊ b ¥ d. 12. Condition for the lines r = a + tb and r = + pd to be coplanar is [ ] = [abd]. 13. The line LM of shortest distance between the lines r = a + tb and r = + pd is parallel to the vector b ¥ d. Length of LM is the projection of the line joining the points with position vectors a and on LM. LM =

(c - a ) ◊ (b ¥ d) b ¥d

=

[cbd ] - [abd ] b ¥d

intersection of the two planes through the given lines and the line LM are (r – a) ◊ b ¥ (b ¥ d) = 0, (r – ) ◊ d ¥ (b ¥ d) = 0 The line of intersection of these planes is the lines. Illustration 11

between the lines r = l (i – j + k) and r = (i – j) + m (–2j + k)

(c - a) ◊ (b ¥ d) b¥d

=

Solution: a = 0,

b=i–j+k

d = – 2j + k

(r – a) ◊ b ¥ (b ¥ d) = 0, (r – ) ◊ d ¥ (b ¥ d) = 0 (b ¥ d) = i – j – 2k

(i - j) ◊ (i - j - 2k ) (1) + (-1) + (-2) 2

2

2

=

2 6

SOLVED EXAMPLES

SINGLE CORRECT ANSWER TYPE QUESTIONS Let P Example 1 with respect to the plane x – y + z plane through P and containing the straight line z is 1

x y = = 1 2

(a) x + y – 3z = 0 (b) 3x + z = 0 (c) x – 4y + 7z = 0 (d) 2x – y = 0 Ans. (c) Solution: x – y + z = 3 and passing through A(3, 1, 7) is x-3 y -1 z-7 = = = r (say) 1 -1 1 Let the coordinates of P be (r + 3, –r + 1, r + 7) r r Êr ˆ M of AP is Á + 3, - + 1, + 7˜ and lies Ë2 ¯ 2 2 on x – y + z = 3 fi

= i – j,

-2

r ◊ N = q is q - a ◊ N . N

vector a

Thus

-1

1

r r r + 3 + - 1 + + 7 = 3 fi r = –4 2 2 2 Thus coordinates of P are (–1, 5, 3) P and containing the

x y z x y z line = = is -1 5 3 = 0 1 2 1 1 2 1 fi x – 4y + 7z = 0

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Example 2 x y z = = and perpendicular to the plane 2 3 4 x y z x y z containing the straight lines = = and = = is 3 4 2 4 2 3 (a) x + 2y – 2z = 0 (b) 3x + 2y – 2z = 0 (c) x – 2y + z = 0 (d) 5x + 2y – 4z = 0 Ans. (c) Solution: Let a, b, c to the plane containing the lines x y z x y z = = and = = , 3 4 2 4 2 3 then 3a + 4b + 2c = 0 and 4a + 2b + 3c = 0 straight line

fi Let a, b, g

r2 = 25 fi r = ± 5/3 5 1 ± = (5 + a) 3 9

fi Thus, fi

± 15 = 5 + a fi a = 10 5 \ r= 3 Hence, coordinates of Q are

(

(a) (c) 2 Ans. (a) Solution:

1 (x – 0) – 2 (y – 0) + 1 (z – 0) = 0 x – 2y + z = 0. If the distance of the point P(1, –2, 1) x + 2y – 2z = a, where a > 0, is 5, then the P to the plane is 8 4 7 4 4 1 , ,,- , (b) 3 3 3 3 3 3

Example 3

(a) (c)

) (

( (

1 2 10 , , 3 3 3

) )

a > 0]

)

The point P is the intersection of the

straight line joining the pionts Q (2, 3, 5) and R (1, –1, 4) with the plane 5x – 4y – z = 1. If S is the foot of the T (2, 1, 4) to QR, then PS is

plane is or

[

5 10 10 8 4 7 + 1, - 2, - + 1 = , , 3 3 3 3 3 3

Example 4

a b g = = . 1 -2 1



PQ2 = r2 + 4r2 + 4r2 = 25

a b c = = 8 -1 -10

8a – b – 10g = 0 2a + 3b + 4g = 0 a b g = = 26 -52 26



Q lies on x + 2y – 2z = a, we get (r + 1) + 2(2r – 2) –2 (–2r + 1) = a 1 fi r – 5 = a fi r = (5 + a). 9 PQ = 5, we get

(d)

( (

Ans. (a) Solution: Let Q P(1, –2, 1) to the plane x + 2y –2z = a. PQ is x -1 y+2 z -1 = = = r (say) 1 2 -2 Let coordinates of Q be (r + 1, 2r – 2, –2r + 1)

2 1 5 ,- , 3 3 2

) )

1 2

2

(d) 2 2 QR is x-2 y-3 z-5 = = 1- 2 -1 - 3 4-5 x-2 = 1

or

(b)

y-3 z-5 = 4 1

Let coordinates of P be (r + 2, 4r + 3, r + 5). P lies on 5x – 4y – z = 1, we get 5(r + 2) – 4 (4r + 3) – (r + 5) = 1 fi r = – 2/3 Thus, coordinates of P are (4/3, 1/3, 13/3). Let coordinates of S on QR be (l + 2, 4l + 3, l + 5). ST is perpendicular to QR, we get (l + 2 – 2) (1) + (4l + 3 – 1) (4) + (l + 5 – 4) (1) = 0 fi l = –1/2 Thus, coordinates of S

Fig. 20.3

Now,

IIT JEE eBooks: www.crackjee.xyz 20.9 2 2 2 Ê 3 4ˆ Ê 1ˆ Ê 9 13 ˆ PS2 = Á - ˜ + Á1 - ˜ + Á - ˜ Ë 2 3¯ Ë 3¯ Ë2 3¯

= fi

1 4 1 1 + + = 36 9 36 2

PS = 1 / 2

Example 5 the line of intersection of the planes x + 2y + 3z = 2 and 2 x – y + z = 3 and at a distance 3 (a) 5x – 11y + z = 17 2x + y = 3 2 – 1

(b)

(c) x + y + z =

3

(d) x – 2 y = 1 – 2 Ans. (a) Solution: intersection of x + 2y + 3z – 2 = 0 and x – y + z – 3 = 0 is (x + 2y + 3z – 2) + l (x – y + z – 3) = 0 or (1 + l)x + (2 – l)y + (3 + l)z – (2 + 3l) = 0

(1+ l )(3) + ( 2 - l )(1) + (3 + l )( -1) - ( 2 + 3l ) (1 + l )2 + (2 - l )2 + (3 + l )2 = 2l



14 + 4l + 3l 2

fi fi

=

2 3 2 3

3l2 = 14 + 4l + 3l2 l = – 7/2

2(x + 2y + 3z – 2) – 7(x – y + z – 3) = 0 fi 5x – 11y + z = 17 Example 6 x + 2 y +1 z = = to the plane x + y + z = 3. The 2 -1 3 feet of perpendiculars lie on the line x y -1 z - 2 x y -1 z - 2 = = = = (a) (b) -13 -13 5 8 5 8 the line

(c) Ans. (d) Solution:

x y -1 z - 2 = = 4 3 -7

(d)

x y -1 z - 2 = = -7 2 5

Coordinates of any point on x + 2 y +1 z = = 2 -1 3

are P (2r – 2, –r – 1, 3r). If Q x + y + z = 3, then PQ

P to the plane

x - (2r - 2) y - (- r - 1) z - 3r = = = s (say) 1 1 1 Let coordinates of Q be (s + 2r – 2, s – r – 1, s + 3r) x + y + z = 3, we get (s + 2r – 2 ) + (s – r – 1) + (s + 3r) = 3. 1 fi 3s + 4r – 6 = 0 fi s = - ( 4r - 6) 3 Thus, coordinates of Q are Ê 2 r , -7 r + 1, 5 r + 2ˆ ÁË ˜¯ 3 3 3 Taking x =

2 7 5 r , y = – r + 1, z = r + 2 , 3 3 3

we get (x, y, z) lie on the line 3x 3( y - 1) 3( z - 2) = =r = 2 -7 5 or on

x y -1 z - 2 = = . 2 -7 5

ABC Example 7 points of its sides are on the x, y and z axes respectively. the triangle on these axes are respectively a, b, g, then the coordinates of the centroid of the triangle ABC are (a) (– a/3, b/3, g /3) (b) (a/3, – b/3, g/3) (c) (a/3, b/3, – g /3) (d) (a/3, b/3, g/3) And. (d) Solution: x y z ABC is + + = 1 a b g a, 0, 0), (0, b, 0) and (0, 0, g ). Let the coordinates of A be (x1, y1, z1) AB lies on the z g) and thus the coordinates of B are (– x1, – y1, 2g – z1) C are (– x1, 2b – y1, – z1) BC = (– x1, b – y1, g – z1) = (a, 0, 0) fi x1 = – a, y1 = b, z1 = g. and thus the coordinates of A are (– a, b, g ) B are (a, – b, g) and those of C are (a, b, – g ). Hence the coordinates of the centroid of the triangle ABC are (a/3, b /3, g /3).

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P(1, 0, 0) in the

Example 8 line

x -1 y +1 z + 10 = = is 2 -3 8

(3, 0, 3). S o PQ =

(a) (3, – 4, – 2) (b) (5, – 8, – 4) (c) (1, – 1, – 10) (d) (2, – 3, 8) Ans. (b) Solution: Coordinates of any point Q on the given line are (2r + 1, – 3r – 1, 8r – 10). So the direction ratios of PQ are 2r, – 3r – 1, 8r – 10. Now PQ is perpendicular to the given line. if 2(2r) – 3(– 3r – 1) + 8(8r – 10) = 0 fi 77r – 77 = 0 fi r = 1 and the coordinates of Q P on the line are (3, – 4, – 2). Let R (a, b, c P in the given lines then Q PR b c a +1 = 3, = – 4, =–2 2 2 2 a = 5, b = – 8, c = – 4

fi fi

Let P(3, 2 ,6) be a point in space and Q Example 9 be a point on the line r = (i – j + 2k) + m (– 3i + j + 5k) Then the value of m for which the vector PQ is parallel to the plane x – 4 y + 3 z = 1 is (a) 1/4 (b) – 1/4 (c) 1/8 (d) – 1/8 Ans. (a) Solution: Let O be the origin, as Q lies on the given line, OQ = i – j + 2k + m (–3i + j + 5k OP = 3i + 2j + 6k. PQ is parallel to the plane x – 4 y + 3 z = 1. PQ ◊ (i – 4j + 3k) = 0 fi

(OQ – OP) ◊ (i – 4j + 3k) = 0



(1 + 4 + 6 ) + m (– 3 – 4 + 15) – (3 – 8 + 18) = 0



11 + 8 m – 13 = 0 fi m = 1/4.

Example 10 passes through the point P 2x+y+z

k + 2, k – 1, k + 2) which lies on the plane 2 x + y + z k + 2) + k – 1 + k fi k = 1, and for this values of k, the coordinates of Q are

Q.

Example 11

(3 - 2) 2 + (0 + 1) 2 + (3 - 2) 2 =

The angle between the lines whose l2 + m2 – n2 = 0, l + m + n = 0 is (b) p/4 (c) p/3 (d) p/2

(a) p/6 Ans. (c) Solution: We also have l2 + m2 + n2 = 1

So that l2 + m2 – n2 = 0 fi 2n2 = 1 fi n = ± 1/ 2 and l + m + n = 0 fi (l + m)2 = n2 = 1/2 = l2 + m2 fi 2 lm = 0 fi l = 0 or m = 0 If

l = 0, m + n = 0 fi m = – n = ± 1/ 2 So direction ratios of one of the lines are 0, ± 1/ 2 ,

1/ 2

and if m = 0, l + n = 0 fi l = – n = ± 1/ 2 So the direction ratios of the other line are ± 1/ 2 , 0,

1/ 2

cos–1 ÈÎ0 ¥ (± 1/ 2 ) + (± 1/ 2 )(0) + (± 1/ 2 )(± 1/ 2 ) ˘˚ = cos–1 1/2 = p/3. Example 12

The shortest distance between the skew

lines l1 : r = a1 + l b1 and l2 : r = a 2 + mb 2 is (a)

(a 2 - a1 ) ◊ b1 ¥ b 2 | b1 ¥ b 2 |

(b)

(a1 - b1 ) ◊ a 2 ¥ b 2 | b1 ¥ b 2 |

(c)

(a 2 - b 2 ) ◊ a1 ¥ b1 | b1 ¥ b 2 |

(d)

(a1 - b 2 ) ◊ b1 ¥ a 2 | b1 ¥ a 2 |

Ans (a) Solution: Let PQ be the line of shortest distance between l1 and l2. Now l1 passes through A1(a1) and is parallel to b1 and l2 passes through A2(a2) and is parallel to b2. Since PQ is perpendicularto both l1 and l2, it is parallel to b1 × b2. b ¥ b2 Let n be the unit vector along PQ then n = 1 . | b1 ¥ b 2 |

PQ (a) 1 (b) 2 (c) 3 Ans. (c) Solution P x-2 y +1 z-2 = = = k (say) 1 1 1

3.

(d) 2

Fig. 20.4

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If d is the shortest distance between the given lines then |PQ| = d and PQ = dn. Next, PQ being the line of shortest distance between l1 and l2 is the projection of the line joining A1(a1) and A2(a2) on n. |PQ| = | A1A 2 ◊ n | fi

d=

(a 2 - a1 ) ◊ b1 ¥ b 2 | b1 ¥ b 2 |

Example 13 between the plane 3x + 4y – 5z – 60 = 0 and the coordinate planes in cubic units is (a) 60 (b) 600 (c) 720 (d) none of these Ans. (b) Solution: x y z + + =1 20 15 -12

Example 15

The lines r = i - j + l (2i + k ) and

r = (2i - j) + m (i + j - k ) intersect for (a) l = 1, m = 1 (b) l = 2, m = 3 (c) all values of l and m (d) no value of l and m. Ans. (d) Solution: The given lines intersect, if the shortest distance between the lines is zero. We know that the shortest distance between the lines r = a1 + l b1 and r = a 2 + m b 2 is | (a1 - a 2 ) ◊ b1 ¥ b 2 | | b1 ¥ b 2 | So the shortest distance between the given lines is zero if (i - j - (2i - j)) ◊ (2i + k) ¥ (i + j - k ) = 0 -1 0 0 = 2 0 1 1 1 -1

L.H.S.

=1 π 0

Hence the given lines do not intersect. A(20, 0, 0), B(0, 15, 0), C(0, 0, – 12) and the coordinates of the origin are (0, 0, 0) \ OABC is 0 0 0 0 1 20 0 0 6 0 15 0 0 - 12

1 1 = 1 1

Example 16 x = ± 1, y = ± 1, z = ± 1 is 10 units. The locus of the point is

1 ¥ 20 ¥ 15 ¥ (-12) 6

= 600

Example 14 If the planes x = cy + bz, y = az + cx and z = bx + ay pass through a line, then a2 + b2 + c2 + 2abc is (a) – 1 (b) 0 (c) 1 (d) none of these Ans. (c) Solution: Let l, m, n be the direction ratios of the line lying in the three planes, then the line is perpendicular to l – cm – bn = 0 – cl + m – an = 0 – bl – am + n = 0 l, m, n 1 - c -b -c 1 -a = 0 -b - a 1 fi fi

1 – a2 + c(– c – ab) – b(ac + b) = 0 a2 + b2 + c2 + 2abc = 1

(a) x2 + y2 + z2 = 1 (b) x2 + y2 + z2 = 2 (c) x + y + z = 1 (d) x + y + z = 2 Ans. (b) Solution: Let P(x, y, z) be any point on the locus, then | x + 1|, |x – 1|, |y + 1|, |y – 1|, |z + 1|, |z – 1| |x + 1|2 + |x – 1|2 + | y + 1|2 + |y – 1|2 + |z + 1|2 + |z – 1|2 = 10 fi 2(x2 + y2 + z2) = 10 – 6 = 4 fi x2 + y2 + z2 = 2. x+3y+4 Example 17 z – 3 = 0 in the plane x – y + z – 3 = 0 is the plane (a) 4x – 3y + 2z –15 = 0 (b) x – 3 y + 2z – 15 = 0 (c) 4x + 3y – 2z + 15 = 0 (d) none of these Ans. (a) Solution a¢ x + b¢ y + c¢ z + d¢ = 0 in the plane ax + by + cz + d = 0 is given by 2 (aa¢ + bb¢ + cc¢) (ax + by + cz + d) = (a2 + b2 + c2) (a¢ x + b¢ y + c¢ z + d¢) 2 (2 + 3 + 4) (x – y + z – 3) = (1 + 1 + 1) (2x – 3y + 4z – 3)

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fi fi

6 (x – y + z – 3) = 2x – 3y + 4z –3 4x – 3y + 2z – 15 = 0

Example 20 A, B, C with position vectors - 6i + 3j + 2k , 3i - 2 j + 4k , 5i + 7 j + 3k is

Note The plane x – y + z

(a) r.(i - j - 7k ) + 23 = 0 (b) r.(i + j + 7k ) = 23 (c) r . (i + j - 7k ) + 23 = 0

Example 18

The plane 2x – y + 3z + 5 = 0 is rotated

5x – 4y – 2z position is

(d) r. (i - j - 7k ) = 23 Ans. (a) Solution: points A, B, C with position vectors a, b, is r ◊ (a × b + b × + × a) = a . So that if a, b, represent the given vectors, then (a × b + b × + × a) = i j k i j k i j k -6 3 2 + 3 -2 4 + 5 7 3 3 -2 4 5 7 3 -6 3 2

(a) 6x y z – 31 = 0 (b) 27x – 24y – 26z – 13 = 0 (c) 43x – 32y – 2z + 27 = 0 (d) 26x – 43y – 151z – 165 = 0 Ans. (b) Solution: intersection of the given planes is

= i (12 + 4 – 6 – 28 + 14 –

2x – y + 3z + 5 + l (5x – 4y – 2z + 1) = 0

+ 10 + 18) + k

or (2 + 5l)x – (1 + 4l)y + (3 – 2l)z + 5 + l = 0 This will be perpendicular to the plane 2x – y + 3z + 5 = 0 if 2 (2 + 5l) + (1 + 4l) + 3(3 – 2l) = 0 fi l 4(2x – y + 3z + 5) – 7(5x – 4y – 2z + 1) = 0 fi 27x – 24y – 26z – 13 =0 Example 19

r = 2i + 3j - k + (i + j + k ) t and the plane

and

a.b ×

2 6 15

(c) cos q = -

11 7 70

2 6 15

(d) sin q = -

Ans. (b) Solution:

11 7 70

The line is parallel to the vector i + j + k and i – 4j+ 5k, so that the angle between these vectors is p/2 – q. (i + j + k ) ◊ (3i - 4i + 5k ) cos (p / 2 – q) = 1 + 1 + 1 32 + 42 + 52 sin q =

3- 4+ 5

r. (i - j - 7k ) + 23 = 0

Example 21 (b) sin q =

2 6 = 15 3 50

-6 3 2 = 3 -2 4 = 299 5 7 3

r. (-13i + 13j + 91k ) fi

r ◊ (3i - 4 j + 5k ) = q, then



= -13i + 13j + 91k

If q is the angle between the line

(a) cos q =

j

between the lines

x y z x-2 y -1 z + 2 = = and = = -5 2 -3 1 3 2

is (a) 3 (x – 21) = 3y z – 32 x - (62/3) y - 31 z + (31/3) = = (b) 1/3 1/3 1/3 (c)

x - 21 y - (92/3) z + (32/3) = = 1/3 1/3 1/3

(d)

x - 2 y + 3 z -1 = = 1/3 1/3 1/3

Ans. (a) Solution:

Let P (2r1, - 3r1, r1) and Q (3r2 + 2, -5r2 +

1, 2r2 - 2) be the points on the given lines so that PQ is the line of shortest distance between the given lines. Now direction ratios of PQ are

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2r1 -3r2 - 2, - 3r1+ 5r2 - 1, r1 - 2 r2 + 2. Since it is perpendicular to the given lines 2 (2r1-3r2 -2) - 3 (-3r1+ 5r2 -1)+ (r1 - 2r2 + 2) = 0



1- x 1 1 1 1- y 1 =0 1 1 1- z



fi P is (62/3, - 31, 31/3) and Q is ( 21, - 92/3, 32/3) and direction ratios of PQ are 1/3, 1/3, 1/3. PQ as it passes through Q are

1- x x x

fi fi

x (0 + y) – z (– y + x y – x) = 0 xy+yz+zx=xyz

x - 21 y + 92/3 z - 32/3 = = 1/3 1/3 1/3



1 1 1 + + =1 x y z

and 3(2r1 -3r2 -2) -5(-3r1 + 5r2 -1) + 2(r1 - 2r2 + 2) = 0 fi

14r1 - 23r2 + 1 = 0, 23r1 - 38r2 + 3 = 0



r1 = 31/3, r2



3(x - 21) = 3y

z - 32

a, b, g with the Example 22 coordinate axes. If a + b = p/2, then (cos a + cos b + cos g)2 (a) 1 + sin 2a (b) 1 + cos 2a. (c) 1 – sin 2a (d) none of these Ans. (a) Solution: We have cos2 a + cos2 b + cos2 g = 1 fi

cos2 a + cos2 (p/2 – a) + cos2 g = 1



cos2 a + sin2 a + cos2 g = 1



cos2 g = 0 fi cos g = 0

So (cos a + cos b + cos g)2 = (cos a + sin a)2 = 1 + 2 sin a cos a = 1 + sin 2a. Example 23 If the points (2 – x, 2, 2), (2, 2 – y, 2) (2, 2, 2 – z) and (1, 1, 1) are coplanar, then 1 1 1 (a) + + = 1 (b) x + y + z = 1 x y z (c)

1 1 1 + + =1 x -1 y -1 z -1

(d) none of these

Ans. (a) Solution



1 1 2- x 2 2 2- y 2 2

1 2 2 2- z

1 1 1 1

=0

1 -y 0

1 0 -z

=0

Example 24 the point of intersection of lines

È R2 Æ R2 - R1 ˘ ÍR Æ R - R ˙ Î 3 3 1˚

x -1 y-2 z-3 = = 3 1 2

x-3 y -1 z-2 = = 1 2 3 the origin is (a) 7x + 2y + 4z = 54 (b) 4x + 3y + 5z = 50 (c) 3x + 4y + 5z x + y + z = 12 Ans. (b) Solution P (3l + 1, l + 2, 2l + 3) and on the second line is Q (m + 3, 2 m + 1, 3m + 2) P and Q l = m = 1 and the pint of intersection of the given lines is P (4, 3, 5). The planes given is (a), (b), (c) and (d) all pass through P. The and

to OP 2 2 2 the origin is 4 + 3 + 5 = the plane in (b).

50 O (0, 0, 0)

Example 25

A (1, 2, 1), B (2, 1, 3) and C (–1, 1, 2). The angle between the faces OAB and ABC is 19 17 (a) cos–1 (b) cos–1 35 31

( )

( )

Ans. (a) Solution ax + by + cz = 0 st a + 2b + c = 0 and 2a + b + 3c = 0 fi

OAB is a b c = = 5 -1 -3

and the direction ratios of the plane are 5,– 1, –3



1 1 1 1- x 1 1 1 1- y 1 1 1 1- z

1 0 0 0

=0

ABC are 1, –5, –3 If q 5 ¥ 1 + (- 1) (- 5) + (- 3) (- 3) 19 = . cos q = 35 52 + 12 + 32 12 + 52 + 32

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MULTIPLE CORRECT ANSWERS TYPE QUESTIONS If the straight lines

Example 26

and direction of L2 is N2 = 2i + 2j + k Thus, direction of L, which is perpendicular to both L1 and L2 is

z x -1 y +1 = = 2 2 k

i N= 1 2

x +1 y +1 z = = are coplanar, then the plane(s) 5 2 k containing these two lines is(are) (a) y + 2z = – 1 (b) y + z = – 1 (c) y – z = – 1 (d) y – 2z = – 1 Ans. (b), (c) and

fi fi

-2 2 5

-1 - ( -1)

0-0

2 5

k 2

2 k

0 k 2





=0

0 2 =0 k

y +1 k 2

z 2 =0 k

(x – 1) (k2 – 22) – (y + 1) (2k – 10) + z(4 – 5k) = 0 Taking k = 2, we get 6 (y + 1) – 6z = 0 or y – z + 1 = 0 Taking k = – 2, we get 14(y + 1) + 14z = 0 y+z+1=0

L passing through the origin is Example 27 perpendicular to the lines L1 : (3 + t)i + (–1 + 2t)j + (4 + 2t)k, –• < t < • L2 : (3 + 2s)i + (3 + 2s)j + (2 + s)k, –• < s < • Then, the coordinate(s) of the point(s) on L2 at a L distance of 17 and L1 is (are) Ê 7 7 5ˆ (a) Á , , ˜ Ë 3 3 3¯

(b) (–1, –1, 0)

(c) (1, 1, 1)

Ê 7 7 8ˆ (d) Á , , ˜ Ë 9 9 9¯

Ans. (b), (d) Solution: Direction of L1 is N1 = i + 2j + 2k

2

1

on L2 is given to be 17 . Thus, AP = 17 fi AP2 = 17 fi (2s + 1)2 + (2s + 6)2 + (s)2 = 17 fi s2 + 28s + 20 = 0 fi s + 10) (s + 2) = 0 fi s

–2 [k2 – 22] = 0 fi k = ± 2 x -1 2 5

k 2

= –2i + 3j – 2k L is r = l (– 2i + 3j – 2k) For point of intersection of L and L1, set 3 + t = –2l, –1 + 2t = 3l, 4 + 2t = –2l fi t = –1, l = –1 Thus, point of intersection of L and L1 is A (2, –3, 2). Distance of A P(3 + 2s, 3 + 2s, 2 + s)

Solution -1 - 1

j 2

(–1, –1, 0). Example 28 x = a,

Two lines L1 : x = 5,

y z = and L2 : 3 - a -2

y z = are coplanar. Then a can take value(s) -1 2 - a (a) 1

(b) 2

(c) 3

(d) 4

Solution: The lines L1 and L2 are x-5 y z = = 0 3 - a -2 and

x -a y z = = . 0 -1 2 - a

0 0 a -5 fi fi fi

3-a -1 0-0

-2 2-a = 0 0-0

(a – 5)[(3 – a) (2 – a) – 2] = 0 (a – 5) (a2 – 5a + 4) = 0 a = 5, 1, 4

Example 29

P(λ, λ, λ) perpendiculars

PQ and PR are drawn respectively on the lines y = x, z = 1 and y = – x, z = –1. If P is such that ∠QPR is a right angle, then the possible value(s) of λ is(are)

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(a)

2

Distance of P3 -1 + k = 1 ⇒ (k – 1)2 = k2 + 2 2 1+ k +1

(b) 1

(c) –1 (d) – 2 Ans. (c) Solution: Let the line L1 be x = y, z = 1. a, a, 1) Q

L1

⇒ k2 – 2k + 1 = k2 + 2 ∴ P3 is 1 (x + z – 1) - y = 0 2



k= -

1 2

or 2x – y + 2z – 2 = 0*

...(1) α, β, γ) is 2,

P

2a - b + 2g - 2

R

4 +1+ 4

L2

Fig. 20.5

Let coordinates of Q be (a, a, 1) Direction ratios of PQ are l – a, l – a, l – 1. PQ ^ L1 1(l – a) + 1(l – a) + 0(l – 1) = 0 fi a = l. \ Coordinates of Q are (l, l, 1). Direction ratios of PQ are 0, 0, l – 1. Let line L2 be x = – y, z L2 is b, – b, – 1). Let coordinates of R be (b, – b, – 1) Direction ratios of PR be l – b, l + b, l + 1 PR ^ L2, we get 1 (l – b) + (– 1) (l + b) + 0 (l + 1) = 0 fi b = 0. Thus coordinates of R (0, 0, – 1) Direction ratios of PR are l, l, l + 1 QP ^ PR, (l + 1) (l – 1) = 0 fi l = – 1, 1 Reject l = 1 as for l = 1, P and Q coincide. Example 30

Let R3, consider the plane P1 : y = 0 and

P1 and P2, P2 : x + z = 1. Let P3 which passes through the intersection of P1 and P2. If the P3 is 1 and the distance of a point (α, β, γ P3 is 2, then which of the following relations is (are) true ? (a) 2α + β + 2γ + 2 = 0 (b) 2α – β + 2γ + 4 = 0 (c) 2α + β – 2γ – 10 = 0 (d) 2α – β + 2γ – 8 = 0 Ans. (b), (d) Solution P1 and P2 is P3 : (x + z – 1) + ky = 0.

=2

Note *Plane P1

P3 P 1.

⇒ 2α – β + 2γ – 2 ± 6 ⇒ 2α – β + 2γ + 4 = 0 or 2α – β + 2γ – 8 = 0 Example 31

In R3, let L be a straight line passing

through the origin. Suppose that all the points on L are at a constant P1 : x + 2y – z + 1 = 0 and P2 : 2x – y + z – 1 = 0. Let M be the locus of the feet of L to the plane P1. Which of the following points lie(s) on M? 5 2ˆ Ê (a) Á 0, - , - ˜ Ë 6 3¯

Ê 1 1 1ˆ (b) Á - , - , ˜ Ë 6 3 6¯

Ê 5 1ˆ (c) Á - , 0, ˜ Ë 6 6¯

Ê 1 2ˆ (d) Á - , 0, ˜ Ë 3 3¯

Ans. (a), (b) Solution L P1 and P2, L is parallel to both P1 and P2. If direction ratios of L are a, b, c then a + 2b – c = 0 2a – b + c = 0 a -b c ⇒ = = 2 -1 1+ 2 -1 - 4 ⇒

a b c = = 1 -3 -5 L is

x y z = = = λ(say) 1 -3 -5

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L is (λ, –3λ, –5λ) Let I (α, β, γ) be foot of perpendicular of P (λ, –3λ, –5λ) on the plane P1 : x + 2y – z + 1 = 0 then IP P1. Therefore a -l b + 3l g + 5l = = = k(say) 1 2 -1 ⇒ α = λ + k, β = 2k – 3λ, γ = –k – 5λ I lies on P1 (λ + k) + 2(2k – 3λ) – (–k – 5λ) + 1 = 0

So fi fi

1 1 1 Thus, α = λ – , β = –3λ – , γ = –5λ + 6 3 6

{

1 1 1ˆ Ê ∴ M = Á l - , -3l - , - 5l + ˜ l Œ R Ë 6 3 6¯

}

Ê 1 1 1ˆ For λ = 0, Á - , - , ˜ Œ M Ë 6 3 6¯ 1 Ê -5 -2 ˆ , Á 0, , ˜ Œ M 6 Ë 6 3¯

Example 32

(a) 1

3, -1

3, 1

3

(b) 1

3 , -1

3 , -1

3

(d) 1

3, -1 3, 1

3, 1

3, 1

6

=

2 (2 + 3l ) + (l - 1) - (l + 1) - 3 - 5l (2 + 3l ) 2 + (l - 1)2 + (l + 1)2

11l2 + 12l + 6 = 6 (l – 1)2 = 6 (l2 – 2l + 1) 5 l2 + 24l = 0 fi l = 0 or l = – 24/5.

2x – y + z – 3 = 0 or 62x

y

z – 105 = 0.

The plane passing through the origin and

tional to 1, – 2, 2 and 2, 3, – 1 passes through the point. (a) (1, – 2, 2) (b) (2, 3, – 1) (c) (3, 1, 1) (d) (4, 0, 7) Ans. (a), (b) and (c) Solution: plane be l, m, n the lines lying in the plane, whose direction cosines are proportional to 1, - 2, 2 and 2, 3, -1; we have l - 2m + 2n = 0 and 2l + 3m - n = 0 m m l fi = = -4 5 7

inclined with the coordinate axes are

(c) - 1

1

Example 34

1 ⇒ 6k = – 1 ⇒ k = – 6

and for λ =

Solution 2x – y + z – 3 + l (3x + y + z – 5) = 0 fi (2 + 3l) x + (l –1) y + (l + 1) z – (3 + 5l) = 0

lx + my + nz = 0 or 4x - 5y - 7z = 0 which clearly passes through the points given in (a), (b), (c). Example 35

3

3

Ans. (a), (b), (c) and (d) Solution: Let the line be inclined at an angle a with each of the three coordinate axes, that the direction cosines of the line are cos a, cos a, cos a. and cos2 a + cos2 a + cos2 a = 1 fi cos2 a = 1/3 fi cos a = ± 1 3

The plane passing through the point

(– 2, – 2, 2) and containing the line joining the points a, b, c respectively on the axes of x, y and z respectively, then (a) a = 3b (b) b = 2c (c) a + b + c = 12 (d) a + 2b + 2c = 0 Ans. (a), (b) and (c) Solution: (-2, -2, 2) is A (x + 2) + B ( y + 2) + C (z – 2) = 0

Example 33 lines 2x – y + z – 3 = 0, 3x + y + z – 5 = 0 and which is at a distance of 1 / 6 (a) 2x – y + z –3 = 0 (b) 3x + y + z – 5 = 0 (c) 62x y z = – 105 = 0 (d) x + 2 y – 2 = 0 Ans. (a), (c)

Since it contains the line joining (1, 1, 1) and (1, – 1, 2) these points also lie on this plane. fi 3A + 3B - C = 0 and 3A + B + 0 = 0 fi

or

A B C = = . 1 -3 -6 (x + 2) - 3 (y + 2) - 6 (z - 2) = 0 x - 3y - 6z + 8 = 0

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2i + 3j + 4k + l (i + j + 3k)

or

x y z + + =1 - 8 8/3 8/6



a = 8, b = 8/3, c = 8/6



a = 3b, b = 2c, a + b + c = 12

and

a + 2b + 2c = 16

Example 36 (a) – 27, 18, 54 (b) 27, – 18, – 54 (c) 27, – 18, 54 (d) – 27, 18, – 54 Ans. (c) and (d) Solution: be a, b, c, then (1) a2 + b2 + c2 = (63)2 a b c = = = l (say) then also 3 -2 6 a = 3l, b = - 2l and c = 6l l2 + 4l2 + 36l2 = (63)2



Example 37

The lines

fi fi

x-2 y-3 z-4 = = 1 1 -k

and

x -1 y - 4 z - 5 = = are coplanar if k 2 1 (a) k = 0 (b) k = – 1 (c) k = 2 (d) k = – 3 Ans. (a) and (d) Solution: The given lines are coplanar if

Example 38

) (

)

a+2 b+3 c+4 5 7 9 , , , , = 2 2 2 2 2 2 (a, b, c) = (3, 4, 5) AB is not perpendicular to BC because 1 × 1 + 1 × 1 + 3 ¥ 5 π 0. ABCD is not a rectangle. The coordinates of a point on the line

x -1 y +1 = = z at a distance 4 14 -3 2 (1, – 1, 0) are

1 0 0 2 1 = 1 k k + 2 1+ k 2(1 + k) - (k + 2) (1 - k) = 0 if k = 0, - 3.

(

Example 39

2 -1 3 - 4 4 - 5 1 -1 -1 -k = 1 1 -k 1 0= 1 k 2 1 k 2 1

or if or

x-2 y-3 z-4 = = 1 1 5 Next, if D is (a, b, c), then since ABCD BD

l2 = (63)2

63 l=± 7



x-2 y-3 z-4 = = 1 1 5 (c) Coordinates of D are (3, 4, 5) (d) ABCD is a rectangle Ans. (a), (b), (c), Solution: Direction ratios of AB are 4 – 2, 5 – 3, 10 – 4 or 1, 1, 3. So AB in parallel to the vector i + j + 3k and passes through B (2, 3, 4), the vector 2i + 3j + 4k 2i + 3j + 4k + l (i + j + 3k) BC passes through the points B (2, 3, 4) and its direction ratios are 2 – 1, 3 – 2, 4 + 1 or 1, 1, 5.

(b) ( 8 14 + 1, – 12 14 – 1, 4 14 ) (c) (– 7, 11, – 4) (d) (– 8 14 + 1, 12 14 – 1, – 4 14 ) Ans. (a) and (c) Solution: The coordinates of any point on the given line are (2r + 1, – 3r - 1, r)

- 1, 0) is given

2

or if k + 3k = 0 to be 4 14 .

The points A (4, 5, 10), B (2, 3, 4) and C

fi fi

(

(2r)2 + (- 3r)2 + (r)2 = 4 14 14r2 = 16 ¥ 14 fi r = ± 4.

)

2

ABCD, then - 13, 4) or (- 7, 11, - 4).

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OPQRS located in x ≥ 0, y ≥ 0, z ≥ 0) with O as origin, and OP and OR along the x y OPQR OP = 3. The point S T of diagonal OQ such that TS = 3. Then p (a) the acute angle between OQ and OS is . 3 Example 40

O to RS be M, and let M divide RS in the ration k : 1. Coordinate of M are Ê 3k /2 , (3k /2) + 3 , 3k ˆ ÁË ˜ k +1 k +1 k + 1¯ Direction ratio of OM are Direction ratio of RS :

P to the 3 2

fi 6k = 3 fi k =

O to the straight line containing RS is

15 2

= S (3/2, 3/2, 3)

= R (0, 3, 0)

3 P

y

T (3, 0, 0)

=

Fig. 20.6

3 3 OQ = 3i + 3j OS = i + j + 3k 2 2 Let q be angle between OQ and OS, then OQ ◊ OS cos q = OQ OS

=

3 3 3Ê ˆ + 3Ê ˆ Ë 2¯ Ë 2¯ 9+9

=

1

9 9 + +9 4 4 =

1

2 3/2 3 O, Q, S lies on x – y = 0, the plane containing triangle OQS is x – y = 0. P to the plane 3-0 3 = containing OQS = 1+1 2

3 k 2 + ( k + 2 )2 + 4 k 2 2 ( k + 1) 3 3 Ê 2 ˆ Ë 2¯

6 4 + +4 4 2

15 2

MATRIX-MATCH TYPE QUESTIONS

Q (3, 3, 0)

x

1 2

2 1 9 2 Ê3 k + Á k + 3ˆ˜ + (3k )2 Ë2 ¯ k +1 4

Thus (OM)2 =

Ans. (b), (c), (d) Solution:

O

3 -3 , , 3 or 1, –1, 2 2 2

OM ^ RS, we get Ê 3k /2 ˆ 1 + Ê (3k /2) + 3 ˆ -1 + Ê 3k ˆ 2 ÁË ˜ ( ) ÁË ˜( ) =0 ˜ ( ) ÁË k + 1¯ k +1 ¯ k + 1¯

angle OQS is x – y = 0. plane containing the triangle OQS is

3k /2 (3k /2) + 3 3k , , k +1 k +1 k +1

Consider the lines L1 :

Example 41

x -1 y z + 3 = = , 2 -1 1

x-4 y+3 z+3 = = and the planes P1: 7x + y + 1 1 2 2z = 3, P2 : 3x + 5y – 6z = 4. Let ax + by + cz = d be and L2 :

intersection of lines L1 and L2, and perpendicular to planes P1 and P2, then Column 1 Column 2 (a) a = (p) 13 (b) b (c) c = (r) 1 (d) d = (s) –2 p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

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Solution:

L1 and L2. L1 l1 + 1, –l1, l1 – 3) and a point on L2 l2 + 4, l2 – 3, 2l2 – 3). For point of intersection of L1 and L2, we set 2l1 + 1 = l2 + 4, – l1 = l2 – 3, l1 – 3 = 2l2 – 3 fi l1 = 2, l2 = 1 Thus, point of intersection of L1 and L2 is (5, –2, –1). A(x – 5) + B (y + 2) + C(z + 1) = 0 This plane will be perpendicular to the planes P1 and P2 if 7A + B + 2C = 0 3A + 5B – 6C = 0 A -B C = = fi -6 - 10 -42 - 6 35 - 3 A B C = = 1 -3 -2



Thus, a = 1, b = – 3, c = – 2 and d = 5a – 2b – c = 13. Example 42 Column 1 Column 2 (a) Centroid of the triangle with vertices (p) (2, 3, 5) A(2, 3, 7), B(6, 7, 5), C(1, 2, 3)

B(- 1, - 1, - 1) x y z = = 2 3 5 (r) (3, 3, 2) (d) Coordinates of the point dividing the join of (5, 5, 0) (s) (4/ 38 ,

(4/ 38, 6/ 38, 10/ 38) & (2, 3, 5) (d) Coordinates of the point are

(

3¥5+ 2¥0 3¥5+ 2¥0 3¥0+ 2¥5 , , 5 5 5 = (3, 3, 2) Example 43

Column 1 Column 2 x-2 y-7 z+5 (a) = = (p) Perpendicular to the 3 4 2 plane 3x + 4y + 2z =1 x +1 y-3 z+7 = = (b) 3 4 2 (2, 7, - 5) x-5 y+2 z-2 = = (r) direction cosines are (c) 1 3 4 2/ 30, 5/ 30, 1/ 30 (d)

6/ 38 ,

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

Solution:

10/ 38 )

x y–2 z+6 = = 2 5 1 p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

A

and (0, 0, 5) in the ratio 2 : 3.

)

(s) lies in the plane 7x - y - z = 35

(a) and (b) pass through (2, 7, -5), are parallel x + 4y + 2z = 1 and hence is perpendicular to it (c) Point (5, - 2, 2) lies on (c) and Solution:

(s) as 1 × 7 - 3 × 1 - 4 × 1 = 0. So (c) lies in the plane (s). 2 5 , , (d) direction cosines are 4 + 25 + 1 4 + 25 + 1 1 4 + 25 + 1 =

2

,

5

,

1

30 30 30 and passes through (2, 7, – 5)

(a) centroid of DABC

as

2 + 6 +1 3+ 7 + 2 7 + 5 + 3 , , 3 3 3 = (3, 4, 5) =

AB =

(

)

7 - 1 9 - 1 11 - 1 , , = (3, 4, 5) 2 2 2

2 7–2 –5+6 = = 2 5 1

Example 44 ax + by + cz + d = 0,

x -a y-b z -g = = l m n

.

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Column 1 Column 2 (a) line is perpendicular to (p) if al + bm + cn = 0 the plane al + bm + cn = 0 plane if and aa + bb + cg + d=0 a b c = = (c) line lies in the plane (r) if l m n (d) line intersects the plane (s) aa + bb + cg + d = 0 p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

Solution:

(a) line is perpendicular to the plane if it is

a b c = = l m n (b) It is parallel to the plane if al + bm + cn = 0 (c) it lies in the plane if it is parallel to the plane and the point (a, b, g) lying on the line lies on the plane. i.e.

x-2 y-3 z-4 = = 3 4 5

Example 45

Column 1 (a) Point on the line at a

Column 2 (p) (- 1, - 1, - 1)

to the plane x+y+z+3=0 (c) Point on the line at a distance (r) (8, 11, 14) 29

p

the plane x + y – z + 3 = 0 q r s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(- 1, - 1, - 1) (c) (3r + 2)2 + (4r + 3)2 + (5r + 4)2 50r2 + 76r = 0 fi r = 0, r = - 76/50 For r = 0, the point is (2, 3, 4). (d) r = – 2 Example 46 If ax + by + gz + d = 0 represents a plane passing through the points (- 6, 3, 2), (3, - 2, 4) and (5, 7, 3) Column 1 Column 2 (a) a (p) 7 (b) b (c) g (r) 23 (d) d (s) - 1 p q r s Ans. a p q r s b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution

distance 10 2 (2, 3, 4)

Ans.

For r = 2, the point is (8, 11, 14), For r = – 2 it is (– 4, – 5, – 6) (b) 3r + 2 + 4r + 3 + 5r + 4 + 3 = 0 fi 12r + 12 = 0 fi r = - 1

Solution r + 2, 4r + 3, 5r + 4) 2 2 (a) (3r + 2 - 2) + (4r + 3 - 3) + (5r + 4 - 4)2 = 200 fi r2 = 200, fi r = ± 2

r ◊ (a ¥ b + b ¥ c + c ¥ a) = a . b ¥ c where a = - 6 i + 3 j + 2k , b = 3 i - 2 j + 4k = 5 i + 7 j + 3k fi r ◊ (i - j + 7 k ) + fi x - y + 7z + 23 = 0 fi a = 1, b = - 1, g = 7, d = 23 Example 47 Column 1 (a) Point (a, b, g ) lies on the plane x + y + z = 2. Let a = a i + b j + g k, k × (k × a) = 0, then g

Column 2 (p) 1

x + y = |a| and ax – y = 1 intersects each other in the interval a Œ (a, •), the value a is (c)

1

0

0

1

2 2 Ú (1 - t ) dt + Ú (t - 1) dt

(r) 4/3

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(d) If sin A sin B sin C + cos A cos B = 1, then the value of 2 sin2 (C/2) is p q r s Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(s)

2

(c) Distance of the points,

(r)

(3 3 / 2, 3 3, 3) plane p in units. (d) Distance of a face of the

(s) 3

Ans.

Solution: 1. (a) We have a + b + g = 2 and (k ◊ a) k – (k ◊ k)a = 0

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution

= g k – (ai + b j + g k) = 0 fi a = 0, b = 0. Thus, g = 2 | a| +1 a| a| -1 and y = (b) x = a +1 a +1 x > 0, y > 0, we get a + 1 > 0 and a | a | – 1 > 0 fi a > 1. Thus, a = 1. 1 0 2 2 and Ú (t 2 - 1) dt = (c) Ú (1 - t 2 ) dt = 0 1 3 3 fi

1

Ú0 (1 - t

2

) dt +

0

Ú1 (t

2

3

- 1) dt =

4 3



A (3, 0, 0), B (0, 3, 0) and C of the tetrahedron 0 0 0 1 1 3 0 0 1 1 9 OABC = = ¥ 27 = 6 0 3 0 1 6 2 0 0 3 1 (b)

C £ 1, 1 = sin A sin B sin C + cos A cos B £ sin A sin B + cos A cos B = cos (A – B) £ 1. fi cos (A – B) = 1. fi A – B = 0 or A = B Thus, 1 = sin C sin2 A + cos2 A fi sin C = 1 fi C = p /2 fi 2 sin2 (C/2) = 2(1/ 2 )2 = 1.

p is A (x – 1) + B (y – 1) + C (z – 1) = 0 wherer A + 0. B – 1. C = 0 and – A + B + 0.C = 0. A=B=C p is x + y + z = 3.

AB = BC = CA = 3 2 , so the area of the face ABC 3 9 3 ¥ (3 2) 2 = OAB, 4 2 OBC, OCA is an isosceles traingle with sides 3, 3 2 so area of each of these faces is 1 9 9 ¥3 2¥ 9- = 2 2 2

p passes through the point Example 48 (1, 1, 1) and is parallel to the vectros b = (1, 0, –1) and a = (–1, 1, 0) p A, B and C tetrahedron OABC. Column 1 Column 2 9 2 OABC is cubic units. 9 3 2

3 3 +3 3 +3-3 9 2 = 2 1+1+1

ABC =

0+0+0-3

1+1+1 distance of A, B, C OCA, OAB

= 3 OBC,

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Example 49 Column 1 (a) The plane 3x + 2y – through the point x y = (b) The line 3 2

Column 2 (p) (1, 1, – 2) z = 7 passes

through the points A (0, 0, 0) and B(3, –1, 2) parallel to the line

x+y=0 a distance

(r) (2, – 2, – 5)

z+2 x +1 and -6 1

perpendicular to the plane x – 3y = 0 and z

plane 2x – 2y – 5 z = 33 is at the point. q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

d

p

q

r

origin. p q r

s

a

p

q

r

s

s

b

p

q

r

s

s

c

p

q

r

s

d

p

q

r

s

Ans.

Solution (a) The plane passes through all the points given (b) both the line and the plane pass through (0, 0, –7) (c) First line passes though (1, 1, –2) which also 1+1 1+1 -2+6 = = lies on the 2nd line as 1 1 2 (d) If (a, b, c) is the foot of the perpendicular plane is a (x – a) + b (y – b) + c (z – c) = 0 or ax + by + cz = a2 + b2 + c2. which representes plane 2x – 2y – 5z = 33 2

Then

14

2

a a +b +c b c a +b +c = = = 2 2 2 = 2 -2 -5 33 2 +2 +5 2

is (2, – 2, – 5)

2

2

y – 11z = 0

bisecting the angle between the planes x + 2y + z + 5 = 0 and 2x – y – z – 5 = 0

intersect at the point (d) Foot of the perpendicular (s) (3, –1, 0)

p

10

x

y +1 z+6 = = 1 2

Ans.

Column 2 x – 3y – 2z – 10 =0

x-4 y-3 z +1 = = . -4 1 7

z+7 = 1 x+y+z=7=0 x -1 y -1 (c) Lines = 2 1 =

Example 50 Column 1

2

x + y + 2z + 10 =0

Solution A be ax + by + cz = 0. Since if passes through B and is parallel to the given line. 3a – b + 2c = 0 and a – 4b + 7c = 0. a c b = = fi 1 - 19 - 11 x y – 11z = 0. (b) The planes is (p) or (s) are at a distance ± 10 3 + 2 +1 2

2

2

=

10 14

tween the given planes are

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x + 2y + z + 5 1+ 4 +1



2x - y - z - 5 4 +1+1

fi x – 3y – 2z – 10 = 0 or 3x + y = 0 (d) The plane in (s) is perpendicular to x – 3y as 3 ¥ 1 – 3 ¥ 1 +2 ¥ z-axis at origin.

ASSERTION-REASON TYPE QUESTIONS Example 51

r = a - c + m (a + 3c) are coplanar. Statement-2: There exist l and m such that the two values of r Ans. (a) Solution: Since a and a and in the two values of r we get. 6 – l = 1 + m, 2l – 1 = 3m – 1 fi l = 3, m = 2. So there exist values for l and m such that the two values of r

Consider the planes 3x – 6y – 2z = 15 and

2x + y – 2z = 5 Statement-1: intersection of the given planes are x = 3 + 14t, y = 1 + 2t, z = 15t Statement-2: The vector 14i + 2j + 15k is parallel to the line of intersection of the given planes. Ans. (d) Solution

Example 54 and

x - 2 y +1 z = = 1 2 3

x - 3 y -1 z - 0 = = =t 14 2 15 If this represents the line of intersection of the given planes

direction ratios of the line of intersection of the planes is (– 6) (– 2) – (– 2) (1), (– 2) (2) – (– 2) (3), (3) (1) – (2) (– 6) i.e. 14, 2, 15 showing that the vector 14 iˆ + 2 ˆj + 15 kˆ is

-1 1 –1 1 =0 = 1 -1 1 2 3

Statement-1: P is a point (a, b, c). Let A, P in yz, zx and xy planes respectively A, B

x y z and C is + + = 1 a b c Statement-2:

x -1 y z +1 1 -1 1 = – (5x + 2 y – 3z – 8) = 0 1 2 3

P in a plane is the foot P on the plane.

Ans. (c) Solution P in a plane is a point M such that PM is perpendicular to the PM lies on the plane. The points A, B, C are respectively (– a, b, c), (a, – b, c) and (a, b, – c) x y z which lie on the plane + + a b c is true. Example 53

x - 2 y +1 z = = is perpendicular 1 2 3 z – 8 = 0 and parallel to the plane

to the plane 3x + 6y x+y–z=0 Ans. (b) 1- 2 0 +1 -1- 0 -1 1 1 Solution: Since 1 2 3

parallel to the line of intersection of the given planes and

B, C

x -1 y z +1 = = -1 1 1

x + 2y – 3 z – 8 = 0 Statement-2: The line

written as

Example 52

Statement-1: The lines

Statement-1: If the vectors a and are r = 6a – + l (2 – a) and

1 2 3 = = and 1 + 2 – 3 = 0 3 6 9 Example 55

Consider the planes

P1 : x – y + z = 1 P2 : x + y – z = – 1 P3 : x – 3y + 3z = 2 Let L1, L2, L3 be the lines of intersection of the planes P2 and P3, P3 and P1, and P1 and P2 respectively.

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L1, L2 and L3 are

Statement-1:

Statement-2: point. Ans. (d) Solution: If l, m, n denote the direction ratios of L1 then l + m – n = 0 and l – 3m + 3n = 0 fi l = 0, m = n fi direction ratios of L1 L2 and L3 0, 1, 1 showing that L1, L2, L3 is False. we get x = 0, y – z = – 1 and y – z = – 2/3 which is not possible. Example 56

Statment-1: The locus of a point which

3i – 2j + 5k and i + 2j – k is r (i – 2j + 3k) = 8. Statement-2: a and b is a+b r. (a – b) = 0 2 Ans. (a) Solution: r be the position vector of the point on the locus, then | r – a| = |r – b| fi

Shortest distance between the lines L1 and

Statement -2: If P is a point is on L1 and Q is a point on L2 then shortest distance between L1 and L2 PQ. Ans. (c) Solution: We can write L1: r = 3i + 8j + 3k + l (3i – j + k) = a + lb L2: r = – 3i – 7j + 6k + m (– 3i + 2j + 4k) = + ld Where a = 3i + 8j + 3k, = – 3i – 7j + 6k b = 3i – j + k, d = – 3i + 2j + 4k a – = 6i + 15j – 3k = 3 (2i + 5j – k) i j k = - 6 i - 15 j + 3k |b × d| = 3 -1 1 = - 3 (2 i + 5 j - k ) -3 2 4 |b × d|= 3 4 + 25 + 1 = 3 30 .

| (a - c) · (b ¥ d) | |b ¥ d| =

on L1 and Q

9 (4 + 25 + 1) 3 30

= 3 30 .

P L2 such that PQ = 18 which is

(r – a)2 = (r – b)2



| r |2 + |a|2 – 2 r · a = |r |2 + | b |2 – 2r · b



2 r · (a – b) + | b|2 – |a|2 = 0



2 r · (a – b) + (b – a)· (b + a) = 0



Statement-2: L2 is 3 30

(

r-

)

a+b · (a – b) = 0 2

Èr - 3i - 2 j + 5k + i + 2 j - k ˘ ÍÎ ˙˚ 2 ◊ (3i – 2j + 5k – (i + 2j – k) = 0 fi

[r – (2i + 2k)] ◊ (2i – 4j + 6k)] = 0



r ◊ (i – 2j + 3k) = 2 ¥ 1 – 0 ¥ 2 + 2 × 3 = 8

Example 57

y-8 x-3 z-3 L1: = = -1 3 1 L 2: =

x+3 y+7 z-6 = = . -3 2 4

Example 58 Statement-1: If two pairs of opposite edges of a tetrahedron are at right angles, then the third pair is also at right angles. Statement-2: The relation of perpendicularity is a transitive relation. Ans. (c) Solution: Let OABC be a tetrahedron, let O be the origin and the coordinates of A, B, C be (xi, yi, zi), i = 1, 2, 3. Suppose that OA, BC and OB, CA are orthogonal. The direction ratios of OA and BC are x1, y1, z1 and x3 – x2, y 3 – y 2, z 3 – z 2. Since OA^BC, x1 (x3 – x2) + y1 (y3 – y2) + z1 (z3 – z2) = 0 (1) x2 (x1 – x3) + y2 (y1 – y3) + z2 (z1 – z3) = 0 x3 (x1 – x2) + y3 (y1 – y2) + z3 (z1 – z2) = 0 Thus OC and AB are at right angles.

(2)

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Example 59 Statement-1: Coordinates of a point on the line passing through P (a, b, g) and having direction ratios a, b, c; at a distance r P are (a + ar, b + b r, g + c r). Statement-2: x the plane x – y + z = 2 y z = is 1. 3 -6 Ans (d) Solution: Statement-1 is true if a, b, c are the direction cosines of the line fi if a2 + b2 + c2 = 1 x -1 y+2 = 2 3 r r z-3 = = = -6 7 4 + 9 + 36

(

)

- 6r 2r 3r + 1, - 2, + 3 which lies on the given plane 7 7 7 if r is true. angle q with each of the x and z If the angle a y-axis, is such that 2 2 2 sin a = 3 sin q, then cos q = 3/5. Statement-2: If a line with direction ratios l, m, n angles a, b, g respectively with x, y and z l m cosa = , cos b = , 2 2 2 2 l +m +n l + m2 + n2 cosg =

n l 2 + m2 + n2

Ans. (a) Solution:

x y z = = 1 0 0

x

so direction ratios of x given line are l, m, n 1. l + 0. m + 0 . n fi cosa = = 1 + 0 + 0 l 2 + m2 + n2 b= and cos g =

l l 2 + m2 + n2

m l +m +n 2

2

l l 2 + m2 + n2 m

and

cos a =



l = n, cos2 q = cos2 a =

l 2 + m2 + n2

n

=

l 2 + m2 + n2

.

l2 , 2 l 2 + m2

m2 . 2 l 2 + m2

sin2 a = 3 sin2 q m2 =3 2 l 2 + m2



1–

fi fi

2l2 = 3 (l2 + m2) l2 = – 3 m2 = n2



cos2 q =

È ˘ l2 Í1 - 2 2˙ Î 2l + m ˚

- 3m 2 3 2 2 = 5 - 6m + m

COMPREHENSION-TYPE QUESTIONS

Statement-1:

Example 60

cos q =

Paragraph for Question Nos. 61 to 64 A (-2, 2, 3) and B (13, -3, 13) L is a line through A Example 61

A point P

3PA = 2PB, then the locus of P is (a) x2 + y2 + z2 + 28x (b) x2 + y2 + z2 - 28x + (c) x2 + y2 + z2 + 28x (d) x2 + y2 + z2 - 28x + Example 62

12y 12y 12y 12y

l 2 + m2 + n2

the direction ratios of the line are l, m, n is then

10z 10z 10z 10z

+ +

247 247 247 247

= = = =

0 0 0 0

Coordinates of the point P which divides

the join of A and B in the ratio 2 : 3 internally are (a) (33/5, (c) (32/5, - 12/5, 17/5) (d) (20, 0, 35) Example 63

L, perpendicular to the

line AB is 2

+ + -

x+2 y-2 z-3 = = -5 15 10 x-2 y+2 z+3 (b) = = 3 13 2 (a)

(c)

x+2 y-2 z-3 = = 3 13 2

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(d)

x-2 y+2 z+3 = = -5 15 10

Example 64 passing through the origin and the points A and B are (a) 15, - 5, 10 (b) 11, - 1, 16 (c) 3, 13, 2 (d) 7, 13,– 4 Ans.

61. (a)

62. (b)

P in the line

Example 66

63. (c) 64. (d)

Solution: 61. Let P(x, y, z) be any point on the locus then 3PA = 2PB fi PA)2 = 4(PB)2 fi x + 2)2 + (y - 2)2 +(z - 3)2] = = 4[(x - 13)2 + (y+ 3)2 + (z - 13)2] fi 5(x2 + y2 + z2) + 140x - 60y + 50z - 1235 = 0 fi x2 + y2 + z2 + 28x - 12y + 10z - 247 = 0 Ê 2 ¥ 13 + 3(- 2) , 2 ¥ (- 3) + 3(2) , 2 ¥ 13 + 3 ¥ 3 ˆ ÁË ˜¯ 2+3 2+3 2+3 = (4, 0, 7) 63. Direction ratios of AB are 13 + 2, - 3 - 2, 13 - 3 i.e. 15, - 5, 10 L be x+2 y-2 z-3 = = l m n then 15l - 5m + 10n = 0

origin be ax + by + cz A(- 2, 2, 3) and B(13, - 3, 13). - 2a + 2b + 3c = 0 13a - 3b + 13c = 0 a b c fi = = 26 + 9 39 + 26 - 26 + 6

r = a + l b is (a) (11, 12, 11) (c) (5, 8, 15) Example 67

(b) (5, 2, - 7) (d) (17, 16, 7)

If A is the point with position vector a D PLA

(a) 3 6 (c)

17

(b) 7 17/2 (d) 7/2

Ans. 65. (c), 66. (c), 67. (b) Solution: 65. Let the position vector of L be a + l b (6 + 3l) i + (7 + 2l) j + (7 - 2l) k So PL = (6 + 3l) i + (7 + 2l) j + (7 - 2l) k - (i + 2j + 3k) (5 + 3l) i + (5 + 2l) j + (4 - 2l) k Since PL is perpendicular to the given line which is parallel to b = 3 i + 2 j - 2 k fi 3(5 + 3l) + 2(5 + 2l) - 2(4 - 2l) = 0 fi l = - 1 and thus the position vector of L is 3i + 5 j + 9k . 66. Let the position vector of Q P in the given line be x1i + y1 j + z1 k, then L of PQ. i + 2 j + 3 k + x1 i + y1 j + z1 k fi 3i + 5 j + 9k = 2 x1 + 1 y +2 z +3 = 3, 1 = 5, 1 2 2 2 fi x1 = 5, y1 = 8, z = 15 fi P in the line is (5, 8, 15) 1 PL AL DPLA = 2 fi

or

a b c = = 35 65 - 20

=

1 2 i + 3 j + 6k - 3i - 2 j + 2k 2

or

a b c = = 7 13 -4

=

1 4 + 9 + 36 2

=

7 17 2

Paragraph for Question Nos. 65 to 67 a = 6 i + 7 j + 7 k , b = 3 i + 2 j - 2 k , P(1, 2, 3).

9+4+4

Paragraph for Question Nos. 68 to 70 The position vector of L, the foot of the P on the line r = a + l b is (a) 6 i + 7 j + 7 k (b) 3 i + 2 j - 2 k

Example 65

(c) 3 i + 5 j + 9 k

(d) 9 i + 9 j + 5 k

P(2, 3, - 4), b = 2i – j +2k Example 68 the point P perpendicular to the vector b is (a) r ◊ (2 i - j + 2 k ) = - 7

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(b) r ◊ (2 i - j + 2 k ) = 7 (c) r ◊ (2 i + 3j - 4 k ) = - 7 (d) r ◊ (2 i + 3j - 4 k ) = 7

fi x – 2y + z = 0

(1)

Ax – 2y + z = d, we get A = 1.

(2)

p passing

Example 69

through the point with position vector b and perpendicular to the vector OP, O being the origin is (a) 2x - y + 2z + 7 = 0 (b) 2x - y + 2z - 7 = 0 (c) 2x + 3y - 4z + 7 = 0 (d) 2x + 3y - 4z - 7 = 0 Example 70

6 , we get |d | 1+ 4 +1 Example 72

=

6 fi|d|=6

If q is the angle between the line

x +1 y -1 z-2 = = and the plane 3 2 4 2x + y - 3z + 4 = 0, then k cosec2 q = 203 if k = Ans. 8 Solution: The line is parallel to the vector b = 3 i + 2 j + 4 k

by the plane p on the coordinate axes is

h = 2 i + j - 3k . b◊h b h

6 + 2 - 12

Ans

fi sin q =

Solution P whose position vector is a= 2 i + 3j - 4 k , perpendicular to b is (r - a) ◊ b = 0

fi 8 cosec q = 203 fi k = 8

fi r ◊ (2 i - j + 2 k ) = (2 i + 3j - 4 k ) ◊ (2 i - j + 2 k ) = 2 × 2 + 3(-1) - 4 × 2 = - 7 p is 2x + 3y - 4z = - 7 p on the coordinate axes = -

7 7 -7 + + -4 2 3

=

91 12

INTEGER-ANSWER TYPE QUESTIONS Example 71

If the distance between the plane

Ax – 2y + z = d and the plane containing the lines x -1 y-2 z-3 x-2 y-3 z-4 = = and = = 2 3 4 3 4 5 is

6 , then | d | is

Ans. 6 Solution x -1 y - 2 z - 3 = = 2 3 4 x-2 y-3 z-4 = = and 3 4 5 is x -1 y - 2 z - 3 2 3 4 =0 3 4 5

=

29 14

=

-4 406

2

Example 73

If d is the distance between the point

(- 1, - 5, - 10) and the point of intersection of the line x-2 y +1 z-2 = = with the plane x - y + z = 5, then 3 4 12 d3 = (10 + a)3 if a = Ans. 3 Solution r + 2, 4r - 1, 12r + 2) which lies on the plane if 3r + 2 - (4r - 1) + 12r + 2 = 5 fi r = 0 and the point of intersection of the line and the plane is (2, - 1, 2) So d2 = (2 + 1)2 + (- 1 + 5)2 + (2 + 10)2 fi d3 = (10 + 3)3 fi a = 3 Example 74

If Q

the point P(4, -5, 3) on the line

x-5 y+2 z-6 = = -4 3 5

then 25(PQ)2 – 450 = Ans. 7 Solution: The given line passes through A(5, - 2, 6) (PQ)2 = (AP)2 - (AQ)2 (AP)2 = (4 - 5)2 + (- 5 + 2)2 + (3 - 6)2 AQ is the projection of AP on the given line 3 (- 4) + (- 2 + 5) \ AQ = (5 - 4) 50 50 Ê -5 ˆ + (6 - 3) Á = Ë 50 ˜¯

6 50

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fi (PQ)2

-

Ans Solution: The direction ratios of RS are 3-1, 5-2, 7-3 i.e. 2 3 4 , , 2, 3, 4 so direction cosines of RS are and 29 29 29 the projection of PQ on RS is

36 457 = 50 25

fi 25(PQ)2 – 450 = 7

s= fi fi

1

s2 k

2

Example 78 Fig. 20.7

Example 75

If p1, p2 are respectively the perpendicular

distances of the points with position vectors a = 3 i - 5 j + 8 k and b = 2 i - 41j + 21k

r ◊ (2 i + 2 j - k ) =

12, then p2 – 4p1 Ans. 5 Solution: p1 = =

p2 = and

12 - (3 i - 5 j + 8 k ) ◊ (2 i + 2 j - k ) 2i + 2 j – k 12 - (6 - 10 - 8) 4 + 4 +1

=

24 =8 3

102 ¥ 102 fid = = (51)2 ¥ 2 2

The lines

x-4 y - 17 z - 11 = = and 15 9 8

x - 15 y-9 z -8 = = intersect at the point P, then 4 17 11 P a, a = Ans. 2 x-4 y - 17 z - 11 = = is Solution 15 9 8 x - 15 (15r r + 17, 8r + 11) and any point on the line 4 y-9 z -8 = = is 17 11 r¢ + 8) r + 17 = 17r¢

15r + 4 = 4r¢

r + 11 = 11r¢ + 8

fi r = r¢ = 1 and the point of intersection is OP)2

P

Example 76 perpendicular to two planes 2x - 2y + z = 0 and x - y + 2z (52, 53, 57) is (51)2 p, p = Ans. 2 Solution: Let the plane be a(x - 1) + b(y - 2) + c(z - 2) = 0 where 2a - 2b + c = 0, a - b + 2c = 0 a b c fi = = 1 1 0 x+y-3=0 d 1+1

2

(4r¢ + 15, 17r¢

1 111 |12 - (4 - 82 - 21)| = = 37 3 3 p2 – 4p1 = 5

52 + 53 - 3

[2(8 - 1) + 3(10 - 2) + 4(11 + 2)]

29

=

2

+ (26)2

fia=2 Example 79

It S denotes the area of the triangle with

vertices A (1, – 1, 2), B (2, 1, – 1) C (3, – 1 , 2) then S2 Ans. 3. Solution: = BC =

AB =

(1 - 2) 2 + (- 1 - 1)2 + (2 + 1)2

1+ 4 + 9 =

14

14 , CA = 2. B (2, 1, – 1)

102 2

2

Example 77

P, Q, R, S are the points (1, 2, - 2),

(8, 10, 11), (1, 2, 3) and (3, 5, 7) respectively. If s denotes the projection of PQ on RS then s2 = 100k2, k =

A (1, – 1, 2)

D Fig. 20.8

C (3, – 1, 2)

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So the triangle ABC is isosceles with base AC. If D is the AC, coordinates of D are

(

)

3 + 1 -1 - 1 2 + 2 , , = (2, – 1, 2) 2 2 2 (2 - 2) 2 + (1 + 1) 2 + (- 1 - 2) 2

and BD =

4+9 =

=

13 1 AC ¥ BD = 2

DABC = S = fi

13

S2 – 10 = 3

Example 80 the origin cuts the coordinate axes at points A, B, C. If the centroid (x, y, z) of the triangle ABC x–2 + y–2 + z–2 = k, then the value of k is Ans. Solultion: x y z + + =1 a b c fi A (a, 0 ,0), B (0, b, 0) C (0, 0, c) fi (x, y, z) = (a/3, b/3, c/3) k = x–2 + y–2 + z–2 a–2 + b–2 + c–2)

4x - 3 2y + 1 7z - 6 = = is 6 7 5 (a) 56 (b) 29 3. l = m = n = 1 represent the direction cosines of the (a) x y (c) z 4. Let N be the foot of the perpendicular of length p l, m, n be the direction cosines of ON (a) px + my + nz = l (b) lx + py + nz = m (c) lx + my + pz = n (d) lx + my + nz = p 5. If the planes bx – ay = n, cy – bz = l and az – cx = m intersect in a line, then al + bm + cn (a) – 1 (b) 0 (c) 1 (d) none of these the axes is (a) 2 (b) 4 (c) 6 (d) 8 7. The line x – 2y + 4z + 4 = 0, x + y + z – 8 = 0 intersects the plane x – y + 2z + 1 = 0 at the point (a) (3, 2, 3) (b) (5, 2, 1) (c) (2, 5, 1) (d) (3, 4, 1)

-1

() () () 1 a

2

+

Hence k

1 b

2

+

1 c

2

= 1 fi a–2 + b–2 + c–2 = 1

.

with position vector i + 4k on the line joining the points having position vector as – 11i + 3k and 2i – 3j + k has the position vector (b) 4i + 5 j - 5k (a) 4i + 5 j + 5k (c) 5i + 4 j – 5k

(d) none of these

EXERCISE ular to two planes 2x – 2y + z = 0 and x – y + 2z LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS plane in (a, b, c x y z + + =1 (a) a b c (b) ax + by + cz = 3 (c) ax + by + cz = a2 + b2 + c2 (d) ax + by + cz = a + b + c 2. Shortest distance between the two straight lines x - 3/ 4 y + 1/ 2 z + 6/ 7 = = and 2 3 4

(1, 2, 2) is (a) 0 (c)

(b) 1

2

(d) 2 2

10. If a = i + 2 j + k , b = i - j + k and = i + j-k projection on is 1/ 3 , is

a and b whose

(a) 4 i - j + 4 k

(b) 3 i + j - 3 k

(c) 2 i + j - 2 k

(d) 4 i + j - 4 k

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS O, and A, B and C, the coordinates of the centroid can be

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1 Ê , 1ˆ˜ (a) Á1, Ë 2 ¯

Ê 1 , 1, 1ˆ (b) Á ˜¯ Ë 2

Ê (c) Á1, 1, Ë

(d) (1, 1, 1)

1 ˆ ˜ 2¯

Column 1

A, B and C, the point of intersection of the planes through A, B, C parallel to the coordinate planes can be (c) (1, 1, – 1)

(d) (2, 3, – 4) x -1 y-2 = 2 3

z-3 and parallel to a coordinate axis is 4 (a) 4y – 3z + 1 = 0 (b) 2x – z + 1 = 0 (c) 3x – 2y + 1 = 0 (d) 2x + 3y + 1 = 0 14. Let A be vector parallel to line of intersection of planes P1 and P2 through origin. P1 is parallel to the vectors 2 j + 3 k and 4 j - 3 k and P2 is parallel =

to j – k and 3 i + 3 j , then the angle between A and 2 i + j - 2 k is (a) p/2 (b) p/4 (c) p/6 (d) 3p/4 15. The plane passing through the origin and containing the lines whose direction cosines are proportional to 1, –2, 2 and 2, 3, –1 passes though the point (a) (1, –2, 2) (b) (2, 3, –1) (c) (3, 1, 1) (c) (4, 0, 7)

MATRIX-MATCH TYPE QUESTIONS 16. If a, b, g positive direction of the axes, then Column 1 (a) sin2 a + sin2 b + sin2 g (b) cos2 a + cos2 b + cos2 g (c) cos 2a + cos 2b + cos 2g (d) cos2a + cos (b + g)cos (b – g) Column 1 (a) passing through (2, 1, - 1)

Column 2 (p) 1 -1 (r) 2 (s) 0

Column 2 (p) 15x - 21y + 35z = 105 x + 21y + 35z = 16

plane 5y - 3z = 0 x + 21y - 35z = 86 lengths 7, 5, 3, on the coordinate axes.

(d) parallel to the plane (s) 15x – 21y + 35z + 26 15x – 21y + 35z = 8 =0 ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0 (a) a + b + c π a2 + b2 + c2 + bc + ca (b) a + b + c a2 + b2 + c2 + bc + ca (c) a + b + c π a2 + b2 + c2 + bc + ca (d) a + b + c a2 + b2 + c2 + bc + ca

Column 2 = ab at a single point π ab

resent the line x = y =z

π ab

resent identical planes

= ab

resent the whole of sional plane

Column 1 (a) The centroid of the triangle with vertices at the points where x y z the plane + + a b c

Column 2 (p) 1

axes is (1, r, r2). The plane passes through the point (4, - 8,15) if r x -1 y+2 = 2 (b) The line r r z is parallel to the -12 plane x + y + z = 3 if r (c) The plane is perpendicular x y z = = to the line 1 r r2 passes through the origin and the point (- 4, 3, 1) if r (d) The lines whose vectorial r = 2i – 3j =

(r) -4

(s) 3

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Statement-2: Two straight lines intersect if the

+ 7x + l (5i + pj + 2 pk ) and r = i + 2 j + 3k + m

23. Statement-1: The plane 5x + 2z – 8 = 0 contains the line 2x – y + z – 3 = 0 and 3x + y + z = 5, and is perpendicular to 2x – y – 5z – 3 = 0

(-i - pj + 3k ) are perpendicular for all values of l and m if p

Statement-2: The plane 3x + y + z x – 1 = y + 1 = z – 1 at the point (1, 1, 1). 24. p: x + y – 2z = 3, P: (2, 1, 6), Q

20. Column 1 •

(a) If

Column 2 Ê 1 ˆ

 tan - 1 ÁË 2 r 2 ˜¯

=q

(p)

r =1

2 3

then tan q (b) Sides a, b, c ABC a cos 2a = b+c b a+c then tan2 a + tan2 b

Statement-2: If the direction ratios of a line are 1 + l, 1 – l

cos 2b =

L is perpendicular to x + 2y + 2z = 0 and passes through (0, 1, 0). The perpendicular distance of L

Statement-1: The line joining PQ is perpendicular p Statement-2: Q P in the plane p. 25. Statement-1: Let PM the point P(1, 2, 3) to the xy plane. If OP angle q with the positive direction of z OM f with the positive direction of x is, where O is the origin and q, f are acute angles, then 3 tan q tan f = 2 5 .

y (r)

5 3

(1, 2, 3) and is perpendicular to two planes x = 0 and y = 0.

ASSERTION-REASON TYPE QUESTIONS x +1 y + 2 z +1 x–2 y+2 z–3 = = = = , L2 : 3 1 2 1 2 3 Statement-1: The distance of the point (1, 1, 1)

21. L1 :

L1 and L2 is 13 5 3 . Statement-2: The unit vector perpendicular to both – i – 7 j + 5k . the lines L1 and L2 is 5 3

l= 2± 5 .

COMPREHESION-TYPE QUESTIONS Paragraph for Question Nos. 26 to 28

are ai x + bi y + ci z + di = 0 (i = 1, 2, 3, 4) is

D3 6D1 D2 D3 D4

where D1, D2, D3, and D4 are cofactors of d1, d2, d3 and d4 a1 b1 c1 d1 a2 b2 c2 d 2 in D = a3 b3 c3 d3 a4 b4 c4 d 4 x + 3y + z = 6, 2x + 3y = 0, 3y + z = 0 and 2x + z = 0 is (a) 10 (b) 12 (c) 24 (d) 36 k 3. The product of lengths of intercepts of the plane on the coordinate axis is (b) 4 k 3 (a) 24 k 3 (c) 6 k 3 (d) 6 ¥ 64 k 3

22. Statement-1: r = i + 2 j + 3k + l ( ai + 2 j + 3k ) and r = 2i + 3j + k + m (3i + aj + 2k ) intersect at a

y + z = 0, z + x = 0, x + y = 0 and x + y + z = 1 2 2 (a) 1 (b) (c) 4 (d) 3 5

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Paragraph for Question Nos. 29 to 31 L1 :

x +1 y - 3 y + 2 x y-7 z+7 = = = , L2 : = 1 2 -3 2 1 -3

L1 and L2 are (a) perpendicular (b) parallel (c) coplanar (d) none of these 30. Lines L1 and L2 intersect at the point (a) (–3, 2, 1) (b) (2, –1, 3) (c) (1, –3, 2) (d) none of these L1 and L2 is (a) x + y + z = 0 (b) 3x – 2y – z = 0 (c) x – 3y + 2z = 0 (d) x + y + z = 42

x–y – z – 4 = 0 x + y + 2z – 4 which is parallel to the line of intersection of the planes 2x + 3y + z = 1 and x + 3y + 2z = 2 is Ax + By + Cz + D = 0, then the value of | A + B + C

nate axes then

S (33)2 ax + by + cz =

d, then a4 + b3 + c2 + d

Paragraph for Question Nos. 32 to 33 x +1 y +1 z +1 = = , p1: x + 2y + 3z = 14 2 3 4 p2: 2x – y + 3z p, at p 32. The line through P perpendicular to p1 passes through the point (a) (1, 1, 1) (b) (0, 1, 0) (c) (0, 0, 0) (d) (0, 0, 1) 33. If the line through P perpendicular to p1 plane p2 in the point Q, then the coordinates of the PQ are

LEVEL 2

L:

(c) (2, 3, 4)

(d) (2, 4, 6)

INTEGER-ANSWER TYPE QUESTIONS 34. If the position vector of the intersection of the line r = (i + 2j + 3k) + l(2i + j + 2k) and plane r ◊ (2i – 6j + 3k) + 5 = 0 is ai + bj + ck, then 1 (50a + 60b + 75c)2 1445 35. If d is the shortest distance between the lines r = 3i + 5j + 7k + l(i + 2j + k) and r = – i – j – k + m(7i – 6j + k), 125 2 d then 1058 36. If the point of intersection of the line r = (i + 2j + 3k) + l(2i + j + 2k) and the plane r ◊ (2i – 6j + 3k) + 5 = 0 lies on the plane r ◊ (i + 75j + 60k) – a = 0, 1 then a 67 x -1 y -1 z -1 = = 37. If the line intersects the curve 15 16 n 6x2 + 5y2 = 1, z = 0 then n2 – 2n + k = 0 where the 1 k value of 263

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. The line of intersection of the planes r◊ (3i - j + k ) = 1 and r◊ (i + 4 j – 2k ) = 2 is parallel to the vector (a) -2i + 7 j + 13k

(b) 2i + 7 j + 13k

(c) -2i - 7 j + 13k

(d) -2i + 7 j - 13k

a, b, g, d with four diagonals of a cube, then cos2 a + cos2 b + cos2 g + cos2 d is (a) 1/3 (b) 2/3 (c) 4/3 (d) 8/3 3. The distance between the planes given by r ◊ (i + 2j – 2k) + 5 = 0 and r ◊ (i + 2j – 2k) – 8 = 0 is (a) 1 unit (b) 13/3 units (c) 13 units (d) none of these 4. The area of the triangle with vertices A(3, 4, –1), B (2, 2, 1) and C (3, 4, –3) is (a) 4 5 (b) 3 (c) 6 (d) none of these k3. The locus of the centroid of the tetrahedron is (a) xyz = 6k3 (b) xyz = 8k3 (c) x + y + z = 6k (d) x3 + y3 + z3 = 64k3 r

(a) 4 3 (i + j + k )

(b) 4 3 (i – j + k )

(c) 6 2 (i + j)

(d) 6 2 (i + j + k )

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7. The perpendicular distance of the point (x, y, z) x (a)

x2 + y 2

(b)

y2 + z2

(c) z 2 + x 2 (d) x 2 + y 2 + z 2 8. The non zero value of a for which the lines ax – 4y + 7z + 16 = 0 = 4x + 3y – 2z + 3 and x – 3y + 4z + 6 = 0 = x – y + z + 1 are coplanar is (a) 35 (b) 37 (c) 47 (d) none of these P (a, b, c) perpendiculars PM and PN are drawn to zx and xy O is the OMN is (a)

x y z =0 - a b c

(b)

x y z (c) =0 + + a b c

x y z =0 - + a b c

x y z (d) =0 + a b c

x–y+z direction cosines are proportional to 2, 3, – 5 is (c) 15/7

(d) none of these

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 11. If q is the angle between the lines whose vector r = 3i + 2 j + 4k + l (i + 2 j + 2k ) and r = 5i – 2k + m (3i + 2 j + 6k ) ; l and m being (a) cos q =

19 21

(b) sin q =

19 21

(c) sin q =

4 5 21

(d) cos q =

4 5 21

12. If l, m, n are the direction cosines of the line of shortest distance between the lines x - 3 y + 15 z - 9 x +1 y -1 z - 9 = = = = and -7 -3 2 5 2 1 then (a) 3l – 15m

n=0

(b) 2l – 7m + 5n = 0

(c) l = m = n = 1

3

join of A

B (11, 0, – 1) lies on the plane

(d) 2l + m – 3n = 0

(a) 2x + y + z = 6 (b) x – y + z = 1 (c) x + y + z = 1 (d) x – y – z = 1 14. If O is the origin and the line OP of length r an angle a with x containing x P are (a) (r cos a, o, r sin a) (b) (r cos a, r sin a, o) (c) (o, r cos a, r sin a) (d) (r sin a, o, r cos a) 15. If a plane p passes through the point (1, 2, 3), p are l, m, n; and it contains the line joining the origin to the point (1, 1, 1), then (a) l + 2m + 3n = 0 (c) l + m – n = 0

(b) l + m + n = 0 (d) l – m + 2n = 0

MATRIX-MATCH TYPE QUESTIONS 16. Column 1 (a) a point on the plane 2x + 4y - 5z + 15 = 0

Column 2 (p) (1, 2, 3)

x +1 y+2 z+2 = = 2 4 5 (c) a point on both 2x + 4y - 5z + 15 = 0 and

(r) (5, 10, 13)

x +1 y+2 z+2 = = 2 4 5

x2 + y2 + z2 = 14 17. If t and p Column 1 Column 2 (a) passing through the (p) r = a + t (b - a) + pc point with position vector a and parallel to the vectors b and r◊(b ¥ c + c ¥ a + a ¥ b) points with position

= [a b c ]

vectors a and b and parallel to the vector (c) passing through three (r) r = a + t b + pc points with position vectors a, b, , (d) passing through the (s) b + t (a – b ) + pc point with position vector b, parallel to the vectors a – b and

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18. p: 2x - y + l z + 4 = 0 Column 1 (a) The angle between the x +1 y -1 z-2 = = line 1 2 2 -1 and the plane p is sin (1/3) (b) The plane p the point (l, l, - 4) (c) The plane p is perpendicular

3a2 – 34a + 31 = 0. Column 2 (p) if l = 4

and (– 2, 3, – 4) l = 5/3 (r) if l

to the plane lx + xy - l z - 4 = 0 (d) The plane p l=2 intercept of length 2 on z P(a, b, g) and Q(1, - 1, 0) be two points such PQ is R(x, y, z). If x of a and b, y b and g and z of g and a, then Column 1 Column 2 (a) a (p) can take any value (b) b (c) g (r) 1 (d) a + b + g (s) 5 x = y + a = z at P and the line x + a = 2y = 2z at Q R of the line joining P and Q lies on the plane x + 2y + z = 14. Then if O is the origin. Column 1 Column 2 2 (p) 41 (a) (OP) (b) (OQ)2 (c) (OR)2 (r) 88 2 (s) 36 (d) (PQ)

ASSERTION-REASON TYPE QUESTIONS 21. Statement-1:

Statement-2: The distance between the parallel lines x–2 y+3 z–4 x+2 y–3 z+4 = = = = and is 1 2 2 1 2 2

x + 2 y +1 = 3 2

23. Statement-1: The sphere x2 + y2 + z2 + 3x – 4z + 1 = 0 intersects the plane 3x – 4z + 1 = 0 in a circle of radius 21 2 . Statement-2: The plane 3x – 4z + 1 = 0 lies outside the sphere x2 + y2 + z2 + 3x – 4z + 1 = 0. 24. Statement-1: The distance of the centre of the sphere x2 + y2 + z2 + 4x – 2y + 6z line x = y = z is 26 3 . Statement-2: radius r on a line is 2a then the distance of the centre r 2 – a2 . 25. L: r = (i + 3j – k ) + t ( j + 2k ) p : r ◊ (i + 4 j + k ) + 6 = 0 Statement-1: The line L intersects the plane p at the point (1, 0, – 7) Statement-2: The angle between the line L and the plane p is (1/2 cos –1 (1/5)).

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 26 to 29

y -1 x z-2 = = 1 3 2 p : 2x + 3y - z = 7 P: (1, 6, 3), L:

P on the line L is

z–3 at a distance 3 2 2 x+7 y+5 z–2 = = . lies on the line 5 4 1

(a)

x -1 y-6 z-3 = = -3 0 2

(b)

Statement-2: The distance between the points (– 2, – 1, 3) and (– 7, – 5, 2) is 42

x -1 y-6 z-3 = = -3 2 0

(c)

x -1 y-6 z-3 = = -3 0 2

(d)

x -1 y-6 z-3 = = 3 0 2

=

22. Statement-1: The integer value of a for which the shortest distance between the lines x–3 y–5 z–7 x +1 y +1 z +1 = = = = and a –2 1 7 –6 a

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P on the line L is (a)

13

(b)

plane r ◊ (2i - 6 j + 3 k ) + 5 = 0 lies on the plane r ◊ (i + 75 j + 60 k ) - a a k,

14

(c) 46 (d) 10/ 14 28. The angle q between the plane p and the line L is given by (a) cos q = 5/14 (b) sin q = 5/14 (c) cos q = 1/14 (d) sin q = 1/14 P and perpen dicular to the plane p is (a) x + y + 5z = 1 (b) 5y + 3z (c) 3x - 2y = 15 (d) 3x + y z = 36

INTEGER-ANSWER TYPE QUESTIONS 30. If a, b, c are the lengths of the intercepts of the plane passing through the intersection of the planes 2x + y + 2z x - 5y - 4z = 1 and the point (3, 2, 1), on the coordinate axes, then (5a + b + c)2 = (21)2k, where k = 31. If the point of intersection of the line r = (i + 2 j + 3 k ) + l (2i + j + 2 k ) and the

where k = x -1 y +1 z -1 = = intersect the curve 15 16 n 6x2 + 5y2 = 1, z = 0; then 10n2 - 20n + 21k3 = 0, where the value of k is 33. If p is the perpendicular distance of an angular point

32. If the line

which does not pass through that angular point, then p 2 3 ÊÁ ˆ˜ Ë 47 ¯ 34. For a point P in the plane, let d1(P) and d2(P) be the distances of the point P x – y =0 and x + y = 0 respectively. The area of the region R consisting of all points P of the plane and satisfying 2 £ d1(P) + d2(P) £ 4, is p passes through the point (1, 1, 1) and is a parallel to the vectors b = (1 , 0, – 1) and = (–1, 1, 0). If p A, B and C, then twice the OABC (in cubic units) is.

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. The value of k for which the line z-k x-4 y-2 = = lies in the plane 2 1 1 2x – 4y + z = 7, is (a) 7 (b) 6 (c) no real value (d) – 7

4.

p is perpendicular to the two planes 2x – 2y + z = 0 and x – y + 2z = 4 and passes through the point (1, – 2, 1). The distance of p (1, 2, 2) is (d) 2 2 [2006] (a) 0 (b) 1 (c) the point P 2x + y + z

[2003]

x -1 y +1 z -1 x-3 2. If the lines = = and = 2 3 4 1 y-k z = intersect at a point then the value of k is 2 1 [2004] cuts the coordinate axes at A, B and C. If the centroid G(x, y, z) of triangle ABC 1 1 1 2 + 2 + 2 = k, then the value of k is x y z [2005]

at point Q

PQ

(c) 3 (a) 1 (b) 2 6. Let P(3, 2, 6) be a point in space and Q be a point on the line r = (i – j + 2k) + m(–3i + j + 5k). Then the value of μ for which the vector PQ is parallel to the plane x – 4y + 3z = 1 is 1 1 1 1 (a) (b) (c) (d) 4 4 8 8

x y z = = and perpendicular to the plane containing 2 3 4 x y z x y z the straight lines = = and = = is 3 4 2 4 2 3

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(a) x + 2y – 2z = 0 (c) x – 2y + z = 0

(b) 3x + 2y – 2z = 0 (d) 5x + 2y – 4z = 0 [2010] 8. If the distance of the point P x + 2y – 2z = a, where a > 0, is 5, then the foot of P to the plane is (a)

( (

) )

8 4 7 , ,3 3 3

(b)

( (

4 4 1 ,- , 3 3 3

) )

1 2 10 2 1 5 , , ,- , (d) [2010] 3 3 3 3 3 2 P is the intersection of the straight line joining the pionts Q (2, 3, 5) and R (1, –1, 4) with the plane 5x – 4y – z = 1. If S is the foot of the per T (2, 1, 4) to QR, PS is 1 2 (a) (b) 2 (c)

(d) 2 2

(c) 2

[2012]

intersection of the planes x + 2y + 3z = 2 and x – y + z 2 = 3 and at a distance 3 (a) 5x – 11y + z = 17 (b)

(d) x –

3

2y = 1 –

2

[2012]

x + 2 y +1 z = = to the plane x + y + z = 3. The feet 2 -1 3 of perpendiculars lie on the line x y -1 z - 2 x y -1 z - 2 = = = = (b) (a) -13 -13 5 8 5 8 (c)

x y -1 z - 2 = = 4 3 7

(d)

x y -1 z - 2 = = -7 2 5 [2013]

12. Let P to the plane x – y + z plane passing through and containing the straight x y z line = = is 1 2 1 (a) x + y – 3z = 0 (c) x – 4y + 7z = 0

1. If the straight lines

z x -1 y +1 x +1 = = and = 2 2 k 5

y +1 z = are coplanar, then the plane(s) containing 2 k these two lines is(are) (a) y + 2z = – 1 (b) y + z = – 1 (c) y – z = – 1 (d) y – 2z = – 1 [2012] l passing through the origin is perpendicular to the lines l1 : (3 + t)i + (–1 + 2t)j + (4 + 2t)k, –• < t < • l2 : (3 + 2s)i + (3 + 2s)j + (2 + s)k, –• < s < • Then, the coordinate(s) of the point(s) on l2 at a distance of 17 l and l1 is (are) Ê 7 7 5ˆ (b) (–1, –1, 0) (a) Á , , ˜ Ë 3 3 3¯ (c) (1, 1, 1) 3. Two lines L1 : x = 5,

Ê 7 7 8ˆ (d) Á , , ˜ Ë 9 9 9¯

[2013]

y z = and L2 : x = a, 3 - a -2

y z = are coplanar. Then a can take value(s) -1 2 - a

2x + y = 3 2 – 1

(c) x + y + z =

MULTIPLE CORRECT ANSWERS TYPES QUESTIONS

(b) 3x + z = 0 (d) 2x – y = 0 [2016]

(a) 1 (c) 3

(b) 2 (d) 4 [2013] P(λ, λ, λ) perpendiculars PQ and PR are drawn respectively on the lines y = x, z = 1 and y = – x, z = –1. If P is such that ∠QPR is a right angle, then the possible value(s) of λ is(are) (a)

2

(b) 1

(c) –1 (d) – 2 [2014] 3 5. Let R , consider the plane P1 : y = 0 and P2 : x + z = 1. P1 and P2, which Let P3 passes through the intersection of P1 and P2. If the P3 is 1 and the distance of a point (α, β, γ P3 is 2, then which of the following relations is (are) true ? (a) 2α + β + 2γ + 2 = 0 (b) 2α – β + 2γ + 4 = 0 (c) 2α + β – 2γ – 10 = 0 (d) 2α – β + 2γ – 8 = 0 [2015]

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6. In R3, let L be a straight line passing through the ori gin. Suppose that all the points on L are at a constant P1 : x + 2y – z + 1 = 0 and P2 : 2x – y + z – 1 = 0. Let M be the locus of the on L to the plane P1. Which of the following points lie(s) on M? 5 2ˆ Ê Ê 1 1 1ˆ (b) Á - , - , ˜ (a) Á 0, - , - ˜ Ë 6 3 6¯ Ë ¯ 6 3 Ê 5 1ˆ (c) Á - , 0, ˜ Ë 6 6¯

Ê 1 2ˆ (d) Á - , 0, ˜ Ë 3 3¯

2. Column 1

Column 2



(a) If

Ê 1 ˆ

 tan - 1 ÁË 2 r 2 ˜¯

=q

(p)

r =1

2 3

then tan q (b) Sides a, b, c ABC a cos 2a = b+c cos 2b =

[2015]

b a+c

then tan2 a + tan2 b

OPQRS tant (x ≥ 0, y ≥ 0, z ≥ 0) with O as origin, and OP and OR along the x y The base OPQR OP = 3. The point S T of diagonal OQ such that TS = 3. Then p . (a) the acute angle between OQ and OS is 3

L is perpendicular to x + 2y + 2z = 0 and passes through (0, 1, 0). The perpendicular distance of L

(r)

5 3

(1, 2, 3) and is perpendicular to two planes x = 0 and y = 0.

angle OQS is x – y = 0. plane containing the triangle OQS is

P to the 3

ax + by + cz = 0 bx + cy + az = 0 cx + ay + bz = 0

2

O to the straight line containing RS is

15 2

[2016]

MATRIX-MATCH TYPE QUESTIONS 1. Column 1 (a) Point (a, b, g ) lies on the plane x + y + z = 2. Let a = a i + b j + gk, k × (k × a) = 0, then g

Column 2 (p) 1

(c)

Ú (1 - t 0

0

2

) dt + Ú (t 2 - 1) dt

Column 2 represent planes single point

(b) a + b + c a2 + b2 + c2 π ab + bc + ca (c) a + b + c π a2 + b2 + c2 π ab + bc + ca (d) a + b + c a2 + b2 + c2 = bc + ca + ab

x + y = |a| and ax – y = 1 intersects each other in the interval a Œ (a, •), the value a is 1

Column 1 (a) a + b + c π a2 + b2 + c2 = ab + bc + ca

represent the line x=y=z represent identical planes. represent the whole of the three

(r) 4/3

1

(d) If sin A sin B sin C + cos A cos B = 1, then the value of 2 sin2 (C/2) is

(s)

2 [2006]

x -1 y z + 3 = = , 4. Consider the lines L1 : 2 -1 1

[2007]

x-4 y+3 z+3 = = and the planes 1 1 2 P1: 7x + y + 2z = 3, P2 : 3x + 5y – 6z = 4. Let

L2 :

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ax + by + cz = d be through the point of intersection of lines L1 and L2, and perpendicular to planes P1 and P2. correct answer using the code given below the lists: Column 1 Column 2 P. a = 1. 13 b= 2. –3 R. c = 3. 1 S. d = 4. –2 Codes: (a) (b) (c) (d)

3 1 3 2

2 3 2 4

4 4 1 1

1 2 4 3

[2013]

ASSERTION-REASON TYPE QUESTIONS 1. Consider the planes 3x – 6y – 2z = 15 and 2x + y – 23 = 5. Statement-1: of intersection of given planes are x = 3 + 14t, y = 1 + 2t, z = 15t Statement-2: The vector 14i + 2j + 15k is parallel to the given planes. [2007] 2. Consider three planes P1 : x – y + z = 1 P2 : x + y – z = – 1 P3 : x – 3y + 3z = 2. Let L1, L2, L3 be the lines of intersection of the planes P2 and P3, P3 and P1, and P1 and P2, respectively. Statement-1: L1, L2 and L3 Statement-2: The three planes do not have a

COMPREHENSION-TYPE QUESTIONS Consider the lines x +1 y+2 z +1 = = L1 : 3 1 2 x-2 y+2 z-3 = = 1 2 3 1. The unit vector perpendicular to both L1 and L2 is L2 :

(c)

-i + 7 j + 7k 99 -i + 7 j + 3 k 5 3

(b) (d)

(a) 0 (c)

(b) 41

5 3

(d)

17 3 17 5 3

passing through the point (–1, – 2, – 1) and whose L1 and L2 is 2 7 (a) (b) 75 75 13 23 (c) (d) [2008] 75 75

INTEGER-ANSWER TYPE QUESTIONS 1. If the distance between the plane Ax – 2y + z = d and the plane containing the lines x -1 y-2 z-3 x-2 y-3 z-4 = = and = = 2 3 4 3 4 5 is 6 , then | d | is [2010] 2. For a point P in the plane, let d1(P) and d2(P) be the distances of the point P x – y =0 and x + y = 0 respectively. The area of the region R consisting of all points P of the plane and satisfying 2 ≤ d1(P) + d2(P) ≤ 4, is [2014]

FILL

IN THE

BLANKS TYPE QUESTIONS

1. The area of triangle whose vetics are A(1, – ,1, 2), B(2, 1, – 1), C(3, – 1, 2) is ______ . by P(1, – 1, 2), Q(2, 0, –1), R(0, 2, 1) is

SUBJECTIVE-TYPE QUESTIONS the points (2, 1, 0), (5, 0, 1) and (4, 1, 1). (b) If P Q such that PQ is perpendicular to the plane in (a) PQ lies on it. [2003]

Paragraph for Question Nos. 1 to 3

(a)

2. The shortest distance between L1 and L2 is

-i - 7 j + 5k 5 3 -7 i - 7 j - k 99

T has one of its face as ABCD. The face opposite to it ABCD is A¢ B¢C¢ D¢. The parallelopiped T S with face ABCD A¢ B¢C¢ D¢ changing to A¢¢ B¢¢C¢¢ D¢¢ S T, show that locus of A¢¢ is a plane. [2004] 3. p1 and p2 are two planes passing through the origin. L1 and L2 are two lines passing through the origin

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such that L1 lies on p1 not on p2 and L2 lies on p2 but not on p1. Show that there exist three points A, B, C A¢, B¢, C¢ can be chosen such that (a) A is on L1, B on p1 but not on L1 and C not on p1 (b) A¢ is on L2, B¢ on p2 but not on L2 and C¢ not on p2. [2004] p passes through the point (1, 1, 1) and is a parallel to the vectors b = (1 , 0, – 1) and = (–1, 1, 0). If p A, B and C of the tetrahedron OABC. [2004] line 2x – y + z – 3 = 0, 3x + y + z – 5 = 0 and which are at a distance of 1/ 6 [2005]

Answers LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. (c) 5. (b)

2. (c) 6. (b)

3. (d) 7. (c)

4. (d) 8. (d)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 11. (a), (b), (c) 13. (a), (b), (c) 15. (a), (b), (c)

12. (a), (b), (c), (d) 14. (b), (d)

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

18.

20.

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

ASSERTION-REASON TYPE QUESTIONS 21. (a) 25. (b)

22. (a)

23. (c)

24. (d)

COMPREHENSION-TYPE QUESTIONS 30. (d)

31. (a)

32. (c)

33. (d)

INTEGER-ANSWER TYPE QUESTIONS 34. 5

35. 2

36. 2

37. 1

MATRIX-MATCH TYPE QUESTIONS p

q

r

s

a

p

q

r

s

LEVEL 2

b

p

q

r

s

c

p

q

r

s

SINGLE CORRECT ANSWER TYPE QUESTIONS

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

16.

17.

1. (a) 5. (a)

2. (c) 6. (a)

3. (b) 7. (b)

4. (d) 8. (c)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 11. (a), (c) 13. (a), (b)

12. (b), (c), (d) 14. (a), (b) 15. (a), (b)

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MATRIX-MATCH TYPE QUESTIONS p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

c

p

q

d

p

16.

17.

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. (a) 5. (c)

2. (b) 6. (a)

1. (b), (c) 5. (b), (d)

2. (b), (d) 6. (a), (b)

3. (a), (d) 4. (c) 7. (b), (c), (d)

MATRIX-MATCH TYPE QUESTIONS q

r

s

a

p

q

r

s

s

b

p

q

r

s

r

s

c

p

q

r

s

q

r

s

d

p

q

r

s

p

q

r

s

p

q

r

s

a

p

q

r

s

a

p

q

r

s

b

p

q

r

s

b

p

q

r

s

c

p

q

r

s

c

p

q

r

s

d

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

p

q

r

s

r

s

a

p

q

r

s

b

p

q

r

s

b

p

q

r

s

c

p

q

r

s

c

p

q

r

s

d

p

q

r

s

d

p

q

r

s

20.

4. (d) 8. (a)

MULTIPLE CORRECT ANSWERS TYPES QUESTIONS

p

18.

3. (a) 7. (c)

1.

2.

3.

ASSERTION-REASON TYPE QUESTIONS

4. (a)

21. (b) 25. (b)

ASSERTION-REASON TYPE QUESTIONS

22. (c)

23. (d)

24. (a)

1. (d)

COMPREHENSION-TYPE QUESTIONS INTEGER-ANSWER TYPE QUESTIONS 30. 4

31. 5

32. 5

33. 2

2. (d)

COMPREHENSION-TYPE QUESTIONS 1. (b)

2. (d)

3. (c)

INTEGER-ANSWER TYPE QUESTIONS 1. 6

2. 6

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FILL 1.

IN THE

13

BLANKS TYPE QUESTIONS 2. ±

1 6

(2i + j + k )

SUBJECTIVE TYPE QUESTIONS 1. (a) x + y – 2z = 3

(b) (6, 5, – 2)

9 2

4.

5. 2x – y + z – 3 = 0; 62x

y

z – 105 = 0

Hints and Solutions LEVEL 1

a(x – a) + b(y – b) + c(z – c) =0 fi ax + by + cz = a2 + b2 + c2 Ê 3 1 6ˆ 2. Both the lines pass through the point Á , - , ˜ . Ë 4 2 7¯ 3. direction cosines of x 0, 1, 0 and of z

p=

-k l 2 + m2 + n2

i + j-k fi [(l1 + l2) i + (2l1 – l2) j + (l1 + l2) k] ◊ 1+1+1 1 = 3 fi 2l1 - l2 = 1 and the vector is r = (3l1 - 1) i + j + (3l1 - 1)k fi r = - 4 i + j – 4 k for l1 = - 1 fi - r = 4i - j + 4k x y z + + = 1, then a b c 1 1 1 1 + 2 + 2 2 a b c

y

=

3 1 1 1 4 fi 2 + 2 + 2 = 2 9 a b c

A(a, 0, 0), B(0, b, 0) and C(0, 0, c) l, m, lx + my + nz = k. Length of

n

a b c = = 1 1 0 x+y+1=0 1+ 2 +1 = 2 2 2 10. Vector lying in the plane of a and b is 1 r = l1 a + l2 b and its projection on is 3 2a - 2b + c = 0, a - b + 2c = 0 fi

fik=p x, y, z

we get al + bm + cn = 0 6. If a coordinate axis, then cos2 a + cos2 a + cos2 a = 1 fi cos2 a = 1/3 and the direction cosines of the lines are 1/ 3 , ± 1/ 3, ± 1/ 3 which give us four lines. y - z = 4. x+z by (2, 5, 1). r = (1 - t) (- 11i + 3 k ) + t (2 i - 3 j + k ) i.e. r = (13t - 11) i - 3 t j + (3 - 2 t ) k which does not represent the vectors in (a), (b) or (c) for any value of t. a (x - 1) + b(y + 2) + c(z - 1) = 0

Centroid of the D s.t x–2 + y–2 + z–2 in (a), (b) and (c).

(a/3, b/3, c/3) = (x, y, z)

x y z + + = 1 which a b c A(a, 0, 0), B(0, b, 0) and C(0, 0, c) planes parallel to coordinate planes, through A, B, C are x = a, y = b, z = c and their point 2 3 4 of intersection is (a, b, c) where + + = 1 as the a b c points in (a), (b), (c) and (d). a (x - 1) + b(y - 2) + c(z - 3) = 0, where 2a + 3b + 4c = 0 (i) If it is parallel to the x are 1, 0, 0, then a ◊ 1 + b ◊ 0 + c ◊ 0 = 0 fi a = 0 b = (- 4/3)c is - 4(y - z) + 3(z - 3) = 0 or 4y - 3z + 1 = 0 y z respectively 2x - z + 1 = 0 and 3x - 2y + 1 = 0 14. Vector A is parallel to

[(2 j + 3 k ) ¥ (4 j - 3 k )] ¥ [( j - k ) ¥ (3 i + 3 j)]

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(b) line is parallel to the plane if r + r 2 – 12 = 0 fi r = 3, – 4

= 54 (j – k) If q cos q = ±

54 + 108 3 ¥ 54 2



1 2

fi q = p/4 or 3p/4 x y z 1 -2 2 2 3 -1

fi 4x – 5y – 7z = 0 which passes through the points in (a), (b), (c) 16. Since cos a, cos b, cos g are the direction cosines of the line cos2 a + cos2 b + cos2 g = 1 fi sin2 a + sin2 b + sin2 g = 2 and cos 2a + cos 2b + cos 2g = 1 - 2 = - 1, cos2a + cos2b – sin2g = 0

x + ry + r2 z = 0, since it passes through (– 4, 3, 1), – 4 + 3r + r2 = 0 fi r = 1, – 4 (d) The lines are perpendicular if 5(– 1) + (p) (–p) + (2p)3 = 0 fi p2 – 6p + 5 = 0 fi p = 5, 1 20. (a) Since

Ê 1 ˆ tan–1 Á 2 ˜ = tan–1 (2r + 1) – tan–1(2r – 1) Ë 2r ¯

17. (a) (2, 1, -



= tan–1 (2n + 1) – tan–1 (1)

x y z - + =1 7 5 3



fi a = b = c π represents three identical planes x + y + z = 0 (b) fi |A| = 0 fi ax + by + cz = 0 fi by + cz = (b + c)x and cy + az = (c + a)x fi (b + c)y + (c + a)z = (b + c)x + (c + a)x fix=y=z (c) fi | A| π x=y=z point. (d) fi a = b = c = 0; x, y, z can take any values

a, 0, 0), (0, b, 0) and (0, 0, c) so the centroid is (a/3, b/3, c/3) = (1, r, r 2) fi a = 3, b = 3r, c = 3r2 and the x y z + + 2 = 1 3 3r 3r 2 2 or r x + ry + z = 3r which passes through (4, – 8, 15) if 4r2 – 8r + 15 = 3r2 fi r2 – 8r + 15 = 0 fi r = 3 or 5



Ê 1 ˆ

 tan - 1 ÁË 2 r 2 ˜¯

r =1

a b c 1 18. | A| = b c a = – (a + b + c) [(a – b)2 + 2 c a b (b – c)2 + (c – a)2] (a) fi (a – b)2 + (b – c)2 + (c – a)2 = 0

Ê 1 ˆ

n

 tan - 1 ÁË 2 r 2 ˜¯

r =1

(b) 5 × 21 - 3 × 35 = 0 so plane perpendicular to 5y - 3z (p) Can be written as

(2 r + 1) - (2 r - 1) 1 = , we get 2 1 - (2 r + 1) (2 r - 1) 2r

p˘ È = lim Ítan - 1 (2n + 1) - ˙ nÆ• Î 4˚

p p p – = 2 4 4 \ q = p /4 fi tan (q) = 1.

=

(b)

1 - tan 2 a a = cos 2a = b+c 1 + tan 2 a fi tan2 a =

b+c-a b+c+a 2

b=

a+b-c . a+b+c

Thus, tan2 a + tan2 b =

2b 2 = 3 a+b+c [ a + c = 2b]

L is x-0 y -1 z-0 = = 1 2 2 Now, AM = projection of OA on AM O (0, 0, 0)

A (0, 1, 0)

M Fig. 20.9

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2 1+ 4 + 4

=

=

Now, OM2 = OA2 – AM2 = 1 – fi OM =

x = r sin q cos f = 1 y = r sin q sin f = 2

2 3

z

4 5 = 9 9

P (x, y, z)

5/3 q

a(x – 1) + b(y – 2) + c(z – 3) = 0. as it is perpendicular to x = 0, y = 0, we get a = 0, b = 0 and the z = 3.

O

r 90 - q

y

f M

x Fig. 20.10

0-3 1

z = r cos q = 3

= 3.

21. L1 and L2 are parallel to the vectors a = 3i + j + 2k and b = i + 2 j + 3k respectively. The unit vector perpendicular to both L1 and L2 is a ¥ b – i – 7 j + 5k = a¥b 1 + 49 + 25 fi

x + 1) – 7(y + 2) + 5(z + 1) = 0 whose 13 5 3 and thus state

1 + 4 + 9 = 14 (Taking the +ve sign)

r = OP = fi cos q = and

3 14

fi tan q =

sin q sin f 2 = fi tan f = 2. sin q cos f 1

Thus tan q tan f = 2 5 1- l (1 + l ) + (1 - l ) + 4 2

[(2 – 1) i + (3 – 2) j + (1 – 3) k ] ◊ (ai + 2 j + 3k ) ¥ (3i + aj + 2k ) = 0 1 fi a 3

1 2 a

2 – 2 = 0 fi 2a + 5a – 25 = 0 fi a = 5 2 or – 5 3 2

2x – y + 2z – 3 + l (3x + y + z – 5) = 0 For l = 1, we get 5x + 2z – 8 = 0 which is perpendicular to 2x – y – 5z – 3 = 0 as 5 ¥ 2 + 0(– 1) + 2(– 5) = 0. 24. Direction ratios of PQ are 6 – 2, 5 – 1, – 2 – 6 i.e. p, so PQ is perpendicular to p PQ is perpendicular to p of PQ lies on p. 25. Let P (x, y, z) = P (1, 2, 3) and OP = r OM = r q) = r sin q

5 3

2

1 2

fi 4 (1 – l)2 = 2(l2 + 3) fi l2 – 4l – 1 fi l = 2± 5

2 2 26. D = 0 2

3 3 3 0

1 -6 2 3 0 0 0 = 6 0 3 1 = 72 1 0 2 0 1 1 0

2 3 0 D1 = – 0 3 1 = –12 2 0 1 2 3 1 D2 = 0 3 1 = 6 2 0 1 2 3 1 D3 = – 2 3 0 = 6 2 0 1

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r – 1, 3r – 1, 4r – 1 which lies on the plane p1 if 2r – 1 + 3 (3r – 1) + 3(4r – 1) = 14 fi r = 1 and the point P is (1, 2, 3) P(1, 2, 3); perpendicu x -1 y - 2 z - 3 = = lar to p1 is 1 2 3

2 3 1 D4 = 2 3 0 = 6 0 3 1 72 ¥ 72 ¥ 72 = |– 24| = 24. 6(-12) ¥ 6 ¥ 6 ¥ 6 27. Let the variable plane be ax + by + cz + d = 0 a 1 D= 0 0

b 0 1 0

c 0 0 1

d 0 = –d 0 0

D1 = 1, D2 = a, D3 = – b, D4 = c -d 3 = 6 ¥ 1 ¥ a ¥ ( -b) ¥ c

64 k3 fi -

d d d - = 6 ¥ 64 k3 a b c

1 0 28. D = 1 1

1 1 0 1

1 -1 1 0 = 2, D1 = 2, D2 = 1, D3 = 1, D4 = 1 1 0 0 0 23 2 = 6 ¥ 2 ¥1¥1¥1 3

=

=0

which is true. L1 is (–3r1 – 1, 2r1 + 3, r1 – 2) and on L2 is (r2, – 3r2 + 7, 2r2 r2 = 2, r1 = –1 and the point of intersection of L1 and L2 is (2, 1, –3)

x +1 y - 3 z + 2 -3 2 1 =0 1 -3 2

(8i + 6 j - 2k ) 64 + 36 + 400

32 + 36 - 160

=

-1 - 0 3 - 7 -2 + 7 -3 2 1 =0 -3 1 2

containing the lines L1 and L2 is

35. d[(3 + 1)i + (5 + 1)j + (7 + 1)k]◊ (i + 2 j + k ) ¥ (7 i - 6 j + k ) |(i + 2 j + k ) ¥ (7i - 6 j + k )| = (4i + 6 j + 8k ) ◊

the lines are coplanar if and only if

-1 -4 5 i.e -3 2 1 1 -3 2

which passes through (0, 0, 0). P perpendicular to p1, is (r + 1, 2r + 2, 3r + 3), if it lies on p2 then 2(r + 1) – (2r + 2) + 3(3r + 3) = 27 fi r = 2 The coordinates of Q PQ are (2, 4, 6). r = (2l + 1)i + (l + 2)j + (2l + 3)k which lies on the plane if 2(2l + 1) – 6(l + 2) + 3(2l + 3) + 5 = 0 fi l = –1 and the position vector of the point of intersection is –i + j + k = ai + bj + ck fi a = –1, b = 1, c = 1 \ (50a + 60b + 75c)2 = (85)2 = 7225

500 92 10 5



=

46 5 5

fi 125d 2 = 46 ¥ 46 = 2116

125d 2 =2 1058

l, 2 + l, 3 + 2l) which lies on the given plane if 2 (1 + 2l) – 6(2 + l) + 3 (3 + 2l) + 5 = 0 fi 4l + 4 = 0 fi l = –1 so, the point of intersection is (–1, 1, 1) which lies on the plane r . (i + 75j + 60k) – a = 0 if –1 + 75 + 60 – a = 0 fi a = 134 a fi = 2. 67 l + 1, 16l – 1, nl + 1) 1 intersects 6x2 + 5y2 = 1, z = 0 if l = n

IIT JEE eBooks: www.crackjee.xyz 20.45 2

Ê 15 ˆ Ê 16 ˆ so 6 Á - + 1˜ + 5 Á - - 1˜ Ë n ¯ Ë n ¯

s is parallel to r, r ◊ s = 0

2

fi (l + 1) – (l – 1) + 2l – 1 = 0 fi l = -

fi n2 – 2n + 263 = 0 fi k = 263

x – 3y – 4z – 4 =0 fi |A + B + C| = 6

1 and k=1 263 x – y – z – 4 + l(x + y + 2z – 4) = 0 2x + 3y + z – 1 and x + 3y + 2z – 2 = 0 is i j k s = 2 3 1 = 3(i – j + k) 1 3 2 r = (l + 1)i + (l - 1) j + (2l - 1)k

1 2

(i)

11(x – 11) + 11(y – 11) + 11 (z –11) = 0 fi x + y + z = 33 S S = 3(33)2 fi =3 (33) 2 5x – 7y + 11z + 4 = 0 fi a4 + b3 + c2 + d

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21 Functions In everyday life, many quantities depend on one or more changing variables eg: (a) the wind chill depends on the speed of the wind (b) amount earned by your investment depends on interest rate (c) plant growth depends on sunlight and rainfall (d) voltage depends on current and resistance (e) test marks depend on attitude, listening in lectures and doing tutorials (among many other variables!!)

If x is an element in the domain of a function f of a function requires that f assign one and only one value to x. This means that a function cannot be multiple-valued. For example, the expression ± x since it assigns two values to each positive x.

x,

Illustration 1 Let f (x) = x3. This tells us that the value of f at x is x3. Thus, f (2) = 23 = 8, f (– 4) = (– 4)3 = –64 and f (0) = 03 = 0.

21.1 DEFINITIONS

A function from a set D to a set R is a rule that assigns a unique element f(x) in R to each element x in D i.e. there is only one output (y-value) for each input (x-value). If the number of elements in D and R are n and m respectively D m to R is n . In general, the sets D and R need not be sets of real numbers. However, we consider only those functions for which D and R are both subsets of the real numbers. We shall denote by R the set of all real numbers. The set D domain of the function. We usually denote it by dom f. If x is an element in the domain of a function f, then the element that f associates with x is denoted by the symbol f (x), and is called the image of x under f, or the value of f at x. The set of all possible values of f (x) as x varies over the domain is called the range of f. If f : D Æ R, then the range of f is a subset of R and the set R is called co-domain of f. A function is depicted pictorially below

Domain

Remark

Range

Illustration 2 Let g(x) = x - 3 . When x < 3, g(x) is imaginary because x – 3 will be a negative number. Thus the domain of g is [3, •) = {x: x ≥ 3}. As x varies over this interval, g(x) varies from 0 to •. Therefore, the range of g is [0, •). 21.2

DOMAINS AND RANGES OF SOME ELEMENTARY FUNCTIONS

1. Constant functions A function that assigns the same value to every member of its domain is called a constant function. The domain of the constant function f (x) = c is R and its range is {c}. 2. Polynomial functions A function of the form c xn, where c is a constant and n is a non-negative integer, is called a monomial in x. Examples are 2x3, 5x4, –6x and x8. The function 4 x1/2 and x–3 are not monomials because the powers of x are not non-negative integers. A function that x is called a polynomial in x. Thus + a1 x + a0 f (x) = an xn + an – 1 xn – 1 + is a polynomial. The domain of a polynomial is R and its range is a subset of R depending on the degree n. 3. The domain of f (x) = loge x (= ln x or log x) is (0, •) = {x Œ R, x > 0} = R+, and its range is (– •, •) (i.e., the whole of R).

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4. The domain of f (x) = etx is R and its range is R+. 5. The domain and range of elementary trigonometric functions are as follows: (i) The domain of f (x) = sin x is R and its range is [– 1, 1] = {x Œ R: –1 £ x £ 1} (ii) The domain of f (x) = cos x is R and its range is [–1, 1]. (iii) The domain of f (x) = tan x is R ~ {(2m + 1)p/ 2, m Œ I} = {x Œ R: x π (2m + 1)p/2, m is an integer} and its range is R. (iv) The domain of f (x) = cot x is R ~ {mp : m Œ I} = {x Œ R, x π mp, m is an integer} and its range is R. (v) The domain of f (x) = sec x is R ~ {(2m + 1)p/2, m Œ I} and its range is ( •, – 1] » [1, •) = {x Œ R, x œ (–1, 1)}, so | sec x | ≥ 1. (vi) The domain of f (x) = cosec x is R ~ {mp, m Œ I} and its range is the same as that of sec x. 6. The domains and ranges of the inverse trigonometric functions are as follows: (i) The domain of f (x) = sin–1 x (also written as arcsin x) is [–1, 1] and its range is [–p /2, p /2]. (ii) The domain of f (x) = cos–1 x is [–1, 1] and its range is [0, p]. (iii) The domain of f (x) = tan–1 x is R and its range is (–p/2, p/2) = {x Œ R, –p /2 < x < p/2} (iv) The domain of f (x) = cot–1 x is R and its range is (0, p). (v) The domain of f (x) = sec–1 x is R ~ (–1, 1) and its range is [0, p/2) » (p/2, p] = {x Œ R, 0 £ x £ p, x π p/2} (vi) The domain of f (x) = cosec–1 x is R ~ (–1, 1) and its range is [–p/2, 0) » (0, p/2] = {x Œ R, – p/2 £ x £ p/2, x π0} 7. For any subset A Õ R. The function jA: R Æ {0, 1} Ï1 if x Œ A jA (x) = Ì Ó0 if x œ A is called the of A. Its domain is R and the range is {0, 1}. The following are some simple properties: (i) jA » B = jA + jB – jA « B (ii) jA « B = jA jB (iii) If A and B are disjoint sets then jA » B = jA + jB

21.3 ALGEBRAIC OPERATIONS ON FUNCTIONS

1. If f and g are two functions, then sum of the functions, f+g x Œ dom f « dom g by (f + g) (x) = f (x) + g(x) 2. If k is any real number and f is a function, then kf x Œ dom f by (k f ) (x) = k f (x). 3. If f and g are two functions, then the pointwise product fg x Œ dom f « dom g by ( f g) (x) = f (x) g(x) 4. If f and g are functions, then f/g xŒ dom (f ) « dom(g) « {x: g(x) π 0} by (f /g)(x) = f(x)/g(x). 5. Composition of functions Let f : A Æ B and g : B Æ C be functions, then gof : A Æ C gof ) (x) = g ( f (x)). Note that if gof fog Let f : R Æ R and g : R Æ R f (x) = cos x 3 and g(x) = x . Then gof (x) = g( f (x)) = g(cos x) = cos3 x and fog (x) = f (x3) = cos x3. Thus even if fog and gof are fog = gof. We have the following formulae for domains of functions. 1. dom ( f + g) = dom f « dom g 2. dom ( f g) = dom f « dom g 3. dom ( f /g) = dom f « dom g « {x : g(x) π 0} 4. dom f = dom f « {x : f (x) ≥ 0} 21.4 SOME CLASSES OF FUNCTIONS

1. Rational function of two polynomials y=

a0 x n + a1 x n - 1 + b0 x + b1 x m

2

m -1

+

+ an + bm

3

For example, y = (x + 3)/(x + 4) is rational function. 2. Irrational function If in the function y = f (x), the operations of addition, subtraction, multiplication, division and raising to a power with rational non-integral exponents are performed on the right-hand side, the function y = f (x) is said to be irrational. Examples are y = (3x2 + x )/ 2 + 4 x and y = x . 3. Algebraic function An algebraic function is any function y = f (x + Pn (x) = 0 P0(x)yn + P1(x)yn – 1 + where P0 (x), P1(x), , Pn(x) are certain polynomials in x. Clearly, every rational and irrational function is an algebraic function. 4. Transcendental function A function which is not an algebraic function is called a transcendental function. Examples are trigonometric, logarithmic and exponential functions.

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5. Explicit function If the dependent variable, say y, is expressible explicitly in terms of the independent variable, the function y = f (x) is called an explicit function. Otherwise it is said to be an implicit function. For example, y = cos3 x + x3 is an explicit function, whereas x5 + y5 – 3y x = 0 is an implicit function. 6. Even function A function y = f (x) is said to be an even function if f (–x) = f (x) " x Œ dom ( f ). Examples are y = cos x and y = x 2n. The graph of an even function y = f(x) is symmetric about y-axis. 7. Odd function A function y = f (x) is said to be an odd function if f (– x) = – f (x) " x Œ dom ( f ). Examples are y = sin x and y = x2n + 1. The graph of an odd function is symmetric about origin. Clearly, y = f (x) + f (– x) is always an even function, and y = f (x) – f (– x) is always odd. Any function y = f (x) can be expressed uniquely as the sum of an even and an odd function as follows: 1 1 ( f (x) + f(–x)) + ( f (x) – f (– x)) f (x) = 2 2 8. Periodic function A function y = f(x) is said to be periodic if there exists a number T > 0 such that f (x + T) = f (x) for all x in the domain of f. The least such T is called the fundamental period or simply period of f. For example, the period of sin x and cos x is 2p, and that of tan x is p. If f (x) is a periodic function with period T, then the function f (a x + b), a > 0, is periodic with period T/a. For example, sin 2x has a period p, cos 3x has a period 2p/3 and tan (2 x + 4) has a period p/2. The period of |sin x| is p, whereas that of |sin x| + |cos x| is p/2. 9. Onto function (or Surjective Function) If a function f : A Æ B is such that each element in B is the f-image of at least one element in A, then we say that f is a function of A ‘onto’ B. Equivalently a function f is an onto function if codomain of f = Range of f.

as f (1) = 1 and f (–1) = 1. Note that a periodic function f : R Æ R cannot be one-to-one as f (x + T) = f (x) for some T > 0. In order to show that a function f : A Æ B is one-to-one, we may take any x, y Œ A such that f (x) = f ( y) and try to show that x = y. Let f: D Æ G and g:C Æ B so that gof:D Æ B (i) If both f and g are one - one then so is gof (ii) If both f and g are onto then so is gof 11. Bijective function (One-to-One and onto). If a function f is both one-to-one and onto, then f is said to be a bijective function. For example an identity function iA (x) = x is trivially a bijective iA : A Æ A function. The function f : [–p/2, p/2] Æ by f (x) = sin x is a bijective function. The function f (x) = tan x is also a bijective f : (– p/2, p/2) Æ R function. The concept can be illustrated by the following diagram (Fig. 21.2).

Fig. 21.2

If A and B are finite sets and f: A Æ B is a bijection then n(A) = the number of elements in A = the number of elements in B. If n(A) = n, then the number of bijections from A to B is the total number of arrangements of n items taken all at a time which is n!. 12. A function I is said bounded on I if there is k > 0 such that | (x)| £ k for all x Œ I. Equivalently there is m 1 and M such that m £ (x) £ M for all x Œ I. E.g. (x) = x is not bounded on (0, 1) and (x) = x2 is bounded on [0, 1]. (x) = sin x or cos x are bounded functions on R whereas (x) = tan x is unbounded on (– p/2, p/2). 21.5 GRAPHS OF SOME FUNCTIONS

Fig. 21.1

For example, f : R Æ f (x) = sin x is an onto function but f : R Æ R f (x) = sinx is not onto since Range of f = [– 1, 1] and codomain of f = R. In order to show that a function f : A Æ B is onto we start with any y Œ B x Œ A such that f (x) = y. 10. One-to-one function (or injective function) A function f is said to be one-to-one if it does not take the same values at two distinct points in its domain. For example, f (x) = x3 is one-to-one, whereas f (x) = x2 is not,

1.

Constant function f (x) = c represents a constant function (Fig. 21.3).

Fig. 21.3

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2. Proportional values. If variables y and x are direct proportional, then the functional dependence between them is represented by the equation: y = kx, where k is a constant a factor of proportionality. A graph of a direct proportionality is a straight line, going through an origin of coordinates and forming with an x- axis an angle a, a tangent of which is equal to k: tan a = k. Therefore, a factor of proportionality is called also a slope. These are shown in three graphs with k = 1/3, k = 1 and k = –3 on Fig. 21.4 k=–3

a hyperbola points is a constant value, equal in this case to 1. In general case this value is k, as it follows from a hyperbola equation: xy = k Y 2 1 –2

–1

X

O

Y

–1

k=1

–2

k = 1/3 a

X

O

Fig. 21.6 Fig. 21.4

3. Linear function. If variables y and x are related by the 1-st degree equation: Ax + By = C, (at least one of numbers A or B is non-zero), then a graph of the functional dependence is a straight line. If C = 0, then it goes through an origin of coordinates, otherwise – not. Graphs of linear functions for different combinations of A, B, C are represented on Fig. 21.5 Y

B = 0, A π 0 C=0

A π 0, B π 0, C π 0

C/B

A = 0, B π 0

X

C/A Fig. 21.5

4. Inverse proportionality. If variables y and x are inverse proportional, then the functional dependence between them is represented by the equation: y = k/x, where k is a constant. A graph of an inverse proportionality is a curve, having two branches (Fig.21.6). This curve is called a hyperbola. These curves are received at crossing a circular cone by a plane. As shown on Fig. 21.6, a product of coordinates of

The main characteristics and properties of hyperbola: the function domain: x π 0, and codomain: y π 0; the function is monotone (decreasing) at x < 0 and at x > 0, but it is not monotone on the whole, because of a point of discontinuity x = 0 the function is unbounded, discontinuous at a point x = 0, odd, non–periodic; there are no zeros of the function. 5. Quadratic function. This is the function: y = ax2 + bx + c, where a, b, c – constants, a π 0. In the simplest case we have b = c = 0 and y = ax2. A graph of this function is a quadratic parabola – a curve going through an origin of coordinates. Every parabola has an axis of symmetry OY, which is called an axis of parabola. The point O of intersection of a parabola with its axis is a vertex of parabola. Y

y = 1/2 x2

3 2 1 –3 –2 –1 O –1

1

2

3

X

Fig. 21.7

A graph of the function y = ax2 + bx + c is also a quadratic parabola of the same shape, that y = ax2, but its vertex is not an origin of coordinates, this is a point with coordinates:

IIT JEE eBooks: www.crackjee.xyz Functions 21.5 Y

Ê b b2 ˆ , c Á 2a 4 a ˜¯ Ë

2

n=1

y = x ,n > 0

The form and location of a quadratic parabola in a coordinate system depends completely on two parameters: a of x2 and discriminant D : D = b2 – 4ac. These properties follow from analysis of the quadratic equation roots. All possible different cases for a quadratic parabola are shown on Fig. 21.8

n = 1/2 n = 1/4 n=0

1

Y a>0, D = 0

n=4 n=2 n

O

1

a>0, D 1 and decreases at 0 < a < 1; the function is unbounded, continuous in everywhere, non-periodic; the function has one zero: x = 1. 9. Absolute-value function is given by Ï x x≥0 y = | x | = max {x, – x} = Ì Ó- x x < 0 It is depicted in Fig. 21.17.

Fig. 21.17

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Greatest Integer Function and Fractional part function f(x) = [x] will denote the integral part of x or greatest integer less than or equal to x. Eg. [5] = 5, [5.2], = –5, [–5.1] = –6, [p] = 3, [e] = 2. The domain of this function is R and the range is I. For positive x, this function might represent, for example, the legal age of a person is a function of his chronological age x. The fractional part of x is the fractional part [x] of x so {x} = x – [x] e.g. (5.04) = 0.04. {–4} = 0, {–1,7} = 0.3. {p} = p–3. The domain of this function is R whereas the range is (0, 1).

8.

Ï x, if 0 < x < 1 Ó0, x ŒI

{ x} = Ì

Ï0, x Œ I 9. {x} + {–x} = Ì Ó1, if x œ I n - 1˘ 1˘ È È = [ nx ]. 10. [ x ] + Í x + ˙ + ... + Í x + n n ˙˚ Î Î ˚ 11. Signum function is given by if x > 0 Ï1 Ô sgn ( x ) = Ì0 if x = 0 Ô-1 if x < 0 Ó and it is depicted in Fig. 21.20

graph of [x]

x'

y' Fig. 21.18 Fig. 21.20

y

graph of [x]

x'

y'

12. The least integer function: The function whose value at any number x is the smallest integer greater than or equal to x is called the least integer function, e.g. if f (x) = [x]¢ = smallest integer, greater than or equal to x. then [1.5]¢ = 2 [1.99]¢ = 1 [1.4]¢ = 2.

Fig. 21.19

Some Properties

1. 2. 3. 4.

x – 1 < [x] £ x [{x}] = 0 = {[x]} If n £ x < n + 1, n Œ I, then [x] = n and conversely. If x Œ I, then [x] = x otherwise [x] < x.

, if x ŒI Ï-[ x ] 5. [–x] = Ì Ó-[ x ] - 1 , if x œI Ï0, if x ŒI in other word, [x] + [–x] = Ì Ó-1, if x œI 6. [x + y] = [x] + [y], if x or y Œ I. if {x} + {y} < I Ï[ x ] + [ y], 7. [ x + y] = Ì + + 1 if {x} + {y} ≥ I [ x ] [ y ] , Ó Moreover, [x + y] ≥ [x] + [y].

Fig. 21.21

Figure 21.21 shows the graphs. For positive value of x, this function might represent for example, the cost of parking x hours in a parking lot which charges Rs. 10 for each hour or part of an hour in this case f(x) = 10 [x]¢.

IIT JEE eBooks: www.crackjee.xyz 21.8 Comprehensive Mathematics—JEE Advanced 21.6 INVERSE OF A FUNCTION 1/3

Illustration 4

3

The function f (x) = x and g(x) = x have the following property: f (g(x)) = f (x3) = (x3)1/3 = x g( f (x)) = g(x1/3) = (x1/3)3 = x Similarly, the functions f (x) = loge x and g(x) = ex cancel the effect of each other. If two functions f and g satisfy f (g(x)) = x for every x in the domain of g, and g( f (x)) = x for every x in the domain of f, we say that f is the inverse of g and g is the inverse of f, and we write f = g–1, or g = f –1. (The symbol f –1 does not mean 1/f.) f, write down the equation y = f (x) and then solve x as a function of y. The resulting equation is x = f –1 ( y). If f is one-to-one, then f range and, conversely, if f has an inverse, then f is onef. If f is one-to-one to-one. f –1 from A to B, and g is one-to-one from B to C, then f og is one-to-one from A to C, and ( fog)–1 = g–1 o f –1 If f : X Æ Y is bijective then f –1 : Y Æ X is bijective and –1 –1 (f ) = f. Moreover, f –1 ({ y }) = { f –1(y)} for all y Œ Y. Illustration 3 Let f(x) = Let y =

ax + b . The domain of f is R – cx + d

Ï d¸ Ì- ˝ . Ó c˛

If f : R Æ R f (x) = | x | and A = (– •, 0), then f –1(A) = {x: f (x) Œ A} = {x: | x | Œ (– •, 0)} = f (the empty set). If B = (–3, 3), then f –1 (B) = (–3, 3). Note that if f is one-to-one, then f –1({x}) is a singleton A, B 1 Y. (i) A 1 B = f –1 (A) 1 f –1 (B) (ii) f –1 (A » B) = f –1 (A) » f –1 (B) (iii) f –1 (A « B) = f –1 (A) « f –1 (B). (iv) f –1 (A ~ B) = f –1 (A) ~ f –1 (B). (v) For S Õ X, f –1 ( f(S))   S (vi) f (f –1 (A)) Õ A. Monotone Functions Let D be a doain of real numbers. 1. A function f : D Æ R is non-decreasing (increasing) if for every x1 < x2 in D, f(x1) £ f(x2) (f(x1) < f (x2)). 2. A function f : D Æ R is non-increasing (decreasing) if for every x1 < x2 in D, f(x1) ≥ f(x2) (f(x1) > f (x2)). 3. A function f : D Æ R is monotone if it is either increasing on D or decreasing on D. One-to-one function need not by monotone but a monotone function is one-one. SOLVED EXAMPLES

ax + b b - dy so cxy + dy = ax + b x = . cx + d cy - a

Hence f –1(x) =

b - dx cx - a

21.7 DIRECT AND INVERSE IMAGES

Let f be a function with domain X and co-domain Y. If A Õ X, then the direct image of A under f is the subset of Y (denoted by f (A f (x) : x Œ A}. For example, let f : R Æ R be a function defined by f (x) = x2. If A = {–3, –1, 0, 1, 3}, then f (A) = {0, 1, 9}. Note that if f (X) = Y, then the function f : X Æ Y is onto. (i) A 1 B fi f (A) 1 f (B) (ii) f (A » B) = f (A) » f (B) (iii) f (A « B) 1 f (A) « f (B). The inclusion may be strict. (iv) f (A ~ B) 4 f (A) ~ f (B) Let f be a function with domain X and range Y and let B be a subset of Y. The inverse image of B under f is a subset x: f (x) Œ B}. of X (denoted by f –1 (B

SINGLE CORRECT ANSWER TYPE QUESTIONS x and g(x) = 7x + b. If the function y = fog(x) passes through (4, 6) then the value of b is Example 1

Consider the functions f(x) =

(a) 8 (c) –25 Ans. (a) Solution:

(b) – 8 (d) 4 – 7 6

y = f (g (x)) = f (7x + b) =

Given that y (4) = 6, so Example 2

7x + b

28 + b = 6 fi 28 + b = 36 fi b = 8.

Let f : P(N) Æ N be defined by

f(S) = min S, the minimum value of S for S Õ N. Then f is (a) one-one and onto (b) onto but not one-one (c) one-one but not onto (d) neither one-one nor onto Ans. (b)

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Solution: Given any n Œ N, consider the set S ={n, n + 1, ....}. S Œ P(N) and min S = n, so f is onto. Consider S1 = {1} and S2 = {1, 2}, min S1 = 1 = min S2 but S1 π S2 so f is not one-one. The domain of the function f(x) =

Example 3 1 sin x

+ 3 sin x is (a) [0, p] [2kp , (2k + 1)p ]

(b) k ŒI

(2kp , (2k + 1)p )

(c)

C . d -1 +1 Example 6

The period of the function f(x) =

cosec2 3x + cot 4x is (a) p/3

(b) p/4

(c) p/6

(d) p

Ans. (d) Solution: f(x) = 2/(1 – cos 6x) + 1/(tan 4x). So the period of f = l.c.m (period of cos 6x, period of tan 4x) = l.c.m (p/3, p/4) = p. Example 7

Which of the following sets of ordered

k ŒI

(a) R = {(x, y); x2 + y2 = 2} on R (b) A = {1, 2, 3}, B = {1, 2, 3, 4, 5} and R = {(x, y): 5x + 2y is a prime number} on A (c) A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5, 6, 7} and R = {(x, y): y = x2 – 3x + 3} on A (d) A = (0, •); R = {(x, g): y = x2} on A

[2kp , (2k + 1)p ]

(d) k ŒN

Ans. (c) Solution: f x in R for which sin x > 0 i.e. x Π(2kp, (2k + 1)p), k ΠI. Example 4

The period of the function f(t) =

p p ) + 2 sin (3pt + ) + 3 sin 5 pt + cos 4pt is 3 4 (a) 1 (b) 2 (c) 4 (d) 6

Ans. (d)

sin (2pt +

Solution:

Ans. (b)

R (1, 1), (2, 1) ΠR in (c). So R to-one function. R in (d) is one-to-one as A = {x : x > 0}

p 2p = 1 . The ) is 2p 3 p 2p 2 = . The period of ) is period of 2 sin (3pt + 3p 3 4 1 2p 2 3 sin 5pt is = . The period of cos 4pt is . The period of 2 5p 5

Solution: The period of sin (2pt +

Ï 2 2 1¸ f(t) is l.c.m Ì1, , , ˝ = 2 . Ó 3 5 2˛ Consider the function h: I × I Æ Q m , then h is by h (m, n) = | n | +1 Example 5

(a) (b) (c) (d)

onto but not one-one one-one and onto one-one but not onto neither one-one nor onto

Ans. (a) Solution:

Unequal element (1,2) and (1, –2) are in I × I 1 but h (1,2) = h(1, –2) = , so h is not one-one. Take any 3 bŒ Q, b =

c for some c, d ΠI, and d > 0. Then h (c, d Р1) = d

Since (1, Р1) and (1, 1) ΠR in (a), so R cannot ΠR in (b). So

Example 8 f (x) =

Let f : R " R

e| x | - e- x ex (a) (b) (c) (d)

. Then + e- x f is both one-one and onto f is one-one but not onto f is onto but not one-one f is neither one-one nor onto.

Ans. (d) Solution: f is not one-one as f (0) = 0 and f (–1) = 0. f is also not onto as for y = 1 there is no x Œ R such that f (x) = 1. If there is such a x Œ R then e| x | – e–x = ex + e–x. Clearly x π 0. For x > 0, this equation gives –e–x = e–x which is not possible. For x < 0, the above equation gives ex = – e–x which is also not possible. Example 9

If f (x) is a polynomial satisfying

f (x) · f (1/x) = f (x) + f (1/x), and f (3) = 28, then f (4) is given by (a) 63 Ans. (b)

(b) 65

(c) 67

(d) 68

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Solution: By considering a general nth degree polynomial and writing the expression f (x) · f (1/x) = f (x) + f (1/x) in of xn, xn–1, and the constant term, that the polynomial satisfying the above equation is either of the form xn + 1 or – xn + 1. Now, from f (3) = 3n + 1 = 28, we get 3n = 27, or n = 3. But f (3) = – 3n + 1 = 28 is not possible, as –3n = 27 is not true for any value of n. Hence f (4) = 43 + 1 = 65. Example 10

Let the function f (x) = x2 + x + sin x –

cos x + log (1 + |x function g(x) on [– 1, 1] satisfying g(– x) = – f (x) is (a) x2 + x + sin x + cos x – log (1 + |x|) (b) –x2 + x + sin x + cos x – log (1 + |x|) (c) –x2 + x + sin x – cos x + log (1 + |x|) (d) none of these Ans. (b) Solution [–1, 1]. Let g(x) = –x2 + x + sin x + cos x – log (1 + | x |). (The choice of g is clear as x2, cos x log (1 + | x | ) are even functions and x, sin x are odd functions). Then g(–x) = –x2 – x – sin x + cos x – log (1 + | x |) = – (x2 + x + sin x – cos x + log (1 + | x |) = – f (x). Example 11

Let f(x) = a x (a > 0) be written as

f (x) = g(x) + h(x), where g(x) is an even function and h(x) is an odd function. Then the value of the g(x + y) + g(x – y) is (a) 2g(x) g(y) (b) 2g(x + y) g(x – y) (c) 2g(x) (d) g(x) – g(y) Ans. (a) 1 x (a + a–x) and Solution: Clearly g(x) = 2 1 h(x) = (ax – a–x) see 21.4(7). Now 2 1 1 g(x + y) + g(x – y) = (ax + y + a–(x + y)) + (ax – y + a–x + y) 2 2 1 (ax ay + ax a–y + a–x ay + a–x a–y) = 2· 4 1 =2· (ax (ay + a–y) + a–x (ay + a–y)) 4 Ê1 x -x ˆ Ê 1 y -y ˆ = 2 ÁË (a + a )˜¯ ÁË (a + a )˜¯ 2 2 = 2 g(x) g(y)

Example 12

Let f be a real valued function such

that for any real x, f(k + x) = f(k – x) and f(2k + x) = – f(2k – x) for some k > 0. Then (a) f is even and periodic (b) f is odd but not periodic (c) f is odd and periodic (d) f is even but not periodic Ans. (c) Solution: f (k + x) = f (k – x) fi f (2k – x) = f (x), (replacing x by k – x) so f (x) = – f (2k + x) ...(i) Changing x to x + 2k, we have f (x + 2k) = – f (x + 4k) So f (x) = f (x + 4k). Thus f is periodic. Again f (2k – x) = f (x) fi f (2k + x) = f (– x). Therefore, (i) fi f (x) = – f (– x). Hence f is odd and periodic. Example 13

Let a function : R Æ R

(x) = 2x + sin x for x Œ R. Then is (a) one-to-one and onto (b) one-to-one not onto (c) onto but not one-to-one (d) neither one-to-one nor onto Ans. (a) Solution: Since f¢(x) = 2 + cos x > 0 for all x Œ R, so f is one-to-one. Moreover, f (x) Æ • as x Æ • and f (x) Æ – • as x Æ – •, hence the range of f is R. Therefore, f is onto as well. Example 14

Suppose f (x) = (x + 1)2 for x ≥ – 1. If g(x)

is the function whose graph is the reflection of f (x) w.r.t. y = x, then g(x) equals 1 (a) - x - 1 , x ≥ 0 (b) ,x>–1 ( x + 1)2 (c) x + 1 , x ≥ – 1

(d)

x -1, x ≥ 0

Ans. (d) Solution: Clearly g(x) = f –1 (x). Let y = f (x) = (x + 1)2 fi y – 1 = x. Hence f –1 (x) = x – 1, x ≥ 0. Example 15

Let n(A) = 4 and n(B) = 6. Then the number

of one-one functions from A to B is (a) 120 (b) 360 (c) 24 (d) 140 Ans. (b) Solution: Let f : A Æ B be any function. Suppose that A = {x1, x2, x3, x4} and B = {y1, y2, , y6}. Now f (x1) y 6 say can be chosen to be as any one of y 1 , y 2 y6 so f (x1) = y1, f (x2) can be chosen as any one of y2, those are five choices for f (x2). This means that we can have

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6.5.4.3 = 360 such functions from A to B. (c) log Remark

Example 16 f (x) =

Ê x - 1ˆ 1 ¥ log0. 4 Á is Ë x + 5 ˜¯ x 2 - 36

(a) (– •, 0) ~ {– 6}

(b) (0, •) ~ {1, 6}

(c) (1, •) ~ {6}

(d) [1, •) ~ {6}.

Ê1+ xˆ (d) tan–1 Á Ë 1 - x ˜¯

x π – 6, 6 and

For f

Ê x - 1ˆ x -1 > 0. Since loga x for 0 < a ÁË x + 5 ˜¯ ≥ 0, x+5 x -1 < 1 is a decreasing function, we have £ 1 x+5 x -1 > 0. For x > – 5, we must have x – 1 £ x + 5 and and x+5 x have x > 1. For x < –5, we have x – 1 ≥ x + 5, x – 1 < 0. These inequalities are not possible. Hence the domain of f is (1, •) ~ {6}.

log0.4

Example 17

The graph of the function

(

)

f(x) = loga x + x 2 + 1 is symmetric about (a) x-axis (c) y-axis

(b) origin (d) the line y = x

(

)

f(x) = loga - x + x 2 + 1 = loga

1

x + x2 + 1 = – loga x + x 2 + 1 = – f(x). Hence f is an odd function, therefore, the graph of f is symmetric about origin. Since a x - a- x , so graph of f, cannot by symmetric f–1(x) = 2 about the line y = x. Non-symmetry about x-axis and y-axis is clear. Ê x-y ˆ and f has domain If f(x) – f(y) = f Á Ë 1 - x y ˜¯ (–1, 1) then a function satisfying the above functional equation is Ê1- xˆ Ê1+ xˆ (b) log ÁË (a) 2 + log Á ˜ ˜ 1+ x¯ Ë1- x¯ Example 18

Ê (1 - x )(1 + y) ˆ Ê 1 - yˆ Ê1- xˆ – log = log Solution: log ÁË ÁË (1 + x )(1 - y) ˜¯ ˜ ˜ Á 1+ x¯ Ë 1 + y¯ x-yˆ Ê 1Á 1 - xy ˜ Ê 1 - xy - ( x - y) ˆ = log Á = log Á ˜ Ë 1 - xy + ( x - y) ˜¯ Á1+ x - y ˜ ÁË 1 - xy ˜¯ Ê1- xˆ Thus f(x) = log ÁË ˜ 1+ x¯ 1+ x = – f(x) so the function in (a) 1- x cannot satisfy the given functional equation. The function x = 0 and taking x = 0 and y = 1, we can verify that the function in (d) does not satisfy the given equation.

equation. Since log

Ans. (c)

Ans. (b) Solution:

1- x

2

Ans. (b)

If n(A) = m and n(B) = m £ n. Then the number of onen! one functions from A to B is n - m ! . ( )

Solution:

2x

Example 19

Which of the following function is non

(a) f (x) = {x}, the fractional part of the number x (b) f (x) = cot (x + 7) sin 2 x cos2 x (c) f (x) = 1 – 1 + cot x 1 + tan x (d) f (x) = x + sin x Ans. (d) Solution: The function in (a) is periodic with period 1. The function in (b) is periodic with period p. The function in (c) on simplication is equal to (1/2) sin 2x, which is periodic with period p. The function in (d) is nonperiodic as g(x) = x is non-periodic. Example 20

Let

then the function g(x) = (9x2 – 1) has domain (a) [0, 2] (b) [–1/3, 1/3] (c) [–3, 3] 1 1 1 1 ˘ ,- ˘»È , (d) ÈÍ Î 3 3 ˙˚ ÍÎ 3 3 ˙˚ Ans. (d) Solution: g is meaningful if 0 £ 9x2 – 1 £ 2 ¤ 1 £ 9x2 £ 3 i.e.

È 1 1 ˘ x Œ [ (- •, - 1 / 3] » [1 / 3, •) ] « Í , ˙ Î 3 3˚ È 1 1 ˘ È 1 1 ˘ = Í,˙»Í , ˙. 3˚ Î 3 3˚ Î 3

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Example 21

If 2 (x2) + 3 (1/x2) = x2 – 1 for all

x ΠR ~ {0}. Then (x2) is equal to (a)

1 - x4

(c)

(

5x2

(1 - x ) (3 + 2 x ) 2

(b)

)(

1 - x2 3 - 2 x2 3x

)

(d)

2

2

5x2 1- x

2

5x2

Ans. (b) Solution:

Replacing x by 1/x in the given equation we get, 1 - x2 . Solving the given 2f (1/x2) + 3f (x2) = (1/x2) – 1 = x2 functional equation and the last equation, we have 9f(x2) – 4f (x2) =

(

3 1 - x2 x

2

)

– 2(x2 – 1)

(1 - x ) (3 + 2 x ) 2

= (1 – x2) ((3/x2) + 2) =

(1 - x ) (3 + 2 x ) 2



2

f(x ) =

2

x2

2

5x2

Example 22

If [x] and {x} represent integral and

fractional part of x then the function defined by f (x) = [x] + 1000

{x + r}

r =1

1000

Â

of f(x) is (a) [–1, 0] (b) [0, 1/2] (c) [–1/2, 0] (d) [0,1] Ans. (b) Solution: Let g(x) = |sin x| + |cos x x and is periodic with period p/2. For xŒ[0, p/2] g(x) = sin x + cos x = 2 sin (x + p/4). Since g(0) = 1 and g(p/4) = 2 and g increases on [0, p/4] and decreases on [p/4, p/2], so the range of g is [1, 2 ]. Hence the range of f is [0, 1/2]. Suppose f(x) = (x + 1)2 for x ≥ – 1. If h(x)

Example 25

is the function whose graph is the reflection of the graph of f(x) with respect to the line y = x and g(x) is the function whose graph is obtained by shifting 3 units to the left the graph of h(x), then g(x) equals. 1 ,x>–1 (a) x - 3 – 1, x ≥ 3 (b) ( x + 4 )2 x-2 , x > 2

(c)

(d)

x + 3 – 1, x > – 3

Solution h(x) = f–1(x) But y = f(x) = (x + 1)2 (b) 4x (d) 4[x] + 1000{x}

Ans. (c) Solution: If x ΠI then [x] = x and {x + r} = 0 for any r ΠI. Thus f (x) = x. If x ΠR ~ I then [x] = integral part of x and {x + r} = {x} for any r = 1, 2, 1000. Thus f (x) = [x] + {x} = x. Example 23

Let f (x) = log2(|sin x| + |cos x|). The range

Example 24

Ans. (d)

is equal to

(a) 2[x] + {x} (c) x

0 < log g(x) £ 1 fi 0 < sin log g(x) £ sin 1 (sin x increases on [0, 1]. Thus cos (sin 1) £ cos (sin log g(x)) < 1 (cos x decrease on [0, sin 1]). Similarly sin (cos 1) £ sin cos (log g(x)) < sin 1. Thus the range of f (x) is [cos (sin 1) + sin cos 1, 1 + sin 1).

The range of

Ê Ê Ê Ê Ê x 2 + e2 ˆ ˆ ˆ x 2 + e2 ˆ ˆ sin cos log + f(x) = cos Á sin Á log 2 Á ˜ Á ˜ Á 2 ˜˜˜ x +1 ¯ ¯ Ë x +1 ¯ ¯ ¯ Ë Ë Ë Ë is (b) [–2, 2] (a) [– 2 , 2 ] (c) [cos (sin1) + sin cos1, 1 + sin1] (d) [–1, 1] Ans. (c) x 2 + e2 is continuous Solution: The function g(x) = x2 +1 with range (1, e]. Since log x increases for all x > 0 so

fix=

y – 1 so f–1(x) =

x – 1. g(x) = h(x + 3) =

x + 3 – 1.

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS Example 26

If f (x) = sin x + cos x + 4,

g(x) = log4 log5 log3 (18x – x2 – 77), then (a) dom g = (18, 10) (b) range f = [– 2, 2] (c) dom f =

n ŒI

p 4p ˘ È Í np + 3 , np + 3 ˙ Î ˚

(d) range f = [2, 6] Ans. (a) and (d) Solution: g is defined if log5 (log3 (18x – x2 – 77) > 0 and 18x –x2 – 77 > 0 fi log3 (18x – x2 – 77) > 5° = 1, x2 – 18x + 77 < 0

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fi fi fi fi

(a) h(x) = cos x4 = f(x4) = f(g(g)x)) = fogog(x) (b) h(x) = cos (cos2x) = f(cos2x) = f (g(f(x)) = fogof(x) (c) h(x) = cos2 (x4+x2) = gof (x4+x2) = gofo (gog+g) (d) h(x)=cos2 x2 = gof (x2) = gofog(x)

18x – x2 – 77 > 31 = 3, (x – 11) (x – 7) < 0 x2 – 18x + 80 < 0 and 7 < x < 11 (x – 8) (x – 10) < 0 and x Œ (7, 11) x Œ (8, 10) « (7, 11) = (8, 10)

Since domain of sin x and cos x is R so dom f = R. Also Ê 3 ˆ 1 f(x) = 2 Á 2 sin x + 2 cos x˜ + 4 = 2 sin (x + p/6) + 4. Since Ë ¯ range of sin x is [– 1, 1] so the range of f = [2, 6]. Example 27

Which of the following functions have

inverse defined on their ranges? (a) f (x) = x2, x Œ R (b) f (x) = x3, x Œ R (c) f (x) = sin x, 0 < x < 2p (d) f (x) = ex, x Œ R Ans. (b) and (d) Solution: The function in (a) is not one-one, so f –1 is not R Æ R given by from R Æ R, hence f –1 f –1(x) = x1/3. The function f (x) = sin x is not one-one in (0, 2p) as f (p/4) = f (3p/4) = 1/ 2 , so it is not invertible. The function f (x) = ex is one-one as ex = ey ¤ x = y. The range f = R+ = {x: x > 0}. So inverse of f viz log x on R+ to R. Example 28 (a)

p is the period of the function

(1 + sin x ) cos x (1 + cosec x )

(b) |sin x| + |cos x| (c) sin 2x + cos 3x (d) cos (sin 2x) + cos (cos 2x) Ans. (a) and (d) Solution: sin x = tan x, so it has period p. The function in form as cos x (b) has perivd p/2. Since the period of sin 2x is p and the period of cos 3x is 2p/3, the period of sin 2x + cos 3x is L.C.M. ( p, 2p/3) = 2p. The period of the function in (d) is clearly p. Example 29 Let f(x) = cos x and g (x) = x2 (a) if h(x) = cos x4 then h = (b) if h(x) = cos (cos2x) then h = (c) if h(x)=cos2 (x4+x2) then h = (gog + g) (d) if h(x)=cos2 x2 then h = Ans. (a), (b), (c), (d) Solution:

Example 30

Which of the following defines a one – one

function? (a) f (x) = ex (– • < x < •) 2 (b) f (x) = ex (– • < x < •) (c) f (x) = cos x (0 £ x £ p) (d) f (x) = ax + b (– • < x < •) (a π 0) Ans. (a), (c) and (d) Solution: If f (x) = ex then f (x1) = f (x2) fi ex1 = ex2 fi ex1 – x2 = 1 fi x1 – x2 = 0 fi x1 = x2. If f (x) = cos x then f (x1) = f(x2) fi cos x1 – cos x2 = 0 fi sin sin

x1 - x2 x1 + x2 x1 - x2 sin = 0 fi sin = 0 or 2 2 2

x1 + x2 = 0 fi x1 = x2 (as x1, x2 Œ [0, p] so x1 π – x2). 2

If f (x) = ax + b then f (x1) = f (x2) fi ax1 + b = ax2 + b fi a(x1 – x2) = 0 fi x1 = x2. So in any of the above case f is 2 1 – 1. If f (x) = ex then f (1) = f (– 1). Hence f is not 1 – 1. Example 31

The range of the function f(x) = cos [x],

for – p/2 < x < p/2 contains (a) {–1, 1, 0} (b) {cos 1, 1, cos 2} (c) {cos 1, – cos 1, 1} (d) [– 1, 1] Ans. (b) Solution: For the given range of x, we have [x] = – 2 for – p/2 < x < – 1 fi f(x) = cos (– 2) = cos 2, [x] = – 1 for – 1 £ x < 0

fi f(x) = cos (– 1) = cos 1,

[x] = 0 for 0 £ x < 1

fi f(x) = cos 0 = 1,

and [x] = 1 for 1 £ x < p/2

fi f(x) = cos 1,

so that the range of f(x) is {cos 1, 1, cos 2}. Example 32

Let y = f(x – 1) + f(x + 1) where f(x)

Ï1 – | x |, | x | £ 1 = Ì Ó0, | x | > 1 The number of points of intersections with y = sin a is (a) 0 (p < a < 2p) (b) 4 (0 < a < p) (c) infinitely many (a = p)

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(d) 2 (0 < a < 2p)

Ans. (b), (c), (d)

Ans. (a), (b), (c) Solution:

Ï0, | x | > 1 Ô f (x – 1) = Ì1 - x, 0 £ x £ 1 Ô1 + x, - 1 £ x < 0 Ó Ï0, x < 0 Ô x, 0 £ x £ 1 Ô f (x + 1) = Ì Ô 2 - x, 1 £ x £ 2 ÔÓ0, x > 2

Ï0, x < - 2 Ô x + 2, - 2 £ x £ - 1 Ô Ô - x, - 1 £ x £ 0 So f (x – 1) + f(x + 1) = Ì Ô x, 0 £ x £ 1 Ô 2 - x, 1 £ x £ 2 Ô Ó0, x > 2

Solution: dom f = R and dom g = {x : x > 1}. Since x2 ≥ 0 for any x, so range f = [4, •) and the range g = (0, •). Hence dom (f + g) = R « (1, •) = (1, •), range f « range g = [4, •) and range f » range g = (0, •) 2x , then Example 35 Let f(x) = 1 + x and g(x) = 2 x +1 (a) dom (f + g) = (– 1, •) (b) dom (f + g) = [0, •) (c) range f « range g = [1] (d) range f » range g – [– 1, •) Ans. (b), (c) , (d) Solution: dom f = [0, •) and dom g = R. Also range f = {x : x ≥ 1} and range g = {x : | x | £ 1} = [– 1, 1]. Hence dom (f + g) = [0, •) « R = [0, •), range f « range g = [1, •) « [– 1, 1] = {1} and range f » range g = [– 1, •).

g(x) = y = sin p

Fig. 21.22

Example 33

Which of the following functions are

periodic (a) f(x) = {x}, the fractional part of the number x (b) f(x) = |cos x| sin 2 x cos2 x (c) f(x) = 1 – 1 + cot x 1 + tan x (d) f(x) = x + sin x Ans. (a), (b), (c) Solution: The function in (a) is periodic with period 1. The function in (b) is periodic with period p. The function x, hence is periodic with period p. The function in (d) is non-periodic. 1 , then Example 34 Let f(x) = x2 + 4 and g(x) = x -1 (a) dom (f + g) = (0, •) ~ (0, 1) (b) range f « range g = [4 •) (c) range g = (0, •) (d) range f » range g = (0, •)

Let f: R Æ R

Example 36

f(x) = [x] and

3 - 2x then 4 (a) f is neither one-one nor onto (b) g is one-one and onto (c) f is one-one and g is onto (d) neither f nor g is onto

Ans. (a), (b) Solution: Since f(3/4) = f(1/2) = 0 so f is not one-one. If y ΠR ~ I, there is no x ΠR such that f(x) = y. Hence f is neither one-one nor onto. g being polynomial of degree one is both one-one and onto. Let g(x) = x2 Р4x Р5 then

Example 37 (a) g (b) g (c) g (d) g Ans. (b), (d)

is is is is

one-one on R one-one on (–•, 2] not one-one on (–•, 4] one-one on (–•, 0)

Solution: If g(x1) = g(x2) then x12 – 4x1 = x22 – 4x2 fi (x1 – x2) (x1 + x2 – 4) = 0. Hence if x1 + x2 π 4 then x1 = x2. Thus g is one-one only on (– •, 2]. Ê p Êp ˆˆ sin Á sin x˜ ˜ , g(x) Example 38 Let f(x) = sin Á Ë3 ¯¯ Ë2 3 =

p p sin x and h(x) = x, then 2 2 È 1 1 ˘ , (a) Range of f is Í˙ Î 2 2˚ È 1 1˘ (b) Range of f o g is Í- , ˙ Î 2 2˚

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(c) There is x Œ R such that g o f (x) = 2 (d) The period of g o f o h is 4 Ans. (a), (d) p p p Solution: Since –1 £ sin x £ 1, so - £ sin x £ 3 3 3 È p p˘ Now sinx is an increasing function in Í- , ˙ Î 2 2˚

Example 39

Let f(x) = sin (p/x) and D+ = {x: f(x) > 0}.

so -

3 3 Ê pˆ Êp ˆ = sin Á - ˜ £ sin Á sin x˜ £ Ë ¯ Ë ¯ 2 3 3 2

Then D+ contains (a) (1/3, 1/2) (b) (1/5, 1/4) (c) (– 1, – 1/2) (d) (– p, – 1/2) Ans. (a), (b), (c) Solution: D+ = {x : sin (p/x) > 0} ÏÔ p ¸Ô ÏÔ 1 ˆ ¸Ô Ê 1 , ˜˝ Ì x : Œ (2np ,(2n + 1) p ) ˝ = Ì x : x Œ Á Ë ¯ nŒ I 2 n + 1 2 n Ô ÓÔ x nŒI ˛Ô ÓÔ ˛

fi-

3 p 3 p p ˆ Êp sin Á sin x˜ £ £ ¯ Ë 2 2 3 2 3 3 2 2 3

This shows that D+ 2 (1/3, 1/2), D+ 2 (1/5, 1/4) and D+ 2 (– 1, – 1/2).

p p Êp ˆ p sin Á sin x˜ £ fi- £ Ë ¯ 4 4 2 3 3 fi-

Ê p 1 ˆ ˆ Êp sin Á sin x˜ £ £ sin Á ˜ ¯ Ë Ë 3 2 2 3 2¯

1

Since sine function is a continuous function, so range 1 ˘ È 1 , f = Í˙ 2˚ Î 2 fog (x) = f (g (x)) Ê p Êp ˆˆ sin Á sin g( x )˜ ˜ = sin Á Ë3 ¯¯ Ë2 3 Ê p Êp Êp ˆˆˆ sin Á sin Á sin x˜ ˜ ˜ = sin Á Ë2 ¯¯¯ Ë3 Ë2 3 since

-

p p p £ sin x £ 2 2 2

Example 40

Let R = {(x, y): x, y Œ R, x2 + y2 £ 25}

R¢ = {(x, y): x, y Œ R, y ≥ (4/9)x2} then (a) dom R « R¢ = [– 3, 3] (b) Range R « R¢ 2 [0, 4] (c) Range R « R¢ = [0, 5] (d) R « R¢ defines a function Ans. (a), (b), (c) Solution: The equation x2 + y2 = 25 represents a circle with centre (0, 0) and radius 5 and the equation 4 y = x 2 represents a parabola with vertex (0, 0) and focus 9 (0, 1/9). Hence R « R¢ is the set of points indicated in the Fig. 21.23. Thus dom R « R¢ = [– 3, 3] and range R « R¢ = [0, 5] 2 [0, 4] Since (0, 0) Œ R « R¢ and (0, 5) Œ R « R¢ Hence R « R¢

Êp ˆ fi -1 £ sin Á sin x˜ £ 1 Ë2 ¯ As above, -

Ê p Êp 1 Êp ˆˆˆ £ sin Á sin Á sin Á sin x˜ ˜ ˜ £ Ë2 ¯¯¯ Ë3 Ë2 3 2 2

1

È 1 1 ˘ , i.e. range of fog is Í˙ Î 2 2˚ p p p p Since g o f (x) = sin f(x), but - £ sin f ( x ) £ < 1 so 2 2 2 2 there is no x Œ R such that g o f (x) = 1 g o f o h (x) = g o f (h (x)) Ê Ê p Êp Ê p ˆˆˆ = g Á sin Á sin Á sin Á x˜ ˜ ˜ Ë 2 ¯¯¯ Ë 3 Ë Ë2 3 So the period of g of o h is

2p =4. p /2

Fig. 21.23

MATRIX-MATCH TYPE QUESTIONS Example 41

Let f(x) = cos–1 g(x) =

1 sin x

x-3 + log5 (6 – x) and 2

+ 3 sin x

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(a) (b) (c) (d)

Column 1 f is defined on g is defined on f is continuous on g is continuous on

(p) (q) (r) (s)

Column 2 [1, 3} [1, 2] [1/2, 4] [2, 5]

Example 43

Let f(x) = a cos x + b sin x (a2 + b2 > 0)

and tan a = b/a. Then Column I (a) f increases on (b) f decreases on (c) f is positive on

Ans.

(d) f(x) ≥

a 2 + b2 2

Column II (p) [7 + a, 8 + a] (q) [–p + a, a] (r) [a, p/2 + a] on (s) [a, a + p/4]

Ans.

Solution: For f to be defined we must have x-3 –1£ £ 1 and 6 – x > 0 i.e. 1 £ x £ 5 and x < 6. So the 2 domain of f is [1, 5]. Hence f kp < x < (2k + 1)p, k Œ I For k = 0, we must have 0 < x < p [1, 3], [1, 2]. f and g are continuous function wherever they Example 42

(

)

Let f(x) = loga x + x 2 + 1 (a > 0, a

π1). Then Column 1 (a) f is defined on (b) f increases on (c) The inverse f is positive on (d) (af(x) – x)2 increases on

Column 2 (p) (0, •) (q) (– •, •) (r) [0, a]

Solution: a

f(x) =

a 2 + b2 cos(x – a) where cos a =

a 2 + b2 and sin a = b

cos x decreases on 2np £ x £ (2n + 1)p, n Œ I and increases on (2n – 1)p £ x £ 2np, n Œ I. Thus cos (x – a) decreases on 2np + a £ x £ (2n + 1)p + a and increases on (2n – 1) p + a £ x £ 2np + a, n £ I. For n = 1, [2np + a,(2n + 1) p + a] = [2p + a, 3p + a] 2 [7 + a, 8 + a]. For n = 0 f increases on [– p + a, a]. For n = 0 f decreases [a, p + a] È p ˘ 2 [a, p/2 + a] and [a, a + p/4]. f(x) > 0 on Ía , + a ˙ Î 2 ˚ 1 and f(x) ≥ a 2 + b2 2 if cos (x – a) ≥ . 2 1+ x 3x + x3 and g(x)= , Example 44 Let f(x) = log 1- x 1 + 3 x2 then Column 1 Column 2

(s) (– •, a]

Ans.

Ê 1 - eˆ (a) fog Á Ë 1 + e ˜¯ Solution:

f(x) = loga(x +

x2 + 1

Ê e - 1ˆ (b) gof Á Ë e + 1˜¯ (c) fog(0)

x,

since x 2 + 1 > |x| and is odd. It increases for all value of x and has an inverse. Solving y = loga(x + so a–y = x –

x 2 + 1 ), ay = x +

x 2 + 1 . Adding we get x =

Ê e - 1ˆ (d) fog Á Ë e + 1˜¯

x2 + 1 1 y (a – a–y). 2

1 x (a – a–x) > 0 for x > 0. (af(x) – x)2 = 2 x2 + 1 which increases on [0, •).

Thus f–1(x) =

a 2 + b2 . It is known that

Ans.

(p) 3 (q) –3 (r) 0 (s) 1

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Solution:

= log

fog(x) = f(g(x)) = log

1 + 3x 2 + 3x + x3

1 + g ( x) 1 - g ( x) 3

1+ xˆ = log ÊÁ = 3f(x) Ë 1 - x ˜¯

1 + 3x 2 - 3x - x3

1- e 1 + e = - 3, 1- e 1+ 1+ e fog (0) = 3f(0) = 0.

1 - eˆ = 3 log Thus fog ÊÁ Ë 1 + e ˜¯

Ans.

3x - 1

(s) f is an odd function

3x + 1

1-

Ê e - 1ˆ = g (log e) = g(1) = 1. and g o f Á Ë e + 1˜¯ Let f: R Æ

Example 45

(d) f(x) = x

f(x) = sin (2x + 1). If domain is restricted to Column 1 Column 2 (a) [–3p/4 –1/2, – p/2 –1/2] (p) f is one-one and onto (b) [– 3p/4 –1/2, –1/2] (q) f is one-one but not onto (c) [p/4 –1/2, 3p/4 –1/2] (r) f is onto but not one-one È 3p 1 p 1 ˘ – ,– – ˙ (s) f is neither one (d) Í – Î 4 2 2 2˚ -one nor onto È 5p 1 3p 1 ˘ »Í – , - ˙ Î 4 2 2 2˚

Solution: f(x)

f(x) = {x} is a period function with period 1,

16 x - 1

= 4x – 4–x so f(– x) = – f(x). Thus f in this case is 4x an odd function. =

If y = log4

(x +

x2 + 1

) then x +

y x 2 + 1 = 4 and

1 (4y – 4–y) fi 4–y = – x + x 2 + 1 , whence x = 2 1 x f–1(x) = (4 – 4–x). It is easy to verify that the function 2 in (d) is even. Example 47 function (a)

Ans.

Column 1 log x - 2 2

Column 2 (p) R ~ [– 3, 2]

x +3

x+2 (b) log (1 - x ) + sin–1 x 10

(q) R ~ [0, 1)

(c) 1/(x – [x]) (d) 1/[x]

(r) [–1, 1) – {0} (s) R ~ I

[x] denotes the greatest integer less than x Solution: The function f(x) = sin (2x + 1) decreases from 1 to 0 on [– 3p/4 – 1/2, – p/2 – 1/2] so f is not onto but one-one. Since f(– p/2, - 1/2) = 0 = f(– 1/2) so f is not one-one on [– 3p/4 – 1/2, –1/2] but is onto. On [p/4 – 1/2, 3p/4 – 1/2] f decreases from 1 to – 1 and is continuous. Hence f is one-one and onto as well. Example 46 Column 1

Column 2

(a) f(x) = {x}, the fractional part of x (b) f(x) = (c) f(x) =

(

16 x - 1 4x

log4 x + x 2 + 1

Ans.

)

(p) f

–1

(x) = (4x – 4–x)

(q) f is an even function (r) f is a periodic function

Solution: f in (a which is true

x-2 x-2 > 0 and π1 x+3 x+3

if x Œ (2, •) or x Œ (– •, – 3). Hence the domain of the function in (a) is R if x ≥ – 2, 1 – x > 0 and 1 – x π 0 and x Œ [– 1, 1]. Hence the domain in this case is [–1, 1) – {0}. x – [x] = 0 if and only x Œ I and [x] = 0 if and only if x Œ[0, 1).

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Example 48

Example 50

have domain (– p/2, p/2) and range (– •, •) Column 1 Column 2 (a) 1 + 2x (p) onto but not one-one (b) tan x (q) one-one but not onto (r) one-one and onto (c) x2 2 (s) neither one-one nor onto (d) f(x) = tanx f(p/2) = 0 Ans.

Column 1 ( x + 3)

(a) f(x) =

Column 2 2

(p) 0 £ f(x) £ 3

,

x2 + 1

–• 0 so 3 £ y y

(d) The domain of the given function is [Рp/4, p/4]. For x Π[Рp/4, p/4],

(p 2 16) - x 2 Π[0, p/4] and sine

function increases on [0, p/4] so 0 £ f(x) £ 3. Example 51 Column 1 (a) The period of sin px + tan

px px + sin 2 +.... + 2 2

Column 2 (p) 22n – 1

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sin

px 2

+ tan

n -1

px 2

n

(b) g(x) = 3 + 4x, the value of

(q) 22n

n

=

g (0) = g o g ....... o g(0) is (c) f(x) = x3 + 2n x2 + bx + c is bijection if and only if 3b ≥ d where d is equal to

(r) 2n

(d) (2

– 1)

Â

r =0

r

(– 1) ncr

n

(s) 2 + 1

1 2

n

+

1 4

n

+

1 8

n

+ ..... =

1 2n 1-1 2

n

=

1 2 -1 n

Example 52

Let f(x) = sin–1

x-3 – log10 (4 – x) 2

Statement-1: The domain of f(x) is [1, 3]

Ê 1 ˆ 3 7 Á r + 2r + 3 r + ....... upto infinity˜ 2 2 Ë2 ¯ r

n

ASSERTION-REASON TYPE QUESTIONS

n

2n

n

3ˆ Ê Ê 7ˆ Ê 1ˆ ÁË 1 - ˜¯ + ÁË 1 - n ˜¯ + ÁË 1 - ˜¯ + ...... 2 8

is

n

r

Ans. (d) Ans.

x| £ 1 and log10 x is

Statement-2: sin–1 x x > 0. Solution: –1£

For sin–1

x-3 2

x-3 £ 1 ¤ 1 £ x £ 4. But log10(4 – x 2

4 – x > 0 ¤ x < 4. Hence the domain of f(x) = [1, 4] « (– •, 4) = [1, 4). Solution: (a) Since the period of sin ax (a > 0) is 2p/a and the period of tan ax (a > 0) is p/a, so the periods of px px sin px, sin 2 , ... sin n -1 are 2, 22 · 2, ... 2n–1 · 2 re 2 2 spectively. The periods of tan px/2, tan px/22, ... tan px/2n are 2, 22, ... 2n. Hence the period of the given function is lcm (2, 22, ... 2n) = 2n. (b) g2(x) = g(g(x)) = g(3 + 4x) = 3 + 4 (3 + 4x) = (42 – 1) + 42x suppose that gk(x) = (4k – 1) + 4k x. Then gk+1 (x) = gk (g(x)) = gk(3 + 4x) = (4k – 1) + 4k (3 + 4x). = 4k(1 + 3) – 1 + 4k+1x = (4k + 1 – 1) + 4k+1 x Hence by induction, gn(x) = (4n – 1) + 4n x for all n Œ N. Thus gn(0) = 4n – 1 = 22n – 1 (c) Note that f(x) Æ – • as x Æ – • and f (x) Æ • as x Æ •, therefore f is onto. Also, f ¢(x) = 3x2 + 2n+1 x + b ≥ 0 for all x Œ R if and only if the discriminant of f ¢(x) = 0 viz. 4 (22n – 3b) £ 0. That is, y = f(x) is non-decreasing (and hence one-to-one) on R if and only if 22n – 3b £ 0. (d) Given series can be written as

Example 53

Statement-1: The range of the function

f(x) = sin x + a sin x + b where |a | > 2 will be real numbers 2

between b –

a2 and b + a + 1. 4

Statement-2: The function g(t) = t2 + at + 1 where t Œ [– 1, 1] and |a| > 2 will attain minimum and maximum values at – 1 and 1. Ans. (d) Solution: f(x) = (sin x + (a/2))2 + b – a2/4 If |a| > 2 so |a/2| > 1 but |sin x| £ 1 so max f(x) = max (sin2 x + a sin x + b) = 1 + a + b, sin x = 1 min f(x) = min(sin2 x + a sin x + b = 1 – a + b, sin x = – 1 range f(x) = [1 – a + b, 1 + a + b] Clearly the statement 2 is true. Example 54

Let f(x) =

1 Ê |sin x | sin x ˆ + 3 ÁË cos x |cos x | ˜¯

Statement-1: The period of f(x) is 2p Statement-2: The period of sin x and cos x is 2p. Ans. (b) Solution: On the interval [0, 2p], the function can be represented, as follows

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Ïtan x Ô0 f(x) = ÔÌ Ô- tan x ÔÓ0

Statement-2: The domain of inverse of y is R.

x Œ[0, p / 2) x Œ(p / 2, p ] x Œ[p , 3p / 2) x Œ[3p / 2, 2p ]

Ans. (b) Solution: The given equation makes sense if x > 1. Moreover y =

Since the period of tan x is p so f is periodic with period 2p. Example 55

Statement-1: Let f(x) be a function such

that f(x Р1) + f(x + 1) = 2 f(x) for all x ΠR then f is periodic. Statement-2: There exists a function with period 8. Ans. (b) Solution:

Replacing x – 1 by x, we get

f(x) + f(x + 2) = 2 f(x + 1) Again replacing x by x + 2 f(x + 2) + (x + 4) =

Adding (1) and (2), we get f(x) + 2f(x + 2) + f(x + 4) =

(2)

2 {f(x + 1) + f(x + 3)}

= 2 f(x + 2) = 2f(x + 2) fi f(x) + f(x + 4) = 0 (3) fi f(x + 4) + f(x + 8) = 0 (4) Substracting (4) and (3) we obtain f(x) = f(x + 8) Thus f is periodic. We know that the period of the function kx – [kx] is 1/k. 1 È1 ˘ A function with period 8 is x – Í x ˙ . 8 Î8 ˚ Example 56 function

Statement-1: The linear fractional

a ax + b (ad – bc π 0) cannot attain the value . c cx + d

Statement-2: The domain of the function b - dy is R ~ {a/c}. g(y) = cy - a Ans. (a) b - dy ax + b Solution: If f(x) = y = then x = cy - a cx + d Thus f–1(y) = g(y) =

b - dy and dom g = R ~ {a/c}. cy - a

Hence f Example 57 2

A function y of x is represented by

equation y – 1 + log2(x – 1) = 0 Statement-1: The domain of y is [1, 3]

(x – 1) ¤ 2 ≥ x – 1 ¤ x £ 3. Hence the domain of y is [1, 3]. Also 1 – y2 = log2(x – 1) 2 fi x = 1 + 21–y . 2 y is 1 + 21–y whose domain is R.

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 58 to 78

(1)

2 f(x + 3)

1 - log2 ( x - 1) makes sense if 1 ≥ log2

A tower has the following shape: a truncated right circular cone (one with radii 2R (the lower base) and R (the upper base), and the height R bears a right circular cylinder whose radius is R, the height being 2R. Finally a semisphere of radius R is mounted on the cylinder. Suppose that the cross-sectional area S of the tower is given by f(x), where x is the distance of the cross-section from the lower base of the cone. Example 58

The domain of the function f(x) is

(a) (– •, •) (c) [0, R]

(b) [0, 4R] (d) [R, 4R]

Example 59 For 0 £ x £ R, the function f(x) is given by (a) p (2R – x)2 (b) p (R – x)2 (c) pR2 (d) 4px2 Example 60

The range of f(x) is

(a) [0, 4pR2] (c) [0, pR2]

(b) [pR2, 4pR2] (d) [pR2, 3pR2]

Example 61 The function f(x) is (a) one-one on [R, 2R] (b) one-one on [R, 3R] (c) one-one on [0, 4R] (d) one-one on [0, R] » [3R, 4R] Ans. 58. (b), 59. (a), 60. (a), 61. (d) Solution 58 to 61 The function f(x) is given by Ïp (2 R - x )2 , 0£ x£R ÔÔ 2 f(x) = Ìp R , R £ x £ 3R Ô 2 2 ÔÓp (6 R x - x - 8 R ) , 3 R £ x £ 4 R f R]. f decreases on [0, R] so f takes values in [pR2, 4pR2] for x Œ [0, R]. On [3R, 4R], f

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decreases so f takes values in [0, pR2]. Hence the range of f is [0, 4pR2]. Since f is strictly decreasing function on [0, R] » [3R, 4R] so it one-one on this interval. Moreover, the function is constant on [R, 3R], thus it cannot be oneone on [R, 3R] or a set containing [R, 3R].

Then F is equal to

Paragraph for Question Nos. 62 to 65

Then G is equal to

A cylinder is inscribed in a sphere of radius R. The volume V of the cylinder is written as V = f(x), where x is the height of the cylinder. f is

Example 62 (a) (0, R)

(b) (0, R2)

(c) (0, 2R)

(d) (R, •)

(c) 2

(d) f 4

Let G be a function such that G o f3 = f6.

(a) f 5 Example 68

(c) f 3

(b) f 4

(c) f 3

(d) f 2

Let H be a function such that f4 o H = f5.

Then H is equal to (b) f 4

(c) f 5

(d) f 6

Let J be a function such that f3 o J o f2 =

f4. Then J is equal to (b) f 5 (a) f 6

(c) f 4

(d) f 3

Ans. 66. (d), 67. (a), 68. (b), 69. (c) Solution: 66 to 69 66. f1 o F = f4 fi F = f1–1 o f4 = f1 o f4. Thus F(x) = f1(1/ (1 – x)) = 1/(1 – x) = f4(x). 67. G o f3 = f6 fi G = f6 o f3–1 = f6 fi f3. So G(x) = f6(1 – x) = (1 – x)/(1 – x) – 1 = x – 1/x = f5(x).

p x13 . 8

(d) 3

Ans. 62. (c), 63. (d), 64. (b), 65. (b) Solutions: 62 to 65 Ê 2 x2 ˆ V = f(x) = px Á R - ˜ 4¯ Ë

Example 67

(b) f 2

It is easy to see that f1–1 = f1, f2–1 = f2, f3–1 = f3, f4–1 = f5.

(d) equal to pR2x

Then number of such x1 is (a) 0 (b) 1

(a) f 1

Example 69

f ( x) p 2 + x is Example 64 The function g(x) = x 4 (a) one-one (b) a constant function p (c) equal to R 2 for all x 4

If x1 in the dom f is such that f(x1) =

Let F be a function such that f1 o F = f4.

(a) f 2

Example 63 The function V/x represents (a) a straight line (b) an increasing function (c) a circle (d) a decreasing function

Example 65

Example 66

for x Π(0, 2R). The

R). V/x = p (R2 – x /4) which is clearly a decreasing function. g(x) = f ( x) p 2 + x = pR2 which is constant. f(x1) = px13/8 for x1 x 4 2

68. f4 o H = f5 fi H = f4–1 o f5 = f5 o f5. Therefore, ( x - 1) x - 1 1 = = f4(x). H(x) = f5(f5(x)) = f5((x – 1)/x) = ( x - 1) x 1- x For Q 69, J = f3–1 o f4 o f2–1 = f3 o f4 o f2. Thus J(x) = f3 o f4 (1/x) = f3 (1/(1 – 1/x)) = f3 (x/x – 1) = 1 – x/x – 1 = 1/1 – x = f4(x). Paragraph for Question Nos. 70 to 73 The function f(x) = mx f(x + y) = f(x) + f(y) and f(x) f(x + y) = f(x) f(y). = ax From the given functional relations, we can determine several things about the functions. At times the function can be determined uniquely from the functional equation. Example 70 f(4) is equal to

= 2 2 3 R, otherwise values satisfying f(x1) = px13/8 don’t lie in (0, 2R). Paragraph for Question Nos. 66 to 69

(a) f(1) Example 71

f4(x) = 1/(1 – x), f5(x) = (x – 1)/x, f6(x) = x/(x – 1) This family of functions is closed under composition that is, the composition of any two of these functions is again one of these.

(b) 4f(1)

(c) 2f(1)

(d) 0

If f(x + y) = f(x) + f(y) for all x, y and

f(1) = 1 then f(– 9/8) is equal to

For x π f1(x) = x, f2(x) = 1/x, f3(x) = 1 – x,

If f(x + y) = f(x) + f(y) for all x, y then

(a) 9/8 Example 72 then

(b) 8/9

(

(c) – 9/8

(d) 1

2 2 If f(x) + f(y) = f x 1 - y + y 1 - x

(a) f(4x2 + 3x) + 3f(x) = 0 (b) f(3x – 4x3) + 3f(x) = 0

)

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(c) f(4x3 + 3x) – 3f(x) = 0

Paragraph for Question Nos. 74 to 77

3

(d) f(4x – 3x) + 3f(x) = 0 Example 73 The function f(x) = log ((1 + x)/(1 – x)) satisfy the equation.

f(x) with domain [0, 2] and range [0, 1]

(a) f(x + 2) – 2f(x + 1) + f(x) = 0 (b) f(x) + f(x + 1) = f(x(x + 1)) (c) f(x + y) = f(x) f(y) Ê x+yˆ (d) f(x) + f(y) = f Á Ë 1 + xy ˜¯

Fig. 21.24

Ans. 70. (b), 71. (c), 72. (d), 73. (d) Solution: 70 to 73 Putting x = 1 = y in f(x + y) = f(x) + f(y), we obtain f(2) = 2f(1). Assuming that f(k) = kf(1). f(k + 1) = f(k) + f(1) = kf(1) + f(1) = (k + 1) f(1). Thus by induction, f(n) = nf(1) for all n Œ N. In particular, f(4) = 4f(1). If n = – m, m Œ N then – f(m) = (– m) f(1) but f(n + (– n)) = f(n) + f(–n) and f(0) = 0 so f(– n) = – f(n). Thus f(n) = nf(1) for all n Œ I. If x = p/q, p, q Œ I, q > 0 then f(p) = f(qx) but f(p) = p f(1) and f(qx) =

f ( x +...... + x)  q times

= qf(x) so f(x) =

p f(1). q

Thus f(4) = 4f(1) and f(– 9/8) = – 9/8 f(1) = – 9/8. 72. putting x = y = 0, we obtain f(0) = 0. Putting y = x, we get

(

)

2 2f(x) = f 2 x 1 - x . Also replacing y by 2 x 1 - x 2 , we

(

2 get f(x) + f 2 x 1 - x

) = f(3x – 4x ) = – f(4x 3

3

Fig. 21.25

Example 74

Fig. 21.26

Fig 21.25 represents the graph of the

function (a) – f(x)

(b) – f(x – 1) + 1

(c) – f(x + 1) – 1

(d) – f(x + 1) + 1

Example 75

[1, 3] and [0, 1] are the domain and range

(respectively) of the function (a) – f(x)

(b) f(x – 1)

(c) – f(x + 1) + 1

(d) – f(x + 1)

– 3x) Example 76

Fig. 21.26 represents the graph of function

(putting y = – x; we can see that f is an odd function)

(a) 2f(x)

(b) f(x – 2)

fi 3f(x) = – f(4x3 – 3x) fi f(4x3 – 3x) + 3f(x) = 0

(c) f(x + 2)

(d) f(x – 2) + 1

73. The equations in (a) and (b) are not meaningful for the function f(x) as the domain of f(x) is (Р1, 1). For x Π(Р1, 1), f(x + 1) or f(x equation in (c) is also not meaningful e.g. if x1 = x2 = 1/2. L.H.S. is meaningful but R.H.S. = f f(x1) + f(x2) = log

(1 + x1 )(1 + x2 ) 1 + x1 1 + x2 + log = log (1 - x1 )(1 - x2 ) 1 - x1 1 - x2

= log

1 + x1 + x2 + x1 x2 1 - x1 - x2 + x1 x2

Ê x + x2 ˆ = fÁ 1 Ë 1 + x1 x2 ˜¯

x + x2 1+ 1 1 + x1 x2 = log x + x2 1- 1 1 + x1 x2

Example 77 The domain and range (respectively) of (a) (– x) are [– 2, 0] and [– 1, 0] (b) (x) – 1 are [0, 2] and [0, 1] (c) (x) + 2 are [0, 2] and [1, 2] (d) – (x + 1) + 1 are [– 1, 1] and [0, 1] Ans. 74. (d), 75. (d), 76. (c), 77. (d) Solution 69 to 72 To shift the graph of a function y = f(x) straight up, we add a positive constant. To shift the graph of a function y = f(x) straight down, we add a negative constant to the right hand side of y = f(x). To shift the graph of y = f(x) to the left, we add a positive constant to x and to shift to the right, we add a negative constant to x. 74. since the Fig. 21.25 is inversion then shifting down and then shifting to the left so it –f(x + 1) + 1. – f(x + 1) + 1 are [– 1, 1] and [0, 1].

IIT JEE eBooks: www.crackjee.xyz Functions 21.23

Paragraph for Question Nos. 78 to 81 The builders of the Trans-Alaska pipeline used insulted pads to keep the heat from the pipeline from melting the permanent frozen soil beneath. To design the pad, it was necessary to take into account the variation in air Ê 2p ˆ ( x - 101)˜ + 25 where f is temperature f(x) = 37 sin Á Ë 365 ¯

) (x) = g(x) = log ( x + (



2 x = log y + y + 1 (Since ex > 0)

\

f –1

= loge501 = 501.

The highest temperature shown was

(a) 37° F

(b) 25° F

(c) 62° F

(d) 12° F

Example 79

Example 80

(b) – 12° F (d) 25° F

The maximum temperature will be in the

month (a) May (c) July Example 81

(b) June (d) Aug. The period of f is

(a) 2p (c) 125

(b) 365 (d) p

Ans. 78. (c), 79. (b), 80. (c), 81. (b) Solutions: 73 to 76 Since the maximum value of sine function is 1 and minimum value is – 1 so the maximum temperature was 62° F and minimum was – 12° F. Moreover, sine function is maximum at x = p/2 so maximum temperature was after 192 days i.e. in the month of July. The period of f is 2p/(2p/365) = 365.

INTEGER-ANSWER TYPE QUESTIONS Example 82

Let f(x) =

e x - e- x and if f(g(x)) = x then 2

1 Ê e1002 - 1ˆ g is equal to 501 ÁË 2 e 501 ˜¯ Ans. 1

e x - e- x fi e2x – 1 = 2yex 2 Therefore, t2 – 2yt – 1 = 0, t = ex Solution:

Let y =



2 y ± 4 y2 + 4 = y ± y2 + 1 2

t=

Example 83

The lowest temperature shown was

(a) – 2° F (c) – 18° F

)

Ê e1002 - 1ˆ Ê e1002 - 1 e1002 + 1ˆ g Á = log + Á ˜ ˜ 2 e501 ¯ Ë 2 e501 ¯ Ë 2 e501

in Fahrenheit and x is the number of the day counting from the begining of the year. Example 78

x2 + 1

The value of

sin3 x

0

cos ec 2 x

Let f(x) = sec x x cot 2

4

tan 2 x

sin 2 x

3

.

2 (Period of f) is p

Ans. 4 Solution:

f(x) = 4 (3 sin3 x – cot

x cosec 2x) 2

– sin 2x (sin3 x tan2 x – sec x cosec 2x) = 12 sin x – 4 cot x/2 cosec 2x – sin 2x sin3 x tan2 x + sec x The period of each term is 2p. Hence period of f(x) is 2p. 2 (Period of f) = 4. Thus p 3

Example 84

If f (x) is a polynomial function satisfying

f(x) f (y) = f (x) + f (y) + f (xy) – 2 for all real x and y and f(3) = 10, then f (4) – 8 is equal to Ans. 9 Solution: Putting x = y = 1 in the given equation, we have (f(1))2 = 3f (1) – 2 fi (f(1) – 2) (f(1) – 1) = 0 fi f(1) = 2 or f(1) = 1. If f(1) = 1 then putting y = 1 in the given equation, we have f (x) = 2f (x) + 1 – 2 fi f (x) = 1 for all x which is not true as f (3) = 10. Thus f (1) = 2. Putting y = 1/x in the given equation we get f (x) f (1/x) = f (x) + f (1/x) + f (1) – 2 = f (x) + f (1/x). Using Example 9, we have f (x) = xn + 1 or f (x) = – xn + 1. If f (x) = – xn + 1 then 10 = f (3) = – 3n + 1 fi 3n = – 9 which is not possible. Thus f (x) = xn + 1 so 10 = f (3) = 3n + 1 fi 3n = 9 fi n = 2. Therefore, f (x) = x2 + 1 fi f(4) = 42 + 1 = 17. 4x . Then the value of Example 85 Let f (x) = x 4 +2 1 È Ê 1 ˆ Ê 2 ˆ fÁ + fÁ + ˜ Í Ë 1997 ˜¯ 499 Î Ë 1997 ¯ Ans: 2

Ê 1996 ˆ ˘ + fÁ Ë 1997 ˜¯ ˙˚ is given by

IIT JEE eBooks: www.crackjee.xyz 21.24 Comprehensive Mathematics—JEE Advanced

Solution:

f (1 – x) =

41 - x 1- x

+2

4

4x

Thus f (x) + f (1 – x) =

4x + 2

Ê 1 ˆ Ê 2 ˆ + fÁ + Now f Á Ë 1997 ˜¯ Ë 1997 ˜¯

= +

4 4 + 2(4 ) x

2

.=

4x + 2

=

2 2+4

4x + 2 4x + 2

x

f (1) + f (1) +  + f (1) f (1 +  + 1)   =   =  Solution: f (n) =  n -times

.

n -times

4n, for every natural number n. =1

n

Â

Ê 1996 ˆ + fÁ Ë 1997 ˜¯

k =1

Ê n ˆ 998 Ê n ˆ 1996 Ê n ˆ f  ÁË 1997 ˜¯ =  f ÁË 1997 ˜¯ +  f ÁË 1997 ˜¯ n =1 n =1 n = 999

n=0

Find

= 998

n =1

60

Example 86

If a = e

then find the value of

1 11

2pi/11

and f(x) = 5 +

Â

k =1

k

Solution:

Ak x ,

10

Â

P n (x) = (1 – 2hx + h 2 ) –1/2 , |x| £ 1, |h| < 1/3.

1 P (10). 355 3 Ans: 7

998

Â1

Let Pn(x) be a function satisfying



 hn

Ê n ˆ 998 Ê 1997 - n ˆ =  f ÁË ˜ ˜+ÂfÁ 1997 ¯ n = 1 Ë 1997 ¯ n =1 Ê Ê n ˆ n ˆˆ Ê + f Á1 = ÂÁfÁ = ˜ Ë 1997 ˜¯ ˜¯ Ë 1997 ¯ n =1 Ë

4 an + 2n (n + 1) = 2n (32 + n)

Example 88

998

998

4(a + k) = 2n (32 + n) fi

fi a = 16

1996

=

The given relation can be written as

=1+

f(a r x).

(1 – 2hx + h2)–1/2 = (1 – h(2x – h))–1/2 h (- 1/ 2) (-3 / 2) h2 (2x – h)2 (2x – h) + 2 2

r=0

+

Ans: 5 Solution: Clearly a is solution of the equation x11 = 1 so that 1 + a + + a10 = 0. 10

Â

f (ar x) = f (x) + f (a x) +

+ f (a10 x)

r=0 60 60 Ê ˆ Ê ˆ k k k = Á 5 + Â Ak x ˜ + Á 5 + Â Ak a x ˜ + Ë ¯ Ë ¯ k =1 k =1

1 1 (3x2 – 1) h2 + (5x3 – 3x) h3 + 2 2 P3(x) = Coefficient of h3 in (1 – 2hx + h2)–1/2 1 = (5x3 – 3x). 2 = 1 + xh +

P3 (10) =

1 1 (5000 – 30) = 2485 fi P (10) = 7 2 355 3

60 Ê ˆ 10 k + Á 5 + Â Ak a x ˜ Ë ¯ k =1 60

= 55 +

Â

k =1

Ak xk [1 + ak +

60

= 55 +

Â

k =1

Example 87

Ak xk

( )

1- ak

1- ak

11

= 55 + 0 = 55 1 a for which 4

n

Â

EXERCISE LEVEL 1

+ a10k]

Find the natural number

f (a + k) = 2n (32 – n) where the function f satisfies

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. The domain of y = x/ x 2 - 3 x + 2 is (a) R ~ {1, 2} (c) (– •, 1) » (2, •)

k =1

the relation f (x + y) = f (x) + f (y) for all natural numbers x, y and further f (1) = 4. Ans. 4

1.3.5 3 h (2x – h)3 + 2.4.6

2. The domain of y = (a) [0, 5] (c) [–1, 2]

(b) (– •, 2) (d) (1, •).

Ê 5 x - x2 ˆ log10 Á is 4 ˜¯ Ë (b) [1, 4] (d) [2, 5]

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3. The domain of y = log10 (a) [4, 6] (c) (2, 3) 4. The domain of y = sin–1 (a) (– •, 4) (c) (– •, 3)

(

)

x - 4 + 6 - x is (b) (– •, 6) (d) [4, •)

x-3 – log10 (4 – x) is 2 (b) [1, 4) (d) [1, •)

5. If f (x) = x3 – x and j (x) = sin 2x, then (a) j ( f (2)) = sin 2 (b) j ( f (1) = 1 (c) f ( j (p/12)) = – 3/8

(d) f ( f (1))= 2

6. Which of the following functions is an odd function? 2 (a) f (x) = cos x (b) y = 2 – x 4

(c) y = 2x – x

(d) x2n cot2n + 1x

7. The inverse of the function y = (a) log10 (2 – x) 1 log10 (2x – 1) 2

(c)

10 x - 10 - x 10 x + 10 - x

(b)

1+ x 1 log10 1- x 2

(d)

1 2x log 4 2-x

(

a

(b) 6 p (d) p

)

( ) (d) ( a ,• ) a3 2

10. The domain of the function y =

log

1 sin x

is

(a) R ~ {–p, p} (b) R ~ {2np, n Œ I} (c) R ~ {n p, n Œ I} (d) (– •, •) 11. The function f: R Æ R given by f(x) = 3 – 2 sin x is (a) one-one (b) onto (c) bijective (d) neither one-one nor onto 12. Which of the following functions are one-one and (a) (b) (c) (d)

f: R Æ R, f(x) = sin x f: R+ Æ R, f(x) = 2 x + 1 f: [0, p] Æ [– 1, 1], f(x) = cos x f :N Æ N, f(x) = 2x

R Æ R, f(x) = 4x + 5 R Æ R+ » {0}, f(x) = 2x2 R+ Æ R+, f(x) = 1/x2 N Æ {n2 : n ŒN}, f(x) = x2

{(x, y): y2 = 4ax, x, y ΠR} {(x, y): y = |x|, x, y ΠR} {(x, y): x2 + y2 = 1, x, y ΠR} {(x, y): x2 Рy2 = 1, x, y ΠR} functions f and g are given by f(x) = (x), the 1 fractional part of x and g(x) = sin [x]p, where [x] 2 denotes the integral part of x, then range of gof is (a) [Р1, 1] (b) {0} (c) {Р1, 1} (d) [0, 1]

(a) (b) (c) (d) 15. The

16. Let g (x + y), g(x).g(y) and g(x – y) are in A.P. For all x, y and g(0) π 0. Then (a) g(2) = 0 (b) the graph of g is symmetry about y-axis (c) g is an odd function (d) g(0) = 1 17. Let f(x) = cos (p/x) and D+ = {x: f (x) > 0}. Then D+ contains (a) (2/5, 1/2) (b) (1/2, 2/3) (c) (– 2/3, – 1/2) (d) (– p, – 1/2) 18. The set of all values of ‘a’ (a > 0) for which the function f : [–3, 3] Æ R defined by f (x) = [x2/a] cosec ax + sec ax is an even function contains (a) (11, •) (b) (7, •) (c) (11, •) (d) (9, 11)

9. The domain of f(x) = log a1 log a2 log a3 log a4 x (ai > 0) is a (a) (0, •) (b) a4 3 ,• (c) a3 4 ,•

f: f: f: f:

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

2t + 3 is 6p

8. The period of the function y = sin (a) 2 p (c) 6 p2

is

(a) (b) (c) (d)

log2 ( x + 3) 19. The domain of definition of f(x) = 2 x + 3x + 2 contains (a) R ~ {– 1, – 2} (b) (– 2, •) ~ {– 1} (c) R ~ {– 1, – 2, – 3} (d) (– 3, •) ~ {– 1, – 2} 1 20. If f: [1, •) Æ [2, •) is given by f(x) = x + then x f –1 (x) equals (a)

x + x2 - 4 2

x (b) 1 + x 2

(c)

x - x2 - 4 2

2 (d) 1 + x - 4

IIT JEE eBooks: www.crackjee.xyz 21.26 Comprehensive Mathematics—JEE Advanced

21. The domain of the function

Column 1

1

x2 - 3x + 2 +

f(x) =

28.

3 + 2 x - x2

contains

3

(a) y = x –

(a) (– 1, 1] (b) [2, 3) (c) (0, •) (d) (0, 2] 22. If y = log3 x and F = (3, 27), the set onto which the set F is mapped contains (a) (0, 3) (b) (1, 3) (c) (1, 2) (d) (0, 2) 23. The interval of variation of x for which cos–1

(b) y =

(a) R

ax + 1 ax - 1

(q) f(0) = 1 and odd function (r) odd function (s) neither even nor odd R

Column 1 Column 2 (a) 7x + 1 (p) onto [– 1, 1] but not one-one [0, p] (b) cos x (q) one-one on [0, p] but not onto R (c) sin x (r) one-one and onto R (d) 1 + log x (s) one-one on (0, •)

(b) R ~ [3, 4] (d) (– •, 3)

(c) [3, 4] 2

25. If f(x – 1) = 2x – 3x + 1 then f(x + 1) is given by (b) 2x2 + 5x + 3 (a) 2x2 + 5x + 1 (d) 2x2 + x + 4 (c) 2x2 + 3x + 5

30. Column 1 (a) Range of f(x) pˆ Ê = 3sinx + 4sin ÁË x + ˜¯ 3 + 7 contains (b) Range of f(x) = log 5 ( 2 (sinx

MATRIX-MATCH TYPE QUESTIONS 26. The period of the function Column 1 (a) sin 2q –

Column 2 3 cos 2q

(p) 4p

3 (b) cosec2 x + cot x 4 (c) 2 sin 3x + 3 sin 2x

(r) 24

pt pt + sin 3 4 27. The domain of the function

(s) 2p

Column 1 3 - 2x 5 (1 – log10

(x2 – 5x + 16)) 2 (c) cos–1 2 + sin x sin x + 16 - x

(d) domain of f(x) = sin–1[1 – x2] contains

Column 2

3 - x +sin–1

(b) log10

– cosx) + 3) contains (c) domain of f(x) Ê 1 - | x |ˆ = sin–1 ÁË ˜+ 3 ¯ Ê | x | - 3ˆ cos–1 ÁË ˜ contains 5 ¯

(q) p

(d) sin

(d)

x x

(p) even function

1 - x 2 = sin–1 x is true is

(a) [0, 1] (b) [– 1, 1] (c) [0, 1/2] (d) [– 1/2, 1] 24. The set of values of x for which |f(x)| + |g(x)| > |f(x) + g(x)| if f(x) = x – 3 and g(x) = 4 – x

(a)

x x + 6 120

a -1 ax - 1 (c) y = x x a +1

(d) y =

Column 2 5

(p)

(2kp,(2k+1)p) k ŒI

(q) [–4, –p] » [0, p]

(r) (2, 3) 2

(s) [–1, 3]

Column 2 (p) [0, 2]

(q) {0}

(r) {1}

(s) {0, 1}

31. Column 1 (a) {x : [sin–1x] > [cos–1x]} contains (b) Range of [|sinx| + |cosx|] contains (c) Range of [x + 1/2] + [x – 1/2] + 2[– x] È 1 ˘ (d) Range of Í ˙ contains Î[ x - 3] ˚

Column 2 (p) {0, 1} (q) {1} (r) [sin1, 1]

(s) {– 1}

IIT JEE eBooks: www.crackjee.xyz Functions 21.27

ASSERTION-REASON TYPE QUESTIONS 32. Statement-1: The period of the function f(x) = p p p sin [x] + cos x + cot [x] is 24 4 2 4 Statement-2: The period of sinx, cosx is 2p and period (f(x) + g(x)) = l.c.m (period of f(x), period of g(x)) e x - e- x 33. Statement-1: The range of f(x) = x e + e- x on [0, •) is [0, •) R is [0, •). Statement-2: Range of ex 34. Statement-1: The set of all values of a for which È x2 ˘ the function f(x) = cot(sinx) + Í ˙ whose domain Îa˚ is [– 4, 4] is an odd function is R ~ [– 16, 16] Statement-2: The set of all zeros of the greatest integer function is (– 1, 1)

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 35 to 38 A right cone is inscribed in a sphere of radius R. Let S = f(x) be the functional relationship between the lateral surface area S of the cone and its generatrix x. 35. The domain by f(x) is (a) (0, 2R) (b) (0, R) (c) R (d) (R, 3R) 36. The value of f(R) is given by 3 p R2 (c) pR2 (d) 2pR 2 37. S2 is a polynomal of degree (a) 4 (b) 5 (c) 3 (d) 6 n 38. If x is such that x , n ≥ 6 is negligible then S is given by p x2 Ê x2 ˆ p x4 1 (a) (b) ˜ 2 R ÁË 2R R2 ¯ (a)

3 p R 2 (b)

Ê x2 ˆ 2 p x 1 (c) Á ˜ Ë 8 R2 ¯

px Ê x2 ˆ 1(d) Á 2 R Ë 8 R 2 ˜¯

40. The function h(x) (a) decreases on [– 2, 2] (b) decreases strictly on [– 2, 1] (c) increases on [– 2, 2] (d) increases on [1, 2] 41. The function h(x) is (a) one-one on its domain (b) one-one on [– 1, 1] (c) a linear function on [– 2, 1] (d) a linear function on [1, 2] 42. Which of the following is true (a) h(x) = |x| on [– 2, 2] (b) h(x) = |x| c[–2, 0] (x) + 2(x – 1) c(1, 2](x) (c) h(x) is nonzero constant on [0, 1] (d) h(x) = |x| c[–2, 0](x) + (x – 1) c(0, 2](x)

INTEGER-ANSWER TYPE QUESTIONS 43. If f(x) = sin2 x + sin2 (x + p/3) + cos x cos (x + p/3) 1 and g(5/4) = 1298 then g o f(x) is equal to. 649 44. Let f(x) = 1 + x2 and g be a function such that f(g(x)) 1 = 1 + x2 – 2x3 + x4. The value of g(18) is 34 45. If f: R Æ R f(x) = x2 + 1, then the num–1 ber of elements in f (17) » f–1(– 3) is 46. If f x satisfying Ê x + 1ˆ . f(x) = f Á Ë x + 2 ˜¯ 47. If f(x) = 1 – x3 – x4 – 2x5 = g(x) + h(x), where g is an even function and h is an odd function. Then 1 h(– 5) is equal to 2125 LEVEL 2

2

Paragraph for Question Nos. 39 to 42 Let f(x f(x) = – 1, – 2 £ x £ 0; f(x) = x – 1, 0 < x £ 2 and g(x) = |x|. Form the composite function h(x) = fog(x) + g o f(x). 39. The range of h(x) is (a) [0, 1] (b) [– 2, 2] (c) [0, 2] (d) [1, 2]

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. If f(x) = cos [2p2]x + cos [– 2p2]x, where [x] stands for the greatest integer function, then (a) f(p/2) = – 1 (b) f(p) = + 1 (c) f(–p) = 0 (d) f(p/4) = 1

IIT JEE eBooks: www.crackjee.xyz 21.28 Comprehensive Mathematics—JEE Advanced

2. If f(x) = sin log

( 4-x

2

)

(1 - x ) , then the domain

of f and range of f are (respectively) (a) [– 1, 1], (– 1, 1) (c) (1, 2), [– 1, 1]

(b) (– 2, 1), (– 1, 1) (d) (– 2, 1), [– 1, 1]

3. Range of the function f(x) = (a) (1, •) (c) (1, 7/3]

11.

x2 + x + 2 x2 + x + 1

, x ΠR is

12.

(b) (1, 3/2) (d) (1, 7/5]

4. If the function f: [2, •) Æ [2, • f(x) = 3x(x – 2), then f–1(x) is

13.

(a) (1/3)x(x – 1)

( (1 -

) x)

(b) 1 + 1 + log3 x (c)

1 + log3

(d) not defined (a) f(x) = {x}, the fractional part of x (b) f(x) = cos x (c) f(x) = x cos x (d) f(x) = sin 1/x Ê 2 x -1 ˆ + 6. The domain of the function f(x) = sin -1 Á Ë 3x - 2 ˜¯ 3x - 4 ˆ cos -1 ÊÁ is Ë 5 ˜¯ (a) [1, 3] (b) [–1/3, 3] (c) [–1/3, 3/5]»[1, 3] (d) [–1/3, 0) 7. The domain of definition of the function y(x) given by the equation ax + ay = a (a > 1) is (a) 0 < x £ 1 (b) 0 £ x < 1 (c) – • < x < 1 (d) – • < x £ 0 8. Let f: X Æ [1, 27] be a function by f(x) = 5 sin x + 12 cos x + 14. The set X so that f is one-one and onto is (a) [– p/2, p/2] (b) [0, p] (c) [0, p/2] (d) none of these 9. If f (x + 2y, x – 2y) = xy then f (x, y) is equal to (a) (1/4) (xy) (b) (1/4) (x2 – y2) (c) (1/8) (x2 – y2) (d) (1/2) (x2 + y2) 10. If g(x) = 1 +

3

x then a function f such that

f (g(x)) = 3 – 3 x + x is (a) f (x) = x3 – 3x2 + x + 5 (b) f (x) = x3 + 3x2 – x – 5

14. f

(c) f (x) = x3 – 3x2 + 2x + 3 (d) f (x) = x3 – 3x2 + 3x + 3 If f (x) = ux + v. The number of points in {(u, v); f (x) = fof(x) for all x Œ R} is are (a) finitely many (b) infinitely many (c) 2 (d) 4 Let f be a function satisfying 2f(x) – 3f (1/x) = x2 for any x π 0, then the value of f (2) is (a) – 2 (b) – 7/4 (c) – 7/8 (d) 4 x x + + 1 is The function f (x) = x 2 e -1 (a) even (b) periodic (c) odd (d) neither even nor odd If 1 x ( x + 1) 2x x ( x - 1) ( x + 1) x (x) = 3 x ( x - 1) x ( x - 1) ( x - 2 ) x ( x - 1) ( x + 1)

then f (50) + f (51) + + f (99) is equal to (a) 0 (b) 1275 (c) 3725 (d) 1025 15. The domain of the function x-5 - 3 x + 5 is f (x) = log10 2 x - 10 x + 24 (a) (– 5, •) (c) (2, 5) » (5, •)

(b) (5, •) (d) (4, 5) » (6, •)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 16. If f is a strictly decreasing function with range [a, b], its domain is (a) [b, a] (b) [f –1(a), f –1(b)] (d) [ f(b), f(a)] (c) [f –1 (b), f –1 (a)] 2 17. Let f(x) = (x + 1) – 1 (x ≥ – 1). Then the set S = {x : f o f(x) = f–1 (x)} contains -3 + i 3 -3 - i 3 Ô¸ ÔÏ , ˝ (b) {0, 1, – 1} (a) Ì0, - 1, 2 2 ÔÓ Ô˛ (c) {0, – 1} 18. If f(x) =

(d) {0}

a-x (a > 0), the domain of f–1(x) contains a+x

(a) (– •, •) (c) (a, •)

(b) (– •, – 1) (d) (0, •)

IIT JEE eBooks: www.crackjee.xyz Functions 21.29

sin ([ x ] p )

19. If f(x) =

where [x] denotes the integral x2 + x + 1 part of x, then (a) is one-one (b) is not one-one and non-constant (c) is a constant function (d) is zero function

20. If f(x) =

2x + 1 2 x + 3x2 + x 3

and S = {x: f(x) > 0} then S

contains (a) (– •, – 3/2) (c) (– 1/4, 1/2)

(b) (– 3/2, – 1/4) (d) (1/2, 3)

e- x , where [x] is the 1 + [ x] 1 + x2 greatest integer less than or equal to x. Then (a) dom (f + g) = R ~ [– 2, 0) (b) dom (f + g) = R ~ [– 1, 0) (c) Range f « Range g = [– 2, 1/2] (d) Range f « Range g = [– 1/2, 1/2] ~ {0} 22. Let the function f: R ~ {– b} Æ R x+a (a π b) then by f(x) = x+b (a) f is one-one but not onto (b) f is onto but not one-one (c) f is both one-one and onto (d) f -1(2) = a - 2b

21. Let f(x) =

x

and g(x) =

MATRIX-MATCH TYPE QUESTIONS 23. The domains of the functions Column 1 Column 2 –1 (a) sin (x/2 – 1) + log (x – [x]) (p) (3 – 2p, 3 – p) » (3, 4] (q) (0, 4) ~ {1, 2, 3} (b) ex + 5sin

p 2 16 - x 2

(c) log10sin(x – 3) +

(r) [– p/4, p/4]

16 - x 2 1 - 2x (d) cos–1 (s) [– 3/2, 5/2] 4 24. The periods of the functions Column 1 Column 2 (a) y = sin (2pt + p/3) + (p) 1/4 2 sin (3pt + p/4) + 3 sin 5pt (b) y = |sin 4px| + |cos 8px| (q) 8

1 Ê |sin (p 4) x | sin (p 4) x ˆ (c) y = 2 ÁË cos(p 4) x + |cos(p 4) x | ˜¯ (r) 2 (d) y = sin pt + cos 2pt (s) 1 –1 –1 25. Let f(x) = sin x, g(x) = cos x and h(x) = tan–1 x. For what interval of variation of x the following are true. Column 1 Column 2 (a) f ( x ) + g ( x ) = p/2 (p) [0, •)

( 1- x ) = 0 2

(b) f(x) + g

Ê 1 - x2 ˆ (c) g = 2h(x) Á ˜ Ë 1 + x2 ¯

(q) [0, 1] (r) (– •, 1)

Ê1+ xˆ (d) h(x) + h(1) = h ÁË 1 - x ˜¯ (s) [– 1, 0] f, g, h are as in Q25 and p(x) = cot–1 x) Column 1 (a) f (b) g

(

1- x

) +f(

( 1- x ) 2

Ê 1 - x2 ˆ (c) g Á ˜ Ë 1 + x2 ¯

Column 2 x ) (p) R ~ {0} (q) [– 1, 1] (r) R

(d) h(x) – p(1/x) (s) [0, 1] 27. Let f(x) = (x + a)/(x + b) whose a π b, b > 1 Column 1 Column 2 (a) dom f contains (p) (0, •) (b) range f contains (q) (b, •) (r) (1, •) (c) dom f –1 contains (s) (–b, •) (d) range of f –1 contains 28. Period of the function Column 1 Column 2 (a) f(x) = |sin(cosx)| + (p) p/2 cos(sinx) is less than or equal to (b) f(x) = sin4x + |cos4x|is (q) 2p (c) f(x) = min (sinx, |x|) (r) p (d) f(x) = 2x – [2x] is less (r) 3/4 than or equal to

ASSERTION-REASON TYPE QUESTIONS 29. Statement-1: max(cA, cB) is a characteristic function Statement-2: max(cA, cB) = cA » B 30. Statement-1: The period of the function f(x) = cos[2p2]x + cos[– 2p2]x + {x} is p, [x] being greatest integer function and {x} is fractional part of x is p.

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Statement-2: cosine function is periodic with period 2p. 31. Statement-1: Êx ˆ f(x) = ex + cos–1 ÁË - 1˜¯ + log x - [ x ] is 2 (0, 4) ~ {1, 2, 3} Êx ˆ Statement-2: Domain of cos–1 Á - 1˜ is (0, 4) Ë2 ¯

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 32 to 35 Let f and g be functions and let D = {x Πdom (g): g(x) Πdom (f)}. The composition fog D by fog(x) = f(g(x)). Similarly it is possible to form the triple composition or four function composition. Suppose then fn(x) = f o f .. o f (n times) (x) 1 32. If fog o h(x) = then | x | +3 (a) h(x) = x + 3 (c) g(x) = 1/x 33. If fog o h(x) =

(b) f(x) = |x| (d) g(x) = x + 3

1 log then |sin x |

(a) h(x) = x (b) g(x) = log |x| (c) f(x) = x , h(x) = sin x (d) g(x) = log |x|, h(x) = sin x 34. If f(f (f(x))) = x, then (b) f –1 (x) = x (a) f –1 (x) = f(x) (d) (f o f)–1 = f (c) f 5(x) = f(x) n

35. If f(x) =

10 - x n , x > 0 then

(a) f 4(x) = x (c) f 4(x) = f(x)

(b) f 3(x) = x (d) f 5(x) = x

Paragraph for Question Nos. 36 to 39 Consider an arbitrary function f: X Æ Y: It induces two important set mappings. If A is a subset of X, then its image f(A) is the subset of Y f(A) = {f(x): x Œ A} Similarly, if B is a subset of Y, then its inverse image f–1(B) is a subset of X f–1(B) = {x: f(x) Œ B} 36. Which of the following is not true (a) A1 Õ A2 fi f(A1) 1 f(A2) n

Ai ) = (b) f ( i« =1 n

n

« f(Ai)

i =1 n

(c) f ( » Ai ) = » f(Ai) i =1 c

i =1 c

(d) f(A) Õ f(A ) if and only if f is onto where A, Ai 1 X.

37. Which of the following is not true (a) A = f –1 (f(A)) (b) f –1(Y) = X (c) f –1 (»i Bi ) = f –1(Bi) (d) f –1 (»i Bi ) = « f –1(Bi) where Bi 1 Y 38. Consider the function Ï3 x -1 + 8 2 for 0 £ x £ 2 Ô f(x) = Ì7 + log2 ( x - 2) for 2 < x £ 4 Ô 2 for 4 < x £ 6 Ó x - 9 x + 21 If A = {1, 3, 5, 7} then f–1(A) has (a) 7 elements (b) 6 elements (c) 5 elements (d) 4 elements 39. Let f(x) = – log2 x + 3 and A = [1, 4] then f(A) is equal to (a) [1, 9] (b) [1, 2] (c) [2, 4] (d) [1, 3]

INTEGER-ANSWER TYPE QUESTIONS x +1 and g(x) = |x| + 1. Then the 2x - 1 number of elements in the set {x: f(x) ≥ g(x)} ~ [(1/2, 3/5) » (3/5, 7/10) » (7/10, 4/5) » (4/5, 1)] is

40. Let f(x) =

x 4 1 then g(256) is equal to 512 42. Let f, g, h x x f(x) = , g(x) = and h(x) = x +1 1- x 1 (fog o h)–1 (12). 216 43. Let f(x) = 1/2 + x - 3 4 and f(x) = f

3

x . Find

–1

(x) then

1 f(30) is equal to. 871 44. Let f(x) = x4 + x tion f(x) so that f is even. If value of the extended 1 K is equal to. function at – 3 is K, then 42 100

45. If a = e2pi/17 and f(x) = 7 +

Â

K =1

value of

1 17

16

Â

r =0

f(ar x).

AK xK

IIT JEE eBooks: www.crackjee.xyz Functions 21.31

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. Let f(x) = |x–1|. Then (a) f(x2) = (f(x)2 (b) f(x+y) = f(x) + f(y) (c) f(|x|) = |f(x)| (d) none of these [1983] 2. If f(x) = cos (log x), then ˘ 1 È Ê xˆ f ( x ) f ( y) - Í f Á ˜ + f ( xy)˙ 2 Î Ë y¯ ˚ has the value 1 (a) –1 (b) 2 (c) –2 (d) none of these [1983] 1 + x + 2 is y= log10 (1 - x )

1 3x - 5 x+5 (b) is given by 3 (c) does not exist because f is not one-one (d) does not exist because f is not onto [1998] (a) is given by

(

8. If g(f(x)) = |sin x| and f(g(x)) = sin x

ÏÔ -3 + i 3 -3 - i 3 ¸Ô , (a) Ì0, - 1, ˝ (b) {0, 1, –1} 2 2 ÔÓ Ô˛ (d) empty [1995]

)

2

, then

(a) f(x) = sin2 x, g(x) = x (b) f(x) = sin x, g(x) = |x| (c) f(x) = x2, g(x) = sin x (d) f and g cannot be determine. 9. If the function f: [1, •) Æ [1, • f(x) = 2x(x – 1), then f–1(x) is 1 (a) ÊÁ ˆ˜ Ë 2¯ (c)

(a) (– 3, – 2) excluding –2.5 (b) [0, 1] excluding 0.5 (c) [– 2, 1] excluding 0 (d) None of these [1983] 4. Which of the following functions is periodic (a) f(x) = x – [x] (b) f(x) = sin 1/x, x ! 0, f(0) = 0 (c) f(x) = x cos x (d) None of these [1983] 5. Let f(x) = sin x and y(x) = ln |x|. If the ranges of the composition functions f o g and g o f are R1 and R2 respectively, then (a) R1 = {u: –1 £ u < 1}, R2 = {v: –• < v < 0} (b) R1 = {u: –• < u < 0}, R2 = {v: –1 £ v £ 0} (c) R1 = {u: –1 < u < 1}, R2 = {v: –• < v < 0} (d) R1 = {u: –1 £ u £ 1}, R2 = {v: –• < v < 0} [1994] 2 6. Let f(x) = (x + 1) – 1, (x ≥ –1). Then the set S = {x: f(x) = f–1(x)} is

(c) {0, –1}

7. If f(x) = 3x – 5, then f –1(x)

x ( x -1)

(b)

[1998]

(

1 1 + 1 + 4 log2 x 2

)

)

(

1 1 - 1 + 4 log2 x (d) not defined [1999] 2 y(x) given x

y

by the equation 2 + 2 = 2 is (a) 0 < x £ 1 (b) 0 £ x £ 1 (c) –• < x £ 0 (d) –• < x 0 Ó Then for all x, f(g(x)) is equal to (a) x (b) 1 (c) f(x) (d) g(x)

[2001]

IIT JEE eBooks: www.crackjee.xyz 21.32 Comprehensive Mathematics—JEE Advanced

14. Let f(x) =

ax , x π 1. Then, for what value of a is x +1

f(f(x)) = x (a)

2

(c) 1

(b) – 2 (d) –1

[2001]

15. Let a function f: R Æ R f (x) = 2x + sin x for x Œ R. Then f is (a) one-to-one and onto (b) one-to-one not onto (c) onto but not one-to-one (d) neither one-to-one nor onto [2002] 2 16. Suppose f (x) = (x + 1) for x ≥ – 1. If g(x) is the f (x) w.r.t. y = x, then g(x) equals (a) – x – 1, x ≥ 0 1 ,x>–1 (b) ( x + 1)2 (c)

x +1 , x > – 1

f(x) =

x –1, x ≥ 0

np , nΠ{0, 1, 2,...}

(b) ±

np , nΠ{1, 2,...}

(b) [– 1/2, 1/9] (d) [– 1/4, 1/4] x2 + x + 2

[2003]

: x Œ R is x2 + x + 1 (a) (1, •) (b) (1, 3/2) (c) (1, 7/3] (d) (1, 7/5] [2003] 2 20. Let f(x) = x – 1 and g(x) = sin x + cos x, then f(g(x)) will be an invertible function if x lies in (a) [0, 3p/2] (b) [0, p/2] (c) [– p/4, p/4] (d) [– p/2, 0] [2004] 21. Let f and g R to R as follows: Ï0 if x is rational f(x) = Ì Ó x if x is irrational and 19. Range of the function f(x) =

p + 2np, nŒ [...., –2, –1, 0, 1, 2....] 2

(d) 2np, nŒ {...., –2, –1, 0, 1, 2....} 24. The function f: [0, 3] Æ –15x2 + 36x + 1, is (a) one-one and onto (b) onto but not one-one (c) one-one but not onto (d) neither one-one nor onto

[2011] f(x) = 2x3

[2012]

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

sin -1 (2 x ) + p 6 is

(a) [– 1/4, 1/2] (c) [– 1/2, 1/2]

(a) ±

(c)

[2002] x 17. If f: [0, •) Æ [0, •) and f(x) = then the 1+ x function f is (a) one-one and onto (b) one-one but not onto (c) onto but not one-one (d) neither one-one nor onto [2003] (d)

Ï0 if x is irrational g(x) = Ì Ó x if x is rational then the function f – g is (a) one-one but not onto (b) onto but not one-one (c) both one-one and onto (d) neither one-one nor onto [2005] 22. Let X and Y be two nonempty sets. Let f: X Æ Y be a function. For A à X and B à Y f(A) = {f(x) : x Œ A}, f –1(B) = {x Œ X | f(x) Œ B}, then (b) f(f –1 (B)) à B (a) f(f –1 (B)) = B (d) f –1(f(A)) à A (c) f –1(f(A)) = A [2005] 2 23. Let f(x) = x and g(x) = sin x for all x Œ R. Then the set of all x satisfying (f c g c g c f) (x) = (g c g c f ) (x), where (f c g)(x) = f(g(x)), is

1. Let g(x of the equilateral triangle with two of its vertices at (0, 0) and (x, g(x)) is

3/4, then the function g(x) is

2 (a) g(x) = ± 1 - x

(b) g(x) =

1 - x2

(c) g(x) = - 1 - x 2

(d) g(x) =

1 + x2 [1989]

2. Let f:(0, 1) Æ R b-x 1 - bx where b is a constant such that 0 < b < 1. Then (a) f is not invertible on (0, 1) 1 (b) f ! f –1 on (0, 1) and f ' (b) = f ¢(0) f ( x) =

IIT JEE eBooks: www.crackjee.xyz Functions 21.33

point Ak, k = 1,2,..., n. If

1 (c) f = f –1 on (0, 1) = f ' (b) = f ¢(0)

n -1

n -1

 k =1 (a k ¥ a k -1 ) =  k =1 (a k ¥ a k +1 ) ,

(d) f –1 is differentiable on (0, 1)

[2011]

Êp pˆ 3. Let f: Á , ˜ Æ R Ë 2 2¯

then the minimum value of n is R. If the normal from the point P(h,1) on the ellipse

3

f(x) = (log(secx + tanx) . Then (a) f(x) is an odd function (b) f(x) is an one-one function (c) f(x) is an onto function (d) f(x) is an even function

x 2 y2 + = 1 is 6 3

perpendicular to the line x + y = 8, then the value of h is [2014]

Êp Êp ˆˆ 4. Let f(x) = sin Á sin ÁË sin x˜¯ ˜ for all x Œ R and Ë6 ¯ 2 p g(x) = sin x for all x Œ R. Let (f c g)(x) denote 2 f(g(x)) and (g c f)(x) denote g(f(x)). Then which of È 1 1˘ (a) Range of f is Í- , ˙ Î 2 2˚ È 1 1˘ (b) Range of f c g is Í- , ˙ Î 2 2˚ f ( x) p = (c) lim x Æ 0 g( x ) 6

S. Number of positive solutions satisÊ 1 ˆ fying the equation tan–1 Á Ë 2 x + 1˜¯ Ê 1 ˆ Ê 2ˆ = tan–1 ÁË = tan–1 Á 2 ˜ is 4 x + 1˜¯ Ëx ¯ P Q R S (a) 4 3 2 1 (b) 2 4 3 1 (c) 4 3 1 2 (d) 2 4 1 3

MATRIX-MATCH TYPE QUESTIONS

[2014]

1. Let f: R Æ R and g : R Æ R be respectively given by f(x) = |x| + 1 and g(x) = x2 + 1. Define h : R R by Ïmax{ f ( x ), g( x )} if x £ 0, h( x ) = Ì Ó min{ f ( x ), g( x )} if x > 0. The number of points at which h(x) is not differentiable is [2014]

domains (– p/2, p/2) and the range (– •, •) Column 1 Column 2 (a) 1 + 2x (p) is onto but not one-one (b) tan x (q) is one-one but not onto (r) is one-one and onto (s) is neither one-one nor onto [1992]

FILL

IN THE

BLANKS TYPE QUESTIONS

Ê p2 ˆ - x 2 ˜ lie in the 1. The value of f(x) = 3 sin Á ÁË 16 ˜¯ interval_______. [1983]

2. Column 2 1. 1

3 . Then x [–1, 1], x ! ± 2 dy( x ) Ô¸ d 2 y( x ) 1 ÏÔ 2 +x ˝ Ì( x - 1) 2 dx Ô˛ equals y( x ) ÔÓ dx Q. Let A1, A2, ...., An (n > 2) be the vertices of a regular polygon of n sides with its centre at the origin. Let ak be the position vector of the

4. 9

INTEGER-ANSWER TYPE QUESTIONS

(d) There is an x ΠR such that (g c f )(x) = 1 [2015]

Column 1 P. Let y(x) = cos (3 cos–1 x),

3. 8

2. 2

2. The domain of the function f(x) = sin given by________.

–1

Ê x2 ˆ log Á 2 2 ˜ is Ë ¯ [1984]

Ê 4 - x2 ˆ 3. If f(x) = sin log Á ˜ , then the domain of f(x) ÁË 1 - x ˜¯ is _____ and its range is _______. [1985] 4. There are exactly two distinct linear functions _____ and ________ which map [–1, 1] onto [0, 2]. [1989]

IIT JEE eBooks: www.crackjee.xyz 21.34 Comprehensive Mathematics—JEE Advanced

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

5. If f is an even function defined on the interval (–5, 5), then four real values of x satisfying Ê x + 1ˆ f ( x) = f Á Ë x + 2 ˜¯ are _____ and ______.

[1996]

pˆ Ê 6. If f(x) = sin2x + sin2 ÁË x + ˜¯ + cos x cos 3 and g = 1 then (g c f) (x) = _______.

pˆ Ê ÁË x + 3 ˜¯ [1996]

16. 18. 20. 22. 24.

26.

x 2 + 4 x + 30

is not one-to-one. [1983] 3. If f1(x) and f2(x) are defined on domain D1 and D2 respectively then f1(x) + f2(x) is defined on D1 » D2. [1988]

2. The function f(x) =

x 2 - 8 x + 18

27.

SUBJECTIVE-TYPE QUESTIONS p¸ Ï p 1. Given A = Ì x : £ x £ ˝ and f(x) = cos x – x(x + 1), 3˛ Ó 6 and f(A). [1980] 2. Let f be a one-one function with domain {x, y, z} and range {1, 2, 3}. It is given that exactly one of the following statements is true and the remaining two are false: f(x) = 1, f(y) ! 1, f(z) ! 2 [1982] Determine f 1(1). 3. Find the natural number a for which

28.

n

 f (a

+ k) = 16(2n – 1) where the function f

k =1

29.

satisfies the relation f(x + y) = f(x) f(y) for all natural numbers x, y and further, f(1) = 2. [1992]

Answers LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13.

(c) (c) (b) (b)

2. 6. 10. 14.

(b) (d) (c) (b)

3. 7. 11. 15.

(a) (b) (d) (b)

4. (b) 8. (c) 12. (c)

(d) (c), (d) (c) (d)

17. 19. 21. 23. 25.

(a), (b), (a), (a), (b)

(b), (c) (d) (b) (c)

MATRIX-MATCH TYPE QUESTIONS

TRUE/FALSE TYPE QUESTIONS 1. If f(x) = (a – xn)1/n where a > 0 and n is positive integer, then f(f(x)) = x. [1983]

(b), (a), (a) (b), (b),

30.

IIT JEE eBooks: www.crackjee.xyz Functions 21.35

31.

24.

25.

ASSERTION-REASON TYPE QUESTIONS 32. (c)

33. (d)

34. (c)

COMPREHENSION-TYPE QUESTIONS 35. (a) 39. (c)

36. (b) 40. (d)

37. (d) 41. (d)

38. (d) 42. (b)

26.

INTEGER-ANSWER TYPE QUESTIONS 43. 2 47. 3

44. 9

45. 2

46. 4

LEVEL 2

27.

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13.

(c) (a) (c) (a)

2. 6. 10. 14.

(d) (c) (c) (a)

3. 7. 11. 15.

(c) (b) (b) (d)

4. (b) 8. (d) 12. (b)

28.

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 16. 18. 20. 22.

(c) (b), (c), (d) (a), (d) (c), (d)

17. (c), (d) 21. (c), (d) 21. (b), (d)

MATRIX-MATCH TYPE QUESTIONS 23.

ASSERTION-REASON TYPE QUESTIONS 29. (a)

30. (d)

31. (c)

COMPREHENSION-TYPE QUESTIONS 32. (d) 36. (b)

33. (c) 37. (a)

34. (d) 38. (c)

35. (a) 39. (d)

INTEGER-ANSWER TYPE QUESTIONS 40. 4 44. 2

41. 2 45. 7

42. 8

43. 1

IIT JEE eBooks: www.crackjee.xyz 21.36 Comprehensive Mathematics—JEE Advanced

PAST YEARS' IIT QUESTIONS

Hints and Solutions

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17. 21.

(d) (d) (b) (b) (b) (c)

2. 6. 10. 14. 18. 22.

(d) (c) (d) (d) (a) (b)

3. 7. 11. 15. 19. 23.

(d) (b) (d) (a) (c) (a)

4. 8. 12. 16. 20. 24.

(a) (a) (a) (d) (c) (b)

MULTIPLE CORRECT ANSWER TYPE QUESTIONS 1. (b), (c) 2. (a) 4. (a), (b), (c)

3. (a), (b), (c)

MATRIX-MATCH TYPE QUESTIONS 1.

LEVEL 1

x2 – 3x + 2 > 0 i.e. (x – 1) (x – 2) > 0 fi x < 1 or x > 2. 2. We must have log10

4.

5. 6. 7.

Ê 5x - x2 ˆ 5x - x2 ≥ Á 4 ˜ ≥0¤ Ë ¯ 4

1 ¤ (x – 4) (x – 1) £ 0 ¤ x Œ [1, 4]. x ≥ 4 and x £ 6. Thus the domain is [4, 6]. £ (x – 3)/2 £ 1 The sin–1 (x x < 4. So we must have and log10 (4 – x 1 £ x £ 5 and x < 4. Thus the domain is [1, 4). f(2) = 6, f(1) = 0, f(p/12) = 1/2, f(f(2)) = sin 12, f(f(1)) = 0, f(f(p/12)) = – 3/8, f(f(1)) = 0. 2 The function f(x) = cos x, 2–x are even functions and 4 y = 2x – x is neither even nor odd. y(10x + 10–x) = 10x – 10–x fi 102x (y – 1) = – 1 – y fix=

1 1+ x 1+ y 1 log10 . Thus f–1(x) = log10 . 2 1- x 1- y 2

8. Period is equal to 2p/(2/6p) = 6p2 9. f

2. (a)

INTEGER-ANSWER TYPE QUESTIONS 1. 3

FILL

IN THE

BLANKS TYPE QUESTIONS

1. [0, 3/ 2 ] 3. (–2, 10); [–1, 1] 1 1 (–3 ± 5 ), (3 ± 5. 2 2 6. 1

2. [–2, –1] [1, 2] 4. x + 1, 1 – x 5)

TRUE/FALSE-TYPE QUESTIONS 1. False

2. True

3. False

SUBJECTIVE-TYPE QUESTIONS È 3 p2 p 1 p2 p ˘ 1. Í 2 - 36 - 6 , 2 - 9 - 2 ˙ ÍÎ ˚˙ 2. [y]

3. 3

log a2 log a3 log a4 x > 0 i.e. loga loga 3 4 a

x >1 i.e. loga4 x > a3 which is true only if x > a4 3 . 1 10. y (x x| ! 0, |sin x | ≥ 1 so x ! np, n Œ I. 11. We have, for instance, f(0) = 3 and f(p) = 3, so f is not one-one. It is not onto, as there is no x Œ R such that 3 – 2 sin x = 0. 12. The function in (a) is neither one-one nor onto, and that in (b) is not onto. The function in (d) is not onto. 13. The function in (b) is not one-one, while those in (a), (c) and (d) are both one-one and onto. 14. The sets of points described by answers (a), (c) and (d) are symmetric in y. That is, if (x, y) is in one of these sets, so is (x, – y). This means the ordered pairs in these answers cannot represent functions. However, answer (b) is correct because it is symmetric in x. 1 sin [(x)]p = 0, for all 15. g o f(x) = g(f(x)) = 2 x Œ R. Hence the range of g o f is {0}. 16. 2 g(x) g(y) = g(x + y) + g(x – y) " x, y. Interchanging x by y, we have 2g(x) g(y) = g(x + y) + g(y – x). Thus

IIT JEE eBooks: www.crackjee.xyz Functions 21.37

g(x – y) = g(y – x) taking y = 0, we get g(x) = g(–x). Thus the graph of g is symmetrical about y-axis. Putting x = y = 0, we get 2 g(0)2 = 2 g(0). Since g(0) ! 0 so g(0) = 1. 17. D+ = {x : cos (p/x) > 0} Ï p Ôx : x Ô = Ì Ô Ô Ó

pˆ¸ Ê ÁË 2np , (4n + 1) ˜¯ Ô 2 Ô n ŒI ˝ p Ê ˆ Ô 4 n 1 , 2 n p ( ) ÁË ˜¯ Ô 2 ˛

Œ

Ï Ê 2 1ˆ Ê 1 2ˆ ¸ ÁË , ˜¯ ÁË , ˜¯ Ô Ôx : x Œ 2 3 Ô Ô n ŒI 5 2 2 Ì ˝ Ê 2 1ˆ Ô Ô ÁË - , - ˜¯ ÔÓ Ô˛ 3 2 È x2 ˘ 18. f(–x) = f(x) tfi 2 Í ˙ cosec ax = 0 Îa˚ È x2 ˘ fi Í ˙ = 0 " x Œ [–3, 3] Îa˚ x2 < 1 " x Œ [–3, 3] a fi 0 £ x2 < a " x Œ [–3, 3] fi a > 9. 21. dom f = {x: x + 3 > 0, x2 + 3x + 2 ! 0} = {x: x > – 3, x ! – 1, – 2} = (– 3, •) ~ {– 1, – 2} fi0£

x + x2 - 4 for x > 2 2 21. dom f = {x : (x – 1) (x – 2) ≥ 0, 3 + 2x – x2 > 0} = {x : x £ 1, x ≥ 2, (x – 3) (x + 1) < 0} = [– 1, 1] » [2, 3) 22. log3 x is an increasing function, so F is mapped onto the set (log3 3, log3 27) = (1, 3).

20. f–1 (x) =

1 - x 2 is meaningful for – 1 £ x £ 1. Range of cos–1 x in [0, p] and that of sin–1 x is [– p/2, p/2]. The given equation is valid only when 0 £ x £ 1. 24. |f(x)| + |g(x)| = 7 – 2x if x < 3; and is equal to 1 if 3 £ x £ 4 and is equal to 2x – 7 if x > 4. Also f(x) + g (x) = 1 so |f(x) + g(x)| = 1. We need those points for which LHS is greater than 1. Clearly, we can exclude values of x between 3 and 4. For x < 3, 7 – 2x > 1 and for x > 4, 2x – 7 > 1. Therefore, the given inequality is true for values if x Œ R ~ [3, 4]. 25. f(x + 1) = f((x + 2 – 1) = 2(x + 2)2 – 3(x + 2) + 1 = 2(x2 + 4x + 4) – 3x – 6 + 1 = 2x2 + 5x + 3 26. Note that period of sin ax is 2p/a, a > 0 23.

27. Use dom sin-1 x is [- p/2, p/2] 28. Check f(- x) = f(x) or f(- x) = - f(x) 29. sin x is one-one on [- p/2, p/2] and cos x is one-one on [0, p]. 30. Range of f in (a) is ÈÎ- 37 + 7, 37 + 7˘˚ Range of f in (b) is [0, 2] domain of f in (c) is [– 4, 4] domain of f in (d) is ÈÎ- 2 , 2 ˘˚ 31. {x : [sin–1x] > [cos–1x]} = [sin 1, 1] Range of [|sinx| + |cosx|] = {0, 1} Range of [x + 1/2] + [x – 1/2] + 2[– x] = {– 3, – 1} È 1 ˘ Range of Í ˙ = {– 1, 0, 1}. Î[ x - 3] ˚ p p (8 + [x]) = sin [x], so its period is 8. The 32. sin 4 4 p p p period of cos x is 4 and cot [3 + [x]) = cot [x] 2 3 3 is 3. Hence period of f(x) is l.c.m (8, 4, 3) = 24. the statement (b) may not be. The l.c.m is a period of f(x) + g(x) but may not be the least period. E.g. |sin x| and |cos x| are periodic with period p but the least period of |sin x| + |cos x| is p/2. 2 . For x ≥ 0, 1 £ e2x < • 33. f(x) = 1 – 2x 1+ e fi0£

2 1+ e

2x

£

1 2 fi0£1– £ 1. 2 1 + e2 x

È x2 ˘ 34. f(– x) = – f(x) fi Í ˙ = 0 " x Œ [– 4, 4] Îa˚ x2 < 1 fi For a > 0, 0 £ x2 < a and for a < a 0, – a < x2 £ 0. Moreover, 0 £ x2 £ 16. Thus a Œ (– •, – 16) » (16, •) = R ~ [– 16, 16]. fi0£

35-38. p x2 S = f(x) = 2R

4 R 2 - x 2 for 0 < x < 2R

otherwise f(x Ê x2 ˆ 2 f(x) = p x Á 1 ˜ 4 R2 ¯ Ë

12

Ê 1 x2 ˆ 6 = px2 Á 1 ˜ + 0(x ) Ë 2 4 R2 ¯

39–42. h(x) = – x if – 2 £ x £ 0; h(x) = 0 if 0 < x £ 1; h(x) = 2 (x – 1) if 1 < x £ 2. 43. On simplifying f(x) = 5/4

IIT JEE eBooks: www.crackjee.xyz 21.38 Comprehensive Mathematics—JEE Advanced

1 1 g o f(x) = g(5/4) = 2.1298 = 2. 649 649 44. f(g(x)) = 1 + (g(x))2. Therefore, 1 + (g(x))2 = 1 + x2 – 2x3 + x4 fi g(x) = |x (x – 1)| so

45. f–1 (17) = {4, – 4}, f–1 (– 3) = f x +1 -x + 1 and – x = x +2 -x + 2 47. h(x) = – x3 – 2x5

46. Equate x =

IIT JEE eBooks: www.crackjee.xyz

22 Limits and Continuity 22.1

LIMITS

Let f

c,

e f (x) – l

= l if x c

d right-hand limit

c f (x x–c

lim f(x) = l

x and c l

l

x

c–d

e.

xÆc +

f (x)

Y

c lim f(x) = l

l1

xÆc

f (x) and l can be

l2

x c

c. e

e

x

xÆ c f (x) = l

f (x) – l

d d

lim f (x) = l1 x ® c–

c x–c

e.

f (x) – l

x Æ c + f (x xÆc a function f (x

y = f (x)

l

f (x

x x Æ c – f (x f(x) = 1/x xÆ

x f (x

ONE-SIDED LIMITS xÆc+

left-hand limit

f (x

xÆc–

xÆc -

c c – h, h

x Æ c+

f (x

Ï 1, x > c f ( x) = Ì Ó-1, x < c xÆc– f (x) = – 1.

lim f (x) = l

can

f (x

xÆc f (x) = 1/x

x

Fig. 22.1

If x

f (x xÆc

xÆ a-d a a+d

f (x

xÆc–

f (x

xÆc +

l-e

22.2

lim f (x) = l2 x ® c+

Fig. 22.2

y l+e

X

c

f (x

f (x) and l lim f ( x )

x Æ c-

x f(x)

c

c 1

c + .1 1

1

c

c –1

c – .1 –1

–1

IIT JEE eBooks: www.crackjee.xyz 22.2 22.3

1.

FREQUENTLY USED LIMITS

lim

xÆ0

sin x = 1 = lim xÆ0 x = lim

xÆ0

2. 3.

1/x

lim (1 + x)

xÆ0

lim

xÆ0

x = lim

xÆ0

sin -1 x tan -1 x = lim . xÆ0 x x

log a (1 + x ) x

a

xÆ0

ax - 1 x

a

lim

xÆa

22.4

log x

lim

xÆ2

f ( x ) f (2 ) 8 = = . g ( x ) g (2 ) 9

xn - an = nan–1. x-a

lim

xÆ0

xÆc

3.

4.

lim (f(x) ± g(x)) = lim f(x) ± lim g(x). xÆc

lim [kf(x)] = k lim f(x

xÆc

f

f(x g

lim

xÆc

xÆc

x

3 + sin x 3 + 0 = =3 x Æ 0 cos x 1

If lim f (x) and lim g (x

2.

3 + sin x cos x

ALGEBRAIC OPERATIONS ON LIMITS

g(x

1.

x3 + 1

Illustration 2

m

xm

x2 + x + 2

f (x) = x2 + x + 2 and g(x) = x3 + f(2) = 4 + 2 + 2 = 8 and g

e = 1.

lim

f (a) and g(a

Illustration 1

xÆ2

6.

g(a) π

f ( x ) f (a ) = g ( x ) g (a )

lim

(a

5.

f (x) g (x)

1. If f, g

lim

(1 + x )m - 1 lim = m, m Œ R. xÆ0 x

7.

lim

xÆa

xÆa

log (1 + x ) = 1. x

ex - 1 lim xÆ0 x

xƕ

COMPUTATION OF LIMITS

a π 1).

e (a

xÆ0

lim

22.5

1ˆ x Ê = e = lim Á 1 + ˜ . xÆ• Ë x¯

a = e, lim 4.

tan x x

xÆc

xÆc

2. If g(a (x – a

f(a

f f and g

k

lim (f(x) ◊ g(x)) = lim f(x) ◊ lim g(x).

xÆc

xÆc

lim f ( x ) f ( x) xÆc lim = x Æ c g( x) lim g ( x ) xÆc

5. If f(x

lim f ( x ) = L

xÆ a

xÆc

Illustration 3 lim g(x) π

xÆc

lim

L≥

lim f(x) = f Ê lim xˆ = f(c), Ë xÆc ¯ xÆc

x Æ1

x3 - 1 x2 - 1

If f(x) = x3 –1 and g(x) = x2 f(1) = g x x3 –1 and x2

( x - 1) ( x 2 + x + 1) x Æ1 x2 - 1 x Æ1 ( x - 1) ( x + 1) lim

x3 - 1

= lim

x and

IIT JEE eBooks: www.crackjee.xyz 22.3

lim (1 + f(x))1/g(x) = lim (1 + f(x))1/f(x).( f (x) /g (x)) = el

x2 + x + 1 = lim x Æ1 x +1 =

xÆ0

Illustration 6

1+1+1 3 = . 1+1 2

x

lim (1 + (x2 + 3x))

p ( x ) – q ( x ) and p(x), q(x), r(x), s(x f (a) = g(a

3. If f and g r ( x ) - s( x )

xÆ0

xÆ0

lim (1 + ( x 2 + 3 x )[1 /( x

2

+ 3 x )] ( x ( x + 3) / sin x )

xÆ0

2

= lim (1 + (x2 + 3x))[(1/(x

+ 3x)] (x(x +3) /

x)

xÆ0

2

= lim (1 + (x2 + 3x))(1/(x

+ 3x)) (x

x) (x + 3)

xÆ0

Illustration 4

= e1.3 = e3 4a + 3 x - x + 6a

lim

= lim

(2 x - 2a ) ( 2a + 5 x + 3a + 4 x ) ( x - a)

xÆa

a

xÆ0

4a + 3 x + x + 6a

4a + 3 x + x + 6a

xÆa

lim

(

2a + 5 x + 3a + 4 x

= 2 lim

4.

22.6

2 a + 5 x - 3a + 4 x

xÆa

f (x)

-1 g (x)

xÆ0

lim

1. ( 0 0 form) If f and g d

)

(i) g¢(x) π =

2 (2 7 a ) 2 7a

= 2.

xÆ0

f(x) Æ



a f (x) - 1 a f (x) - 1 f ( x ) = lim ◊ xÆ0 g (x) f (x) g (x) f (x) g (x)

asin x - 1 lim x Æ 0 x cos x

5.

xÆ0

lim

xÆ0

f ¢( x ) =L g ¢( x )

lim

f ( x) = L. g( x ) x Æ •). If f and g a, •

a f(x) Æ

lim f(x) = lim g(x

xƕ

3. (•/• form). If f and g d (i) g¢(x) π xŒ (ii) f(x) Æ •, g(x) Æ •

d) xÆ

lim

xÆ0+

a xÆ

xƕ

xƕ

sin x

lim (1 + f (x))1/g(x)

lim g(x)

Ê f ¢( x ) ˆ (ii) lim Á ˜ =L x Æ • Ë g ¢( x ) ¯

(iii)

- 1 sin x 1 ◊ x cos x sin x

d)

xÆ0+

lim

xÆ0+

(i)

Illustration 5

xÆ0

lim f(x

2. (L¢

xÆ0

a



xÆ0+

xÆ0+

a lim

= lim

(ii) (iii)

f (x) g (x)

and lim

L’HÔPITAL’S RULES FOR CALCULATING LIMITS

f ¢( x ) =L g ¢( x )

lim

lim

xÆ0+

f ( x) = L. g( x )

f ( x) =L g( x )

Illustration 7

f (x) =l g (x) lim

x Æ1

x3 + 3x - 4 2 x2 + x - 3

3x2 + 3 6x 3 = lim = . x Æ1 4 x + 1 x Æ1 4 2

= lim

IIT JEE eBooks: www.crackjee.xyz 22.4

3x2 + 3 4x + 1 In fact

0 0

(

x Æ 1.

x) = x –

Ê ˆ x2 x3 + + ˜ x) = – Á x + 2 3 Ë ¯

)

lim 3 x + 3 3x2 + 3 x Æ1 6 = . = x Æ1 4 x + 1 lim ( 4 x + 1) 5 lim

2

x Æ1



0 • and . 0 •



, ••, • – •

x= x-

x3 x 5 + - 3! 5!

x = 1-

x2 x4 + - 2! 4!

8. tan x = x + –1

x= x+

1 x3 1 3 x 5 + ◊ + 2 3 2 4 5

–1

x= x-

x3 x 5 + - 3 5

x

xÆ0 +

. If L L = lim x

x = lim

x Æ 0+

= lim

x Æ 0+

x Æ 0+

log sin x 1/ x

Ê •ˆ ÁË ˜¯ •

x cos x - cos x 0 = lim = 1 x Æ 0 1 (sin x ) / x sin x ÊÁ 2 ˆ˜ Ëx ¯

Illustration 10

lim

(1 + x )1 / x - e

xÆ0

x

Let y = (1 + x)1/x fi

y=

1 x

x) =

1 x

xÆ0

If L xÆ0

4 log (1 - 3 x ) Ê 0 ˆ 4 ( -3) = – 12 ÁË ˜¯ = lim x Æ 0 1 - 3x x 0

–12

L=e 22.7

È ˘ x x2 - ˙ Í1 - + 2 3 ˚ eÎ

fiy=

= e◊e

p ( p - 1) 2 p ( p - 1) ( p - 2 ) 1. (1 + x)p = 1 + px + x + 3! 2! x3 + º x pŒQ x2 x3 + + 2. ex = 1 + x + 2! 3! e a) +

x2 2!

x x2 + - 2 3

2 ˘ ˆ 1 Ê x x2 ˙ + +   ˜¯ 2 ÁË 2 3 ˙˚

y-e x

È 1 x = e Í- + + terms containing Î 2 3 x 2 and higher powers of x˘˙ ˚ \ lim

xÆ0

2 e a) + º

-

È Ê x x2 ˆ = e Í1 + Á - + - ˜ + ¯ Î Ë 2 3

USE OF SERIES EXPANSION IN FINDING LIMITS

3. ax = 1 + x

È ˘ x2 x3 x + - ˙ Í 2 3 Î ˚

È ˘ x x2 = Í1 - + - ˙ Î 2 3 ˚

lim (1 – 3x)4/x (1•

L = lim

x

x

L=1 Illustration 9

x

x3 2 x 5 + + 3 15

Illustration 8 lim (sin x )

x2 x3 x 4 + + x 2 3 4

(1 + x )1 / x - e x

=-

e . 2

IIT JEE eBooks: www.crackjee.xyz 22.5 22.8

a h(x) £ f(x) £ g(x x ŒD and lim h(x) = lim g(x) = l lim f(x) = l.

SEQUENTIAL LIMITS

xÆa

f (1) = f1, f (2) = f2 ,

NÆR

f (n) = fn, 

3.

Œ

lim an = a

nƕ

Œ

N n > N.

an – a

xÆa

lim ex = 1, lim ex = •, lim e–x

xÆ0

xƕ

lim ex

Œ

xÆa

lim

xÆa

xÆ-•

xƕ

e x - ea = ea. x-a

Illustration 13 Illustration 11 1 n

lim

nƕ

lim

nƕ

1 n

1 1 , 2 n n

2

2n - 1 xn = 2n + 1

xÆ 0

1 =0. x £x

x

lim xn = 1.

nƕ

lim f(x) = l

xn

xÆa

of x

lim x sin

lim xn = a

Illustration 14

lim f(xn) = l.

n Æ•

nƕ

lim f(x

Find lim

x Æa

f

xÆ 0+

xn

lim xn = a

nƕ

xn and x¢n

lim xn = a = lim x¢n

nƕ

nƕ

x

Hence lim

xÆ 0+

but lim f(xn) π lim f(x¢n). nÆ•

[ x] x



lim f(xn

nƕ

1 £ x and lim x xÆ 0 x 1 lim x sin = 0 . xÆ 0 x

nƕ

22.10

[ x] x

[ x] x

CONTINUITY

Illustration 12 1 x -1

lim

xÆ1

1 and ( x - 1)

xn =1 +

f (x) =

lim xn = 1 = lim x¢n. Now f (xn)

nƕ

lim f(xn

nƕ

lim f(x¢n

nƕ

22.9

nƕ

np

4n + 1 p = 1. Hence 2

f(x¢n lim f(x

xÆc

nƕ

lim

x Æc



f xÆ c f (x) = f (c

f interval (a, b f interval [a, b] if (i) f

SOME USEFUL RESULTS ON LIMITS

1. If lim f(x) = l

c

1 2 , xn¢ = 1 + , ( 4n + 1)p np

(ii)

a, b),

lim f (x) = f (a

xÆa+

at a)

lim g(x

xÆc

(iii)

f (x) g (x)

continuous at x = c if h Æ f (c + h) = f (c). continuous on an open a, b). continuous on a closed

lim f (x) = f (b

xÆb-

at b) f, g and h D

f (x f be discontinuous at c

IIT JEE eBooks: www.crackjee.xyz 22.6

f

discontinuous on an interval if

function f

4. If f I and x ΠI

c

lim f(x) and lim f(x

x Æ x0 +

(i) f

c xÆc

(iii) f

f (x) = 1/x at x

lim f(x) £ f(x )£ lim f(x).

f (x

x Æ x0 -

c f (x) π f (c xÆc

xÆc

f (x

I

c and c – d, c + d ) f(x

x Æ1

f and g f + g, a f, f g

a

–1

f

6. Let f f(x) π

lim f ( x ) = lim( x 2 + x + 1) = 3 π f (1) x Æ1

x Æ x0 +

5. If f

Ï x3 - 1 Ô x π1 f(x) = Ì x - 1 ÔÓ 4 x =1

But

x Æ x0 -

x = c and f, g)

f, g

7. Let f f(b c Π(a, b

a, b]. If f(a) and f(c R

g (c) π

c f/g If g

c. c and f

g (c

fog

SOLVED EXAMPLES

c.

SINGLE CORRECT ANSWER TYPE QUESTIONS Illustration 15

(a) Ï| x | , xπ0 Ô f(x) = Ì x ÔÓ 1 , x = 0

x Æ0 -

x x

5 2

(c) 1

(d) 2

1 - cos5 x (1 - cos x)(1 + cos x + cos 2 x + cos3 x + cos 4 x) = 1 - cos x (1 - cos x)(1 + cos x)

f

x

a, b a, b k £ f(x) £ K

k, K ΠR

x Π[a, b].

2. f(x 3. Intermediate Value Theorem If m M f (x a£x£b m£A£M [a, b f (x ) = A, i.e., a

1 + cos x + cos 2 x + cos3 x + cos 4 x 1 + cos x 5 1 - cos x 5 lim = xÆ xÆ x Æ 0 1 - cos x 2 Ï 3 ÔÔ x + 1 + x + 11 , x π -2 Let f(x) = Ì Example 2 x+2 Ô k, x = -2 ÓÔ =

FUNCTIONS CONTINUOUS ON A CLOSED INTERVAL

f (x 1. f(x

(b)

Solution:

|x| |x| x = lim = 1 but lim x Æ0 + x x Æ0 + x x Æ0 - x = – lim

3 2

Ans. (b)

lim

22.11

Ê 1 - cos5 x ˆ lim Á ˜ x Æ 0 Ë 1 - cos 2 x ¯

Example 1

k

a, b]. f (x) and (a) 2

A x Œ

Ans. (d)

f (b)

11 4

x (c)

17 4

(d) -

17 6

IIT JEE eBooks: www.crackjee.xyz 22.7

3 + x + 11 x π –2, x + 1 x+2

Solution:

ˆ Ê n Ê a + n b - 2 ˆ n a + n b -2 ˜ = lim Á Á1 + ˜ ˜ x Æ• Á Ë 2 ¯ ˜¯ ÁË 2

2

Ê 3 ˆ ˜ - ( x + 11) ÁË x + 1¯ = ˆ Ê 3 - x + 11˜ ( x + 2) ÁË ¯ x +1

1

-(2 + x3 + 13x 2 + 23x) = È 3 ˘ ( x + 1) 2 ( x + 2) Í - x + 11˙ x + 1 Î ˚

=

-( x + 2)( x 2 + 11x + 1) È 3 ˘ ( x + 1) 2 ( x + 2) Í - x + 11˙ ˚ Î x +1

=

x 2 + 11x + 1 x Æ-2 È 3 ˘ - x + 11˙ ( x + 1) 2 Í Î x +1 ˚ 4 - 22 + 1 -17 = = . -(-3 - 3) 6

Example 3

a,

(a) 3 ab a+ b

lim Bn (a, b) = 4 ab - ab = 3 ab . lim log tan 2 x (tan 2 2 x)

Example 4

x Æ0

1 2 Ans. (c) Solution:

log tan 2 x (tan 2 2 x) =

Example 5

sin 2 x

x 1 Ï 0, ÔÔ x, 0 £| x |< 1 f(x) = Ì Ô 1 . x = 1, -1 ÔÓ 2

x2n Æ

x = 1 and x = 2 x = 1 and x = 2 x = 1 but not at x = 2 x = 2 but not at x = 1

lim f ( x)

lim f ( x)

x Æ-1-

f

lim f(x

x

xÆ2

nƕ

f

If lim

(d) lim f(x) = 1/4.

(c) n

x

Ê x ˆ Ê Ê -3 ˆ -3 ˆ ÁË 3 + x ˜¯ = ÁË1 + 3 + x ˜¯ = ÁË1 + 3 + x ˜¯

x2

Ê -3 ˆ lim Á1 + xÆ• Ë 3 + x ˜¯

(d) n + 1/n

lim

and

xƕ

xÆ •

3+ x -3

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS nƕ

x x2n + 1

t Æ0

-3 ◊ x = –3 3+ x

1

Ê 1 ˆ 1 = lim f(x) = Á ˜ x Æ1 4 Ë 3 + 1¯

a = n + 1/n Example 45 (a)

If f (x) = x

lim f (x) = –1

xÆ0+

3 + x Ê -3 ˆ x -3 ÁË 3 + x ˜¯

= lim (1 + t)1/t = e

f(x) = e–3

fi (1) (n) ((a – n) n

Let f(x) = lim

x Æ1

Ans. (a) and (d)

Ê sin nx ˆ tan x ˆ Ê (n) Á ( a - n) n lim Á ˜ Ë x Æ 0 Ë nx ˜ x ¯ ¯

Example 43

x

(c) lim f(x) = e–5

Solution:



1 2

Solution:

xÆ0

Ans. (d)

x Æ1-

xƕ

xƕ

(( a - n) nx - tan x ) sin n x a n n +1

x

(b) lim f(x) = 2

x

nŒR

fi (a – n) n

f(1) =

xƕ

x Example 42

Æ

(a) lim f(x) = e–3

2

Ê 2ˆ +6Æ 1 and f (x¢n ) = Á1 + n ˜¯ Ë

2n

lim f ( x) = lim x = 1

x Æ1-

Ê x ˆ If f(x) = Á Ë 3 + x ˜¯

Example 44

1 2 and x¢n = 1 + xn Æ 1 f (2) = 5 ¥ xn = 1 + n n 1ˆ Ê n Æ • and f (x n ) = 5 Á1 + ˜ Æ nÆ• x¢n Æ Ë n¯

Ê 1ˆ ÁË ˜¯ x

x

x Æ1+

Ans. (d) Solution:

3 2

(b) lim f ( x) + lim f ( x) + f (1) =

x

lim (aeax – ex – 1)

\a

x Æ1-

e[ x] + x - 2 [ x] + x

IIT JEE eBooks: www.crackjee.xyz 22.17

(b) (c)

lim f (x

xÆ0-



Ï0 if |sin x| < 1 x)2n = Ì Ó1 if |sin x| = 1

f(x) = lim

nƕ

lim f (x

xÆ0+

f

(d) lim f (x xÆ0

Ans. (a), (b) and (d) Solution:

x

e

f (x) = x

lim f (x f (x) = x

lim

-1- x

-2 -1- x

e-1 - 2 ¥ -1

xÆ0-

x

p x Æ (2 k + 1) 2

ex - 2 = e x – 2. x

lim f(x) = 1 – 2 = – 1. If f

lim f (x

xÆ0

È1˘ lim x 8 Í 3 ˙ xÆ0 Îx ˚

x

pˆ Ê π 1 = f Á (2k + 1) ˜ Ë 2¯

f(x

Hence f(x p

Let f (x

Example 48

xÆ0+

Example 46

x x = (2k + 1) p 2 (k ΠI

x

Ï x + a 2 sin x ,0£ x£p /4 Ô , p /4< x£p /2 f(x) = Ì 2 x cot x + b Ôa cos 2 x - b sin x , p / 2 < x £ p Ó p (a) a = p/ 6 (b) b = – p/ 12 (c) a = p/ 6 and b – p/ 12 (d) a = p b – p/ 12

Ans. (a), (b), (c), (d) Solution: Ans. (b), (c) and (d)

f (p /4) =

x – 1 £ [x] £ x

Solution:

f (p / 4) =

xŒR o

lim =

x Æ p /4 +

=

1 È1˘ 1 – 1 £ Í 3˙£ 3 3 x Îx ˚ x fi

f (p /2) = lim f (x) = lim (a xÆ

xÆ0

È1˘ lim x 8 Í 3 ˙ xÆ0 Îx ˚ Example 47

p + 2



p + 2

x–b

x)

= –a–b b = – a Hence a = p /6 and b = – p /12.

Œ I Õ Q. If f(x) = lim

nƕ

x)2n

x=p x= p 2 x = –p 2 Ans. (a), (b), (c) and (d) Solution:

p +b 2

f (p / 2) = b and

x

lim (x5 – x8

xÆ0

lim (2 x cot x + b)

x Æ p /4 +

a – b = p/4

Ê 1 ˆ È1˘ x Á 3 - 1˜ £ x 5 Í 3 ˙ £ x5 Ëx ¯ Îx ˚ 8

But lim x5

p + a. 4

Ï0 if | x| < 1 lim x2n = Ì nÆ• Ó1 if | x| = 1

f

Example 49

Let f (x) = g ¢(x)

ea/ x - e- a/ x ea/ x + e- a/ x

g

a

lim f (x

xÆ0

(a) (b) (c) (d)

g (x g(x) = x g (x) = x 2 g(x) = x 3 h(x

Ans. (c) and (d)



h(x

IIT JEE eBooks: www.crackjee.xyz 22.18

Solution:

and lim

xÆ0-

lim

Æ +

ea/ x

e

a/ x

ea/ x

e

a/ x

ea/ x - e- a/ x ea/ x + e- a/ x

xÆ0+

= lim

xÆ0-

Hence lim f (x

1 - e- 2a / x

= lim

1 + e- 2a / x

e2a / x - 1 e2a / x + 1

=1

= –1

g(x) = x

xÆ0

g (x) = x h(x

Let f (x) = [x] + [–x

n and k ΠR ~ I (a) lim f(x

= lim

(b) lim f (x (c) f (d) f

x=n x=k

(a) a = c (c) a = 2

f (n) = [n] + [– n] = n – n

lim f (x) = lim [n – h] +[– n + h]= n – 1+ (– n)= – 1

xÆn-

a–b a = b = 6 and c

Example 53

Ans. (a), (b) and (d)

hÆ0 +

= lim x] and [– x

k.

xÆ0

(a) (b) (c) (d)

Let f (x) = lim

nƕ

f (x f (x f (x f(x

x 2n - 1 x 2n + 1

x

= lim

xÆ0

x x x

lim x2n = • 1- 1 nÆ•

x x

lim x

1+ 1

2n

xÆ0

x 2 n = 1. x

2n

x

nƕ

x 2n

= lim

x

nƕ

f (x) = lim

ae x - b cos x + ce - x x sin x

Ê ˆ x2 + c Á1 - x + + ˜ 2! Ë ¯ 3 Ê ˆ x xÁ x- + ˜ 3! Ë ¯ ( a - b + c) + ( a - c) x +

Ans. (a), (b) Solution: x

(b) c = 1 (d) b = 2

Ê ˆ Ê x2 ˆ x2 a Á1 + x + + ˜ - b Á1 - + ˜ 2! Ë ¯ Ë 2! ¯

xÆk

Example 51

xÆ0

ae x - b cos x + ce - x x sin x

Solution:

hÆ0 +

If k œ I Hence lim f (x

If lim

c

Ans. (a), (b), (d)

lim f (x) = lim [n + h] + [– n – h]= n +(– n – 1) = –1

xÆn+

ˆ 3 4 ˜¯ - 2 x + x

2 x5 3 - x6 2 +

and 1 – a

xÆk

- bx + cx 2 + x 3

( a - b) x + cx 2 + (1 - a 6) x 3 + ax 5 120 +

xÆ0

xÆn

Solution:

(b) c (d) c = 2

Ê ˆ x3 x5 + - ˜ aÁ x 6 120 Ë ¯ lim 2 3 xÆ0 Ê x x 2x2 Á x + 2 3 Ë

2

h(x

Example 50

2 x 2 log (1 + x ) - 2 x 3 + x 4

Solution:

lim f(x) = A and

xÆ0+

lim f(x) = – A. Hence lim f (x 3

a sin x - bx + cx 2 + x 3

Ans. (a), (b), (c)

If g (x) = Ax + B, A π xÆ0-

xÆ0

(a) a = b (c) a = b



xÆ0

If lim

Example 52

n

f(x) = –1. f (x

Ê a b cˆ x 2 Á + + ˜ + 0( x 3 ) Ë 2! 2! 2!¯ Ê ˆ x2 x4 + - ˜ x 2 Á1 3! 5! Ë ¯

a– b+c a–c 2 and if (a + b + c)/2 = 2, i.e. a + b + c a = 1, c = 1, and b = 2.

IIT JEE eBooks: www.crackjee.xyz 22.19

f (x

Example 54 [ x2 ] - 1

x 2 π 1,

x -1 2

( - 4)(8) 16 + 16 + 16 + 16 + 16

f(x) =

f (1) = f

Ê 2ˆ ÁË - ˜¯ 5

(a) lim f (x xÆ1+

Hence b

(b) lim f(x) = 1 xÆ1-

If f (x) =

Example 56

(c) lim f (x

a=b a sin 3x + A sin 2 x + B sin x x5

xÆ1-

xπ (a) f (c) A = – 4 Ans. (a), (c) Solution: Let g (x g h(x) = x5

(d) lim f(x xÆ1

Ans. (a), (c), (d) Solution:

x

lim f(x) = lim

[ x2 ] - 1 1-1 = lim 2 2 xÆ 1 + x -1 x -1

lim f (x) = lim

[ x2 ] - 1 x2 - 1

xÆ 1+

and

2 , [x 2 x x2 xÆ 1+

xÆ 1-

xÆ 1-

lim x 4

Ïae -1 Ô 1 | x + 2| Ô 2-e Ô If f (x) = Ìb Ô 4 Ôsin Ê x - 16 ˆ Ô ÁË x 5 + 32 ˜¯ Ó x b b=–

x Æ0

3 + 2A + B 3cos 3x + 2 A cos 2 x + B cos x

x Æ -2

x Æ -2

= – lim

x Æ0

a

x Æ -2 +

=

2 - e1 | x + 2|

= if f A = – 4 and B

Ê 1 ˆ ÁË | x + 2| Æ • as x Æ - 2 -˜¯

Example 57

x Æ -2

lim f (x) =

x Æ0

a e1 | x + 2| - 1

a e -1 | x + 2| - 1

lim

x Æ -2 +

lim

x Æ -2 +

2 e -1 | x + 2| - 1

27 cos 3x + 8 A cos 2 x + B cos x 120 x

27 + 8A + B

= –a

= lim -

20 x 3

x Æ0

x = -2 -2 < x < 0

9sin 3x + 4 A sin 2 x + B sin x

= – lim

= lim

Solution: lim - f (x) = lim -

5 x4

x Æ0

-3 < x < -2

x and

lim f (x

x Æ0

1 | x + 2|

x

5 x4

= lim

(a) a (c) a Ans. (b), (c), (d)

x+A h

x Æ0

0 -1 -1 = lim 2 = lim 2 xÆ 1- x - 1 xÆ 1- x - 1

Example 55

A=4 (d) B = – 5

3cos 3x + 2 A cos 2 x + B cos x

lim f (x) = lim

x Æ0

x

Ê x 4 - 16 ˆ Á 5 ˜ Ë x + 32 ¯ ( x - 2)( x 2 + 4) ( x 4 - 2 x 3 + 4 x 2 - 8 x + 16)

81sin 3x + 16 A sin 2 x + B sin x 120 x

81 ¥ 3 32 1 + A+ B =f 120 120 120

(a) A > 3 (d) A Ans. (a), (b), (d)

x f If A = lim

xÆ- 2

(b) A > 4

tan p x 1ˆ Ê + lim Á1 + 2 ˜ xÆ• Ë x+2 x ¯ (c) A

x

IIT JEE eBooks: www.crackjee.xyz 22.20

Solution:

lim

xÆ-2

= lim

yÆ0

and

f

tan p x tan p ( x + 2) = lim x + 2 xÆ-2 x+2

1ˆ Ê lim Á1 + 2 ˜ Ë x ¯

p tan y = p [y = p (x + 2)] y

x 2 (1/ x )

xƕ

1ˆ Ê = lim Á1 + 2 ˜ xÆ• Ë x ¯

x 2 lim

xƕ

1 x

e 4

9 p

Solution: x

= p + 1 > 4 and

lim

x Æ1

(log(1 + x ) - log 2)(3.4 x -1 - 3x ) {(7 + x )1/3 - (1 + 3x )1/2 }sin p x (log (2 + y ) - log 2)(3.4 x -1 - 3x )

= lim

yÆ0

Let

Example 58

Ï (1 - cos 4 x) tan x Ô Ô x3 Ô f(x) = Ì2a2 Ô x Ô Ô 16 + x - 4 Ó a

yˆ Ê log Á1 + ˜ .(3(4 y - 1) - 3 y ) Ë 2¯ = – lim 1/3 1/2 yÆ0 ÏÔÊ yˆ Ê 3 ˆ ¸Ô 2 ÌÁ1 + ˜ - Á1 + y˜ ˝ sin p y Ë Ë 4 ¯ Ô 8¯ ÓÔ ˛

if x < 0 if x = 0 if x > 0

= – lim

yÆ0

(b) 2

Solution: 1 - cos 4 x tan x 2sin 2 2x tan x = lim x Æ 0x Æ 0x x x2 x2

= -

lim f (x) = lim

=

and

log (1 + y / 2) ÔÏ 3(4 y - 1) Ô¸ - 3˝ ¥ .Ì y y ÓÔ ˛Ô 2

py y y 3 sin p y È 2 ˘ ÍÎ1 + 24 - 1 - 8 y + O ( y ) ˙˚

Ans. (b), (c)

x Æ 0-

{(7 + x )1/3 - (1 + 3x )1/2 }sin x [ y = x – 1]

f

(a) 4

x

Ans. (c), (d)

=e =1 tan p x 1ˆ Ê \A = lim + lim Á1 + 2 ˜ xÆ-2 x + 2 xÆ• Ë x ¯

(d) -

f

lim f (x) =

x Æ 0+

= =

lim

x Æ 0-

lim

x Æ 0+

lim

x Æ 0+

lim

x Æ 0+

tan x sin 2 2 x .4 2 x (2 x )

Example 60

x

x . 16 + x + 4 16 + x - 16

x (1 + a cos x ) - b sin x x3

xÆ0

)

x + 4 = 8,

lim

xÆ0

{(7 + x )1/3 - (1 + 3x )1/2 }sin p x

x (1 + a cos x ) - b sin x x3

Ê Ê ˆˆ x2 x4 + - ...˜ ˜ x Á1 + a Á1 2! 4! Ë ¯¯ Ë

Let (log(1 + x ) - log 2)(3.4 x -1 - 3x )

(b) a = – 5/2 (d) b = 3/2

Solution:

a = 8fia=±2

f (x) =

If lim

Ans. (b), (c)

2

Example 59

9 p

(a) b – a = 5/6 (c) a – b = – 1

16 + x - 4

( 16 +

1 p

,xπ1 = lim

xÆ0

Ê ˆ x3 x5 - bÁ x + - .....˜ 3! 5! Ë ¯ x3

4 e

IIT JEE eBooks: www.crackjee.xyz Limits and Continuity 22.21

È -a b ˘ x (1 + a - b) + x 3 Í + ˙ + 0 ( x 4 ) Î 2! 3!˚ = lim xÆ0 x3 This limit exists if 1 + a – b = 0 and is equal to 1 and if – a/2 + b/6 = 1. Solving these two equations, we get a = – 5/2 and b = – 3/2.

MATRIX-MATCH TYPE QUESTIONS Example 61

(a) (b) (c) (d) Ans.

Let f(x) = lim

xƕ

Column 1 f = 1 on f = –1 on f ≥ 0 on f(x) = sgn (| x| – 1) on p q r s a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution:

x 2n - 1 x 2n + 1 Column 2 (p) (1, •) (q) (2, •) (r) [2, •) (s) (– 1/2, 1/2)

(r) (1, 2)

(d) The function

(s) (4, 8)

f(x) = x - [ x ] is continuous on p q r s Ans. a p q r s b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: The function in (a) can be written as Ï-1 if x œ I f (x) = Ì Ó0 if x Œ I So f is discontinuous on I The function in (b) is continuous on R ~ {0}. The function in (c) can written as Ï 1 if sin x > 0 f (x) = Ì Ó-1 if sin x < 0

If |x| > 1, then lim x 2 n = •, so xÆ•

f (x) = lim

xƕ

If |x| < 1 then lim x xƕ

1 - x -2 n 1 + x -2 n 2n

So has discontinuities at x = kp, k ΠI. The function in (d) has discontinuous at x = 1, 4, 9 ..., n2, ...

=1

Example 63

= 0, therefore f(x) = – 1. If x = ± 1

2n

then x = 1 for any n, therefore f (x) = 0. Thus Ï1 if | x | > 1 i.e. x Œ( - •, - 1) » (1, •) Ô f (x) = Ì-1 if | x | < 1 i.e. - 1 < x < 1 Ô0 if x = ± 1 Ó or

(c) The function | sin x | is f(x) = sin x continuous on

f(x) = sgn (|x| – 1).

Example 62 Column 1 (a) The function f (x) = [x] + [– x] is continuous on (b) The function 1/ x ÔÏe , x π 0 f(x) = Ì is x=0 ÔÓ0, continuous on

Column 2 (p) (– 1, 0)

(q) (3, 4)

(a) (b) (c) (d) Ans.

Column 1 |sin x – cos x| |x| sin (1/x) ax2, 0 < a < 1 p q r

(p) (q) (r) (s) s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Column 2 no limit at x = 0 is always decreasing continuous everywhere is decreasing on (–p /4, p/4)

Solution: The function f (x) = |x| is continuous everywhere since sin (1/x) oscillates so lim sin(1/x) doesn’t exist. xÆ 0 Also |sin x – cos x| Ïsin x - cos x, sin x ≥ cos x = Ì Ócos x - sin x sin x < cos x

IIT JEE eBooks: www.crackjee.xyz 22.22 Comprehensive Mathematics—JEE Advanced

For x Π(p/4, p/4), cos x > sin x so f (x) = |sin x Рcos x| = cos x Рsin x = 2 cos (x + p/4) since 0 < x + p /4 < p/2 and cosine function decreases on (0, p /2) so f (x) decreases on (Рp /4, p /4). By the graph of a x 2, it is clear that it decreases on R for 0 < a < 1. Example 64 Column 1 (a) f (x) = x sgn (x Р1)

(

1 + tan -1 3x - 3 1 - sin -1 3x 1 - sin -1 2 x - 1 + tan -1 2 x

xÆ 0

Ê 1 -1 ˆ Ê 1 -1 ˆ ÁË1 + tan 3x + ....˜¯ - ÁË1 - sin 3x + ....˜¯ 3 3 = lim xÆ 0 Ê 1 -1 ˆ Ê 1 -1 ˆ ÁË1 - sin 2 x - ....˜¯ - ÁË1 + tan 2 x + ....˜¯ 2 2

Column 2 (p) lim f (x)

1 tan -1 3x 1 sin -1 3x 1 + + ( higher power of 3 3 x x x sin -1 3x and tan -1 3x )

xÆ1

2

sin sin (tan( x 2)) (b) f(x) =

3

lim

log cos 3x

)

doesn’t exist

= lim

xÆ 0

(q) lim f (x)

-

xÆ 0

doesn’t exist 3

-1

1 + tan 3x 3

(c) f (x) =

1 - sin -1 2 x -

(r) lim f (x) xÆ 0

= – 1/9

-1

1 + tan 2 x

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(s) lim f (x) = –1 xÆ 0

Solution: For function f in (a) Ï x, x > 1 Ô f (x) = Ì0, x = 1 so lim f (x) does not exist. xÆ1 Ô Ó- x, x < 1 lim

xÆ 0

sin (sin (tan ( x 2 2 ))) log cos 3 x

= lim

xÆ 0

sin (sin (tan ( x 2 2 ))) sin (tan x 2 2) ×

=

¥

sin (tan x 2 2) tan x 2 2

1 2x 1 lim = 2 xÆ 0 -3tan 3x 9

e1/ x - 1

lim

xÆ 0 -

e1/ x + 1

= – 1 and lim

xÆ 0 +

sin (sin (tan ( x 2 2 )))

×

sin (tan x 2 2) tan ( x 2 2) x2 2

×

x2 2 log cos 3x

e1/ x - 1 e1/ x + 1

= 1.

The equation have at least one root on

Example 65

e1/ x - 1 (d) f (x) = 1/ x e +1 Ans.

1+1 =–1 -1 - 1

=

1 - sin -1 3x

1 sin -1 2 x 1 tan -1 2 x 1 + ( higher powers of 2 2 x x x sin -1 2 x and tan -1 2 x )

the interval Column 1 (a) sin x – x + 1 = 0 (b) x2/4 – sin p x + 2/3 = 0 (c) (x 3/4) – sin p x + 2/3 (d) 2 x – 3x p q r s Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(p) (q) (r) (s)

Column 2 [– 2, 1/2] [0, 1] [0, 3p /2] [– 2, 2]

Solution: Let f (x) = sin x Рx + 1, f(0) = 1 > 0 and f(3p /2) = sin (3p/2) Р3p/2 + 1 = Р3p/2 < 0. Thus by Intermediate value theorem there is x Π(0, 3p/2) such that f (x) = 0. Let g(x) = x2/4 Рsin p x + 2/3, g (Р2) = 1 + sin 2p + 2/3 > 0 and g (1/2) = 1/16 Р1 + 2/3 < 0. Thus again applying Intermediate value theorem there is x Π(Р2, 1/2) such that g(x) = 0. Let f be as in (iii) f is continuous on [Р2, 2], f (Р2) = Р4/3 < 0, f (2) > 0 apply again intermediate value theorem on [Р2, 2]. If f is as in (iv) then f(0) = Р1 and f (1) = 1.

IIT JEE eBooks: www.crackjee.xyz Limits and Continuity 22.23

Example 66 Column 1 1 (a) f (x) = x-2 x - sin x x + sin x (c) f (x) = x sin p/x, f(0) = 0 (d) f(x) = tan–1 1/x (b) f (x) =

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

Column 2 (p) lim f(x) = 1 xÆ0

(q) f is continuous on R

Solution: If a Œ I then f (a) = [a – 1] + [1 – a] = a – 1 + 1 – a = 0. If x œ I then there is n Œ I such that n < x < n + 1 so f (x) = [x – 1] + [1 – x] = n – 1 + (– n) = – 1. Hence for any a Œ R lim f (x) = – 1. Thus f is continuous xÆa at a if a Œ R ~ I. lim

a[ x ]+ x - 1 a x -1 - 1 = lim = 1 – a–1 but x Æ 0x -1 [ x] + x

lim

a[ x ]+ x - 1 acot x - acos x = log a. Also lim x Æp 2 cot x - cos x [ x] + x

x Æ 0-

(r) lim f(x) = 0 xƕ

(s) lim f(x) does not

x Æ 0+

exist

Example 68 Extending the functions in col. 1 by continuity we have Column 1 Column 2 1 - cos (7( x - p )) (p) f (p) £ – 7/2 (a) f (x) = x -p 1 - cos (7( x - p ))

(b) f(x) =

( x - p )2

nƕ

Also lim x sin p /x = 0 = f (0)

sin x 1 - x2 p 2

(r) f (p) £ 0

(d) f (x) =

sin 7 x sin 2 x

(s) f (p) £ 49/2

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

xÆ 0

So f (x) = x sin p/x is continuous on R. tan–1 1/x Æ p /2 as x Æ 0+ whereas tan–1 1/x Æ – p /2 as x Æ 0–.

Ans.

Example 67 Column 1 Column 2 (a) Let f (x) = [x – 1] + (p) continuous at a [1 – x], [x] is the greatest integer function, a is an integer (b) Let f be as in (i) a is (q) lim f (x) does xÆ 0

not an integer

not exist

Solution:

1 - cos 7( x - p ) x -p

lim

x Æp

[ x ]+ x

(c) f (x) =

a -1 ,xπ0 [ x] + x

(r) f (a) = 0

(d) f(x) =

a -a cot x - cos x

(s)

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

= lim

x Æp

cos x

lim f (x) =

-2sin 2 ((7 2)(p - x ) ) 49 . (p - x ) = 0 4 ((7 2) (p - x ))2

So for the function f in (i) f (p) = 0

x Æp 2

log a

(q) f (p) £ p/2

(c) f (x) =

|sin x| £ 1 and 1/x Æ 0 as x Æ • so lim f (x) = 1.

Ans.

cot x - cos x

x Æp 2

f (x) =

cot x

) = log a

acos x acot x - cos x - 1

= lim

1 is trivially continuous except x-2 1 - (sin x ) x x - sin x at x = 2. Also f(x) = = . Since 1 + (sin x ) x x + sin x

Solution:

(

xÆ0

lim

x Æp

1 - cos 7( x - p ) ( x - p )2

= lim

x Æp

2sin 2 (7 2)(p - x ) 49 49 = 2 2 ((7 2) (p - x )) 4

So for the function f in (ii) f (p) = 49/2 lim

x Æp

sin x 1- x p 2

2

= lim

x Æp

sin (p - x ) p .p 2 = 2 (p - x )(p + x )

Thus for the function f in (iii) f (p) = p /2

IIT JEE eBooks: www.crackjee.xyz 22.24 Comprehensive Mathematics—JEE Advanced

lim

x Æp

sin 7 x sin 7(p - u) - sin 7u -7 = lim = lim = uÆ 0 sin 2(p - u ) uÆ 0 sin 2u sin 2 x 2

3+ 2 x

= lim

(1 + x -1 )(1 + 2 x -1 ) + 1

xƕ

Therefore, for the function f in (iv) f (p) = – 7/2

For f to be continuous f (0) is given by

Example 70 xƕ

Column 1 (a) f (x) =

3

( x + 1) - ( x - 1)

(

3

x +1 - x -1 3

(p) £ 1

2

3

)

(q) ≥ 3/2

(c) f (x) =

x2 - 2x - 1 - x2 - 7x + 3

(r) ≥ 0

(d) f (x) =

( x + 1)( x + 2) – x

(s) £ 5/2

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

lim

xƕ

3

=

lim x

xƕ

= lim

xƕ

lim

xƕ

(

xƕ

= lim

xƕ

lim

1 1 sin x tan x

(r) 4

(d) f(x) =

1 - cos3 x x sin 2 x

(s) 3/4

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

)

x + 1 - x - 1 = lim 3

2 1 + x -3 + 1 - x -3

xƕ

2 x3 2 x3 + 1 + x3 - 1

)

2

= lim

3 1 Ê sin x 2 ˆ (1 + cos2 x + cos x ) = Á ˜ 4 4Ë x 2 ¯ (sin 2 x ) 2 x

If a and b are positive and x Èb˘ b [x] denotes the greatest integer £ x, then lim = . x Æ0 + a Í Î x ˙˚ a {x} Statement-2 = 0, where {x} denotes fractional lim x Æ• x part of x. Example 71

5x - 4 x - 2x - 1 + x2 - 7x + 3 5-4 x 1- 2 x -1 x + 1- 7 x + 3 x

xƕ

xÆ 0

ASSERTION-REASON TYPE QUESTIONS

2

2

log (4 + x ) - log 4 log (1 + x 4) = lim = 1/4 xÆ 0 x 4 x

(1 - cos x )(1 + cos2 x + cos x ) 1 - cos3 x = lim xÆ 0 x sin 2 x 2 x 2 (sin 2 x ) 2 x

xÆ 0

=1

( x + 1)( x + 2) - x = lim

xÆ 0

log (1 + 4 x ) log (1 + 4 x ) = lim .4 xÆ 0 4x x

1 ˆ 1 - cos x Ê 1 lim =0 f (0) = lim Á ˜ = xÆ xÆ 0 Ë sin x 0 tan x ¯ sin x lim

3

f (0) = lim

xÆ 0

x2 - 2x - 1 - x2 - 7x + 3

= lim

xƕ

(

(c) f (x) =

f (0) = lim

0 =0 1+1+1 32

(q) 0

Solution: =4

(1 + x -1 ) 4 3 + (1 - x -1 ) 4 3 + (1 - x -2 )2 3

xƕ

(p) 1/4

log (4 + x ) - log 4 x

( x + 1)2 - 3 ( x - 1)2

4 x -1 3

Column 2

(b) f (x) =

Ans.

( x + 1)2 - ( x - 1)2 = lim xƕ ( x + 1) 4 3 + ( x - 1) 4 3 + ( x 2 - 1)2 3 = lim

Column 1 log (1 + 4 x ) (a) f (x) x

Column 2 2

(b) f (x) = x3/2

Solution:

3 2

Match the lim f (x) in col. 1 with col. 2

Example 69

Ans.

=

2

=

5 2

3x + 2 ( x + 1)( x + 2) + x

Statement-1

Ans. (a) Solution:

lim

x Æ0

x Èb˘ x Ê b Ï b ¸ˆ = lim Á - Ì ˝˜ Í ˙ x Æ 0 + a Îx˚ a Ë x Ó x ˛¯

IIT JEE eBooks: www.crackjee.xyz Limits and Continuity 22.25

Ê b Ê b ˆ {b / x} ˆ = lim Á - Á ˜ x Æ 0 + Ë x Ë a ¯ {b / x} ˜ ¯ = = Since, 0 £ {x} < 1 so Hence

lim

x Æ•

= lim f ( x ) x Æ 1-

Hence there is no break point at x = 1 but there is a break point at x = – 1.

b b { y} - lim a a x Æ• y

Example 73

b a

lim

{x} 1 £ for x > 0. x x

Statement-2 degree, then

= 100

If p(x) and q(x) are polynomials of same

lim

Ans. (a) Solution:

Let p(x) = a0 x n + a1xn– 1 + ... + a n and q(x) = b0 x n + b1xn– 1 + ... + b n then

a + a / x + a2 / x 2 + ... + an / x n p( x ) = lim 0 1 x Æ • q( x ) x Æ • b + b / x + b / x 2 + ... + b / x n 0 1 2 n lim

=

a0 b0

(x + 1)10 + ... + (x + 100)10 is a polynomial of degree 10 x10 + 910 is a polynomial

f is a discontinuous function f has a break point at x = 1.

follows from statement-2.

Ans. (c) Solution:

x10 + 910

p( x ) leading coefficients of p(x ) = x Æ • q( x ) leading coefficients of q(x )

{x} =0 x

Ï x 2 - 1, x £ - 1 Ô Ô p - cos -1 x ÔK , -1 < x < 0 Ô x +1 = Ì Ô 1 - x2 , 0 £ x < 1 Ô 1 1 Ô , x > 1, ÔÓ sin( x - 1) tan( x - 1) Statement-1 Statement-2

( x + 1)10 + ( x + 2)10 + .... + ( x + 100)10

xƕ

For K > 0, let f (x) be given by

Example 72

Statement-1

lim

x Æ -1 +

p - cos -1 x f (x) = K lim x Æ -1 + x +1 = K lim yÆp

= K lim yÆp

= K

p- y 1 + cos y

Example 74

(cos -1 x = y )

1 p-y 2 cos y / 2

1 p +

2 p /2- y/2 1 lim ¥ p+ 2 y Æ p sin(p / 2 - y / 2)

y

K π 0 = f ( -1) 2p So f is discontinuous at x = – 1 =

Ê ˆ 1 1 lim f ( x ) = lim Á x Æ1 + x Æ 1 Ë sin( x - 1) tan( x - 1) ¯˜ 1 ˆ Ê 1 = lim Á ˜ u Æ 0 Ë sin u tan u ¯ 1 - cos u = 0 = f (1) u Æ 0 sin u

= lim

then | f | is continuous at x = 0 Statement-2 If f is continuous then | f | is continuous Ans. (a) Solution:

y

Statement-1: If |f (x)| £ |x| for all x Œ R

Put x = 0 in | f (x)| £ |x| so | f (0) £ 0

Thus |f (0)| = 0 fi f(0) = 0. We can write | f (x)| £ |x| as | f (x) – f(0)| £ |x – 0| Thus f is continuous at 0. So applying statement-2 | f | is continuous at x = 0. Of course, if f is continuous then | f | is continuous function. Example 75

1+ x x is said to nƕ

convergent. The following are well known (i) lim 1/n p = 0 ( p > 0) nƕ

(ii) lim x n = 0 (| x| < 1) nƕ

Æ•

nƕ

Example 76

(c) lim g o f (1/n) = 0 nƕ

[x] + lim

= 1 + 1 + 1 = 3. So f is continuous at x = p/2. Thus f is continuous on R ~ {1}. Also [x] is not continuous at every x ΠI.

(iii) If lim s n = l, then lim

Let f (x) = – 1 + |x – 1|, g(x) = 2 – |x + 1| then

s1 + s2 + ...... + sn =l n

Which of the following sequences do not

(d) lim g o f (1/n) = – 1 nÆ•

nx

Ê a 1/ x + a21/ x + ...... + an1/ x ˆ Example 79 If f (n) = lim Á 1 ˜ xÆ•Ë n ¯ then f (n) is equal to (b) a1 + a2 + ... + an (a) log a1, a2 ... an (c) n (d) a1 a2 ... an Ans. 76. (c), 77. (b), 78. (b), 79. (d) Solution: 76– 79 1

Æ 0 as n Æ •. 3 a + a + .... + an Hence lim 1 2 = 0. Clearly 1/n! Æ 0 as n Æ •. nÆ• n 1 (1 + 1/n)n Æ e as n Æ •. n + 1 - n = Æ n +1+ n 0 as n Æ •.

For 76. Let an =

n-1

For 77. Let x n = 1/n, xn¢ =

2 . lim xn = 0 = lim xn¢ nÆ• 2n + 1 nÆ•

g(x n ) = < n>* = 0, g(xn) =

2n + 1 2

converge to zero. (a)

1È 1 1 1 ˘ 1 + + + ..... + n-1 ˙ n ÍÎ 3 32 3 ˚

(b)

1 n!

(c) (1 + 1/n) (d)

n

n +1- n

Example 77 < x>*

Ï< 1 / x > * x π 0 , where Let g(x) = Ì x=0 Ó0

= < n + 1/2 >* = 1/2

Hence g is not continuous at O. *

n +1 Ê n ˆ Ê n ˆ = = * = 1/n so lim g Á gÁ ˜ ˜ =0 Ë n + 1¯ nÆ• Ë n + 1¯ n Ï-( x + 3) , x < -2 Ô x + 1 , - 2 £ x £ -1 Ô For 78. f o g(x) = Ì Ô-1 - x , - 1 £ x £ 0 ÔÓ x - 1 , x>0 and

is the distance from x to the integer nearest to x then (a) g is continuous at x = 0 (b) g is not continuous at x = 0 Ê 2 ˆ =2 (c) lim g Á Ë 2n + 1˜¯ nÆ•

*

Ï1 + x , x < -1 Ô g o f (x) = Ì1 - x , - 1 £ x < 1 Ô3 - x , x ≥ 1 Ó f o g(1/n) = 1/n – 1 Æ – 1 as n Æ • and g o f (1/n) = 1 – 1/n Æ 1 as n Æ •

IIT JEE eBooks: www.crackjee.xyz Limits and Continuity 22.27

For 80. Clearly f (0) = 0, let I = (0, •). Since f is continuous on I, and f(x) π 0 for any x Œ I, so f (x) = x for all x Œ I or f(x) = – x for all x Œ I. Similarly, if J = (– •, 0) f(x) = x for all x Œ J and f(x) = – x. Combining these possibilities, we have f(x) = x for all x Œ R; f(x) = – x for all x Œ R; f(x) = |x|; f (x) = – |x|.

For 79. 1/ x 1/ x ÔÏ a + ........ + an Ô¸ lim Ì 1 ˝ xÆ• Ô n Ô˛ Ó

ÏÔ a y + a2y ........ + any = lim Ì 1 yÆ 0 Ô n Ó n Ï Ê Â aiy Ô Á Ô i =1 = lim Ì1 + Á ( -1) + yÆ 0 n Á Ô Á ÔÓ Ë

= e

log a1 + log a2 +.......+ log an

ˆ¸ ˜ ÔÔ ˜˝ ˜Ô ˜¯ Ô ˛

= e

nx

¸Ô ˝ ˛Ô

n/y

- n + a1y

)

(

(

)

a1 y -1 +......+ an y -1 n ¥ y y +......+ an

log a1 a2 .......an

= a1 a2 ... an

Paragraph for Questions Nos. 80 to 83 Among various properties of continuous, we have if f is continuous function on [a, b] and f (a) f(b) < 0, then there exists a point c in (a, b) such that f (x) = 0 equivalently if f is continuous on [a, b] and x ΠR is such that f (a) < x < f (b) then there is c Π(a, b) such that x = f (c). It follows from the above result that the image of a closed interval under a continuous function is a closed interval. Example 80

The number of continuous functions on R

which satisfy ( f (x))2 = x 2 for all x ΠR is (a) 1 Example 81

(b) 2

(c) 4

(d) 8

Suppose that f (1/2) = 1 and f is continuous

on [0, 1] assuming only rational value in the entire interval. The number of such function is

Example 82

Let f be a continuous function on [–1, 1]

satisfying ( f(x)) + x = 1 for all x Œ [–1, 1] 2

2

The number of such functions is (a) 2 (b) 1 Example 83 Let f (x) = x, x π 0 and f (0) = 1. Then (a) f is a continuous function (b) Range of f is an interval (c) Range of f is R (d) hypothesis of the result in the comprehension is violated Ans. 80. (c), 81. (d), 82. (a), 83. (d) Solution: 80–83

For 81. We claim that f (x) = 1 for all x Œ [0, 1]. Let if possible there is c Œ [0, 1] such that f (c) π 1 so f(c) < 1 or f(c) > 0. Suppose that c < 1/2. Consider the interval [c, 1/2], by above result (comprehension) f takes all values between f(c) and 1 but f takes only rational values. This contradiction shows there is no c Œ[0, 1] such that f(c) π 1. For 82. f (x) = 1 - x 2 , or f(x) = - 1 - x 2 satisfy the given condition. Apply the result (comprehension) to show that these are the only possibilities. For 83. f is not continuous at x = 0. Range of f is not an interval for 1, – 1 Œ Range f but 0 Œ Range f.

Paragraph for Questions Nos. 84 to 87 x n as x Æ 0. This follows from the Binomial theorem for rational indices. Taking advantage of this fact, we can approximately calculate the roots of the numbers.

The function n 1 + x - 1 and

Example 84 The value of (a) 10.2 (c) 10.12

3

1047 is (b) 10.16 (d) 10.1

Example 85 The value of (a) 20.2 (c) 20.12

3

8144 is (b) 20.16 (d) 20.4

Example 86 (a) 1.02 (c) 1.06

5

Example 87 (a) 4.02 (c) 4.08

5

1.1 is equal to (b) 1.04 (d) 1.03 1080 is equal to (b) 4.04 (d) 4.07

Ans. 84. (b), 85. (c), 86. (a), 87. (b) Solution: 84-87 For Q.84

3

1047 = 3 1000 + 47 = 10

3 1+

47 47 ˆ Ê = 10 Á1 + ˜ Ë 1000 3000 ¯

= 10 (1 + .016) = 10.16

IIT JEE eBooks: www.crackjee.xyz 22.28 Comprehensive Mathematics—JEE Advanced

For Q.85 write 8144 = 8000 + 144

\

For Q.86 1.1 = 1 + .1 For Q.87 1080 = 1024 + 56

lim

xÆ0

| x| x4 + 4 x2 + 7

(

)

For Q.89, Since –1 £ sin 1 / 3 x £ 1, so – x 4 £ x4

Paragraph for Questions Nos. 88 to 91

(

)

sin 1 / 3 x £ x 4. But lim x4 = 0, xÆ0

If f, g and h are functions having a common domain D and h (x) £ f (x) £ g(x), x Œ D and if lim h(x) = lim g (x) =

Hence 0 £ x4 sin 1/3 x £ 0,

l then lim f(x) = l. This is known as Sandwich Theorem.

\ lim x4 sin 1/3 x = 0.

xÆa

xÆa

xÆa

Using this result, compute the following limits. xÆ0

x + 4 x2 + 7 (b) 0 (d) doesn’t exist

(a) 1 (c) 1/2 4

(

)

xÆ0

(a) 0 (c) 1/3

(b) 1 (d) does not exist Let f (x) = x 2

Example 90 f (0) = 1 then

xÆ0

For Q.90, 0 £ x2

-2/ x e1/ x - e -1/ x 2 1- e = x £ x2 for x > 0 e1/ x + e -1/ x 1 + e -2/ x

4

lim x sin 1 / 3 x is

Example 89

(a)

| x|

The value of lim

Example 88

=0

e1/ x - e -1/ x , x π 0 and e1/ x + e -1/ x

So lim f (x) = 0. Also lim f(x) = lim x 2 x Æ 0+

x Æ 0-

x Æ 0-

e 2/ x - 1 =0 e 2/ x + 1

Hence lim f (x) = 0 xÆ0

Q.91, since x – 1 £ [x] £ x for all x Œ R 1 È1˘ 1 -1£ Í 3 ˙ £ 3 3 x Îx ˚ x fi

Ê 1 ˆ x 5 Á 3 - 1˜ £ x5 [1/x3] £ x2 for x > 0 Ëx ¯

and x2 £ x5[1/x3] £ x5((1/x3) – 1) for x < 0

lim f (x) doesn’t exist

x Æ 0+

so lim x5[1/x3] = 0 xÆ0

(b) lim f (x) doesn’t exist xÆ0

INTEGER-ANSWER TYPE QUESTIONS

(c) lim f(x) exist Æ

(d) f is a continuous function

Example 92

Let f (x) = x [1/x ] x π 0 and f (0) = 0 5

Example 91

3

512

(a) lim f (x) doesn’t exist

f(x) =

xÆ0

(b) f is not continuous at x = 0 (c) lim f (x) = 1 xÆ0

(d) lim f (x) = 0

(

The value of x +4 -2

sin 2 x

For Q.88, since 0£

) is continuous to

x +4 -2

xÆ0

Ans. 88. (b), 89. (a), 90. (c), 91. (d); = 512 lim

Solution: 88–91

xÆ0

x 4 + 4 x 2 + 7 ≥ 1 , so | x|

x4 + 4 x2 + 7

Hence 0 £ lim

xÆ0

| x| x4 + 4 x2 + 7

sin 2 x -4 + ( x + 4) 1 ◊ sin 2 x x +4 +2

= 512

£ |x|. But lim |x| = 0, xÆ0

£ lim |x| = 0. xÆ0

f(0) for which

Ans. 2 Solution: f (0) = 512 lim

xÆ0

1 32

1 2x 1 ◊ lim 2 x Æ 0 sin 2 x 2 + x + 4

1 1 = 512 (1) = 64 2 2+ 4

IIT JEE eBooks: www.crackjee.xyz Limits and Continuity 22.29

lim (p) (1 – x) tan

Example 93

xÆ1

px is equal to. 2

Ans. 2

= p lim

xÆ1

=

Example 94

1- x Êp ˆ cos Á (1 - x )˜ Ë ¯ p 2 Ê ˆ sin Á (1 - x )˜ Ë2 ¯

2 y È p ˘ p lim cos y = 2 Í y = (1 - x )˙ 2 p yÆ 0 sin y Î ˚ lim

- log (1 + 2h) + 2 log (1 + h) h2

hÆ0

is equal to

Ans. 1 Solution: The required limit is equal to -1 h2

Let f be a continuous function on R

= – lim

hÆ0

Example 95

h2 (1 + h)2

1- tan x

¥ -

=

1 (1 + h)2

= 1.

Solution: Since f(1) is a rational number and f is a continuous function, f takes all values between f(1) and f(2) = 10. When f(1) π irrational numbers between f(1) and 10, so the only possibility is that f is a constant function. In this case f(1.5) = f(2) = 10. Given the function f(x) = lim n(x1/n – nÆ•

7 f (1/x) + k f (x) = 0 then k is

nƕ

= e2

.

2 tan x x (1- tan x )

tan x Ê ˆ = 1, lim (1 + x )1/ x = e˜ ÁË xlim ¯ Æ0 x xÆ0

Example 99

ABC is an isosceles triangle inscribed in

a circle of radius r. If AB = AC and h is the altitude from A to BC. If the triangle ABC has perimeter P and area D then D lim 512 r 3 is equal to hÆ 0 P Ans. 4 Solution: We have BC = 2BD, AD = h and OD = h – r. \BC = 2 r 2 - ( h - r )2 = 2 2hr - h2 fi AB =

2hr - h2 + h2 = 2hr

so that P = 2AB + BC 2 ÈÍ 2hr - h2 + 2hr ˘˙ Î ˚ Also the area of DABC is =

Ans. 7 f (x) = lim

2 tan x ˆ 2 tan x Ê lim Á1 + ˜ xÆ 0 Ë 1 - tan x ¯

1/ x

Thus the reqd. limit is 1.

198

Solution:

xÆ 0

2 tan x ˆ Ê lim tan (p/4 + x)1/x = lim Á1 + ˜ xÆ 0 xÆ 0 Ë 1 - tan x ¯

Let f(x

1), x > 0. Suppose f equal to

The value of (e–2 ) lim tan (p/4 + x)1/x is

Solution: We have

for 1 £ x £ 3. If f(x) takes rational values for all x and f(2) = 198, then f(1.5) =

Example 96

Solution: Putting x = y = 0, we obtain f (0) = ( f (0))2 so f(0) = 0 or f (0) = 1. If f(0) = 0 then f(1) = f(1 + 0) = f (1) f(0) = 0 not true. Hence f(0) = 1. Replacing y by x/2 and x by x/2, we have f (x) = ( f (x/2))2 > 0 for all x. (1 = f (0) = f (x) f (–x) so f (x) π 0 for all x g(x) = log f (x) so g (x + y) = g(x) + g ( y) and g is continuous so g(x) = g(1)x = (log f (1)) x fi f(x) = ekx = ax where a = ek = elog f (1) = f (1) = 4. Thus f(x) = 4x fi f (3) = 43 = 64.

Ans. 1

[log (1 + 2h) + log(1 + h)2]

Ê h2 ˆ log Á1 (1 + h)2 ˜¯ Ë

Ans. 4

Example 98

1 + 2h 1 = – lim 2 log hÆ0 h (1 + h)2

Ans.

Substituting in the given equation, we obtain k = 7 satisfying f (x + y) = f (x) f (y) for all x, y ΠR and f (1) = 4 1 f (3) is equal to then 16

Ê p p xˆ Reqd. limit = p lim (1 – x) cot Á ˜ Ë2 xÆ1 2 ¯

lim

f(1/x) = log(1/x) = –log (x) = –f(x)

Example 97

Solution:

hÆ0

Hence

x1/ n - 1 xu - 1 = lim = log x uÆ0 1/ n u

D = BD ¥ AD = h 2hr - h2 .

IIT JEE eBooks: www.crackjee.xyz 22.30 Comprehensive Mathematics—JEE Advanced

Solution:

Reqd. limit is equal to tan x( 3 - tan x )(tan x + 3) Ê1 ˆ ÁË cos x 3 - tan x ˜¯ 2

= lim

(

x Æ p /3

2 tan x(tan x + 3) cos x

= lim

x Æ p /3

Fig. 22.3

2 =

D \ 3 = P

8

= 8 fi

(

h 2hr - h2

(

2hr - h2 + 2hr 2r - h 2r - h + 2r

)

)

3

x Æp

)

3

Solution:

=4

r =1

kp for 2

k

any k ≥ 1 and A = Â ( b r ) r , then

xÆ A

- (1 - 2 x )

1/4

x + x2

2 lim

25 2 - ( (sin x + cos x )2 )5 2 1 - sin 2 x

=

2 lim

25 2 - (1 + sin 2 x )5 2 2 - (1 + sin 2 x )

=

2 lim

=

5 2 .23 2 2

is equal to

x Æp 4

x Æp 4

k

r =1

cos–1b r = p /2 ¤ b r = 0. k

A = Â ( br )r = 0

uÆ 2

= lim

(1 + x 2 / 3 + 0( x 4 )) - (1 - x / 2 + 0( x 2 )) x (1 + x )

= lim

1 2 + x 3 + 0 ( x2 ) = 1/2. 1+ x

xÆ 0

xÆ 0

Thus the reqd. limit is 2. Example 101 Ans. 4

1 3 tan x - tan3 x lim is equal to x Æ p /3 6 cos ( x + p 6 )

(u = 1 + sin 2x)

Example 103 tan x tan( x + h) tan( x + 2h) D(x) = tan( x + 2h) tan x tan( x + h) tan( x + h) tan( x + 2h) tan x

r =1

(1 + x 2 )1/3 - (1 - 2 x )1/4 Thus lim xÆ 0 x + x2

25 2 - u 5 2 2-u

= 10

Since maximum value of cos–1 x in [0, 1] =

p /2, Â cos -1 b r = kp/2 is possible if and only if each

x Æp 4

4 2 - (cos x + sin x )5 1 - sin 2 x

=

Ans. 2 Solution:

1 8 - 2 (cos x + sin x )5 45 1 - sin 2 x

2 lim

Reqd. limit =

r =1

(1 + x ) lim 4.

= 24

12

Ans. 2

k

2 1/3

)

Example 102 The value of lim

Let 0 £ b r £ 1 and  cos -1 b r =

Example 100

3+ 3

is

(

hÆ0

( 3 )(

3

D 2r = 512 r 3 P 8 2 2r

lim 512 r

)

Find the value of lim

hÆ 0

3 D (p 3) 24h2

Ans. 6 Solution: Using C3 Æ C3 – C2 and C2 Æ C2 – C1, we can write tan x [tan( x + h) - tan x ] / h D = tan( x + 2h) [tan x - tan( x + 2h)] / h h2 tan( x + h) [tan( x + 2h) - tan( x + h)] / h [tan( x + 2h) - tan( x + h) / h [tan( x + h) - tan x ] / h [tan x - tan( x + 2h)] / h

IIT JEE eBooks: www.crackjee.xyz Limits and Continuity 22.31

Ans. 2

But

tan( x + h) - tan x lim hÆ 0 h

sin x

Ê sin x ˆ x -sin x lim Á ˜ xÆ 0 Ë x ¯

Solution:

1 sin ( x + h) cos x - cos( x + h)sin x = lim hÆ 0 h cos( x + h) cos x sin h 1 = sec2x h cos( x + h) cos x

= lim

hÆ 0

or

d (tan x) = sec2 x) dx

(use

hÆ 0

1

hÆ 0

D ( x) h2

sec2 x

tan x

sec2 x

= tan x -2sec2 x

sec2 x

tan x

sec2 x

-2sec2 x

tan x

sec2 x

sec2 x

0

-3sec x

0

0

0

-3sec2 x

=

2

ÈUsing R2 Æ R2 - R1 ˘ = 9 tan x sec4 x Í ˙ Îand R3 Æ R3 - R1 ˚ 3 D (p 3)

So lim

24h2

hÆ 0

x Æ tan 3

9.24 =6 24

1 tan6 x - 2 tan5 x - 3tan 4 x is 81 tan 2 x - 4 tan x + 3

xÆ1

–1

EXERCISE LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS

(tan x - 2 tan x - 3) 2

lim

x Æ tan -1 3

= = Example 105

tan 4 x

lim

Æ tan 3

lim

x Æ tan -1 3

tan x - 4 tan x + 3 2

(tan x - 3)(tan x + 1) tan x (tan x - 3)(tan x - 1) 4

tan 4 x

(tan x + 1) 3 +1 = 34 = 162 tan x - 1 3 -1

The value of

sin x È 1 ˘ - sin x x sin x Ê ˆ Í 1 + lim x x ˙ e lim Á Í x Æ 0 Ë x ˜¯ ˙ x Æ1 ÍÎ ˙˚

x2 + 1 - 1

1. The value of lim

xÆ0

(a) 4

x2 + 9 - 3

(b) 3

is

(c) 1

(d) 2

x-a - b-a

2. The value of lim

x 2 - b2

xÆb

(b > a) is

(a)

1 4b

(b)

1 b b-a

(c)

1 2b b - a

(d)

1 4b b - a

(

2

log e x + 2 x 3. If f(x) =

Ans. 2 Solution:

xÆ1

The value

Example 104 lim-1

=

1

lim x 1- x = lim [1 + ( x - 1) ]1- x = e–1 So their sum = 2e

tan x - tan ( x + 2h) = –2 sec2 x h

Thus, lim

sin x x

= e–1

tan( x + 2h) - tan( x + h) Similarly, lim = sec2 x hÆ 0 h and lim

1 È ˘ Ê Ê sin x ˆ ˆ sin x x -1 ˙ Í - 1˜ ˜ = lim Á1 + Á ˙ Ë x ¯¯ xÆ 0 ÍË ÍÎ ˙˚

tan x

), x

0. The value of

lim f(0) is Æ

1 2

(b)

2

(c) 2

(d)

1 2

(a)

4. The value of lim

( x + 1)20 + ( x + 2)20 + ... + ( x + 100)20

xƕ

(a) 100 (c) 10

x 20 + 1020 (b) 1 (d) 20

is

IIT JEE eBooks: www.crackjee.xyz 22.32 Comprehensive Mathematics—JEE Advanced x

Ê x 2 + 4 x + 3ˆ 5. If f(x) = Á 2 f(x) is ˜ , then xlim Æ• Ë x +x+2¯ (a) e3

(b) e4

6. Let f(x) =

( tan (p

(c) e2

(d) 24

4 - x ) cot 2 x ) ( x π p 4 ) . The

value which should be assigned to f at x = p 4 , so that it is continuous everywhere, is (a) 1/2 (b) 1 (c) 2 (d) 1/4 x tan 2 x - 2 x tan x , x π 0. Then the value 7. Let f(x) = (1 - cos 2 x )2 f(0) so that f is continuous (a) 2 (b) – 2 (c) 1/2 (d) – 1/2 8. lim

1 + log x - x

x, y ΠR and f (2) = 5 then lim f (x) is (b) 17

(c) – 5

1 30 1 (c) 60 11. Let f be a continuous f (x + y) = f(x) + f(y)

(b) 0

1 - cos (1 - cos x )

x4 is continuous everywhere is (a) 1/8

function on satisfying for all x, y ΠR and

xÆ0+

f(x) =

1 + sin x - cos x 1 - sin x - cos x

x = 0. The value of f (0) so that f (x) is continuous at x = 0, is (a) 1 (b) –1 (c) 0 (d) none of these

(d)

(a) 1

lim f ( x ) = 1

xÆ0+

Ï x 2 + 2 cos x – 2 ,x0 Ô 6x2 Ó

)

cos x - sin x cos 2 x

x = p 4 . The value of f (p 4 ) so that f(x) is continuous everywhere, is (b) –1

2

(c)

(

(d) 1

19. The value of f(0), for f(x) = 1 + tan 2

x

2

)

1/ 2 x

, so

that f(x) is continuous everywhere, is (a) e (b) 1/2 (c) e1/2 (d) 0 (1 + x )1/ 4 - (1 - x )1/ 4 is xÆ0 x (b) 0 (c) –1 (d) –1/2

20. The value of lim

xÆ0+

(

(d) none of these

log(1 + ax ) - log (1 - bx ) x x = 0. The value which should be assigned to f at x = 0 so that it is continuous there, is (a) a – b (b) a + b (c) log a + log b (d) none of these

(d) none of these

(a) 4 (b) 80 (c) 0 (d) none of these 12. Let f (x) = e x sgn (x + [x]), where sgn is the signum function and [x] is the greatest integer function. Then (b) lim f ( x ) = – 1 (a) lim f ( x ) = 0 xÆ0+

(c) 1/4

16. The function

f(x) =

xÆ4

lim f ( x ) = 1

(b) 1/2

18. The function

f(1) = 5 then lim f (x) is equal to

(c)

f (x) =

(d) 21

Ê 2 + cos x 3 ˆ 10. The value of lim Á 3 – 4 ˜ is x Æ 0 Ë x sin x x ¯ (a)

xƕ

(a) –1 (b) 1 (c) 0 (d) none of these 15. The value of f(0) so that the function

17. The function

xÆ4

(a) 5

1 1 (d) 12 6 1/x 14. If f(x) = x(e – 1), then lim f (x) is (c)

f(x) =

equals

1 - 2x + x2 (a) 1 (b) 0 (c) –1 (d) – ½ 9. Let f be a continuous function satisfying f (x) f(y) = f (x) + f(y) + f (x y) – 2 for all xÆ1

The value of f(0), so that f is continuous is 1 1 (a) (b) 5 3

(a) 1/2

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 21. The function y = 1 + log [x] is ([x] is the greatest integer function)

IIT JEE eBooks: www.crackjee.xyz Limits and Continuity 22.33

(a) continuous nowhere (b) continuous everywhere

22. The function f (x) = [sin x] is not continuous at ([x] is the greatest integer less than or equal to x) (a) x = p /2 (b) x = p/4 (c) x = 0 (d) x = p 23. If a lim x1/(1 – x) + b = e– 1 (a ≥ 0, b ≥ 0) then xÆ2

(b) a = 1, b = e – 1 (d) none of these

(a) a = 1, b = 0 (c) a = 2, b = e – 1

24. The set of all points of discontinuities of the function tan x log x f (x) = contains 1 - cos 4 x (a) {np /2 : n Œ I} (b) {n p /2, n Œ Q} (c) (– •,0) » {np/2, n Œ N} (d) (– •,0] » {np/2, n Œ N} 25. Let f (x) = (a) (b) (c) (d)

f f f f

is is is is

(

log 1 + x

2

)

x - 26 x + 25 4

continuous continuous continuous continuous

2

. Then

on [6, 10] on [– 2, 2] on [– 6, 6] [2, 4]

a – a2 – x 2 – 26. Let L = lim

x Æ0

x

4

30. Which of the following functions are continuous on (0, 1) 1 (a) x – [x] (b) x - [ x] (c) (–1) [x]

(d) sin [x]

MATRIX-MATCH TYPE QUESTIONS 31. lim f (x), where f(x) is as in column 1 and [x] is the xÆ 0

greatest integer function is Column 1 tan[e 2 ] x 2 - tan[-e 2 ] x 2 (a) f (x) = sin 2 x (b) f (x) =

3

(c) f (x) =

x 4 ,a>0

(b) a = 1

1 (c) L = 64

1 (d) L = 32

Ï1 2 Ô 5 (2 x + 3) for - • < x £ 1 Ô 27. If f (x) = Ì6 – 5 x for 1 < x < 3 Ôx – 3 for 3 £ x < • Ô Ó then f is (a) continuous at x = 1 (b) discontinuous at x = 1 (c) continuous at x = 3 (d) discontinuous at x = 3 (1/2) - cos2 x is continuous at (b) x = 3p/4 (d) x = 7p /4

(d) f(x) = 32. Let f (x) =

Column 2 (p)

[5 2 + tan x + tan 2 x ] - [5 2] tan x

2 8

(q) 15

1 + x2 - 4 1 - 2 x

(r) 0

x + x2

2

If L (a) a = 2

28. The function at y = (a) x = p /4 (c) x = 5p /4

ÏÔ x 2 if x is rational 29. If f (x) = Ì 2 ÔÓ- x if x is irrational then f is (a) continuous at x = 0 (b) discontinuous at x = 1/2 (c) discontinuous at x = 0 (d) continuous at x = 1/2

2 - 1 + cos x

(s) 1/2

sin 2 x a0 x m + a1 x m+1 + ..... + ak x m+ k b0 x n + b1 x n+1+ ..... + bl x n+ l

where a0 π 0, b0 π 0 then lim f(x) is equal xÆ 0

Column 1 (a) m > n (b) m = n (c) m < n and n – m is even, a0 /b0 > 0

Column 2 (p) • (q) – • (r) a0/b0

(d) m < n and n – m is even, a0 /b0 < 0

(s) 0

33. lim f(x) is less than equal to, where xÆ 0

Column 1

Column 2

(a) f (x) =

e -e x

(b) f(x) =

e x - e- x sin x

x

2x

(p) e (q) – 2

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e2 x - e 4 x x

(c) f (x) =

(d) (1 + sin x)

(r) –1

cosec x

(s) 2

ASSERTION-REASON TYPE QUESTIONS 36. Statement-1: lim ( cos p / n ) = 1 n

nƕ

Statement-2: The above limit is equal to

34. lim f (n) is

lim ( - p tan p / n )

Æ•

Column 1 (a) f (n) =

(b) f(n) =

( n + 1)2

(p) £ 1

2 n2 n3 - 100n2 + 1

(q) £ 0

100 n2 + 15 n 3

(c) f (n) = (d) f (n) =

Column 2

n + 2n - 1 n+2 3

(r) ≥ 1/2

n! ( n + 1)!- n!

Column 2

35. (a) If f (x) = x – 3, 2 < x < 3 and f (x) = 2x + 5, 3 < x < 4 the quadratic equation whose roots are

(p) 8x2 – 6x + 1 = 0

nƕ

1 n +1 4

xÆc

COMPREHESION-TYPE QUESTIONS Paragraph for Question Nos. 39 to 42

xÆc

x Æ3+

(b) If a = lim

38. Statement-1: If a continuous function on [0, 1] satisfy 0 £ f (x) £ 1 then there is c Œ [0, 1] such that f (c) = c. Statement-2: lim f ( x ) = f (c)

The continuity on an interval has a geometric interpretation, namely, a function f I is continuous on I if its graph has no ‘holes’ or ‘jumps’. f is said to have a removable discontinuity at c if f (x) has a limit at c but lim f (x) π f (c). If lim f(x) and lim f (x) exist

lim f(x) and lim f (x)

x Æ3-

37. Statement-1: Let g(x) = [x) + x 2 then g is continuous from right at every integer where [x) is greatest integer less than x Statement-2: g(x) is discontinuous from right at every integer.

(s) = •

Column 1 2

nƕ

(q) x2 – 11x + 10 = 0

8 n3 and + ... n4 + 1 n4 + 1 tan 4 x then n Æ 0 tan8 x the quadratic equation whose roots are a, b is

b = lim

(c) If f (0) is determined (r) x2 – 17x + 66 = 0 1 2 = x x e -1 is continuous at 0 then x – f (0) is a factor of quadratic equation given by so that f (x) =

xÆc+

xÆc-

but are not equal then c is called jump discontinuity. If lim f(x) or lim f (x) fail to exist then c xÆc+

xÆc-

discontinuity. 39. Let f (x) = sgn x then x = 0 (a) is a point of continuity (b) is a jump discontinuity (c) is a removable discontinuity Ï x2 + 5 , x < 2 Ô , x=2 40. Let g(x) = Ì10 Ô 3 Ó1 + x , x > 2 then x = 2 is (a) a point of continuity (b) is a removable discontinuity (c) is a jump discontinuity

(d) If f (a, b, c, d) =(s) x2 – 20x + 19 = 0 c + dx

1 ˆ Ê lim Á1 + ˜ nÆ• Ë a + bx ¯ then x – f (1, 1, 1, 0) is a factor of

41. Let g(x) = x + 7, x < –3; g(x) = |x – 2|, – 3 £ x < –1; g(x) = x2 – 2 x, – 1 £ x < 3 and g(x) = 2x – 3, x ≥ 3. then (a) x = – 1 is a jump discontinuity (b) x

IIT JEE eBooks: www.crackjee.xyz Limits and Continuity 22.35

(c) x = – 3 is a jump discontinuity (d) x = – 1 is a removable discontinuity Ï1 , x £ - 2 Ô Ô1 42. g(x) = Ì x , - 2 < x < y Ô2 ÓÔ x , x ≥ 4 then (a) g is a continuous function (b) all the discontinuities are removable discontinuities (c) all the discontinuities are jump

1 x + x 2 + ...... + x 99 - 99 . xÆ1 2475 x -1 48. Let f be a continuous function and g a bounded function such that f

47. Find the lim

xÆ1 nÆ•

49. If f (u) = lim

t Æ•

(1 + sin p u)t - 1 (1 + sin p u )t + 1

u=x 50. Let f (x) =

x + x - 16 x + 20 if x π 2. Find the value ( x - 2)2 2

of f(2) so that f is a continuous function. 51. Find the value of lim

Let f [a, b] then

(3tan 3x - 4 tan 2 x - tan x )

xÆ 0

(i) f is bounded on [a, b] and (ii) f attains both its maximum value M and its minimum value m on [a, b] i.e. there is c, d Π[a, b] such that f (c) = M and f (d ) = m. 43. Which of the following functions are not bounded 2x (a) f (x) = , [Р2, 2] 1 + x2 1 - cos x (b) f (x) = [Р2, 2] x2 x3 - 8x + 6 , [0, 5] 4x + 1

(d) none of these 44. Let g(x) = 1/x 2, x > 0; g(0) = 0 then (a) g is a continuous function (b) g is a bounded function (c) g is bounded on [1, •) (d) g has a minimum value on [1, •).

46. Find the value of f(0) for which 64 ( x + 4 - 2) f (x) = is continuous. sin 2 x

4 x 2 tan x

.

52. Find the number of integral discontinuities of f(x) =

tan x tan -1 (1 ( x - 1)) x ( x - 2)( x - 4)

53. Find the value of f (2) so that f (x) = is continuous.

(2 x +8 - 1024) 4 x +1 - 64

LEVEL 2

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. The number of points at which the function f(x) = 1/(x – [x]) is not continuous is (a) 1 (b) 2 (c) 3

2.

INTEGER-ANSWER TYPE QUESTIONS 1 45. Find the value of f (2), so that the function f(x) = 4 2x - 4 , x π 2 is continuous everywhere. (4 + 2 x )1 3 - 2

is not continuous at

x. 3

Paragraph for Question Nos. 43 to 44

(c) f (x) =

1 f ( x) en x + g ( x) . f (1) en x + 1

lim lim

lim

x cos x - log (1 + x ) x2

xÆ0

(a) 1/2 (c) 1

is equal to (b) 0 (d) none of these

3. If f(x) = 1/(2 – x), then the points of discontinuity of the composite function y = f( f(f(x))) are (a) 2, 3/4 (b) 1, 2 (c) 2, 3 (d) 2, 3/2 4.

lim

xÆ0

sin x n (sin x ) m

(a) 1 (c) n m

( m < n) is equal to (b) 0 (d) none of these

IIT JEE eBooks: www.crackjee.xyz 22.36 Comprehensive Mathematics—JEE Advanced

Ï 2 ÔÔ x + 5 5. If g(x) = Ì 10 Ô 3 ÔÓ 1 + x then

(

)

if x < 2 if x = 2

(1 - x )

if x > 2

Ê 1 1ˆ -Á + Ë | x | x ˜¯

and f(0) = 0 10. Let f (x) = (x + 1) 2 (a) f is continuous at x = 0 (b) lim f (x) exists xÆ0

(c)

(a) lim g(x) = –9 xÆ2+

(d)

(b) lim g(x) = 9 xÆ2

(c) lim g(x) = 3 xÆ2+

(d) g is continuous at x = 2 6. If Ï - x 2 if x < - 1 Ô if x = - 1 Ô3 g(x) = Ì Ô2 - x if -1 < x £ 1 Ô x2 if x > 1 Ó then (a) g is not continuous at x = –1 ex - 1 (b) lim g ( x ) < lim x Æ1 x Æ 0 sin x (c) g is not continuous at x = 1 (d) g is continuous at x = –1

lim f(x) does not exist

xÆ0+

lim f(x) π lim f(x)

xÆ0+

xÆ0-

ÏÔ x 2 if x is rational 11. If f (x) = Ì 2 ÓÔ- x if x is irrational then (a) f is continuous at x = 0 (b) f is continuous at x = 1/2 (c) f is not continuous at x = 0 (d) f is continuous at x = 1/2 sin x

Ê sin x ˆ x - sin x is 12. The value of lim Á ˜ xÆ0 Ë x ¯ (a) e

(b) 1

13. Let f (x) =

(c) e 2

1 - cos 2 ( x - 2) x-2

(d) e–1

, x π 2.

Then lim f (x) xÆ2

7. If f(x) =

( a + x )2 sin ( a + x ) - a2 sin a ,xπ0 x

the value of f(0) so that f is continuous at x = 0 is (a) a2 cos a + a sin a

(b) a2 cos a + 2a sin a

(c) 2a2 cos a + a sin a

(d) none of these

8. Given that y = 1/(u2 + u – 2), where u = 1/(x – 1), the function is discontinuous at (a) x = 1, 2, 3 (b) x = 1/2, 3 (c) x = 2, 1/2 (d) x = – 2 Ï1 2 Ô 5 (2 x + 3) for - • < x £ 1 Ô 9. If f(x) = Ì6 - 5 x for 1 < x < 3 Ôx - 3 for 3 £ x < • Ô Ó then (a) f is continuous at x = 1 (b) f is discontinuous at x = 1 (c) f is continuous at x = 3 (d) f is discontinuous at x = 2

(a) exists and it is equal to

2

(b) does not exist because lim f (x) doesn’t exist xÆ2+

(c) equals to 1 (d) doesn’t exist because lim f (x) π lim f (x) xÆ2+

14. lim

xÆ2-

(sin x - tan x )2 + (1 - cos 2 x )4 + x5

xÆ0

7 tan 7 x + sin6 x + 2 sin5 x

(a) 0

(b) 1

(c) 1/2

is equal to

(d) 2

15. Let f (x) = x – 1 and g(x) = 1/x. Then the set of points where gofog is continuous is (a) R ~ {0}

(b) R ~ {1}

(c) (– •, •)

(d) (– •, •) ~ {0, 1}

È x 2 ˘ + È( 2 x )2 ˘ + ... + È( n x )2 ˘ Î ˚ Î ˚ Î ˚ then 16. Let f (x) = lim nÆ• n the set of all points of continuity of f ([x] denotes the greatest integer function) (a) (– •, •) ~ {0}

(b) (– •, •) ~ I

(c) (– •, •)

(d) (– •, •) ~ {0, 1}

IIT JEE eBooks: www.crackjee.xyz Limits and Continuity 22.37

x is rational Ï1 17. Let f (x) = Ì . Then x is irrational Ó0 (a) f is discontinuous for every real x (b) f is continuous on R (c) f is continuous at the points where x is rational (d) f is continuous at the points where x is irrational. 18. Let f : R Æ R x irrational Ï x f (x) = Ì x is rational Ó1 - x then f is (a) discontinuous at x = 1/2 (b) continuous at x = 1/2 (c) continuous everywhere (d) discontinuous everywhere

(b) –1 (d) none of these

sin [ cos x ] , (where [x] denote the greatest x Æ 0 1 + [ cos x ] lim

integer less than or equal to x) is (a) 1 (b) 0 (c) – 1

(d) 1/2

21. If x 2 + x3 £ x + f (x) £ x 5 – x 3, for values of x near 0, f ( x) then lim is xÆ0 x (a) –1 (b) 0 (c) 1 (d) none of these 22. Let f (x) = [x 2 + 1], ([x] is the greatest integer less than or equal to x). Then f is continuous (a) on [1, 3] (b) for all x is [1, 3] except four points (c) for all x in [1, 3] except seven points (d) for all x in [1, 3] except eight points 23. The value of f (0) so that f (x) = (a) 1/6 24. Let f (x

1+ x - 31+ x x (b) 1/4

2x - 1 p , where [ ] denotes 2 the greatest integer function is discontinuous at (a) all x (b) all integer points (c) no x (d) x which is not integer

25. The functionf (x) = [x] cos

26. If lim (cos x + a sin bx)1/x = e2 then the values of

xÆ0

20.

Ê 1ˆ (b) f Á ˜ Æ 0 as x Æ 0 Ë x¯ (c) xf (x) Æ 1 as x Æ 0 (d) f(x) = log x

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

sin - 1 ([ x ] + x ) , [x] π 0 19. If f (x) = [ x] = 0, [x] = 0 where [x] denote the greatest integer less than or equal to x, then lim f (x) is (a) 1 (c) 0

(a) f (x) is bounded

xÆ 0

a and b are (a) a = 1, b = 2 (b) a = 2, b = 1/2 (c) a = 2 2 , b = 1 (d) a = 4, b = 2

Ï x2 - 4 x - 5 if x > 5 Ô 27. Let f (x) = Ì x - 5 Ô2 x - 4 if x £ 5 Ó then (a) lim f (x) = 6 xÆ5+

(b) lim f(x) = 6 xÆ5-

(c) f is not continuous at x = 5 (d) f is a continuous function 28. If lim

x Æ- a

(c) 1/3

(d) 3

x > 0 and be continuous. Ê xˆ Let f(x) satisfy f Á ˜ = f (x) – f (y) for all x, y and f(e) = 1. Then Ë y ¯

x 7 + a7 = 7, then the value of a is x+a

(a) 1 (c) 7

(b) –1 (d) none of these

29. Let f (x) = [x /(x + 3)]3x. Then (a) lim f (x) = e–3

(b) lim f(x) = 1/64

(c) lim f (x) = 64/625

(d) none of these

xƕ

xÆ 2

is continuous is

2

xÆ1

30. The set of all points of discontinuities of the function tan x log x contains f (x) = 1 - cos 4 x (a) {np /2; n Œ Z} (b) {np/2 ; n Œ Q} (c) (–•, 0) » {np/2; n Œ N} (d) none of these

IIT JEE eBooks: www.crackjee.xyz 22.38 Comprehensive Mathematics—JEE Advanced

31. Points of discontinuities of the function 38. If f (x) =

f (x) = 4x + 7 [x] + 2 log (1 + x) are (a) 0 (b) 1 (c) – 3/2 (d) – 5/4

(a) (b) (c) (d)

x2 + 4 x + 1 contains the sets 1 -| x|

39. Let f and g then (a) f o g is (b) g o f is (c) f o g is (d) none of

(1, •) (– 1, •) (– •, – 1) (– •, •) ~ {–1, 0, 1}.

(b) B = 5 (d) A = – 3 f (x) =

x and g(x) = x – 1

continuous on [0, •) continuous on [0, •) continuous on [1, •] these 1/ x 2

(e x - 1)4 , x π 0; f(0) = 8 sin ( x 2 k 2 ) log{1 + ( x 2 2)}

may be continuous function is (a) 1 (b) –1 (c) 2

xÆ0

(c) f is discontinuous at x = 0 (d) – 2

(d) lim f (x) = 0 xÆ0

34. The function Ïx a ÔÔ f (x) = Ìa Ô 2 2 ÓÔ(2b - 4b) x 2

0 £ x – 1. Then lim a (a) and lim b (a) are 3

5 and 1 2 7 (c) - and 2 2

aÆ 0 +

1 and – 1 2 9 (d) - and 3 [2012] 2 (b) -

Questions on Continuity (b) p (d) 1

9. The integer n for which lim

10. If lim

Ê 1ˆ fÁ ˜ Ë 2¯

(a) -

equals

(a) – p (c) p /2

(a) 1

(b)

aÆ 0 +

[2000]

2 f(2) p

f (2)

13. If lim ÈÎ1 + x ln(1 + b2 )˘˚ x = 2b sin 2 q , b > 0 and q x Æ0 (– p, p], then the value of q is p p (b) ± (a) ± 4 3 p p (c) ± (d) ± [2011] 6 2

(

(b) – 2 (d) – 1/2 x

8. lim

8 p 2 (c) p

(a)

f (t ) dt

x 2 - p 2 / 16

(c) 1

equals

(d) 2 [2006]

log (1 + a x ) - log (1 - b x ) is not x x = 0. The value which should be assigned to f at x = 0 so that it is continuous at x = 0, is (a) a – b (b) a + b (c) log a + log b (d) none of these [1983] Ê 2 x - 1ˆ p, where [ . ] 17. The function f (x) = [x] cos Á Ë 2 ˜¯ denotes the greatest integer function, is discontinuous at (a) all x (b) all integer points (c) no x (d) x which is not an integer [1995] 18. Let f (x x > 0 and be continuous. Let f(x) satisfy f(x/y) = f(x) – f(y) for all x, y and f(e) = 1. Then (a) f (x) is bounded (b) f (1/x) Æ 0 as x Æ 0 (c) x f (x) Æ 1 as x Æ 0 (d) f (x) = log x [1995] 16. The function f (x) =

IIT JEE eBooks: www.crackjee.xyz 22.42 Comprehensive Mathematics—JEE Advanced

19. The function f (x) = [x]2 – [x2] (where [y] is the greatest integer less than or equal to y), is discontinuous at (a) all integers (b) all integers except 0 and 1 (c) all integers except 0 (d) all integers except 1 [1999]

max {f(x):x [0, 1]} = max {g(x):x [0, 1]}, the correct statement(s) is(are): (a) (f(c))2 + 3f(c) = (g(c))2 + 3g(c) for some c [0, 1] (b) (f(c))2 + f(c) = (g(c))2 + 3g(c) for some c [0, 1] (c) (f(c))2 + 3f(c) = (g(c))2 + g(c) for some c [0, 1] (d) (f(c))2 = (g(c))2 for some c [0, 1] [2014]

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS Questions on Limits 1. Let L = lim

xÆ0

a - a2 - x 2 x4

x2 4 , a > 0.

If L (a) a = 2 (b) a = 1 1 1 (d) L = [2009] (c) L = 64 32 2. For every integer n, let an and bn be real numbers. Let function f: R Æ R be given by Ïa + sin p x, for x Œ[ 2n, 2n + 1] f(x) = ÔÌ n ÔÓbn + cos p x, for x Œ( 2n - 1, 2n) for all integers n. If f is continuous, then which of the following holds(s) for all n? (b) an – bn = 1 (a) an–1 – bn–1 = 0 (d) an–1– bn = – 1 (c) an – bn+1 = 1 [2012] Questions on Continuity 3. If f (x) = (1/2)x – 1, then on the interval [0, p] (a) tan ( f (x)) and 1/f (x) are both continuous (b) tan (f (x)) and 1/f (x) are both discontinuous (c) tan (f (x)) and f –1 (x) are both continuous (d) tan ( f (x)) is continuous and 1/f (x) is not [1989] 4. Which of the following functions are continuous on (0, p)? 1 (a) tan x (b) Ú x t sin dt 0 t 1 if 0 < x £ 3 p /4 Ï Ô (c) Ì 3p 2 ÔÓ2 sin 9 x, if 4 < x < p if 0 < x £ p / 2 Ï x sin x (d) ÔÌ p [1991] ÔÓ 2 sin (p + x ) if p / 2 < x £ p 5. For every pair of continuous functions f, g:[0, 1] R such that

INTEGER-ANSWER TYPE QUESTIONS 1. The largest value of the non-negative integer a for which 1– x

Ï – ax + sin( x – 1) + a ¸1– x 1 is [2014] = lim Ì ˝ x Æ1 Ó x + sin( x – 1) – 1 ˛ 4 2. Let m and n be two positive integers greater than 1. Ê cos(a n ) ˆ e –e m e If lim Á ˜ = – ÁÊ ˜ˆ then the value of m Ë ¯ a Æ0 Á n 2 a ˜¯ Ë is [2015] 3. Let a, b Œ R be such that x 2 sin b x = 1. Then 6(a + b) equals [2016] x Æ 0 a x - sin x lim

FILL

IN THE

BLANKS TYPE QUESTIONS

Questions on Limits p xˆ = _______. 1. lim(1 – x) tan ÊÁ Ë 2 ˜¯ xÆ1

[1984]

f (x) = sin x, x π np, n = 0, ± 1, ± 2, ...... =0 otherwise and g (x) = x2 + 1, x π 0, 2 =4 x =0 =5 x =2 then lim g( f (x)) is _______. [1986]

2. If

xÆ 0

3.

Ê x 4 sin (1 / x ) + x 2 ˆ lim Á ˜ = _______. x Æ- • Ë 1 + | x3 | ¯

[1987]

4. ABC is an isosceles triangle inscribed in a circle of radius r. If AB = AC and h is the altitude from A to BC, then the triangle ABC has perimeter P=2

(

)

2 h r - h2 + 2 h r and area A = _______.

Also lim

hÆ 0

A = _______. P3

[1989]

IIT JEE eBooks: www.crackjee.xyz Limits and Continuity 22.43

Ê x + 6ˆ 5. lim x Æ• Á Ë x + 1 ˜¯

x+4

Ê 1 + 5 x2 ˆ 6. lim Á ˜ x Æ0 Ë 1 + 3 x 2 ¯ 7. lim

1/ x

= _______.

[1990]

1/ x 2

= _______.

[1996]

log (1 + 2h) - 2 log (1 + h)

= _______. [1997]

h2

hÆ 0 x2

cos t 2 dt 8. lim Ú0 = _______.

[1997] x sin x Questions on Continuity Ï( x 3 + x 2 - 16 x + 20) / ( x - 2)2 if x π 2 9. Let f (x) = ÔÌ k if x = 2 ÔÓ If f (x) is continuous for all x, then k = _______. x Æ0

[1981] 10. A discontinuous function y = f (x) satisfying x2 + y2 = 4 is given by f (x) = _______. [1982] 11. Let f(x) = [x] sin (p/[x + 1]) where [.] denotes the greatest integer function. The domain of f is _____ and the points of discontinuities of f in the domain are _______. [1996] 12. Let f(x £x£ 3. If f(x) takes rational values for all x and f(2) = 10 then f (1.5) = _______. [1997]

TRUE/FALSE TYPE QUESTIONS 1. If lim ( f (x) g(x)) exists, then both lim f (x) and xÆa

xÆa

lim g (x) exist.

[1981]

xÆa

SUBJECTIVE-TYPE QUESTIONS Questions on Limits a + 2x - 3x , 1. lim xÆa 3 a + x - 2 x

aπ0

[1978]

p xˆ 2. lim(1 – x) tan ÊÁ Ë 2 ˜¯ xÆ1

[1978]

p 6. Find lim ÏÌtan ÊÁ + xˆ˜ ¸˝ [1993] Ë4 ¯˛ xÆ 0 Ó Questions on Continuity 7. Let f (x + y) = f(x) + f (y) for all x and y. If the function f (x) is continuous at x = 0, then show f (x) is continuous at all x. [1981] 8. Determine the values of a, b, c for which the function Ï sin ( a + 1) x + sin x for x < 0 Ô x Ô f (x) = Ô c for x = 0 Ì Ô ( x + b x 2 )1/2 - x1/2 Ô for x > 0 b x 3/2 ÔÓ is continuous at x = 0 [1982] 9. Let f (x) be a continuous and g(x) be a discontinuous function. Prove that f (x) + g(x) is a discontinuous function. [1987] 10. Find the values of a and b so that the function Ï x + a 2 sin x 0 £ x < p /4 Ô f (x) = Ì 2 x cot x + b p /4 £ x £ p /2 Ôa cos 2 x - b sin x p /2 < x £ p Ó is continuous for 0 £ x £ p [1989] Ï 1 - cos 4 x x0 Ô ÔÓ 16 + x - 4 Determine the value of a, if possible, so that the function is continuous at x = 0. p Ï a /|sin x| , - n then f (x) =

24 x 3 a2 – x 2

sin15 x 2

Thus for f in (b) f (x) = 0. Hence lim f (x) = 0 f in xÆ 0 (c) can be written as

+ x 2 a2 – x 2

3

x Æ0

2 . So y

Also [5/2 + tan x + tan2 x] = 2 if x near 0.

x 2 (%)

a – 1/2 = 0 (as the numerator tends to 1/a – 1/2 as x 0), so a = 2. In this case the last limit is equal to 2 –3/2

cos 5p /4 = - 1

2 = cos 7p /4 while cos 3p/4 =

Ê x ˆ ¥15 Á So lim f (x) = 15. xÆ 0 Ë sin x ˜¯

12 x 2

xÆ0

2, cos p/4 = 1

2 £ cos x £

2

i.e. x π ± 5, ± 1

2

1

7 < e 2 < 8. Thus f (x) =

25. f is continuous if x – 26x + 25 π 0 4

f (x) =

a0 x m- n + a1 x m+1- n + ..... + ak x m+ k - n b0 + b1 x + ..... + bl x l

so lim f(x) = 0. If m = n then xÆ 0

24 f(x) =

a0 + a1 x + ..... + ak x k b0 + b1 x + ..... + bl x l

so lim f (x) = a0 /b0. xÆ 0

IIT JEE eBooks: www.crackjee.xyz 22.48 Comprehensive Mathematics—JEE Advanced

If m < n and n – m is even then f (x) Æ • or – • according as a 0 /b 0 > 0 or a0 /b 0 < 0 respectively. 34. Divide the numerator and denominator by n2 in (a); n in (b); n! in (c). 35. For (a) lim f ( x ) = 6, lim f ( x ) = 11, for (b) x Æ 3-

1 4

a =

x Æ3+

1 . For (c) f (0) = 1 and for (d) 2

b =

43. f in (a) is continuous on [– 2, 2] so is bounded. The maximum value of f on [– 2, 2] is 1/2. Again the function in (c) is continuous on [0, 5] so bounded. Moreover for f in (b) lim f (x) = 1/2. xÆ 0

44. As x Æ 0+, 1/x Æ • so g is not bounded and not continuous on [0, •) on the other hand g is bounded on [1, •) since 0 £ g(x) £ 1 but g has no minimum on [1, •). 2

45. f (2) = lim f (x) = 2 lim xÆ 2

f(a, b, c, d) = ed/b. 36. u = ( cos p / x )

x

xÆ 2

(L’Hopital Rule)

log(cos p / x ) logu = x log cosp/x = . 1/x

(-p / x ) tan p / x

46. f (0) = lim f (x) = lim

2

So lim log u = lim

xÆ 0

47. Reqd. limit = lim

37. The function [x) is discontinuous from right as [n) = n – 1 and lim [x) = n. x Æ n+

xÆ1

=

en x + 1

xÆ1 nÆ•

lim f (x) = 1 and lim f (x) = – 1 xÆ 0 -

f ( x ) + e - nx g ( x )

= lim lim

xÆ1 nÆ•

Thus x = 0 is a jump discontinuity. 40.

lim f(x) = 1 + 23 = 9; lim f(x) = 4 + 5 = 9

xÆ 2 +

xÆ 2

t Æ•

Thus x = 2 is a removable discontinuity. 41.

t Æ•

lim

xÆ-1+

= lim

f(x) =

lim

2

xÆ-1+

x – 2x = 3,

lim

xÆ-1-

f (x) =

lim |x – 2| = 3 so lim f (x) exists and is equal to

xÆ-1

xÆ-1

1 + e- n x

= lim f (x) = f (1) = 297 xÆ1

(1 + sin p u)t - 1

49. f (u) = lim

xÆ 2 -

so lim f (x) exists but is not equal to f(2) = 10

1 (1 + 2 x + 3x 2 + ... + 99x 98) 2475

f ( x ) e nx + g ( x )

48. lim lim xÆ 0 +

2 cos 2 x

)) = 8

99(100) 1 (1 + 2 + 3 + ... + 99) = = 2. 2475 2 ¥ 2475

38. Apply intermediate value theorem to g(x) = x – f (x) Ï1 x > 0 Ô 39. sgn x = Ì0 x = 0 Ô Ó-1 x < 0

((

(64) 1 2 x + 4

xÆ 0

=0

-1 / x 2

xƕ

xƕ

1 = 12 (1 3) (4 + 2 x)-2 3 (2)

(1 + sin p u)t + 1

1 - (1 + sin p u )- t 1 + (1 + sin p u )- t

= 1 for 0 < u < 1

If u > 1 then f (u) = – 1. Hence f is not continuous at u = 1. 50. f (2) = lim

xÆ 2

x 3 + x 2 - 16 x + 20 ( x - 2)2

f (–1) = 3. Thus f is continuous at x = – 1. lim f(x) xÆ- 3+

= lim |x – 2| = 5, lim xÆ- 3+

xÆ- 3 -

f (x) = lim

xÆ- 3 -

x + 7 = 4.

So x = – 3 is a jump discontinuity.

= lim

xÆ 2

( x - 2)2 ( x + 5) ( x - 2)2

= lim (x + 5) = 7 xÆ 2

42.

lim g (x) = lim g (x) = 2 so f is continuous at x

xÆ 4 +

xÆ 4 -

51. Reqd. limit

= 4. lim g(x) = 1 but lim g(x) = –1 so x = – 2 xÆ- 2 -

xÆ- 2 +

is a jump discontinuity and there is no other discontinuity.

= lim

xÆ 0

3(3tan x - tan3 x ) 8 tan x - tan x 2 1 - 3tan x 1 - tan 2 x 4 x 2 tan x

IIT JEE eBooks: www.crackjee.xyz Limits and Continuity 22.49

= lim

˘ ˆ 83 1 È Ê1 8 3 + -12 Í Á 2 ˜ 2 ˙ 1 - tan x ˚˙ 4x ÎÍ Ë 3 1 - 3tan x ¯

= lim

˘ 2 tan 2 x 8 È ˙ =4 2 Í 2 2 4 x Î (1 - tan x )(1 - 3tan x ) ˚

xÆ 0

xÆ 0

x = 0, 2, 4 and at x = (2 n p + 1) , n ΠI. Except these points the function is 2 continuous, so the required set is

p Ï ¸ Ì(2n + 1) : n ŒI ˝ » {0, 2, 4}. Thus the integral 2 Ó ˛ number of discontinuities is 3. 53. f (2) = lim

xÆ 2

(2 x - 2 - 1) 210 1 26 log 2 = = 23 = 8 4 log 4 (4 x - 2 - 1) 24 4

IIT JEE eBooks: www.crackjee.xyz

23 23.1 DIFFERENTIABILITY

= Let f x 0. then f is said to be differentiable at x0 If lim

DxÆ0

f ( x0 + D x ) - f ( x0 ) Dx

exists. If this limit exists we call it the derivative of f (x) at x0 and denote it by f ¢ (x0) or by dy d x x = x0 If this limit does not exist, we say that f is not differentiable at x = x0. f ( x ) - f ( x0 ) Equivalently lim x Æ x0 x - x0 f has a non vertical tangent line at x = x0. right-hand derivative, denoted by f ¢(x0+), at x = x0 as lim

D x Æ 0+

f ( x0 + D x ) - f ( x0 ) Dx

and the left-hand derivative, denoted by f ¢ (x0 – ), at x = x0 as f ( x0 + D x ) - f ( x0 ) lim D x Æ 0Dx Thus, a function f (x) is derivable at x = x0 If f ¢ (x0+) = f ¢(x0 –). Every differentiable function is continuous but the converse may not be true. That is, a continuous function need not be differentiable. e.g.

f ' (0 –) =

lim

Dx Æ 0 +

Dx =1 Dx

| Dx | - Dx = lim Dx Æ 0 - Dx Dx Æ 0 - Dx lim

(as Dx Æ 0 –, we have Dx< 0) = –1 Thus f is not differentiable at x = 0. 4

y = |x|

3 2 1 -3

-2

-1

0

1

2

3

Fig. 23.1

The absolute value function is not differentiable at x = 0. Clearly f is a continuous function. Similarly If f (x) = |x – a|, then f is not differentiable at x = a, f ¢ (a +) = 1 and f ¢(a –)= – 1. f ¢(x) = lim

hÆ0

f ( x + h) - f ( x ) h

x, f (x)). The line through (x, f (x (x, f (x)). Thus, If there is no tangent line at a certain tangent line, the function is not differentiable

Illustration 1 words, a function is not differentiable at a

f (x) = |x| at x = 0, f ¢ (0+) =

lim

Dx Æ 0 +

f (0 + Dx ) - f (0) Dx

curve suddenly changes direction (Fig. 23.2).

Fig. 23.2

IIT JEE eBooks: www.crackjee.xyz 23.2 23.2 DIFFERENTIABILITY ON AN INTERVAL

f (x a, b], is said to be differentiable a, b] If it is differentiable at every c Π(a, b) and both 23.4 SOME FORMULAE OF DIFFERENTIATION

f (a + h) - f (a) f (b + h ) - f (b ) and lim h Æ 0h h

lim

h Æ 0+

exist. Continuously Differentiable

f is said to be continuously differentiable If the derivative f ¢(x) exists and is itself a continuous function. e.g.

3. Product Rule. d d d ( f (x)g (x)) = g(x) ( f (x)) + f (x) (g(x)) dx dx dx

Illustration 2 1 Ï 2 , xπ0 Ô x sin f (x) = Ì x ÔÓ 0 , x=0 lim

hÆ0

f (0 + h ) - f (0 ) = lim hÆ0 h

For x π 0 f ¢(x) = 2x sin

Let f (x) and g(x) be differentiable functions and a Œ R. 1. Sum and Difference Rule d d d ( f (x) ± g(x)) = ( f (x)) ± (g(x)) dx dx dx 2. Scalar Multiple Rule d d (a f (x)) = a f (x) dx dx

4. Quotient Rule

1 h = lim h sin 1 = 0 . hÆ0 h h

h 2 sin

1 1 - cos x x

d Ê f ( x) ˆ = d x ÁË g( x ) ˜¯

5. Chain Rule If y = h(u) and u = f (x), then d y d y du = d x du d x

x Æ0

not continuously differentiable.

23.3 HIGHER ORDER DIFFERENTIATION

Let y = f (x) be a differentiable function such that z = f ¢(x) is also differentiable. Then the second derivative of y = f (x) is denoted by y2(x), f "(x) or d2y/d x2 d y dx

2

=

d Ê d y ˆ Ê dz ˆ Á= ˜ d x ÁË d x ˜¯ Ë dx ¯

d Ê d2 y ˆ d4 y (iv) = = f ¢¢¢ (x) and f (x) = d x 3 d x ÁË d x 2 ˜¯ d x4 d3 y

or, in general, Physically

dn y d xn

=

acceleration and

d Ê dn - 1y ˆ for any n Œ N, n > 1. d x ÁË d x n - 1 ˜¯ d2 y

dy dx

dx 2 d3 y dx3

d d ( f ( x )) - f ( x ) ( g( x )) dx dx 2 ( g( x )) g (x) π 0

Since lim f ¢( x ) does not exist. So f is differentiable but

2

g( x )

6. Test of constancy val, f ¢(x) = 0, then the function f (x) has a constant value within this interval. f (x) are differentiable function in the table below f (x) f ¢(x) C, a constant xn (f (x))n

0 nxn–1 n(f (x))n–1f ¢(x)

log x

1 x

log f (x)

f ¢ ( x) f ( x)

a x, a > 0 ex ef (x) sin x cos x tan x cot x

ax log x ex ef (x).f ¢(x) cos x – sin x sec2 x – cosec2 x

IIT JEE eBooks: www.crackjee.xyz Differentiability and Differentiation 23.3

sec x cosec x

sec x tan x – cosec x cot x 1

sin–1 x

1- x -

cos–1 x

tan

–1

1 - x2

1 + x2 -

cot–1 x

, –1 < x < 1

1 + x2

| x | x2 - 1 -

cosec–1 x

x = et sint, y = et cos t dy dx dy and . dt dt dx dx = et cos t + et sin t = et (sin t + cos t) dt dy = – et sin t + et cos t = et (cos t – sin t) dt

,xŒR

1

sec–1 x

, |x| > 1

1 | x | x2 - 1

Ê d y ˆ dt ÁË d x ˜¯ d x

Illustration 4

,xŒR

1

y

d y d y dt d2 y d = = . and d x dt d x d x 2 dt

, –1 < x < 1

1

1

x

2

by y(x) = y (j –1 (x

So, , |x| > 1

dy dy dt cos t - sin t ◊ = = . dx dt dx cos t + sin t

23.7 LOGARITHMIC DIFFERENTIATION 23.5 DIFFERENTIATION OF IMPLICIT FUNCTIONS

y/dx when a differentiable function y = y (x) F (x, y) = 0, we differentiate F with x, considering y as a function of x, and solve the resulting equation to obtain for dy/dx. Illustration 3 dy If y x3 + y3 – 3 axy = 0. dx Differentiating w.r.t. x, we have 3x2 + 3y2 fi

dy È dy ˘ - 3a Í x + y ˙ = 0 Î dx ˚ dx

f 1(x) log f 2(x) + log f 3(x)

Illustration 5 y = (sin x)cos x

dy = 3ay - 3 x 2 [3 y 2 - 3ax ] dx dy ay - x = dx y 2 - ax 2



differentiating them is called logarithmic differentiation. It is useful in the following cases: 1. If the given function consists of three or more f actors which are functions of x. 2. If the given function is of the form ( f (x))h(x) 3. If the given function is of the form h1(x)h2(x)/h3(x)h4 (x). If h(x) = (f 2 (x)) f 1(x) f 3(x

log y = cos x log sin x Differentiating, we get d d 1 dy = cos x (log sin x ) + (log sin x ) cos x dx dx y dx

23.6 DIFFERENTIATION OF FUNCTIONS REPRESENTED PARAMETRICALLY

Let there be given two functions x = j (t) and y = y (t), where t Œ (a, b ). If the function x = j (t) has an inverse, t = j –1 (x) on the interval (a, b

= cos x fi

cos x - (log sin x )sin x sin x

2 ˘ dy cos x È cos x (sin x ) - (log sin x )sin x ˙ = ÍÎ ˚ sin x dx

IIT JEE eBooks: www.crackjee.xyz 23.4

Ans. (a)

Illustration 6

x sin x if x ≥ 0 f (x) = sin x| x | = ÏÌ Ó- x sin x if x < 0

Solution: y=

n

x( x 2 + 1) ( x 2 - 1)2

Since x sin x and –x sin x are differentiable functions, f (x x = 0.

1 [log x + log( x 2 + 1) - 2 log( x 2 - 1)] n Differentiating, we get 1 dy 1 È1 2x 4x ˘ + = n ÎÍ x x 2 + 1 x 2 - 1 ˚˙ y dx log y =

dy = dx



n

x( x 2 + 1)

1 È1 2x 4x ˘ ¥ Í + 2 - 2 2 2 n Î x x + 1 x - 1 ˙˚ ( x - 1)

23.8 LEIBNITZ THEOREM AND nTH DERIVATIVES

Let f (x) and g(x nth order. Then, dn dx n

n

( f (x) g(x)) = f n(x) g(x) + nC1 f n – 1 (x) g1(x) +

C2 f n–2(x) g2(x) + … + nCr f n – r (x) gr (x) + … + n

Cn f (x) gn(x).

where for any k > 1, f k(x) = of f (x) dn d xn

dk dx k

(xn) = n!;

(f (x)), i.e., kth derivative

d Ê 1 ˆ (- 1)n n! Á ˜ = n +1 ; dx Ë x¯ x

pˆ Ê (sin x ) = sin Á x + n ˜ , Ë 2¯ dx dn

n

pˆ Ê (cos x ) = cos Á x + n ˜ ; Ë 2¯ dx dn

n

dn d xn

(emn) = mn emx.

SOLVED EXAMPLES

SINGLE CORRECT ANSWER TYPE QUESTIONS Example 1 f (x) = | x | sin x is differentiable is (a) (– •, •) (b) (– •, 0) » (0, •) (c) (0, •) (d •).

Now f ¢(0+) = lim

h Æ 0+

f (0 + h) - f (0) f (h ) = lim h Æ 0+ h h f (0) = 0]

= lim

h Æ 0+

h sin h = lim sin h = 0 h Æ 0+ h

f (0 + h) - f (0) h Æ 0h f (h) - f (0) = lim h Æ 0h - h sin h = lim - sin h = 0 = lim h Æ 0h Æ 0h Hence f is differentiable everywhere. and f ¢(0–) = lim

If

Example 2

the

function

f

is

and strictly increasing in a neighborhood

differentiable of 0, then

È f ( x ) - f ( x ) f ( x ) - f ( x) ˘ lim Í + ˙ is equal to 2 xÆ0 f ( x ) - f ( 0) ˚ Î f ( x ) - f ( 0) 4

2

3

(a) –1 (b) –2 (c) 0 (d) 1 Ans. (b) Solution f is strictly increasing so f ¢(x) > 0 for all x in a neighborhood of 0 È f ( x 4 ) - f ( x 2 ) f ( x3 ) - f ( x) ˘ lim Í + ˙ 2 xÆ0 f ( x ) - f ( 0) ˚ Î f ( x ) - f ( 0) È f ( x 4 ) - f ( 0) f ( x 3 ) - f ( 0) ˘ 1 + = lim Í-1 + ˙ xÆ0 f ( x) - f (0) ˚ f ( x 2 ) - f ( 0) Î ˘ È f ( x 4 ) - f (0) f ( x 3 ) - f ( 0) 2 ¥ ¥ x2 ˙ x Í 3 4 x -0 x -0 ˙ + = lim Í-2 + 2 f ( x) - f (0) ˙ xÆ0 Í f ( x ) - f (0) Í ˙ x-0 x2 - 0 Î ˚ f ¢ (0) f ¢ (0) ¥0+ ¥0 = -2 + f ¢ (0) f ¢ (0) = –2. Example 3 f (x) =

1 - e- x

2

is differentiable is

(a) (0, •) (c) (– •, •) ~ {0}

(b) (– •, •) (d) (–1, •).

IIT JEE eBooks: www.crackjee.xyz Differentiability and Differentiation 23.5

Ans. (c) Solution:

For x π 0, we have

[1 + 1 + º + 1] = – 2 n. n Times

=–2

1 1 x e- x - x2 = [ ( 2 x ) e ] 2 2 1 - e- x2 1 - e- x 2

f ¢(x) =

f ¢(0+) = lim

h Æ 0+

f (h) - f (0) 1- e = lim h Æ 0+ h h

Ê e - h - 1ˆ lim = Á ˜ h Æ 0+ Ë - h2 ¯ 2

g¢(a) = 2, the value of

- h2

lim

xÆa

1/ 2

Êe - 1ˆ and f ¢(0–) = lim - Á ˜ 2 h Æ 0- Ë - h ¯ - h2

1 - e- h 1 - e- h =h h

2

=-

Ans

g( x ) f ( a ) - g( a ) f ( x ) xÆa x-a

Solution:

= –1

1 - e- h

lim

xÆa

xÆa

1/ 2

p Êp ˆ (c) – n (d) – n Ë ¯ 4 4 Ans. (a) Solution: For x Œ (0, p/4), tan x > x > sin x fi 1– tan n x < 1 – xn < 1 – sinn x, thus f (x) = 1– tann x f (p/4)

xÆa

= lim

hÆ0 -

f (p / 4 + h ) - f (p / 4 ) h

lim

xÆa

hÆ0 -

g( x ) - g( a ) x-a

g( x ) f ( a ) - g( a ) f ( x ) = - g( a ) f ¢( a ) + f ( a ) g ¢( a ) x-a = - (- 1)1 + 2 ¥ 2 = 5

(a) (b) (c) (d) Ans. (d)

g1 g0 g2 g3

x Æ0 n

is is is is

differentiable at x = 0 continuous at x = 0 continuously differentiable continuously differentiable

Ïsin(1 / x), x π 0 g0(x) = Ì x=0 Ó 0,

lim x sin

1 doesn’t exist, so g0 is not continuous at x = 0 x

1 Ï Ô x sin , x π 0 g1(x) = Ì x ÔÓ 0, x=0

h (1 - tan h)n (1 + tan h ) + ... + (1 + tan h)n -1 ] h

Ï n Ê 1ˆ Ô x sin Á ˜ , x π 0 Ë x¯ then Let gn(x) = Ì Ô = , x 0 0 Ó

Example 6

- 2 tan h[(1 - tan h)n - 1 + (1 - tan h)n - 2 + lim

xÆa

Since g¢(a) exists, g is continuous at x = a f ¢(a) exists, so that f is continuous at x = a. Hence, limx Æ a g(x) = g(a) = –1 and limx Æ a f (x) = f (a) = 2. Thus.

Solution:

1 - tan n (p / 4 + h) h (1 - tan h) - (1 + tan h)

f ( x ) - f (a) x-a

n -1

n n Ï Êp ˆ ¸ Ê 1 ˆ , 1 = min Ì0, 1 - Á Ë 4 ¯ ˛˝ = 0 Ë 2 ˜¯ Ó

n

xÆa

+ lim f ( x ) lim

1 Рsinn x, 1 Рxn}, x Π(Рp/2, p/2). The left hand derivative of f at x = p/4 is (a) Р2n (b) Р2 (n + 1)

hÆ 0 -

g( x )( f (a) - f ( x )) + f ( x ) ( g( x ) - g(a)) x-a

= - lim g( x ) lim

Hence f is not differentiable at x differentiability are (– •, •) ~ {0}. For n Œ N, let f (x) = min {1 – tann x, Example 4

= lim

= lim

2

h2 2

hÆ0 -

(b) 1/5 (d) 2/5

(c)

1/ 2

Ê e - h - 1ˆ =-Á 2 ˜ Ë -h ¯

f ¢ (p/4 –) = lim

g( x ) f ( a ) - g( a ) f ( x ) is x-a

(a) – 5 (c) 5

=1

because, as h Æ 0–, h is a negative number, so that 2

If f (a) = 2, f ¢(a) = 1, g(a) = –1 and

Example 5

lim

xÆ0

g1 ( x) - g1 (0) 1 = lim sin , which does not exist so g1 x Æ0 x-0 x

IIT JEE eBooks: www.crackjee.xyz 23.6

(b) (c) (d) Ans. (b) Solution:

is not differentiable at x = 0 1 Ï 2 Ô x sin , x π 0 g2(x) = Ì x ÔÓ 0, x=0 g 2 ( x) - g 2 (0) 1 = lim x sin = 0 xÆ0 x Æ0 x-0 x lim

1 1 Ï Ô- cos + 2 x sin , x π 0 g 2¢ (x) = Ì x x ÔÓ 0, x=0

So

1 1ˆ Ê lim g 2¢ ( x) = lim Á - cos + 2 x sin ˜ xÆ0 Ë x x¯

xÆ0

which does not exist so g2 is not continuously differentiable. 1 1 Ï 2 Ô3x sin - x cos , x π 0 g¢3 (x) = Ì x x ÔÓ 0, x=0

f (x) is continuous at x = 0 f (x) is not differentiable at x = 0 f ¢(0) = 1. f (x

lim f (x

at x = 0. Being a constant function f is differentiable at x = 0 and f ¢(0) = 0. Example 9

= lim

hÆ0

f (5) f (h) - f (5) f (0) h hÆ0

If f ¢ is differentiable function and f ¢¢(x) is Example 7 continuous at x = 0 and f ¢¢(0) = 5, the value of lim

2 f ( x ) - 3 f (2 x ) + f (4 x ) x2

xÆ0

(a) 5 (c) 15 Ans. (c) Solution:

is

(b) 10 (d) 20

lim

xÆ0

f ( x) f ¢( x ) f ¢(0) = lim = If g( x ) x Æ 0 g ¢( x ) g ¢(0)

Example 10

lim

x2

2 f ¢( x ) - 6 f ¢(2 x ) + 4 f ¢(4 x ) xÆ0 2x

= lim

= lim

xÆ0

f ¢( x ) - 3 f ¢(2 x ) + 2 f ¢(4 x ) x

= lim ( f ¢¢(x) – 6 f ¢¢(2 x) + 8 f ¢¢(4x)) xÆ0

= f ¢¢ (0) – 6 f ¢¢(0) + 8 f ¢¢(0) = 5 – 30 + 40 = 15. Example 8 and f (x

2

x]. Then

(a) lim f (x) does not exist xÆ0

Let f (x x] and Ï0 if x is an integer g(x) = Ì 2 Ó x otherwise

Ï0 if x is an integer Ô + f o g (x) = Ìn if x Œ R and n < x < n + 1 or Ô – Ó x Œ R and – n +1 < x < – n

2 f ( x ) - 3 f (2 x ) + f (4 x )

xÆ0

f (h) - f (0) = f (5) f ¢(0) = 2◊3 = 6 h

then (a) (g o f )¢ (1) = 1 (b) g of is not continuous (c) f o g is differentable on R ~ I (d) f o g is a differentiable function Ans. (c) Solution: g o f (x) = 0 for all x ŒR and

f (0) = g(0) = 0. \

Let f (x + y) = f (x) f ( y) for all x and y.

If f (5) = 2 and f ¢(0) = 3, then f ¢(5) is equal to (a) 5 (b) 6 (c) 0 (d) 3 Ans. (b) f (5 + h) - f (5) Solution: f ¢(5) = lim hÆ0 h

= f (5) lim

so g3 is continuously differentiable.

x] = 0 for –p /4 < x < p /4. Thus

xÆ0

lim g3¢ ( x) = 0,

xÆ0

2

f o g is differentiable on R ~ I. Let Example 11

So,

Ï x2 if x £ x0 f (x) = Ì Óa x + b if x > x0 a and b for which the function is continuous and has a derivative at x0, are (b) a = 2x0, b = – x 20 (a) a = x0, b = – x0 (c) a = x 20, b = – x0 (d) a = x0, b = – x 20. Ans. (b) Solution: For f to be continuous everywhere, we must have

IIT JEE eBooks: www.crackjee.xyz Differentiability and Differentiation 23.7

x 20 = f (x0) =

lim

x Æ x0 +

f has a derivative at x0 If f ¢(x0 +) = f ¢(x0 –). f ( x0 + h ) - f ( x0 ) Now, f ¢(x0+) = lim h Æ 0+ h

Solution:

h Æ 0-

( x + h) = lim 0 h Æ 0h

where u = x 02 = ax0 + b]

f ( x0 + h ) - f ( x0 ) h 2

1/ x

Æ

f (1 + x ) - f (1) f (1 + x ) - f (1) x = ¥ 1+ x -1 3 f (1)

1 f ¢(1) ¥ 0 = 0 as x Æ 0. 3 = lim (1 + u )

1 f (1+ x ) - f (1) 1 ◊ ¥ u 1+ x -1 f (1)

uÆ0

x02

= e f ¢(1)/f (1) = e6/3 = e2

x02 + h 2 + 2 x0 h - x02 h Æ 0h

¢

= lim

= lim (h + 2 x0 ) = 2 x0

Ï tan - 1 x if |x | £ 1 Ô f (x) = Ì 1 ÔÓ (| x | - 1) if |x | > 1 2

That is, a = 2 x0, and hence b = x20 – a x0 = x 20 – 2 x 20 = – x 02. Given f ¢(2) = 6 and f ¢(1) = 4, lim

hÆ0

(

) 2 f ( h - h + 1) - f (1)

f 2h + 2 + h 2 - f ( 2 )

(a) 3/2 (c) 5/2 Ans. (b) Solution:

= lim

hÆ0

is equal to

(b) 3 (d) – 3

(

f 2h + 2 + h 2 - f ( 2 )

¥

(

p p , f (1) = and lim f (x) = – 1, x Æ1 4 4 lim f (x) = 0 so f is not continuous at – 1, 1, hence not

x Æ1 +

(h - h

2

)

+1 -1

)

f h - h + 1 - f (1) 2

2+h 1 1 ¥ =6¥2¥ = 3. h Æ 0 1- h f ¢ (1) 4

= f ¢(2) ¥ lim

Let f : R Æ R be such that f (1) = 3 and

Ê f (1 + x ) ˆ f ¢(1) = 6. Then lim Á xÆ0 Ë f (1) ˜¯

1/ x

equals

(b) R ~ {1} (d) R ~ {– 1, 1}

Since f (– 1) = –

h (1 - h )

2h + 2 + h 2 - 2

Example 13

(a) R ~ {0} (c) R ~ {– 1} Ans. (d) Solution: We have

Ï (1/2 ) ( - x - 1) if x < - 1 Ô -1 if - 1 £ x £ 1 f (x) = Ì tan x Ô (1/2 ) ( x - 1) if x > 1 Ó

) h Æ 0 f ( h - h 2 + 1) - f (1) f ( 2h + 2 + h 2 ) - f ( 2 ) h ( 2 + h ) ¥ lim

The domain of the derivative of the function

Example 14

h Æ 0-

Example 12

(d) e3

1 f (1+ x ) - f (1) 1 ◊ ¥ f (1) 1+ x -1

xÆ0

Ê a x + b - x02 ˆ + a˜ = lim a = a = lim Á 0 h Æ 0+ Ë h ¯ h Æ 0+

f ¢(x0–) = lim

Ê f (1 + x ) ˆ lim Á xÆ0 Ë f (1) ˜¯

(c) e2

= lim (1 + u ) u

a( x0 + h) + b - x02 = lim h Æ 0+ h

and

(b) e1/2

(a) 1 Ans. (b)

f (x) = ax0 + b

Ï - 1/2 Ô Ô 1 f ¢(x) = Ì 2 Ô1+ x ÔÓ 1/2

if x < - 1 if - 1 < x < 1 if x > 1

Thus the domain of f ¢ is R ~ {– 1, 1}. Example 15 If f (1) = 1, f ¢(1) = 3 then the derivative of y = f (f ( f ( f (x))) at x = 1 is (a) 256 Ans. (c)

(b) 16

(c) 81

(d) 27

IIT JEE eBooks: www.crackjee.xyz 23.8

Solution: y¢(x) = f ¢( f ( f (f (x))) f ¢(f ( f (x))) f ¢( f (x)) f ¢(x) so y¢(1) = f ¢( f (f (f (1))) f ¢(f (f (1))) f ¢(f (1)) f ¢(1) = f ¢(f ( f (1))) f ¢( f (1)) f ¢(1) f ¢(1) = (f ¢(1))4 = 34 = 81 x 1 Ë 2¯ 4 1

9 ˆ Ê 2 ÁË x + x + + cos p x˜¯ 4 1

n

9 ˆ Ê 2 ÁË x + x + + cos p x˜¯ 4

n

IIT JEE eBooks: www.crackjee.xyz Differentiability and Differentiation 23.9

Since x < tan x so

2 tan n x + x n tan n x

Hence f (x) = 2 for all x Therefore f is a differentiable function for all x Œ R. Ï 3 p Ô x cos 2 x , x π 0 ÔÔ Let f (x) = Ì 4 p Ô + x sin x Ô 0, x=0 ÔÓ

Example 21

f ¢(2) exists.

n

Ê x ˆ Æ2 = 2+Á Ë tan x ˜¯

then f is (a) differentiable at x = 0 and x = 1 (b) differentiable at x = 0 not differentiable at x =2 (c) differentiable at x = 2 and not differentiable at x = 1 (d) differentiable neither at x = 0 nor at x = 2.

For

1 1 1 1 < x £1, 1£ < 2 , £ 0 then 0 £ x3 sin £ x3 and 2x x

3 3 If x < 0 then x £ x sin

cos

p ≥0 2x

3 Thus f (x) = x cos

2 f ¢(x) = 3x cos

p p +p = 2 2

f ¢(1–) = -

Example 22

p 2

Ê x x - x- x ˆ If f (x) = cot–1 Á ˜¯ , then f ¢(1) 2 Ë

equals

1 1 1 p p p < < fi < < 6 2x 4 3 x 2

Similarly for 1 £ x £ 2 and 2 < x < 3; sin

1 p p Ï 3 4 ÔÔ- x cos 2 x - x sin x , 2 < x £ 1 f (x) = Ì Ô x3 cos p + x 4 sin p , 1 < x < 2 ÔÓ 2x x f (1 + h) - f (1) hÆ 0 + h p p (1 + h)3 cos + (1 + h) 4 sin 2(1 + h) 1+ h = lim hÆ 0 + h p p p lim 3(1 + h)2 cos - (1 + h)sin hÆ 0 + 2(1 + h) 2 2(1 + h) = p p - p (1 + h)3 cos +4(1 + h)3 sin 1+ h 1+ h

p p + x3 sin 2x x

x π 0, 0 £ cos

1 1 1 1 1 < < 1, < < 2 x 4 2x 2 p p cos > 0 , sin > 0 2x x

f ¢(1+) = lim

Ans. (c) x3 cos

p p £ 0 , sin < 0 2x x



p ≥0 x

p p + x 4 sin , 2x x

p p p p p - x sin + 4 x3 sin - p x 2 sin x x 2x 2 2x

(a) –1 (b) 1 (c) log 2 (d) – log 2. Ans. (a) Solution: Putting u = x x and using logarithmic differentiation, we get log u = x log x. fi

1 du 1 du = x ◊ + log x fi = x x (1 + log x ) u dx x dx

Similarly, If v = x –x, we get dv = –x–x (1 + log x) dx \

f ¢(x) = –

d Ê x x - x- x ˆ Á ˜¯ 2 Ê x x - x- x ˆ d x Ë 1+ Á ˜¯ 2 Ë 1

2

IIT JEE eBooks: www.crackjee.xyz 23.10

= fi

4 x

2x

+x

f ¢(1) = –

- 2x

È1 [(x + 2 ÍÎ 2

x

˘ + x - x )(1 + log x )]˙ ˚

2 1+1+ 2

tan–1

If f (x) = (1 + x)n, then the value of

Example 23

f ¢¢(0) + f (0) + f ¢(0) + 2!

+

(0)

is

f (x) = (1/2) 52x+1 and g(x) = 5x + 4x log 5 is (a) (1, •) (b) (0, 1) •) (d) (0, •) Ans. (d) Solution: f ¢(x) = (1/2) 52x + 1 (log 5) ¥ 2 = (log 5)◊52x+1 g¢(x) = 5x log 5 + 4 log 5. So {x: f ¢(x) > g¢(x)} = {x: log 5◊52x + 1 > (log 5) 5x + 4 log 5} = {x: 52x+1 > 5x + 4} = {t = 5x : 5t2 – t – 4 > 0} = {t = 5x : (5t + 4) (t – 1) > 0} = {t = 5x : t > 1 or t < – 4/5} = {t = 5x : t > 1} = (0, •) Example 25 Let f (x) = sin x; g(x) = x2 and h(x) = log x.

(a) 2 – p (c) p – 2 Ans. (b)

d 2u d x2

x= p2

is

(c)

+ tan -1

1 x + 5x + 7 2

+

to

Solution:

= tan–1

1 + n2 n

n2 1 + n2

(d) none of these.

1 + n2 1

y = tan–1

1+ x + x

2

+ tan -1

1 x + 3x + 3 2

+ + n terms ( x + 1) - x ( x + 2) - ( x + 1) + tan -1 1 + x(1 + x ) 1 + ( x + 1)( x + 2)

+ + n terms = tan (x + 1) – tan x + tan (x + 2) – tan–1 (x + 1) + + tan–1 (x x + n) – tan–1 (x + (n – 1)) = tan–1 (x + n) – tan–1 x y¢(x) =

–1

1 1 + ( x + n)

y ¢(0) =



1 1 + n2

Example 27

2

-

-1=

–1

1 1 + x2 - n2 1 + n2

Let f and g be functions satisf ying f (x) =

ex g(x), f (x + y) = f (x) + f (y), g(0) = 0, g¢(0) = 4, g and g¢ are continuous at 0. Then (a) f (x) = 0 for all x (b) f (x) = x for all x (c) f (x) = x + 4 for all x (d) f (x) = 4 x for all x Ans. (d)

hÆ0

u (x) = h( f (g(x))) = h( f (x )) = h (sin x ) Solution: = log sin x2 Hence, u ¢(x) = 2 x cot x2 and u¢¢(x) = 2 cot x2 – 4x2 cosec2 x2, so

(b) -

–1

= lim 2

1

Ans. (b)

Solution:

(b) 2 – 2p (d) 2p – 2 2

u¢¢( p /2) = 2 – 2p.

x + 3x + 3

(a) -

+ + nC n = 2 n. The solution set of f ¢(x) > g¢(x) where

If u (x) = h( f (g(x))), then

1 2

+

1 + x + x2

n terms, then y¢(0) is equal to

(a) n (b) 2n n–1 (c) 2 (d) none of these. Ans. (b) Solution: f (0) = 1, f ¢(x) = n(1 + x)n – 1, f ¢¢(x) = n(n – 1) (1 + x)n – 2, , f n(x) = n(n – 1) 1(1 + x)n – n. So f ¢(0) = n, f ¢¢(0) = n(n – 1), , f n(0) = n is equal to n(n - 1) n! + + 1+n+ = nC 0 + nC 1 + nC 2 2! n! Example 24

1

If y = tan–1

Example 26

f ¢ (x) = lim

hÆ0

f ( x + h) - f ( x ) h

f ( x ) + f (h ) - f ( x ) e h g (h ) Ê 0 ˆ = lim hÆ0 h Ë 0¯ h

= lim eh g¢(h) + ehg(h) hÆ 0

= g¢ (0) + g(0) = 4 Hence

f (x) = 4 x + c but f (0) = g (0) = 0

So f (x) = 4 x.

IIT JEE eBooks: www.crackjee.xyz Differentiability and Differentiation 23.11

Example 28

Let f : R Æ R 3

f (x) is

by f (x) = max {x, x not differentiable is

(a) {– 1, 1} (c) {0, 1} Ans. (d) Solution:

(b) {– 1, 0} (d) {– 1, 0, 1} y = x and y = x3 are given in the

Note that If x £ – 1 Ï x 3 If – 1 £ x < 0 f (x) = ÔÔ x Ì x If 0 £ x £ 1 Ô 3 ÔÓ x If x ≥ 1 Clearly f is not differentiable at x = – 1, 0, 1 as there are x = – 1, 0, 1. y = x3

y=

x

1

–1

Fig. 23.3

Example 29 Which of the following functions is differentiable at x = 0? (a) cos (|x|) + |x| (b) cos (|x|) – |x| (c) sin |x| + |x| (d) sin (|x|) – |x| Ans. (d) Solution: cos |x| = cos x or cos (– x). Thus, n any case cos |x| = cos x for all x Œ R. Since h(x) = |x| is not differentiable at x = 0, so cos |x| + |x| = cos x + |x| is not differentiable at x = 0. Ï - (sin x + x ) if x < 0 f (x) = sin |x| + |x| = Ì if x ≥ 0 Ó sin x + x Now f ¢(0 –) = – 2, f ¢(0 +) = 2 so f is not differentiable at x = 0. Finally Ï - sin x + x if x < 0 g(x) = sin ( x ) – |x| = Ì Ó sin x - x if x ≥ 0

hÆ0

(a) 4 sin (log 4)) (b) 15 sin log 5 (c) 12 sin (log 5) (d) 9 sin log 7 Ans. (c) cos ( log x ) d = (sin (log x) + C) Solution: f ¢¢(x) = x dx So f ¢(x) = sin (log x) + C but f ¢(1) = 0 so C = 0. Thus f ¢(x) = sin (log x). Now dy Ê 2 x + 3ˆ d Ê 2 x + 3ˆ = f ¢Á Ë 3 - 2 x ˜¯ d x ÁË 3 - 2 x ˜¯ dx Ê Ê 2x + 3 ˆ ˆ 2 (3 - 2 x ) - ( - 2 )( 2 x + 3) = sin Á log Á Ë Ë 3 - 2 x ˜¯ ˜¯ ( 3 - 2 x )2 =

In this case g¢(0 +) = 0 and g¢(0 –) = 0. Thus sin ( x ) – |x| is differentiable at x = 0. Example 30 If f and f ¢(a) π 0 then which of the following is true.

f (a) - f (a - h) h f ( a + 2h ) - f ( a + h ) 1 f ¢ ( a ) = lim (b) hÆ0 2 2h f ( a + 2h ) - f ( a ) (c) f ¢(a) = lim hÆ0 h f ( a + 3h ) - f ( a + h ) (d) 3f ¢(a) = lim hÆ• h Ans. (b) f ( a ) - f ( a - h) f ( a - h) - f ( a ) = lim Solution: lim hÆ0 h hÆ0 -h = f ¢(a) f ( a + 2h ) - f ( a + h ) lim hÆ0 2h f ( a + 2h ) - f ( a ) - ( f ( a + h ) - f ( a ) ) = lim hÆ0 2h f ( a + 2h ) - f ( a ) 1 f (a + h) - f (a) - lim = lim hÆ0 2h 2 hÆ0 h 1 1 = f ¢(a) – f ¢(a) = f ¢(a) 2 2 f ( a + 2h ) - f ( a ) f ( a + 2h ) - f ( a ) lim = 2 lim hÆ0 hÆ0 2h h = 2f ¢(a) f ( a + 3h ) - f ( a + h ) Similarly lim = 2 f ¢ (a ) hÆ 0 h cos ( log x ) , f ¢(1) = 0 and If f ¢¢(x) = Example 31 x dy Ê 2 x + 3ˆ y= fÁ then is equal to ˜ dx x = 1 Ë 3 - 2x ¯ (a) – f ¢(a) = lim

a’

Ê Ê 2x + 3 ˆ ˆ sin Á log Á Ë Ë 3 - 2 x ˜¯ ˜¯ (3 - 2 x ) 12

2

dy = 12 sin log 5 dx x = 1

IIT JEE eBooks: www.crackjee.xyz 23.12

Which of the following could be not true

Example 32

(a)

If f ¢¢(x) = x–1/3 (a) f (x) =

3 2/3 x –3 2

(c) f ¢¢¢(x) = –

(b) f (x) =

1 –4/3 x 3

(d) f ¢(x) =

9 5/3 x –7 10

(b)

3 2/3 x +6 2

2n x n - 1 x2n + 1 2n x n

(

So f ¢(x) = fi f (x) =

f ¢¢(x) = x–1/3 =

d dx

(c) -

Ê 3 2 /3 ˆ ÁË x + C ˜¯ 2

(d)

3 2/3 d Ê 9 5/3 ˆ x + Const = ÁË x + Const ˜¯ d x 10 2

f (x

(a) f (b) there is a differentiable function on (– •, •) whose derivative is f (x). (c) f (d) f Ans. (c)

=–

Solution: f is not differentiable at 1, 2, 3 (in f act not continuous) so f f differentiable function on (– •, •) whose derivative is f (x). f is not continuous at x = 1 so f is not differentiable on f (x f x

+ e-

x

then xy2 + (1/2)y1 is

equal to

(

(a) y

(b) x e

(c) (1/4)y

(d)

x

+ e-

x

)

y1 =

fi 2 x y1 = e 2 x y2 + fi

1

1 2 x x

-e

y1 =

e

x

- x

1

-

2 x

e-

e

If y = cos–1

x

+

1

x

2 x

x2n - 1 x2n + 1

e-

x

=

=–

1 Ê x 2 n - 1ˆ 1 - Á 2n Ë x + 1˜¯

2

d dx

Ê x 2 n - 1ˆ ÁË x 2 n + 1˜¯

x2n + 1

( x2n + 1) - ( x 2n - 1) 2n x 2 n - 1 ( x 2 n + 1) - 2 n x 2 n - 1 ( x 2 n - 1) ¥ 2 ( x2n + 1) 2

2

4n x 2 n - 1

1

=-

2n x 2 n - 1

( x 2n - 1) | x n | ( x 2n + 1) n 2n (| x |2 ) 2n | x |n = x | x |n ( x 2 n + 1) x ( x 2 n + 1) 4 x2n

Ï 2n x n - 1 Ô x2n + 1 Ô = Ì 2n x n ÔÔ | x | x2n + 1 Ó

(

Example 36

if n is even if n is odd

)

2

n

If y(n, x) = ex ex … ex , 0 < x < 1

Then lim

. Differentiating again, we get

x 2 x xy2 + (1/2)y1 = (1/4)y.

Example 35

1

( x2n + 1)

dy(n, x ) at x = 1/2 is nƕ dx (a) e (b) 4e (c) 2e Ans. (b)

xy

Ans. (c) Solution:

If n is odd

)

2n x n - 1

x]¢, the least integer =–

If y = e

(

| x | x2n + 1

y ¢(x) = –

Solution:

function then

Example 34

2n x n

Ans. (c)

9 5/3 x + const. Thus, (a) cannot be true. 10

Example 33

If n is odd

)

| x | x2n + 1

Ans. (a) Solution:

If n is even

1 2 x

Solution: y(n, x) = e

x + x 2 +...+ x n

=

x (1- x n ) e 1- x

x (1 - x ) n

y

(d) 3e

dy(n, x ) d Ê x(1 - x n ) ˆ So = e 1- x ¥ dx dx ÁË 1 - x ˜¯ dy(n, x ) For 0 < x < 1, lim n Æ• dx

then y¢(x) is equal to = lim n

x x n +1 1 e x 1- x

d dx

È x x n +1 ˆ lim Í ˜ Î n Æ• 1 - x 1 - x ¯

IIT JEE eBooks: www.crackjee.xyz Differentiability and Differentiation 23.13

x

= e1- x = dy(n, x ) n Æ• dx lim

x 1 e x

Ans. (b)

1

Solution:

(1 - x )2

Putting y = 0 in the given functional equation, f ( x ) + f (0) 1 = 2 2 f (x) = 2f (x/2) – 1

We get f (x/2) =

= 4e. x = 1/ 2

Example 37

p at t = is If x = a cos t, y = a sin t then 3 4 dx d3 y

(a) 3/a2 (c) – 3/a2

(b) – 12/a2 (d) 12/a2



f ¢(0) = –1 we get lim

Since

hÆ0



lim

hÆ0

Solution: Clearly x2 + y2 = a2 and y(p/4) = a / 2 , x(p/4) = a / 2 . Differentiating we get 2x + 2yy1 = 0 fi y1 = –

f ¢(x) = lim

hÆ0

= lim

1 + ( y1 (p /4 )) -2 2 = fi y2 (p/4) = y (p /4 ) a 2

hÆ0

f (0 + h) - f (0) = –1 h

f (h ) - 1 = –1 h

Example 38

3 ( - 1)( - 2 ) 2 a◊a/ 2

=-

12 a2

3y1 y2 y

.

Ê x 2 - y2 ˆ dy = log a then is If cos-1 Á 2 2˜ dx Ëx +y ¯

using (i)

1 f (x) – 1 + f (h) – f (x)] h f (h ) - 1 = –1. = lim hÆ0 h Thus f ¢(x) = –1, so we get f (x) = –x + C. But fi Thus,

f (0) = 1, therefore, 1 = f (0) = –0 +C C = 1. f (x) = 1 – x f (2) = 1 – 2 = –1.

equal to (a) y/x (c) x2/y2

(b) x/y (d) y2/x2

Ans. (a) x 2 - y2

= cos log a = k (say) x 2 + y2 Putting u = y/x we have (y/x)2 = u2 = (1 - A)/(1 + A) y/x =

1 È1 {2 f ( x ) - 1 + 2 f (h) - 1} - f ( x )˘˙ Í h Î2 ˚

hÆ0

2y1 y2 + y1y2 + yy3 = 0 fi y3 = – y3(p/4) = -

f ( x + h) - f ( x ) h

= lim

Differentiating again, we have



(i)

f (2 x ) + f (2h) - f ( x) 2 = lim hÆ0 h

x , so y1(p/4) = – 1. y

Now x + yy1 = 0 fi 1 + y12 + yy2 = 0

Solution:

f (x)] (f (0) = 1)

Let x ΠR, then

Ans. (b)



(b) - 1 (d) 2

(a) 1 (c) 1/2

d Ê 1 ˆ -1 + Ë dx 1- x¯

(1 - A)/(1 + A) fi x dy/dx - y = 0

fi dy/dx = y/x 1 Êx + yˆ = f (x) + f (y)] for real Example 39 Let f Ë ¯ 2 2 x and y. If f ¢(0) exists and equals - 1 and f (0) = 1 then the value of f (2) is

Example 40

If f : R Æ R is a function such that f (x) = x3

+ x2 f ¢(1) + xf ≤(2) + f ≤¢(3) for x Œ R then the value of f (2) is (a) 5 (b) 10 (c) 6 (d) - 2 Ans. (d) Solution: Putting x = 0 in the given equation, We have f (0) = f ¢¢¢(3) and x = 1, we get f (1) = 1 + f ¢(1) + f ¢¢(2) + f ¢¢¢(3). Thus, f (1) – f (0) = 1 + f ¢(1) + f ¢¢(2). f ¢(x) = 3x2 + 2x f ¢(1) + f ¢¢(2)

(i)

f ¢¢(x) = 6x + 2 f ¢(1), f ¢¢¢(x) = 6 Thus f ¢¢¢(3)= 6 and f ¢¢(2) = 12 + 2 f ¢(1). Putting x = 1 in (i), we have f ¢(1) = 3 + 2 f ¢(1) + f ¢¢(2) = 3 + 2 f ¢(1) + 12 + 2 f ¢(1) = 15 + 4 f ¢(1)

IIT JEE eBooks: www.crackjee.xyz 23.14



f ¢(1) = – 5 and so f ¢¢(2) = 12 – 10 = 2. f (2) = 23 + 22 f ¢(1) + 2 f ¢¢(2) + f ¢¢¢(3)

Example 43

= 8 + 4(–5) + 2 (2) + 6 =-2 Example 41

( x - 1)

n

log cos ( x - 1) m

; 0 < x < 2 and

m and n are integers, m π 0, n > 0, and let p derivative of | x – 1| at x = 1. If lim g( x ) = p then x Æ1

(a) n = 1, m = 1 (c) n = 2, m = 2 Ans. (c) Solution: Clearly p = – 1 and

(b) n = 1, m – 1 (d) n > 2, m = n

for x = - 3

hÆ0

hÆ0 m

n

h log cos h

n -1

n h Ê0 ˆ form¯ = – lim m h Æ 0 tan h Ë 0 =–

(a) f is continuous at x = –3 (b) f is not derivable at x = –3 (c) f is not continuous at x = –3 (d) f is continuous at x = –3 but not derivable. Ans. (b) and (c) | x + 3| lim f ( x ) = lim Solution: x Æ - 3+ x Æ - 3+ tan -1 ( x + 3) = lim

lim g( x ) = lim g (1 + h) = lim

n h lim ¥ hn - 2 h Æ 0 tan h m

The last limit is nonzero only If n = 2 and so n/m = 1 fi m = 2.

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS Example 42

for x π - 3

then

Let g(x) =

x Æ1

Ï | x + 3| Ô If f (x) = Ì tan -1 ( x + 3) Ô4 Ó

Let f be a differentiable function satisfying

f (x + y) = f (x) + f (y) + xy(x + y) for all x, y Œ R. If f ¢ (0) = 2, then (a) f (b) f (c) f is thrice differentiable (d) f 0, ±i 2 Ans. (a), (c), (d) Solution: Differentiating w.r.t. x y constant, f ¢(x + y) = f ¢(x) + 0 + y(x + y) + xy Put x = 0 f ¢(y) = f ¢(0) + y2 = y2 + 2 fi f (y) = y3 + 2y + c Putting x = 0 = y f (0) = f (0) + f (0) + 0 fi f (0) = 0 Thus f (y) = y3 + 2y = y(y2 + 2)

x+3

x Æ - 3+

-1

tan ( x + 3)

= lim

y Æ 0+

y tan - 1 y

=1

where y = x + 3, and lim f ( x ) = lim

x Æ - 3-

x Æ - 3-

| x + 3| -1

tan ( x + 3)

= lim

y Æ 0-

-u tan -1 u

= –1,

u=x+3 Hence, f is neither continuous nor derivable at x = –2. Example 44

If x + | y | = 2y, then y as a function of x is

x (b) continuous at x = 0 (c) differentiable for all x (d) such that dy/dx = 1 3 for x < 0 Ans. (a), (b) and (d) Solution: If x + | y | = 2y, then y can be written in terms of x as if x ≥ 0 Ïx Ô y = Ì1 ÔÓ 3 x if x < 0 Therefore, y as a function of x x and is continuous at x = 0, but it is not differentiable at x = 0, because f (0 + h) - f (0) lim =1 h Æ 0+ h f (0 + h) - f (0) 1 lim = and h Æ 03 h Example 45 (a) (b) (c) (d)

The function f (x) = 1 + | sin x | is

continuous nowhere continuous everywhere differentiable nowhere not differentiable at x = 0

IIT JEE eBooks: www.crackjee.xyz Differentiability and Differentiation 23.15

Ans. (b), (d) and (e) Solution: Since g(u) = | u | and u(x) = sin x are continuous functions, we have f (x) = 1 + | u (x) | = 1 + g o u (x). Therefore, f is continuous everywhere, and we can write it as Ï1 + sin x for sin x ≥ 0 f (x) = Ì Ó1 – sin x for sin x < 0 So, f where sin x = 0, i.e., x = n p, n Œ I. Hence f is also not differentiable at x = 0. 1 - 1 - x 2 , then

If f (x) =

Example 46

(a) f (b) f is continuous at x = 0 (c) f is not differentiable at x = 0 (d) f is differentiable everywhere Ans. (a), (b) and (c) Solution: interval – 1 £ x £ 1, because, for f have 1 – x2 ≥ 0. 1 1 ◊ (2 x ) \ f ¢(x) = 2 2 1 - 1 - x2 2 1 - x at x π 0 and x π x Æ 1 – or x Æ –1 + , we have f ¢(x) Æ• f ¢(x) exists at the x = 0, we note that f ¢(0 +) = lim

h Æ 0+

1 - 1 - h2 h 1 - (1 - h 2 )1/ 2 h

= lim

h Æ 0+

= lim

h Æ 0+

= lim

h Æ 0+

1 1 - ÊÁ 1 - h 2 + powers of h 2 ˆ˜ Ë ¯ 2 2 h 1 1 + powers of h 2 = 2 2

f ¢(0–) = –

Similarly

1

2 Hence, f has no derivative at x = 0. Since for a continuous function h(x) with h(x) ≥ 0, the function g(x) = h ( x ) is continuous. So f is continuous on x = 0. Example 47

If

Ï x log cos x for x π 0 Ô f (x) = Ì log (1 + x 2 ) Ô 0 for x = 0 Ó

then (a) f (x) is continuous at x = 0 (b) f (x) is continuous at x = 0 but not differentiable at x = 0 (c) f (x) is derivable at x = 0 and f ¢ (0) = – 1/2 (d) f (x) is not continuous at x = 0. Ans. (a) and (c) f ( x ) - f (0) log cos x lim = lim Solution: xÆ0 x Æ 0 log (1 + x 2 ) x-0 = lim

xÆ0

=

log [1 - (1 - cos x )] 2 sin 2 ( x /2) x2 ◊ ◊ 1 - cos x 4 ( x /2)2 log (1 + x 2 )

1 log (1 - y) Ê sin ( x /2) ˆ ◊ lim Á lim x Æ 0Ë 2 yÆ0 y x /2 ˜¯

2

◊ lim

xÆ0

x2 log (1 + x ) 2

=-

1 2

where y = 1 – cos x. So f is derivable at x = 0 and hence also continuous. x is differentiable Example 48 The function f (x) = 1 + |x| on (a) (0, • •) (c) (– •, 0) (d) (– •, •). Ans. (a), (b), (c) and (d) Solution: The given function can be written as Ï x ÔÔ1 + x if x ≥ 0 f (x) = Ì Ô x for x < 0 ÓÔ1 - x Since x/(1 + x) (x > 0) and x/(1 – x) (x < 0) have non-zero x that f ¢(0+) = lim

h Æ 0+

= lim

h Æ 0+

f ¢(0–) = lim

h Æ 0-

= lim

h Æ 0-

h (1 + h ) - 0 f (h) - f (0) = lim h Æ 0+ h h-0 1 =1 1+ h h (1 - h ) - 0 f (h) - f (0) = lim h Æ 0 h h-0 1 =1 1- h

Thus, f is derivable at x = 0 also, and hence f is derivable everywhere.

IIT JEE eBooks: www.crackjee.xyz 23.16

Example 49 then

1 Ï 2 Ô x cos If f (x) = Ì x ÔÓ 0

for x π 0 for x = 0

(a) f and f ¢ are continuous at x = 0 (b) f is derivable at x = 0 (c) f is derivable at x = 0 and f ¢ is not continuous at x=0 (d) f is derivable at x = 0 and f ¢ is continuous at x = 0. Ans. (b) and (c) Solution: For x π 0, we have f ¢(x) = 2x cos

1 Ê 1ˆ Ê 1 ˆ + x2 sin Á ˜ Á 2 ˜ Ë x¯ Ë x ¯ x

1 1 = 2xcos + sin x x and for x π 0, we have f ( x) - f (0) x 2 cos (1/ x) 1 = = x cos fi f ¢(0) = 0 x-0 x x 1 1 lim f ¢ ( x) = lim ÁÊ 2 x cos + sin ˜ˆ xÆ0 x Æ 0Ë x x¯ Let us write 1 1 1 1 sin = 2 x cos - ÊÁ 2 x cos - sin ˆ˜ Ë x x x x¯ Now limx Æ 0 (2xcos (1/x)) = 0, so that if limx Æ 0 f ¢(x) were to exist, then lim xÆ0 sin (1/x) would also exist, which is not true. Hence f ¢ is not continuous at x = 0. 2x , then Example 50 If f (x) = sin–1 1 + x2 (a) f is derivable for all x, with | x | < 1 (b) f is not derivable at x = 1 (c) f is not derivable at x = – 1 (d) f is derivable for all x, with | x | > 1. Ans. (a), (b) (c) and (d) Solution: 1 2 (1 + x 2 ) - 4 x 2 ◊ f ¢(x) = 2 2 (1 ) 1 - 2x 1 + x =



Example 51

Let f be a differentiable function on R lim

xÆ0

satisfy

f ¢ ( x ) exists and x

f (x) f (y) = f (x1) f (y1) For all x, y, x1, y1 such that x2 + y2 = x12 + y12 such that f (0) = 1, then (a) f (b) f (c) lim f ( x) = 1 x Æ0

(d) f Ans. (b), (c), (d) Solution log f (x) + log f (y) = log f (x1) + log f (y1) Let u = x2, v = y2, u¢ = x12, v¢ = y12 fi log f ( u ) + log f ( v ) = log f ( u1 ) + log f ( v1 )

Thus, f is derivable at x

( (

Since sin–1 x is not differentiable for |x| = 1, we have f is 2x = 1 which is same as x = ± 1. not differentiable for 1 + x2

))

2 (1 - x 2 ) (1 - x 2 )2 (1 + x 2 ) Ï 2 Ô1 + x 2 Ô f ¢(x) = Ì Ô- 2 ÔÓ 1 + x 2

for | x | < 1 for | x | > 1

Let

g(u) = log f ( u ) , so

u + v = u¢ + v¢ fi g(u) + g(v) = g(u¢) + g(v¢) Setting u¢ = 0, v¢ = u + v, g(u) + g(v) = g(0) + g(u + v) = log f (0) + g(u + v) = log 1 + g(u + v) = g(u + v) g ( x + h) - g ( x ) g ( x ) + g ( h) - g ( x ) lim = lim hÆ 0 h Æ 0 h h g ( h) = lim = g¢(0) = C(say). Note that g¢(0) exists hÆ0 h g¢(x) = c for all x Œ R fi g(x) = cx + b But g(0) = 0 so b = 0 fi g(x) = x f ( x ) = ecx fi

cx f (x) = e

2 2

lim f ¢ ( x) = lim 2cxecx = 0

xÆ0

x Æ0

Clearly, f x] denote the greatest integer less Example 52 than or equal to x. If f (x x sin p x], then f (x) is (a) continuous at x = 0 (b) continuous in (–1, 0) (c) differentiable at x = 1 (d) differentiable in (– 1, 1)

IIT JEE eBooks: www.crackjee.xyz Differentiability and Differentiation 23.17

(e) none of these Ans. (a), (b) and (d) Solution: x], we have f (x x sin p x] = 0 for – 1 £ x £ 1, because 0 £ x sin p x £ f (x x sin p x] = –1 for 1 < x < 1 + h h > 0, because sin p x is negative and ≥ – 1 for 1 < x < 1 + h. Thus f (x) is constant and equal to 0 in the interval

Hence f is differentiable at x = 1 and so also continuous. Since g(x) = | x | is continuous everywhere but not differentiable at x = 0, f is continuous at x = 3 but not differentiable there. xn sin x - cos x If f (x) = n! sin ( n p 2 ) cos ( n p 2 )

Example 54

a2

a f (x) is continuous at x = 0 and in the interval x = 1, lim xÆ1+ f (x) = –1 and limxÆ1– f (x) = 0. Hence f is not continuous at x differentiable at x = 1. Example 53

(c) a

for x ≥ 1 Ï| x - 3| Ô 2 f (x) = Ì x 3 x 13 + for x < 1 ÔÓ 4 2 4 (a) continuous at x = 1 (b) continuous at x = 3 (c) differentiable at x = 1 (d) differentiable at x = 3 Ans. (a), (b) and (c) Solution: The given function can be written as

d xn dn d xn

n! sin ( x + np 2 ) - cos ( x + np 2 ) sin ( np 2 ) cos ( np 2 ) ( f (x)) = n! a2

a

a3

n! sin ( n p 2 ) - cos ( n p 2 ) = n! sin ( n p 2 ) cos ( n p 2 )

( f ( x )) x=0

a2

a

a3

a

a3 0 0

= 0.

a3

(n is odd) Example 55

f (1 + h) - f (1) h Æ 0+ h 3 - (1 + h) - 2 = lim = -1 h Æ 0+ h f (1 + h) - f (1) and f ¢(1–) = lim h Æ 0h

+

fi f ¢(1+) = lim

x + 2 2x - 4

If f (x) =

x - 2 2 x - 4 then (a) f

x=4 (b) f is differentiable on (2, •) ~ {4} (c) f is differentiable on (– •, •) (d) f ¢(x) = 0 for all x Œ Ans. (a), (b) and (d) Solution: The domain of f •). Put t = 2 x - 4

(1 + h)2 3(1 + h) 13 + -2 4 2 4 = lim h Æ 0h 2

Ê 1 - 3 + 13 - 2ˆ + Ê h - 3 h ˆ + h ÁË ˜¯ ÁË ˜ 4 2 4 2 2¯ 4 = lim h Æ 0h

h

a

n! sin ( np 2 ) = n! sin ( np 2 )

Ï Ôx - 3 for x ≥ 3 Ô 3 x for 1£ x £ 3 f (x) = Ì Ô 2 Ô x - 3 x + 13 for x < 1 ÔÓ 4 2 4

h Æ 0-

(b) 0

6

Solution: dn

= lim

( f (x)) at x = 0 for n = 2m + 1 is

Ans. (b) and (d)

is

h2 4 = -1

d xn

(a) –1

The function

-h +

dn

the value of

a3

f (x) = =

t 2 /2 + 2 + 2 t + t 2 /2 + 2 - 2 t 1 2

(t + 2) +

1 2

Ï 1 ¥ 4 if t < 2 Ô = Ì 2 Ô 2t if t ≥ 2 Ó

| t – 2| ÏÔ2 2 =Ì ÔÓ2 x - 2

if x Œ[2, 4) if x Œ[4, •)

IIT JEE eBooks: www.crackjee.xyz 23.18

Ï0 Ô Hence, f ¢(x) = Ì 1 Ô x-2 Ó

if x Œ[2, 4) if x Œ(4, •)

Let h(x) = min {x, x2} for every real

number x. Then (a) (b) (c) (d)

q

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

and f ¢(4) doesn’t exist. Example 56

p

x-b a-b r s

(d) y = 2 cos-1

h is continuous for all x h is differentiable for all x h¢(x) = 1, for all x > 1 h is not differentiable at two values of x

Solution:

Solution: Thus h

=

1 a -b 1 ¥ 1 - ( x - b )/(a - b ) 2 x - b a - b 2

1

=

(a - x )/( x - b )

1 a -x a -b ◊ 1 + ( x - b )/(a - x ) 2 x - b (a - x )2

x = 0, 1 h(0 + k ) - h(0) k2 = lim = 0. k Æ0 + k Æ0 + k k

h(0 + k ) - h(0) k = lim = 1. h¢ (0 –) s= m lim k Æ0 k Æ 0 k k Hence, h is not differentiable at x = 0. Similarly h¢(1 +) lim h (x) = 0 = h (0) = = 1 and h¢ xÆ 0 +

. If y is as in (b) then

4

y¢(x) =



2

=

h¢(0 +) = lim

dy 1 =– dx (a - x ) ( x - b )

If y is as in (a) then y¢(x)

Ans. (a), (c) and (d) Ï x, - • < x £ 0 Ô 2 h(x) = Ì x , 0 < x £ 1 Ô Ó x, 1 £ x < •

(s)

(a - x )( x - b ) 1-

¥

2 1 a -b 2

4 (a - x )/( x - b )

lim h(x) and lim h(x) = 1 = h(1) = lim h(x),

xÆ 0 -

xÆ1+

xÆ1-

Hence h is a continuous function and is not differentiable at two values of x.

- x + b + a - x] =

-1

4 (a - x )( x - b ) (a - b )2

4 (a - x )/( x - b )

-1

Since sin x + sin x = p /2 so (d) is clear. Example 58

function h(x).

MATRIX-MATCH TYPE QUESTIONS Example 57 Column 1

Column 1 (a) f (x) = x2 cos (1/x), f

(b) f (x) =

Column 2 x-b a-b

(b) y = 4 sin-1

x-b a-x

dy 4 = dx (a - x ) ( x - b ) (q)

dy = dx

dy (r) = dx

where q = 2 (a - x )( x - b ) a -b

1 (a - x ) ( x - b ) 2 (a - x ) ( x - b )

Column 2 not derivable

(x - 4x )

4 1/2

(a) y = 2 sin-1

(c) y = 4 sin-1q

4

If y is as in (c), we have y¢(x) =

- ( x/4)

1/3

1 - (8 x )

3 1/4

f (1/2) = 2/9 (c) f (x) = x cos 1/x, f (0) = 0 (d) f (x) = x3 sin 1/x, f (0) = 0

(q) f is differentiable everywhere but f ¢ is not continuous (r) f is not continuous (s) f ¢ is continuous but not derivable

IIT JEE eBooks: www.crackjee.xyz Differentiability and Differentiation 23.19

Ans. a

p

q

r

s

p

q

r

s

b

p

q

r

s

c

p

q

r

s

p

d

Solution:

q

r

dy d2 y 1 + x 2 /y2 1 =0fi = = 2 y dx dx a sin3 t

fix+y Hence

s

d2 y dx

t =p /2

2

If f is as in (a) then for x π 0, f ¢(x) = x = 0 lim

x Æ1/2

1 ( x - 4 x 4 )- 1/2 (1 - 16 x3 ) - 1 ( x/4)- 2/3 Ê 0 ˆ 2 form¯ = lim 2 Ë 1 x Æ1/2 0 - (8 x 3 ) - 3/4 24 x 2 4

( )

1 1 1 2 2 4 =

- 1/2

Hence, f is not continuous at x = 1/2 f ( x ) - f (0) = lim cos 1/x which If f is as in (c) then lim x Æ0 xÆ 0 x-0 does not exist so f is continuous function but not derivable. If f is as in (iv) f ¢(x) = 3x2 sin 1/x - x cos (1/x), x π 0, f ¢(0) f ¢( x) - f ¢(0) = 0 but lim = lim (3x sin 1/x - cos 1/x) x Æ0 xÆ 0 x-0 which does not exist. Example 59 Column 2 d2 y

(a) x = a cos t, y = a sin t

dx 2

t =p /3

(c) x = a cos2 t, y = a sin2 t (d) x = a (t - sin t), y = a (1 - cos t) p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(r)

d y dx 2

t =p /6

d2 y dx

2

t =p /2

3

(s)

d y dx

3

cos2 t - 4 sin 2 t

=

For (d)

9a 2 cos7 t sin3 t

d2 y dx

= -

2

. For (c) x + y = a so

= -

4 a

t =p/4

=-

=0

a (1 - cos t )2

Column 2 16 xy y2 = - 2 ( y - 2)3 2 ( 3 y 4 + 8 y 2 + 5)

(b) x2 + y2 = 4

(q) y3 =

(c) y = tan (x + y)

(r) y3 = -

(d) x3 + y3 - 6xy = 0 (d)

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

For (a), 8x + 18y

y2 = -

12x y5 16 9y3

dy dy 4x =0fi = 9y dx dx

2 d2 y Ê dy ˆ + 9y =0fi Ë dx ¯ dx 2

1 a

d2 y

16 3a

dx 2

1

Column 1

=0 = -

d2 y

Example 60

Solution:

2

(b) x = a cos3 t, y = a sin3 t (q)

1 a

(a) 4x2 + 9y2

1 (1 - 2) - (1/8) - 2/3 2 = 2. - 3/4 - 6 (1) ◊ (1/4)

Column 1

= -

dx dy d3 y and . Show that dt dt dx 3

h2 cos(1/h) - 0 =0 hÆ• h so f is differentiable everywhere but f ¢ is not continuous at x = 0. If f is as in (b) then lim f ( x ) 2x cos (1/x) + sin (1/x

Ans.

For (a) x2 + y2 = a2

Solution:

dx 2

2

= -

4 (9 y2 + 4 x 2 )

For (b) x + y

81y

3

= -

=-

4 + 16 x 2 /9 y 2 9y

4 ¥ 36 81y

3

=-

16 9y3

dy dy x =0fi =y dx dx

2 d2 y d2 y 1 + x 2 /y2 Ê dy ˆ 1+ Ë ¯ + y 2 =0fi = y dx dx dx 2

IIT JEE eBooks: www.crackjee.xyz 23.20

y3 =

y3 4

y

4

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

4

= -

¥3

dy 12 = - 5x dx y

(c) and (d) can be done similarly. Example 61 Column 1

Column 2

Solution

| log(1 + h) | log(1 + h) = lim =1 hÆ0+ hÆ0 h h f ¢(1–) = – 1.

f ¢(1+) =

1 1 xy2 - y1 - y 2 4 =0

(a) xy - log y

2x - x 2

Ans.

x (d) y = e + e p q r s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(s) y3y2 + 1 = 0

and

lim

f ¢(1+) = 3 and f ¢(1–) = – 1 If f as in (d) then f ¢(1+) = 1 and f ¢(1–) = – 3 x] denotes the greatest

Example 63

integer less than or equal to x Column 1 Column 2 (a) x|x

Solution: If xy - log y = 1 then xy1 + y – y1/y = 0 so y2 + (xy - 1) y1 = 0 yx 2 + y1 1 - x 2 If y 1 - x = sin-1 x then 2 1- x =

| h | | (1 + h)2 + (1 + h ) + 1 | =3 hÆ0+ h f ¢(1–) = – 3. If f is as in (c) then

f ¢(1+) =

(r) y2 + (x y - 1)y1 x

lim

If f (x) = |x 3 – 1| = | x – 1| |x2 + x + 1|

2 (b) y = (sin-1 x)/ 1 - x (q) (1 - x2) y1 - xy = 1

(c) y =

If f (x) = |log x| then

(b)

|x|

(q) differentiable in (– 1, 1)

(c) x x] (d) |x – 1| + |x + 1|

(r) strictly increasing in (– 1, 1) (s) not differentiable at least

p

q

r

s

a

p

q

r

s

1- x so - xy + y1 (1 - x2) = 1

b

p

q

r

s

For (c) y2 = 2x - x2 fi yy1 = 1 - x

c

p

q

r

s

d

p

q

r

s

1

Ans.

2

fi y12 + yy2 = - 1 so

(1 - x )2 y

2

+ y y2 = - 1

Solution:

x - 2x + y ] = - 1 fi y y2 = (d) can be obtained similarly. Example 62 3

2

Column 1 (a) If f (x) = |log x (b) If f (x) = | x 3 – 1| then

2

ÏÔ - x 2 - 1 < x < 0 For (a) f (x) = x|x| = Ì 2 0 £ x 0

if

x 0 for all x. Statement-2: f in statement is of the form x2 + x sin f + 1. Ans. (a) Solution: Putting y = 0 in the given equation, we have f (x) = f (x) + f (0) - 1 fi f (0) = 1 f ( x + h) - f ( x ) f ¢(x) = lim hÆ0 h f ( x ) + f (h ) + 2 hx - 1 - f ( x ) = lim hÆ0 h f (h ) - 1 f (h) – f (0) + 2x + 2 x = lim = lim hÆ0 hÆ0 h h

Example 66 (a) 1/p cm/min (c) 1/2p cm/min

Example 67 The amount of water when the height of water is 3 cm is (in cm3) (a) 4p (b) 3p (c) 27p (d) 2p

sin f ˆ cos f 3 Ê + + >0 = Ëx + 2 ¯ 4 4 Example 65

Statement-1: f (x) = e–|x| is differentiable

for all x. Statement-2: f (x) = e-|x| is continuous for all x Ans. (d) Solution:

ÏÔe- x f (x) = Ì x ÓÔ e

if

x≥0

if

x 0, and let p be the lef t hand derivative of | x – 1 | at x = 1. If lim g(x) = p, then x Æ1

(a) n = 1, m = 1 (c) n = 2, m = 2

(b) n = 1, m = – 1 (d) n > 2, m = n

30. Let g(x) = log (f (x)) where f (x) is a twice differentiable •) such that f (x + 1) = xf (x). Then g≤ (N + 1/2) – g≤ (1/2), N = 1, 2, 3, ... Ï 1 1 ¸ 1 + ... + (a) – 4 Ì1 + + ˝ 2 (2 N – 1) ˛ Ó 9 25 Ï 1 1 ¸ 1 + ... + (b) 4 Ì1 + + 2˝ 9 25 (2 N – 1) ˛ Ó Ï 1 ¸ 1 (c) – 4 Ì1 + + ... + ˝ 2 (2 N + 1) ˛ Ó 9 Ï 1 1 ¸ 1 + ... + (d) 4 Ì1 + + ˝ 2 (2 N + 1) ˛ Ó 9 25 p Ï 2 Ô x cos , x π 0 x 31. Let f (x) = Ì , x ŒR , ÔÓ 0, x = 0 then f is (a) differentiable both at x = 0 and at x = 2

(b) differentiable at x = 0 but not differentiable at x=2 (c) not differentiable at x = 0 but differentiable at x=2 (d) differentiable neither at x = 0 nor at x = 2

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 1. If x + | y | = 2y, then y as a function of x is x (b) continuous at x = 0 (c) differentiable for all x dy 1 = for x (d) such that dx 3 2. The function f (x) = 1 + | sin x| is (a) continuous nowhere (b) continuous everywhere (c) differentiable (d) not differentiable at x = 0

x ] denote the greatest integer less than or equal to x. If f (x x sin p x], then f (x) is (a) continuous at x = 0 (b) continuous in (– 1, 0) (c) differentiable at x = 1 (d) differentiable in (– 1, 1) (e) none of these 4. The function | x – 3| if x ≥ 1 Ï Ô 2 f (x) = Ì x 3 x 13 – + if x < 1 ÔÓ 4 2 4 is (a) continuous at x = 1 (b) differentiable at x = 1 (c) continuous at x = 3 (d) differentiable at x Ï 0 if x < 0 5. Let f (x) = Ì 2 Ó x if x ≥ 0 (a) f ¢ is differentiable (c) f ¢ is continuous

(b) f is differentiable (d) f is continuous

6. Let g (x) = x f (x) where Ï x sin (1/ x ) x π 0 f (x) = Ì 0 x=0 Ó x=0

IIT JEE eBooks: www.crackjee.xyz Differentiability and Differentiation 23.39

(a) (b) (c) (d)

g is differentiable but g¢ is not continuous g is differentiable while f is not both f and g are differentiable g is differentiable and g¢ is continuous

7. The functions f (x) = max {(1 – x), (1 + x), 2}, x Œ (– •, •) is

x = 1 and x =–1 x = 1 and x 8. Let h (x) = min {x, x2}, for every real number x. Then (a) h is continuous for all x (b) h is differentiable for all x (c) h¢ (x) = 1, for all x > 1 (d) h is not differentiable at two values of x 9. If f (x) = min {1, x2, x3} then (a) f (x) is continuous for all x (b) f ¢(x) > 0, " x > 1 (c) f (x) is not differentiable at x = 1. (d) f (x) is not differentiable for two values of x 10. Let f : R → R be a function such that f (x + y) = f (x) + f (y), " x, y ∈ R. If f (x) is differentiable at x = 0, then (a) f (x containing zero. (b) f (x) is continuous " x ∈ R (c) f ′(x) is constant " x ∈ R (d) f (x p p Ï Ô- x - 2 , x £ - 2 Ô p Ô 11. If f (x) = Ì- cos x, - < x £ 0 2 Ô < £1 x x 1, 0 Ô Ôln x, x > 1 Ó then

p 2 (b) f (x) is not differentiable at x = 0 (c) f (x) is differentiable at x = 1 3 (d) f (x) is differentiable at x = – 2 12. Let g : R → R be a differentiable function with g(0) = 0, g’(0) = 0 and g’(1) ≠ 0. Let (a) f (x) is continuous at x = -

Ï x g( x ), x π 0 Ô f ( x) = Ì | x | x=0 ÓÔ 0, and h(x) = e|x| for all x ∈ R. Let (f º h) (x) denote f (h(x)) and (h º f )(x) denote h(f (x)). then which of the following is (are) true ? (a) f is differentiable at x = 0 (b) h is differentiable at x = 0 (c) f º h is differentiable at x = 0 (d) h º f is differentiable at x H : x2 – y2 = 1 and a circle S with center N(x2 H and S P(x1, y1) with x1 > 1 and y1 > 0. The common tangent to H and S at P intersects the x M. If (l, m) is the centroid of the triangle ΔPMN is(are) dl 1 = 1 – 2 for x1 > 1 (a) dx1 3 x1 x1

(b)

dm = dx1 3

(c)

dl 1 = 1 + 2 for x1 > 1 dx1 3 x1

(d)

dm 1 = for y1 dy1 3

(

x12 – 1

)

for x1 > 1

È 1 ˘ È 1 ˘ 14. Let f : Í- , 2˙ Æ R and g : Í- , 2˙ Æ R be funcÎ 2 ˚ Î 2 ˚ x2 – 3] and g(x) = |x| f (x) + |4x – 7| f (x y] denotes the greatest integer less than or equal to y for y Œ R. Then (a) f È 1 ˘ Í - 2 , 2˙ Î ˚ (b) f È 1 ˘ Í - 2 , 2˙ Î ˚ (c) g Ê 1 ˆ is Á - , 2˜ Ë 2 ¯ f (x

(d) g Ê 1 ˆ in Á - , 2˜ Ë 2 ¯ 15. Let a, b Œ R and f : R Æ R f (x) = a cos (|x3 – x|) + b |x| sin (|x3 + x|). Then f is

IIT JEE eBooks: www.crackjee.xyz 23.40

(a) (b) (c) (d)

differentiable at x = 0 If a = 0 and b = 1 differentiable at x = 1 If a = 1 and b = 0 not differentiable at x = 0 If a = 1 and b = 0 not differentiable at x = 1 If a = 1 and b = 1

Ï| x | if f 1(x) = Ì x if Óe

(b) (c)

Ê 1ˆ lim f ¢ Á ˜ = 1 Ë x¯ xÆ0+

f 2(x) = x ; Ïsin x if x < 0, f 3(x) = Ì if x ≥ 0 Óx and if x < 0, Ï f2 ( f1 ( x )) f 4(x) = Ì Ó f2 ( f1 ( x )) - 1 if x ≥ 0. P. Q. R. S.

Ê 1ˆ lim xf Á ˜ = 2 Ë x¯

xÆ0+

lim x 2 f ¢ ( x) = 2

xÆ0+

(d) |f (x)| £ 2 for all x Œ 17. Let f : R Æ R, g : R Æ R and h : R Æ R be differentiable function such that f (x) = x3 + 3x + 2, g(f (x)) = x and h(g(g(x)) = x for all x Œ R. Then 1 15 (c) h(0) = 16 (a) g¢(2) =

MATRIX-MATCH TYPE QUESTIONS 1. The following functions are Column 1 Column 2 (a) sin (p x where (b) sin (p (x x]) (q) nowhere differentiable (r) not differentiable at 1

x]

(d) |x – 1| + |x + 1|

P

Q

R

S

3 1 3

1 3 1

4 4 2

2 2 4

List II onto but not one-one neither continuous nor one-one differentiable but not one-one continuous and one-one

c in

2. Let f and g interval (– 1, 1) such that g≤(x) is continuous, g(0) π 0, g¢(0) = 0, g≤(0) = 0 and f (x) = g (x) sin x. Statement-1: lim g(x) cot x – g(0) cosec x] = f ¢(0) xÆ0

Statement-2: f ¢(0) = g

x] denotes the greatest integer less

(c) x

1. 2. 3. 4.

1. Let f (x) = 2 + cos x for all real x. Statement-1: For each real t t, t + p] such that f ¢(c) = 0. Statement-2: f (t) = f (t + 2p) for each real t.

(d) h(g(3)) = 36

COMPREHENSION-TYPE QUESTIONS Paragraph 1 to 3: If a continuous function f real line R the equation f (x) = 0 has a root in R f on R

Column 2 (q) differentiable in (–1, 1) (r) strictly increasing (– 1, 1) (s) not differentiable at

(a) (b) (c)

List I f 4 is f 3 is f 2 o f 1 is f 2 is

ASSERTION-REASON TYPE QUESTIONS

(b) h¢(1) = 666

than or equal to x Column 1 (a) x |x |x| (b)

x ≥ 0;

2

16. Let f : (0, •) Æ R be differentiable function such f ( x) for all x Œ (0, •) and f (1) π 1 that f ¢(x) = 2 x then (a)

x < 0,

in

R then

f (x) = 0 has a root in R. Consider f (x) = kex – x for all real x where k is a real constant. 1. The line y = x meets y = kex for k £ 0 at

(– 1, 1) 3. Let f 1: R Æ R, f 2 Æ •

•) Æ R, f 3: R Æ R and f 4: R

k for which kex – x = 0 has only one root is (a) 1/e (c) e

(b) 1 (d) log e

IIT JEE eBooks: www.crackjee.xyz Differentiability and Differentiation 23.41

3. For k > 0, the set of all values of k for which ke x – x = 0 has two distinct roots is Ê 1ˆ (a) Á 0, ˜ Ë e¯

Ê1 ˆ (b) Á , 1˜ Ëe ¯

Ê1 ˆ (c) Á , •˜ Ëe ¯

(d) (0, 1).

x 2

x –3 equals 8. If f (x) = | x – 2| and g(x) = f (f (x)), then g¢(x) = for x 9. If f (x) = x| x f (x) is twice differentiable is

–1

1. If the function f (x) = x + e and g(x) = f (x), then the value of g¢ Ê -1 Ê sin q ˆ ˆ p p 2. Let f (θ) = sin Á tan Á ˜ ˜ , where - < q < . cos 2 q 4 4 Ë ¯¯ Ë Then the value of

FILL

IN THE

d ( f (q )) d (tan q )

Ï Ê 1 ˆ 2 – | x | if x π 1 Ô ( x – 1) sin Á Ë x – 1˜¯ 1. Let f (x) = Ì Ô if x = 1 –1 Ó where f (x) is not differentiable is

10. If xe xy = y + sin2 x, then at x = 0,

. 11. Let F (x) = f (x) g(x) h(x) for all real x, where f (x), g(x) and h(x x0 F ¢(x0) = 21 F (x0), f ¢(x0) = 4f (x0), g¢(x0) = – 7g(x0) h¢(x0) = kh (x0). Then k =

1. The derivative of an even function is always an odd

SUBJECTIVE-TYPE QUESTIONS

.

Ê 2 x – 1ˆ dy 2. If y = f Á 2 and f ¢(x) = sin x2, then dx Ë x + 1˜¯

1. Find the derivative of f (x) = sin (x2 + 1) with x 2. Find the derivative of x –1 Ï Ô 2 f (x) = Ì 2 x – 7 x + 5 Ô – 1/3 Ó

= 3. For the function f (x) =

, x π 0, f (0) = 0, the

1 + e1/ x

right hand derivative f ¢(0+) = derivative f rom the lef t, f ¢(0–),

and the .

4. If f r(x), gr (x), hr (x), r such that f r (a) = gr(a) = hr(a), r = 1, 2, 3 and f1 ( x ) f2 ( x ) f3 ( x ) F (x) = g1 ( x ) g2 ( x ) g3 ( x ) h1 ( x ) h2 ( x ) h3 ( x ) then F ¢(x) at x = a is 5. If f (x) = logx (ln x), then f ¢(x) at x = e is Ê 6. The derivative of sec–1 Á Ë2 1 – x 2 at x = 1/2 is

ˆ – 1˜¯

dy = dx

TRUE/F ALSE TYPE QUESTIONS

BLANKS TYPE QUESTIONS

x

f ( x) – 3

lim

xÆ9

INTEGER-ANSWER TYPE QUESTIONS 3

7. If f (9) = 9, f ¢(9) = 4, then

x

when x π 1 when x = 1

at x 3. Given y=

5x (1 – x )2 /3

+ cos2 (2x + 1)

dy/dx 4. Let f be a twice differentiable function such that f ≤(x) = – f (x) and f ¢(x) = g(x) f (x)]2

h (x

g(x)]2

h (10) if h 5. If (a + bx)ey/x = x x3

d2 y dx 2

2 Ê dy ˆ – y˜ = Áx Ë dx ¯

ÏÔ x 2 /2 0 £ x 0

Find the continuous function f (x) satisf ying f ¢(1) =f 11. Let R be the set of real numbers and f : R Æ R be such that for all x and y in R, |f (x) – f (y)| £ |x – y |3. Prove that f (x y

1ˆˆ ˜˜ x¯¯

if x π 0 if x = 0

Test whether (a) f (x) is continuous at x = 0 (b) f (x) is differentiable at x 18. Determine the values of x for which the following function f ails to be continuous or differentiable 1– x if x < 1 Ï Ô f (x) = Ì (1 – x ) (2 – x ) if 1 £ x £ 2 Ô 3– x if x > 2 Ó

9. Let f (x Ï – 1 if – 2 £ x £ 0 f (x) = Ì Ó x – 1 if 0 < x £ 2 and g(x) = f (| x|) + | f (x)| Test the differentiability of g(x 10. Let g(x

Ï Ê Ê 1 + Ô x exp ÁË – ÁË |x| 17. Let f (x) = Ì ÔÓ 0

x] + |1 – x|, – 1 £ x £ 3

13. If x = sec q – cos q, y = secn q – cosn q, then show that 2

Ê dy ˆ (x + 4) Á ˜ = n2 (y2 Ë dx ¯

19. If y =

ax 2 bx + + ( x – a ) ( x – b) ( x – c ) ( x – b) ( x – c ) c x–c

b c ˆ y¢ 1Ê a + + = Á y x Ë a – x b – x c – x ˜¯ 20. Let f (x), x ≥ 0 be a non-negative continuous x

function, and let F (x) =

Ú f (t ) dt, x ≥ 0. If for some 0

c > 0, f (x) £ c F (x) for all x ≥ 0, then show that f (x) = 0 for all x ≥ 21. Let a Œ R. Prove that a function f : R Æ R is differentiable at a if and only if there is a function g : R Æ R which is continuous at a f (x) – f (a) = g(x) (x – a) for all x Œ R

2

f: R Æ R f (x + y) = f (x) f (y) for all x, y in R and f (x) π 0 for any x in R. Let the function be differentiable at x = 0, and f ¢(0) = 2. Show that f ¢(x) = 2f (x) for all x in R. Determine f (x dy at x = – 1, when 15. Find dx (sin y)sin ((p /2)x) +

3 sec–1 2x + 2x tan (log (x + 2)) 2

1 Ê x + yˆ = f (x) + f (y)] for all real x and y. If 16. Let f Á Ë 2 ˜¯ 2 f ¢(0) exists and equals – 1 and f f (2).

Ï x + a if 22. Let f (x) = Ì Ó | x – 1 | if

x P (x) for all x ≥ 1, then dx

P (x) > 0 for all x

IIT JEE eBooks: www.crackjee.xyz Differentiability and Differentiation 23.43

24. If a function f a, 2a] Æ R is an odd function such that f (x) = f (2a – x) for x Œ a, 2a] and the lef t hand derivative at x = a derivative at x = – a 25. Let f Æ R be a function such that f (0) = 0. Using the f act that f ¢(0) = lim n f (1/n nÆ•

È2 ˘ –1 Ê 1 ˆ value of lim Í (n + 1) cos Á ˜ – n ˙ Ë n¯ n Æ • Îp ˚ 26. Let Ï –1 Ê x + c ˆ Ôb sin ÁË 2 ˜¯ Ô 1/2 f (x) = Ì Ô 1 Ô (eax –1) x Ó

if

–1/2 < x < 0

if

x=0

if

0 < x < 1/2

If f (x) is differentiable at x b2 = 4 Рc2 27. Let f and g that for each x, y ΠR

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 31. 33. 35. 37. 39. 41. 43.

a R be such

g¢ (0).

28. f (x) is a differentiable function and g(x) is a double differentiable function such that | f (x)| £ 1 and f ¢(x) = g(x) " x Œ R. If (f (0))2 + (g(0))2 that there exists at least one c Œ (– 3, 3) such that g(c) g≤(c

2. 6. 10. 14. 18. 22. 26. 30.

(c) (b) (b) (d) (a) (c) (c) (c)

3. 7. 11. 15. 19. 23. 27.

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p p

q q

r r

s s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p p

q q

r r

s s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p p

q q

r r

s s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p p

q q

r r

s s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

47.

SINGLE CORRECT ANSWER TYPE QUESTIONS (b) (a) (a) (c) (d) (a) (b) (d)

r

46.

Answers

1. 5. 9. 13. 17. 21. 25. 29.

(c) (c) (b) (d) (d) (b) (a)

4. 8. 12. 16. 20. 24. 28.

(d) (b) (b) (c) (d) (a) (a)

(c)

q

45.

LEVEL 1

32. 34. 36. 38. 40. 42.

p

44.

g (x – y) = g (x) g(y) + f (x) f (y) Rf ¢

(c) (d) (b), (c)

(a), (b), (a), (a), (a), (d)

(b), (c), (b), (c) (b),

(c), (d) (d) (c) (c), (d)

MATRIX-MATCH TYPE QUESTIONS

f (x – y) = f (x) g (y) – g (x) f (y) If

(b), (a), (a), (a) (b), (b) (c)

48.

IIT JEE eBooks: www.crackjee.xyz 23.44

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

p

q

r r

s s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p p

q q

r r

s s

LEVEL 2

a

p

q

r

s

SINGLE CORRECT ANSWER TYPE QUESTIONS

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

ASSERTION-REASON TYPE QUESTIONS 49. (b)

52.

50. (a)

COMPREHENSION-TYPE QUESTIONS 51. (b) 55. (c) 59. (d)

52. (d) 56. (a)

53. (c) 57. (c)

54. (b) 58. (b) 53.

INTEGER-ANSWER TYPE QUESTIONS 60. 1 64. 2 68. 0

61. 8 65. 2 69. 8

62. 1 66. 2

63. 4 67. 2 54.

1. 5. 9. 13. 17. 21. 25. 29. 33.

(c) (c) (c) (c) (b) (a) (b) (d) (d)

2. 6. 10. 14. 18. 22. 26. 30. 34.

(d) (d) (a) (c) (b) (b) (b) (b) (c)

3. 7. 11. 15. 19. 23. 27. 31. 35.

(d) (c) (a) (a) (a) (a) (c) (a) (d)

4. 8. 12. 16. 20. 24. 28. 32.

(d) (b) (d) (a) (b) (a) (a) (b)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 36. 38. 40. 42. 44. 46. 48. 50.

(a), (a), (d) (b) (b), (b) (b), (a),

(d) (d)

37. 39. 41. 43. 45. 47. 49.

(c) (d) (b), (d)

(a), (a), (b), (b), (b), (a), (b)

(c) (d) (c) (d) (c), (d) (c)

MATRIX-MATCH TYPE QUESTIONS p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

51.

ASSERTION-REASON TYPE QUESTIONS 55. (a)

56. (d)

COMPREHENSION-TYPE QUESTIONS 57. (c) 61. (a)

58. (b) 62. (b)

59. (d)

60. (c)

INTEGER-ANSWER TYPE QUESTIONS 63. 0 67. 2 71. 2

64. 1 68. 1

65. 3 69. 2

66. 3 70. 4

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17. 21. 25. 29.

(d) (b) (a) (b) (b) (a) (d) (c)

2. 6. 10. 14. 18. 22. 26. 30.

(a) (a) (d) (a) (d) (d) (b) (a)

3. 7. 11. 15. 19. 23. 27. 31.

(d) (c) (d) (d) (b) (b) (a) (b)

4. 8. 12. 16. 20. 24. 28.

(c) (b) (c) (d) (c) (b) (d)

IIT JEE eBooks: www.crackjee.xyz Differentiability and Differentiation 23.45

MULTIPLE CORRECT ANSWERS TYPES QUESTIONS 1. 3. 5. 7. 9. 11. 13. 15. 17.

(a), (a), (b), (a), (a), (a) (a), (a), (b),

(b), (d) (b), (d) (c), (d) (c) (c)

2. 4. 6. 8. 10. 12. 14. 16.

(b), (d) (b) (c)

(b), (a), (a), (a), (b), (a), (b), (a)

p

q

r

s

2.

p

q

r

s

p

q

r

s

p

q

r

s

b

p

q

r

s

b

p

q

r

s

c

p

q

r

s

c

p

q

r

s

d

p

q

r

s

d

p

q

r

s

3. (d)

ASSERTION-REASON TYPE QUESTIONS 2. (b)

3 1ˆ Ï Ê2 Ô – ÁË 3 log 2 + 9 ˜¯ x if Ô 10. f (x) = Ì 1/ x Ê1+ xˆ Ô if ÁË 2 + x ˜¯ ÓÔ 12. 0, 1, 2 3 15. p p2 – 3

FILL

1. {0} 3. 0; 1 7. 4 11. 24

BLANKS TYPE QUESTIONS Ê 2 x – 1ˆ È 2 2 x – 2 x ˘ 2. sin Á 2 Í ˙ Ë x + 1˜¯ Î ( x 2 + 1)2 ˚ 4. 0 8. 1

5. 1/e 9. 2 | x |

6. 4 10. 1

TRUE/F ALSE TYPE QUESTIONS 1. False

SUBJECTIVE-TYPE QUESTIONS 1. 2x cos (x2 + 1)

2. – 2/9

14.

e2x

16.

–1

LEVEL 1

2. 1 IN THE

x>0

Hints and Solutions

3. (a)

INTEGER-ANSWER TYPE QUESTIONS 1. 2

x£0

17. (a) continuous at x = 0 (b) not differentiable at x = 0 18. f is continuous on (– •, 2) » (2, •) and is differentiable on R – {1, 2}. 22. a = 1, b = 0 24. 0 25. 1 – 2/p

COMPREHENSION-TYPE QUESTIONS 2. (a)

f ≤ is not

–8 32 + log 2 16 + p 2 8. g is continuous on (0, 2) and differentiable on (0, 1) » (1, 2). 9. differentiable on (– 2, 2) – {0, 1}

a

1. (b)

– 2 sin (4x + 2)

7.

a

1. (b)

3 (1 – x )5/3

4. 1 6. f , f ¢

(d), (e) (b), (c) (b) (c), (d) (c) (d) (c)

MATRIX-MATCH TYPE QUESTIONS 1.

5 (3 – x )

3.

1. lim

f (5 + h2 ) - f (5 - h2 )

hÆ 0

2 h2

=

1 f (5 + h 2 ) - f (5) lim 2 hÆ0 h2

=

1 f ¢(5) + f ¢(5)] = 4 2

Ê 1 - 3 log x ˆ Ê 3 + 3 log x ˆ + tan-1 Á 2. y = tan-1 Á Ë 1 + 3 log x ˜¯ Ë 1 - 9 log x ˜¯ = p/4 - tan-1 (3 log x) + tan-1 3 + tan-1 (3 log x) = p/4 + tan-1 3 dy d2 y Thus, = 0 and so is also zero. dx dx 2 3. f (x) =

tan 2 x + sec 2 x + 2 tan sec x - 1 (1 + tan x )2 - sec 2 x

IIT JEE eBooks: www.crackjee.xyz 23.46

=

tan 2 x + sec2 x + 2 tan x sec x - 1 = tan x + sec x 2 tan x

Thus, f ¢(x) = sec2 x + sec x tan x = sec x (sec x + tan x) p f ¢ ÊÁ ˆ˜ = 2 + 2 = 2 2 + 1 Ë 4¯

(

4. log f (x) =

)

1 log x. Differentiating both sides we x

have f ¢ ( x) 1 log x 1 - log x = 2 - 2 fi f ¢(x) = f (x) f ( x) x x x2 Differentiating again both the sides, we have 1 - log x f ≤(x) = f ¢(x) + f (x) x2 È x (- 1/ x ) - (1 - log x )2 x ˘ Í ˙ Î ˚ x4

t˘ È 3t ÍÎsin 2 cos 2 ˙˚

Ê d yˆ and Á ˜ = - 2 sin t + 2 sin 2t = 2 Ë dt ¯

Ê d2 y ˆ Ê d yˆ Hence, Á ˜ = cot (t/2) and Á 2 ˜ Ë d x¯ Ëdx ¯ = -

1 dt cosec2 (t/2) × dx 2

Ê d2 y ˆ 1 Therefore, Á 2 ˜ = - cosec2 (t/2) × 2 Ëdx ¯ 1 4 sin 3t /2 sin t /2 and

d2 y dx

2

= t =p

1 1 1 ¥1¥ = 2 4 ( -1) .1 8

9. Since g(x) = |x| is a continuous function and lim

hÆ1 +

2

thus f ≤(e) = 5.

f (e) e3

= e- 3 + 1/e

dy du 3x + 4 = f ¢(u) , where u = dx dx 5x + 6 But

du 3(5 x + 6) - (3 x + 4)5 2 = = 2 dx (5 x + 6) (5 x + 6)2

2 dy 1 Ê 3x + 4 ˆ Thus = - 2 tan Á ˜ Ë 5 x + 6 ¯ (5 x + 6)2 dx d 6. (f 2(x) + g2(x f (x) f ¢(x) + g(x) g¢(x)] = dx

f (x) - g(x) f (x)] = 0 Hence f 2(x) + g2(x) is constant. Thus f 2(8) + g2(8) = f 2(2) + g 2(2) = f 2(2) + ( f ¢(2))2 = 16 + 16 = 32. sin 2 x(e3/ x - e-3/ x ) f ( x) - f (0) 7. lim = lim xÆ0+ xÆ0+ x-0 x(e3/ x + e-3/ x ) =

lim x

sin 2 x(1 - e -6 / x )

xÆ0 +

lim

xÆ0-

(

x 2 1 + e -6 / x

)

xÆ0-

hÆ1 -

f is continuous at x = 1 and x = 4. f is clearly not differentiable at x = 4. Since g(x) = |x| is not differentiable at x = 0. Now f (1 + h) - f (1) f ¢(1 +) = lim hÆ0 + h -3+h -3 = lim = - 1, hÆ0 + h

f ¢(1 -) = lim

hÆ 0 +

- (h2 + 2h) + 3h = 5/2. h

= lim

hÆ 0 -

Hence, f is not differentiable at x = 1. 10. For x = 0, f (0) = 0 and for x π 0, 1 1 È ˘ + + º˙ f (x) = x2 Í1 + 2 2 2 1 x (1 x ) + + Î ˚ = x2

1 1 - 1/(1 + x 2 )

=

x 2 (1 + x 2 ) x2

= 1 + x 2,

since lim f (x) = 1, so f is not continuous at x = 0 hÆ 0

and hence non differentiable at x = 0.

sin x Ê e6 / x - 1ˆ ◊ x ÁË e6 / x + 1˜¯

11. For x > 2,

= 0.1. (–1) = 0 f is differentiable at x = 0 and f ¢(0) = 0. Ê d xˆ 8. Á ˜ = 2 cos t - 2 cos 2t Ë dt ¯

(1/2)(1 + h )3 - (1 + h )2 + 3(1 + h ) + 1/2 - 3 h (1/2)(h3 + 3h2 + 3h)

= 0.1.1 = 0

sin 2 x(e6 / x - 1) f ( x) - f (0) = lim xÆ0x-0 x(e6 / x + 1)

= lim x ◊

f (x) = 3 = lim f (x) so f is a continuous function.

t/2) sin t/2]

x

x

Ú1 {1 + |1 - t|}d t = 1 + x2/2

1

Ú0 {1 + |1 - t|}dt = Ú0 {1 + |1 - t|} d t +

IIT JEE eBooks: www.crackjee.xyz Differentiability and Differentiation 23.47

+ f (0) + 0 fi f (0) = 0. Thus, C = 0 and f (x) = 5x + x2/2.

Ï1 + x 2 /2 x > 2 Thus, f (x) = Ì Ó 5x - 7 x £ 2 lim f (x) = 1 + 4/2 = 3 = f (2) = lim f (x)

xÆ 2 +

xÆ 2 -

1 + (1/2)(2 + h )2 - 3 xÆ 0 + h

= lim

xÆ a

xÆ 0

2 f ≤( x ) - 12 f ≤ (2 x ) + 16 f ≤ (4 x ) = 15. 2

14. f ¢(x) = (1/2)52x+1 (log 5) (2) = log 5 ◊ 52x+1 also g¢(x) = 5x log 5 + 4 log 5. So {x: f ¢(x) > g¢(x)} = {x: log 5 ◊ 52x+1 > log 5 5x + 4 log 5} = {x : 52x + 1 > 5x + 4} = {t = 5x : 5t2 - t - 4 > 0} = {t = 5x : (5t + 4) (t - 1) > 0} = {t = 5x : t > 1 or t < - 4/5} x = {t = 5 : t > 1} = (0, •) f (2 + h) - f (2) 15. f ¢(2) = lim hÆ 0 h f (2 + h) But f (2) = lim f (2 + h) = lim h = (0) (3) = 0 hÆ 0 hÆ 0 h f (2 + h) = 3. hÆ 0 h f ( x + h) - f ( x ) 16. f ¢(x) = lim hÆ 0 h Hence f ¢(2) = lim

= lim h hÆ 0

f ( x ) + f (h ) + xh - f ( x ) h

1 f (h) + x = f ¢(0) + x = 5 + x hÆ 0 h Hence, f (x) = 5x + x2/2 + C. Putting x = y = 0 in the given equation, we have f (0) = f (0 + 0) = f (0) = lim

= lim

xÆ a

f ( x ) - f (a) ¥ f ( x ) - f (a)

f ( x) - a2 f ( x) - b

2

¥

f ( x) + b f ( x) + a

f ( x) + b f ( x) + a =

f ¢ (a) ¥ f ¢ (a)

f (a ) + b f (a) + a

=

5b a

1 + sin x + 1 - sin x

18.

Hence, f is continuous but not differentiable at x = 2. 12. Since g (x) = |x| is continuous but not differentiable at x = 0, so f is continuous but not differentiable for log5 (x3 + 10x2 + 11x – 69) = 0 i.e., x3 + 10x2 + 11x – 69 = 1 fi x3 + 10x2 + 11x – 70 = 0 fi (x – 2) (x + 7) (x + 5) = 0 fi x = 2, –7, –5 2 f ¢( x ) - 6 f ¢ (2 x ) + 4 f ¢ (4 x ) lim xÆ 0 2x = lim

f ( x) - b

xÆ a

f ¢(2 +) = lim

(1/2) h 2 + 2h lim = = 2, xÆ 0 + h 5(2 + h) - 7 - 3 =5 f ¢(2 -) = lim xÆ 0 h

f ( x) - a

17. lim

1 + sin x - 1 - sin x =

cos( x/2) + sin ( x/2) + cos( x/2) - sin ( x/2) cos( x/2) + sin ( x/2) - cos( x/2) + sin ( x/2)

=

x cos( x/2) = cot sin ( x/2) 2

È 1 + sin x + 1 - sin x ˘ x \ y = cot-1 Í ˙ = 2 Î 1 + sin x - 1 - sin x ˚ fi 19.

dy 1 d = fi dx

=0

dx dy dy = - sin q, = 3 sin2 q cos q so that = dq dq dq - 3 sin q cos q. Differentiating again, we have = - 3 cos 2q

d2 y d x2

dq 3cos 2q = . dx sin q

2

d2 y 3cos 2q Ê d yˆ 2 2 3 + y ÁË d x ˜¯ 2 = 9 sin q cos q + sin q sin q dx = 9 sin2 q cos2 q + 3 sin2 q cos 2q q = p/4

is 9/4. 20. y = tan-1

1 1+ x + x

2

+ tan-1

1 x + 3x + 3 2

+ ... + n

terms = tan-1

( x + 1) - x ( x + 2) - ( x + 1) + tan-1 + ... 1 + x ( x + 1) 1 + ( x + 1)( x + 2)

n terms = tan-1 (x + 1) - tan-1 x + tan-1 (x + 2) - tan-1 (x + 1) + ...+ tan-1 (x + n) - tan-1 (x + (n - 1)) = tan-1 (x + n) - tan-1 x. 1 1 y¢(x) = 2 1 + ( x + n) 1 + x2

IIT JEE eBooks: www.crackjee.xyz 23.48

fi y¢¢(x) =

y¢¢(0) =

– 2 ( x + n)

+

(1 + ( x + n) )

2 2

-2 x

+

2x

(1 + x ) (1 + x2 ) 2 2

2

2x

(1 + x2 )

2

=0

21. Differentiating both the sides, we have cos x cos y - sin x sin y d y/dx = 0. Putting x = y = p/4, we d y˘ have = 1. Differentiating again, we get d x ˙˚ ( p /4, p /4) - sin x cos y - cos x sin y d y/dx - cos x sin y d y/ d x - sin x sin y (dy/d x)2 - sin x sin y d2 y/d x2 = 0. Putting x = y = p/4, we have d2 y ˘ =-4 ˙ d x 2 ˚ ( p /4, p /4) 22.

24. f (g(x)) = x, so f ¢(g(x)) g¢(x) = 1. Putting x = a, we get f ¢(g(a)) g ¢(a) = 1 fi f ¢(b) = 1/2. 25. y1 = nxn-1 (a cos (log x) + b sin (log x)) + b cos(log x ) ˆ Ê a x n Ë - sin (log x ) + ¯ x x fi xy1 = ny + xn (- a sin (log x) + b cos (log x)) fi xy2 + y1 = ny1 + nxn-1 (- a sin (log x) + Ê a cos(log x ) b sin (log x ) ˆ b cos (log x)) + xn Ë ¯ x x 2 fi x y2 + xy1 = nxy1 + n(xy1 - ny) - y = nxy1 + ny1x - n2y - y fi x2y2 + (1 - 2n)xy1 + (1 + n2)y = 0 26. For f to be continuous

dx dy = cos t, = k cos k t, i.e., dt dt



dy k cos kt = . Thus, dx cos t (1 - x = k (1 - y ) 2 fi (1 - x )2y1y2 - 2xy12 = - k2(2yy1) fi (1 - x2) y2 - xy1 + k2y = 0 2

) y12

2

lim (ax + b) =

2

xÆ1+

xÆ1-

if 0 £ x < 1 Ï x f ¢(x) = Ì Ó4 x - 3 if 1 < x £ 2 2(1 + h )2 - 3(1 + h) + 3/2 - 1/2 fi f ¢(1+) = lim hÆ0 + h 2h 2 + h =1 hÆ0 + h Similarly, f ¢(1-) = 1. Hence = lim

if 0 £ x £ 1 Ï x f ¢(x)= Ì Ó4 x - 3 if 1 < x £ 2 So f ¢ Ï1 if 0 £ x £ 1 f ≤(x) = Ì Ó4 if 1 < x £ 2 f ¢(1 + h) - f ¢(1) hÆ0 + h 4(1 + h) - 3 - 1 = lim =4 hÆ0 + h Similarly, f ≤(1-) = 1. Hence f ≤ x = 1, so it is not continuous there. fi

f ≤(1+) = lim

6

ap +b 6

Ï2 cos 2 x if 0 < x < p /6 f ¢(x)= Ì if p /6 < x < 1 Ó a f ¢(p/6+) = a and f ¢(p/6-) = 1 Thus a= 1, b =

23. We have lim f (x) = 1/2 and lim f (x) = 1/2, so that f

p + 6

3 = f (p/6) = lim f (x) = p 2 xÆ +

27. f

3 p 2 6

x = 1 and x = - 1 lim f (x) = a + b and lim f (x) = 1 so a + b = 1

xÆ1 -

xÆ1 +

f ¢(1+) = - 1, f ¢(1-) = 2a. Hence a = - 1/2, b = 3/2. 28. Put x = tan q so u = tan-1 Put x = sin q so v = tan-1

1 + x2 - 1 1 = tan-1 x. x 2 2 x 1 - x2 1 - 2 x2

Ê 1 - x2 ˆ du 1 du = ˜fi 2 Á 2 dv dv 2 (1 + x ) Ë ¯

= 2 sin-1 x

= x=0

1 4

-1

29. Put x = cos q so y = sin (tan q/2) and dy dy dq 1 ◊ = = dx dq dx 4 sin (q /2)cos2 (q /2) cos q when x = cos q = 1/2, we have q = p/3 so -2 2 y¢(1/2) = 3 30. Put x = sin (cos q ) so y = tan q /2. Thus dy Ê1 ˆ 2 = Á sec q /2˜ Ë2 ¯ dx

1 Ê ˆ ÁË - sin q (cos q ) ˜¯

IIT JEE eBooks: www.crackjee.xyz Differentiability and Differentiation 23.49

fi y¢(0) = - 1. 31. f ¢(x) = eax (a sin (bx + c) + b cos (bx + c)) 2 2 Put a = r cos q1, b = r sin q1, so that fi r = a + b and q1 = tan-1 b/a, we have f ¢(x) = eax r sin (bx + c + q1) f ≤(x) = reax a sin (bx + c + q1) + b cos (bx + c + q1)] = r2 eax sin (bx + c + 2q1)

a 2 + b2 and q = c + 2 tan-1 b/a.

r=

Thus

f ( x + h) - f ( x ) f ( x )[ f (h) - 1] = lim = hÆ 0 hÆ 0 h h f (x) f ¢(0) = f (x) Let g (x) = f (x) e-x, g¢(x) = - f (x) e-x + f ¢(x) e-x = - f (x)e-x + f (x)e-x = 0. Hence f (x) e-x = c, f (x) = cex but f (0) = 1, so c = 1 and thus f (x) = ex which is differentiable for all x and f ¢(x) = f (x). 33. f (x) = x + |x| + cos 19x and g(x) = cos 9x, so f and g are continuous function. Clearly f is not differentiable at x = 0 so f + g is not differentiable at x f g(x) = (x + |x| + cos 19x) (cos 9x) 32. f ¢(x) = lim

x 1 Ó x

Similarly f ¢(- 1+) π f ¢(- 1-)

g(x) =

Ú -1d t = - x

if x £ 1

0

(

)

1

2 46. If y = tan-1 x + 1 + x then

g(x) = Ú (-1)d t + 0

dy = dx

1

(

1 + x + 1 + x2

)

2

¥

x + 1+ x 1+ x

2

2

lim g( x ) = –1

x Æ1

g¢(1–) = –1 50. lim

hÆ0

x

Ú f (t )d t = – x + (x–) = –1 for x > 1 0

so g is continuous at x =1 and g¢(1+) = 0

f (0 + h) - f (0) f (h ) =1. = lim hÆ0 h h

IIT JEE eBooks: www.crackjee.xyz Differentiability and Differentiation 23.51

u = e2x and v = x4 y10 = (e2x)10 x4 + 10C1 (e2x)9 4x3 + 10C2 (e2x)8 12x2 + 10C3 (e2x)7 24x + 10C4 (e2x)6 24 y10(0) = 10C4 (e2x)6 (0) ◊ 24 = 315 × 210. pˆ pˆ Ê Ê y8 = x2 cos Ëx + 8 ¯ + 8C1 2x cos Ëx + 7 ¯ 2 2 pˆ Ê + 8C2 ◊ 2 cos Ëx + 6 ¯ 2 8¥7 Ê pˆ ◊ 2 cos Á 6 ˜ = - 56. Ë 2¯ 2 2 2 -1 2 53. (1 - x ) y = (sin x) fi - 2xy2 + (1 - x2)2yy1 so y8(0) =

=

2 sin

-1

1- x

x

2

= 2y

fi - xy + (1 - x2) y1 = 2 fi (1 - x2 )y2 - 2xy1 - xy1 - y = 0 fi (1 - x2)y2 - 3xy1 - y = 0 Differentiating n times, we obtain (1 - x2) yn+2 -(2n + 3)x yn+1 - (n + 1)2 yn = 0. Putting x = 0, we get yn+2 (0) = (n + 1)2 yn(0) 54– 56. D* f (x) = lim

hÆ 0

104 58. 40° =

u = cos x, v = x2

f 2 ( x + h) - f 2 ( x ) h

( f ( x + h) - f ( x )) ( f ( x + h) + f ( x )) = lim hÆ 0 h f ( x + h) - f ( x ) = lim ( f (x + h) + f (x)) lim hÆ 0 hÆ 0 h = 2f (x) f ¢(x) So D* f g(x) = 2f g(x) ( f g)¢ (x) = 2 f (x) g(x) (f ¢(x) g(x) + f (x) g ¢(x)) = (g(x))2 (2f (x) f ¢(x)) + (f (x))2 (2g(x) g ¢(x)) = g2(x) D*f (x) + f 2(x) D*g (x). D*( fog) = 2f (g(x)) f ¢(g(x)) g¢(x) 2 log x + x (log x)2 2 sec2 x tan x D* (tan x log x) (1) = 0. If f (x) = sin x and g (x) = cos x then fog(x)) = sin (cos x). Thus D*(fog) (x) = 2 sin (cos x) cos (cos x) (- sin x). D* (tan x log x) = tan2 x

57. Let f (x) = x when x increases f rom 100 to 104, h 1 and df = f ¢(x)h = with x = 100, h f ¢(x) = 2 x 2 x

1 = 0.02 5

= 4, d f =

100 + 0.2 = 10.2. p p radians, let f (x) = cos x and x 4 36

decreases f rom p/4 to p/4 - p/36 d f = f ¢(x)h = h sin x. When x = p/4, h = - p/36 we have df = p p p 2 sin = so 72 36 4 1

cos 40 =

2

p

+

236

=

1 2

(1 + p/36).

4 pr3 dV = 4 pr2 h: when r = 10 and h = 0.02, 3 we have dV = 4p (10)2 (0.02) = 8p.

59. V =

60. f ¢(2+) = lim

hÆ 0 +

(2 + h)2 - 5(2 + h) + 6 h

h2 - h = lim |h - 1| = 1. hÆ 0 + hÆ 0 + h 2 61. Put x = 4 tan q then f (x) = tan-1 x2/4 8x so f ¢(x) = 4 x + 16 =

lim

62. f (x) = 1 + tan-1 = 1 + 2 tan-1 63. y =

4x 4 - x2

x 4x so f ¢(x) = . 2 4 + x2

1È 1 1 ˘ + so 2 ÍÎ x + 1 x - 1 ˙˚

y5(x) =

(- 1)5 5! È 1 1 ˘ + 6 Í 2 ( x - 1)6 ˙˚ Î ( x + 1)

64. y = 1 - 2 sin2 x cos2 x = 1 so y4(x) = 65. y1 =

x = 2 tan q, we get f (x)

1 (1 - cos 4x) 4

pˆ 44 Ê cos Ë4 x + 4 ¯ fi y4(0) = 43. 4 2

4 cos(4 sin - 1 x ) 1 - x2

fi (1 - x2) y12 - 16 (1 - sin2 (4 sin-1 x)) = 16 (1 - y2) Differentiating again (1 - x2) 2y1y2 - 2xy12 = - 32yy1 fi (1 - x2)y2 - xy1 + 16y = 0.

IIT JEE eBooks: www.crackjee.xyz 23.52

66.

dy dx = - et sin t + et cos t, = et cos t - 1 - et sin t, dt dt

f1 ( x ) Df1 ( x )

thus dy cos t + sin t d2 y = fi dx cos t - sin t d x2 =

D =

d Ê cos t + sin t ˆ e- t . Á ˜ d t Ë cos t - sin t ¯ cos t + sin t

67. Since | f (x) – f (y) | £ | x – y | is true for all x, y Œ R, we have for x π y | f ( x ) - f ( y)| £ | x – y |2 | x - y| yÆ x

f ( x ) - f ( y) x-y

f ( x ) - f ( y) fi lim yÆ x x-y

a1 f1 ( x ) + a2 f2 ( x ) + a3 f3 ( x ) a1Df1 ( x ) + a2 Df2 ( x ) + a3 Df3 ( x )

f3 ( x ) Df3 ( x )

D 2 f1 ( x ) a1 D 2 f1 ( x ) + a2 D 2 f2 ( x ) + a3 D 2 f3 ( x ) D 2 f3 ( x )

(C2 Æ C2 + a1 C1 + a3 C3) 0 f3 ( x ) f1 ( x ) 1 Df1 ( x ) 0 Df3 ( x ) = 0. = a2 2 D f1 ( x ) 0 D 2 f3 ( x )

3

fi lim

1 a2

| x – y |2

69. Putting x = y = 0 in (i), we have 2f (0) = (f (0))2 fi f (0) = 0 or f (0) = 2. If f y = 0 in (i), we have 2f (x) = 2f (x + 0) = f (x) f (0) = 0. f (x) = 0 for all x Œ R but by (ii) lim f (x) xÆ0

£0 fi

| f ¢(x) | £ 0

fi f ¢(x) = 0 Hence f (x) is a constant function. Hence f (7) = f (4) = 192 68. Differentiating twice the equation a1 f 1(x) + a2 f 2(x) + a3 f 3(x) = 0, we get a1 Df 1(x) + a2 Df 2(x) + a3 Df 3(x) = 0 and a1 D2 f 1(x) + a2 D2 2 f 2(x) + a3 D f 3(x) = 0. Since not all ai’s are zero so at least one a1, a2, a3 is nonzero say a2 π 0.

2

= 2 + lim x g(x) = 2 + 0 = 2 which contradicts f (x) xÆ0

= 0 for all x ΠR. Hence f (0) = 2. lim

xÆ0

f ( x + h) - f ( x) f ( x ) f (h) - f ( x ) = lim (1 / 2) x Æ 0 h h

f (x) lim

xÆ0

f ( x) f (h) - 2 = 2 2h

lim (4 + h g(h))

xÆ0

(using (ii)) = 2 f (x) Thus f is differentiable at every x and f ¢(x) = 2f (x). fi f (x) = Ce 2x but f (0) = 2 so C = 2. Thus f (x) =2e2x. Ê1 ˆ Thus log Ë f (4)¯ = 8. 2

IIT JEE eBooks: www.crackjee.xyz

24 Applications of Derivatives 24.1

THE DERIVATIVE AS A RATE OF CHANGE

In case of a linear function y = mx + b the graph is a straight line and the slope m measures the steepness of the line by giving the rate of climb of the line, the rate of change of y with respect to x. As x changes from x0 to x1, y changes m times as much: y1 – y0 = m(x1 – x0) Thus, the slope m = (y1 – y0)/(x1 – x0) gives the change in y per unit change in x. In more general case of a differentiable function y = f (x), the difference quotient f ( x + h) - f ( x ) = f ( x + h) - f ( x ) , h π 0 x+h-x h gives the average rate of change of y (or f ) with respect to x. The limit as h approaches zero is the derivative dy/ dx = f ¢(x), which can be interpreted as the instantaneous rate of change of y with respect to x. Since the graph is a curve, the rate of change of y can vary from point to point. Illustration 1 Find the rate of change of volume of a sphere with respect to its radius when r = 4 cm V= when

4 3 pr 3

r = 4,

dV = 4p r 2 dr dV = 64p dr

Illustration 2 A point is in motion along a curve 12y = x3. Which of its coordinate change faster? Differentiating both the coordinates with respect to t we have

12

dy dx = 3x2 dt dt yt¢



xt¢

=

x2 4

Hence, if (i) –2 < x < 2, then

yt¢

< 1, i.e., the rate of change of xt¢ the ordinate is less than that of the abscissa. y¢ (ii) For x > 2 or x < –2, t > 1, i.e., the rate of change xt¢ of the ordinate is greater than that of the abscissa. (iii) For x = ± 2, the rate of change of ordinate is equal to that of the abscissa. Velocity and Acceleration

Suppose that an object is moving along a straight line and that, for each time t during a certain time interval, the object has (coordinate) x(t). Then at time t + h, the position of the object is x(t + h), and x(t + h) – x(t) is the change in position that the object experienced during the time period t to t + h. The ratio x (t + h ) - x (t ) = x (t + h ) - x (t ) t+h-t h gives the average velocity of the object during this time period. If x(t + h) - x(t ) = x¢(t) lim hÆ0 h exists, then x¢(t) gives the (instantaneous) rate of change of position with respect to time. This rate of change of position is called the velocity of the object. If the velocity function is itself differentiable, then its rate of change with respect t time is called the acceleration, in symbols a(t) = v¢(t) = x≤(t) speed at time t = |v(t)|

IIT JEE eBooks: www.crackjee.xyz 24.2 Comprehensive Mathematics—JEE Advanced

If the velocity and acceleration have the same sign then the object is speeding up, but if the velocity and acceleration have opposite signs then the object is slowing down. A sudden change in acceleration is called a jerk. Thus, Jerk is the derivative of acceleration. If a body’s position at the time t is x(t), the body’s jerk at time t is j=

da d3 x = dt dt 3

Illustration 3 A point moves in a straight line so that its distance from the 1 start in t sec is equal to s = t 4 - 4t 3 + 16t 2 . What will be 4 acceleration and at what times is its velocity equal to zero? v=

a=

ds 3 = t - 12t 2 + 32t dt d2s

= 3t 2 - 24t + 32

dt 2 v = 0 fi t3 – 12t2 + 32t = 0 fi t(t2 – 12t + 32) = 0 fi t(t – 4) (t – 8) = 0 at t = 0, t = 4, t = 8 the velocity will be zero and the corresponding acceleration will be 32, – 16, 32 respectively.

The change in f from x to x + h can be approximated by f ¢(x) h; f (x + h) – f (x)  f ¢(x) h. Illustration 5 Find the increment and differential of the function y = 3x3 + x – 1 at x = 1, Dx = 0.1. dy = (9x2 + 1) Dx At x = 1, Dx = 0.1, dy = (9 +1) (0.1) =1 3 Dy = [3(x + Dx) +(x + Dx) – 1] – (3x3 + x – 1) = 9x2 Dx + 9x Dx2 + 3Dx3 + Dx Dy – dy = 9x Dx2 + 3Dx3 At x = 1, Dx = 0.1 Dy – dy = 0.09 + 0.003 = 0.093. 24.2

TANGENT AND NORMAL

Let y = f(x) be the equation of a curve, and let P(x0, y0) be a point on it. Let PT be the tangent, PN the normal and PM the perpendicular to the x-axis (Fig. 24.1). Y

P ( x0, y0)

Illustration 4 A body whose mass is 3 kg performs rectilinear motion according to the formula s = 1 + t + t2, where s is measured Ê mv2 ˆ in cm and t in secs. Determine the kinetic energy Á 2 ˜ Ë ¯ of the body in 5s after the start. ds ds = 1 + 2t , v = = 11 cm /s dt t = 5 dt K.E. =

mv 2 3000 ¥ 121 = 181500 gm cm2/s2 = 2 2 = 181.5 ¥ 103 erg.

Differentials Let y = f (x) be a differentiable function. Let h π 0. The difference f (x + h) – f (x) is called the increment of f from x to x + h, and is denoted by Df. Df = f (x + h) – f (x) The product f ¢(x) h is called the differential of f at x with increment h, and is denoted by df df = f ¢(x) h

O

T

N

M

X

Fig. 24.1

The slope of the tangent to the curve y = f(x) at P is given by f ¢ (x0) Thus the equation of the tangent to the curve y = f(x) at (x0, y0) is y – y0 = f ¢(x 0) (x – x0) Since PN is perpendicular to PT, it follows that, if f ¢ (x 0) π 0, the slope of PN is –1 / f ¢ (x0). Hence the equation of the normal to the curve y = f(x) at (x0, y0) is f ¢ (x0) (y – y0) + (x – x0) = 0 An equation of the normal parallel to the x-axis is y = y0. The length of the tangent at (x0, y0) is PT, and it is equal to

y0

1+

1 ( f ¢( x0 ))2

.

The length of the normal is PN and it is equal to

IIT JEE eBooks: www.crackjee.xyz Applications of Derivatives 24.3

y0

1 + ( f ¢( x0 ))2 .

TM is called subtangent and the length of the subtangent is equal to 1 y f ¢ ( x0 ) 0 MN is called subnormal and the length of the subnormal is equal to |y0 f ¢(x0)|. Illustration 6 Derive the equation of tangent and normal at (x0, y0) of the curve y = log x. dy 1 = x dx



dy dx

=

x = x0

1 x0

24.4

THE ROLLE’S AND LAGRANGE’S THEOREMS

Rolle’s theorem Let f (x interval [a, b], such that (i) f (x) is continuous on [a, b], (ii) f (x) is derivable on ]a, b[, and (iii) f(a) = f(b). Then there exists a c Œ ]a, b[ such that f ¢(c) = 0. Rolle’s theorem for polynomials If f (x) is any polynomial, then between any pair of roots of f(x) = 0 lies a root of f¢(x) = 0. If a polynomial equation P (x) = 0 has at least n real roots, then P¢ (x) = 0 has at least (n–1) real roots, P"(x) = 0 has at least (n–2) real roots and so on. Lagrange’s Mean Value theorem Let f(x) be a function a, b], such that (i) f(x) is continuous on [a, b], and (ii) f(x) is derivable on ]a, b[. Then there exists a c Œ]a, b[ such that

Hence, equation of tangent at (x0, y0) is 1 ( X - x0 ) x0 Equation of normal is Y – log x0 = –x0(X – x0) y – log x0 =

f ¢(c) = Illustration 7

Length of tangent = log x0

1 + x02

Length of normal = log x0

1+

1 x02

Check the validity of Rolle's theorem for y = 4sin x in the interval [0, p]. Being exponential function, y is a continuous function x and y¢ = (4 sin x) (log 4) cos x. Also y(0) = 4° = 1 and y(p) = 4 sin p = 4° = 1. Hence, all the hypothesis of Rolle's theorem

Length of subtangent = x0 log x0 Length of subnormal = | yo / xo 24.3

ANGLE BETWEEN TWO CURVES

angle between the tangents to the two curves at their point of intersection. Let y = f(x) and y = g(x) be two curves, and let P(x0, y0) be their point of intersection. Also, let y and j be the angles of inclination of the two tangents at P with the x-axis, and let q be the angle between the two tangents. Then ± tan q =

f (b ) - f ( a ) b-a

g ¢ ( x0 ) - f ¢ ( x0 ) tan j - tan y = 1 + tan j tan y 1 + f ¢( x0 ) g ¢( x0 )

Orthogonal Curves

If the angle of intersection between two curves is a right angle then the two curves are said to be intersecting orthogonally. Two curves y = f (x) and y = g(x) cut orthogonally if f ¢(x) g¢(x) = – 1.

y¢ (p /2) = 0

and

p/2 Π[0, p]

Illustration 8 Show that the equation x3 – 3x + C = 0 cannot have two different roots in (0, 1). Consider f (x) = x3 – 3x + C. Suppose f has two distinct roots say x1 and x2 in (0,1). f is a polynomial so continuous on [x1, x2] and differentiable in (x1, x2) and f (x1) = f (x2) = 0 so by Rolle's theorem there is a Œ (x1, x2) à (0, 1) such that f ¢(a) = 0



3a 2 – 3 π 0



a = ± | œ (0, 1).

3

So x – 3x + c = 0 cannot have two different roots in (0, 1). 24.5

MONOTONICITY

A function f(x D is said to be nondecreasing, increasing, non-increasing and decreasing respectively, if for any x1, x2 Œ D and x1 < x2, we have f(x1) £ f(x2), f(x1) < f(x2), f(x1) ≥ f(x2) and f(x1) > f(x2), respectively (Figs 24.2 and 24.3). The function f(x) is said to be monotonic if it possesses any of these properties.

IIT JEE eBooks: www.crackjee.xyz 24.4 Comprehensive Mathematics—JEE Advanced Y

y¢(x) =

y = f (x)

=

f (x2) f (x1) x2 x1 An increasing function

X

-2 + 2 x 2 (1 + x + x 2 )2

=

2( x 2 - 1) (1 + x + x 2 )2

Fig. 24.2

24.6

f (x1) f (x2)

O

Thus,

x1 x2 A decreasing function

X

Fig. 24.3

As an immediate consequence of Lagrange’s Mean theorem, we have Testing monotonicity Let f (x) be continuous on [a, b] and differentiable on ]a, b[. Then (i) for f (x) to be non-decreasing (non-increasing) on [a, b f ¢ (x) ≥ 0 ( f ¢(x) £ 0) for all x Œ ]a, b[. (ii) for f (x) to be increasing (decreasing) on [a, b] it is f ¢(x) > 0 ( f ¢(x) < 0) for all x Œ]a, b[. Illustration 9 Show that the function y = 2 x - x 2 increases in the interval (0, 1) and decreases in (1, 2) 1 2 - 2x 1- x = x π 0, 2. 2 2 2x - x (2 - x ) x x = 0, x = 2.

y¢(x) > 0 if 1 – x > 0 so y increases on (0, 1) and y¢(x) < 0 for x Œ (1, 2), so y decreases on (1, 2). Illustration 10 Find the interval of monotonicity of y =

y¢(x) > 0 ¤ x2 – 1 > 0 ¤ x > 1 or x < –1

Hence, y increases on (1, •) » (– •, –1) and decreases on (–1, 1).

Y



(1 + x + x 2 )2

Since the denominator is always positive so the sign of y¢(x) according to the sign of x2 –1.

O

y¢(x) =

(1 + x + x 2 )(-1 + 2 x ) - (1 - x + x 2 )(1 + 2 x )

1 - x + x2 1 + x + x2

MAXIMA AND MINIMA

A function has a local maximum at the point x0 if the value of the function f (x) at that point is greater than its values at all points other than x0 of a certain interval containing the point x0. In other words, a function f (x) has a maximum at x0 if it a, b) containing x0, i.e., with a < x0 < b, such that for all points different from x0 in (a, b), we have f (x) < f (x0). A function f (x) has a local minimum at x0 if there exists an interval (a, b) containing x0 such that f (x) > f (x0) for x Œ (a, b) and x π x0. One should not confuse the local maximum and local minimum of a function with its largest and smallest values over a given interval. The local maximum of a function is the largest value only in comparison to the values it has at Similarly, the local minimum is the smallest value only in ciently close to the local minimum point. The general term for the maximum and minimum of a function is extremum, or the extreme values of the function. A necessary condition for the existence of an extremum at the point x0 of the function f (x) is that where f ¢(x0) = 0, or f ¢ (x0) does not exist. The points at which f ¢ (x) = 0, or where f ¢(x) does not exist, are called critical points of the function f. First derivative test (i) If f ¢(x) changes sign from positive to negative at x0, i.e., f ¢(x) > 0 for x < x0, and f ¢(x) < 0 for x > x0, then the function attains a local maximum at x0. (ii) If f ¢(x) changes sign from negative to positive at x0, i.e., f ¢(x) < 0 for x < x0, and f ¢(x) > 0 for x > x0, then the function attains a local minimum at x0. (iii) If the derivative does not change sign in moving through the point x0, there is no extremum at that point. Second derivative test Let f be twice differentiable, and let c be a root of the equation f ¢(x) = 0. Then

IIT JEE eBooks: www.crackjee.xyz Applications of Derivatives 24.5

(i) c is a local maximum point if f ¢¢ (c) < 0 (ii) c is a local minimum point if f¢¢ (c) > 0. However, if f¢¢(c) = 0, then the following result is applicable. Let f ¢(c) = f¢¢ (c) = º = f (n – 1) (c) = 0 (where f r denotes the rth derivative), but f (n)(c) π 0. (i) If n is even and f (n) (c) < 0, there is a local maximum at c, while if f (n) (c) > 0, there is a local minimum at c. (ii) If n is odd, there is no extremum at the point c.

Illustration 13 Find the greatest and least value of y = x3 – 3x2 + 6x – 2 on [–1, 1]. y is a differentiable function of x and y¢(x) = 3x2 – 6x + 6 = 3(x2 – 2x + 2) = 3((x – 1)2 + 1) > 0. Hence y increases on [–1, 1]. Thus the greatest value = y(1) = 1 – 3 + 6 – 2 = 2, least value = y(–1) = –1 – 3 – 6 –2 = –12. A point x = c = f (x) if f ≤ (c

y f ≤¢(c) π 0

Illustration 11 Find the extrema of y = 2x3 – 3x2. y is differentiable function and y¢ = 6x2 – 6x = 6x(x – 1). So, points extremum are 0 and 1 y¢¢ = 6 (2x – 1) y¢¢(1) = 6 > 0, y¢¢(0) = – 6 < 0. Hence, ymax = y(0) = 0 and ymin = y(1) = 2 – 3 = –1 Greatest/Least Value (Absolute Maximum/Absolute Minimum) Let f be a function with domain D. Then f has a greatest value (or absolute maximum) at a point c Œ D if f (x) £ f (c) for all x in D and a least value (or absolute minimum) at c if f (x) ≥ f (c) for all x in D.

SOLVED EXAMPLES

SINGLE CORRECT ANSWER TYPE QUESTIONS The point on the parabola 2y = x2, which Example 1 is nearest to the point (0, 3) is 1ˆ Ê (a) (±4, 8) (b) Á ±1, ˜ Ë 2¯

Ans. (c) Let the point on the parabola 2y = x2 be

Solution:

Illustration 12 Function

Domain

Greatest/least value

y = x2

(– •, •)

No greatest value, least value at x = 0

y = x2

[0, 2]

greatest value at x = 2 least value at x = 0

y = x2

(0, 2]

greatest value at x = 2 no least value

y = x2

(0, 2)

no greatest value, no least value

If f is continuous at every point of D and D = [a, b], a closed interval, then f assumes both a greatest value M and a least value m, i.e., there are x1, x2 Œ [a, b] such that f (x1) = M and f (x2) = m and m £ f (x) £ M for every x in [a, b]. The greatest (least) value of continuous function f (x) on the interval [a, b] is attained either at the critical points or value of the function, we have to compute its values at all the critical points on the interval (a, b), and the values f (a), f (b) of the function at the end-points of the interval, and choose the greatest (least) out of the numbers so obtained.

9ˆ Ê (d) Á ±3, ˜ Ë 2¯

(c) (±2, 2)

Ê x2 ˆ Á x, 2 ˜ . If L is distance of this point from (0, 3) then Ë ¯ Ê x2 ˆ S = L = x + Á - 3˜ Ë 2 ¯ 2

2

2

Ê x2 ˆ1 dS = 2x + 2 Á - 3˜ ◊ 2 x dx Ë 2 ¯2 Ê x2 ˆ = 2 x Á - 2˜ Ë 2 ¯ dS = 0 fi x = 0, ±2 dx d 2S dx 2

È 3x 2 ˘ d 2S = 2Í - 2˙ so 2 Î 2 ˚ dx

=8>0 x =±2

S is minimum when x = ±2, hence required point is (±2, 2). A dynamite blast blows a heavy rock Example 2 straight up with a launch velocity of 160 m/sec. It reaches a height of s = 160t – 16t2 after t sec. The velocity of the rock when it is 256 m above the ground on the way up is

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(a) 98 m/s (c) 104 m/s

(b) 96 m/s (d) 48 m/s

x2/8 + y2/18 = 1 so that the area of the triangle formed by the tangent at P and the coordinate axes is the smallest, are given by

Ans. (b) Solution:

v=

ds = 160 – 32t dt

t

for which s(t) = 256. So 160t – 16t2 = 256 fi

16(t – 10t + 16) = 0



(t – 2) (t – 8) = 0

Ans.

fi So

t = 2, t = 8. v (2) = 160 – 32.2 = 96 v(8) = 160 – 256 = – 96. So the velocity on the way up in 96 m/s.

Let f be a twice differentiable function Example 3 such that f ≤(x) > 0, x Œ R Ê x2 ˆ h(x) = 2 f Á ˜ + f (6 – x2), x Œ R Ë 2¯ increases increases increases increases

on on on on

(–2, •) (–•, •) (0, •) (–2, •) ~ [0, 2]

(b) ( 8, 0)

(c) ( 18, 0)

(d) none of these

(a)

2 x 2 dy =0 + y 8 18 dx

Now dy dx (



8 cos q , 18 sin q

È Ê x2 ˆ ˘ h¢(x) = 2 x Í f ¢ Á ˜ - f ¢ (6 - x 2 )˙ ÍÎ Ë 2 ¯ ˙˚ f ≤(x) > 0, x Œ R fi f ¢ is an increasing function

y–

= -

9 8 cos q 4 18 sin q

= -

9 cot q 2

(

9 cot q x - 8 cos q 2

18 sin q = -

)

Êx x > 6 - x 2 , i.e., x2 > 4 then f ¢ Á ˜ > f ¢(6 – x2), 2 Ë 2¯ hence for x > 0, h¢(x) > 0 and for x < 0, h¢(x) < 0. (i) h¢(x) > 0, x > 0, x > 2

=

Ê 8 ˆ Á cosq , 0˜ Ë ¯

6 = 12 cosec 2q cos q sin q

But cosec 2q is smallest when q = p 4 . Therefore A is smallest when q = p 4 . Hence the required point is 1 1 ˆ Ê , 18 ◊ ˜ = (2, 3) ÁË 8 ◊ 2 2¯

(ii) h¢(x) < 0, x < 0, x < –2 Thus, h increases on (2, •) and decreases on (–•, –2) Ê x2 ˆ x2 < 6 - x 2 , i.e., x2 < 4, f ¢ Á ˜ < f ¢(6 – x2) Also if 2 Ë 2¯ 2 So, for x > 0, x < 4, h¢(x) < 0

and

Thus the area of the triangle formed by this tangent and the coordinate axes is 1 1 A= 18 ◊ 8 ◊ 2 cos q sin q



If

dy 9x =dx 4y

Hence the equation of the tangent at ( 8 cos q, 18 sin q) is

Ê 18 ˆ Á 0, sinq ˜ Ë ¯

Solution:

2

)



Therefore, the tangent cuts the coordinate axes at the points

Ans. (d)



(a) (2, 3)

Solution : Any point on the ellipse is given by ( 8 cos q, 18 sin q)

2

(a) (b) (c) (d)

The coordinates of the point P(x, y) lying

Example 4

The point(s) on the curve y3 + 3x2 = 12y Example 5 where the tangent is vertical is/are

and for x < 0, x2 < 4, h¢(x) > 0

(a) (± 4/ 3 – 2)

(b) (±

(c) (0, 0)

(d) (± 4/ 3 , 2)

Thus h increases on (–2, 0)

Ans. (d)

Hence h increases on (–2, 0) » (2, •) and decreases on (–•, –2) » (0, 2).

Solution: 3y2

11/3 , 1)

Differentiating the given curve w.r.t. x, we get

dy dy dy 2x + 6x = 12 fi = - 2 dx dx dx y -4

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At point where the tangent(s) is(are) vertical,

dy is not dx

d 2 y 2a 2 2b2 d2 y = + so dx 2 x 2 ( a - x )3 dx 2

x=

y2 – 4 = 0 fi y = ± 2 y

4 when y = 2, 8 + 3x2 = 12(2) fi 3x2 = 16 fi x = ± 3 2 2 when y = – 2, – 8 + 3x = – 24 fi 3x

y

=

x=

a2 and a+b

=

( a + b)2 a

y = ax3 + bx2 + cx

Example 8 x-axis at P

(b) y = 2x + 1 (d) y = x + 2

(a) 3y = 9x + 2 (c) 2y = x + 8

>0

b 2 ( a + b) a 2 ( a + b) Èb ˘ + = ( a + b) Í + 1˙ 2 ab Îa ˚ a

3 Example 6 y2 = 8x and xy = – 1 is

a2 a+b

y-axis at a point Q where its a, b, c are respectively

Ans. (d) Solution: (t t

xy

y= -

dy dy 1 1 1 = 2 fi = 2 fi x dx d x (t , - 1/ t ) t x

\

xy = – 1 at (t

Ans t) is

Solution:

1 1 1 2 = 2 (x – t) fi y = 2 x – t t t t For this line to be tangent to the parabola y2 = 8x it 2 y = mx + so m y+

m= \

–t=

Example 7

dy = 0 fi 12a – 4b + c = 0 d x ( - 2, 0) dy =3fic=3 d x (0, 8)

-2 1 2 and or m = – t = 2 m t t 1 fi – t3 = 1 fi t = – 1 2 t y=x

0 = – 8a + 4b – 2c + 8 fi – 8a + 4b – 2 = 0 (ii) Solving (i) and (ii) a

(c)

(a) (4, 4) (b)

1 1 + a b

(d) none of these

ab a+b

Ans.

Ans. (d) dy -b a dy = ( -1) - 2 ◊ =0fix= a±b dx ( a - x )2 x dx 2

Solution:

Since 0 < x < a so x =

2y

dy dy = (2 – x)2 – 2x (2 – x) , so = dx d x (1, 1) y–1= –

2

a2 ab and a – x = a+b a+b

1 2

P, then P is (b) (–1, 2)

(c)

Solution: –

b=0

If the tangent at (1, 1) on y2 = x(2 – x)2

Example 9

If a, b

a+b a

(i)

y-axis at (0, 8) so

b2 a2 y= + , 0 < x < a is a-x x (a)

dy = 3ax2 + 2bx + c dx x-axis at (– 2, 0) so



y=

1 (x – 1) 2

-x + 3 2

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(1/4) (–x + 3)2 = x(4 + x2 – 4x) fi

x2 – 6x + 9 = 16x + 4x3 – 16x2



4x3 – 17x2 + 22x – 9 = 0



(x – 1) (4x2 – 13x + 9) = 0



(x – 1)2 (4x – 9) = 0

3p p – 3 cos q = 0 fi q = or . Corresponding to these 2 values of q, we have 2 p p = – 1, y = 3 + 2 cos = 3; 2 2 3p 3p = 2 + 3 = 5, y = 3 + 2 cos =3 x = 2 – 3 sin 2 2 Thus the required points are (– 1, 3), (5, 3). x = 2 – 3 sin

Since x = 1 is already the point of tangency, x = 9/4 and 2

y2 =

9Ê 9ˆ 9 . Thus the required point is (9/4, 3/8). Á 2 - ˜¯ = 4Ë 4 64

Example 10

The tangent to the curve x = a cos 2q

cos q, y = a cos 2q sin q at the point corresponding to q = p /6 is (a) parallel to the x-axis (b) parallel to the y-axis (c) parallel to line y = x (d) parallel to the line y = –x Ans.

x x and g(x) = where sin x tan x 0 < x £ 1, then in this interval (a) both f(x) and g(x) are increasing functions (b) both f(x) and g(x) are decreasing functions (c) f(x) is an increasing function (d) g(x) is an increasing function Ans. (c) sin x - x cos x f ¢(x) = , Solution: sin 2 x Example 12

If f(x) =

(a)

Solution:

dx - a cos q sin 2q = –a cos 2q sin q + dq cos 2q =–a =

cos 2q

- a sin 3q cos 2q

dy =a dq =

(cos 2q sin q + cos q sin 2q )

cos 2q cos q – a

cos 2q

Example 13 function



Ê 5 - 27 ˆ (a) Á - 3, » (2, •) 2 ˜¯ Ë È 3 - 21 ˘ (b) Í- 4, ˙ » (1, •) 2 ˚ Î (c) (– •, •) (d) [1, •).

Example 11 The points of contact of the vertical tangents to x = 2 – 3 sin q, y = 3 + 2 cos q are (a) (2, 5), (2, 1) (b) (– 1, 3), (5, 3)

Ans.

(c) (2, 5), (5, 3)

Solution:

Ans. (b) Solution:

For vertical tangents

dx = 0 so, we have dq

The set of all values of a for which the

decreases for all real x is

So the tangent to the curve at q = p 6 is parallel to the x-axis.

(d) (– 1, 3), (2, 1)

tan 2 x

Ê a+4 ˆ 5 - 1˜ x – 3x + log 5 f (x) = Á Ë 1- a ¯

cos 2q dy = 0. d x q = p /6

tan x - x sec2 x

Let u(x) = sin x – x cos x, so that u¢(x) = x sin x > 0 for 0 < x £ 1. So u(x) > u(0) = 0. Also f ¢(x) > 0 for 0 < x £ 1. Hence f increases on (0, 1]. Let v(x) = tan x – x sec2 x, so that v¢(x) = – 2x sec2 x tan x < 0 for 0 < x £ 1. Thus v(x) < v(0), i.e., g¢(x) < 0 for 0 < x £ 1. So g decreases on (0, 1].

a cos 3q

dy = – cot 3q dx

Hence

sin q sin 2q

g¢(x) =

(b) Differentiating, we get Ê a+4 ˆ - 1˜ 5x4 – 3 f ¢(x) = Á 1 a Ë ¯

For f (x) to be decreasing for all x, we must have f ¢(x) < 0 for all x.

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(a) (b) (c) (d)

Ê a+4 ˆ 4 3 - 1˜ x < " x Á 5 Ë 1- a ¯



This is possible only if a+4 1- a

decreases on (–•, 0) increases on (0, •) x = –10 is a point of minimum x = 0.1 is a point of maximum

Ans. (c)

-1£ 0

Solution f ¢(x) = 12x3 + 120x2 – 0.12x – 1.2

This inequality is always true if a > 1, i.e., a Œ (1, •).

= 12(x3 + 10x2 – 0.01x – 0.1)

Moreover, we must have a ≥ – 4 for Therefore, we have

= 12(x2 – 0.01) (x + 10)

a+4 1- a

a + 4 to be real.

= 12(x – 0.1) (x + 0.1) (x + 10) £1 fi a + 4 £1- a

So

[∵ we consider only a < 1] fi

a + 4 £ 1 + a2 – 2a





Thus



on (–10, –0.1) so f has a point of minimum at x = –10, x = 0.1 is a point of local minimum and x = –0.1 is point of local maximum. Example 16

2 log (p + x ) (x≥ 0, Example 14 Thefunction g(x) = eax log (e + x ) a > 0) is (a) increasing on [0, •) (b) decreasing on [0, •) (c) increasing on [0, p /e) and decreasing on [p /e, •) (d) decreasing on [0, p /e) and increasing on [p /e, •) Ans. (b) 2

Since eax increases on [0, •) so it is log (p + x ) enough to consider f (x) = log (e + x )

f ¢(x) = =

log (e + x ) ¥ (e + x ) - (p + x ) log (p + x ) (p + x ) (e + x ) (log (e + x ))2

Solution

f ¢(x) = ex(1 – x) + x(1 – 2x) ex(1 – x) = (1 + x – 2x2) ex(1 – x) = – (x – 1) (2x + 1) ex(1 – x)

Since ex(1 – x) > 0 for all x, so f ¢(x) > 0 if and only if (x – 1) (2x + 1) < 0 i.e. – 1/2 = min (1, – 1/2) < x < max (1, – 1/2) = 1

The equation ex–1 + x – 2 = 0 as (b) two real roots (d) four real roots

Ans. (a)

Thus (e + x) log (e + x) < (e + x) log (p + x) < (p + x) log (p + x) for all x > 0. Thus, f ¢(x) < 0 for " x > 0 fi f (x) decreases on (0, •). The function f (x) = (3x + 40x – 0.06x

Ans. (a)

(a) one real root (c) three real roots

Since log function is an increasing function and e < p, log (e + x) < log (p + x).

Example 15 – 1.2x)

increasing on [– 1/2, 1] decreasing on R increasing on R decreasing on [– 1/2, 1]

Example 17

(log (e + x ))2

3

(a) (b) (c) (d)

Let f (x) = xex(1 – x), then f (x) is

Thus f increases on [–1/2, 1].

1 1 - log (p + x ) log (e + x ) ¥ p+x e+x

4

f ¢(x) < 0 on (–•, –10) » (–0.1, 0.1) Since f ¢(x) < 0 on (–•, –10) and f ¢(x) > 0

0 £ a2 – 3a – 3

3 - 21 2 È 3 - 21 ˘ a Œ Í- 4, ˙ » (1, •) 2 ˚ Î

Solution:

f ¢(x) > 0 on (–10, –0.1) » (0.1, •)

2

Solution: Clearly, x Assume that f (x) = ex – 1 + x – 2 = 0 has a real root a other than x = 1. We may suppose that a > 1 (the case a < 1 is exactly similar). Applying Rolle’s theorem on [1, a] (if a < 1 apply the theorem on [a, 1]), we get b Œ (1, a) such that f ¢(b) = 0. But f ¢(b) = eb – 1 + 1, so that eb – 1 = –1, which is not possible. Hence there is no real root other than 1.

IIT JEE eBooks: www.crackjee.xyz 24.10 Comprehensive Mathematics—JEE Advanced

Example 18 The function f satisfying f ¢(x) for any x Œ (a, b) is

f ( b) - f ( a) π b-a

(a) f (x) = x1/3, a = – 1, b = 1 x =1 Ï2 Ô 2 (b) f (x) = Ì x 1 < x < 2 , a = 1, b = 2 Ô1 x=2 Ó

Ï6 - 3x Ô10 - x Ô = Ì Ôx + 4 ÔÓ3x - 6

(c) f (x) = x|x|; a = – 1, b = 1 (d) f (x) = 1/x; a = 1, b = 4 Ans. (b)

(b) is differentiable on (1, 2) but not continuous at x = 1 and x = 2 as lim f (x) = lim x2 = 1 π f (1) and lim f (x) x Æ1 +

= lim x2 = 4 π f (2). Moreover, xÆ2-

xÆ2-

f (2) - f (1) = – 1 and 2 -1

f ¢(x) = 2x > 0 on (1, 2) so there is no x Œ (1, 2) such that f (2) - f (1) = f ¢(x). 2 -1 Example 19 Suppose f is differentiable on R and a £ f ¢(x) £ b for all x Œ R where a, b > 0. If f (0) = 0, then (a) f (x) £ min(ax, bx) (b) f (x) ≥ max(ax, bx) (c) a £ f (x) £ b (d) ax £ f (x) £ bx. Ans. (d) Solution: For x > 0. Applying Lagrange’s theorem on [0, x] we have c Œ (0, x) such that f ( x ) f ( x ) - f (0) = = f ¢( c) x x-0 f ( x) £ b fi ax £ f (x) £ bx, x > 0. x Similarly for x < 0 applying Lagrange’s theorem for [x, 0], we have ax £ f (x) £ bx. But a £ f ¢(c) £ b so a £

Example 20 The minimum value of f(x) = |3 – x| + |2 + x| + |5 – x| is (a) 0 (b) 7 (c) 8 (d) 10 Ans. (b) Solution:

f can be written as

if x £ - 2 if - 2 < x < 3 if 3 £ x < 5 if x ≥ 5

if x £ - 2 if - 2 < x < 3 if 3 £ x < 5 if

x≥5

By drawing the graph of f (x) we see that minimum value of f is 7.

Solution: The functions in (a), (c), (d) satisfying hypothesis of Lagrange’s Mean Value theorem so there is f ( b) - f ( a) = f ¢(c). The function in c Œ (a, b) such that b-a

x Æ1 +

Ï3 - x - (2 + x ) + 5 - x Ô3 - x + 2 + x + 5 - x Ô f(x) = Ì Ôx - 3 + 2 + x + 5 - x ÔÓ x - 3 + 2 + x + x - 5

Example 21 (a) (b) (c) (d)

Ï 4Ê 1ˆ Ô x Á 2 + sin ˜ , x π 0 then Let f (x) = Ì Ë x¯ Ô x=0 0, Ó

f is not differentiable at x = 0 f has a maximum at x = 0 The range of f is [0, •) f increases on (–•, •) ~ (–1, 2).

Ans. (c) Solution:

lim

xÆ0

1ˆ f ( x) - f (0) 3Ê = lim x ÁË 2 + sin ˜¯ xÆ0 x x-0

Ï 3Ê 1ˆ 1 2 Ô4 x Á 2 + sin ˜ - x cos , x π 0 Ë Thus f ¢(x) = Ì x¯ x Ô x=0 0, Ó Ï 2È 1 1˘ Ô x Í8 x + 4 sin - cos ˙ , x π 0 = Ì Î x x˚ Ô x=0 0, Ó For x > 0, f ¢(x) > 0 and for x < 0, f ¢(x) < 0. Thus f decreases on (–•, 0) and increases on (0, •). Therefore, f has a minimum at x = 0. Since f (x) > 0 for all x π 0 and f is continuous so range of f is [0, •). Example 22 The coordinates of the point on the parabola y2 = 8x which is at minimum distance from the circle x2 + (y + 6)2 = 1 are (a) (2, – 4) (b) (18, –12) (c) (2, 4) (d) (1, 8) Ans. (a) Solution: Let P (2t 2, 4t) be any point on the parabola. The centre of the given circle is O(0, –6) and the radius is 1. OP2 = 4t4 + (4t + 6)2 = 4[t4 + 4t2 + 9 + 12t] = 4x, where x = t4 + 4t2 + 12t + 9

IIT JEE eBooks: www.crackjee.xyz Applications of Derivatives 24.11

dx = 4t3 + 8t + 12 = 4(t3 + 2t + 3) dt = 4(t + 1) (t2 – t + 3) dx =0 dt

So



t=–1

= – 4 sin x sin (3x 2) sin ( x 2) Hence x = 0, 2p/3, p and 2p are the critical points. Also, f (0) = 1+ 1/2 – 1/3 = 7/6, f (2p /3) = – 13/12, f (p) = –1/6 and f (2p) = 7/6. Hence the greatest value is 7/6 and the least value is –13/12. Thus the difference is

(other roots are imaginary) 2

7 Ê 13 ˆ 27 9 - Á- ˜ = = 6 Ë 12 ¯ 12 4

2

d x d x = 4(3t2 + 2), > 0. 2 dt dt 2 t = -1 Hence OP2 is minimum at t = –1. But if A is any point on the circle and on OP (min), then AP will be minimum when OP is minimum as AP = OP – (radius of circle). Thus the required point is P(2 (– 1))2, 4(–1)) = (2, – 4). Example 23 The image of the interval [– 1, 3] under the maping f (x) = 4x3 – 12x is (a) [–2, 0] (b) [–8, 72] (c) [–8, 0] (d) [–4, 36] Ans. (b) Solution: f (x) for x Œ [–1, 3]. By virtue of the continuity of f (x), the image is the interval È ˘ f ( x ), max f ( x )˙ Í x Œmin 1, 3 x Œ 1, 3 [ ] [ ] Î ˚ The critical points of f (x) are given by f ¢(x) = 12x2 – 12 = 12(x2 – 1) = 0. That is, x = ± 1, so that f (1) = 4 ◊ 1 – 12 = – 8, f (– 1) = – 4 + 12 = 8 and f (3) = 4 ◊ 27 – 12 ◊ 3 = 108 – 36 = 72. max f(x) = f(3) = 72 \ x Œ[ - 1, 3]

min

and

x Œ[ - 1, 3]

Example 25 If y = a log |x| + bx2 + x has its extremum values at x = – 1 and x = 2, then (a) a = 2, b = –1 (b) a = 2, b = – 1 2 (c) a = –2, b = 1 2 (d) a = 1, b = –2 Ans. (b) Solution: We have y¢ = a/x + 2bx + 1 and y¢ (– 1) = 0 and y¢(2) = 0. \ – a – 2b + 1 = 0 a 2 + 4b + 1 = 0 and a 2 – 2a + 2 + 1 = 0 fi Hence a = 2 and b = –1 2 . Example 26 If q is the angle (semi-vertical) of a cone of maximum volume and given slant height, then tan q is given by (a) 2 (b) 1 (c) Ans.

f(x) = f(1) = – 8

Hence the image of [–1, 3] under the mapping f(x) is [–8, 72].

2

(d)

(c)

Solution: See Fig. 24.4. Let OB = l, OA = l cos q and AB = l sin q (0 £ q £ p/2). Then

The difference between the greatest and 1 least values of the function f (x) = cos x + cos 2x – 2 cos 3x is (b) 8 7 (a) 2 3 Example 24

(c) 9 4

(d) 3 8

Ans. (c) Solution: The given function is periodic, with period 2p. So the difference between the greatest and least values of the function is the difference between these values on the interval [0, 2p ]. We have f ¢(x) = – (sin x + sin 2x – sin 3x)

3

Fig. 24.4

IIT JEE eBooks: www.crackjee.xyz 24.12 Comprehensive Mathematics—JEE Advanced

V=

p p 3 (AB)2 (OA) = l sin2 q cos q 3 3

For minimum value of S,

p 3 dV l sin q (3 cos2q – 1) = 3 dq



So from dV/dq = 0, we get q = 0 or cos q = 1 V(0) = 0, V (p 2) = 0 and 1 ˆ 2p l 3 Ê V Á cos - 1 ˜= Ë 3¯ 9 3 Hence V is maximum when cos q = 1 2

3 . Also,

2.

2

2

2

Ans. (d) Solution: fi

We have f (x) = x2 + 2bx + 2c2 = (x + b)2 + 2c2 – b2

min f (x) = 2c2 – b2 g(x) = – x2 – 2cx + b2

Also

= b2 + c2 – (x + c)2 So

max g(x) = b2 + c2.

Since min f (x) > max g(x), So

2c2 – b2 > b2 + c2



c2 > 2b2 fi |c| >

2 |b|.

x2 + y2 = 1 at 27 ( 3 3 cos q, sin q) (where q Π(0, p/2)). Then the value of q such that sum of intercepts on axes made by this tangent is least is (a) p /3 (b) p /6 (c) p /8 (d) p /4 Example 28

Tangent is drawn to ellipse

Ans. (a) Solution:

Equation of tangent at ( 3 3 cos q, sin q) is x cos q 3 3

-

cos q sin 2 q

Since q Π(0, p/2), so tan q = d2 S 2

= 0 fi tan3 q = 3 3 3 fi q = p/3.

= 3 3 sec q tan2 q + 3 3 sec3 q + cosec2 q

dq cot q + cosec3 q

3 , i.e., tan q =

(d) |c| > |b|

2

cos q 2

Also

Example 27 If f (x) = x + 2bx + 2c and g(x) = – x – 2cx + b2 are such that min f (x) > max g(x), then the relation between b and c is (a) |c| < |b| (b) 0 < c < b/2 (c) |c| < |b|

3 3 sin q



dS =0 dq

+ y sin q = 1

Its intercept on coordinates axes are 3 3 cos q and cosec q. If S denotes their sum, then S = 3 3 sec q + cosec q dS = 3 3 sec q tan q – cosec q cot q dq



d2 S dq 2

> 0. Thus S is least when q = p/3. q = p /3

Example 29 (a) (b) (c) (d)

Let f (x) = (x – 3 )5 (x + 1)4 then

x = 7/9 is a point of maxima x = 3 is a point of minimum x = –1 is a point of maxima f has no point of maximum or minimum

Ans. (c) Solution: f ¢(x) = (x – 3)4 (x + 1)3 (9x – 7) and f ¢¢(x) = 8(x – 3)3 (x + 1)2 (9x2 + 14x +1) f¢¢(x) = 24(x – 3)2 (x + 1) (21x3 – 49x2 + 7x + 13) f iv(x) = 24(x – 3) (3x – 1) (21x3 – 49x2 + 7x + 13) + 168(x + 3)2 (x + 1) (9x2 – 14x + 1) f v(x) = 48(3x – 5) (21x3 – 49x2 + 7x + 13) + 336(x – 3) (9x2 – 14x + 1) (3x – 1) + 336(x – 3)2 (x + 1) (9x – 7) f ¢(3) = f ¢(– 1) = f ¢(7/9) = 0 At x = –1, f iv(x) < 0 so x = –1 is a point of maxima At x = 3, f ¢¢(x) = f¢¢¢(x) = fiv(x) = 0 but f v(x) π 0 So f has neither maximum nor minimum at x = 3. At x = 7/9, f ¢¢(x) > 0 and therefore is a point of minima. Example 30 Let f (x) = (1 + x)n – (1 + nx), x Œ [– 1, •). Then f (a) has an absolute maximum at x = 0 (b) has neither absolute maximum nor absolute minimum at x = 0 (c) has an absolute minimum at x = 0 1

(d) In = Ú0 f ( x ) dx is bounded for all n Ans. (c) Solution:

f (x) = (1 + x)n – (1 + nx)

fi f ¢(x) = n[(1 + x)n – 1 – 1]. Now for x = 0, f ¢(0) = 0 and for x > 0, f ¢(x) > 0. Thus f increases on [0, •) i.e. f (x) ≥ f (0) for x Œ [0, •). For – 1 £ x < 0, 0 £ 1 + x < 1 so (1 + x)n

IIT JEE eBooks: www.crackjee.xyz Applications of Derivatives 24.13

– 1 < 0. So f decreases on [– 1, 0) i.e. f (x) ≥ f (0) = 0 for x Œ [– 1, 0). Hence f has an absolute minimum at x = 0. 1 n+2 In = Ú f ( x ) dx = [2n+1 - 1] Æ • as n Æ • n +1 2 1 0

Example 31 The tangent to the curve y = e x drawn at the point (c, e c) intersects the line joining the points (c – 1, ec–1) and (c + 1, ec + 1) (a) on the left of x = c (b) on the right of x = c (c) at no point (d) at all points Ans. (a) Solution: Equation of straight line joining A(c + 1, ec+1) and B(c – 1,e c–1) is ec + 1 - ec - 1 y – ec+1 = (x – c – 1) (1) 2 A(c + 1, ec–1) c–1 )

e – 1, B(c c– 1

1

P c

Fig. 24.6

Equation of tangent at (c, e c ) is (2)

Subtracting (1) from (2), we get 1 (e – e–1) (x – c) + 2 1 (e – e–1)] 2 1 1 fi (e + e –1) – 1 = (x – c)[1– (e – e –1)] 2 2 ec (e – 1) = e c [(x – c) –

fi fi

e + e -1 - 2 2 and 2 + e–1 – e < 0] x 1, y = log x – (x inequality (a) x – 1 > y (b) x2 – 1 > y x -1 (c) y > x – 1 (d) 1, we have f (x) < f (1). Thus log x < x – 1. Also, x2 – 1 > x – 1, so that x2 – 1 > log x. Similarly, (x – 1)/x < log x. Example 35 Let f be a differentiable function with range (0, •) and g(x) = (f (x))2 – (f (x))3 + (f (x))4 for every x Œ R. Then (a) g is increasing whenever f is increasing (b) g is increasing whenever f is decreasing (c) g is decreasing whenever f is decreasing (d) critical points of g are same as of that of f. Ans. (a), (c), (d) Solution: f ¢(x)

g¢(x) = 2 f (x) f ¢(x) –3 (f (x))2 f ¢(x) + 4(f (x))3

3 1˘ È = 4f (x) f ¢(x) Í( f ( x)) 2 - ( f ( x)) + ˙ 4 2˚ Î

IIT JEE eBooks: www.crackjee.xyz 24.14 Comprehensive Mathematics—JEE Advanced 2 ÈÊ 3ˆ 1 9˘ f ( x ) Í ˜¯ + - ˙ = 4 f (x) f ¢(x) ÁË 8 2 64 ˙˚ ÍÎ 2 ÈÊ 3ˆ 23 ˘ = 4 f (x) f ¢(x) ÍÁË f ( x) - 8 ˜¯ + 64 ˙ ˙˚ ÍÎ 2

3˘ 23 È Since f (x) > 0 and Í f ( x) - ˙ + >0 8˚ 64 Î For all x so g¢(x) > 0 if f ¢(x) > 0 g¢(x) < 0 if f ¢(x) < 0 Therefore, g increases (decreases) whenever f increases (decreases). Moreover, g¢(x) = 0 ¤ f ¢(x) = 0, so g and f have same critical points. Example 36

If

2 ÔÏ3x + 12 x - 1 , -1 £ x £ 2 f(x) = Ì , 2< x£3 ÔÓ37 - x

Then (a) (b) (c) (d)

f(x) is increasing on [–1, 2] f(x) is continuous on [–1, 3] f ¢(2) doesn’t exist f(x) has the maximum value at x = 2.

Ans. (a), (b), (c) and (d) Solution: For x Œ [–1, 2), f ¢(x) = 6x + 12 > 0 on (–1, 2), so f increases on [–1, 2]. f is trivially continuous on [–1, 3] except possibly x = 2. At point x = 2, f(2) = 35 and lim f(x) = lim (37 – x) = 35,

xÆ2+

xÆ2+

lim f(x) = lim (3x2 + 12x – 1)

xÆ2-

xÆ2-

= 3 ¥ 4 + 12 ¥ 2 – 1 = 35. Thus, f is continuous at x = 2 as well. Now f ¢(2+) = =

lim

f (2 + h) - f (2) h

lim

37 - (2 + h) - 35 = –1 h

h Æ 0+

h Æ 0-

Similarly f ¢(2–) = 24 π f ¢(2 +) As f increases on [–1, 2] and decreases on [2, 3] so f has a maximum at x = 2. Example 37 If the line aX + bY + c = 0 is a normal to the curve xy = 1. Then (a) a > 0, b > 0 (b) a > 0, b < 0 (c) a < 0, b > 0

(d) a < 0, b < 0 Ans. (b) and (c) Solution: We have

Differentiating the equation of curve xy = 1, dy +y= 0 x dx

dy = –y/x . dx Hence the slope of normal = x/y. Morever the slope of the line aX + bY + c = 0 is –a/b. So we have x/y = –a/b, i.e., bx + ay = 0 solving this with xy = 1, we have x2 = –a/b. So we must have a/b < 0, i.e., a > 0, b < 0 or a < 0, b > 0. fi

Example 38 The function f (x) = 2 log (x – 2) – x2 + 4x + 1 increases in the interval (a) (1, 2) (b) (2, 3) (c)

(5 2, 3)

(d) (2, 4)

Ans. (b) and (c) Solution:

f ¢(x) =

Ê 1 - ( x - 2)2 ˆ 2 Á x - 2 ˜¯ Ë

2 2 – 2x + 4 = – 2(x – 2) = x-2 x-2

Therefore f ¢(x) > 0 only if (i) 1 – (x – 2)2 > 0 and x – 2 > 0, or (ii) 1 – (x – 2)2 < 0 and x case, we have (x – 3) (x – 1) < 0 and x > 2. That is, 1 < x < 3 and x > 2, so that x Œ (2, 3). The second case is not possible as the domian of f (x) is {x : x > 2}. Hence f (x) increases on (2, 3) and, since (5 2, 3) à (2, 3), it increase on (5 2, 3) as well. Example 39 The equations of the tangents to the curve y = x4 from the point (2, 0) not on the curve, are given by (a) y = 0 (b) y – 1 = 5(x – 1) 4098 2048 Ê 8ˆ (c) y – = Á x - ˜¯ 81 27 Ë 3 (d) y –

32 80 Ê = Áx 243 81 Ë

2ˆ ˜ 3¯

Ans. (a) and (c) Solution: Let (x0, x40) be the point of tangency. Then the equation of the tangent will be y – x40 = y¢(x0) (x – x0). Since this tangent passes through the point (2, 0), we have –x40 = 4x30 (2 – x0), or 3x40 – 8x30 = 0. That is, x0 = 0 or x0 = 8/3, so that the points of tangency are (0, 0) and (8/3, 4096/81). Therefore, the equations of the tangents are y = 0 and y –

4096 2048 Ê 8ˆ = ÁË x - ˜¯ 81 27 3

IIT JEE eBooks: www.crackjee.xyz Applications of Derivatives 24.15

Example 40 then f (x) has (a) (b) (c) (d)

Ï x 3 + x 2 - 10 x - 1 £ x < 0 Ô sin x 0 £ x < p /2 Let f (x) = Ì Ô 1 + cos x p /2 £ x £ p Ó

local maxima at x = p/2 local minima at x = p/2 absolute maxima at x = 0 absolute maxima at x = p/2

Ê 1 1ˆ Ê 1 1ˆ ÁË , ˜¯ and increases on ÁË , ˜¯ . 4 3 7 6 Example 42 The angle at which the curve y = kekx intersects the y-axis is

Ans. (a) and (c) Solution: The function f is not differentiable at x = 0, x = p/2 as f ¢(0 –) = – 10, f ¢(0 +) = 1; f ¢(p/2 –) = 1, f ¢(p/2 +) = – 1. The function f ¢(x) is given by Ï3x 2 + 2 x - 10 - 1 £ x < 0 Ô cos x 0 < x < p /2 f ¢(x) = Ì Ô - sin x p /2 < x £ p Ó The critical points of f are given by f ¢(x) = 0 or x = 0, p/2. Thus critical points are x = p/2, x = 0. Since f ¢(x) > 0, for 0 < x < p/2and f ¢(x) < 0, for p/2 < x < p so f has local maxima at x = p/2. Also f ¢(x) < 0 for – 1 £ x < 0 and f ¢(x) > 0 for 0 < x < p/2 so f has local minima at x = 0. Since f (– 1) = – 9, f (p/2) = 1, f (0) = 0 and f (p) = 0. Thus f has absolute minimum at x = – 1 and absolute maximum at x = 0. Ï p Ôcos , x π 0 then the Example 41 Let y = f (x) = Ì x Ô x=0 function y Ó 1, Ê 1 1ˆ (a) decreases on Á , ˜ Ë 4 3¯ Ê 1 1ˆ (b) increases on Á , ˜ Ë 8 7¯ Ê 1 1ˆ (c) increases on Á , ˜ Ë 7 6¯

and

sin

Ans. (b)

(d) sec-1

1 + k4

dy = k2 ekx. The curve intersects y-axis at dx

Solution: (0, k) so

)

(b) cot-1 (k2)

dy dx

= k2. If q is the angle at which the given (0, k )

k2 - 0 curve intersects the y-axis then tan (p/2 - q) = 1 + 0.k 2 = k2. Hence q = cot-1 k2. Example 43 If f ¢(x) = g(x) (x - a)2, where g(a) π 0 and g is continuous at x = a then (a) (b) (c) (d)

f f f f

is is is is

increasing near a if g(a) > 0 increasing near a if g(a) < 0 decreasing near a if g(a) > 0 decreasing near a if g(a) < 0

Ans. (a), (d) Solution: Since g is continuous at x = a, if g(a) > 0 there exist an open interval I containing a so that g(x) > 0 " x Œ I fi f ¢(x) ≥ 0 " x Œ I. Therefore, f is increasing near a. Similarly f is decreasing near a if g(a) < 0.

(d) f (6) £ 5

Ans. (b), (c) p x

2

sin

p x

So the sign of y¢ coincide with sign of sin Now sin

(

(c) sin-1 1/ 1 + k 4

(c) f (6) > 5

Ans. (a), (c), (d) For x π 0, y¢ =

(a) tan-1 (k2)

Example 44 Let f be differentiable for all x. If f (1) = - 2 and f ¢(x) ≥ 2 for x Œ [1, 6], then (a) f (6) < 8 (b) f (6) ≥ 8

1 1 ˆ (d) decreases on » ÊÁ , ˜ k ŒI Ë 2k + 2 2k + 1¯

Solution:

1ˆ Ê 1 Hence the function increases on » Á , ˜ and k ŒI Ë 2k + 1 2k ¯ 1 ˆ Ê 1 , decreases on » Á ˜ . Thus y decreases on k ŒI Ë 2k + 2 2k + 1¯

Solution:

p x

p p > 0 if 2kp < < (2k + 1)p , k ΠI x x p p < 0 if (2k + 1)p < < (2k + 2)p , k ΠI x x

By Lagrange’s mean value theorem there is f (6) - f (1) c Œ (1, 6) such that = f ¢(c) 6 -1 f (6) + 2 fi = f ¢(c) i.e. f (6) = 5f ¢(c) - 2 ≥ 5.2 - 2 = 8 5 so f (6) > 5 also. Example 45

Let f (x) = cos x sin 2x then

(a) min f (x) > - 7/9, x Π[- p, p]

IIT JEE eBooks: www.crackjee.xyz 24.16 Comprehensive Mathematics—JEE Advanced

(b) min f (x) > - 9/7, x Π[- p, p] (c) min f (x) > - 1/8, x Π[- p, p] (d) min f (x) > - 2/9, x Π[- p, p]

Solution:

Solution: f (x) = cos x sin 2x = 2 sin x - 2 sin x, min f (x) = min g(t) where g(t) = 2t - 2t3, x Œ [- p, p], t Œ [- 1, 1]. g ¢(t) = 2 - 6t2 = 0 fi t = ± 1/ 3 g≤(t) = - 12t, so g≤ (1/ 3) < 0 and g¢¢ ( - 1/ 3) >0 Hence min g(t) = g ( - 1 3) = -

4 7 9 >- >9 7 3 3

If f (x) = tan-1 x - (1/2) log x. Then

Example 46

(c) (d)

e1/x - 1

x Æ 0 + e1/ x

3

(b)

xÆ0

continuous. But f ¢(0+) = lim

Ans. (a), (b)

(a)

If f is as in (a) then lim f (x) = 0 = f (0) so f is

the greatest value of f (x) on ÈÎ1/ 3, 3 ˘˚ is p/6 + (1/4) log 3 the least value of f (x) on ÈÎ1/ 3, 3 ˘˚ is p/3 - (1/4) log 3 f (x) decreases on (0, •) f (x) increases on (- •, 0)

Ans. (a), (b), (c) Solution: The domain of f (x) is (0, •). For x > 0,

(

2x - 1 + x2 1 1 f ¢(x) = = 1 + x2 2x (1 + x 2 )2 x

)

= -

Thus f ¢(x) < 0, i.e. f (x) decreases on (0, •). Also f ¢(x) = 0 if

Example 47 Column 1 (a) For x π 0, f (x) =

Column 2 (p) y2 - m2y = 0

Ê e1/x - 1ˆ x Á 1/x ˜ , f (0) = 0 Ë e + 1¯ (b) y = aemx + be-mx (c) y = x + cos x + x3 + 2x2 + 6x + 17 (d) y = tan-1 x - x p q r s Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(q) continuous at x = 0 but not differentiable (r) always decreasing (s) always increasing.

xÆ0+1 +

e - 1/x

Example 48 Let f (x) = (2x - 1) (2x - 2) and g(x) = 2 sin x + cos 2x Column 1 Column 2 (a) f increases on (p) (p, •) (b) f decreases on (q) (- •, p) (c) g decreases on (r) (p/2, 5p/6) (d) g increases on (c) (0, p/6) p q r s Ans.

(1 + x 2 )2 x

MATRIX-MATCH TYPE QUESTIONS

1 - e - 1/x

= 1 and f ¢(0-) = - 1. If y = aemx + be-mx then y1 = m (aemx - be-mx ) and y2 = m2 (aemx + be-mx ) = m2y. If y = x + cos x + x3 + 2x2 + 6x + 7 then y¢ = 1 - sin x + 3x2 + 4x + 6. Since the discriminant of 3x2 + 4x + 6 is negative and 1 - sin x ≥ dy 0 for all x so y¢ > 0 for all x, so y increases. For y in (d) dx - x2 = < 0 for all x. 1 + x2

(1 - x )2

x = 1 and f (1) = p/4, f (1/ 3 ) = p/6 + (1/4) log 3, f ( 3 ) = p/3 - (1/4) log 3. Thus the greatest value is p/6 + log 3 and the least value is p/3 - (1/4) log 3.

+1

= lim

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution y¢(x) = 2x (2x + 1 - 3) log 2. Since 2x > 0 and log 2 > 0 so y¢(x) > 0 if 2x + 1 - 3 > 0, log 3 = log2 3. Thus y¢(x) > 0 for x > log2 3/2. i.e. x + 1 > log 2 Hence y increases on (log2 3/2, •) and decreases on (- •, log2 3/2) … (- •, - p). The period of g is 2p so it is enough to consider g on [0, 2p]. g¢(x) = 2 cos x - 2 sin 2x = 2 cos x (1 - 2 sin x) x Œ (0, p/6) » (p/2, 5p/6) » (3p/2, 2p)

>0

if

0 for any x π - 1. Thus the least f (x) ( x + 1)2 = f (0) = - 1 and the greatest f (x) = f (4) = 3/5. (iii) and (iv) can be done similarly. Example 50 domain (0, p) Column 1 (a) x2 + 2 cos x (b) 8 log x + x + 16/x

(

1 + x2 - x

)

Column 2 (p) increasing (q) decreasing

(r) neither increasing nor decreasing 2 Ê x ˆ (d) log ( 1 + x) – Á x - ˜ 2¯ Ë (s) increases on (0, p/2) (c) log

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

Solution: Let f (x) = x + 2 cos x so f ¢(x) = 2 (x - sin x) since x ≥ sin x on (0, p) so f is increasing. If f (x) = 8 log x x 2 + 8 x - 16 16 8 16 then f ¢(x) = + 1 - 2 = +x+ x x x2 x 2

=

(x + 4 + 4 2) (x + 4 - 4 2) . x

2

Hence f neither increases nor decreases. If f (x) = log 1 1 + x 2 - x then f ¢(x) = < 0 for all x. For f in 1 + x2

(

)

(d) f ¢(x) =

x2 > 0 for x > 0. 1+ x

Example 51 domain (0, p/2) then Column 1 (a) x2 + 2 cos x + 2 on (0, p/2) has (b) 9x - 4 tan x on

Column 2 (p) local maximum at cos-1 (2/3) (q) maximum at x = 1/2

(0, p/2) has (c) (1/2 - x) cos x +

(r) no local extremum

x -x 4 2

sin x -

Ê1 ˆ (d) Á - x˜ cos p (x + 3) (s) minimum at x = 1 Ë2 ¯ + (1/p ) sin p (x + 3) on (0, 4) p q r s Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: Let f (x) = x2 + 2 cos x + 2, then f ¢(x) = 2(x sin x). Now f ¢(x) = 0 iff sin x = x but sin x π x for any x Œ (0, p/2). Hence f has no extremum. If f (x) = 9x - 4 tan x then f ¢(x) = 9 - 4 sec2 x. Thus, f ¢(x) = 0 iff sec2 x = 9/4 i.e. x = cos-1 (2/3) or x = cos-1 (- 2/3). Also f ≤(x) = - 8 sec2 x tan x. 9 2 f ≤(cos-1 2/3) = 8 ◊ ◊ 0 (a) If f (x) = Ì if x = 0 Ó0 then f ¢(x) vanishes at

Column 2

point of (0, 1)

IIT JEE eBooks: www.crackjee.xyz 24.18 Comprehensive Mathematics—JEE Advanced 2 ÔÏ x sin (p / x ) at x > 0 (b) If f (x) = Ì (q) at least 10 at x = 0 ÔÓ0 points of (0, 1) then f ¢(x) vanishes at x3 , (r) no point (0, 1) (c) If f (x) = tan –1 x – x + 6 0 < x £ 1 f (x) vanishes at (d) If f (x) = sin x – x, (s) exactly one f¢ vanishes at point of [0, 1]

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

Solution: The function f in (a) vanishes at points where sin (p/x) = 0, p /x = kp, x = 1/k, k = 1, 2, 3, ... Since the function has derivative at any interior point of the interval [0, 1], the Rolle’s theorem is valid to anyone of the interval [1/2, 1], [1/3, 1/2], .. [1/(k + 1), 1/k], ... Consequently there exist ck Œ (1/(k + 1), 1/k) so that f ¢(ck) Similarly the conclusion holds for the function in (b). x 2 ( x 2 - 1) 1 x2 = . If f is as in (c) then f ¢(x) = 1 + 2 2(1 + x 2 ) 1 + x2 So for x Œ (0, 1), f ¢(x) < 0. Therefore, f decreases on [0, 1] and so f (x) < f (0) = 0. Thus f is never zero in (0, 1). For f in (d) f ¢(x) = cos x – 1 and cos x decreases on (0, 1). f ¢ is zero exactly at one point i.e. x = 0. Example 53 Column 1

Column 2

10 4 x - 9x2 + 6x decreases on

(a) f (x) =

3

(b) y = 2x2 – log x increases on (c) y = x – 2 sin x decreases (d) y = x 2 e– x decreases on p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

(p) (–•, 0)

(q) (0, 1/2) (r) (2, •) (s) (0, 2)

Let g(x) = 4x 3 – 9x 2 + 6x

Solution: fi

g¢(x) = 12x2 – 18x + 6 = 6(2x2 – 3x + 1)

= 6(2x – 1) (x – 1) Then g increases on {x : x > 1} » {x : x < 1/2} and decreases on (1/2, 1). Hence the function f decreases on (1, •) » (– •, 1/2) and increases on (1/2, 1). In particular f decreases on (– •, 0), (0, 1/2) and (2, •). 4 x2 - 1 1 = . Thus If f is as in (b) then f ¢(x) = 4x – x x f ¢(x) > 0 if x Œ (– 1/2, 0) » (1/2, •) so f increases on (– 1/2, 0) » (1/2, •) in particular on (2, •). If f is as in (c) then f ¢(x) = 1 – 2 cos x. Thus f ¢ is periodic with period 2p and f decreases on (0, p /3) » (5p/3, 2p) and increases (p/3, 5p /3). In particular f decreases on (0, 1/2). The function in (d) decreases on (– •, 0) » (2, •) and increases on (1, 2). In particular decreases on (– •, 0), (2, •). Example 54 Column 1 (a) If y = 2x 3 – 3x2, then 2 (b) If y = x 2 3 6 x - 7 , then 3

(

(c) If y =

)

1 Ê 2 1ˆ –1 Á x - ˜¯ sin x 2Ë 2

p 1 x 1 - x 2 - x 2 , then 4 12 (d) If y = x sin x + cos x 1 2 – x 2 (– p/2 £ x £ p /2) then p q r s Ans. p q r s a

Column 2 (p) ymax = 0 (q) ymin = –1 (r) ymin = 1

+

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(s) ymin = –

2 3

Solution: If y is as in (a) then y is differentiable everywhere and y¢ = 6x 2 – 6x = 6x (x – 1). For x < 0, y¢ > 0 and for 0 < x < 1, y¢ < 0, so ymax = y(0) = 0. Also for x > 1, y¢ > 0, so ymin = y (1) = – 1. If y is an in (b) then y is differentiable except at x = 7/6 4 4 otherwise y¢ = x 3 6 x - 7 + (6x – 7)–2/3 x2 3 3

(

)

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=

f ¢(x) = 0 fi cos 2x = cos p/3 fi x = p /6

4 7 x( x - 1) . 3 (6 x - 7)2/3

f (– p/2) = 1, f (p/2) = – 1 and f (p/6) =

For x < 0, y¢ > 0 and for 0 < x < 1, y¢ < 0 so ymax = y(0) = 0. Also x > 1, y¢ > 0 so ymin = y(1) = – 2/3.

=

If y is as in (c) then ymax = y(0) = 0 ymin = y(1/2) =

and

3 3 - 2p (verify) 48

If y is as in (d) then ymin = y (0) = 1 ymax = y ( ± p / 3) =

and

Example 56

6p 3 - p + 18 (verify) 36

(a) If f (x) = on [0, 1]

Column 2

1 - x + x2

(p) greatest value of f = 1

1 + x - x2

2 (sin 2x – x) p on [–p /2, p/2] 1 (d) f (x) = (x 3 – 3x2 + 2 6x – 2) on [–1, 1] p q r s Ans.

(r) least value of f = – 1

1 + 3x

(d) f (x) = (x 2 – 2x) log x

p

q

r

s

a

p

q

r

s

b

p

q

r

s

Ans.

q

r

s

c

p

q

r

s

b

p

q

r

s

d

p

q

r

s

c

p

q

r

s

d

p

q

r

s

So f ¢(x) > 0 if x Œ (– •, – 1) » (4, •) and is negative if x Œ (– 1, 4). Hence f increases on (4, •), (– •, – 1).

(1 + x - x 2 )(2 x - 1) - (1 - x + x 2 )(1 - 2 x ) (1 + x - x 2 )2 (1 + x - x )

2 2

(s) (e, •)

Solution: If f is as in (a) then f ¢(x) = 6x 2 – 18x – 24 = 6(x 2 – 3x – 4) = 6(x + 1) (x – 4).

If f is as in (a) above then

2(2 x - 1)

(q) (–•, – 1)

3 2 x + 4x increases on 2

p

=

(p) (4, •)

(r) (–•, 0)

4 + 5x2 increases on

a

f ¢(x) =

3

(b) f (x) = 4x 3 – 21x2 + 18x + 20 increases on

– (s) least value of f = – 6

Column 2

(a) The function f (x) = 2x – 9x 2 – 24x + 7 increases on

(q) least value of f = 3/5

(c) f (x) =

Solution:

Column 1

(c) f (x) =

(b) If f (x) = 2 tan x – tan 2 x on [0, p/2)

3 1 - 0 on (– •, 1/2) » (3, •) and is negative on (1/2, 3). Hence f increases on (– •, – 1); (4, •); (– •, 0). For f as in (c), f ¢(x) = (4 + 5 x 2 )3 - (1 + 3x )5 x 12 - 5 x = . So f ¢(x) > 0 on 2 3/2 (4 + 5 x ) (4 + 5 x 2 )3/2 (12/5, •) and is negative on (– •, 12/5). In particular, it increases on (4, •) and (e, •) If f is as in (d), one can see easily that f increases (– •, 1) » (e, •) and decreases on (1, e).

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ASSERTION-REASON TYPE QUESTIONS

x2 - 6x + 5

Let f (x) =

Example 57

x – 5x + 6 2

Column 1

Column 2

(a) If – 1 < x < 1, then f (x

f (x) < 1

(b) If 1 < x < 2, then f (x

f (x) < 0

(c) If 3 < x < 5, then f (x

f (x) > 0

(d) If x > 5, then f (x) p q r s Ans. a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution:

f (x) < 1

Ï2 x + 1, - 1 £ x < 0 ÔÔ x=0 Example 58 Let f (x) = Ì 2 x , Ô x ÓÔ2 - 1, 0 < x £ 1 Statement-1: f is bounded but never reaches its maximum and minimum. Statement-2: f has a discontinuity at 0. Ans. (a) Solution: In the interval [-1, 0), the function increases from 3/2 to 2 and in (0, 1] it increases from 0 to 1. It does not attain either the value 2 or 0. Therefore the function is bounded but never reaches its maximum or minimum. This is because there is a discontinuity at the point x = 0.

f (x) = 1 –

2( x + 2) x 2 - 10 x + 24

Example 59 so

È ( x + 2)2 - 48 ˘ f ¢(x) = 2 Í 2 2˙ Î ( x - 10 x + 24) ˚

Statement-2: The function x1/x (x > 0) has local maximum at x = e. Ans. (b)

Also

È (1 + 2 / x ) (1 - 10 / x ) ˘ f (x) = Í ˙ Æ 1 as x Æ ± •. Î (1 - 4 / x ) (1 - 6 / x ) ˚

(a) For – 2 < x < 2, then f (x) > 0, f (x) < 1 so 0 < f (x) < 1 (b) For 2 < x < 4, f (x) < 0 and so f (x) < 1 (c) For 6 < x < 10, f (x) < 0 and so f (x) < 1 (d) For x > 10, f (x) > 0, f (x) < 1 and 0 < f (x) < 1. The graph of y = f (x) is as follows

Let f (x) = x1/x fi log f (x) =

Solution:

\ f ¢(x) > 0 if x < – 2 – 4 3 or 4 < x < – 2 + 4 3 or x > 6 and f ¢(x) < 0 if – 2 – 4 3 < x < 4 or – 2 + 4 3 < x < 6

Statement-1: ep > p e



log x x

f ¢ ( x) - log x + 1 Ê - log x + 1ˆ = fi f ¢(x) = x1/x Á ˜¯ 2 Ë f ( x) x2 x

For x > e, f ¢(x) < 0 and for 0 < x < e, f ¢(x) > 0 Thus f has local maximum at x = e. Since p > e and f decreases on (e,•) so f (p) < f (e) fi p1/p < e1/e fi p e < ep. Example 60

Let f (x) = tan-1 x and g(x) = x -

x3 6

Statement-1: f (x) < g (x) for 0 < x £ 1 Statement-2: h(x) = tan-1 x - x +

x3 decreases on [0, 1]. 6

Ans. (a) 1

Solution:

The derivative of the function

f (x) = tan-1 x - x + –2

2

4

6

10

is equal to f ¢(x) =

Fig. 24.11

x3 6

x 2 ( x 2 - 1) 1 x2 = 1 + 2 1 + x2 2(1 + x 2 )

The function is continuous throughout R and in particular it is continuous on [0, 1] and f ¢(x) < 0 for x Œ [0, 1]. Thus f decreases on [0, 1] and, consequently, for any point x, x3 < 0. 0 < x £ 1, f (x) < f (0) = 0 or tan-1 x - x + 6

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Example 61 Statement-1: For 0 £ p £ 1 and for any positive a and b the inequality (a + b) p £ a p + bp is valid. Statement-2: For 0 £ p £ 1 , the function f (x) = 1 + x p (1 + x) p decreases on [0,•)

f1(x) has minimum at x = - 3 and x = 1. f (x) has maximum at x = - 3 Also f (x) has maximum at x = 1 and f (1) = 1/53. So the absolute maximum of f is 1/53.

Ans. (c) Solution: Let f (x) = 1 + xp - (1 + x) p, x ≥ 0 The derivative of this function 1 È 1 ˘ f ¢(x) = px p-1 – p(1 + x)p-1 = p Í 1 - p 1- p ˙ (1 + x ) ˚ Îx Since for x > 0, x < 1 + x and 1 - p ≥ 0 so x1-p < (1 + x)1-p

Example 64 Statement-1: The critical points of the function x cos x will occur between p /4 and p/3.

Hence f ¢(x) > 0 for x > 0. Thus, the function f increases in [0, •) i.e. f (x) = 1 + x p - (1 + x) p > f (0) = 0, whence 1 + x p > (1 + x) p. Put x = a/b, we get (a + b) p < a p + b p.

Solution:

Statement-1: The absolute minimum Example 62 value of |x - 1| + | x - 2| + |x – 3| is 2.

Statement-2: The function x tan x increases on (0, p/4) and decreases on ( p/4, p/2). Ans. (c) Let f (x) = x cos x fi f ¢(x) = cos x (1 - x tan x)

= cos x (1 - g(x)), where g (x) = x tan x Since g¢(x) = tan x + x sec2 x > 0 for x Œ (0, p /2), hence g increases on (0, p /2). Also p /4 < 1 so 1 - (p/4) tan p /4 > 0 fi f ¢(p /4) > 0 and p/3 > 1 so 1 - (p/3) tan (p/3) = 1 - (p/3) 3 0 for all x.

,2< x£3 ,x>3

Ans. (a)

The function has absolute minimum at x = 2. So the absolute minimum is equal 2. f ¢(1-) = – 3, f ¢(1+) = – 1 f ¢(2+) = 1, f ¢(2-) = - 1, f ¢(3-) = 1, f ¢(3+) = 3. Hence f is not differentiable at x = 1, 2, 3. Example 63 value of f (x) =

Statement-1: The absolute maximum 1 1 is . 2 53 3x + 8 x - 18 x + 60 4

3

Statement-2: The minimum value of f1(x) = 1/f (x) is at x = 1 and x = - 3. Ans. (a) Solution:

f 1 (x) = 3x4 + 8x3 - 18x2 + 60. Since

f ¢1(x) = 12x3 + 24x2 - 36x = 12 x(x2 + 2x - 3) = 12 x (x - 1) (x + 3) f ≤1 (x) = 12(3x2 + 4x - 3) The critical points of f1(x) are x1 = - 3, x2 = 0, x3 = 1 f 1≤(- 3) > 0, f 1≤(0) = - 36, f 1≤(1) = 48 > 0.

Solution: Since g(x) is polynomial of odd degree so it has at least one real root. If there are two roots say x1 < x2. Then in the interval [x1, x2], the function g(x) = 3x5 + 15x continuous, vanishes at the end-points and has a derivative at all points. So for some c Œ (x1, x2), g¢(c) = 0 but g¢(x) > 0 for all x. This contradiction shows that the equation in question has only one real root. 1- x 1+ x Statement-1: The difference of the greatest and smallest value of f (x) on [0, 1] is p /4. Statement-2: If a function g is decreases on [a, b] then greatest value of g = g(a) and least value of g is g(b). Example 66

Let f (x) = tan-1

Ans. (a) f (x) = p/4 - tan-1 x fi f ¢(x) =

-1

π 0 for all 1 + x2 x and f ¢(x) < 0 for all x. Hence greatest value of f = f (0) = p /4 and the least value = f (1) = 0.

Solution:

Example 67

Let f (x) = 2 sin x + tan x - 3x

Statement-1: f (x) has no maximum and minimum on (- p/2, p/2).

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Statement-2: f increases on (- p/2, p /2)

Example 69 Statement-1: |tan x - tan y| £ |x - y| for all x, y Œ (– p/2, p/2)

Ans. (a) Solution: The derivative of f 1 -3 f ¢(x) = 2 cos x + cos2 x =

Statement-2: If f is differentiable on an open interval and | f ¢(x)| £ M then | f (x) - f (y)| £ M |x - y| Ans. (d) Solution: Applying Lagrange’s Mean value on [x, y] we have c Œ (x, y) such that

(1 - cos x ) (1 + cos x - 2 cos2 x ) cos2 x

f ¢(c) =

3

=

4 sin ( x 2) sin (3 x 2) cos2 x

>0

| f (y) - f (x)| = | f ¢(c)| |y - x| £ M | y - x| (if | f ¢(x)| £ M)

in the interval (- p /2, 0) and (0, p/2) and vanishes only at x = 0. Hence f increases on (- p /2, p/2) so f has no maximum and minimum. Example 68 Statement-1: The equation of tangents drawn to the curve y2 - 2x3 - 4y + 8 = 0 from the point (1, 2) are y - (2 + 2 3 ) = 2 3 (x - 2) and y - (2 - 2 3 ) = - 2 3 (x - 2). Statement-2: Two tangents can be drawn from (1, 2) to the curve y2 - 2x3 - 4y + 8 = 0 Ans. (b) Solution: Let the tangent drawn from (1, 2) to the curve y2 - 2x3 - 4y + 8 = 0 meets the curve at the point (h, k). Slope of tangent at (h, k) =

dy dx

= ( h, k )

3 h2 k -2

and | f (y) - f (x)| ≥ M | y - x | (if | f ¢(x)| ≥ M) Putting f (x) = tan x, f ¢(x) = sec2 x so | f ¢(x)| ≥ 1 Hence |tan x - tan y| ≥ |x - y|.

COMPREHENSION-TYPE QUESTIONS Paragraph for Questions Nos 70 to 72 If f ≤(x) < 0 (> 0) on an interval (a, b) then the curve y = f (x) on this interval is convex (concave), i.e. it is situated below (above) any of its tangent lines. If f ≤(x0) = 0 or does not exist but f ¢(x0) does exist and the second derivative changes sign when passing through the point x0 then the y= point (x0, f (x0 f (x). Example 70

3 h2 Equation of tangent is y - k = (x - h). This passes k -2 through (1, 2) so 3 h2 2-k= (1 - h) fi 3h3 - 3h2 - k2 + 4k - 4 = 0 k -2 ...(1) Also (h, k) is a point on y - 2x - 4y + 8 = 0 2



f ( y) - f ( x) y-x

If y = x4 + x3 - 18x2 + 24x - 12 then

(a) (- 2, (b) (1ˆ Ê (c) Á 3/2, - 8 ˜ Ë 16 ¯ (d) y is convex on (3/2, •)

3

k2 - 2h3 - 4k + 8 = 0

Example 71 ...(2)

Adding (1) and (2), we get h3 - 3h2 + 4 = 0 (h + 1) (h - 2)2 = 0 fi h = - 1, h = 2 For h = - 1, k2 - 4k + 10 = 0 whose roots are not real. Hence h = 2 fi k - 4k - 8 = 0 fi k = 2 ± 2 3 . 2

Thus the points of tangency are (2, 2 + 2 3 ) and (2, 2 - 2 3 ). Hence the equation of tangents are y - (2 + 2 3 ) = 2 3 (x - 2), y - (2 - 2 3 ) = – 2 3 (x - 2)

(a) (b) (c) (d)

If y = x sin (log x) then

y y p/ y

3 2 x + 1 is concave along Example 72 y = x4 + ax3 + 2 the entire number scale then (a) |a| ≥ 1 (b) |a| £ 1 (c) |a| £ 2 Ans. 70. (c), 71. (d), 72. (c)

(d) |a| > 2

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Solution 70.

y¢ = 4x3 + 3x2 - 36x + 24

x Ê ˆ y ≤ = 12x + 6x - 36 = 12 Á x 2 + - 3˜ Ë ¯ 2 whence y≤ = 0 at x1 = - 2, x2 = 3/2 2

Also y≤ > 0 on (- •, - 2) and (3/2, •): y≤ < 0 on (- 2, 1ˆ Ê3 3/2) so the points (- 2, - 124) Á , - 8 ˜ are points of Ë2 16 ¯ y is concave on (3/2, •). Solution: 71. y≤ = 12x2 + 6ax + 3. The curve will be concave along the entire number scale of y≤ ≥ 0 for all values of x i.e. when 4x2 + 2ax + 1 ≥ 0 for all x 4a2 - 16 £

a| £ 2.

(a) y =

p 6 + 3p /2

(b) y =

3p 12 + 3p /2

(c) y =

3p 12 + 5p /2

(d) y =

2p 11 + 5p /2

Ans. 73. (b), 74. (a), 75. (c) Solution: Let y be the length and x be the breadth of the rectangular portion Total perimeter of the window is 2 x + 2y + (1/2) p y = p (say). Let amount of light per square metre for the coloured glass be m. If L is the total light transmitted then L = 3 m ¥ Area of rectangular portion + m ¥ Area of semi-circular portion

y¢ = sin (log x) + cos (log x) and 1 y≤ = [cos (log x) - sin (log x)] = x 2 Êp ˆ sin Á - log x˜ Ë4 ¯ x p/4 + kp y≤ = 0 at xk = e , k = 0, ± 1, ± 2, ...

= 3mxy +

Example 73 The ratio of the sides y : x of the rectangle so that the window transmit the maximum light is (a) 3 : 2 (b) 6 : 6 + p (c) 6 + p : 6 (d) 1 : 2 Example 74 If m is the amount of light per square metre for the coloured: glass and L is the total light transmitted then dL dy

is equal to y =1

(a)

mÈ 5p ˘ 3 p - 12 Í 2Î 2 ˙˚

3p ˘ È (b) m Í2 p - 6 2 ˙˚ Î

˘ ˆ 1 y˜ + p y 2 ˙ ¯ 8 ˚

dL mÈ pˆ p Ê = 3 p - 6Á 2 + ˜ y + Í Ë dy 2Î 2¯ 2

passing through each point xk. Consequently, the points xk

glass. The clear glass transmits three times as much light per square metre as the coloured glass. Suppose that y is the length and x is the breadth of the rectangular portion and p the perimeter.

1 m p y2 8

È3 Ê pˆ Ê = m Í yÁ p - Á2 + ˜ Ë 2¯ Î2 Ë

Êp ˆ The function sin Á - log x˜ and so y≤, changes sign when Ë4 ¯

arch) is in the form of a rectangle surmounted by a semi-

Critical points of L are

Example 75

Solution 72.

Paragraph for Questions Nos 73 to 75

p˘ È (d) 2m Í3 p - ˙ Î 2˚

p˘ È (c) m Í p - 2 - ˙ Î 2˚

dL dy

= y =1

m 2

dL =0 dy

˘ y˙ ˚

p˘ È ÍÎ3 - 12 - 2 ˙˚ fi

y=

3p 12 + 5 p /2

5p ˆ m Ê d2 L = ÁË - 12 ˜ < 0. 2 2 2 ¯ dy Therefore, for y = 3p /[12 + (5p/2)], there will be maximum light transmission. 3p 2◊ 12 + 5p /2 y = 3p pˆ x Ê p - Á2 + ˜ Ë 2 ¯ 12 + 5p /2 =

6 6 = 6+p (12 + 5p /2) - 3(2 + p /2)

Paragraph for Questions Nos 56 to 78 A straight line is called an asymptote to the curve y = f (x) if the distance from the variable point M of the curve to the straight line approaches zero as the point M recedes kinds of asymptotes; vertical, horizontal and inclined.

IIT JEE eBooks: www.crackjee.xyz 24.24 Comprehensive Mathematics—JEE Advanced

Vertical asymptotes: If at least one of lim f (x) or lim xÆa+

f (x If

xÆa-

x = a is a vertical asymptote. lim f (x) = A then y = A is a horizontal asymptote.

xƱ•

f ( x) = k2, lim [ f (x) - k1x] = b2 then y = xƱ• x k2 x + b2 is an inclined asymptote.

e1/x - 1 =1 x Æ ± • 1/ x xƱ• y = x + 1 is an inclined asymptote to curve 2. For Example 77 b=

If limits lim

lim y = lim

xƱ•

Example 76 be curve 2 then

3x + 3x be curve 1 and y = xe1/x Let y = x -1

(a) curve 1 has no horizontal asymptote and curve 2 has no vertical asymptote (b) y = 3x + 3 and y = x + 2 inclined asymptotes to curve 1 curve 2 (c) y = 3x + 3 and y = x + 1 are inclined asymptotes to curve 1 and curve 2 (d) y = x + 1 and y = 3x + 3 are inclined asymptotes to curve 1 and curve 2 5x then Let y = x-3

Example 77 (a) (b) (c) (d)

There There There x=3

Let y =

Example 78 (a) (b) (c) (d)

are no vertical asymptotes are no horizontal asymptotes are no inclined asymptotes and y = 5 are the only asymptotes

There There There There

x 2 + 1 sin 1/x

are no horizontal asymptotes is only one horizontal asymptote are only two horizontal asymptotes is one vertical asymptote.

xÆ3

xÆ3

y =

b = lim

So

x = 1 is a vertical asymptote y k = lim = 3 and b = lim ( y - kx) = 3 xƱ• x xƱ•

So

y = 3x + 3 is an inclined asymptote

lim

y = x

et =• 1 t = Æ0 t lim x 0

lim e1/x = e = 1

xƱ•

xÆ •

|x|

1 + 1/x 2 sin 1/x

as x Æ • Ï1 = Ì Ó- 1 as x Æ - • So curve has two horizontal asymptote.

INTEGER-ANSWER TYPE QUESTIONS If the value of greater of sin x + tan x 1 and 2x (0 < x < p/2) at p/4 is g(p/4) then g (p/4) 2 is equal to Example 79

Ans. 1 Solution: Let f(x) = sin x + tan x – 2x. Then g(x) = f¢(x) = cos x + sec2 x – 2. fi

g¢(x) = – sin x + 2 sec2 x tan x

Ê 1 - cos3 x + 1ˆ = sin x Á ˜ cos3 x ¯ Ë

Ê 3x ˆ lim y = lim Á + 3x˜ = - • x Æ1 + Ë x - 1 ¯

xƱ•

( y - kx) = lim

Ê 2 ˆ = sin x Á - 1 + ˜ cos3 x ¯ Ë

x Æ1 +

k=

5x =∓ • x-3

lim

xÆ •

For curve 1 in Example 76 Ê 3x ˆ lim y = lim Á + 3x˜ = - • x Æ1 x Æ1 - Ë x - 1 ¯

xÆ+0

lim

For Example 78, the curve has no vertical asymptotes since it continuous at x = 0 1 x 1 + 1/ x 2 sin y x = ± 1.0 = 0 k = lim = lim xÆ • x xÆ • x

Solution:

xÆ0+

(xe1/x - x) =

5x = 5. So x = 3 and y = 5 are xÆ • xÆ • x - 3 the only asymptotes. lim

Ans. 76. (c), 77. (d), 78. (c)

For curve 2, lim y = lim e1/x =

lim

Since for 0 < x < p 2 , we have 0 < cos3 x < 1, g is an increasing function. Hence g(x) > g(0), i.e., cos x + sec2 x – 2 > 0. Therefore, f is an increasing function, so f(x) > f (0) for 0 < x < p 2 . Hence sin x + tan x > 2x. 1 Thus g(p/4) = + 1. 2 Example 80

If the greatest value of y = x/log x on

[e, e3] is u then e3/u is equal to Ans. 3

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Solution:

log x - 1

y¢ =

Since (log x)2 is always positive, we have y¢ > 0 if log x > 1. That is, if x > e. Therefore, y increases on [e , e3] Hence u = e3/3. Example 81 If (u, v) are the coordinates of the point on the curve x3 = y (x - 4)2 where the ordinate is minimum 1 then uv is equal to 81 Ans. 4 Solution: by y =

The ordinate of any point on the curve is given x3

( x - 4)2 dy = dx

3x 2

( x - 4 )2

dy =0 dx

-

2 x3 ( x - 4)3



x=0

x 2 ( x - 12)

=

( x - 4)3 or

x = 12

( x - 4)3 (3x 2 - 24 x ) and d2 y d x2 and

d y d x2

dy > 0 for x < 1 dx

8 ¥ 64 (9 ¥ 16) - 27 ¥ 16 ¥ 0 (2 ¥ 4)6

Solution:

\

= 3

)(

dy < 0 for x > 1 dx

3.

3 or when A = p 6 and

Let f(q) = 64 sec q + 27 cosec q

f ¢(q) = 0



sin q

64

= 27

cos q 3 tan q = 27/64, i.e., tan q = 3/4 2

cos q sin 2 q

sin q = 3/5 and cos q = 4/5

(3 ¥ 4)3 (2 ¥ 4)2

min f (q ) = 64 ¥ (5/4) + 27 ¥ (5/3) = 125.

>0

= 27

Example 84 If A is the area formed by the positive x-axis, and the normal and tangent to the circle x2 + y2 = 4 at (1, 3 ) then A/ 3 is equal to Ans. 2

y = 3 tan A tan B = 3 tan A tan ( p 3 – A)

(

3

f ¢(q) = 64 sec q tan q – 27 cosec q cot q



1 + 3x dy = 3 dx

)

If f (q) = 64 secq + 27 cosecq when q 1 lies in (0, p /2) then min f (q) is equal to 25 Ans. 5

Example 82 If A > 0, B > 0 and A + B = p/3 then the maximum value of 3 tan A tan B is

1 + 3x

3x

Example 83

Solution:

3-x

3,

Thus, y has a maximum at x = 1 ymax = 1/3.

Thus (u, v) = (12, 27)

(

dy/dx = 0 if x = 1

Thus, Also, as x > 0,

i.e.,

y=

3x

(1 +

)

2

\ f has minimum value when tan q = 3/4,

x = 12

=

)(

3x - 1 x + 3

+ 27 cosec q cot2 q > 0 if q Π(0, p /2)

Hence y is minimum at x = 12 and its value is

Ans. 1 Solution:

)

(

-

f ¢¢(q) = 64 sec3 q + 64 sec q tan2 q + 27 cosec3q

=0

=

3x

2

=3

0 < A < p 3 , x = tan A > 0.

( x - 4)6

x=0

2

(1 +

As



- 3x 2 ( x - 12) ( x - 4)2

d2 y = d x2

3 - 2 x - 3x 2

= 3

(log x )2

)

where x = tan A

) (

3 - 2x - x

3-x

) ( 3)

(1 + 3x )2

3 + 3x - 2 x - 2 3x 2 - 3x + 3x 2

(1 +

3x

)

2

dy dx

2x + 2y

(1, 3 )

=–

dy =0 dx



dy x =– y dx

1 3

( 3)

)

Therefore, the equation of the tangent at 1, 3 is y–

(

3 = -1

(x – 1)

and the point of intersection of this tangent with the

(

)

x-axis is (4, 0). The equation of the normal at 1, 3 is y – 3 = 3 (x – 1), and the point of intersection of this normal with the x-axis is (0, 0). Hence the required area is 1 ◊4 3 =2 3 2

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Example 85 A curve passes through the point (2, 0) and the slope of the tangent at any point (x, y) is x2 - 2x for all values of x then 6ymax is equal to

The corresponding ordinate is given by y=

u2 (sin a – cos a) cos a tan a a

Ans. 8



dy = x2 – 2x dx ˆ d Ê x3 - x 2 + c˜ = Á dx Ë 3 ¯

Solution:

=

x3 fi y= – x2 + c 3 Since the curve passes through (2, 0), we get 0 = 8/3 – 4 + c, i.e, c = 4/3. Hence the equation of the curve is 4 x3 – x2 + 3 3 Now, from dy/dx = 0, we get x = 0 or 2. Also, d2 y = 2x – 2 d x2 fi

d2 y d x2

and

u2 È 2 1 ˘ sin a - sin a cos a - (1 - 2 sin a cos a ) ˙ Í a Î 2 ˚

u2 (sin2 a – 1/2) a u2 (sin2 a – 1/2) = u2/4a According to given condition a fi sin2 a = 3/4. =

If p1 and p2 are lengths of the Example 87 perpendiculars from origin on the tangent and normal to 1 the curve x2/3 + y2/3 = 62/3 then the value of (4p21 + p22) is 6 equal to

y=

d2 y d x2

Ans. 6 Solution: Any point on the curve x2/3 + y2/3 = 6 2/3 is of the form (6 cos3 q, 6 sin3q). Now dy/dx = – y1 3 x1 3 , so

= –2 x=0

=2

dy dx

x=2

Hence at x = 0, y has a maximum and ymax = 4/3 Example 86

If the point on y = x tan a -

ax

dy ax = tan a – 2 dx u cos2 a

the tangent is parallel to y = x if i.e., tan a –

ax u 2 cos2 a



dy = 1, dx

=1,

(tan a - 1) u cos a a 2

x=

(6 cos3 q , 6 sin3 q )

sin q = – tan q. cos q

Y – 6 sin3 q = – tan q (X – 6 cos3 q )

2



X sin q + Y cos q = 6 sin3 q cos q + 6 sin q cos3 q = 6 sin q cos q. The equation of normal at (6 cos3 q, 6 sin3 q ) is

Ans. 3



2

=–

Thus the equation of tangent at (6 cos3 q, 6 sin3 q ).

2

24 cos a (a > 0), where the tangent is parallel to y = x has an ordinate u2/4a, then 4 sin2 a is equal to

Solution:

a u4 (sin a – cos a)2 cos2 a 2u 2 cos2 a a2

x=

Y – 6 sin3 q = cot q (X – 6 cos3 q ), i.e. cos q X – sin qY = 6 (sin2 q – cos2 q ) = – 6 cos 2q

p2 = 6 cos 2q

and 4p21

2

u2 (sin a - cos a) cos a a

p1 = 6 sin q cos q

Therefore,

Example 88

+ p22 = 36 sin2 2q + 36 cos2 2q = 36. Let q (x) denote an angle measured in

on the y-axis at a point x on the positive x-axis. If q0 is the maximum value of q (x). then p /q0 is equal to Ans. 6

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Solution: Since the slope AP is tan–1 (– 1/x) and the slope of BP is tan–1 (– 3/x) so q (x) = tan–1 (– 1/x) – tan–1 (– 3/x) = tan–1 (3/x) – tan–1 (1/x) = tan–1

2/ x 1 + 3/ x

2

= tan–1

2x x +3 2

Since tan x is an increasing function so tan q (x) is maximum when q (x) is maximum. But 2x A = tan q (x) = 2 x +3 dA x2 + 3 - 2 x2 - x2 + 3 = 2 = 2 dx ( x 2 + 3)2 ( x 2 + 3)2 dA = 0 fi x = ± 3. dx dA dA > 0 for x2 < 3 and < 0 for x2 > 3. Thus A is Also dx dx 2 3 maximum when x = 3 . Hence max q (x) = tan–1 3+3 = tan–1 1/ 3 = p /6.

Fig. 24.7

Fig. 24.8

3sin q cos3 q (1 + sin q )2 fi

- (1 + sin q )3 (cos3 q - 2sin 2 q cos q ) dV 1 = pr3 dq 3 sin 2 q cos 4 q =

(1 + sin q )2 1 pr3 [3sin q (1 – sin2 q) 3 sin 2 q cos3 q – (1 + sin q) (1 – 3 sin2 q )]

=

(1 + sin q )3 1 p r3 (3 sin q – 3 sin2 q 3 sin 2 q cos3 q – 1 + 3 sin2 q)

=

(1 + sin q )3 (3 sin q - 1) 1 pr3 3 sin 2 q cos3 q

So from dV/dq = 0, we get sin q = –1 or sin q = 1 3 . That is, q = 3p/2 or q = sin–1 1 3 . Now, since q = 3p /2 is not possible, we must have q = sin–1 (1 3) and 0 < q < p 2

A cone is circumscribed about a sphere

and d V d q < 0 for 0 < q < sin - 1 (1 3) and d V d q > 0 for sin–1 (1/3) < q < p /2, V is minimum when q = sin–1 (1/3).

Solution: Let K be the centre of the sphere and let CAB be the cone. Let O be the centre of the base of the cone and q its semi-vertical angle (Fig. 24.8). The altitude of the cone is given by h = r + r cosec q and the radius of the base of the cone is R = r (1 + cosec q ) tan q. Therefore, the volume of the cone is

Example 90 A soft drink manufacturer wants to fabricate cylindrical cans for its products. The can is to have a volume of 343.75 cu cm. If r is the radius for can 8p r 2 is equal to require least amount of material then 275 to

Example 89 of radius r. The volume of the cone is minimum when 3 sin q is equal to Ans. 1

V= =

1 p r3 (1 + cosec q )3 tan2 q 3 (1 + sin q )3 1 p r3 3 sin q cos2 q

Ans. 5 Solution: Let r be the radius and h be the height of the can. Total surface area S = 2pr2 + 2prh We need minimize S so that V = pr2 h = 343.75 fi

h=

343.75 p r2

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Ê 343.75 ˆ S(r) = 2pr2 + 2pr Á Ë p r 2 ˜¯ = 2pr2 +

set

4 1 + = a has at least one solution in sin x 1 - sin x the interval (0, p/2) is equation

687.5 r

dS 687.5 = 4pr dr r

The least value of a for which the

Example 92

S,

Ans. 9 Solution:

dS 687.5 = 0 fi r2 = dr 4p



d2S 687.5 8p r 2 Since = 4p + > 0 so for minimum 275 d r2 r2 S is equal to 5 cm2. Example 91 Two roads OA and OB intersect at an angle of 60°. A car driver approaches O from A where OA = 800 m at a uniform speed of 20 m/sec. Simultaneously another car moves from O towards B at a uniform speed of 25m/sec. If t is the time when the two cars are closest, then [t]2 is equal to Ans.256

Let y =

dy cos x 4 = - 2 cos x + dx sin x (1 - sin x )2 Ê ˆ 4 1 = cos x Á - 2 + 2˜ Ë sin x (1 - sin x ) ¯

So from dy/dx = 0, we get cos x = 0 fi

or

x= p 2



s2 = (800 – 20t)2 + 625t2 – (40 – t)250t

For s to be minimum, we have fi

ds = 0. dt

– 40(800 – 20t) + 1250t – (40 – t) 250 + 250t = 0



– 42000 + 2550t = 0 840 fit= sec so [t] = 16 51

It is easy to check d 2 s dt 2 > 0.

8 ± 64 - 48

x = sin–1

6 2 3

=

8±4 2 = 2, 6 3

Ê ˆ d2 y 4 1 = – sin x Á - 2 + 2˜ 2 dx Ë sin x (1 - sin x ) ¯ Ê 8 cos x 2 cos x ˆ + + cos x Á ˜ 3 Ë sin x (1 - sin x )3 ¯

Fig. 24.9



4(1 – sin x)2 = sin2 x

or

sin x =



(800 - 20t )2 + (25t )2 - s2 1 = cos 60° = (800 - 20t ) 25t 2

4 1 = 2 sin x (1 - sin x )2

Since x Π( 0, p 2) , we have from 3 sin2 x Р8 sin x + 4 = 0



Solution: Let the distance between two cars after t seconds be s t seconds is 20 t, so its distance from O will be (800 – 20t m) and the distance covered by second car will be 25t m.

4 1 + on ( 0, p 2) sin x 1 - sin x



d2 y d x2

x = sin -1

2 3

2 = - ( -9 + 9) 3

ˆ 4ˆ Ê 8 2 Ê + Á1 - ˜ Á = 35 > 0 + Ë 9 ¯ Ë 8/9 [1 - (2/3)]3 ˜¯ Hence a has a minimum at x = sin–1 2 3 in ( 0, p 2) , and its minimum value is a=

4 1 + =9 2/3 1 - ( 2/3)

Thus a = 9. Example 93 For the circle x2 + y2 = r2, the value of r for which the area enclosed by the tangents drawn from the point (6, 8) to the circle and the chord of contact is maximum is Ans. 5

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Solution:

p( x ) ˆ Ê Ê p( x ) ˆ lim Á1 + 2 ˜ = 2 fi lim Á 2 ˜ = 1 x Æ0 Ë x ¯ x ¯ Let p(x) = a0 x4 + a1x3 + a2x2 + a3x + a4. The given limit is a3 = a4 = 0 and is 1 if a2 = 1. Thus p(x) = a0x4 + a1x3 + x2 p(x) has extremum at x = 1, 2 if p¢(1) = 0, p¢(2) = 0 fi 4a0 + 2a1 = – 2, 32a0 + 12a1 = –4.

Solution:

x Æ0 Ë

Solving we get, a0 = 1/4, a1 = –1

Fig. 24.10

Let O be the centre of the circle x2 + y2 = r2 and P(6, 8) be the point outside the circle. Let q be the angle between the tangent AP and OP. Clearly –OAM = q so that BM = r cos q and OM = r sin q (0 < q < p /2). Also OP = = 10 and r = 10 sin q. If A is the area of the triangle PAB, then A = 2 (area of DPBM )

36 + 64

= 2 ((1/2) PM ¥ BM) = (OP – OM) BM = (10 – r sin q) r cos q

\ p(x) =

f (x) = 2x – 15x + 36x – 48 on the set A = {x| x2 + 20 £ 9x} is 3

= 50 sin 2q cos2 q 25 [sin 4q + 2 sin 2q ] 2 dA = 50 (cos 4q + cos 2q ) So dq = 100 cos 3q cos q

2

Ans. 7 Solution:

x2 + 20 £ 9x fi x2 – 9x + 20 £ 0



(x – 4)(x – 5) £ 0



A = [4, 5] f ¢(x) = 6x2 – 30x + 36

= (10 – 10 sin q) (10 sin q cos q) = 100 cos3 q sin q

= 6(x – 2) (x – 3) > 0 on [4, 5] Thus f increases on [4, 5], Therefore, max f ( x ) = f (5) = 7. x ŒA

=

For maximum value of q, fi cos 4q + cos 2q fi 2 cos 3q cos q fi 3q fi q

= = = =

dA =0 dq

0 0 p /4 (0 < q < p /2) p /6

d2 A = – 100 (2 sin 4q + sin 2q) dq 2 Ê 3ˆ = – 100 Á 3 + 0 upon the coordinate axes is (a) 2a (b) a

x+ y

(c) a/2 (d) a 4. Let x and y be two real numbers such that x > 0 and xy = 1. The minimum value of x + y is (a) 1 (b) 1/2 (c) 2 (d) 1/4

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x2 - x + 1

5. The maximum value of 2 for all real valx + x +1 ues of x is (a) 1/2 (b) 1 (c) 2 (d) 3 6. If y = 2x + cot-1 x + log

(

)

1 + x 2 - x , then y

(a) decreases on (- •, •) (b) decreases on [0, •) (c) neither decreases nor increases on [0, •) (d) increases on (- •, •) 7. Let g(x) = (log (1 + x))-1 - x-1, x > 0 then (a) 1 < g(x) < 2 (b) - 1 < g(x) < 0 1 (c) 0 < g(x) < 1 (d) < g( x) < 1 2 8. Given n real numbers a1, a2, ... an, the value of x for which sum of the square of all the deviations is least is (a) a1 + a2 + ... + an (b) 2(a1 + a2 + ... + an ) (c) a12 + a22 + ... + an2 a1 + a2 + º + an n 9. The number of solutions of the equation a f (x) + g (x) = 0, where a > 0, g(x) π 0 and g(x) has minimum value 1/4, is (a) one (b) two (d)

10. Suppose that 4 1 + =a sin x 1 - sin x has at least one solution on the interval (0, p/2). Then a has minimum value of x = (b) sin–1 1/4 (a) sin–1 2/3 –1 (d) 1 (c) cos 4/5 11. If the tangent at (1, 1) on y2 = x(2 - x)2 meets the curve again at P, then P is (a) (4, 4) (b) (-1, 2) (c) (9/4, 3/8) (d) (3/4, 7/4) sin ( x + a ) has no maximum or min12. The function sin ( x + b ) imum if (k an integer) (a) b - a = kp (b) b - a π kp (c) b - a = 2k p (d) none of the above 13. The two curves x3 - 3xy2 + 2 = 0 and 3x2y - y3 - 2 = 0 (a) cut at right angles (b) touch each other (c) cut at an angle p/3 (d) cut at an angle p/4

14. If x cos a + y sin a = p touches x2 + a2 y2 = a2, then (a) p2 = a2 sin2 a + cos2 a (b) p2 = a2 cos2 a + sin2 a (c) 1/p2 = sin2 a + a2 cos2 a (d) 1/p2 = cos2 a + a2 sin2 a 15. The set of all values of the parameters a for which the points of minimum of the function y = 1 + a2x x2 + x + 2 £ 0 is x3 satisfy the inequality 2 x + 5x + 6 (a) an empty set (b) (c) (d)

( - 3 3, - 2 3 ) (2 3, 3 3 ) ( - 3 3, - 2 3 )

» ( 2 3, 3 3 )

16. Three normals are drawn to the parabola y2 = 4x from the point (c, 0). These normals are real and distinct when (a) c = 0 (b) c = 1 (c) c = 2 (d) c = 3 17. The function f (x) = (log (x - 1))2 (x - 1)2 has (a) local extremum at x = 1 x=1 (c) local extremum at x = 2 x=2 18. Given the function f (x) = x2 e-2x, x > 0. Then f (x) has the maximum value equal to (b) (2e)-1 (a) e-1 (c) e-2 (d) none of these 19. Let f (x) = (x - 4) (x - 5) (x - 6) (x - 7) then (a) f ¢(x) = 0 has four real roots (b) three roots of f ¢(x) = 0 lie in (4, 5) » (5, 6) » (6, 7) (c) the equation f ¢(x) = has only two roots (d) three roots of f ¢(x) = 0 lie in (3, 4) » (4, 5) » (5, 6) 20. The points on the curve 5x2 - 6xy + 5y2 = 4 that are the nearest the origin are (a) (1/2, - 1/2), (- 1/2, 1/2) (b) (0, 2/ 5 ), (0, - 2/ 5 ) (c) ( 2/ 5 , 0), (- 2/ 5 , 0) Ê 2 ˆ 3, 0 , Á ,1 Ë 5 ˜¯ 21. A given right circular cone has a volume p, and the largest right circular cylinder that can be inscribed in the cone has a volume q. Then p : q is (a) 9 : 4 (b) 8 : 3 (c) 7 : 2 (d) 5 : 3 (d)

(2/

)

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22. The point of intersection of the tangents drawn to the curve x2y = 1 - y at the points where it is met by the curve xy = 1 - y is given by (a) (0, - 1) (b) (1, 1) (c) (0, 1) (d) (1, –1) 23. The equation of the tangent to the curve y = (2x - 1) e2(1 - x) at the point of its maximum is (a) y = 1 (b) x = 1 (c) x + y = 1 (d) x - y = - 1 24. The distance of the point on y = x4 + 3x2 + 2x which is nearest to the line y = 2x - 1 is (b) 3/ 5 (a) 4/ 5 (c) 2/ 5 (d) 1/ 5 2 25. If the function f (x) = x + a/x has a local minimum at x = 2, then the value of a is (a) 8 (b) 16 (c) 18 (d) 12

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 26. The coordinates of the point on the curve (x2 + 1) ( y - 3) = x where a tangent to the curve has the greatest slope are given by (a)

(

3, 3 + 3/4 )

(b)

(c) (0, 3)

(d)

(-

(

3, 3 - 3/4 )

3, 3/4

)

27. The critical points of the function f (x) = (x - 2)2/3 (2x + 1) are (a) - 1 and 2 (b) 1 (c) 1 and - 1/2 (d) 1 and 2 28. The function f (x) = sin x cos2 x has extremum at (a) x = p/2

(b) x = cos

(

(c) x = cos-1 - 2/3

)

-1

(1/ 3 )

(

(d) x = cos-1 - 2/3

)

29. The function f (x) = 2 log (x - 2) - x2 + 4x + 1 increases in the interval (a) (1, 2) (b) (2, 3) (c) (5/2, 3) (d) (2, 4) 30. The critical points of the function f ¢(x), where f (x) x-2 are = x3 (a) 0 (b) 1 (c) 3 (d) - 1 31. y = log x x > 1, the equality (a) x - 1 > y (b) x2 - 1 > y (c) y > x - 1 (d) (x - 1)/x < y 32. The interval (s) of decrease of the function f (x) = x2 log 27 - 6x log 27 + (3x2 - 18x + 24) log (x2 - 6x + 8) is

(a) (3 -

1 + 1/3e , 2)

(c) (3, 4 +

1 + 1/3e )

(b) (4, 3 +

1 + 1/3e )

(d) none of these

33. If the line, ax + by + c = 0 is a normal to the curve xy = 2, then (a) a < 0, b > 0 (b) a > 0, b < 0 (c) a > 0, b > 0 (d) a < 0, b < 0 34. The normal to the curve given by x = a (cos q + q sin q), y = a (sin q - q cos q) at any point q is such that it (a) makes a constant angle with x-axis (b) is at a constant distance from the origin (d) passes through the origin log (p + x ) is log (e + x ) (a) increasing on [0, •) (b) decreasing on [0, •) (c) increasing on [0, p/e) and decreasing on [p/e, •) (d) decreasing on [0, p/e) and increasing on [p/e, •) x 36. e > 10 x (a) for all x (b) for all x > 3 (c) for x only if x > 20 (d) for all x > 18 37. An extremum value of the function f (x) = (sin-1 x)3 + (cos-1 x)3 (- 1 < x < 1) is (b) p3/8 (a) 7p3/8 (d) p3/16 (c) p3/32 35. The function f (x) =

38. Let f (x) = (1 - x 2 )(1 + 2 x 2 ) then (a) the greatest value of f (x) is 1

- 1, 1]

(b) the greatest value of f (x) is 3/ 8 (c) the least value of f (x) is 0 (d) the least value of f (x) is - 1. 39. Let f (x) = x4 - 4x3 + 6x2 - 4x + 1 then (a) f increases on [1, •) (b) f decreases on [1, •) (c) f has a minimum at x = 1 (d) f has neither maximum nor minimum y = ax, y2 + x2 = c2 are perpendicular for (a) a = 2, c = 4 (b) a = - 2, c = 3 (c) a = 3, c = 2 (d) a = 3, c = - 2

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MATRIX-MATCH TYPE QUESTIONS 41. Match the maxima of the functions f (x) on L.H.S (a) x2 e-x 4x (b) 2 x +4

(p) - 1

(c) - x2 5 ( x - 2)2

(r) f (2)

14 (d) 4 x - 8x2 + 2

(s) 1

(q) 0

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos 48 to 50

42. Match the difference of greatest and least value of the functions f on L.H.S (a) sin x sin 2x on (- •, •) (p) p/6 (b) 2x3 - 3x2 - 12x + 1 on [- 2, 5/2]

(q) 8/ 3 3

(c) cos-1 x2 on ÈÎ- 2/2, 2/2˘˚

(r) 6

(d) x + x on [0, 4] 43. For the function on L.H.S (a) cos x - 1 x 2 x3 + 2! 3!

(s) 27

(p) minimum value is - 4

x2 (q) there is no extremum 2! at x = 0 4 -x2 (c) x e (r) the minimum is f (0) = 0 (d) sin 3x - 3 sin x (s) the functions reaches (b) cos x - 1 +

maximum at x = 2 - 1, 1] and has values in [–12, 8] (a) 2x3 - 3x2 - 12x + 1 (p) is one-one but not onto

(1 - x 2 )(1 + 2 x 2 )

(b) (c)

1 - x + x2 1 + x + x2 3

2

(d) x – 3x + 6x – 2

Statement-2: f decreases from p/4 to 0. 47. Statement-1: ex + e–x > 2 + x2, x π 0 Statement-2: f (x) = ex + e–x – 2 – x2 is an increasing function.

(q) is one-one and onto (r) is neither one-one nor onto (s) is not one-one but onto

ASSERTION-REASON TYPE QUESTIONS 20 45. Statement-1: Let f (x) = then the 3 4 x 9 x2 + 6x range f = [6, 20] Statement-2: f increases on (1/2, 1) and decreases on (1, •) » (– •, 1/2). 1- x on [0, 1] then 46. Statement-1: Let f (x) = 2 tan–1 1+ x the range of f = [0, p/2].

A manufacturing plant has a capacity of 25 articles per week. Experience has shown that x articles per week can be sold at a price of ` p each where p = 110 - 2x and the cost of producing x articles is 600 + 10x + x2 `. (a) straight line (b) a parabola with vertex at (20, 700/3) Ê 50 700 ˆ and focus (c) a parabola with vertex at Á , Ë 3 3 ˜¯ Ê 50 2199 ˆ ÁË , ˜ 3 3 ¯ Ê 50 700 ˆ (d) a parabola with vertex at Á , and focus Ë 3 3 ˜¯ Ê 50 2099 ˆ ÁË , ˜ 3 3 ¯ 49. Integral value of x maximum (a) 24 (b) 22 (c) 17 (d) 16 50. If P (a) P increases on [8, 25] (b) P decreases on [8, 25] È 2 ˘ (c) P decreases on Í16 , 25˙ Î 3 ˚ (d) P increases on [8, 22] Paragraph for Question Nos 51 to 53 Among several applications of maxima and minima is an> be a sequence. Consider f (x) obtained by replacing x by n e.g. n x let an = consider f (x) = on [1,•) f ¢(x) = n +1 x +1 1 > 0 for all x. Hence max f (x) = lim f (x) = 1, so xÆ• ( x + 1)2 the largest term of < an> is 1. 51. The largest term of an = n2/(n3 + 200) is (a) 29/453 (b) 49/543 (c) 43/543 (d) 41/451 52. The largest term of the sequence an = n/(n2 + 10) is

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(a) 3/19 (b) 2/13 (c) 1 (d) 1/7 53. If f (x Q.51 then (a) f increases for all x (b) f decreases for all x (c) f has a maximum at x = 3 400 (d) f increases on [0, 9].

INTEGER-ANSWER TYPE QUESTIONS 54. If A is the area of the triangle formed by positive x-axis and the normal and the tangent to the circle x2 + y2 = 4 at (1, 3 ) then A/ 3 is equal to 55. The minimum value of

2

e x - 1 is

Ï x -1 + a , x £1 56. Let f (x) = Ì , x >1 Ó 2x + 3 If f (x) has local minimum at x = 1 and a ≥ 5 then a is equal to x2 y2 + = 1, 57. Let P be a variable point on the ellipse 25 9 with foci F1 and F2. If A is the area of the triangle 1 PF1F2, then the maximum value of A is 3 58. The maximum value of e|x log x| for 0 < x £ 1 is 59. If f (x) = logx 1/9 - log3 x2 (x > 1) then max f (x) is equal to 60. Let f (x) = cos2 x + cos x + 3 then greatest value of f (x) + 4 least value of f (x) – 9 is equal to 2 1 log (greatest 61. If f (x) = ex - 4x + 3 on [- 5, 5] then 12 value of f (x)) is

62. If (u, v) is a point on 4x2 + a2y2 = 4a2, where 4 < a2 < 8, that is farthest from the point (0, - 2) then u + v is equal to 63. Let f (x f ( x) ˆ Ê lim Á1 + 3 ˜ x ¯

xÆ0Ë

1/ x

= e2 and has local maximum at x

5 f (3) = 1 and local minimum at x = 0 and 2 then 112 is equal to 64. If y = f (x) is represented as x = g(t) = t5 - 5t3 - 20t + 7 and y = h(t) = 4t3 - 3t2 - 18t + 3 (- 2 < t < 2) then max f (x) – 9 is equal to 65. If f (x) = x2 log x on [1, e] then log (greatest of f (x) - least of f (x)) is equal to

LEVEL 2

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. Let f f (x) = 2x2 – log |x|, x π 0 then (a) f increases on [– 1/2, 0] » [1/2, •) (b) f decreases on (– •, 0) (c) f increases on (– •, – 1/2) (d) f decreases on [1/2, •) 2. The smallest value of M such that | x2 – 3x + 2| £ M for all x in [1, 5/2] is (a) 1 (b) 1/2 (c) 1/4 (d) 3/4 3. A cone is made from a circular sheet of radius 3 by cutting out a sector and giving the cut edges of the remaining piece together. The maximum volume attainable for the cone is (a) p/3 (b) p/6 (d) 3 3 p

(c) 2p/3

point of f (x) = x4 – 6x3 + 12x2 – 8x + 3 is (a) y = 3x + 4 (b) y = 4 (c) y = 3x + 2 (d) none of these 5. The dimensions of the rectangle of maximum area that can be inscribed in the ellipse (x/4)2 + (y/3)2 = 1 are (a)

8, 2

(c) 2 8, 3 2 6. The maximum value of (a) (1/e) (c) 1

e

(b) 4, 3 (d)

( 1 x)

x

8, 3 is

(b) e1/e (d) e–1/e

7. Let f (x) = 6x4/3 – x1/3 (a) maximum value of f is 7 (b) maximum value of f is 5 (c) maximum value of f is 9 (d) minimum value of f is – 3/2 8. The shortest distance of (0, 0) from the curve e x + e- x is y= 2 (a) 1/2 (b) 1/3 (c) 2 (d) 1 9. The normal to the circle x2 + y2 – 2x – 2y = 0 passing through (2, 2) is

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(a) x = y (b) 2x + y – 6 = 0 (c) x + 2y – 6 = 0 (d) x + y – 4 = 0 10. If f (0) = 0 and f ≤(x) exists for all x > 0 then f (x)/x (a) decreases on (0, •) (b) increases on (0, •) (c) decreases on (1, •) (d) neither increases nor decreases on (0, •) 11. The value of k so that the equation x3 – 12x + k = 0 has distinct roots in [0, 2] is equal to (a) 4 (b) 2 (c) – 2 (d) no real number 12. The number of real roots of the equation e–|x|(2x3 – 3x2 + 6x + 6) = 0 is (a) 1 (b) 2 (c) 3 (d) 4 13. Let f (x) = (x – 2) (x4 – 4x3 + 6x2 – 4x + 1) then value of local minimum of f is (a) – 2/3 (b) – (4/5)4 (c) – 44/55 (d) – (4/5)5 2 14. Let f (x) = x – 2|x| + 2, x Œ [– 1/2, 3/2] then (a) (b) (c) (d)

min

f ( x ) = 1/2

min

f ( x) = 1

min

f ( x ) = 3/2

min

f ( x ) = 3/4

x Œ[ -1/ 2, 3/ 2] x Œ[ -1/ 2, 3/ 2] x Œ[ -1/3, 3/ 2] x Œ[ -1/ 2, 3/ 2]

x xˆ Ê f (x) = (a2 – 3a + 2) Á cos2 - sin 2 ˜ Ë 4 4¯ + (a – 1)x + sin 1 does not possess critical points is (a) (1, •)

(b) (–2, 4)

(c) (1, 3) » (3, 5) (d) (0, 1) » (1, 4). 19. Let f (x) = (x – 2) (x – 3) (x – 4) (x – 5) (x – 6) then (a) f ¢(x (b) four roots of f ¢(x) = 0 lie in (2, 3) » (3, 4) » (4, 5) » (5, 6) (c) the equation f ¢(x) has only three roots (d) four roots of f ¢(x) = 0 lie in (1, 2) » (2, 3) » (3, 4) » (4, 5) 20. Let f (x) = (x – 3)5 (x + 1)4 then (a) x = – 1 is point of minima (b) x = 1 is point of maxima (c) x = 7/9 is a point of maxima (d) x = – 1 is neither a point of maxima and minima. 21. The equations of those tangents to 4x2 – 9y2 = 36 which are perpendicular to the straight line 2y + 5x = 10, are Ê 117 ˆ (a) 5( y – 3) = 2 Á x 4 ˜¯ Ë (b) 5( y – 2) = 2(x – (c) 5( y + 2) = 2 (x – (d) y + 2 = x - 18

x 15. Let f (x) = tan x and g(x) = , x > 0 then 1 + x2 –1

(b) f (x) £ g(x) on [1, •) (d) none of these 1 16. The greatest value of f (x) = tan–1 x – log x on 2 [1/ 3, 3] is (a) f (x) < g(x), x > 0 (c) g(x) < f (x) on (0, •)

(a)

p 1 + log 3 3 2

(b)

p 1 + log 3 6 4

(c)

p 1 + log 3 6 2

(d)

p 1 + log 3 3 4

17. The condition for y = ax4 + bx3 + cx2 + dx + e to (b) 3b2 – 8ac = 0 (a) b2 – 4ac > 0 2 (d) 3b2 – 8ac < 0 (c) 3b – 8ac > 0 18. The set of all values of a for which the function

18 ) 18 )

22. The value of a for which the equation x3 – 3x + a = 0 has two distinct roots in [0, 1], is given by (a) –1 (b) 1 (c) 3 (d) 0 23. The point of intersection of the tangents drawn to the curve x2y = 1 – y at the points where it is intersected by the curve x y = 1 – y, is given by (a) (0, 1) (b) (1, 1) (c) (0, –1) (d) (–1, –1) 24. If the sum of the squares of the intercepts on the axes cut off by the tangent to the curve x1/3 + y1/3 = a1/3 (a > 0) at (a/8, a/8) is 2, then a has the value (a) 1 (b) 2 (c) 4 (d) 8 25. The value of m for which the area of the triangle included between the axes and any tangent to the curve xm y = bm is constant, is (a) 1 2

(b) 1

(c) 3 2

(d) 2

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26. If the tangent at any point on the curve x4 + y4 = a4 cuts off intercepts p and q on the coordinate axes, the value of p–4/3 + q–4/3 is (b) a–1/2 (c) a1/2 (d) 0 (a) a–4/3 27. A channel 27 m wide falls at a right angle into another channel 64 m wide. The greatest length channels is (a) 120 (b) 125 (c) 100 (d) 110 28. For a Œ [p, 2p], the function 1 sin a tan3 x f (x) = 3 a-2 + (sin a – 1) tan x + 8-a has (a) x = n p (n Œ I) as critical points (b) no critical points (c) x = 2n p (n Œ I) as critical points (d) x = (2n + 1)p (n Œ I) as critical points. 29. The greatest value of y = (x + 1)1/3 – (x – 1)1/3 on [0, 1] is (a) 1 (b) 2 (c) 3 (d) 21/3 30. The coordinates of the point M(x, y) of y = e–|x| so that the area formed by the coordinates axes and the tangent at M is greatest, are (a) (e, 1) (c) (– 1, e–1)

(b) (1, e–1) (d) (0, 1)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 31. An interval of increase of the function y = x – 2 sin x if 0 £ x £ 2p, is (a) (p /3, p) (b) (0, p ) (c) (p /2, p) (d) (p/3, 5p /3) | x - 1| are 32. The critical points of the function f (x) = x2 (a) 0 (b) 1 (c) 2 (d) – 1 33. The interval of increase of the function y = x – ex + tan (p /7) is (a) (– •, – 1) (b) (0, •) (c) (– •, 0) (d) (1, •) 34. The critical points of the function f (x) = (x – 2)2/3 (2x + 1) are (a) – 1 and 2 (b) 1 (c) 1 and – 1/2 (d) 1 and 2 35. The function f (x) = x3/4 – sin p x + 3 on [–2, 2] takes the value

(a) 1 (b) 7 3 (c) 6 (d) 3 36. The normal to the curve represented parametrically by x = a(cos q + q sin q) and y = a(sin q – q cos q) at any point q, is such that it (a) makes a constant angle with the x-axis (b) is at a constant distance from the origin (d) passes through the origin. 37. The value of a for which the function f (x) = (4a – 3) (x + log 5) + 2(a – 7) cot ( x 2) sin 2 ( x 2) does not possess critical points is (a) ( - •, - 4 3) (b) (– •, – 1) (c) [1, •) (d) (2, •) 38. The interval to which b may belong so that the function Ê 21 - 4 b - b2 ˆ 3 f (x) = Á1 ˜ x + 5x + 6 ÁË ˜¯ b +1 is increasing at every point of its domain is (a) [–7, –1] (b) [– 6, –2] (c) [2, 2.5] (d) [2, 3]. 39. Which of the following is correct (b) e3 > 3e (a) e2 > 2e (c) ep > p e (d) ep > (p /2) e 40. Let f (x) = (x2 – 1)n (x2 + x – 1) then f (x) has local extremum at x = 1 when (a) n = 2 (b) n = 3 (c) n = 4 (d) n = 5

MATRIX-MATCH TYPE QUESTIONS 41. Match the interval of monotonicity Column 1 Column 2 (p) increases on (- •,•) (a) y = x - ex

(

(b) y = log x + 1 + x 2 (c) y = x 4 x - x 2

(d) y =

10 4 x - 9x2 + 6x 3

) (q) increases on (0, 3) (r) increases on (- •, 0) and decreases on (0, •) (s) increases on (1/2,1) and decreases on (- •, 0) » (0, 1/2) » (1, •)

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42. Match the greatest value of the functions Column 1 Column 2 (a) y = 100 - x 2 on [- 6, 8] (b) y = 2 tan x - tan2 x on [0, p/2) 1- x on (c) y = tan-1 1+ x [0, 1]

(p) no greatest value (q) 10

COMPREHENSION-TYPE QUESTIONS (r) p/4

a2 b2 + on (s) 1 x 1- x (0, 1) a > 0, b > 0 43. Match the extrema of the functions Column 1 Column 2 2 3x + 4 x + 4 (a) y = (p) ymax = 4 x2 + x + 1 (d) y =

(b) y =

3

x - 3x + 64 3

(q) ymin = -1/3

2

(c) y = - x 2 x 2 + 2 (d) y =

(r) ymax = 2

1 log ( x + 4 x 3 + 30) 4

Column 1 (a) y = x3 - 5x2 + 3x - 5 (b) y = (x + 1)4 + ex

(s) ymax = 0

Column 2 (p) (2, 62) and (4,206)

tion point (r) (1, - 1)

(c) y = x - 12x + 48x2 - 50 (d) y = x + 36x2 (s) (5/3, - 250/27) 3 4 2x - x 45. Match the extrema of the functions in Column 1 Column 2 4

(a) y = x +

3

1- x

(p) ymin = 0

(b) y = x2 e-x

(q) ymin = 1

(c) y = x 2 - x 2

(r) ymax = 5/4

(d) y =

e an + e - ax 2

2 sin x £ < 1 for 0 < |x| £ p/2 p x sin x Statement-2: f (x) = , 0 < x £ p/2 is x increasing.

47. Statement-1:

(s) ymin = - 1

ASSERTION-REASON TYPE QUESTIONS 46. Statement-1: Between any two roots of e x sinx = 1, there is at least one real root of ex cos x + 1 = 0. Statement-2: ex sinx – 1 is an increasing function on R.

Paragraph for Question Nos. 48 to 50 Using monotone property, we can establish certain inequalities. Some of inequalities can also be established using Lagrange’s theorem or Rolle’s theorem. Suppose we have to establish ex > 1 + x (x π 0). Consider the function f (x) = ex - (1 + x) so f ¢(x) = ex - 1 > 0 for all x > 0 then f increases on R. So f (x) > f (0) = 0 for all x > 0 fi ex > 1 + x. Similarly, f ¢(x) < 0 for x < 0 so f (x) < f (0) for x < 0 i.e. ex > 1 + x. 48. Let f (x) = 2 x , g(x) = 3 - 1/x, h(x) = ex + e-x and u(x) = 2 + x2 then (a) f (x) < g(x) for x > 1 (b) f (x) £ g(x) for x > 1 (c) h(x) > u (x) for x π 0 (d) h(x) £ u (x) for x π 0 49. Let f (x) = 2x tan-1 x, g(x) = log (1 + x2), h(x) = x3 x4 + then sin x, u(x) = x 6 120 (a) f (x) < g(x) (b) f (x) ≥ g (x) (c) h(x) > u(x), x > 0 (d) h(x) ≥ u(x), x > 0 tan - 1 x x , r(x) = 50. Let u(x) = log (1 + x), v(x) = + 1 + x 1 x then (a) u (x) £ v(x), x > 0 (b) r(x) ≥ u(x), x > 0 (c) r(x) > u(x), x > 0 (d) u (x) > v(x), for x > 0 Paragraph for Question Nos. 51 to 53 One can use Rolle’s theorem to show that a function f (x) has exactly one zero, two zeros etc. in an interval. We can also establish that certain equation has at least one root inside an interval using Rolle’s theorem. 4 51. Consider f (x) = x3 + 2 + 7, (- •, 0), then x (a) f has exactly three zeros in (- •, 0) (b) f has exactly two zeros in (- •, 0) (c) f has exactly one zero in (- •, 0) (d) f has no zero in (- •, 0) 52. Let f (x) = x4 + 3x + 1, [- 2, - 1] then (a) f has exactly two zeros in [- 2, - 1] (b) f has exactly one zero in [- 2, - 1]

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(c) f has at least one zero in [- 2, - 1] (d) f has no zero in [- 2, - 1] 53. Consider the equation x log x + x = l, l > 1 then this equation (a) has at least one solution in [1, l] (b) has no solution in (1, l) (c) has two solutions in (1, l) (d) has three solutions in (1, l)

INTEGER-ANSWER TYPE QUESTIONS 54. If y = 1 - (x - 2)4/5, then ymax is 55. The greatest value of the function y = sin2 x 20 cos x + 1 is 56. If y = a log x + bx2 + x has its extremum values at x = - 1 and x = 2 then the value of - 2ab is 57. If the tangent at the point (16, 64) on the curve y2 = 1 x3 meets the curve again at Q(u, v) then uv is 16 3 -1 58. If u(x) is larger of x - (x /6) and tan x for 0 < x £ 48 1 then u(1/2) is equal to 23 59. If v(x) is larger of ex - 1 and (1 + x) log (1 + x) for x Œ (0, •) then log (V(8) + 1) is equal to 1 cu metres, open 7 at the top is to be manufactured from a sheet of met-

60. A cylindrical vessel of volume 25

al. If r and h are the radius and height of the vessel so that amount of metal is used in the least possible then rh is equal to 61. The altitude of a cylinder of the greatest possible volume which can be inscribed in a sphere of radius 3 3 is 62. If a and b are two parts of 8 such that sum of their 1 cubes is the least possible then ab is equal to 4 63. A straight line L with negative slope passes through the point (8, 2) and cuts the positive coordinate axes at points P and Q. The absolute minimum value of 1 (OP + OQ) as L varies, where O is the origin is 9 3 x=0 Ï Ô 2 64. If f (x) = Ì- x + 3x + a 0 < x < 1 Ô mx + b 1£ x £ 2 Ó then the value of (a + b)m is 65. The circle x2 + y2 = 1 cuts the x-axis at P and Q. Another circle with centre at Q and variable radius R above the x-axis and the line segment PQ at S. If A is the maximum area of the triangle QSR then 3 3 A is equal to _____.

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS Questions on Tangents and Normal 1. The normal to the curve x = a (cos θ + θ sin θ ), y = a (sin θ – θ cos θ) at any point θ is such that (a) it makes a constant angle with the x-axis (b) it passes through the origin (c) it is at a constant distance from the origin (d) none of these [1983] 2. If the normal to the curve y = f (x) at the point (3, 4) makes an angle 3π/4 with the positive x-axis, then f (3) = (a) – 1 (b) – 3/4 (c) 4/3 (d) 1 [2000] 3. The point(s) on the curve y3 + 3x2 = 12y where the tangent is vertical, is/are (a) (±4 / 3 , - 2)

(b) (± 11 / 3 , 1)

(c) (± 11 / 3 , 1)

(d) (± 4 / 3 , 2) [2002]

4. Tangent to the curve y = x2 + 6 at a point P (1, 7) touches the circle x2 + y2 + 16x + 12y + c = 0 at a point Q. Then the coordinates of Q are (a) (– 9, – 13) (b) (– 10, – 15) (c) (– 6, – 7) (d) (6, – 7) [2005] 5. The tangent to the curve y = ex drawn at the point (c, ec) intersects the line joining the points (c – 1, ec – 1) and (c + 1, ec + 1) (a) on the left of x = c (b) on the right of x = c (c) at no point (d) at all point [2007] Questions on Increasing and Decreasing 6. Let f and g be increasing and decreasing functions, respectively from [0, ) to [0, ). Let h (x) = f (g (x)). If h (0) = 0, then h (x) – h (1) is (a) always zero (b) always negative (c) always positive (d) strictly increasing

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(e) none of these [1987] 7. The function f f (x) = (x + 2)e– x is (a) decreasing for all x (b) decreasing in (– , – 1) and increasing in (–1, ) (c) increasing for all x (d) decreasing in (– 1, ) and increasing (– , – 1) [1994] log (p + x ) is 8. The function f (x) = log (e + x ) (a) increasing on [0, ) (b) decreasing on [0, ) (c) increasing on [0, π /e] and decreasing on [π /e, ) (d) decreasing on [0, π /e] and increasing on [π /e, ) [1995] x x and g (x) = , where 0 < x 1, 9. If f (x) = sin x tan x

10.

11.

12.

13.

then in this interval (a) both f (x) and g (x) are increasing functions (b) both f (x) and g(x) are decreasing functions (c) f (x) is an increasing function (d) g (x) is an increasing function [1997] The function f (x) = sin4 x + cos4 x increases if (a) 0 < x < π /8 (b) π/4 < x < 3π /8 (c) 3π/8 < x < 5π /8 (d) 5π /8 < x < 3π /4 [1999] For all x (0, 1) (b) loge (1 + x) < x (a) ex < 1 + x (c) sin x > x (d) loge x > x [2000] If f (x) = x e x(1 – x), then f (x) is (a) increasing on [– 1/2, 1] (b) decreasing on R (c) increasing on R (d) decreasing on [– 1/2, 1] [2001] Let the function g: (– , ) (– π /2, π/2) be given by g (u) = 2 tan–1 (eu) – π /2, then g is (a) even and is strictly increasing in (0, ) (b) odd and is strictly decreasing in (– , ) (c) odd and is strictly increasing in (– , ) (d) neither even nor odd, but in strictly increasing in (– , ). [2008]

14. The number of points in (– , ), for which x2 – x sin x – cos x = 0 is (a) 6 (b) 4 (c) 2 (d) 0 [2013] Questions on Maxima and Minima 15. Two towns A and B are 60 km apart. A school is to be built to serve 150 students in town A and 50 students in town B. If the total distance to be travelled by all 200 students is to be as small as possible, then the school should be built at

(a) town B (c) town A

16.

17.

18.

19.

(b) 45 km from town A (d) 45 km from town B [1982] If y = a log x + bx2 + x has its extremum values at x = – 1 and x = 2, then (a) a = 2, b = – 1 (b) a = 2, b = – 1/2 (c) a = – 2, b = 1/2 (d) none of these [1983] 2 4 2n Let P(x) = a0 + a1x + a2x + …+a n x be a polynomial in a real variable x with 0 < a0 < a1 < a2 < … < an. The function P(x) has (a) neither a maximum nor a minimum (b) only one maximum (c) only one minimum (d) only one maximum and only one minimum (e) none of these [1986] 25 75 In the interval [0, 1], the function x (1 – x) takes its maximum value at the point (a) 0 (b) 1/4 (c) 1/2 (d) 1/3 [1995] The function f (x) = |px – q| + r|x|, x (– , ), where p > 0, q > 0, r > 0 assumes its minimum value only at one point if (a) p q (b) r q (c) r p (d) p = q = r [1995]

20. If f (x) =

x2 - 1 x2 + 1

for every real number x, then the

minimum value of f (a) does not exist because f is bounded (b) is not attained even though f is bounded (c) is equal to 1 (d) is equal to – 1 [1998] Ï| x | for 0 < | x | £ 2 21. Let f (x) = Ì for x = 0 Ó1 Then at x = 0, f has (a) a local maximum (b) no local maximum (c) a local minimum (d) no extremum [2000] x2 + y2 = 1 at ( 3 3 22. Tangent is drawn to ellipse 27 cos θ, sin θ) (where θ (0, π /2)). Then the value at θ such that sum of intercepts on axes made by this tangent is least is (a) π /3 (b) π /6 (c) π/8 (d) π /4 [2003] 23. The total number of local maxima and local minima of the function ÏÔ(2 + x )3 , - 3 < x £ - 1 f (x) = Ì 2/3 ÓÔ x , - 1 < x < 2 (a) 0

(b) 1

(c) 2

(d) 3 [2008]

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24. Let f, g and h 2 2 2 the interval [0, 1] by f (x) = ex + e–x , g(x) = xex 2 2 2 + e–x and h(x) = x2ex + e–x . If a, b and c denote, respectively, the absolute maximum of f, g and h on [0, 1], then (a) a = b and c

b

(b) a = c and a

(c) a

b

(d) a = b = c [2010]

b and c

b

Questions of Mean Value Theorem 25. For which of the following functions [0, 1] the Lagrange’s Mean Value theorem is NOT applicable to Ï1 1 x< Ô - x, 2 Ô2 (a) f (x) = Ì 2 1 ÔÊ 1 ˆ ÔÁË 2 - x˜¯ , x ≥ 2 Ó Ï sin x , xπ0 Ô (b) f (x) = Ì x ÔÓ1, x=0 (c) f (x) = x|x| (d) f (x) = |x|

(b) lim f (x) = 1 xƕ

(c) for all x in the interval [1, ), f (x + 2) – f (x) > 2 (d) f (x) is strictly decreasing in the interval [1, ) [2009] Questions on Maxima and Minima

[2003]

ÏÔ xa log x if x > 0 26. Let f (x) = Ì if x = 0 ÔÓ 0 If Rolle’s theorem can be applied to f on [0, 1], then value of α can be (a) –1 (b) –1/2 (c) 0 (d) 1/2 [2004]

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS Questions on Tangent and Normal 1. If the line ax + by + c = 0 is a normal to the curve xy = 1, then (a) a > 0, b > 0 (b) a > 0, b < 0 (c) a < 0, b > 0 (d) a < b, b < 0 [1986] Questions on Increasing and Decreasing x+2 then 2. If y = f (x) = x -1 (a) x = f (y) (b) f (1) = 3 (c) y increases with x for x < 1 (d) f is a rational function x [1984] ÏÔ3x 2 + 12 x - 1 , - 1 £ x £ 2 3. If f (x) = Ì then ,2< x£3 ÔÓ37 - x (a) f (x) is increasing on [– 1, 2]

(b) f (x) is continuous on [– 1, 3] (c) f (2) does not exist (d) f (x) has the maximum value at x = 2 [1993] 2 3 4. Let h(x) = f (x) – (f (x)) + (f (x)) for every real number x. Then (a) h is increasing whenever f is increasing (b) h is increasing whenever f is decreasing (c) h is decreasing whenever f is decreasing (d) nothing can be said in general [1998] 1 5. For function f (x) = x cos , x 1, x (a) for atleast one x in interval [1, ), f (x + 2) – f (x) < 2

6. f (x) is a cubic polynomial which has local maximum at x = – 1. If f (3) = 18, f (–1) = 2 and f ′ has local minimum at x = 0 then (a) the distance between (– 1, 2) and (a, f (a)) where x = a is the point of local minima is 2 5 . (b) f (x) is increasing for x [1, 2 5 ] (c) f (x) is has local minima at x = 1 (d) the value of f (0) = 5

[2006]

x, 0 £ x £ 1 Ï ÔÔ x 7. f (x) = Ì2 - e x - 1 , 1 £ x £ 2 ; g ( x ) = Ú0 f (t ) dt Ô ÔÓ x - e , 2 < x £ 3 x [1,3] then g (x) has (a) local maxima at x = 1 + log 2 (b) local maxima at x = 1 and local minima at x = 2 (c) local maxima at x = e (d) local minima at x = e [2006] having their lengths in the ratio 8 : 15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100, the resulting box has maximum volume. Then the lengths of the sides of the rectangular sheet are (a) 24 (b) 32 (c) 45 (d) 60 [2013] 9. The function f (x) = 2| x | + | x + 2 | – | | x + 2 | –2| x | has a local minimum or a local maximum at x = -2 (a) – 2 (b) 3

IIT JEE eBooks: www.crackjee.xyz 24.40 Comprehensive Mathematics—JEE Advanced

(b) 2

(d)

2 3 R be given by

[2013]

10. Let a R and let f : R f (x) = x5 – 5x + a then (a) f (x) has three real roots if a > 4 (b) f (x) has only one real roots if a > 4 (c) f (x) has three real roots if a < –4 (d) f (x) has three real roots if –4 < a < 4

MATRIX-MATCH TYPE QUESTIONS Questions on Increasing and Decreasing

[2014] x ˆn

Ê nˆ Ê nˆ Ê n n ( x + n) Á x + ˜  Á x + ˜ ˜ Á Ë 2¯ Ë n¯ ˜ , 11. Letf (x) = lim Á 2 x Æ• Á Ê 2 n2 ˆ ˜ 2 2 Ê 2 n ˆ Á n !( x + n ) Á x + ˜  Á x + 2 ˜ ˜ ÁË 4¯ Ë n ¯ ˜¯ Ë

for all x > 0. Then Ê 1ˆ (a) f Á ˜ ≥ f (1) Ë 2¯

Ê 1ˆ Ê 2ˆ (b) f Á ˜ ≥ f Á ˜ Ë 3¯ Ë 3¯

f ¢ (3) f ¢ (2) ≥ f (3) f (2) [2016] 12. Let f : R Æ (0, •) and g : R Æ R be twice differentiable functions such that f≤ and g≤ are continuous function of R. Suppose f ¢(2) = g(2) = 0, f ≤(2) π 0 f ( x) g ( x) and g¢(2) π 0. If lim = 1, then xÆ 2 f ¢ ( x) g ¢ ( x) (c) f ¢(2) £ 0

(a) (b) (c) (d)

(d)

f has a local minimum at x = 2 f has a local maximum at x = 2 f ≤(2) > f (2) f (x) – f ≤(x) = 0 for at least one x Œ R [2016]

Questions on Mean Value Theorem 13. Let f, g : [–1, 2] → R be continuous functions which are twice differentiable on the interval (–1, 2). Let the values of f and g at the points –1, 0 and 2 be as given in the following table: f (x) g (x)

x = –1 3 0

x=0 6 1

(c) f '(x) – 3g'(x) = 0 has exactly one solution in (0, 2) (d) f '(x) – 3g'(x) = 0 has exactly two solutions in (–1, 0) and exactly two solutions in (0, 2) [2015]

x=2 0 –1

In each of the intervals (–1, 0) and (0, 2) the function (f – 3g)" never vanishes. Then the correct statement(s) is(are) (a) f '(x) – 3g'(x) = 0 has exactly three solutions in (–1, 0) » (0, 2) (b) f'(x) – 3g'(x) = 0 has exactly one solution in (–1, 0)

Ê p pˆ domain Á - , ˜ . Ë 2 2¯ Column 1 (a) x + sin x (b) sec x

Column 2 (p) is increasing (q) is decreasing (r) is neither increasing nor decreasing [1992] 2. Match the statements given in Column I with the intervals/union of intervals given in Column II. Column I Column II (a) The (p) (– , –1) (1, ) Ï Ê 2iz ˆ ÌRe ÁË ˜ : z is a complex number 1 - z2 ¯ Ó | z |= 1, z π ±1} is (b The domain of the function

(q) (– , 0)

(0, )

Ê 8(3) x - 2 ˆ f (x) = sin -1 Á ˜ is Ë 1 - 32( x -1) ¯ 1 (c) If f (q) = - tan q -1 then the set

tan q 1 - tan q

p¸ Ï Ì f (q ) : 0 £ q < ˝ is 2˛ Ó

1 tan q , 1

(r) [2, )

(d) If f (x) = x3/2 (3x – 10), (s) (– , –1] [1, ) x 0, then f (x) is increasing in (t) (– , 0] [2, ) [2011]

ASSERTION-REASON TYPE QUESTIONS Questions on Increasing and Decreasing 1*. Consider the following statements S and R. S : Both sin x and cos x are decreasing functions in the interval (π /2, π). * Originally a question of Straight Objective Type

IIT JEE eBooks: www.crackjee.xyz Applications of Derivatives 24.41

R : If a differentiable function decreases in an interval (a, b), then its derivative also decreases in (a, b). Which of the following is true? (a) Both S and R are wrong. (b) Both S and R are correct, but R is not the correct explanation for S. (c) S is correct and R is the correct explanation for S. (d) S is correct and R is wrong. [2000]

COMPREHENSION-TYPE QUESTIONS Paragraph for Questions 1 to 2 Let f : [0, 1] Æ R (the set of all real numbers) be a function. Suppose the function f is twice differentiable f (0) =f f ≤(x) – 2f ¢(x) + f (x) ≥ ex, x Œ [0, 1] 1. Which of the following is true for 0 < x < 1? (a) 0 < f (x) < • (c) - 1 < f ( x) < 1 4

(b) - 1 < f ( x) < 1 2 2 (d) –• < f (x) < 0

2. If the function e–x f (x) assumes its minimum in the 1 interval [0, 1] at x = , which of the following is 4 true? (a) f ¢(x) < f (x), 1 < x < 3 4 4 1 (b) f ¢(x) > f (x), 0 < x < 4 1 (c) f ¢(x) < f (x), 0 < x < 4 3 (d) f ¢(x) < f (x), < x < 1 [2013] 4

INTEGER-ANSWER TYPE QUESTIONS

[2009]

4. Let f R (the set of all real numbers) such that f (x) = 2010(x – 2009) (x – 2010)2 (x – 2011)3 (x – 2012)4, for all x R. If g interval (0, ) such that

R with values in the

f (x) = ln (g (x)), for all x

R,

then the number of points in R at which g has a local maximum is [2010] 5. Let f : R R f (x) = | x | + | x 2 - 1| . The total number of points at which f attains either a local maximum or a local minimum is [2012] 6. Let p(x) be a real polynomial of least degree which has a local maximum at x = 1 and a local minimum at x = 3. If p (1) = 6 and p(3) = 2, the p (0) is [2012] 7. A cylindrical container is to be made from certain solid material with the following constraints: It has V mm3, has a 2 mm thick solid wall and is open at the top. The bottom of the container is a solid circular disc of thickness 2 mm and is of radius equal to the outer radius of the container. If the volume of the material used to make the container is minimum when the inner radius of V the container is 10 mm, then the value of is 250p [2015]

FILL

IN THE

BLANKS TYPE QUESTIONS

Questions on Tangent and Normal 1. Let C be the curve y3 – 3xy + 2 = 0 If H is the set of points on the curve C where the tangent is horizontal and V is the set of points on the curve where the tangent is vertical, then H = ___________ and V = ___________. [1994] Questions on Increasing and Decreasing

Questions on Tangent and Normal 1. The slope of the tangent to the curve (y – x5)2 = x (1 + x2)2 at the point (1, 3) is

Then the value of p(2) is

[2014]

Questions on Maxima and Minima 2. The maximum value of the function f (x) = 2x3 – 15x2 + 36x – 48 on the set A = {x | x 2 + 20 9x} is [2009] 3. Let p(x) be a polynomial of degree 4 having extremum at x = 1, 2 and p( x ) ˆ Ê lim Á1 + 2 ˜ = 2. x Æ0 Ë x ¯

2. The function y = 2x2 – log |x| is monotonically increasing for values of x (π0) satisfying the inequalities ___________ and monotonically decreasing for values of x satisfying the inequalities __________. [1983] 3. The set of all x for which log (1 + x) £ x is equal to __________. [1987] Questions on Maxima and Minima p 4. If A > 0, B > 0 and A + B = , then maximum value 3 of tan A tan B is ___________. [1993]

IIT JEE eBooks: www.crackjee.xyz 24.42 Comprehensive Mathematics—JEE Advanced

SUBJECTIVE-TYPE QUESTIONS Questions on Tangent and Normal 1. Find all the tangents to the curve y = cos (x + y), – 2π x 2π that are parallel to the line x + 2y = 0. [1985] 2. Tangent at a point P1 (other than (0, 0)) on the curve y = x3 meet the curve again at P2. The tangent at P2 meets the curve at P3, and so on. Show that the abscissae of P1, P2 … Pn ratio [area ( P1 P2 P3)]/[area ( P2 P3 P4)] [1993] 3. Find the equation of the normal to the curve y = (1 + x)y + sin–1 (sin2 x) at x = 0. 3

[1993]

2

4. The curve y = ax + bx + cx + 5 touches the x-axis at P(– 2, 0) and cuts the y-axis at a point Q where the gradient is 3. Find a, b, c. [1994] 5. If a function f (x x, y R | f (x) – f (y)| (x – y)2 Find an equation of tangent to the curve y = f (x) at point (1, 2). [2005] Questions on Increasing and Decreasing 6. Use the function f (x) = x1/x, x > 0 to determine the [1981] larger of the two numbers eπ and πe. 7. Show that 1 + x ln ( x + x 2 + 1) ≥ x 2 + 1 for all x 0. [1983] 8. Show that 2 sin x + tan x 3x, 0 x < π /2. [1990] 9. Let ÏÔ xe ax , x£0 f (x) = Ì 2 3 ÓÔ x + ax - x , x > 0 where a is a positive constant. Find the interval in which f (x) is increasing. [1996] 10. Let – 1 p 1. Show that the equation 4x3 – 3x – p = 0 has a unique root in the interval [1/2, 1] and identify it. [2001] 11. Using the relation 2(1 – cos x) < x2, x otherwise prove that sin (tan x)

x

x

[0, π /4]

0 or [2003]

12. Justifying all the inequalities used, show that f (x) = sin x + 2x (3/π) x (x + 1) for 0 x π 2. [2004] Questions on Maxima and Minima 13. Let x and y be two real numbers such that x > 0 and xy = 1. Find the minimum value of x + y. [1981]

14. A swimmer S is in the sea at a distance d km from the closest point A on a straight shore. The house of the swimmer in on the shore at a distance L km from A. He can swim at a speed of U km/hr and walk at a speed of V km/hr (V > U). At what point on the shore should be land so that he reaches his house in the shortest possible time? [1983] 15. Find the coordinates of the point on the curve x where the tangent to the curve has the y= 1 + x2 greatest slope. [1984] 3 2 16. Let f (x) = sin x + λ sin x, – π/2 < x < π/2. Find the intervals in which λ should lie in order that f (x) has exactly one minimum and exactly one maximum. [1985] 17. Let A ( p2, – p), B (q2, q), C(r2, – r) be the vertices of the triangle ABC. A parallelogram AFDE is drawn with vertices D, E and F on the line segments, BC, CA and AB respectively. Using calculus, show that maximum area of such a parallelogram is: (1/4) (p + q) (q + r) (p – r) [1986] 18. Find the point on the curve 4x2 + a2y2 = 4a2, 4 < a2 < 8 that is farthest from the point (0, –2). [1987] 19. A point P is given on the circumference of a circle of radius r. Chord QR is parallel to the tangent at P. Determine the maximum possible area of the triangle PQR. [1990] the arch) is in the form of a rectangle surmounted by

with clear glass. The clear glass transmits three times as much light per square metre as the coloured glass does. What is the ratio of the sides of the rectangle so that the window transmits the maximum light? [1991] 2 21. What normal to curve y = x forms the shortest chord? [1992] 22. Let Ï 3 (b3 – b2 + b - 1) , 0 £ x< 1 Ô- x + ( b 2 + 3b + 2) Ô f (x) = Ì Ô 2x - 3 1 £ x £ 3. Ô Ó Find all possible real values of b such that f (x) has the smallest value at x = 1. [1993] 2 2 23. The circle x + y = 1 cuts the x-axis at P and Q. Another circle with center at Q and variable radius

IIT JEE eBooks: www.crackjee.xyz Applications of Derivatives 24.43

8 metres apart. The speed of the object after t seconds is given by ds = (2t + 1) metres per second. dt Let α and β be the angles subtended by the object at A and B, respectively after one and two seconds. Find the value of cos (α – β ). [1987]

R above the x-axis and the line segment PQ at S. Find the maximum area of the triangle QSR. [1994] 24. Determine the points of maxima and minima of the function f (x) = (1/8) log x – bx + x2, x > 0, where b 0 is a constant. [1996] 25. Suppose f (x) is a function satisfying the following conditions (a) f (0) = 2, f (1) = 1 (b) f has a minimum value at x = 5/2 and (c) for all x 2ax f (x) =

2ax - 1

b b +1 2(ax + b) 2ax + 2b + 1

LEVEL 1

2ax + b + 1

SINGLE CORRECT ANSWER TYPE QUESTIONS

-1 2ax + b

where a, b are some constants. [1998] Determine the constants a, b and the function f (x). 26. Find the coordinates of all the points P on the x2 y2 ellipse 2 + 2 = 1, for which the area of the a b triangle PON is maximum, where O denotes the origin and N, the foot of the perpendicular from O to the tangent at P. [1999] r for 27. For the circle x2 + y2 = r2 which the area enclosed by the tangents drawn from the point P (6, 8) to the circle and the chord of contact is maximum. [2003] 28. P(x) is a cubic polynomial satisfying P(–1) = 10, P(1) = –6. If P (x) has a local maximum at x = –1 and P (x) has a local minimum at x distance between maximum and minimum points on the curve y = P (x). [2005] Questions on Mean Value Theorem 29. For all x [0, 1], let second derivative f (x) of a function f (x) exist for and satisfy |f (x)| 1. If f (0) = f (1), then show that |f (x)| < 1 for all x [0, 1]. [1981] 30. Using Rolle’s theorem or otherwise, show that f (x) = 51x101 – 2323 x100 – 45x + 1035 = 0 has a root lying in the interval ( 45

Answers

1 100

1. 5. 9. 13. 17. 21. 25.

from the midpoint of two locations A and B,

2. 6. 10. 14. 18. 22.

(a) (d) (a) (b) (c) (c)

3. 7. 11. 15. 19. 23.

(b) (c) (c) (c) (b) (a)

4. 8. 12. 16. 20. 24.

(c) (d) (b) (d) (a) (d)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 26. 29. 32. 35. 38.

(a), (b), (a), (b) (b),

(c) (c) (b)

27. 30. 33. 36. 39.

(c)

(d) 28. (a), (c) 31. (a), (a), (b) 34. (b), (c), (d) 37. (c) (a), (c) 40. (a),

(c), (d) (d) (c) (b), (c), (d)

MATRIX-MATCH TYPE QUESTIONS p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

41.

, 46).

[2004] 31. If f (x) is a twice differentiable function such that f (a) = 0, f (b) = 2, f (c) = – 1, f (d) = 2, f (e) = 0 where a < b 0 for x < - 1 and S¢(x) < 0 for x > - 1. So S is maximum when x = - 1. Hence Smax = 1 + 2/1 = 3. x 1 1 d Ê ˆ + 6. y¢(x) = 2 ¥ -1 2 Á ˜ 2 2 x d 1+ x 1+ x - x Ë 1+ x ¯ =2-

(



log (1 + x ) x 1 > fi log (1 + x) > 1+ x 1+ x x

fi g (x) < 1 8. Sum of the squares of the deviation n

S = Â (x - ai)2 n

dS dS = 2 Â (x - ai). For S to be minimum =0 dx dx i =1 fi

)

1 + x2 - 1 + x2 + x2 1 1 = ≥0 1 + x2 1 + x2 1 + x2

(x - ai) = 0

a1 + a2 º + an d2S . Also = 2(1 + ... + 1) n dx 2 1 = 2n > 0. So S is minimum when x = Â ai . n 9. a f (x) = - g(x) £ - 1/4, since g(x) ≥ 1/4 for all x. But this is not possible as a f (x) > 0 for all x. Thus the number of solution is zero. 10. See Example 92. fi

11. 2y

2

=-2

log (1 + x) < x fi g(x) > 0 1 1 > Also x Π(0, x) fi x < x fi 1+ x 1+ x

i =1

So the sum of intercepts = x + y + 2 x+



. Equation of

(X - x). So x intercepts of X-axis and Y-axis are x + x y and y + xy .

(

log (1 + x ) 1 = 0. Hence x = 2 is a point of extremum. 18. f ¢(x) = 2xe-2x - 2x2 e-2x = 2(1 - x)x e-2x, f ¢(x) = 0 ¤ x = 1, 0. Also f ≤(x) = 2(1 - x)e-2x - 2xe-2x - 4(1 - x)xe-2x, so f ≤(1) = - 2e-2 < 0 and f ≤(0) > 0. Thus max f (x) = f (1) = e-2. 19. Since f (4) = f (5) = f (6) = f (7) = 0, so by Rolle’s theorem applied on the interval [4, 5], [5, 6], [6, 7] there are x1 Œ (4, 5), x2 Œ (5, 6), x3 Œ (6, 7) such that f ¢(x1) = f ¢(x2) = f ¢(x3) = 0. Since f¢ is a polynomial of degree 3 so cannot have four roots. 20. Differentiating the given equation, we have 10x - 6y - 6y y¢(x) + 10y y¢(x) = 0 (3y - 5x)/(5y - 3x)

dy 2 dy = so dx y dx

= (t 2 , 2t )

2 1 = . 2t t

y¢(x) =

6 È 5 y2 - 5x2 ˘ Í ˙ 5 Î 5 y - 3x ˚

S¢(x) = 0 ¤ .y = ± x. If y = x then 5x2 - 6xy + 5y2 = 4 fi 10x2 = 6x2 = 4 fi x = ± 1 If y = - x fi 16x2 = 4 fi x = ± 1/2. Thus the extremum points are (1, 1), (- 1, - 1), (1/2, - 1/2) (- 1/2, 1/2). dy Ê ˆ - 10 x˜ (5 y - 3 x ) Á10 y dx Ë ¯

16. Any point on y2 = 4x is (t2, 2t). Since



The square of the distance of (x, y) from origin is S = x2 + y2 = 2(2 + 3xy)/5 ((x, y) is a point on the given curve)

=

we must have x2 + 5x + 6 < 0. If x = - a/ 3 , we have 2

Thus the roots are real and distinct if c > 2 so c = 3 is the correct choice. 17. f ¢(x) = (x - 1)2 2 log (x - 1) 1/(x - 1) + 2 (log(x - 1))2 (x - 1) = 2 (x - 1) [log (x - 1) [1 + log (x - 1)] f ¢(x) = 0 fi x = 2 or 1 + e-1, since the domain of f is {x : x > 1}. f ≤(x) = 2 log (x - 1) (1 + log (x - 1))

dS (3 y - 5 x ) ˘ 6 6 dy 6È = Íy + x = y+ ¥ 5 5 dx 5 Î 5 y - 3 x ˙˚ dx

dy d2 y = a2 - 3x2 = 0 ¤ x = ± a/ 3 . Since dx d x2 = - 6x so y is minimum for x = - a/ 3 . Since x2 + x + 2 > 0 for all x so for

t = 0, t2 = c - 2

d2 S

6 = 2 5 dx

Ê dy ˆ - (5 y 2 - 5 x 2 ) Á 5 -3 Ë d x ˜¯ (5 y - 3 x )2

IIT JEE eBooks: www.crackjee.xyz 24.48 Comprehensive Mathematics—JEE Advanced

d2 S d x2

= ( - 1, 1)

6 2( - 20) < 0 etc. Thus the required 5 4

points are (1/2, - 1/2) and (- 1/2, 1/2). 21. Let H be the height of the cone and a be its semivertical angle. Suppose that x is the radius of the inscribed cylinder and h be its height h = QL = OL - OQ = H - x cot a V = volume of the cylinder = px2 (H - x cot a) 1 (i) Also p = p (H tan a)2 H 3 dV = p (2Hx - 3x2 cot a) dx So

dV =0 dx

¤

d2 V d x2

2 x = H tan a 3

when x = p

x = 0, x =

2 H tan a, 3

= - 2pH < 0. So, V is maximum

2 H tan a and q = Vmax = 3

4 2 1 4 4 p 3 p tan 2 a H tan2 a H = = p. 9 3 9 27 p tan 2 a

Hence p : q = 9 : 4.

1 . Thus the equation of tangents are Y = 1 and 2 Y - 1/2 = (- 1/2) (X - 1). Their intersection is given by (0, 1). 23. y¢(x) = - 2 (2x - 1) e2(1 - x) + 2e2(1 - x) = 2e2(1 - x) (- 2x + 2) Thus y¢(x) = 0 fi x = 1. Since, y≤(1) < 0 so (1, 1) is the point of maximum and the equation of tangent is y - 1 = 0 (x - 1), i.e. y = 1. 24. Distance of any point (x, y) from y = 2x - 1 is y - 2x + 1 . If (x, y) is on y = x4 + 3x2 + 2x then this 5 x 4 + 3x2 + 1 distance is S = 5 = -

d S 4 x3 + 6 x = dx 5

Also, S ¢(x) < 0 for x < 0 and S¢(x) > 0 for x > 0. Thus S is minimum when x = 0, and min. S is 1/ 5 . 25. f ¢(x) = 2x - a/x2, f ¢(x) = 0 ¤ 2x = a/x2 ¤

x = (a/2)1/3.

Clearly for x = (a/2)1/3 we have f ≤(x) > 0. So 2 = (a/2)1/3 fi a = 16. 26. Solving for y the given equation we have y=3=

dy 1 - x2 x = f (x) (say) fi = d x 1 + x2 2 x2 + 1

(

Now f ¢(x) =

Fig. 24.12

22. The intersection of x2y = 1 - y and xy = 1 - y are given by x2y = xy. Since y π 0 on any of two curves so we must have x = 0, 1. Thus the curves intersects at (0, 1); (1, 1/2). dy 2x y = . Now, differentiating x2 y = 1 - y, we have d x 1 + x2 dy dy = 0 and Thus d x (0, 1) d x (1, 1/ 2)

dS = 0 fi x = 0. dx



)

- 2 x (3 - x 2 )

(1 + x 2 )

2

For extremum of f (x), we have f ¢(x) = 0 fi x = 0, x= ± 3. At x = 0, f ¢(x) changes sign from positive to negative fi f (x) has maxima at x = 0. Thus the required point is (0, 3). 2 (x - 2)-1/3 (2x + 1) + 2(x - 2)2/3 = 27. f ¢(x) = 3 10(x - 1) (x - 2)-1/3. f ¢(x) = 0 fi x = 1 also f ¢(x) does not exist at x = 2 hence the critical points are x = 1 and x = 2. 28. f ¢(x) = cos3 x + 2 sin2 x (- cos x) = cos x (3 cos2 x - 2). Since the function is differentiable everywhere so the points of extremum are given by f ¢(x) = 0. f ¢(x) = 0 So

¤

x = p/2 or cos x = ± 2/3

x = p/2, cos-1

(

)

(

2/3 , cos-1 - 2/3

)

IIT JEE eBooks: www.crackjee.xyz Applications of Derivatives 24.49

29. Note that f (x x > 2. For x > 2, f ¢(x) = (2/(x - 2)) - 2(x - 2) = 2[1 - (x - 2)2]/(x - 2). As x > 2, f ¢(x) > 0, if 1 - (x - 2)2 > 0 or (x - 2))2 < 1 or 1 < x < 3. As x > 2, we get f (x) increases on (2, 3).

Similarly + 8)e > 0

Ïx - 2 x >1 ÔÔ x 3 , 30. f (x) = Ì Ô 2 - x , x < 1, x π 0 ÔÓ x 3

Hence

f ≤(x) doesn’t exist at x = 3. Thus critical point of f ¢(x) is 3. 31. Consider f (x) = log x - (x - 1) fi f ¢(x) = 1/x -1 So f ¢(x) < 0 for x > 1. Thus f (1) > f (x) for x > 1 fi 0 > log x - (x - 1) fi x - 1 > log x x -1 - log x fi g¢(x) Consider g(x) = x >0 x2 for all x. Hence g(x) > g(1), x > 1 fi (x - 1)/x > log x (b) is clearly not true. This can ascertained by taking x = 2. 32. f ¢(x) = 2x log 27 - 6 log 27 + (6x - 18) log (x2 - 6x + 8) (3 x 2 - 18 x + 24) (2 x - 6)

For f (x

= 6(x - 3) log (3(x2 - 6x + 8)e) x2 - 6x + 8 > 0

fi x > 4 or x < 2 If x > 4, then f ¢(x) < 0 if log 3(x2 - 6x + 8) e < 0 i.e. 3(x2 - 6x + 8)e < 1 i.e. x2 - 6x + (8 - 1/3e) < 0 i.e. ¤

( x - (3 +

3-

Hence

1 + 1/3e

)) ( x - (3 -

1 + 1/3e < x < 3 +

(

x Π4, 3 + 1 + 1/3e

))

1 + 1 / 3e < 0

1 + 1/3e

)

x Π3 - 1 + 1/3e , 2

1 + 1/3e

)

dx ˘ = a q cos q ˙ dq ˙ fi d y = tan q dy ˙ dx and = a q sin q ˙ dx ˚ so the slope of the normal is - cot q. Thus the equation of normal is y - a (sin q - q cos q) = - cot q [x - a (cos q + q sin q)]. x cos q + y sin q = a, in which form we see that the distance of the normal from the origin is |a| and that it touches the circle x2 + y2 = a2. log (e + x ) ¥

( x - 1/2)2 + 1/4

x2 - 6 x + 8

(

x>3+

or

So the slope of normal is x/y. Hence we must have - a/b = x/y = x2/xy = x2/2 > 0 so a/b < 0, i.e. either a > 0, b < 0 or a < 0, b > 0. 34. We have

Ï 6 ( x - 4) , x Œ( - •, 0) » (0, 1) ÔÔ x 5 f ≤(x) = Ì Ô 6 (4 - x ) , x Œ(1, 3) » (3, •) ÔÓ x 5

+

1 + 1/3e

x e + x for x > 0, and log x is an increasing function so log (p + x) > log (e + x). Thus (p + x) log (p + x) . (p + x) log (e + x) > (e + x) log (e + x). Hence f ¢(x) < 0 for x > 0. Thus f decreases on [0, •). 36. Let f (x) = ex - 10x so f ¢(x) = ex - 10. Hence f increases for x > log 10. In particular, f (x) = ex - 10x > f (18) = e18 - 180 > 0. So ex > 10x for x > 18. In particular ex > 10x for x > 20. However, ex > 10 x is not true for all x > 3. 37. f ¢(x) = 3[(sin-1 x)2 - (cos-1 x)2]/

(

)

1 - x2 .

For extremum value, we have f ¢(x) = 0, so (sin-1 x)2 - (cos-1 x)2 = 0 i.e. sin-1 x - cos-1 x = 0, or sin-1 x + cos-1 x = 0 but since sin-1 x + cos-1 x = p/2 so the equality is not possible. Hence x = 2 . Thus f (p/4) = (p/4)3 + (p/4)3 = p2/32. 38. The points of extremum of f (x) is same as that of g(x) = (1 - x2) (1 + 2x2), x Œ [- 1, 1]. But g ¢(x) =

IIT JEE eBooks: www.crackjee.xyz 24.50 Comprehensive Mathematics—JEE Advanced

39.

40.

41.

42.

- 2x (1 + 2x2) + 4x (1 - x2) = 2x [- 1 - 2x2 + 2 - 2x2] = 2x (1 - 4x2). g ¢(x) = 0 ¤ x = 0, x = ± 1/2. Since f (0) = 1, f (- 1) = 0, f (1) = 0, f (- 1/2) = f (1/2) = 3/4 ¥ 3/2 = 3/ 8 . f ¢(x) = 4(x3 - 3x2 + 3x - 1) = 4(x - 1)3 > 0 for x > 1. Hence f increases on [1, •). Moreover, f ¢(x) < 0 for x < 1. Hence f has a minimum at x = 1. dy/dx = a y/dx = - x/y from the equation of the circle. So a × (- x/y) = - ax/y = - 1 as y = ax, so the curves are perpendicular for every value of a and c. If f (x) is as in (i) then f ¢(x) = 2xe-x - x2e-x = x [2 - x]e-x. Since e-x π 0 for all x, therefore critical points are x = 0, x = 2. For 1 < x < 2, f ¢(x) > 0 and 2 < x < 3, f ¢(x) < 0. Hence x = 2 is a point of maxima. Max. f (x) = f (2) = 4e-2. Similarly other functions in (ii), (iii) and (iv) If f is as in (i), then f (x) = 1/2 (cos x - cos 3x). Since period of f is 2p cient to seek the greatest and least value among the interval [0, p]. f ¢(x) = (1/2) (3 sin 3x - sin x) The derivative vanishes at x1 = 0, x2 = cos-1 1/ 3 , x3 = cos-1 ( - 1/ 3 ) , x4 = p 4 y (0) = y(p) = 0 y(cos-1 ± 1/ 3 ) = ± 3 3 Hence the greatest value is 4/3 3 and the least value is - 4/3 3 . For function in (ii) greatest value of f (x) = f (2) = 8 and least value is f (2) = - 19. For function in (iii) greatest value is f (0) = p/2 and least value is f ( ± 2/2) = p/3. For (iv) greatest value is y(4) = 6 and the least value is y(0) = 0.

43. For f in (i), f ¢(x) = - sin x + x + x2/2, f ¢(0) = 0 f ≤(x) = - cos x + 1 - x, f ≤(0) = 0 f ≤¢(x) = sin x - 1, f≤¢(0) = - 1 π 0 so there is no extremum at x = 0. Functions in (ii), (iii), (iv) can be treated similarly. 44. Let f (x) = 2x3 - 3x2 - 12x + 1 so f ¢(x) = 6x2 6x - 12 = 6(x3 - x - 2) = 6(x - 2) (x + 1). Thus f decreases on [- 1, 2], in particular on [- 1, 1]. Hence max f (x) = f (- 1) = - 2 - 3 + 12 + 1 = 8 and min f (x) = f (1) = 2 - 3 - 12 + 1 = - 12. So the range of f = [- 12, 8] and hence f is onto. f is one-one on [- 1, 1] as if f (x1) = f (x2) then (x1 - x2) [2(x21 + x22 + x1 x2) - 3(x1 + x2) - 12] = 0. This equation is possible

for x1, x2 Œ [- 1, 1] only if x1 = x2. If f (x) = g ( x ) where g(x) = (1 - x2) (1 + 2x2) then clearly f is not one-one as f (1) = f (- 1). g¢(x) = - 2x (1 + 2x2) + (1 - x2)4x = x[2 - 8x2]. So g ¢(x) = 0 fi x = 0, x = ± 1/2. Now g(0) = 1, g(1/2) = g(- 1/2) = 9/8 and g(1) = g(- 1) = 0. Hence the greatest value of f (x) on [- 1, 1] is 3/ 8 and the least value is 0. Thus f in this case is not onto. 45. f decreases on (– •, 0) » (0, 1/2) » (1, •) and increases on (1/2, 1). 46. Greatest value of f is p/2 and least value is 0. 47. f decreases on (– •, 0) and increases on (0, •). 48– 50. P - Cost = xp - (600 + 10x + x2) = x (110 - 2x) - (600 + 10x + x2) = 100x - 600 - 3x2 = - 3[(x - 50/3)2 - 700/9] 2

700 50 ˆ Ê = - 3 Á x - ˜ . This represents Ë 3 3¯ a parabola with vertex at (50/3, 700/3) and focus (50/3, 2099/12). Now P is negative if x < 8 or greater 25. Thus it is enough to consider

So

P -

P (x) = 100x - 600 - 3x2, 8 £ x £ 25 2 P¢(x) = 100 - 6x = 0 fi x = 16 . Since f ¢(x) > 0 on 3 2ˆ 2ˆ Ê Ê ÁË 8, 16 ˜¯ , f increases on ÁË 8, 16 ˜¯ and P¢(x) < 0 on 3 3 Ê 2 ˆ Ê 2 ˆ ÁË16 , 25˜¯ , f decreases on ÁË16 , 25˜¯ . The largest 3 3 value of P corresponding to an integer value of x will therefore occur at x = 16 or x = 17. Direct calculation of P(16) and P(17) show the choice x = 17 is correct. 51–53. For Q.51, consider the function f (x) = x2/(x3 + 200) in the interval [1, •). Since the derivative x (400 - x 3 ) f ¢(x) = ( x 3 + 200)2 is positive when 0 < x < 3 400 and negative when x > 3 400 , the function increases when 0 < x < 3 400 and decreases when x > 3 400 . So f (x) has a maximum at x = 3 400 . Further, since 7 < 3 400 < 8, it follows that the largest term in the sequence is either

IIT JEE eBooks: www.crackjee.xyz Applications of Derivatives 24.51

a7 or a8. Since a7 = 49/543 > 8/89 = a8, the largest term in the given sequence is a7 = 49/543. Similarly for Q.52, consider f (x) = x/(x2 + 10). 54. Differentiating x2 + y2 = 4, we have dy/dx = - x/y. So y¢(1,

3 ) = - 1/ 3 . Thus the equation of tan-

gent at (1, 3 ) is Y - 3 = ( - 1/ 3 ) (X - 1). Its intersection with X-axis is (4, 0). The equation of normal at (1, 3 ) is Y - 3 = 3 (X - 1). Its intersection with X-axis is (0, 0). Hence A is 2 3 .

55. The points of extremum of f (x) = x2

e

x2

- 1 is same

x2

as that g(x) = e - 1. Now g¢(x) = 2x e so g ¢(x) = 0 22 ¤ x = 0 also g≤(x) = 2ex (1 + 2x2) and g ≤(0) = 2 > 0. Therefore g is minimum at x = 0. Hence min f (x) = 0. Ï1 - x + a x £ 1 56. f (x) = Ì Ó 2x + 3 x > 1 If f (x) has local minimum at x = 1, then f (1) £ f (1 + h), h > 0 i.e. a £ 2(1 + h) + 3. Hence a £ 5. Also 1 - (1 - h) + a ≥ f (1) = a for h > 0 which is trivially true. Hence, a = 5. 57. The coordinates of F1 and F2 are (- 4, 0) and (4, 0) respectively. Let P(x, y) be any point on the given ellipse. The area A of the triangle PF, F2 is given by x y 1 1 | - 4 0 1 | = 4y = 12 1 - x 2 /25 2 4 0 1 So A is maximum when x = 0, i.e max A = 12. 58. Since for 0 < x < 1, log x < 0 so f (x) = |x log x| = - x log x and f ¢(x) = - x ◊ (1/x) - log x = - (1 + log x). Therefore, f ¢(x) = 0 ¤ x = e-1. Clearly f (1/e) = - e < 0. Thus max f (x) = f (e-1) = e-1. log 9 2 log x 59. Let f (x) = logx (1/9) - log3 x2 = log x log 3 so

f ¢(x) =

=

2 log 3 x (log x )

2

-

2 x (log 3)

2 ÈÎ(log 3)2 - (log x )2 ˘˚ x (log x )2 log 3

f ¢(x) = 0 fi log x = log 3 or log x = - log 3 fi x = 3 or x = 1/3. Since x > 1 so we must have x = 3. If 1 < x < 3 then f ¢(x) > 0 and for x > 3, f ¢(x) < 0, hence max f (x) = f (3) = - log 9/log 3 - 2 = - 4. 60. f (x) = (1/2) (1 + cos 2x) + cos x + 3. Since f is periodic with period 2p, the greatest and the least value are the values on [0, 2p]

f ¢(x) = - sin 2x - sin x = - 2 sin (3x/2) cos (x/2). Hence x = 0, p, 2p/3, 4p/3 are the critical points. Now f (0) = 5, f (2p/3) = 11/4, f (p) = 3, f (4p/3) = 11/4, f (2p) = 5. Hence the greatest value is 5 and the least value is 11/4. 2 61. f ¢(x) = 2(x - 2) ex - 4x + 3. Thus x = 2 is a critical point. Since f (- 5) = e48, f (2) = e-1 and f (5) = e8 so the greatest value is e48. 62. The equation of the given curve can be written as x2 y2 =1 + 4 a2 Any point on this ellipse is of the form (a cos q, 2 sin q ). Hence the distance z from the point (0, –2) is given by z=

( a cos q - 0)2 + (2 sin q + 2)2

Since the critical points of z and z2 are the same, we only need to consider S = z2 = a2 cos2 q + 4 sin2 q + 8 sin q + 2 dS = – 2a2 cos q sin q + 8 sin q cos q + 8 cos q dq = [2(4 – a2) sin q + 8] cos q



Therefore, from d S/d q = 0, we get cos q = 0 or 4 sin q = 2 > 1 (because 4 < a2 < 8). a -4 Hence the only possibility is cos q = 0, i.e., q = p 2 . d2S = – 2 sin q [(4 – a2) sin q + 4] dq 2 + 2 cos q (4 – a2) cos q

Also

d2S dq 2



= – 2(4 – a2 + 4) = – 2(8 – a2) < 0 q =p 2

So the farthest point is ( a cos (p 2) , 2 sin (p 2)) = (0, 2). 63. The given limit will exist only if there is no constant term, one degree, two degree and three degree terms in f (x). Let f (x) = a x4 + b x5 + c x6. In this case, the given limit can be written as lim (1 + a x + b x2 + c x3)1/x xÆ0

(

= lim 1 + x ( a + bx + cx 2 ) xÆ0

= ea

so

)

1 x ( a + bx + cx 2 )

a=2

Thus f (x) = 2x4 + b x5 + c x6 fi

f ¢(x) = x3 (8 + 5b x + 6c x2)

( a + bx + cx 2 )

IIT JEE eBooks: www.crackjee.xyz 24.52 Comprehensive Mathematics—JEE Advanced

Since f (x) has local extremum at x = 0, 1, 2 . So f ¢(0) = f ¢(1) = f ¢(2) = 0 fi c = 2/3, b = – 12/5. 12 5 2 6 Hence f (x) = 2 x4 – x + x . 5 3 224 fi f (3) = 5 64. g¢(t) = 5t4 - 15t2 - 20. In the interval (- 2, 2), Q ¢(t) π 0. Now h¢(t) = 12t2 - 6t - 18 so h≤(t) = 0 fi t1 =

- 1 and t2 = - 3/2. Furthermore h≤(t) = 24t - 6, h≤ (-1) = - 30 < 0, h≤(3/2) = 30 > 0. Consequently the function y = f (x) has a maximum y = 14 at t = - 1. 65. f ¢(x) = x (1 + 2 log x) so f ¢(x) never vanishes inside [1, e]. Hence there is no critical point inside [1, e]. Since f (1) = 0, f (e) = e2. So greatest f (x) = e2 and least f (x) = 0.

IIT JEE eBooks: www.crackjee.xyz

25 A function F(x) is an antiderivative of a function f(x) if F¢(x) = f(x) for all x in the domain of f. The set of all antiderivatives of f f with respect to x, denoted by

Ú f ( x) dx

Ú f ( x) dx

= F(x) + C

The constant C constant.

1.

0

2.

xn (n π – 1)

4.

1 x ex

5.

ax (a > 0, a π 1)

6. 7. 8. 9. 10. 11. 12.

cos x sin x sec2 x cosec2 x sec x tan x cosec x cot x 1

Ú f(x) dx* C 1 n +1 x n +1 x

a -x

2

(a π 0)

1

19.

x + a2

20.

x2

21.

a 2 - x 2 (a π 0)

22.

x2 + a2

23.

x2 - a2

2

a2

ex ax log a sin x – cos x tan x – cot x sec x – cosec x

24.

25. 26.

1 x - a2 (a π 0) 1 ( a π 0) 2 a - x2 2

1 ( a π 0) a2 + x2

sin–1 x or – cos–1 x

14.

1 - x2 1 1 + x2 tan x

15.

cot x

x

16.

sec x

x x + tan x

13.

1 2

TABLE OF BASIC FORMULA

S.No. f(x)

3.

x – cot x tan

x is the

have found one antiderivative, the other antiderivative of f differ from F by a constant. We indicate this by

cosec x or

18.

The function f

25.1

17.

by parts –1

–1

tan x or – cot x x x

Ê x pˆ tan Á + ˜ Ë 2 4¯

u(x) v(x)

x 2

Ê xˆ sin–1 ÁË ˜¯ a x + x2 + a2 x + x2 - a2 x 2 1 a - x2 + a2 2 2 x Ê ˆ sin -1 Á ˜ Ë a¯ x 2 x + a2 2 1 + a 2 log x + x 2 + a 2 2 x x2 - a2 2 1 - a 2 log x + x 2 - a 2 2 1 x-a log x+a 2a 1 a+x log a-x 2a 1 Ê xˆ tan -1 Á ˜ Ë a¯ a v(x) Ú u(x)dx È dv ˘ Ú ÍÎÚ dx Ú u ( x) dx˙˚ dx

TIP

Ú (integrand) dx = Note: Use a2 or – a2

x a2 dx (integrand) ± Ú 2 2 integrand

IIT JEE eBooks: www.crackjee.xyz 25.2 Comprehensive Mathematics—JEE Advanced 25.2

A LIST OF EVALUATION TECHNIQUES

S.No.

Form of Integrate

Value/Evaluation Technique

1.

Ú f ¢(x + a) dx

f(x + a) + C

2.

Ú f ¢(ax + b)dx, a π 0

1 f(ax + b) + C a

3.

Ú

4.

Ú (f(x))n f ¢(x) dx, n π – 1

1 (f(x))n+1 + C n +1

5.

Ú f ¢(g(x)) g¢(x) dx

f(g(x)) + C

6.

Ú cos mx cos nx dx

Use product to sum identities

f ¢( x ) dx f ( x)

f (x

C

cos A cos B =

7.

1 [cos (A – B) + cos (A + B)] 2

Ú sin mx sin nx dx

sin A sin B =

1 [cos (A – B) – cos (A + B)] 2

Ú sin mx cos nx dx

sin A cos B =

1 [sin (A – B) + sin (A + B)] 2

Ú sinm x cosn x dx

If m is odd put cos x = t

where m, n ΠN

If n is odd put sin x = t If both m and n are odd, put sin x = t if m ≥ n and cos x = t otherwise. If both m and n sin2 x =

1 (1 – cos 2x) 2

cos2 x =

1 (1 + cos 2x) 2

If m + n put tan x = t 8.

a sin x + b cos x + c

Ú p sin x + q cos x + r dx ae x + be- x

Ú pe x + qe- x

dx

(p, q π 0 and at least one of a, b π 0)

Write Num = a (DEN) + b

d (DEN) + g dx

Find a, b sin x and constant term (or ex, e– x d ( DEN ) dx + g a Údx + b Ú dx DEN

dx

Ú DEN

x,

IIT JEE eBooks: www.crackjee.xyz 25.3

S.No.

Form of Integrate

9.

Ú p tan x + q

a tan x + b

ae2 x + b

Ú pe2 x + q

Value/Evaluation Technique

a sin x + b cos x

Ú p sin x + q cos x

dx

or

dx

(p, q π 0, at least one of a, b π 0) 10.

Ú

dx

or

ÚQ

or

Ú

or

Q

dx

or

Q or

Ú be x + qe- x

dx

and use (8) above Use a = r cos a, b = r sin a

dx

Ú a sin x ± b cos x

1 dx . Now, use formula for Ú r sin ( x ± a ) Ú cosec x dx

a, b π 0)

11.

ae x + be- x

dx

L

Ú

Q L

ÚQ

ÚL

dx

b Write Q = a ÊÁ x 2 + x + Ë a

dx

and put t = x +

cˆ ˜ a¯

b 2a

Q

where L = px + q, p π 0 and Q = ax2 + bx + c, a π 0 12.

13.

L1 dx L2

Ú

Write as

L1

where L1 = ax + b, a π 0

Ú

L2 = px + q, p π 0

to reduce to form

dx

ÚL

Q

L 1L 2

Put L =

dx

Ú

L Q

dx

1 to reduce to form t

Ú

dx Q

where L = px + q, p π 0 and 14.

Ú

Q = ax2 + bx + c, a π 0 Q

Q dx L

L Q

where L = px + q, p π 0 Q = ax2 + bx + c,

Divide Q by L to obtain

aπ0

where L1 is a linear expression in x and a is a constant.

Q = L1 (L) + a

Ú

L1 Q

dx + a Ú

dx L Q

.

IIT JEE eBooks: www.crackjee.xyz 25.4 Comprehensive Mathematics—JEE Advanced S.No.

Form of Integrate

15.

Ú

Q1

Value/Evaluation Technique

Write Q1 ∫ a Q2 + b

dx

Q2

d [Q2] + g dx

Find a, b, g

where Q1 = ax2 + bx + c, a π 0

a

Ú

Q2 dx + b

Ú

d dx [Q ] dx + g 2 Q2

Ú

dx Q2

Q2 = px2 + qx + r, p π 0 16.

dx

ÚL

1

Put L2 = t2

L2

where L1 = ax + b and L2 = px + q, with a, p π 0, 17.

a π p.

dx

ÚQ

1

Q2

where Q1 = ax2 + b, Q2 = px2 + q with

a, p π 0, p π 0

First put x = 1/t and simplify then put p + qt2 = u2

18.

ÚR(x x1/q, x1/r ...) dx where R is a rational function.

Let l = lcm (p, q, r, ...) and put x = t l

19.

Ú ex (f(x) + f¢(x)) dx

Write as Ú ex f (x) dx + Úex f ¢(x) dx

1/p,

ex as the second function to obtain e f(x) - Úe f ¢(x) dx + Úe f ¢(x) dx x

20.

x

eax as the second function to obtain

Ú eax cos (bx + c) dx or Ú eax sin (bx + c) dx

x

1 ax b e cos (bx + c) + Ú eax cos (bx + c) dx a a eax as the second function and Ú e cos (bx + c) dx to the left. ax

21.

P( x )

P(x) ≥

Ú Q( x) dx where P(x) and Q(x) are polynomials in x

Q(x

A(x) when

P(x) is divided by Q(x). Now,

P( x ) B( x ) = A(x) + Q( x ) Q( x ) B(x

22.

P( x )

Ú Q( x) dx

Q(x)

Write

An A1 A2 P(x Q(x) and Q(x) is a P ( x ) = +...+ + product of distinct linear factors, that is, x - an x - a1 x - a 2 Q( x )

IIT JEE eBooks: www.crackjee.xyz 25.5

S.No.

Form of Integrate

Value/Evaluation Technique

Q(x) = A(x – a1) (x – a2) . . . (x – ar)

Evaluate Ai¢ s and use

dx

Úx-a

where ai¢ s are distinct and A π 0 23.

24.

P( x)

Ú ( x - a)n dx

Put x – a = t,

where n ≥ 1 and P(x) is a polynomial in x.

express P(x) in terms of ts

P( x) dx

Ú ( x - a ) m ( x - b) n Am A1 A2 + + ... 2 x - a ( x - a) ( x - a)m

where m, n ≥ 1, a π b,

+ P(x) < m + n.

25.

Evaluate Ai¢ s and B ¢j s

P( x) dx

Ú ( x - a) ( x 2 + bx + c) where b2 – 4c

26.

Bn B1 B2 + + ... + 2 x - b ( x - b) ( x - b) n

P(x) < 3

x2 + 1

Ú x4 + kx2 + 1 dx

A Bx + C + 2 x - a x + bx + c Divide the numerator by x2 to obtain

(1 + 1 / x 2 ) dx

Ú x2 + 1/ x2 + k Now, put x – 1/x = t 27.

x2 - 1 Ú x4 + kx2 + 1 dx

Divide the numerator by x2 to obtain

(1 - 1 / x 2 ) dx Ú x2 + 1 / x2 + k Now, put x + 1/x = t 28.

xP( x 2 ) Ú Q( x2) dx

Put x2 = t

where P and Q are polynomials in x 29.

P( x 2 ) Ú Q( x2 ) dx

Put x2 = t

where P and Q are polynomials in x

for partial fractions not for integration

x–a

IIT JEE eBooks: www.crackjee.xyz 25.6 Comprehensive Mathematics—JEE Advanced

S.No.

Form of Integrate

Value/Evaluation Technique

30.

Ú ( x2 + a 2)n dx = An

1

An–1 An–1 = Ú 1.

where n > 1

31.

32.

ÚR (sin x, cos x) dx where R is a rational function Special Cases (a) If R (– sin x, cos x) = – R (sin x, cos x) (b) If R (sin x, – cos x) = – R (sin x, cos x) (c) If R (– sin x, – cos x) = R (sin x, cos x)

1 dx ( x + a 2 )n - 1 2

Universal substitution tan (x/2) = t. Put cos x = t Put sin x = t Put tan x = t

Úxm (a + bxn) p dx where m, n, p are rational numbers. Expand (a + bxn) p Put x = tk

(a) p (b) p m=

a c ,n= , b d

where k = l cm (b, d)

a, b, c, d ΠI, b > 0, d > 0

33.

(c)

m +1 n

Put a + bxn = ts where p = r/s, r, s ΠI, s > 0.

(d)

m +1 +p n

Put a + bxn = xnts where p = r/s, r, s ΠI, s > 0.

Ú

Pn ( x) ax 2 + bx + c

where Pn(x

Write

dx n in x.

Ú Ú

Pn ( x) 2

ax + bx + c

dx = Qn–1 (x) ax 2 + bx + c + k

dx ax 2 + bx + c

where Qn–1 (x

n – 1) in x and k is a

constant. Differentiate both the sides w.r.t. x by

ax 2 + bx + c to 1 Qn–1 (x) (2ax + b) + k 2 Qn–1 (x) and k.

Pn(x) = Qn¢ - 1 (x) (ax2 + bx + c) +

IIT JEE eBooks: www.crackjee.xyz 25.7 25.3

SOME TIPS FOR SIMPLIFYING COMPUTATIONS

1

1. In case of

Q

, Q , or

the substitution x +

25.4

SOME TIPS FOR PARTIAL FRACTIONS

1. Partial fractions

1 , where Q = ax2 + bx + c, Q

f(x) =

P(x) < n and a1, a2 ... an are distinct) is

b = t may be used instead of 2a Q.

If Q = (x – a) (x – b) with a b, then x – 1 (a + b) 2

= t can be used. L L 2. In case of , L Q and Q Q

An A1 A2 + + ... + x - a1 x - a2 x - an where Ai =

2

where L = px + q and Q = ax + bx + c, p 0, q 0, the following steps in unnecessary Express d [Q] + b L=a dx

P( x) ( x - a1) . . . ( x - an )

P(ai ) (ai - a1) . . . (ai - ai - 1) (ai - ai + 1) . . . (ai - an )

= value of the expression (x – ai) f(x) at x = ai. In other words, Ai x = ai everywhere in f(x) except in the factor x – ai. For instance, 2x - 1 2(1) - 1 1 ∫ ( x - 1) ( x + 1) ( x - 2) (1 + 1) (1 - 2) x - 1

b =t 2a 3. To evaluate

+

x+

Ú R ( x,

+

)

2

ax + bx + c dx ∫ -

where R is a rational function of x and ax 2 + bx + c ,

(i)

ax 2 + bx + c = t ± x a if a > 0

(ii)

ax 2 + bx + c = tx ±

(iii)

if c > 0

2

ax + bx + c = t(x – a)

where a is a real root of ax2 + bx + c = 0 4. To evaluate dx I= Ú a + b sin x dx

Ú a - b cos (p / 2 + x)

and put p/2 + x = q. This will lead to simpler computation then direct evaluation.

2(2) - 1 1 (2 - 1) (2 + 1) x - 2

1 1 1 1 1 + 2 x -1 2 x +1 x - 2

x = 1, 2(– 1) – 1 or – 3 at x = – 1 and 2(2) – 1 or 3 at x = 2. Indeed, it is the polynomial* 1 ( x + 1) ( x - 2) (- 3) ( x - 1) ( x - 2) + (1 + 1) (1 - 2) (-1 - 1) (-1 - 2) +

3( x - 1) ( x - 2) (2 - 1) (2 + 1)

2. If P(x) = b0 + b1x + ... + bnxn, the partial fraction of P( x) ( x - a1) . . . ( x - an )

we rewrite I=

2(-1) - 1 1 (-1 - 1) (-1 - 2) x + 1

where, a1, a2, ... an bn +

An A1 A2 + + ... + x - a1 x - a2 x - an

where Ai¢s b, c are distinct

a,

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x3 a3 1 ∫1+ ( x - a ) ( x - b) ( x - c ) ( a - b) ( a - c ) x - a

=

y3

(1 – 2y + 3y2 – ...)

+

b 1 (b - a) (b - c) x - b

Next, put x + 2 = z 1 1 1 = 2 = - 2 (1 - z ) -3 3 2 3 ( x + 1) ( x + 2) z ( z - 1) z

+

c3 1 (c - a ) (c - b ) x - c

=–

3

1

z2 Thus,

3. To resolve

(1 + 3z + ...) 1

1

3

( x + 1) ( x + 2)

( x - a ) r ( x - b)

=

3

=

3 2 1 + x + 1 ( x + 1)2 ( x + 1)3 –

where a b, put x – a = y, so that 1 1 = r r y ( y + a - b) ( x - a ) ( x - b) y ˘ È 1+ ˙ r Í ( a - b) y Î a - b ˚ 1

( x - a ) ( x - b) ( x 2 + c 2 ) where a b.

-1

Ê y ˆ Now expand Á1 + upto yr–1 and divide by a - b ˜¯ Ë yr to obtain Cr + 1 C0 C + r 1+ 1 + . . . + . r y y y 1

The value Ck

( x - a)r - k

( x - 1) ( x - 2)

=

1 3

y ( y - 1) 1 y3

=

1 y3

be obtained as explained in (1). For instance, we can write 1 2

x( x + 1) ( x + 2)

(1 - y ) -1

(1 + y + y 2 + ....)

SINGLE CORRECT ANSWER TYPE QUESTIONS

1

1 1 = 3 ( x - 1) ( x - 2) x - 2 x - 1 -

1 ( x - 1)

2

Illustration Resolve 1 3

( x + 1) ( x + 2)2 into partial fraction. First, put x + 1 = y, ( x + 1)3 ( x + 2)2

=

1 1 Ax + B + 2 x 3( x + 1) x 2 + 2

SOLVED EXAMPLES

Thus,

1



+ (Ax + B) x (x + 1) Now, A and B x.

For instance, if x – 1 = y, then =

1 1 and can x-b x-a

To obtain A and B, we rewrite above expression as 1 1 2 2 1 ∫ (x + 1) (x + 2) – x (x + 2) 3 2

.

1 1 is x-b (b - a)r 1

3 1 x + 2 ( x + 2)2

4. To obtain partial fractions of 1

-1

3

1

1 y 3 ( y + 1)2

-

1 ( x - 1)

Example 1 3

Let I =

Ú

x 4 + 4 x3 + 12 x 2 + 9 x dx ( x + 3)5 - x5 - 243

(a)

1 [ x - 2 ln( x + 3)] + c 15

(b)

1 [ x + 2 ln ( x + 3)] + c 3

1 [ x + 2 ln ( x + 3)] + c 15 1 (d) [2 x - ln ( x + 3)] + c 3 (c)

IIT JEE eBooks: www.crackjee.xyz 25.9

Ans. (a) Solution:

I=Ú

3

(d) - cot x +

2

x ( x + 4 x + 12 x + 9) 15 ( x 4 + 6 x3 + 18 x 2 + 27 x)

1 ( x + 1) ( x 2 + 3x + 9) dx = Ú 15 ( x + 3) ( x 2 + 3x + 9) 1 È 2 ˘ dx 1Ú Í 15 Î x + 3 ˙˚ 1 È x - 2 ln ( x + 3)˚˘ + c = 15 Î Let I = Ú

(a)

x 2 sec2 x

( x tan x + 1)2



then I

-x + tan x + c cos x ( x sin x + cos x )

- x sec x + tan x sec x + c x sin x + cos x x sec x (c) - tan x + c x sin x + cos x -x (d) - tan x + c x sin x + cos x

=Ú =Ú

( x tan x + 1)2 x

(1 - cos x) (1 + cos x) 2

( x sin x + cos x )2

dx

x x cos x dx cos x ( x sin x + cos x )2

Example 3

-x + tan x + c cos x ( x sin x + cos x ) Let I = Ú

(a) cot x -

dx , then sin x + tan 2 x 2

1

Ê tan x ˆ +C tan -1 Á Ë 2 ˜¯ 2

1 1 Ê tan x ˆ (b) - cot x +C tan -1 Á Ë 2 ˜¯ 2 2 1 1 Ê tan x ˆ (c) - cot x +C tan -1 Á Ë 2 ˜¯ 2 2 2

dx

dx

1 È 1 1 ˘ dx Ú 2 Í 2 Î1 - cos x 1 + cos 2 x ˚˙

=

1 sec2 x 2 cosec x dx Ú 2 + tan 2 x dx 2Ú

Let I = Ú

dx x + 2 x2 - 1

(

)

(

)

(a) ln x + x 2 - 1 -

1 x + 2 x2 - 1 +c ln 2 x - 2 x2 - 1 1

(d)

ln

2 2

)

(

, then I

x2 - 2 x - 2 x2 - 1

2 ln x + x 2 - 1 - ln x + 2 x 2 - 1 + c

(c)

2

)

=

Example 4

dx

x Ê -1 cos x + x sin x ˆ sec2 x dx ÁË ˜¯ + Ú cos x x sin x + cos x x sin x + cos x =

2

(b) ln x + x 2 - 1 -

d As (x sin x + cos x) = x cos x dx I=

(

sin 2 x 1 + cos 2 x

Ê tan x ˆ 1 1 = - cot x +c tan -1 Á Ë 2 ˜¯ 2 2 2

Ans. (a) I=Ú

cos 2 x cos 2 x

(b)

x 2 sec2 x

Ê1 ˆ tan -1 Á tan x˜ + C Ë ¯ 2 2

Ans. (c) Solution: Write I = Ú

=

Example 2

1

(

2

)

(2 - x ) ( x + 2

2 ln x + x - 1 + ln

2 x2 - 1

x - 2 x2 - 1

) +c

Ans. (c) Solution:

I=Ú =Ú

(

x - 2 x2 - 1

)(

x + 2 x2 - 1 x - 2 x2 - 1

x - 2 x2 - 1

(

)

x2 - 2 x2 - 1

)

dx

dx x2 - 1



x dx - 2 2 - x2

=-

1 2 - x2 - 1 dx ln 2 - x 2 + 2 Ú 2 2 - x2 x2 - 1

=-

1 ln 2 - x 2 + 2 ln x + x 2 - 1 2

Ú

(2 - x ) 2

(

)

(

- 2 I1

x2 - 1

)

dx

IIT JEE eBooks: www.crackjee.xyz 25.10 Comprehensive Mathematics—JEE Advanced

Solution

Where I1 = Ú Put x =

= -Ú

\

(2 - x ) 2

1 , so that t I1 = Ú

Put

1 2

x -1

2

1 ˆ Ê 1ˆ Ê ÁË 2 - 2 ˜¯ ÁË ˜¯ - 1 t t

( 2t

Put

=

=

=

2 2 1 2 2 1 2 2 1 2 2

)

( 1+ x

= (u + 1)

)

=

1 + 2u 1 - 2u x + 2 x -1 x - 2 x2 - 1

(x +

ln

2

2

2 x -1

(

)

2

2 x2 - 1 2 - x2

)

2

)

(

I = 2 ln x + x 2 - 1 - ln x + 2 x 2 - 1

(

If I = Ú x10 + 1 + x 20

)

21 10

1 20

19 2 (a) u (1 + u ) - (1 + u ) 20 + C 9

(b) 2u (1 + u )

19 20

4 -1 - u 20 + C 9

19 4 (c) u (1 + u ) - (1 + u ) 20 + C 9 1 19 40 (d) 2u (1 + u ) 20 (1 + u ) 20 + C 9 Ans. (d) -

1 20

(u + 1)

-1

1 20

1 20

-

21 20

dx

2udu

( 2u ) - 2Ú (u + 1) -

-

1 20

du

19 2 ¥ 20 (u + 1) 20 + c 19

sin x

1 - 2 sin x

1 + sin x

+

+C

1 1 (d) A = , B = 8 4 2 Ans. (c)

dx and

Solution:

I=

u = x -20 + 1 then I

-

)

21 10

-

dx

1 1 (a) A = , B = 8 4 2 1 1 (b) A = - , B = 8 4 2 1 1 (c) A = - , B = 8 4 2

x - 2 x -1

(x + ln

-

2u

1 + 2 sin x

B

)

2

)

Ú sin 4 x dx = A log 1 - sin x

Example 6

2

ln

-20

Ê 1ˆ I = Ú Á - ˜ (u + 1) Ë 20 ¯

1- t2

-1

ln

Thus,

Example 5

=Úx

-21

21 10

-

- 1 ˘˙ ˚

-20 x -21 dx = 2udu

t dt

(

1

( 1+ x

-20

1 + x -20 = u fi x -20 = u 2 - 1 , so thus

1 - t 2 = u, so that – tdt = udu 4du du =Ú I1 = Ú 2 1 - 2u 2 4 ÈÎ 2 1 - u - 1˘˚ =

I = Ú ÈÍ x10 Î

Ê 1ˆ ÁË - 2 ˜¯ dt t

2

)

(

divide by - x10 + 1 + x 20 ,we have

dx

= =

sin x

1

cos x

Ú sin 4 x dx = 4 Ú cos2 x cos 2 x dx

cos x 1 dx 4 Ú (1 - sin 2 x)(1 - 2 sin 2 x) dt

Ú (1 - t 2 )(1 - 2t 2 )

(where t = sin x)

=

1 È 2 1 ˘ Í ˙ dt Ú 4 ÎÍ1 - 2t 2 1 - t 2 ˚˙

=

1 + 2t 1 1+ t 1 1 ◊ log - log +C 2 2 2 1- t 1 - 2t 8

IIT JEE eBooks: www.crackjee.xyz 25.11

1

=

1 + 2 sin x

log

4 2

1 - 2 sin x

-

1 + sin x 1 log +C 8 1 - sin x

1 1 A = - ,B = 8 4 2



Ú(

If I =

Example 7

(

)

tan x - cot x + C

2

(b)

2

x + cos x +

(c)

2

x – cos x +

(d)

2

x + p/4) +

Ans. (b) Solution:

I=

Ú

cos x - sin x cos x sin x

sin 2x

C

2 sin x cos x 2 sin x cos x

2

I=

t2 - 1

2

=

dx

Ú

1 2

[sin

–1

(b)

2 [sin

2

=

(c)

2 [sin

I1 =

and

I2 =

Ú( Ú(

(d)

1 log x3/2 + 1 - x3 + C 3

x3

sin 2x

(sin x + cos x

(cos x – sin x

Ú x3

1- x

and 1 – x3 = t2, so that

3

( -2/3) t dt = - 2

Ú (1 - t 2 ) t

I=

==

x sin 2x

x 2 dx

– 3x dx = 2t dt and

C

(sin x – cos x

x sin 2x ] + C

1 - x3 + 1

1 log 3

1 - x3 - 1

x] + C 2x + C

) tan x ) dx

and use Example 7 and sin x + cos x I1 = Ú dx . Put sin x – cos x = t, so sin x cos x dt that I1 = 2 Ú = 2 sin -1 (sin x - cos x ) + C 2 1- t

(

)

(

)

2 1 + x3 9 2 (b) 1 + x3 9

5/2

3/ 2

1 + x3

dx, then I

(

)

(

)

3/2 2 1 + x3 +C 3 1/ 2 2 - 1 + x3 +C 3

+

x + 1 + x3 + C

(c) log (d) x2 Ans. (b) Solution:

+C

x5

Ú

If I =

Example 10 (a)

dt

3 Ú 1 - t2

2 Ê 1ˆ 1- t log +C Á ˜ 3 Ë 2¯ 1+ t

C

tan x + cot x dx cot x -

C

2

tan x dx , then I

(d) sin–1 (cos x – sin x Ans. (a) 1 Solution: Write I = ( I1 - I 2 ) 2 where

2 3

t + t -1 + C

– cos x + –1

+C

1 - x3 - 1

(c)

2

+ cos x + –1

+C

1 - x3 + 1

Solution: Write I =

x + cos x +

If I =

Example 8 (a)

dt

Ú

1 - x3 + 1

, then I

Ans. (b)

Put sin x + cos x = t, so that 2 sin x cos x = t2 – 1 \

1 (b) log 3

C C

1 - x3

1 - x3 - 1

cot x - tan x dx , then I

(a)

dx

1 log 3

(a)

)

Úx

If I =

Example 9

x 3) + C

Put 1 + x3 = t2, 3x2 dx = 2t dt, so that I=

=

Ú

(

(t

2

) ◊ 2 t dt = 2 Ê t

-1 t

2 1 + x3 9

ˆ - t˜ + C Á 3Ë 3 ¯

3

)

3/2

-

3

(

2 1 + x3 3

)

1/2

+C

IIT JEE eBooks: www.crackjee.xyz 25.12 Comprehensive Mathematics—JEE Advanced

(a) (b) (c) (d)

( x - 2) ( x + 2) (2 - x ) (2 + x )

1- x

Ú

If I =

Example 11

1+ x

dx , then I

=

Ú

1 - x + sin -1 x + C 1 - x + cos

-1

x+C

1 - x - sin -1 x + C

I=

Ú

=–

Ú

=2 =2

Ú

1- t t

Ú

2

1 - x2 1 + x x

dt

dt - 2

1 - t2

Ú

1 - t2

+2

Ú

2

1 - t dt

Example 13 value of I is

(

)

x-2

Ê 2x + b - a ˆ (c) sin–1 Á Ë a + b ˜¯

1 - x - sin -1 x + C

)

Ê x -a - bˆ +C (d) sin -1 Á Ë b - a ˜¯ Ans. (a)

1 - x + cos x + C '

where C¢ = p/2 + C Example 12

I=

1 + 1 - x2

+ sin–1

x

1 - 1 - x2

+ tan–1 x + C

x

1 + 1 - x2

– sin–1 x + C

1 1- x

Úx

1- x

2

1 (x – a + x – b) 2

1 (a + b), so that 2 È 1 ˘ (x – a) (b – x) = Ít + (a + b ) - a ˙ Î 2 ˚

x +C

1 (b – a)2 – t2 4 Ê 2t ˆ I = sin–1 Á +C Ë b - a ˜¯

= x

dx

Put t = =x–

x

Ans. (c) Solution:

Solution:

1 1- x dx, then I If I = Ú x 1+ x

1 + 1 - x2

dx

Ú ( x – a ) ( b - x) , (b < a) then

Ê x +a + bˆ (b) sin–1 Á Ë b - a ˜¯

= (t - 2) 1 - t 2 - sin -1 t + C x-2

– sin–1 x + C

Ê 2x - a - b ˆ +C (a) sin–1 Á Ë b - a ˜¯

1 Èt ˘ + 2 Í 1 - t 2 + sin -1 t ˙ + C 2 2 Î ˚

(

If I =

– sin–1 x + C

1 + 1 - x2

x dt

1 , so that t

–1 1 1 – sin x 1- 2 t t dt – sin–1 x 2 t -1

Ú

t - 1 + 1 - t2

1 - x2

t + t 2 - 1 – sin–1 x + C

= -2 1 - t 2 - 2 sin -1 t

=

(-1 / t 2 ) dt

t (1 - t ) 1- t dt ( 2t ) dt = 2 1+ t 1 - t2

Ú

I=

dx

-Ú x=

x = t or x = t2, so that dx = 2t dt and

Put

=

x 1 - x2

1 - x + cos -1 x + C

Ans. (a) Solution:

dx

–1

x+C

\

Ê 2x - a - b ˆ +C = sin–1 Á Ë b - a ˜¯

1 È ˘ Íb - 2 (a + b ) - t ˙ Î ˚

IIT JEE eBooks: www.crackjee.xyz 25.13

If I =

Example 14

( ( ( (

(a) x (b) x (c) x (d) x Ans. (c)

Ú log (

)

1 - x + 1 + x dx , then I is

) 12 x + C 1 1 + x ) + sin x + C 2 1 1 1 + x ) + sin x - x + C 2 2 1 1 1 + x ) + sin x + x + C 2 2

1- x + 1+ x + 1- x + 1- x + 1- x +

(

Solution: I = x

1- x + 1+ x

Ú

(

1+ x - 1- x

x

1+ x + 1- x

1 - x2

1+ x - 1- x

)

Ú (1 + x) - (1 - x)

2

x 1 - x2

\

dx

dx

(a)

Ú

x–a

(

) 12 (sin

1- x + 1+ x +

If I =

Ú

If I =

I

1 +C log sin ( a - b ) cos ( x - b )

1 + sin x dx , then the domain of

(– •, •) (0, •) (– •, 0) (– p/2, p/2)

Ú

1 + sin x 1 - sin x 1 - sin x cos x

Ú

1 - sin x

dx

dx = -2 1 - sin x + C

Since sin x £ 1, so domain of I is R.

-1

)

x- x +C

If I =

Example 17

(b) (c) -

sin ( x - a ) 1 (c) log +C sin ( a + b ) cos ( x - b )

(d) -

Ú

x+a

(a) -

sin ( x - a ) 1 (b) log +C cos ( a - b ) cos ( x - b )

sin ( x - a ) 1 (d) log +C cos ( a + b ) cos ( x - b )

Ú

C

+C

cos ( x - b )

Example 16

=

dx , then I is sin ( x - a ) cos ( x - b ) sin ( x - a )

x–b

sin ( x - a )

= log

(a) (b) (c) (d) Ans. (a)

dx

[cot (x – a) + tan (x – b)] dx

Solution: I =

Ê ˆ 1 Á - 1˜ dx = sin–1 x – x ÁË 1 - x 2 ˜¯

I=x

Example 15

=

2

Ú

Ú

sin ( x - a ) cos ( x - b )

)

)

dx

cos ( x - b ) cos ( x - a ) + sin ( x - b ) sin ( x - a )

-1

1 1+ x +1- x - 2 1- x dx 2 1 - x2 =

Ú

-1

(

I1 =

cos {( x – b ) – ( x – a )} Ú sin ( x – a ) cos ( x – b)

cos (a – b) I =

x

Ú

=

Solution cos (a – b), so that

-1

1 Ô¸ ÔÏ -1 + Ì ˝ dx 1 - x + 1 + x ÓÔ 2 1 - x 2 1 + x Ô˛ 1 =x 1 - x + 1 + x + I1 2

where

Ans. (b)

2ax + x 2

2

2

1 3

3/2

, then I

+C

2ax + x 2

1

a

(2ax + x )

x+a

1 a

a

dx

x+a 2ax + x 2 x+a 2ax + x3

+C

+C

+C

Ans. (c) Solution: Write 2ax + x2 = (x + a)2 – a2, and put x + a = a sec q, so that dx = a sec q tan q dq

IIT JEE eBooks: www.crackjee.xyz 25.14 Comprehensive Mathematics—JEE Advanced

\

Ú

a sec q tan q

1

Ú sin 2 q dq = - a2 sin q + C

I= =

a

=-

3

3

a tan q

dq

cos q

2

If I =

Ú

(

)

2 1 x +1 (a) 3 x3

x2 + 1 x4

x+a

+C

2ax + x 2

x2 + 1 x2

dx , then I

+C

)

(

)

If I =

Ú

1Ê 1 (b) sin x - cos x log cosec x + cot 2 ÁË 2 (c)

1 Ê 1 ˆ ÁË sin x + cos x + 2 log cosec x - cos x ˜¯ + C 2

(d) none of these Ans. (d) sin x cos x dx Solution: I = sin x + cos x

Ú

, then I

Ê x pˆ Ê x pˆ 2 log cosec Á - ˜ + cot Á - ˜ + C Ë 2 4¯ Ë 2 4¯

(d)

Ê x pˆ 2 log tan Á - ˜ + C Ë 4 8¯ 1 + sin x = 1 – cos (p/2 + x) = 2 sin2 (p/4 + x/2) I=

+C

ˆ x˜ +C ¯

1 + sin x

(c)

Ans. (b) Solution:

Example 21 =A

dx

Ê x pˆ Ê x pˆ 2 log cosec Á + ˜ - cot Á + ˜ + C Ë 2 4¯ Ë 2 4¯

=

ˆ 1Ê 1 (a) cos x + sin x log ( cosec x - cos x )˜ + C 2 ÁË ¯ 2

Ú

(b)

3/2

dx , then I sec x + cosec x

˘ ˙ dx 2 sin ( x + p /4) ˙˚ 1

Ê x pˆ Ê x pˆ 2 log cosec Á + ˜ + cot Á + ˜ + C Ë 2 4¯ Ë 2 4¯

\

1 sec3 q 1 x +1 + C = 3 tan 3 q 3 x3

dx

(a)

3/2

2

È Ísin x + cos x ÍÎ

If I =

Example 20

1 1 +C = 3 sin 3 q

Example 19

Ú

3/2

2 1 x +1 (d) +C 3 x2 Ans. (a) Solution: Put x = tan q so that x 2 + 1 = sec q, dx = sec2 q dq sec q sec2 q cos q dq = Ú 4 dq \ I= Ú 4 tan q sin q

= -

1 2

sin x + cos x

1 [sin x - cos x ] 2 1 log cosec ( x + p /4) - cot ( x + p /4) + C 2

+C

(

=

Ú

(sin x + cos x )2 - 1

=

(b) x3 (x2 + 1)–1/2 + C (c)

1 2

1

1 sec q 1 +C =- 2 2 tan q a a

Example 18

=

1

Êx



cosec Á + ˜ dx Ë 2 4¯ 2Ú

Ê x pˆ Ê x pˆ 2 log cosec Á + ˜ - cot Á + ˜ + C Ë 2 4¯ Ë 2 4¯ If I =

x +1 5 + 2 x + x2

Ú

dx

(

15 + 2 x + x 2

)

,

+ C then A

(a)

1 2

(b) 2

(c)

1 4

(d)

Ans. (c) Solution:

3

4 3

5 + 2x + x2 = 4 + (x + 1)2. Put x + 1 = t dt I=Ú 3 . Put t = 2 tan z, so that 4 + t2 4 + t2 = 2 sec z and dt = 2 sec2z dz,

(

)

IIT JEE eBooks: www.crackjee.xyz 25.15

2 sec2 z dz 1 = Ú cos z dz 4 23 sec3 z

(a)

1 È1 2 1 a + x2 - 3 4 Íx a Î 3x

=

1 1 tan z sin z + c = +C 4 4 1 + tan 2 z

(b)

1 È1 2 1 a + x2 - 3 a2 + x2 4 Íx a Î 3x

=

1 t /2 +C 4 1 + t 2 /4

(c)

1 È1 2 1 a + x2 a2 + x2 4 Íx a Î 2 x

=

1 x +1 +C 4 5 + 2 x + x2

(d)

2 a4

I=Ú

If I =

Example 22 (a) (b) (c) (d) Ans. (a) Solution: I=

Ú

Ú

cos x dx , then I sin ( x - a )

(

3/2 ˘

(

˙+C ˚

3/2 ˘

)

˙+C ˚

È 1 a2 + x2 - 1 a2 + x2 ˘ + C ÍÎ x ˙˚ x3 2

Ê aˆ a 2 + x 2 = x5 Á ˜ + 1 , so that Ë x¯

Solution: Write x4

I=

dx / x3

Ú x2

( a / x )2 + 1

Put a2/x2 = t, so that (– 2a2/x3) dx = dt Put x – a = q, so that

sin q

dq =

= (cos a)

Ú

Example 23 then I

I =-

cos a cos q - sin a sin q dq sin q

q x–a

If I =

Ú

a)q + C a)x + C ¢

[1 + 2 tan x (tan x + sec x)]1/2dx

x + tan x tan ( x / 2 + p / 4) cos x

x

C

Example 24 If I =

Ú x4

dx a2 + x2

1 2a

4

Ú

t +1-1

If I =

t +1

dt

3/2 ˘

)

˙+C ˙ ˚

dx

Ú sin 4 x + cos4 x , then I

Ê tan 2 x ˆ +C tan -1 Á Ë 2 ˜¯ 2

1

(a)

Ê 1 + cos 2 x ˆ tan -1 Á ˜ +C Ë 2 2 ¯

1

Ê tan x + cot x ˆ tan -1 Á ˜¯ + C Ë 2 2

1

Ê tan x + cot x ˆ 2 tan -1 Á ˜ +C 2 Ë ¯

(d) Ans. (a)

Ú (sec x + tan x) dx Ê x pˆ tan Á + ˜ Ë 2 4¯

t +1

dt = -

(

Example 25

Ans. (b)

I=

t

2 2 1 ÈÍ a + x 1 - 3 a2 + x2 4 Í x a 3x Î

(c)

1 + 2 tan x (tan x + sec x) = sec2 x – tan2 x + 2 tan2 x + 2 tan x sec x = (sec x + tan x)2

Ú

=

(b) C C

2a

4

1 È 1 3/2 ˘ t + 1 - (t + 1) ˙ + C 4 Í 3 a Î ˚

+C

x/2 + p/4) cos x p/4 – x/2) cos x

1

=

Ú cot q dq – (sin a) Ú dq

= (cos a = (cos a where C¢ = C + a sin a

\

)

Ans. (b)

cos a x–a x sin a + C cos a x–a x sin a + C – cos a x–a x sin a + C x sin a – cos a x–a C

cos ( a + q )

Solution:

˘ a2 + x2 ˙ + C ˚

Solution: x

C

1 sin4 x + cos4 x = [(1 – cos 2x)2 + (1 + cos 4 2 2x) ] =

, then I

1 (1 + cos2 2x) 2

IIT JEE eBooks: www.crackjee.xyz 25.16 Comprehensive Mathematics—JEE Advanced

I =2

\

=2 =

dx

sec2 2 x

Ú 2 + tan 2 2 x dx dt

Ú 2 + t2

where t = tan 2x

(c)

x + 2 5 - x + 5 sin -1

x+2 +C 5

(d)

10 + 3x - x 2 + 3sin -1

x+2 +C 5

Solution:

Ê tan 2 x ˆ tan -1 Á ˜ +C Ë 2 2 ¯ If I =

Example 26

Ú sin 4 x + cos4 x

˘ ˙+C ˚˙

2 + sin 2 x ˘ ˙+C 2 - sin 2 x ˙˚

1 + 2 sin 2 x 1È 1 log Íx + 2 ÍÎ 2 2 1 - 2 sin 2 x

˘ ˙+C ˙˚

Ans. (c) 1 2 x= (1 – cos 2x) and 2

Solution

1 cos x = (1 + cos 2x), 2 we can write

=

Ú

1È 1 log = Íx 2 ÍÎ 2 2 Example 27 If I =

(a)

Ú

2 + sin 2 x ˘ ˙+C 2 - sin 2 x ˙˚

5- x dx , then I 2+ x

x + 2 5 - x + 3sin -1

x+2 +C 3

t

( 2t ) dt = 2Ú

7 - t 2 dt

x+2 dx , then I x-2

Ú ( x + 1)

If I =

(a)

1 ( x + 6) x 2 - 4 + 2 log x + x 2 - 4 + C 2

(b)

1 ( x + 6) x 2 - 4 + 4 log x + x 2 - 4 + C 2

(c)

1 ( x + 6) x 2 - 4 + 6 log x + x 2 - 4 + C 2

(d) ( x + 6) x 2 - 4 + 2 log x + x 2 - 4 + C Ans. (b) I=

(1 - cos 2 x )2

1 È 2 cos 2 x ˘ Í1 ˙ dx 2 ÎÍ 2 - sin 2 2 x ˚˙

Ú

Example 28

Solution:

Ú 2 (1 + cos2 2 x) dx

7 - t2

Ê x+2ˆ = x + 2 5 - x + 7 sin -1 Á ˜ +C 7 ¯ Ë

2

I=

+C

Ê t ˆ = t 7 - t 2 + 7 sin -1 Á +C Ë 7 ˜¯

dx , then I

2 + sin 2 x ˘ ˙+C 2 - sin 2 x ˙˚

1 + sin 2 x 1È 1 log Íx 2 ÎÍ 1 - sin 2 x 2 2

1È 1 (c) log Íx 2 ÍÎ 2 2

7

Put 2 + x = t2, so that dx = 2t dt and I=

sin 4 x

1È 1 (a) log Íx + 2 ÍÎ 2 2

(d)

x + 2 5 - x + 7 sin -1

1

=

x+2

(b)

Ans. (b)

Ê t ˆ +C tan -1 Á Ë 2 ˜¯ 2

1

=

(b)

sec2 2 x

Ú 1 + cos2 2 x Ú sec2 2 x + 1dx =2

=

Ú

=

Ú

=

Ú

(x

( x + 1) ( x + 2) x2 - 4 2

dx

)

3 (2 x) + 6 2 dx x2 - 4

-4 +

x – 4 dx +

Ú

x dx 2

x -4

+6

Ú

dx x2 - 4

x 2 x - 4 - 2 log x + x 2 - 4 2

+ 3 x 2 - 4 + 6 log x + x 2 - 4 =

1 ( x + 6) x 2 - 4 + 4 log x + x 2 - 4 + C 2

IIT JEE eBooks: www.crackjee.xyz 25.17

If I =

Example 29

dx

Ú x3

(b)

, then I

x2 - 1

2 ˆ 1 Ê x -1 Á (a) + tan -1 x 2 - 1˜ + C x 2 ÁË ˜¯ 2 ˆ 1 Ê x -1 2 -1 Á ˜ +C (b) x + tan 1 2 ÁË x 2 ˜¯ 2 1 Ê x -1 Á (c) + tan -1 2 ÁË x3

(d)

ˆ x 2 - 1˜ + C ˜¯

2 ˆ 1 Ê x -1 2 -1 Á ˜ +C + x x tan 1 2 ÁË x 2 ˜¯

a 2 a 2 - b2 x2 ax

(c)

ab

x dx

Ans. (b) Solution: Write I =

Ú

I=

\

I=

Ú

(t

2

)

2

+1 t

Ú

dt =

(t

dt 2

=

=

)

+1

2

t = t2 + 1

=

=

t t2 + 1 t 2

t +1

I=

\

dt

1

Ú t 2 + 1 = Ú 1. t 2 + 1dt Ú

+ t

+2

(t

2t 2

)

+1

2

Ú

(

)

t2 + 1

2

2a

(a)

x a 2 - b2 x2

Ú

(a

2

+C

- b2 x2

Út

-3/2

dt =

1

(

a 2 a 2 / x2 - b2 x

(

a 2 a 2 - b2 x2 If I =

1 a

)

1/2

)

1/2

2

t

+C

+C

+C

dx

Ú x1/2 + x1/3 , then I x1/6) + C x1/6) + C

x + 3x1/3 – 6x1/6

x1/6) + C

1/6 x + 6x

x1/6) + C dx

Ú x1/3 (1 + x1/6 )

Put x1/6 = t or x = t6, so that

1 t 1 + tan -1 t + C 2 t2 + 1 2

If I =

2

Solution: Write I =

I =6

2 ˆ 1 Ê x -1 2 -1 ˜ +C x + tan 1 = Á 2 ÁË x 2 ˜¯

Example 30

3/2

t 3/2

(d) 3x1/3 – Ans. (a)

+ 2 tan -1 t - 2 I

)

(a) 2 x – 3x1/3 + 6x1/6 (c)

dt

dx

/ x2 - b2

(b) 3x1/3 – 6x1/6

dt

t2 + 1 - 1

2

2

Example 31 tan -1 t =

(a

dx

(-1/2a ) dt 1

2t dt and t

Ú x3

and put a2/x2 = t + b2, so that (– 2a2/x3) dx = dt

and put x2 – 1 = t2, so that 2x dx =

x2 - 1

+C

a 2 - b2 x2

= -

Ú x4

+C

+C

a 2 - b2 x2

(d)

Ans. (b) Solution Write I =

x

)

3/2

, then I

= 6

Ú

t 5 dt

Ú t 2 (1 + t ) Ú =6

t3 dt t +1

t3 + 1 - 1 dt t +1

È 1 ˘ = 6 Í t2 - t + 1 ˙ dt t + 1˚ Î

Ú(

)

È t3 t 2 ˘ = 6Í + t - log t + 1 ˙ + C 2 ÎÍ 3 ˚˙ = 2 x – 3x1/3 + 6x1/6

x1/6) + C

IIT JEE eBooks: www.crackjee.xyz 25.18 Comprehensive Mathematics—JEE Advanced

Ú 5 - cos2 q - 4 sin q dq , then I

If I =

Example 32

I=

I=

q

4 +C 2 + sin q

Ú ÍÎ 3 1 + 2t - 3 t + 2 ˙˚ dt

=

q) –

2 +C 2 + cos q

1 3

=

1 + 2 tan ( x /2) 1 log +C 3 tan ( x /2) + 2

2 +C 2 - cos q

(3x - 2)

Ú 5 - (1 - x2 ) - 4 x Ú

3 (t + 2) - 2 t2 4 +C t

t

q

(a)

(b)

(c)

(d)

If I =

Ú

È2

dx =

3x - 2

Ú ( x - 2)2

dx

Ê3 4ˆ dt = Á + 2 ˜ dt Ët t ¯

1 1 ˘

If I =

t

C

x3 + x

Ú x4 - 9 dx , then I

(a)

1 4

x4

1 12

x2 + 3

(b)

1 4

x4

1 12

x2 - 3

(c)

1 4

x4

1 12

x-3 +C x+3

(d)

x2 - 3 1 1 log x 4 - 9 - log 2 +C 2 6 x +3

Ú

4 +C sin q - 2

1

t

Example 34

4 +C q) + 2 - sin q Example 33

dt

=

x – 2 = t, so that

Put

dt

Ú 2t 2 + 5t + 2 = Ú ( 2t + 1)(t + 2)

4 +C 2 - sin q

Put sin q = x, so that I=

\

q) +

q) – Ans. (a) Solution:

dx = 2dt/(1 + t2)

(3 sin q - 2) cos q

Ans. (b) Solution: Write I =

dx , then I 4 + 5 sin x

I=

1 2

Ú

(

x2 + 3

) dx , put x

x x2 + 1

( ) x

2

t +1

2

+C

x2 - 3

2

-9 1

t

+C

= t, so that 1

tan ( x /2) + 5 1 log +C 3 3

=

2 tan ( x /2) + 1 1 log +C 3 tan ( x /2) + 2

t-3 1 1 log t 2 - 9 + log +C t+3 4 12

=

x2 - 3 1 1 log x 4 - 9 + log 2 +C 4 12 x +3

2 tan ( x /2) + 1 1 log +C 3 tan ( x /2) - 2 1 3

log

2 tan ( x /2) - 1 tan ( x / 2) + 1

+C

Ans. (b) Solution: Put tan (x/2) = t, so that Ê 2t ˆ 2 = (2 + 2t2 + 5t) and 4 + 5 sin x = 4 + 5 Á 2 2˜ 1 + t + 1 t Ë ¯

Example 35

If I =

Ú e (x cos x + sin x)dx, then I x

1 x e (x sin x – cos x) + C 2 1 x e (x sin x + cos x) + C (b) 2 1 x e (x cos x – sin x) + C (c) 2 1 x e (x (sin x + cos x) – cos x) + C (d) 2 Ans. (d) (a)

dt

Ú t 2 - 9dt = 2 Ú t 2 - 9 dt + 2 Ú t 2 - 9

IIT JEE eBooks: www.crackjee.xyz 25.19

Solution: We have

Úe

x

cos x dx =

therefore, x x I= e (sin x + cos x) 2

1 x e (sin x + cos x), 2

1 ex (sin x + cos x)dx + ex sin x dx 2 1 1 = x ex (sin x + cos x) – ex (cos x – sin x)dx 2 2 1 1 x x ex (sin x + cos x) – e cos x + C = 2 2

Ú



Ú

Ú

(Ú e

x

( f ( x ) + f ¢ ( x )) dx = e x f ( x ))

1 x = e (x sin x + x cos x – cos x) + C 2 If I =

Example 36 (a)

Ú

+ ae

x

Ê a + b tan 2 tan–1 Á ÁË b-a b-a

xˆ ˜ +C ˜¯

(d)

Ê b-a tan–1 Á 2 b-a Ë a + b tan

ˆ ˜ +C x¯

x/2

t

(c)

e2x + ae x + a

(d)

e2 x + ae x + log e x / 2 + e x + e x / 2 + C

+a

e

\

+a )+C ex + a ) + C

ex/2 +

(

)

1 + ae- x dx , put ex = t2, so that

I=

Ú

aˆ Ê ÁË1 + 2 ˜¯ (2t) dt t

=2

Ú

t 2 + a dt

=

Ê t ˆ tan -1 Á ˜ +C b- a Ë b-a¯

=

Ê a + b tan 2 tan -1 Á ÁË b-a b- a

I=

Ú

tan x a + b tan 2 x

Ê a + b tan 2 sin–1 Á b-a ÁË b-a

xˆ ˜ +C ˜¯

(b)

Ê a + b tan 2 Á ÁË b-a

xˆ ˜ +C ˜¯

1

1 b-a

cos–1

1

xˆ ˜ +C ˜¯

sin x

Ú

a cos 2 x + b sin 2 x

dx

–1

- dt

I=

Ú

=

Ê b-a ˆ cos -1 Á cos x˜ + C b b-a Ë ¯

b - (b - a ) t 2 1

–1

x = tan

dx (a < b), then I

(a)

1

Put cos x = t, so that

I= If I =

dt

Ú t ◊ t 2 + (b - a)

= e2 x + ae x + a log e x /2 + e x + a + C

Example 37

t

)

I=

2 2 = t t + a + a log t + t + a + C x /2 e x + a + a log e x /2 + e x + a + C = e

(

Alternative Solution

Ans. (c) x

b

dt fi tan x dx = b b + t2 - a

ex

e

Úe

1

ex + ae2x) + C

(b)

Solution: Write I =

1

Ans. (c) Solution: Put a + b tan2 x = t2 fi 2b tan x sec2 x dx = 2t dt

e2x + ae x dx , then I

e2x + ae x + a 2x

(c)

=

=

Ê 1 - x2 Á x ÁË

ˆ ˜ , therefore ˜¯

Ê tan–1 Á ÁË b-a 1

1 b-a

–1

tan

ˆ 1 ˜ +C (b - a ) cos2 x ˜¯ b

(

Ê a + b tan 2 tan ÁÁ b-a b-a Ë 1

–1

)

Ê b 1 + tan 2 x - b + a ˆ Á ˜ +C Á ˜ b a ÁË ˜¯ xˆ ˜ +C ˜¯

IIT JEE eBooks: www.crackjee.xyz 25.20 Comprehensive Mathematics—JEE Advanced

If I =

Example 38

Ans. (b)

dx

Ú sin x (3 + 2 cos x) , then I x

x) x) + C

x

dx and put 4 cos ) 3 + 4 cos x = t, so that – 4 sin x dx = dt and

a) x) + C

x

I=

x) x) + C

x

x+C

=

sin x dx

Ú sin 2 x (3 + 2 cos x)

and put

2

1 Ê t - 3ˆ 2 sin2 x = 1 - Á = È22 - (t - 3) ˘ Î ˚ Ë 2 ¯˜ 4 1 (t – 1) (5 – t) 4

I= 2

\

16 t

2 = 5

(a)

(b)

(c)

(d)

x)

x) + C¢

If I =

1 10 sin 2 x

Ú (3 + 4 cos x)3

3 cos x + 8

(3 + 4 cos x )2

(3 + 4 cos x )2

Úe

2

16 (3 + 4 cos x )

2

+C

ex + 1) dx, then I

–x

ex + 1) + C ex + 1) + C

(d) none of these

ex + 1) +

dx, then I

= – e–x

ex + 1) +

= – e–x = – e–x = – (e–x

ex ex

If I =

Example 41

+C

e- x e x

Ú e x + 1dx e- x

Ú e- x + 1dx

e–x + 1) + C ex) + x + C ex + 1) + x + C dx

Ú x3 + 1 , then I

(a)

x +1 Ê 2 x - 1ˆ 1 1 + +C log tan -1 Á 3 Ë 3 ˜¯ 3 x2 - x + 1

(b)

Ê x + 1ˆ 1 +C log x 2 - x + 1 + tan -1 Á 3 Ë 3 ˜¯

(c)

1 1 log x 2 - x + 1 + log x + 1 3 3

(

)

(

)

Ê 2 x - 1ˆ + tan -1 Á +C Ë 3 ˜¯

+C

+C

3 - 8 cos x

2

(c) x – (e–x

(d)

16 (3 + 4 cos x ) 3 + cos x

16 (3 + 4 cos x )

+C

3 + 8 cos x

1 Ê1 3 1 ˆ +C 8 ÁË t 2 t 2 ˜¯

8 cos x + 3

Solution: I = – e–x

1 x) – 2

1 10

Example 39

dt =

ex + 1) + C

1 1 È1 ˘ =2 Í log | t | - log | t - 1| + log | t - 5 |˙ + C 5 4 20 Î ˚

1 2

=

3

(b) x + (ex

Ú

where C¢ = C –

t

If I =

Example 40

È1 1 1 1 1 1 ˘ = 2 Í + ˙ dt Î 5 t 4 t - 1 20 t - 5 ˚

+

2

(t - 3)

Ans. (c)

dt t 1 ( ) ( t - 5) t

Ú

Ú

(a) x + (e–x

3 + 2 cos x = t, so that – sin x dx = dt/2 and

=

-1 8

2t - 3

Ans. (c) Solution: Write I =

2 sin x cos x

Ú (3

Solution: Write I =

(

)

1 1 log x 2 - x + 1 + log x + 1 3 6 Ê 2 x - 1ˆ + tan -1 Á +C Ë 3 ˜¯

Ans. (a) Solution:

1 3

x +1

=

1

( x + 1) ( x 2 - x + 1)

IIT JEE eBooks: www.crackjee.xyz 25.21

=



x-2 ˘ 1È 1 - 2 Í ˙ 3 ÎÍ x + 1 x - x + 1 ˚˙ 1 1 log | x + 1| - I1 3 3

I=

where I1 =

=

x-2

Ú x2 - x + 1 dx

Ú = =

t + 1/2 - 2 t 2 + 3/4

Ê 2 x - 1ˆ 1 log x 2 - x + 1 - 3 tan -1 Á 2 Ë 3 ˜¯

(

1 log 3

Example 42

(a)

(b)

(c)

(d)

If I =

)

x +1 x2 - x + 1

2 +t

3 2

2 -t

1

2 -t

2 +t

3 2

2 -t

1

2 +t

3 2

2 -t

-

x -1 + C

Ú

Úq

q (2 sec2 q tan q) dq

(

d sec2 q dq

= q sec2 q –

= q sec2 q – = x sec–1 Example 44

)dq

Ú sec

2

q dq

2 tan–1 (t) + C 3

(sin x + cos x )2 (1 - sin x cos x )

sec2 q - 1 + C

x - x -1 + C

If I =

dx

Ú sin ( x - p /3) cos x , then I

x + sin (x – p x – p/3) sec x x – sin (x – p (d) none of these

2 tan–1 (t) + C 3

sin x + cos x

Put sin x – cos x = t, so that 1 – 2 sin x cos x = t2, and

x - x -1 + C

= q sec2 q – tan q + C

C C C

Ans. (b) Solution

where t = sin x + cos x Ans. (c) 3 Solution x + cos3 x = (sin x + cos x) (sin2 x + cos2 x – sin x cos x), we can write

I=Ú

=

+ tan–1 (t) + C

+

x +C

, then I

1 + tan–1 (t) + C 3 2 +t

1

x dx then I

2 x = sec q or x = sec q

I=

Ê 2 x - 1ˆ +C tan -1 Á Ë 3 ˜¯ 3

Ú sin3 x + cos3 x

x – tan–1

Solution

1

dx

1

3 2

+

-1

x) + C

(d) x sec–1 x + Ans. (c) \

1 ˘

˘ + tan -1 (t )˙ + C 2 -t ˙˚

x

(c) x sec–1

dt

1

2 +t

Ú sec

If I =

(b) sec–1

Ê 2t ˆ 1 3ˆ 3 2 Ê log Á t 2 + ˜ - ◊ tan -1 Á Ë 2 4¯ 2 3 Ë 3 ˜¯

I=

\

È 1 log Í ÍÎ 2 2

(a) x sec–1

2 3 1 1 \ x2 – x + 1 = ÁÊ t + ˜ˆ - ÁÊ t + ˜ˆ + 1 = t 2 + 4 Ë Ë 2¯ 2¯

Thus, I1 =

2 3

Example 43

b 1 Put t = x + =x2a 2

2 È

2dt

Ú (2 - t 2 )(1 + t 2 ) = 3 Ú ÍÎÍ 2 - t 2 + 1 + t 2 ˙˚˙ dt

I=

I=

1 cos (p /3)

cos (p /3)

Ú sin ( x - p /3) cos x dx

cos ÈÎ x - ( x - p /3)˘˚

=2

Ú sin ( x - p /3) cos x dx

=2

Ú

dx

cos x cos ( x - p /3) + sin x sin ( x - p /3) sin ( x - p / 3) cos x

= 2 ÈÎcot ( x - p /3) + tan x ˘˚ dx

Ú

dx

IIT JEE eBooks: www.crackjee.xyz 25.22 Comprehensive Mathematics—JEE Advanced

= 2 ÈÎlog sin ( x - p / 3) | + log sec x ˘˚ + C x – p/3) sec x

If I =

Example 45

(a)

Ú

(

1 + x2

)

(

(d)

1+ 1 (1 + nx n ) n + K n (n - 1)

(d)

1+ 1 (1 + nx n ) n + K n -1

1

Ans. (a) Solution: f o f (x) =

)

2 1È 2 ˘ (b) x + + x log 1 ˙˚ + C 2 ÍÎ

)+C x log ( x + 1 + x ) + C

(c) log x + 1 + x

(c)

1

) dx, then I

1 log x + 1 + x 2 + C 2

(

11 (1 + nx n ) n + K n -1

C

log x + 1 + x 2

(

1

(b)

Ú

1

Solution

1 + x2

) ( 1+ x ) dx

I = log x + 1 + x 2 log x + 1 + x 2

Ú

x n - 2 g ( x ) dx =

1 n

)

)

(

Ú

)

2 1È 2 ˘ x + + x log 1 ˙˚ + C 2 ÎÍ

Example 46

Let f(x) =

x (1 + x n )1/ n

f o f ... o f (x) (n times). Then

Úx

n-2

1

(a)

for n

g(x) dx

11 (1 + nx n ) n + K n (n - 1)

(1 + nx n )1/ n

.

Ú (1 + nxn )1/ n dx . Put (1 + nx )

n 1/n

= t, we

tn - 1 1 dt = t n -1 + K t n ( n - 1)

Let I =

ex

Ú e4 x + e2 x + 1 dx, e- x

Ú e- 4 x + e- 2 x + 1 dx

Then, for an arbitrary constant C, the value of J-I

Ú

(

x

J=

)

=

1/ n

(1 + 3x n )1/ n

2

1È log x + 1 + x 2 ˘˙ + C 2 ÍÎ ˚ Alternative Solution 1 x + 1 + x 2 = t, so that dx = dt and 1 + x2 1 I = t dt = t 2 + C 2

(

n

1

Example 47

2

)

(1 + 2 )

11 (1 + nx n ) n + K n (n - 1)

=

1 + x2

I=

x

x

as 2nd

2

(

=

xn - 1

2 I = Ê log x + 1 + x 2 ˆ + C1 Ë ¯



(1 + ( f ( x)) )

f o f ... o f (n times) (x) =

Ans. (b)

-

n 1/ n

Similarly f o f o f (x) =

2

2

( log ( x +

f ( x)

(a)

1 2

Ê e 4 x - e 2 x + 1ˆ Á 4x ˜ +C Ë e + e 2 x + 1¯

(b)

1 2

Ê e 2 x + e x + 1ˆ Á 2x ˜ +C Ë e - e x + 1¯

(c)

1 2

Ê e 2 x - e x + 1ˆ Á 2x ˜ +C Ë e + e x + 1¯

(d)

1 2

Ê e 4 x + e 2 x + 1ˆ Á 4x ˜ +C Ë e - e 2 x + 1¯

2 and g(x) = Ans. (c) Solution J-I =

e3 x

Ú e4 x + e2 x + 1

dx -

ex

Ú e4 x + e2 x + 1 dx

.

IIT JEE eBooks: www.crackjee.xyz 25.23

= =

e x (e2 x - 1)

Ú e4 x + e2 x 1 - 1/ t

dx = +1

t2 - 1

Ú t4 + t2 + 1

2

5Ê 11 ˆ = - Á x 22/5 + x12/5 + 11x 2/5 ˜ Ë ¯ 88 3

du

Ú (t + 1/ t )2 - 1dt = Ú u 2 - 1 , u = t + 1 / t u -1 1 +C = 2 u +1

1 2

=

e2 x - e x + 1

1 = 2

e2 x + e x + 1

e +e

-x

-1

e +e

-x

+1

x x

λ=+C

I=Ú

(

= λ x + αx + 11x a

b

)

c -k

(x

)

+1

Ê 113/ 25 11 63/ 25 ˆ + x + 11x13/ 25 ˜ ÁË x ¯ 3

+C

5 88

b=

12 5

a+α=

k

c

1 + ln x dx x ln x

Ú

5

d

a

Ans. t = 1 + ln x fi t 2 = 1 + ln x

Solution



2tdt =

\

I=Ú

112 15

1 dx x

t.2t 2 ˘ È dt = Ú Í2 + 2 ˙dt 2 t -1 t - 1˚ Î

= 2t +

2 t -1 ln +k 2 t +1

= 2 1 + ln x + ln

Solution

(

= x3/ 25

)

5

5

11 Ê ˆ = x3/5 Á x 22/5 + x12/5 + 11x 2/5 ˜ Ë ¯ 3 I=Ú



(

+k

f ( x ) = 1 + ln x = ln ( ex )

a

b

c

Example 50

x17/5 + 2 x 7/5 + x -3/5

x

+k

d

)

Ê 22/5 11 12/5 ˆ + x + 11x 2/5 ˜ ÁË x ¯ 3

22/5

1 + ln x + 1

f ( x) + 1

5

5

x -3/5 x 4 + 2 x 2 + 1

Ê 22/5 11 12/5 2/5 ˆ ÁË x + x + 11x ˜¯ 3

1 + ln x - 1

f ( x) - 1

= 2 f ( x ) + ln

È 110/ 25 11 60/ 25 ˘ + x + 11x10/ 25 ˙ Íx 3 Î ˚

b

f ( x ) = ln ( ex )

f ( x ) = ln x

dx

Ans. Ê 113/ 25 11 63/ 25 ˆ + x + 11x13/ 25 ˜ ÁË x ¯ 3

+C

f ( x) + c b a f ( x ) + ln +k f ( x) + d 2

2

Then λ=-

I

+C

2

-4

5 22 11 12 2 , a = , α =, , b = , c = , k = 4 88 5 3 5 5

Example 49

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS Example 48

5 dt 5 -5 5 = t dt = - t -4 + C Ú Ú 5 22 t 22 88

I=

dt , t = e x

5

5

dx

dx

11 + x12/5 + 11x 2/5 = t 3

44 7/5 22 -3/5 ˆ Ê 22 dt = Á x17/5 + x + x ˜ dx Ë 5 ¯ 5 5

f ( x) =

( 2 x - 1) x 2 + x + 1 - ( 2 x + 1) x 2 - x + 1 , x π 0 ( 2 x - 1)2 ( x 2 + x + 1) - ( 2 x + 1)2 ( x 2 - x + 1) xf ( x )

I = 6Ú

x4 + x2 + 1

2

(

dx

I

)

x2 + x + 1 - x2 - x + 1 + C 4x

x + x + 1 + x2 - x + 1 2

+C

IIT JEE eBooks: www.crackjee.xyz 25.24 Comprehensive Mathematics—JEE Advanced

8x

(c)

( 2 x + 1)

2

+3+

( 2 x - 1)

2

+3

+C

(a) 2ab = – 1

(2x – 1)2 (x2 + x + 1) – (2x + 1)2 (x2 – x + 1) = (4x2 + 1) (x2 + x + 1 – x2 + x – 1) – 4x (x2 + x + 1 + x2 – x + 1)

\

Solution

x4 + x2 + 1 = (x2 + 1)2 – x2

Ê x + 7ˆ I = ÚÁ Ë x + 5 ˜¯

= (x2 – x + 1) (x2 + x + 1)

Ê x + 5ˆ = ÚÁ Ë x + 7 ˜¯

È 2x + 1 I =ÚÍ 2 ÎÍ x + x + 1

Put t =

= 2 =

(

˘ ˙ dx x 2 - x + 1 ˚˙ 2x - 1

)

x2 + x + 1 - x2 - x + 1 + C

2 ÈÎ x 2 + x + 1 - x 2 + x - 1˘˚

=

x2 + x + 1 + x2 - x + 1 4x x2 + x + 1 + x2 - x + 1

Example 51 sin (sin x) sin2

+C

dt = \

I=

(b) u¢(x) = h(x) (d) u (2p) = 0

Ans. (a), (b), (d) Solution: We have 2f (x) = cos (sin x) (1 + cos x) + sin (sin x) (1– cos x) = cos (sin x) – cos x sin (sin x) + sin (sin x) + cos x cos (sin x) \ I = I1 + I2 È ˘ d where I1 = Ú e x Ícos ÊÁ (sin x ) + ( cos (sin x ))ˆ˜ ˙ dx Ë ¯ dx Î ˚ x = e cos (sin x) d Similarly, I2 = Ú e x ÈÍsin (sin x ) + (sin (sin x )) ˘˙ dx dx Î ˚ x = e sin (sin x) \ I = ex [cos (sin x) + sin (sin x)] + C Thus h (x) = cos x, u (x) = sin x

x+5 x+7

5

dx

( x + 7 )2

-5

dx

( x + 7 )2

2 x+5 =1x+7 x+7 2

( x + 7 )2

dx

1 -5 1 t dt = - t -4 + C Ú 2 8 4

Now, f ( x ) =

For x ΠR, let f(x) = cos (sin x) cos2

(u(x))] + C, then (a) h(x) = cos x (c) u (p/2) = – 1

(d) f(x) =

1 Ê x + 7ˆ =- Á ˜ +C 8 Ë x + 5¯

+C

x + 2 x , and if 2Ú e x f ( x)dx = ex [h(u(x)) + u 2

= a(f (x))b + C

Ans. (a), (b), (c)

= 8x3 + 2x – 4x (2x2 + 2) = – 6x Also,

3

(b) 32a + b = 0

x+7 x+5

(c) f(x) =

Ans. (a), (b), (c), (d) Solution:

( x + 5) ( x + 7 ) 5

then

( 2 x + 1)2 + 3 - ( 2 x - 1)2 + 3 + C

(d)

dx

If I = Ú

Example 52

Example 53

If In =

x+7 1 ,a = - ,b = 4 x+5 8

Ú cot

n

x dx and I0 + I1 + 2(I2 + …

Ê u2 u9 ˆ + I8) + I9 + I10 = A Á u + + º + ˜ + constant where 2 9¯ Ë u = cot x then (a) A is constant (b) A = – 1 (c) A = 1 (d) A is dependent on x Ans. (a), (b) Solution: In = =

Ú cot

Thus

n–2

Ú cot

x (cosec2

In + In–2 = -

Ú cot x – 1) dx = Ú cot n

x dx =

n–2

x cot2 x dx

n–2

x cosec2 x dx – In–2

cot n -1 x + constant n -1

(i)

I0 + I1 + 2 (I2 + … I8) + I9 + I10 = (I2 + I0) + (I3 + I1) + (I4 + I2) + (I5 + I3) + (I6 + I4) + (I7 + I5) + (I8 + I6) + (I9 + I7) + (I10 + I8) Ê cot x cot 2 x cot 9 + + º + = Á 2 9 Ë 1 Hence A

xˆ ˜¯ + constant

IIT JEE eBooks: www.crackjee.xyz 25.25

If

Example 54

sin x

x – cos x 2

(b) A = 1/ 2

(c) f (x) = sin x Ans. (b), (d)

ln

(

)

x-a + x-b +C b – a))

)

x-a – x-b ) Ê 1ˆ -Á ˜ Ë 2¯

(d) f (x) = x

sin x

Ú sin ( x - p /4) dx = Ú

Solution:

1 [2x – (a + b (b) 2

C then

(a) A =

(

(a) [2x – (a + b

Ú sin ( x - π /4) dx = A (f(x) +

sin (u + p /4) du sin u

(

1 [2x – (a + b 2

(c)

(u = x – p/4) =

Ú

= = = = =

1 1

2 1

u

2

(x – p/4) +

2

Solution: Write I = Ú 1 ◊ log

C

2

x–p

2

C

If f ¢(x) =

then f

1 2

-x + x + 1

2 – 1)

and f(0) = -

1+ 2 2

(

)

2

x x 2 + x + 1 + log x + x 2 + 1 + C 2 2 x = 0, we have f (0) = C so C = –1/2 – 1/ 2 =

Ê 1 1 1 1 ˆ + 2 + log |1 + 2 | + Á - ˜ Ë 2 2 2 2¯ 2

Example 56

If I =

Ú log (

2 – 1).

)

x - a + x - b dx , then I

)

x-a + x-b

ˆÊ 1 + ˜Á x- a + x-b¯ Ë x- a 1

(

)

x-a + x-b -

ˆ ˜ dx x-b¯ 1

1 I1 2

x

Put t = x -

1 1 ( a + b) , and k = (a – b) 2 2 I1 = Ú

t + ( a + b ) /2

(t - k ) (t + k )

I1 = t 2 - k 2 + I=x

f ( x) = Ú x + x + 1 dx

)

x - a + x - b dx

Ú ( x - a ) ( x - b) dx

\ Thus, 2

(

where I1 =

2)

Ans. (a), (d) Solution:



Ú 2 ËÁ

=x

(b) 1

2

(c) 1 +

and f (1) =

-

( x + log | (sin x - cos x) | ) + C

Thus A = 1/ 2 and f (x) = x Example 55

(

=x

1

1 Ê 1 ˆ ÁË x + log | sin x - cos x | + log ˜ +C 2 2¯ 1

( x - a ) ( x - b) + C

Ans. (b), (c)

Ú cot u du 2

1

u+

1 2

)

[2 x - (a + b)] log ( x - a + x - b ) + C

(d)

1

Ú du + 2

x-a + x-b -

sin u cos (p /4) + cos u sin (p /4) du sin u

( x - a )( x - b) + C

-

(

dt

a+b log t + t 2 - k 2 2

)

x-a + x-b -

a+b 1 + ( a + b) log x 4 2 = x log

(

)

( x - a ) ( x - b)

( x - a ) ( x - b)

x-a + x-b -

(

1 2

1 2

( x - a )( x - b)

)

2 1 ( a + b) log x - a + x - b + C 4 1 = ÈÎ2 x - ( a + b )˘˚ log x - a + x - b 2 1 ( x - a ) ( x - b) + C 2

-

(

+ C1

)

IIT JEE eBooks: www.crackjee.xyz 25.26 Comprehensive Mathematics—JEE Advanced

(a) (b)

x 2 + 20

Ú ( x sin x + 5 cos x)2

If I =

Example 57

Ans. (a), (d) dx , then I

Solution: We can write I=

x + tan x + C cos x ( x sin x + 5 cos x )

x + cot x + C sin x ( x sin x + 5 cos x )

=

(c) (x sin x – 5 cos x) sin x + 7x + C x + 2 tan x + C cos x ( x sin x + 5cos x )

Ans. (a) Solution: We can write I=

Put t =

(

x8 x 2 + 20

Ú

)

2

4

5

dx

3

5

1 x

5

5

1 x7

Ú

(b)

cos x x5 sin x + 5 x 4 cos x

Ú ( x5 sin x + 5x4 cos x) sec

2

If I =

Ú

x7 2 x7 + x 2 + 1 x5 2

x + 1 + 2x

7

(x

5 x8 + 7 x 6 2

+ 1 + 2 x7

)

2

p( x) q ( x)

+C

2

dx

+ 2 , so that - dt t

2

1 = +C t +C

If

Ú cos (q / 2)

sin 3 (q /2) cos3 q + cos 2 q + cos q

d q then I

q q Ê ˆ (c) tan–1 Á tan + sec + 1˜ + Const Ë 2 2 ¯

x dx

(d) tan–1 (cos q + sec q + 1) + Const Ans. (b), (d) Solution: I = to =

Ú

sin (q /2) cos (q /2) sin 2 (q /2)

(cos q /2) 2

cos3 q + cos 2 q + cos q

I =-

= q(x) = 7

dq

sin q (1 - cos q ) 1 dq Ú 2 (1 + cos q ) cos3 q + cos 2 q + cos q

Put cos q = x, so that

+C

p(x

1

(a) cot–1 (tan q + sec q) + Const (b) cot–1 (cos q + sec q + 1) + Const

)

dx then I

-1 (c) 2 x 7 + x 2 + 1 + C (d)

I=

x + tan x + C cos x ( x sin x + 5 cos x )

Example 58 (a)

x5

5 x 4 cos x + x5 sin x

= -

x8

x 2 + 1 + 2 x7

Example 59

d [x5 sin x + 5x4 cos x] = (x5 + 20x3) cos x dx

+

dx

2

4

(

2

7

x7

Note that

I =-

+

6

ˆ ÁË 5 + 7 + 2˜¯ x x

+

=

x

ÚÊ 1

I=

( x sin x + 5x cos x) ( x + 20 x ) cos x Ê x ˆ dx =Ú Á ˜ ( x sin x + 5x cos x) Ë cos x ¯ 5

1 Ê 1 ˆ x Á 5 + 7 + 2˜ Ëx ¯ x 14

5

–1

(d)

Ú

5 x8 + 7 x 6

1 2

Ú

(1 - x ) 1 dx Ú 2 (1 + x ) x3 + x 2 + x x2 - 1

( x + 1)

2

1 x x + +1 x

dx

IIT JEE eBooks: www.crackjee.xyz 25.27

=

1 2

Ú

x2 - 1 1 1 Ê ˆ x 2 Á x + + 2˜ x + + 1 Ë ¯ x x

Ú

Ê 1 ˆ Thus, I = exsinx + cosx Á x +C x cos x ˜¯ Ë

1 + 1 = t2 x so that (1 – 1/x2)dx = 2t dt and 2t dt 1 I= 2 = tan–1 t + C 2 t +1 t Put x +

Ú(

If I =

Example 61

)

I=

Ú

2x + 3 + x + 2

u=

+C

x + 2 - 2x + 3

Ê x 4 cos3 x - x sin x + cos x ˆ e x sin x + cos x Á ˜ dx x 2 cos 2 x Ë ¯

(

, then I

2 x + 3, v = x + 2

x + 2 + 2x + 3

If

)

x + 2 + 2x + 3 + C

(d) is transcedental function in u and v, u =

then I sec x ˆ Ê (a) ex sin x + cos x Á x +C Ë x ˜¯ x sin x + cos x

(b) e (c) e

Ans. (a), (d) Solution

cos x ˆ Ê ÁË x sin x - x ˜¯

I=

where I1 =

Ú

I1 = dx

Ú (e

x sin x + cos x

)

x cos x x dx

Ú

Ú

ex sin x + cos x dx

I 2 = e x sin x + cos x

= ex sin x + cos x

(cos x - x sin x ) dx ( x cos x )2

Ê 1 ˆ ÁË - x cos x ˜¯

Ú

2x + 3 x +1

I1 =

x ¸Ô ˝ dx ˛Ô

dx and I2 =

Ú

x+2 x +1

dx

2t ◊ t

1 ˘

È

Ú t 2 - 1 dt = 2Ú ÍÎ1 + t 2 - 1˙˚ dt

È 1 t -1 ˘ = 2 Ít + log ˙ t + 1 ˚˙ ÎÍ 2 In I2, put x + 2 = y2, so that

(x sin x + cos x) = x cos x

I1 = xex sin x + cos x –

dx

x +1

Put 2x + 3 = t2, in I1, so that

Solution ÏÔ cos x - x sin I = e x sin x + cos x Ì x 2 cos x + ( x cos x )2 ÓÔ = I1 + I2

2x + 3 - x + 2

Ú

= I1 – I2

cos x – x sin x ˆ Ê (d) xe x sin x + cos x – Ú e x sin x + cos x Á1 – ˜ dx Ë x 2 cos 2 x ¯ Ans. (a), (d)

where

2x + 3

x+2

v=

Ê x sec x ˆ ÁË tan x - x ˜¯ + C

x sin x + cos x

and

dx

Ú

u –1 v –1 + log +C u +1 v +1

(a) 2(u – v

= tan–1 (cos q + sec q + 1) + C Example 60

dx x cos x

+ e x sin + cos x ( x cos x )

dx

2 y2

I2 =

Ú y 2 - 1dy = 2 y + log

I=2

(

y -1 y +1

Thus, 2x + 3 - x + 2

)

2x + 3 - 1 2x + 3 + 1 x + 2 -1 x + 2 +1

+C

IIT JEE eBooks: www.crackjee.xyz 25.28 Comprehensive Mathematics—JEE Advanced

cot x

Ú

Example 62 If I =

a + b cot 2 x

= – 2 sin (5x/2) sin (x/2) = cos 3x – cos 2x

dx (0 < a < b), then

Ú (cos 3x - cos 2 x) dx =

I 1

Ê b-a ˆ sin–1 Á sin x˜ + Const b-a b Ë ¯

(a)

–1

b - a sin

(b)

(

(c) sin–1

(

)

b - a sin x + Const

)

b - a x + Const

Ê b-a ˆ cos –1 Á sin x˜ + Const b b–a Ë ¯

Ans. (a), (d) Solution: I =

cos x

Ú

2

2

a sin x + b cos x

dx

Put sin x = t, I=

b - (b - a ) t

2

Ê b-a ˆ t +C sin -1 Á b ˜¯ b-a Ë Ê b-a ˆ sin -1 Á sin x ˜ + C n b-a Ë ¯ 1

=

cos 8 x – cos 7 x dx is expressed as 1 + 2 cos 5 x K sin 3x + M sin 2x + C then (a) K = – 1/3 (b) K = 1/3 (c) M = – 1/2 (d) M = 1/2 Ans. (b), (c) Example 63

Solution:

f(x) if

Example 64 (a) f ( x ) = (b) f ( x ) = (c) f ( x ) =

1

(p)

2

( x + 1) x 2 + 2

If

Ú

cos 8 x - cos 7 x cos 8 x - cos 7 x 2 sin 5 x = ◊ 1 + 2 cos 5 x 1 + 2 cos 5 x 2 sin 5 x

=

sin 13x - sin 3x - sin 12 x + sin 2 x 2 (sin 5 x + sin 10 x)

=

1 sin 13x + sin 2 x - (sin 3x + sin 12 x) ◊ 2 sin 5 x + sin 10 x

15 x 11x 15 x 9x - 2 sin 2 sin cos cos 1 2 2 2 2 = 15 x 5x 2 2 sin cos 2 2 Ê 11x ˆ Ê 9x ˆ cos Á - cos Á ˜ ˜ Ë ¯ Ë 2 ¯ -2 sin 5 x sin ( x /2) 1 2 = = 2 2 cos (5 x /2) Ê 5x ˆ cos Á ˜ Ë 2¯

)

5/2

+C

x +1

Ë ( x + 2)

x 4 + x8

(r ) - 2

(1 – x 4 )7/2

ˆ + C˜ 2 ¯

1 - x + cos -1 x

+ x 1– x +C

1– x

(s) - tan -1 1 + 2 / x 2 + C

1+ x p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

(

5 1 - x4

(q ) sin -1 Á

( x + 2) x 2 + 6 x + 7

(d) f ( x ) =

Column 2 x5

Ê

1

dt

Ú 1

=

MATRIX-MATCH TYPE QUESTIONS Column 1

–1

(d)

sin 3x sin 2 x +C . 3 2

x = 1/t, I = Ú f ( x ) dx, where f(x) is as in

Solution (a) we have

– 1 / t 2 dt

I=

Ú (1 / t 2 + 1)

=–

1 2

=–Ú

Ú

2

z 2 – 1ˆ + z 1 Á 2 ˜¯ Ë

dz 2

z +1



1 t2

-tdt



(1 + t ) ( z = 1 + 2t )

1/ t + 2

z dz

ÚÊ

2

1 + 2t 2

2

= - tan -1 z = - tan -1 1 +

x + 2 = 1/t in I =

We have I =

2

2 x2

+C

Ú f ( x) dx where f(x) is as in (b)

dt

1 1 + 2t – t t t2

2

= -Ú

dt 2 - (t - 1)

2

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Solution

x +1 Ê t - 1ˆ = sin–1 +C = – sin–1 Á Ë 2 ˜¯ ( x + 2) 2

I = 2Ú

For (c) As (1 – x4)7/2 = x7(1/x2 – x2)7/2, we can write I=Ú –2

2

x –3 + x (x

–2

2 7/2

–x )

dx

x/2 = t dt

u + vt 2

2 v

If v > 0 then I =

dt

Ú t2 + u / v

=

2 2 t+C= tan x/2 + C. u u 2 v

v tan–1 u

Ê vˆ Á t u ˜ + C. Ë ¯

2

Put x – x = t , so that – 2(x–3 + x)dx = 2t dt dt 1 and I = – Ú 6 = 5 + C 5t t =

x

1- x 1+ x

+C

dx = Ú

1 - cos q ( - sin 2q ) dq 1 + cos q

q sin 2q d q 2 q = - 4Ú sin 2 cos q dq 2

(a)

= - 2 1 – x + cos -1 x + x 1 - x + C

(b)

a + b = u, a – b = v Column 1 Column 2 1 Ê u + tan x /2 ˆ (a) v = 0 (p) Á ˜ +C u Ë u – tan x /2 ¯

(c) v < 0

(r)

(d) v = –1

(s)

uv 1

(

– uv

p

q

2 tan x/2 + C u r s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

u – – v tan ( x /2)

+C

+C

The antiderivative of Column 2 sin x – 2 +C sin x – 1

2

cos x (sin x – 1)(sin x – 2) 2x

1 (sec x + tan x)–2 + C 2 (r) 2x tan–1 x

1 + x2

tan x + cot x

(s)

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

)

u – – v tan x /2

u + – v tan ( x /2)

log

(sec x + tan x)

(c) sin–1 (d)

tan–1 tan x /2 v / u + C u + – v tan x /2

u +t –v +C u / –v – t

sec x

dx

Ú a + b cos x , where a, b > 0 and

(b) v

dt u – t2 –v

Column 1

= – 2 sin q + q + sin q cos q + C

2

– uv

Example 66

=–2

I=

1

=

Ú (1 – cos q) cos q dq = – 2 cos q dq + Ú (1 + cos 2q) dq Ú

Ú

2 –v = log –v 2 u

= - Ú tan

Example 65

2 –v

If v < 0 then I =

5

5(1 – x 4 )5/2

I=Ú

Ê vˆ tan–1 Á (tan x / 2) +C u ˜¯ vu Ë

2

=

For (d) Put x = cos q, i.e., x = cos2 q, so that dx = – 2 cos q sin q dq. Then

Ans.

. If v = 0 then I =

x2) + C

Ê tan x – 1 ˆ tan–1 Á a +C Ë 2 tan x ˜¯

Solution: sec x

Ú (sec x + tan x)2 dx = Ú =

dt

Ú t3

sec x(sec x + tan x)

(t = sec x + tan x)

(sec x + tan x)3

dx

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p

q

r

s

a

p

q

r

s

Ú (sin x – 1) (sin x – 2) dx = Ú (t – 1) (t – 2) (t = sin x)

b

p

q

r

s

p

q

r

s

sin x - 2 Ê 1 1 ˆ = ÚÁ +C – dt = log ˜ sin x - 1 Ë t – 2 t – 1¯

c d

p

q

r

s

= –

Ans.

1 (sec x + tan x)–2 + C 2 dt

cos x

Ú

sin–1

2x 1+ x

Ú

dx = 2

2

= 2 [t tan t – log |sec t| + C] = 2 x tan–1 x – log (1 + x2) + C. Putting tan x = y2, so that dx = 2y dy/(1 + y4), we have

Ú(

=2 Ú

)

= Ú

Ê 1 ˆ 2y y +1 dy = 2Ú dy = ÚÁ y + ˜ 4 y¯ 1+ y Ë 1 + y4 2

1 + 1 / y2 y2 + 1 / y2

dy = 2 Ú

= 2Ú

1 + 1/ y 2

( y - 1/ y )2 + 2

= dy

1

tan -1

2

= 2 tan

-1

du u +2 2

= Ú

2

=

+C

Column 1 1 ( a + b2 ) – 2 2

=

Column 2 1 xˆ Êa (p) tan–1 Á tan ˜ + C Ëb ab 2¯

a sin x + 2

b cos x

(d) f(x) =

2

sec2 x dx b + a tan x 2

2

2

= Ú

dt b + a2t 2 2

1 a 1 xˆ Ê at ˆ Êa ◊ tan–1 Á ˜ = tan–1 Á tan ˜ + C 2 Ë ¯ Ë ¯ ab b 2 b b a

1 a + b2 2

For (d), write f(x) =

1 a cos x + b sin x

cosec (q + x).

sec2 x . Now put tan x = t. a2 (1 + tan 2 x ) – b2

The antiderivative of

Example 68

Column 1 1 Ê tan x ˆ (q) 2 tan–1 Á Ë sin a ˜¯ a sin a

2

2

(c) f (x) =

dx a sin x + b2 cos2 x 2

2

1 2

dt 1 a = tan–1 t+C 2 ab b a t +b 2 2

1 = r sin(q + x )

( a – b ) cos x (b) f(x) =

2

For (c) put a = r sin q, b = r cos q so that

Ê tan x – 1 ˆ ÁË ˜ +C. 2 tan x ¯

The antiderivative of

Example 67

(a) f (x) =

u

dt (tan x/2 = t) ( a + b ) (1 + t 2 ) – ( a2 – b2 ) 2

1 Êa ˆ tan–1 Á tan x / 2˜ + C Ëb ¯ ab

Ú

[where u = y – 1/y] = 2◊

dx ( a2 + b2 ) – ( a2 – b2 ) cos x

(1 – t 2 )

tan x + cot x dx

= 2Ú

Ú

Solution:

t sec2 t dt (x = tan t)

+ C, a = cos–1

x

(a) b a

1 1 Êa ˆ (r) tan–1 Á tan x˜ + C Ëb ¯ ab a cos x + b sin x 1 a2 – b2 cos2 x

(a2 > b2) log

(s)

e (2 – x ) (1 – x ) 1 – x 2

(b) 2

(c)

1

Column 2 2

x ex 1+ e

x

x +1 x 2

x(1 + x e )

(p)

ex +C x+2

(q) e x

1+ x 1– x

(r) (x – 2)u – log

+ C, u =

a2 + b2 1Ê aˆ tan Á x + tan –1 ˜ + C 2Ë b¯

u –1 u +1

(d)

e x ( x + 1) ( x + 2)

2

(s) log

x ex 1+ xe

x

+

ex + 1

1 +C 1 + x ex

IIT JEE eBooks: www.crackjee.xyz 25.31

p

Ans.

q

r

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution:

Now

1– x 1 ¥ 1 + x (1 – x )2 =

2

Thus Ú e x

(2 – x ) (1 – x ) 1 – x 2

dx = ex

x -1

Ú

( x 2 + 1) x 4 + 1

Statement-2: Ú

dt t t -a 2

dx = sec–1

Thus, I = (x – 2)

2t dt 2

t –1

( x + 1) e x

x 2

1

Solution

(1 – x ) 1 – x 2

Ú

1+ x +C 1- x

= 2t + log

x2 - 1

dx = Ú

( x 2 + 1) x 4 + 1

= Ú

t –1 t +1

Ê e x + 1 – 1ˆ e x + 1 – log Á ˜ +C ÁË e x + 1 + 1˜¯

xe (1 + xe )

dx = Ú

t +C a

Ans. (d)

For (c) the given integral can be written as x

+C

+C

x 2

1 sec–1 a

=

e x + 1 = t2, so that I1 = Ú t ◊

x2 + 1

2

I1 = Ú e + 1 dx

Put

1 + xe x

Statement-1:

Example 69

x

where

Ú

( x + 2)2

I = x e x + 1 – I1

For (b)

1

ASSERTION-REASON TYPE QUESTIONS

d Ê 1+ xˆ 1 1 – x (1 – x ) + (1 + x ) ¥ = Á ˜ dx Ë 1 – x ¯ 2 1+ x (1 – x )2 =

1 + xe

+

x

Ê 1 –1 ˆ = ex Á + ˜ Ë x + 2 ( x + 2)2 ¯

e x ( x + 1)

For (d), write

È 1+ x ˘ 1 ˙ + = ex Í 1– x˙ Í (1 – x ) 1 – x 2 Î ˚

(1 – x ) 1 – x 2

xe x

= log

In (a) the integrand can be written as

1 + (1 – x 2 )

ex

s

dt (t – 1) t 2

(t = 1 + xex, dt = (x + 1)ex dx) 1˘ 1 È 1 = Ú Í - ˙ dt Ît - 1 t ˚ t Ê 1 1ˆ = ÚÁ – 2 ˜ dt Ë t (t – 1) t ¯ Ê 1 1 1ˆ = ÚÁ – – 2 ˜ dt Ët –1 t t ¯ 1 +C = log |t – 1| – log |t | + t

(

( x + 1 / x) dt t t2 - 2

x2 + 1 / x2

dx

[t = x + 1/x]

=

1 t sec–1 +C 2 2

=

x2 + 1 1 sec–1 +C 2 x 2

Statement-1: Ú x - x 2 - 4 dx

Example 70 È1 = Í Î3

1 - 1 / x2

x+2- x-2

)

3

+2

(

)

˘ x+2+ x-2 ˙ +C ˚

Statement-2: The integral in statement-1 can be computed by substituting

(

x +2 - x -2

)

2

= 2t

Ans. (d) Solution: fi fi

Putting x–

(

x +2 - x -2

)

2

= 2t

x2 - 4 = t fi

(x – t)2 = x2 – 4 1Ê 4ˆ t2 + 4 = 2tx fi x = Á t + ˜ Ë 2 t¯

Thus dx =

1Ê 4ˆ Á1 – 2 ˜¯ 2Ë t

Therefore, the given integral =

1 4ˆ Ê t Á1 - 2 ˜ dt Ú Ë 2 t ¯

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1 Ê 2 3/ 2 -1/ 2 ˆ Á t + 8t ˜¯ + C 2 Ë3

= È = 1Í 1 2 Î3 2

(

x+2- x-2 +

1 È1 2 2 ÍÎ 3

(

x+2-

=

)

3

x-2

)

3

8 2 + 16

Example 73

˘ ˙ +C x+2 - x-2 ˚

=–

1

(

x+2+

)

˘ x - 2 ˙ +C ˚

u ( x) v( x ) sin (6x + 2) + cos (6x + 2) 72 72 1 1 + x3 + x2 – x + C 2 4 (a) u(x) = 3x 2 + 6x – 13 (b) u(x) = 18x2 + 2x – 13 (c) v(x) = 3x + 1 (d) v(x) = – (6x + 1)

COMPREHENSION-TYPE QUESTIONS Example 74 Paragraph for Question Nos 71 to 74 In calculating a number of integrals we had to use the method of integration by parts several times in succession. The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts

Ú u( x ) v( x ) dx = u(x) v1(x) – u¢(x) v2(x) + u ≤(x) v3(x)

For Example 71, the given integral is equal to sin 2 x cos 2 x ˆ – (3x 2 – 4x + 3) ÊÁ – (x 3 – 2x2 + 3x – 1) ˜ Ë 2 4 ¯ cos 2 x Ê sin 2 x ˆ + (6x – 4) Á – +C ˜ –6 Ë 8 ¯ 16 sin 2 x cos 2 x [2x 3 – 4x2 + 3x] + [6x2 – 8x + 3) + C 8 4 For Example 72, applying the above formula in the comprehension, the given integral is equal to

=

(2x3 + 3x2 – 8x + 1)

If

3 2 Ú (x – 2x + 3x – 1) cos 2x dx

+ (12x + 6)

sin 2 x cos 2 x u(x) + v(x) + C 4 8

= u(x) u(x) v(x) v(x)

Example 72 2x + 6 5.7.9 (a) (b) (c) (d)

= = = =

3

2

x – 4x + 3x 2x3 – 4x2 + 3x 3x2 – 4x + 3 6x2 – 8x

If Ú (2x3 + 3x 2 – 8x + 1)

2 x + 6 dx

(2x + 6) f (x) + C then f(x) is equal to x3 – 6x2 – 91x + 297 7x 3 – 3x2 – 132x + 597 70x 3 – 45x2 – 396x + 897 70x3 – 45x 2 – 132x + 597

(2 x + 6)3/2 (2 x + 6)5/2 – (6x2 + 6x – 8) 3 3◊5

(2 x + 6)7/2 (2 x + 6)9/2 – 12 +C 3◊5◊ 7 3◊5◊ 7 ◊9

2x + 6

(2x + 6) (70x 3 – 45x2 – 396x + 897) + C 5◊7◊9 For Example 73 1 (3x 2 + x – 2) (1 – cos (6x + 2)). Now applying the Ú 2 formula in comprehension, the last integral can be written as

then

=

9 È3 2 2˘ ÍÎ 4 ( x – 7 x + 1) – 40 (2 x – 7)(2 x + 1) + K (2 x + 1) ˙˚ + C. Then K is equal to (a) 27/320 (b) 3/4 (c) 17/93 (d) 9/320 Ans. 71 (b), 72 (c), 73 (d), 74 (a)

– (– 1)n–1 Ú u n ( x ) vn ( x ) dx

Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration by parts is especially useful when calculating Ú Pn(x) Q(x)dx, where Pn(x) is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n + 1 times.

(a) (b) (c) (d)

x2 - 7x + 1 dx = (2x + 1)2/3 3 2x + 1

Solutions of Examples 71–74

v1(x) = Ú v( x ) dx , v2(x) = Ú v1 ( x ) dx ..., vn(x) = Ú vn-1 ( x ) dx

=

If Ú

+ ... + (–1)n–1 un–1(x) vn (x)

where

Example 71

If Ú (3x 2 + x – 2) sin2 (3x + 1) dx

˘ 1 È 3 x2 – 2x˙ Íx + 2Î 2 ˚ –

1 2

sin(6 x + 2) cos(6 x + 2) È 2 + (6 x + 1) ÍÎ(3x + x – 2) 6 36 sin (6 x + 2) ˘ –6 ˙+C 6 ¥ 36 ˚

IIT JEE eBooks: www.crackjee.xyz 25.33

˘ 18 x 2 + 6 x – 13 1 È 3 x2 sin (6x + 2) – 2x˙ – Íx + 2Î 2 72 ˚

=

1 (6x + 1) cos (6x + 2) + C. 72 The integral in Example 74 can be shown to be equal to 3 9 (x 2 – 7x + 1) (2x + 1)2/3 – (2x – 7) (2x + 1)5/3 4 40 27 + (2x + 1)8/3 + C. 320 –

Paragraph for Question Nos 75 to 78 can be found by the method of reduction and recursion. Reduction formulas make it possible to reduce an integral dependent on the index n > 0, called the order of the integral, to an integral of the same type with a smaller index. Integration by parts helps us to derive reduction formulas. (Add a constant in each of Examples 75–78.) If I n = Ú

Example 75

dx then In + 1 + ( x + a2 )n 2

1 – 2n 1 I n is equal to ◊ 2n a 2 x ( x + a2 )n

(c)

1 x 1 x (d) ◊ 2 2 2 n 2 2 2n a ( x + a ) 2n a ( x + a2 ) n –1

In, – m +

(b)

2

If I n,– m = Ú

1 1 2 2 2n a ( x + a2 ) n –1

sin n x dx then, cos m x

(c)

(b)

1 sin n –1 x ( n – 1) cos m –1 x

(d)

If un = Ú

1 sin n –1 x ( m – 1) cos m –1 x

(a) x (c)

xn ax + 2bx + c 2

(d)

1 (x3m + x2m + xm)(m+1)/m 6(2m + 1)

Ans. 75 (c), 76 (b), 77 (d), 78 (c) Solution of Examples 75–78 For Example 75, using integration by parts In =

x x2 + 2n dx Ú ( x 2 + a2 )n ( x 2 + a2 )n + 1

=

x 1 + 2n Ú 2 dx – 2 n (x + a ) ( x + a2 )n 2

1 dx ( x + a2 )n + 1 1 – 2n 1 1 x I = Whence I n +1 + ◊ 2 2 n 2 2n a 2n a ( x + a 2 ) n For Example 76 In, –m =

2

n – 1 sin n – 2 x sin n –1 x dx – Ú ( m – 1) cos m –1 x m – 1 cos m – 2 x n –1 sin n –1 x – In–2, 2–m m –1 m –1 ( m – 1) cos x

un +1 = Ú

x n+1 ax 2 + 2bx + c

dx

=

1 x n (2ax + 2b) - 2bx n dx Ú 2a ax 2 + 2bx + c

dx, then

=

1 x n (2ax + 2b) b dx - un Ú a 2a ax 2 + 2bx + c

xn ax 2 + 2bx + c

ax + 2bx + c (b) 2

1 (2x3m + 3x2m + 6x m)(m +1)/m 6( m + 1)

n – 1 sin n –1 x m – 1 cos m –1 x

(n + 1)au n+1 + (2n + 1)bun + nc u n–1 is equal to n–1

(c)

For Example 77, consider

sin n –1 x cos m –1 x

Example 77

1 (2x2m + 3x m + 6)(m+1)/m ( m + 1)

=

n –1 In–2, 2–m is equal to m –1

(a)

(b)

2na 2 Ú

(a)

Example 76

The value of I m = Ú (x3m + x2m + xm) (2x2m + 3x m + 6)1/m dx is 1 (a) (2x2m + 3x m + 6)(m+1)/m 6( m + 1) Example 78

x

n –2

ax 2 + 2bx + c

= In – In =

(d) x n ax 2 + 2bx + c =

b un, where a

1 x n (2ax + 2b) dx Ú 2a ax 2 + 2bx + c 1 2a

È x n 2 ax 2 + bx + c ÍÎ

(i)

IIT JEE eBooks: www.crackjee.xyz 25.34 Comprehensive Mathematics—JEE Advanced

- Ú n x n-1 2 ax 2 + 2bx + c ˘˙ dx ˚

Ê 1 1 1 ˆ = ÚÁ+ + dx Ë 3( x + 1) x - 1 3( x - 2) ˜¯

xn ax 2 + 2bx + c a

=

= – (1/3) log | x + 1| + log | x –1| + (1/3) log | x – 2 | + const. = (1/3) log ( | x + 1|)–1 + (1/3) log | x – 1|3

n x n-1 ( ax 2 + 2bx + c) dx - Ú a ax 2 + bx + c

+ (1/3) log | x – 2 | + (1/3) log C

aIn = x n ax 2 + 2bx + c – na un+1

= (1/3) log C

–2bn un– nc un–1 Putting this value in (i) we have

Hence

a un+1= xn ax 2 + 2bx + c – na un +1 – 2bn un – ncu n – 1 – b un fi (n + 1) aun +1 + (2n + 1) bun + ncun–1

A 1 = fi A = 3. 9 3

Example 80

For Example 78 I = Ú xm (x2m + xm + 1) (2x2m + 3xm + 6)1/m dx = Ú xm – 1 (x2m + xm + 1) (xm (2x2m + 3xm + 6))1/m dx

I=

1 2 (t + t + 1) (2t3 + 3t2 + 6t)1/m dt Ú m

Put 2t3 + 3t2 + 6t = um fi (6t2 + 6t + 6) dt = mum –1 du 1 2 1 (t + t + 1) dt = um–1 du m 6



I=

1 m–1 1 m 1 m+1 Ú u u du = 6 Ú u du = 6 (m + 1) u 6

1 = (2t 3 + 3t 2 + 6t ) 6( m + 1) =

1 6( m + 1)

m +1 m

m +1 3m 2m m m (2 x + 3x + 6 x )

INTEGER-ANSWER TYPE QUESTIONS Example 79

If

x2 – 3

Ú

x3 – 2 x 2 – x + 2 then A is equal to.

dx =

Ê ( x – 2)( x – 1)3 ˆ A log Á C ˜ 9 x +1 Ë ¯

Ans. 43 Solution:

Ú

x2 - 3 x3 - 2 x 2 - x + 2

x2 – 1 ( x 2 + 1) x 4 + 1

dx is equal to

ˆ x 2 + 1 / x 2 ˜ + C then A is equal to. ¯

Ans. 2

Ú

Solution:

= Ú xm – 1 (x2m + xm + 1) (2x3m + 3x2m + 6xm)1/m dx Put xm = t fi mxm–1 dx = dt \

If Ú

2 Ê 1 A tan–1 Á Ë 2 4

= xn ax 2 + 2bx + c

( x - 2)( x - 1)3 x +1

x2 - 1 ( x 2 + 1) x 4 + 1

dx = Ú

= Ú = Ú

dt t t2 - 2

= Ú

y dy ( y + 2) y 2

=

1 - 1/ x 2 ( x + 1/ x ) x 2 + 1/ x 2

t dt t2 t2 - 2 1 y +C tan -1 2 2

Ê ÁË t = x +

dx 1ˆ ˜ x¯

( y2 = t2 – 2)

( ( x + 1 x) - 2 ) + C

=

1 1 tan -1 2 2

=

1 Ê 1 ˆ x 2 + 1 x 2 ˜ + C. tan -1 Á Ë ¯ 2 2

2

Thus A = 2 Example 81 If the graph of the antiderivative F(x) of f(x) = log (log x) + (log x)–2 passes through (e, 7 – e) then the term independent of x in F(x) is Ans. 7 Solution: An antiderivative of f (x) = F(x) = Ú (log (log x) + (log x)–2) dx + C = x log (log x) – Ú

x dx + Ú (log x)–2 dx + C x log x

= x log (log x) – [x (log x)–1 + Ú (log x)–2 dx] dx = Ú

x2 - 3 dx ( x + 1) ( x - 1)( x - 2)

+ Ú (log x)–2 dx + C (again integrating by parts)

IIT JEE eBooks: www.crackjee.xyz 25.35

= x log (log x) – x(log x)–1 + C Putting x = e, we have 7 – e = e.0 – e + C. Thus C = 7. If Ú

Example 82 –

cos2 x + sin 2 x (2 cos x – sin x )2

–A dx = x 25

2 1 log |2 cos x – sin x| + + C then A is equal to 5 2 – tan x

= tan–1 (e x/2 ) – Ans. 2

I = Ú = Ú = Ú = Ú

1 + 2 tan x (2 - tan x )2

dx

sec2 x - tan 2 x + 2 tan x (2 - tan x )2 sec2 x - tan x (tan x - 2) (2 - tan x )2

Ú

sec2 x (2 - tan x )

where I1 = Ú = and I2 = Ú

2

dx + Ú

tan x d x = I1 + I2 (say) 2 - tan x [t = tan x]

2

1 1 = 2 - t 2 - tan x tan x sin x dx=Ú dx 2 - tan x 2 cos x - sin x

x and cos x, we get 0 = 2a – b and 1 = – a – 2b. Solving we get a = –1/5 and b = – 2/5. 1 2 cos x - sin x 2 - 2sin x - cos x dx - Ú dx \ I2 = - Ú 5 2 cos x - sin x 5 2 cos x - sin x 1 2 = - x - log |2 cos x – sin x| + const 5 5

that

1 1 2 - x - log |2 cos x – sin x| + C so 2 - tan x 5 5

–A –1 = i.e. A = 5. 25 5

(e 2 x + 4)(e 2 x – 1)2

(t 2 + 4) (t 2 - 1)2 dt t +4 2

+

t 3 + 4t (t 2 + 4) (t 2 - 1)2

dt + Ú

dt

t (t 2 + 4) (t 2 + 4) (t 2 - 1)2

dt

2t 1 dt Ú 2 2 (t - 1)2

1 t – +C 2 2 2(t - 1)

(

)

Example 84

1 2(e

If Ú

2x

- 1)

+C

z x3 A dx = tan–1 – 16 128 2 4+x

= x 4 / 2 then A is equal to. Ans. 4 Solution: Putting x 4 = t in the given integral, we have dt dt x3 1 1 dx = Ú = Ú Ú 4 4 + t 4 16 1 + (t / 2) 4 4 + x16

dx

( y = t / 2)

=

2 dy Ú 16 1 + y 4

=

y 2 + 1 - ( y 2 - 1) 1 dy Ú 16 2 y4 + 1

=

˘ 1 - 1/ y 2 1 È 1 + 1/ y 2 d d y y ÍÚ 2 ˙ Ú 16 2 Î y + 1/ y 2 y 2 + 1/ y 2 ˚

=

˘ 1 + 1/ y 2 1 - 1/ y 2 1 È d d y y ÍÚ ˙ Ú 16 2 Î ( y – 1/ y )2 + 2 ( y + 1/ y )2 - 2 ˚

= 2e5 x + e 4 x – 4e3 x + 4e 2 x + 2e x

(t 2 - 1)2

= tan–1 e x 2 -

Example 83 If Ú

(t 2 + 4) (t 2 - 1)2

dt + Ú

dt

u– 2 1 log + C, where u = y + 1/y and z = y – 1/y, y 64 u+ 2

Putting sin x = a(2 cos x – sin x) + b (– 2 sin x – cos x)

Thus, I =

2t 4 - 4t 2 + 2

= tan–1

dt

(2 - t )

= Ú

(t 2 + 4) (t 2 - 1)2

=2 Ú

dx

+ C then K is equal to.

– 1)

2t 4 + t 3 - 4t 2 + 4t + 2

=2 Ú dx

4(e

Solution: (i) Putting e x = t in the given integral and taking ex common in the integrand, we can write the given integral as

Ans. 5 Solution: (i) Dividing the numerator and denominator by cos2 x, the given integral can be written as

K 2x

du ˘ 1 È dz -Ú 2 ÍÚ 2 ˙ 16 2 Î z + 2 u - 2˚ [z = y – 1/y, u = y + 1/ y]

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+ C.

=

2 5 2a 3 u – u 5 3

where u = y + 1/y, z = y – 1/y and y = x4/ 2 .

=

2 2a (t + a)5/2 – (t + a)3/2 5 3

=

1 tan–1 32

u- 2 z 1 – log 64 2 u+ 2

Thus Example 85

cos2 x sin x A dx = log |sin x – cos x| If Ú 16 sin x – cos x

1 (sin 2x + cos 2x) + C then A is equal to. 8 Ans. 4

I=

+

Solution

cos2 x sin x 1 cos x sin 2 x Ú sin x - cos x d x = 2 Ú sin x - cos x d x



Ú

1 (1 - (sin x - cos x ) ) cos x dx Ú 2 sin x - cos x

cos x 1 1 = Ú d x - Ú (sin x - cos x ) cos x d x 2 sin x - cos x 2 1 - (sin x - cos x ) + (cos x + sin x ) = Ú dx 4 sin x - cos x 1 Ú (sin 2 x - 1 - cos 2 x ) d x 4 1 1 = [– x + log | sin x – cos x | ] + cos 2x 4 8 -

+ =

2 [4 (tan x + 4)3/2 – 3(tan x + 3)3/2] + C 3

Example 87

2

=

2 [(tan x + 4)5/2 – (tan x + 3)5/2] 5

cos3 x dx (sin 4 x + 3sin 2 x + 1) tan –1 (sin x + cosec x )

= – A log |tan–1 (sin x + cosec x)| + C then A is equal to. Ans. 1 Solution:

Put sin x = t, so that

Example 86

Ê 1ˆ (t 4 + 3t 2 + 1) tan –1 Á t + ˜ Ë t¯ Divide numerator and denominator of the integrand by t 2. – (1 – 1 / t 2 ) dt 1ˆ Ê 2 1 ˆ –1 Ê ÁË t + 2 + 3˜¯ tan ÁË t + ˜¯ t t Put t + 1/t = u, so that I=Ú

sin 2 x 1 x+ +C 4 8

I=Ú

sin x + 3cos x + sin x + 4 cos x

A ((tan x + 4)5/2 – (tan x + 3)3/2) 30 2 [4 (tan x + 4)3/2 – 3 (tan x + 3)3/2] + C – 3 then A is equal to Ans. 6 (sin x ) (cos x ) –3 dx Solution: I = Ú tan x + 3 + tan x + 4

where Ia = Ú t t + a dt Put t + a = u2, so that Ia = Ú (u2 – a) u(2u) du

(u + 1) tan –1 u

= – log |tan–1 (sin x + cosec (x)| + C Example 88 A log 9

If I = Ú

2 x esin x + 1 – 1 2 xesin x + 1 + 1

x cos x + 1 2 x 3esin x + x 2

dx =

+ C then A is equal to.

Ans. 9 Solution: We can write I= Ú

Put tan x = t, so that t I= Ú dt = I 4 – I3 t +3+ t +4

– du 2

= – log |tan–1 (u)| + C

sin x (cos x ) –5/2 dx

=

(1 – t 2 ) dt

I=Ú

1 1 log | sin x – cos x | + (sin 2x + cos 2x) + C. 4 8 If I = Ú

If I =

( x cos x + 1) dx x 2 x esin x + 1

Put 2xesin x + 1 = t 2 fi 2esin x (x cos x + 1) dx = 2tdt \

I= Ú

2t dt 2

(t – 1)t

= log

t –1 +C t +1

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2 xesin x + 1 – 1

= log

If I = Ú

Example 89

2 xesin x + 1 + 1

Thus

dx 2x 1 – x

I =–

2 – x + 1– x

1 1 =– log u + + u 2 + u + + 2 3 2 3 1 K log v – + v 2 – v + 1 + C where u = 16 2 1 then K is equal to 1– x +1 Put 1 – x = t 2, so that

I= Ú

1 1 log v – + v 2 – v + 1 + C 2 2 1 1 where u = and v = 1– x –1 1– x +1

1 1– x –1

Example 90

2(1 – t 2 )t 1 + t 2 + t

Solution:

1 = (I1 – I2) 2

and I2 = Ú

= Ú

1 dt t – 1 1 + t2 + t

In I1, put t – 1 = 1/u, so that

= –

= –

= –

Ê 1 + Á1 + Ë 1 Ú 3

1ˆ Ê ˜¯ + ÁË1 + u

1ˆ ˜ u¯

du u2 + u +

1 3

1 du Ú 2 3 1ˆ 1 Ê ÁË u + ˜¯ + 2 12 1 1 1 log u + + u 2 + u + 2 3 3

In I2 put t + 1 = 1/v, so that I2 = Ú

= Ú

Ê –1ˆ ÁË 2 ˜¯ du u 2

If I = Ú

dx 1 + x2 + 2x + 2

=

x2 + 2x + 2 – 1 x +1

+C

I= Ú

dx

( x + 1)2 + 1 + 1

( x + 1)2 + 1 - 1 dx ( x + 1)2

Put (x + 1)2 + 1 = t2, so that t -1 t I= Ú 2 dt 2 t -1 t -1

1 dt t + 1 1 + t2 + t

I1 = Ú u

1 1 + u2 + u + 2 3

then A is equal to. Ans. 7

– 2t dt

where I1 = Ú

2 3

log u +

A log x + 1 + x 2 + 2 x + 2 – 7

Ans. 8 Solution:

1

1 + v2 – v + 1 2

+

1

v=

= – log v –

+C

dt dt t dt = Ú -Ú 2 t + 1 t2 - 1 t -1 (t + 1) t 2 - 1

= log t + t 2 - 1 – I1 In I1, put t + 1 = 1/u, so that Ê 1ˆ ÁË - 2 ˜¯ du du u = -Ú I1 = Ú 2 1 - 2u Ê 1ˆ Ê 1 ˆ ÁË ˜¯ ÁË - 1˜¯ - 1 u u = Thus,

1 - 2u =

t -1 t +1

I = log t + t 2 - 1 -

t -1 t2 - 1

+C

= log x + 1 + x 2 + 2 x + 2

dv 2

1ˆ 3 Ê ÁË v – ˜¯ + 2 4

-

x2 + 2x + 2 - 1 x +1

+C

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EXERCISE LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS È ex ˘ 1. If I = Ú Íe x ln (sin x) + ˙ dx then I is equal to sin 2 x ˚ Î (a) ex [ln (sin x) + cot x] + C (b) ex [ln (sin x) – cot x] + C (c) ex [ln (cos x) – sec x] + C (d) ex [ln (cos x) + tan x] + C cosec x - cot x sec x 2. Let I = Ú ◊ dx cosec x + cot x 1 + 2 sec x Ê 2 Ê xˆ ˆ (a) - 2 sin Á 1 - tan 2 Á ˜ ˜ + C Ë 2¯ ¯ Ë 5

(c) I =

1 ln( x 6 + 1) + C 3

(d) I =

1 ln( x 6 + 1) + C 6

5. Let I = Ú

Ê 5 - xˆ ˜ 2 Á (5 - x) Ë 5 + x ¯

Ê 5 - xˆ (a) 4 Á Ë 5 + x ˜¯

1/ 4

Ê 5 - xˆ (c) 2 Á Ë 5 + x ˜¯

1/ 4

6. Let I = Ú (a)

-1

(b)

(c)

Ê 2 Ê xˆ ˆ 1 + tan 2 Á ˜ ˜ + C 2 sin Á Ë 2¯ ¯ Ë 5 2 xˆ 2 tan Á tan ˜ + C 2¯ Ë 5

Ê 2 xˆ (d) - 2 tan -1 Á tan ˜ + C 2¯ Ë 5 2

3. Let I = Ú (a) (b) (c) (d)

(c)

-1

-1 Ê

1 2

sin x 1 + cos 2 x

dx

tan -1 ( 2 tan x) - x + C

Ê Ê xˆˆ tan -1 Á tan Á ˜ ˜ - x + C Ë 2¯ ¯ Ë 2 2

tan -1 (2 tan x) - x + C

ˆ Ê 1 tan -1 Á tan x˜ - x + C ¯ Ë 2 2

1

4. Let I = Ú

dx x x6 + 1

x6 + 1 - 1

1 ln 6

x +1 -1

(b) I =

x6 + 1 + 1 6

x6 + 1 + 1

x5 + x + 1

(

1/ 4

Ê 5 - xˆ (d) 4 Á Ë 5 + x ˜¯

7/ 4

+C

+C

(b)

+C

(d)

(

x2

(1 + x 2 ) 1 + 1 + x 2

)

+C +C

dx , then I is equal to x5 + x x5 + x + 1 x5 - x x5 + x + 1

+C +C

)

dx is equal to

(a) ln x + x 2 + 1 - tan -1 x + C (b)

(

)

1 ln x + x 2 - 1 - 2 tan -1 x + C 2

( ln (1 +

) 1 + x ) + tan

(d)

2

-1

x+C

8. For a > 1, let È Ê xx ˆ ˘ I = Ú a x Íln x + (ln a ) ln Á x ˜ ˙ dx then I is equal to Ë a ¯ ˙˚ ÍÎ 1 ˘ È Ê xˆ +C (a) a x Í x ln Á ˜ + 1 Ë ¯ a ln a ˙˚ Î 1 ˘ È Ê xˆ +C (b) a x Í x ln Á ˜ Ë ¯ a a ˙˚ ln Î

, then

1 (a) I = ln 3

x5

7. Let I = Ú

Ê 5 + xˆ (b) 4 Á Ë 5 - x ˜¯

( x5 + x + 1) 2

x5 + x + 1

dx , then I is equal to

+C

4 x5 + 5 x 4 x3

3/ 4

(c) ln 1 + 1 + x 2 - tan -1 x + C

1 1

10

+C

È Ê xˆ ˘ (c) a x Í x ln Á ˜ + 1˙ + C Ë a¯ ˚ Î

+C

1 ˘ È Ê xˆ +C (d) a x Íln Á ˜ + x ln a ˙˚ Î Ë a¯

IIT JEE eBooks: www.crackjee.xyz 25.39

Ê x 4- xˆ 9. Let I = Ú Á dx , then x ˜¯ Ë 4- x

(c)

(d) none of these

(a) (4 - x)3/ 2 - x3/ 2 + 4 - x - x + C (b) x

3/ 2

- (4 - x)

3/ 2

x – x1/4 + 6 log (x1/4 + 1) – 3x3/4 + C cos3 x

13. If I = Ú

+ x +C

sin11 x

dx, then I equals

(c) - 4 x - x 2 + C

(a) -

2 2 (cot x)5/2 – (cot x)7/2 + C 5 7

Ê x - 2ˆ +C (d) - sin -1 Á Ë 2 ˜¯

(b) -

7 3 (cot x)7/2 – (cot x)11/3 + C 2 11

È ( x + a) x + a ˘ 10. Let a, b > 0. For x > 0, let f(x) = ln Í x +b ˙ Î ( x + b) ˚ and g(x) = (x + a) (x + b) ln (x + a) ln (x + b), then f ( x) Ú g ( x) dx is equal to

5 11 (cot x)5/2 + (cot x)2/11 + C 2 2 (d) none of these (c)

14. If I = Ú [1 + cot (x – a) cot (x + a)]dx, then I equals cot x - cot a cot x + cot a

+C

ln( x + b) +C (a) ln( x + a)

(a) log

Ê ln( x + b) ˆ +C (b) ln Á Ë ln( x + a) ˜¯

(b) cot 2a log

(c) ln (ln (x + b) – ln (x + a)) + C

(c) cosec 2a log

Ê x + a ˆ ln( x + b) +C (d) ln Á Ë x + b ˜¯ ln( x + a)

(d) log |tan x| – x log |tan a| + C

11. Let I = Ú (a) (b)

(

) dx , then

1 + x + x2 - 1

( 2 1x(

x2 1 + x + x2

2

15. If I = Ú

) 1+ x + x )

2 1 - 1 + x + x2 + C x

(c)

-2(1 + x) 1+ x + x +1 2

)

(

+ ln x + x 2 + x + 1 + C

(

)

Ê log (1 + x1/4 ) ˆ 1 12. If I = Ú Á 1/2 + ˜ dx, then I equals x1/2 + x1/4 ¯ Ë x - x1/4 (a) 2 x + 4x1/4 + log |x1/4 – 1| + C (b) 2(x

1/2

1/4

1/4

– 6x + 3) log (x + 1) + C – (x1/2 – 10x1/4 + 7) + 2{log(x1/4 + 1)}2

tan x - cot a tan x + cot a

+C

1 log (x2 + a2) dx, then I equals 2 x 1 Ê xˆ log (x2 + a2) + 2 tan–1 Á ˜ + C Ë a¯ x

(b) -

1 2 Ê xˆ log (x2 + a2) + tan–1 Á ˜ + C Ë a¯ x a

(c) -

1 2 Ê xˆ log (x2 + a2) + cot–1 Á ˜ + C Ë a¯ x a

(d) – log (x2 + a2) + 16. If I = Ú

ˆ 1Ê 1 2 (d) - Á1 + ˜ + ln x + x + x + 1 + C 2¯ xË 1+ x + x

+C

(a) -

2

1 Ê ˆ + ln Á x + + x 2 + x + 1˜ + C Ë ¯ 2

1 - cot x tan a 1 + cot x tan a

2 x tan -1 ÊÁ ˆ˜ + C Ë a a¯

1 tan–1 (ex) dx, then I equals ex

(a) – e–x tan–1 (ex) + log (1 + e2x) + C (b) x – e–x tan–1 ex – (c) x – e–x tan–1 (ex) – (d) ex + tan–1 x +

1 log (1 + ex) + C 2 1 log (1 + e2x) + C 2

1 log (1 + e2x) + C 2

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17. If I = Ú (a) (b) (c) (d)

log (cos x )

tan tan tan tan

18. If I = Ú

cos2 x x log cos x log cos x log cos x log cos

22. If I = Ú cos q log (tan q/2) dq, then I equals (a) sin q log (tan q/2) + q + C (b) cos q log (tan q/2) + q + C (c) sin q log (tan q/2) – q + C (d) cos q log (tan q/2) – q + C dx 23. If I = Ú , then I equals x x4 + 1

, then I equals x x x x

+ – – +

tan tan cot cot

log ( x + x 2 + a2 ) x 2 + a2

( (b) (1/2) log ( x + (c) (1/2) log ( x + (d) (1/2) Èlog ( x + ÍÎ

x x x x

– + + –

x+C x2 + C x+C x+C

dx (a) (1/4) log

) +C x +a ) – x + C x +a ) + x + C x + a )˘ + C ˙˚

(a) (1/2) log x + x 2 + a2

19. If I = Ú

2

2

2

(b) (1/4) log

2

2

x-2 x e +C x+2

(b)

2- x x e +C x+2

(c)

1+ x x e +C x+2

(d)

x -1 x e +C x+2

x2 + 1

(a) (x + 4)

dx, then 2I equals

(

)

x 2 + 1 + log x + x 2 + 1 + C

(

)

(b) x x 2 + 1 + 2 log x + x 2 + 1 + C

(

)

21. If I = Ú

(

)

x 2 + 1 + log x + x 2 + 1 + C

x2 dx, then I equals ( x - a) ( x - b)

(a) x +

(b) x +

1 log a-b 1 log a-b

x-a +C x-b x-a x-b

(a) (b) (c) (d)

25. If I = Ú

+C

sin x - cos x sin 2 x

dx, then I equals

(a) log sin x – cos x +

sin 2x

+C

(b) log sin x + cos x –

sin 2x

+C

(c) log |sin x + cos x| + C (d) log |sin x – cos x| + C cos x dx a + b cot 2 x

(a > b > 0), then I equals

(a)

1 a sin 2 x + b cos2 x + C a-b

(b)

1 a + b cot 2 x + C a-b

(c)

1 a-b

a2 + b2

1 {a2 log |x – a| – a-b b2 log |x – b|} + C 1 x-a +C (d) x 2 + log a-b x-b (c) x +

+C

dx , then I equals tan x log cosec x log |log cosec x| + C log |log cos x| + C – log |log (cosec x)| + C log |log sin x| + C

26. If I = Ú

(c) x x 2 + 1 + log x + x 2 + 1 + C (d) (x – 3)

x4 + 1 - 1

24. If I = Ú

x2 ex dx, then I equals ( x + 2)2

( x + 1)2

x4 + 1 + 1

+C

(d) (1/4) (x4 – 1)–3/2 + C

(a)

20. If I = Ú

x4 + 1 + 1

(c) (1/4) (x4 + 1)3/2 + C

2

2

x4 + 1 - 1

(d)

( 1 a-b(

27. If I = Ú (a)

) a + b cot x – x ) + C a + b cot 2 x + x + C 2

dx , then I equals (e + 2)3 x

ex + 3 1 1 x - log (ex + 2) + +C 8 8 4(e x + 2)2

IIT JEE eBooks: www.crackjee.xyz 25.41

1 1 x+ 8 8

ex

1 1 x+ 8 8

ex

ex ex +

x-

C

x +1

ex ex +

C

Ú

I

1

x

+

x

-

x

+

x

-

x

1

x

I

I

+

x x

C

x

+

x x

C

Ú

x -a x +a

1 a

Ê ax ˆ Á ˜ Ëx -a ¯

1 a

Êx a ˆ Á ax ˜ Ë ¯

C

Ê ÁË

I

Ú

Ú

C

x

C

Ê ÁË

xˆ ˜¯

C C I

C

(

)

x-

I

C

)

C

x-x

x

a- xˆ

I

Ê xˆ ÁË ˜¯ a

C

a

Ê xˆ ÁË ˜¯ a

C

Ê xˆ ÁË ˜¯ a

C

a

Ê xˆ ÁË ˜¯ a

C

I

1+ x

1 x

-

-

x

)

1- x

1+ x

x

1 x +

C

C x 1+ x

C

x +x +C x 1+ x x +a + x dx x -a a a

Ú

a

xˆ ˜¯

(

dx

x

ax +C x -a

C

Ê a+x

1+ x -

I

Ú Á a - x + a + x ˜ dx Ë ¯

x

I

xˆ ˜¯

x-x

-x

C

Ê ÁË

dx

x

C

x+

xˆ ˜¯

Ê ÁË

- x

x +1

x

dx x

x-

I

I

x+ x +a +

Ú

(

C C

-x

x

x-x

1- x

dx

x+ x -a

I

1+ x

C

x

x +a

( ) ( x+ )

C

1- x

dx

Ú

-x

x

C

-x

x I

x dx x

-x

x

x -1

C

a a

a

a

a

I x

a

a a x C x a a a x C x a a a x C x a a a x C

IIT JEE eBooks: www.crackjee.xyz 25.42 Comprehensive Mathematics—JEE Advanced

36. If I =

Úe

x

x3 + 3 x 2 + 4 dx, then I equals ( x + 1)3

Ê x 2 - x + 1ˆ (a) e Á ˜ +C Ë ( x + 1)2 ¯ x

Ê x 2 - x + 1ˆ (b) e Á ˜ +C Ë ( x + 1)2 ¯ x

2 Ê x2 + 2x - 2ˆ x Ê x + 2 x - 1ˆ (c) ex Á + C (d) e Á ˜ +C ˜ Ë ( x + 1)2 ¯ Ë ( x + 1)2 ¯

(a) log |cos x| + (1/2) log |cos 2x| + (1/3) log |cos 3x| + C (b) log |cos x| – (1/2) log |cos 2x| – (1/3) log |cos 3x| + C (c) log |cos x| + (1/2) log |cos 2x| – (1/3) log |cos 3x| + C (d) none of these dx , then I equals 38. If I = Ú (2 sin x + sec x) 4

(b) (c)

1 1 2 +C + 5 6 5 tan x 3 tan x 7 tan 7 x

1 1 1 +C + 5 6 5 tan x 3 tan x (2 sin x + sec x )3

-1 (d) – tan–1 3 tan x + C 3 (2 sin x + sec x )3 Ê x + 2ˆ 39. If I = Ú e Á Ë x + 4 ˜¯

)

2

x

dx, then I equals

(a)

x ex + C x+4

(b)

x-2 x e +C x+4

(c)

x +1 x e +C x+4

(d)

x -1 x e +C x+4

Ê ˆ 1 x tan–1 Á +C 4 1/4 ˜ 2 Ë (1 + x ) ¯

(b)

1 - (1 + x 4 )1/4 1 log 4 1 + (1 + x 4 )1/4

(c)

)

)

(b)

(2

)

(

)

2 + 1 tan–1 1 + x + x

(

)

– log x + 2 x + 2 + C (c)

(

x -1

)

2

(

)

tan–1 1 + x - x

(

)

+ log x +1 + C (d) none of these 2x + 3 dx, then I equals 42. If I = Ú 2 ( x + 2 x + 3) x 2 + 2 x + 4 (a) log

(b) log

x2 + 2x + 4 - 1 x2 + 2x + 4 + 1

+C

x2 + 2x + 4 - 1

Ê x + 2ˆ + tan–1 Á +C Ë 3 ˜¯ x2 + 2x + 4 + 1

Ê x + 3ˆ (c) log tan–1 Á ˜ +C Ë 2 ¯ Ê a2 - ax + x 2 ˆ 43. If I = Ú cot–1 Á ˜ dx, then I equals a2 Ë ¯ Ê x - aˆ Ê xˆ (a) x tan–1 Á ˜ – (x – a) tan–1 Á +C Ë a¯ Ë a ˜¯ a (b) log (2a2 – 2ax + x2) 2 a – log (x2 + a2) + C 2 Ê x - aˆ Ê xˆ (c) x tan–1 Á ˜ + (x – a) tan–1 Á Ë a¯ Ë a ˜¯

+C

(1 + x 4 )1/4 1 tan–1 2 x 1 - (1 + x 4 )1/4 1 - log 4 1 + (1 + x 4 )1/4

(

(d) none of these

dx 40. If I = Ú , then I equals (1 + x 4 )1/4 (a)

(

+ log x + 2 x + 2 + C

)

(

)

+C

x + (1 + x 4 )1/4

41. If I = Ú tan–1 1 + x dx, then I equals

-1 + tan–1 3 tan x + C 3 (2 sin x + sec)3

(

(

x - (1 + x 4 )1/4

(a) x2 tan–1 1 + x - x

37. If I = Ú tan x tan 2x tan 3x dx, then I equals

(a) -

(1 + x 4 )1/4 1 tan–1 2 x 1 - log 4

(d)

+

+C

(d) tan–1

a log (2a2 – 2ax + x2) + C 2

x-a x + x tan–1 a a + log (2a2 – 2ax + x2) + C

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44. If I = Ú

cos x sin3 x - cos3 x

(c) f (x) = x2/3 dx, and f(x) =

51. If I = Ú (tan x)1/3 dx

1 1 Ê 2 tan x + 1ˆ log (tan x – 1) + tan–1 Á ˜¯ , g(x) Ë 3 3 3 1 = log (tan2 x + tan x + 1) then I equals 6 (a) f(x) – g(x) + C (b) f(x) g(x) + C (c) f(x)/g(x) + C (d) f(x) + g(x) + C 3/2

dx, and f(x) = (cot x) , g(x) = sin5 x (cot x)5/2, then I equals

(2

(a)

)

(

cos 4 x 2

sin x

dx = K cot x + M sin 2 x + L

(a) L = – 3 (c) M = – 1/4 3x + 4 x3 – 2 x – 4

x + C, then 2

(b) K = – 2 (d) none of these dx = log |x – 2| + K log f (x) + C, then

(a) K = – 1/2 (b) f (x) = x 2 + 2x + 2 2 (c) f (x) = |x + 2x + 2| (d) K = 1/4 dx xˆ Ê = K tan–1 Á M tan ˜ + C, then 49. If Ú Ë 5 + 4 cos x 2¯ (a) K = 1 (c) M = 1/3 50. Let Ú

(b) K = 2/3 (d) M = 2/3

x1/ 2 1– x

dx =

3

(a) A = 1/4

3/2

(b) B =

(d) B = 2 3

(c) A = 1/4 52. If I = Ú

sin x + sin x dx = A cos x + B log | f (x)| + C, cos 2 x

then

(a) A = 1/4, B = – 1/2, f(x) =

3

2 cos x – 1 2 cos x + 1

2

2 gof (x) + C (C being the 3

53. The value of the integral Ú

(b) f(x) = x 3/2

2 cos x + 1 2 cos x – 1 2 cos x + 1 2 cos x – 1

log ( x + 1) – log x dx is x ( x + 1)

(a) – (1/2) (log x + 1)2 – (1/2) (log x)2 + log (x + 1) log x + C 2 (b) – [(log (x + 1) – (log x)2] + log (x + 1) log x + C (c) C – (1/2) (log (1 + 1/x)2 (d) none of these 54. If Ú

xex 1+ e

x

dx = f (x) 1 + e x – 2 log g(x) + C, then

(a) f (x) = x – 1 (c) g(x) =

1+ ex –1 1+ ex +1

(b) f (x) = 2(x – 2) (d) g (x) =

1+ ex +1 1+ ex –1

2

55. If Ú xe –5x sin 4x2 dx = Ke–5x2 (A sin 4x2 + B cos 4x2) + C. Then (a) K = – 1/82 (b) K = 1/82 (c) A = 5 (d) A = 3 3 3 56. If Ú f (x) dx = tan5 x (5 tan2 x + 11) + C then 55 f (x) is equal to tan 2 x(1 + tan 2 x )6

(a)

3

sin 2 x cos –14 x

(b)

3

(c)

3

cos2 x sin –14 x

(d)

73 2 sin x cos –14 x 3

constant of integration) then (a) f (x) =

+C

x then

(d) A = 1/2, B = 3 / 4 2 , f(x) =

x dx 46. If Ú 2 = K tan–1 x + L tan–1 + C, 2 2 ( x + 1) ( x + 4) then (a) K = 1/3 (b) L = 2/3 (c) K = – 1/3 (d) L = – 1/6

48. If Ú

)

2 3

(c) A = – 1/2, B = 3 / 5 , f (x) =

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

47. If Ú

where t = tan

1/3

2

(b) A = 1/2, B = – 3/4

)

(1 / 2 3 ) f(x) + C (2 2 / 3) f (x) + 1/5 g(x) + C

(d)

(

t2 + 1

2t 2 – 1

+ B tan–1

3 / 3 f(x) – (1/5) g(x) + C

(b) – 4 2 / 5 g(x) + C (c)

t4 – t2 + 1

= A log

3

sin3 2 x

45. If I = Ú

(d) g(x) = sin–1 x

IIT JEE eBooks: www.crackjee.xyz 25.44 Comprehensive Mathematics—JEE Advanced

57. If the primitive of

x5 + x 4 – 8

is

3

x – 4x log | f (x) + C then (a) A = 1 (b) A = 4 (c) f (x) = x 2(x – 2)5 (x + 2)–3 (d) f (x) = x 2(x – 2)3 (x + 2)–2 x

58. If the primitive of

1–e

2x

h ( x) +

is f o g(x) –

x3 – 1

(q) log

x

(1 + x ) 10

– 1 5

(c) f (x) =

(d) f(x) =

4x – x

(a) f (x) =

x+ x

2

10

5 10 5 – +C + 10 5 3 x 3 x 2 x2

5

)

(r) 2 x – 3 3 x – 8 4 x

)

2 x 2 – 3x + 3 x – 2x + x 3

2

171 7 (d) f (x) =

(

6

tan–1

)

x – 12 x + 2 – 212 x + 1 7

+C

1 3 (s) log (1 + x2/3) + C 4 2 x + x +2 x 3

–9 16

log |2x + 1| + C x2 – 2

+C

x2 – 1

(p) 3 log |x| –

(b) f (x) =

x3 + 1 x( x – 1)3

2 – x –1

(q) log |x – 1| – 2 log |x| x

– 2–x x( x – 1)2

( x – 1)2

x +C ( x – 1)2



(d) f (x) =

– x3 + 5 x 2 – 4x + 2 x( x – 1)3

+C

(r) 2 log |x – 1| – log |x|

(s) 2 log |x| – 2 log | x – 1| + C

64. The antiderivative of 1 1 sec5 x + sec3 x + C 5 3

(a) cos 3 x sin –6 x

(p)

(b) sin 2 x cos–8 x

(q) –

(c) cos–8 x sin x (sin5 x + cos3 x (1 + sin 2 x ))

(r)

3 log 1 + 12 x + 33 log 2

+C

Ê ( x – 2)2 ˆ x 1 (s) log Á ˜ +C 5 x 4 – 3x 2 + 2 Ë 2x + 1¯

+ 6 6 x + 4812 x +

(

( x + 2)3

1 –7 x + log |x| 4 16

(r) log

x3 – 4 x

x 2 ( x – 2)5

log | x – 1| + C

(c)

10 + 10 x

x5 + x 4 – 8

x3 x2 + + 4x + 3 2

63. If I = Ú f (x) dx, where

(p) 2x1/2 – 3x1/3 + 6x1/6 – 6 log (x1/6 + 1) + C

1 (b) f (x) = x + x1/3

(p)

(q)

3

2x

61. The antiderivative of 1 (a) f (x) = 1/2 x + x1/3

(

x 2 2 x – 3x – 2

log |2x – 1|

MATRIX-MATCH TYPE QUESTIONS

x

(a) f (x) =

(b) f(x) =

(b) g(x) = e (d) h(x) = 1 – e2x 1 59. If Ú cos 2 3x cosec 3x dx = (log | f (x)| + g(x)| ) + C 3 then 3x (a) f (x) = tan (b) f(x) = cot 3x 2 (c) g(x) = sin 3x (d) g(x) = cos 3x 1 1 60. If the primitive of 6 is f (x) + + g(x) x–3 4 x x +x + C then (b) f(x) = 2 tan–1 x (a) f(x) = tan–1 x (c) g(x) is a constant (d) g(x) = – 1/3 function

(c) f (x) =

62. Ú f (x) dx, where f (x) is

log

x

e (1 + e )

C then (a) f (x) = sin–1 x (c) g(x) = ex

x3 x 2 + + Ax + 3 2

1 1 cosec 5 x + 3 5 3 cosec x + C

tan3 x 2 tan5 x + 3 5

+

tan 7 x +C 7

IIT JEE eBooks: www.crackjee.xyz 25.45

(d)

sin x (1 + cos2 x ) cos6 x

(s)

tan 2 x 2 tan 4 x + 2 4 +

tan 7 x +C 7

ASSERTION-REASON TYPE QUESTIONS 2 65. Let F(x x. Statement 1: The function F(x F(x + p) = F(x) for all real x. Statement 2: sin2 (p + x) = sin2 x for all real x. 66. Let F(x

cos3 x sin 2 x + sin x

◊ x π np.

Statement 1: The function F(x) is 1 – 1 on (0, p /2]. Statement 2: log x increases from – • to 0 on (0, 1] and sin x increases from 0 to 1 on (0, p /2]. 67. Let F(x 1 , F(0) = 0. (sin x + cos x )2 Statement 1: lim F(x) = 1 x Æp /4

Statement 2:

cos x is a continuous at x = p /4. sin x + cos x

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos 68 to 71 If the integrand is a rational function of x and fractional ax + b powers of a linear fractional function of the form , cx + d then rationalization of the integral is affected by the substiax + b tution = t m, where m is l.c.m. of fractional powers cx + d of

ax + b . cx + d

68. If I = Ú

(2 x – 3)1/2 (2 x – 3)1/3 + 1

dx

1 1 È1 ˘ (2 x – 3)7/6 – (2 x – 3)5/6 + (2 x – 3)1/2 ˙ 5 3 = 3 Í7 +C Í ˙ 1/6 – (2 x – 3) + g ( x ) ˙˚ ÍÎ then g(x) is equal to (b) (2x – 3)1/2 (a) tan–1 (2x – 3)1/6 –1 1/6 (c) 3 tan (2x – 3) (d) 4(2x – 3)1/6

69. If I = Ú

dx 4

( x – 1)3 ( x + 2)5

A is equal to (a) 1/3 (c) 3/4 70. If I = Ú 3 =K

3

14

+ C then

(b) 2/3 (d) 4/3 dx

( x + 1)2 ( x – 1) 4

1+ x +C 1– x

then K is equal to (a) 2/3 (c) 1/3 71. If Ú

Ê x – 1ˆ =A Á Ë x + 2 ˜¯

dx (1 – x ) 1 – x 2

(a) 1/2 (c) 1/3

(b) –3/2 (d) 1/2 =K

x +1 + C then K is equal to 1– x (b) 1 (d) 2/3

Paragraph for Question Nos 72 to 75 We can derive reduction formulas for the integral of the form Ú sin n x dx, Ú cos n x dx Ú tan n x dx, n Ú cot x dx and other integrals of these form using integration by parts. In turn these reduction formulas can be used to compute integrals of higher power of sin x, cos x etc. 1 72. If Ú sin 5 x dx = – sin 4 x cos x + A sin2 x cos x 5 8 – cos x + C then A is equal to 15 (a) – 2/15 (b) – 3/5 (c) – 4/15 (d) – 1/15 1 tan 5 x + A tan3 x + tan x – x + C 73. If Ú tan 6 x dx = 5 then A is equal to (a) 1/3 (b) 2/3 (c) – 2/3 (d) – 1/3 1 tan 3 x + A tan 3 x + tan x + C 74. If I = Ú sec 6 x dx = 5 then A is equal to (a) 1/3 (b) 2/3 (c) – 1/3 (d) – 2/3 75. If Ú cosec n x dx =

– cos ecn – 2 x cot x + n –1

A Ú cosecn–2 x dx then A is equal to

IIT JEE eBooks: www.crackjee.xyz 25.46 Comprehensive Mathematics—JEE Advanced

(a)

1 n–2

(b)

n –1 (c) n–2

n n–2

n–2 (d) n –1

INTEGER-ANSWER TYPE QUESTIONS 76. Let F(x) be the antiderivative of f (x) = 3 cos x – 2 sin x whose graph passes through the point (p/2, 1). Then F(0) is equal to. 77. Let F(x) be the antiderivative of f (x) = 1/(3 + 5 sin x + 3 cos x) whose graph passes through 1 8 log + 5 is the point (0, 0). Then F(p /2) – 10 3 equal to 78. Let f be a function satisfying f ≤(x) = x–3/2, f ¢(4) = 2 and f (0) = 0. Then f (784) – 2238 is equal to 1 A dx = tan–1 e2x + C then A is equal to 79. Ú 2 x –2 x 8 e +e dx 1 x A = tan–1 x + + C then A is 80. If Ú 2 2 4 2 x2 + 1 ( x + 1) equal to 81. If Ú

x2 + x + 1 f ( x) dx = log + x –1 1 - x3

2x + 1 A tan–1 + C then A = _____, where f(x) 3 3 is a polynomial of second degree in x such that f (0) = f(1) = 3f(2) = 3. 82. If Ú

sin –1 x – cos –1 x sin –1 x + cos –1 x

dx =

A [–(sin–1 x ) (1 – 2x) + p A is equal to 83. If Ú

cos7 x + cos5 x sin 2 x + sin 4 x

x 1 – x ˘˚ – x + C then

dx

2 = –A sin3 x + 5 sin x – – 12 tan–1 (sin x) + C sin x then A is equal to 1 A tan x – 84. If Ú sec 4 x cosec 4 x dx = tan 3 x + 3 2 cot x + C then A is equal to sin x A dx = sec x – tan x + x + C 85. If Ú 266 1 + sin x then A is equal to

dx

86. If Ú

(1 + x )

x–x

2

=

A x –1 +C 3 1– x

then A is equal to cos x – sin x ÊA ˆ 87. If Ú dx = sin–1 Á (sin x + cos x )˜ + C Ë ¯ 9 8 – sin 2 x then A is equal to LEVEL 2

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. If I = Ú

sin (2 x ) - sin (2a ) dx, then I (sin x - sin a ) + (cos x - cos a )

equals (a) sin x – cos x +

2 x cos (p/4 + a) + C

(b) sin x – cos x +

2 x cos (a – p/4) + C

(c) sin x + cos x +

2 x cos (p/4 + a) + C

(d) sin x – cos x +

2 x cos (p/4 – a) + C

2. If I = Ú sin3 x sin

x dx, then I equals 2

(a)

16 x 16 x sin5 sin7 +C 5 2 7 2

(b)

xÊ xˆ 16 sin Á1 - cos 4 ˜ + C 5 2Ë 2¯

16 x 16 x cos5 cos7 +C 2 7 5 2 (d) none of these dx then I equals 3. If I = Ú 1/3 1/3 x ( x - 1) (c)

x2/3 + x1/3 log |x1/3 – 1| + C 3x1/3 + 3 log |x1/3 – 1| + C 3x2/3 + 3 log |x1/3 – 1| + C 3x1/3 + log |x1/3 – 1| + C x -1 dx, then I equals 4. If I = Ú x+3 (a) (b) (c) (d)

Ê x - 1ˆ (a) 2 x - 1 + 4 tan–1 Á ˜ +C Ë 2 ¯ Ê x - 1ˆ (b) 2 x - 1 – tan–1 Á ˜ +C Ë 2 ¯

IIT JEE eBooks: www.crackjee.xyz 25.47

(c)

Ê x - 1ˆ 2 x - 1 + 4 tan–1 Á ˜ +C Ë 2 ¯

1 È ˘ 2 Í x - 1 - 2 tan -1 x - 1˙ + C 2 Î ˚ dx 5. If I = Ú , then I equals x + 1- x (d)

1

+

log

2 2

( )( ) ( 2 x + 1) ( 2 (1 - x) ) - 1 2x - 1

(b)

x + 1- x +

(c)

x - 1- x +

1

1 2 2

6. If I = Ú (a) (b) (c) (d)

x

(e (ex (ex (ex

2x - 1 2x + 1

2 (1 - x ) - 1 2x - 1 2x + 1

+C

+ – – –

e ) e–x) e–x) e–x)

tan (x/2) + cot (x/2) + sec (x/2) + cosec (x/2)

+C

(a) x 1 - x

+ (1/2) sin

1 log tan x + a + b tan 2 x + C a-b dx cos ( x + a ) sin ( x + b ) 3

(b) 2 sec (b – a)

tan ( x + a ) sin ( b - a ) + C

(c) 2 cosec (a – b)

cot ( x + a ) + sin ( b - a ) + C

(d) 2 cosec (a + b)

cot ( x + a ) + sin ( b - a ) + C

cos ( x - a ) dx, then I equals cos ( x + a )

(b) – sin a log cos x + cos2 x - sin 2 a

(d) sin a log

a + b tan x + 1 - 2 cos a a + b tan x a + b tan x + 1 + 2 cos a a + b tan x + cos a tan–1

x+C 11. If I = Ú ( x + 1) (a) (x + 8)

(d) (2 + x)

1 - x 2 – (3/2) sin–1 x + C

(b) (1/2) (x + 8)

a + b tan 2 x

a cos x + b sin x - a - b cos x 2

log

2

a cos2 x + b sin 2 x + a - b cos x

+C

+C

+C

x2 - 9

9 8 - log x + x 2 - 9 + log 2 3

1 (a) 2 a-b

a + b tan x

x 2 - 9 + 25 log x - x 2 - 9

1 - x 2 – (1/2) sin–1 x + C

(a > b > 0), then I equals

a + b tan x - 1

x+3 dx, then I equals x-3

(c) (2 – x)

tan x dx

+C

Ê sin x ˆ +C (c) cos a sin–1 Á Ë cos a ˜¯

1È (b) ( x - 2) 1 - x 2 - sin -1 x ˘˙ + C ˚ 2 ÎÍ

8. If I = Ú

, then I equals

(a) 2 sin a log |cos (x + a) – cos (x – a)| + C

C C C +C

–1

+C

(a) 2 sec (b – a)

10. If I = Ú

+C

1- x 7. If I = Ú x dx , then I equals 1+ x 2

a - b log cos x + a + b tan 2 x

tan ( x + a ) cos (b - a ) + sin (b - a ) + C

e x + e - x + (e x - e - x ) sin x dx 1 + cos x –x

(c)

9. If I = Ú

+C

2(1 - x ) + 1

log

x + 1 - x + log

(d)

2 (1 - x + 1

log

2 2

cos x + a sin 2 x + b cos2 x 1 log +C a-b cos x - a sin 2 x + b cos2 x

(d)

x - 1- x

(a)

(b)

x-3+ x+3 + C

(c) (x + 8)

x 2 - 9 – 2 log x + x 2 - 9

(d) (x + 8)

x 2 - 9 + log x + x 2 - 9

+C +C

12. If u = Ú eax sin (bx) dx, and v = Ú eax cos (bx) dx, then (u2 + v2) (a2 + b2) equals

IIT JEE eBooks: www.crackjee.xyz 25.48 Comprehensive Mathematics—JEE Advanced

(a) 2eax (c) 2e2ax

(b) e2ax (d) bxeax

Ê uˆ 13. If u and v are as in problem 12, then tan–1 Á ˜ + Ë v¯ Ê bˆ tan–1 Á ˜ equals Ë a¯ (a) bx (c) b2 x2

(b) 2bx (d) bx

14. If I = Ú 2 x sin (2x) dx, then I equals (a)

2x [(log 2) sin 2x – 2 cos 2x] + C (log 2)2 + 4

(b)

2x [2 sin 2x + (log 2) cos 2x] + C (log 2)2 + 4

(c)

2x [2 cos 2x – (log 2) sin 2x] + C (log 2)2 + 1

(d)

2x [2 sin 2x – (log 2) cos 2x] + C (log 2)2 + 4

15. If I = Ú xe x sin x dx, then I equals (a) (1/2) e x {x(sin x + cos x) + cos x} + C (b) (1/2) e x {x(sin x + cos x) – sin x} + C (c) (1/2) e x {x(sin x – cos x) + cos x} + C (d) ex {x(sin x – cos x) + cos x} + C 16. If I = Ú e x (x cos x + sin x) dx (a) (1/2) x(sin x + cos x)ex + (1/2) e x cos x + C (b) (1/2) x(sin x + cos x)ex + (1/2) ex cos x + C (c) (1/2) x(sin x – cos x)ex + (1/2) ex cos x + C (d) (1/2) x(sin x – cos x)ex – (1/2) ex cos x + C dx , then I equals 17. If I = Ú x (1 + 4 x 3 + 3x 6 ) 3 ˘ (a) 1/3 ÎÈ3log | x| + 1/ 2 log | x 3 + 1| - log |3x 3 + 1|˙ + C 2 ˚ (b) 1 / 3 È2 log | x| - log | x 3 + 1| + 1/ 6 log | x 3 + 3 |˘˚ + C Î (c) 1/3 È2 log | x| - log | x 3 + 1| - log | x 3 + 3 |˘˚ + C Î (d) Èlog | x| + log | x 3 + 1| - log | 3x 3 + 11|˘˚ + C Î 18. If I =

Ú

1 + log x x2 x - 1

(c) log x x - x 2 x - 1 + C (d) tan–1 xx + C 19. If I = Ú sec2 x + 1 dx, then I equals

( (b) log ( tan x + (c) log ( tan x +

(b) log x x + x 2 x - 1 + C

2

–1

2

–1

Ê tan x ˆ ÁË ˜ +C 2 ¯ Ê tan x ˆ ÁË ˜ +C 2 ¯

2 - sin 2 x Ê 1 ˆ sin x˜ - tan -1 +C (d) sin -1 Á Ë 2 ¯ sin x 20. If I = Ú

cot x - 3 cot 3x dx, then I equals 3 tan 3x - tan x

(a) x –

1 log 3

3 + tan x

(b) x +

1 log 3

3 + tan x

(c) x –

3 - tan x 2 log 3 + tan x 3

+C

(d) x +

3 - tan x 2 log 3 + tan x 3

+C

21. If I = Ú

3 - tan x 3 - tan x

+C

+C

x cos x dx

then I equals

( x - 1) sin 2 x + x sin 2 x 2

(a) sin–1 (2x sin x + cos x) + C (b) log | 2 x sin x + cos x +

x cos x - 1 | + C

(c) log |x sin x + 2 cos x +

x cos x + 1 | + C

(d) log |sin x + cos x| + 22. If I = Ú (a)

dx, then I equals

(a) sec–1 (xx) + C

) sec x + 1) + tan tan x + 2 ) – tan

(a) log tan x + sec2 x + 1 + C

(b) -

( x 2 - 1)sin 2 x + x sin 2 x

1 tan–1 x dx, then I equal 4 x Ê x 2 + 1ˆ 1 ˘ 1È tan–1 x + Ílog Á 2 ˜ - 2 ˙ + C 6 ÍÎ Ë x ¯ x ˙˚ 2 1 1 –1 –1 Ê x + 1ˆ x + tan tan Á x ˜ +C 6x 3x 3 Ë ¯

IIT JEE eBooks: www.crackjee.xyz 25.49

(c)

(d)

1 3x

cot–1 x –

3

1 1

23. If I = Ú (a) (b) -

(c) -

(d) -

Ê x 2 + 1ˆ 1 tan–1 Á ˜ +C 2x Ë x ¯

tan–1 x –

x3

x3

sin–1

2x 2 1

1

sin

2x 2 1

+

2x 2

–1

1- 1- x

+C

x

2 1 1- x x– +C 2 x

1 - x2 x

(c) tan–1 (d) sin

–1

( ( (

+C

x + 1+ x (1 + x )1/3

(b) (1 + x)

(c) (1 + x)2/3

dx then I equals (c) 3˘ +C 4 ˙˚

6 3 È3 2 ÍÎ 8 (1 + x ) - 5 x + 10 6 ˘ + 1+ x˙ + C 7 ˚ 7˘ È 2 1 ÍÎ x - 5 x + 8 ˙˚ + C

26. If I = Ú

x -1 1+ x

2

dx 1 + x2 + x4

1 + x2

)

15

dx, then I equals

1 + x2

(

(

(

x + x2 + 1

)

)

)

15

+C

15

+C

14

+C

(d) none of these 29. If I = Ú cos 2x log (1 + tan x) dx (a) (1/2) sin 2x log (1 + tan x) – (1/2)x + (1/2) log |sec x| + C (b) (1/2) (sin 2x + 1) log (1 + tan x) + (1/2)x + (1/2) log |sec x| + C (c) (1/2) (sin 2x + 1) log (1 + tan x) – (1/2)x – (1/2) log |sec x| + C (d) none of these 30. If I = Ú

1 + x2 x 1 + x4

, then I equals

1 Ê ˆ (a) tan–1 Á x 2 + 2 + 1˜ + C Ë ¯ x

1 - tan 2 x

+ (1/2) sin a sin –1 (tan x) + C (b) sin (tan x) cos a + tan a cos–1 (sec x – cosec x) + C (c) cos–1 (cos x – sin x) + tan–1 (sec x – cosec x) + C 1 (d) 1 - tan 2 x (–(cos a + sin a) + tan x) 2 1ˆ Ê + Á sin a + ˜ cos–1(tan x) + C Ë 2¯

(b) log x + 1 + x 2

6 3˘ È3 (d) (1 + x)2/3 Í (1 + x )2 - x + ˙ + C 5 4˚ Î8 2

cos x

–1

+C

6 È3 (a) (1 + x)2/3 Í (1 + x )2 - (1 + x ) + 8 5 Î 2/3

cosec x + sec x dx then I equals cosec x - sec x

3

(a) (1/15) x + x 2 + 1

2

25. If I = Ú

sin ( x + a )

(x + 28. If I = Ú

) 2 sin x ) + C 2 sin x ) + C

2 sin x + C

(2 sin x )

1 4 x + x2 + 1 + C x

(a) – (1/2) (tan x + 2) cos a

2

24. If I = Ú 1 + sec 2x dx, then I equals

(b) cos–1

(d) tan–1 27. If I = Ú

2 1 1- x x+ +C 2 x

sin–1 x + log

2

(a) sin –1

1 Ê ˆ (b) sec–1 Á x 2 + 2 + 1˜ + C Ë ¯ x 1 Ê ˆ (c) cosec–1 Á x 2 + 2 + 1˜ + C Ë ¯ x

sin–1 x dx, then I equals

1

2x

Ê x 2 + 1ˆ 1 tan–1 Á ˜ +C 6x Ë x ¯

dx

(a) log x -

1 1 + x2 + 2 x x

+C

(b) log x +

1 1 + x2 + 2 x x

+C

IIT JEE eBooks: www.crackjee.xyz 25.50 Comprehensive Mathematics—JEE Advanced

1ˆ Ê (c) sec–1 Á x - ˜ + C Ë x¯ (d) log x -

(a)

1 + x3

dx , then I equal

4 2 2 1 + x3 + C (1 + x3 )5/ 2 - (1 + x3 )3/ 2 + 15 9 3

4 8˘ 2 È1 1 + x3 Í x 6 + x3 + ˙ + C 3 15 15 ˚ Î6

(a) (b) (c) (d)

sec-1 x - cosec-1 x sec-1 x + cosec-1 x

4 ( x sec-1 p 4 ( x sec-1 p 4 ( x sec-1 p 4 (2 x sec-1 p

Ê x 2 - 1ˆ 1 sec–1 Á ˜ + 2 Ë 2x ¯

(c)

1 log 2

(d)

dx , then I equals 35. If I = Ú

(a)

x - x - 1) - x + C x + x - 1) - x + C

(b)

x + x - 1) - x + C

3 tan x

px e

+ 4x

pxe

Ê 1ˆ (a) Á ˜ log Ë 2¯

dx, then I equals

2

tan x

+4 -2

tan x

+4+2

pxe

(c) tan

(d) cos

34. If I = Ú

–1

Ê p x e tan x + 4 ˆ Á ˜ 2p Ë ¯

x

2

(1 - x ) 1 + x 4 4

È 1 (c) Í log Í2 2 Î

36. If I =

Ê 2x ˆ 2 sin–1 Á 2 ˜ Ë x + 1¯

x4 + 1 + 2 x

Ê x 2 + 1ˆ + sec–1 Á ˜ +C Ë x 2¯

x4 + 1 + 2 x x4 + 1 - 2 x 1 x4 + 1 ˘ ˙ +C tan -1 2 2 x ˙˚

dx, then I equals

dx, then I equals (upto to a constant)

x4 + 1 - 2 x x2 - 1

x4 + 1 - 2x x 4 +1 + 2 x x4 + 1 - 2 x

+

+

1 tan–1 2

x3 Ú (ax3 + b)7/3 dx then I equals

(b)

3 x3 +C 4b ( ax 3 + b) 4/3

(c) -

3 x4 +C 4b ( ax 3 + b) 4/3

2 x4 +C 3b ( ax 3 + b) 4/3

x4 + 1 ˘ ˙ 2 x ˚˙

1 x4 + 1 ˘ ˙ tan -1 2 2 x ˙˚

Ê 2 xˆ x2 + 1 2 + sin–1 Á 2 ˜ + x 3 2 Ë x + 1¯

1 x4 +C 4b ( ax 3 + b) 4/3

(d)

Ê x4 + 1ˆ + sin–1 Á ˜ ÁË x ˜¯

x4 + 1 - 2x

(a)

1/2

Ê p x e tan x + 4 ˆ Á ˜ +C ÁË p x e tan x ˜¯

1 - x4

1È 1 log 2 ÍÎ 2 2

+C

+C

1 + x4

1 log 2

(d) sec–1

Ê p x e tan x + 4 ˆ (b) sin–1 Á ˜ +C ÁË ˜¯ 8

Ê x 2 + 1ˆ sin–1 Á ˜ 2 2 Ë 2x ¯ 1

x4 + 1 - 2 x

È 1 Í 1 log 4 Í2 2 Î

x - x - 1) + x + C

x sec x + 1

–1

2x

+

2

33. If I = Ú

( x 2 - 1)2 + x 2

(b)

2 1 + x3 (3x 6 - 4 x3 + 8) + C (d) 45 32. If I = Ú

2 2

log

-

2 2 È1 ˘ (b) 1 + x3 Í (1 + x3 ) 2 + (1 + x3 ) + 1˙ + C 3 3 Î5 ˚ (c)

1

1 1 + x2 + 2 + 1 + C x x

x8

31. If I = Ú

(a)

IIT JEE eBooks: www.crackjee.xyz 25.51

x 5/6 - x -7/6

37. If I = Ú

x1/3 ( x 2 + x + 1)1/2 - x1/2 ( x 2 + x + 1)1/3

dx

then I equals 1 ˆ Ê (a) 2x log Á x + + 1˜ Ë x ¯ (b) tan

–1

1/6

Ê x + 1/ xˆ – tan–1 Á ˜ +C Ë 3 ¯

1 ˆ Ê x + 1/ xˆ Ê ÁË ˜¯ + log ÁË x + + 1˜¯ x 2

-3/2

Ê x - 1ˆ (c) log (x + 1/x + 7)1/6 – tan–1 Á ˜ Ë x ¯ 2

+C

38. If I = Ú (a)

1 - x4 x

(c) tan–1

39. Let I = Ú (a)

(b)

Ê x ˆ (b) sin–1 Á ˜ ÁË 1 - x 4 ˜¯

x

1- x x

x

(c)

(e x + 1) log(e x + 1) e +1 e log(1 + e x ) e +1 x

x

e (log(1 + e x )) 2

+C

log | t + t 2 - a 2 | +C ,

42. If I = Ú

x ( x + 1) x 2 + 2 x + 2

tan x dx = A log |cos x| + x – tan( x + p /3) 3 log |sin x + B cos x| + C Then (a) A > B (b) B > A (c) A = B (d) A – B = 1 dx (0 < a < p/2) and 41. If I = Ú sin x sin (2 x + a ) f (x) = log cot x + cot a +

(cot x )2 - cosec2 a

+C

1 + log x 2 – log 1 + 2 x 2 + x + 1 + C (c) log |x| – log |x + 1|

(b) log x +

+

2 log 1 + x 2 + 2 x + 2

+C

(

)

1 log x + log(x + 1) – log 1 + x 2 + x + 2 2 1 log( x + 2) + 2 x 2 + 2 x + 2 2 dx , then I equals ( x - 1) ( x + 2)5/4 3/4

Ê x - 1ˆ (c) 4 Á Ë x + 2 ˜¯

+C

, then I equals

(a) log |x + 1| – log 1 + x 2 + 2 x + 2

4 Ê x - 1ˆ (a) 3 ÁË x + 2 ˜¯

40. If Ú

then I equals

2sin a

1 t = cot x + cot a 2 dx

dx is equal to

+C

x

(d)

-1

(d)

43. If I = Ú

ex

x

2

+C

x

(c) 1/2 f(x) + C

(d)

ex

(1 + e )(log(1 + e )) 1 + e

log(1 + e x )

1 f(x) + C cos a

(d) none of these

4

log(1 + e x ) - e x ex

(b)

+C

dx, then I equals up to a constant

(1 - x 4 )3/2

1 f(x) + C sin a

-3/2

(d) 2t3 + 3t2 + 6t + 6 log |t – 1| + C, t = (x + 1/x + 1)6 1 + x4

(a)

5/4

+C

- 3/4

+C

4 Ê x - 1ˆ (b) 3 ÁË x + 2 ˜¯

1/4

Ê x - 1ˆ (d) 4 Á Ë x + 2 ˜¯

+C 1/4

+C

44. If I = Ú (x7m + x2m + xm) (2x6m + 7xm + 14)1/m dx then I equals 1 (a) (x7m + x2m + xm)(m + 1)/m + C m +1 (b)

2 (2x7m + 7x2m + 14)m+1 + C m +1

1 (2x7m + 7x2m + 14xm)(m+1)/m + C 14 ( m + 1) 1 (2x7m + 7x2m + 14xm)1/m + C (d) 14m cos 7 x - cos 8 x 45. If I = Ú dx, then I equals 1 + 2 cos 5 x (c)

IIT JEE eBooks: www.crackjee.xyz 25.52 Comprehensive Mathematics—JEE Advanced

(a)

1 1 sin 2x + sin 3x + C 2 3

(b)

1 1 sin 2x – sin 4x + C 2 4

(c)

1 1 sin 2x – sin 3x + C 2 3

1 1 [log (x + 1)]2 – (log x)2 2 2 + log (x + 1) log x + C 2 (b) – [{log (x + 1)} – (log x)2] + log (x + 1) log x + C (c) C – (1 2) [log (1 + 1/x)]2 (a) –

(d) none ot these.

1 1 (d) sin 3x + sin 4x + C 3 4

52. The antiderivative of

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

(a) f (x) = log ( 1 - x + 1 + x ) (c) M = – 48. If Ú

2 3 dx

(x + 4)( x 2 + 7) 2

= K tan–1

1 (b) L = – 2 1 (d) M = – 2 x x + L tan–1 + C, 2 7

then (a) K = – 1 / 7

(b) L = 1/6

(c) L = – 1/3 (d) K = – 1 / 3 7 4 cos x x 49. If Ú dx = K cot x + M sin 2x + L + C, then 2 2 sin x (a) L = 1 (c) M = – 50. If Ú

(b) K = – 2 1 4

xe x 1+ e

x

d x = f ( x ) 1 + e x –2 log g(x) + C, then (b) g(x) = x

(c) g(x) =

1+ e +1 1 + ex - 1

51. The value of the integral Ú

(d) C +

1 3

Ú

1 + ex - 1 1 + ex + 1

(d) f (x) = 2(x – 2) log ( x + 1) - log x dx is x ( x 1)

) ˘˙˚

(

) ˘˚˙

( 1 - 9x

2

)

3/2

is

3

2

(

)

3 + cos -1 3x ˘˙ ˚

(

- cos -1 3x

)

2

)

dx is equal to cos x sin3 x

(a) – (1/2) cot2 x + log |tan x| + C (b) – (1/2) cot2 x – log |cot x| + C 1 (c) + log |tan x| + C 2 tan 2 x (d) 54. If Ú

1 – log |cot x| + C 2 tan 2 x dx

( x - a ) (b - x)

= A sin–1 f (x) + C then

(a) A = 1 b +a 2 b -a 2

x(b) f (x) =

(d) none of these

(a) f (x) = x – 1

53.

(

1È 1 - 9x2 3 ÍÎ

2

(

1È 1 - 9 x 2 + cos -1 3x 9 ÍÎ 1 (b) C + ÈÍ 1 - 9 x 2 + cos -1 3x 9Î (c) C –

)

1 - 9x2

(a) C –

46. If Ú tan4 x dx = K tan3 x + L tan x + f (x), then 1 (a) K = (b) L = – 1 3 2 (c) f (x) = x + C (d) K = 3 47. If Ú log ( 1 - x + 1 + x ) dx = x f (x) + L x + M sin–1 x + C, then

(

x + cos -1 3x

(c) f (x) = ( x - a ) ( b - x ) (d) A = 1/2 55. If Ú cos7 x dx = A sin7 x + B sin5 x + C sin3 x + sin x + k then (a) A = 1/7 (b) A = – 1/7 (c) B = 1/5 (d) B = 3/5

MATRIX-MATCH TYPE QUESTIONS 56. Ú f (x) dx when

IIT JEE eBooks: www.crackjee.xyz 25.53

f x

1 e +1

f x

ee

ex

e–x

ex x e +1

C

x

x x

ex – 1

f x

1 e 4

e +1 x

fx

e

x

ee

x

Ú f x dx

C

1– x 1+ x

x

C

1

f x

a +x

–1

a

x

f x

1 a

C

–1

x 1 x

a

x x –a

ex

e

x

f x

C

x

x

x

x

e

x

x

fx

C

x x 4

x

x

x

x

p p

x n

p /a

ax

n

Ú

C

x

ˆ ˜ ¯

x

1

x

x

n

bx

n –1

Ú C

x

n n

n a n –1 b

C

– AÚ

n –1

dx ax + b

A n n

C

C

ax + b

dx ax + b

n

m

x

1

x

Ê xÁ Ë

x

n

x

Statement 1: F x p Statement 2:

C

x 4

f x

f x

1

x

n

Paragraph for Question Nos. 63 to 66

Ú f x dx f x

1+ x

x

x

x x

C,

COMPREHENSION-TYPE QUESTIONS

C

x x

C

x x

f x

u -1 u +1

Fx x x Statement 1: F Statement 2: Fx

C

x x

x +C

ASSERTION-REASON TYPE QUESTIONS

x –a

x

e

C

x –1

u

x

)

x

u

x f x

x

1+ x

x +a

a

x

x

C

Ú f x dx f x

(

1 x

x

C

1+ 1- x

x

x

1

fx

–1

ˆ ˜¯

x

a –x

C

a

a |x|

x a

a –x

f x

x+ ˆ Ê -Á ˜ Ë x- ¯

1– x 1 1+ x x



Ú f x dx

Ê + Á Ë

x+ x-

+C

x

1



n

x x A

n n m +1

dx

n

a b m

x

n –1

a b

x

A Ú

n

x dx x

IIT JEE eBooks: www.crackjee.xyz 25.54 Comprehensive Mathematics—JEE Advanced

(a)

m–n+2 n –1

m – n +1 (d) n –1

m–n (c) n–2 m

65. If Ú AÚ (a) (c)

cos x n

sin x

m–n+2 n–2

(b)

dx =

cos

m –1

x

( m – n) sin

n –1

x

+

cos m – 2 x dx + C then A is equal to sin n x m m+n

m –1 (b) m+n

m m + n –1

(d)

66. If I n = Ú eax sinn x dx =

m –1 m–n

ax

e

a +n 2

2

sinn–1 x

(a sin x – n cos x ) + A I n–2 then A is equal to (a) (c)

n( n – 1) a +n 2

(b)

2

( n – 1) ( n – 2) a +n 2

(d)

2

n –1 a 2 + n2 a 2 + n2

=

A 2 –1 x sin 8

2a – a

x Ô¸ ˝ dx Ô˛

where u =

70. If I = Ú (2 cos 2 x - 1)(4 cos3 x - 3 cos x)dx 1 A sin 5 x + sin x + C then A equal to 30 2 2 cos x - sin x + k 71. If Ú dx = A log |cos x + sin x – 2| cos x + sin x - 2 + Bx + C then A + B + |k| is equal to

A 1863

x – 1 then A is equal to.

cos2 x sin x dx sin x - cos x = A log |sin x – cos x| + B(sin2x + cos2x) + C then 4(A + 4B) is equal to

74. If I = Ú

x2 - 2 x2 + 2 ˆ ( x + 5 x + 4) tan Á ˜ Ë x ¯ 4

–1Ê

2

dx

= A log |tan–1 (x + 2/x)| + C then A equals. 75. If I = Ú (log x) sin–1 x dx A log = (sin–1x) x(log x – 1) + (log x – 2) 1 - x 2 + 16 1+ t + C, 1– t 1 - x 2 then A is equal to.

t= 76. If Ú

sin x dx = sin 4 x

1+ 2 u 1 1+ u ˘ A È 1 log – log Í ˙ + C, 36 ÎÍ 2 1 – u ˚˙ 1– 2 u 2 where u = sin x then A is equal to

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS

-

then k is equal to

)

x – 1 dx = (u2 + 1)2 tan–1 u –

73. Ú

ÏÔ 1 2 - x ¸Ô 1 1 ÏÔ t 5/ 2 + Ì + 4 a2t1/ 2 Ì ˝+ 2 a 2 a a 2 5 Ô˛ ÓÔ ÓÔ

4a 3/ 2 ¸ t ˝ + C , t = 2a + x then A is equal to. 3 ˛ A 3 1/ 3 68. If Ú ( x + (tan x)1/ 3 )dx = x 4 / 3 - log(1 + t 2 ) + 4 2 2 1 -1 2t - 1 1/3 tan + C t = tan x then A is equal to 3 3 x +1 k x -1 69. If Ú +C dx = 3 x2 + x + 1 ( x 2 + x + 1) x 2 + x + 1

(

u3 – u + C

n ( n – 2)

INTEGER-ANSWER TYPE QUESTIONS ÔÏ 1 67. If I = Ú x sin–1 Ì ÔÓ 2

72. If I = Ú tan –1

1. Let f (x) = Ú e x ( x - 1)( x - 2) dx then f decreases in the interval, (a) (–•, –2) (b) (–2, –1) (c) (1, 2) (d) (2, •) [2000] 2. Ú

(a)

=

(c)

x2 - 1 x3 2 x 4 - 2 x 2 + 1 2x4 - 2x2 + 1 x2 2x4 - 2x2 + 1 x

dx =

+c

(b)

+c

(d)

2x4 - 2x2 + 1 x3 2x4 - 2x2 + 1 2x2

+c

+c

[2006]

IIT JEE eBooks: www.crackjee.xyz 25.55

3. Let f (x) = g(x) = ( f

[2012]

x for n ≥ 2 and (1 + x n )1/ n

ASSERTION-REASON TYPE QUESTIONS

f ... f )( x )

F(x + p) = F(x) for all real x.

1

11 (a) (1 + nx n ) n + K n( n - 1)

Statement–2: sin2 (p + x) = sin2 x for all real x. [2007]

1

11 (1 + nx n ) n + K n -1

FILL

1

1+ 1 (c) (1 + nx n ) n + K n( n - 1) 1

4. Let I = Ú

e 4 x + e2 x + 1

[2007]

BLANKS TYPE QUESTIONS

4e x + 6e - x 9e x - 4e - x

dx

= Ax + B log (9e2x – 4) + C then A = _____, B = _____ and C = _____ [1990]

dx, J = Ú

e- x e- 4 x + e- 2 x + 1

dx.

Then for an arbitrary constant C, the value of J–I equals Ê e 4 x - e 2 x + 1ˆ log Á 4 x ˜ +C Ë e + ex + 1 ¯

(a)

IN THE

1. If I = Ú

1+ 1 (d) (1 + nx n ) n + K n +1

ex

SUBJECTIVE-TYPE QUESTIONS 1. I = Ú

sin x dx dx or I = Ú dx 1 - cot x sin x - cos x

2. I = Ú

x dx 1 + x4

[1979]

x2 dx ( a + bx )2

[1979]

(b)

Ê e 2 x + e x + 1ˆ 1 log Á 2 x ˜ +C 2 Ë e - e x + 1¯

3. I = Ú

(c)

Ê e 2 x - e x + 1ˆ 1 log Á 2 x ˜ +C 2 Ë e + e x + 1¯

Ê xˆ 4. I = Ú 1 + sin Á ˜ dx Ë 2¯

(d)

Ê e 4 x + e 2 x + 1ˆ 1 log Á 4 x ˜ +C 2 Ë e - e 2 x + 1¯

5. The integral Ú

sec2 x

(sec x + tan x )9/2

[2008]

5. I = Ú

dx equals (for some

7. I = Ú

(a) –

(b)

1

(sec x + tan x )11/2

(c) –

(d)

(sec x + tan x )

11/2

(sec x + tan x )11/2 1

(sec x + tan x )11/2

Ï1 1 2¸ Ì - (sec x + tan x ) ˝ + K Ó11 7 ˛

Ï1 1 2¸ Ì - (sec x + tan x ) ˝ + K 11 7 Ó ˛

1

x2 dx 1- x

Ï1 1 2¸ Ì + (sec x + tan x ) ˝ + K Ó11 7 ˛

Ï1 1 2¸ Ì + (sec x + tan x ) ˝ + K Ó11 7 ˛

8. I = Ú 9. I = Ú 10. I = Ú 11. I = Ú

( x - 1)

[1978]

[1980]

[1980]

6. I = Ú (e log x + sin x ) cos x dx

arbitrary constant K) 1

x.

Statement–1: The function F(x

Then Ú x n - 2 g(x)dx equals

(b)

2

1. Let F(x) b

f occur n times

[1981]

ex dx

[1983]

dx x ( x + 1)3/4

[1984]

( x + 1)3 2

4

1- x 1+ x

dx

sin - 1 x - cos - 1 x sin - 1 x + cos - 1 x cos 2 x sin x

dx

[1985]

dx

[1986]

[1987]

IIT JEE eBooks: www.crackjee.xyz 25.56 Comprehensive Mathematics—JEE Advanced

12. I = Ú

(

)

tan x + cot x dx

[1989]

Ê log (1 + x1/6 ) ˆ 1 13. I = Ú Á 1/3 + ˜ dx x1/3 + x ¯ Ë x + x1/ 4

[1992]

Ê cos q + sin q ˆ dq 14. I = Ú cos 2q log Á Ë cos q - sin q ˜¯

[1994]

x +1

15. I = Ú

x(1 + xe x )2

[1996]

( x 2 + 1)2 ( x + 1)

dx

[1999]

Ê ˆ 2x + 2 18. I = Ú sin - 1 Á ˜ dx ÁË 4 x 2 + 8 x + 13 ˜¯

(

)(

19. I = Ú x 3m + x 2 m + x m 2 x 2 m + 3x m + 6 m>0

[2001]

)

1/ m

[2002]

SINGLE CORRECT ANSWER TYPE QUESTIONS 2. 6. 10. 14. 18. 22. 26. 30. 34. 38. 42.

(a) (c) (b) (b) (d) (d) (c) (c) (a) (a) (d)

3. 7. 11. 15. 19. 23. 27. 31. 35. 39. 43.

(d) (a) (b) (b) (a) (b) (a) (b) (b) (a) (d)

4. 8. 12. 16. 20. 24. 28. 32. 36. 40. 44.

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 46. (a), (d) 48. (a), (b), (c)

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

63.

LEVEL 1

(b) (b) (c) (a) (a) (a) (a) (b) (b) (c) (a) (b)

51. 53. 55. 57. 59.

q

62.

dx,

Answers

1. 5. 9. 13. 17. 21. 25. 29. 33. 37. 41. 45.

(d) (d) (c) (b) (c), (d) (c), (d)

p

61.

[1997]

x 3 + 3x + 2

(b), (b), (b), (a), (a), (a),

(a), (a), (a), (b), (a),

(b) (c) (c) (c) (d)

MATRIX-MATCH TYPE QUESTIONS

dx

1 - x dx 16. I = Ú 1+ x x 17. I = Ú

50. 52. 54. 56. 58. 60.

47. (a), (c) 49. (b), (c)

(b) (a) (d) (c) (c) (a) (a) (b) (c) (d) (a)

64.

ASSERTION-REASON TYPE QUESTIONS 65. (d)

66. (a)

67. (d)

COMPREHENSION-TYPE QUESTIONS 68. (a) 72. (c)

69. (d) 73. (d)

70. (b) 74. (b)

71. (b) 75. (d)

IIT JEE eBooks: www.crackjee.xyz 25.57

INTEGER-ANSWER TYPE QUESTIONS 76. 0 80. 2 84. 1

77. 5 81. 2 85. 3

78. 2 82. 2 86. 6

p

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58.

79. 4 83. 3 87. 3

LEVEL 2

59.

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17. 21. 25. 29. 33. 37. 41. 45.

(b) (a) (a) (a) (a) (d) (b) (c) (a) (d) (d) (c)

2. 6. 10. 14. 18. 22. 26. 30. 34. 38. 42.

(a) (a) (d) (a) (a) (a) (d) (a) (d) (a) (d)

3. 7. 11. 15. 19. 23. 27. 31. 35. 39. 43.

(b) (b) (b) (c) (d) (c) (d) (a) (b) (b) (b)

4. 8. 12. 16. 20. 24. 28. 32. 36. 40. 44.

(d) (a) (b) (a) (a) (a) (a) (b) (a) (c) (c)

(a), (b), (b), (a) (a),

ASSERTION-REASON TYPE QUESTIONS 61. (d)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 46. 48. 50. 52. 54.

60.

(b), (c) (d) (d)

47. 49. 51. 53. 55.

(b)

(a), (a), (a), (c), (b),

(d) (c) (c) (d) (d)

62. (a)

COMPREHENSION-TYPE QUESTIONS 63. (b)

64. (a)

65. (d)

66. (a)

INTEGER-ANSWER TYPE QUESTIONS 67. 4 71. 8 75. 8

MATRIX-MATCH TYPE QUESTIONS

68. 3 72. 9 76. 9

69. 4 73. 3

70. 3 74. 1

PAST YEARS’ IIT QUESTIONS

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ASSERTION-REASON TYPE QUESTIONS

a

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1. (d)

b

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c

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p

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56.

57.

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. (c) 5. (c)

FILL

2. (d)

IN THE

3. (a)

4. (b)

BLANKS TYPE QUESTIONS

1. A = –3/2, B = 35/36, C = an arbitrary constant

IIT JEE eBooks: www.crackjee.xyz 25.58 Comprehensive Mathematics—JEE Advanced

SUBJECTIVE TYPE QUESTIONS 1 1. x + log |sin x – cos x| + C 2

17.

1 2. tan–1 (x2) + C 2 È a2 ˘ + bx 2 a log | a bx | Í ˙+C a + bx ˚ Î 4. 4[sin (x/4) – cos (x/4)] + C

3.

16. 2 log

1 b3

1

( x + 1)

2

10.

11.

(

x -2

)

19.

LEVEL 1

1. I = Ú e x ln(sin x)dx + Ú e x cosec2 xdx = ex ln (sin x) – Ú e x

2 cot x - cot 2 x - 1

2.

2 sin–1(sin x – cos x) + C

– 4x1/4 + 6x1/6 – 12x1/12 + 12 log (x1/12 + 1) + [2(x1/6 + 1)3 – 9(x1/6 + 1)2 + 18(x1/6 + 1)] log (1 + x1/6) – [

2 1/6 9 (x + 1)3 – 3 2

(x1/6 + 1)2 + 18(x1/6 + 1)] – 3[log (x1/6 + 1]2 + C

+ xe x 1 + +C x 1 + xe 1 + xe x

cosec x - cot x 1 - cos x x = = tan 2 2 cosec x + cot x 1 + cos x \ I = Ú tan( x / 2)

(1 + tan( x / 2))

1 - tan 2 ( x / 2)

1 - tan 2 ( x / 2) 3 + 2 tan 2 ( x / 2)

Put tan (x/2) = t t I = 2Ú 1- t2

dt 3 + 2t 2

1 - t 2 = u fi – t dt = u du u du du = -2Ú \ I = -2Ú 5 - 2u 2 u 3 + 2(1 - u 2 )

Put

Ê 2u ˆ = - 2 sin -1 Á ˜ +C Ë 5¯ Ê 2 Ê xˆ ˆ = - 2 sin -1 Á 1 - tan 2 Á ˜ ˜ + C Ë 2¯ ¯ Ë 5

Ê cos q + sin q ˆ 1 sin (2q) log Á 2 Ë cos q - sin q ˜¯

15. log

+ Ú e x (-cosec2 x)dx + Ú e x cosec 2 xdx = ex (ln (sin x) – cot x) + C

2 cot x + cot 2 x - 1

3 2/3 12 7/12 12 5/12 13. x – x +2 x – x + 3x1/3 2 7 5

14.

cos x dx + Ú e x cosec2 xdx sin x = ex ln (sin x) – ex cot x

1 - x - sin - 1 x + C

cot x + cot 2 x - 1 1 – log +C 2 cot x - cot 2 x - 1 12.

1 [2x3m + 3x2m + 6xm)1+1/m + C 6( m + 1)

Hints and Solutions

4Ê 1ˆ 2 -1 x+ x(1 - x ) - x + C Á x - ˜¯ sin pË 2 p 1 log 2

x +C

( x + 1)2 1 3 x log 2 + tan–1 x + 2 +C 4 2 x +1 x +1

ex + C

8. – (1 + x–4)1/4 + C 9.

– 2 sin–1

Ê 2x + 2ˆ 3 – log (4x2 + 8x + 13) + C 18. (x + 1) tan–1 Á Ë 3 ˜¯ 4

2 1 3/2 5/2 ˘ È 5. –2 Í 1 - x - (1 - x ) + (1 - x ) ˙ + C 3 5 Î ˚ 1 6. x sin x + cos x – cos 2x + C. 4 7.

1- 1- x x

1 log |cos 2q| + C 2

3. I = Ú

2 - (1 + cos 2 x) 1 + cos 2 x

where I1 = Ú

dx = 2I1 – x

dx 1 + cos 2 x

= Ú

sec2 x

dx sec2 x + 1

dx

IIT JEE eBooks: www.crackjee.xyz 25.59

= Ú

tan x + 2 2

tan x ˆ tan Á ˜ - x+C Ë 2 2 ¯

x5 x6 x6 + 1

\ I=

=

5. Let t = dt =

dx

and 3 x5

x6 + 1 + 1

= ax (x ln x) – I1 – Ú a x dx fi I1 + I2 = a x ( x ln x) Also,

+C

5+ x 10 = –1 5- x 5- x (5 - x)

2

\ 9. I = Ú

dx

1/ 4

+C

= Ú

4 x5 + 5 x 4 x10 (1 + x -4 + x -5 ) 2 4 x -5 + 5 x -6 (1 + x -4 + x -5 ) 2

dx

Put 1 + x–4 + x–5 = t, so that (–4x–5 – 5x–6) dx = dt dt 1 \ I = -Ú 2 = + C t t = 7. I = Ú

1 1 + x -4 + x -5

=

x - (4 - x) 4x - x

2

dx = Ú

2x - 4 4 x - x2

dx

È ln( x + b) ˘ Thus, I = ln Í ˙+C Î ln( x + a) ˚

x5 + x + 1

(1 + x 2 )[1 + 1 + x 2 ]

1 ˘ È Ê xˆ +C I = a x Í x ln Á ˜ + 1 ¯ Ë a a ˙˚ ln Î

= ln (ln (x + t))

x5

x2

I3 = ax (x ln a) – Ú a x (1) ln a dx = ax (x ln a) – ax.

= -2 4 x - x 2 + C 10. f(x) = (x + a) ln (x + a) – (x + b) ln (x + b) \ I = I(b) – I(a) Where dx I(t) = Ú ( x + t ) ln( x + t )

6. Write I= Ú

ax ln a

Put t = 4x – x2, dt = – (2x – 4) dx dt = -2 t + C \ I = -Ú t

\ I = Ú t -3/ 4 dt = 4t1/4 + C Ê 5 + xˆ = 4Á Ë 5 - x ˜¯

I3 = Ú a x (ln a)( x ln a) dx I2 = ax (x ln x) – Ú a x (1 + ln x) dx

dx = dt

x6 + 1

x6 + 1 - 1

10

8. Write I = I1 + I2 – I3 where I1 = Ú a x ln x dx ,

1 dt 1 t -1 +C = ln Ú 2 3 t -1 6 t +1 1 ln 6

)

(

= ln x + x 2 + 1 - tan -1 x + C

I2 = Ú a x (ln a)( x ln x) dx

x6 + 1 = t fi

Put

Ê tan x ˆ +C tan -1 Á Ë 2 ˜¯ 2

1

dx =

-1 Ê

1

\ I= 4. I = Ú

sec2 x

dx

x 2 ÈÍ 1 + x 2 - 1˘ ˙˚ Î = Ú dx 2 2 (1 + x )(1 + x - 1) Ê 1 1 ˆ = ÚÁ dx 2˜ Ë 1 + x2 1 + x ¯

+C

1 + x + x2 - 2 1 + x + x2 + 1 dx x2 1 + x + x2 = I1 – I2 + I3, where 2+ x 1 dx I1 = Ú 2 dx , I2 = 2Ú 2 , 2 x x 1+ x + x

11. I = Ú

I3 = Ú Now, I1 = Ú

dx 1 + x + x2 2 x -3 + x -2 x -2 + x -1 + 1

dx

IIT JEE eBooks: www.crackjee.xyz 25.60 Comprehensive Mathematics—JEE Advanced

= -2 x -2 + x -1 + 1 2 = - 1 + x + x2 x 2 I2 = x dx I3 = Ú 2 2 Ê 3ˆ 1ˆ Ê ÁË x + ˜¯ + Á ˜ 2 Ë 2 ¯

and

È ˘ 1 14. I = Ú Í1 + ˙ dx Î tan ( x - a ) tan ( x + a ) ˚ È (1 + tan x tan a ) (1 - tan x tan a ˘ = Ú Í1 + ˙ dx (tan x - tan a ) (tan x + tan a ) ˚ Î È tan 2 x - tan 2 a + 1 - tan 2 x tan 2 a ˘ = ÚÍ ˙ dx tan 2 x - tan 2 a Î ˚

1 Ê ˆ = ln Á x + + x 2 + x + 1˜ Ë ¯ 2 \

I=

2 1 ˆ Ê 1 - 1 + x + x 2 + ln Á x + + x 2 + x + 1˜ + C ¯ Ë x 2

12. We can write I = I1 + I2 where I1 = Ú

log (1 + x1/4 ) dx and I = dx Ú 2 x1/2 - x1/4 x1/2 + x1/4 3

\

4t dt t -t 2

I1 = 4 Ú

15. I = -

and I2 = Ú

log (1 + t ) t2 + t

4t3 dt

= 2 x + 4x1/4 + log |x1/4 – 1| log (t + 1) 2 t dt I2 = 4 Ú t +1 Put t + 1 = u log u (u – 1)2 du I2 = 4 Ú u 1ˆ ˜ log u du u¯

– 10x

1/4

+ 7) + 2 (log (x

sec4 x (tan x )11/2

dx = Ú

1 + t2 t11/2

dt

=–

1 1 1 tan–1 (t) + Ú 2 dt t t t +1

=–

È1 1 t ˘ tan–1 (t) + Ú Í - 2 ˙ dt t Î t t + 1˚

(

18. Put log x + x 2 + a2

1/4

+ 1))

2

3 11 = – 4 is a negative even integer, we put 2 2 tan x = t. I= Ú

1 1 2x log (x2 + a2) + Ú dx x x 2 + a2 x

1 log (e2x + 1) + C 2

17. I = Ú sec2 x log cos x dx = tan x log cos x – Ú tan x (– tan x) dx = tan x log cos x + tan x – x + C

= 2(x1/2 – 6x1/4 + 3) log (x1/4 + 1) 1/2

tan x - tan a +C tan x + tan a

= – e–x tan–1 (ex) + x –

= (2u2 – 8u) log u – (u2 – 8u) + 2 (log u)2

13.

1 - tan 2 a tan x - tan a log +C 2 tan a tan x + tan a

1 2 Ê xˆ log (x2 + a2) + tan–1 Á ˜ + C Ë a¯ x a x x 16. Put e = t, e dx = dt 1 I = Ú 2 tan–1 (t) dt t

= 2t2 + 4t + log |t – 1|

Ê = 4 ÚÁu - 2 + Ë

dx

= -

Ê t2 1 ˆ dt = 4 Ú Á t + 1 + dt t -1 t 1˜¯ Ë

– (x

=

tan 2 x - tan 2 a

= cot 2a log

Put x1/4 = t or x = t4, so that I1 = Ú

(1 - tan 2 a ) sec2 x

= Ú

)

(

2 -5/2 2 -7/2 - t +C t 5 7

= -

) =t

1 2 t +C 2 19. Put x + 2 = t, so that I = Ú t dt =

I = Ú et - 2

(t - 2)2 t2

dt

4 4ˆ Ê = Ú et - 2 Á1 - + 2 ˜ dt Ë t t ¯

IIT JEE eBooks: www.crackjee.xyz 25.61

4e t - 2 +C t

= et–2 – = 20. I = Ú

x-2 x e +C x+2

x2 + 1 + 2x x2 + 1

27. Put ex + 2 = t, I = Ú Now, 1

dx

t (t - 2) 3

= Ú x 2 + 1 dx + 2 Ú =

x x +1 2

dx

(

\

1 1 x x 2 + 1 + log x + x 2 + 1 2 2

)

1 (x + 4) 2

x2 + 1 +

(

)

=

I=

28. I = Ú

+ 2 x2 + 1 + C =

1 1 dt = t+c Ú a-b a-b

I=

t-2 1 1 1 + + 2 +C log t 4t 4 t 8

tan x (1 - 3 tan 2 x )

tan x (3 - tan 2 x ) Put t = tan x, so that

1 log x + x 2 + 1 + C 2

I= Ú

x ( x - a) ( x - b) =1+ \

(1 + tan 2 q /2) qˆ Ê dq 22. I = sin q log Á tan ˜ - Ú sin q Ë 2¯ 2 tan (q /2)

29. I = Ú

24. I = – Ú

cot x dx log sin x

dt = – log |t| + C t = – log |log sin x| + C 25. Put sin x + cos x = t, sin 2x = t2 – 1 dt = – log t + t 2 - 1 + C \ I = –Ú 2 t -1 Put log sin x = t, I = – Ú

= log sin x + cos x – 26. I = Ú

sin x cos x a sin 2 x + b cos2 x

dx

Put a sin2 x + b cos2 x = t2 fi (a – b) sin x cos x dx = t dt

sin 2x + C

(3 - t 2 ) (1 + t 2 )

1 + a2 / x 2 x 2 + a 4 / x 2 - a2

dt

1 log 3

3+t 3-t

+C

dx

Put x – a2/x = t, so that I= Ú

23. Put x4 + 1 = t2, so that t dt t -1 1 1 +C = log Ú 2 (t 2 - 1) t 4 t +1

1 - 3t 2

= tan–1 t –

qˆ Ê = sin q log Á tan ˜ – q + C Ë 2¯

I=

dx

Ê 1 2 ˆ = ÚÁ 2 ˜ Ë t + 1 3 - t2 ¯

1 Ê a2 b2 ˆ a - b ÁË x - a x - b ˜¯

1 I =x+ {a2 log |x – a| – b2 log |x – b|} + C a-b

t (t - 2)

1 1 1 1 - - 2 - 3 8 (t - 2) 8t 4t 2t

2

21.

dt 3

=

dt 1 Êtˆ = tan–1 Á ˜ + C 2 Ë a¯ a t +a 2

Ê x 2 - a2 ˆ 1 tan–1 Á ˜ +C a Ë ax ¯

30. cos4 x – cos2 x sin2 x + sin4 x = cos2 2x + (1/4) (1 + cos 2x) (1 – cos 2x) = (1/4) [4 cos2 2x + 1 – cos2 2x] = (1/4) (1 + 3 cos2 2x) Thus, 4 sec2 2 x 4 dx = dx I= Ú Ú 2 1 + 3 cos2 2x sec 2 x + 3 Put tan 2x = t dt Êtˆ I = 2Ú 2 = tan–1 Á ˜ + C Ë 2¯ t +4 31. Write I = Ú

dx x (2 - x ) 3

3

that I= Ú

4 sin q cos q dq 8 sin3 q cos3 q

and put x = 2 sin2 q, so

IIT JEE eBooks: www.crackjee.xyz 25.62 Comprehensive Mathematics—JEE Advanced

= 2Ú =



tan x tan 2x tan 3x = tan 3x – tan 2x – tan x 1 log |cos 2x| Thus I = log |cos x| + 2 1 – log |cos 3x| + C 3 sec4 x 38. I = Ú dx (2 tan x + 1 + tan 2 x ) 4 Put tan x + 1 = t, so that

dq = – cot 2q + C sin 2 2q

2 sin 2 q - 1 +C 2 sin q cos q x -1

=

+C

x (2 - x )

32. Put x = sin2 q, 2 sin q cos q dq I= Ú (1 + sin q ) sin q cos q = 2 (tan q – sec q) + C = 2 33. I = Ú 34. I = Ú

I= Ú

(t - 1)2 + 1 8

t 39. Put x + 4 = t, so that

a -x 2

2

2

(

)

Ê t - 4t + 4 ˆ = e–4 Ú et Á ˜¯ dt Ë t2

Ê xˆ dx = 2a sin–1 Á ˜ + C Ë a¯

4 ˘ È = e–4 Íet - et ˙ + C t ˚ Î x = ex + C x+4

dx x5 1 + 1 / x 2

Put 1 + 1/x2 = t2, so that (– 2/x3) dx = 2t dt and t -1 I=– Ú t dt t x2 + 1

=

x

40. I = Ú

I = –Ú

35. Put x – a = q, so that sin (q + 2a ) + cos (q + a ) dq I= Ú sin q = Ú [(sin 2a + cos a) cot q + (cos 2a – sin a)] dq = (sin 2a + cos a) log |sin q| + (cos 2a – sin a)q + C 36. Put t = x + 1, so that

=

1 tÏ Ê 1 2 ˆ¸ Ú e Ì1 - 3 ÁË 2 - 3 ˜¯ ˝ dt e t t ˛ Ó 1 e

È t 3e ˘ Íe - 2 ˙ + C t ˚ Î t

Ê x2 + 2x - 2ˆ = ex Á ˜ +C Ë ( x + 1)2 ¯ 37. We have tan 3x =

tan x + tan 2 x 1 - tan x tan 2 x

t3 dt (t 4 - 1) t

=

1 Ê 1 1 ˆ + 2 ˜ dt Ú Á 2 2 Ë t + 1 t - 1¯

=

1- t 1 1 tan–1 t – log +C 2 4 1+ t

41. Put 1 +

x = tan q, so that

I = Ú q 2(tan q – 1) sec2 q dq = (tan q – 1)2 q – Ú (sec2 q – 2 tan q) dq

ÏÔ (t - 1)3 + 3 (t - 1)2 + 4 ¸Ô I = Ú et - 1 Ì ˝ dt t3 ÓÔ ˛Ô =

dx x -5 = Ú -4 -4 1/4 dx x ( x -4 + 1)1/4 x ( x + 1)

Put x–4 + 1 = t 4,

1Ê 1 ˆ - 1˜ + C Á ¯ 2 Ë x2

-

1 1 2 + 6 - 7 +C 5 5t 3t 7t

Ê t - 2ˆ dt I = Ú et–4 Á Ë t ˜¯

2 x -1 (sin q - 1) +C= +C cos q 1- x

a+x+a-x

dt = –

( ) - (1 + x ) + log ( x + 2

= x2 tan–1 1 + x

42. I = 2I1 + I2 where I1 = Ú and

I2 = Ú

( x + 1) dx È( x + 1) + 2˘ ( x + 1)2 + 3 Î ˚ dx 2

È( x + 1)2 + 2˘ ( x + 1)2 + 3 Î ˚ 2 2 In I1, put (x + 1) + 3 = t , so that I1 = Ú

t -1 tdt 1 = log 2 t +1 (t - 1) t 2

)

x +2 +C

IIT JEE eBooks: www.crackjee.xyz 25.63

=

1 log 2

x2 + 2x + 4 + 1

44. I = Ú

In I2 put x + 1 = 1/t, so that ( -1 / t ) t dt 2

I2 = Ú

3

(1 + 2t 2 ) 1 + 3t 2

2

a Ê a - xˆ = (x – a) tan–1 Á + log (2a2 – 2ax + x2) Ë a ˜¯ 2

x2 + 2x + 4 - 1

sec2 x

Put tan x = t, so that I= Ú

2

Put 1 + 3t = u , so that u du I2 = – Ú (2u 2 + 1) u =–

= –

(

1 tan–1 2

Ê 2 ( x 2 + 2 x + 4) ˆ Á ˜ ÁË ˜¯ x +1

a - ax + x 2

43. Since

a

=

1 tan–1 2

2u

2

2

=1–

)

Ê a2 - ax + x 2 ˆ cot–1 Á ˜ a2 Ë ¯

= tan

–1

È x a-x ˘ + Í ˙ a Í a ˙ Í1 + x Ê a - x ˆ ˙ Á ˜ ÍÎ a Ë a ¯ ˙˚

Ê a - xˆ Ê xˆ = tan–1 Á ˜ + tan–1 Á Ë a¯ Ë a ˜¯ \ I = I1 + I2 where

Ê xˆ I1 = Ú tan–1 Á ˜ dx Ë a¯ xa Ê xˆ dx = x tan–1 Á ˜ - Ú 2 Ë a¯ x + a2 Ê xˆ a = x tan–1 Á ˜ - log (x2 + a2) Ë a¯ 2

In I2,

put

where

a-x = t, so that a

t -1

=

t -1 ˆ 1 Ê 1 dt - 2 Ú Á 3 Ë t - 1 t + t + 1¯˜

t -1 (t + 1/2)2 + 3/4

dt

Put t + 1/2 = u \

I1 = Ú =

45. I = 2 2 Ú

u - 3/2 u 2 + 3/4

du

1 3 2 Ê 2u ˆ log (u2 + 3/4) – ◊ tan–1 Á ˜ Ë 3¯ 2 3 2

(cos x )3/ 2 (sin x )7/ 2

dx

= 2 2 Ú (cot x)3/2 cosec2 x dx cot x = t, then

Put

I = –2 2 Ú

t3/2 dt

4 2 (cot x)5/2 + C 5 46. We can rewrite the integrand as follows: =–

1 1Ê 1 1 ˆ = Á 2 – 2 2 3 Ë x + 1 x + 4 ˜¯ ( x + 1) ( x + 4) 2

The integral can now be easily evaluated to give x 1 1 tan–1 x – tan–1 + C. 2 3 6 Therefore K = 1/3 and L = – 1/6. 47. The given integral equals

Ú

(1 – sin 2 x )2 2

sin x

dx = Ú (cosec x – sin x)2 dx

= Ú (cosec 2 x + sin2 x – 2) dx

I2 = – a Ú tan–1 t dt 1 È ˘ = – a Ít tan -1 t - log (t 2 + 1)˙ 2 Î ˚

dt 3

1 1 log |t – 1| – I1 3 3

I1 = Ú

x Ê a - xˆ Á ˜ aË a ¯

Ê 1ˆ and cot–1 (x) = tan–1 Á ˜ , Ë x¯ We get

dx

tan3 x - 1

1 – cos 2 x Ê ˆ – 2˜ d x = Ú Á cos ec2 x + Ë ¯ 2 = – cot x +

sin 2 x ˆ 1Ê Áx – ˜ – 2x + C 2Ë 2 ¯

IIT JEE eBooks: www.crackjee.xyz 25.64 Comprehensive Mathematics—JEE Advanced

1 3 sin 2x – x + C, 4 2 So that K = – 1, M = – 1/4 and L = – 3. 48. Since x 3 – 2x – 4 = (x – 2) (x2 + 2x + 2) and the integrand is a proper rational function, we use the method of partial fractions to get Bx + C 3x + 4 A = + 2 3 x – 2 x + 2x + 2 x – 2x – 4

3x + 4

x +1 1 dx – Ú 2 dx x–2 x + 2x + 2 = log | x – 2| – (1/2) log |x2 + 2x + 2| + C

dx = Ú

3

x – 2x – 4

Hence K = – 1/2, and f (x) = |x 2 + 2x + 2 | = x 2 + 2x + 2, because x2 + 2 x + 2 > 0. 49. Since the integrand is a rational function of cos x, we put tan (x/2) = t. Then dx = Ú I= Ú 5 + 4 cos x

2 dt Ê 4(1 – t 2 ) ˆ (1 + t 2 ) Á 5 + 1 + t 2 ˜¯ Ë

= 2 Ú

dt 5(1 + t 2 ) + 4(1 – t 2 )

= 2 Ú

t dt 2 = tan–1 +C 3 3 t +9 2

xˆ 2 Ê1 tan –1 Á tan ˜ + C Ë3 2¯ 3 Hence K = 2/3 and M = 1/3. =

50. Putting x3/2 = t, we have (3/2) 1/2

So Ú

x

1 – x3

Thus

dx =

x dx = dt

2 dt 2 = sin–1 t + C Ú 3 1 – t2 3

= (2/3) sin–1 x 3/2 + C. f (x) = x3/2 and g(x) = sin–1 x.

51. Put tan x = t3, so that I= 3 Ú

t3 3 y dt = dy (y = t 2 ) Ú 6 2 1 + y3 1+ t

1 (I1 – I2) where 2 y +1 dy I1 = Ú 2 dy and I2 = Ú y +1 y - y +1 =

3

52. Since the integrand is odd w.r.t. sine we put cos x = t so cos 2x = 2t2 – 1 and dt = – sin x d x. Thus t2 – 2 1 2t 2 – 1 – 3 I= Ú 2 dt = Ú dt 2 2t – 1 2t 2 – 1

fi 3x + 4 = A(x2 + 2x + 2) + (Bx + C) (x – 2) Putting x = 2, A of x 2, we get 0 = A + B, i.e., B = – 1. Putting x = 0, we have 4 = 2 A – 2C, i.e., C = – 1. Thus

Ú

2y -1

I1 = log(y2 – y + 1) 2 3 tan -1

= – cot x –

=

1 3 dt dt – Ú 2 Ú 2 2 2t – 1

=

1 3 1 t– ¥ log 2 2 2 2

=

1 3 cos x – log 2 4 2 A = 1/2, B = –

So,

2 t –1 2 t +1

3 4 2

2 cos x – 1 2 cos x + 1

3 4 2

, f (x) =

2 cos x + 1 2 cos x – 1

+C

, 2 cos x – 1

f (x) =

B=

+C

2 cos x + 1

or

.

53. Put t = 1 + (1/x). Then dt = – dx/x 2, and the integral becomes 1ˆ Ê log Á1 + ˜ Ë log t 1 x¯ 2 Ú Ê 1 ˆ dx = – Ú t dt = – 2 (log t) + C x 2 Á1 + ˜ Ë x¯ 2 1È Ê 1ˆ ˘ = – Ílog Á1 + ˜ ˙ + C 2Î Ë x¯˚ 1 1 [log (x + 1)]2 – (log x)2 =– 2 2 + log (x + 1) log x + C. 54. Making the substitution 1 + e x = t 2, we have e x dx = 2t dt and x = log (t 2 – 1). Then the given integral becomes È log(t 2 – 1) ˘ 2 ˙ 2t d t = 2 Ú log (t – 1)dt ÚÍ t Î ˚ È ˘ t2 dt ˙ + C = 2 Ít log (t 2 – 1) – 2 Ú 2 t –1 ˚ Î È Ê t – 1ˆ ˘ 1 =2 Ít log (t 2 – 1) – 2 Á t + log ˙ +C 2 t + 1¯˜ ˚ Ë Î = 2x

1 + ex – 4 1 + ex

IIT JEE eBooks: www.crackjee.xyz 25.65

1 + ex – 1

– 2 log 55. Ú e au sin bu du =

1 + ex + 1

So the primitive is equal to Ú

+ C.

e au (a sin bu – b cos bu) + C a2 + b2

(t = e x, u = e2x) = sin –1 t –

d dx

Ê 3 3 ˆ tan5 x (5 tan 2 x + 11)˜ ÁË ¯ 55

=

3 d (5 tan11/3 x + 11 tan5/3 x) sec2 x 55 dx

=

3 Ê 55 8/5 55 2/3 ˆ 2 Á tan x + tan x˜¯ sec x 55 Ë 3 3

= (tan 8/3 x + tan 2/3 x ) sec 2 x = tan 2/3 x (1 + tan2 x ) 2 =

3

sin 2 x cos –14 x .

57. The integrand can be written as x2 + x + 4 +

4( x 2 + 4 x – 2) 2

x( x – 4)

60.

Hence the primitive of the given function is x3 x 2 + + 4x + 2 log | x| + 3 2 5 log |x – 2| – 3 log | x + 2| + C x3 x 2 + + 4x + log |x2 (x – 2)5 (x + 2) –3 | + C = 3 2 ex e2 x + 58. The integrand can be written as 1 – e2 x 1 – e2 x

1 1 log |cosec 3x – cot 3x| + cos 3x + C 3 3

=

1Ê 3x ˆ + cos 3x˜ + C log tan ¯ 3 ÁË 2 3x = cosec 3x – cot 3x, 2 g(x) = cos 3x.

1 1 1 È1 1 ˘ = 4 2 = 2Í 2 – 2 ˙ 4 x +x x ( x + 1) x Îx x + 1˚ 1 1 1 = 4 – 2 + 2 x x x +1 Hence the primitive of

1 1 –3 1 is – x + 4 3 x x +x 6

tan–1 x + C Thus f (x) = tan –1 x, g(x) = – 1/3 which is a constant function. dx =6 1/2 x + x1/3

t3 6 Ú t + 1 d t (x = t )

È 1 ˘ = 6 Ú Ít 2 – t + 1 – ˙ dt = t + 1˚ Î

x( x 2 – 4)

2 5 3 = x2 + x + 4 + + – x x–2 x+2

=

6

4( x – 4 + 4 x + 2)

4 16 8 = x2 + x + 4 + + 2 + 2 x x –4 x ( x – 4)

1 – e2 x + C

1 – u + C = sin –1 e x –

f (x) = tan

Thus

2

+

1 – t2

1 du Ú 2 1– u

Ú (cosec 3x – sin 3x) dx

61. Ú

= x2 + x + 4

+

Hence f (x) = sin–1 x, g (x) = e x and h (x) = 1 – e 2x.

If u = x22, then du = 2xdx, a = – 5 and b = 4, –5x 2 Ú x e sin 4x dx 1 –5u 1 e 5u = e sin 4u du = 2 2 25 + 16 (– 5 sin 4u – 4 cos 4u) + C 1 =– (5 sin 4x2 + 4 cos 4x2 ) 2 82 e –5x + C. 1 Thus K = – , A = 5 and B = 4. 82 56. f (x) =

dt

Ê t3 t2 ˆ 6 Ú Á – + t – log | t + 1|˜ + C Ë3 2 ¯ = 2x1/2 – 3x1/3 + 6x1/6 – 6 log (x1/6 + 1) + C

Ú

dx t2 3 2t d t =3 Ú 3 dt (t3 = x) = Ú 2 1/3 2 x+x t +t t +1

3 3 log (t 2 + 1) + C = log (1 + x 2/3) + C. 2 2 In (iii), put x = t 10 and in (iv) put x = t12. =

62. Factorize the denominator and use integration of rational functions. 63. Since

x 3 – 2x2 + x = x (x – 1)2 and

2 x 2 – 3x + 3 x( x – 1)2

2 x 2 – 3x + 3 3 2 1 – so dx = + Ú x ( x – 1)2 x – 1 x( x – 1)2

=

IIT JEE eBooks: www.crackjee.xyz 25.66 Comprehensive Mathematics—JEE Advanced

3 log |x | – Also

Thus

2 – log |x – 1| + C. x –1

x3 + 1 x( x – 1)

Ú

3

x3 + 1 x( x – 1)3

2 1 1 2 – + + 2 x – 1 x ( x – 1) ( x – 1)3

=

dx =

2 log |x – 1| – log |x| –

1 1 +C – x – 1 ( x – 1)2

= 2 log | x – 1| – log | x| –

x +C ( x – 1)2

F(x) = log sin x Рsin x + C, x Π(0, p /2] which is one-one 1 , F(0) = 0 fi C = 1 67. F(x) = C Р1 + tan x \

–6

64. If I = Ú cos x sin x dx, put sin x = t so that I= Ú =

1 – sin 2 x sin6 x

cos x dx = Ú (t–6 – t–4) dt

1 1 cosec 5 x + cosec3 x + C 5 3

and Ú sin2 x cos–8 x dx = Ú t2 (1 + t2)2 dt (t = tan x) tan3 x tan 7 x 2 tan5 x + + +C 3 7 5 Put tan x = t in (iii) and put sec x = t in (iv) 1 65. F(x) = Ú sin 2 x dx = Ú (1 - cos 2 x ) dx 2 1 (2x – sin 2x) + K = 4 =

1 F(x + p) – F(x) = [2 (x + p) – sin (2x + 2p) – 2x 4 + sin 2x)] 1 = p π0 2 Thus the statement 1 is false. The statement 2 is true, since sin2 (x + p) = (– sin x)2 = sin2 x 66. F(x) = Ú

cos x (1 - sin 2 x ) (1 - t ) =Ú dt (t = sin x) sin x (1 + sin x ) t = log | sin x| – sin x + C

F(x) = F(p /4) = 1/2.

68. Put (2 x – 3)1/6 = t x+2 =t 69. Put x –1 70. Put

x +1 = t3 1– x

71. Put

x +1 = t2 1– x

Use partial fraction in the remaining two cases. 3

lim

x Æ p /4

sin5 x = (1 – cos2 x)2 sin x tan6 x = (sec2 x – 1) tan4 x sec6 x = sec4 x sec2 x, integrate by parts cosecn x = cosecn – 2 x (cot2 x + 1), integrate by parts F (x) = 3 sin x + 2 cos x – 2 1Ê 1 x ˆ 77. F (x) = Á log 5 tan + 3 ˜ ¯ 5Ë 3 2

72. 73. 74. 75. 76.

78. F(x) = – 4 x + 3x 79. Put e 2 x = t 1 dx, integrate by parts 80. tan–1 x = Ú 1 + x2 81. Determine the polynomial f(x), f(x) = – x2 + x + 3 82. Put sin –1 x = q so that cos–1 x = p/2 – q and integrate by parts. 83. Put sin x = t, the integrand reduces to – t 2 + 5 2 12 + 2 – 1 + t2 t 84. Reduce the integrand to tan 2 x sec2 x + 2 sec2 x + cosec2 x. 85. Reduce the integrand to sec x tan x – sec2 x + 1. 86. Put x = sin 2 t 87. Write 8 – sin 2x = 9 – (sin x + cos x)2

IIT JEE eBooks: www.crackjee.xyz

26 26.1 GEOMETRICAL INTERPRETATION OF THE DEFINITE INTEGRAL

If we graph the integrand y = f (x), then for f (x) ≥ 0, the definite integral

b

Úa

f ( x ) dx is numerically

equal to the area bounded by the curve y = f (x), the x-axis and the straight lines x = a and x = b (Fig. 26.1).

26.3

DEFINITE INTEGRAL AS THE LIMIT OF A SUM

Let f (x interval [a, b]. Then In particular, r=n

lim

Y

nƕ

y = f (x)

1 Ê rˆ

Ú

b

a

n

f ( x ) dx = lim h  f (a + rh ). n Æ• , h Æ 0 nh = b - a

r =1

1

 n f ÁË n ˜¯ = Ú0 f ( x)dx

r =1

Illustration 1

x=a

x=b

2 n˘ È1 lim Í 2 + 2 + ... + 2 ˙ nÆ• Î n n n ˚

X

Fig. 26.1

In general

b

Úa

f ( x ) dx represents an algebraic sum of the

bounded by the graph of the function y = f (x), the x-axis and the straight lines x = a and x = b. The areas above the x-axis enter into this sum with a plus sign, while those below the x-axis enter it with a minus sign (Fig. 26.2).

1

x2 1r = lim  = Ú xdx = nÆ• 2 r =1 n n 0 n

1

= 0

Illustration 2 È 12 22 ( n - 1)2 ˘ lim Í 3 + 3 + ... + ˙ nÆ• n n n3 ˚ Î 2

1

x3 1 Ê rˆ = lim  Á ˜ = Ú x 2 dx = Ë ¯ nÆ• n 3 r =1 n 0 26.4 Fig. 26.2

26.2 THE NEWTON-LEIBNITZ FORMULA

If F (x) is one of the antiderivatives of a continuous function f (x) on [a, b] i.e., F¢(x) = f (x) (a < x < b), then we have the following formula due to Newton and Leibnitz: (FUNDAMENTAL THEOREM OF CALCULUS) b

Úa

= F ( b) - F ( a) a

= 0

1 3

PROPERTIES OF DEFINITE INTEGRALS a

f ( x )dx = 0

b

f ( x )dx = - Ú f ( x )dx

1.

Úa

2.

Úa

3.

1

b

Úa

a

b

b

f (u )du = Ú f (t )dt a

4. If f (x) > 0 on the interval [a, b], then

b

Úa

f ( x ) dx b

b

f ( x ) dx = F ( x )

1 2

> 0. If (x) > 0 for all points of [a,b]. Then Ú f ( x ) a dx > 0

IIT JEE eBooks: www.crackjee.xyz 26.2 Comprehensive Mathematics—JEE Advanced b

b

5. If f(x) < g(x) on [a, b], then Úa f ( x ) dx < Úa f ( x ) dx. b

6.

b

b

Úa ( f1 ( x) + f2 ( x)) dx = Úa

f1 ( x ) dx + Ú f 2 ( x )dx

18. If the function j(x) and y(x) are defined on [a, b] and differentiable at every point x Œ (a, b), and f (t) is continuous for j(a) < t < j(b), then y ( x) ˆ d Ê dy dj . - f (j ( x )) Á Ú f (t ) dt ˜ = f (y ( x )) ˜¯ dx ÁË j ( x ) dx dx

a

and b

b

Úa a f ( x) dx = a Úa b

Úa

7.

b

f ( x )dx

b

f ( x )dx £ Ú f ( x ) dx a

c

b

a

c

8.

Ú

9.

Ú0

f ( x )dx = Ú f ( a - x )dx

b

b

10.

a

a

Úa

f ( x )dx = Ú f ( x )dx + Ú f ( x )dx.

a

a

Ú0

b

f ( x ) dx = Ú f ( a + b - x )dx

b

a

a/2

0

f ( x) + Ú

a/2

0

Ï0 Ô f ( x )dx = Ì a / 2 f ( x )dx 2 ÓÔ Ú0

Úa

f ( a - x )dx if f ( a - x ) = - f ( x ) if f ( a - x ) = f ( x )

12. If f (–x) = f (x) (i.e., f is an even function), then a

a

Ú- a f ( x) dx = 2Ú0

f ( x )dx = Ú f (j (t )) j ¢(t ) dt t1

20. Let a function f (x, a) be continuous for a £ x £ b and c £ a £ d. For any a Œ [c, d], if I(a)

0

f ( x )dx = Ú

t2

Úa

a

11. In particular

Ú0

19. Change of variables If the function f (x) is continuous on [a, b] and the functions x = j(t) is continuously differentiable on the interval [t1, t2] and a = j(t1), b = j (t2), then

f ( x ) dx

13. If f (–x) = – f (x) (i.e., f is an odd function), then

b

f ( x, a )dx, then I ¢(a ) = Ú f ¢( x, a dx where a

(a) is the derivative of I(a) w.r.t. a, and f ¢(x, a) is the derivative of f (x, a) w.r.t. a, keeping x constant. 21. If f 2(x) and g2(x) are integrable on [a, b], then

Ú

b

a

f ( x )g( x ) dx £



b

a

) (Ú 1/ 2

f 2 ( x ) dx

b

a

)

1/ 2

g 2 ( x ) dx

22. Let f be continuous and non-negative at all points of [a, b] and f (x0) > 0 at least at one point x0 within this interval then

a

Ú- a f ( x) dx = 02.

b

Úa

f ( x ) dx > 0.

Illustration 3

14. If f is continuous on [a, b], then the integral function x

g

g(x) =

Úa f (t)

dt for x Π[a, b] is

derivable on [a, b], and g¢(x) = f (x) for all x Œ [a, b]. 15. If m and M are the smallest and greatest values of a function f (x) on an interval [a, b], then a

16. If f (x) is continuous on [a, b], then there exists a point c Π(a, b) such that The number f (c) =

b

Úa

f ( x ) = f ( c ) ( b - a) .

b 1 f ( x )dx called the mean Ú b-a a

ˆ Ê x2 Ú0 ÁË 4 - 7 x + 5˜¯ dx 2

Evaluate

The given integral is equal to 2 2 1 2 2 x dx - 7Ú xdx + 5Ú dx Ú 0 0 4 0

b

m(b - a) £ Ú f ( x )dx £ M (b - a).

1 x3 = 4 3

2

0

x2 -7 2

2

2

+ 5x 0

0

1 È8 ˘ 7 - 0 ˙ - [4 - 0] + 5[2 - 0] Í 4 Î3 ˚ 4 2 10 = - 14 + 10 = 3 3

=

value of the function f (x) on the interval [a, b]. The Illustration 4 for integrals. 17. If f (x) is periodic with b

Úa

f ( x )dx = Ú

In particular,

b + nT

a + nT nT

Ú0

period T then

f ( x ) dx, where n is an integer. T

f ( x )dx = nÚ f ( x ) dx. 0



Find the average value of f (x) = 4 – x2 on [0, 3]. b 1 f ( x )dx average (f) = Ú a b-a

IIT JEE eBooks: www.crackjee.xyz 26.3

=

1 3 ( 4 - x 2 ) dx 3 - 0 Ú0

1 x3 = 4x 3 3

3

0

where

I1, q =

1È 27 ˘ = Í12 - ˙ = 1 3Î 3˚

I 0, q

Illustration 5 The value of

1

Ú sin x dx 2

0

0< 26.5

Ïq -1 q - 3 2 ÔÔ q q - 2 ◊ ◊ ◊ 3 if q is odd =Ì Ô q - 1 ◊ q - 3 ◊ ◊ ◊ 1 ◊ p if q is even ÔÓ q q - 2 2 2

Note that if p is odd then it does not matter whether q is even or odd.

cannot be 2

For x Π[0, 1], 0 < sin x2 < 1, so 1

1 and q +1

Areas by integration

1

Ú sin x dx £ Ú 1. dx = 1 2

0

0

INTEGRALS WITH INFINITE LIMITS

26.7

If a function f (x) is continuous for a £ x < •, then by

Ú



For the curve y = f (x) > 0,

b

f ( x )dx = lim Ú f ( x )dx

(1)

bƕ a

a

AREA BETWEEN A CURVE AND THE X-AXIS b

Úa

f ( x )dx is the area bounded

by the curve y = f (x), the x-axis and the straight line x = a, x = b

(1), then the improper integral is said to be convergent; otherwise it is divergent. Geometrically, the improper integral (1) for f (x) > 0, is y = f (x), the straight line x = a and the x-axis. Similarly, we x=a b

b

Ú- • f ( x)dx = a Ælim- • Úa and



Ú- •

f ( x )dx = Ú

-•

p /2

Ú0 26.6 p /2

Ú0

a



f ( x )dx + Ú f ( x )dx a

sin x dx and = Ú n

Fig. 26.3

f ( x ) dx

p /2

0

p

x=b

q

sin x cos x dx

Illustration 6 Find the area between the curve y = x (x – 3) and the ordinates x = 3 and x = 5.

REDUCTION FORM[ULAE FOR

cos n xdx =

p /2

Ú0

sin n x dx

1 p Ïn -1 n - 3 ÔÔ n ¥ n - 2 ¥ ◊ ◊ ◊ ¥ 2 ◊ 2 if n is even =Ì Ô n - 1 ¥ n - 3 ¥ ◊ ◊ ◊ 2 if n is odd ÔÓ n n-2 3 I p, q =

p /2

Ú0

x=3 Fig. 26.4 5

5

3

3

Required area = Ú y dx = Ú x ( x - 3)dx 5

x3 x2 = -3 3 3 2

sin p x cos q xdx

2 p-3 Ï p -1 ÔÔ p + q ¥ p + q - 2 ¥ ◊ ◊ ◊ ¥ 3 + q ◊ I1,q if p is oddd =Ì Ô q - 1 ¥ p - 3 ¥ ◊ ◊ ◊ 1 I if p is even 0, q 2+q ÓÔ p + q p + q - 2

x=5

5

3

2 =8 . 3 26.8

AREA BETWEEN TWO CURVES

The area between y = f (x) and y = g(x) on the interval [a, b] provided f (x) > g(x) on [a, b] is

IIT JEE eBooks: www.crackjee.xyz 26.4 Comprehensive Mathematics—JEE Advanced b Ê uppar

b

Ê lower

ˆ

ˆ

Úa [ f ( x) - g ( x)dx = Úa ÁË function˜¯ - ÁË function˜¯ dx

x=a

Illustration 8 Find the area between the curve y = x(x – 3) and the ordinates x = 0 and x = 5.

x=3

x=b

x=5

Fig. 26.5

Illustration 7

Fig. 26.8

Determine the area between y = 2x2 + 10 and y = 4x + 16 We first find the points of interscction 2x2 + 10 = 4x + 16 fi 2x2 – 4x – 6 = 0 fi 2(x + 1) (x – 3) = 0 The first curve is a parabola and the second curve is a straight line y = 4x +

16

Area between y = x (x – 3) between x = 0 and x = 3 is 3 9 2 Ú0 ( x - 3x)dx = - 2 Area between y = x (x – 3) between x = 3 and x = 5 is 5 26 2 Ú3 ( x - 3x)dx = 3 Required area = 26 - ÊÁ - 9 ˆ˜ = 79 . 3 Ë 2¯ 6 26.9 INTEGRATION ALONG THE Y-AXIS

10

x = -1 0

x=3 Fig. 26.6

3

Required area = Ú (( 4 x + 16) - (2 x + 10))dx = Ú ( -2 x 2 + 4 x + 6)dx -1

2 64 3 = - x 3 + 2 x 2 + 6 x -1 = 3 3 If the curve crosses x-axis at x = x0, then the area bounded by y = f (x) and the ordinates x = a, x = b is given by

Úa =Ú

x0

a

f ( x )dx +

b

Úx

f ( x )dx

b

f ( x )dx - Ú f ( x )dx x0

x=b

Fig. 26.7

( f ( y) - g( y))dy

Illustration 4 Find the area of the region in the first quadrant enclosed by x = y2 and x = y + 2 y2 = y + 2 fi y = –1 or y = 2 Required area 2

= Ú (( y + 2) - y 2 ) dy 0

y2 y3 + 2y 2 3

2

y=2 x = y2

x=y+2

0

10 = ◊ 3

y = f (x)

x = x0

d

c

=

0

x=a

Ú

2

-1 3

x0

If f (y) and g(y) are continuous function on [c, d] and f (y) ≥ g(y) for all y in [c, d], then the area of the region bounded by x = f(y) and x = g(y) and horizontal lines y = c and y = d is

When calculating the area under a curve y = f (x) follow the steps below: Fig. 27.9 1. Stretch the curve. 2. Determine the boundaries x = a and x = b or y = c and y = d.

IIT JEE eBooks: www.crackjee.xyz 26.5

(i) Lines parallel to the coordinate axes. 4. Integrate. Following will be useful for tracing the curve. 1. Symmetry: Find out whether the curve is symmetrical about any line with the help of the following points: (i) The curve is symmetrical about the x-axis if the equation of the curve remain unchanged when y is replaced by – y i.e. if the equation of the curve contains only even powers of y, e.g. x2 + y2 = 4, y2 = 8x. (ii) The curve is symmetrical about y-axis if the equation of the curve remain unchanged when x is replaced by – x i.e. if the equation of the curve contains only even powers of x e.g. x2 + y2 = 4, x2 = 4y. (iii) The curve is symmetrical in opposite quadrants if the equation of curve remain unchanged when both x and y are replaced by –x and –y e.g. x2 + y2 = 2. (iv) The curve is symmetrical about the line y = x if the equation of the curve remains unchanged when x and y are interchanged e.g. x2 + y2 = 8. 2. Origin: Find out whether the origin lies on the curve y2 = 4x of tangents at origin by equating to zero the lowest degree terms. For examples, x = 0 is tangent at origin to the the curve y2 = 4x. 3. Intersection with origin: Find out the points of intersection of the curve with the coordinate axes e.g. y2 = 2x + 1, put x = 0, y2 = 1, y = ±1 put y = 0, x = - 1 . Thus points of intersection with coordinate 2 Ê 1 ˆ axes are (0, 1) (0, –1) and Á - , 0˜ . Ë 2 ¯ 4. Regions where no part of the curve lies: Find the regions of the plane where no part of the curve lies. For examples, y2 = 4x. x < 0 no part of the curve lies. The curve line straight like, circle, parabola, describe them here. (a) Straight Line: General equation of straight line is of the form ax + by + c = 0. To plot a straight line we put y x. Also we put x y. We plot these two points and join them to get the required line.

y

y y=k x=h

O

Fig. 26.10

(ii) Lines passes through origin y

y

y = mx m>0

y = mx m 0, the value of

Example 2

Solution

Ans. (c)

f ( x )g ¢( x ) - f ¢( x )e- g ( x ) f ( x )e- g ( x ) + 1



e- g ( b ) f ( b )

e- g ( a ) f ( a )

dt , t = f ( x )e- g ( x ) t +1 e- g ( b ) f ( b )

= log(t + 1) e- g ( a ) f ( a ) Ê e - g ( b ) f (b ) + 1 ˆ = log Á - g ( a ) ˜ f ( a ) + 1¯ Ëe

dx

IIT JEE eBooks: www.crackjee.xyz 26.8 Comprehensive Mathematics—JEE Advanced

The value of

Example 5

I =Ú

p

0

(

x sin x dx is cos 4 x + 4 cos 2 x + 11

))

p p - log 1 + 2 2 32 p p - log 3 - 2 2 (b) 32 2 (c) p p - log 3 + 2 2 16 p (d) p - log 3 - 2 2 16 Ans. (b) p (p - x ) sin x Solution: I = Ú0 cos 4 x + 4 cos 2 x + 11 dx p sin x dx - I = pÚ 0 cos 4 x + 4 cos 2 x + 11 p p sin x I= Ú dx 4 2 0 2 8 cos x - 8 cos x + 1 + 4 (2 cos2 x - 1) + 11 (a)

(

(

( (

-p 2

))

(

)) ))

( (

-1

dt Ú1 8 t 4 + 8 p -1 dt p 1 dt = Ú 4 = Ú 4 16 -1 t + 1 8 0 t + 1 =

p 16

=

(t = coos x )

11 + t 2

Ú0

+1- t2 dt 1+ t4

˘ È Í 1 ˙ 2 2 1 1 + 1/t 1 + 1/t p Í ˙ = dt dt Ú0 Ê 1ˆ 2 ˙ 16 Í Ú0 Ê 1ˆ 2 t + ˜ -2 ˙ Í Át - ˜ + 2 Á Ë t¯ Î Ë t¯ ˚ 1 1˘ È 1 1 tt+ - 2 ˙ p Í 1 t - 1 log t ˙ = Í tan -1 1 16 Í 2 2 2 2 t+ + 2 ˙ ˙ Í t 0 0˚ Î =

1 È 2- 2 t 2 + 1 - 2t ˘ ˘ p È p log lim log Í ˙˙ Í 16 ÍÎ 2 2 2 2 Î 2 + 2 t Æ0 t 2 + 1 + 2t ˙˚ ˚˙

2- 2 ˘ p È p 1 log Í ˙ 16 ÍÎ 2 2 2 + 2 ˙˚ 2 p Èp - log (3 - 2 2 ) ˘ = ˚ 32 2 Î

=

Example 6

equals

p + 4 x5 If I = Ú dx, then I -p / 6 1 - sin ( x + p / 6) p /6

(a) 4p

(b) 2p + 1/ 3

(c) 2p - 3

(d) 4p + 3 - 1/ 3

Ans: (a) 4 x5 is an odd function, and 1 - sin ( x + p / 6 ) p is an even function, we get 1 - sin ( x + p / 6 )

Solution

As

I = 2p Ú

p /6

0

dx 1 - sin( x + p /6)

Put x + p/6 = t, dx = dt I = 2p Ú

p /3

p /6

p /3 1 - sin t dt = 2p Ú dt p /6 cos 2 t 1 - sin t

= 2p (tan t + sec t )]pp /3 /6 Ï = 2p Ì Ó

2 ˆ¸ Ê 1 + 3+2 -Á ˜ ˝ = 4p . Ë 3 3¯ ˛

(

)

Ú

If I =

Example 7

2

(|x + 1| + |x + 2| + |x - 1|) dx,

-3

then I equals 31 (a) 2 37 (c) 2 Ans. (a)

35 2 39 (d) 2

(b)

Solution We can write I = I1 + I2 + I3 2

I1 = Ú |x + 1| dx etc.

where

-3

x + 1 = t, so that

Put

3

I1 =

Ú-2

|t| dt =

1 = - t2 2 Similarly,

I2 = I3 = I =

Thus,

0

1 + t2 2 -2

0

(- t) dt + -2 3 0

=

Ú

3

0

t dt

13 2

9 2

31 . 2 1.7

Example 8

Ú

If I =

Ú0

È x 2 ˘ dx , then I equals Î ˚

(a) 2◊4 +

2

(b) 2◊4 -

(c) 2◊4 -

2

(d) 2◊4 - 1/ 2

Ans. (b) Solution:

Put x2 = t

2

IIT JEE eBooks: www.crackjee.xyz 26.9

or

x=

\

I=

1

t or dx = [t ]

2.89

Ú0 Ú

=

dt

2 t [t ]

2.89

0

2 t

2

dt + Ú1

[t ]

2.89

dt + Ú2

2 t 2 t 2 1 2.89 2 =0+Ú dt + Ú dt 1 2 t 2 2 t 2

t˘ +2

=

˚1

Ú

0

-2

1/ 2 0

[t ] 2 t

dt

e cos t dt = 12 et cos t ˘˚

+ 12Ú

2

4 If I = Ú-2 1 - x dx, then I equals

Example 11

2.89

(a) 6 (c) 12

˚2

(b) 8 (d) 21

= 2◊4 - 2

Solution As |1 - x4| is an even function, I = 2Ú

1

(a) - 4

(b) - 3

(c) - 2

(d) - 1

Ú

1

I=

Ú

1

=

-1

Example 12

(- 1)dt = - t]1–1 = - 2

as t3 + t cos t is an odd function.

The value of

2

Ú-1 [ x ] - {x} dx , where [x]

is the greatest integer less then or equal to x and {x} is the fractional part of x is (a) 7/2 (b) 5/2 (c) 1/2 (d) 3/2 Ans. (a) For any x ΠR, x = [x] + {x} so

Solution

If I =

Example 10

2

Ê Ê x5 ˆ˘ x5 ˆ ˘ = 2 Á x - ˜ ˙ + 2 Á - x˜ ˙ 5 ¯ ˙˚ Ë Ë 5 ¯ ˙˚1 0 1 ˆ Ê 32 Ê 1ˆ = 2 Á1 - ˜ + 2 Á - 2 - + 1˜ Ë 5 Ë 5¯ 5 ¯ = 12.

[(x + 1)3 - 1 + (x + 1) cos (x + 1)]dx

[t3 - 1 + t cos t]dt -1

0

1

Solution: We can write

Put x + 1 = t, so that

1

0

Ans. (c)

-2

|1 - x4| dx

= 2Ú (1 - x 4 )dx + 2Ú ( x 4 - 1)dx

[x3 + 3x2 + 3x + (x + 1) cos (x + 1)] dx, is

0

2

0

The value of

Ú

e sin t dt

1/ 2 È Ê 1ˆ ˘ = 12 Í e cos Á ˜ - 1˙ + 12 et sin t ]10/ 2 - 12Ú et cos t dt 0 ¯ Ë 2 Î ˚ 1 1ˆ ˆ Ê Ê fi 2 I = 12 Á e Á cos + sin ˜ - 1˜ . Ë Ë 2 2¯ ¯

Ans. (c)

I=

1/ 2 t

0

= ( 2 - 1) + 2(1.7 - 2)

Example 9 I=



1/ 2 t

0

I = 12Ú

dt

[x] – {x} = 2 [x] – x. Thus

Ú

p

0

(1/ 2 ) cos x

e

Ï Ê1 ˆ¸ Ê1 ˆ Ì2 sin Á cos x˜ + 3 cos Á cos x˜ ˝ sin x dx, Ë2 ¯˛ Ë2 ¯ Ó

then I equals (a) 7 e cos (1/2) (c) 0 Ans. (d) Solution

|t|

1 cos x = t, so that - sin x dx = 2dt and 2

Put I=

(b) 7 e [cos (1/2) - sin (1/2)] 1 1ˆ ˆ Ê Ê (d) 6 Á e Á cos + sin ˜ - 1˜ Ë Ë 2 2¯ ¯

Ú

-1/ 2

1/ 2

e|t| (2 sin t + 3 cos t) (- 2) dt |t|

As e sin t is an odd function, and e cos t is an even function,

2

Ú-1 [ x ] - {x} dx = 0

1

2

Ú-1 2 [ x ] - {x} dx + Ú0 2[ x] - x dx + Ú1 0

1

2

-1 0

0

1 2

2[ x ] - x dx

= Ú 2 + x dx + Ú x dx + Ú 2 - x dx =Ú

-1

1

(2 + x ) dx + Ú0 xdx + Ú1 (2 - x )dx

3 5 1ˆ 1 Ê = - Á -2 + ˜ + + 2 - = Ë 2 2 2¯ 2 3

Example 13 If I = Ú2

2 x5 + x 4 - 2 x3 + 2 x2 + 1 dx, then I ( x 2 + 1)( x 4 - 1)

equals (a)

1 1 log 6 + 2 10

(b)

1 1 log 6 2 10

IIT JEE eBooks: www.crackjee.xyz 26.10 Comprehensive Mathematics—JEE Advanced

(c)

1 1 log 3 2 10

(d)

1 1 log 2 + 2 10

\

Ans. (b) Solution: Note that 2x5 + x4 - 2x3 + 2x2 + 1 = 2x3 (x2 - 1) + (x2 + 1)2

3 dx 2 x 3 dx +Ú 2 2 2 2 ( x + 1) 2 x -1 3



Solution

3

x -1 ˘ 1 = I1 + log x + 1 ˙˚2 2

x

( x + 1)2 2



p /2

0

1

-1

1

= 2 Ú (1 - t 2 ) (1 + 5t 2 )dt

(2 x )dx

[∵ remaining part of integrand is an odd function] 1

= 2 Ú (1 - 4t 2 - 5t 2 ) dt 0

10

10 t

If I = Ú

p /2

0

cos n x sin n x dx =

sin n x dx then l equals (a) 2-n+1 (c) 2-n

using

Ú0

a

f ( x ) dx = Ú [ f ( x ) + f (2a - x )] dx 0

1

Ê 4 ˆ˘ = 2 Á t + t3 - t5 ˜ ˙ Ë 3 ¯ ˚0 4 ˆ 8 Ê = 2 Á 1 + - 1˜ = . Ë 3 ¯ 3 Example 16 If I = (a) 3 /2 (c) 0

1 Ê1 ˆ sin Á - x˜ dx, then I equals 1/ 3 x Ëx ¯ (b) p + 3 /2 (d) p – 3 /2 3

Ú

Ans. (c) (b) 2-n-1 (d) 2-1

Ans. (c) 1 p /2 Solution: I = n Ú (2 sin x cos x ) n dx 2 0 1 p /2 = n Ú (sin 2 x )n dx 2 0 Put 2x = q, so that 1 p 1 I = n Ú (sin n q ) dq 0 2 2 1 È p /2 = n+1 Ú [(sin q )n + (sin(p - q )n ] dq 2 ÍÎ 0 2a

Put cos x = t, so that

= Ú (1 - t 2 ) (1 + 4t + 5t 2 + 2t 3 )dt

-1 1ˆ ˘ Ê Ú5 t 2 dt = ÁË log t + t ˜¯ ˙˚ 5 1 = log 2 10 1 1 I = log 6 2 10

Example 14

(d) 2

1

Put x2 + 1 = t, 2x dx = dt

Thus,

2 3

0

3

I1 =

(b)

-1

2

2

\

x(1 + 2 cos x) (1 + cos x)2 dx,

I = Ú (1 - t 2 ) (1 + 2t ) (1 + t )2 dt

1Ê 1 1ˆ = I1 + Á log - log ˜ 2Ë 2 3¯ where I1 = Ú

3

then I equals 4 (a) 3 8 (c) 3 Ans. (c)

- 1) + ( x 2 + 1)2 dx ( x 2 + 1)2 + ( x 2 - 1)

2

p

Ú0 sin

If I =

Example 15

3 2 x3 ( x 2

I =Ú

\

1 p /2 n Ú sin q d q 2n 0 l = 2- n

Thus, I =

Solution:

Put x = 1/t, so that 1/ 3 Ê 1ˆ (-1) I = Ú t sin Á t - ˜ 2 dt 3 Ë t¯ t = -Ú

3

1/ 3

1 Ê1 ˆ sin Á - t ˜ dt = - I Ët ¯ t

fi 2I = 0 or I = 0. Example 17 If I = Ú

p /2

0

(a) - p/2 (c) - 1/3

sin 8 x log (cot x ) dx, then I equals cos 2 x (b) p/3 (d) 0

Ans. (d) Solution: Using

a

Ú0

a

f ( x )dx = Ú f ( a - x )dx, 0

IIT JEE eBooks: www.crackjee.xyz 26.11

( 4p - 8 x ) log cot (p /2-x ) dx cos(p - 2 x ) p / 2 ( - sin 8 x )(log tan x ) dx =Ú 0 - cos 2 x

I =Ú

p / 2 sin

The natural number n (£ 5) for which

Example 20

0

1

I n = Ú e x (x - 1) n dx = 16 - 6e is 0

=-I [∵ log tan x = - log cot x] fi 2I = 0 or I = 0.

(a) 2 (c) 4

(b) 3 (d) 5

Ans. (b) 1

If I = Ú

Example 18

e

l /e

(a) 2 (c) 2(1 - 1/e)

and for n > 1,

e

1/ e

\

e

1 1 - + e - e +1 e e = 2(1 - 1/e). If for k ΠN,

0

(b) 0 (d) p

p /2

0

2[cos x + cos 3 x + ... + cos(2k - 1x ] cos x dx [(1 + cos 2x ) + (cos 4x + cos 2x ) + ...... + I (cos 2kx + cos (2k - 2)x )]dx p /2

1 1 Ê ˆ˘ = Á x + sin 2 x + sin 4 x + ... + (sin 2 kx˜ ˙ Ë ¯ ˚0 2 2k =

p . 2

sin 3 x cos x dx, sin 4 x + cos4 x

(1)

(b) p/4 (d) p

I=Ú

p /2

0

Ú

a

0

a

f ( x ) dx = Ú f (a - x ) dx, we get 0

cos3 x sin x dx cos4 x + sin 4 x

(2)

Adding (1) and (2), we get p /2 sin x cos x 2I = Ú dx 0 cos4 x + sin 4 x Using cos4 x + sin4 x = (cos2 x)2 + (sin2 x)2 = (1/4) (1 - cos 2x)2 + (1/4) (1 + cos 2x)2

and using the given identity, we can write

I=Ú

If I = Ú0

Solution: Using

Ans. (c) Solution: Writing p /2 sin 2 kx I= Ú0 sin x cos x dx p /2

= 16 - 6e n = 3.

Ans. (a)

sin 2kx = 2[cos x + cos 3x + ... + cos (2k - 1)x], sin x p /2 then value of I = Ú sin 2kx cot x dx is

0

\

then I equals (a) p/8 (c) p/2

=1-

I=Ú

I3 = 1 - 3I2 = 1 - 3(2e - 5)

p /2

1

1

(a) - p/2 (c) p/2

and

Example 21

= ( -t log t + t )]1/e +(t + log t - t ) ]1

Example 19

I1 = 1 - (1)I0 = 1 - (e - 1) = 2 - e

e

= Ú ( - log t ) dt + Ú (log t ) dt 1/e

0

I2 = - 1 - 2I1 = - 1 - 2 (2 - e) = 2e - 5

- log t dt

1

1

0

= - (- 1)n - nIn - 1

1 = t , so that x 1/e 1 I = Ú log ( -1) dt e t =Ú

1

I n = e x ( x - 1)n ˘˚ - n Ú e x ( x - 1)n-1 dx

Ans. (c) Solution: Put

1

x x Solution: We have I 0 = Ú0 e dx = e ˘˚ 0 = e - 1

dx log x 2 , then I equals x (b) 2/e (d) 0

= (1/2) (1 + cos2 2x) p /2 2 sin x cos x \ dx 2I = Ú 0 1 + cos2 2 x Put cos 2x = t, so that -1 dt -1 -1 dt p = -Ú = 2I = 2 2 Ú 0 1+ t 4 2 1 1+ t fi I = p /8 Example 22 such that then

Ú

p

0

Ú

p

0

Let f : [0, 1] " R be a continuous function

f (sin x ) dx = 2018,

xf (sin x ) dx is equal to

IIT JEE eBooks: www.crackjee.xyz 26.12 Comprehensive Mathematics—JEE Advanced

(a) 1009p (c) 1008p Ans. (a) Solution

Let I =

(b) 2016p (d) 2017p p

p

Ú0 xf (sin x) dx = Ú0 (p - x)sin(p - x)dx

(a)

p

= Ú (p - x ) f (sin x ) dx

(c)

0

p

fi 2 I = p Ú f (sin x )dx = 2018 p

Ans. (c)

0

Solution Thus,

fi I = 1009 p For a > 0, let

Example 23

x( x + x + 1) 3 a dx, Ú 0 2 x + 1 x4 + x2 + 1 then I (a) is equal to I ( a) =

(c) ln Ans. (c) Solution:

(

(

a +1 + a

)

a +1 + a 3

(b) ln

3

)

(d) 2ln

(

I ( a) =

3 a 2 Ú0

=

3 a 2 Ú0

x

( x + 1) ( x 2 - x + 1) x1/ 2

( x + 1) 3

x p + = q , so that x = 2q - p/2 and dx = 2dq 2 4

Put

5p /4 2q -p /2 e cos q p /4

where

)

a +1 + a 2

5p /4 2q e p /4

I1 = Ú

dx

fi fi Thus,

dx



3 x dx = dt 2 dt a3 / 2 I ( a) = Ú0 t2 +1

( = ln (

I=

-3 2 2p (e + 1). 5

Example 26

a3 / 2

= ln t + t 2 + 1 ˘ ˚˙0

Example 24

If I =

Ú

2p

0

a3 + 1 + a3

)

Ê x pˆ e x /2 sin Á + ˜ dx, then I equals Ë 2 4¯

(a) p (b) 0 Ans. (b) Solution Put x/2 = q, so that

(c) - p/2 (d) 2p

p pˆ Ê I = 2 Ú eq sin Á q + ˜ dq 0 Ë 4¯ p q e 0

= 2Ú

c

x4 + 1

b

x6 + 1

I = ea Ú

)

cos q dq

5 1È 1 ˆ 1 ˆ˘ Ê 2 Ê -2 I1 = Íe5p /2 Á - ep /2 Á + ˜ ˜˙ Ë 2 Ë 2 4 4Î 2¯ 2¯˚ -3 5p /2 I1 = (e + ep /2 ) 5 2

Put x3/2 = t, Thus

dq

5p /4 1 1 5p /4 2q e sin q dq = e 2q cos q ˘˚ + Ú p /4 2 2 p /4 1 1 5p /4 = e 2q (2 cos q + sin q )]p /4 - I1 4 4

)

x4 + x2 +1 = (x2 + 1)2 – x2 = (x2 + 1 + x) (x2 + 1 – x)

\

(d) 0

= 2e-p/2 I1 a2 + 1 + a

(

-3 2 2p (e + 1) 5

I = 2Ú

2

(a) ln

2p Ê x pˆ x If I = Ú0 e cos ÁË + ˜¯ dx, then I equals 2 4 -3 2 -3 2p (e - 1) (b) (e2p - 1) 5 5

Example 25

(sin q + cos q ) dq p

= 2 eq sin q ˘˚ = 0 0

[∵ Ú ex(f (x) + f ¢(x)) dx = ex f (x)]

Let a, b, c, > 0 and b > c, If dx, then

Ê cˆ (a) I < Á ˜ Ë b¯

a

(c) I < eac Ans. (a) Solution For x > 0,

(b) I > eab (d) I < ea(c – b)

1 x 4 + 1 x6 + 1 - x5 - x = x x6 + 1 x( x 6 + 1) =

(1 - x ) - x 5 (1 - x ) (1 - x )(1 - x 5 ) = >0 x( x 6 + 1) x( x 6 + 1)

x4 + 1 1 < , for all x > 0 x6 + 1 x c x4 + 1 c1 \ Ú 6 dx < Ú dx = log c - log b b x +1 b x fi

IIT JEE eBooks: www.crackjee.xyz 26.13

fiI 0 and Ë x¯ 2 +1 dx I=Ú 2 2 -1 1 + x 1 + a f ( x)

(

)(

)

(a) independent of a (b) independent of function f (c) tan–1 ( 2 + 1) – I = p/4 (d) tan–1 ( 2 – 1) – I = p/4 Ans. (a), (b), (c), (d)

(1)

IIT JEE eBooks: www.crackjee.xyz 26.23

Solution:

Put x = 1/t, then dx = (–1/t2) dt.

Also, when x = t=

2 – 1, t = 2 –1 2 -1

I =Ú

\

2 +1

=Ú =Ú

2 +1 2 -1

2 + 1 and when x =

(-1/t ) dt (1 + 1/t ) ÈÎ1 + a

2 -1

2 +1,

(

a

- f ( x ) - g( x ) (b - cos x )m h( x ) + k ( x )(- sin x )2 n f ( x ) + g( x ) =m (b - cos) h( x ) + k ( x )sin 2 n x = – F (x)

f (1/ t ) ˘

˚

)

f ( x) f ( x)

2

dx

(2)



2 +1 2 -1

f ( x)

2

dx ˘ = tan -1 x ˙ 2 -1 1 + x 2 ˚ 2 +1

= tan

-1

( (

) ( Ê 2 + 1) - tan Á Ë 2 + 1 - tan -1 -1

(

F is an odd function

\

I = Ú F ( x ) dx = 0

a

-a

)

(

Example 64 Let f (x) =

dx

Ê 1 - sin x ˆ sin x - 2 log Á Ë cos x ˜¯

2 +1

)

2 -1

and let I = Ú

1 ˆ ˜ 2 + 1¯

(

)

)

F: [–a, a] " R by F ( x) =

)

2 +1

(

)

2 +1

(c) I = log cot (p / 8) (d) I = log cot (p/8) Ans. (b), (c) Solution:

Write I = I1 – I2 where p / 4 sin x 1 dx I1 = Ú 2 0 2 cos x Ê 1 + sin x ˆ log Á Ë cos x ˜¯

(sec x ) log ÊÁË 1 -cossinx x ˆ˜¯ 2

and

I2 = Ú

p /4

0

Ê 1 + sin x ˆ log Á Ë cos x ˜¯

f ( x) + g( x) (b - cos x ) m h( x ) + k ( x )(sin x )2 n

dx

Now, note that d È Ê 1 + sin x ˆ ˘ d log Á ˜ = dx ÍÎ Ë cos x ¯ ˙˚ dx

ÈÎlog (sec x + tan x )˘˚

sec x tan x + sec2 x sec x + tan x = sec x

=

Ans. (a), (b), (c) Solution:

(

(b) I = log

fi I + tan–1 ( 2 – 1) = p/4 Example 63 Let a Œ (0, p/2), b > 1, m, n ŒN and let f, g, h, k: [–a, a] " R be four continuous functions such that f(–x) = –f(x), g(–x) = –g(x), h(–x) = h(x), k(–x) = k(x), and a f ( x ) + g( x ) let I = Ú dx - a (b - cos x )m h( x ) + k ( x )sin 2 n x (a) I is independent of a (b) I is independent of f and g (c) I is independent of h and k (d) I depends only on b.

p 4

f ( x ) dx, then

(a) I = log

)

(

p /4

0

˘ 2 +1 ˙ ˚

,0£ x£

Ê 1 + sin x ˆ log Á Ë cos x ˜¯

(1 + cos 2 x )

2 -1

Èp 2 + 1 - Í - tan -1 Î2 p I = tan -1 2 + 1 4 Ê 1 ˆ p = tan -1 Á Ë 2 - 1˜¯ 4 p p = - tan -1 2 - 1 2 4

= tan -1 fi

1 + a f ( x)

(1 + x ) (1 + a )

= tan -1



This shows that I is independent of a, b, f, g, h, and k.

Adding (1) and (2), we get

2I = Ú

m

=

dx 2 x + 1 ÈÎ1 + a - f ( x ) ˘˚

( x + 1) (1 + a )

f (- x) + g(- x) (b - cos( - x ) h( - x ) + k ( - x )(sin( - x ))2 n

F (- x) =

2

2

2 +1

Then

IIT JEE eBooks: www.crackjee.xyz 26.24 Comprehensive Mathematics—JEE Advanced

\

1 p /4 sec x tan x dx Ú 2 0 log (sec x + tan x )

I1 =

)

= tan x

(

log (sec x + tan x ) ˘˙ ˚0

= log

(

2 + 1 - I3

p /4

p /4

- I3

)

sec2 x log (sec x + tan x ) dx

where

I3 = Ú

Next,

Èlog (sec x + tan x ) ˘ sec x Í ˙ p /4 ÍÎ+ log (sec x - tan x )˙˚ dx I2 + I3 = Ú 0 log (sec x + tan n x)

0

=

p p p A +B= .A= , B = 0 and A = , 2 4 2 sin a sina -p B= satisfy the last equation. 4 sin a Thus



p / 4 sec

(

x log sec x - tan x 2

2

log (sec+ tan x )

0

) dx

1

Ans. (b), (c) Solution:



=Ú =Ú

p



p

0

0

dx

0

Ê 2x Ê 2x 2 xˆ 2 xˆ ÁË cos + sin ˜¯ - cos a ÁË cos - sin ˜¯ 2 2 2 2

0

a

0

(p - x ) sin 2 x sin ((p / 2) cos x ) dx p - 2x



p

(2 x - p ) sin 2 x sin ((p / 2) cos x ) dx 2x - p

p

ˆ Êp 2 sin x cos x sin Á cos x˜ dx ¯ Ë2

p

ˆ Êp sin x cos x sin Á cos x ˜ dx ¯ Ë2

2I =

Ú0

=

Ú0

I=

Ú0

= -Ú

dx x x (1 - cos a ) cos 2 + (1 + cos a ) sin 2 2

=–

a

= sec2 a/2 cot a/2 [tan–1 (t/tan a/2)]0tana/2

(2)

-1

1 Êp ˆ Êp ˆ t sin Á t ˜ dt = 2Ú t sin Á t ˜ dt 0 Ë2 ¯ Ë2 ¯

[t = cos x, dt = – sin x dx]

0

Put tan x/2 = t, we have 2 tan a / 2 sec a / 2 dt I=Ú 2 0 t + tan 2 a / 2

(1)

Adding (1) and (2) we get

2

dx x x 2 sin 2 a / 2 cos2 + 2 cos2 a / 2 sin 2 2 2 2 2 a 1 (sec a / 2) sec x / 2 dx = Ú 2 0 tan 2 a / 2 + tan 2 x / 2 =Ú

x sin 2 x sin ((p / 2) cos x ) dx 2x - p

(p - x ) sin 2 x sin ((– p / 2) cos x ) dx p - 2x

1



p

(d) 2/ p 2

(p - x ) sin (2p - 2 x ) sin ((p / 2) cos (p - x )) dx 2(p - x ) - p

p

dx 1 - cos a cos x

a

Ú

Let I =

Ans. (a), (b) 0

(b) 8/p 2

(c) 2 Ú0 t sin (p/2) t) dt

a 0

a

x sin 2 x sin((p / 2) cos x ) dx 2x - p

(a) 4/p 2

0

Solution: I = Ú

p

Ú0

is

= 0. dx A = + B (a π 0) . Example 65 If = Ú 1 - cos a cos x sin a Then possible values of A and B are p p p ,B=0 (b) A = ,B= (a) A = 4 sin a 2 4 p p p (c) A = ,B= (d) A = p, B = sin a 6 sin a

The value of

Example 66

2

2

2 p p ◊ = sin a 4 2 sin a

=–

4 Êp ˆ cos Á t ˜ ◊ t Ë2 ¯ p

4 p

+ 0

Ú

1

0

Êp ˆ cos Á t ˜ dt Ë2 ¯

4 8 Êp ˆ (0 - 0) + 2 sin Á t ˜ Ë2 ¯ p p

Example 67 (a)

1

The value of

Ú

p /2

0

= 0

8 p2

x sin x cos x dx is sin 4 x + cos4 x

Ê 5p /4 ˆ sin 2 x dx˜ Á Úp 4 4 cos x + sin x ¯ Ë

(b) p 2 /16

1

2

IIT JEE eBooks: www.crackjee.xyz 26.25

(c) 3p 2 /4

F ¢¢ (x) = 0 implies that x = 2, 4/3. As F ¢¢ (x) changes sign around x = 2 and 4/3. So x = 2 and x = 4/3 are points of F (2) = – 4/3 and F (4/3) = – 112/81. So

(d) p 2 /2 Ans. (a), (b) Solution: I=Ú

p /2

0

p = 2 p fi 2I = 2 fiI =

p 4

p /2

Ú 0

Ú

sin x cos x dx sin 4 x + cos4 x

p /2

0

p /2

Ú 0

x sin x cos x dx sin 4 x + cos4 x

Column 1

p p / 2 sin 2 x sin 2 x = Ú dx 0 1 4 0 1 + cos2 2 x 1 - sin 2 2 x 2 2 -1 -p - p [tan–1 (–1) – tan–1 1] = p dt = = 16 8 Ú1 1 + t 2 8

Ú

p /2

1 ◊ 2

Ú

5p / 4

p

sin 2 x = = f ( x) 4 cos x + sin 4 x p /4 sin 2 x sin 2 x dx dx = Ú 4 4 4 0 cos x + sin x cos x + sin 4 x = 2Ú

p /4

0

2

tan x sec x dx 1 + tan 4 x



0

Example 68

Let F (x) =

Ú

0

(b) I2 (c) I3

(q) is more than 1

(d)

( )

(s) 0

lim I m

m Æ•

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans

Solution Put ( x + 1) e x = t dt = dx

e x + ( x + 1)

=

1 ( x + 3) e3 2

Im = 2 Ú

\

¤

Ans. (a), (b), (c)

2 e

1

Êm+2 eˆ 1 ln Á ˜< Ë m +1 ¯ 2

¤ ¤

m

¤

m>

(

)

e -1 > e e e -1

2 e

dt ˘ = 2 ln (t + m)˙ t+m ˚1

m+2 e < e m +1 m+2 e < m e + e

¤

1 x e 2

Êm+2 eˆ = 2 ln Á ˜ Ë m +1 ¯

Now, Im < 1

Solution F ¢ (x) = (x – 1) (x – 2)2 so F ¢ (x) = 0 implies that x = 1 and x = 2. Around x = 1, F ¢ (x) changes sign from negative to positive. Hence F (x) is minimum at 1 1 5 x = 1 and F (1) = Ú0 (t3 – 5 t2 + 8t – 4)d t = - + 4 4 3 17 – 4 = - . Hence (1, – 17/12) is a point of minimum. 12 Moreover, F ¢¢(x) = 3x2 – 10x + 8 = (3x – 4) (x – 2) and

dx

(r) 2ln 2 e

(a) (1, – 17/12) is a point of minimum

(d) (1, - 3) is a point of minimum

( x + 1) e x + m

(p) is less than 1



(t – 1) (t – 2)2 d t, then

0

(a) I0

p 2t dt = tan -1 t 2 = 4 4 1+ t 0 x

( x + 3) e x

Column 2

1

1

1

For m > 0, let I m = Ú

Example 69

sin x cos x dx sin 4 x + cos4 x

The integrand in (a) is a periodic function with period p, since sin 2( x + p ) f (x + p) = 4 cos ( x + p ) + sin 4 ( x + p )

\

MATRIX-MATCH TYPE QUESTIONS

p / 2 (p / 2 - x )sin x cos x x sin x cos x dx dx=Ú 4 4 0 sin x + cos x sin 4 x + coss4 x

IIT JEE eBooks: www.crackjee.xyz 26.26 Comprehensive Mathematics—JEE Advanced

It is not true for m = 1, 2, but true for m > 3. Ê 1 + 2 e / mˆ lim I m = lim ln Á ˜ =0 mÆ• Ë 1+1/ m ¯

È I 2016 ˘ È 2016 p ˘ Í ˙=Í ˙=3 Î I 515 ˚ Î 515 p ˚

mƕ

p

Let I n = Ú0

Example 70

Ê nx ˆ sin 2 Á ˜ Ë 2¯ dx, n Œ N. Ê xˆ sin 2 Á ˜ Ë 2¯

Column 1 1 I2017 – 2015 p Value of n such that In + I2017 £ I2022

(a) (b) (c)

(b)

x2 + 2

a

x6 + 1

Column 1 (a)

I (0)

(b)

I (1)

Value of n so that I2021 – In ≥ I 2018

(r) 3 (s) 4

(c)

I 1/ 3

È I 2016 ˘ Í ˙ Î I 515 ˚

(t) 5

(b)

I

(

Ans.

)

( 3)

(r)

1 p + ln 2 - 3 4 3

(s)

p 2 -1 Ê 2 ˆ 1 - tan Á ˜ ln (7) Ë 3¯ 2 3 6 3

p

q

r

s

a

p

q

r

s

t

a

p

q

r

s

b

p

q

r

s

t

b

p

q

r

s

c

p

q

r

s

t

c

p

q

r

s

d

p

q

r

s

t

d

p

q

r

s

Jn = Ú

p / 2 sin

)

x2 + 2 x2 + 2 = x6 + 1 x2 + 1 x4 - x2 + 1

(

=

)(

)

1 1 1 x2 - 5 3 x2 + 1 3 x4 - x2 + 1

(

) (

)

2 2 1 1 1 3 x -1 - 2 x +1 = 3 x2 + 1 3 x4 - x2 + 1 2 2 x2 + 1 x -1 1 1 = + 3 x2 + 1 x4 - x2 + 1 3 x4 - x2 + 1

0

< In> is an A.P. with common difference p. Also, I1 = p

(

Solution: Write

(2n + 1)q

dq sin q p / 2 sin q cos (2n + 2)q J n+1 - J n = Ú dq 0 sin q 1 p /2 = ÈÎsin (2n + 2)q ˘˚0 = 0 2n + 2 p Jn + 1 = Jn = ........ = J1 = J0 = 2 In + 1 – In = p

dx.

Column 2 p 2 1 Ê 2 ˆ (p) + tan -1 Á ˜ ln (7) Ë 3¯ 2 3 3 3 5p (q) 6

(q) 2

x = q , so that 2 2 p /2 sin ( n q ) I n = 2Ú dq 0 sin 2 q 2 2 p / 2 sin ( n + 1)q - sin ( nq ) dq I n+1 - I n = 2Ú 0 sin 2 q p / 2 sin (2n + 1)q sin q dq = 2Ú 0 sin 2 q = 2 Jn





I ( a) = Ú

(p) 1

Solution: Put

fi \

For a > 0, let

Example 71

Column 2

where [x] greatest integer < x Ans. p q r s t

where

"nŒN

Thus, In = np

\

x2 + 2 1 -1 2 Ú x6 + 1 dx = 3 tan x - I1 + 3 I 2

where I1 = Ú =Ú

(1 - 1 x2 ) x2 - 1 dx = Ú 2 2 x - x +1 x + 1 x2 - 1 4

(1 - ) dx 1

(x + 1x)

2

x2

x4 - x2 + 1

dx

IIT JEE eBooks: www.crackjee.xyz 26.27

Put x +1/x = t, I1 = Ú

dt t -3 2

=

t- 3 1 ln 2 3 t+ 3

1

F0 (x) = ln x and Fn ( x ) = Ú Fn-1 ( x ) dx " n Œ N 0

Column 1

x 2 - 3x + 1 ln 2 = 2 3 x + 3x + 1 1

I2 = Ú

and

1 n

Ú0 x

(a)

Put

x – 1/x = u, (1 + 1/x ) dx = du

\

I2 = Ú

1

2 -1 -1 Ê x - 1ˆ tan u = tan = ( ) Á x ˜ u2 + 1 ¯ Ë

dx

2 3

ln

1Ê 1 1ˆ ÁË1 + +  + ˜¯ 2 n! n

ln x dx (p) (q) 0

1 Ê1 1 1ˆ + + .... + ˜ Á n! Ë 2 3 n¯

(b)

Fn (1)

(r) -

(c)

Fn (e)

(s) e – 1

lim Fn ( x )

(t) -

(b)

xÆ0 +

x 2 + 2 1 -1 2 -1 Ê x 2 - 1ˆ tan tan Á = x + ˜ Ú x6 + 1 3 3 Ë x ¯ -

Column 2

(1 + 1 x2 ) dx x2 + 1 dx = Ú x2 + 1 2 - 1 x4 - x2 + 1 x 2

\

For x > 0, let

Example 72

1

(n + 1)2

x a ln x = 0 " a > 0] [Use lim xÆ 0 +

x 2 - 3x + 1 x 2 + 3x + 1

p

q

r

s

t

a

p

q

r

s

t

b

p

q

r

s

t

c

p

q

r

s

t

d

p

q

r

s

t

Ans.

Thus, 1Êp ˆ I ( a) = Á - tan -1 (a )˜ ¯ 3Ë 2 Ê a - 1ˆ ˆ 2Êp + Á - tan -1 Á ˜˜ 3Ë 2 Ë a ¯¯ 2

+

=

+ fi

I ( 0) = I (1) =

1 2 3

ln

t

˘ 1 t n+1 Ê 1 ˆ x n+1 ln x ˙ (a) Ú0 x ln x dx = Ú x ÁË x ˜¯ dx n +1 ˚0 n + 1 0 t

a2 - 3 a + 1

=

Ê a2 - 1ˆ p 2 - tan -1 ( a ) - tan -1 Á ˜ 2 3 Ë a ¯ 1 2 3

ln

n

a2 + 3 a + 1

a2 + 3 a + 1

1

1

0

(n + 1)2

x

x

0

0

(b) F1 ( x ) = Ú F0 (t ) dt = Ú ln t dt

p 2 p 5p + = , 2 32 6

= t ln t ] 0x - Ú dt = x ln x - x x

Ê 2 - 3ˆ p 1 1 p + = + ln Á ln 2 - 3 ˜ 4 2 3 Ë 2 + 3¯ 4 3

(

1 p 2 -1 Ê 2 ˆ - tan Á ˜ ln (7) Ë 3¯ 2 3 6 3

1 n+1 1 t n+1 t ln t n +1 (n + 1)2

fi Ú x n ln x dx = -

a2 - 3 a + 1

2 1 Ê 1 ˆ p 2 I Á ˜ = + tan -1 ln 7 Ë 3¯ 3 3 3 2 3 I ( 3) =

Solution:

0

)

F2 ( x ) = Ú [t ln t - t ] dt x

0

x

x

1 1 ˆ˘ ˘ Ê1 = t 2 ln t ˙ - Á t 2 - t 2 ˜ ˙ Ë 2 4 2 ¯ ˚0 ˚0 =

1 Ê 1ˆ 1 2 x ln x - Á1 + ˜ x 2 2! Ë 2 ¯ 2!

IIT JEE eBooks: www.crackjee.xyz 26.28 Comprehensive Mathematics—JEE Advanced

Proceeding likewise, we get 1 1Ê 1 1ˆ Fn ( x ) = x n ln x - Á1 + +  + ˜ x n Ë 2 n! n¯ n! Fn (1) = -

1Ê 1 1ˆ Á1 + +  + ˜¯ n! Ë 2 n

a I(a) = – 16 cos ÊÁ ˆ˜ Ë 2¯

Thus,

Ê p aˆ Ê aˆ I(a + p) = – 16 cos Á + ˜ = 16 sin Á ˜ Ë 2 2¯ Ë 2¯ Ê aˆ I(p – a) = – 16 sin Á ˜ Ë 2¯



(c) Fn (e) = - 1 ÊÁ 1 + 1 +  + 1 ˆ˜ n! Ë 2 3 n¯ lim Fn ( x ) = 0

(d)

I(2p – a) = – 16 cos (p – a/2) = 16 cos (a/2)

x Æ0 +

I ( a) = Ú

3p + a

-p + a

Column 1 I (a + p) I (p – a) I (2p + a) I (2p – a) p

q

a

p

Ê xˆ x - a - p sin Á ˜ dx Ë 2¯ Column 2 (p) – 16 sin (a/2) (q) – 16 cos (a/2) (r) 16 sin (a/2) (s) 16 cos (a/2)

r

s

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Ans.

Solution:

Example 74

For a > 0,

Example 73

(a) (b) (c) (b)

I(2p + a) = – 16 cos (p + a/2) = 16 cos (a/2)

Put x – a – p = q, then

I ( a) = Ú

2p



2p

Ê p q + aˆ q sin Á + ˜ dq Ë2 2 ¯

-2p

Ê q + aˆ q cos Á dq Ë 2 ˜¯

-2p

2p Ê q + aˆ Ê q + aˆ dq + Ú q cos Á dq q cos Á ˜ 0 Ë 2 ˜¯ Ë 2 ¯ 2p In the first integral put q = – f to obtain 0 2p Ê q + aˆ Ê -f + a ˆ + 1 I ( a) = Ú f cos Á ( ) d f q cos Á dq ˜ Ú Ë 2 ˜¯ Ë 2 ¯ 2p 0

= -Ú



Column 1 a 2 1 a sin2 t e dt - Ú esin t dt (a) xlim Ú + y x y Æ0 x (b)

(c)

(d)

( (Ú lim Ú

Ú0

È Ê q + aˆ Ê a -qˆ˘ + cos Á q Ícos Á dq ˜ Ë 2 ˜¯ ˙˚ ¯ Ë 0 2 Î

2p

0

Êqˆ q cos Á ˜ dq Ë 2¯ 2p

2p Êqˆ Êqˆ˘ = 2q sin Á ˜ ˙ - 2Ú sin Á ˜ dq 0 Ë 2¯ Ë 2 ¯ ˚0 2p

Êqˆ˘ = 4 cos Á ˜ ˙ = -8 Ë 2 ¯ ˚0

2y

dt

sin x dx y

x

(r) 0

cos t 2 dt

y

xÆ0

(s) 2/3

p

x q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

lim

1 x

(Ú e

xÆ0

1 xÆ0 x

= lim

a

sin 2 t

y

x+y

Úy

xƕ

dt - Ú

e

2

)

2

x + y t2 e dt 0

= lim

xƕ

e( x + y )

2 e( x + y ) e

esin

( x + y )2

2

2

) 2

( x + y)

1

(2 Ú = lim

ˆ Ê0 ÁË form˜¯ 0

x + y t2 e dt 0

xƕ

dt

(2 Ú = lim xÆ•

2

esin t dt

esin t dt = lim

x + y t2 e dt 0 x + y 2t2

Ú0

a x+y

xÆ0

(Ú lim

Ê aˆ = 2 cos Á ˜ I1 , where Ë 2¯ I1 = Ú

(q) esin

x3

Ú lim

Solution

(p) 1

x2

0

2p

e

xÆ0

Ans.

)

2 x + y t2 e dt 0 x + y 2t 2

xƕ

lim

)

Column 2

)

)e

e2 ( x + y )

( x + y )2

2

Ê• ˆ ÁË form˜¯ • Ê• ˆ ÁË • form˜¯

1 1 = lim =0 2 ( x + y) x Æ • x + y

The limits in (c) and (d) can be obtained similarly.

IIT JEE eBooks: www.crackjee.xyz 26.29

lim f ( n ) , where f (n) is

Example 75

nƕ

Column 1 Ê 1 + (a) Á ÁË n 2

Column 2

(b) Ê n + 1 + n + 2 + .. 1 ˆ Á n 2 + 12 n 2 + 22 n ˜¯ Ë 1 n

Ans.

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(r) p /6

ˆ ˜ (s) p /2 4 n 2 - n 2 ˜¯ 1

+ .. +

4 n 2 - 22

p

Solution: (a)

1

p 1 + log 2 4 2 (q) 1/2

È 2 p np ˘ 2p + sin 2 + .. + sin 2 Ísin ˙ 2n 2n ˚ 2n Î

Ê 1 + (d) Á ÁË 4 n 2 - 1

1

=Ú 0

Ú0

(1 - cos p r ) dx =

1 1 È si n p x ˘ ˘ 1 ˙ 2 ÍÎ p ˙˚0 ˙˚

1 = . 2 The limit in (d) is equal to Ú

1

0

dx 4 - x2

= sin -1

x 2

1

= p /6 0

Example 76 The area bounded by curves Column 1 Column 2 2 (a) y = x + 2, y = – x, x = 0 and x = 1 (p) 4/3

(d) x = – 2 y2, x = 1 – 3y2 p q r s Ans. a p q r s

1 n -1 1 = lim  2 2 nƕ n r =0 n -r 1 - (r 2 / n2 )

r =0

1

dx 1- x

1

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

(q) 2 2 (r) p – 2/3 (s) 17/6

1

Â

nƕ

1 2

(c) y = sin x, y = cos x (one of the region)

Ê 1 ˆ 1 1 1 lim Á + + + ... + ˜ 2 nÆ •Á n2 - 1 n2 - 22 n2 - ( n - 1)2 ˜¯ Ë n = lim

=

(b) y = x2, y = 2/(1 + x2)

For (a)

n -1

1

1 n 2p p sin 2 = Ú sin 2 x dx  nÆ• n 2n 0 2 r =1

= lim

ˆ 1 1 + .. + ˜ n2 - 1 n 2 - (n - 1) 2 ˜¯ (p)

(c)

np ˘ 1È 2 p 2p sin + sin 2 + ... + sin 2 (c) nlim Æ• n Í n n 2 2n ˙˚ 2 Î

2

= sin -1 x

=

p 2

Solution Let g(x) = – x and f (x) = x2 + 2. Then g (x) £ f (x) for all x in [0, 1]. Moreover, y = g(x) represents a straight line and y = f (x) represents a parabola. So the required area is (Fig. 26.7)

0

È ˘ 1 Ê rˆ j Á ˜ = Ú j ( x) d x ˙ Í nlim  ƕ Ë n¯ 0 r =1 n Î ˚

Ú

1

n

n n+r Ê n +1 1ˆ n+2 lim + + ◊ ◊ ◊  (b) nlim = 2 2 2 2 2 2 ˜ nÆ• Æ•Á n¯ Ë n +1 n +2 r =1 n + r

1

0

=

Ú

1

0

[(x2 + 2) – (–x)]dx x3 x2 + 2x + (x + 2 + x)dx = 3 2

1 n 1 + (r / n)  2 2 nƕ n r = 1 1 + (r / n )

= lim

1+ x 1 1 2x dx = Ú dx + Ú dx 2 2 2 0 1 + x2 + + 1 x 1 x 0 0 1

1

1



1

= tan -1 x 0 +

1 log (1 + x 2 ) 2

1 0

=

1

=

2

p 1 + log 2 4 2 Fig. 26.16

0

17 . 6

IIT JEE eBooks: www.crackjee.xyz 26.30 Comprehensive Mathematics—JEE Advanced

For (d), solving the system of equations we get the points of intersection y1 = – 1, y2 = 1. Since 1 – 3y2 ≥ – 2y2 for – 1 £ y £ 1, the required area is equal to (Fig. 26.19) 1 4 Ú-1 [(1 – 3y2) – (– 2 y)2] dy = 3 . 2t + 1

x

For (b) Clearly y = x2 is a parabola. The curve y = 2/(1 + x2) passes through (0, 2) and as x Æ • y Æ 0. Moreover y is never greater than 2(if y > 2 then x2 < 0). So the shape of y = 2/(1 + x2) is a bell-shaped curve. The points of intersection of these curves is given by x2 = 2/(1 + x2), i.e., (x2)2 + x2 – 2 = 0 which implies x2 = – 2, 1 so x = ± 1, and hence y(1) = 1, y(–1) = 1. Required area is (Fig. 26.17) È 1 2 ˘ 1 = 2 ÍÚ d x - Ú x2 d x˙ 0 0 1 + x2 Î ˚ 2 1˘ Ê p 1ˆ È = 2 Í2 [tan -1 1] - ˙ = 2 Á - ˜ = p - . ¯ Ë 3 3 2 3 ˚ Î For (c) The points of intersection of f (x) = sin x and g(x) = cos x are given by x = p /4 and 5p/4. Since sin x ≥ cos x on the interval [p/4, 5p/4], the area of the required region is (Fig. 26.18) 5p / 4

5p / 4

(sin x – cos x)dx = – cos x – sin x

Column 1 (a) F (x) < 2 for x in 4 - 3p (b) F (x) > for x in 4

Column 2 (p) [–1, 0)

(c) F increases on (d) F decreases on p q r

(r) [–1/2, 1] (s) [–1, 1]

Ans.

Area OPSQO = 2[area (ORQS) – area (OQR)]

p /4

dt, x Œ [–1, 1]

then

Fig. 26.17

Ú

Ú0 t 2 - 2t + 2

Let F (x) =

Example 77

p /4

È 2 2 Ê 2 2ˆ˘ =Í + - Á˙ = 2 2. 2 Ë 2 2 ˜¯ ˙˚ ÍÎ 2

Solution

(q) [–1, –1/2]

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

2x + 1

F¢ (x) =

x2 - 2 x + 2

=

2x + 1 x2 - 2 x + 2

Since x2 –2x + 2 = (x – 1)2 + 1 > 0 so F¢ (x) > 0 for x > –1/2 and F¢ (x) < 0 for x < –1/2. Hence F increases on (–1/2, 1) and decreases on [–1, –1/2]. The greatest value of F is F (1) and least value is F (–1/2). But 1

2t + 1

0 t2

- 2t + 2

F (1) = Ú

dt = Ú

1

2t - 2 + 3

0 t2

- 2t + 2

= log |t2 – 2t + 2 10 + 3

dt

1

Ú0 (t - 1)2 + 1

= 0 – log 2 + 3 tan–1 (t – 1) = – log 2 + 3

dt

1 0

p 2 – 1 – 3 tan–1 1 = 1 – Fig. 26.19

-1/ 2 0

3p 4

IIT JEE eBooks: www.crackjee.xyz 26.31

ASSERTION-REASON TYPE QUESTIONS Statement-1: For a, b > 0, a2 > b

Example 78 Ê 1 + x a+ ln Á • Ë 1 + x a-

Ú0

ˆ b˜ ¯ b

(1 + x ) ln ( x ) 2

dx =

p/4 < x < 3p/4

p b 2

Statement-2: If f is an odd function, on (–a, a), then

Ú

Ú0

(

ln 1 + x m

(1 + x ) 2

)

Ans. (a)

Put x = q + p/2, then

(

2

and

1

(

ln 1 + t - m

0

)

1ˆ Ê ÁË1 + 2 ˜¯ ln (1 / t ) t

1

\ g( q) is an odd function. Thus,

Ú

I = mÚ

1

np

Ú0

1 sin x 2 1 1 dx ≥ ÊÁ1 + + + ◊ ◊ ◊ + ˆ˜ n¯ x pË 2 3 sin x 2 ≥ (0, p /2) p x p sin x 2p sin x sin x dx = Ú dx + Ú dx 0 p x x x

np

Next, let

Solution ˆ b˜ ¯ b

Ú0

+ ◊◊◊ + Ú

2

Ú

3p /4

p /4

np

( n - 1)p

dx

(1 + x ) ln ( x ) = I (a + b ) - I (a - b ) p p = È(a + b ) - (a - b )˘ = ˚ 2 4Î Statement-1:

g (q ) dq = 0

Ans. (b)

01+

Example 79

p /4

Statement-1: If n is a positive integer then

Statement-2:

dx mp = 2 4 x

f ( x ) dx = Ú

-p / 4

Example 80

2

Ê 1 + x a+ ln Á • Ë 1 + x a-

3p / 4

p /4

Ê -1ˆ ÁË 2 ˜¯ dt t

m

0

0

g(q )dq

g(– q) = – g (q)

(1 + x ) - m lnn ( x ) dx = -Ú (1 + x ) ln ( x ) 1 ln

J =Ú

p /4

-p / 4

Replacing q by -q, we get

2

Put x = 1/t in I2 to obtain

\

f ( x )dx = Ú

ˆ Ê ˜ Á (-1) sin q 1 Ê 1 ˆ g(q ) = cos Á ˜ cos Á ˜¯ Ë 1 cos q ˆ˜ Ê 1 ˆ Á sin Ê cos2 q sin 2 Á ÁË ÁË cos q ˜¯ ˜¯ Ë cos q ˜¯

dx = I1 + I 2

m



I2 = Ú

3p / 4

p /4

2

0

I2

Ú

where

m

1

I1

)

(1 + x ) ln ( x ) ln (1 + x ) =Ú dx (1 + x ) ln ( x ) ln (1 + x ) =Ú dx (1 + x ) ln ( x ) 0

where

= 0.

Solution: Statement-2 is true. See theory

ln 1 + x m



f ( x )dx

Ans. (a)

mp dx = 4 ln ( x )

Solution: Let I ( m) = Ú

a

-a

Statement-2: For m > 0 •

ˆ Ê ˜ Á 1 cos x Ê 1 ˆ , f ( x ) = cos Á ˜ cos Á ˜ Ë sin x ¯ 2 2Ê 1 ˆ Á sin ÊÁ 1 ˆ˜ ˜ sin x sin Á ÁË Ë sin x ¯ ˜¯ Ë sin x ˜¯



p

0

b

f ( x ) dx = 0, where

sin x x

dx

p sin u p sin u sin x dx + Ú du + Ú du + 0 p +u 0 p + 2u x p

Ú0

sin u du u + (n - 1)p

(Putting x – p = u in 2nd integral, x – 2p = u in 3rd integral,...)

IIT JEE eBooks: www.crackjee.xyz 26.32 Comprehensive Mathematics—JEE Advanced p

Ú0 \

p sin u sin u 2 ≥Ú = 0 p + ( r - 1)p u + (r - 1)p rp

np

Ú0

sin x 2 dx ≥ x p

È 1 Í1 + 2 + ◊ ◊ ◊ + Î

r = 1, 2, ◊◊◊, n

log t

1/ x

1 + t + t2

dt

log (1/ u ) Ê du ˆ Ê 1ˆ u= ˜ 1 1 ÁË u 2 ˜¯ ÁË t¯ 1+ + 2 u u x log u =Ú 2 du = F ( x) 1 u + u +1



x

F (x) + F (1/x) =

Ú1

x

=

Ú1

x

=

Ú1

x

=

Ú1

x

=

Ú1

log t dt + t +1

x

Ú1

Ê 3ˆ dt (x > 0) at x = 1 is ÁË ˜¯ cos 1 2 d j ( x) f (t ) dt = f (j(x)) – f (y(x)) Statement-2: dx Úy ( x ) Ans. (c) d j ( x) Solution: f (t ) dt = f (j(x)) j¢(x) – f (y(x)) y¢(x) dx Úy ( x ) x

Ú1/ x cos t

=

log (1/ u ) Ê du ˆ Á- ˜ 1 + 1/ u Ë u 2 ¯

Statement-1: If I =

cos x

Ú0 1 + x2

dx

–1

then

-1 Úa f ( x ) dx + Ú f (a) f ( x ) dx = b f (b) - a f (a) .

Example 84

p /2

Ú-p / 2 2

sin x

dx + Ú

4

5/ 2

(a)

5p 4

(b) p

(c)

3p 4

(d)

sin -1 log2 ( x - 2) dx is

p 4

Example 85 If f (x) = x3 + x then dx then

Ú f ( x ) dx + 2 Ú 2

1

5

1

f -1 ( x - x ) dx is

(a) 21

(b) 9

(c) 8

(d) 18

Example 86 If f (x) = ex + 2x + 1 then

b

Úa j ( x)

f (b )

b

log t (log x) 2 dt = t 2

f ( x) j (x) dx = f (c)

( x )¢ – cos (1/x)2 (1/x)¢

If f is an invertible function and its inverse is f

Statement-2: If f and j continuous function on [a, b] then

Úa

2

Paragraph for Question Nos. 84 to 86

I < (p/4) cos 1

b

( x)

1 1 cos x 2 + (cos (1/x )) 2 x 2 x 1 Ê 3ˆ F¢ (1) = cos 1 + cos 1 = ÁË ˜¯ cos 1 2 2

Thus

Ê 1 1 ˆ + 2 log t Á ˜ dt Ë t +1 t + t¯ log t Ê 1ˆ 1 + ˜ dt t + 1 ÁË t¯

for some c Π(a, b)

2

=

log t dt t +1

1

Example 82

p (for some c, 0 < c < 1) 4

COMPREHESION-TYPE QUESTIONS

1/ x

Ú1

0

Statement-1: The derivative of F (x)

Example 83

F¢ (x) = cos

So statement-1 is not true. x log t If F (x) = Ú1 dt then t +1 log t dt + t +1

Ú0 1 + x2

1

= cos c tan–1 x

But the function cos x decreases on (0, 1) so cos c > cos 1. p Thus I > (cos 1) . Statement-2 is known as generalized 4 mean value theorem.

1

x

Ú0 1 + x2

dx

1

dx = cos c

= cos c

Ans. (d)

Ú1

cos x

1˘ n ˙˚

F (x) + F (1/x) = (1/2) (log x)2

F (1/x) =

1

Solution

Statement-2 is true but is not a correct reason for the statement-1. x log t Example 81 Statement-1: F (x) = Ú1 dt 1 + t + t2 then F (x) = – F (1/x) x log t Statement-2: If F (x) = Ú1 t + 1 dt then

Solution

Ans. (a)

(a) 4 (c) 2

(b) 1/6 (d) 8

Ú

e+3

2

f -1 ( x ) dx is

IIT JEE eBooks: www.crackjee.xyz 26.33

Ans. 84. (a), 85. (d), 86. (c) Solution:

(a) 1 + 2

È p p˘ 84. f (x) = 2 sinx is invertible over Í- , ˙ with inverse Î 2 2˚ sin–1 (log2 x)

Ú

p /2

-p / 2

2

sin x

2

dx + Ú sin

-1

1/ 2

(log2 z ) dz

Example 89

f ( x ) dx + Ú

10

Ú

2

1

5

f ( x ) dx + 2 Ú f (2 x ) dx -1

1

f -1 ( x ) dx = 2 ¥ 10 – 1 ¥ 2 = 18

2

e +3

Ú2 \

Ú0

f is an increasing function on [0, 1] so invertible

86.

1

f -1 ( x ) dx + Ú f ( x ) dx = 1(e + 3) - 0 (2)

e +3

(b)

1.22

1.2

(d)

1.4

Ans. 87. (a), 88. (b), 89. (c) Solutions: 1 sin x dx = sin c 87. Ú0 1 + x2

1

(

0

)

Ú

b

a

Ú

g(x) f (x) dx = f (c)

a

g(x) dx. This

b f ( a) + f ( b) (b – a) < Ú f ( x ) dx < (b – a) f (b) a 2

Úa

f ( x) g ( x)d x £



a

f 2 ( x ) dx

)(Ú

b

a

g 2 ( x ) dx

sin x dx is 1 + x2 (a) (p/4) sin 1

(b) p sin 1

(c) (p/2) sin 1

(d) (p/4) sin 1/2

upper bound of

Example 88 bound of

Ú

1

0

)

Using Mean-Value Theorem, the best

Example 87

Ú

p sin c (0 < c < 1) 4

sin x p sin 1. 2 dx < 1+ x 4 88. f (x) = 1 + x 4 is concave on [0, 1] Hence

Ú

1

0

b

be used for estimation are (i) If f increases and has a concave graph in the interval [a, b] then b f ( a ) + f (b ) (b – a) f (a) < Úa f (x) dx < (b – a) 2 (ii) If f increases and has a convex graph in the interval [a, b] then

b

=

dx 1 + x2

Since the function sin x increases on [0, 1] so sin c < sin 1.

is known as Mean-Value Theorem. This result can be used

2

0

0

If functions f (x) and g(x) are continuous on the interval [a, b] and g(x) retain the same sign on [a, b] then there is c Π(a, b) such that

1

Ú

1

= sin c tan x

Paragraph for Question Nos. 87 to 89

(iii)

(c)

–1

f -1 dx = e + 3 - Ú e x + 2 x + 1 dx = 2.

b

Using (iii) above the best upper bound of

1 + x dx is (a) 1.2

0

Ú2

2 -1

4

f (x) = x + x is invertible on [1, 2] with inverse f (x)

Ú1

)

(

–1

whose domain is [2, 10] \ =

1

3

2

(d) 2

(Putting x – 2 = z)

Ê p ˆ Ê 1ˆ = 2 Á ˜ - Á ˜ (-p /2 ) = 5p /4 Ë 2 ¯ Ë 2¯ 85.

2 -1 2

(c)

1+ 2 2

(b)

1

since f ¢¢(x) = So 1
0

f ¢¢(p

Division by x2 gives

y–1/3 = v fi y -4 /3 – 3x

dy dv =–3 dx dx

dv + 6v = 3x dx

dv 2 - v =–1 dx x which is a linear equation whose integrating factor is x–2. 1 x–2 v = +C x fi v = x + Cx2 fi



y(x) =

(x

Cx 2 )3

IIT JEE eBooks: www.crackjee.xyz Differential Equations 27.19

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 48 to 50

dx + k1x = k1m0. This is a linear dt equation with integrating factor ek1 t. So the solution is

the rate of generation of the product is proportional to the

given by xe k1 t = k1 m0

e k1t + const. But for t k1

x=0

so const = – m0. Hence x = m0 (1 – e–k1 t). The equation (ii) A substance whose initial quantity is m0 turns into another substance and from the product thus obtained a second product begins to generate immediately. Both reactions are k1 for k2 for the second reaction. If x is the t units y is the quantity of the second product. y at t = 2 is given by m0 (a) m0 + (2k1 e– k1 – 2k1 e– k 2) k1 - k2

Example 48

dy + k2 y = k2 m0 (1 – e– k 1t dt linear equation. Hence reduces to

È e k2t e( k2 - k1 )t ˘ ye = k2 m0 Í ˙ + const. k2 - k1 ˙˚ ÎÍ k2 But y = 0 at t = 0 so k1 t

k2 m0 . Hence k2 - k1

const. = – m0 +

m0 (k1 e–k2t – k2 e–k 1t ) k2 - k1

(b) m0 +

2 m0 (k1 e– k1 – k1 e–k 2) k1 - k2

(c) m0 +

m0 (k1 e–2k1 – k2 e–2k2) k1 - k2

At

(d) m0 +

m0 (k2 e–2k1 – k1 e–2k2) k1 - k2

and lim y = m0 and Ú0 t Æ•

Example 49

(b) m0 +

y = m0 + 1

m0 (k1 e–2k2 – k2 e–2k 1) k2 - k1 ydt = m0

Ê - k2 e - k1 t k1 e - k2 t ˆ m0 + + k1 - k2 ÁË k1 k2 ˜¯

m0 k1 - k2 (d) k1m0 + k2

(b) m0 +

2m0 (c) m0 + k1 - k2

(a) m0 +

t

lim y is equal to

t Æ•

(a) m0

Example 50

y = m0 +

= m0 +

Ú y dt is equal to

m0 k È k2 - k1 ˘ e - 1 + 1 e - k2 - 1 ˙ Í k1 - k2 Î k1 k2 ˚

(

)

(

)

A chemical manufacturing company has a 1000 k1 holding

m0 [e–k2 – e–k1] k1 - k2

kl of water k1 of pollutant per k1 1

kl per hour

+ k12 e–k2 + k22 – k12]

(c) m0 +

m0 [e–k1 – e–k2] k1 - k2

(d) m0 +

m0 k2 [e– k2 – e– k1] k1 - k2

of 40 kl per hour. If P (t t. Example 51

The value of P(10) is approximately

(a) 2000 (c) 2109

Ans. 48. (d); 49 (a); 50. (b) Solution of Examples 48-50 According to the given conditions dx dy = k1(m0 – x) and = k2(x – y) dt dt

0

Paragraph for Question Nos. 51 to 53

1 0

m0 1 [– k 22 e k1 - k2 k1 k2

1

Example 52

(b) 3100 (d) 4321 1

The value of Ú0 P(t) d t is

(a) 1140 - log 5/3 (ii)

(b) 1140 - 3240 log

10 9

IIT JEE eBooks: www.crackjee.xyz 27.20 Comprehensive Mathematics—JEE Advanced

(c) 2140 - 1140 log 8/9 10 (d) 2140 - 3240 log 9 Example 53

dP dt

t = 10

who have heard it and the number who have not. Suppose that 100 people initiate the rumour and that a total of 500

is approximately equal to

(a) 158 (c) 129

Let y (t rumour at time t Example 54 The time required for half the population to hear rumor is

(b) 149 (d) 212

(a) 4 days (c) 4.58 days

Ans. 51. (c); 52. (b); 53. (c) Solutions to Examples 51–53: dP is given by dt

Example 55

¥

Thus we have fi

Ú

dt 9+t

= e log (9 + t) = 9 + t. Thus

d (P(9 + t)) = 240 (9 + t) dt 240 (9 + t)2 + C fi P(9 + t) = 2 C fi P = 120 (9 + t) + 9+t Since the amount of pollutant at t = 0 is 2 ¥

Example 56

3240 40080 = 19 19

(a)

5000 1 + 50e -5000 kt

(b)

5000 1 + 49e -5000 kt

(c)

5000 1 - 50e -5000 kt

(d)

500 1 + 49e -5000 kt

54. (c); 55. (a); 56. (b)

Solutions of Example 54–56: Let y(t) denote the number of people who t. Maximum value of y(t) is y(0) = 100 and y



C = 720 fi C = – 3240 9 3240 P(t) = 120 (9 + t) – 9+t

Now P (10) = 120 ¥ 19 –

The function y(t) is given by

dy dy μy(5000 – y) fi = k y(5000 – y) dt dt Separating the variables and integrating we have

We have P(0) = 120 ¥ 9 +

Ê ˆ Ú0 P (t) dt = 120 ÁË 9 + 2 ˜¯ - 3240 (log 10/9) 1

= 1140 – 3240log (10/9) Paragraph for Question Nos. 54 to 56 A rumour spreads through a population of 5000 people at a rate proportional to the product of the number of people

dy

Ú y 5000 - y = kt + C ( )



1 È 1 dy ÍÚ d y + Ú 5000 Î y 5000 -



1 y = kt + C log 5000 5000 - y



y = Const e5000 kt = Ke5000kt 5000 - y

~ (approximately)

1

(d) 5000 log (1 – e–5000k )

Ans.

equation whose I.F. = e

Thus

(b) 5000 log (1 + e5000k ) (c) 5000 log (1 + e500k )

dP P = 240 – 9+t dt

dP P + dt 9 + t

1

Ú0 y (t) dt is equal to

Ê 49 + e5000 k ˆ (a) 5000 log Á ˜ 49 Ë ¯

dP = (rate in) – (rate out) dt

kl/hour and is leaving at the rate of 40 kl amount of water at time t is 360 + 40 t kl. Hence amount P(t ) and the rate at which of pollutant at time t is 360 + 40t P(t ) P(t ) = . 360 + 40t 9+t

(b) 6.2 days (d) 3.32 days

˘ ˙ = kt + C y˚

fi y(1 + Ke5000 kt) = 5000 Ke5000kt Put y K) = 5000 K fi K = 1/49. Using y k k ) = 5000 Ke 500 (1 + Ke 49 k k fi 1 = 9 Ke fie = 9

(i)

IIT JEE eBooks: www.crackjee.xyz Differential Equations 27.21



fiy=

2500 = fi

5000/49 e - 5000 kt + 1/49

=

5000 49e - 5000 kt + 1

5000 1 + 49 e - 5000 kt

1 + 49 e–5000 kt = 2 fi e–5000kt = 5000 kt = log 49 fi t =



(a) x2 = 2y + k (c) y2 = 2x + k

1 49 k= log 10, 000 9

t=

1 49

log 49 5000 k

2 log 49 2 2 = = log 49 - log 9 1 - log 9 1 - 129 229 log 49

Example 59 The orthogonal trajectories for y = Ce–x represent (a) exponential curve (b) family of ellipses (c) family of circles (d) family of parabolas Ans.

Ú

e5000 kt y(t) dt = 5000 Ú dt 49 + e5000 kt 1 0

Ê 49 + e5000 k ˆ =5000 log Á ˜. 49 Ë ¯ Paragraph for Question Nos. 57 to 59 Any curve which cuts every member of a given family of curve at right angle is called an orthogonal trajectory of y = mx passing

f (x y c) = y – cx2 = 0

Solution: fi fi

x 2 y ¢ - 2 xy y = c. Differentiating =0 x2 x4 2y y¢ = x

Replacing

= 4.58 days 1 0



dy dx by – dy dx

–x 2y

1 2 x = constant 2 which represent a family of ellipses yx–3/2 = C and differentiate this 3 dy dx by – equation y¢ x–3/2 – x–5/2y = 0. Replace dy 2 dx have fi

y2 +

1+

3 –1 dy x y =0 2 dx 2 xdx 3 1 y2 = – x2 + const 2 3 2x2 + 3y2 = const ydy = –

(i) Let f (x y z c is an arbitrary parameter. (ii) Differentiate the given equation w.r.t. x and then eliminate c. dy dx by – in the equation obtained (iii) Replace dy dx in (ii).



(iv) Solve the differential equation in (iii).

we have 1 – y

Example 58 The orthogonal trajectories for y = Cx3/2 where C is arbitrary constant is given by

y¢ =

2y dy + x dx = constant

of the circles x2 + y2 = a2.

Example 57 The orthogonal trajectories for y = Cx2 where C is arbitrary constant represent (a) a family of circles (b) a family of straight lines (c) a family of ellipses (d) a family of hyperbolas

(b) 2x2 + 3y2 = k (d) 3x2 + 2y2 = k



yex = C and differentiate y¢ex + exy = 0. Replace y¢ by –

dx dy

dy y2 = 0 fi ydy = dx fi =x+C 2 dx which represent a family of parabolas.

INTEGER-ANSWER TYPE QUESTIONS Example 60

If y = e4x + 2e– x

y3 + Ay1 + By = 0 then the value of Ans.

4

Solution:

1 AB is 39

y1 = 4e4x – 2e–x fi y2 = 16e4x + 2e–x fi y3

= 64e4x – 2e–x

IIT JEE eBooks: www.crackjee.xyz 27.22 Comprehensive Mathematics—JEE Advanced

Putting these values in y3 + Ay1 + By 64e

4x

– 2e

–x

+ A(4e

4x

–x

– 2e ) + B(e

4x



–x

+ 2e ) = 0

-

(64 + 4A + B)e4x + (–2 – 2A + 2B )e–x = 0 This is true for all x only if 64 + 4A + B = 0 and 2 + 2A – 2B = 0. Solving we get A = –13 and B = –12. i.e. AB = 156. y=

Example 61 A sin (98t + w0) + B cos (98 t + w0) (w0 1 is y2 + Cy = 0 then the value of C - 1 is 2401 Ans. 3 Solution: fi

Ans. 5 Solution: The given family of lines can be represented by the equation x cos a + y sin a = 10 (1) where a is an arbitrary constant. Differentiating we have (2) cos a + sin a y1 = 0 Multiplying (2) by x and subtracting it from (1) dy y sin a – x sin a = 10 dx (3) fi ( y – xy1) sin a = 10 Multiplying (1) by y1 and (2) by y and subtract xy1 cos a – y cos a = 10y1 fi (xy1 – y) cos a = 10 y1 (4) y – xy1)2 = 100 (1 + y12). Let

yf' ( x ) - y 2 dy = dx f ( x) f

The given equation can be written as dy f ¢( x ) - y 2 -y = f ( x) f ( x) dx

dx

= e log f ( x ) = f ( x )

have d f ( x) [u f (x)] = 1 fi u f (x) = x + constant fi y = dx x+C 1 f ( x) fi y(1) = fi C = 0. Thus y = 1+ C x Hence y(4) =

1296 = 324 4

Example 64 trajectory of x2 + y2 – ay = 0 then the orthogonal trajectory is a circle with radius Ans. 5 Solution: Differentiating w.r.t. x 2x + 2y

dy dy -a = 0. fi a = 2 dx dx

Ê dx ˆ ÁË x d y + y˜¯ .

Ê dx ˆ dx = 0. x2 + y2 – 2 Á x + y˜ y = 0 fi x2 – y2 – 2xy dy Ë dy ¯ dy dx Replacing by (see paragraph for Q57 to 59) d x d y we have y2 - x2 dy dy = 0. fi = x2 – y2 + 2xy 2 xy dx dx This is a homogeneous equation in x and y. Putting y = Vx

f (x) is a

f (4) = 1296. If y (1)

1 = 1 then y(4) is equal to 81 Ans. 4 Solution:

This is a linear equation in u so its integrating factor is

If the differential equation of all straight

(y - xy1)2 = 20A(1 + y12 ) then A is equal to ... units.

Example 63

du f ¢( x ) 1 +u = dx f ( x) f ( x) f ¢( x)

d2 y = – 9604 A sin (98t + w0) – 9604 B cos (98t + w0) d t2 = – 9604 y fi y2 + 9604y = 0. Hence C = 9604.

(i)

1 d y du 1 = u so that - 2 = . Now (i) becomes y y dx dx

eÚ f( x)

dy = 98A cos (98t + w0) - 98B sin (98 + w0) dt

Example 62

Put

1 d y 1 f ¢( x ) 1 + = 2 y d x y f ( x) f ( x)

V+x

V 2 x2 - x2 dV V2 -1 = = 2 x ◊V x dx 2V

dV V 2 - 1 - 2V 2 1+V2 = = dx 2V 2V dx – 2V fi = dV x 1+V 2 x

fi log x + log (1 + V2) = const fi

x(1 + y2/x2) = const = b

fi x2 + y2 – bx = 0 is required orthogonal trajectories.

IIT JEE eBooks: www.crackjee.xyz Differential Equations 27.23

b = 10. Thus x2 + y2 – 10x = 0 which is circle of radius 5. Example 65 The value of y x(1 – y y (0) = 4/3 is

( 8 ) - 13 if (1 + x ) dd xy = 2

Solution: Let the height of the water level above the base at time t be h(t) cm and let it fall dh(t) in time dt. Let the decrease in volume be dv. Let r be the radius of the surface of water at time t r2 = (200)2 – (200 – h)2 = 400h – h2 cm and dV = – p r2 dh

Ans. 3 Solution: The given equation can be written as dy x x + y= d x 1 + x2 1 + x2

dh dV = – p r2 dt dt



This is a linear equation with I.F. 1 + x 2 . So y 1 + x 2 = Ú x 1 + x 2 dx + C. = (1/3) (1 + x2)3/2 + C fi

y = (1/3) (1 + x2) + C (1 + x 2)–1/2

4 1 = y(0) = + C fi C = 1. Hence y 3 3

( 8)

1 1 1 = 3. ◊ 9 + fi y ( 8) – 3 3 3 1 y(log 4) if y2 – 7y1 + Example 66 The value of 160 12y = y y1(0) = 7 is

=

Fig. 27.2

V=

Ans. 2

dV dt

Solution: The given equation can be written as Ê d ˆ Êdy ˆ ÁË d x - 3˜¯ ÁË d x - 4 y˜¯ = 0

(i)

dy du – 4y = u then (i) reduces to – 3u = 0 dx dx du dy = 3dx fi u = C1 e3x fi dx u 3x – 4y = C1 e which is a linear equation whose I.F. is e–4x. d So (ye–4x) = C1 e–x fi ye– 4x = – C1 e–x + C2 dx If

fi y = C1 e + C2 e . So 2 = y(0) = C1 + C2. y1(0) = 3C1 + 4C2 = 7 fi C1 = C2 = 1. 3x



Example 67 is initially full of water and has an outlet of 12 cm2 cross sectional area at the bottom. The outlet is opened at some 2gh (t )



g is the

36 2 gh 5

h.

dh 36 = dt 5p

2g

h 400h - h2

36 5p

2g

1 400 h - h3/2

(400

)

h - h3/2 dh = -

36 5p

2 g dt

800 3/2 2 5/2 36 h – h =– 3 5 5p

2g t + C

At t

h = 200 cm 800 2 56 C= (200)3/2 – (200)5/2 = 2 ¥ 105 cm 15 3 5 t h = 0. Let t = T when h 0= -

V(t) and h(t) are respectively

of water level above the outlet at time t acceleration due to gravity. If t 27 T g - 1 10 is equal to 2p Ans. 7

d h 36 = 2g dt 5

= -

4x

Hence y = e3x + e4x. Thus y(log 4) = 43 + 44 = 320.

V(t) = 0.6

Thus – p r 2

3 2 gh . Rate 5

36 5p

2g T +

2 ¥ 56 ¥ 5p



T=

Thus

27T g -3 10 = 1400 p

15 ¥ 36 ¥ 2g

2

56 ¥ 105 15

105 =

14 p 105 units 27 g

IIT JEE eBooks: www.crackjee.xyz 27.24 Comprehensive Mathematics—JEE Advanced

Example 68 An object falling from rest in air is subject not only to the gravitational force but also to air resistance. Assume that the air resistance is proportional to the velocity with constant of proportionality as k acts in a direction opposite to motion (g = 9.8 m/S2). If the 10 Ak is velocity cannot exceed A m/s then the value of 49 Ans. 2 Solution: Let V(t) be the velocity of the object at time t. dV We have dt fi

dV = dt 9.8 - kV kt + C fi

e–kt

But V

Putting y = v x

d v v( v sin v + cos v) = dx v sin v - cos v d v v( v sin v + cos v) 2 v cos v fi x = -v= dx v sin v - cos v v sin v - cos v v sin v - cos v dx dx 1ˆ Ê p1 fi dv = 2 fi Á tan v - ˜ d v = 2 Ë ¯ x x v cos v v v+x



log sec v – log v = log x2 + constant sec 1 = constant fi sec = constant fi xy x fi xy cos(y/x) = constant. Since y(1) = 2p p. Hence xy cos y/x = 2p y(4) y(4) 4 cos = 2. Putting x p 4

e– kt 9.8 9.8 (1 – e–kt) < k k 9.8 10 for all t. Hence V(t) cannot exceed . Thus Ak is 2. k 49 fi

If x sin ( y/x) dy = ( y sin ( y/x) – x) d x and 1 y(1) = p /2 then the value of cos ( y/e12) is 3 Ans. 4. Solution: The given equation can be written as Example 69

d y y sin ( y x ) - x = dx x sin ( y x ) Put y/x = v v+x

d v v sin v - 1 = = v – cosec v dx sin v



dx x cos v = log x + constant



cos ( y x ) = log x + constant



EXERCISE

e–kt ) fi V(t) =

– sin v dv =

y(1) = p/2 fi constant = 0. Hence cos y/e12 = log e12 = 12. Example 70

If x cos ( y/x) ( y dx + x dy) = y sin ( y/x)

(x dy – y dx y(1) = 2p then the value of y(4) Ê y (4) ˆ cos Á is Ë 4 ˜¯ p Ans. 2 4

Solution:

The given equation can be written as d y y ( y sin ( y x ) + x cos ( y x )) = d x x ( y sin ( y x ) - x cos ( y x ))

LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. The particular solution of t(1 + t2) dx = (x + xt2 – t2) dt x|t=1 = - p is 4 (b) x = – t cot–1 t (a) x = –t tan–1 t (c) x = –t tan t (d) x = –t cot t 2 2 2. General solution of y + x y¢ = xyy¢ is (b) e–y/x = Cx (a) ex/y = Cx (c) ey/x = Cy (d) ex/y = Cy x- y x+ y = sin y(0) = p then 3. If y ¢ + sin 2 2 (a) x = tan–1 e–sin y/2 (b) y2 = 4tan–1 e–2sin x/2 (c) y = 2 tan–1 e–sin y/2 (d) y = 4 tan–1 e–2sin x/2 4. The orthogonal trajectories of the family of curves an-1 y = xn are given by (b) ny2 + x2 = const (a) xn + n2y = const (d) n2x - yn = const (c) n2x + yn = const 5. The equation of the curve in which subnormal varies as the square of the ordinate is (k is constant of proportionality) (b) y = ekx (a) y = Ae2kx (c) y2/2 + kx = A (d) y2 + kx2 = A y 2 - 2 xy - x 2 and pass6. The curve satisfying y1 = 2 2 y + 2 xy x (a) a straight line (b) a circle (c) an ellipse (d) a parabola

IIT JEE eBooks: www.crackjee.xyz Differential Equations 27.25

7. The solution of differential equation dy f ( y/ x ) y = +2 is dx x f ¢ ( y/ x ) (b) y2 f (y/x) = k (a) x2 f (y/x) = k 2 (d) f ( y/x) = ky2 (c) f ( y/x) = kx 8. The degree of the differential equation of all curves having normal of constant length C is (a) 1 (b) 3 (c) 4 (d) 2 dy = sin (x - y) is 9. The solution of the equation dx x- y Ê x - yˆ = c (b) x + cot Á =c (a) y + cot Ë 2 ˜¯ 2

( )

Êx(c) x + tan Á Ë 2

yˆ ˜¯ = c

(d) none of these

10. Equation of the curve passing through the origin dy = sin(10x + 6y) is satisfying dx 5 tan(5 x + 3 y ) + 3 ˆ -1 3 (a) tan -1 ÊÁ ˜¯ = 4 x + tan Ë 4 4 –1 2 (b) sin (5x + 3y) = x (c) tan–1(10x + 6y) = 5x + 3y 5 tan(5 x + 3 y ) + 2 ˆ -1 1 (d) tan -1 ÁÊ ˜¯ = 5 x + tan Ë 4 2 dy 11. The particular solution of log = 3x + 4y, y(0) = dx 0 is

(b) 4e3x - e- 4y = 3 (a) e3x + 3e-4y = 4 3x 4y (d) 4e3x + 3e-4y = 7 (c) 3e + 4e = 7 12. The solution of the differential equation dy 1 is = 2 dx x y [ x sin y 2 + 1]

(a) x2 (cos y2 - sin y2 - 2 ce- y ) = 2 2 (b) y2 (cos x2 - sin y2 - 2ce- y ) = 2 2 (c) x2 (cos y2 - sin y2 - e-y ) = 4 2 2 2 (d) x (cos y - sin y ) = 2 ce- y/2 13. The differential equation corresponding to the family of curves y = ex (a cos x + b sin x), a and b being arbitrary constants is (a) 2y2 + y1 - 2y = 0 (b) y2 - 2y1 + 2y = 1 (d) y2 - 2y1 + 2y = 0 (c) 2y2 - y1 + 2y = 0 2

dy 14. The solution of y x + y - x = 0 is dx (a) x4/4 + 1/5 (x/y)5 = C (b) x5/5 + (1/4) (x/y)4 = C

(c) (x/y)5 + x4/4 = C (d) (x y)4 + x5/5 = C 15. The equation of the curve passing through (3, 9) y/dx = x + 1/x2 is (a) 6xy = 3x2 - 6x + 29 (b) 6xy = 3x2 - 29x + 6 (c) 6xy = 3x3 + 29 x - 6 (d) 6xy = 3x3 – 29 x + 7 x dy y Ê ˆ - 1˜ dx is = Á 2 16. The solution of 2 2 x + y x + y2 Ë ¯ (a) y = x cot (c - x) (b) cos-1 y/x = -x + c (c) y = x tan (c - x) (d) y2/x2 = x tan (c - x) 17. The solution of ( y(1 + x-1) + sin y) dx + (x + log x + x cos y) dy = 0 is (a) (1 + y-1 sin y) + x-1 log x = C (b) ( y + sin y) + xy log x = C (c) xy + y log x + x sin y = C (d) xy + y log x – x sin y = C 18. If f (x) is a differentiable function then the solution of dy + (y f ¢(x) - f (x) f ¢(x)) dx = 0 is (a) y = (f (x) - 1) + Ce-f /x ) (b) yf (x) = (f (x))2 + C (c) yef (x) = f (x) e f (x) + C (d) ( y - f (x)) = (f (x)) e-f (x) 19. Which of the following differential equation is not of degree 1 (a) x3 y2 + (x + x2) y12 + ex y3 = sin x (b) y = 2y1 + 3 1 - y 12 (c) y21/2 + sin xy1 = xy = x (d)

y1 + y = x + 1

20. The solution of the equation (2x + y + 1)dx + (4x + 2y - 1)dy = 0 is (a) log (2x + y - 1) = C + x + y (b) log (4x + 2y - 1) = C + 2x + y (c) log (2x + y + 1) + x + 2y = C (d) log (2x + y - 1) + x + 2y = C

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 21. The solution of

x dx + y d y = x d y - y dx

a2 - x 2 - y 2 x2 + y2

(a)

x 2 + y 2 = a (sin (tan-1 y/x) + C)

(b)

x 2 + y 2 = a (cos (tan-1 y/x) + C)

(c)

x 2 + y 2 = a (tan (sin-1 y/x) + const)

5

1 Ê (d) y = x tan Á const + sin - 1 Ë a

ˆ x2 + y2 ˜ ¯

is

IIT JEE eBooks: www.crackjee.xyz 27.26 Comprehensive Mathematics—JEE Advanced

dy d3 y d2 y - 2y = 0 is 4 +5 3 2 dx dx dx (a) y = C1 ex + C2 e2x + C3 e3x (b) y = (C1 + C2 x + C3 x2 ) ex (c) y = (C1 + C2x) ex + C3 e2x (d) y = (C1 + C2 x2 ) ex + C3 e2x 23. The curve for which sum of the reciprocals of the radius vector and the polar subtangent is k is (b) k r2 = C eq + 1 (a) kr = C r eq + 1 (c) r = k r eq + C (d) k2 r2 = k eq + C dy - 3y = sin 2x is 24. The solution of the equation dx 1 -3x e (2 cos 2x + 3 sin 2x) + C (a) ye-3x = 13 1 (2 cos 2x + sin x) + C e3x (b) y = 13 1 (c) y = cos (2x - tan-1 3/2) + C e3x 13 1 (d) y = sin (2x + tan-1 2/3) + C e3x 13 22. The solution of

2

du Ê du ˆ + 25. The solutions of u = u , where u = y and d u ÁË d u ˜¯ u = xy are (a) y = 0 (b) y = - 4x (d) x2y = cy + c2 (c) xy = cy + c2 2

dy x Ê d yˆ 26. The solution of Á ˜ (e + e-x ) + 1 = 0 dx Ë dx¯ (b) x = log (c + y) (a) y + e-x = const -x (d) y = ex + const (c) y = e + const 27. The solution of (3x + 3y - 4)dy + (x + y)dx = 0 is given by (a) (x + y) + log |x + y - 4| = C (b) 3x + 3y - 4 + log |x + 4y| = C 3 (x + y) + log |x + y - 2| = x + const (c) 2 (d) e1/2(x + 3y) |x + y - 2| = C 28. The solutions of p + cos px sin y = sin px cos y d yˆ Ê ÁË p = d x ˜¯ is (a) y = 0 (b) cx2 - y = sin-1 x (c) cx - y = sin-1 c (d) y =

x 2 - 1 - sin-1

x2 - 1 x

dy 29. The solution of y + px = p2 x4, p = is dx 2 4 (a) y + Cx = C x (b) xy = C2 x - C (c) obtained by putting p = C/x in the given equation (d) obtained by putting p = C/x2 in the given equation dy 30. The solution of + x = x e(n-1)y dx (a)

Ê e ( n - 1) y - 1ˆ 1 2 log Á ˜ = x /2 + C n -1 Ë e ( n - 1) y ¯

(b) e (n - 1)y = Ce (n-1) y + (n - 1)x2/2 + 1 Ê e ( n - 1) y - 1 ˆ (c) log Á = x2 + C ( n - 1) y ˜ ( n 1) e Ë ¯ (d) e (n-1)y = ce(n-1)y2/2+x + 1

MATRIX-MATCH TYPE QUESTIONS 31. The differential equation of Column 1 Column 2 (a) Parabolas having (p) is a equation whose their vertices at variables are separable origin and foci on the x-axis (b) Straight lines which (q) is a equation of order 2 ce p from origin (c) All conics whose (r) is a equation of axes coincide with degree 2 the axes of coordinates (d) Curve passing (s) is a equation whose through the origin, variable are separable all the normals to and is of order 1 which pass through x 0, y 0) 32. If p = dy/dx then the solution of Column 1 Column 2 (p) ( y + c) ( y + x2 + c) (a) p2 - 7p + 12 = 0 (cy + xy + 1) = 0 (q) y = 2cx + c2y2 (b) p3 + 2x p2 - y2p2 - 2xy2p = 0 2 3 (r) (y - 4x + c) (y (c) y = 2px + y p 3x + c) = 0 (s) ( y + c)2 = x (x (d) 4xp2 = (3x - 2)2 2)2 33. The orthogonal trajectories of Column 1 Column 2 (p) y2/3 - x2/3 = c (a) a y2 = x3 (q) x2 + 2y2 = c2 (b) y = a x2

IIT JEE eBooks: www.crackjee.xyz Differential Equations 27.27

(c) x2/3 + y2/3 = a2/3 (r) 3y2 + 2x2 = c2 (d) y2 = 4x (s) 2x2 + y2 = c2 34. The solution of the differential equations Column 1 Column 2

( ( -1 +

) dx + + y ) y dy

(a) 1 + x x 2 + y 2 x2

(p) xesin y/x = C

2

=0 y2 1 2 + (x 2 3

(b) (x - 1) dy + y dy

(q) x -

= x(x - 1) y1/3 dx (c) ( y (1 + x-1) + sin y)dx

+ y2)3/2 = C (r) y2/3 (x - 1)2/3 ( x - 1)8/3 + 4 (2/5)(x - 1)5/3 + C (s) xy + y log x + x sin y = C

+ (x + log x + x cos y) dy

=

(d) x cos y/x dy = ( y cos y/x - x) dx

35. The solution of the differential equations Column 1 Column 2 tan x + sin x (a) (x cos y/x + (p) y = C + sin x y sin y/x) y dx + (x cos y/x y sin y/x)xdy = 0 x - tan y (b) y1 = cos y (c) y - y1 cos x = y2cos x (1 - sin x) x2 + y2 x

(d) 2yy1 = e x2 + y2 - 2x x

+

(q) sin y = x - 1 + Ce-x (r) log |C x| = - e

-

x2 + y2 x

(s) xy cos y/x = C

ASSERTION-REASON TYPE QUESTIONS 36. Let a solution y = y (x) of the differential equation x

x 2 - 1 dy - y

y 2 - 1 dx = 0 satisfy y(2) =

2 3. Statement-1: y (x) = sec (sec-1 x - p /6) Statement-2: y(x) is given by

1 2 3 1 = - 1- 2 y x x

37. Let a solution y = y(x) of the differential equation y sin x + y¢ cos x = 1 satisfy y (0) = 1. Statement-1: y(x) = sin (x + p /4) Statement-2: The integrating factor of the given differential equation is sec x.

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 38 to 40 A right circular cone with radius R and height H contains a liquid which evaporates at a rate proportional to its surface area in contact with air (proportionality constant = k > 0). Suppose that r (t) is the radius of liquid cone at time t. 38. The time after which the cone is empty is (a) H/2k (b) H/k (c) H/3k (d) 2H/k 39. The radius of water cone at t = 1 is (a) R [1 - k/H] (b) R [1 - H/k] (c) R [1 + H/k] (d) R[1 = k /H] 10

40. The value of  r(i) is equal to i =1

k˘ È (a) 10 R Í2 - ˙ H˚ Î

11 k ˘ È (b) 5 R Í2 + Î H ˙˚

11 k ˘ È (c) 5 R Í2 Î H ˙˚

11 k ˘ È (d) 4 R Í2 Î H ˙˚

Paragraph for Question Nos. 41 to 43 Newton’s law of cooling states that the rate of change of the temperature T of an object is proportional to the difference between T and the (constant) temperature T of the surrounding medium. We can write it as dT = - k (T - t), k > 0 constant dt A cup of coffee is served at 185∞ F in a room where the temperature is 65°F. Two minutes later the temperature of the coffee has dropped to 155°F. (log 3/4 = 0.144, log 3 = 1.09872). 41. The temperature of any object at t = 2 is (a) t e-k + [T (0)]e-2k (b) t ek + (T(0) + t)e-2k (c) t + [T(0) - t] e-2k (d) t + 2[T(0) - t ] e-k 42. Time required for coffee to have 105°F temperature is (a) 6 min (b) 6.43 min (c) 7.23 min (d) 7.63 min 43. Temperature of the coffee at time t is given by (b) 75 + 110e-kt (a) 65 + 120e-kt -2kt (c) 65 + 140e (d) 75 + 140e-kt

IIT JEE eBooks: www.crackjee.xyz 27.28 Comprehensive Mathematics—JEE Advanced

INTEGER-ANSWER TYPE QUESTIONS 44. The differential equation of parabola with axis parallel to y-axis and length of latus rectum as 4a is of order 45. The degree of the differential equation whose primitive is c2 + 2cy + a2 - x2 = 0, where c is an arbitrary and a 2 46. The largest value of c such that there exists a p differentiable function h(x) for - c < x < c that is a solution of y1 = 1 + y2 with h(0) = 0 is 47. If the solution of

y x Ê ˆ dy = Á 2 - 1˜ 2 2 x +y Ëx + y ¯ 2

16 dx y (0) = 1, then the value of y(p/4) p is 48. If the curve satisfying (xy4 + y) dx - xdy = 0 passes through (1, 1) then the value of - 41 ( y(2))3 – 29 is 49. If the curve satisfying (1 + ex/y)dx + ex/y (1 - x/y)dy = 0 passes through (1, 1) then 9 + y (2) e2/y(2) - e is equal to 50. If the curve satisfying ydx + (x + x2y)dy = 0 pass1 es through (1, 1) then 4log y(4) is equal y (4) to

LEVEL 2

1. The degree of the differential equation satisfying

(

(b) x =

x- y+a a log +c x- y-a 2

(c) y2 = a log

x- y+a +c x- y-a

a x- y+a log +c x+ y-a 2 5. The curve y = f (x) ( f (x) ≥ 0, f (0) = 0, f (1) = 1) bounding a curvilinear trapezoid with the base [0, x], whose area is proportional to the (n + 1)th power of f (x) is (b) yn + 1 = x (a) yn = x n (d) xn + 1 = y (c) x = y 6. The general solution of yy2 = y 21 is (a) y = C1 x + C2 (b) y = C2 eC1x (c) y = C2 + eC1x (d) y = eC2x + eC1x 7. Solution of the differential equation (d) y =

dy = sin (x + y) + cos (x + y) dx is Ê x + yˆ =y+c (a) log 1 + tan Á Ë 2 ˜¯ Êx+ (b) log 2 + sec Á Ë 2

SINGLE CORRECT ANSWER TYPE QUESTIONS 1 + x2 + 1 + y2 = A x 1 + y2 - y 1 + x2

dy = 4. Solution of the differential equation (x – y)2 dx a2 is x- y-a a +c (a) y = log x- y+a 2

) is

(a) 2 (b) 3 (c) 4 (d) 1 2. A curve passing through (2, 4) satisfying the integral x

equation Ú0 t 2 y (t )dt = x3 y(x) is (a) xy2 = 32 (b) x2y = 16 (c) xy = 8 (d) y = 2x 3. The differential equation of the family of circles a, 0) and (– a, 0) is (a π 0) (a) y1 ( y2 – x2) + 2 xy + a2 = 0 (b) y1 y2 + xy + a2 x2 = 0 (c) y1( y2 – x2 + a2) + 2xy = 0 (d) y1( x2 – y2 + a2) + 2xy = 0

yˆ ˜¯ = x + c

(c) log |1 + tan (x + y) | = y + c Ê x + yˆ =x+c (d) log 1 + tan Á Ë 2 ˜¯ 8. Equation of the curve through the origin satisfying dy = (sec x + y tan x) dx is (a) y sin x = x (b) y cos x = x (c) y tan x = x (d) x sin y = y dy y = 2 when 9. y = ae–1/x + b is a solution of dx x (a) a is any real number, b = 0 (b) a = 3, b = 0 (c) a = 1, b = 1 (d) a = 2, b = 2 10. The solution of y – xy1 = 2(x + yy1), y(1) = 1 is (a) tan–1 y/x + log (x2 + y2) = log 2 1 2 (x + y2) = p/4 (b) tan–1 y/x + log 2

IIT JEE eBooks: www.crackjee.xyz Differential Equations 27.29

1 2 (x + y2) = 0 3 x 1 p (d) tan -1 + log ( x 2 + y 2 ) = 2 4 y 11. The curve such that the area of the trapezoid formed by coordinates axes, ordinate at an arbitrary point and the tangent at this point equals half the square of the abscissa is (a) y = Cx2 (b) y = (1/2)x + Cx2 (c) 3y = x + Cx2 (d) y = x + Cx2 12. A curve passing through origin, all the normals to which pass through (x0, y0) is (a) yy0 = x2 + y2 (b) a circle with centre (x0, y0) (c) x2 + y2 = x 20 + y 20 (c) tan–1 y/x + log

1 2 ( x0 + y02 ) 2 13. The trajectories orthogonal to (2a – x) y2 = x3 is (a) x2 + y2 = C(y2 + 2x2) (b) (x2 + y2)2 = Cxy (c) (x2 + y2)2 = C(y2 + 2x2) (d) (x2 + y2) = x3 + xy2 (d) x2 + y2 =

14. The solution of

x 3d x + yx 2 d y x2 + y2

(a)

x 2 + y 2 = ( 2 - 1)x

(b)

x +y

(c)

x 2 + y 2 + y/x2 =

2

2

= ydx – xdy, y(1) = 1 is

(d) x2 + tan–1 y/x = 1 +

Ê dn y ˆ 17. The general solution of xy5 = y4 Á yn = n ˜ is dx ¯ Ë given by (a) y = C1 x5 + C2 x3 + C3 x2 + C4 x + C5 (b) y = C1 x5 + C2 x4 + C3 x2 + C4 x + C5 (c) y = C1 + C2 x + C3 x2 + C4 x3 + C5 x4 (d) xy = C1 + C2 x + C3 x2 + C4 x4 (C1, C2, C3, C4, C5 being arbitrary constant) x-axis, and two ordinates, one of which is constant, the other variable, is equal to ratio of the cube of the variable ordinate to the variable abscissa. The curve passing through (1, 1)is given by (b) (2y2 - x2)3 = x2 (a) (2y2 - x2)2 = x2 2 2 3 2 (c) (y - x ) = 8x (d) (2y - x2)3 = x2 19. The solution x y y ˆ Ê1 ÁË y sin y - x 2 cos x + 1˜¯ dx + y x x 1ˆ Ê1 ÁË x cos x - y 2 sin y + y 2 ˜¯ dy = 0 is given by (a) sin y/x - cos (b) cos x/y - sin (c) sin y/x - cos (d) sin y/x + cos 20. The solution of

(d) (x2 + y2)2 + xy2 = 2 15. The solution of the equation (x – y)2 dx + 2xy dy = 0 is (a) y = x (C – log |x|) (b) x2 = y(C – log |x|) (c) y3 = x (C – log |x|) (d) log x1/3 (3y2 – 3xy + x2)1/3 - 2 3y - x = +C tan -1 3 2 16. The solution of (1 + y + x2 y) dx + (x + x3) dy = 0, y(1) = 1 p (a) y + tan–1 x = 1 + 4 p –1 (b) xy + tan x = 1 + 4 p (c) y2 + tan–1 x = 1 + 4

y/x y/x x/y x/y

y + sin x cos2 ( x y ) (a) (b) (c) (d)

+ + + -

x x x x

+

1/y 1/y 1/y 1/y

= = = =

C C C C

x dy + sin y dy cos ( x y ) cos ( x y ) = 0 is given by sin (xy) - cos x - cos y = C tan (xy) + (cos x + cos y) = C cos (xy) - (sin x + sin y) = C tan (xy) - (cos x + cos y) = C 2

+ y/x = 1 + 2 2 -1

p 4

dx +

2

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 21. The differential equation of the curve such that the initial ordinate of any tangent is equal to the corresponding subnormal is (a) a linear equation (b) a homogeneous equation (c) an equation with separable variables (d) a second order differential equation 22. The solutions of (x + y + 1) dy = dx are (b) x + y + 4 = C log y (a) x + y + 2 = Cey (c) log (x + y + 2) = Cy (d) log (x + y + 2) = C + y

IIT JEE eBooks: www.crackjee.xyz 27.30 Comprehensive Mathematics—JEE Advanced

23. The solutions of x2 y12 + xyy1 – 6y2 = 0 are (b) x2 y = C (a) y = Cx2 1 (c) log y = C + log x 2 (d) x3 y = C Ê d yˆ is 24. The solution of p2 + 2y cot x p = y2 Á p = d x ˜¯ Ë (a) y (1 + cos x) = C

(c) x2 – y2 = a2 (d) system of circles with centre on x-axis dy = y - x is given by 28. The solution of dx (a) x + C = 2 y - x + 2 log (b) x2 + C =

(c) x = 2 sin

(

)

y - x -1

)

y - x -1

(c) x + C = (y - x) + log (y - x - 1) 2

(b) y (1 – cos x) = C –1

y - x + log

(

c /2 y

(d)

(d) x = 2cos–1 c / y 25. The solution of the differential equation (x + y) dy - (x - y) dx = 0 is (a) y2 + 2xy + x2 = k (b) y2 + 2xy - x2 = k 2 2 (d) y2 - 2xy + x2 = k (c) y + 2xy + x = 0 26. The solution of the differential equation (x2 sin3 y - y2 cos x) dx + (x3 cos y sin2 y - 2y sin x) dy = 0 is (a) x3 sin3 y = 3y2 sin x + C (b) x3 sin3 y + 3y2 sin x = C (c) x2 sin3 y + y3 sin x = C (d) 2x2 sin y + y2 sin x = C 27. The orthogonal trajectories of the family of curves xy = k2 are (a) rectangular hyperbola (b) y2 = 4a (x + a)

y- x - 1 = C¢ e

x/2 - y - x

Ê d yˆ 29. The solution of Á ˜ (x2 y3 + xy) = 1 is Ë dx¯ (a) 1/x = 2 - y2 + C e-y /2 (b) the solution of an equation which is reducible to linear equation. (c) 2/x = 1 - y2 + e-y/2 2

(d)

2 1 - 2x = - y2 + Ce-y /2 x

dy x2 + y2 + 1 = satisfying y(1) = 1 30. The solution of dx 2 xy is given by (a) a system of hyperbola (b) a system of circles (c) y2 = x (1 + x) - 1 (d) (x - 2)2 + ( y - 3)2 = 5.

MATRIX-MATCH TYPE QUESTIONS 31. The solution of the differential equations Column 1 (a)

Column 2

1 + x2 - y2 xdx - yd y = xd y - ydx x2 - y2

xdx + yd y = (b) xd y - ydx (c)

(

1 - x2 + y2 x +y 2

1 + x 2 + y 2 + x 2 y 2 + xy

(d) xy1 - y =

(p)

)

2

dy =0 dx

x2 + y2

1 1 + x 2 + log 2

(q) sin-1

32. The solution of the differential equations Column 1 (a) (x2 + y)dx - xdy = 0

(p)

(b) y (1 + xy)dx - xdy = 0 (c) y1 (y - x - 4) = x + y - 2

(q) (r)

(d) xd y + ydx + y2 (xdy - yd x) = 0

(s)

1 + x2 + 1

+ 1 + y2 = C

x 2 + y 2 = tan-1 y/x + C

(r) Cx2 = y + (s)

1 + x2 - 1

x2 + y2

x2 - y2 + 1 + x2 - y2 = C

x Ê + Á 2 2 Ë x -y

ˆ ˜ x -y ¯

Column 2 y-3 1 log( x 2 + y 2 + 2 x - 6 y + 10) = tan -1 +C 2 x +1 x - y/x = C y2 - 1 + C xy = 0 2x x2 + =C y

y

2

2

IIT JEE eBooks: www.crackjee.xyz Differential Equations 27.31

Column 1 (a) y = x tan (x + 2) 1 (b) y = tan (log 2x) x (c)

Column 2 (p) xdx = 2(y – y3) dy (q) y¢ + 2xy = 2x3y3

2 1 1 = 4 e2x + x2 + 2 2 y

(r) (x2 + y2 + y)dx = xdy

(d) x2 = y2(2 - y2) 34. The curves in column 1 are represented by column 2 Column 1 (a) The curve such that the length of the normal is double the square of the ordinate. (b) The curve such that the area of the x-axis, two ordinates and arc MM¢ of this curve is equal to the arc length, MM¢ at any choice of points M and M ¢ (c) The curve such that the initial ordinate at any tangent is less than the abscissa of the point of tangency by two scale units (d) The curve such that subtangent is double to the abscissas of the point of tangency

ASSERTION-REASON TYPE QUESTIONS 35. Let y(x) be a solution of xdy + ydx + y2 (xdy – ydx) = 0 satisfying y(1) = 1 Statement 1: The range of y(x) has exactly two points. Statement 2: The constant of integration is zero. 36. Let y1 and y2 be two different solutions of the equady tion + P(x) y = Q(x) dx Statement 1: y = y1 + C(y2 - y1) is the general solution of the same equation Statement 2: If y1, y2, y3 are three different soluy - y1 tions then 2 is constant. y3 - y1

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos 37 to 39 a population of 25,000. The disease is spreading at a rate proportional to the product of the number of people who have it and the number who don’t. Suppose that 100 people had the disease initially and 400 had it after 10 days.

(s) (1 - xy + x2y2)dx = x2dy Column 2 (p) y = Cx - x log |x| - 2

(q) 4y = const . e2x + const. e–2x

(r) Catenary

(s) y2 = Cx

log

498 = 0.1382, log 249 = 5.528 123

37. The number of days so that half the population have (a) 50 (c) 45

(b) 40 (d) 25

(a) 1400 (b) 1540 (c) 1498 (d) 1492 39. The population any time t is 25000 25000 (b) (a) 498 Ê ˆ 1 - 249 e 0.1382 t logÁ t Ë 123 ˜¯ 1 - 149 e 25000 25000 (c) (d) 0.2764 t 1 + 149 e 1 + 249 e - 0.1382 t Paragraph for Question Nos. 40 to 42 A sky driver equipped with parachute falls from rest toward earth. The total weight plus the equipment is 160 kg. Before the parachute opens, the air resistance is equal 1 to v , v is the velocity. The parachute opens 5 sec after 2

IIT JEE eBooks: www.crackjee.xyz 27.32 Comprehensive Mathematics—JEE Advanced

the fall begins, after it opens, the air resistance is equal 5 194 Ê ˆ to v 2 . Á e-1/ 2 = , g = 32˜ Ë ¯ 8 320 40. The velocity at the instant the parachute open is (a) 124 m/s (b) 102 m/s (c) 126 m/s (d) 174 m/s 41. An expression for the velocity of the sky driver at time t after the parachute opens is Ê 110 20- 4t ˆ Ê 110 20- 4t ˆ 16 Á e + 1˜ 8Á e + 1˜ ¯ ¯ Ë 142 Ë 142 (b) (a) 110 20- 4t 110 20- 4t -1 1e e 142 142 Ê 110 10- 2t ˆ 4Á e + 1˜ ¯ Ë 142 (c) 110 10- 2t 1e 142 42. Velocity when t = 8 is

Ê 110 10- 2t ˆ 16 Á e + 1˜ ¯ Ë 142 (d) 110 10- 2t 1e 142

Ê 110 -6 ˆ 16 Á e + 1˜ ¯ Ë 142 (a) 110 -6 1e 142

Ê 110 -12 ˆ 16 Á e + 1˜ ¯ Ë 142 (b) 110 -12 1e 142

Ê 110 -12 ˆ 8Á e + 1˜ ¯ Ë 142 (c) 110 -12 1e 142

Ê 110 -6 ˆ 4Á e + 1˜ ¯ Ë 142 (d) 110 -6 1e 142

INTEGER-ANSWER TYPE QUESTIONS 43. If

dy - x tan (y - x) = 1, y (0) = p/2 then the value dx

of sin ( y(4) - 4)e-8 is 44. If

dy + dx

y x +a 2

y ( 3a ) 8 - 3 3 a2

2- 3

2

= 3x, y(0) = a2 then the value of

– 30 is equal to

dy = cos x (2 cos y - sin2 x) and y(0) dx = p/2 then e- 2 + 4 cos y(p/2) is equal to 46. If the curve satisfying yy1 sin x = cos x (sin x - y2) passes through (p /2, 2) then the value of 3( y (p/6))2 –38 is 47. If curve satisfying x(x + 1) y1 - y = x(x + 1) passes 5 through (1, 0) then the value of y (4) - log 4 is 4

45. If sin y

48. If a curve satisfying xy1 - 4y - x 2

y = 0 passes

2

through (1, (log 4) ) then value of y(2)/(log 32)2 is Ê x2 ˆ 49. If the curve satisfying xdx = Á - y 3 ˜ dy passË y ¯ es through (0,2) then the value of (4 - (y(4))2) is equal to

1 ( y(4))2 4

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. The order of the differential equation whose general solution is given by y = (C1 + C2) cos (x + C3) – C4 e x +C5 , where C1, C2, C3, C4, C5 are arbitrary is (a) 5 (b) 4 (c) 3 (d) 2 [1998] 2. A solution of the differential equation 2

dy Ê dy ˆ + y = 0 is ÁË ˜¯ – x dx dx (a) y = 2 (c) y = 2x – 4

(b) y = 2x (d) y = 2x2 – 4

[1999] 3. If y(t) is a solution of the differential equation dy (1 + t) – ty = 1 and y(0) = – 1, then y(1) is equal to dt (a) – 1/2 (b) e + 1/2 (c) e – 1/2 (d) 1/2 [2003] 4. If y = y(x dy (2 + sin x) = – (1 + y) cos x, y(0) = 1 then y(p/2) is equal to dx (a) 1

(b) 2/3

(c) 1/3

(d) 5/3

[2004] 5. The solution y = y(x) of the differential equation (x2 + y2)dy = xy dx satisfying the conditions y(1) = 1, y(x0) = e, then the value of x0 is (a)

3e

(b)

2(e 2 - 1)

(c)

2(e 2 + 1)

(d)

(e 2 + 1) / 2

[2005] 6. Suppose y = y(x ydx + y2dy = xdy. If y(x) > 0 " x ΠR and y(1) = 1, then y(Р3) equals (a) 1 (b) 2 (c) 3 (d) 5 [2005]

IIT JEE eBooks: www.crackjee.xyz Differential Equations 27.33

7. Let f (x) be a differentiable on the interval (0, •) such that f (1) = 1, and t f ( x ) - x f (t ) = 1 for each x > 0. Then f (x) is tÆx t-x 2

1 2x2 + 3 3x

(c) -

F (x) =

2

lim

(a)

x2

(b) –

1 2 + 2 x x

(d)

1 4 x2 + 3 3x

1 x

[2007]

1 - y2 dy 8. The differention equation = determines y a family of circles with dx

Ú

f ( t ) dt

0

for x Œ [0, 2]. If F ¢(x) = f ¢(x) for all x Œ (0, 2), then F (2) equals (b) e4 – 1 (a) e2 – 1 [2014] (c) e – 1 (d) e4 12. The function y = f (x) is the solution of the differential equation x4 + 2x dy xy + 2 = dx x – 1 1 – x2 in (– 1, 1) satisfying f (0) = 0. Then 3 2

Ú

f ( x) dx

3 2

x-axis. y-axis.

[2007]

Ê pˆ 9. A curve passes through the point Á1, ˜ . Let the slope Ë 6¯ y Ê yˆ of the curve at each point (x, y) be + sec Á ˜ , x > 0. Ë x¯ x Then the equation of the curve is

(a)

p 3 3 2

(b)

p 3 3 4

(c)

p 3 6 4

(d)

p 3 6 2

[2014]

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

1 Ê yˆ (a) sin Á ˜ = log x + Ë x¯ 2 Ê yˆ (b) cosec Á ˜ = log x + 2 Ë x¯ Ê 2yˆ (c) sec Á ˜ = log x + 2 Ë x¯ 1 Ê 2yˆ (d) cos Á ˜ = log x + Ë x¯ 2

[2013]

È1 ˘ 10. Let f : Í ,1˙ Æ R (the set of all real numbers) be Î2 ˚ a positive, non-constant and differentiable function Ê 1ˆ such that f ¢(x) < 2f (x) and f Á ˜ = 1. Then the value Ë 2¯ 1

of Ú1/2 f (x)dx lies in the interval (a) (2e – 1, 2e)

(b)

(e – 1, 2e – 1)

Ê e -1 ˆ (c) Á , e - 1˜ Ë 2 ¯

(d)

Ê e - 1ˆ ÁË 0, ˜ 2 ¯

[2013]

11. Let f : [0, 2] Æ R be a function which is continuous on [0, 2] and is differentiable on (0, 2) with f (0) = 1. Let

1. The differential equation representing the family of curves y2 = 2c (x + c ) where c is a parameter is of (a) order 1 (b) order 2 (c) degree 3 (d) degree 4 [1999] 2. Let f (x) be a function of x such that a tangent at a point P on the curve y = f (x) intersects the x-axis at A and y-axis at B. If AP : BP = 3 : 1 and f (1) = 1, then (a) f (2) = 1/8 (b) differential equation of curve is dy + 3y = 0 x dx (c) equation of normal at (1, 1) is x – 3y + 2 = 0 (d) differential equation of curve is dy – 3y = 0 [2006] x dx 3. If y(x y¢ – y tan x = 2x sec x and y (0) = 0, then p2 Êpˆ (a) y Á ˜ = Ë 4¯ 8 2

p2 Êpˆ (b) y¢ Á ˜ = Ë 4¯ 9

IIT JEE eBooks: www.crackjee.xyz 27.34 Comprehensive Mathematics—JEE Advanced

(d) If y

2 Êpˆ p (c) y Á ˜ = Ë 3¯ 9

Êpˆ (d) y¢ Á ˜ = Ë 3¯

4p 2p 2 [2012] + 3 3 3 4. Let y(x) be a solution of the differential equation (1 + ex)y¢ + yex = 1. If y (0) = 2, then which of the following statement is (are) true? (a) y(–4) = 0 (b) y(–2) = 0 (c) y(x) has a critical point in the interval (–1, 0) (d) y(x) has no critical point in the interval (–1, 0) [2015] 5. Consider the family of all circles whose centres lie on the striaght line y = x. If this family of circles is represnted by the differential equation Py≤ + Qy¢ + 1 = 0, where P, Q are functions of x, y and y¢ (here dy d2 y y ¢ = , y ≤ 2 ), then which of the following dx dx statement is (are true)? (a) P = y + x (b) P = y – x (c) P + Q = 1 – x + y + y¢ + (y¢)2 (d) P – Q = x + y + y¢ – (y¢)2 [2015] 6. A solution curve of the differential equation (x2 + xy + 4x + 2y + 4) dy - y 2 = 0, x > 0 passes through dx (1, 3). Then the solution curve (a) intersects y = x + 2 exactly at one point (b) intersects y = x + 2 exactly at two points (c) intersects y = (x + 2)2 (d) does not intersect y = (x + 3)2 [2016]

MATRIX-MATCH TYPE QUESTIONS

(b) Values(s) of k for which the (r) 3 planes kx + 4y + z = 0, 4x + ky + 2z = 0 and 2x + 2y + z (s) 4 = 0 intersect in a straight line (c) Value(s) of k for which | x –1| + (t) 5 |x – 2 | + | x + 1| + |x + 2| = 4k has integer solution(s) (d) If y ¢ = y + 1 and y(0) = 1, then value(s) of y (ln 2) [2009] 3. Match the statements/expressions given in Column I with the open intervals in Column II. Column I (a) Interval contained in the

Column II Ê p pˆ (p) Á - , ˜ Ë 2 2¯

solutions of the differential equation (x – 3)2 y¢ + y = 0 (b) Interval containing the value of the integral 5

Ú ( x -1)( x - 2)( x - 3)

1

Ê pˆ (q) Á 0, ˜ Ë 2¯ Ê p 5p ˆ (r) Á , ˜ Ë8 2 ¯

( x - 4)( x - 5)dx

(q) 0

1. Let a solution y = y(x) of the differential equation

0

¥ [cos x cot x – log(sin x)sin x]dx (b) Area bounded by – 4y2 = x and x = 1 – 5y2 (c) Tangent of the angle of intersection of curves y = 3x – 1 log x and y = xx – 1 equals

2. Match the statements/expressions given in Column I with the values given in Column II. Column I Column II (a) The number of solutions of (p) 1 sinx the equation xe – cos x = 0 (q) 2 Ê pˆ in the interval Á 0, ˜ Ë 2¯

(p) 1

p /2

cos x Ú (sin x)

[2006]

Ê pˆ (c) Interval in which at least (s) Á 0, ˜ Ë 8¯ one of the point of local 2 maximum of cos x + sin x lies (d) Interval in which (t) (– p, p) tan–1 (sin x + cos x) is increasing [2009]

1. Match the statements/expressions given in Column I with the values given in Column II. Column I Column II (a)

dy 2 = , y(1) = 1 then x+y dx (x + y + 2)2 e–y equals

equation

ASSERTION-REASON TYPE QUESTIONS x x 2 - 1 dy – y y 2 - 1 dx = 0 satisfy y(2) = 2 / 3 . (r) 16/e

Statement 1: y(x) = sec (sec–1 x – p/6) Statement 2: y(x) is given by 1 2 3 1 = – 1- 2 y x x

[2008]

IIT JEE eBooks: www.crackjee.xyz Differential Equations 27.35

INTEGER-ANSWER TYPE QUESTIONS 1. Let f be a real-valued differentiable function on R (the set of all real numbers) such that f (1) = 1. If the y-intercept of the tangent at any point P(x, y) on the curve y = f (x) is equal to the cube of the abscissa of P, then the value of f (– 3) is equal to [2010] 2. Let y (x) + y(x)g (x) = g(x) g (x), y(0) = 0, x R, where df ( x ) and g(x) is a given non-constant f (x) denotes dx differentiable function of R with g(0) = g(2) = 0. The the value of y(2) is [2011]

FILL

IN THE

BLANKS TYPE QUESTIONS

1. A spherical rain drop evaporates at a rate proportional to its surface at any instant t. The differential equation giving the rate of change at its radius of the rain drop is _____.

SUBJECTIVE-TYPE QUESTIONS 1. A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant length k, then show that the differential equation describing such curves is dy = ± k 2 - y2 y dx Find the equation of such a curve passing through (0, k). [1994] 2. Let y = f (x) be a curve passing through (1, 1) such that the triangle formed by the coordinates axes and the tangent at any point of the curve lies in the 1st quadrant and has area 2. Form the differential equations and determine all such possible curves. [1995] 3. Determine the equation of the curve passing through the origin, in the form y = f (x dy = sin (10x + 6y). differential equation dx [1996] 4. Let u(x) and v(x) satisfy the differential equations du dv + p(x)u = f (x) and + p(x)v = g(x), where dx dx p(x), f (x) and g(x) are continuous functions. If u(x1) > v(x1) for some x1 and f (x) > g(x) for all x > x1, prove that any point (x, y) where x > x1 does not satisfy the equations y = u(x) and y = v(x). [1997] 5. A and B are two separate reservoirs of water. Capacity of reservoir A is double the capacity of

reservoir B with water, their inlets are closed and then the water is released simultaneously from both the reservoirs. any instant of time is proportional to the quantity of water in the reservoir at the time. One hour after the water is released, the quantity of water is reservoirs 1 A is 1 times the quantity of water in reservoir B. 2 After how many hours do both the reservoirs have the same quantity of water? [1997] 6. A curve C has the property that if the tangent drawn at any point P on C meets the coordinate axes at A and B then P is the mid-point of AB. The curve passes through the point (1, 1). Determine the equation of the curve. [1998] 7. A curve passing though the point (1, 1) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of P from the x-axis. Determine the equation of the curve. [1999]

that of the last year. Assuming at the average food requirement per person remains constant, prove that n years, where n is the smallest integer greater than or equal to ln 10 - ln 9 ln (1.04 ) - 0.03

[2000]

9. A hemispherical tank of radius 2 metres is initially full of water and has an outlet of 12 cm2 crosssectional area at bottom. The outlet is opened at some to the law V(t) = 0.6

2gh(t ) , where V(t) and h(t)

outlet and the height of water level above the outlet at time t, and g is the acceleration due to gravity. Find the time it takes to empty the tank. [2001] 10. A right circular cone with radius R and height H contains liquid which evaporates at a rate proportional to its surface area in contact with air (proportionality constant = k > 0). Find the time after which the cone is empty. [2003] ( x + 1)2 + y - 3 at point (x, y) and x +1 passes through (2, 0). Find equation of the curve and the area enclosed by the curve and the x-axis in the fourth quadrant. [2004]

11. A curve has slope

IIT JEE eBooks: www.crackjee.xyz 27.36 Comprehensive Mathematics—JEE Advanced

12. If length of tangent at any point on the curve y = f (x) intercepted between the point and the x-axis is of length 1. Find the equation of the curve. [2005]

Answers LEVEL 1

(a) (b) (d) (d) (c)

2. 6. 10. 14. 18.

(c) (a) (c) (b) (a)

3. 7. 11. 15. 19.

(d) (c) (d) (c) (b)

4. 8. 12. 16. 20.

(b) (d) (a) (c) (d)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 21. 24. 26. 28. 30.

(a), (a), (a), (a), (a),

(d) 22. (c) (b), (c), (d) (b), (d) (c), (d) (b)

23. 25. 27. 29.

(a) (a), (b), (c) (c), (d) (b), (d)

MATRIX-MATCH TYPE QUESTIONS p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

31.

32.

33.

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

35.

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17.

p

34.

ASSERTION-REASON TYPE QUESTIONS 36. (c)

37. (d)

COMPREHENSION-TYPE QUESTIONS 38. (b) 42. (d)

39. (a) 43. (a)

40. (c)

41. (c)

INTEGER-ANSWER TYPE QUESTIONS 44. 2 48. 3

45. 2 49. 5

46. 1 50. 4

47. 4

LEVEL 2

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17.

(d) (a) (a) (c) (a)

2. 6. 10. 14. 18.

(b) (b) (b) (b) (b)

3. 7. 11. 15. 19.

(c) (d) (d) (d) (c)

4. 8. 12. 16. 20.

(a) (b) (b) (b) (d)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 21. 24. 27. 29.

(a), (a), (a), (a),

(b) 22. (a), (d) 23. (a), (c), (d) (b), (c) 25. (b) 26. (a) (c) 28. (a), (d) (b), (d) 30. (a), (c)

IIT JEE eBooks: www.crackjee.xyz Differential Equations 27.37

MATRIX-MATCH TYPE QUESTIONS

9. (a)

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

a

c

p

q

r

d

p

q

r

32.

33.

34.

11. (b)

12. (b)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

p

31.

10. (d)

1. (a), (c) 4. (a), (c)

2.(a), (b), (c) 5. (b), (c)

3. (a), (d) 6. (b)

MATRIX-MATCH TYPE QUESTIONS p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

t

p

q

r

s

t

b p

q

r

s

t

p

q

r

s

t

d p

q

r

s

t

p

q

r

s

t

p

q

r

s

t

s

b p

q

r

s

t

s

c

p

q

r

s

t

d p

q

r

s

t

1.

2. a c

3.

ASSERTION-REASON TYPE QUESTIONS 35. (b)

ASSERTION-REASON TYPE QUESTIONS

36. (b)

COMPREHENSION-TYPE QUESTIONS 37. (b) 39. (d)

38. (c) 40. (c)

INTEGER-ANSWER TYPE QUESTIONS 41. (a)

42. (b)

INTEGER-ANSWER TYPE QUESTIONS 43. 1 47. 3

44. 7 48. 4

45. 1 49. 4

3. (a) 7. (a)

2. 0

FILL 1.

SINGLE CORRECT ANSWER TYPE QUESTIONS 2. (c) 6. (c)

1. 9

46. 3

PAST YEARS’ IIT QUESTIONS

1. (c) 5. (a)

1. (c)

IN THE

BLANKS TYPE QUESTIONS

dr = -k dt

SUBJECTIVE-TYPE QUESTIONS 1. x2 + y2 = k2 2. x + y = 2, xy = 1, x > 0, y > 0

4. (c) 8. (c)

3. y =

1 tan–1 3

3˘ È4 -1 Ê 3 ˆ Í 5 tan (4 x + tan ÁË 4 ˜¯ ) - 5 ˙ Î ˚

IIT JEE eBooks: www.crackjee.xyz 27.38 Comprehensive Mathematics—JEE Advanced

5.

log 2 log (4/3)

6. xy = 1.

7. x2 + y2 = 2x

3. y¢ = sin 9.

14 p 27 g

(105) units

12. log |y| +

1- y

2

2 log tan y/4 = -4 sin

Hints and Solutions LEVEL 1

1. t(1 + t2) dx = x(1 + t2) dt – t2 dt dx x -t - = dt t 1 + t2 I.F. = e

1 - Ú dt t

=

1 t

1 d Ê 1ˆ Á x. ˜ = dt Ë t ¯ 1+ t2

fi 2.

1 x. = – tan–1 t + C t p p Putting t = 1, - = - + C fi C = 0 4 4 x = –t tan–1 t

dy y2 = . Put y = ux, so that dx xy - x 2 du dy = u+x dx dx



x dy = -2 cos dx sin y /2 2

– log (1 + 1 - y )

2

=±x+C



x y = -2 cos sin 2 2

11. y = x2 – 2x; 4/3

10. H/k

x- y x+ y - sin 2 2

x +C 2

Putting x = 0, y = p, we have C = 0, so log tan y/4 = –2 sin x/2. fi y = 4 tan–1 e–2 sin x/2 4. Differentiating, we have dy dx = nxn-1 fi a n-1 = nxn-1 an-1 dx dy Putting this value in the given equation, we have dx y = xn nxn-1 dy dy dx dx Replacing by , we have ny = - x dx dy dy 2 2 fi ny dy + x dx = 0 fi ny + x = const. which is the required family of orthogonal trajectory. dy = ky2 5. According to the given condition y dx dy = kdx fi log y = kx + C fi y = Aekx where k fi y is the constant of proportionality. 6. Put y = vx. Then dy/dx = v + x (dv/dx), and the given equation becomes Ê v 2 + 2v - 1 ˆ dx = –Á 3 fi ˜ dv x Ë v + v2 + v + 1¯

du u2 u+x = u -1 dx

È v 2 + 2v - 1 ˘ = - Í 2 ˙ dv Î ( v + 1) + ( v + 1) ˚

du u 2 - u (u - 1) u x = = u -1 dx u -1

2v ˆ Ê -1 dv. = Á + 2 Ë v + 1 v + 1˜¯

u -1 dx fi u – log u = log x + const du = u x u = log ux + const



1 ux = eu C



Cy = ey/x

Integrating both sides of this, we now get log x - log (v + 1) + log (v2 + 1) + log c = 0. This can be rewritten Ê y2 ˆ x ÁË 2 + 1˜¯ 1 È x v2 + 1 ˘ x ˙ =1fi= c Í = y c ÎÍ v + 1 ˚˙ +1 x fi c(y2 + x2) = y + x.

(

)

IIT JEE eBooks: www.crackjee.xyz Differential Equations 27.39

If this curve passes through (1, -1), we get 2c = 0, i.e., c = 0. Hence the required curve is y + x = 0, which is a straight line. dy du =u+x . The given 7. Putting y/x = u we have dx dx

Putting x = 0, y = 0, we have C = Thus the required curve is tan–1 ÊÁ Ë

differential equation can be written as du f (u) u+x =u+2 dx f¢ (u ) du f (u) fix =2 fi dx f¢ (u)

11. ¢( ) dx du = 2 ( ) x

Integrating, we get log f (u) = log x2 + log k, so f (u) = kx2 i.e. f (y/x) = kx2. 8. According to the given condition Ê dxˆ y 1+ Á ˜ Ë dy¯

2

2

Ê dxˆ = C fi y2 + y2 Á ˜ = C2 Ë dy¯

The degree of this equation is 2. du dy 9. Putting u = x - y, we have = 1. The givdx dx du = sin u fi en equation can be written as 1 dx du Ú 1 - sin u = dx

dy 1 du ˆ = ÊÁ - 10˜ ¯ dx 6 Ë dx The given differential equation reduces to 1 Ê du ˆ - 10˜ = sin u Á ¯ 6 Ë dx



du = dx 6sin u + 10 1 du =x+C Ú 2 3sin 4 + 5

Now, Ú

du du = 2Ú 2 (tan u/2 = t) 5t + 6t + 5 3sin u + 5 =

= Hence

2 du Ú 5 Ê 3ˆ 2 Ê 4 ˆ 2 ÁË t + ˜¯ + ÁË ˜¯ 5 5 1 -1 Ê 5 tan(u /2) + 3 ˆ tan Á ˜¯ + Const Ë 2 4

1 -1 Ê 5 tan(u /2) + 3 ˆ tan Á ˜¯ = x + C Ë 4 4

5 tan(5 x + 3 y ) + 3 ˆ -1 3 ˜¯ = 4 x + tan 4 4

dy = e3x + 4y = e3x e4y fi e- 4y dy = e3x dx dx Thus

e - 4 y e3 x = const. But y(0) = 0, 3 -4 -1 1 - = C. 4 3

so,

Hence e-4y/(- 4) - e3x/3 = - 7/12 fi 3e-4y + 4e3x = 7. 12. The given differential equation can be written as dx 1 dx 1 - 2 y = y sin y2. = xy [x2 sin y2 + 1] fi 3 dy x dy x This equation is reducible to linear equation, so putting -1/x2 = u, the last equation can be written du as + 2uy = 2y sin y2 dy 2 The integrating factor of this equation is e y So required solution is 2

uey = Ú 2y sin y2 ◊ ey dy +C 2

fi tan u + sec u = x + C fi tan (x - y) + sec(x - y) = x + C. 10. Put 10x + 6y = u, so

1 3 tan–1 4 4

1 y2 e (sin y2 - cos y2) + C 2 2 2u = (sin y2 - cos y2) + Ce-y =

fi fi

2 = x2 [cos y2 - sin y2 - 2 Ce-y ]. 2

13. y1 = ex (- a sin x + b cos x) + ex (a cos x + b sin x) fi

y1 = ex (- a sin x + b cos x) + y



y2= y1 + ex (- a cos x - b sin x) + ex (- a sin x + b cos x)



y2 - 2y1 + 2y = 0.

14. The given differential equation can be written as y5 xdx + ydx - xdy = 0. Multiplying by x3/y5, we have x3 Ê y d x - x d y ˆ x4 dx + 3 Á ˜¯ = 0. Integrating, we get y Ë y2 x5/5 + (1/4) (x/y)4 = C. 15. The given differential equation has variable separable so integrating, we have y = x2/2 - 1/x + C. This will pass through (3, 9) if 9 = 9/2 - 1/3 + C fi C = 29/6. Hence the required equation is 6xy = 3x3 + 29x - 6.

IIT JEE eBooks: www.crackjee.xyz 27.40 Comprehensive Mathematics—JEE Advanced

16. The given equation can be written as xd y - ydx xd y - ydx 1 ¥ = - dxfi = - dx 2 2 x +y x2 1 + y 2 /x 2 fi

1 d ( y/x) = - dx. Integrating we have 2 2 1 + y /x d x tan



-1

(y/x) = - x + c

y = x tan (c - x).

17. The given equation can be written as y(1 + x-1)dx + (x + log x)dy + sin y dx + x cos y dy = 0 fi d( y(x + log x)) + d(x sin y) = 0 fi y(x + log x) + x sin y = C fi x (y + sin y) + y log x = C 18. The given equation can be written in the linear form as follows: dy + yf ¢(x) = f (x) f ¢(x) dx The integrating factor of this equation is f ¢( x) d x eÚ = ef (x). Integrating, we have yef (x) = t t t Ú te dt + C, (where t = f (x)) = te - e + C. Hence

y = ( f (x) - 1) + C e-f (x) 19. The degree of the equation in (a) is clearly 1. The equation in (b) can be written as ( y - 2y1)2 = 9(1 y12) which is of degree 2. The equations in (c) and (d) are also of degree 1. d y dX . Therefore, the 20. Put 2x + y = X fi 2 + = dx dx given equation is reduced to dX X +1 dX 3( X - 1) = -2= =fi dx 2X - 1 dx 2X - 1 21. Taking, x = r cos q and y = r sin q , so that x2 + y2 = r2 and y/x = tan q, we have x dx + y dy = r dr and x dy - y dx = x2 sec2 q dq = r2 dq. The given equation can be transformed into r dr r dq 2

a2 - r 2

=

r

2



dr = dq

a2 - r 2

x +y

2

= a sin (const + tan

-1

y ◊ e-2x = Ú (C¢1x + C¢2)e-x dx + C3 = -(C¢1x + C¢2)e-x - (C ¢1x + C ¢2)e-x - C ¢1 e-x + C3 fi y = (C1x + C2)ex + C3e2x 1 dr 23. The reciprocal of polar subtangent is = 2 . r dq According to the given condition, dr 1 1 dr + = k fi dq = fi kr - 1 = creq. r r2 d q r ( kr - 1) 24. The given equation is a linear equation with I.F. e-3x. So ye-3x = Ú e-3x sin 2x dx + c 1 -3x e (2 cos 2x + 3 sin 2x) + c = 13 1 (2 cos 2x + 3 sin 2x) + ce3x 13 Putting 2 = r cos q and 3 = r sin q, we can get (c) and substituting 2 = r sin q and 3 = r cos q the answer (d) can be obtained. 25. Differentiating w.r.t. u, we obtain fi

y= -

dv dv d v d2v d v ˆ d2v d2v Ê +u 2 +2 fi + = =0 u 2 ÁË du d u d u d u2 d u ˜¯ d u 2 du fi

dv dv u = const or = 2 du du dv = c in the given equation du

we get v = cu + C2 fi xy = cy + C2. dv Again putting = - u/2 in the given equation, we du

y/x)

22. The given equation can be rewritten as Ê d ˆÊ d ˆÊ d ˆ ÁË d x - 1˜¯ ÁË d x - 1˜¯ ÁË d x - 2˜¯ y = 0

dy ˆ Ê d ˆ Êdy x ÁË d x - 1˜¯ ÁË d x - 2 y˜¯ = C ¢1 e . Putting V = d x dV - V = C ¢1 ex whose I.F. is e-x so 2y we have dx dy -x Ve = C ¢1x + C ¢2 fi V = (C ¢1x + C ¢2)ex. Hence dx x - 2y = (C ¢1x + C ¢2)e , which is again a linear equation with I.F. e-2x. Hence

Putting

fi C + sin-1 r/a = q = tan-1 y/x 2

Ê d ˆÊ d ˆ Let Á -1 - 2˜ y = U, then (i) reduces to Ë d x ˜¯ ÁË d x ¯ dU - U = 0 fi U = C ¢1 ex. Therefore, we have dx

(i)

obtain v = - u2/4 fi y2 = - 4xy fi y = 0 or y = - 4x.

IIT JEE eBooks: www.crackjee.xyz Differential Equations 27.41

26. The given equation can be factored as Êdy xˆ ÁË d x - e ˜¯



Êdy -xˆ ÁË d x - e ˜¯ = 0



dy - ex = 0 fi c + y = ex fi x = log (c + y). dx dy - e-x = 0 fi y + c = -e-x fi y + e-x + C = 0 dx -x

fi y + e = const. 27. Rewriting the given equations as =-

dy dx

x+y and putting x + y = u we obtain 3( x + y ) - 4

du u 2 ˆ Ê3 fiÁ + du = dx -1 = 3 u - 4 Ë 2 2u - 4 ˜¯ dx fi e (1/2) (x + 3y) |x + y - 2| = C. 28. The given equation can be written as px - y = sin-1 p

(i)

Differenting w.r.t. x, both the sides 1 ˆ dp Ê x=0 Á ˜ 1 - p2 ¯ d x Ë fi

p = Const or x =

x=

1 1 - p2

1 - p2

fip=

x2 - 1 x

.

Putting this value of p in (i), we get y =

x 2 - 1 - sin-1

e

-1

= x dx

( n - 1) e( n - 1) y 1 dy = x2/2 + C Ú n - 1 e( n - 1) y - 1 e( n - 1) y

fie

(

(n – 1)y

= Ce

)

( n -1) y + ( n -1)

x2 2

+1

31. Equation of parabolas having their vertices at the origin and foci on x-axis is of the form y2 = 4ax where a is parameter. Differentiating, we get 2y (dy/dx) = 4a fi a = (y/2) (dy/dx). Putting this value in y2 = 4ax, we have y2 = 2xy (dy/dx) i.e. y = 2x (dy/dx) whose variables are separable. The required straight lines can be represented by x cos a + y sin a = p (i). Differentiating we have cos a + sin a dy/dx = 0 (ii). Multiplying (ii) by x and subtract it form (i), we get y sin a - x sina y1= p (iii) Multiplying (i) by dy/dx and (ii) by y and subtracting, we have (xy1 - y) cos a = py1 (iv) Squaring and adding (iii) and (iv), we get (y - xy1)2 = p2(1 + y12). The degree of this equation is two. The equation ax2 + by2 = 1 represents the family of all required conies. The differential equation satis-

1

Putting p = c in the equation (i), we have cx - y = sin-1 c Also

dy (n - 1) y

x2 - 1

. x Clearly y = 0 is a solution. 29. Differentiating w.r.t. x, we have dy dp dp +p+x = 4x3 p2 + 2x4p dx dx dx dp = 0 fi px2 = C fi p = C/x2 fi 2p + x dx Substituting this value of p, in the given differential equation, we obtain y = - C/x + C2 fi xy = C2x - C. 30. Rewriting the given equation, we get dy = x (e (n-1)y - 1) dx

are two arbitrary constants a and b. (d) part is clear. 32. The equation in (i) can be written as dy dy ( p - 4) (p - 3) = 0 fi = 4 or =3 dx dx fi y - 4x + C = 0 or y - 3x + C = 0 So the general solution is ( y - 4x + C) (y - 3x + C) = 0. The equation in (ii) can be written as p(p2 - y2p + 2xp - 2xy2) = 0 fi p (p - y2) (p + 2x) = 0 If p = 0 then y + C = 0. If p + 2x = 0 then y + x2 + C = 0. If p - y2 = 0 then 1/y + x + C = 0. Hence the general solution is (y + C) (y + x2 + C ) (Cy + xy + 1) = 0. the value of p and separate the variables. 33. Differentiating the equation in (b) we get 2ayy1 = 3x2 fi 2

x3 yy1 = 3x2 fi 2xy1 = 3y. y2

Replacing y1 by fi 2x2 + 3y2 = c2.

dx we have 2x dx + 3y dy = 0 dy

IIT JEE eBooks: www.crackjee.xyz 27.42 Comprehensive Mathematics—JEE Advanced

Differentiating the equation in (a) we have dx y1 = 2ax = 2y/x. Replacing y1 by , we obtain dy x dx + 2y dy = 0 fi x + 2y = c 2

2

2

dy dy dx +y = 0. Replacing by x dx dx dy dx we obtain x-1/3 = y-1/3 fi y2/3 - x2/3 = c dy (d) is similar. -1/3

34. The equation in (a) can be written as dx - y dy + x fi

x 2 + y 2 dx +

x 2 + y 2 y dy = 0

fi (y dx + x dy) + (yx-1 + dx + log x dy) fi d(xy) + d (y log x) + d(x sin y) = 0 Integrating both the sides, we have xy + y log x + x sin y = C. For (d), put y/x = V. 35. Put y/x = V in (a); put sin y = u in (b) and 1/y = z in (c) and x = r cos q, y = r sin q in (d). 36. Separating the variable of given equation and then integrating, we have

Ú

dx - y dy + x 2 + y 2 (x dx + y dy) = 0

1 fi dx - y dy + x 2 + y 2 d(x2 + y2) = 0 2 Integrating both the sides we have y2 1 + Ú 2 2

x-

(y + yx-1 + sin y)dx + (x + log x + x cos y)dy = 0

+ (sin y dx + x cos y dy) = 0

Differentiating the equation in (c), we have -1/3

For (c), we can write the given equation as

y2 1 2 + (x + y2)3/2 = C 2 3 For (b), divide both the sides by (x - 1)y1/3 the given equation can be written as 1 d y y 2/3 =x (1) + y1/3 d x x - 1 2 -1/3 d y d u y . Then (1) bePut y = u so that = 3 dx dx comes du 2u 3 du u 2 + + = x =xfi 2 dx x -1 d x 3( x - 1) 3

= Ú

dy y

y2 - 1

sec–1 x = sec–1 y + C. Putting x = 2, we have

C=

p p p p - = . So sec–1 x = sec–1 y + 3 6 6 6



pˆ 1 Ê y = sec Á sec-1 x – ˜ . Also cos–1 Ë 6¯ x

x-

2/3

x x -1 2



where u = x2 + y2

u du = C

dx

= cos–1 fi

Ê 3ˆ 1 p 1 + cos–1 Á ˜ + = cos–1 Ë 2 ¯ y 6 y

2 3 1 = - 1- 2 y x x

tan x dx 37. It is linear equation with I.F. e Ú = sec x. Required solution is y = sin x + cos x.

Solution of Question 38–40 Let q be the semi vertical angle of the cone so that tan q = R/H

which is a linear equation in u and x whose integrating factor is (x - 1)2/3. Therefore, d 2 (u (x - 1)2/3) = x (x - 1)2/3 dx 3 2 2/3 fi u(x - 1)2/3 = ...(i) Ú x (x - 1) dx + C 3 Putting x - 1 = t3 in the last integral, it reduces to Ê t8 t5 ˆ 2 Ú (t3 + 1)t4 dt = 2 Á + ˜ Ë 8 5¯ Thus (1) can be written as y

2/3

(x - 1)

2/3

=

( x - 1)8/3 + 2 4

5

(x - 1)5/3 + C

Fig. 27.3

IIT JEE eBooks: www.crackjee.xyz Differential Equations 27.43

Let the radius and height of water cone at time t be r and h respectively. So r h If V is the volume of water and S is the surface of the cone in contact with air at time t, then tan q =

1 2 1 pr h = pr 3 cot q and S = p r 2 V= 3 3 dV We are given that μS dt fi

dV = – kS (V is decreasing) dt



dr 1 p(3r2) cot q = – k(pr2) dt 3



dr = – k tan q dt

Integrating, we get r = – (k tan q )t + C when t = 0, r = k

\C=R

Thus r = (– k tan q )t + R When cone is empty, r = 0. If T is the time taken for the cone to be empty, then 0 = (– k tan q )T + R R R H = = fiT= k tan q k ( R / H ) k Hence, the cone will be empty in time H/k. kˆ R Ê r(1) = - R tan q + R = - k + R = R Á1 - ˜ Ë H H¯ 10

Rk R 10 55  r (i) = 10R - k  i = 10R H i =1 H i =1 Solution Question 41–43 dT + kT = kt The given equation be written as dt fi T = t + Ce-kt. Putting t = 0, we have T(0) = t + C so C = T(0) - t. Thus T(t) = t + (T(0) - t)e-kt T(2) = t + (T(0) - t)e-2k We have t = 65, T(0) = 185 so T(t) = 65 + 120e-kt 3 fi k = 0.144 4 log 3 = 7.63 min. 65 + 120e-0.144t = 105 fi t = 0.144 T(2) = 155 fi e-2k =

44. Let (u, v) be the vertex of the parabola. The equation of required parabolas are of the form (x - u)2 = 4a (y - v). So there are two unknown constants. 45. Differentiating the given equation, we have 2c

dy – 2x = 0 dx



c=

(d y

x

d x)

Putting this value in the given equation, we have x2

(d y d x ) fi

2

+2

x d y dx

x2 + 2 xy

46. y1 = 1 + y2 fi

y + a2 – x2 = 0

Ê d yˆ dy + (a2 – x2) Á ˜ dx Ë dx¯

dy 1 + y2

2

=0

= dx

fi tan-1 y = x + C But y (0) = 0, so C = 0, i.e., tan-1 y = x. Since the range of tan-1x is (- p/2, p/2), the largest value of c such that y = h(x) is differentiable is p/2. 47. The given equation can be written as xd y - ydx d ( y/ x ) = - dx fi = - dx. 2 2 x +y 1 + y 2 /x 2 So tan-1 y/x + x = C. Putting x = 0, we get C = p/2. Thus y/x = tan (p/2 - x) = cot x. 48. The presence of y dx – x dy terms suggests a factor of the form 1/y2 f (x/y). Dividing both sides by y4, we have xdx +

ydx - xd y

= 0 fix3dx +

y4

x2 y d x - x d y =0 y2 y2 (multiplying by x2)

2

Ê xˆ Ê xˆ x3 dx + Á ˜ d Á ˜ = 0. Ë y¯ Ë y¯ Integrating this last equation, we have x4 1 Ê x ˆ + 4 3 ÁË y ˜¯

3

= C. Since y(1) = 1, so

7 8 41 . Thus = 3 12 4 ( y (2)) 49. Putting x = v y, we have C=

dx dv =v+y dy dy The given equation reduces to Ê dv ˆ (1 + ev ) Á v + y ˜ + ev (1 – v) = 0 d y¯ Ë

IIT JEE eBooks: www.crackjee.xyz 27.44 Comprehensive Mathematics—JEE Advanced



50. The presence of y dx + x dy suggests a factor of the form f (xy). y dx + xd y 1 + dy = 0 ydx + x dy + x2 y dy = 0 fi y x2 y2

dy 1+ ev =dv y v + ev

Integrating we have Êx ˆ log y Á + e x / y ˜ = constant Ëy ¯ Since y(1) = 1, so C = 1 + e.



x + yex/y = C



d( xy ) ( xy )

2

+

dy 1 =0fi + log y = C y xy

y (1) = 1 fi C = - 1.

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28 Vector Algebra A vector is a quantity having both magnitude and direction, such as displacement, velocity, force and acceleration. Any portion of a straight line, where the two end-points are distinguished as initial and terminal, is called a directed line segment. The directed line segment with initial point A and terminal point B is denoted by the symbol AB. Graphically, a vector is represented by a directed line segment. We denote it by a letter with an arrow over it, as in a, or in bold type, as in a. A scalar is a quantity having magnitude only but no direction, such as mass, length, time, temperature and any real number. A vector whose initial and terminal points are the same is called a zero vector, 0. The position vector r or any point P with respect to the origin of reference O is a vector OP. A point with position vector r will be written as P(r). Two vectors a and b are said to be collinear if they are supported on the same or parallel lines. For such vectors, b = xa for some scalar x. A set of vectors is said to be coplanar if they lie on the same plane, or the planes in which the different vectors lie are all parallel to the same plane. Three vectors a, b and c are coplanar if c = xa + yb for some scalars x and y. Equal vectors Two vectors (a and b) are said to be equal if they have equal magnitude, a and b lie on the same line or on parallel lines and, a and b have the same direction. Unit vectors A vector whose magnitude is of unit length. a A unit vector in the direction of a is . |a| 28.1 LINEAR COMBINATIONS

The vector r = a1 a1 + a2 a2 + + an an, where a1, a2, , an are scalars, is called the linear combination of a1, a2, , an. The following results are useful in determining coplanar and collinear vectors: 1. Let a, b be two non-zero, non-collinear vectors and r any vector coplanar with a and b. Then r can be represented uniquely as a linear combination of a and b.

2. If a and b are non-collinear vectors, then xa + yb = x¢a + y¢b fi x = x¢, y = y¢ 3. Fundamental theorem. If a and b are non-collinear vectors, then any vector r, coplanar with a and b, can be expressed uniquely as a linear combination of a and b. That is, there exist unique x and y Œ R such that r = xa + yb. 4. If a, b and c are non-coplanar vectors, then xa + yb + zc = x¢a + y¢b + z¢c fi x = x¢, y = y¢, z = z¢ In other words, a, b and c are linearly independent. 5. Fundamental theorem in space. If a, b and c are non-coplanar vectors in space, then any vector r can be uniquely expressed as a linear combination of a, b and c. That is, there exist unique x, y, z Œ R such that r = xa + yb + zc. If i, j and k are three unit vectors along the x-axis, y-axis and z-axis respectively, then any vector r can be represented uniquely as r = a1i + a2 j + a3k, where a1, a2 and a3 are the coordinates of r. 6. Section formula. The position vector of a point P which divides the line joining the points A and B with position vectors a and b respectively in the ratio m : n, is na + mb (m π – n) m+n 1 (a + b). The position vector of mid-point M of AB, is 2 The point A with position vector a is written A(a). If A(a), B(b) and C(c) are the vertices of a triangle ABC, then the 1 centroid of this triangle is (a + b + c). 3 7. Test of collinearity. Three points A(a), B(b), and C(c) are collinear if and only if there exist scalars x, y and z, not all zero, such that xa + yb + zc = 0, where x + y + z = 0. 8. Test of coplanarity. Four points A(a), B(b), C(c) and D(d) are coplanar if and only if there exist scalars x, y, z and w, not all zero, such that xa + yb + zc + wd = 0, x + y + z + w = 0.

IIT JEE eBooks: www.crackjee.xyz 28.2 Comprehensive Mathematics—JEE Advanced 28.2 THE UNIT VECTOR i, j, k

Let OX, OY and OZ be three mutually perpendicular straight lines so that they form right handed system. Let i, j, k denote unit vectors along OX, OY, OZ. Let OP represent a vector r. With OP as diagonal, construct a rectangular parallelopiped whose three coterminous edges OA, OB and OC lie along OX, OY and OZ respectively. Let OA = x, OB = y and OC = z then r = xi + yj + zk

= Projection of b on a a ◊b = projection of a on b. OM = |b| Component of a in the direction of b is the vector a ◊b OM = a. | b |2 5. (a + b) ◊ (a – b) = a2 – b2, (a + b)2 = a2 + b2 + 2a ◊ b, (a – b)2 = a2 + b2 – 2a ◊ b. 6. If i, j and k are three unit vectors along three mutually perpendicular lines, then i ◊ i = j ◊ j = k ◊ k = 1 and i◊j=j◊k=k◊i=0 7. If a = a1i + a2 j + a3k and b = b1i + b2 j + b3 k, then a ◊ b = a 1b 1 + a 2b 2 + a 3b 3, |a| = and

Fig. 28.1

If a, b, g are the angles which OP makes with the coordinate axes then l = cos a, m = cos b, n = cos g are the direction cosines. In case x = rl = r cos a, y = rm = r cos b, z = rn = r cos g. A unit vector r along r is r = li + mj + nk 28.3 SCALAR OR DOT PRODUCT

The scalar product of two vectors a and b is given by |a| |b| cos q, where q (0 £ q £ p) is the angle between the vectors a and b. It is denoted by a ◊ b.

Fig. 28.2

Properties of the scalar product a ◊ a = |a|2 = a2. a◊b=b◊a 2. a ◊ (b + c) = a ◊ b + a ◊ c. 3. Two non-zero vectors a and b make an acute angle if a ◊ b > 0, an obtuse angle if a ◊ b < 0 and are inclined at a right angle if a ◊ b = 0. a ◊b 4. a ◊ b = (projection of a on b)|b|. OL = |a|

1.

cos q =

a12 + a22 + a32 a1b1 + a2b2 + a3b3 a12

+ a22 + a32 b12 + b22 + b32

8. Component of a vector r in the direction of a given r ◊a vector a in the plane of r and a is a. | a |2 9. Component of a vector r perpendicular to a vector a Ê r ◊a ˆ in the plane of r and a is r – Á a. Ë | a |2 ˜¯ 28.4

VECTOR OR CROSS PRODUCT

The vector product of two vectors a and b, denoted a ¥ b, is the vector c with |c| = |a| |b| | sin q |, where q is the angle between a and b, with 0 £ q £ p. c is supported by the line perpendicular to a and b, and the direction of c is such that a, b and c form a right-handed system. Properties of the vector product 1. a ¥ b = – b ¥ a. 2. a ¥ a = 0. 3. a ¥ (b + c) = a ¥ b + a ¥ c. 4. (a ¥ b)2 = a2b2 – (a ◊ b)2. 5. i ¥ i = j ¥ j = k ¥ k = 0 and i ¥ j = k, j ¥ k = i, k ¥ i = j. 6. Two non-zero vectors a and b are collinear if and only if a ¥ b = 0. 7. If a = a1i + a2j + a3k and b = b1i + b2 j + b3k, then i j k a ¥ b = a1 a2 a3 b1 b2 b3 = (a2b3 – a3b2)i + (a3b1 – a1b3)j + (a1b2 – a2b1)k 8. The area of the parallelogram whose adjacent sides are represented by the vectors OA = a and OB = b,

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is |a ¥ b|, and the area of the triangle OAB is 1 |a ¥ b|. The vector area of the above parallelogram 2 is a ¥ b. 9. A unit vector perpendicular to the plane of a and b is a¥b |a ¥ b| and a vector of magnitude l perpendicular to the plane of a and b is l (a ¥ b) ± |a ¥ b| 10. If a = a1 i + a2 j + a3 k, b = b1 i + b2 j + b3 k and q is the angle between a and b then

sin2 q =

(a2b3 - a3b2 )2 + (a3b1 - a1b3 )2 2 + ( a1b2 - a2b1 )

( )( ) Sai2

Sbi2

12. Perpendicular distance of a point P(p) from a line l which is passing through a point A(a) and is parallel to a unit vector b is |(p – a) ¥ b|. 28.5 SCALAR TRIPLE PRODUCT

The scalar triple product of the three vectors a, b and c is denoted by [a, b, c a ¥ b ◊ c. Properties of the scalar triple product 1. (a × b) ◊ c = a ◊ (b × c). 2. [a b c] = [b c a] = [c a b] = – [b a c] = – [c b a] = – [a c b]. 3. If a = a1i + a2 j + a3k and b = b1i + b2j + b3k, then (a ¥ b) ◊ c = [a b c] =

10.

a ◊u b ◊u c ◊u [a b c] [u v w] = a ◊ v b ◊ v c ◊ v a◊w b◊w c◊w

11. Four points with position vectors a, b, c and d will be coplanar if [d b c] + [d c a] + [d a b] = [a b c] or equivalently [b – a

c–a

d – a] = 0

12. [b + c c + a a + b] = 2 [a b c] 13. [b – c c – a a – b] = 0

11. Two nonzero vector a and b are parallel if and only a a a if a ¥ b = 0 or equivalently 1 = 2 = 3 . b1 b2 b3

a1 a2 b1 b2 c1 c2

6. Any three vectors a, b and c are coplanar if and only if [a b c] = 0. 7. [a + b c d] = [a c d] + [b c d]. 8. Three vectors a, b and c form a right-handed or lefthanded system, according as [a b c] > or < 0. a ◊c b ◊c 9. (a ¥ b) ◊ (c ¥ d) = a ◊d b ◊d

a3 b3 . Thus c3

[pa qp rc] = pqr [a b c], where p, q, r, are scalars 4. The volume of the parallelopiped whose adjacent sides are represented by the vectors a, b and c, is (a ¥ b) ◊ c. In particular, If any two vectors among a, b, c are equal then [a b c] = 0. 5. The volume of a tetrahedron ABCD is equal to 1 |AB ¥ AC ◊ AD|. 6

28.6 VECTOR TRIPLE PRODUCT

The vector triple product of three vectors a, b and c is the vector a ¥ (b ¥ c), where a ¥ (b ¥ c) = (a ◊ c)b – (a ◊ b)c and (a ¥ b) ¥ c = (a ◊ c)b – (b ◊ c)a Clearly, a ¥ (b ¥ c) π (a ¥ b) ¥ c in general. In fact, (a ¥ b) ¥ c = a ¥ (b ¥ c) if and only if the vectors a and c are collinear. Also, (a ¥ b) ¥ (c ¥ d) = [a b d] c – [a b c] d = [a c d] b – [b c d] a and [a ¥ b b ¥ c c ¥ d] = [a b c]2 28.7 GEOMETRICAL AND PHYSICAL APPLICATIONS

Bisector of an angle If a and b are unit vectors along the sides of an angle, then a + b and a – b are the vectors along the internal and external bisectors of the angle, respectively. The bisectors of the angles between the lines r = xa and r = yb are given by Ê a bˆ r = l Á + ˜ (l Œ R) Ë |a| |b| ¯ Reciprocal systems of vectors Let a, b and c be a system of three non-coplanar vectors. Then the system a¢, b¢ and c¢ a ◊ a¢ = b ◊ b¢ = c ◊ c¢ = 1 and a ◊ b¢ = a ◊ c¢ = b ◊ a¢ = b ◊ c¢ = c ◊ a¢ = c ◊ b¢ = 0 is called the reciprocal system to the vectors a, b and c.

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If [a b c] π 0, a¢ =

c×a a×b b×c , b¢ = , c¢ = [a b c] [a b c] [a b c]

Vectorial equation of a line Let a be the position vector of any point P on the line, and let b be a vector parallel to a given line. Then an equation of this line is r = a + tb, t being a parameter.

1 2 b j, OB = bi + 2 1 2 2 ab then OA ◊ OB = ab + 4 1 = ÊÁ ab - 1ˆ˜ ¯ Ë2

2

+1≥1

Vector equation of a line passing through points with position vectors a and b is r = a + t(b – a), t is a parameter. Shortest distance between two non-intersecting and nonparallel lines passing through the points whose position vector are a and b and are parallel to the vectors c and d is (a - b) ◊ (c ¥ d) and the vector determined by the short |c ¥ d| (a - b) ◊ (c ¥ d) distance is (c ¥ d). | c ¥ d |2

OA ◊ OB is least when ab = 2. 1 Area of DOAB = | OA ¥ OB | 2

Equation of a plane. The equation of a plane passing through the point with position vector a and parallel to the plane containing b and c, is r = a + lb + mc or [r – a b c] = 0.

Example 2 that

l and m being parameters. The equation of a plane through three points a, b and c is r = (1 – l – m)a + lb + mc, l and m being parameters.

1 (b + c) 2 (b) a + b + c = 0 (c) one of a, b, c must be unit vector (d) all three a, b, c must be unit vectors. Ans. (b) Solution: From (1), to get a, b, c are perpendicular to the same vector. \ a, b, c are coplanar. Let a = bb + g c for some scalars b and g. fi a ¥ b = b(b ¥ b) + g (c ¥ b)

or

r ◊ (b ¥ c + c ¥ a + a ¥ b) = [a b c]

Equation of a plane which is at a distance d from the origin having a unit normal n is r. n = d. Equation of a plane passing through a point with position vector a having a unit normal n is (r – a) ◊ n = 0. Work done. If a force F acts at a point A and displaces it to the point B, then the work done by the force F is F ◊ AB. The moment of a force F applied at B about the point A is the vector AB ¥ F. SOLVED EXAMPLES

SINGLE CORRECT ANSWER TYPE QUESTIONS Let A, B be two distinct points other than 1 2 the origin on the curve y = x . Least value of area of 2 DOAB when OA ◊ OB is minimum is (a) 2 (b) 1 Example 1

(c) 2 Ans. (a) Solution:

Let OA = ai +

(d) 1 2 a j, 2

3

=

1 | ab2 + a2b | 4

=

a+b ≥ 2

ab =

2

If a, b, c are three non-zero vectors, such a¥b=b¥c=c¥a

(1)

then (a) a =

= – g (b ¥ c) = – g (a ¥ b) fi Similarly, b = –1 \

g = –1. a+b+c=0

A vector c, directed along the internal biExample 3 sector of the angle between the vectors a = 7i – 4j – 4k and b = –2i – j + 2k, with |c| = 5 6 , is (a)

5 (i – 7j + 2k) 3

(b)

5 (5i + 5j + 2k) 3

(c)

5 (i + 7j + 2k) 3

(d)

5 (–5i + 5j + 2k). 3

Ans. (a) Solution:

The required vector c is given by

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Ê a bˆ l Á + ˜. Ë |a| |b| ¯ Now fi

a 1 b 1 = (–2i – j + 2k) = (7i – 4j – 4k) and |a| |b| 3 9 7 2 ˆ 27 Ê1 c = l Á i - j + k ˜ fi |c|2 = l2 ¥ ¥2 Ë9 9 9 ¯ 81

Therefore, l2 = 225 or l = ± 15. 5 fi c = ± (i – 7j + 2k) 3 If d = a ¥ (b ¥ c) + b ¥ (c ¥ a) Example 4 + c ¥ (a ¥ b), then (a) d is a unit vector (b) d = a + b + c (c) d = 0 (d) a, b c and d are coplanar. Ans. (c) Solution: a ¥ (b ¥ c)= (a ◊ c)b – (a ◊ b)c b ¥ (c ¥ a) = (b ◊ a)c – (c ◊ b)a c ¥ (a ¥ b) = (c ◊ b)a – (c ◊ a)b Since the dot product is commutative, we have a ¥ (b ¥ c) + b ¥ (c ¥ a) + c ¥ (a ¥ b) = 0 Example 5

Let a = i + j + k, b = i – j + k and c = i –

j Рk be three vectors. A vectors v in the plane of a and b, 1 , is given by whose projection on c is 3 (a) i Р3j + 3k (b) Р3i Р3j Рk (c) 3i Рj + 3k (d) i + 3j Р3k Ans. (c) Solution: As v lies in the plane of a and b, let v = a a + b b, where a, b ΠR. fi v = (a + b)i + (a Рb)j + (a + b )k 1 We are given projection of v and c = 3 fi fi

|c◊ v| 1 = |c| 3 | (a + b ) - (a - b ) - (a + b )

3 fi | a – b | = 1, a – b = ±1 Thus, only possible answer is (c)

=

1 3

If p, q, r are three mutually perpendicular Example 6 vectors of the same magnitude and if a vector x p ¥ ((x – q) ¥ p) + q ¥ ((x – r) ¥ q) + r ¥ ((x – p) ¥ r) = 0, then vector the x is

(a) (1/2) (p + q – 2r) (b) (1/2) (p + q + r) (c) (1/3) (p + q + r) (d) (1/3) (2 p + q– r). Ans. (b) Solution: p ¥ ((x – q) ¥ p) = (p ◊ p) (x – q) – (p ◊ (x – q)) p = |p|2 (x – q) – (p ◊ x) p (p ◊ q = 0) Similarly q ¥ ((x – r) ¥ q) = |q|2 (x – r) – (q ◊ x) q and r ¥ ((x – p) ¥ r) = |r|2 (x – p) – (r ◊ x) r. Without loss of generality we can assume that p = |p| i, q = |p| j, r = |p| k (|p| = |q| = |r|) Therefore, L.H.S. of the given expression = 3 |p|2 x –|p|2 (p + q + r) – |p|2 [(i ◊ x)i + (j ◊ x)j + (b ◊ x) k] 2 2 2 = 3 |p| x – |p| (p + q + r) – |p| x (see Example 1) 2 2 = 2 |p| x – |p| (p + q + r) This expression will be equal to 0 if 1 (p + q + r) x= 2 Example 7

If a and b are two unit vectors, then the

vector (a + b) ¥ (a ¥ b) is parallel to the vector (a) a – b (b) a + b (c) 2a – b (d) 2a + b. Ans. (a) Solution: (a + b) ¥ (a ¥ b) = a ¥ (a ¥ b) + b ¥ (a ¥ b) = (a ◊ b)a – |a|2 b + |b|2 a – (a ◊ b) b = a ◊ b (a – b) – b + a (|a| = |b| = 1) = ((a ◊ b) – 1) (a – b). Hence (a + b) ¥ (a ¥ b) is parallel to a – b. Example 8

Let a = i + 2j + k, b = i – j + k and c =

i + j – k. A vector in the plane of a and b whose projection 1 on c is is 3 (a) – 4i + j – 4k (b) 3i + j – 3k (c) i + j – 2k (d) 4i + j – 4k Ans. (a) Solution. A vector d in the plane of a and b is of the form la + mb = (l + m)i + (2l – m)j + (l + m)k. The projection c.d 1 l + m + ( 2l - m ) - ( l - m ) = = if of d on c is c 3 2l – m = 1. Thus d is of the form (3l – 1)i + j + (3l – 1)k so – 4i + j – 4k is d for l = –1.

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If a = l (i + j – k), b = m (i – j + k), and c Example 9 are unit vectors perpendicular to the vector a and coplanar with a and b, then a unit vector d perpendicular to both a and c is 1 1 (a) (2i – j + k) (b) (j + k) 6 2 1

(c)

6

Ans. (b) Solution: so

(i – 2j + k)

(d)

1 2

Thus

c=

1 6

Thus

2

a1 a2 = b1 b2 c1 c2

1 6

,

l 1 (0i + 3j + 3k) = (j + k). 6 | a ¥ c| 2

2

3 2 (a1 + a22 + a32 ) (b12 + b22 + b32 ) 4 (c12 + c22 + c32 )

a3 b3 c3

a1b1 + a2b2 + a3b3 b12

a1c1 + a2 c2 + a3c3

is equal to (a) 0 (b) 1 1 2 (a 1 + a 22 + a 32 ) (b12 + b22 + b32 ) (c) 4

Ans. (c)

c1 c2 c3

= a1b1 + a2b2 + a3b3

Example 10 Let a = a1i + a2 j + a3k, b = b1i + b2j + b3k and c = c1i + c2j + c3k be three non-zero vectors such that c is a unit vector perpendicular to both a and b. If the angle between a and b is p /6, then

(d)

b1 b2 b3

a12 + a22 + a32

+

b22

+

a1 a2 a3

= a1b1 + a2b2 + a3b3 0

+

b22

+

c1 c2 c3

a1c1 + a2 c2 + a3c3

a1b1 + a2b2 + a3b3 b12

b1 b2 b3

b1c1 + b2 c2 + b3c3

b32

b1c1 + b2 c2 + b3c3

a12 + a22 + a32

b32

0

c12 + c22 + c32

0 0 1

= (a12 + a22 + a32) (b12 + b22 + b32) =

(a12

+

a22

+

a32) –

=

1

a3 b3 c3

a1c1 + a2c2 + a3c3 = 0, b1c1 + b2c2 + b3c3 = 0 and

a1 a2 a3

(–2i + j – k) and

a1 a2 b1 b2 c1 c2

+ a22 + a32 b12 + b22 + b32

Now,

i j k l 1 1 a¥c= 1 1 -1 d = | a ¥ c| 6 | a ¥ c| -2 1 -1 =

a12

3 2 (a1 + a22 + a32 )1/2 (b12 + b22 + b32 )1/2 2 = a 1b 1 + a 2b 2 + a 3b 3

0 ◊ x – 2y – 2z = 0

fi y + z = 0. Also c is perpendicular to a, so x + y – z = 0. x y z Therefore and x 2 + y 2 + z 2 = 1 = = -2 1 -1 2 (c being a unit vector). We have x = – , y = 6 1 . z=– 6

a1b1 + a2b2 + a3b3

3 p = cos = 2 6

( j – k)

Let c = xi + yj + zk. Since a, b, c are coplanar,

x y z lm 1 1 -1 = 0 fi 1 -1 1

Solution: According to the given conditions, c12 + c22 + c32 = 1, a◊ c = 0, b ◊ c = 0 and

2

(b1 +

– (a1b1 + a2b2 + a3b3)2 + b32)

b22

3 (a12 + a22 + a32 ) (b12 + b22 + b32 ) 4

1 (a12 + a22 + a32 ) (b12 + b22 + b32 ) 4

Example 11

The values of k for which the points

A(1, 0, 3), B(–1, 3, 4), C(1, 2, 1) and D(k, 2, 5) are coplanar, are (a) 1 (b) 2 (c) 0 (d) –1. Ans. (d) Solution: Let a = (1, 0, 3), b = (–1, 3, 4), c = (1, 2, 1) and d = (k, 2, 5). Since A, B, C and D are coplanar, we have [d b c] + [d c a] + [d a b] = [a b c] fi fi fi

k 2 5 k 2 5 k 2 5 1 0 3 -1 3 4 + 1 2 1 + 1 0 3 = -1 3 4 1 2 1 1 0 3 -1 3 4 1 2 1 –5k – 15 + 6k – 14 – 9k + 1 = – 20 – 8k – 28 = – 20 fi k = –1

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Example 12 Let a, b and c be three non-zero vectors, no two of which are collinear. If the vector a + 2b is collinear with c, and b + 3c is collinear with a, then a + 2b + 6c is equal to (a) l a

(b) lb

(c) lc

(d) 0

Therefore, the given expression is equal to 1 + 0 + 1 + 0 + 1 + 0 = 3. Example 15 The value of a for which the volume of parallelopiped formed by the vectors i + aj + k, j + ak and a i + k is minimum is (a) – 3

(l being some non-zero scalar),

(c) 1 / 3

(b) 3

(d) - 3

Ans. (d)

Ans. (c)

Solution: Let a + 2b = xc and b + 3c = ya. Then a + 2b + 6c = (x + 6) c and also, a + 2b + 6c = (1 + 2y) a. So (x + 6)c = (1 + 2y)a. Since a and c are non-zero and non-collinear, we have x + 6 = 0 and 1 + 2y = 0, i.e., x = –6 and y = – 1/2. In either case, we have a + 2b + 6c = 0.

Solution: Volume of the parallelopiped formed by i + aj + k, j + a k, ai + k is 1 a 1 V = 0 1 a = 1 + a3 – a a 0 1

Example 13 A vector a has components 2p and 1 with respect to a rectangular Cartesian system. The system is rotated through a certain angle about the origin in the counterclockwise sense. If a has components p + 1 and 1 with respect to the new system, then (a) p = 0 (b) p = 1 or p = –1/3 (c) p = –1 or p = 1, 3 (d) p = 1 or p = –1. Ans. (b) Solution: Since the rotation of axes does not affect the distance between the origin and the point, we have 4p2 + 1 = (p + 1)2 + 1 fi

p + 1 = ± 2p

Example 14



Let a, b and c be three non-coplanar

Then the value of the expression (a + b) ◊ p + (b + c) ◊ q + (c + a) ◊ r is equal to (a) 0 (b) 1 (c) 2 (d) 3.

Solution:

For minimum value of V, we have fi a = ± 1 / 3 Also

d 2V da2

dV =0 da

= 6a > 0 for a = 1 / 3 . Thus V

is minimum when a = 1 / 3 . Example 16

Let V = 2i + j – k and W = i + 3k. If U is

a unit vector, then the maximum value of the scalar triple product [U V W] is (a) – 1 (b)

p = 1 or –1/3

vectors, and let p, q and r relations b¥c c¥a a¥b p= ,q= and r = [a b c] [a b c] [a b c]

Ans. (d)

dV = 3a 2 – 1 da

10 + 6

(c)

59

(d)

60

Ans. (c) Solution:

V ¥ W = 3i – 7j – k [U V W] = U ◊ (3i – 7j – k) = |U| |3i – 7j – k| cos q = 59 cos q £ 59 The maximum 59 is attained when q = 0 i.e. taking V¥W U= . |V ¥ W| Example 17

If a and b are two unit vectors such that

a + 2b and 5a – 4b are perpendicular to each other then the angle between a and b is (a) 45° (b) 60° (d) cos–1 (2/7) (c) cos–1 (1/3)

a ◊ (b ¥ c) [a b c] = =1 a◊ p = [a b c] [a b c]

b ◊ p=

0 b ◊ (b ¥ c) = =0 [a b c] [a b c]

b ◊ q=

b ◊ (c ¥ a) (b ¥ c) ◊ a a ◊ (b ¥ c) [a b c] = = = =1 [a b c] [a b c] [a b c] [a b c]

c ◊ q=

c ◊ (c ¥ a) =0 [a b c]

c◊ r=

c ◊ (a ¥ b) (a ¥ b) ◊ c = =1 [a b c] [a b c]

Ans. (b)

and

a◊ r = 0

Solution: Since the vectors a + 2b + and 5a – 4b are perpendicular so (a + 2b) ◊ (5a – 4b) = 0 fi 5|a|2 + 6a ◊ b – 8|b|2 = 0 fi 6a ◊ b = 3 (| a| = |b| = 1) fi a ◊ b = 1/2 fi cos q = 1/2 (| a| = |b| = 1| fi q = 60°

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Example 18

If a, b and c are unit vectors, then | a – b|2

+ |b – c|2 + |c – a|2 does not exceed (a) 4

(b) 9

(c) 8

If a, b are nonzero vectors and a Example 21 is perpendicular to b then a nonzero vector r satisfying r ◊ a = a , for some scalar a, a ¥ r = b is a a + (a + b) aa + a ¥ b (b) (a) 2 |a| | b |2

(d) 6

Ans. (b) Solution: |a – b|2 + |b – c|2 + |c – a|2 = 2(| a|2 + |b|2 + |c|2 – (a ◊ b + b ◊ c + c ◊ a) = 2(3 – (ab + bc + ca))

(i)

But, 0 £ |a + b + c| = | a| + |b| + | c| + 2(a ◊ b + b ◊ c + c ◊ a) = 3 + 2 (a ◊ b + b ◊ c + c ◊ a) fi a ◊ b + b ◊ c + c ◊ a ≥ – 3/2 (ii) Putting (ii) in (i), we get | a – b|2 + |b – c|2 + |c – a|2 £ 6 + 3 = 9 The bound 9 is attained if a + b + c = 0. Example 19 Let a, b, c be unit vectors with r = 2

2

2

(a) 2 2 (c) 1 2

2 then the max value of (b) 6 (d) 3/2

Ans. (d) Solution: Taking dot product with a on both the sides, we have r ◊ a = a b ¥ c ◊ a = a [a b c] 1 1 r ◊ a. Similarly b = r ◊ b and so a = 2 2 1 1 r ◊ a. Hence a + b + g = r.(a + b + c) 2 2 1 | r| |a + b + c| cos p/4 = 2 1 1 £ 2 (| a| + | b | + |c|) 2 2 = 3/2.

g=

r ¥ (i + 2j + k) Example 20 If r = i – k, then for any scalar a, r is equal to (a) i + a (i + 2j + k) (c) k + a (i + 2j + k)

(c)

2

a b ¥ c + b c ¥ a + g a ¥ b, [a b c] = 2 and the angle between r and a + b + c is p/4 with | r | = a+b+g

be seen that r = j + a (i + 2j + k) satisfy r ¥ (i + 2j + k) = i – k for any scalar a .

(b) j + a (i + 2j + k) (d) i – k + a(i + 2j + k)

Ans. (b) Solution: Let r = a i + b j + g k then r ¥ (i + 2j + k) = (b – 2g ) i + (g – a)j + (2a – g )k. Thus b – 2g = 1, g = a . So r = ai + (1 + 2a)j + ak = j + a (i + 2j + k). Also it can

a a - (a ¥ b) 2

|a|

(d)

a a - (a ¥ b) | b |2

Ans. (c) Solution: Taking cross product with a on both the sides of a ¥ r = b, we get a ¥ (a ¥ r) = a ¥ b fi (a ◊ r) a – (a ◊ a) r = a ¥ b fi a a – (a ¥ b) = | a|2 r a a - (a ¥ b) fi r= | a |2 clearly this r satisfy a ¥ r = b and r ◊ a = a Example 22 The vectors a = i + j + m k, b = i + j + (m + 1)k and c = i – j + mk are coplanar if m is equal to (a) 1 (b) 4 (c) 3 (d) none of these Ans. (d) Solution: a, b, c are coplanar if 1 1 1 0 m m 1 1 m +1 = 0 fi 1 0 m +1 1 -1 1 -2 m m fi

=0

(C2 Æ C2 – C1 ) 2=0

So there is no value of m for which the vectors are coplanar. Example 23 The position vector of a point P is r = x i + yj + zk, where x, y, z Œ N and u = i + j + k. If r ◊ u = 10 then the possible positions of P are (a) 72 (b) 36 (c) 60 (d) 108 Ans. (b) Solution: r ◊ u = 10 fi x + y + z = 10. The number of positive integral solution of this equation is 9C2 = 36. Example 24 If a, b, c be three unit vectors such that a ¥ (b ¥ c) = (1/2)b, b and c being non-parallel. If q1 is the angle between a and c and q2 is the angle between a and b then (b) q1 = p /2 (a) q1 = p /3 (d) q2 = p /6 (c) q2 = p /3

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Ans. (a)

1 2 3 1 | 3 1 -2 | = 2 1 -3 -4

Solution: (1/2) b = a ¥ (b ¥ c) = (a ◊ c)b – (a ◊ b)c fi (a ◊ c – (1/2)) b = (a ◊ b) c The last relation is possible only if a ◊ c = 1/2 and a ◊ b = 0 as b and c are non-parallel. hence q2 = p /2 and q1 = p/3.

W

Example 25 The distance of the point B with position vector i + 2j + 3k from the line passing through the point A whose position vector is 4i + 2j + 2k and parallel to the vector 2i + 3j + 6k is (a)

10

(b)

5

(c)

6

(d)

T

U S Q Fig. 28.3

Solution: AB = – 3 i + k. Since AB. (2i + 3j + 6k) = – 6 + 6 = 0. Hence AB is perpendicular to the given line. Thus the required distance is equal to |AB| = 9 + 1 = 10 .

Using R2 Æ R2 – 3R1 and R3 Æ R3 Æ R3 – R1, we get 1 2 3 1 V = | 0 -5 -11 | 2 0 -5 -15

If a and b are vectors such that | a + b | =

29 and a ¥ (2i + 3j + 4k) = (2i + 3j + 4k) ¥ b, then a possible value of (a + b) ◊ (–7i + 2j + 3k) is (a) 0 (b) 3 (c) 4 (d) 8 Ans. (c) Solution: a ¥ (2i + 3j + 4k) = (2i + 3j + 4k) ¥ b = – b ¥ (2i + 3j + 4k) fi (a + b) ¥ (2i + 3j + 4k) = 0 fi (a + b) is parallel to (2i + 3j + 4k) fi (a + b) = a (2i + 3j + 4k) for some a Œ R fi |a + b| = | a |

R

P

8

Ans. (a)

Example 26

V

4 + 9 + 16

fi 29 = a 29 fi | a | = 1 fi a=±1 Also, (a + b) ◊ (–7i + 2j + 3k) = a (2i + 3j + 4k) ◊ (– 7i + 2j + 2k) = a (–14 + 6 + 12) = 4a Taking a = 1, we get one of the possible values as 4. Example 27 Let PR = 3i + j – 2k and SQ = i – 3j – 4k determine diagonals of a parallelogram PQRS and PT = i + 2j + 3k be another vector. Then the volume of the parallelepiped determined by the vectors PT, PQ and PS is (a) 5 (b) 20 (c) 10 (d) 30 Ans. (c) Solution: Volume of the parallelepiped formed by PT, PQ and PS is V = [PT PQ PS] = (PT) ◊ (PQ ¥ SQ) 1 (PT) ◊ (PR ¥ SQ) = 2

=

1 -5 -11 1 -5 -11 | |= | | 2 -5 -15 2 0 -4

= 10 Suppose x, y, z > 0 and let

Example 28

pˆ Ê pˆ Ê a = ÁË x - y cos ˜¯ i - ÁË y sin ˜¯ j 6 6 pˆ Ê pˆ Ê b = ÁË y - z cos ˜¯ i - ÁË z sin ˜¯ j , 3 3

and

then least value of | a | + | b | is (a)

x2 + z 2

(b)

y2 + z2

(c)

x2 + y 2

(d)

x2 + y 2 + z 2

Ans. (a) Solution:

We have 2

pˆ pˆ Ê Ê | a | = Á x - y cos ˜ + Á y sin ˜ Ë ¯ Ë 6 2¯ 2

= x2 –

2 3 xy + y



|a| =

x 2 - 3xy + y 2

and

|b| =

y 2 - yz + z 2 O p/6 p/3

x

z

y |a|

B

|b| C

A Fig. 28.4

2

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Refer Fig. 28.2 AB = | a |, BC = | b |. Also | a | + | b | = AB + BC ≥ AC = [Equality holds when x = y = z] Example 29 such that

x2 + z 2

Suppose a, b, c are three non-zero vectors a + 4b + 2c = 0 and | a – b | = | b – c | = | c – a |.

(1)

then (a) a ◊ b = 0 (b) b ◊ c = 0 (b) a ◊ c = 0 (c) a ◊ (b + c) = – 6 Ans. (c) Solution: Let a = | a |, b = | b |, c = | c |. Let | a – b | = | b – c | = | c – a | = k (say) Also,

b= -

1 (a + 2c) 4

Also, | 2a – 9d |2 = | – 6b – 7c |2 fi 4| a |2 + 81| d |2 = 36 a ◊ d = 36 | b |2 + 49| c |2 – 84 b ◊ c fi 3 (cos a cos d – sin a sin d) = 7 (cos b cos g – sin b sin g) fi 3cos (a + d) = 7 cos (b + g) Example 31

a = 3l i + (2l + 1)j + (l + 1)k b = (2l – 1)i + (2l – 1)j + (l – 2)k c = (4l – 1)i + 3lj + 2lk Product of all the values of l is for which a, b, c are coplanar is (a) 1 (b) 0 (c) –1 (d) –2 Ans. (c) Solution: a, b, c are coplanar

Now, 2

k = = k2 = =

1 |a – b| = |5a + 2c|2 16 1 (25a2 + 4c2 + 20 a ◊ c), (2) 16 1 | b – c |2 = | a + 6c |2 16 1 2 (a + 36c2 + 12 a ◊ c) (3) 16 | a – c |2 = a2 + c2 – 2 a ◊ c (4)

Example 30

3l 2l + 1 l + 1 D = 2l - 1 2l - 1 l - 2 = 0 4l - 1 3l 2l

2

and k2 = From (2), (4), we get 25a2 + 4c2 + 20 a ◊ c = 16a2 + 16c2 –32 a ◊ c (5) fi 9a2 – 12c2 + 52 a ◊ c = 0 From (3), (4) a2 + 36c2 + 12 a ◊ c = 16a2 + 16c2 – 32 a ◊ c (6) 15a2 – 20c2 – 44 a ◊ c = 0 5 Multiply (5) by and subtract from (6) to obtain 3 – (44 + 65) a ◊ c = 0 fi a ◊ c= 0 Let a = cos a i + sin a j,

b = cos bi – sin bj, c = cos g i + sin g j, d = cos d i – sin d j. If 2a + 6b + 7c – 9d = 0, then (a) 3 cos (a + d) = 5 cos (b + g ) (b) 3 cos (a + d) = 7 cos (b + g ) (c) cos (a + d) + cos (b + g ) = 0 (d) sin (a + d ) + sin (b + g ) = 0 Ans. (b) Solution: Note that | a | = | b | = | c | = | d | =1.

Let

Use C3 Æ C3 – C2

and

C2 Æ C2 – C1 to obtain

3l -l + 1 -l D = 2l - 1 0 -l - 1 = 0 4l - 1 - l + 1 -l R1 Æ R1 – R3 gives

-l + 1 0 0 -l - 1 = 0 0 D = 2l - 1 4l - 1 - l + 1 -l

D = (–l + 1) (–l + 1) (l + 1) = 0 fi l = 1, 1, –1 Thus, product of values of l is –1. Example 32 A vector equation of the line of intersection of the planes r = b + l1 (b – a) + m1 (a + c) r = c + l2 (b – c) + m1 (a + b) a, b, c being non-coplanar vectors is. (a) r = a + m1 (b + c) (b) r = b + m1 (a + 2 c) (c) r = a + m1 (b + 2c) (d) r = b + m1 (a + c) Ans. (d) Solution: At points of intersection of the two planes, we have b + l1 (b – a) + m1 (a + c) = c + l2 (b – c) + m2 (a + b) fi

(–l1 + m1 – m2) a + (1 + l1 – l2 – m2) b + (m1 – 1 + l2) c = 0.

As a, b, c are non-coplanar, we have

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–l1 + m1 – m2 = 0, 1 + l1 – l2 – m2 = 0, m1 – 1 + l2 = 0. Eliminating l2, m2, i.e., writing l2 = 1–m1 from the last equation, in the second equation we have l1 = 0, m2 = m1. So the required line is r = b + m1(a + c)

Example 35 If r ◊ a = 0, r ◊ b = –1 and [r a b] = 1, a ◊ b π 0, | a ¥ b | = 1 the value of r in terms of a and b is (a) a × (a × b) + a × b a¥b (b) a × (a × b) + a¥b2

Example 33 An equation of a plane containing the lines r = a1 + t b1 and r = a2 + t b2 where [a2 - a1, b1, b2] = 0 is (a) [r - a1, b1, b2] = 0 (c) [r - a2, a1, b2] = 0

(b) [r - a2, b1, b2] = 0 (d) [r - a, a2, b2] = 0

Ans. (a) Solution: The given lines passes through the points with position a1 and a2 and a normal to both the lines is b1 ¥ b2 (the given condition [a2 – a1, b1, b2] = 0 ensures that lines are either parallel or intersecting). Hence the required equation of the plane is r◊(b1 ¥ b2) = a1 ◊(b1 ¥ b2) or [r – a1 b1 b2] = 0. Example 34 The equation of the plane which passes through the line of intersection of the planes r ◊ n1 = q1, r ◊ n2 = q2 and is parallel to the line of intersection of the line of intersection of the planes r ◊ n3 = q2 and r ◊ n4 = q4 given that [n1 n3 n4] = {n2 n3 n4] is (a) r ◊ n2 - q2 = r ◊ n3 - q3 (b) r ◊ n1 - q1 = [n1 n2 n3] (r ◊ n2 - q2) (c) r ◊ n1 - q1 = r ◊ n2 - q2 (d) r ◊ n3 - q3 = r ◊ n1 - q1, Ans. (c) Solution: Equation of any plane through the intersection of r ◊ n1 = q1, r2 ◊ n2 = q2 is of the form (1) r ◊ n1 + l r ◊ n2 = q1 + l q2 where l is a parameter. So n1 + l n2 is normal to the plane (1). Any plane containing planes r ◊ n3 = q3, r ◊ n4 = q4 is of the form r ◊ (n1 + mn4) = q3 + m q4. Hence we must have n1 + l n2 = k (n3 + m n4) for some k fi

[n1 + l n2]◊[n3 ¥ n4] = 0

fi [n1, n3, n4] + l [n2, n3, n4] = 0 fi l = – 1. Putting this value in (1), we have equation of required plane as r ◊ n1 – q1 = r ◊ n2 – q2 or

[n2 n3 n4] (r ◊ n1 – q1) = [n1 n3 n4] (r ◊ n2 – q2)

(c) a × (b × a) + (d)

b¥a b¥a2

a ¥ (a ¥ b) +a×b a ¥ (a ¥ b)

Ans. (b) Solution : Writing r as linear combination of a, b and a ¥ b, we have r = xa + yb + z (a ¥ b) For scalars x, y, z 0 = r ◊ a = x |a|2 + ya ◊ b –1 = r ◊ b = xa ◊ b + y |b|2 2 Solving we get x = a · b, y = - | a |

|a ¥ b|2 = |a|2|b|2 – (a◊b)2

since

Also 1 = [r a b] = z |a ¥ b|2 fi

z=

1 | a ¥ b |2

Thus r = ((a ◊ b) a – |a|2 b) + (a ¥ b) = a ¥ (a ¥ b) + (a ¥ b). Example 36 If a ¥ b and c ¥ d are perpendicular satisfying a ◊ c = l, b ◊ d = l, (l > 0) a ◊ d = 4, b ◊ c = 9 then l is equal to (a) 2 (b) 3 (c) 6 (d) 36 Ans. (c) Solution. We know that a◊c a◊d (a ¥ b).(c ¥ d) = b◊c b◊d =

l 9

4 l

= l2 – 36

Since a ¥ b and c ¥ d are perpendicular so l = 6. Example 37 Let a, b, c be unit vectors such that a + b + c = 0. Which one of the following is correct? (a) a ¥ b = b ¥ c = c ¥ a = 0 (b) a ¥ b = b ¥ c = c ¥ a π 0 (c) a ¥ b = b ¥ c = a ¥ c π 0 (d) a ¥ b, b ¥ c, c ¥ a are mutually perpendicular Ans. (b) Solution: a + b + c = 0 fi a, b, c are coplanar. Also no two of a, b, c can be parallel, for if a is parallel to b, then

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a = kb but a, b are unit vectors so k = ± 1. If k = 1, we get a = b and so 2a + c = 0. fi |c| = 2|a| = 2, which is contradiction the assumption that c is a unit vector. If k = – 1, we get a + b = 0 fi c = 0 which contradicts the assumption that c is a unit vector. Therefore a is not parallel to b fi a ¥ b π 0. Also a + b + c = 0 fi a ¥ (a + b + c) = 0 fi a ¥ b= c ¥ a Similarly a ¥ b = b ¥ c Thus a ¥ b = b ¥ c = c ¥ a π 0. Example 38 Let two non-collinear unit vectors a and b form an acute angle. A point P moves so that at any time t the position vector OP (where O is the origin) is given by a cos t + b sin t. When P is farthest from origin O, let M be the length of OP and u be the unit vector along OP. Then a+b and M = (1 + a.b)1/2 (a) u = |a + b| (b) u =

a-b and M = (1 + a.b)1/2 |a - b|

(c) u =

a+b and M = (1 + 2a.b)1/2 |a + b|

(d) u =

a-b and M = (1 + 2a.b)1/2 |a - b|

1 1/ 2 1/ 2 = 1/ 2 1 1/ 2 1/ 2 1/ 2 1

Example 40 If a, b, c, d are unit vectors such that (a ¥ b). (c ¥ d) = 1 and a ◊ c = 1/2 then (a) a, b, c are non-coplanar (b) b, c, d are non-coplanar (c) b, d are non-parallel (d) a, d are parallel and b, c are parallel Ans. (c) Solution: Let q1 be the angle between a and b, q2 be the angle between c and d and f be angle between a ¥ b and c ¥ d then (a ¥ b).(c ¥ d) = 1 fi |a ¥ b| |c ¥ d|cosf = 1 fi sinq1 sinq2 cos f = 1 fi q1 = p/2, q2 = p/2, f = 0 As f = 0 so a ¥ b and c ¥ d are parallel fi a, b, c, d are coplanar Also a ◊ c = 1/2 fi cosq3 = 1/2 (where q3 is the angle between a and c) q3 = p/2 fi angle between a and c is p/3 fi angle between b and d is p/3 \ b and d are non-parallel.

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

Ans. (a) Solution: OP = a cos t + b sin t so M2 = |OP|2 = (a cos t + b sin t) . (a cos t + b sin t) = (cos2 t + sin2 t + 2 cos t sin t a.b) = (1 + sin2t a.b) |OP| max = (1 + a.b)1/2, t = p 4 u=

a+b OP a/ 2 +b/ 2 = = . | a+b| | OP | a b

2 Example 39

+

2

The edges of a parallelopiped are of unit

length and are parallel to non-coplanar unit vectors a, b, c such that a . b = a . c = b . c = 1/2. Then the volume of the parallelopiped is (a) 1 / 2 (c) 3 / 2 Ans. (a) Solution: =

(b) 1 / 2 2 (d) 1 / 3

(Volume) 2 = [a b c] [a b c] a ◊a b ◊a c◊c a ◊b b ◊b c◊b a◊c b ◊c c◊c

= 1. 2

Example 41

Let a = cos a i + sin a j,

b = cos bi + sin bj, c = cos g i + sin g j, 0 £ g, b,g < 2p. If a ◊ c = b ◊ c = – 3 a ◊ b, then a ◊ b is equal to 1 1 1 1 (b) (c) (d) (a) 3 2 2 3 Ans. (a), (c) Solution: a ◊ c = cos a cos g + sin a sin g = cos (a – g) b ◊ c = cos (b – g), a ◊ b = cos (a – b) Let A = a – g, B = b – g, a – b = A – B. Now, a ◊ c = b ◊ c fi cos A = cos B fi B = 2mp ± A where m Œ I. But – 2p < A, B < 2p \ B = A, B = – A, B = 2p – A or B = – 2p + A When B = A, or B = – 2p + A, we get cos fi When

A=–

3 cos 0

or

- 3 cos (–2p)

cos A = - 3 . Not possible. B = – A or B = 2p – A, we get cos A = - 3 cos (2A)

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fi 2 3 cos2 A + cos A –

=

p = (sin A)i + (sin B)j, q = (cos B)i + (cos A)j be two vectors such that p ◊ q = sin 2c. If sin A, sin C, sin B are in A.P., and AC ◊ (BA – CA) = 32, then (a) –BCA = p /3 (b) DABC is equilateral D. let

-1 ± 1 + 4(2 3 )( 3 )

fi cos A =

2(2 3 ) -1 ± 5

=

1

3 2

,-

4 3 3 Now, a ◊ b = cos (2A) = 2 cos2 A – 1 Ï 1 ÔÔ - 3 when cos A = = Ì Ô 1 when cos A = ÔÓ 2 Example 42

3 3 2

a = i + (cos q)j + (cos 2q)k b = (cos q)i + (cos 2q)j + (cos 3q)k c = (cos 2q)i + (cos 3q)j + (cos 4q)k and let f (q) = [a b c], then Ê 3p ˆ Êp ˆ (b) f Á ˜ = 0 (a) f Á ˜ = 0 Ë 2¯ Ë 2¯

cos q cos 2q cos 3q

1 f (q) = cos q cos 2q

1 z+z 8 2 z + z2

cos 2q cos 3q cos 4q

z+z

z2 + z 2

z2 + z 2

z3 + z 3

z3 + z 3

z4 + z 4

2

1 = (D + D ) 8 where z = cos q + i sin q and 1

z+z

z2 + z 2

z

z2 + z 2

z3 + z 3

z2

z3 + z 3

z4 + z 4

Using C2 Æ C2 – z C1 followed by C3 Æ C3 – z C2, we get 1 D=

B

(d) f (p + q) = – f (p – q)

(c) f (p) = –1 Ans. (a), (b), (d) Solution:

D=

4 3 3 8 (d) circumradius of DABC is . 3 Ans. (a), (b), (c) Solution: p ◊ q = sin A cos B + sin B cos A fi sin 2C = sin (A + B) = sin (p – C). fi 2 sin C cos C = sin C sin A, sin C, sin B are in A. P. sin A + sin B = 2 sin C = 2 sin (p/3) = 3 (c) in radius of DABC is

1

Let a, b, c be the vectors

=

Let ABC be a triangle, and

Example 43

3 =0

z 2

z

z

z2

z3

z2 z

3

C

A

Fig. 28.5

By the law of sines. AB AC BC AC + BC = = sin C sin B sin A sin B + sin A AB 3/2

=

AC + BC 3

fi AC + BC = 2AB we have AC ◊ (BA – CA) = 32 fi AC ◊ (BA + AC) = 32 fi AC ◊ BC = 32 fi CA ◊ CB = 32 Let M be foot of perpendicular from B to CA, then CA ◊ CB = (CA) (CM) Thus, (CA) (CM) = 32 Let As

32 CM = x, the CA = x –BCA = p/3,



CM Êpˆ 1 = cos Á ˜ = Ë 3¯ BC 2 BC = 2x.

z4

Using C3 Æ C3 – z2 C1, we get D = 0 \ f (q) = 0 " q ŒR.

M

(1)

(2)

(3)

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fi l = sin (a/2) Let point P be (cos q, sin q), then OP = cos q i + sin q j We have | AP |2 = (cos q – 1)2 + (sin2 q) = 2(1 – cos q) 2 Also | BP | = (cos q – cos a)2 + (sin q – sin a)2 = 2[1 – cos (q – a)]

From (1), (2), (3) 16 AB = +x x Also, AB2 = BM + MA2 = BC 2 – CM 2 + (AC – CM)2 2

16 ˆ Ê 32 ˆ Ê 2 2 ÁË x + ˜¯ = 4x – x + ÁË - x˜¯ x x



2

16 32 fi ÊÁ x + ˆ˜ - ÊÁ - xˆ˜ Ë x ¯ Ë x¯

2

(

2 As | AP| | BP| = 2 1 - 1 - l

2

= 3x

2

fi | a | = cos a | b | + sin a | c | + (2 cos a sin a) b ◊ c Let | a | = | b | = | c | = R, therefore, R2 = (cos2a + sin2a)R2 + (sin)(2a)) b ◊ c fi b ◊ c = 0 fi –BOC = p/2 Now, AB = b – a = (1 – cos a) b – sin a c AC = c – a = – cos a b + (1 – sin a)c \ AB ◊ AC = – cos a (1 – cos a) – sin a (1 – sin a) < 0 [ 0 < a < p /2] fi –BAC is an obtuse angle \ –BAC = 3p/2 [ –BOC = p/2] 2

2

2

2

2

Example 44 Let AB be a chord of the unit circle with centre O and let length of AB = 2l, where 0 < l < 1. If point

(

)

P on the unit circle is such that | AP | | BP | = 2 1 - 1 - l 2 , then (a) P can be mid-point of arc (AB) (b) –POA = sin–1 (l) (c) arc (AB) = 2 sin–1 (l) (d) arc (AB) = 2 cos -1 1 - l 2 Ans. (a), (b), (c), (d) Solution: Suppose OA is the x-axis, the OA = i. Let –AOB = a, then OB = cos a i + sin a j, P a O

a˘ È fi | AP |2 | BP |2 = 4 Í1 - cos ˙ 2˚ Î

A

2

aˆ Ê fi 4(1 – cos q) [1 – cos (q – a)] = 4 ÁË1 - cos ˜¯ 2

2

Thus, a possible value of q is a/2. \ q = sin–1 (l) fi x2 +

256 x

2

Ê 1024 ˆ + 32 - Á 2 - 64 + x 2 ˜ = 3x2 Ë x ¯

fi x4 – 32x2 + 256 = 0 fi (x2 – 16)2 = 0 fi x = 4 Thus, AB = 8, BC = 8, CA = 8 fi DABC is equilateral. r= Also R = 2r =

8 4 D = = 3 s 2 3 3

8 3 3

Example 45 Let n Œ N, n ≥ 2. Given (n + 2) points O A0, A1,…, An in the XY-plane such that –Ak–1OAk = p/n, (1 £ k £ n), –OAk–1 Ak = –OA0 A1 (2 £ k £ n). If | OA0 | = 1, | OA1| = 1 +

1 , then n

Ê 1ˆ (a) |OAk| = Á1 + ˜ Ë n¯

B

)

k

(O £ k £ n)

Ê 1ˆ 2 Ê p ˆ 1 (b) |A0A1|2 = 4 Á1 + ˜ sin Á ˜ + 2 Ë n¯ Ë 2n ¯ n n

Fig. 28.6

We have, cos a = fi

OA2 + OB 2 - AB 2 = 1 – 2l2 2(OA)(OB)

2l2 = 1 – cos a = 2 sin2 (a/2)

(c) lim  | Ar -1 Ar | = (e - 1) p 2 + 1 nƕ r =1

(d) lim | OAn | = e nƕ

Ans. (a), (b), (c), (d) Solution: Note that D OA0A1 ~ D OA1A2 ~ D OA2A3 etc.

IIT JEE eBooks: www.crackjee.xyz Vector Algebra 28.15 Ak

n

lim  | Ar -1 Ar | = (e - 1) p 2 + 1

Ak - 1

a

nƕ r =1

lim | OA n | = e

nƕ

A2

Example 46 a p/n p/n

2i – j + l k are coplanar if (a) l = –2 (c) l = 1 – 3 Ans. (a), (b) and (c)

A1

p/n |+|/n a

A0

Solution:

Fig. 28.7

Let ai = |OAi| (0 £ i £ n) OAk -1 OAk = OA0 OA1 fi OAk = a1(OAk–1)

[

OA0 =1]

k

Let br = | Ar–1 Ar | (1 £ r £ n) We have b2r = a2r + a2r–1 – 2ar ar–1 cos (p/n)

1 = ÊÁ1 + ˆ˜ Ë n¯ 1 = ÊÁ1 + ˆ˜ Ë n¯

2r

Ê 1ˆ + Á1 + ˜ Ë n¯

2r - 2

2r - 2

Ê 1ˆ - 2 Á1 + ˜ Ë n¯

2 r -1

Êpˆ cos Á ˜ Ë n¯

ÈÊ 1 ˆ 2 Ê 1ˆ Êp ˆ˘ ÍÁ1 + ˜ + 1 - 2 Á1 + ˜ cos Á ˜ ˙ Ë n¯ Ë n¯˙ ÍÎË n ¯ ˚

2r - 2

b12 r -1

b1 , where

pˆ 1 Ê 1ˆ Ê b21 = 2 Á1 + ˜ Á1 - cos ˜ + 2 Ë n¯ Ë n¯ n Ê 1ˆ 2 Ê p ˆ 1 = 4 Á1 + ˜ sin Á ˜ + 2 Ë n¯ Ë 2n ¯ n fi

Ê 1ˆ  br = b1  ÁË1 + ˜¯ n

n

r =1

r =1

r -1

n

ÈÊ 1 ˆ n ˘ b1 ÍÁ1 + ˜ - 1˙ ÍÎË n ¯ ˚˙ = 1 n 2 n Ê ˆ È ˘ Ê 1 ˆ Ê sin(p / 2n) ˆ 2 Ê 1ˆ  br = ÁÁ ÁË1 + n ˜¯ ÁË p / 2n ˜¯ p + 1˜˜ ¥ ÍÁË1 + n ˜¯ - 1˙ ˙˚ r =1 Ë ¯ ÎÍ n

3

The given vectors are coplanar if and only if

fi l(l2 – 1) – (l + 2) + 2(–1 – 2l) = 0 fi l3 – 6l – 4 = 0 fi (l + 2) (l2 – 2l – 2) = 0 fi

l=–2

Example 47

2± 4+8 =1± 3 2 The vectors a = xi – 2j + 5k and or

l=

b = i + yj – zk are collinear if (a) (b) (c) (d)

x x x x

= = = =

1, y = –2, z = –5 1/2, y = –4, z = –10 – 1/2, y = 4, z = 10 –1, y = 2, z = 5.

Ans. (a), (b), (c) and (d) br = ÊÁ1 + 1 ˆ˜ Ë n¯



(b) l = 1 + (d) l = 0.

l 1 2 1 l -1 = 0 2 -1 l

1 fi ak = a 21 OAk–2 = … = ak1 = ÊÁ1 + ˆ˜ Ë n¯

Ê 1ˆ = Á1 + ˜ Ë n¯

The vectors l i + j + 2k, i +l j – k and

Solution: The given vectors are collinear if a = kb for some k, i.e., xi – 2j + 5k = ki + kyj – zkk fi (x – k)i + (–2 – ky)j + (5 + zk)k = 0 Hence x = k, y = –2/k and z = –5/k. Taking k = 1, 1/2, –1/2 and –1, we get the values given by (a), (b), (c) and (d), respectively. Example 48 The values of x for which the angle between the vectors a = xi – 3j – k and b = 2xi + xj – k is acute, and the angle between the vector b and the axis of ordinates is obtuse, are (a) 1, 2 (b) –2, –3 (c) all x < 0 (d) all x > 0. Ans. (b), (c) Solution: According to the given conditions, a ◊ b > 0 and b ◊ c < 0 , where c = (0, 1, 0). Thus 2x2 – 3x + 1 > 0 for x < 0. But 2x2 – 3x + 1 is greater than zero for all x < 0.

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Example 49 The vectors a, b and c are of the same length and, taken pairwise, they form equal angles. If a = i + j and b = j + k, the coordinates of c are (a) (1, 0, 1) (c) (–1, 1, 2)

(b) (1, 2, 3) (d) (–1/3, 4/3, –1/3)

Ans. (a), (d) Solution: Let c = (c1, c2, c3). Then |c| = |a| = |b| c12 + c22 + c32 = 2 = . It is given that the angles between the vectors are identical, and equal to f (say). Then a ◊b 0 + 1 + 0 1 = = cos f = |a| |b| 2 2 2 a ◊ c c1 + c2 1 = = |a| |c| 2 2 2

and

b ◊ c c2 + c3 1 = = 2 2 |b| |c|

Hence c1 + c2 = 1 and c2 + c3 = 1. That is, c1 = 1 – c2 and c 3 = 1 – c 2. fi 2 = c12 + c22 + c32 = (1 – c2)2 + c22 + (1 – c2)2 = 3c22 + 2 – 4c2 Therefore, c2 = 0 or c2 = 4/3. If c2 = 0, then c1 = 1 and c3 = 1, and if c2 = 4/3, then c1 = –1/3, c3 = –1/3. Hence the coordinates of c are (1, 0, 1) or (–1/3, 4/3, –1/3). Example 50

Let a = 2i – j + k, b = i + 2j – k and c =

i + j – 2k be three vectors. A vector in the plane of b and c whose projection on a is of magnitude

2 / 3 is

(a) 2i + 3j – 3b (b) 2i + 3j + 3k (c) –2i – j + 5k (d) 2i + j + 5k. Ans. (a), (c) Solution: Let the required vector be d = xi + yj + zk. For this to be coplanar with b and c , we must have x y z 1 2 -1 = 0 1 1 -2 fi fi

x (–4 + 1) + y(–1 + 2) + z(1 – 2) = 0 –3x + y – z = 0 |a ◊ d| . So The projection of d on a is |a| 2 1 = |2x – y + z| 3 6



2x – y + z = ± 2

(i)

(ii)

The choices (a) and (c) satisfy the equations (i) and (ii). Let a = 4i + 3j and b be two vectors Example 51 perpendicular to each other in xy-plane. The vectors c in the same plane having projections 1 and 2 along a and c are

2 11 i+ j 3 2 2 11 (c) - i + j 5 5 Ans. (b), (c) (a) -

(b) 2i – j (d)

2 11 i+ j 3 2

Solution:

Let b = xi + yj. Since a is perpendicular to b so c◊a . Let c = ui + vj be the required 4x + 3y = 0. Thus b = |a| vector. According to the given condition c◊b fi 4u + 3v = 5. Also 1= |b| c◊b ux - (4 / 3) vx fi = 2 fi 3u – 4v = ± 10 |b| x 2 (1 + (16 / 9) Solving these equations we have u = 2 and v = – 1 or u = – 2/5, v = 11/5. 2=

Example 52

The vectors AB = 3i – 2j + 2k and

BC = – i – 2k are the adjacent sides of a parallelogram. The angle between its diagonals is (a) p /4 (b) p /3 (c) 3p /4 (d) 2p /3 Ans. (a), (c) Solution: The diagonals are given by AB – BC = 4i – 2j + 4k, AB + BC = 2i – 2j These vectors have magnitudes 6 and 2 2 , respectively, and their dot product is 12. Therefore the angle between them is 12 1 p cos–1 = cos–1 = or p minus this value ( 6) ( 2 2 ) 2 4 i.e. 3p /4. Example 53

For non-coplanar vectors A, B and C,

|[A B C]| = | A| | B| |C| holds if and only if (a) A ◊ B = B ◊ C = C ◊ A = 0 (b) A ◊ B = 0 = B ◊ C (c) A ◊ B = 0 = C ◊ A (d) B ◊ C = 0 = C ◊ A Ans. (a), (b), (c), (d) Solution: Let q be the angle between A and B and f the angle between C and A ¥ B i.e. the angle between C and the perpendicular to the plane containing A and B. Then | (A ¥ B) ◊ C| = |A ¥ B | |C| cos f = | A| |B| |C| sin q cos f so if the given relation holds, we have sin q cos f = 1 since |sin q |, |cos f | £ 1 so we must have sin q = 1, cos f = 1 i.e. q = p /2, f = 0. The former implies that A and B are perpendicular so that A ◊ B = 0. On the hand, if f is zero, C must be perpendicular to both A and B so that B ◊ C = 0 = A ◊ C.

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Example 54 Let the unit vectors A and B be perpendicular and the unit vector C be inclined at an angle q to both A and B. If C = aA + b B + g (A ¥ B) then (a) a = b (b) g 2 = 1 – 2a 2 1 + cos 2q (c) g 2 = – cos 2q (d) b 2 = 2 Ans. (a), (b), (c), (d) Solution: We have |A| = | B| = |C| = 1. It is also given that the angle q between A and C equals that between B and C i.e. A◊C B◊C = A ◊ C = cos q = =B◊C | A| |C| |B| |C| Since A ◊ B = 0, we get from the given values of C. A ◊ C = aA ◊ A + b A ◊ B + g A ◊ (A ¥ B) = a i.e. a = cos q and similarly B ◊ C = cos q = b i.e. a = b = cos q, so answer (a) is correct. Next 1 = C ◊ C = 2a 2 + g 2 |A ¥ B |2 = 2a 2 + g2 [| A |2 | B|2 – (A ◊ B)2 = 2a 2 + g 2 2 fi g = 1 – 2a 2 = 1 – 2 cos2 q = – cos 2q 1- g 2 1 + cos 2 q = 2 2 which, proves (b), (c) and (d).



i a a ¥ (a + b) = a ¥ b = a ¥ (i – 3j – 4k) = 1 1

= (–4a2 + 3a3) i + (a3 + 4a1) j + (–3a1 – a2) k. Hence

2i + 2j – k = (–4a2 + 3a3) i + (a3 + 4a1) j + (–3a1 – a2) k.

So 2 = – 4a2 + 3a3, 2 = a3 + 4a1, –1 = –3 a1 – a2. Solving these equations, we have 3k - 2 2-k a3 = k, a2 = , a1 = , k being any real 4 4 numbers. Also

2- k 2+ k , = 4 4 -10 - 3k 3k - 2 =– b2 = –3 – a2 = – 3 – 4 4 b3 = – 4 a3 = – 4 – k. b1 = 1 – a1 = 1 –

Hence

a =

a2 = b 2 =

Example 55 If a, b, c are three non-coplanar vectors such that a + b + c = a d and b + c + d = ba then a + b + c +d is equal to (a) 0 (b) a a (c) b b (d) (a + b )c Ans. (a) Solution: a + b + c = a (b a – b – c) fi (1 – ab )a + (1 + a)b + (1 + a)c = 0 fi 1 = ab and a = – 1 (a, b, c are non-coplanar so linearly independent) Putting the value a = – 1, we get a + b + c + d = 0. Example 56 then

If a × b = 2i + 2j - k, a + b = i - 2j + 4k

1 1 3 13 i + j + k; b = i - j - 5k 4 4 4 4 (b) a = j + 2k; b = i - 4j - 6k (a) a =

(c) a = (d) a =

1 7 5 19 i + j + 2k; b = i - j - 7k 4 4 4 4

1 1 3 13 i - j ; b = i - j - 4k 2 2 4 4

Ans. (a) (b) (c) Solution: Let a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k . Taking cross product with a on both sides of a + b = i – 3j + 4k, we get ,

j k a2 a3 -3 -4

2-k 3k - 2 i+ j+kk 4 4

2+k (10 + 3k ) i– j – (4 + k) k. 4 4 Let u = u1i + u2j + u2k be a unit vector in

b = Example 57 R3 and w =

1 6

(i + j + 2k ) . Given that there exists a vector

v in R3 such that |u ¥ v| = 1 and w.(u ¥ v) = 1. Which of the following statement(s) is(are) correct? (a) There is exactly one choice for such v v (c) If u lies in the xy-plane then |u1| = |u2| (d) If u lies in the xz-plane then 2|u1| = |u3| Ans. (b), (c) Solution: w.(u – v) = 1 fi |w| |u – v| cos a = 1 where a is angle between w and u – v. fi (1) (1) cos a = 1 fi cos a = 1 fi a = 0 fiu¥v=w (1) Let v = v1i + v2j + v3k, then (1) gives (u2v3 – u3v2)i + (u3v1 – u1v3)j + (u1v2 – u2v1)k 1 (i + j + 2k ) = 6 Thus, u2v3 – u3v2 = u 3v 1 – u 1v 3 =

1 6 1 6

(2)

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u 1v 2 – u 2v 1 =



As a is orthogonal to x, we get 0 = a. x = ay ◊ x + bz ◊ x

2 6

Ê v1 ˆ Ê 1ˆ A Á v2 ˜ = 1 Á 1˜ , where Á ˜ 6Á ˜ Ë v3 ¯ Ë 2¯

fi fi Also,

Êpˆ = a ( 2 ) 2 + b ( 2 )( 2 ) cos Á ˜ Ë 3¯

As det (A) = 0, A is not invertible, the system of equation u1, u2 in terms

Given u3 of u3 number of solutions.

1 = –u1v3 If u lies in the xy-plane, u3 = 0, and u2v3 = 6 u2 = – u1 fi |u1| = |u2| Also, if u lies in the xz-plane, then u2 = 0, and (2) gives

fi fi

1

, u 1v 2 =

6 u1v2 = – 2u3v2 |u1| = 2|u3|.

2 6

Example 58 The vector(s) which is/are coplanar with vectors i + j + 2k and i + 2j + k, and perpendicular to the vector i + j + k is/are (a) j – k (b) –i + j (c) i – j (d) – j + k Ans. (a), (d) Solution: A vector coplanar with i + j + 2k and i + 2j + k is a = a(i + j + 2k) + b(i + 2j + k) = (a + b)i + (a + 2b)j + (2a + b)k a is perpendicular to i + j + k gives, (a + b) + (a + 2b) + (2a + b) = 0 fi a + b = 0 Thus, a = bj – bk = b(j – k) Thus, possible answers are (a) and (d) Example 59

a + b = 0. a ◊ y= a y ◊ y + b z ◊ y

Ê 0 1 -1ˆ A = Á -1 0 1 ˜ Á ˜ Ë 1 -1 0 ¯

–u3v2 =

Êpˆ 0 = ( 2 )( 2 ) cos Á ˜ (a + b ) Ë 3¯

Let x, y and z be three vectors each of

magnitude 2 and the angle between each pair of them p is . If a is a nonzero vector perpendicular to x and y ¥ z 3 and b is a nonzero vector perpendicular to y and z ¥ x, then (a) b = (b.z) (z – x) (b) a = (a.y) (y – z) (c) a.b = – (a.y) (b.z) (d) a = (a.y) (z – y) Ans. (b), (c) Solution: We can write a as a linear combination of y, z and y ¥ z. As a is orthogonal to y ¥ z, we can write a = ay + bz.

= 2a + b = a Similarly, a ◊ z = b. Also, 0= Thus, a= Next, we write b= We have 0= Also, b ◊ z= and b ◊ x= As 0= Thus, b= Next, a ◊ b= where u= =

[

a + b = 0]

a+bfia=–b=a◊y (a ◊ y) (y – z) a1 x + b1 z b ◊ y = a1 + b1

a1 + 2b1 = b1 2a1 + b1 = a1 a1 + b1 fi a1 = –b1 = – b ◊ z b ◊ z (z – x) (a ◊ y) (b ◊ z) u (y – z) ◊ (z – x) y◊z–z◊z–y◊x+z◊x Êpˆ = ( 2 ) 2 cos Á ˜ - ( 2 ) 2 Ë 3¯ Êpˆ Êpˆ -( 2 )2 cos Á ˜ + ( 2 ) 2 cos Á ˜ Ë 3¯ Ë 3¯

= –1 Thus, a ◊ b = – (a ◊ y) (b ◊ z) Example 60 Let DPQR be a triangle. Let a = QR, b = RP and c = PQ and If |a| = 12, and b.c = 24 then which of the following is (are) true? 2 (a) | c | - | a | = 12 2 2 (b) | c | - | a | = 30 2 (c) |a ¥ b + c ¥ a| = 48 3 (d) a.b = –72 Ans. (a), (c), (d) Solution: a + b + c = QR + RP + PQ = 0 fi b + c= – a

(1)

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fi fi

|b + c|2 = |–a|2 |b| + |c| + 2b.c = |a|2 2

The two lines intersect at the tip of this vector

2

Example 62

P

If a, b, c are three unit vectors such that

1 b and c being non parallel then 2 (a) angle between a and b is p/2 (b) angle between a and c is p/4 (c) angle between a and c is p/3 (d) angle between a and b is p/3 Ans. (a), (c) 1 Solution: a × (b × c) = (a ◊ c)b - (a ◊ b)c = b 2 fi (a ◊ c - 1/2)b = (a ◊ b)c

a × (b × c) = c

Q

b

a

R

Fig. 28.8

fi fi

2

48 + |c| + 2(24) = 144 |c|2 = 48 1 2 | c | - | a | = 12 fi 2 1 2 fi | c | + | a | = 24 + 12 = 36 2 From (1) a ¥ (a + b + c) = a ¥ 0 fiO+a¥b+a¥c=0 fi a¥b=c¥a Similarly, a¥b=b¥c Thus, |a ¥ b + c ¥ a|2 = (b ¥ c + b ¥ c)2 = 4|b ¥ c|2 = 4[|b|2 |c|2 – (b.c)2] = 4[16(3) (48) – (24)2] = (4) (24)2 (3) fi |a ¥ b + c ¥ a| = Finally, as a ¥ b= |a ¥ b|2 = fi |a|2|b|2 – (a.b)2 = 2 fi (12) (16 ¥ 3) – (a.b)2 = fi (a.b)2 = 2 = (12) (3) (16 – 4) = fi a.b =

2(24) 3 = 48 3 b¥c |b ¥ c|2 (24)2 (3) (24)2 (3) (12)2(16)(3) – (24)2(3) (12)2(36) ± (12) (6) = ±72

Example 61 The point of intersection of the lines l1: r(t) = (i - 6j + 2k) + t(i + 2j + k) l2: R(u) = (4j + k) + u(2i + j + 2k) (a) at the tip of r(7) (c) (8, 8, 9)

(b) at the tip of R (4) (d) at the tip of R(2)

Ans. (a), (b), (c) Solution: Equation r (t) = R(u) we obtain (1 + t - 2u)i + (- 10 + 2t - u)j + (1 + t - 2u)k = 0 Thus 1 + t - 2u = 0 ; - 10 + 2t - u = 0 Solving, we obtain t = 7, u = 4, so r (7) = 8i + 8j + 9k = R(4)

But b and c are non parallel so a ◊ c - 1/2 = 0, a ◊ b = 0 fi angle between a and c is p/3 and the angle between a and b is p/2. Example 63 If a, b and c be non-coplanar unit vectors equally inclined to one another at an acute angle q. If a × b + b × r = pa + qb + rc then (a) p = r 1 2 cos q ,q= (b) p = 1 + 2 cosq 1 + 2 cos q (c) r =

1

(d) p =

1 + 2 cosq

2 cos q 1 + 2 cos q

Ans. (a), (b), (c) Solution: We are given that a ¥ b + b ¥ c = pa + qb + rc Taking dot product with a, we have [a b c] = pa ◊ a + qa ◊ b + ra ◊ c = p + q cos q + r cos q since a ◊ a = | a |2 = 1, a ◊ b = | a | | b |cos q = cos q a ◊ c = cos q. Similarly, taking dot product with b and c we get 0 = p cos q + q + r cos q and [a b c] = p cos q + q cos q + r Therefore, from (2), (4) we get p= r Adding (2), (3) and (4) and we get 2 [a b c] = (2 cos q + 1) (p + q + r) fi

2 [a b c ] =p+q+r 2 cosq + 1

(1)

(2) and

(3) (4)

(5)

Multiplying (5) by cosq and subtracting from (2) we get [a b c] – fi

2 cos q [a b c] = p (1 – cos q) 2 cos q + 1

[a b c ] (2 cos q + 1) (1 - cos q )

=p

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q=

Similarly, Now,

- 2 [a b c] cos q (2 cos q + 1) (1 - cos q )

a ◊a a ◊b a ◊c 1 [a b c] = b ◊ a b ◊ b b ◊ c = cos q c ◊a c ◊b c ◊c cos q 2

1 = (1 + 2 cos q ) cos q cos q

1 1 cos q

cos q 1 cos q

Ans.

cos q cos q 1

1 cos q 1

1 0 0 = (1 + 2 cos q ) cos q 1 - cos q 0 cos q 0 1 - cos q fi

= (1 + 2 cos q) (1 – cos q)2 [a b c] = 1 + 2 cosq (1 – cos q)

(

)

acute, 0 < cos q < 1.) Hence

p = r=

1 1 + 2 cos q

q= -

(since q is

2 cos q 1 + 2 cos q

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: (a) |[a b c]| = 2 [given] Therefore, |[2(a ¥ b) 3(b ¥ c) c ¥ a]| = 6|[a ¥ b b ¥ c c ¥ a]| = 6|[a b c]2| = 6(4) = 24. (b) |[a b c]| = 5 given Therefore, |[3(a + b) (b + c) 2(c + a)]| = 6|2[a b c]| = 6(2)(5) = 60 1 | a ¥ b | = 20 2 \ area of triangle with vectors 2a + 3b and a – b is 1 | (2a + 3b) ¥ (a - b) | 2

(c) We are given .

MATRIX-MATCH TYPE QUESTIONS Example 64 Match statement in column 1 with appropriate statement in column 2. Column 1 Column 2 (a) Volume of parallelepiped determined (p) 100 by vectors a, b and c is 2. Then the volume of the parallelepiped determined by vectors 2(a ¥ b), 3(b ¥ c) and (c ¥ a) is (b) Volume of parallelepiped determined (q) 30 by vectors a, b and c is 5. Then the volume of the parallelepiped determined by vectors 3(a + b), (b + c) and 2(c + a) is (c) Area of a triangle with adjacent sides (r) 24 determined by vectors a and b is 20. Then the area of the triangle with adjacent sides determined by vectors (2a + 3b) and (a – b) is (d) Area of a parallelogram with adjacent (s) 60 sides determined by vectors a and b is 30. Then the area of the parallelogram with adjacent sides determined by vectors (a + b) and a is

1 | -2(a ¥ b) + 3(b ¥ a) | 2 5 = | a ¥ b | = 5(20) = 100 2 =

(d) We are given |a ¥ b| = 30 \

|(a + b) ¥ a| = |b ¥ a| = 30

Example 65 If a and b are two units vectors inclined at angle a to each other then Column 1 Column 2 2p 8

(r) DABC is a right triangle (s) for every DABC

(d) |a – b|2 + |b – c|2 + |c – a|2 = 8

Ans.

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

Solution: Note that |a| = |b| = |c| = 1 Also, BC2 = |b – c|2 = |b|2 + |c|2 – 2b.c = 1 + 1 – 2cos 2A = 4 sin2 A etc. A Æ Æ

b

a

Æ

O

c

B

C

Fig. 28.10

Thus, AB2 + BC2 + CA2 = 4(sin2A + sin2B + sin2C) 2 Now, 2sin A + 2sin2 B + 2 sin2 C = 4 – [cos (2A) + cos (2B)] – 2 cos2 C = 4 – 2 cos (A + B) cos (A – B) – 2 cos2 C = 4 + 2cos C cos (A – B) – 2cos2 C £ 4 + 2 cos C – 2cos2 C 1 = 4 - {4 cos 2 C - 4 cos C + 1 - 1} 2 =

9 1 - (1 - 2 cos C )2 2 2

\ AB2 + BC2 + CA2 £ 9 Also, AB2 + BC2 + CA2 = 9 ¤ cos (A – B) = 1 and 1 – 2 cos C = 0 ¤ A = B and C = p/3 ¤ A = B = C = p/3 Next, assume AB2 + BC2 + CA2 > 8 fi sin2A + sin2B + sin2C > 2 fi cos2 A + cos2B – sin2C < 0 fi cos2 A + cos(B + C) cos(B – C) < 0 fi cos2 A – cos A cos(B – C) < 0 fi cos2 A [–cos (B + C) – cos (B – C)] > 0 fi 2cos A cos B cos C > 0 (1) As at most one of cos A, cos B, cos C is less than 0, (1) gives cos A, cos B, cos C > 0 fi A, B, C are acute angles. Next, AB2 + BC2 + CA2 = 8 fi 2cos A cos B cos C = 0 fi one of cos A, cos B, cos C is 0 fi one of A, B, C is right angle

ASSERTION-REASON TYPE QUESTIONS Example 71 Statement-1: Let f1(x), f2(x), f3(x) be three real polynomials and let a(x) = f1(x)i + f2(x)j + f3(x)k and b(x) = i + (sin x)j + (sin 2x)k, x Œ R. If a(x) ◊ b(x) = 0 " x Œ R then f1(x), f2(x), f3(x) are zero polynomials. Statement-2: If f(x) is a real polynomial of degree m and f (x) = 0 for (m + 1) distinct real numbers x, then f (x) ∫ 0. Solution: Statement-2 is true. see Theory. As f1(x) + (sin x) f2(x) + (sin 2x) f3(x) = 0

(1)

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Suppose deg (f1(x)) = m, we have f1(kp) + (sin (kp))f2 (kp ) + (sin (2kp))f3 (kp) = 0 for k = 1, 2, …, m + 1. \ f1(x) ∫ 0 Therefore, (1) becomes (sin x) f2(x) + (sin (2x)) f3(x) = 0 (2) Let deg (f2(x)) = r, put (2r + 1)p p 3p 5p x= , , , (2) 2 2 2 2 to obtain Ê 2k - 1 ˆ (-1) k +1 f 2 Á p ˜ = 0 for k = 1, 2, … r + 1 Ë 2 ¯ fi f2(x) = 0 for (r + \ f2(x) ∫ Now, (2) becomes (sin (2x)) f3(x) = Let deg (f3(x)) = put

(2 s + 1)

)

7- 3 £x£

1 2

(

7+ 3

)

[

x > 0]

)

(

)

)

Example 73 Statement-1: Let P, Q, R be the feet of perpendiculars drawn from the circumcentre of a DABC to the sides BC, CA and AB respectively. If (t + 1) OP + (t – 1) OQ – t(t + 1) OR = O for some t Œ (0 ,1), then to –BAC = 3p/4 Statement-2: If a, b are two non-zero vectors, and a ◊ b < 0, then angle between a, b is more than p/2. Ans. (a) Solution: We have a ◊ b = | a | | b | cos q where q is (0 £ 0 £ p) angle between a and b. If | a | | b | cos q = a ◊ b < 0 then cos q < 0 fi q > p/2 \ Statement-2 is true. Let OA = a, OB = b, OC = c, A Q

R O

a 2 + 4b 2 + 2ab

p

B

4a 2 + b 2 + 2ab

fi (4 + t2 + 2t)x2 = 1 + 4t2 + 2t fi (x2 – 4)t2 + 2(x2 – 1)t + (4x2 – 1) = 0

(

2

required range as [1/2, 2]

is [1/2, 2]. Statement-2: If a > 0, b2 – 4ac > 0, then ax2 + bx + c > 0 if 1 1 -b - b 2 - 4ac < x < -b + b 2 - 4ac 2a 2a Ans. (c) Solution: a = | a |, b = | b |. Now, | a + 2b |2 = a2 + 4b2 + 4 a ◊ b = a2 + 4b2 + 4ab cos (p/3) = a2 + 4b2 + 2ab and | 2a + b |2 = 4a2 + b2 + 2ab | a + 2b | Let x = . Note that x > 0 | 2a + b | x2 =

1 2

(

Example 72 Statement-1: Let a and b be two nonzero vectors such that an angle between a and b is p/3, then range of | a + 2b | | 2a + b |



)

Also, (1) has both negative roots if x2 – 4 > 0, x2 – 1 > 0 and 4x2 – 1 > 0, that is, if x > 2 or if x2 – 4 < 0, x2 – 1< 0, 4x2 – 1< 0 that is, if 0 < x < 1/2. Thus, (1) has atleast one positive root if 1/2 £ x £ 2 1 1 1 As > 7 + 3 , we get the 7 - 3 as 2 > 2 2 2

p 4

(

(

1 1 5 - 21 £ x 2 £ (5 + 21) 2 2



fi f3(x) = 0 for (s + 1) distinct values, we get f3(x) ∫ 0.

)



2

Ê 2s - 1 ˆ (-1) s +1 f3 Á p˜ = 0 Ë 4 ¯

(

2 21 5 25 fi ÊÁ x 2 - ˆ˜ £ -1 = Ë 4 2¯ 4

Ê 7 - 3ˆ Ê 7 + 3ˆ fi Á £ x2 £ Á ˜ ˜ 2 2 Ë ¯ Ë ¯

1) distinct value 0.

0 s, p 3p x= , , 4 4

As t is real 4(x2 – 1)2 – 4(x2 – 4) (4x2 – 1) ≥ 0 fi (x4 – 2x2 + 1) – (4x2 – 17x2 + 4) ≥ 0 fi – 3x4 + 15x2 – 3 ≥ 0 fi x4 – 5x + 1 £ 0

C

Fig. 28.11

(1)

then OP =

1 1 (b + c), OQ = (c + a) 2 2

IIT JEE eBooks: www.crackjee.xyz Vector Algebra 28.25

1 OR = (a + b) 2 As (t + 1) OP + (t – 1) OQ – t(t + 1) OR = 0 fi (t + 1) (b + c) + (t – 1) (c + a) – t(t + 1) (a + b) = 0 fi (t – 1 – t2)a + (t + 1 – t2 – t) b + (t + 1 + t – 1)c = 0 fi

a=

1- t2 1+ t2

b+

2t 1 + t2

c

Let t = tan (a/2), 0 < a < p/2. \ a = (cos a) b + (sin a) c R2 = | a |2 = (cos2a) | b |2 + (sin2a) | c |2 + (2 cos a sin a) b ◊ c fi R2 = (cos2 a + sin2 a)R2 + (sin 2a) b ◊ c fi b◊c =0 [ sin (2a) π 0] fi –BOC = p/2 (1) Also, AB ◊ AC = (b ◊ a) ◊ (c – a) = [(1 – cos a)b – sin a] ◊ [– cos a b + (1 – sin a)] = – cos a (1 – cos a) – sin a (1 – sin a) p/2 (2) Now (1), (2) gives –BAC = 3p/4. Example 74 Statement-1: If vectors a and c are non collinear then the lines r = 6a – c + l(2c – a) and r = a – c + m(a + 3c) are coplanar Statement-2: There exist l and m such that the two values of r become same. Ans. (a) Solution: If the lines have a common point then there exist l and m such that 6 – l = 1 + m and –1 + 2l = –1 + 3m fi l = 3, m = 2.

= cos2 a/2 fi Unit vector along AD =

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 76 to 78 Let a, b and c be a system of three non-coplanar vectors. Then the system a¢, b¢ and c¢ a ◊ a¢ = b ◊ b¢ = c ◊ c¢ = 1 and a ◊ b¢ = a ◊ c¢ = b ◊ a¢ = b ◊ c¢ = c ◊ a¢ = c ◊ b¢ = 0 is called the reciprocal system to the vectors a, b and c. Example 76 If a¢, b¢, c¢ is a reciprocal system of a, b, c then (a) [a b c] [a¢ b¢ c¢] = |a × a¢| |b × b¢| |c × c¢| (b) [a b c] [a¢ b¢ c¢] = 1 (c) a × a¢ = b × b¢ = c × c¢ (d) [a b c] [a¢ b¢ c¢] < 1 Example 77 a × a¢ + b × b¢ + c × c¢ is a (a) zero vector (b) a nonzero vector 1 ((a × a¢) × b) (c)

[a b c]3 (d) is a scalar multiple of a¢ + b¢ + c¢

Example 78 (a) a + b + c =

Example 75 Statement-1: If u and v unit vectors inclined at an angle a and a is a unit vector bisecting the 1 angle between them then a = (u + v) 2 cos(a / 2) Statement-2: If ABC is an isosceles triangle with AB = AC = 1 then vector representing bisector of angle A is 1 ( AB + AC) . 2 Ans. (a) Solution: In an isosceles triangle ABC in which AB = AC, the median and bisector from A must be same line so statement 2 is true 1 AD = (u + v) 2 1 | AD |2 = [| u |2 + | v |2 + 2uv] 4 =

1 [2 + 2 cos a] 4

1 (u + v) 2 cos(a / 2)

(b) a =

b ¢ ¥ c¢ [a¢ b¢ c¢ ]

(c) b =

a ¢ ¥ c¢ [a¢ b¢ c¢ ]

(d) a + b + c =

1 [b¢ × c¢ + c¢ × a¢] [a¢ b¢ c¢ ]

1 [a¢ × b¢ + b¢ × c¢] [a¢ b¢ c¢ ]

Ans. 76. (b), 77. (a), 78. (b) Solution: b ◊ a¢ = c ◊ a¢ = 0, a¢ is perpendicular to both b and c, and hence parallel to b ¥ c. so a¢ = t(b ¥ c). Also, a ◊ a¢ = 1, so 1 = a ◊ a¢ = ta ◊ (b ¥ c) fi Thus

t=

1 1 = a ◊ (b ¥ c) [a b c]

a¢ =

b¥c . Similarly [a b c]

b¢ =

c¥a and [a b c]

IIT JEE eBooks: www.crackjee.xyz 28.26 Comprehensive Mathematics—JEE Advanced

a¥b [a b c] a ¥ a¢ + b ¥ b¢ + c ¥ c¢ a ¥ (b ¥ c) + b ¥ (c ¥ a) + c ¥ (a ¥ b) = [a b c]

c¢ = fi

Now and \

a ¥ (b ¥ c) = = c ¥ (a ¥ b) = a ¥ (b ¥ c) +

= = =

Ans. 79. (c), 80. (d), 81. (b)

(a◊c)b – (a ◊ b)c, b ¥ (c ¥ a) (b◊a)c – (b ◊ c)a (b◊c)a – (a ◊ c)b b ¥ (c ¥ a) + c ¥ (a ¥ b) = 0

So a ¥ a¢ + b ¥ b¢ + c ¥ c¢ = 0. Next [a¢ b¢ c¢] = (a¢ ¥ b¢) ◊ c¢ 1 [(b ¥ c) ¥ (c ¥ a)] ◊ (a ¥ b) = [a b c]3 1 = [(e◊a) c – (e◊c)a]◊(a ¥ b) [a b c]3 [e = b ¥ c] =

Example 81 The value of [a b c]2 is (b) (1/2)k2 (a) k2 (c) (1/3)k2 (d) k3

1 [a b c]3 1 [a b c]3 1 [a b c]3

[{(b ¥ c)◊a}c]◊(a ¥ b) [(b ¥ c) ◊ a] [c ◊ (a ¥ b)] [a◊(b ¥ c)] [(a ¥ b)◊c]

1 [a b c]

Fig. 28.12

Solution: Equation of the face ABC is r ◊ (b ¥ c + c ¥ c + a ¥ b) = [a b c] and that of edge OA is r = t a. If q denotes the angle between the face and the edge, then (b ¥ c + c ¥ a + a ¥ b) ◊ a sin q = |b ¥ c + c ¥ a + a ¥ b| ◊|a| =

[a b c] |b ¥ c + c ¥ a + a ¥ b| ◊|a|

a ◊a a ◊b a ◊c Now [a b c]2 = b ◊ a b ◊ b b ◊ c c ◊a c ◊b c ◊c

We can also show that a=

b ¢ ¥ c¢ c¢ ¥ a ¢ a¢ ¥ b ¢ ,b= ,c= [a¢ b¢ c¢ ] [a¢ b¢ c¢ ] [a¢ b¢ c¢ ]

=k

6

=k

6

Paragraph for Question Nos. 79 to 91 Let k be the length of any edge of a regular tetrahedron. (A tetrahedron whose edges are all equal in length is called a regular tetrahedron. The angle between a line and a plane is equal to the complement of the angle between the line and the normal to the plane whereas the angle between two planes is equal to the angle between the normals. Let O be the origin of reference and A, B and C vertices with position vectors a, b and c respectively of the regular tetrahedron. Example 79 The angle between any edge and a face not containing the edge is (b) cos-1 1/4 (a) cos-1 (1/2) (c) cos-1 1 / 3

The angle between any two faces is

Example 80 (a) cos (c) p/3

(d) p/3

-1

1/ 3

(b) cos-1 1/4 (d) cos-1 1/2

1 cos 60∞ cos 60∞ cos 60∞ 1 cos 60∞ cos 60∞ cos 60∞ 1 1 1/ 2 1/ 2 1/ 2 1 1/ 2 1/ 2 1/ 2 1

= k6 (3/4 – 1/8 – 1/8) = (1/2) k6. Also |b ¥ c + c ¥ a + a ¥ b| is twice the area of the triangle ABC which is equilateral with each side k so that this is 3 2 k. 2 k3 2 2 = Hence sin q = 3 2 6 k ◊k 2 1 fi cos q = . 3 We have | a| = |b| = |c| = k |a ◊ b| = |b ◊ c| = | c ◊ a | = k 2 cos 60° = k2/2 (ABC is an equilateral triangle)

IIT JEE eBooks: www.crackjee.xyz Vector Algebra 28.27

Equation of OAB and OBC are r◊ (a ¥ b) = 0, r ◊ (b ¥ c) = 0. The angle between two planes is equal to the angle between the normals. If q is the angle then a¥b b¥c ◊ cos q = |a ¥ b | |b ¥ c | (a ¥ b) ◊ (b ¥ c) =

a◊b a◊c b◊b b◊c

= -

1 4 k 4

INTEGER-ANSWER TYPE QUESTIONS Ê kp ˆ Ê kp ˆ ak = cos Á ˜ i + sin Á ˜ j Ë 11 ¯ Ë 11 ¯ then the value of the expression 12

 | ak +1 - a k |

3 2 k = |b ¥ c| 2 cos q = 1/2. Thus the angle between any

| a ¥ b| = |a| |b| sin 60° = \

two plane faces is cos-1(1/2) Paragraph for Question Nos. 82 to 84

Let a = cxi – 6j – 3k b = xi + 2j + 2cxk. Let q be angle between a and b. Example 82

If 0 < q < p/2, " x ΠR then

(a) c > 0 (c) 0 < c < 4/3 Example 83

(b) 0 < c < 1/2 (d) no value of c

If q > p/2 " x ΠR, then

(a) –4/3 < c £ 0 (c) 0 < c < 4/3 Ans. 82. (d) 83. (d) 84. (a) Solution. a ◊ b = cx2 – 12 – 6cx

 | a6 j - a6 j -1 | j =1

Ans. 3 Solution: | ak + 1 – ak |2 = 2 2 È Ê k +1 ˆ Ê kp ˆ ˘ È Ê k + 1 ˆ Ê kp ˆ ˘ + cos p cos sin p sin ˜ ÁË ˜¯ ˙ Í ÁË ÁË ˜¯ ˙ Í ÁË 11 ˜¯ 11 ˚ Î 11 ¯ 11 ˚ Î k ˆ Êpˆ Ê k +1 = 2 - 2 cos Á p - p ˜ = 4 sin 2 Á ˜ Ë 22 ¯ Ë 11 11 ¯ fi

| ak + 1 – ak | = 2| sin (p/22)| = b(say) 12b =3 4b

S=

Thus,

fi cx2 Р6cx Р12 > 0 " x ΠR fi c > 0 and 36c2 + 48c < 0 fi c > 0, and 36c(c + 4/3) < 0 fi c > 0 and c < Р4/3 No value of c. 83. cx2 Р6cx Р12 = 0 " x ΠR Not possible for any value of c. cx Р6cx Р12 < 0 " x ΠR. 2

If c = 0, then cx2 – 6cx – 12 = –12 < 0 For c < 0 36 c2 + 48 c < 0 fi 36 c (c + 4/3) < 0 fi c + 4/3 > 0. \ – 4/3 < c £ 0

Let n ΠN and qk =

Example 86

2k -1 p 2n + 1

Suppose ak = tan (qk)i + j and bk = i – cot (qk)j

(b) c > 0 (d) No value of c

82. If 0 < 0 < p/2, then a ◊ b > 0 " x Œ R

84.

k =1 4

S =

If q = p/2 " x ΠR, then

(a) c > 1/2 (c) 1/2 < c < 2 Example 84

(b) – 4/3 < c < 0 (d) no value of c.

For k ΠN, let

Example 85

n

2 Â | a k |2 k =1 n

then value of

is

 | b k |2

k =1

Ans. 6 Solution: qn + 1 =

Note that qk + 1 = 2qk and 2 p n

2 +1 n

=p–

1 2 +1 n

p = p – q1

fi cot (qn + 1) = cot (p – q1) = – cot q1 Also, cot2 (qk + 1) – cot2 (qk) = cot2(2qk) – cot2(qk) Ê 1 - tan 2 q k ˆ = Á ˜ Ë 2 tan q k ¯

2

– cot2 qk

=

1 [cot2 qk + tan2 qk – 2 – 4 cot2 qk] 4

=

1 [| ak |2 – 3 | bk |2] 4

IIT JEE eBooks: www.crackjee.xyz 28.28 Comprehensive Mathematics—JEE Advanced n

n

fi  (|ak | – 3|bk | ) =  [cot (qk + 1) – cot (q1)] = 0 2

2

k =1

2

2

k =1

n



2 Â | a k |2 k =1 n

=6

 | bk |

2

k =1

If a, b and c are unit vectors satisfying

Example 87

| a – b |2 + | b – c |2 + | c – a |2 = 9, then | a + 2b + 2c | is Ans. 1 Solution: We have 9. = | a – b |2 + | b – c |2 + | c – a |2 £ | a – b |2 + | b – c |2 + | c – a |2 + | a + b + c |2 = | a |2 + | b | – 2a ◊ b + | b |2 + | c |2 – 2b ◊ c + | c |2 + | a |2 – 2c ◊ a + | a |2 + | b |2 + | c |2 + 2a ◊ b + 2b ◊ c + 2c ◊ a =9 [ | a | = | b | = | c | = 1] Thus, 9 = | a – b |2 | | b – c |2 + | c – a |2 + | a + b + c |2 fi | a + b + c | = 0 or a + b + c = 0 Also, | a + 2b + 2c | = | – a + 2(a + b + c) | = | – a + 2(0) | = |–a| = 1 Consider the set of eight vectors

Example 88

V = {ai + bj + ck : a, b, c Œ {–1, 1}}. These non-coplanar vectors can be chosen from V in 2p ways. Then p is Ans. 5 Solution: Let these vectors be OA, OB, OC, OD, OE, OF, OG, OH H

G

E

A

Example 89 Suppose that p, q and r are three noncoplanar vectors in R3. Let the components of a vector s along p, q and r be 4, 3 and 5, respectively. If the components of this vector s along (– p + q + r), (p – q + r) and (– p + q + r) are x, y and z, respectively, then the value of 2x + y + z is Ans. 9 Solution: We are given s = 4p + 3q + 5r (1) Also, s = x(– p + q + r) + y (p – q + r) + z (– p – q + r) (2) = (– x + y – z) p + (x – y – z)q + (x + y + z)r As p, q, r are non-coplanar vectors, from (1) sand (2), we get –x + y – z = 4 (3) x – y – z= 3 (4) x + y + z= 5 (5) From (4) and (5), we get 2x = 8 fi x = 4 Now, 2x + y + z = x + (x + y + z) = 9 Example 90 Let a, b and c be three non-coplanar unit vectors such that the angle between every pair of them is p . If a ¥ b + b ¥ c = pa + qa + qb + rc, where p, q and r 3 p 2 + 2q 2 + r 2 are scalars, then the value of is q2 Ans. 4 Solution: We have a◊a=b◊b=c◊c=1 a◊b=b◊c=c◊a Êp ˆ 1 = (1)(1) cos Á ˜ = Ë 3¯ 2 Now, a ◊ b + b ◊ c = pa + qb + rc fi a ◊ (a ¥ b + b ¥ c) = p(a ◊ a) + q(a ◊ b) + r(a ◊ c)

F O D

We can choose 3 vectors out of 8 in 8C3 = 56 ways, out of these 24 sets are coplanars. \ There are 56 – 24 = 32 = 25 ways to choose 3 noncoplanar vectors.

C



1 1 p+ q+ r=k 2 2

where Similarly,

k = [a b c]

B Fig. 28.13

There are six planes in which these vectors lie. These are OABHG, OACGE, OADFG, OBCEH, OCDEF and OBDHF. We can choose any of the six planes and select 3 vectors out of 4 in 4C3 = 4 ways. Thus, there (6) (4) set of coplanar vectors.

and

(1)

1 1 p+q+ r = k 2 2

(2)

1 1 p+ q+r = k 2 2

(3)

IIT JEE eBooks: www.crackjee.xyz Vector Algebra 28.29

From (1), (2), (3) we get p=q=r 2 2 Êrˆ Ê pˆ p 2 + 2q 2 + r 2 + + 2 Now, = Á ˜ ÁË q ˜¯ = 4. Ë q¯ q2 If d = l(a × b) + m(b × c) + n(c × a), [a b c] 1 = 1/8 and d. (a + b + c) = 8 then (l + m + n) is equal to. 16 Ans. 4 Example 91

Solution:

d ◊ c = l (a ¥ b) ◊ c + m (b ¥ c) ◊ c + n (c ¥ a) ◊ c = l[a b c] + 0 + 0 = l [a b c] = l/8

Hence

l = 8(d ◊ c). Similarly, m = 8 (d ◊ a) and n = 8(d ◊ b)

\ l + m + n = 8d ◊ c + 8d ◊ a + 8d ◊ b = 8d ◊ (a + b + c) = 64. =

1 (l + m + g ) = 4 16

Example 92 If a = (0, 1, - 1) and c = (1, 1, 1) are given a × b + c = 0 and a ◊ b vectors, then |b|2 = 3 is. Ans. 6 Solution: Let b = xi + yj + zk. So a ¥ b = (z + y) i – xj – xk a ¥ b + c = 0 fi z + y + 1 = 0, –x + 1 = 0 fi x=1 a◊b = 3 fi y – z = 3 Solving these equations we have y = 1, z = –2. Thus b = (1, 1, –2). i.e. |b|2= 6 Example 93

Let b = 4i + 3j and c be two vectors

perpendicular to each other in the xy-plane. If ri, i = 1, 2 ... n are the vectors in the same plane having projections 1 and 2 along b and c respectively then

1 10

n

 | ri |2

is equal to.

i =1

Ans. 2 Solution: Let r = lb + mc and c = ± (xi + yj). Since c and b are perpendicular, we have 4 4x + 3y = 0 fi c = ± x ⎛⎜ i − j ⎞⎟ 3 ⎠ ⎝ r ◊b ( l b + m c) ◊ b l b ◊ b = ± 1 = proj. of r on b = = |b| |b| |b| [ b ◊ c = 0] = l|b| = 5l. Hence l = ± 1/5. r ◊c Also, ± 2 = proj. of r on c = |c|

( l b + m c) ◊ c 5 = m |c| = m x |c| 3 Thus, mx = ± 6/5. Therefore, 6⎛ 4 ⎞ 1 r= (4i + 3j) + ⎜ i − j ⎟ = ± (2i – j,) 5⎝ 3 ⎠ 5 Ê 2 11 ˆ 1 6⎛ 4 ⎞ r= (4i + 3j) - ⎜ i − j ⎟ = ±, ÁË - i + j˜¯ 5 5 5 5⎝ 3 ⎠ =

Thus there are four such vectors 1 10

4

Â

| ri |2 = 1 |2i - j|2 + 1 5 5 i =1

2 11 - i+ j 5 5

2

=2

Example 94 If A = (1, 1, 1) and C = (0, 1, - 1) are given vectors and B is a vector satisfying A × B = C and 9 A ◊ B = 3 then |B|2 is equal to. 11 Ans. 3 Solution: Let B = (x, y, z), so that x + y + z = 3. Also, i j – k = C= 1 x = (z –

j 1 y y)i

k 1 z + (x – z)j + (y – x)k

Hence z = y, x – z = 1 and x – y = 1. That is, z = y and x = 1 + y. Therefore, 1 + y + y + y = 3. Thus y = 2/3, z = 2/3 and 9 x = 1 + 2/3 = 5/3. So |B|2 = (25 + 4 + 4) 11 = 3 11 Example 95 If [b c d] = 24 and (a × b) × (c × d) + (a × c) × (d × b) + (a × d) × (b × c) + ka = 0 then k/16 is equal to. Ans. 48 Solution:

We know that

(a ¥ b) ¥ c = (a ◊ c) b – (b ◊ c)a. Putting c ¥ d = e, we get (a ¥ b) ¥ (c ¥ d) = (a ¥ b) ¥ e = (a ◊ e)b – (b ◊ e)a = {a ◊ (c ¥ d)}b – {b ◊((c ¥ d)}a = [a c d] b – [b c d] a (1) Similarly, (a ¥ c) ¥ (d ¥ b) = [a d b] c – [c d b] a = [a d b] c – [b c d] a (2) Also (a ¥ d) ¥ (b ¥ c) = – (b ¥ c) ¥ (a ¥ d) = (b ¥ c) ¥ (d ¥ a) = [b d a] c – [c d a] b = – [a d b] c – [a c d] b (3) From (1), (2) and (3), we get u = (a ¥ b) ¥ (c ¥ b) + (a ¥ c) ¥ (d ¥ b) + (a ¥ d)

IIT JEE eBooks: www.crackjee.xyz 28.30 Comprehensive Mathematics—JEE Advanced

From (1), (2) and (3), we get u = (a ¥ b) ¥ (c ¥ b) + (a ¥ c) ¥ (d ¥ b) + (a ¥ d) ¥ (b ¥ c) = [a c d] b – [b c d ] a + [a d b] c – [b c d] a – [a d b] c – [a c d] b = – 2 [b c d] a. = - 48 a. fi k/16 = 3

Example 99

If a = i + 2j - 3k, b = 2i + j - k and u is

a vector satisfying a × u = a × b and a ◊ u = 0 then 2 |u|2 is equal to. Ans. 5 Solution: a × u = a × b fi a × (u - b) = 0 fi

u - b = t a for some t

Example 96 The volume of the tetrahedron whose vertices are (0, 1, 2) (3, 0, 1) (4, 3, 6) (2, 3, 2) is equal to. Ans. 6 1 |[AB AC AD]|. But Solution: The required volume = 6 AB = 3i - j - k; AC = 4i + 2j + 4k; AD = 2i + 2j



u = b + ta



u = (2 + t)i + (1 + 2t)j - (1 + 3t)k

Now a ◊ u = 0



2 + t + 2 (1 + 2t) + 3(1 + 3t) = 0

3 -1 -1 1 So volume of the tetrahedron = | 4 2 4 | = 6 6 2 2 0

vectors with | r | = | x0| and | r¢| = |x1| then r ◊ r¢ is equal to Ans. 4 Solution. f ¢(x) = x(x + 1)(x + 2)(x + 3) – 24 Clearly f ¢(1) = 0, Dividing f ¢(x) by x – 1, we get f ¢(x) = (x – 1)(x + 4)(x2 + 3x + 6). Thus the critical points are x = 1, – 4. Hence r ◊ r ¢ = |x0| | x1 | = 1 ¥ 4 = 4.

Example 97 If q is the angle between the lines AB, AC where A, B, C are the three points with coordinates (1, 2, - 1), (2, 0, 3) (3, - 1, 2) respectively then 462 cos q is equal to. 10 Ans. 2 Solution: It is easy to see AB = i - 2j + 4k; AC = 2i - 3j + 3k AB ◊ AC = 20 AB ◊ AB = 21, AC ◊ AC = 22 Thus cos q =

AB ◊ AC = AB AC

20 462

462 cos q = 2 10 Example 98

Suppose that a, b, c do not lie in the same

plane and are nonzero vectors such that |a| = 1, |b| = 2, |c| = 2 a ◊ b = 1, b ◊ c = 2 and the angle between a - b and b - c is p/6. If d is any vector such that d ◊ a = d ◊ b = d ◊ c and |d|2 = |a|2 k for any scalar a, then k is equal to. Ans. 3 Solution: We have d ◊ (a - b) = 0 and d ◊ (b - c). Therefore d is perpendicular to a - b and b - c = 0. So vector d is scalar multiple of (a - b) × (b - c). |d|2 = |a|2 |a - b|2 |b - c|2 sin2 p/6 = |a|2 (|a|2 + |b|2 - 2a ◊ b) (|b|2 + |c|2 - 2b ◊ c) = |x|2 ◊ 3 ◊ 4

1 = 3|a|2 4

1 4



t = - 1/2. Hence u = (3/2)i + (1/2)k and 2|u|2 = 5.

Example 100 =

x

Ú1

Let x0 and x1 be the critical points of f(x)

(t(t + 1) (t + 2) (t + 3) – 24)dt and r and r¢ be parallel

EXERCISE LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. The volume of the parallelopiped whose sides are given by OA = 2i - 3j, OB = i + j - k, OC = 3i - k is (a) 4/13 (b) 4 (c) 2/7 (d) 1/17 2. The points with position vectors 60i + 3j, 40i - 8j, ai - 52j are collinear if (a) a = - 40 (b) a = 40 (c) a = 20 (d) a = –20 3. Let p, q, r be three mutually perpendicular vectors of the same magnitude. If a vector x equation p ¥ ((x - q) ¥ p) + q ¥ ((x - r) ¥ q) + r ¥ ((x - p) ¥ r) = 0 then x is given by 1 1 (p + q - 2r) (b) (p + q + r) (a) 2 2 (c)

1 (p + q + r) 3

(d)

1 (2p + q - r) 3

4. If |a| = 2, |b| = 3 |c| = 4 and a + b + c = 0 then the value of b ◊ c + c ◊ a + a ◊ b is equal to

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(a) 19/2 (b) - 19/2 (c) 29/2 (d) - 29/2 5. If A, B, C, D are four points in space and |AB ¥ CD + BC ¥ AD + CA ¥ BD| = l (area of the triangle ABC). Then the value of l is (a) 1 (b) 2 (c) 3 (d) 4 6. Given a = i + j - k, b = - i + 2j + k and c = - i + 2j - k. A unit vector perpendicular to both a + b and b + c is 2i + j + k (b) j (a) 6 (c) k

(d)

i + j+k 3

7. Let P, Q, R be points with position vectors r1 = 3i 2j - k, r2 = i + 3j + 4k and r3 = 2i + j - 2k relative to an origin 0. The distance of P from the plane OQR is (magnitude) (a) 2 (b) 3 (c) 1 (d) 11/ 3 8. A vector a which is collinear with the vector 6i 15 8j k of magnitude 50 making on obtuse angle 2 with z-axis is (a) 24i - 32j - 30k (b) - 24i + 32j + 30k (c) 24i + 32j - 30k (d) none of these 9. Let there be two points A, B on the curve y = x2 in the plane OXY satisfying OA ◊i = 1 and OB◊ i = - 2 then the length of the vector 2OA - 3OB is (b) 2 51 (c) 3 41 (d) 2 41 (a) 14 10. If A, B, C, D are four points in space satisfying AB ◊ CD = K[|AD|2 + |BC|2 - |AC|2 - |BD|2] then the value of K is (a) 2 (b) 1/3 (c) 1/2 (d) 1 11. For unit vectors b and c and any vector a, the value of {{(a + b) ¥ (a + c)} ¥ (b ¥ c)} ◊ (b + c) is (b) 2|a|2 (c) 3|a|2 (d) 0 (a) |a|2 12. The distance of the point B with position vector i + 2j + k from the line passing through the point A with position vector 4i + 2j + 2k and parallel to the vector 2i + 3j + 6k is (a)

10

(b)

5

(c)

6

(d) 2

13. Three non-coplanar vector a, b and c are drawn from a common initial point. The angle between the plane passing through the terminal points of these vectors and the vector a ¥ b + b ¥ c + c ¥ a is (a) p/4 (b) p/2 (c) p/3 (d) p/6

14. If pth, qth, rth term of a G.P. are the positive numbers a, b, c, then the angle between the vectors log a2 i + log b2 j + log c2k and (q - r)i + (r - p)j + ( p - q)k is (a) p/3 (b) p/2 1 p (d) (c) sin-1 4 a 2 + b2 + c2 15. Given three unit vectors a, b, c no two which are 1 collinear satisfying a ¥ (b ¥ c) = b. The angle 2 between a and b is (a) p/3 (c) p/2

(b) p/4 (d) 2p/3

16. The volume of the tetrahedron with vertices P (- 1, 2, 0), Q(2, 1, - 3), R (1, 0, 1) and S(3, - 2, 3) is (a) 1/3 (b) 2/3 (c) 9/14 (d) 2/3 17. Consider the parallelopiped with sides a = 3i + 2j + k, b = i + j + 2k and c = i + 3j +3k then the angle between a and the plane containing the face determined by b and c is (b) cos-1 (9/14) (a) sin-1 (1/3) (d) sin-1 (2/3) (c) sin-1 (9/14) 18. A unit vector n perpendicular to the plane determined by the points A(0, - 2, 1), B(1, - 1, - 2) and C(- 1, 1, 0) 1 1 (2i + j + k ) (b) (a) ( 2i + j + 2k ) 3 6 (c)

1 3

(i - j + k )

(d)

1 14

(3i + j + k )

19. The acute angle between the lines x = - 2 + 2t, y = 3 - 4t, z = - 4 + t and x = - 2 - t, y = 3 + 2t, z = - 4 + 3t is 1 1 (b) cos-1 (a) sin-1 3 6 (c) cos-1

1 5

(d) cos-1

2 3

20. If the vectors AB = -3i + 4k and AC = 5i - 2j + 4k are the sides of a triangle ABC. Then the length of the median through A is (a)

14

(b)

18

(c)

29

(d) none of these

21. A vector a = (x, y, z) of length 2 3 which makes equal angles with the vectors b = (y, - 2z, 3x) and c

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= (2z, 3x, - y) and is perpendicular to d = (1, - 1, 2) and makes an obtuse angle with y-axis is (a) (- 2, 2, 2) (c) (2, - 2, - 2)

(b) (1, 1, 10 ) (d) none of these

22. If a ¥ b = c and b ¥ c = a, then (a) (b) (c) (d)

a, b, c are orthogonal in pairs but |a| π |c| a, b, c are orthogonal but |b| π 1 a, b, c are not orthogonal to each other a, b, c are orthogonal in pairs and |a| = |c|, |b| = 1

23. If a + b + c = 0 and |a| = 3, |b| = 5 and |c| = 7 then the angle between a and b is (a) p/6 (c) p/3

(b) 2p/3 (d) 5p/3

24. a ◊ ((b ¥ c) ¥ (a + b + c)) is equal to (a) 0 (c) [a b c]

(b) 2[a b c] (d) 3[a b c]

25. If a and b are two unit vectors and f is the angle 1 between them, then |a - b| is equal to 2 (a) 0 (b) p/2 (c) |sin f/2| (d) |cos f/2|

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 26. If in a right angle triangle ACB, the hypotenuse AB = p, then AB ◊ AC + BC ◊ BA + CA ◊ CB is equal to (b) p2/2 (a) 2p2 27.

28.

29.

30.

(c) p2 (d) 0 If a, b, c are three non-coplanar vectors, then (a + b + c) ◊ ((a + b) ¥ (a + c)) equals (a) 0 (b) [a b c] (c) 2 [a b c] (d) - [a b c] The vectors ai + 2a j - 3a k, (2a + 1)i + (2a + 3)j + (a + 1)k and (3a + 5)i + (a + 5)j + (a + 2)k are non-coplanar for a in (a) {0} (b) (0, •) (c) (– •, 1) (d) (1, •) For non zero, non-collinear vectors p and q, the value of [i p q ]i +[j p q]j + [k p q]k is (a) 0 (b) 2(p ¥ q) (c) (q ¥ p) (d) (p ¥ q) The position vectors a, b, c and d of four distinct points A, B, C and D lie on a plane are such that

|a - d| = |b - d| = |c - d| then the point D is the (a) centroid of DABC (b) orthocentre of DABC (c) circumcentre of D ABC (d) incenter of DABC 31. The components of vector i + j + k along vector, i + 2j + 3k is (a) (3/7) (i + 2j + 3k) (b) (2/7) (i + 2j + 3k) (c) (1/7) (i + 2j + 3k) (d) (4/7) (i + 2j + 3k) 32. The vector i ¥ ((a ¥ b) ¥ i) + j ¥ [(a ¥ b) ¥ j] + k ¥ [(a ¥ b) ¥ k] equals (a) 0 (b) (a ◊ b)b (c) b (d) 2(a ¥ b) 33. The vectors AB = 3i - 2j + 2k and BC = - i - 2k are the adjacent sides of a parallelogram. The angle between its diagonals is (a) p/4 (b) p/3 (c) 3p/4 (d) 2p/3 34. Let the unit vectors A and B be perpendicular and the unit vector C be inclined at an angle q to both A and B. If C = aA + bB + g (A ¥ B) then (b) g2 = 1 - 2a2 (a) a = b 1 + cos 2q 2 35. If the unit vectors a and b are inclined at an angle 2q and |a - b| < 1, then if 0 £ q £ p, q lies in the interval (a) [0, p/6) (b) (5p/6, p] (c) g2 = - cos 2q

(d) b2 =

(c) [p/6, p/2] (d) (p/2, 5p/6] 36. For non-coplanar vectors A, B and C, |(A ¥ B) ◊ C| = |A| |B| |C| holds if and only if (a) A ◊ B = B ◊ C = C ◊ A = 0 (b) A ◊ B = 0 = B ◊ C (c) A ◊ B = 0 = C ◊ A (d) B ◊ C = 0 = C ◊ A 37. If K is the length of any edge of a regular tetrahedron then the distance of any vertex from the opposite face is 2 2 K (a) 3/2 K (b) 3 2 K (d) 3 K 3 38. Let a = cos a i + sin a j, b = cos b i + sin b j, c = cos g i + sin g j and a + b + c = 0, then (a) cos 2a + cos 2b + cos 2g = 0 (b) sin 2a + sin 2b + sin 2g = 0 (c)

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(c) cos (b + g) + cos (g + a) + cos (a + b) = 0 (d) sin (b + g) + sin (g + a) + sin (a + b) = 0 39. Let a, b, be two non-zero vectors such that |a| = |b| = |a + b| If q (0 £ q £ p) is angle between a and b, then (a) a.b = 0 (b) cos q = –1/2 (c) a.b < 0 (d) sin q = 3 / 2 40. Let a = OA, b = OB, and c = OC be three distinct vectors such that |a| = |b| = |c|. If a + b = 0, then (a) DABC is a right triangle (b) –ACB = p/2 (c) –ABC = p/3 (d) DABC is an isosceles triangle

MATRIX-MATCH TYPE QUESTIONS 41. An equation of the plane Column 1 Column 2 (a) containing the line (p) r = (1 - t - p) a r = a + t b and passing tb + pc through a point c (b) containing the line (q) r = a + t(c - a) r = a + tb and perpenpb dicular to the plane r◊c=q (c) through points with (r) r = a + t (b – a) position vector a, b, c pb (d) through two point (s) r = (1 - t)a + ta whose position vectors pc are a, b and parallel to c

+

+

(a) triangle with vertices whose position vector w.r.t O is - i + 2j + 3k, 2i - j - k, i+j-k (b) tetrahedra with vertices 0, i + j - k, i - j + k, -i+j+k

+ +

Column 2 (p)

3/2

(q) 2/3

ASSERTION-REASON TYPE QUESTIONS 46. In each of the three planes determined by two of the lines OA, OB, OC, (O being origin) a straight line is drawn through O perpendicular to the third line.

42. If a = i + j + k; b = i - j + k; c = i + j - k then Column 1 Column 2 (a) a ◊ b + b ◊ c + c ◊ a (p) 8 (b) ((a ◊ c)c + (c ◊ b)a) ◊ b (q) - 2 (c) (a + 2b) ◊ (a + (a ◊ c)b) (r) 7 (d) (a + b) (a + b + c) (s) 1 43. The area / volume of Column 1

(c) tetrahedra with vertices (r) 89/2 - i + k, 2i - j, i + 2j + 5k, i + 2j + k (d) triangle with vertices (s) 6 i, j, i + j + k 44. Match the equation in column 1 with curve in column 2. (r = xi + yj + zk, x, y, z, ŒR) Column 1 Column 2 (a) |r| = |2i – r| (p) x2 + y2 + z2 = 6 (b) |r – 2i| = |r + 2j| (q) x2 + y2 + z2 = 1 (c) r.r = 4 (r) x + y = 0 (d) (r – i) . (r + i) = 5 (s) x = 1 45. Match the functions in column 1 with their domains in column 2. (r = xi + yj + zk, x, y, z Œ R) Column 1 Column 2 (p) 0 < |r| £ 1 (a) tan–1 (|r|) (q) |r| £ 1 (b) cos–1 (|r|) (r) R3 (c) ln |r| + sin–1 (|r|) (d) ln |r – i| + cosec–1(|r|) (s) |r| ≥ 1, r π i

Statement-1: The three lines so determined are coplanar Statement-2: (a ¥ b) ¥ c + (b ¥ c) ¥ a + (c ¥ a) ¥ b = 0 where OA = a, OB = b and OC = c . 47. If a . c = 3/2, b . d = 2, a . d = 3 and b . c = 1/2 Statement-1: a ¥ b, c, d are non-coplanar Statement-2: (a ¥ b).(c ¥ d) = (b . c) (a . d) – (a . c) (b . d) 48. Let the vectors PQ, QR, RS, ST, TU and UP represent the sides the regular hexagon Statement-1: PQ ¥ (RS + ST) π 0 Statement-2: PQ ¥ RS = 0 and PQ ¥ ST π 0 49. Consider the planes 3x – 6y – 2z = 15 and 2x + y – 2z =5 Statement-1: The parametric equation of the line of intersection of the given planes are x = 3 + 14t, y = 1 + 2t, z = 15t Statement-2: The vector 14i + 2j + 15k is parallel to the line of intersection of given planes. 50. Let A, B, C, D are four points in space with position vectors a, b, c, d respectively.

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Statement-1: AC2 + BD2 + AD2 + BC2 ≥ AB2 + CD2 Statement-2: |a – b| + |b – c| + |c – d| ≥ |a – d|

COMPREHENSION-TYPE QUESTIONS Paragraph for Question Nos. 51 to 53 If b π 0, then every vector a can be written in a unique manner as the sum of a vector a11 parallel to b and a vector a^ perpendicular to b. If a is parallel to b, then a11 = a and a^ = 0. If a is perpendicular to b, then a11 = 0 and a^ = a. The vector a11 is called the projection of a on b and is denoted by projb a. Since projb a is parallel to b, it is a scalar multiple of the unit vector in the direction of b i.e. Projb a = lub The scalar l is called the component of a in the direction of b and is denoted by compb a. In fact, Projb a = (a ◊ ub)ub and compb a = a ◊ ub 51. If a = - 2i + j + k and b = 4i - 3j + k then projb a 5 (a) 4i - 3j + 2k (b) - (4i - 3j + k) 13 (c)

5 (4i - 3j + k) 13

(d) -

4 (2i - j + 2k) 11

52. If a and b are as in Example 41 and a = a11 + a^ then a1 is equal to 1 1 (a) (- 3i + j + 9k) (b) (- 3i - j + 4k) 13 13 2 2 (- 3i - j + 9k) (d) (3i + j - 9k) (c) 13 13 53. If a and b are as in Example 41 then compb a is (a) (- 5/13) 26

(b)

(5/13)

26

5 17 11 Paragraph for Question Nos. 54 to 56 (c) 5 26

(d) -

Let C: r(t) = x(t)i + y(t)j + z(t)k

r (t + h) - r ( h) exist xÆ0 h

be a differentiable curve i.e. lim

for all t. The vector r¢(t) = x¢(t)i + y¢(t)j + z¢(t)k if not 0, is tangent to the curve C at the point P(x(t), y (t), z(t)) and r¢(t) points in the direction of increasing t. 54. The point P on the curve r(t) = (1 - 2t) i + t2 j + 2e2(t - 1) k at which the tangent vector r ¢(t) is parallel to the radius vector r(t) is

(a) (- 1, 1, 2) (b) (1, - 1, 2) (c) (- 1, 1, - 2) (d) (1, 1, 2) 55. A parameterized tangent vector to r (t) = ti + t2j + t3k at (2, 4, 8) is (a) R (u) = 2i + 4j + 8k + u(i + j + 4k) (b) R(u) = i + 2j + 4k + u(i + 4j + 12k) (c) R(u) = i + 4j + 12k + u(2i + 4j + 8k) (d) R(u) = 2i + 4j + 8k + u(i + 4j + 12k) 56. The tangent vector to r(t) = 2t2 i + (1 - t) j + (3t2 + 2) k at (2, 0, 5) is (a) 4i + j - 6k (b) 4i - j + 6k (c) 2i - j + 6k (d) 2i + j - 6k

INTEGER-ANSWER TYPE QUESTIONS 57. If D is the area of the triangle whose vertices A, B and C have position vectors i - j + 2k, 2i + j - k and 3i - j + 2k w.r.t. origin of reference O then D2/13 is 58. If A, B, C are the non-coplanar vectors, then A◊B¥C B◊A ¥C + is equal to C¥ A◊B C◊A ¥ B 59. If the vectors ai + j + k, i + bj + k and i + j + ck, ( a π b π c π 1) are coplanar, then the value of 1 1 1 is + + 1- a 1- b 1- c a a 2 1 + a3 60. If

b b 2 1 + b3 c

c

2

1+ c

= 0 and the vectors A =

3

(1, a, a2), B = (1, b, b2), C = (1, c, c2) are non-coplanar then the product |abc| is 61. The value of |a ¥ i|2 + |a ¥ j|2 + |a ¥ k|2 when |a| = 1, is 62. A nonzero vector a is parallel to the line of intersection of the plane determined by the vectors i, i + j and the plane determined by the vectors i - j, i + k. If the acute angle between a and the vector i - 2i + 2k then q

2 cos q is

63. Let OA = a, OB = 10a + 2b and OC = b, where O, A and C are non-collinear points. Let p denote the area of the quadrilateral OABC, and let q denote the area of the parallelogram with OA and OC as adjacent sides. If p = Kq then K is. 64. Let a = i - j, b = i + 2j + 2k, c = 2i + j + 2k and d = 2i - j + k. If p is the shortest distance between the lines r = a + t b and r = c + p d then 2p2 is 65. Let a = i + j + k, b = i + 2k and c = 2i + j + 2k. If the equation of the plane through points with position

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vectors a and b and parallel to the vector c is r . (- 3i + 2j + 2R) = l, then l is 66. If a, c, d are non-coplanar vectors satisfying d ◊ (a ¥ (b ¥ (c ¥ d))) = k[a c d] and b ◊ d = 14 then k/7 is LEVEL 2

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. The position vectors of the points A, B and C are i + j + k, i + 5j – k and 2i + 3j + 5k, respectively. The greatest angle of the triangle ABC is (a) 3p/4 (b) p/2 (c) cos–1 2 (d) cos–1 5 . 3 7 2. Let a, b and c be three vectors such that a + b + c = 0 and |a| = 3, |b| = 5, |c| = 7. Then an angle between a and b is (a) p/12 (b) cos–1 2 (c) p/6 (d) p/3. 3 3. If r ◊ a = 0, r ◊ b = 0 and r ◊ c = 0 for some non-zero vector r, then the value of [a b c ] is (a) 0 (b) 1/2 (c) 1 (d) 2. 4. The position vectors of three consecutive vertices of a parallelogram are i + j + k, i + 3j + 5k and 7i + 9j + 11k. The position vector of the fourth vertex is (a) 6 (i + j + k) (b) 7 (i + j + k) (c) 2j – 4k (d) 6i + 8j + 10k. 5. The volume of the parallelopiped whose sides are given by OA = 2i – 3j, OB = i + j – k and OC = 3i – k is (a) 4/13 (b) 4 (c) 2/7 (d) none of these. 2 6. The value of |(a ¥ i) ¥ j| + 1|(a ¥ j) ¥ k|2 + |(a ¥ k) ¥ i|2 is (b) 2a2 (a) a2 2 (c) 3a (d) none of these. 7. If a, b and c are any three vectors, then a ¥ (b ¥ c) = (a ¥ b) ¥ c if and only if (a) (b) (c) (d)

b and c are collinear a and c are collinear a and b are collinear a, b, c are collinear

8. The value of a for which the vectors 2i – j + k, i + 2j + a k and 3i – 4j + 5k are coplanar is (a) 3 (b) –3 (c) 2 (d) 5. 9. The area of a parallelogram having diagonals a = 3i+ j – 2k and b = i – 3j + 4k is

(b) 2 3

(a) 4

(c) 4 3 (d) 5 3 .

10. If r r ¥ (i + 2j + k) = i – k, then for any scalar t, r is equal to (a) i + t(i + 2j + k) (b) j + t(i + 2j + k) (c) k + t(i + 2j + k) (d) i – k + t(i + 2j + k). 11. The vectors a, b and c are equal in length and taken pairwise, make equal angles. If a = i + j, b = j + k and c make an obtuse angle with the base vector i, then c is equal to (a) i + k (b) –i + 4j – k 1 4 1 1 4 1 (d) i + j - k . (c) - i + j - k 3 3 3 3 3 3 12. A line makes angles a, b, g and d with diagonals of a cube. The value of cos2 a + cos2 b + cos2 g + cos2 d is (a) 1 (b) 1/3 (c) 8/3 (d) 4/3 13. If (a ¥ b) ¥ (c ¥ d) = [a b d] c + kd then the value of k is (a) [b a c] (b) [a b c] (c) [b c d] (d) [c b d] 14. A value of x for which the angle between c = xi + j + k and d = i + xj + k is p/3 is (a) 1 +

2

(b) 4

(c) 3 + 2 (d) 4 2 15. The line x = – 2, y = 4 + 2t, z = – 3 + t intersect (a) the xy-plane in (–2, 0, –5) (b) the xz-plane in (– 2, 0, – 4) (c) the yz-plane (d) the yz-plane in (– 2, 0, 7) 16. Let u = 2i – j + 3k and a = 4i – j + 2k. The vector component of u orthogonal to a is (a) (1/7) (20i – 5j + 10k) (b) (1/7) (4i + 24j + 4k) (c) (1/7) (11i + 2j + 6k) (d) (- 1/7) (6i + 2j – 11k) 17. If a, b, c, d lie in the same plane then (a ¥ b) ¥ (c ¥ d) is equal to (a) c + d (b) 0 (c) [a b c] a + 2b (d) [b c d] c + d 18. A vector of length 7 which is perpendicular to 2j – k and – i + 2j – 3k and makes an obtuse angle with y-axis is

( ) (1/ 3 ) (4i – j – 2k)

(a) 1/ 5 (4i – j + k) (b)

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( ) (1/ 3 )

(c) 1/ 3 (– 4i + j + 2k) (d)

(– 4i – j + 2k)

19. The angle between two diagonals of a cube is 1 2 (a) cos–1 (b) cos–1 3 3 1 2 (c) cos–1 (d) cos–1 3 3 20. The point of intersection of the lines r ¥ a = b ¥ a, r ¥ b = a ¥ b is (a) a (b) b – a (c) a – b (d) (a + b) 21. Let |a| = |b| = 2 and p = a + b, q = a – b. If |p ¥ q| = 2(k – (a ◊ b)2)1/2 then (a) k = 16 (b) k = 8 (c) k = 4 (d) k = 1 22. The value of |a ¥ (i ¥ j)|2 + |a ¥ (j ¥ k)|2 + |a ¥ (k ¥ i)|2 is (b) 2 |a|2 (a) |a|2 (c) 3 |a|2 (d) none of these 23. Let the vector a, b, c and d be such that (a ¥ b) ¥ (c ¥ d) = 0. Let P1 and P2 be planes determined by the pairs of vectors a, b; c and d respectively. Then the angle between P1 and P2 is (a) 0 (b) p/4 (c) p/3 (d) p/2 24. Let a = 2i + j – 2k and b = i + j. If c is vector such that a ◊ c = |c|, |c – a| = 2 and the angle between a ¥ b and c is 30°, then |(a ¥ b) ¥ c| is equal to (a) 2/3 (b) 3/2 (c) 2 (d) 3 25. Let a, b, g be distinct real numbers. The points A, B, C with position vectors a i + b j + g k, b i + g j + a k, g i + a j + bk respectively (a) are collinear (b) form an equilateral triangle (c) form an isosceles triangle (d) form a right angled triangle. 26. Let a = j, b = j – k, c = k – i. If d is a unit vector such that a ◊ d = 0 = [b c d] the vector d equals i-k i + j-k (b) (a) ± 2 3 (c) ±

i+k 2

28. If A(4, 7, 8), B(2, 3, 4), C(2, 5, 7) are the position vectors of the vertices of DABC. Then the length of angle bisector of angle A is (a) (3/2)

27. The shortest distance between r = (1 + l) i + j + k and r = (i – j – k) + m (i + 2j + 3k) is (a) 2 / 3

(b) 3/ 3

(c) 4 / 6

(d) 5 / 6

(b) (2/3)

34

(c) (1/2) 34 (d) (1/3) 34 29. A triangle ABC of the vertices A(1, – 2, 2), B(1, 4, 0) and C(– 4, 1, 1). The vector BM, where M is the foot of the altitude drawn from B to AC is 20 10 (a) - i - 10 j + k 3 3 (b) (c)

10 30 10 ij+ k 7 7 7

20 10 i + 5j- k 7 7

30 10 - 20 ij+ k 7 7 7 30. If the vector a, b and c form the sides BC, CA and AB respectively of a triangle ABC, then (a) a ◊ b + b ◊ c + c ◊ a = 0 (b) a ¥ b = b ¥ c = c ¥ a (c) a ◊ b = b ◊ c = c ◊ a (d) a ¥ b + b ¥ c + c ¥ a = O (d)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 31. Two sides of a triangle are formed by the vectors a = 3i + 6j – 2k and b = 4i – j + 3k. Acute angles of the triangle are (a) cos–1 (c) cos–1

7 75 3

(b) cos–1 (d) cos–1

26 75 2 . 3

15 32. A vector of magnitude 5 perpendicular to the vectors 2i + j – 3k and i – 2j + k is 5 3 5 3 (a) (i + j + k) (b) (i + j – k) 3 3 (c)

(d) ± k

34

5 3 5 3 (i – j + k) (d) – (i + j + k). 3 3

33. Let |a| = 3 and |b| = 4. The value of m for which the vectors a + mb and a – mb are orthogonal is (a) 3/4 (b) 2/4 (c) – 3/4 (d) – 2/3. 34. Suppose a, b, c are three non-zero vectors such that a ¥ b = c, b ¥ c = a and c ¥ a = b, then (a) a . b = b . c = c . a = 0

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(b) [a b c] = 1 (c) |a| = |b| = |c| = 1 (d) (a ¥ b) ¥ (c ¥ a) + a = 0 35. If b, c, d are non-coplanar vectors then (a ¥ b) ¥ (c ¥ d) + (a ¥ c) ¥ (b ¥ d) + (a ¥ d) ¥ (b ¥ c) is (a) a ¥ b (b) b ¥ c (c) d ¥ c (d) 0 36. If the unit vectors a and b are inclined at an angle q and |a – b| < 1, then if 0 £ q £ p, q lies in the interval ⎡ π⎞ ⎛ 5π ⎤ (b) ⎜ , 2π ⎥ (a) ⎢0, ⎟ ⎠ ⎝ 3 3 ⎣ ⎦ ⎡π ⎤ ⎡ 5π ⎤ (c) ⎢ , π ⎥ (d) ⎢π , ⎥. ⎣3 ⎦ ⎣ 3 ⎦ 37. If in a right angled triangle ACB, the hypotenuse AB = p then AB ◊ AC + BC ◊ BA + CA ◊ CB is equal to (a) p2

(b) p2/2

(c) 2p2

(d) p

38. The volume of the tetrahedron whose vertices are the points A, B, C and D with position vectors i – 6j + 10k, –i – 3j + 7k, 5i – j + lk and 7i – 4j + 7k respectively is 11 cubic units if the value of l is (a) –1 (b) 1 (c) –7 (d) 7. 39. Let a, b, c be three unit vectors such that |a – b|2 + |b – c|2 + |c – a|2 ≥ 9 then (a) a ¥ b = b ¥ c = c ¥ a (b) a + b + c = 0 (c) a , b, c are coplanar (d) a, b, c form sides of an equilateral triangle. m m 40. Let am = ( 3 - 1) i + ( 3 + 1) j " m Œ N, then

(a) (b) (c) (d)

|a2019|2 is an integer gcd (|a2019|2, |a2018|2) is a prime |am+2|2 + |am|2 = 4|am+1|2 " m. |am|2 is an even integer " m ≥ 2

MATRIX-MATCH TYPE QUESTIONS 41. Let r = xi + yj. Match the expression in column 1 with name of curve on in column 2. Column 1 Column 2 (a) |r – 2i| = 2 (p) Straight line with slope – 1 (b) |r – 2i| = |r + 2j| (q) Circle through origin | r - 2i | =2 (r) an ellipse (c) | r + 2j |

(d) |r – 2i| + |r + 2j| = 8 (s) circle with centre at (–2/3)i + (–8/3)j (t) a line segment joining 2i and –2j 42. If a ¥ b = b ¥ c = c ¥ a and |a| = 5. Match the expressions in column 1 with values in column 2 Column 1 Column 2 (a) |a + 2b + 2c| (p) 20 (b) |a – 2b + c| (q) 15 (c) |a + b + 3c| – |a| (r) 5 (d) |a + b + c| + 2|a| (d) 10 43. Let a, b, c be three vectors, such that [a b c] = 2. Match the expressions in column 1 with values in column 2. Column 1 Column 2 (a) [a + b + c a + b b] (p) 60 (b) [3a + 4b + 5c + a + 2b b] (q) 4 (c) [a ¥ b b ¥ c c ¥ a] (r) 2 (d) [b + c c + a a + b] (s) 10 44. Let a = (x + y)i + xj + xk, b = (5x + 4y)i + 4xj + 2xk, and c = (10x + 8y)i + 8xj + 3xk let f(x, y) = [a b c]. Match expressions in column 1 with values in column 2 Column 1 Column 2 (a) f(2, 2017) (p) 8 (b) f(3, 2018) (q) 27 (c) f(2, 3) (r) 4034 (d) f(3, 2) (s) 6054 (t) 6

ASSERTION-REASON TYPE QUESTIONS 45. If a, b, c Are three non-coplanar vectors such that |a| = 1, b ¥ c = i + j – k, c ¥ a = i + 3j, a ¥ b = i – j + 2k, a . b = 1 = a . c and [a b c] = 2 Statement-1: a can be determined explicitly 1 [(b . c) (b ¥ c) + (a . b) (c Statement-2: a = [a b c] ¥ a) + (a . c) (a ¥ b)] 46. If a = i + j – k b = 2i + j – 3k and r is a vector satisfying 2r + r ¥ a = b 1 Statement-1: r = [7i + 5j – 9k + (a ¥ b)] 7 Statement-2: r can be expressed in terms if a, b, a ¥b

IIT JEE eBooks: www.crackjee.xyz 28.38 Comprehensive Mathematics—JEE Advanced

COMPREHENSION-TYPE QUESTIONS Paragraph for Question No. 47 to 50 Equation of a line can be obtained as the intersection of two planes, or passing through a point and parallel to given plane. Similarly equation a plane can be obtained having different condition e.g. passing through three points or through a point and perpendicular to two planes. 47. The line through the point c, parallel to the plane r . n = 1 and perpendicular to the line, r = a +tb is (a) r = c + t(a ¥ n) (b) r = c + t(b ¥ n) (c) r = c + tn (d) r = a + t(c ¥ n) 48. The line through the point a and parallel to the planes r . n1 = q1, r . n2 = q2 is (b) r = a + t(n1 – n2) (a) r = a + tn1 (d) r = a + t(n1 ¥ n2) (c) r = a + tn2 49. An equation of plane which passes through the two points a and b and is perpendicular to the plane r . n = q is (a) r . ((b – a) ¥ n) = q (b) r . (a – b) = q (c) r . ((b – a) ¥ n) = [a b n] (d) r . ((b – a) ¥ n) = [a n b] 50. An equation of plane which passes through a and is perpendicular to the plane r . n = q and is parallel to the line r = b + tc is (a) r . b = [a n c] (b) [r n c] = [a n c] (c) r . a = [b n c] (d) [r c n] = [a n c]

INTEGER-ANSWER TYPE QUESTIONS 51. Let a = cos2(p/8)i + cos2(p3/8)j + k, b = i Рcos2(p5/8) j + cos2(p7/8)k then 2(|a|2 + |b|2) is 52. Let a = xi + (x + 1)j + (x Р1)k b = (x + 1)i + xj + (x + 2)k c = (x Р1)i + (x + 2)j + xk. If a, b, c are coplanar, then 4x + 3 is equal to 53. Let n ΠN and a =

Êpˆ Êpˆ sin Á ˜ i - cos Á ˜ j . If |a| = Ë n¯ Ë n¯

1 n , then n is 2

a b c 54. Suppose a, b, c > 0 and let D = b c a . c a b Let A = bc – a2, B = ac – b2, C = ab – c2 Let x = Ai + Bj + Ck, y = Bi + Cj + Ak and z = Ci + Aj + Ck. Ê 1ˆ If [x y z] = 36, then Á - ˜ D is equal to Ë 3¯ 55. Let a, b, c > 0 and x = ai + bj + ck. Suppose A = a+b+c abc Let q = tan–1 ((Ax) . i) + tan–1 ((Ax) . j) + tan–1 ((Ax) . k), 3 then q is p

PAST YEARS’ IIT QUESTIONS SINGLE CORRECT ANSWER TYPE QUESTIONS 1. The scalar A · {(B + C) × (A + B + C)} equals (a) 0 (b) [ABC] + [BCA] (c) [ABC] (d) none of these [1981] 2. For non zero vectors a, b, c, |(a × b) · c| = |a| |b| |c| holds if and only if (a) a ◊ b = 0, b ◊ c = 0 (b) b ◊ c = 0, c ◊ a = 0 (c) c ◊ a = 0, a ◊ b = 0 (d) a ◊ b = b ◊ c = c ◊ a = 0 [1982] 3. The points with position vectors 60i + 3j, 40i – 8j, ai – 52j are collinear if

(a) a = – 40 (c) a = 20

(b) a = 40 (d) none of these [1983] 4. The volume of the parallelopiped whose sides are given by OA = 2i – 3j, OB = i + j – k, OC = 3i – k is (a) 4/13 (b) 4 (c) 2/7 (d) none of these [1983] 5. Let a = a1i + a2j + a3k b = b1i + b2j + b3k and c = c 1i + c 2j + c 3k be three non-zero vector such that c is a unit vector perpendicular to both the vectors a and b. If the

IIT JEE eBooks: www.crackjee.xyz Vector Algebra 28.39

a1 p angle between a and b is , then b1 6 is equal to c1 (a) 0 (b) 1

(c) (d)

a2 b2 c2

a3 b3 c3

2

1 2 (a1 + a22 + a32 ) (b12 + b22 + b32 ) 4 3 2 (a1 + a22 + a32 ) (b12 + b22 + b32 ) (c12 + c22 + c32 ) 4

(e) none of these [1986] 6. A vector a has components 2p and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to the new system, a has component p + 1 and 1, then (a) p π 0 (b) p = 1 or p = – 1/3 (c) p = – 1 or p = 1/3 (d) p = 1 or p = – 1 (e) none of these [1986] 7. The number of vectors of unit length perpendicular to vectors a = (1, 1, 0) and b = (0, 1, 1) is: (a) one (b) two (e) none of these [1987] 8. Let a, b, c be three non-coplanar vectors and p, q, r

p=

c¥a a¥b b¥c ,q= ,r= [a b c] [a b c] [a b c ]

then the value of the expression (a + b) · p + (b + c) · q + (c + a) · r is equal to (a) 0 (b) 1 (c) 2 (d) 3 [1988] 9. Let a, b, c be distinct non-negative numbers. If the vectors ai + aj + ck, i + k and ci + cj + bk lie in a plane, then c is (a) the arithmetic mean of a and b (b) the geometric mean of a and b (c) the harmonic mean of a and b (d) equal to zero [1993] 10. Let p and q be the position vectors of P and Q respectively, with respect to O and |p| = p, |q| = q. The points R and S divide PQ internally and externally in the ratio 2 : 3 respectively. If OR and OS are perpendicular then (b) 4p2 = 9q2 (a) 9p2 = 4q2 (c) 9p = 4q (d) 4p = 9q [1994] 11. Let a, b, g be distinct real numbers. The points with position vectors ai + bj + g k, b i + g j + a k, g i + a j + bk (a) are collinear (b) form an equilateral triangle

(c) form a scalene triangle (d) form a right angled triangle [1994] 12. Let a = i – j, b = j – k, c = k – i. If d is a unit vector such that a · d = 0 = [bcd], then d equals i + j - 2k i + j-k (b) ± (a) ± 6 3 (c) ±

i + j+k

(d) ± k

3

[1995]

13. If a, b, c are non coplanar unit vectors such that 1 (b + c) , then the angle between a ¥ (b ¥ c) = 2 a and b is (a)

3p 4

(b)

p 4

(c)

p 2

(d) p [1995]

14. Let u, v and w be vectors such that u + v + w = 0. If |u| = 3, |v| = 4 and |w| = 5, then u ◊ v + v ◊ w + w ◊ u is (a) 47 (b) – 25 (c) 0 (d) 25[1995] 15. If a, b and c are three non coplanar vectors, then (a+ b + c) ◊ {(a + b) × (a + c)} equals (a) 0 (b) [a b c] (c) 2 [a b c] (d) –[a b c] [1995] 16. Let p, q, r be three mutually perpendicular vectors of the same magnitude. If a vector x equation p ¥ ((x – q) ¥ p) + q ¥ ((x – r) ¥ q) + r ¥ ((x – p) ¥ r) = 0 then x is given by 1 (p + q – 2r) 2

(b)

(c)

1 (p + q + r) 3

(d)

(a)

2 3

(a)

1 (p + q + r) 2

1 (2p + q – r) 3 [1997] 17. If a = i + j + k, b = 4i + 3j + 4k and c = i + aj + bk are linearly dependent vectors and | c| = 3 , then (a) a = 1, b = 1 (b) a = 1, b = ±1 (c) a = – 1, b = ±1 (d) a = ± 1, b = 1 [1998] 18. For three vectors u, v, w which of the following expression is not equal to any of the remaining three (a) u ◊ (v ¥ w) (b) (v ¥ w) ◊ u (c) v ◊ (u ¥ w) (d) (u ¥ v) ◊ w [1998] 19. Let a = 2i + j – 2k and b = i + j. If c is a vector such that a ◊ c = |c|, |c – a| = 2 2 and the angle between (a ¥ b) and c is 30°, than |(a ¥ b) × c| = (b)

3 2

(c) 2

(d) 3 [1999]

IIT JEE eBooks: www.crackjee.xyz 28.40 Comprehensive Mathematics—JEE Advanced

20. Let a = 2i + j + k, b = i + 2j – k and a unit vector c be coplanar. If c is perpendicular to a, then c = 1 1 (–j + k) (b) (–i – j – k) (a) 2 3 (c)

1 5

(–i – 2j)

(d)

1 3

(i – j – k)

[1999] 21. If the vectors a, b and c from the sides BC, CA and AB respectively of a triangle ABC, then (a) a ◊ b + b ◊ c + c ◊ a = 0 (b) a × b = b × c = c × a (c) a ◊ b = b ◊ c = c ◊ a (d) a × b + b × c + c × a = 0 [2000] 22. Let the vectors a, b, c and d be such that (a × b) × (c × d) = 0. Let P1 and P2 be the two planes determined by the pair of vectors a, b and c, d respectively, then the angle between P1 and P2 is (a) 0 (b) p/4 (c) p /3 (d) p/2 [2000] 23. If a, b and c are unit coplanar vectors, then the scalar triple product [2a – b 2b – c 2c – a] = (a) 0

(b) 1

(c) – 3

(d)

(b)

(b) 3

(c) 1/ 3 (d)

3

[2003] 29. If a = i + j + k, a◊b = 1 and a×b = j – k, then b is (a) i – j + k (b) 2j – k (c) i (d) 2i – k [2004]

1 10

(3j – k)

(d)

1 69

(2i – 8j + k) [2004]

31. Let a, b, c be three non coplanar vectors and a ·b a ·b b1 = b – a 2 a, b2 = b + |a| | a |2

32.

33.

34.

10 + 6

(c) 59 (d) 60 [2002] 28. The value of a so that the volume of parallelopiped formed by i + aj + k, j + ak and ai + k becomes minimum is (a) –3

(c)

3

[2000] 24. If a, b and c are unit vectors, then |a – b|2 + |b – c|2 + |c – a|2 does not exceed (a) 4 (b) 9 (c) 8 (d) 6 [2001] 25. Let a = i – k, b = xi + j + (1 – x)k and c = yi + xj + (1 + x – y)k, then [a b c] depends on (a) only x (b) only y (c) neither x nor y (d) both x and y [2001] 26. If a and b are two unit vectors such that a + 2b and 5a – 4b are perpendicular to each other then the angle between a and b is (a) 45° (b) 60° (d) cos–1(2/7) [2002] (c) cos–1(1/3) 27. Let v = 2i + j – k, w = i + 3k and u is a unit vector then the maximum value of [uvw] is (a) –1

30. The unit vector which is orthogonal to the vector 5i + 2j + 6k and is coplanar with the vectors 2i + j + k and i – j + k is 1 1 (2i – 6j + k) (b) (2i – 5j) (a) 41 29

35.

c1 = c –

c ·a b ·c b1 2 a + |a| | c |2

c2 = c –

b1 ·c c ·a b1 2 a – |a| | b1 |2

c3 = c –

c ·a b ·c b1 2 a + |c| | c |2

c4 = c –

c ·a b ·c b1 2 a – |c| | b1 |2

then the triplet of pairwise orthogonal vectors is (b) (a, b1, c2) (a) (a, b1, c1) (d) (a, b1, c3) (c) (a, b2, c2) [2005] Let a = i + 2j + k, b = i – j + k and c = i + j – k. A vector in the plane of a and b whose projection on c is 1/ 3 is (a) 4i – j + 4k (b) 3i + j + 3k (c) 2i + j + 2k (d) 4i + j – 4k [2006] The number of distinct real values of l, for which the vectors –l2i + j + k, i – l2j + k and i + j – l2k are coplanar, is (a) zero (b) one (c) two (d) three [2007] Let a, b, c be unit vectors such that a + b + c = 0. Which one of the following is correct? (a) a¥ b = b ¥ c = c ¥ a = 0 (b) a ¥ b = b ¥ c = c ¥ a π 0 (c) a ¥ b = b ¥ c = a ¥ c π 0 (d) a¥ b, b¥ c, c¥ a are mutually perpendicular. [2007] Let two non-collinear unit vectors a and b form an acute angle. A point P moves so that at any time t the position vector OP (where O is the origin) is given by a cos t + b sint. When P is farthest from origin O, let M be the length of OP and u be the unit vector along OP. Then,

IIT JEE eBooks: www.crackjee.xyz Vector Algebra 28.41

(a) u =

a+b and M = (1 + a ◊b)1/2 |a + b|

(b) u =

a-b and M = (1 + a◊b)1/2 |a - b|

(c) u =

a+b and M = (1 + 2a◊b)1/2 |a + b|

(d) u =

a-b and M = (1 + 2a◊b)1/2 |a - b|

40. Let a = i + j + k, b = i – j + k and c = i – j – k be three vectors. A vector v in the plane of a and b, whose 1 projection on c is , is given by 3 (a) i – 3j + 3k (b) –3i – 3j – k (c) 3i – j + 3k

(d) i + 3j – 3k [2011] 41. If a and b are vectors such that a + b = 29 2008]

36. The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors a,b,c such that a◊ b = b◊ c = c◊a = 1/2. Then the volume of the parallelopiped is (a)

1 2

(b)

1 2 2

(c)

3 2

(d)

1 3 [2008]

37. If a, b, c and d are unit vectors such that (a ¥ b) ◊ (c ¥ d) = 1 1 and a¥c= , 2 then (a) a, b, c are non-coplanar (b) b, c, d are non-coplanar (c) b, d are non-parallel (d) a, d are parallel and b, c are parallel [2009] 38. Let P, Q, R and S be the points on the plane with position vectors – 2i – j, 4i, 3i + 3j and – 3i + 2j respectively. The quadrilateral PQRS must be a (a) parallelogram, which is neither a rhombus nor a rectangle (b) square (c) rectangle, but not a square (d) rhombus, but not a square [2010] 39. Two adjacent sides of a parallelogram ABCD are given by AB = 2i + 10j + 11k and AD = –i + 2j + 2k The side AD is rotated by an acute angle a in the plane of the parallelogram so that AD becomes AD¢. If AD¢ makes a right angle with the side AB, then the cosine of the angle a is given by (a)

8 9

(b)

17 9

(c)

1 9

(d)

4 5 9

[2010]

and a ¥ ( 2i + 3j + 4k ) = ( 2i + 3j + 4k ) ¥ b , then a possible value of (a + b ) ◊ ( -7 i + 2 j + 3k ) is (a) 0 (b) 3 (c) 4 (d) 8 [2012] 42. Let PR = 3i + j – 2k and SQ = i – 3j – 4k determine diagonals of a parallelogram PQRS and PT = i + 2j + 3k be another vector. Then the volume of the parallelepiped determined by the vectors PT, PQ and PS is (a) 5 (b) 20 (c) 10 (d) 30 [2013]

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 1. Let a = 2i – j + k, b = i + 2j – k and c = i + j – 2k be three vectors. A vector in the plane of b and c whose 2 projection on a is of magnitude is 3 (a) 2i + 3j – 3k (b) 2i + 3j + 3k (c) –2i – j + 5k (d) 2i + j + 5k [1993] 1 2. The vector d = (2i – 2j + k) is 3 (a) a unit vector (b) makes an angle p/3 with the vector (2i – 4j + 3k) 1 (c) parallel to the vector - i + j - k 2 (d) perpendicular to the vector 3i + 2j– 2k [1994] 3. Which of the following expression are meaningful? (a) u·(v¥ w) (b) (u·v)· w (c) (u·v)w (d) u¥ (v·w) [1998] 4. Let a and b be two non-collinear unit vectors. If u = a – (a·b)b and v = a×b, then |v| is (a) |u| (b) |u| + |u·a| (c) |u| + |u·b| (d) |u| + u·(a +b) [1999]

(

)

IIT JEE eBooks: www.crackjee.xyz 28.42 Comprehensive Mathematics—JEE Advanced

5. Let a be a vector parallel to line of intersection of planes P1 and P2. Plane P1 is parallel to the vectors 2j + 3k and 4j + 3k and that P2 is parallel to j – k and 3i + 3j, then the angle between vector a and vector 2i + j – 2k is (a)

p 2

(b)

p 4

p 3p (d) [2006] 6 4 6. The vector(s) which is/are coplanar with vectors i + j + 2k and i + 2j + k, and perpendicular to the vector i + j + k is/are (a) j – k (b) – i + j (c) i – j (d) – j + k [2011] (c)

7. Let x, y and z be three vectors each of magnitude p 2 and the angle between each pair of them is . 3 If a is a nonzero vector perpendicular to x and y × z and b is a nonzero vector perpendicular to y and z × x, then (a) b = (b.z) (z – x) (b) a = (a.y) (y – z) (c) a.b = – (a.y) (b.z) (d) a = (a.y) (z – y) [2014] 8. Let PQR be a triangle. Let a = QR, b = RP and c = PQ and If |a| = 12, and b.c = 24 then which of the following is (are) true? (a)

| c |2 - | a | = 12 2

(b)

| c |2 - | a | = 30 2

(c) | a ¥ b + c ¥ a | = 48 3 (d) a.b = – 72 [2015] 3 9. Let u = u1i + u2j + u3k be a unit vector in R and w 1 = (i + j + 2k). Given that there exists a vector v 6 in R3 such that |u ¥ v| = 1 and w.(u ¥ v) = 1. Which of the following statement(s) is(are) correct? (a) There is exactly one choice for such v v (c) If u lies in the xy-plane then |u1| = |u2| (d) If u lies in the xz-plane then 2|u1| = |u3| [2016]

MATRIX-MATCH TYPE QUESTIONS 1. Match the statements in Column 2 with the values in Column 2

Column 1 (a) Root(s) of the equation

Column 2 (p) =

p 6

(q) =

p 4

2 sin2 q + sin2 2q = 2 (b) Points of discontinuity of the function È 6x ˘ È 3x ˘ f(x) = Í ˙ cos Í ˙ p Î ˚ Îp ˚ where [y] denotes the largest integer less than

(r)

p 3

(s)

p 2

or equal to y (c) Volume of the parallelopiped with its edges represented by the vectors i + j, i + 2j and i + j + pk (d) Angle between vectors

(t) p

a and b where a, b and c are unit vectors satisfying [2009] a + b + 3c = 0 2. Match Column 1 with Column 2 and select the correct answer using the code given below the lists Column 1 Column 2 P. Volume of parallelepiped determined by 1. 100 vectors a, b and c is 2. Then the volume of the parallelepiped determined by vectors 2(a ¥ b), 3(b ¥ c) and (c ¥ a) is Q. Volume of parallelepiped determined by 2. 30 vectors a, b and c is 5. Then the volume of the parallelepiped determined by vectors 3(a + b), (b + c) and 2(c + a) is R. Area of a triangle with adjacent sides 3. 24 determined by vectors a and b is 20. Then the area of the triangle with adjacent sides determined by vectors (2a + 3b) and (a – b) is S. Area of a parallelogram with adjacent sides 4. 60 determined by vectors a and b is 30. Then the area of the parallelogram with adjacent sides determined by vectors (a + b) and a is Codes: P Q R S (a) 4 2 3 1 (b) 2 3 1 4 (c) 3 4 1 2 (d) 1 4 3 2 [2013]

IIT JEE eBooks: www.crackjee.xyz Vector Algebra 28.43

3. Column 1 (a) In R2, if the magnitude of the projection vector of the vector αi + βj on

3i + j is

Column 2 (p) 1 3

and if α = 2 + 3 β, then possible value(s) of |α| is (are) (b) Let a and b be real numbers such that the function ÏÔ-3ax 2 - 2, x < 1 f ( x) = Ì is ÔÓ bx + a 2 , x ≥ 1 differentiable for all x R. Then possible value(s) of a is(are) (c) Let ω 1 be a complex cube root of unity. If (3 – 3ω + 2ω2)4n +3 + (2 + 3ω – 3ω2)4n +3 + (–3 + 2ω + 3ω2)4n +3 = 0, then possible value(s) of n is (are) (d) Let the harmonic mean of two positive real numbers a and b be 4. If q is a positive real number such that a, 5, q, b is an arithmetic progression, then the value(s) of |q – a| is (are)

(q) 2

[2014] 6. Suppose that p, q and r are three non-coplanar vectors in R3. Let the components of a vector s along p, q and r be 4, 3 and 5, respectively. If the components of this vector s along (–p + q + r), (p – q + r) and (–p + q + r) are x, y and z, respectively, then the value of 2x + y + z is [2015]

(r) 3

FILL

(s) 4

(t) 5 [2015]

ASSERTION-REASON TYPE QUESTIONS

INTEGER-ANSWER TYPE QUESTIONS

and b =

2i + j + 3k 14

i - 2j 5

, then the value of

(2a + b) .[(a × b) × (a – 2b)] is

IN THE

BLANKS TYPE QUESTIONS

1. Let A, B, C be three vectors of length 3, 4, 5 respectively. Let A be perpendicular to B + C, B to C + A and C to A + B. Then the length of the vector A + B + C is ________ . [1981] 2. A, B, C, D are four points in a plane with position vectors a, b,c,d respectively, such that (a – d)·(b – c) = (b – d) · (c – a) = 0. The point D, then, is the ______ of the triangle ABC [1984] a a 2 1 + a3

1. Let the vectors PQ, QR, RS, ST, TU and UP represent the sides of a regular hexagon. Statement-1: PQ ¥ (RS + ST) π 0. Statement-2: PQ ¥ RS = 0 and PQ ¥ ST π 0 [2007]

1. If a and b are vectors is space given by a =

coplanar vectors can be chosen from V in 2p ways. Then p is [2012] 5. Let a, b and c be three non-coplanar unit vectors such that the angle between every pair of them is p . If a × b + b × c = pa + qb + rc, where p, q and 3 p 2 + 2q 2 + r 2 r are scalars, then the value of is q2

[2010]

2. Let a = – i – k, b = –i + j and c = i + 2j + 3k be three given vectors. If r is a vector such that r × b = c × b and r . a = 0, then the value of r . b is [2011] 3. If a, b and c are unit vectors satisfying |a – b|2 + |b – c|2 + |c – a|2 = 9, then |2a + 5b + 5c| is [2012] 4. Consider the set of eight vectors V = {ai + bj + ck : a, b, c Œ {-1, 1}} . Three non-

2 3 3. If D = b b 1 + b = 0 and the vectors

c

c2

1 + c3

(1, a, a2) (1, b, b2), (1, c, c2) are non coplanar then the product abc = ________ . [1985] 4. If A, B,C are three non coplanar vectors then A · (B ¥ C) B · ( A ¥ C) + x= = ________ . [1985] (C ¥ A) · B C · ( A ¥ B) 5. If A = (1, 1, 1), C = (0, 1, –1) are given vectors then the vector B satisfying the equations A ¥ B = C and A·B = 3 is ________ . [1985] 6. If the vectors ai + j + k, i + bj + k and i + j + ck (a, b, c π 1) are coplanar, then the value of x=

1 1 1 + + ________ . 1- a 1- b 1- c

[1987]

7. Let b = 4i + 3j and c be two vectors perpendicular to each other in the xy-plane. All vectors in the same plane having projections 1 and 2 along b and c respectively, are given by ________ . [1987]

IIT JEE eBooks: www.crackjee.xyz 28.44 Comprehensive Mathematics—JEE Advanced

8. The components of a vector a along and perpendicular to a non-zero vector b are ________ and ________ respectively. [1988] 9. Given that a = (1, 1, 1), c = (0, 1, – 1), a·b = 3 and a ¥b = c, then b = ________ . [1991] 10. A unit vector coplanar with i + j + 2k and i + 2j + k and perpendicular to i + j + k is ________ . [1992] 11. A unit vector perpendicular to the plane determined by the points P(1, – 1, 2), Q(2, 0, – 1) and R(0, 2, 1) is ________ . [1994] 12. A non zero vector a is parallel to the line of intersection of the plane determined by the vectors i, i + j and the line determined by the vectors i – j, i + k. The angle between a and the vector i – 2j + 2k is ________ . [1996] 13. If b and c any two non-collinear unit vectors and a is any vector, then a · (b ¥ c) (b × c) = ________ . (a·b)b + (a·c)c + |b ¥ c| [1996] 14. Let OA = a, OB = 10a + 2b and OC = b where O, A and C are non-collinear points. Let p denote the area of the quadrilateral OABC and let q denote the area of the parallelogram with OA and OC as adjacent sides. If p = kq then k = ______ . [1997] 15. Let a, b, and c be three vectors having magnitudes 1, 1 and 2 respectively. If a ¥ (a ¥ c) + b = 0, then the acute angle between a and c is ______ . [1997]

TRUE-FALSE TYPE QUESTIONS 1. Let A, B, C be unit vectors. Suppose that p A·B = A·C = 0 and the angle between B and C is . 6 Then A = ±2(B ¥ C) [1981] 2. If X · A = 0, X · B = 0, X·C = 0 for some non zero vector X then [ABC] = 0 [1983] 3. The point with position vectors a + b, a – b, a + kb are collinear for all real values of k. [1984] 4. For any three vectors a, b and c (a – b) · (b – c) ¥ (c – a) = 2a · b ¥ c [1989]

SUBJECTIVE-TYPE QUESTIONS 1. A1, A2, ..... An are the vertices of a regular polygon with n sides and O is its centre. Show that

n -1

 ( OAi ¥ OAi+1) = (1 – n) (OA2 ¥ OA1)

i =1

[1982]

2. Find all values of l such that (x, y, z) π (0, 0, 0) and x(i + j + 3k) + y(3i – 3j + k) + z(– 4i + 5j) = l(xi + yj + zk) where i, j, k are unit vectors along the coordinate axes. [1982] 3. If c be a given non zero scalar and A and B be given non zero vectors such that A ^ B X A·X = c, A ¥ X = B [1983] 4. A vector A has components A1, A2, A3 in a right handed rectangular cartesian coordinate system OXYZ. The coordinate system is rotated about the x-axis through an angle p/2. Find the components of A in the new coordinate system in terms of A1, A2, [1983] A3. 5. The position vectors of the points A, B, C and D are 3i – 2j – k, 2i + 3j – 4k, – i + j + 2k and 4i + 5j + lk respectively. If the points A, B, C and D lie on a l. [1986] 6. If A, B, C, D are four points in space, prove that |AB ¥ CD + BC ¥ AD + CA ¥ BD| = 4 (area of D ABC) [1987] 7. Let OACB be a parallelogram with O at the origin and OC a diagonal. Let D be the midpoint of OA. Using vector methods prove that BD and CO intersect in same ratio. Determine this ratio. [1988] 8. If vectors a, b, c are coplanars show that a b c a ·a a ·b a ·c b ·a b ·b b ·c

=0

[1989] 9. In a triangle OAB, E is the midpoint of BO and D is a point on AB such that AD : DB = 2 : 1. If OD and AE intersect at P, determine the ratio OP : PD using vector methods. [1989] 10. Let A = 2i + k, B = i + j + k and C = 4i – 3j + 7k. Determine a vector R satisfying R¥B = C¥B and R·A = 0 [1990] 11. Determine the value of c so that for all real x, the vector cxi – 6j – 3k and xi + 2j + 2cxk make an obtuse angle with each other. [1991] 12. In a triangle ABC, D and E are points on BC and AC respectively, such that BD = 2DC and AE = 3EC. Let P be the point of intersection of AD and BE. Find the BP using vector methods. [1993] ratio PE

IIT JEE eBooks: www.crackjee.xyz Vector Algebra 28.45

13. If the vectors a, b, c, d are not coplanar then prove that the vector x = (a¥b) × (c¥d) + (a¥c) × (d¥b) + (a¥d) ¥ (b¥c) is parallel to a [1994] 14. The position vectors of the vertices A, B, C of a tetrahedron ABCD are i + j + k, i and 3i respectively. The altitude from vertex D to the opposite face ABC, meets the median line through A of the DABC at E. If the length of the side AD is 4 and the volume of the 2 2 , E for tetrahedron is 3 all its possible positions. [1996] 15. If A, B and C are vectors such that |B| = |C|. Prove that [(A + B) ¥ (A + C)] ¥ (B ¥ C) · (B + C) = 0 [1997] 16. Let a, b and c be non coplanar unit vectors, equally inclined to one another at an angle q. If a¥b + b¥c = pa + qb + rc Find scalars p, q and r in terms of q. [1997] 17. Prove, by vector methods or otherwise, that the point of intersection of the diagonals of a trapezium lies on the line passing through the mid-points of the parallel sides. (You may assume that the trapezium is not a parallelogram) [1998] 18. For any two vectors u and v, prove that (i) (u · v)2 + |u ¥ v|2 = |u|2 |v|2 and (ii) (1 + |u|2) (1 + |v|2) = |u + v + (1 – u·v)2 + u¥v|2 [1998] 19. Let u and v be unit vectors. If w is a vector such that w + (w¥u) = v, then prove that |(u¥v) ·w | £ 1/2 and that the equality holds if and only if u is perpendicular to v. [1999] 20. Show, by vector methods, that the angle bisectors of for the position vector of the point of concurrency in terms of the position vectors of the vertices. [2001] 21. Find 3-dimensional vectors v1, v2, v3 satisfying v1 . v1 = 4, v1. v2 = – 2, v1 . v3 = 6, v2 . v2 = 2, v2 . [2001] v3 = 5, v3 . v3 = 29. 22. Let A(t) = f1(t)i + f2(t)j and, B(t) = g1(t)i + g2(t)j, t Œ [0, 1], where f1, f2, g1, g2 are continuous functions. If A(t) and B(t) are nonzero vectors for all t and A(0) = 2i+ 3j, A(1) = 6i + 2j, B(0) = 3i + 2j and B(1) = 2i + 6j, then show that A(t) and B(t) are parallel for some t. [2001] 23. Let V be the volume of the parallelopiped formed by the vectors a = a1i + a2j + a3k, b = b1i + b2 j + b3k,

c = c1i + c2j + c3k. If ar, br, cr, where r = 1, 2, 3 are 3

non-negative real numbers and

 (ar + br + cr ) =

r =1

[2002] 3L. Show that V £ L3. 24. If u, v, w are three non-coplanar unit vectors and a, b, g are the angle between u and v, v and w and w and u respectively and x, y, z are unit vectors along the bisectors of the angle a, b, g respectively. Prove that [x ¥ y y ¥ z z ¥ x] =

a b g 1 [uvw]2 sec2 sec2 sec2 2 2 2 16

[2003]

25. If a, b, c, d are distinct vectors satisfying relation a¥b = c¥d and a¥c = b¥d. Prove that a·b + c·d π a·c + b·d [2004] 26. Let v be a unit vector along the incident ray, w be a a be a unit vector along the outward normal to the plane mirror at the point of incidence P. Express w in terms a and v. [2005]

Answers LEVEL 1

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17. 21. 25.

(b) (d) (d) (b) (c) (c) (c)

2. 6. 10. 14. 18. 22.

(a) (c) (c) (b) (b) (d)

3. 7. 11. 15. 19. 23.

(b) (b) (d) (c) (b) (c)

4. 8. 12. 16. 20. 24.

(d) (a) (a) (b) (b) (a)

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS 26. 29. 32. 34. 36. 38. 40.

(c) (d) (d) (b), (a), (a), (a),

(c), (d) (b), (c), (d) (b), (c), (d) (b)

27. 30. 33. 35. 37. 39.

(d) 28. (b), (d) (c) 31. (a) (a), (c) (a), (b) (c) (b), (c), (d)

IIT JEE eBooks: www.crackjee.xyz 28.46 Comprehensive Mathematics—JEE Advanced

MATRIX-MATCH TYPE QUESTIONS p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

41.

42.

43.

57. 1 61. 2 65. 1

1. 5. 9. 13. 17. 21. 25. 29.

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

s

d

p

q

r

s

p

q

r

s

a

p

q

r

s

b

p

q

r

c

p

q

d

p

q

31. 33. 35. 37. 39.

2. 6. 10. 14. 18. 22. 26. 30.

(d) (a) (b) (b) (b) (b) (a) (b)

3. 7. 11. 15. 19. 23. 27.

(a), (b) (a), (c) (d) (a) (a), (b), (c), (d)

32. 34. 36. 38. 40.

q

r

s

t

a

p

q

r

s

t

s

b

p

q

r

s

t

r

s

c

p

q

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s

t

r

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d

p

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t

p

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a

p

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b

p

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s

c

p

q

r

s

d

p

q

r

s

41.

48. (c)

49. (d)

COMPREHENSION-TYPE QUESTIONS 52. (c) 56. (b)

(a) (b) (c) (a) (c) (a) (a)

4. 8. 12. 16. 20. 24. 28.

(b) (b) (d) (d) (d) (b) (b)

(a), (a), (a), (b), (a),

(d) (b), (c), (d) (b) (d) (d)

MATRIX-MATCH TYPE QUESTIONS

ASSERTION-REASON TYPE QUESTIONS

51. (b) 55. (d)

(b) (b) (d) (a) (b) (a) (b) (d)

p

47. (c)

60. 1 64. 1

MULTIPLE CORRECT ANSWERS TYPE QUESTIONS

r

46. (b) 50. (b)

59. 1 63. 6

SINGLE CORRECT ANSWER TYPE QUESTIONS

q

45.

58. 0 62. 1 66. 2

LEVEL 2

p

44.

INTEGER-ANSWER TYPE QUESTIONS

53. (a)

54. (a)

42.

IIT JEE eBooks: www.crackjee.xyz Vector Algebra 28.47

p

q

r

s

a

p

q

r

s

b

p

q

r

s

c

p

q

r

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d

p

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s

p

q

r

a

p

q

b

p

c d

43.

44.

5. (b), (d) 7. (a), (b), (c) 9. (b), (c)

MATRIX-MATCH TYPE QUESTIONS p

q

r

s

t

p

q

r

s

t

s

b p

q

r

s

t

r

s

c

p

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d p

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2. (c) 3. p

q

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a

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b

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c

p

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d

p

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t

1. a

ASSERTION-REASON TYPE QUESTIONS 45. (c)

46. (a)

COMPREHENSION-TYPE QUESTIONS 47. (b)

48. (d)

49. (c)

50. (b)

INTEGER-ANSWER TYPE QUESTIONS 51. 7 55. 3

52. 1

53. 6

54. 2

ASSERTION-REASON TYPE QUESTIONS 1. (c)

INTEGER-ANSWER TYPE QUESTIONS PAST YEARS’ IIT QUESTIONS

1. 5 5. 4

SINGLE CORRECT ANSWER TYPE QUESTIONS 1. 5. 9. 13. 17. 21. 25. 29. 33. 37. 41.

6. (a), (b) 8. (a), (c), (d)

(a) (c) (b) (a) (d) (b) (c) (c) (c) (c) (c)

2. 6. 10. 14. 18. 22. 26. 30. 34. 38. 42.

(d) (b) (a) (b) (c) (a) (b) (c) (b) (a) (c)

3. 7. 11. 15. 19. 23. 27. 31. 35. 39.

(a) (b) (b) (d) (b) (a) (c) (b) (a) (b)

4. 8. 12. 16. 20. 24. 28. 32. 36. 40.

MULTIPLE CORRECT ANSWERS TYPES QUESTIONS 1. (a), (c) 3. (a), (c)

2. (a), (c), (d) 4. (a), (c)

2. 9 6. 9

FILL (b) (d) (a) (b) (a) (b) (c) (c) (a) (c)

IN THE

3. 3

4. 5

BLANKS TYPE QUESTIONS

1. 5 2 4. 0

2. Orthocentre 5. (5/3, 2/3, 2/3) 1 7. ± (2i – j) and ± (2i– 11j) 5 8.

a ·b b ¥ (a ¥ b) b, 2 |b| | b |2

10. ±

1 2

(j – k)

12. p /4, 3p 14. 6

9.

3. – 1 6. 1

5 2 2 i+ j+ k 3 3 3

11. ±

1 6

(2i + j + k)

15. p/6, 5p/6

TRUE-FALSE TYPE QUESTIONS 1. True

2. True

3. True

4. False

IIT JEE eBooks: www.crackjee.xyz 28.48 Comprehensive Mathematics—JEE Advanced

SUBJECTIVE-TYPE QUESTIONS 2. l = 0, – 1. 3. X =

1 c (A × B) 2 – | A |2 |A|

4. – A2, A1, A3

5. l = –146/17

7. 2 : 1

9. 3 : 2 11. – 4/3 < c £ 0.

10. – i – 8j + 2k 12. 8 : 3 16. p = r = 20.

14. 3i – j – k; – i + 3j + 3k 1 - 2 cos q ,q= 1 + 2 cosq 1 + 2 cos q

|b - c|a +|c - a|b +|a - b|c |b - c|+|c - a|+|a - b|

26. w = v – 2(v · a)a

Hints and Solutions

a ◊ a + a ◊ b + a ◊ c = 0, b ◊ a + b ◊ b + b ◊ c = 0, c◊a+c◊b+c◊c=0 Adding these equations, we have |a|2 + |b|2 + |c|2 + 2 (a ◊ b + b ◊ c + c ◊ a) = 0 fi 29 + 2 (a ◊ b + b ◊ c + c ◊ a) = 0 fi (a ◊ b + b ◊ c + c ◊ a) = – 29/2. 5. Let D be the origin of reference and DA = a, DB = b, DC = c |AB ¥ CD + BC ¥ AD + CA ¥ BD| = |(b – a) ¥ (– c) + (c – b) ¥ (– a) + (a – c) ¥ (– b)| = |b ¥ c – a ¥ c + c ¥ a – b ¥ a + a ¥ b – c ¥ b| = 2 |a ¥ b + b ¥ c + c ¥ a| = 2 (2 area of DABC) = 4 area of DABC. Hence l = 4. 6. (a + b) ¥ (b + c) = 3j ¥ (– 2i + 4j) = 6k Hence the required unit vector is k. 7. Equation of the plane OQR is r = lr2 + mr3, i.e. r ◊ (r2 ¥ r3) = 0 So the distance of P from the plane OQR is r1 ◊ (r2 ¥ r3 ) . r2 ¥ r3 Since r2 ¥ r3 = – 10i + 10 j – 5k so |r2 ¥ r3| = 15

LEVEL 1

and the perpendicular distance = 1. The volume of the parallelopiped 2 -3 0 = |[OA, OB, OC]| = 1 1 - 1 = 4. 3 0 -1 2. Denoting a, b, c by the given vectors respectively. These vectors will be collinear if there is some constant K such that c – a = K(b – a) fi a – 60 = – 20 K and – 55 = – 11 K fi a = – 100 + 60 = – 40 3. Let |p| = |q| = |r| = K. Let p, q, r be unit vectors along p, q, r respectively. Clearly p, q, r are mutually perpendicular vectors, so any vector x can be written as a1 p + a2 q + a3 r. p ¥ ((x – q) ¥ p) = (p ◊ p) (x – q) – (p ◊ (x – q))p = K2 (x – q) – (p ◊ x) p( p ◊ q = 0) = K2 (x – q) – K p ◊ (a1 p + a2 q + a3 r) K p = K2 (x – q – a1 p) Similarly q ¥ ((x – r) ¥ q) = K2 (x – r – a2 q) and r ¥ ((x – p) ¥ r) = K2 (x – p – a3 r) According to the given condition K2 (x – q – a1 p + x – r – a2 q + x – p – a3 r) = 0 fi K2 {3x – (p + q + r) – (a1 p + a2 q + a3 r)] = 0 fi K2 [2x – (p + q + r) = 0 fi x = (1/2) ( p + q + r) (K π 0) 4. Taking dot product on both sides by a, b and c, we have

- 30 - 20 + 5 = 3. 15

8. A unit vector along b = 6 i – 8j –

15 k, is 2

2 Ê 15 ˆ ÁË 6i - 8 j - k ˜¯ , so a vector of length 50 along 25 2 15 ˆ Ê b is ± 4 Á 6 j - 8 j - k ˜ . Since a makes obtuse anË 2 ¯ ±

gle with z-axis so we must have a ◊ k < 0. Thus a = 24i – 32j – 30k. 9. Let OA = x1 i + y1 j and OB = x2 i + y2 j. Since 1 = OA. i = x1 and – 2 = OB. i = x2. Moreover, y1 = x12 = 1 and y2 = x22 = 4, so OA = i + j and OB = – 2i + 4j. Hence |2OA – 3OB| = |8i – 10j| = 164 = 2 41 . 10. Let A be the origin of reference and the position vector of B, C, D be b, c, d, w.r.t. A. So AB = b, CD = d – c, AD = d, BC = c – b, AC = c and BD = d – b. The L.H.S. is equal to b ◊ (d – c). The R.H.S. is K [|d|2 + |c – b|2 – |c|2 – |d – b|2] = K [d ◊ d + c ◊ c + b ◊ b – 2c ◊ b – c ◊ c – d ◊ d – b ◊ b + 2d ◊ b] = 2K[b ◊ (d – c)]. Hence K = 1/2. 11. The given expression = {{a ¥ c + b ¥ a + b ¥ c} ¥ (b ¥ c)} ◊ (b + c) = {(a ¥ c) ¥ (b ¥ c) + (b ¥ a) ¥ (b ¥ c)} ◊ (b + c)

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= [(a ◊ (b ¥ c)) c – (c ◊ (b ¥ c)) a + (b ¥ (b ¥ c)) a – (a ◊ (b ¥ c)b)] ◊ (b + c) = [(a ◊ (b ¥ c)) (c – b) ◊ (b + c)] = (a ◊ (b ¥ c)) [|c|2 – |b|2] = 0 [ |b| = |c| = 1]. 12. AB = – 3i + k. Since AB ◊ (2i + 3j + 6k) = – 6 + 6 = 0. Hence AB is perpendicular to the given line. Thus required distance is equal to |AB| = 9 + 1 = 10 . 13. Let the terminal points be A, B, C and the common initial point be the origin of reference so that AB = b – a and AC = c – a. The vector AB ¥ AC is perpendicular to the plane ABC. AB ¥ AC = (b – a) ¥ (c – a) = b ¥ c + c ¥ a + a ¥ b. Hence the angle between the plane and the given vector is p /2., x0 and the common p – 1 ratio be x. So a = x0 x , i.e. log a = log x0 + (p – 1) log x. Similarly, log b = log x0 + (q – 1) log x and log c = log x0 + (r – 1) log x. If a = log a2 i + log b2 j + log c2 k and b = (q – r) i + (r – p)j + ( p – q )k then a ◊ b = 2[(log a) (q – r) + (log b ) (r – p) + (log c) ( p – q)] = 2 [(q – r) [log x0 + ( p – 1) log x] + (r – p) [log x0 + (q – 1) log x] + ( p – q) [log x0 + (r – 1) log x]]. = 2 [(q – r + r – p + p – q) log x0 + (qp – pr – q + r + qr – pq – r + p + pr – qr – p + q] log x = 0. Hence the angle between a and b is p/2. 15. The given equality implies that 1 (a ◊ c) b – (a ◊ b) c = b 2 1 fi (a ◊ c – )b = (a ◊ b)c 2 But since b and c are non-parallel so the only possibility is a ◊ c = 1/2 and a ◊ b = 0. Hence the angle between a and b is p /2. 16. PQ = 3i – j – 3k, PR = 2i – 2j + k and PS = 4i – 4j + 3k. The volume of the tetrahedron = (1/6) |PQ ◊ (PR ¥ PS)| 3 -1 - 3 1 2 -2 1 = 6 4 -4 3

=

2 . 3

17. b ¥ c = – 3i – j + 2 k. If q is the angle between a and the plane containing b and c a ◊ (b ¥ c) then cos (90 – q) = a b¥c =

1

1

14

14

| (– 9 – 2 + 2)| =

9 . 14

Hence

q = sin–1 (9/14)

18. AB = i + j – 3k and AC = – i + 3j – k and AB ¥ AC = 8i + 4j + 4k. Hence n = ± (2i + j + k ) / 6 . 19. The direction vector of these lines are d = 2i – 4j + k and d¢ = – i + 2j + 3k Now d ◊ d¢ = – 7. We change the direction vector of d to obtain a positive dot product, corresponding to an acute angle q (- d) ◊ d ¢ 7 1 . cos q = = = - d d¢ 21 14 6 20. Let A be the origin of reference so that the position vectors of B and C – 3i + 4 k and 5i – 2 j + 4k respectively. The position vector of mid point of BC is i – j + 4k. Thus the length of the median is 1 + 1 + 16 = 18 . 21. Since a is perpendicular to d, so x – y + 2 z = 0 (i). Moreover, |b| = |c| so a ◊ b = a ◊ c as a makes equal angles with b and c. Thus xy – 2yz + 3xz = 2xz + 3xy – yz fi xz – 2xy – yz = 0 (ii) Also x2 + y2 + z2 = 12 (iii) and y < 0, Substituting the value of y from (i) in (ii) we get x2 + 2xz + z2 = 0 so x = – z and y = z. Again substituting these value in (iii) we get z2 = 4 i.e. z = ± 2 but y < 0 and y = z So z = –2 = y and x = 2 22. a ¥ b fi c is perpendicular to a and b b ¥ c = a fi a is perpendicular to b and c Thus a, b, c are orthogonal is pairs. Since |c| = |a ¥ b| = |a| |b| and similarly |a| = |b| |c| so |c| = |c| |b|2. Hence |b| = 1 and |c| = |a|. 23. a + b + c = 0 fi a + b = – c fi (a + b) ◊ (a + b) = |c|2. Thus |a|2 + |b|2 + 2|a| |b| cos q = |c|2, where q is the angle between a and b. Therefore, cos q =

49 - 9 - 25 1 = 2 ◊ 3◊ 5 2



q = p /3.

24. a ◊ (b ¥ c) ¥ (a + b + c) = a ◊ ((b ¥ c) ¥ a + (b ¥ c) ¥ b + (b ¥ c) ¥ c) = a ◊ ((b ¥ c) ¥ a + b + c) = a ◊ ((a ◊ b)c – (a ◊ c)b) = (a ◊ b) (a ◊ c) – (a ◊ c) (a ◊ b) = 0 25. |a – b|2 = (a – b) ◊ (a – b) = |a|2 + |b|2 – 2 |a| |b| cos f = 2(1 – cos f) = 4 sin2 (f/2). Hence (1/2)|a – b| = |sin f /2|.

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26. The given expression = |AB| |AC| cos q + |BC| |AB| cos (p/2 – q) = p(b cos A + a cos B) = p2 (using projection formula)

These vectors have magnitudes 6 and 2 2 , respectively, and their dot product is 12. Therefore the angle between them is p 12 1 = cos-1 = or p cos-1 ( ) 4 2 (6) 2 2 minus this value, i.e. 3p/4.

27. The given expression is equal to (a + b + c) ◊ (a ¥ a + a ¥ c + b ¥ a + b ¥ c) = b ◊ (a ¥ c) + c ◊ (b ¥ a) + a ◊ (b ¥ c) = [a b c] – [a b c] – [a b c] = – [a b c] a 2a - 3a 28. 2 a + 1 2 a + 3 a + 1 = 3a + 5 a + 5 a + 2 1 2 -3 a 2a + 1 2a + 3 a + 1 3a + 5 a + 5 a + 2

π 0, for a π 0. Hence the three vector are coplanar only for a = 0. 29. We can write the given expression = (i ◊ ( p ¥ q))i + (i ◊ ( p ¥ q))j + (k ◊ ( p ¥ q))k = p ¥ q. Since for any vector a, a = (a ◊ j) i + (a ◊ j)j + (a ¥ k)k. 30. a – d = – AD, b – d = – BD and c – d = – CD. According to the given condition |AD| = |BD| = |CD|. Thus D is the circumcentre of D ABC. 31. Component of i + j + k along i + 2j + 3k = |proj of i + j + k on i + 2j + 3k | 6 14

¥

1 14

(i + 2j + 3k) =

Since A ◊ B = 0, we get from the given value of C, A ◊ C = aA ◊ A + bA ◊ B + gA ◊ (A ¥ B) = a, i.e., a = cos q, and similarly, B ◊ C = cos q = b. This is, a = b = cos q, so that answer (a) is correct. Next, we have 1 = C ◊ C = 2a2 + g2 |A ¥ B|2 = 2a2 + g2 [|A|2 |B|2 (A ◊ B)2] = 2a2 + g2 fi

-3 1 2 7 a + 4 = a (15a2 + 31a + 37) = a 0 -2a + 1 0 - 5(a + 1) 10 a + 17

=

34. Since they are unit vectors, we have |A| = |B| = 1 = |C|. It is also given that the angle q between a and c equals that between b and c, i.e., A◊B B◊C = A ◊ C = cos q = =B◊C A C B C

1 14

(i + 2j + 3k)

3 (i + 2j + 3k) 7

32. i ¥ ((a ¥ b) ¥ i) = (i ◊ i) (a ¥ b) = (i ◊ (a ¥ b))i j ¥ ((a ¥ b) ¥ j) = ( j ◊ j) (a ¥ b) – (j ◊ (a ¥ b))j k ¥ ((a ¥ b) ¥ k) = (k ◊ k) (a ¥ b) – (k ◊ (a ¥ b))k The given expression = 3(a ¥ b) – (i ◊ (a ¥ b)i + ( j ◊ a ¥ b)j + k ¥ (a ¥ b))k = 3 (a ¥ b) – a ¥ b = 2 (a ¥ b). 33. The two diagonals are given by AB - BC = 4i - 2j + 4k and AB + BC = 2i -2j.

g 2 = 1 - 2a2 = 1 - 2 cos2 q = - cos 2q

1- g 2 1 + cos 2q = 2 2 which proves answers (b), (c) and (d). 35. Since a and b are unit vector, we have fi

a2 = b2 =

|a - b| =

(a - b)2 =

(a - b) ◊ (a - b)

=

a 2 + b 2 - 2a ◊ b

=

1 + 1 - 2 cos 2q =

=

2 (2sin 2 q ) = 2 |sin q|

2 (1 - cos 2q )

Therefore, |a - b| < 1 implies |sin q| < 1/2



q Π[0, p/6) or (5p/6, p].

36. Let q be the angle between A and B, and f the angle between C and the vector A ¥ B, that is, the angle between C and the perpendicular to the plane containing A and B. Then |(A ¥ B) ◊ C| = |A ¥ B| |C| cos f = |A| |B| |C| sin q cos f, so if the given relation holds, we must have sin q cos f = 1, which means both sin q and cos f must be 1. That is, we must have q = p/2 and f = 0. The former implies that A and B are perpendicular, so that A ◊ B = 0. On the other hand, if f is zero, C must be perpendicular to both A and B, so that B ◊ C = 0 = A ◊ C. 37. Let OABC be a regular tetrahedron and O be the origin of reference. Let the position vectors of A,

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B, C be a, b, c. so |a| = |b| = |c| = k and |a ◊ b| = 1 2 |b ◊ c| = |c ◊ a| = K2 cos 60º = K,a◊a=b◊b= 2 c ◊ c = K 2. a◊a a◊b a◊c 1 6 b [a, b, c] = ◊ a b ◊ b b ◊ c = K 2 c◊a c◊b c◊c Also |a ¥ c + c ¥ a + a ¥ b| is twice the area of the triangle ABC so that this is ( 3/2) K2 The equation of the plane ABC is r ◊ [b ¥ c + c ¥ a + a ¥ b] = [a b c] So that the distance of the vector O, from this plane (1/ 2 ) K 3 [a b c ] = = b¥c+c¥a+a¥b ( 3/2) K 2 2 K. 3 38. a + b + c = 0 fi cos a + cos b + cos g = 0, sin a + sin b + sin g = 0 Now see solution to Example 66 of Chapter 1. 39. |a + b|2 = |a|2 fi |a|2 + |b|2 + 2a.b = |a|2 fi 2|a| |a| cos q = – |a|2 [ |a| = |b|] fi cos q = –1/2 = cos (2p/3) fi q = 2p/3. =

\ sin q = sin (2p/3) = 3 / 2 . Also, a . b = – |a|2 < 0 40. OA + OB = O fi OB = – OA \ BA is diameter of the unit circle |r| = 1. Thus, DABC is a right triangle with right angle at C, also –ACB = p/2. 41. A plane containing r = a + tb passes through a and vector r - a and b are collinear. Hence the equation of plane passing through c and containing line r = a + tb has r - a, c - a and b as coplanar vector. Thus the required equation is r = a + t (c - a) + pb. Again for the plane containing the line r = a + tb and perpendicular to r ◊ c = a, the vectors r - a, b and c are coplanar so we have r = a + tb + pc. For (iii) the vectors r - a, b - a, c - a being coplanar, the equation of the plane is r = a + t(b - a) + p(c - a) = (1 - t - p) a + tb + pc For (iv) again r - a, b - a and c are coplanar, so the required equation is r = a + t(b - a) + pc = (1 - t) a + tb + pc.

42. a ◊ b + b ◊ c + c ◊ a = 1 + (- 1) + 1 = 1 Similarly other dot products. 43. Use the formula of area and volume given in text. 44. (a) |r| = |2i – r| represents plane x = 1 (b) |r – 2i| = |r + 2j| ¤ (x – 2)2 + y2 + z2 = x2 + (y + 2)2 + z2 ¤x+y=0 (c) |r| = 2 represents sphere x2 + y2 + z2 = 2 (d) |r|2 = 6 or |r| = 2

2

6 represents sphere

2

x +y +z =6 r, so domain R3 45. (a) tan–1(|r (b) cos–1 (|r r| £ 1 (c) ln |r r| > 0 and sin–1(|r|) is r| £ 1, so required domain is 0 < |r| £ 1 (d) cosec–1(|r r| ≥ 1 and ln |r – i| r – i| > 0, thus, |r| ≥ 1 and r π i. Hence, required domain is |r| ≥ 1, r π i 46. The plane containing OA and OB is r . (a ¥ b) = 0 Any line in this plane is perpendicular to the vector a ¥ b. Thus the line in this plane perpendicular to OC is parallel to the vector (a ¥ b) ¥ c so the equation of three lines are r = t(a ¥ b) ¥ c, r = p(b ¥ c) ¥ a and r = r(c ¥ a) ¥ b But (a ¥ b) ¥ c + (b ¥ c) ¥ a + (c ¥ a) ¥ b = 0 (use 20.6) so that three vectors (a ¥ b) ¥ c, (b ¥ c) ¥ a, (c ¥ a) ¥ b are coplanar. 47. (a ¥ b) . (c ¥ d) = (a . c) (b . d) – (b . c) (a . d). 48. We have RS + ST = RT. As PQ and TR are not parallel to each other, PQ ¥ (RS + ST) = PQ ¥ RT π 0 Thus, the statement 1 is true. Since PQ and RS are not parallel, PQ ¥ RS π 0. Therefore, the statement 2 is false. 49. We have 3(3 + 14t) – 6(1 + 2t) – 2(15t) = 9 + 42t –6 – 12t – 12t – 30t = 3 π 0 Thus the line x = 3 + 14t, y = 1 + 2t, z = 15t does not lie on the plane 3x – 6y – 2z = 15 \ x = 3 + 14t, y = 1 + 2t, z = 15t cannot be the line of intersection of the planes 3x – 6y – 2z = 15 and 2x + y – 2z = 5. Since 3(14) – 6(2) – 2(15) = 0 and 2(14) + 1(2) – 2(15) = 0, so the vector 14i + 2j + 15k is perpendicular to the line of intersection of the two planes. 50. Statement-2 is true in view of triangle in equality. For statement-1, note that |a – c|2 + |b – d|2 + |a – d|2 + |b – c|2 – |a – b|2 – |c – d|2

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= 2(|a|2 + |b|2 + |c|2 + |d|2) – 2a.c – 2b.d – 2a.d – 2b.c – (|a|2 + |b|2 – 2a . b + |c|2 + |d|2 – 2c . d) = |a|2 + |b|2 + |c|2 + |d|2 + 2a.b + 2c.d – 2a.c – 2b.d – 2a.d – 2b.c = |a + b – c – d|2 b 1 (4i - 3j + k) 51–53. |b| = 26 , ub = = |b| 26 5 26 . 13 5 (4i - 3j + k) Projb a = (compb a) ub = 13 2 (- 3i - j + 9k) a^ = a - (a ◊ ub)ub = 13

compb a = a ◊ ub = -

54. r¢(t) = - 2i + 2tj + 4e2(t-1)k Since r¢(t) is parallel to r (t) so r(t) = a r¢(t) fi 1 - 2t = - 2a ; t2 = at; 2e2(t - 1) = 4ae2(t - 1) fi a = 1/2. The only value of t three equations is t = 1. So r(1) is the required point i.e. (- 1, 1 2), 55. r¢(t) = i + 2tj + 3t2k. Since (2, 4, 8) corresponds to r(2) so r¢(2) = i + 4j + 12k is tangent to the curve at (2, 4, 8). Required equation of line R (u) = 2i + 4j + 8k + u(i + 4j + 12k) 56. (2, 0, 5) corresponding to r(1) and r¢(t) = 4ti - j + 6tk. Required tangent vector is r¢(1) = 4i - j + 6k. 1 57. Area of DABC = |BC ¥ BA| 2 But BA = i + 2j - 3k and BC = i - 2j + 3k, so i j k BC ¥ BA = 1 - 2 3 = 6j + 4k. Thus 1 2 -3 1 36 + 16 = 13 . 2 58. A ◊ B ¥ C = [A B C], C ¥ A ◊ B = [A B C] and A ◊ B ¥ C = [A B C] A ◊ (B ¥ C) B ◊ ( A ¥ C) + so (C ¥ A) ◊ B C ◊ ( A ¥ B) area of D ABC =

=

[ A B C] - [ A B C] =0 [ A B C] [ A B C]

59. Since ai + j + k, i + bj + k and i + j + ck are coplanar, there exist scalars x, y, z (not all zero) such that x (ai + j + k) + y (i + bj + k) + z(i + j + ck) = 0 fi

(ax + y + z)i + (x + by + z)j + (x + y + cz)k = 0



ax + y + z = 0, x + by + z = 0, x + y + cz = 0

Since at least one of x, y, z is different from zero, the above system must have a non-zero solution. a 1 1 a 1 1 1 b 1 = 0 fi 1- 2 b -1 0 = 0 \ 1 1 c 1- a 0 c -1 fi a(b - 1) (c - 1) - (1 - a) (b - 1) - (c - 1) (1 - a) = 0 Dividing by (1 - a) (1 - b) (1 - c), we get a 1 1 + + =0 1- a 1- c 1- b fi

1 1 1 1 a + + = = 1. 1- a 1- b 1- c 1- a 1- a

60. Since (1, a, a2), (1, b, b2) and (1, c, c2) are non-coplanar, 1 a a2 2 D= 1 b b

1 c

c

π 0. We have

2

a a2 1 + a2

a a2

a3

2 2 2 2 0 = b b 1+ b = b b 1 + b b

b3

c

c2

1 + c2

a a2 1 c2 1

c

c

c2

c3

1 a a2 fi

2 0 = (1 + abc) 1 b b

1 c

c

= (1 + abc) D.

2

As D π 0, we get 1 + abc = 0 or abc = - 1. 61. Let a = a1 i + a2 j + a3 k a ¥ i = a2 j ¥ i + a3 k ¥ i = - a2 k + a3 j fi |a ¥ i|2 = a22 + a32. Similarly |a ¥ j|2 = a12 + a32 and |a ¥ k|2 = a21 + a22. So |a ¥ i|2 + |a ¥ j|2 + |a ¥ k|2 = 2(a12 + a22 + a32 ) = 2. 62. Equation of the plane containing i and i + j is [r - i, i, i + j] = 0 (r - i) ◊ (i ¥ (i + j)) = 0 fi ((x - 1) i + yj + zk) ◊ k = 0 fi z=0 (i) Equation of the plane containing i - j and i + k is [r - (i - j), i - j, i + k] = 0 (r - i + j) ◊ ((i - j) ¥ (i + k)) = 0 (x - 1)i + (y + 1)j + 2k) ◊ (- i - j + k) = 0 fi - (x - 1) - (y + 1) + z = 0 fi x+y-z=0 (ii) Let a = a1 i + a2 j + a3 k. Since a is parallel to (i)

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and (ii) a3 = 0 and a1 + a2 - a3 = 0 fi a1 = - a2, a3 = 0. Thus a unit vector in the direction of 1 1 1 ¥ 1 + (- 2) ¥ (- 1) a= (i - j). So cos q = 2 2 1◊ 1 + 4 + 4 =

1 2

63. q = area of parallelogram with OA and OC as adjacent sided = |OA ¥ OC| = |a ¥ b| and p = area of quadrilateral OABC = (1/2) |OA ¥ OB| + (1/2) |OB ¥ OC| = (1/2) |a ¥ (10a + 2b)| + (1/2) |(10a + 2b) ¥ b| = |a ¥ b| + 5 |a ¥ b| = 6 |a ¥ b| = 6q. Thus K = 6. 1 -1 1 1 -1 0 64. [c b d] = 1 2 2 = - 4 [a b d] = 1 2 2 = 1 2 -1 1 2 -1 1 and b ¥ d = 4i + 3j - 5k so |b ¥ d| = 5 2 .

Hence the shortest distance = =

5

=

[c b d ] - [a b d ] b¥d

1

5 2 2 65. b ¥ c = - 2i + 2j + k, c ¥ a = - i + k and [a b c] = 1 1 1 1 0 2 = 1. 2 1 2 The required equation of the plane is r ◊ [b ¥ c + c ¥ a] = [a b c] fi r ◊ (- 3i + 2j + 2k) = 1 66. The given expression can be written as d ◊ (a ¥ c (b ◊ d) c - (b ◊ c) d) = d ◊ ((b ◊ d) (a ¥ c) - (b ◊ c) (a ¥ d )) = (b ◊ d) d ◊ (a ¥ c) - (b ◊ c) d ◊ (a ¥ d) = (b ◊ d) [a c d]. Thus K = b ◊ d = 14

IIT JEE eBooks: www.crackjee.xyz JEE (Advanced)—2017: Mathematics Paper-I  PYP.1

JEE (Advanced)—2017 Mathematics Paper-I SECTION I • • • •

This section contains SEVEN questions. Each question has FOUR options (a), (b), (c) and (d). One or More Than One of these four option(s) is/are correct. For each question, marks will be awarded in one of the following categories: Full marks : + 4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened Partial marks : + 1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened Zero marks : 0 If none of the bubbles is darkened Negative Marks : – 2 In all other cases For example, if (a), (c) and (d) are all the correct options for a question, darkening all these three will get +4 marks; darkening only (a) and (d) will get +2 marks; and darkening (a) and (b) will get –2 marks, as a wrong option is also darkened.

1 1. Let X and Y be two events such that P(X) = , 3 2 1 P(X | Y ) = and P(Y | X ) = . Then 2 5 1 4 (a) P(Y) = (b) P(X ¢|Y ) = 2 15 2 1 (c) P(X » Y) = (d) P(X « Y) = 5 5 2. Let f : R Æ (0, 1) be a continuous function. Then, which of the following function(s) has(have) the value zero at some point in the interval (0, 1)? (a) ex –

Ú (b) f(x) + Ú

(c) x –

x

0

f (t )sin t dt

p /2 0

p /2 - x

Ú0

f (t )sin t dt f (t ) cos t dt

9

(d) x – f (x) 3. Let a, b, x and y be real numbers such that a – b = 1 and y π 0. If the complex number z = x + iy Ê az + b ˆ satisfies Im ÁË ˜ = y, then which of the z +1 ¯ following is(are) possible value(s) of x?

(a) 1 –

1 + y2

(b) –1 –

1 – y2



(c) 1 +



(d) –1 +

1 – y2

4. If 2x – y + 1 = 0 is a tangent to the hyperbola x2 y2 – = 1, then which of the following CANa 2 16 NOT be sides of a right angled triangle? (a) a, 4, 1 (b) 2a, 4, 1 (c) a, 4, 2 (d) 2a, 8, 1 5. Let [x] be the greatest integer less than or equal to x. Then, at which of the following point(s) the function f (x) = x cos (p(x + [x])) is discontinuous? (a) x = –1 (b) x = 1 (c) x = 0 (d) x = 2 6. Which of the following is(are) NOT the square of a 3 × 3 matrix with real entries? È 1 0 0˘ È 1 0 0˘ Í ˙ Í0 1 0 ˙ (a) Í ˙ (b) Í0 -1 0˙ ÍÎ0 0 -1˙˚ ÍÎ0 0 -1˙˚ 1 0 0   −1 0 0  0 1 0   0 −1 0  (c)     (d) 0 0 1   0 0 −1 7. If a chord, which is not a tangent, of the parabola y2 = 16x has the equation 2x + y = p, and midpoint (h, k), then which of the following is(are) possible value(s) of p, h and k? (a) p = –1, h = 1, k = – 3 (b) p = 2, h = 3, k = – 4 (c) p = –2, h = 2, k = – 4 (d) p = 5, h = 4, k = – 3

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SECTION II • • •

This section contains Five questions The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive For each question, marks will be awarded in one of the following categories: Full Marks : +3 If only the bubble corresponding to the correct answer is darkened Zero Marks : 0 In all other cases

8. For a real number È1 a Í Ía 1 Í 2 Îa a

a, if the system a 2 ˘ È x˘ È 1 ˘ ˙ a ˙ ÍÍ y ˙˙ = ÍÍ-1˙˙ ˙ 1 ˚ ÎÍ z ˚˙ ÎÍ 1 ˚˙ of linear equations, has infinitely many solutions, then 1 + a + a2 = 9. The sides of a right angled triangle are in arithmetic progression. If the triangle has area 24, then what is the length of its smallest side? 10. Let f : R Æ R be a differentiable function such Êpˆ that f (0) = 0, f ÁË ˜¯ = 3 and f ¢(0) = 1. If 2

g(x) =

p 2

Ú [ f ¢(t) cosec t – cot t cosec t f(t)] dt x

p ], then lim g (x) = 2 xÆ0 11. For how many values of p, the circle x2 + y2 + 2x + 4y – p = 0 and the coordinate axes have exactly three common points? 12. Words of length 10 are formed using the letters A, B, C, D, E, F, G, H, I, J. Let x be the number of such words where no letter is repeated; and let y be the number of such words where exactly one letter is repeated twice and no other letter is rey peated. Then, = 9x For x Œ (0,

SECTION III • This section contains Six questions of matching type • This section contains Two tables (each having 3 columns and 4 rows) • Based on each table, there are Three questions • Each question has Four options (a), (b), (c) and (d). Only One of theses four options is correct • For each question, marks will be awarded in one of the following categories: Full marks : + 3 If only the bubble corresponding to the correct option is darkened Zero marks : 0 If none of the bubbles is darkened Negative marks : –1 In all other cases Answer Q. 13, 14 and 15 by appropriately matching the information given in the three columns of the following Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively Column 1

Column 2

(I)

x2 + y2 = a2

(i)

my = m2x + a

(II)

x 2 + a 2y 2 = a 2

(ii)

2 y = mx + a m + 1

Column 3 Ê a 2a ˆ (P) ÁË 2 , ˜¯ m m a ˆ Ê – ma , (Q) Á Ë m 2 + 1 m 2 + 1 ˜¯

(III) y2 = 4ax

(iii) y = mx +

a 2 m2 – 1

Ê –a2m ˆ 1 , (R) Á 2 2 ˜ Ë a m + 1 a 2 m2 + 1 ¯

(IV) x2 – a2y2 = a2

(iv)

y = mx +

a 2 m2 + 1

Ê –a2m ˆ -1 , (S) Á 2 2 ˜ Ë a m – 1 a 2 m2 – 1 ¯

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13. For a = 2 , if a tangent is drawn to a suitable conic (Column 1) at the point of contact (–1, 1), then which of the following options is the only CORRECT combination for obtaining its equation? (a) (I) (ii) (Q) (b) (I) (i) (P) (c) (III) (i) (P) (d) (II) (ii) (Q) 14. The tangent to a suitable conic (Column 1) at 1 ( 3 , ) is found to be 3x + 2y = 4, then which 2 of the following options is the only CORRECT combination? (a) (IV) (iv) (S) (b) (II) (iv) (R) (c) (IV) (iii) (S) (d) (II) (iii) (R)

Column 1

Column 2

(I)

f (x) = 0 for some x Œ(1, e2)

(i)

(II)

f ′(x) = 0 for some x Œ(1, e)

(ii)

(III)

f ′(x) = 0 for some x Œ(0, 1)

(iii)

(IV)

f ″(x) = 0 for some x Œ(1, e)

(iv)

(a) (I) (iii) (P)

lim f (x) = – ∞

(Q) f is decreasing in (e, e2)

lim f ¢(x) = – ∞

(R) f   ¢ is increasing in (0, 1)

lim f ¢¢(x) = 0

(S)

xƕ xƕ

xƕ

(b) (II) (iv) (Q)

(a) (I) (ii) (R)

(b) (III) (iv) (P)

(c) (II) (iii) (S) (d) (IV) (i) (S) 18. Which of the following options is the only CORRECT combination?

(a) (III) (iii) (R)

(b) (IV) (iv) (S)



(c) (II) (ii) (Q)

(d) (I) (i) (P)

Answers Section-I

f is increasing in (0, 1)

(P)

(c) (II) (iii) (P) (d) (III) (i) (R) 17. Which of the following options is the only CORRECT combination?

Column 3

lim f (x) = 0

xƕ

16. Which of the following options is the only INCORRECT combination?

15. If a tangent to a suitable conic (Column 1) is found to be y = x + 8 and its point of contact is (8, 16), then which of the following options is the only CORRECT combination? (a) (III) (i) (P) (b) (I) (ii) (Q) (c) (II) (iv) (R) (d) (III) (ii) (Q) Answer Q. 16, Q. 17 and Q. 18 by appropriately matching the information given in the three columns of the following table. Let f(x) = x + loge x – x logex, x Œ(0, •), • Column 1 contains information about zeros of f (x), f ′(x) and f ′′(x). • Column 2 contains information about the limiting behaviour of f (x), f ′(x) and f ′′(x) at infinity. • Column 3 contains information about increasing/ decreasing nature of f (x) and f ′(x).

1. (a), (b) 2. (c), (d) 3. (b), (d) 4. (a), (c), (d) 5. (a), (b), (d) 6. (a), (c) 7. (b)

f ¢ is decreasing in (e, e2)

Section-II

8. 1 11. 2

9. 6 12. 5

10. 2

Section-III

13. (a) 16. (d)

14. (b) 17. (c)

15. (a) 18. (c)

Hints and Solutions 1. P(X|Y) =

P( X « Y ) P( X « Y ) and P(Y |X) = P(Y ) P( X )

2 2 P( X « Y ) = fi P(X « Y) = 5 15 1/3 2/15 P( X « Y ) 4 Now, P(Y) = = = 1/2 P( X |Y ) 15 \

P(X ¢|Y) = 1 – P(X |Y) = 1 –

1 1 = 2 2

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and P (X » Y) = P(X) + P(Y) – P (X « Y)

i.e.

1 4 2 7 = + – = . 3 15 15 15



x

2. Let g1(x) = ex – Ú0 f (t ) sin t dt,

1 = a2 (22) –16 a2 = 17/4

As, 42 + 12 = 17 > a2; a, 4, 1 cannot form a right triangle.

then    g1¢(x) = ex – f (x) sin x

Next, (2a)2 = 4a2 = 17 = 42 + 1, therefore 2a, 4, 1 are sides of a right triangle.

As 0 < f (x) < 1 and 0 < sin x < 1 " x Œ (0, 1) g1¢ (x) > 0 " x Œ (0, 1)

fi a, 4, 2 cannot be sides of a right triangle.

Also, lim g1(x) = 1 xÆ 0 +

fi  g1(x) > 1 " x Œ (0, 1) Let   g2(x) = f (x) +

p /2

Ú0

f (t ) sin t dt,

Note that g2(x) > f (x) > 0 " x Œ(0, 1) Next, let p /2 - x g3(x) = x – Ú f (t ) cos t dt 0



g3(0) = – Ú

p /2 0

f (t ) cos t dt < 0

Lastly 1 + (2a)2 = 1 + 4a2 = 1 + 17 = 18 < 82 fi 2a, 8, 1 cannot be sides of a right triangle. 5. f(x) = x cos (p (x + [x])) Ï x cos (p ( x - 2)), -2 £ x < -1 Ô x cos (p ( x - 1)), -1 £ x < 0 ÔÔ 0 £ x 0

[Q 0 < f (t), cos t < 1] As g3(x) continuous on R, we get g3(x) = 0 at least for one value of x Œ (0, 1) Finally, let g4(x) = x9 – f(x) g4(0) = – f (0) < 0 g4(1) = 1 – f (1) > 0 \  g4(x) = 0 at least for one value of x Œ (0, 1)

a( z + 1) + b − a 1 az + b 3. = = a− z +1 z +1 z +1 Now,  az + b   1  y = Im   = – Im    z +1   z +1 z +1   = – Im    zz + z + z + 1  y y = zz + z + z + 1 fi x2 + y2 + 2x + 1 = 1 fi (x + 1)2 = 1 – y2 fi

x = – 1 ±

and

2 2 4. y = 2x + 1 is a tangent to the hyperbola x – y = 1 a 2 16 if c2 = a2m2 – b2

if - 2 £ x < -1 if - 1 £ x < 0 if 0 £ x < 1 if 1 £ x < 2 if 2 £ x < 3

lim f (x) = (–1) cos (–p) = 1

xÆ –1–

lim f (x) = –(–1) cos (p) = –1

xÆ –1+

lim

xÆ 0 –

lim f (x) f (x) = 0 = xÆ 0+

lim f (x) = (1) cos (p) = –1

xÆ1–

lim f (x) = – (1) cos p = 1

xÆ1+

lim f (x) = – 2cos (2p) = –2

xÆ 2–

lim f (x) = 2cos (2p) = 2

xÆ 2 +

Thus, f is discontinuous at x = –1, 1 and 2. 6. If A is a 3 × 3 matrix, then det (A2) = (det (A))2 ≥ 0 1 0 0 -1 0 0 0 1 0 = –1 = 0 -1 0 As 0 0 -1 0 0 -1

   [∵ y π 0]

1 – y2

Ï x cos (p x) Ô- x cos (p x) ÔÔ = Ì x cos (p x) Ô- x cos (p x) Ô ÓÔ x cos (p x)



È1 0 0 ˘ Í0 1 0 ˙ Í ˙, ÍÎ0 0 -1˙˚

 −1 0 0   0 −1 0     0 0 −1

cannot be square of a 3 × 3 matrix.

IIT JEE eBooks: www.crackjee.xyz JEE (Advanced)—2017: Mathematics Paper-I  PYP.5 2 1 0 0  È1 0 0 ˘   Í ˙ Also, 0 0 -1 = 0 −1 0  Í ˙ 0 0 −1 ÍÎ0 1 0 ˙˚ 2 1 0 0  È1 0 0 ˘   Í ˙ and = 0 1 0  Í0 1 0 ˙ 0 0 1  ÍÎ0 0 1˙˚ 7. Equation of chord of y2 = 16x which has (h, k) as mid point is ky – 8(x + h) = k2 – 16h 8x – ky = 8h – k2 Comparing it with 2x + y = p, we get

–k 8h – k 2 8 = = 1 2 p



fi k = – 4, 4p = 8h – k2 fi k = – 4, 8h – 4p = 16 fi k = – 4, 2h – p = 4 ...(i) So only possible choices are (b) and (c). For p = 2, h = 3, k = –4, (i) is satisfied and (–4)2 π 16(3) For p = –2, h = 2, k = – 4, (1) is not satisfied. \  only choice is (b). 8. As the system of equations has infinite number of solutions 1 D= a a2

a a2 1 a =0 1

1

We have (a + d)2 = a2 + (a – d)2 fi 4ad = a2 fi a = 4d 1 Area of triangle = a (a – d) 2 1 fi 24 = (4d) (4d – d) 2 fi 24 = 6d 2 fi d = 2 [∵ d > 0] \  smallest side = a – d = 3d = 6. 10. g(x) = f (t )cosec t ]x

p /2







a2

p /2 x

p /2 x

f (t )(-cosec t cot t )dt

(cosec t cot t ) f (t )dt

Êpˆ Êpˆ = f ÁË ˜¯ cosesc ÁË ˜¯ – f (x) cosec x 2 2 f ( x) = 3 – sin x f ( x) lim g(x) = lim g(x) = 3 – lim xÆ 0 + sin x xÆ 0 + xÆ 0 f ¢ (0) lim f ¢ ( x) = 3 – = 3 – xÆ 0 + cos x 1 = 3 – 1 = 2. 11. Write the equation as (x + 1)2 + (y + 2)2 = p + 5 ...(i) We must have p + 5 > 0. (i) If 0 < p + 5 < 1, (i) does not intersect any of the axes. y

Using R1 Æ R1 – aR2, we get 1- a2 a D=



0 0 1 a =0 1

1

fi (1 – a2)(1 – a) = 0 fi (1 + a) (1– a)2 = 0 fi a = – 1, 1 For a = 1, first two equation become x + y + z = 1 x + y + z = –1 which is not possible \ a = –1 Thus, 1 + a + a2 = 1 9. Let sides of triangle be a – d, a, a + d, where a >d>0

O

x'

x

00 x'

x

2 2 a m – 1 = 2 In this case option(S), column 3 is not possible as –1 1 = does not hold. 2 2 a m2 – 1

For column 2, (iv) option m = – 3/2 , 2 =

(vi) For p = 0, (i) meets the axes in exactly three points.



=

y

p=0 x

y'



–a2m 2

=

2

a m +1

3,

1 2

P10 = 10!

Ê 10!ˆ y = (10C1) (9C8) ÁË ˜¯ 2!

 10!  = (10) (9)    2! 

2

a m +1

=

1 2

– a 2 (– 3 / 2) = 3 fi a2 = 4 fi a = 2 2 Thus, correct option is (b), that is, (II), (iv), (R) 15. When y = x + 8, m = 1 For option Q, set a = 16 fi a = 16 2 m2 + 1 fi

x'

12. x =

a 2 m2 + 1

In this for case option (R)

y'

10

3 x+2 2

\ m = – 3 /2 For column 2, (iii) option



2

m2 + 1 fi (m + 1)2 = 2(m2 + 1) fi (m –1)2 = 0 fi m = 1 if

Ê – 2 2 ˆ , Point of contact Á = (–1, 1) Ë 1 + 1 1 + 1 ˜¯

p = –1 x'

2

For  a =

1 0 1 and f ¢(e) = – 1 < 0 e \ f ¢(x) = 0 for some x Œ (1, e) Also, f ¢¢(x) > 0 " x Œ (0, 1) \ f ¢(x) = 0 for some x Œ (0, 1) is incorrect

Lastly, f ¢¢(x) < 0 V x Œ (1, e) \ f ′′(x) = 0 for some x Œ (1, e) is incorrect. lim f(x) = lim [1 – (1 – x) (1 – log x)] We have, xÆ• e xÆ• = lim [1 – (x – 1) (logex –1)] = – ∞ xÆ•

1 lim f ¢(x) = lim  – log e x  = – ∞ x→∞ xÆ• x 

Ê 1 1ˆ lim f  ¢¢(x) = lim Á – – 2 ˜ = 0. xÆ• Ë x xÆ• x ¯ For column 3, f ¢(x) > 0 " x Œ (0, 1) fi f increases on (0, 1) 1 – x log e x Next, f ¢(x) = < 0    " x Œ (e, e2) x fi f (x) decreases on (e, e2) We have 1   f ¢¢(x) = –   < 0 " x Œ (0, 1) x   fi f  ¢ decreases on (0, 1) Also, f ¢¢ (x) < 0   " x Œ (e, e2) fi f ¢ decreases on (e, e2) 16. In option (d), (III), (i), (R) is only incorrect option. 17. In option (c), (II), (iii), (S) is the only correct combination 18. In option (c), (II), (ii), (Q) is the only correct combination.

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IIT JEE eBooks: www.crackjee.xyz JEE (Advanced)—2017 Mathematics Paper-II SECTION I • • •

This section contains Seven questions. Each question has Four options (a), (b), (c) and (d). Only One of these options is correct. For each question, marks will be awarded in one of the following categories: Full marks : + 3 If only the bubble corresponding to the correct option is darkened.



Zero marks

:

0

If none of the bubbles is darkened



Negative marks

:

–1

In all other cases

1. How many 3 × 3 matrices M with entries from 5. If f : R → R is a twice differentiable function such {0, 1, 2} are there, for which the sum of the Ê 1ˆ 1 that f  ¢¢(x) > 0 for all x ∈ R, and f Á ˜ = , diagonal entries of MTM is 5? Ë 2¯ 2

(a) 135

(b) 198

f(1) = 1, then

(c) 162 (d) 126 (a) f ¢ (1) ≤ 0 2. Three randomly chosen nonnegative integers 1 (b) < f ¢ (1) ≤ 1 x, y and z are found to satisfy the equation x + y 2 + z = 10. Then the probability that z is even, is (c) f ¢ (1) > 1 5 6 (a) (b) 1 11 11 (d) 0 < f  ¢ (1) ≤ 2 1 36 (c) (d) 6. Let S = {1, 2, 3, ... ,9}. For k = 1, 2, ... ,5, let Nk 2 55 be the number of subsets of S, each containing five 3. The equation of the plane passing through the elements out of which exactly k are odd. Then N1 point (1, 1, 1) and perpendicular to the planes + N2 + N3 + N4 + N5 = 2x + y – 2z = 5 and 3x – 6y – 2z = 7, is (a) 126 (b) 252 (a) 14x + 2y –15z = 1 (c) 210 (d) 125 (b) 14x – 2y + 15z = 27 7. Let O be the origin and let PQR be an arbitrary (c) –14x + 2y + 15z = 3 triangle. The point S is such that (d) 14x + 2y + 15z = 31 OP◊ OQ + OR ◊ OS = OR ◊ OP + OQ ◊ OS 4. If y = y(x) satisfies the differential equation     = OQ ◊ OR + OP◊ OS –1 Then triangle PQR has S as its 8 x 9 + x dy = 4 + 9 + x dx, x > 0 (a) circumcentre (b) incentre and y (0) = 7 , then y (256) = (c) centroid (d) orthocenter

(

)

)

(



(a) 3

(b) 16



(c) 9

(d) 80

• • •

SECTION – 2 This section contains Seven questions. Each question has Four options (a), (b), (c) and (d). One or More than One of these four options is(are) correct. For each question, marks will be awarded in one of the following categories: Full marks : + 4 If only the bubble(s) corresponding to all the correct option (s) is (are) darkened. Partial marks : + 1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened

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Zero marks : 0 If none of the bubbles is darkened Negative marks : – 2 In all other cases For example, if (a), (c) and (d) are all the correct options for a question, darkening all these three will get +4 marks; darkening only (a) and (d) will get +2 marks; and darkening (a) and (b) will get –2 marks, as a wrong option is also darkened.

8. If I =

k +1

 k =1 Úk 98

k +1 dx, then x( x + 1)

49 49 (a) I < (b) I > 50 50 (c) I < loge 99 (d) I > loge 99 9. If the line x = a divides the area of region R = {(x, y) ∈ R2 : x3 ≤ y ≤ x, 0 ≤ x ≤ 1} into two equal parts, then (a) a4 + 4a2 –1 = 0 1 (b) 0 < a ≤ 2 (c) 2a4 – 4a2 + 1 = 0 1 (d) < a < 1 2 sin(2 x ) 10. If g(x) = sin -1 (t )dt , then

Úsin x



p (a) g¢ ÊÁ ˆ˜ = 2p Ë 2¯ Ê pˆ (b) g¢ Á – ˜ = 2p Ë 2¯



Êpˆ (c) g¢ Á ˜ = –2p Ë 2¯



Ê pˆ (d) g¢ Á – ˜ = –2p Ë 2¯

cos(2 x) cos(2 x) sin(2 x) - sin x , then 11. If f(x) = - cos x cos x sin x sin x cos x

(a) f(x) attains its maximum at x = 0



(b) f(x) attains its minimum at x = 0

(c) f ′(x) = 0 at exactly three points in (–p, p)



(d) f ′(x) = 0 at more than three points in (–p, p) 12. Let a and b be nonzero real numbers such that 2(cos b – cos a) + cos a cos b = 1. Then which of the following is/are true? Êβˆ =0 Ë 2 ˜¯



Êaˆ (a) tan Á ˜ + Ë 2¯



Êaˆ Êβˆ (b) tan Á ˜ – 3 tan Á ˜ = 0 Ë 2¯ Ë 2¯



(c)

3 tan Á ˜ – tan Á ˜ = 0 Ë 2¯ Ë 2¯



(d)

3 tan Á ˜ + tan Á ˜ = 0 Ë 2¯ Ë 2¯

13. Let f(x) =

3 tan Á

Êaˆ

Êβˆ

Êaˆ

Êβˆ

Ê 1 ˆ 1 – x(1 + |1 – x |) ˜ , for x ≠ 1. cos ÁË 1 – x¯ |1 – x |

Then

(a) lim- f(x) does not exist



(b) lim- f(x) = 0



(c) lim+ f(x) = 0



(d) lim+ f(x) does not exist

x Æ1 x Æ1 x Æ1 x Æ1

14. If f : R → R is a differentiable function such that f ¢(x) > 2f  (x) for all x ∈ R, and f(0) = 1, then

(a) f ¢(x) < e2x in (0, ∝)



(b) f (x) is increasing in (0, ∞)



(c) f (x) is decreasing in (0, ∞)



(d) f (x) > e2x in (0, ∞)

SECTION – 3 • • • •

This section contains Two paragraphs. Based on each paragraph, there are Two questions. Each questions has Four options (a), (b), (c) and (d). Only One of these four options is correct. For each question, marks will be awarded in one of the following categories: Full marks : +3 If only the bubble corresponding to the correct option is darkened. Zero marks : 0 In all other cases

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Paragraph-1 Let O be the origin, and OX, OY, OZ be three unit vectors in the directions of the sides QR, RP, PQ respectively, of a triangle PQR. 15. If the triangle PQR varies, then the minimum value of cos (P + Q) + cos (Q + R) + cos (R + P) is

Èa1 Í M M = Í b1 ÍÎ c1 T

a2 b2 c2

a3 ˘ È a1 b3 ˙˙ ÍÍa2 c3 ˙˚ ÍÎ a3

b1 b2 b3

c1 ˘ c2 ˙˙ c3 ˙˚

Sum of the diagonal elements of MTM is d = a12 + a22 + a32 + b12 + b22 + b32 + c12 + c22 + c32 where a 1, a2, a3, b1, b2, b3, c1, c2, c3 ∈ {0, 1, 2} 3 5 (a) (b) Now, d = 5, if exactly 5 of 2 3 a1, a2, a3, b1, b2, b3, c1, c2, c3 are 1’s and rest 4 3 5 (c) – (d) – of them are zeros 2 3 This 5 numbers can be chosen in 9C5 = 126 ways 16. | OX × OY| = Also, d = 5, if exactly one of (a) sin (P + R) (b) sin (Q + R) a1, a2, a3, b1, b2, b3, c1, c2, c3 is 2 and exactly one (c) sin (P + Q) (d) sin 2R of it is 1. We can choose two numbers in 9C2 and arrange Paragraph-2 them in two ways, that is, in (9C2)(2) = 72 ways Let p, q be integers and let a, b be the roots of the Hence, the number of ways is 198. equation, x2 – x – 1 = 0, where a ≠ b. For n = 0, 1, 2,..., let 2. Number of non-negative integral solutions of x + an = pan + qb n. y + z = 10 is 12C2= 66. FACT : If a and b are rational numbers and a + b 5 = 0, We now count the number of ways in which z is then a = 0 = b. even. 17. If a4 = 28, then p + 2q = For 0 ≤ k ≤ 5, (a) 12 (b) 14 x + y = 10 – 2k (c) 7 (d) 21 has exactly (10 – 2k) + 1 non-negative integral solutions 18. a12 = \ number of ways in which z is even is (a) a11 + 2a10 (b) a11 – a10

(c) 2a11 + a10

(d) a11 + a10

Section-1 1. (b) 4. (a) 7. (d)

2. (b) 5. (c)

3. (d) 6. (a)

Section-2

8. 10. 11. 14.

(b), (c) 9. (c), (d) None of the given answer is correct (a), (c) 12. (a), (b) 13. (b), (d) (b), (d) Section-3

15. (d) 18. (d)

16. (c) Hints and Solutions

È a1 Í 1. Let M = Ía2 ÎÍ a3

b1 b2 b3

c1 ˘ c2 ˙˙ c3 ˚˙





 (11 – 2k ) k =0

= 62 = 36.



Answers

5

17. (a)

Thus, probability of required event 36 6 = = 66 11 3. Equation of any plane through (1, 1, 1) is a (x –1) + b (y –1) + c(z – 1) = 0 .....(i) As (i) is perpendicular to the given planes 2a + b – 2c = 0 3a – 6b – 2c = 0 a c –b ⇒ = = –2 – 12 –12 - 3 –4 + 6 a b c or = = 14 2 15 Thus, equation of required plane is 14(x – 1) + 2(y – 1) + 15(z – 1) = 0 or 14x + 2y + 15z = 31 4. Write the differential equation as 1 dy = = f(x) (say) dx 8 x 9 + x 4 + 9 + x

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Put





y =

P

Ú f ( x)dx M

4 + 9 + x = t 9 + x = t2 – 4





2tdt =



\

y =







Now,







\





1

Ú dt = t + C

= 7 = y(0) =

4+ 9+ x +C 4+ 9+0 +C

C = 0. y(256) = =

Q

1

◊ dx 2 9+ x 2 x

4 + 9 + 256 4 + 9 + 16

R

L

8. For k < x < k + 1, k ∈ N 1 1 1 1 < < < x +1 k +1 x k k +1 k +1 1 ⇒ 0 " x ∈ R, f ′ is a strictly increasing function on R. By the first mean value theorem there exists some a ∈ (1/2, 1) such that f (1) – f (1/2) = f ¢(a) 1 – 1/2 ⇒ f ¢(a) = 1 for some a ∈ (1/2, 1) As f ¢ is strictly increasing on R, f ¢(x) > f ¢(a) " x > a ⇒ f ¢(1) > 1 [∵ 1 > a] 5 4 6. N k = ( C k) ( C 5 – k) ⇒ N1 + N2 + N3 + N4 + N5 = (5)(1) + (10)(4) + (10)(6) + (5)(4) + (1)(1) = 126 7. Let OP = p, OQ = q, OR = r, OS = s We are given p ◊ q + r ◊ s = r ◊ p + q ◊ s = q ◊ r + p ◊ s From first two expressions, (r – q)◊p + (q – r)◊s = 0 ⇒ (r – q) ◊ (p – s) = 0 ⇒ (QR)◊(SP) = 0 ⇒ SP ⊥ QR ...(1) Similarly, from last two expressions, ⇒ (PQ)◊(SR) = 0  ⇒ SR ⊥ PQ ...(2) From (1), (2) we get S is orthocentre of DPQR. See figure in the next column.

S





k +1

 Úk k =1

k +1 1

Úk

x

dx = ln (k + 1) – ln k

k +1 dx < ln (99) – ln (1) ( x + 1) x

= In (99) Next, for k < x < k + 1, 1 ≤ k ≤ 98 k +1 k +1 1 1 > = > x( x + 1) (k + 1)( x + 1) x + 1 100 k +1

k +1 1 dx > x( x + 1) 100

k +1

k +1 dx > 98 = 49 x( x + 1) 100 50



Úk



 Úk

98

k =1

9. Let 

A1 =





A2 =



Now, A1 = A2

Ú0

a

( x - x3 )dx

1

Úa ( x - x )dx 3

B(1, 1)

x O



and



\







a

× × ××× × ×× × × ××× × × ××× × ×× × ×

A a

x 1

A1 + A2 =

Ú0 ( x - x )dx

2A1 =

1 1 ⇒ A1 = 4 8

( x - x3 )dx = 1 8 1 2 1 1 a – a4 = 2 4 8

Ú0

3

=

1 1 1 – = 2 4 4



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⇒ 4a2 – 2a4 = 1 4 ⇒ 2a – 4a2 + 1 = 0 Let f (x) = 2x4 – 4x2 + 1 We have 1 Ê 1ˆ Ê 1ˆ Ê 1ˆ f Á ˜ = 2Á ˜ – 4Á ˜ + 1 = > 0 Ë 2¯ Ë 4¯ Ë 16 ¯ 8

f(1) = 2 – 4 + 1 = – 1 < 0 f(x) = 0 has a root in (1/2, 1)

and ⇒



and tan (a/2) –

13. f (x) =







Since f (1/2) = 1/4 > 0

⇒ f (x) > 0 for 0 < x ≤ 1/2. 10. g′(x) = 2 cos(2x) sin–1 (sin 2x) – cosx sin–1 (sin x) Ê pˆ g ¢ Á – ˜ = 0 ≠ 2p, –2p Ë 2¯ Êpˆ and g¢ Á ˜ = 0 ≠ 2p, –2p Ë 2¯

11. Using C1 → C1 – C2, we 0 cos(2 x) f (x) = –2 cos x cos x 0 sin x

get sin(2 x) – sin x cos x

= 2cos x (cos 2x cos x – sin x sin 2x) = 2cos x cos 3x = cos 2x + cos 4x f (x) attains its maximum value 2 at x = 0 f ¢(x) = – 2 sin 2x – 4sin 4x  π π  f ¢  –  = 0, f ′(0) = 0, f ′   = 0  2 2 12. 2 (cos b – cos a) + cos a cos b = 1 ⇒ (cos a + 2) (cos b – 2) = –3 3 ⇒ 2 + cos a =          (1) 2 – cos b



Let a = tan (a/2) and b = tan (b/2), now (1) can be written as



2 +





2

3(1 + b ) 1 – a2 = 2 1+ a 2(1 + b 2 ) – (1 – b 2 )

3 + a2

=

3(1 + b 2 )



1 + a2 1 + 3b 2 ⇒ (3 + a2) (1 + 3b2) = 3 (1 + a2) (1 + b2) ⇒ 3 + (a2 + 9b2) + 3a2b2 = 3 + 3a2 + 3b2 + 3a2b2



⇒ a2 = 3b2 ⇒ a = ± 3 b





3 tan (b/2) = 0

1 – x(1+ |1 – x |)  1  cos  , x ≠ 1 1– x  |1 – x |

Ï (1 – x) 2 Ê 1 ˆ cos Á Ô Ë 1 - x ˜¯ Ô 1– x = Ì 2 Ô1 - x cos Ê 1 ˆ ÁË ˜ 1- x¯ ÓÔ x - 1

8x(x2 – 1) < 0   for 0 < x < 1/2

⇒    f (x) decrease in (0, 1/2]

3 tan (b/2) = 0

1 – x(1 + (1 – x))  1  cos    − x 1  1− x  =  1 – x(1 + x –1) cos  1     x –1 1− x 

We have f ¢(x) = 8x – 8x





⇒ tan (a/2) +

3

=







if x > 1

if x < 1 if x > 1

Ï Ê 1 ˆ Ô(1 - x) cos ÁË 1 – x ˜¯ Ô = Ì Ô-( x + 1) cos Ê 1 ˆ ÁË ˜ ÔÓ 1- x¯



if x < 1

if x < 1 if x > 1



Ê 1 ˆ For x < 1, –1 ≤ cos ÁË ˜ ≤1 1 – x¯



Ê 1 ˆ ⇒   – (1 – x) ≤ (1– x) cos ÁË ˜ ≤ (1– x) 1 – x¯



Since







or



As





lim (1– x) = 0, we get

xÆ1–

Ê 1 ˆ lim (1– x) cos ÁË ˜=0 1 – x¯

xÆ1–

lim f(x) = 0

xÆ1–

lim cos ÊÁ 1 ˆ˜ doesn’t exist, Ë1- x¯

x Æ1+

lim (– x – 1) cos ÊÁ 1 ˆ˜ doesn′t exist Ë1 – x¯

x Æ1+

14. f ¢(x) > 2f (x) " x ∈ R Let g(x) = e–2x f(x), then g¢(x) = e–2x f  ¢(x) – 2e–2x f(x) = e–2x (f ¢(x) – 2f(x)) > 0 " x ∈ R ⇒ g is strictly increasing on R Also, g(0) = f (0) = 1 ⇒ g(x) > g(0) " x > 0 ⇒ e–2x f (x) > 1 " x > 0 ⇒ f(x) > e2x " x > 0 As f ¢ (x) > 2f (x) > e2x > 0 " x > 0 ⇒ f(x) is increasing on (0, ∞) Also, f ¢ (x) > 2f(x) > 2e2x > e2x " x ∈ (0, ∞)

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15. Let E = cos (P + Q) + cos (Q + R) + cos (R + P) = cos (p – R) + cos (p – P) + cos (p – Q) = – (cos P + cos Q + cos R) But in DPQR, cos P + cos Q + cos R ≤ 3/2 therefore E ≥ –3/2 16. |OX × OY| = |OX| |OY| sin (p – R) P

OY OX

Q

R

= sin (P + Q) 17. a + b = 1, ab = –1, 1 1 a= ( 5 +1), b = (– 5 +1) 2 2 an = pan + qbn for n = 0, 1, 2, 3, ..... For n ≥ 1 an + 1 = pan + 1 + qbn+1 = (a + b) (pan+1 + qbn+1)     – ab (pan–1 + qb n–1)

an + 1 = an. + an–1 " n ≥ 1 (1)









a0 = p + q





a1 = pa + qb





a2 = a1 + a0





a3 = a1 + a2





a4 = a2 + a3 = a2 + (a1 + a2)





= a1 + 2a2





=





= 3a1 + 2a0





a1 + 2(a1 + a0)

28 = 3 (pa + qb) + 2(p + q)

7 3 5 (p + q) + (p – q) 2 2 As p, q are integers, we get 7 28 = (p + q), p – q = 0 2 ⇒ p + q = 8, p = q ⇒ p = q = 4 \ p + 2q = 12 18. From (1) a12 = a11 + a10.



=

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