• Author / Uploaded
• Hugo Chávez

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O X

F O

R

D

I B

D

I P L O

M

A

P

R

O

G R

A

M

M

E

ANSWERS

M at h e M at i c s

s ta N D a R D C O U R S E

C O M PA N I O N

Laurie Buchanan

Jim Fensom

Ed Kemp

Paul La Rondie

Jill Stevens

le v e l

WORKED SOLUTIONS

1

Functions

Answers

b

y 16 14 12 10 8 6 4 2

Skills check 1 a

y 4 F 3 2 1

D

–4 –3 –2 –1–10 –2 –3 –4

2

A

C

–2 –1 0

1 2 3 4 5 x E

b

A(0, 2), B(1, 0), C(−1, 0), D(0, 0), E(2, 1), F(−2, −2), G(3, −1), H(−1, 1)

a

4x + 3y = 4(4) + 3(6)

y

c

B

6 4 2 –3

–2

= 16 + 18 z 2 − 3y = (−10)2 − 3(6)

–1 0 –2

5

a

b

y − z = 6 − (−10)

c

= 16 =

=− a

85 60 13 60

3x − 6 = 6

3x = 12 x= b

= x 2 + x − 20

Investigation – handshakes Represents one person. 1

4 So 4 people require 6 handshakes.

5x = −3 − 7 = −10 x= c

b

Number of people

12 = 22 x=

4 a

−2

+ 6 = 11

x+

10

y 4 3 2 1 –4 –3 –2 –1–10 –2 –3 –4

Represents one handshake

a

5x + 7 = −3

x 2

(x + 5)(x − 4) = x 2 − 4x + 5x − 20

2(4)  5 6  ( 10)

=− 3

(x − 1)(x − 3) = x 2 − 3x − x + 3 = x 2 − 4x + 3

= 6 + 10 d

3 x

= x 2 + 9x + 20

= 82

2x  5 yz

2

(x + 4)(x + 5) = x 2 + 5x + 4x + 20

= 100 − 18 c

1

–4

= 34 b

1 2 3 4 5 x

1 2 3 4 5 6 7 x

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Number of handshakes

2

1

3

3

4

6

5

10

6

15

7

21

8

28

9

36

10

45 Worked solutions: Chapter 1

1

WORKED SOLUTIONS c Number of handshakes

y

4

45

3 2

25

1

15

–2 –1 0 –1

5 2 4 6 8 Number of people 1

2

x

0

Function. All x values are different

b

Function. All x values are different

c

Relation. The domain contains more than one 4

d

Relation. The domain contains two ones

e

Relation. The domain contains two – 4s and two –3s

0 –1

f

Function. All x values are different.

–2

a

The domain is {0, 1, 2, 3, 4} The range is {0, 1, 2} It is a function because the domain has exactly one of each value.

4x

3

y

a

b

2

Relation. Crosses twice d

H = 2 n (n – 1)

1

–2

10 x

Exercise 1A 1

y

35

0

d

c

Function. Crosses only once e

y 2 1 1

x

2

Relation. Crosses twice f

y

The domain is {–1, 0, 1, 2, 3}

x

0

The range is {–1, 0, 1, 2} Not a function as domain contains two −1s. 3

It is a function because the domain has exactly one of each value. g

Exercise 1B 1

a

Function. Crosses only once y 2

y

1 0 –1

Function. Crosses only once b

y

2

3

4

5 x

–2

x

0

1

Relation. Crosses twice y

h

3 2 1

0

x

–5 –4 –3 –2 –1 0 –1

1

2

3x

–2

Function. Crosses only once

Function. Crosses only once

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 1

2

WORKED SOLUTIONS 4

y

i 2

y2 = 4 − x 2, y = ± 4 − x 2

1 –2 –1 0 –1

1

There are two possible values of y for any given x. For example, When x = 1, y = 3, − 3. The same value in the domain has two possible values in the range. Therefore x 2 + y 2 = 4 is not a function.

2 x

–2

Function. Crosses only once 2

a

Manipulate the equation to make y the subject:

y=x

Exercise 1C 1

y

y=0

3

y 18 16 14 12 10 8 6 4 2

2 1 –2 –1 0 –1

1

2

3x

–2

b

y=x+2 y 3 2 1

–1 –0.5 0 0.5 1 1.5 2 2.5 x

–3 –2 –1–10 –2

c

1 2 3 4

2

x

y

y 2 1 1

y = 0, x = 0. 8 6 4 2

y = 2x – 3

–2 –1 0 –1

f(x) = 3x

2

3

4

–8 –6 –4 –2–20 –4 –6 –8

x

–2

2 4 6 8 x

–3

d

3

y=4 5

y

y

8 6 4 2

4 3 2

–8 –6 –4 –2–20 –4 –6 –8

1 –2 –1 0 –1

1

2

3 x

–2

3

e

Yes. A vertical line will only cross them once.

f

No, vertical lines such as x = 3 are not functions. 3 2 1

y = 0, x = –1.

4

y = 2, x = – 2. y 8

y

–2 –1–10 –2 –3

2 4 6 8 x

6 4 1 2 3 4x

2 –5 –4 –3 –2 –1 0 –2

Not a function as a vertical line crosses the region in many places

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

1

2

3

4

5 x

–4 –6 –8

Worked solutions: Chapter 1

3

WORKED SOLUTIONS 5

y = 2, x = 1.

y

b

16

y 8 6 4 2

14 12 2 4 6 8 x

–8 –6 –4 –2–20 –4 –6 –8

6

y = x2

10 8 4

y = 0, x = – 3, x = 3.

–4

y

–2

0

4 x

2

Domain x ∈ R, Range y ≥ 0

8 6 4 2

c

y 20 y = x2 + 5x + 6

15

2 4 6 8 x

–8 –6 –4 –2–20 –4 –6 –8

10 5 a –6

Exercise 1D 1

It is a function. Domain of {2, 3, 4, 5, 6, 7, 8, 9, 10} has no value repeated.

–4

Domain x ∈ R, Range y ≥ – 0.25 d

y 10

Range {1, 3, 6, 10, 15, 21, 28, 36, 45}. 2

3

–2

a

Domain {x : −4 < x ≤ 4}, Range { y : 0 ≤ y ≤ 4}

b

Domain {x : 1  x  5}, Range { y : 0  y  4}

c

Domain {x : −∞ < x < ∞}, Range {y : 0 ≤ y < ∞}

d

Domain{x : −2 ≥ x >2}, Range { y : 3 ≥ y ≥ 4}

e

Domain{x : −5 ≤ x ≤ 5}, Range { y : −3 ≤ y ≤ 4}

f

Domain {x : −∞ < x < ∞}, Range {y : −1 ≤ y ≤ 1}

8

g

Domain{x : −2 ≤ x ≤ 2}, Range { y : −2 ≤ y ≤ 2}

6

h

Domain{x :    x  }, Range { y :    y  }

i

Domain x ∈ R, x ≠ 1, Range y ∈ R, y ≠ 0.

1

x

2

–10 a

–15

Domain x ∈ R, Range y ∈ R e

y 10 y = √x

4 2 0

20

40

80 100 x

60

Domain x ≥ 0, Range y ≥ 0

12 y = 2x – 3 10 8 6 4 2 2 4 6 8

0

–1

–5

y

–2 –1–20 –4 –6 –8

y = x3 – 4

5

Note that domain and range can be expressed in many ways.

a

2x

0

–2

y

f a

10 y = √4 – x

8 6

x

Domain x ∈ R, Range y ∈ R

4 2 100 80

60

40

20

0

x

Domain x ≤ 4, Range y ≥ 0

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 1

4

WORKED SOLUTIONS g

l

y 8

y=

4 –8

0

–4

1 x

4

Exercise 1E

Domain x ∈ R, x ≠ 0, Range y ∈ R, y ≠ 0 h

y 8 7 6 5 4 3 2 1

1

a

y = ex

b

0

–1

1

2

x

Domain x ∈ R, Range y > 0

c

y 10 8 6 4 2

1 y= x+2

2 4 6x

–8 –6 –4 –2–20 –4 –6 –8 –10

d

Domain x ∈ R, x ≠ – 2, Range y ∈ R, y ≠ 0 j

y

e

10 8 6 4 2

y=

x+4 x–2

10

x

20

2

c d

y 4 –3

a b

Domain x ∈ R, x ≠ 2, Range y ∈ R, y ≠ 1 k

i

f (7) = 7 – 2 = 5

ii

f (–3) = –3 – 2 = – 5

iii

f ( 2 ) = 2 – 2 = –1 2

iv

f (0) = 0 – 2 = – 2

v

f (a) = a – 2

i

f (3) = 3(7) = 21

ii

f (–3) = 3(–3) = – 9

iii

f ( 1 ) = 3( 1 ) = 1 2

iv

f (0) = 3(0) = 0

v

f (a) = 3(a) = 3a

i

f (7) = 1 × 7 =

ii

f (–3) = 1 × − 3 = − 3

1

y=

–10 1 –4

x2 – 9 x+3 3

e 5

7

x

–8

Domain x ∈ R, x ≠ –3, Range y ∈ R, y ≠ – 6

3

a

1

1

1

2

2

4

4 1 4

1 2

7 4 4

1 2

1 8

iii

f( )= × =

iv

f (0) = 1 × 0 = 0

v

f (a) = × a =

i

f (7) = 2(7) + 5 = 19

ii

f (–3) = 2(–3) + 5 = –1

iii

f ( 1 ) = 2( 1 ) + 5 = 6

iv

f (0) = 2(0) + 5 = 5

v

f (a) = 2(a) + 5 = 2a + 5

i

f (7) = 72 + 2 = 51

ii

f (–3) = (–3)2 + 2 = 11

iii

f ( 1 ) = ( 1 )2 + 2 = 2 4

4 1 4

2

a 4

2

1

2

2

f (0) = (0)2 + 2 = 2 v f (a) = (a)2 + 2 = a2 + 2 f (–a) = (–a)2 – 4 = a2– 4 f (a + 5) = (a + 5)2 – 4 = a2 + 10a + 25 – 4 = a2 + 10a + 21 f (a – 1) = (a – 1)2 – 4 = a2 – 2a + 1 – 4 = a2 – 2a – 3 f (a2 – 2) = (a2 – 2)2 – 4 = a4 – 4a2 + 4 – 4 = a4 – 4a2 f (5 – a) = (5 – a)2 – 4 = 25 – 10a + a2 – 4 = a2 – 10a + 21 g(x) = 3, so 4 x – 5 = 3 4x = 8 x=2 iv

–10 –20 –4 –6 –8 –10

2 x2 + 1

Domain x ∈ R, Range 0 < y ≤ 2

–8

i

y=

–5 –4 –3 –2 –1 0 1 2 3 4 5 6 x

x

8

–4

–2

y 2 1.5 1 0.5

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 1

5

WORKED SOLUTIONS b

c

h(x) = –15 7 – 2x = –15

a

The initial velocity occurs when t = 0. V(0) = (02 – 9) ms –1 = –9 ms–1

2x = 22

b

V(4) = (42 – 9) ms–1 = 7 ms-1

x = 11

c

V(10) = (102 – 9) ms–1 = 91 ms–1

d

The particle comes to rest when V(t) = 0.

7

g(x) = h(x), so 4 x – 5 = 7 – 2x

t 2 – 9 = 0 ⇒ t 2 = 9 ⇒ t = 3s.

4x + 2x = 7 + 5 6x = 12

8

x=2 4

5

6

1 =−1 −3 − 6 9

a

f (2 + h) = (

(2 + h ) + h ) − (2 + h )

b

f (3  h) 

3  h   h   3  h  

a

h(−3) =

b

x = 6, as the denominator is zero and h(x) is undefined.

Exercise 1F

a

f (5) = 5 = 125

1

b

The volume of a cube of side 5

a

b

c d

3

i

3 6 +1 g(6) = ( ) = 19 (6) − 2 4

ii

3 −2 + 1 g(−2) = ( ) = − 5 = 1.25 ( −2 ) − 2 − 4

iii

3 0 +1 g(0) = ( ) = 1 = − 0.5 ( 0 ) − 2 −2

iv

⎛ 1⎞ 3⎜ − ⎟ +1 ⎛ 1⎞ g ⎜ − ⎟ = ⎝ 3 ⎠ = 07 ⎝ 3⎠ ⎛− 1⎞ −2 − ⎜ ⎟ 3 ⎝ 3⎠

= 4.75

h

= 2 + h + h − 2 − h = h =1 h

h

3  h  h  3  h h 1 h

a

(f

 g )( 3 ) = 3 ( 3 + 1) = 12

b

(f

 g )( 0 ) = 3 ( 0 + 1) = 3

c

(f

D g ) ( − 6 ) = 3 ( − 6 + 1) = −15

d

(f

 g )( x ) = 3 ( x + 1) = 3 x + 3

e

( g  f )( 4 ) = (3 ( 4 ) ) + 1 = 13

f

( g  f )( 5 ) = (3 ( 5 ) ) + 1 = 16

h

( g D f ) ( − 6 ) = (3 ( − 6 ) ) + 1 = −17 ( g  f )( x ) = (3 ( x ) ) + 1 = 3x + 1

i

(f

j

( h  f )( 2 ) = (3 ( 2 ) )

k

iii

3 1 .9 + 1 g(1.9) = ( ) = 6.7 = − 67 (1.9 ) − 2 − 0.1

(f

l

( h  f )( x ) = (3 ( x ) )

iv

3 1.99 + 1 g(1.99) = ( ) = 6.97 = − 697 (1.99 ) − 2 − 0.01

m

i

3 1 +1 g(1) = ( ) = 4 = − 4 (1) − 2 −1

ii

3 1 .5 + 1 g(1.5) = ( ) = 5.5 = −11 (1.5 ) − 2 − 0.5

v

3 1.999 ) + 1 g(1.999) = ( = 6.997 = − 6997 (1.999 ) − 2 − 0.001

vi

3 1.9999 ) + 1 g(1.9999) = ( = 6.9997 = − 69997 (1.9999 ) − 2 − 0.0001

The value of g(x) is getting increasingly smaller as x approaches 2. 2 because the denominator equals zero when x = 2. Division by zero is undefined.

e

y

–8

o p 2

a b c

e

10 –4 0 –10

n

d

20 f

g

=0

4

8

12

16

x

–20

There is a vertical asymptote at x = 2, as x = 2 makes the denominator zero and g(x) is undefined.

f g

(

h

h

)

 h )( 2 ) = 3 ( 2 ) + 2 = 18 2

(

2

+ 2 = 38

)

 h )( x ) = 3 ( x ) + 2 = 3 x 2 + 6

( g  h )(3 ) = ( (3 )

2

2

2

+ 2 = 9x 2 + 2

)

+ 2 + 1 = 12

( h  g )(3 ) = (3 + 1) ( g  h )( x ) = ( ( x )

2

2

+ 2 = 18

)

+ 2 +1= x2 + 3

( h  g )( x ) = ( x + 1)

2

( g  f )(1) = 3 − ( (1)

+ 2 = x 2 + 2x + 3

)

2

−1 = 3

( g  f )(2) = 3 − ( (2)2 − 1) = 4 − 2 2 = 0

( g  f )( 4 ) = 3 − ( ( 4 ) − 1) = −12 2 ( f  g )(3 ) = (3 − (3 ) ) − 1 = −1 2

( g D f )(3 ) = 3 − ( (3 ) (f (f

2

)

−1 = −5

D g ) ( − 4 ) = ( 3 − ( − 4 ) ) − 1 = 48 2

 g ) ( x + 1) = ( 3 − ( x + 1) ) − 1 = (2 – x)2 – 1 2

= (4 – 2x + x 2) – 1 = 3 – 2x + x 2 h

 g  x  2    3   x  2    1 = (1 – x)2 – 1 = (1 – 2x + x 2) – 1 = x 2 – 2x

f

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

2

Worked solutions: Chapter 1

6

WORKED SOLUTIONS 3

4

a

(f

 g )( x ) = ( x + 2 ) = x 2 + 4x + 4

b

(f

 g )( 3 ) = ( ( 3 ) + 2 ) = 25

(f

 g )( x ) = 5 ( x 2 + 1) = 5 x 2 + 5

a

5

b c

x – 8x + 19 = x – 1

a

No inverse function. Horizontal line crosses the graph more than once.

2

( g  f )( x ) = ( 5x ) + 1 = 25x 2 + 1 2 ( g  h )( x ) = ( x − 4 ) + 3 = x2 – 8x + 19 ( h  g )( x ) = ( x 2 + 3 ) − 4 = x 2 – 1

b

d

2

y 2 1

2

2

–4 –3 –2 –1 0 1 2 3 4 5 x

2

a

y 8

2

4

–8x + 19 = –1 –8x = –20

–8

–4

( r  s )( x ) = ( x )

2

b

No inverse function. Horizontal line crosses the graph twice. 7 6 5 4 3 2 1

–8

–4

0

x

4

8

x

4

8

x

–4 –8 c

y 8 4

1 2 x

–5 –4 –3 –2 –1 0

–8

Has an inverse function. Any horizontal line crosses the graph once only.

–4

7 6 5 4 3 2 1 –6 –5 –4 –3 –2 –1–10 –2 a –3

0 –4 –8

y

d

y 8 4 –8

1 2 3 4 x

Has an inverse function. Any horizontal line crosses the graph once only.

–4

0 –4 –8

e

y 8

y

4

3 2 1 0

8

4

y

c

4

8

The following have inverse functions. b, c.

b

x

y

Exercise 1G a

8

–8

− 4 = x2– 4

Domain x ∈ R, Range y ≥ – 4

1

4

–4

x = 2.5 6

0

–4 –3 –2 –1 0 1 2 3 4 5 6 x

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

1 2 3 4 x

–4

Worked solutions: Chapter 1

7

WORKED SOLUTIONS y

f

xy = y= 1 2 3 4 5 x

–3 –2 –1 0

f

–4

4

Exercise 1H 1

a

 f  g  1 ii



2

 

iii

( f  g )( x ) = (

2(x ) − 4) + 4

iv

( g  f ) ( x ) = 2 ⎛⎜⎝ x +2 4 ⎞⎟⎠ − 4 = x

2

a

x = 3 y −1 x +1 = 3 y f (x ) =

x +1 3

x = 1 y +5 4

d

4( x − 5) = y h −1( x ) = 4( x − 5)

( x + 3)

3

f f

y

y=

g (x ) = x=

y 3+ y

b

( x ) = ( x + 3)

3

h −1( x ) = 3 h

x=

2y 5− y

5x 2+ x

x=1–y y+x=1 y=1–x f –1(x) = 1 – x x=y f

(x) = x

–1

5x x +2

x −2 y +1 y −2

x ( y − 2) = y + 1 xy − 2 x = y + 1 xy − y = 2 x + 1 y ( x −1) = 2 x + 1 y = 2x + 1 x −1

f (x ) = 2x + 1 −1

=y

f −1( x ) =

4⎝x ⎠ 1 ⎞ 1 17 17 ⎛2 −1 (5) = ⎜ + 3 ⎟ = × = 4⎝5 ⎠ 4 5 20

f (x ) = x + 1 x=

x −3 2

3x 1− x

3x 1− x

f 5

x (5 − y ) = 2 y 5x − xy = 2 y 5x = 2 y + xy 5x = y (2 + x )

=y

f −1( x ) = 1 ⎛⎜ 2 + 3 ⎞⎟

x = 2 y3 + 3 x − 3 = 2 y3

x (3 + y ) = y 3x + xy = y 3x = y − xy 3x = y (1 − x ) f −1( x ) = a

1 x +2

2 4y −3 4 y −3 = 2 x 4y = 2 +3 x 1⎛2 y = ⎜ + 3 ⎞⎟ 4⎝x ⎠

=y

x −3 = y3 2 x −3 3 =y 2

1 x +2

−1

−1

2 4x − 3

x=

x = 3 y −3 x +3= 3 y

y

3

3

x −5 = 1 y

x = 1 −2

f (x ) =

c

g (x ) = x + 2

x +2= 1

g

x +2 = y −1

4

e

x = y3 − 2 3

10 x +7

10 y +7 y + 7 = 10 x 10 y = −7 x −1 f ( x ) = 10 − 7 x 10 −1 f (5) = − 7 = − 5 5

x + 2 = y3

y

−1

c

f (x) = x=

they are inverses of each other

x +1 = 3

1 x

b

=x

b

b

(x) =

–1

f (x) = 6 – x x=6–y x – 6 = –y 6–x=y f –1(x) = 6 – x f –1(5) = 6 – 5 = 1

 2 1  4 4

2  4  1 2 2 −3+ 4 1 f (−3) = = and 2 2  34  g  f 3  2 2  1   4  3 4   2  2

1

1 x

a

g(1) = 2(1) – 4 = –2 and

i

1 y

x=

c

6 4 2

6

x −1

x –3 –2 –1 0 1 2 3 4 5 6 f (x) 0.125 0.25 0.5 1 2 4 8 16 32 64 y 8 4

–2

0

2

4

x

–4

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 1

8

WORKED SOLUTIONS f (x). Domain x ∈R, Range y > 0

d

f 7

–1

2

(x). Domain x > 0, Range y ∈R.

Changing the x-coefficient alters the gradient of the line. y

x has domain x ≥ 0.

g(x) =

First, find g –1(x):

5

y, x ≥ 0

x=

3

x 2 = y, x ≥ 0 g –1(x)

1

= x 2, x ≥ 0

–6

The graph of g (x) is shown below.

–2 0 –2

–4

–1

2

x

4

y

3

y = |x + h| is a translation of –h along the x-axis y

g–1(x) 0

y =|x – 3|

y =|x + 2|

7 5

x

y =|x|

3

You can see that g –1(x) has domain x ≥ 0 range g –1(x) ≥ 0. Now, the graph of f (x) = x 2 is shown below: y

1 –5

4

–3

–1 0 1

3

5

x

The negative sign reflects the graph in the x-axis. Increasing the value of a means the graph increases more steeply.

f(x)

y 4

0

x

3 2

You can see that f (x) has domain x ∈R range f (x) ≥ 0. Hence, f (x) and g –1(x) are different. 8

m

f (x) = –1

1 x m

–3 –2 –1 0 –1

1 m

1

should be –1 but m ×

1 m

a

y 8 6 4 2

= 1.

Changing the constant term translates y = x along the y-axis.

–4

b

y 6 4 2

5 3 1 –4

–2 0 –2

2

4

6 x

2 4 6 8 x

–8 –6 –4 –2–20

y

–6

3 x

Exercise 1I

−c m

Investigation – functions 1

2

–3

For graphs of f(x) and f –1(x) to be perpendicular, m×

1

–2

Let f (x) = mx + c. x = my + c x – c = my x c =y m

1

–8 –6 –4 –2–20 –4 –6

2 4 6 8 x

–4

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Worked solutions: Chapter 1

9

WORKED SOLUTIONS c

a

y 6 4 2 –8 –6 –4 –2–20 –4

2 4 6 8

y

x

8 6 4 2

y

d

6 4 2

–8 –6 –4 –2–20 –4

–8 –6 –4 –2–20

2 4 6 8 x

y 6

e

b

2 –8

–4 0 –4

f

4

8

10 x

3

4

y 10 8 6 4 2

4 2

–6

–4

–f (2x) + 3 has domain –3 ≤ x ≤ 1, and has range 0 ≤ y ≤ 5. 5

a

f (x + 1) is a horizontal translation of f (x) by –1 units. y 4

–8 –6 –4 –2–20 –4 –6

2 4 6 8 x

g is a vertical translation of 2 units, so g(x) = f (x) + 2. h is is a vertical translation of – 4 units, so h(x) = f (x) – 4. q is a horizontal stretch of scale factor 2, so 1 q(x) = f ( 2 x). q is a horizontal translation of – 4 and a vertical translation of –2, so q(x) = f (x + 4) – 2 s is a horizontal translation of – 4, so s(x) = f (x + 4). t is a horizontal translation of 2, so t(x) = f (x – 2). y

2

g –4

f (x)

0

–2

2

4

x

–2 –4

b

f (x) + 1 is a vertical translation of f (x) by +1 unit. y 4 A1

2

–4 f –2

1 2

A

A1

f

g

3

0 –6 –4 –2 –1

2 x

0

–2

–2

–8 –6 –4 –2–20 2 4 6 8 x –4 –6 –8 y

2

–f (2x) + 3 is a horizontal stretch of scale factor 1 followed by reflection in the 2 x-axis, followed by a vertical translation of 3. y

6 4 2

2

2 4 6 8 x

2f (x – 5) has domain –1 ≤ x ≤ 7, and has range –4 ≤ 2f (x – 5) ≤ 6.

–4

g

2f (x – 5) is a horizontal translation of 5, followed by a vertical stretch of scale factor 2.

x

A 0

2

4 x

–2 –4

–2

This is the graph of f (x). It has domain –6 ≤ x ≤ 2 It has range −2 ≤ f (x) ≤ 3 © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 1

10

WORKED SOLUTIONS c

f (–x) is a reflection of f (x) in the y-axis.

3

y

f ( x ) = 3x + 17

a

g f –4

A 2

–2

4

x

2 x − 17 3

f −1( x ) = 2 x − 17

–4

d

2f (x) is a vertical stretch of f (x) by scale factor 2.

4

4 A1

2 g 0 –2

–2

A 2

4

5 1 x = − y −1 5 x +1 = − 1 y 5

f

f

–8 –4 –2–20 –4

6

a b c

−1

( x ) = −5 x − 5 y 6 4

y

g

–6

b

–4 –3 –2 –1 0

2 4 6 8 10 x

5

a

f ( x ) = 3x + 5 x = 3y + 5 x − 5 = 3y x −5 = 3

f (x ) =

1 − (1 − x )

–2

a

b

8

(f

D g )( x ) = 2 (1 − x

) + 7 = 2 − 2x

2

y2

x3  2  y

x −5 3

) x3  2 f 1 ( x b

a

0 –2 –4

2

4 x

y 4 3 2 1 –1 –10 –2 –3

1

2

x

Domain x ∈R,Range y ≥ 0 Domain x ∈R, x ≠ 3, Range y ∈R, y ≠ 0 Reflect in the y-axis. f (x) = –x Vertical stretch scale factor 2. f (x) = –2x 1 3

f (x – 3) = 2(x – 3)2 – 3(x – 3) + 1 = 2x 2 – 12x + 18 – 3x + 9 + 1 = 2x 2 – 15x + 28 2

a b

x

x 2

x  y2

2

7

3

3

3

y

1 2 3 4 x

g(a – 2) = 4(a – 2) –5 = 4a – 8 – 5 = 4a – 13. 1 + (1 − x ) 2 − x = b h (1 − x ) =

f ( x ) x 

4 (0,–1)

a

6 x

Reflect each graph in the line y = x a

Review exercise

2

b

y

−1

6

a

4

f

–6

f

1

2

–4

g(x) is a horizontal translation of f (x) by –3 units, followed by a vertical translation of –2 units.

✗

–2 0 –2

–4

A

y 5 4 3 2 1

2

(0,–1) A1

Reflection in the x-axis. Horizontal translation of 3 units. A vertical stretch of scale factor 2 followed by a reflection in the x-axis and then a vertical translation of 5 units.

7

x +4 5

f ( x ) = − 1 x −1

f (x – 2) + 3 is a horizontal translation of f (x) by 2 units, followed by a vertical translation of 3 units. 6 4 2

3

−5( x + 1) = y −5x − 5 = y

x

–4

e

g −1( x ) =

3

y

f –4

=y

x +4 = y3 5 x +4 3 =y 5

2x = 3 y + 17 2x −17 = 3 y

2 0

–2

2

x = 3 y + 17

4 A1

g ( x ) = 5x 3 − 4 x = 5 y3 − 4 x + 4 = 5 y3

b

2

Horizontal stretch scale factor . f (x) = –2(3x) Translate 3 units left. f (x) = –2(3x + 3) Translate 2 units up. f (x) = –2(3x + 3) + 2

+ 7 = −2 x + 9 2

Expand and simplify. f (x) = –6x – 4 = –2(3x +2)

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Worked solutions: Chapter 1

11

WORKED SOLUTIONS b

Reflect in the x axis. f (x) = – (x 2 ) Stretch vertically by scale factor 1 . 1

f (x ) = − 4 ( x )

13 a

4

2

b

9

a

b

f (x ) Translate 5 units right. 

2

Translate 1 unit down. f ( x)

1 1    x 5 43 

2

c

1

1

2 4 6 8

Domain: x ≥ – 2 Range: y > 0

2

y 16 14 12 10 8 6 4 2 0 –4 –3 –2 –1 –2 –4

(f

x

Domain: x ∈R Range: y ≥ – 4 3

y 12 10 8 6 4 2

 g )( 0 ) = 2 ( −2 ) + 3 = −16 + 3 = −13 3

y3 x −3 2

1 2 x

0 –6 –5 –4 –3 –2 –1 –2 –4 –6 –8

f ( x ) = 2x 3 + 3 x = 2 y3 + 3 x − 3 = 2 y3 f −1 ( x ) = 3

Domain: x ∈R, x ≠ – 2 Range: y ∈R, y ≠ 0 4

a

y 8 6 4 2

f (–x) is a reflection of f (x) in the line x = 0. y 5 4 3 2 1

b

12 a b

–3

–4 –3 –2 –1 0 1 2 x 1 g(x) = 2 f (x – 1) describes

the transformation: Horizontal translation by 1 unit, followed by 1 vertical stretch, scale factor 2 . so P is (4, 1) ( f ° g) (x) = 3(x + 2) = 3x + 6 x f –1(x) = 3 and g –1(x) = x – 2

f (12) = –1

12 3

b 5

–2

0 –2 –4 –6 –8

–1

2 x

1

x-intercept –1.5, y-intercept 3. y

a 6 4 2

=4

g (12) = 12 – 2 = 10 f –1(12) + g –1(12) = 4 + 10 f –1(12) + g –1(12) = 14 –1

x

1 2 3 4

3

g (0) = 3(0) –2 = –2

x −3 = 2

11 a

2

y=x 1

6x − 3 =0 2x − 3

Review exercise

y = 2x – 2

–8 –6 –4 –2–20 –4 –6 –8

b

y = 2x + 3

6x − 3 2x − 3

x=1

The graph of an inverse function is the reflection of the graph of the original function in the line y = x. Graph a line with a y-intercept of 3 and slope of 2. Draw the line y = x. To graph its inverse, sketch the mirror image of the original line. y

10 a

⎠

1 1    x 5 43 

8 6 4 2

=

( 2 x − 1) − 2

6x − 3 = 0 6x = 3

Stretch horizontally scale factor 3. 2 f ( x ) = − 1 ⎜⎛ 1 x ⎞⎟ 4⎝3

3(2 x − 1)

(h ° g) (x) =

–6

b c

–4

–2

0

2

4

6

x

0 Domain: x ∈R, x ≠ 0. Range: y > 0.

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Worked solutions: Chapter 1

12

WORKED SOLUTIONS 6

a

x = –2, y = 2.

10 a

y 16

b

12 4

f

–12 –8 –4 0 –4

4

8

12 x

–8

c 7

y

a

(x) =

x +2 3

c

( f −1  g ) = ( x − 33) + 2 = x 3− 1

d

( f −1  g ) ( x ) = ( g −1  f ) ( x ), so

–1

x −1 = 3x + 1 3

x −1 = 3(3x + 1) x −1 = 9 x + 3 8x = −4

4 2 –1 0 –2

3

( g ° f ) (x) = (3x − 2) + 3 = 3x + 1

6

–2

−1

b

(2.5, 0), (0, –2.5)

–3

x +2

y=

8 f

f (x) = 3x − 2 x = 3y − 2 x + 2 = 3y

1

2

3 x

x =−1

2

–4

e

–6

8

b

± 2

a

f (x) = x3 − 3 x = y3 − 3

–6 –4 –2–20 –4

x+3=y

3

y = 3 x +3 f

−1

f

(x) = 3 x + 3

b

y 12 10 8 6 y=3 4 2

x=3 2 4 6 8 x

x = 3, y = 3

y 6 4 2 –6

–2 0 –2

–4

2

4

6 x

–4

c

1.67

9

y 20 x = –1

10

y=0 –5 –4 –3 –2 –1 0

1 2 x

–10

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Worked solutions: Chapter 1

13

WORKED SOLUTIONS

2

Quadratic functions and equations

Answers Skills check 3a − 5 = a + 7 2a = 12 a = 6

d

(x + 5) (x − 5) = 0 x = ±5

1 a

4 x 2 + 1 = 21 4x 2 = 20 x 2 = 5 x = ± 5 b

3(n − 4) = 5(n + 2) 3n − 12 = 5n + 10 2n = −22 n = −11 c

2 a 2k(k

− 5) b 7a(2a2 + 3a − 7) 2 c 2 x + 4 xy + 3 x + 6 y 2 x ( x + 2 y ) + 3 ( x + 2 y )

x 2 − 25 = 0

e

x 2 + 2x − 48 = 0



x 2

+ 8x − 6x − 48 = 0

x(x + 8) − 6(x + 8) = 0 (x + 8) (x − 6) = 0 x = −8 or x = 6 f

b 2 + 6b + 9 = 0

b 2 + 3b + 3b + 9 = 0 b(b + 3) + 3(b + 3) = 0 (b + 3)2 = 0 b = −3 2 a 6x 2

+ 5x − 4 = 0

6x  + 8x − 3x − 4 = 0 2

2x(3x + 4) − 1(3x + 4) = 0

( 2 x + 3 ) ( x + 2 y )

(3x + 4)(2x − 1) = 0

5a 2 − 10a − ab + 2b 5a ( a − 2 ) − b ( a − 2 ) ( 5a − b ) ( a − 2 ) e (n + 1)(n + 3) f (2x − 3)(x + 1) g (m + 6)(m − 6) h (5x + 9y)( 5x − 9y)

x = − 3 or x =

d

Exercise 2A 1 a

x 2 − 3x + 2 = 0 x 2 − x − 2x + 2 = 0 x(x − 1) − 2(x − 1) = 0 (x − 1)(x − 2) = 0 x = 1 or x = 2 b a 2 + a − 56 = 0 a 2 + 8a − 7a − 56 = 0 a(a + 8) − 7(a + 8) = 0 (a + 8)(a − 7) = 0 a = −8 or a = 7 c m 2 − 11m + 30 = 0 m 2 − 5m − 6m + 30 = 0 m(m − 5) − 6(m − 5) = 0 (m − 5) (m − 6) = 0 m = 5 or m = 6

4

b



1 2

5c 2 + 6c − 8 = 0 5c 2 + 10c − 4c – 8 = 0 5c (c + 2) − 4 (c + 2) = 0 (c + 2) (5c − 4) = 0 4

c = −2 or c = 5 2h 2 − 3h − 5 = 0 2h 2 − 5h + 2h − 5 = 0 c

h(2h − 5) + 1(2h − 5) = 0 (h + 1) (2h − 5) = 0 5

h = −1 or h = 2 4x 2 − 16x − 9 = 0 4x 2 − 18x + 2x – 9 = 0 2x(2x – 9) + 1(2x – 9) = 0 (2x + 1) (2x − 9) = 0 d

9

x = − 1 or x = 2 2

3t 2 + 14t + 8 = 0 3t 2 + 12t + 2t + 8 = 0 3t (t + 4) + 2(t + 4) = 0 (t + 4) (3t + 2) = 0 e

2

t = −4 or t = − 3

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Worked solutions: Chapter 2

1

WORKED SOLUTIONS 6x 2 + x − 12 = 0 6x 2 + 9x − 8x − 12 = 0 3x(2x + 3) − 4(2x + 3) = 0 (2x + 3) (3x − 4) = 0 f

4

3 x = − 2 or x = 3

Exercise 2B 1 a

x 2 + 2x − 7 = 13 + x x 2 + x − 20 = 0 x 2 + 5x − 4x − 20 = 0 x(x + 5) − 4(x + 5) = 0 (x + 5) (x − 4) = 0 x = −5 or x = 4 b 2n 2 + 11n = 3n − n2 − 4 3n 2 + 8n + 4 = 0 3n 2 + 6n + 2n + 4 = 0 3n(n+ 2) + 2(n + 2) = 0 (n + 2) (3n + 2) = 0 n = −2 or

2 n=−3

3z 2 + 12z = −z 2 − 9 4z 2 + 12z + 9 = 0 4z 2 + 6z + 6z + 9 = 0 2z(2z + 3) + 3(2z + 3) = 0 (2z + 3) (2z + 3) = 0 c

z = − 3

2a 2 − 50 = 21a 2a 2 − 21a − 50 = 0 2a 2 − 25a + 4a – 50 = 0 a(2a – 25) + 2(2a – 25) = 0 (a + 2) (2a − 25) = 0 d

a = −2 or a =

25 2

x 2 + 5x = 36 x 2 + 5x − 36 = 0 x 2 + 9x − 4x – 36 = 0 x(x + 9) − 4(x + 9) = 0 (x + 9) (x − 4) = 0 x = −9 or x = 4 f 4x 2 − 2x = x + 1 4x 2 − 3x − 1 = 0 4x 2 − 4x + x – 1 = 0 4x(x – 1) + 1(x – 1) = 0 (4x + 1) (x − 1) = 0 e

x

or x = 1

x 2 − x = 12 x 2 − x − 12 = 0 x 2 − 4x + 3x − 12 = 0 x(x – 4) + 3(x − 4) = 0 (x + 3) (x − 4) = 0 x = −3 or x = 4 2



2

If x = , the leg 5x − 3 would have a negative length. 5 Therefore, the only answer is x = 3.

Investigation – perfect square trinomials 1

2

3

4

5 6



2

=−1 4

(x + 2) 2 + (5x − 3) 2 = (4x + 1) 2 x 2 + 4x + 4 + 25x 2 −30x + 9 = 16x 2 + 8x + 1 10x 2 −34x + 12 = 0 2(5x 2 −17x + 6) = 0 2(5x 2 – 15x – 2x + 6) = 0 2(5x(x – 3) – 2(x – 3) = 0 2(5x − 2) (x − 3) = 0 x=3 3

x 2 + 10x + 25 = 0 → (x + 5) (x + 5) = (x + 5) 2 = 0 x = −5 x 2 + 6x + 9 = 0 → (x + 3) (x + 3) = (x + 3) 2 = 0 x = −3 x 2 + 14x + 49 = 0 → (x + 7) (x + 7) = (x + 7) 2 = 0 x = −7 x 2 − 8x + 16 = 0 → (x − 4) (x − 4) = (x − 4) 2 = 0 x=4 x 2 − 18x + 81 = 0 → (x − 9) (x − 9) = (x − 9) 2 = 0 x=9 x 2 − 20x + 100 =  0 → (x − 10) (x − 10) = (x − 10) 2 = 0 x = 10

Exercise 2C x 2 + 8x + 16 = 3 + 16 (x + 4) 2 = 19 x + 4 = ± 19 x = −4 ± 19 1

25 4

2

x 2 − 5x +



5 ⎞ 37 ⎛ ⎜x − ⎟ = 2⎠ 4 ⎝



x−



x=

=3+

25 4

2

37 5 ± 37 =± = 2 4 2 5 ± 37 2

x 2 − 6x = −1 x 2 − 6x + 9 = −1 + 9 (x − 3) 2 = 8 x −3 = ± 8 = ± 2 2 3



x =3±2 2

4

x 2 + 7x = 4 49 x 2 + 7x + 4 = 4 +



2



7⎞ 65 ⎛ ⎜x + ⎟ = 2⎠ 4 ⎝



x+



x = −7 ±

7 2

=±

65 4

=

49 4

± 65 2

65

2

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Worked solutions: Chapter 2

2

WORKED SOLUTIONS x 2 − 2x = 6 x 2 − 2x + 1 = 6 + 1 (x − 1) 2 = 7 x −1 = ± 7 5

x =1± 7

6

x 2 + x = 3 x2 + x + 1 = 3 +



4

⎛ ⎜x ⎝

6

1 4

2

1⎞ 13 ⎟ = 2⎠ 4

+

x +

1 2

13 4

x=

−1 ± 13 2

=±

=

± 13 2

Exercise 2D x 2 + 6x = 3 x 2 + 6x + 9 = 3 + 9 (x + 3) 2 = 12 x + 3 = ± 12 = ±2 3 x = −3 ± 2 3 1

C

M

Y

CM

MY

CY

CMY

K

x  − 2x = 1 x 2 − 2x + 1 = 1 + 1 (x − 1) 2 = 2 x −1 = ± 2 2

2



x =1± 2

3

5 ( x 2 − 2 x ) = −2



x 2 − 2x = − 2

x 2 − 2x + 1 = − 2 + 1 5

4



x2 + 3 x = 2



3 ⎞ x⎟ 2 ⎠

+

x2 + 3 x + 2 ⎛ ⎜x ⎝

+

3 5

3 5

4 ⎛⎜ x 2 ⎝

=5

5 4 9 16

=

5 4

9

+ 16

2

3⎞ 29 ⎟ = 4⎠ 16

x

3 4



x=

−3 ± 29 4

5

2 ⎛⎜ x 2 − 1 x ⎟⎞ = 6 2



x2 − 2 x = 3



x 2 − 2 x + 16 = 3 + 16



⎛ ⎜x ⎝

⎝

29 16

 29



 

4

⎠

1

1

1

2

− 4 ⎟⎞ = 1

⎠

x=

1 7 ± 4 4

49 16

=±

7

=±4

49 16

x = −2, 2 10 ⎛⎜ x 2 + 2 x ⎟⎞ = 5 ⎝

1

⎠ 1 2 1 25

5



x2 +



x2 +



⎛ ⎜x ⎝

1⎞ ⎟ 5⎠



x +5=±



x=

+

= +

2

=

1

=2+

1 25

27 50

1

27 50

=±

3 3 5 2

±3 6 10

=

−2 ± 3 6 10

Exercise 2E 2  9   4  4  7   24 

9 

1 x 



x=

2

x=



x



x

9 

81  112 8

−9 ± 193 8

2 −2 ± ( 2 ) − 4 ( 3 ) ( −8 ) 2 (3) −2 ± 10 = 6 = − 2, 4 3

3 x 

2 ( x − 1) = 35

x =1±

1 4

2 x 5 2 x 5

5

x −1= ±

x−

3







=

 6 2  4  5  1

6 

−2 ± 4 + 96 6

6 

=

−2 ± 100 6

36  20

  2 5 10

−6 ± 4 10 1 −1, − 5

6  16 10



x=



x=

4

x 2 − 6x + 4 = 0



x=



x=

5

x 2 − x + 3 = 0



x =1±



no solution since 11 is undefined.

6

3x 2 + 10x − 5 = 0



x=



x

7

2x 2 − 3x − 1 = 0



x = 3±



x = 3±

2 6 ± ( −6 ) − 4 (1)( 4 ) 2 (1)

6±2 5 2

6 ± 36 − 16 2

=

6 ± 20 2

=3± 5

( −1)2 − 4 (1) (3 ) 2 (1)

−10 ±

=

=1±

(10 ) − 4 (3 ) ( −5 ) = 2 (3 ) 2

10  4 10 6



1 − 12 1 ± −11 = 2 2

−10 ± 100 + 60 6

=

−10 ± 160 6

 5  2 10 3

( −3 )2 − 4 ( 2 ) ( −1) 3 ± 9 + 8 = 2 (2) 4 17 4

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 2

3

WORKED SOLUTIONS 8

2x 2 − 9x − 4 = 0



x = 9±



x=9±

( −9 )2 − 4 ( 2 ) ( −4 ) 2 (2)

=

9 ± 81 + 32 4

   2 2 2



x

9 



x=

−9 ± 129 4



9  4 2 6   

9 

10

(5 x − 2)( x ) = ( x + 3)( x + 1)



5x 2 − 2 x = x 2 + 4 x + 3



4 x 2 − 6x − 3 = 0 2 6 ± ( −6 ) − 4 ( 4 ) ( −3 ) 2 (4)



x =6±

84 8

=

6 ± 2 21 8

=

81  48 4

6 ± 36 + 48 8

=3±

21 4

Let the two numbers be x and y. x + y = 50 → y = 50 − x xy = 576 → x (50 − x) = 576 50x − x 2 = 576 → x 2 − 50x 2 + 576 = 0 (x − 18)(x − 32) = 0 The possible values for x are 18 and 32. Since y = 50 – x, the two numbers are 18 and 32.

1

This quadratic equation could also be solved using completing the square or the quadratic formula.

Let l represent the length, and w represent the width. 2l + 2w = 70 → l + w = 35 → l = 35 − w lw = 264 → (35 − w)w = 264 35w − w 2 = 264 → w 2 − 35w + 264 = 0 (w − 11)(w − 24) = 0 The dimensions are 24 m and 11 m. 2

(x + 6) 2 + (3x)2 = (4x − 6)2 x 2 + 12x + 36 + 9x 2 = 16x 2 − 48x + 36 6x 2 − 60x = 0 6x (x − 10) = 0  x = 10 (we cannot have x = 0, since this would mean one side has zero length). 3

(23 − x)(16 + x) = 378 368 + 7x − x 2 = 378 → x 2 − 7x + 10 = 0 (x − 5)(x − 2) = 0 x = 2, 5 The dimensions are 18 cm and 21 cm. 4

−14 ±

(14 ) − 4 ( −4.9 )( 2 ) = 2 ( −4.9 ) 2

−14 ± 235.2 −9.8

Investigation – roots of quadratic equations 1 a

Exercise 2F



t=

approximately 2.99 seconds. We cannot have ‘negative time’, so we take only the positive value.

6 − 2x 2 = 9x 2x 2 + 9x − 6 = 0

x=



113 4

9





h = 2 + 14t − 4.9t 2 The ball hits the ground when h = 0. 0 = 2 + 14t − 4.9t 2

5

x = 8±

b

x=

c

x=

2 a

x=

( −8 )2 − 4 (1) (16 ) 2 (1)

=

2 12 ± ( −12 ) − 4 ( 4 ) ( 9 ) 2(4)

= 12 ±

(10 ) − 4 ( 25 )(1) = 2 ( 25 ) 2

−10 ±

b

2 −5 ± ( 5 ) − 4 (1) ( −14 ) 2 (1)

x =8±

x c

3 a

=

2

0

=

= 12 = 8

−10 ± 0 50 −5 ± 81 2

=8±

40 6

= =

3 2

−10 50

= −1 5

−5 ± 9 2

=8±2

10

6

4 ± 10 3

x =3± x=

( −8 )2 − 4 (3 ) ( 2 ) 2 (3 )

=8=4

8

x = −7, 2

8± 0 2

( −3 )2 − 4 ( 5 ) ( −4 ) 2 (5)

2 −3 ± (3 ) − 4 (1) ( 6 ) 2 (1)

=3±

89

10

=

−3 ±

2 4 ± ( −4 ) − 4 ( 2 ) ( 5 ) 2 (2)

=

4±

( 2 )2 − 4 ( 4 )(1) 2(4)

=

−2 ±

−15 2

no solution b

x=

no solution c

x=

−2 ±

no solution

−24 4 −12 8

Exercise 2G Δ = 52 − 4(1)(−3) = 37 two different real roots b Δ = 42 − 4(2)(1) = 8 two different real roots c Δ = (−1)2 − 4(4)(5) = −79 no real roots d Δ = (8)2 − 4(1)(16) = 0 two equal real roots e Δ = (−3)2 − 4(1)(8) = −23 no real roots f Δ = (−20)2 − 4(12)(25) = −800 no real roots 2 a 42 − 4(1)( p) > 0 16 > 4p p < 4 b 52 − 4( p)(2) > 0 25 > 8p 1 a

p
0 p2 > 32 ⇒ p < − 32 or p > 32, so p > 32 p >4 2 c

Investigation – graphs of quadratic functions a

Δ = (−3)2 − 4(1)(−5) = 29 y

(3p)2 − 4( 1)(1) > 0 9p2 > 4 2 −2 4 2 p > ⇒ p < or p > , so d

p >

9 2 3

3

− 4( 1)(k) = 0 100 = 4k k = 25

x



3 a 102

b

0

3

b

Δ = (−6)2 − 4(3)(4) = −12 y

(−3)2 − 4( 2)(k) = 0 9 = 8k 9

k = 8 (−2k)2 − 4( 3)(5) = 0 4k 2 = 60 k 2 = 15 k = ± 15 c

c

0

x

Δ = (2)2 − 4(1)(7) = −24

(−4k)2 − 4( 1)(−3k) = 0 16k 2 + 12k) = 0 4k(4k + 3) = 0 d

y

3

k = 0, − 4 4 a (−6)2

− 4( 1)(m) < 0 36 < 4m m > 9

d

0

x

Δ = (3)2 − 4(4)(5) = −71

(5m)2 − 4( 1)(25) < 0 25m2 < 100 m2 < 4 −2 < m < 2 b

y

(−8)2 − 4( 3m)(1) < 0 64 < 12m m > 16 c

3

(6)2 − 4(1)( m − 3) < 0 36 < 4m − 12 48 < 4m m > 12 d

(−4q)2 − 4( q)(5 − q) < 0 16q2 − 20 q + 4q2 < 0 20q 2 − 20q < 0 20q (q − 1) < 0 0 0: (0, ∞) decreasing when f ′(x) < 0 : (−∞, 0) f (x ) = x 4 − 2x 2 f ′( x ) = 4 x 3 − 4 x 4 x ( x 2 − 1) = 0 4 x ( x + 1)( x − 1) = 0 x = 0, −1, 1

f

( x − 3)

=

1 (t + 1)2

a(t) = = + as t > 0 a(t) > 0 therefore velocity is never decreasing.

In questions 1-3, if y increases as x increases the function is increasing. If y decreases as x increases the function is decreasing. y 1 decreasing (−∞, ∞)

+

–

–1

+

0

1

−5 ( x − 3) 2

signs of f ' x

f

′(x) ≠ 0

f

′(x) undefined when x = 3

–

– –3

decreasing when f ′(x) < 0 : (−∞, 3) and (3, ∞) f (x ) =

7

1 x

=x

f ′( x ) = − 1 x 2

=

4 3 2 1

−1 2

−3 2

signs of f ' x 0

–

−1 3

2x 2

f

–4 –3 –2 –1–10 –2 –3 –4

decreasing when f ′(x) > 0 : (0, ∞) 8

y

1 2 3 4 5 6 x

′(x) ≠ 0

domain x > 0

1 2 3 4 x

1 –3 –2 –1–10 –2 –3 –4 –5 –6

–

0

x −3 ( x − 3)(1) − ( x + 2)(1) ′( x ) = ( x − 3) 2

= ( x − 3)(1) − ( x2 + 2)(1)

Exercise 7Q

2

signs of f ' x

+

f (x ) = x + 2

6

t2 + t − 2 = 0 (t + 2)(t − 1) = 0 when t = 1 b

2 x

increasing when f ′(x) > 0: (−1, 0) and (1, ∞) decreasing when f ′(x) < 0: (−∞, −1) and (0, 1)

particle at rest when v(t) = 0

1 t 2

1

–2

1 s(t) = 4 t2 − ln (t + 1) ds 1 1 v(t) = dt = 2 t − t + 1

dv dt

0

–1

–1

Since the signs of v (0.3) and a (0.3) are different the particle is slowing down at 0.3 seconds. 4

increasing (−1, 1) decreasing (−∞, −1) and (1, ∞)

y 2

increasing (−∞, 2) decreasing (2, ∞)

f ( x ) = x 3e x

signs of f ' x

f ’( x ) = ( x 3 )(e x ) + (e x )(3 x 2 )

–

+ –3

+ 0

e x x 2 ( x + 3) = 0 x = 0, − 3 increasing when f ′(x) > 0: (−3, ∞) decreasing when f ′(x) < 0: (−∞, −3)

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 7

14

WORKED SOLUTIONS 9

f (x ) =

x3 x2 −1

f ′( x ) = ( x

2

5

3

− 1)(3 x 2 ) − ( x 3 )(2 x ) ( x 2 − 1)2

− 3x ( x 2 − 1)2

=x

4

2

signs of f ' x

5 3

f ′( x ) = x

+

–

–√3

–

–

–1

0

– 1

+ 5 3

√3

( x − 3) ( x 2 − 1)2

=x

2

2

decreasing when f ′(x) < 0: ( − 3 , − 1) (−1, 1) (1, 3 ) f is increasing when f ′(x) > 0: (−∞, −2) and (4, ∞)

f (x ) = 2x 2 − 4 x − 3 f ′( x ) = 4 x − 4 4x − 4 = 0 4 ( x − 1) = 0 x =1 f (1) = −5

4

4 3 2 1

4x3 − 4x = 0 4 x ( x + 1)( x − 1) = 0 + – + signs of f ' – x = −1, 0, 1 x 0 1 –1 f (0 ) = 0 f ( −1) = −1 f (1) = −1 Since f ′(x) changes from negative to positive at x = −1 and x =1 there are relative minimums at x = −1 and x = 1

y = f'(x)

–4 –3 –2 –1–10 –2 –3 –4

f (x ) = x 4 − 2x 2 f ʹ( x ) = 4 x 3 − 4 x

1 2 3 4 5 x

The relative minimum points are(−1, −1) and (1, −1) signs of f ' x

–

+

Since f ′(x) changes from positive to negative at x = 0 there is a relative maximum at x = 0. The relative maximum point is (0, 0)

1

5

Since f ′(x) changes from negative to positive at x = 1 there is a relative minimum at x = 1

= ( x + 3) 2 ( 4 x + 3) ( x + 3) 2 ( 4 x + 3) = 0

f ( x ) = x 3 − 12 x − 5

x = −3, − 3

f ′( x ) = 3 x 2 − 12

4

3 x 2 − 12 = 0 + – + signs of f ' 3( x + 2)( x − 2) = 0 x –2 2 x = −2, 2 f ( −2) = 11 f (2) = −21 Since f ′(x) changes from negative to positive at x = 2 there is a relative minimum at x = 2

signs of f ' x

⎛ ⎞ f ⎜ − 3 ⎟ = − 2187 256 ⎝ 4⎠

–

– –3

+ 3

–4

Since f ′(x) changes from negative to positive at x = − 3 there is a relative minimum at x = − 3 4

4

The relative minimum point is

2187 ⎞ ⎛ 3 ⎜− , − ⎟ 256 ⎠ ⎝ 4

There is no relative extremum at x = −3 since f ′(x) does not change signs at x = −3

The relative minimum point is (2, −21) Since f ′(x) changes from positive to negative at x = −2 there is a relative maximum at x = −2. The relative maximum point is (−2, 11)

f ( x ) = x ( x + 3 )3 f ′( x ) = ( x ) ⎡⎣3( x + 3)2 (1) ⎤⎦ + ( x + 3)3 (1)

The relative minimum point is (1, −5) 2

+ 0

Since f ′(x) does not change signs there are no relative maximum or minimum points.

y

f is

1

x =0

+

signs of f ' x

for x > 0, f ′(x) > 0

increasing when f ′(x) > 0: ( −∞, − 3 ) and ( 3 , ∞ )

Exercise 7R

2 3

for x < 0, f ′(x) > 0

f ′( x ) = 0 when x 2 ( x 2 − 3) = 0 or x = 0, − 3 , 3.

decreasing when f ′(x) < 0: (−2, 4)

2 3

x =0

f ′( x ) undefined when ( x 2 − 1)2 = 0 or x = −1, 1.

10

f (x ) = x 3

6

f ( x ) = x 2e − x f ′( x ) = ( x 2 ) ⎡⎣(e − x )( −1) ⎤⎦ + (e − x )(2 x ) = xe − x ( − x + 2)

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 7

15

WORKED SOLUTIONS xe − x ( − x + 2) = 0 x = 0, 2 f (0 ) = 0 f (2) =

–

signs of f '' x

+

–

0

2

2

4 x 3  12 x 2 f ( x )  f ( x )  12 x 2  24 x f ( x )  0 for inflexion points

4 e

2

12 x 2  24 x  0

Since f ′(x) changes from negative to positive at x = 0 there is a relative minimum at x = 0

12 x ( x  2)  0 x  0,2

The relative minimum point is (0, 0)

The relative maximum point is ⎛⎜ 2, ⎝

f (x ) =

1 ( x + 1)2

= f

−2 ( x + 1)3

signs of f '' x

4⎞ ⎟ e2 ⎠

+

3

– –1

f

− 2x + 1 x +1 ( x + 1)(2 x − 2) − ( x 2 − 2 x + 1)(1) ′( x ) = ( x + 1)2 2

− 2) − ( x 2 − 2 x + 1) ( x + 1)2 signs of f '' 2 x = x + 2 x −2 3 ( x + 1) = ( x + 3)( x 2− 1) ( x + 1)

= (2 x

+

– –3

– –1

+ 1

′(−3) = −8

Since f ′(x) changes from negative to positive at x = 1 there is a relative minimum at x = 1 The relative minimum point is (1, 0) Since f ′(x) changes from positive to negative at x = −3 there is a relative maximum at x = −3. The relative maximum point is (−3, −8)

Exercise 7S f (x )  2x 2  4 x  3 f ( x )  4 x  4 f ( x )  4 Since f ″(x) > 0 for all x, f is concave up on (−∞, ∞)

There are no inflexion points.

2

–

+ 2

concave down when f ″(x) < 0: (−∞, 2)

2

f (1) = 0

1

0

f (2)  8 concave up when f ″(x) > 0: (2, ∞)

f ′( x ) undefined when ( x + 1)2 = 0 or x = −1 f ′(x) = 0 when ( x + 3)( x − 1) = 0 or x = −3, 1 f

–

f ( x ) x 3  6 x 2  12 x f ( x )  3 x 2  12 x  12 f ( x ) 6 x  12 f ( x )  0 for inflexion points 6 x  12  0 6( x  2)  0 signs of f '' x x 2

Although f changes signs at x = −1, there is no relative extremum since f is undefined at x = −1 f (x ) = x

+

concave down when f ″(x) < 0: (−∞, 0) and (2, ∞) inflexion points: (0, 0) and (2, 16)

′(x) ≠ 0 f ′( x ) undefined when ( x + 1)3 = 0 or x = −1

8

–

f (2)  16 concave up when f ″(x) > 0: (0, 2)

= ( x + 1)−2

f ′( x ) = −2( x + 1)−3 (1)

signs of f '' x

f (0)  0

Since f ′(x) changes from positive to negative at x = 2 there is a relative maximum at x = 2.

7

f (x )  x 4  4 x 3

inflexion point: (2, 8) 4

f (x )  x 4 4x3 f ( x )  f ( x )  12 x 2 f ( x )  0 for inflexion points + signs of f '' 12 x 2  0 x x 0 concave up when f ″(x) > 0: (−∞, ∞)

+ 0

There are no inflexion points since f ″(x) does not change sign either side of x = 0. f ( x )  2 xe x 5 f ( x ) (2 x )(e x )  (e x )(2)  2e x ( x  1)  f ( x ) (2e x )(1)  ( x  1)(2e x )  2e x ( x  2) f ( x )  0 for inflexion points 2e x ( x  2)  0 x  2

signs of f '' x

–

+ –2

f ( 2)  4 e

2

concave up when f ″(x) > 0: (−2, ∞) concave down when f ″(x) < 0: (−∞, −2) inflexion point: ⎛⎜ −2, − 42 ⎞⎟ ⎝

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

e ⎠

Worked solutions: Chapter 7

16

WORKED SOLUTIONS 6

f (x ) =

1 = ( x 2 + 1)−1 x2 +1

− 4) ( x + 12)3

f ′′( x ) = 1442( x

ii

f ′( x ) = −( x 2 + 1)−2 (2 x ) = f ′′( x ) = =

f ′′( x ) = 0 when 144( x 2 − 4 ) = 0. x = −2, 2

−2 x ( x 2 + 1)2 2

2

⎡⎣( x 2 + 1)2 ⎤⎦

2

f

−2( x 2 + 1) ⎡⎣( x 2 + 1) − 4 x 2 ⎤⎦

f

( x 2 + 1)4

= −2(−23x

+ 1) ( x + 1)3 2

x  1 or  3

⎝

+

–

3 3

⎛ 3⎞ 3 ⎜⎜ − ⎟⎟ = 3 ⎝ ⎠ 4 ⎛ 3⎞ 3 ⎜⎜ ⎟⎟ = ⎝ 3 ⎠ 4

3⎞ ⎟ 3 ⎟⎠

⎝

⎛

concave down when f ″(x) < 0: ⎜⎜ − ⎝

⎛

inflexion points: ⎜⎜ − ⎝

a

f

3 3⎞ , ⎟ 3 4 ⎟⎠

+

√3 –3

concave up when f ″(x) > 0: ⎛⎜⎜ −∞, −

7

and

8

√3 3

and

⎛ 3 ⎞ , ∞ ⎟⎟ ⎜⎜ 3 ⎝ ⎠

⎞ 3 , 3 ⎟⎟ 3 3 ⎠

2

The graph of the second derivative of f is positive for x < −2 and x > 4, so f is concave up on (−∞, −2) and (4, ∞). The graph of the second –4 derivative of f is negative for −2 < x < 4, so f is concave down on (−2, 4).

⎛ 3 3⎞ , ⎟⎟ ⎜⎜ ⎝ 3 4⎠

2

= 48( x

2

1 2 3 4 5 x

⇒ (3 x − 2)( x + 4 ) = 0 ⇒ x = 2 , − 4 3

+ 12)[−( x 2 + 12) + 4 x 2 ] ( x 2 + 12)4

⎛ ⎞ The x -intercepts are ⎜ 2 , 0 ⎟ and ( −4, 0 ). 3 ⎝ ⎠ f ′( x ) = 6 x + 10

+ 12)(3 x − 12) ( x 2 + 12)4 2

2 + 12)( x 2 − 4 ) ( x 2 + 12)4

f ′( x ) = 0 ⇒ 6 x + 10 = 0 ⇒ x = − 5 3

2 = 1442( x − 43 ) ( x + 12)

i

–3 –2 –1–10 –2 –3 –4

y = f''(x)

f ( x ) = 0 ⇒ 3 x 2 + 10 x − 8 = 0

= 144( x

b

y 4 3 2 1

f ( x ) = 3 x 2 + 10 x − 8 f (0 ) = −8 ⇒ the y -intercept is (0, −8).

1

⎡⎣( x 2 + 12)2 ⎤⎦ 2 2 2 2 = ( x + 12) (−482 ) + 1924 x ( x + 12) ( x + 12) 2

2⎠

⎝

2⎠

Exercise 7T

+ 12)2 ( −48) − ( −48 x )[2( x 2 + 12)(2 x )]

= 48( x

+ 2

The inflexion points occur at x = −2, 4 since f ″(x) changes sign at x = −2, 4.

′( x ) = −2 48x 2 ( x + 12)

f ′′( x ) = ( x

–

–2

inflexion points are ⎛⎜ −2, 3 ⎟⎞ and ⎛⎜ 2, 3 ⎞⎟ .

x

f

+

signs of f '' x

Since f ′(x) changes signs at x = −2, 2 the

f ( x )  0 for inflexion points 2(3x 2  1)  0 signs of f '

f

24 x 2 + 12 (−2) = 3 2 3 (2) = 2

f (x ) =

( x + 1) ( −2) − ( −2 x ) ⎡⎣2( x + 1)(2 x ) ⎤⎦ 2

2

signs of f ' x

–

+

f ( x ) 2 48x

⎛ ⎞ f ⎜ − 5 ⎟ = − 49 3 ⎝ 3⎠

f ( x )  0 for relative extrema 48x  0 signs of f ' x x 0

⎛ ⎞ f has a relative minimum point at ⎜ − 5 , − 49 ⎟ . 3 3 ⎝ ⎠ f ′′( x ) = 6 > 0 ⇒ f is always concave up.

( x  12)2

f (x ) 

+

– 0

24 x 2  12

f (0)  2 Since f ′(x) changes from positive to negative at x = 0 there is a relative maximum at x = 0.The relative maximum point is (0, 2).

5 –3

y (–4, 0)

4

( 23 , 0)

–4 –3 –2 –1 0 1 2 3 4 x –4 –8 (0, –8) –12 5 49 – ,– –16 3 3

(

)

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 7

17

WORKED SOLUTIONS 2

f ( x ) = x 3 + x 2 − 5x − 5 f (0 ) = −5 ⇒ the y -intercept is (0, − 5). f ( x ) = 0 ⇒ x 3 + x 2 − 5x − 5 = 0 ⇒

f ′( x ) is undefined when ( x − 4 )2 = 0 or when x = 4 f ′ is decreasing on (−∞, 4 ) and (4, ∞) There are no relative extrema.

x 2 ( x + 1) − 5( x + 1) = 0 ⇒

f ′′( x ) = 12( x − 4 )−3 (1) =

( x + 1)( x 2 − 5) = 0 ⇒ x = −1, ± 5 ⇒

f ′′( x ) is undefined ( x − 4 )3 = 0 ⇒ x = 4 f is concave down on (−∞, 4 ) and concave up on (4, ∞) There are no inflection points.

(

the x -iintercepts are ( −1, 0 ) , − 5 , 0

(

and − 5 , 0

)

)

f ′( x ) = 0 ⇒ 3 x 2 + 2 x − 5 = 0 ⇒ (3 x + 5)( x − 1) = 0 ⇒ signs of f ' x

3

⎛ ⎞ f ⎜− 5 ⎟ = ⎝ 3⎠

40 27

+

–

+

5

and f (1) = −8

⎛ ⎞ f has a relative maximum point at ⎜ − 5 , 40 ⎟ 3 27 ⎝ ⎠ and a relative minimum point at (1, − 8 ) . f ′′( x ) = 6 x + 2 f ′′( x ) = 0 ⇒ 6 x + 2 = 0 ⇒ x

–

signs of f '' x =−1 3

4

–2

( 3

+

–4

signs of f ' x

+

x = 3 ⇒ x -intercept is ( 3, 0 ) .

1

–3

– 3

f ′( x ) = 4 (3 − x )3 ( −1) = −4 (3 − x )3 f ′(( x ) = 0 ⇒ −4 (3 − x )3 = 0 ⇒ x = 3 f has a relative minimum point at ( 3, 0 ) .

⎛ ⎞ at ⎜ − 1 , − 88 ⎟ 27 3 ⎝ ⎠

(–√5, 0)

–

f ( x ) = (3 − x )4 f (0 ) = 81 ⇒ the y -intercept is (0, 81). f ( x ) = 0 ⇒ (3 − x )4 = 0 ⇒

+

⎛ ⎞ f ⎜ − 1 ⎟ = − 88 ⇒ f has an inflexion point 27 ⎝ 3⎠

f ″( x ) = −12(3 − x )2 ( −1) = 12(3 − x )2 f ″( x ) = 0 ⇒ 12(3 − x )2 = 0 ⇒ + signs of f '' x =3 x 3 f ″ do oes not change signs at x = 3, so f does not have an inflection point.

y

(– 53 , 4027 )

signs of f '' x

2 4 6 8 10 12 x

–6 –4 –2 0 –2 –4 –6 –8

1

–3

≠0

y x=4 8 6 (0, –2) 4 2 y=1

f ′( x ) = 3 x 2 + 2 x − 5 x = − 5 ,1

12 ( x − 4 )3

(–1, 0) (√5, 0) 0 –1 1 2 –2 1 , 88 –4 (0, –5) 3 27 –6 –8 (1, –8) –10

x

)

y 10 8 6 4 2

f (x ) = x + 2 x −4

f ( x ) is undefined when x − 4 = 0 ⇒ vertical asymptote at x = 4 1

f (0) = − 1 ⇒ the y -intercept is 2

signs of f '' x 1⎞ ⎛ 0 , − ⎜ ⎟ 2⎠ ⎝

f ( x ) = 0 when x + 2 = 0 ⇒ x = −2 ⇒ signs of f ' the x − intercept is (−2, 0) (x − 4)

−6 ( x − 4 )2

x

(3, 0)

0

Horizontal asymptote at y = 1 or y = 1

f ′( x ) = ( x − 4 )(1) − ( x2 + 2)(1) =

+

–

+ –4

5

1 2 3 4 5 6 x

f (x ) = e

x

− e −x 2

= 1 (e x − e − x ) 2

f (0) = 0 ⇒ the y -intercept is (0, 0) f ( x ) = 0 ⇒ 1 (e x − e − x ) = 0 ⇒ x = 0 ⇒ 2

–

– –4

x -intercept is ( 0, 0 ) f ′( x ) = 1 (e x − e − x (−1)) = 1 (e x + e − x ) 2

2

f ′( x ) ≠ 0 for any x ⇒ f has no relative extrema.. f ″( x ) = 1 (e x + e − x (−1)) = 1 (e x − e − x ) f

2

2

″( x ) = 0 ⇒ 1 (e x 2

−e ) = 0⇒ x = 0

signs of f '' x

–

−x

+ 0

f has an inflection point at (0, 0) © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 7

18

WORKED SOLUTIONS When y = f ′(x) is increasing and y = f (x) is concave up the graph of y = f ″(x) is positive and when y = f ′(x) is decreasing and y = f (x) is concave down the graph of y = f ″(x) is negative.

y 10 8 6 4 2 (0, 0) –4 –3 –2 –1–20 –4 –6 –8

6

y 1 2 3 4 x

y = f ''(x)

y = f '(x)

–3 –2 –1 0

f (x ) = x 2 − 1

1 2 3 x

2

x +1

y = f(x)

Horizontal asymptote at y = 1 or y = 1

2

1

f (0) = −1 ⇒ the y -intercept is ( 0, −1) f ( x ) = 0 when x 2 −1 = 0 ⇒ x = −1, 1 ⇒ the x − intercepts are (−1, 0) and (1, 0) f ′( x ) = ( x

+ 1)(2 x ) − ( x 2 − 1)(2 x ) = 24 x 2 ( x + 1) ( x 2 + 1)2 ⇒ 4 x = 0 ⇒ x = 0 signs of f ' x

2

–

f ′( x ) = 0 f has a relative minimum m point at (0, −1) f ″( x ) = =

x =± ⎛

f ⎜⎜ − ⎝

0

2

4 ( x 2 + 1) ⎡⎣( x 2 + 1) − 4 x 2 ⎤⎦ ( x 2 + 1)4

= 4(12− 3x 3 ) 2

( x + 1)

f ″( x ) = 0 ⇒ 4(1 − 3x 2 )=0 ⇒ 1 3

When the graph of y = f ′(x) has a relative maximum or minimum the graph of y = f ″(x) has zeros. The graph of y = f ″(x) is positive when the graph of y = f ′(x) is increasing and the graph of y = f (x)is concave up. The graph of y = f ″(x) is negative when the graph of y = f ′(x) is decreasing and the graph of y = f (x) is concave down.

+

( x 2 + 1)2 ( 4 ) − ( 4 x ) ⎡⎣2( x 2 + 1)(2 x ) ⎤⎦ ⎡⎣( x 2 + 1)2 ⎤⎦

or ±

3⎞ 1 ⎟=− 3 ⎟⎠ 2

signs of f '' x

3 3

and f

–

+ √3

–3

–

y = f '(x)

y = f(x)

–3 –2 –1 0

⎛ ⎞ 3 , − 1 ⎟⎟ ⎜⎜ − 3 2 ⎝ ⎠

⎛ 3 1⎞ , ⎟⎟ . ⎝ 3 2⎠

and ⎜⎜

3

1

(–√33 , – 12 )

–1 (0, –1)

(1, 0) 1 2 3 4 x

( √33 , – 12 )

Exercise 7U 1

1 2 3 x

y = f ''(x)

y

(–1, 0) –4 –3 –2 –1 0

y

√3 3

⎛ 3⎞ 1 ⎜⎜ ⎟⎟ = − 2 3 ⎝ ⎠

f has inflection points at

When the graph of y = f ′(x) has a zero and changes from positive to negative the graph of y = f (x) has relative maximums and when the graph of y = f ′(x) has a zero and changes from negative to positive the graph of y = f (x) has a relative minimum.

When y = f (x) has a relative minimum or maximum, f ′(x) = 0 and so y = f ′(x) has an x-intercepts. When y = f (x) is increasing the graph of y = f ′(x) is positive and when y = f (x) is decreasing the graph of y = f ′(x) is negative. When y = f ′(x) has a relative minimum, f ″(x) = 0 and so y = f ″(x) as an x-intercept.

When the graph of y = f ″(x) has a zero and changes from negative to positive the graph of y = f ′(x) has a relative minimum and when the graph of y = f ″(x) has a zero and changes from positive to negative the graph of y = f ′(x) has a relative maximum. When the graph of y = f ′(x) has a zero and changes from positive to negative the graph of y = f (x) has relative maximums and when the graph of y = f ′(x) has a zero and changes from negative to positive the graph of y = f (x) has a relative minimum. When the graph of y = f ″(x) is positive the graph of y = f (x) is concave up and when the graph of y = f ″(x) is negative the graph of y = f (x) is concave down.

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Worked solutions: Chapter 7

19

WORKED SOLUTIONS y y = f(x)

–8 –6 –4 –2 0

5 y = f '(x)

f ( x ) = ( x − 1)4 f ′( x ) = 4 ( x − 1)3 (1) = 4 ( x − 1)3

2 4 6 8

f ′( x ) = 0 ⇒ 4 ( x − 1)3 = 0 ⇒ x = 1

x

f ″(( x ) = 12( x − 1)2 (1) = 12( x − 1)2 f ″(1) = 0 ⇒ second derivative test faails ⇒ use first derivative test. f ′(x) changes sign from negative to positive at x = 1 ⇒ relative minimum. signs of f ' – + f (1) = 0 x 1 relative minimum: (1, 0)

y = f ''(x)

Exercise 7V 1

2

3

4

f ( x ) = 3 x 2 − 18 x − 48 f ′( x ) = 6 x − 18 f ′( x ) = 0 ⇒ 6 x − 18 = 0 ⇒ x = 3 f ″( x ) = 6 ⇒ f ″(3) = 6 > 0 ⇒ relative minimum f (3) = −75 relative minimum : (3, − 75) f ( x ) = ( x 2 − 1)2 f ′( x ) = 2( x 2 − 1)(2 x ) = 4 x ( x 2 − 1) f ′( x ) = 0 ⇒ 4 x ( x 2 − 1) = 0 ⇒ x = 0, −1,1 f ″( x ) = ( 4 x )(2 x ) + ( x 2 − 1)( 4 ) = 12 x 2 − 4 f ″(0 ) = −4 < 0 ⇒ relativee maximum at x = 0 f ″( −1) = 8 > 0 ⇒ relative minimum at x = − 1 f ″(1) = 8 > 0 ⇒ relative minimum at x = 1 f (0 ) = 1; f ( −1) = 0; f (1) = 0 relative maximum: (0, 1) relative minimums: ( −1, 0 ) and (1, 0 ) f (x ) = x 4 − 4x 3 f ′( x ) = 4 x 3 − 12 x 2 f ′( x ) = 0 ⇒ 4 x 3 − 12 x 2 = 0 ⇒ 4 x 2 ( x − 3) = 0 ⇒ x = 0, 3 f ″( x ) = 12 x 2 − 24 x f ″(0 ) = 0 ⇒ second derivative test fails ⇒ usse first derivative test Since f ′( x ) does not change signs at x = 0 there is no relative extremum at x = 0. f ″(3) = 36 > 0 ⇒ relative minimum at x = 3 – – signs of f ' f (3) = −27 x 0 3 relative minimum: (3, − 27 ) f ( x ) = xe

x

f ′( x ) = ( x )(e ) + (e )(1) = e ( x + 1) x

x

x

f ′( x ) = 0 ⇒ e ( x + 1) = 0 ⇒ x = −1 x

f ″( x ) = (e x )(1) + ( x + 1)(e x ) = e x ( x + 2)

6

f (x ) =

1 = ( x 2 + 1)−1 x2 +1

f ′( x ) = −1( x 2 + 1)−2 (2 x ) = f ′( x ) = 0 ⇒ f ″( x ) = =

−2 x ( x 2 + 1)2

= 0 ⇒ −2x = 0 ⇒ x = 0

( x 2 + 1)2 ( −2) − ( −2 x ) ⎡⎣2( x 2 + 1)(2 x ) ⎤⎦ ⎡⎣( x 2 + 1)2 ⎤⎦ ( x 2 + 1)4

2 − 1) ( x + 1)3

= 2(32x

f ″(0) = −2 < 0 ⇒ relative maximum f (0) = 1 relative maximum : (0, 1)

Exercise 7W 1

2

3

Neither A nor C can be relative extrema because relative extrema do not occur at endpoints. A is neither an absolute nor a relative A extrema. B is an absolute minimum. C is an absolute maximum. Neither A nor D can be relative extrema because relative extrema do not occur at endpoints. A is A neither an absolute nor a relative extrema. B is a relative minimum C is an absolute maximum. D is an absolute minimum.

C

B C

B D

 f ( x )  ( x  2) for 0  x  4 3

f ( x )  3( x  2)2 (1)  3( x  2)2 f ( x )  0  3( x  2)2  0  x  2 f (0)  8 f (2)  0 f (4)  8

f ( −1) = − 1e

abolute maximum: 8

⎛ ⎞ relative minimum:⎜ −1, − 1e ⎟ ⎝ ⎠

2

−2( x 2 + 1) ⎡⎣( x 2 + 1) − 4 x 2 ⎤⎦

f ″( −1) = 1 > 0 ⇒ relative miniimum e

−2 x ( x 2 + 1)2

abolute minimum:  8

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Worked solutions: Chapter 7

20

WORKED SOLUTIONS 4

5

P   50  4  0  relative maximum

f ( x ) = 8 x − x 2 for − 1 ≤ x ≤ 7 f ′( x ) = 8 − 2 x f ′( x ) = 0 ⇒ 8 − 2 x = 0 ⇒ x = 4 f ( −1) = −9 f ( 4 ) = 16 f (7 ) = 7 absolute maximum: 16 absolute minimum: − 9

x 200  2 y  x 200  2(50) 100 The numbers are 100 and 50 3

y = width A = xy 2 x + 3 y = 400 ⇒ x = 200 − 3 y 2

⎛ ⎞ A = ⎜ 200 − 3 y ⎟ ( y ) = 200 y − 3 y 2 2 2 ⎝ ⎠ A′ = 200 − 3 y

f ( x ) = x − 3 x for − 1 ≤ x ≤ 2 3

2

2

f ′( x ) = 3 x 2 − 3 x = 3 x ( x − 1) f ′( x ) = 0 ⇒ 3 x ( x − 1) = 0 ⇒ x = 0, 1

x

3

A″ = −4

f (0 ) = 0 f (1) =

y

A′ = 0 ⇒ 200 − 3 y = 0 ⇒ y = 200

−5 2

f ( −1) =

x = length

⎛ ⎞ A″ ⎜ 200 ⎟ = −4 < 0 ⇒ relaative maximum 3 ⎝ ⎠

−1 2

⎛ ⎞ x = 200 − 3 y ⇒ x = 200 − 3 ⎜ 200 ⎟ = 100 2 2⎝ 3 ⎠

f (2) = 2 absolute maximum: 2

The dimensions are 100 ft by

ab bsolute minimum: − 5

200 3

ft.

2

Exercise 7Y

Exercise 7X 1

x = the first positive number y = the second positive number S =x+ y x + y = 20 ⇒ x = 20 − y S = 20 − y + y = 20 − y + y −1

S ʹ= −1 + 1 y 2 = 2

Sʹ = 0 ⇒

1 2 y

S ʺ ⎛⎜ 1 ⎞⎟ = ⎝4⎠

⎛1⎞ 4 ⎜ ⎟ ⎝4⎠

3

The numbers

x

S ″ ( 40 ) = 6 > 0 ⇒ relative minimum h = 32002 0 ⇒ h = 32000 = 20 2 x

40

The dimensions are 40 m by 40 m by 20 m. 2

C ( x ) = x 3 − 3 x 2 − 9 x + 30

 x  the first positive number     

C ′( x ) = 3 x 2 − 6 x − 9

y  the second positive number P  xy x  2 y 200  x 200  2 y

C ′( x ) = 0 ⇒ 3 x 2 − 6 x − 9 = 0 ⇒ 3( x + 1)( x − 3) = 0 ⇒ x = −1, 3 The only ciritcal number in [0,10] is 3. C ( 0 ) = 30 C (3) = 3 C (10 ) = 640 Three items should be produced to minimizze the cost.

P (200  2 y )( y )  200 y  2 y 2 P  200  4 y  P   0  200  4 y  0  y  50 P   4

x

S ″ = 2 + 256,000 x −3 = 2 + 256,3000

= −2 < 0 ⇒ relative maximum = 20 − 1 = 79 4 4 79 are and 1 4 4

x

2x 3 = 128,000 ⇒ x 3 = 64000 ⇒ x = 40

4 y3

x = 20 − y ⇒ x

2

4

−1

−1

h

2

−1 2

−3 2

x = length of square base h = height S = x 2 + 4 xh x 2h = 32000 ⇒ h = 32000

x2 ⎛ 32000 ⎞ S = x + 4 x ⎜ 2 ⎟ = x 2 + 128,000 x −1 ⎝ x ⎠ S ′ = 2x −128,000 x −2 = 2 x − 128, 2000 x S ′ = 0 ⇒ 2x − 128, 2000 = 0 ⇒ x

1 2

−1 = 0 ⇒ 1 = y ⇒ y = 1

Sʺ = − 1 y = 4

1 2 y

1

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 7

21

WORKED SOLUTIONS 3

s (t ) = t 3 − 12t 2 + 36t − 10 for 0 ≤ t ≤ 7 s ′(t ) = 3t 2 − 24 t + 36 s ′(t ) = 0 ⇒ 3t 2 − 24 t + 36 = 0 ⇒ 3(t − 2)(t − 6) = 0 ⇒ t = 2, 6 s (0 ) = −10 s (2) = 22 s (6) = −10 s (7 ) = −3 The maximum distance is 22

4

a

c

d ⎛ 3 ⎞ d ⎜ ⎟= dx ⎝ x 4 ⎠ dx

d

d dx



6x 4x

 2 x 3  x 2  6 x 2  2 x  1

4 4

 2x 3  2x 2 

e

d ⎡ x − 4 ⎤ ( x + 7 )(1) − ( x − 4 )(1) = = 11 2 dx ⎢⎣ x + 7 ⎥⎦ ( x + 7 )2 ( x + 7)

ABC and ADE are similar triangles, so

f

⎡⎣e 4 x ⎤⎦ = (e 4 x )( 4 ) = 4 e 4 x

10 − h r

d dx

g

d dx

⎡⎣( x 3 + 1)4 ⎤⎦ = 4 ( x 3 + 1)3 (3 x 2 ) = 12 x 2 ( x 3 + 1)3

h

d⎡ ln(2x + 3) ⎤⎦ = ⎜⎛ 1 ⎞⎟ (2) = 2 x⎣ 2x + 3 ⎝ 2x + 3 ⎠

i

d ⎡ ln x ⎤ = dx ⎢⎣ x 2 ⎥⎦

j

d ⎡ 4 x − 2x ⎤ d ⎡1 4 1 ⎤ 1 2 ⎢ ⎥= ⎢ (4 x − 2 x )⎥⎦ = 6 (8x − 2) = 3 x − 3 dx ⎣ 6 ⎦ dx ⎣ 6

k

d dx

= 10 ⇒ 10 − h = 5 ⇒ r = 30 − 3h 6

3

r

5

V= π r 2 h ⇒V= π ⎛⎜ 30 − 3h ⎞⎟ h or 5

⎝

⎠

A

10 cm 9π (100h − 20h2 + h3 ) 25 9π (100 − 40h + 3h2 ) 25 9π (−40 + 6h) = 18π (3h − 20) 25 25

= V dV = dh d 2V = dh 2

d

( x 2  1)(2 x 3  x 2  x ) 

 10 x 4  4 x 3  3 x 2  2 x  1

9π (100h − 20h2 + h3 ) 25

c

10 – h B

r

C

h D

E 6 cm

l m

3

dV dh 2

r

h =10

= 180 25

⎛1⎞ ( x 2 ) ⎜ ⎟ − (ln x )(2 x ) ⎝x⎠ = x − 2 x4 ln x ( x 2 )2 x

⎛ 10 ⎞ 30 − 3 ⎜ ⎟ ⎝ 3 ⎠ = 5

⎡⎣(3 x 2 + 1)(e x ) ⎤⎦ = (3 x 2 + 1)(e x ) + (e x )(6 x )

x d ⎡ 2e x ⎤ (e x − 3)(2e x ) − (2e x )(e x ) = −x 6e 2 ⎢ x ⎥= x 2 dx ⎣ e − 3 ⎦ ( e − 3) ( e − 3)

d ⎡ 3 dx ⎢⎣

2 x − 5 ⎤⎥ = ⎦

= n

=4

d dx

o

3

b

p( x ) = 4 x − 2x 2 = 4 x 2 − 2x 2

1

dp dx

−1

= 2x 2 − 4 x =

2 x

− 4x

−3

−1

−4

d2 p dx 2

c

dp dx

= −x 2 − 4 =

=0⇒

d2 p dx 2

2 x

x3

2

1

f ′( x ) = lim [2( x + h )

= −6 < 0 ⇒ relative maximum

(4x

d dx

b

d ⎛3 ⎜ dx ⎝

x

3

4

4 ⎞ = d ⎜⎛ x 3 ⎟ ⎠ dx ⎜⎝

⎞ 4 1 ⎟ = x3 ⎟ 3 ⎠

− 6( x + h )] − (2 x 3 − 6 x ) h→0 h 2 x 3 + 6 x 2 h + 6 xh 2 + 2h3 − 6 x − 6h − 2 x 3 + 6 x = lim h→0 h 6 x 2 h + 6 xh 2 + 2h3 − 6h = lim h→0 h h ( 6 x 2 + 6 xh + 2 h 2 − 6 ) = lim h→0 h 3

= lim(6 x 2 + 6 xh + 2h 2 − 6) h→0

= 6x 2 − 6 c

+ 3 x 2 − 2 x + 6 ) = 12 x 2 + 6 x − 2

d ⎡ ⎛ 1 ⎞⎤ ln ⎜ ⎟ = d ⎡ ln x −1 ⎤⎦ = ⎛⎜ 1−1 ⎞⎟ (− x −2 ) dx ⎢⎣ ⎝ x ⎠ ⎥⎦ dx ⎣ ⎝x ⎠ = ( x ) ⎜⎛ −12 ⎞⎟ = −1 x ⎝x ⎠

b

x ≈0.630

a

−1

⎦

( x + h )3 = x 3 + 3 x 2 h + 3 xh 2 + h3

0.630 thousand units or 630 units maximize the profit.

✗

1⎤

)2 ⎥⎥ = 32 (2x − 5) 2 (2)

a

− 4 x = 0 ⇒ x ≈ 0.630

Review exercise

(

= 2 xe 2 x ( x + 1)

radius is 4in and the height is 10 in. p( x ) = 4 x − 2x 2

d ⎡ ⎢3 2 x − 5 dx ⎢⎣ 3 2x − 5

⎡⎣ x 2 e 2 x ⎤⎦ = ( x 2 ) ⎡⎣(e 2 x )(2) ⎤⎦ + (e 2 x ) ( 2 x )

Maximum volume occurs when the

a

x

2

> 0 ⇒ relative minimum

= 30 − 3h ⇒ r 5

= 1 − 2 3ln x

= e x (3 x 2 + 6 x + 1)

dV = 0 ⇒ 9 (100 − 40h + 3h2 ) = 0 ⇒ dh 25 9 (10 − 3h)(10 − h) ⇒ h = 10 , 10 25 3 −180 d 2V = < 0 ⇒ relative maximum 25 dh 2 h=10 2

5

5

 ( x 2  1)(6 x 2  2 x  1)  (2 x 3  x 2  x )(2 x )

2

b

(3x −4 ) = −12x −5 = −x12

f ′( x ) = 6 x 2 − 6 f ′( x ) = 0 ⇒ 6 x 2 − 6 = 0 ⇒ 6( x + 1)( x − 1) = 0 ⇒ x = −1, 1 signs of f ' x

+

– –1

+ 1

f ( x )  0 for  1  x  1, so p 1 and q 1.

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 7

22

WORKED SOLUTIONS d

f ″( x ) = 12 x

e

f ″( x ) = 0 ⇒ 12 x = 0 ⇒ x = 0 signs of f '' x

iii

–

+

f ( x ) = 4 xe x

2

−1

2

−1

= 0 ⇒ 12x ( x − 2) = 0 ⇒ x = 0, 2

)(2 x ) ⎤ + (e x ⎦

2

−1

c

2

y−4 =− 4

− 1)

f ( x ) = 2 x 3 − 3 x + 1; tangent line parallel to y = 5x − 2 f ′( x ) = 6x2 − 3 6x − 3 = 5 ⇒ x = ± 2

f

2 3     3 

=

9−2 3 9

, f

7

5

b

 −2 3     3 

=

9+2 3 9

6

a

and

 −2 3   3

,

i

i

ii

s (t ) = 20t − 100 ln t , t ≥ 1. v (t ) = s ′(t )

v (t ) = 0 ⇒ 20 − 100 = 0 ⇒ t = 5 t

signs of v t 0

c

y 4 3 2 1

5

0 –1 –2

v (t ) 20  100 20  100t 1 t 100 2 t

Since 100  0 and t 2  0, v (t )  0 for all t  1. Therefore velocity is always increasing. 1 2 3 4 5 x

Review exercise 1

a

lim x →2

1 x −2 y

dy = 4 x 3 − 12 x 2 dx d2y = 12x 2 − 24 x dx 2

x 3 ( x − 4 ) = 0 ⇒ x = 0, 4 ⇒ x -intercepts are (0, 0 ) and ( 4, 0 ) dy = 4 x 3 −12 x 2 = 4 x 2 ( x − 3) dx dy = 0 ⇒ 4 x 2 ( x − 3) = 0 ⇒ x = 0, 3 dx d2 y = 12 x 2 − 24 x = 12x ( x − 2) dx 2 d2 y =0 dx 2 x = 0

= 36 > 0 ⇒ relative minimum x =3

+

v (t )  a (t )  100t 2 

y = f(x)

y = x 3 (x − 4) = x 4 − 4 x 3

d2 y dx 2

–

The particle is moving to the left on the interval (1, 5)

9+2 3    9

f ″(2) > 0 since the graph of f is concave up, f (2) = 0 and f ′(2) < 0 since the graph of f is decreasing

ii b

9−2 3    9

f ″(2) > f (2) > f ′(2)

a b

,

a

v (t ) = 20 − 100 t

2 3 3

The tangent line is parallel to y = 5 x + 2 at the points 2 3   3

+ 2

(4, 0) (0, 0) 0 –4 –3 –2 –15 1 2 3 4 x 10 15 20 (2, –16) 25 (3, –27)

1 12

1 (x 12

– 0

y 20 15 10 5

)( 4 )

= 2e x −1 ( 4 x 2 + 2) mtangent = f ′(1) = 12 mnormal = −

+

y (0)  0; y (2)  23 (2  4)  16 inflexion points: (0,0) and (2,  16)

; normal line at (1, 4 )

f ′( x ) = ( 4 x ) ⎡(e x ⎣

= 12x 2 − 24 x = 12 x ( x − 2)

signs of f '' x

0

f ″( x ) > 0 for x > 0 so f is concave up on (0, ∞ ). 3

d2 y dx 2 d2 y dx 2

y (3) = 33 (3 − 4 ) = −27 relative minimum point: (3, − 27)

4 3 2 1 –5 –4 –3 –2 –1–10 –2 –3 –4

1 2 3 4 5 x

x 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5

f(x) −2.0000 −2.5000 −3.3333 −5.0000 −10.0000 10.0000 5.0000 3.3333 2.5000 2.0000

lim 1 does not exist at 2 since the function x →2 x − 2 approaches −∞ from the left side of 2 and ∞ on the right side of 2.

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Worked solutions: Chapter 7

23

WORKED SOLUTIONS b

lim 1 x →3 x − 2 y 4 3 2 1 –5 –4 –3 –2 –1–10 –2 –3 –4

1 2 3 4 5 x

lim 1 = 1 = 1 x →3 x − 2 3−2 c

2 − 16 x −4

lim x x →4

y 8 6 4 2 –10 –8 –6 –4 –2–20 –4 –6 –8  16 x 4

lim x x 4

2

2 4 6 8 10 x

 lim ( x  4)( x  4) x 4

x 4

 lim( x  4)  4  4  8 x 4

d

+3 x −1

lim x x →1

2

y 16 12 8 4 –16–12 –8 –4–40 –8 –12 –16

+3 x −1

lim x x →1

2

4 8 12 16 x

x 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4 3.5

f(x) 2.0000 1.6667 1.4286 1.2500 1.1111 1.0000 0.9091 0.8333 0.7692 0.7143 0.6667

x 3.5 3.6 3.7 3.8 3.9 4.0 4.1 4.2 4.3 4.4 4.5

f(x) 7.5000 7.6000 7.7000 7.8000 7.9000

x 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5

b

i

−1

L′( x ) = 1 ( x 2 + 100) 2 (2 x ) + 2

1

− 1 2 ( x − 60 x + 1525) 2 (2 x − 60) 2 x x − 30

=

x 2 + 100

25 ft

30 – x

x

ii

x 2 − 60 x + 1525

z

y

10 ft

+

L′( x ) = 0 ⇒

x x 2 + 100

= 0 ⇒ x ≈ 8.57

+

x − 30 x 2 − 60 x + 1525

Signs of L′(x) signs of L'(x)

– 0

+ 8.57

L(x) has a minimum at x = 8.57 The stake should be placed 8.47 ft from the 10 foot post.

8.1000 8.2000 8.3000 8.4000 8.5000

f(x) −6.5000 −8.4000 −11.6333 −18.2000 −38.1000 42.1000 22.2000 15.6333 12.4000 10.5000

does not exist at 1 since the function

approaches −∞ from the left side of 1 and ∞ on the right side of 1. 2

a

i

x 2 + 10 2 = y 2 ⇒ y = x 2 + 100

ii

(30 − x )2 + 252 = z 2 ⇒ z = (30 − x )2 + 625 ⇒ z = x 2 − 60 x + 1525

iii

L ( x )  y  z  x 2  100  x 2  60 x  1525 1

1

 ( x 2  100) 2  ( x 2  60 x  1525) 2 © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 7

24

WORKED SOLUTIONS

8

Descriptive statistics

Answers

b

c

Frequency

y

4

12 10 8 6 4 2

a b

0 Red Blue Pink Purple Black x

2

Mean = 4 + 7 + 7 + 8 + 6 = 32 = 6.4 5 5 b The number that occurs most often is 8 a

c

Arrange the data in order of size. 2, 4, 4, 6, 7, 8, 11. The median is the middle member, 6 5, 7, 9, 11, 13, 15. The middle member is in between 9 and 11. 1 (9 + 11) = 10

i

ii

2

6, 8, 11, 11, 14, 17. The middle member is between the two number elevens. 11

iii

a c

2

Discrete. Continuous.

c

1 2

b d

Continuous. Discrete

a

2

b

17

c

2 3

Number of teachers

3

a

10 20 30 40 50 60 70 x Age (years)

Continuous

50

14

f 1 2 4 4 2 2 1 1

b

9 1 2

c

18 and 24

and 2.

170 ≤ t < 180

∑x 66 + 57 + 71 + 69 + 58 + 54 375 = = 6 = 62.5 kmh −1 n 6 ∑ x 1.79 + 1.61 + 1.96 + 2.08 7.44 = Mean = n = = $1.86 4 4

Mean =

a

Discrete Calls per day (x) 2 3 4 5 6 7 8 9 Totals

Mean = a c

5 4 3 2 1

18

b

b

4 y

0

24

5 mins

1

15 30 45 60 75 90 x Time in minutes

Continuous

8

10 15 20 25 30 35 40 45

a

y

a

< < < < < < <
0

= 3 x 2 ; 1 ⎛⎜ du ⎞⎟ = x 2 3 ⎝ dx ⎠

⌠ ⎡ 1 ⎛ du ⎞ 4 ⎤ 4 1 1⎛1 5 ⎞ ⎮ ⎢ ⎜ ⎟ u ⎥ dx = 3 ∫ u du = 3 ⎜ 5 u ⎟ + C ⎝ ⎠ ⌡ ⎣ 3 ⎝ dx ⎠ ⎦

t 2 − 9 dt ≈ 119 m

distance travelled =

7

x3 +C

+C

du

c

3 7

( x 3 + 1)4 dx ⇒ u = x 3 + 1;

3

3

7

x 3 + C=

⌠ 5x 4 − 3x ⎛ 5 2 1 ⎛ 1 ⎞⎞ dx ⌠ ⎮ 6 x 2= ⎮ ⎜ 6 x − 2 ⎜ x ⎟ ⎟ dx ⎝ ⎠⎠ ⌡ ⌡⎝ 5⎛1 3⎞ 1 = ⎜ x ⎟ − ln x + C 6⎝3 ⎠ 2 5 3 1 = x − ln x + C , x 18 2

s( t ) = ∫ v ( t ) dt = ∫ ( t 2 − 9 ) dt = 1 t 3 − 9 t + C

1

dt ≈ 240 cm3

20

2

0

18.4 e 20 dt ≈ 239 billions of barrels

1.5

= 1 (6)(6) + 1 ( 4 + 2)(2) = 24 m 2

t

∫ ∫ (1375t

1

g

⌠ 1= dx ⎮ ⌡ 2x + 3

h

 ln x dx   u   x   1    ln x  x   dx   

2 4 6 8 10 12 14 16 18

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

1 5 u 15

+C =

1 ln(2 x + 3) + C , 2

+ 1)5 + C

x >−3

2

ln x ; du 1 dx x    du      u    dx    dx  

  ud u 

1 (x 3 15

1 2 u C 2

1 (ln x )2  C , 2

x 0

Worked solutions: Chapter 9

12

WORKED SOLUTIONS i

u j

u 3x 2 1; du 6 x

(3x 2 1)(6 x )dx du dx

dx

1 2 u 2

u du

dx 1 (3x 2 2

C

∫

e

1)

 2e dx  u  e x  3; du  e x  x dx  e 3 x  2e dx  2  1  (e x )dx  x  x   e 3  e 3

12

2

f

 2ln u  C 2ln(e x  3)  C

∫3

1 2

[ ln(2 x + 1)]0

=

1 2

[ ln(2(2) + 1)] − 21 [ ln(2(0) + 1)]

2 x − 5dx= 3∫ (2 x − 5) dx

=

3 ⎡ ⎛ ⎞⎤ = 3 ⎢ 1 ⎜ 2 (2 x − 5) 2 ⎟ ⎥ + C ⎢⎣ 2 ⎝ 3 ⎠ ⎥⎦

3

3 2

= (2 x − 5) + C l

⌠ 1 dx ⎮ = ⌡0 2 x + 1

1 2

2 xe dx 2  xe  2x

2

2x

2

2

1 (ln 5) − 1 ln1= 2 2

2  xe

y 3

Area of region B

a

dx 

2

∫

=

1

2

( x 2 − 1)dx

1

2

a

1 2

 e du

1 eu 2

u

1 2x e 2

C

2

∫

=

C

2

1

2

⎡ ⎤ ( x − 1)dx = ⎢ 1 x 3 − x ⎥ 3 ⎣ ⎦1 2

⎡ ⎤ ⎡ ⎤ = ⎢ 1 (2)3 − 2 ⎥ − ⎢ 1 (1)3 − 1⎥ 3 3 ⎣ ⎦ ⎣ ⎦ ⎛ ⎞ = 2 −⎜− 2 ⎟ = 4 3 ⎝ 3⎠ 3

2

2 ⎡ ⎛1 3⎞ ⎤ 2 3 ∫0 (3x − 6)dx =⎢⎣3 ⎜⎝ 3 x ⎟⎠ − 6 x ⎥⎦ 0 =⎣⎡ x − 6 x ⎦⎤ 0

16

16 1 16 − ⎡ 12 ⎤ ⌠ 4 2 = = d t 4 t d t 4 ⎮ ⎢ 2t ⎥ ∫4 ⌡4 t ⎣ ⎦4

e

4

1

( x 2 − 1)dx − ∫ ( x 2 − 1)dx −1

2

π ∫ ( x 2 − 1)2 dx 1

f ′( x ) = 3 x − 2; (2, 6)

e2 ⌠ ⌠ = ⎮ 4x dx 4= ⎮ 1x dx 4 [ ln x ]1 ⌡1 ⌡1

1

∫ 6 xe

3 x 2 +3

0

dx ⇒ u = 3 x 2 + 3;

∫ 6 xe 0

2

3 x +3

5

⌠ ⎛ ⎞ dx = ⎮ ⎜ d u ⎟ e u dx ⌡ x = 0 ⎝ dx ⎠ =

∫

u =3

u

3 2

6

6

3

x 2 − 2x + 4

f ( x )dx = 20 5

Given ∫ f ( x )dx = 20 . 1

5

e du= ⎡⎣e ⎤⎦= e − e 3 u

5

1

a

x =1

u =6

∫

x 2 − 2x + C

− 2(2) + C =6 ⇒ 6 − 4 + C =6 ⇒ C =4

f (x ) =

= 6x ;

when x = 0 then u = 3(0)2 + 3 = 3 and when x = 1 then u = 3(1)2 + 3 = 6 1

3 2

= 3 (2)2 2

du dx

⎛ ⎞ 3⎜ 1 x 2 ⎟ − 2x + C 2 ⎝ ⎠

∫ (3x − 2 ) dx=

f ( x )=

2

= 4 ⎡⎣ ln e 2 ⎤⎦ − 4 [ ln1] = 4(2) − 4(0) = 8 d

2

1

d

16

e

∫

c

1 1 ⎡ 12 ⎤ ⎡ ⎤ ⎡ ⎤ 2 2 = ⎢= 8t ⎥ ⎢8(16) ⎥ − ⎢8(4) ⎥ ⎣ ⎦4 ⎣ ⎦ ⎣ ⎦ = 32 − 16 = 16

c

3 x

Area of region B

b

2

2

2

–2

3 3 = −4 ⎣⎡ (2) − 6(2) ⎦⎤ − ⎣⎡ (0) − 6(0) ⎦⎤ =

b

B

–3 –2 –1 0 A 1 –1

   dx  2  1  du  e u dx  4  dx  

ln 5 2

f (x) = x 2 − 1

1  du  u 2 x 2 ; du 4 x ;    x dx 4  dx  2x2

1

1

⎤ ⎤ ⎡ − 1) ⎥ = ⎢ 1 (3 x − 1)4 ⎥ 12 ⎦ −1 ⎣ ⎦ −1 4

= 16 − 256 = −20 12

u

 u  dx 

1 ⎡1 (3 x 3⎢ ⎣4

⎤ ⎤ ⎡ ⎡ = ⎢ 1 (3(1) − 1)4 ⎥ − ⎢ 1 (3( −1) − 1)4 ⎥ 12 12 ⎦ ⎦ ⎣ ⎣

x

 1  du  1  2    dx 2 du

k

(3 x − 1) dx = 3

−1

C

2

1

⌠ ⎮ ⌡1

1 4

f ( x )dx =

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

1 4

∫

5

1

f ( x )dx = 41 (20 ) = 5;

Worked solutions: Chapter 9

13

WORKED SOLUTIONS 5

5

dx ∫ ∫1 [ f ( x ) + 2]= 1

b

2

5

f ( x ) dx + ∫ 2 dx

a

1

a (t ) = v ′(t ) = 4 t − 11

= 20 + [ 2 x ]1 = 20 + (10 − 2) = 28 5

6

t0

v (t ) = 4 e + 2; s (0 ) = 8 b

1 2x − 1

c 3

dx = ln 5

a

1   1 dx ln(2 x  1)     2 1 2 x  1  1

b

      1 ln(2(k )  1)    1 ln(2(1)  1)  2 2      1 ln(2k 2

1 ln(2k 2

Particle moves left for a < t < b

∫

5

2

2t 2 − 11t + 12 dt ≈ 7.83 m

f ( x ) =x 3 − 2 ⇒ f ( −1) =−3

Use a GDC to solve : x 3 − 2 = 3 x x =2 f ( 2 ) = 23 − 2 = 6

 1)

The point is (2, 6) c

 1) ln 5  ln 2k  1 ln 5 

y 8 6 4 2

2k  1  5  2k  1  25  k  13

Review exercise f (x ) = 4 − x

y

2

2

−2

–2 –1 (–1, –3)

4

V = ∫  ( 4 − x 2 ) dx ≈ 107.2 2

right

f ′( x ) = 3 x 2 ⇒ m = f ′( −1) = 3 y + 3= 3( x + 1) or y= 3 x

k

k

4 left

2t 2 − 11t + 12 = 0 ⇒ x = 1.5, 4 a = 1.5 and b = 4

s ( t )= 2e 2 t + 2t + 6 k

1.5

+

v (t ) = 2t 2 − 11t + 12

2e 2(0) + 2(0) + C =8 ⇒ 2 + C =8 ⇒ C =6

⌠ ⎮ ⌡1

–

right

⎛ ⎞ s (= t ) ∫ ( 4 e + 2 )= dt 4 ⎜ 1 e 2 t ⎟ + 2 t + C 2 ⎝ ⎠ 2t = 2e + 2 t + C

1

+

signs of v

2t

2t

7

v (t ) = 2t 2 − 11t + 12 , t ≥ 0

0 –2 –4

y = 3x

(2, 6) y = f(x)

1

2

x

3 2

d

1 –4 –3 –2 –1 0 –1

1

2

3

4

∫

2

−1

⎡⎣3 x − ( x 3 − 2) ⎤⎦ dx = 6.75

x

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 9

14

WORKED SOLUTIONS

10

Bivariate analysis

Answers Skills check 4

Evaluate 25 = 32

2

33 = 27

3

73 = 343

4

⎛1⎞ ⎜ ⎟ ⎝2⎠

5

⎛3⎞ ⎜ ⎟ ⎝4⎠

6

0.0013 = 0.000 000 001

7

4

Rainfall (cm)

1

y 60

a

40 20

= 1 128 =

0 1999 2001 2003 2005 2007 2009 x

81 256

State the value of n in the following equations

5

b

Strong, negative.

c

As the years increase the rainfall decreases

a

y

n=4

80

2

3 = 243

n=5

60

3

7n = 343

n=3

5 = 625 5 (−4)n = −64

n=4 n=3

4

6

⎛1⎞ ⎜ ⎟ ⎝2⎠

n

=1 8

Science

2n = 16

n

0

3

20

n=3 b

40 60 80 100 x Mathematics

Strong, positive, linear.

6

y

Scatterplot of lean vs year

750

a

Positive

Strong

b

Negative

Weak

c

Negative

Strong

675

d

Positive

Weak

650

e

No correlation

a

i

positive,

ii

linear,

iii

strong,

a

Strong, positive.

b

i

negative,

ii

linear,

iii

strong,

b

The lean is increasing as the years increase.

c

i

positive,

ii

linear,

iii

Moderate.

d

i

No association, ii

iii

zero.

e

i

positive,

f

i

Negative, ii

a

If the independent and dependent variables show a positive correlation then as the independent variable increases the dependent variable increases.

b

lean

725

1975 1977 1980 1982 1985 1987 year

Non linear,

linear,

iii

weak.

non linear,

iii

strong.

If the independent and dependent variables show a negative correlation then as the independent variable increases the dependent variable decreases.

x

Exercise 10B 1

ii

700

a

Mean point = (mean of x, mean of y) = (96.7, 44.1)

b

y 70

Relationship between leaf length and width

60 Width (mm)

2

40 20

Exercise 10A 1

Scores

100

1

n

Rainfall in Tennessee

50

M

40 30 20 10 0

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

40

120 80 Length (mm)

160 x

Worked solutions: Chapter 10

1

WORKED SOLUTIONS 2

a

182 +173 +162 +178 +190 +161+180 + 172 +167 +185 10

i

2

a

= 175 cm

73 + 68 + 60 + 66 +75 + 50 + 80 + 60 + 56 +72 10

y 80 75 70 65 60 55 50 45

Weight (kg)

b

0

3

a

= 66 kg

b

M

Increase

sum of sales number of sales

= 528 = 75.4 7

Note the values of m and b in the equation y = mx + b are approximate.

y 120

150

160 170 180 Height (cm)

190

200

x

100 80 60 40 0 160

14 12 10 8 6 4 2

e Mean Point

4

6

x

8

Strong, positive

d

An increase in the number of hours spent studying mathematics produces an increase in the grade.

280

x

Approximately 70 houses.

70

72

74

76

78

80

12.3

9.5 7.7 6.1

4.3

2.3

The slope is –0.3. As a student plays one more day of sport per year they do 18 mins less homework per week. The y-intercept is 40, which means that the average student who does no sport does 40 hours of homework per week.

2

Exercise 10C Temperature, (x) °F Percentage of diseased leaves, (y)

240 200 Price ($000)

Exercise 10D 1

2

y = –x + 300

20

c

The slope is 6. For every criminal a person knows, they will generally have been convicted of 6 more crimes. The y-intercept is 0.5, which means that people who do not know any criminals will, on average, have been convicted 0.5 times.

3

The slope is 2.4. For every pack of cigarettes smoked per week a person will, on average, take 2.4 more sick days per year. The y-intercept is 7, which means that the average person that does not smoke has 7 sick days per year.

y Percentage diseased

7

140

Hours

1

Mean sales =

c and d

y

0

= 1540 = 220

The mean number of sales is estimated at 75.4

Mean point = (mean of x, mean of y) = (4, 6.67)

b

sum of prices number of prices

The mean house price is $220 000

Sales figure

ii

Mean house price =

12.3

4

2.3 80

70

The y-intercept is –5, which means that –5 people visited his shop in year zero, the y-intercept is not suitable for interpretation.

x

Temperature

a

( x , y ) = (75, 7.03)

b

y = –0.96x + 79

c

% diseased = (–0.96 × Temperature) + 79 % diseased = (–0.96 × 79) + 79 = 3.2%

The slope is 100. 100 more customers come to his shop every year.

5

The slope is 0.8. Every 1 mark increase in mathematics results in a 0.8 increase in science. The y-intercept is –10 which is not suitable for interpretation as a zero in mathematics would mean a –10 in science.

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 10

2

WORKED SOLUTIONS

Exercise 10E 1

a

fairly constant and therefore extrapolating with a linear function is unsuitable.

y

5

Concentration

14

Revisit the data from the leaning tower of Pisa.

12

a

(1981, 694)

10

b

y

8 6

725 lean

4 2 1

2 3 4 5 Time (hours)

6

x

650 1975 1977 1980 1982 1985 1987 year

b

y = 1.84 x + 1.99

c

Concentration after 3.5 hours: y = 1.99 + 1.84 × 3.5 = 8.43

a

y = 9.32x – 17 767

e

Lean = (9.32 × year) – 17 767 = (9.32 × 1990 ) – 17 767 = 780

Cost ($1000)

25

Exercise 10F

20

1

r = 0.863. There is a strong, positive correlation.

2

a

0.789

b

Strong, positive correlation.

c

The income increases as the number of years of education increases.

a

0.907

15 10 5 0

1

2

3 4 Age (yrs)

5

6

7

x

b

y = –2.67x + 28.1

c

Cost = (–2.67 × Age) + 28.1 = (–2.67 × 4.5) + 28.1 = MYR16 085

b

The stopping distance increases, as the car gets older.

d

The relationship may not be linear. Old cars are often more expensive after 50 yrs than when new.

c

Strong positive correlation.

a

–0. 887

b

Strong, negative correlation.

c

Yes, Kelly’s grade would increase if the chat time decreased.

a

0.026

b

Positive, weak correlation.

c

No. Mo’s grade would not increase if the game time decreased.

6

r = 0.994. Strong, positive correlation.

a

3

4

y

Hours of exercise

10 8

5

6 4 2 0

2 4 6 8 10 12 14 Months of membership

x

b

y = –0.665x + 9.86

c

Hours of exercise = (–0.665 × months of membership) + 9.86 = (–0.665 × 3) + 9.86 = 7.865 hrs.

✗

Review exercise 1

a

ii

2

a

b

c

iii

d

i

y 60 50

No. The equation gives –6.1 hrs of exercise!

Fifty years = 600 months, and the line would predict Sarah’s height at 50 years to be about 302 cm = 3.02 meters. Clearly there is a major difficulty with extrapolation. In fact, most females reach their maximum height in their mid to late teens, and from then on, their height is

v

b

Fuel

d 4

x

d

y 30

3

700 675

0

2

Scatterplot of lean vs year

750

40 30 20 10 0

c

200

600 400 Distance

800

x

32. See the dotted lines on the graph.

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 10

3

WORKED SOLUTIONS 3

5

y

a, c

a, c, f

13.6 12.8 Height (cm)

Time (seconds)

13.2 12.4 12.0

Mean point

11.6 11.2 10.8

b

d

20

50

30 40 Age (years)

x

Mean Age =

sum of ages number of policemen

Mean time =

sum of times number of policemen

(4, 30) d i r = 0.986 ii (very) strong positive correlation e y = 1.83x + 22.7 g Height = (1.83 × week number) + 22.7 = (1.83 × 4.5) + 22.7 = 30.9 cm h Not possible to find an answer as the value lies too far outside the given set of data.

10

=

120 10

= 12

Mean age = 34 years. Mean time = 12 secs. Approximately 11.7 secs. y

6

c d 2

a b

3

a c

4

a

Behaviour problems

35

x

Time, minutes

As the time increases, the number of push-ups decreases. y = –1.29x + 9 r = –0.929. There is a strong, negative correlation. w = –22.4 + 55.5h w = –22.4 + 55.5 × 1.6 = 66.4 kg r = 0.785 b y = 30.7 + 0.688x IQ = 30.6 + (0.688 × 100) = 99.4. This should be reasonably accurate since the product moment correlation coefficient shows fairly strong correlation.

a

b c d e f g

Test 2

a b

20 10 0

c d e

20

40 Test 1

60

80

x

Positive, strong. Students with a high score on test 1 tend to have a high score on test 2. y = 0⋅50x + 0⋅48 Test 2 Score = (0.50 × Test 1 score) + 0.48 = (0.50 × 40) + 0.48 = 20.48

20 15

0

7

30

25

5

y 40

30

10

50

b

y 40

0

b

1 2 3 4 5 6 7 8 x Week

b

= 340 = 34

Number of push ups

a

L

0

Review exercise 1

y 38 36 34 32 30 28 26 24 22

c d

1

2 3 4 5 Agreeableness

6 x

Behavior problems decrease –0.797 Strong, negative correlation. Teenagers who were more agreeable tended to have fewer behavior problems. y = –10.2x + 51.0 Number of behavior problems = (–10.2 × Agreeableness score) + 51.0 = (–10.2 × 4.5)+51.0 = 5.1 y = 10.7x + 121 (3sf) i Every coat on average costs $10.66 to produce, ii When the factory does not produce any clothes then x = 0, it has to pay costs of $121. Cost = (10.7 × 70) + 121 = $870 19.99x > 10.66x + 121 9.33x > 121 x > 12.969 13 coats should be produced in one day in order to make a profit.

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 10

4

WORKED SOLUTIONS

11

Trigonometry

Skills check 1

x° + 49° + 41° = 180° x = 90° x° + 2x° + (x − 20)° = 4x° − 20° = 180° x = 50° x° + 56° + 56° = 180° x = 68° (4x)° + (x + 20)° + (x + 20)° = 6x° + 40° = 180° x = 70 °

a b c d

3

e

x 2 = (2.4)2 + (5.6)2 = 37.12

f

x = 37.12 ≈ 6.09 x 2 + (19)2 = (24)2

7

+ ( 5 x − 1) = ( x 2 + 1) 4 x 2 + 25 x 2 − 10 x + 1 = x 4 + 2 x 2 + 1 x 4 − 27 x 2 + 10 x = 0 using GDC, x = 5

(2x )

2

tan B =

2

1

cos40 = 3

4

→B=

→c=

37 cos 40

4 .5 b  →b 4.5

cos B =

2

2

sin 45 =

≈ 53.1°

sin 30 = sin 60 =

sin B = 5

b 11 c

 c →

tan 35 11 sin 35

2

tan A =

8 .5 9 .7

→ A = tan

tan B = 9.7 → B 8 .5

−1 ⎛ 8.5 ⎞

⎜ ⎟ ≈ 41.2° ⎝ 9 .7 ⎠ = tan −1 ⎛⎜ 9.7 ⎞⎟ ≈ 48.8° ⎝ 8 .5 ⎠

9 9 = sin 45 ⎛ 1 ⎞ ⎜ ⎟ ⎝ 2⎠

a 4 .5 b 4 .5

(5 2 ) + b 2

e

→ b = 4.5 sin 60 = 9 ⎛⎜⎜

)

2

(

2

= (10 ) 2

(

tan A = 5

2 5 2

3⎞ ⎟ 2 ⎝ 2 ⎟⎠

→ a2 = 4 3 6 ⎞ ⎟ ⎝4 3⎠

)

2

=9

3 4

cm

− 62 = 12

= 30°

2

→ b 2 = (10 ) − 5 2 b = 50 = 5 2 cm

2

= 9 2 cm

→ a = 4.5 sin 30 = 2.25 cm

B = 180 − 90 − 30 = 60°

 19.2 cm

c = 166.34 ≈ 12.9 cm

c=

4 3

53.1°

2

= 60°

⎜ ⎟ ⎝2⎠

a = 12 = 2 3 cm cos A = 6 → A = cos−1 ⎛⎜

2

c 2   8.5    9.7   166.34

9 9 → c

(

B = 180 − 35 − 90 = 55° 11 ≈ 15.7 cm tan35 = 11 → b = sin35 =

6

B=

⎝ 60 ⎠ sin −1 ⎛⎜ 48 ⎟⎞ ≈ ⎝ 60 ⎠

→ B = cos

a 2 + 62 = 4 3

d

 a  1296 36 cos A = 48 → A = cos−1 ⎛⎜ 48 ⎞⎟ ≈ 36.9° 60 48 → 60

=

⎝2⎠

−1 ⎛ 1 ⎞

A = 180 − 60 − 90 = 30 °

c

≈ 48.3 cm

2

1 2

a tan 45  tan 45 9 cm → a 9

4.5sin 55  3.69 cm

2

2

12 24

B = 180 − 45 − 90 = 45°

b

a + 48 = 60 → a = 60 − 48 = 1296 2

≈ 67.4°

sin A = 12 = 1 → A = sin −1 ⎛⎜ 1 ⎞⎟ = 30° 24

⎝ 20 ⎠ cos−1 ⎛⎜ 12 ⎞⎟ ⎝ 20 ⎠

⎜ ⎟ ⎝ 10 ⎠

b = 432 = 12 3 cm

A = 180 − 55 − 90 = 35° cos 55 = a → a = 4.5 cos 55 ≈ 2.58 cm sin55

→ B = tan

122 + b 2 = 24 2 → b 2 = 24 2 − 122 = 432

a

2

B = 180 − 40 − 90 = 50° tan 40 = a → a = 37 tan 40 ≈ 31.0 cm 37 37 c

−1 ⎛ 24 ⎞

of the questions in this section can be answered in your head, if you remember the patterns of these special right triangles.

sin A = 12 → A = sin −1 ⎛⎜ 12 ⎟⎞ ≈ 36.9° cos B

⎝ 24 ⎠

24 10

Note: many

122 + b2 = 202 → b2 = 202 − 122 = 256 b = 256 = 16 cm 20 = 12 20

24

Exercise 11B

Exercise 11A 1

2

2

tan A = 10 → A = tan −1 ⎛⎜ 10 ⎞⎟ ≈ 22.6°

x = (24) − (19) = 215 x = 215 ≈ 14.7 2

2

)

2

= 100 − 50 = 50

= 1 → A = tan −1 (1) = 45°

B = 180 − 90 − 45 = 45° 2 2 x = ( 8 ) + ( 8 ) = 128 2

x = 128 = 8 2 cm tan30 =

1 3

=

8 y +8

→ y +8=8 3

y = 8 3 − 8 cm sin30 = 1 = 8 → z = 16 cm z 2

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 11

1

WORKED SOLUTIONS 3

a

3 2

sin60 =

x +2 x2 − 4

=

=

1 x −2

x 3 −2 3 =2 → x 3 =2+2 3 → x = b

→

22 3 3

3 AC  x  2 

2+4 3 ⎛ 1 ⎞ ⎜ ⎟ 3 ⎝ 3⎠

AC =

tan 45 = 1 =

 2

22 3 3



2+2 3 3 2 3 3

4x − 1 x2 + 2

=

2+4 3 3

3

cm

y

→ x 2 + 2 = 4x − 1

B

i 2

AB

2

If x = 1, AB = 2 ( 4 (1) − 1) = 2 ( 3 ) = 3 2 cm If x = 3, AB = 2 (4(3) −1) = 2 (11) = 11 2 cm

w 2 = 4 2 + 9 2 = 16 + 81 = 97 → w = 97 w  x  w 2 97 2 194 x y

sin65

=

194 z

x  z

→ z=

x = 3 sin 35 ≈ 1.7207

cos35

y = 3 cos 35 ≈ 2.457456 2

194 → y = tan 65

194 y

A x → 3 y = → 3

sin35 =

AC 2 = x 2 + ( y + 2 ) → AC ≈ 4.778

sin 45

tan65 =



35°

→ x 2 − 4x + 3 = 0 x =1, 3 1 4x  1 → AB = 2 ( 4 x − 1) sin 45  

5

→

9 12.5

 tan 1  9  2  12.5  9    1 →  2tan     71.5  12.5  tan     12.5 →   tan 1  12.5  9 2 2  9  12.5    1 →  2tan     108.5  9  5 x C 2

24 3 3



4

x +2 AC

tan60 = 3 =

tan    

→ 3 ( x − 2) = 2

sin 

194 sin 65

x AC

 1.7207 →   21.1 4.778

4.78 km, N21.1°W

w ≈ 9.8 cm, x ≈ 13.9 cm, y ≈ 6.5 cm, z ≈ 15.4 cm

6

Exercise 11C 1

a b

40° A

a

ˆ  tan AED

28 8

7

ˆ  tan EBA

28 20

ˆ = → EBA

A BX AB

cos47 =

AX = AX → AX = 35 cos 47 ≈ 23.8699 35 AB YC YC = → YC = 15 sin15 ≈ 3.882 15 BC

cos15 =

9°

x

120 tan 9

≈ 758 m

Ship

Z

sin 47 =

sin15 =

120

tan9 = 120 → x =

C

47°

54.5°

Observer

x

15° Y

35 km

ˆ  EBA ˆ  51.5 180  AED

4

B

15 km

ˆ  AED ˆ (alternate angles), so EAB ˆ  EBA ˆ 180  EAB ˆ AEB 3

→ x = 70 tan 40 ≈ 58.737

X

ˆ = tan −1 ⎛⎜ 28 ⎞⎟ ≈ 74.1° → AED ⎝ 8 ⎠ −1 ⎛ 28 ⎞ tan ⎜ ⎟ ≈ ⎝ 20 ⎠

x 70

height = x + 12 ≈ 70.7 m

AE = 282 + 82 = 848 ≈ 29.1 cm BE = 282 + 20 2 = 1184 ≈ 34.4 cm

b

E 12

70

12

tan 40 =

ˆ  38.9  180  2 BAC 2

B

x

h = 152 − 52 = 200 = 10 2 cm ˆ  5 → BAC ˆ = cos−1 5 ≈ 70.5° cos BAC 15 15 ˆ  BCA ˆ ˆ 180  BAC ABC

BY BC

=

=

BX 35

BY 15

→ BX = 35 sin 47 ≈ 25.597

→ BY = 15 cos15 ≈ 14.4889

AZ = BX + BY ≈ 40.086 CZ = AX − YC ≈ 19.988

AC = AZ 2 + CZ 2 ≈ 44.793

b a

18

ˆ = tan −1 ⎛⎜ AZ ⎟⎞ ≈ 63.498° ACZ ⎝ CZ ⎠

44.8 km, bearing approx. (180 + 63.5)° = 243.5° 25

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 11

2

WORKED SOLUTIONS 8

tan17 = a

X

Y

tan55 =

t=

b

≈ 4.01 s

ˆ tan HAD

b

ˆ tan ABE

c

HA = 9 2 + 24 2 = 657 18 ˆ tan GAH

→ a = 95 tan 35 ≈ 66.5197 → b = 95 tan 55 ≈ 135.674

d

C

10 tan 5 tan 17 − tan 5

12 a

9 24 9 18

ˆ = tan −1 9 ≈ 20.6° → HAD 24 ˆ = tan −1 → ABE

9 18

≈ 26.6°

657

X is 135.7 m tall, Y is (135.7 + 66.5) = 202.2 m tall 9

10 and t in the diagram represent 10v and tv (the distance travelled in 10 seconds and in t seconds).

t ( tan17 − tan 5 ) = 10 tan 5

55° 95

tan35 =

→ h = t tan17

→ ( t + 10 ) tan 5 = t tan17

35°

a 95 b 95

h t

ˆ = tan −1 ⎛⎜ → GAH

18 ⎞ ⎟ ⎝ 657 ⎠

d

T

x

≈ 35.1°

DG = 9 2 + 182 = 405 24 ˆ tan AGD 405

B

ˆ = tan −1 ⎛⎜ → AGD

24 ⎞ ⎟ ⎝ 405 ⎠

24°

≈ 50.0°

Exercise 11D 1 240

a b c e

(cos74, sin74) → (0.276, 0.961) (cos90, sin90) → (0, 1)

a

cos−1 0.408 ≈ 66° or sin−1 0.913 ≈ 66°

b

cos−1 0.155 ≈ 81° or sin−1 0.988 ≈ 81°

c

cos−1 0.707 ≈ 45° or sin−1 0.707 ≈ 45°

d

cos−1 0.970 ≈ 14° or sin−1 0.242 ≈ 14°

a

A=

b

A=

c

A=

d

A=

d 72°

2

A

tan72 = tan24 =

x + 240 → x = d tan72 d x → x = d tan24 d

− 240

d tan 24 = d tan 72 − 240 → d ( tan 72 − tan 24 ) = 240 d=

240 tan 72 − tan 24

≈ 91.2 m

3

10 h

A

20

x

B

1

tan 40  h → x = h − 20 tan 40 x  20 h h tan55 = → x = x tan 55 h h = − 20 → h ⎛⎜ 1 − 1 ⎞⎟ tan 55 tan 40 tan 55 ⎠ ⎝ tan 40

h=

20 ⎛ 1 ⎜ ⎝ tan 40

−

1 2 1 2 1 2

(cos70)(sin70) ≈ 0.161

1 2

(cos30)(sin70) ≈ 0.217

(cos38)(sin70) ≈ 0.243 (cos24)(sin70) ≈ 0.186

Investigation – obtuse angles

55°

40°

(cos20, sin20) → (0.940, 0.342) (cos17, sin17) → (0.956, 0.292) (cos60, sin60) → (0.5, 0.866)

1 ⎞ ⎟ tan 55 ⎠

≈ 40.7 cm

(–0.766, 0.643)

(0.766, 0.643) 140°

40°

2 (–0.906, 0.423)

= 20

155° 25°

(0.906, 0.423)

3 (–0.375, 0.927)

(0.375, 0.927) 68°

11 5° A

10

tan5 =

h t + 10

B

112°

h

17° t

→ h = ( t + 10 ) tan 5

Exercise 11E 1

a

B (cos30, sin30) → B (0.866, 0.5), C (− 0.866, 0.5)

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 11

3

WORKED SOLUTIONS b c d e 2

a b

4

B (0.545, 0.839),

sin 84.056 c

B (0.087, 0.996),

sin 40 a

cos (−0.332) ≈ 109.4° → 180 − 109.4 = 70.6° cos−1 (−0.955) ≈ 162.7° → 180 − 162.7 = 17.3°

sin15 ≈ 0.2588, 180 − 15 = 165°

b

sin36 ≈ 0.5878, 180 − 36 = 144°

c

sin81 ≈ 0.9877, 180 − 81 = 99°

d

sin64 ≈ 0.8988, 180 − 64 = 116°

a

sin 0.871 ≈ 60.6°, 180 − 60.6 = 119.4°

b

sin−1 0.436 ≈ 25.8°, 180 − 25.8 = 154.2°

c

sin−1 0.504 ≈ 30.3°, 180 − 30.3 = 149.7°

d

sin−1 0.5 ≈ 30°, 180 − 30 = 150°

2

=

sin 125 c

2

sin 55 4.5

→c=

4.5 sin 84.056 sin 55

sin 15 60

→a=

→c=

sin 15 60

=

60 sin 40 sin 15

sin 110 5 .8

→ a = 5.8 sin 27 ≈ 2.80 cm

sin 43 b

=

sin 110 5 .8

→ b = 5.8 sin 43 ≈ 4.21 cm

sin 110

sin 110

ˆ = 180 − 68.2 − 68.2 = 43.6° YXZ =

sin 68.2 XY

→ XY =

20 sin 68.2 sin 43.6

3

A

2

40°

tan117.5 ≈ − 1.92

c

tan137.7 ≈ − 0.910

C

d

tan45 = 1

a

tan  0.738 → y = 1.09x, θ ≈ 48°

ˆ = 180 − 75 = 105° ABT ˆ = 180 − 40 − 105 = 35° ATB

b

tan

sin 35 2

=

sin 105 AT

c

tan

sin 35 2

=

sin 40 BT

d

tan

e

y = − 0.75x, θ = 180° − α,

B 75°

0.674 0.882  → = 1.87x, θ ≈ 62° 0.471  0.942 → y = −2.80x, θ ≈ 110° 0.336  1.64 → y = −1.21x, θ ≈ 129° 1.35

b

Bˆ = 180 − 40 − 72 = 68° sin 40 sin 72 = a 2 .5

→ a = 2.5 sin 40 ≈ 1.69 cm

sin 68 sin 72 = b 2.5

→ b = 2.5 sin 68 ≈ 2.44 cm

sin 72

sin 72

≈ 2.24 km

K

35

ˆ = 180 − 50 − 36 = 94° AFK sin 94 35

=

sin 36 FK

sin50 =

→ a = 24 sin 47 ≈ 17.7 cm ≈ 18.5 cm

2 sin 40 sin 35

≈ 3.37 km

50°

36°

Cˆ = 180 − 47 − 83 = 50°

→c

→ BT =

2 sin 105 sin 35

F

A

Exercise 11G sin 83

→ AT =

h

1.59

= 24 sin 50 sin 83

T

4

tan  3.76 → y = 2.36x, θ ≈ 113°

sin 50 sin 83 = c 24

≈ 26.9

XY = XZ ≈ 26.9 cm

b

sin 47 sin 83 = a 24

≈ 189 9 190.cm

=

tan56.3 ≈ 1.50

a

≈ 149 cm

60 sin 125 sin 15

a

f

≈ 5.46 cm

sin 27 a

sin 43.6 20

−1

=

Cˆ = 180 – 27 – 43 = 110°

e

tan   0.6 ⇒ α = 36.9 0.8 ⇒ θ = 180 − 36.9 = 143°

1



Aˆ = 180 – 15 – 125 = 40°

d

−1

a

4.5



Cˆ ≈ 180 – 55 – 40.9 ≈ 84.1°

B (0.974, 0.225),

d

3.6 sin 55 → sin Bˆ  4.5

sin Bˆ 3.6



→ Bˆ sin 1  3.6 sin 55   40.9 

Exercise 11F 1

sin 55 4.5

c

B (0.707, 0.707),

cos−1 (−0.903) ≈ 154.6° → 180 − 154.6 = 25.4° cos−1 (−0.769) ≈ 140.3° → 180 − 140.3 = 39.7°

c 3

B (cos57, sin57) → C (− 0.545, 0.839) B (cos45, sin45) → C (− 0.707, 0.707) B (cos13, sin13) → C (− 0.974, 0.225) B (cos85, sin85) → C (− 0.087, 0.996)

h FK

→ FK = 35 sin 36 ≈ 20.6227 sin 94

→ h = FK sin 50 ≈ 15.8 m

Investigation – ambiguous triangles 1

sin 32 3

=

sin C 5

5 sin 32 → sin C = 3

→ Cˆ = sin −1 ⎛⎜ 5 sin 32 ⎞⎟ ≈ 62.0° 1 ⎝

3

⎠

Cˆ2 = 180 − Cˆ1 ≈ 118° The angles are supplementary.

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 11

4

WORKED SOLUTIONS 2

Bˆ1 = 180 − 32 − Cˆ1 ≈ 86.0° 3 sin B1 sin 32 ≈ 5.65 cm = sin B1 → AC = 3

C 2 = 180 − 70 − B2 ≈ 7.8°

sin 32

AC

Bˆ 2 = 180 − 32 − Cˆ2 ≈ 30.0° sin 32 3

=

sin B2 AC

g

3 sin B → AC = sin 322 ≈ 2.83 cm

1

a

=

sin C1 7

7 sin 30 4

→ sin C1 = ⎝

h

4

sin B1 b1

b

sin 56 45

sin 30

sin 56 45

2

→ a1

17sin A1 sin50

a

b

19.0 cm

sin 50 17



sin A2 a2

→ a2

sin 20 2 .5

=

sin B1 6 .8

→ sin B1 = ⎝

17 sin A2 sin 50

2 .5

sin A1 a1

→ a1 =

8.0 cm

d

= =

sin A2 a2 sin C 25

⎠

≈ 7.3 cm

2.5 sin A → a2 = sin 20 ≈ 5.5 cm 2

→ sin C =

c

3

33

sin A2 a2

=

33 sin 70 25

=

b1 sin B 28

f

sin 70 25

=

sin B1 26

8 10

45 sin A2 sin 56

≈ 10.4 cm

⎝ 10 ⎠

→ BCE = sin −1 ⎛⎜ 10 ⎟⎞ ≈ 53.1° 8

⎝

⎠

= sin BDC → BDC = sin −1 ⎛⎜ 10 ⎝

B

10 sin BCD ⎞ ⎟ ≈ 28.1° 17 ⎠

A 40°

⎠

230° 20 km

sin 42

L

→ sin B1 =

26 sin 70 25

b

B

25

X

→ c1 =

230°

16

⎠

25 sin C1 sin 70

B2 = 180 − B1 ≈ 102.2°

A 40° 20

C1 = 180 − 70 − B1 ≈ 32.2° sin C1 c1

→ a2 =

ABD = 180 − EAB − BDC ≈ 98.8° CBD 180  BCD  BDC  25.1 Given side BD = 17 m in triangle ABD and angle Dˆ = 28.1°, and side AB = 10, then there are 2 possible triangles fitting this data, namely DBA and DBC.

→ sin B ≈ 1.05 → triangle does not exist

⎝

=

≈ 45.5 cm

DE 2 + 82 = 172 → DE = 172 − 82 = 15 m AEB = 90° cos EAB = 6 → EAB = cos−1 ⎛⎜ 6 ⎞⎟ ≈ 53.1°

→ B1 = sin −1 ⎛⎜ 26 sin 70 ⎞⎟ ≈ 77.8° sin 70 25

45 sin A1 sin 56

BE 2 + 62 = 102 → BE = 102 − 62 = 8 m

B1  180  42  30.5  107.5 33sin B1 sin 42 47.0 cm  sin B1 → b1

e

→ a1 =

a

25 sin 42 33

→ C = sin −1 ⎛⎜ 25 sin 42 ⎞⎟ ≈ 30.5° ⎝

sin A1 a1

=

sin BCD 17

B2 = 180 − B1 ≈ 111.5° A2 = 180 − 20 − B2 ≈ 48.5° sin 20 2 .5 sin 42 33

→ c=

BCD = 180 – BDC ≈ 126.9°

6.8 sin 20 2 .5

2.5 sin A1 sin 20

=

sin BCE =

A1 = 180 − 20 − B1 ≈ 91.5° =

22 sin C ≈ 29.5 cm sin 45 50 sin 56 → sin C1 = 45 sin −1 ⎛⎜ 50 sin 56 ⎞⎟ ≈ 67.1° ⎝ 45 ⎠

sin C c sin C1 50

=

BCE = EAB ≈ 53.1°

→ B1 = sin −1 ⎛⎜ 6.8 sin 20 ⎞⎟ ≈ 68.5° sin 20 2 .5

 26.7

10

C 2 = 180 − C1 ≈ 108.9° Aˆ 2  180  50  Cˆ2  21.1

c

 

22

CE 2 + 82 = 102 → CE = 102 − 82 = 6 m

Aˆ 1  180  50  Cˆ1  58.9 sin A1 a1

 

A2 = 180 − 56 − C 2 ≈ 11.1°

sin B2 b2



22

1  14 sin 45 

C 2 = 180 − C1 ≈ 112.9°

= → b2 = 4 sin B2 ≈ 4.1 cm sin 30 sin C1 → sin 50 sin C1 = 21sin 50 = 17 17 21 21 sin 50 ⎞ ⎛ → C1 = sin −1 ⎜ ⎟ ≈ 71.1° ⎝ 17 ⎠ sin 50 17

→ sin B1  14 sin 45

A1 = 180 − 56 − C1 ≈ 56.9°

→ b1 = 4 sin B1 ≈ 8.0 8.00cm

C 2 = 180 − C1 ≈ 119.0° Bˆ 2 = 180 − 30 − Cˆ2 ≈ 31.0° sin 30 4

sin 45 22 sin 56 45

→ C1 =

⎠

Bˆ1 = 180 − 30 − Cˆ1 ≈ 89.0° =

=

25 sin C → c 2 = sin 70 2 ≈ 3.6 cm

C = 180 − 45 − B ≈ 108.3°

→ C1 = sin −1 ⎛⎜ 7 sin 30 ⎞⎟ ≈ 61.0° sin 30 4

sin C 2 c2 sin B 14

=

→ B1 sin 

Exercise 11H sin 30 4

sin 70 25 sin 45 22

L

≈ 14.2 cm sin 40 16

= sin AXL → AXL  sin 1  20sin 40  20



16



 180  53.464  126.536 © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 11

5

WORKED SOLUTIONS sin 31 sin A = b 10

ALX = 180 − 40 − AXL ≈ 13.464° sin 40 16

= sin ALX → AX = 16 sin ALX ≈ 5.80 km

c

e B

X

230°

16

= sin ABL 20

→ 16

⎝

f ⎠

a

sin ABL

⎛ ⎝

2 B

= sin B → B = sin −1 ⎛⎜ ⎝

43

43 sin 64 ⎞ ⎟ ≈ 36.0° a ⎠

+ 412 − 20 2 2 ( 33 )( 41)

cos A = 33

→ A = cos

⎜⎜ ⎝

+ 412 − 20 2 ⎞ ⎟ ≈ 28.9° 2 ( 33 )( 41) ⎟⎠

8

A

→d = 52 + 82 − 2 ( 5 ) ( 8 ) cos135 ≈ 12.1 km

3

A

B

2

4.5 E

+ 412 − 332 → 2 ( 20 )( 41)

cos B = 20

2

D

AD = BC = 4.5 + 3 − 2 ( 4.5 )( 3 ) cos 62 ≈ 4.07 cm

+ 2 .4 2 − 3 .6 2 ⎞ ⎟ ≈ 44.4° 2 ( 4.9 )( 2.4 ) ⎟⎠

⎝

2

+ 2 .4 2 − 4 .9 2 2 ( 3.6 )( 2.4 )

cos B = 3.6

2

⎛

2 + 2 .4 2 − 4 .9 2 ⎞ ⎟ ≈ 107.8° 2 ( 3.6 )( 2.4 ) ⎟⎠

→ B = cos−1 ⎜⎜ 3.6 ⎝

C = 180 − A − B ≈ 27.8° d

b 2 = 102 + 14 2 − 2 (10 )(14 ) cos 31

2

ˆ = 180 − 62 = 118° AEB AB = CD = 4.52 + 32 − 2 ( 4.5 )( 3 ) cos118 ≈ 6.48 cm

cos A = 4.9

→ A = cos−1 ⎜⎜ 4.9

C 2

C = 180 − A − B ≈ 98.4° 2 + 2 .4 2 − 3 .6 2 2 ( 4.9 )( 2.4 )

62° 3

2 2 ⎛ 2 ⎞ = cos−1 ⎜⎜ 20 + 41 − 33 ⎟⎟ ≈ 52.8° 2 20 41 ( )( ) ⎝ ⎠

⎛

C

d

ˆ = 270 − 32 −103 = 135° ABC 2 d = 52 + 82 − 2 ( 5 ) ( 8 ) cos135

2

−1 ⎛ 33

103°

32° 58° 5

C = 180 − 64 − B ≈ 80.0°

c

+ 582 − 50 2 ⎞ ⎟ ≈ 56.4° 2 ( 45 )( 58 ) ⎟⎠ 2

C = 180 − A − B ≈ 75.0°

a2 = 432 + 722 − 2 ( 43 )(72 ) cos 64

B

2

→ B = cos−1 ⎜⎜ 45

→ a = 432 + 722 − 2 ( 43 )( 72 ) cos64 ≈ 65.7 m

b

2

+ 582 − 50 2 2 ( 45 )( 58 )

cos B = 45

bearing = 90 + ABL ≈ 143.5°

sin 64 a

+ 582 − 452 ⎞ ⎟ ≈ 48.6° 2 ( 50 )( 58 ) ⎟⎠

⎝

Exercise 11I 1

2

⎛

= sin BLX → BX = 16 sin BLX ≈ 19.1 km BX

+ 582 − 452 2 ( 50 )( 58 )

cos A = 50

→ A = cos−1 ⎜⎜ 50

BLX = 180 − 2 ( ABL ) ≈ 73.07°

d

75 sin 70 = sin A → A = sin −1 ⎛⎜ c ⎞⎟ ≈ 49.4° ⎝ ⎠ 75

B = 180 − 70 − A ≈ 60.6°

ABL = sin −1 ⎛⎜ 20 sin 40 ⎞⎟ ≈ 53.464°

sin ABL 16

c 2 = 752 + 862 − 2 (75 ) ( 86 ) cos 70

sin 70 c

20

L sin 40 16

10 sin 31 ⎞ ⎟ ≈ 43.5° b ⎠

→ c = 752 + 862 − 2 (75 ) ( 86 ) cos 70 ≈ 92.8 m

A 40°

16

⎝

C = 180 − 31 − A ≈ 105.5°

sin 40

AX

→ A = sin −1 ⎛⎜

4

C

B 20 15

36° 27° A

ˆ = 36 + 27 = 63° BAC BC = 152 + 202 − 2 (15 )( 20 ) cos 63 ≈ 18.8 km

→ b = 102 + 14 2 − 2 (10 )(14 ) cos 31 ≈ 7.5 m © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 11

6

WORKED SOLUTIONS

Exercise 11J

B1

5 49 28

P

1

36

a b

A

c

Area = 1 ( 6.8 )( 9.4 ) sin 56.5 ≈ 26.7 cm2

2 1 Area = 10 2 A

( )( 9 ) sin115 ≈ 40.8 cm2

36

49

13.4 25.1

B2

ˆ  49 cos APB

2  282  362 2  49  28 

ˆ = cos−1 ⎛⎜ 49 → APB ⎜

B

+ 282 − 362 ⎞ ⎟ ≈ 46.5° 2 ( 49 )( 28 ) ⎟⎠

⎝

2

C sin 32 sin B = 13.4 25.1

E46.5°N or E46.5°S which is a bearing of 043.5° or 136.5° (since 90 − 46.5 = 43.5 and 90 + 46.5 = 136.5). 6

a

A

32°

A 7.88

8.47

cos A =

Triangle ABD is an isosceles right triangle, so angle ABD = 45°

C

10.98

B C

8.74 2 + ( 7.88 ) − 10.982 2 ( 8.74 )( 7.88 ) 2

2 2 2 ⎞ ⎛ → Aˆ = cos−1 ⎜ 8.74 + (7.88) − 10.98 ⎟ ⎟ ⎜ 2 ( 8.74 )( 7.88 ) ⎠ ⎝ ≈ 82.524°

E

24

⎟ ≈ 83.03° ⎠

2

15

b

13.4

Area = 1 (13.4 )( 25.1) sin A ≈ 152 cm2 d

D

⎜ ⎝

A = 180 − 32 − B ≈ 64.97°

B

15

→ B = sin

−1 ⎛ 25.1 sin 32 ⎞

Area = 1 ( 8.74 )( 7.88 ) sin A ≈ 34.1 cm2 2

24

e

A 46

15

D

C

ˆ  24 cos EDC

41

 152  24 2 2  24 15  2

ˆ = cos−1 ⎛⎜ 24 → EDC ⎜ ⎝

+ 152 − 24 2 ⎞ ⎟ ≈ 71.8° 2 ( 24 )(15 ) ⎟⎠ 2

B sin 58 sin C = 46 41

E

c

C

58°

⎛ 41 sin 58 ⎞ ⎟ ≈ 49.10° 46 ⎠

→ C = sin −1 ⎜⎝

A = 180 − 58 − C ≈ 72.899° 24

Area = 1 ( 41)( 46 ) sin A ≈ 901 cm2 2

24

f

A

A

AC = 152 + 152 = 15 2 ˆ  24 cos EAC

2

⎛



 15 2

⎜ ⎝

2

C 46°

 24 2

B

2  24  15 2



C = 180 − 86 − 46 = 48°

(

)

sin 48 sin 86 = BC 30

ˆ = cos−1 ⎜ 24 EAC ⎜





86°

30

C

− 24 2 ⎞⎟ ⎟ ≈ 63.8° 2 ( 24 ) 15 2 ⎟ ⎠

2

+ 15 2

(

2

)

→ BC =

30 sin 86 ≈ 40.27 sin 48

Area = 1 ( 30 ) ( BC ) sin 46 ≈ 435 cm2 2

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 11

7

WORKED SOLUTIONS 2

→  sin 1   1 (x ) 2

100

O

324 ≈ 22.7 ⎛1 ⎞ . sin . 33 9 57 4 ( ) ⎜ ⎟ ⎝2 ⎠

B

cm

ˆ  cos AOP

+ 10.22 − 17.22 2 (16.4 )(10.2 )

cos A = 16.4

a

2

5

2

1 2

( 2x + 3)( 4 x + 5) sin 30 = 30

1 2

( 8x 2 + 22x + 15) ⎛⎜ 1 ⎞⎟ = 30 → 8x 2 + 22x + 15 = 120

1 2

area of sec tor APB

2.5

( 8)(11) sin A ≈ 20

→ x = 112 + 82 − 2 (11)( 8 ) cos 27.0357 ≈ 5.31 mm x 2 = 112 + 82 − 2 (11)( 8 ) cos152.9643

Exercise 11L   5   180  12

b

240 

c

80 

d

330 

a

56 

b

107 

perimeter = radius + radius + arc length

c

324 

perimeter = 50 + 50 + 2.4 ( 50 ) = 220 cm

d

230 

a

5 6

b

5 3

1

Exercise 11K 1.7 ( 5.6 ) = 9.52 cm

3.25 (12 ) = 39 cm 3   2.5   12.5 ⇒ θ = 5 rad

2

( 2.4 ) ( 50 ) = 3000 cm

2

2

area = 1 ( 5.1) ( 32 ) = 22.95 cm2 2

perimeter = 3 + 3 + 5.1( 3 ) = 21.3 cm

1  2

 

3

 27.2 r

c

 r 2   217.6 →   435.2 2 r

27.2 435.2 = 2 r r 27.2  16

→ 27.2r = 435.2 → r =16 cm

1.7 rad

(1.01072 ) ( 62 ) ≈ 18.19296

≈ 7.96 cm2

75 

2

≈1 2

overlapping area ≈ 13.004 + 18.19296 − 23.2379

a

  r   27.2 → 



overlapping area = area of sec tor AOB + area of sec tor APB −area of quadrilateral OAPB

→ x = 112 + 82 − 2 (11)( 8 ) cos152.9643 ≈ 18.5 mm

6

42

2

x 2 = 112 + 82 − 2 (11)( 8 ) cos 27.0357

5

2

8 62 2 4 8

2

⎝ 44 ⎠

4

42

 23.2379 1 area of sec tor AOB ≈ (1.6255 ) ( 4 2 ) ≈ 13.004

⎝2⎠

area = 1 2

1

ˆ  area of quadrilateral OAPB  2  1  6  8  sin OPA

→ A = sin −1 ⎛⎜ 20 ⎟⎞ ≈ 27.0357° or 152.9643°

1

ˆ → AOP cos

2 2 2 2 62 8 ˆ  6   8   4 → OPA ˆ cosOPA cos 1 2  6  8  2 6 8 0.50536 rad ˆ P 2OPA 1.01072 rad

→ 8x 2 + 22x −105 = 0 6

2

ˆ 1.6255 rad O 2 AOP

Area = 1 (16.4 )(10.2 ) sin A ≈ 81.4 cm2

x=

4 2   8   62 2  4  8 

0.812756 rad

2 2 2 ⎛ ⎞ → Aˆ = cos−1 ⎜⎜ 16.4 + 10.2 − 17.2 ⎟⎟ ⎝ 2 (16.4 )(10.2 ) ⎠ ≈ 76.7°

b

P

8

(33.9 ) sin 57.4 ≈ 324

→x= 4

6

4

   47.8 1     15 18     2

3

A

7

15 18 sin  100

1 2

d 4

a b

  4  3  180 

  4  9  180 

  11  6  180 

    0.977  180 

rad

    1.87  180 

rad

    5.65 rad  180 

    4.01  180 

rad

 180     150 °   

 180     300 °    3  180     270 ° 2    5 4

 180     225 °    1.5  180   85.9°    0.36  180   20.6 °   

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 11

8

WORKED SOLUTIONS c d

Review exercise

2.38  180   136 °

   3.59  180   206°   

1 h

Exercise 11M 1

a b c

2

3

sin  sin 45  4 2 cos  3

1  2

cos120  



2 2



1  3 3 2

tan  tan30  6



36°

1 2

h tan36 == → h 100tan36 ≈ 72.7 m 100

3 3

2

d

sin  sin60  

a

0.892

b

0.949

c

–1.12

d

0.667

a

( 4.5)( 4.5) sin1.3 ≈ 9.76 cm2 BC = 4.52 + 4.52 − 2 ( 4.5 )( 4.5 ) cos1.3 ≈ 5.45 cm

b

3

1.3(4.5)2 2

4

= 50.5 cm2 (3 sf ) 4

a

= sin B 3

→ B = sin

−1 ⎛ 3 sin 0.94 ⎞

⎜ ⎝

11

( cos32 ,sin32 ) → (0.848, 0.530)

b

ˆ ,sin AOC ˆ ⎞⎟ C ⎛⎜ cos AOC

⎟ ≈ 0.222 ⎠

rad

5

ˆ = AOC ˆ + 54 ≈ 126.96° AOD ⎛ ˆ ˆ ⎞⎟ = (−0.600, 0.800) D ⎜ cos AOD ,sin AOD

a

Yˆ =180 − 42.4 − 82.9 =54.7 °

b

sin 82.9 13.2

→ XZ = sin 54.7 =

a

PR =

9.5 + 11.52 − 2 ( 9.5 )(11.5 ) cos118 ≈ 18.03 m

b

sin118 sin Pˆ = = → Pˆ PR 11.5

a

1 2

)( )

shaded area ≈ 15.141 − 4.23 ≈ 10.9 m 1 5 a Area of ∆ POQ = ( 6 )( 6 ) sin1.25 ≈ 17.1 cm2 2 2 2 ⎛ 2 ⎞ ˆ = cos−1 ⎜⎜ 6 + 6 − 11.2 ⎟⎟ ≈ 2.407 rad b QOR 2 6 6 ( )( ) ⎝ ⎠ 1 ˆ ≈ 12.1 cm2 Area of ∆ QOR = ( 6 )( 6 ) sin QOR 2 ˆ ≈ 2.63 rad c θ =2π − 1.25 − QOR 2

7 AB

6

ˆ = sin XZY

b

ˆ = tan30 tan XZY =

8 = 16

1 2

→ YZ = 8 3 cm

ˆ = 30° → XZY

7

tanθ

4

1 2

5

a

2.5 (10 ) = 25 cm

b

1 2

( 2.5= ) (10 ) ( 2.5= )(100 ) 2

1 2

24 + 38 − 2 ( 24 )( 38 ) cos120 ≈ 54.1 km 2

2

→ x sin

1

8sin 82 15

a

sin 82 sin x = 15 8

b

ˆ = 180 − 82 − x ≈ 66.12° ADC ˆ

31.9°

ˆ

= sin ADC → AC = 15sin ADC ≈ 13.9 cm sin 82

AC 2

+ 9 − AC 2 ( 7 )( 9 ) 2

2

⎞ ⎟⎟ ≈ 119° ⎠

(7 ) ( 9 ) sin y ≈ 27.6 cm2

d

1 2

a

Bˆ =− π 1.75 − 0.93 ≈ 0.4616 rad sin Bˆ 12

1 = 8 3 YZ

10 ) sin30 1= ( 4 )(= ( 4 )(10 ) ⎛⎜⎝ 12 ⎞⎟⎠ 2

⎠

ˆ = 170 − 50 = 120° APB

sin 82 15

8

b

=2 5

3

PR

a

AB=

→ AB = 7 2 cm

a

⎝

2 2 AB = 4 + 5.83 − 2 ( 4 )( 5.83 ) cosCˆ ≈ 8.60 cm

⎝

2

sin −1 ⎛⎜ 11.5sin118 ⎞⎟ ≈ 34.3°

( 4 )( 5.83) sinCˆ = 10

⎛

sin 45 =

≈ 10.9 cm

2

7 = c y cos−1 ⎜⎜

Review exercise 1

13.2 sin 54.7 sin 82.9

b

θ ( 6 ) ≈ 15.8 cm

1 = 2

XZ

10 ⎞ 121° ⎟≈ 1 ⎛ ⎜⎜ ⎜ ( 4 )( 5.83 ) ⎞⎟ ⎟⎟ ⎠⎠ ⎝⎝2

( )( )

✗

⎠

→ Cˆ = sin −1 ⎛⎜

Area of ∆ OAB = 1 3 11 sin A ≈ 15.141 2 2 1 Area of sector = 0.94 3 = 4.23 2

d

⎠

c

A = π − 0.94 − B ≈ 1.9795 rad

(

⎝

⎝

3

Shaded area = area of circle − area of sector = π(4.5)2 −

a

ˆ → AOC = cos−1 0.294 ≈ 72.9°

These values are found by using your GDC in radians mode.

1 2

sin 0.94 11

100

O

= sin 0.93 = → BC BC

sin 0.93 sin1.75 = BC AB

12 sin 0.93 ≈ 21.6 sin Bˆ

→ AB =

cm

BC sin1.75 ≈ 26.512 sin 0.93

DB = AB −12 ≈ 14.5 cm

10 cm2

50 = ( 2.5) 125 cm

c

0.93 (12 ) = 11.16 ≈ 11.2 cm

d

DB + arcDEC + BC ≈ 47.3 cm

2

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 11

9

WORKED SOLUTIONS

12

Vectors

Skills check

z

1

2

a

A = (3, 0, 0)

b

B = (3, 4, 0)

H D

c

E = (3, 0, 2)

2

d

F = (3, 4, 2)

O

e

H=

3  , 2

y

4 E

⎛3⎞ c = 3i + 8j = ⎜ ⎟ ⎝8⎠ 0 d 6 j    6

B

3 A

x 2 = 32 + 62

x

= 9 + 36

 3  e  3i  6 j     6 

= 45  x 3

a

45  6.71

X = 180 − 110 = 70 cos X = z ⇒ z = 15 cos 70 15

sin X =

y 15

C

≈ 5 .13 ⇒ y = 15 sin 70

15

≈ 14.1 ( AC )  y  (9  z )2 2

110°

2

 (14.1)  (9  5.13) 2

A

9

 21cm  to the nearest centimetre 

3     4

b

 1    3 

c

2i  5 j 

d

 2.8     4.5 

e

2i  5 j

32  4 2 

25  5

2 12  ( 3)

2 2  52 

 2.8 

2

10  3.16 29  5.39

 (4.5)2  5.3

2 22  ( 5)

29  5.39

3    2  5  

ˆ ABC = cos−1 ⎡⎢ (8.6)

b

 4    1   3   

c

2i  2 j  k 

22  22  12 

9 3 49  7

≈ 101.4

Exercise 12A

2

B

y

a

Using the Cosine Rule ˆ ) 5 a (AC )2 = (AB )2 + (BC )2 − 2(AB ) (BC ) cos (ABC ˆ ) (9.7)2 = (8.6)2 + (3..1)2 − 2(8.6) (3.1) cos (ABC ⎣

1

X

4

2

AC  432.5  20.8 b

 3  a  3i  5 j     5 

⎛ −2 ⎞ b = − 2i + 4 j = ⎜ ⎟ ⎝ 4⎠

F

C

2  

4,

3

G

+ (3.1)2 − (9.7 )2 ⎤ ⎥ 2(8.6)(3.1) ⎦

2

32  2 2  52 

38  6.16

2 4 2  ( 1)2  ( 3)

a

x  2i  3 j

b

y  7j

c

zi jk

d

a

2    AB =  3 

 3    2 2 2  2   ( 3)  2  6   6  

e

j  k

  0    1      6    1  

b

 CD =

c

0    EF =  0    1   

12  ( 1)2

26  5.10

2  1.41

Exercise 12B 2  2  1 a  , b   4  1  6   2 c   3b    3    3  1 1  2 1    d   1a  2  4  2 2    

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 12

1

WORKED SOLUTIONS  2  − 10  e =  = −5  − 1  = −5b    5 

3

a

⎛ −4 ⎞ f =⎜ ⎟ ⎝ −8 ⎠

4

k = 144 =2

We must have s & t so that

7

t = −12k

−5 = 2 s + 2t

(1)

8  4s  t 2  (2)   16 8 s  2t

(2)

= −12 × 2

(3)

= −24 b

−21 10 −84 −t 10 8 − 84 10 −2 5

s = from (2): − 8 = t = =

so f = −2 a  

2 4

2a

For parallel vectors, a  k b for some k   t 7   k    8   10       8 10k



8 10  4. 5

k

2 so t  7k

−1

 7  4 

 0.1       0.7    1  1   10  7    1 i 7j 10

7

7

(1) + (3) : − 21 = 10 s

2

For parallel vectors, r = k s for some k (4 i + t j) = k (14 i − 12 j) 4 = 14k

5 28  5

4



a is parallel to i  7 j with

1 10

the magnitude.

 1  b   7    1 i  7 j 

s

b is parallel to  i  7 j  with opposite direction.  0.05    is not parallel to  i  7 j   0.03   10   is not parallel to i  7 j d      70    c

e = 60 i + 420 j = 60 ( i + 7 j )

ude. e is parallel to ( i + 7 j ) with 60 times the magnitu f  6i  42 j  is not parallel to  i  7 j 

For parallel vectors, v k w for some k t i 5 j 8k k (5i j s k) 5 k so t 5k t 5( 5) 25 8 sk ( 5)s

5

a b c d

6

a b c d

8 5

 OG = j + k  BD = − i − j + k  AD = − i + k 

1 OM = 2 i + j + k  OG = 4 j + 3k  BD = −5i − 4 j + 3k  AD = −5i + 3k 

5

OM = 2 i + 4 j + 3k

g   i  7 j  is not parallel to  i  7 j 

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 12

2

WORKED SOLUTIONS

Exercise 12C 1

6

 ⎛ 7 ⎞  ⎛ 2 ⎞ OP = ⎜ ⎟ OQ = ⎜ ⎟ ⎝4⎠ ⎝3 ⎠    ⎛ 2 ⎞ ⎛ 7 ⎞ PQ = OQ − OP = ⎜ ⎟ − ⎜ ⎟ ⎝3 ⎠ ⎝ 4 ⎠ ⎛ −5 ⎞ =⎜ ⎟ ⎝ −1 ⎠

b

c

d

3

a

b

4

5

(1)

y−3−4=0⇒y−7=0

(2)

−2 + z −(x + y) ⇒ −x − y + z − 2 = 0

(3)

Exercise 12D 1

AB = OB − OA = (−2i + 3j−k) − (i−2j+3k) = ( −2 − 1)i + (3 − ( −2)) j + ( −1 − 3) k = −3i + 5 j − 4 k AC = OC − OA = ( 4 i − 7 j + 7 k ) − ( i − 2 j + 3k )

= ( 4 − 1)i + ( −7 − ( −2)) j + (7 − 3) k = 3i − 5 j + 4 k     we see AB = − AC , so AB and AC are parallel. Since they contain a common point A, they must lie on the same line. ⎛5⎞ ⎛ 2⎞ ⎛ 3⎞    ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 2 a AB = OB − OA = 1 − ⎜ ⎟ ⎜ 3 ⎟ = ⎜ −2 ⎟ ⎜ 5 ⎟ ⎜ −3 ⎟ ⎜ 8 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ 1⎞ ⎜ ⎟ vector is − ⎜ −5 ⎟ = −i + 5j − 6k ⎜ 6⎟ ⎝ ⎠

c

⎛ 1⎞ ⎛ 2 ⎞ ⎛ −1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ vector is ⎜ 2 ⎟ − ⎜ −3 ⎟ = ⎜ 5 ⎟ = −i + 5 j − 6 k ⎜ −1⎟ ⎜ 5 ⎟ ⎜ −6 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

d

⎛ 2 ⎞ ⎛ 1⎞ ⎛ 1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ vector is ⎜ −3 ⎟ − ⎜ 2 ⎟ = ⎜ −5 ⎟ = i − 5 j + 6 k ⎜ 5 ⎟ ⎜ −1⎟ ⎜ 6 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ 8⎞ ⎛ 2⎞ ⎛ 6⎞    ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ b AC = OC − OA = −1 − ⎜ ⎟ ⎜ 3 ⎟ = ⎜ −4 ⎟ ⎜ 13 ⎟ ⎜ −3 ⎟ ⎜ 16 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠     we see AC = 2 AB , so AC and AB are parallel. Since they contain a common point A, then A, B, & C are collinear. ⎛ −2 ⎞ ⎛ 1 ⎞ ⎛ −3 ⎞    ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 3 P1 P2 = OP 2 − OP1 = 1⎟ ⎜ 1⎟ − ⎜ 2 ⎟ = ⎜ −1 ⎜ 4⎟ ⎜4⎟ ⎜ 0⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ −5 ⎞ ⎛ 1⎞ ⎛ −6 ⎞    ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ P1 P3 = OP 3 − OP1 = ⎜ 0 ⎟ − ⎜ 2 ⎟ = ⎜ −2 ⎟ ⎜ 4⎟ ⎜4⎟ ⎜ 0⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠   we see P1 P3 = 2 P1 P2 . Since they contain a common point, they are collinear.

⎛ 1⎞ ⎛ 4 ⎞ ⎛ 5 ⎞    ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ LM = LN + NM = ⎜ −2 ⎟ + ⎜ −2 ⎟ = ⎜ −4 ⎟ ⎜ 0 ⎟ ⎜ −3 ⎟ ⎜ −3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

T

U S

1 + 2x − 1 = 0 ⇒ 0 + 2x = 0

z=9

 ⎛ 1 ⎞ ⎛ 5 ⎞ ⎛ −4 ⎞ AB = B − A = ⎜ ⎟ − ⎜ ⎟ = ⎜ ⎟ ⎝ −3 ⎠ ⎝ 1 ⎠ ⎝ −4 ⎠   ⎛ 4 ⎞ BA = − AB = ⎜ ⎟ ⎝4⎠  ⎛ −2 ⎞ ⎛ 5 ⎞ ⎛ −7 ⎞ AC = C − A = ⎜ ⎟ − ⎜ ⎟ = ⎜ ⎟ ⎝ 4 ⎠ ⎝1 ⎠ ⎝ 3 ⎠  ⎛ 1 ⎞ ⎛ −2 ⎞ ⎛ 3 ⎞ CB = B − C = ⎜ ⎟ − ⎜ ⎟ = ⎜ ⎟ ⎝ −3 ⎠ ⎝ 4 ⎠ ⎝ −7 ⎠ ⎛ 2⎞  ⎜ ⎟ OP = ⎜ −3 ⎟ = 2i − 3 j + 5k ⎜ 5⎟ ⎝ ⎠

From the diagram, we see    US = −TU + TS = −( i − 4 j + 2 k ) + (3i + 4 j − k ) = ( −1 + 3)i + ( 4 + 4 ) j + ( −2 − 1)k = 2i + 8 j − 3k

C B

(1) ⇒ x = 0 (2) ⇒ y = 7 (3) ⇒ −2 + z − 7 = 0

⎛ 1⎞ ⎛5⎞ ⎛ −2 ⎞ A=⎜ ⎟ B =⎜ ⎟ C =⎜ ⎟ ⎝ −3 ⎠ ⎝1 ⎠ ⎝ 4⎠ a

A

1⎞ ⎛ 0 ⎞ ⎛ 1⎞ ⎛ 2 x ⎞ ⎛ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 4⎟ = ⎜0⎟ ⎜ y ⎟ + ⎜ −3 ⎟ − ⎜ ⎜ −2 ⎟ ⎜ z ⎟ ⎜ x + y ⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

  ⎛ 5 ⎞ QP = − PQ = ⎜ ⎟ ⎝1 ⎠ 2

From  the  diagram,  AB + BC − AC = 0

Since P4 collinear with P1, P2, P3, we have   P1 P4 = k P1 P2 for some k ∈ R 1⎞ ⎛ 2 ⎞ ⎛1 ⎞ ⎛  ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ P1 P4 = ⎜ s ⎟ − ⎜ 2 ⎟ = ⎜ s − 2 ⎟ for some s & t ⎜t ⎟ ⎜4⎟ ⎜t − 4⎟ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 12

3

WORKED SOLUTIONS Distance BC = 22 + 10 2 + 52 = 129

1⎞ ⎛ ⎛ −3 ⎞ ⎜ ⎟ Now ⎜ s − 2 ⎟ = k ⎜⎜ −1⎟⎟ ⎜ t − 4⎟ ⎜ 0⎟ ⎝ ⎠ ⎝ ⎠ 1 = −3k ⇒ k =

Distance AB = Distance BC, so ABC is isosceles. B

−1 3

√129 1

7

s − 2 = −k ⇒ s = 2 − k = 2 + 3 = 3 t −4=0⇒t =4 ∴ P4 = ⎛⎜ 2, 7 , 4 ⎞⎟ ⎝ 3 ⎠    4 OA = 3i + 4 j, OB = xi, OC = i − 2 j    AB = OB − OA = ( x − 3)i − 4 j    AC = OC − OA = (1 − 3)i + ( −2 − 4 ) j = −2i − 6 j   If A, B, C are collinear, AB = k AC for some k ∈ R

A

3

x=9−4=5

−4 3

2

4 + 9 + t 2 = 49 t 2 = 36 t = ±6 4

a = xi + 6 j − 2 k 2

a = x 2 + 62 + ( −2 ) = 3x x2 + 36 + 4 = 9x2 8x2 = 40 x2 = ± 5 5

3 

3 3 −4 so AB = 3 i − 4 j   

u = v ,so a 2 + ( −a ) + ( 2a ) = 22 + ( −4 ) + ( −2 ) 2

= ( i − 2 j) − 5 i = 3

AB : BC =

−2 i 3

2

a2 = 4 a = ±2

⎛ −4 ⎞ ⎛ −2 ⎞ ⎜ 3 ⎟:⎜ 3 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ −4 ⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎝ ⎠

6

a

b

b = −3a Then |a + b|=|−2a|=2|a|=10

⎛ 4 ⎞ ⎛ −1⎞ ⎛ 5 ⎞    ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AB = OB − OA = ⎜ 5 ⎟ − ⎜ 5 ⎟ = ⎜ 0 ⎟ ⎜ −1⎟ ⎜ 1⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Distance AB = 52 + ( −2 )

b = 2a Then a + b = 3a = 3 a = 15

Exercise 12E

2

2

6a 2 = 24

− 2j

= 2 :1

1

2

a 2 + a 2 + 4 a 2 = 4 + 16 + 4

BC = OC − OB

 

129 + 242 − 129 2 129 242

CÂB = 46.8°

C

√242

)

a = 7,so 22 + ( −3 ) + t 2 = 7

∴ ( x − 3)i − 4 j = k ( −2i − 6 j) j components ⇒ − 4 = − 6k ⇒ k = 2 3 so x − 3 = −2k =

(

cos CÂB =

√129

c

a+b

b

2

a Using Pythagoras |a|2 + |b|2 = |a + b|2 Hence |a + b| = 52 + 122 = 13

= 29 ≈ 5.39 ⎛ 6 ⎞ ⎛ −5 ⎞ ⎛ 11⎞  ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AB = ⎜ 0 ⎟ − ⎜ 2 ⎟ = ⎜ −2 ⎟ ⎜6⎟ ⎜ 4 ⎟ ⎜ 2⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Exercise 12F

Distance AB = 112 + 22 + 22 = 129

1

j=

⎛ 8 ⎞ ⎛ −5 ⎞ ⎛ 13 ⎞  ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AC = ⎜ 10 ⎟ − ⎜ 2 ⎟ = ⎜ 8 ⎟ ⎜ 1⎟ ⎜ 4 ⎟ ⎜ −3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

3 i+ 4 5 5

2

1 i+2 3 3

j + 2k =

2

3

= 2

Distance AC = 132 + 82 + ( −3 ) = 242 ⎛ 8⎞ ⎛ 6⎞ ⎛ 2⎞  ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ BC = ⎜ 10 ⎟ − ⎜ 0 ⎟ = ⎜ 10 ⎟ ⎜ 1⎟ ⎜ 6 ⎟ ⎜ −5 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2

9 16 ⎛3⎞ ⎛4⎞ + ⎜ ⎟ +⎜ ⎟ = 25 25 ⎝5⎠ ⎝5⎠

= 1 =1

1 22 22 + + 32 32 32 1 4 4 + + 9 9 9

= 1 =1 3

4 i − 3 j = 42 + ( −3 ) = 25 = 5 2

So unit vector is

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

1 5

( 4i − 3 j ) = 45 i − 35 j Worked solutions: Chapter 12

4

WORKED SOLUTIONS

4

⎛ −1⎞ ⎟ ⎜ ⎜ −5 ⎟ ⎟ ⎜ ⎟ ⎜ 4 ⎠ ⎝

2

2

= ( −1) + ( −5 ) + 4 = 42

⎛3⎞ ⎛1⎞ ⎛ 2⎞ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ P1P2 = ⎜⎜ 2 ⎟⎟ − ⎜⎜ 0 ⎟⎟ = ⎜⎜ 2 ⎟⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝ 1 ⎠ ⎝ −1⎠  2

Exercise 12G 1

a

= 5i + j b

= 2i + 3 j

= 5a

7

2i − j = 2 + ( −1) 2

=

1 5

5

×

5

=

5

c

5

2

= 2i + 4 j d

1 5

8

 1  3  2

a + b + d = ( 2 i − j ) + ( 3i + 2 j ) + ( 3i + 3 j ) = ( 2 + 3 + 3 ) i + ( −1 + 2 + 3 ) j

( 2i − j )

Vector of magnitude 5 is  1    3    2  

c + d = ( − i + j ) + ( 3i + 3 j )

= ( −1 + 3 ) i + (1 + 3 ) j

5

= 5 So unit vector is

b + c = ( 3i + 2 j ) + ( − i + j ) = ( 3 − 1) i + ( 2 + 1) j

2

ai + 2aj = a 2 + ( 2a ) = 5a 2 1

a + b = ( 2 i − j ) + ( 3i + 2 j )

= ( 2 + 3 ) i + ( −1 + 2 ) j

⎛ 2⎞ ⎜ ⎟ 1⎜ ⎟ 2 ⎟ 3⎜ ⎜ ⎟ − 1 ⎠ ⎝

Now 5a = 1, so a =

2

= 8i + 4 j

5 5

( 2i − j ) =

5 ( 2i − j )

e

a − b = ( 2 i − j ) − ( 3i + 2 j )

= ( 2 − 3 ) i + ( −1 − 2 ) j = −i − 3 j

 22  14 f

d − b + a = ( 3i + 3 j ) − ( 3i + 2 j ) + ( 2 i − j ) = ( 3 − 3 + 2 ) i + ( 3 − 2 − 1) j

 1   unit vector is  3  14    2 1

 1  7  and vector magnitude7 is 3   14    2 9

a

sec2α

1 cosα 1  1   cosα  So uniit vector is = secα  tanα  sinα 

P1P2 = 22 + 22 + ( −1) = 9 = 3

6

12 + tan2α

= secα =



So unit vector is

 1    =  tanα  =

⎛ −1⎞ ⎟ ⎜ 1 ⎜ −5 ⎟⎟ ⎜ 42 ⎟ ⎜ ⎝ 4⎠

So unit vector is

5

b

2

= 2i + 0 j = 2i  1 14   3   2  2  

⎛ 2 cos  ⎞ 2 2 2 2 ⎜ ⎟ = 2 cos  + 2 sin  ⎝ 2 sin  ⎠ = 4 ( cos 2  + sin 2  ) = 2 (1) = 2 ⎛ 2 cos  ⎞ ⎛ cos  ⎞ So uniit vector is 1 ⎜ ⎟ or ⎜ ⎟ 2 2 sin  ⎝ ⎠ ⎝ sin  ⎠

2

a

⎛ 2 ⎞ ⎛ −4 ⎞ ⎛ 2 −4 ⎞ ⎛ −2 ⎞ a + b=⎜ ⎟+⎜ ⎟=⎜ ⎟=⎜ ⎟ ⎝ −3 ⎠ ⎝ 5 ⎠ ⎝ −3 +5 ⎠ ⎝ 2 ⎠

b

⎛ −4 ⎞ ⎛ −5 ⎞ ⎛ −4 −( −5) ⎞ ⎛ 1 ⎞ b−c =⎜ ⎟−⎜ ⎟ =⎜ ⎟=⎜ ⎟ ⎝ 5 ⎠ ⎝ −3 ⎠ ⎝ 5 −( −3) ⎠ ⎝ 8 ⎠

c

⎡ ⎤ ⎛ 1 1 ⎢ ⎛⎜ 2 ⎞⎟ ⎛⎜ −5 ⎞⎟ ⎥ 1 ⎜ 2 + = a + c = ( ) 2 2 ⎢ ⎜ −3 ⎟ ⎜ −3 ⎟ ⎥ 2 ⎜ −3 ⎠⎦ ⎠ ⎝ ⎝ ⎣⎝

d

−5 ⎞⎟ 1 ⎛⎜ −3 ⎞⎟ = +(−3) ⎟⎠ 2 ⎜⎝ −6 ⎟⎠

⎛ −4 ⎞ ⎛ −5 ⎞ ⎛ 2⎞ a + 3b − c = ⎜ ⎟ + 3 ⎜ ⎟ − ⎜ ⎟ ⎝ 5 ⎠ ⎝ −3 ⎠ ⎝ −3 ⎠ ⎛ 2 + 3(−4) − (−5) ⎞ =⎜ ⎟ ⎝ −3 + 3(5) − (−3) ⎠ ⎛ −5⎞ =⎜ ⎟ ⎝ 15 ⎠

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 12

5

WORKED SOLUTIONS e

⎛ −5 ⎞ ⎛ −4 ⎞ ⎛ 2 ⎞ 3c − 2 b + 5a = 3 ⎜ ⎟ − 2 ⎜ ⎟ + 5 ⎜ ⎟ ⎝ −3 ⎠ ⎝ 5 ⎠ ⎝ −3 ⎠ ⎛ 3( −5) −2( −4 ) +5(2) ⎞ =⎜ ⎟ ⎝ 3( −3) −2(5) +5( −3) ⎠ ⎛ −15 +8 +10 ⎞ =⎜ ⎟ ⎝ −9 −10 − 15 ⎠ ⎛ 3⎞ =⎜ ⎟ ⎝ −34 ⎠

3

a

 3   z  0 2  +  1  =    −5   z2   0  6 + z1 = 0 ⇒ z1 = − 6 −10 + z2 = 0 ⇒ z22 = 10 z = − 6i + 10 j 5

a + b = ( 3i − j − 2 k ) + ( 5i − k )

= ( 3 + 5 ) i + ( −1) j + ( −2 − 1) k = 8i − j − 3 k

b

= − i + 2 j + 3k

d

4

a

y = 21 2

2a − b = 2(3i − j − 2k) − (5i − k) = (6 − 5)i + (−2)j + (−4 + 1)k = i − 2j − 3k 4(a − b) + 2(b + a) = 4((3 − 5)i − j + (−2 + 1)k) + 2(8i − j − 3k) from Q3a = −8i − 4j − 4k + 16i − 2j − 6k = (−8 + 16)i − (4 + 2)j + (−4 − 6)k = 8i − 6j − 10k

x = 6 − y = 6 − ⎛⎜ 21 ⎟⎞ = −9 ⎝ 2 ⎠

6

2x − 3p = q

 12  −7   −20  −  28  = 3y      19  −48  = 3y   1 So y = 3 (19i − 48j)

⎛3⎞ ⎛t − s ⎞ ⎜ ⎟ ⎜ ⎟ 3a = 2 b ⇒ 3 ⎜ t ⎟ = 2 ⎜ 3 s ⎟ ⎜u⎟ ⎜t + s ⎟ ⎝ ⎠ ⎝ ⎠

(2) ⇒ t = 2s (1) ⇒ 9 = 2 ( 2s − s ) = 2s s =9

2

(2) ⇒ t = 9

 4  x=   −5.5 

3  −1  4   − 3(y) = 7    −5  4  −7   12  −20  − 3y =  28     

2

(1) 9 = 2 ( t − s ) (2) 3t = 6s (3) 3u = 2 ( t + s )

 3   −1  2x − 3   =    −5   4   −1 +9   8  2x =  =   4 −15   −11 

b

⎛ x ⎞ ⎛ 6− y ⎞ a=b⇒⎜ ⎟=⎜ ⎟ ⎝ x + y ⎠ ⎝ −2 x − 3 ⎠ x =6− y (1) x + y = − 2x − 3 (2) y = − 3x − 3 Sub (1) into (2) y = −3 ( 6 − y ) − 3 y = −18 + 3 y − 3 −2y = −21

b − 2a = ( 5i − k ) − 2 ( 3i − j − 2 k ) = ( 5 − 6 ) i − 2 ( −1) j + ( −1 − 2 ( −2 ) ) k

c

2p + z = 0

c

(3) ⇒ 3u = 2 ⎛⎜ 9 +

9⎞ ⎟ 2⎠ ⎝ 27 u= =9 3 t = 9, s = 9 , u = 9 2

Exercise 12H P

1 A

N

a O

a b

b

Q

B





AP = OA = a







AB = − OA + OB = −a + b =b−a

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 12

6

WORKED SOLUTIONS c

d

e











PQ = − AP − OA + OB + BQ = − a − a + b + 3b = 4b b − 2a

ii





PN = 1 PQ = 1 ( 4 b − 2a ) 2

2

= 2b − a     ON = OA + AP + PN

iii

= a + a + (2b − a ) = a + 2b

f







AN = AP + PN = a + ( 2b − a ) iv



  





a = OA, b = OB, AC :CB = 3 :1 a

b



AB = − OA + OB = −a + b =b−a

 AC = 3 AB 4 =3 b− 4  1 

c

a)

CB = AB =

d

v



(



4 1 b 4 

(

− a)

3



OA = a, OC = c, CB = 3a a b

c





5

1 2

d



a

4

a









i







AB = − FA + FB = −a + b



3



2 3

AP =

b

(b − a)

M is mid point of OA, so 



MA = 1 OA = 1 a 2 



2 

MP = MA + AP = 1a +

CD = CB − AB

FA = a, FB = b



AB = − OA + OB = −a + b =b−a

2 1 

(



FD = FC + CD   By symmetry, CD = − FA = − a  so FD = 2 ( b − a ) − a = 2b − 3a

since AP = 2 AB

( c + 2a )

2 1 = 3a − c + 2a 2 = 2a − 1 c 2  







= 2a + 1 c 





BC = BE + EC    = BO + OC + EC    = − OB + OC + EC = −a − a + b = b − 2a

OA = a, OB = b

2

=a +



FD = 2b − 3a AC = AF + FC (see iii) = −a + 2(b − a)



OB = OC + CB = c + 3a    AB = − OA + OC + CB = − a + c + 3a = c + 2a    OD = OA + 1 AB



= −3a + 2b FD = AC ∴ FD and AC are parallel

= a ⎛⎜ 1 − 3 ⎞⎟ + b 3 4



c

(b − a)

⎝ 4⎠ 1 = a + 3b 4 4  



FC = FO + OE + EC   By symmetry, OE = BO = − a   and EC = FB = b  so FC = ( b − a ) − a + b

AB is parallel to and half the length of FC



3 4



b

OC = OA + AC =a +





FO = FB + BO   By symmetry, OB = FA = a    so FO = FB − OB =b −a

= 2( b − a )

= 2b 2



=

)

= c



(

)





2 b−a 2 3 2 ⎛1 2⎞ ⎜ − ⎟a + b 3 ⎝2 3⎠ 2 b − 1a 3 6 

MX = MP + PB + BX 



PB = 1 AB = 1 ( b − a ) 3 



3

BX = OB = b 

so MX = ⎛⎜ 2 b − 1 a ⎞⎟ + = 2b

−1 2

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

⎝3

a

6

⎠

1 3

(b − a) + b

Worked solutions: Chapter 12

7

WORKED SOLUTIONS d

MP

2 3

MX

2b −

b

1 a 6 1 a 2

d

MX = 3 MP ∴ MX is parallel to MP Since MX and MP share the common point M, MPX is a straight line

e

2 u ⋅ w = 2 ( −29 ) = − 58

( u − v )⋅( u + w ) =

Exercise 12I 1

a

b

c

=

a ⋅ b = ( 2i + 4 j ) ⋅ ( i − 5 j )

= ( 2 × 1) + ( 4 × −5 ) = 2 − 20 = −18 b ⋅ c = ( i − 5 j ) ⋅ ( −5i − 2 j )

3

= (1 × − 5 ) + ( −5 × − 2 ) = − 5 + 10 =5 a ⋅ a = ( 2i + 4 j ) ⋅ ( 2i + 4 j )

a

b

2

c

= ( −3 × 1) + ( 2 × −5 ) = − 3 −10 = −13

b

c

− 9 − ⎡⎣1 +

− 30 ⎤⎦

0 = = − 9 + 29 = 20

2

= − 32 − 2 − 2 = − 36 2

u v = 18 × 72 = 36 = − u ⋅ v ⇒ paarallel. d

a = 3i − 2 j + k b = 3i − 2 j − k a ⋅ b = ( 3 × 3) + ( −2 × − 2 ) + (1 × −1) = 9 + 4 −1 = 12 a = 32 + 22 + 12 = 9 + 4 + 1 = 14

= −4 + 0 − 5 = −9 ⎛ −1⎞ ⎛ 4 −(−1) ⎞ ⎛ −1⎞ ⎛ 5 ⎞ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ u ⋅ ( v − w ) = ⎜⎜ 0 ⎟⎟ ⋅ ⎜⎜ −3 −(3) ⎟⎟ = ⎜⎜ 0 ⎟⎟ ⋅ ⎜⎜ −6 ⎟⎟ ⎜ ⎟ ⎜⎜ 5 ⎟⎟⎠ ⎜⎝ −1 −(−6) ⎟⎠ ⎜⎜⎝ 5 ⎟⎟⎠ ⎜⎜⎝ 5 ⎟⎟⎠ ⎝

= − 9 − ⎡⎣⎢ ( −1) × ( −1) + 0 × 3 + 5 × ( −6 ) ⎤⎦⎥

= ( −8 × 4 ) + ( 2 × −1) + ( 2 × −1)

v = 4 2 + 12 + 12 = 16 + 1 + 1 = 18

= ( −1 × 4 ) + ( 0 × − 3) + ( 5 × −1)

= ( −1 × 5 ) + ( 0 × − 6 ) + ( 5 × 5 ) = − 5 + 0 + 25 = 20 ⎛ −1⎞ ⎛ −1 ⎞ ⎟ ⎟ ⎜ ⎜ u ⋅ v − u ⋅ w = − 9 − ⎜⎜ 0 ⎟⎟ ⋅ ⎜⎜ 3 ⎟⎟ ⎜⎜ 5 ⎟⎟⎠ ⎜⎜⎝ −6 ⎟⎟⎠ ⎝

u ⋅v =

⎛ −8 ⎞ ⎛ 4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 2 ⎟ ⋅ ⎜ −1⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ 2 − 1 ⎝ ⎠ ⎝ ⎠

u = 8 + 2 + 2 = 64 + 4 + 4 = 72

( c + a ) ⋅ b = ⎡⎢⎣ ( −5i − 2 j ) + ( 2i + 4 j ) ⎤⎥⎦ ⋅ ( i − 5 j )

u ⋅v =

⎠

2

= ⎡⎣ −3i + 2 j ⎤⎦ ⋅ ( i − 5 j )

a

= ( 2 × 1) + (1 × 2 ) = 4

|c|| |d| = ⎛⎜ 5 ⎞⎟ = 5 ≠ c ⋅ d

= ( −5 × 3) + ( −2 × −1) = 15 + 2 = − 13

2

⎛2⎞ ⎛1⎞ ⎟ ⋅⎜ ⎟ ⎜1⎟ ⎜2⎟ ⎝ ⎠ ⎝ ⎠

c ⋅d = ⎜

So neither parallel, nor perpendicular.

c ⋅ ( a + b ) = ( −5i − 2 j) ⋅ ( 2i + 4 j) + ( i − 5j ) 

⎛ −1⎞ ⎛ 4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⋅ ⎜ −3 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ 5 − 1 ⎝ ⎠ ⎝ ⎠

⎛ −5 ⎞ ⎛ −2 ⎞ ⎟ ⎟ ⎜ ⎜ ⎜ 3 ⎟ ⋅⎜ 3 ⎟ ⎟ ⎟ ⎜ ⎜ ⎟⎟ ⎟⎟ ⎜⎜ ⎜⎜ 6 − 1 ⎠ ⎝ ⎝ ⎠

= 10 + 9 − 6 = 13 a ⋅ b = (2 × 4 ) + ( 4 × − 2) =8−8 = 0 ⇒ perpendicular.

⎝

= ( − 5i − 2 j) ⋅ 3i − j 

e

−4 ⎤⎥ ⎡⎢ −1 +(−1) ⎤⎥ − (−3) ⎥⎥ ⋅ ⎢⎢ 0 +3 ⎥⎥ ⎥ ⎢ ⎥ −(−1) ⎥⎦ ⎢⎣ 5 +(−6) ⎥⎦

|c| = 22 + 12 = 5 |d| = 5

= (2 × 2) + ( 4 × 4) = 4 + 16 = 20

d

⎡ −1 ⎢ ⎢ ⎢ 0 ⎢ ⎢⎣ 5

b = 32 + 22 + 12 = 9 + 4 + 1 = 14 a b=

(

14

)

2

= 14 ≠ a ⋅ b

⇒ neither parallel, nor perpendicular e

 

OX ⋅OZ =

⎛1⎞ ⎛0⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⋅⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎝0⎠ ⎝1⎠

= (1 × 0 ) + ( 0 × 0 ) + ( 0 × 1) = 0

⇒ perpendicular f

n ⋅ m = ( 2i − 8 j )( −i + 4 j )

= ( 2 × −1) + ( −8 × 4 ) = − 2 − 32 = −34

n = 22 + 82 = 4 + 64 = 68 m = 12 + 4 2 = 1 + 16 = 17

− 2 j ) ⋅  ( 2i + 4 j ) + ( −i − 5 j ) 

 University Press 2012: this may be reproduced for class use solely for the purchaser’s institute − 2 j ) ⋅  i −©jOxford 

× 1) + ( −2 × −1)

Worked solutions: Chapter 12

8

WORKED SOLUTIONS n m = 17 × 68 = 34 = − n ⋅ m ⇒ parallel. g

b

  AB ⋅ CD = ⎛ 2 ⎞ ⋅ ⎛ −1⎞ = ( 2 × −1) + ( 2 × −1) ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ −1⎠ = −2 − 2  | AB |  22  22  8  |CD |  12  12  2   | AB | ⋅ |CD | 28 

 4   3        4   3  0  1   12 0  1  4 42 4    0    3  2 2    3 1  1   12  4 10 cos 

= −4

 3   cos 1     161.6  10  c  2 i  5 j    2 i  5 j   2      5   

  16 = 4 = − AB ⋅ CD

⇒ parallel

4

vectors a  3 b   i  j  2 k   3  3i  2 j  k 

 4  25   21

 1  3  3  i  1  3  2  j   2  3   1 k  10 i  7 j  k 2a  b 2  i  j  2 k    3i  2 j  k  

 2 1  3 i   2 1  2  j   2  2    1  k

 i  5k (a + 3b) · (2a − b) = (10i + 7j − k) · (−i + 5k)

= (10 × −1) + (−1 × 5) = −10 − 5 = −15 5

Let d  d1i  d2 j  d3 k

a  d 3d1   5  d3  9 b  d  2d1  7d 2  11 c  d  d1  d 2  d3  6 using GDC, d1 = 2, d2 = 1, d3 = 3 ⎛2⎞ ⎜ ⎟ So, d = ⎜⎜ 1 ⎟⎟

 2i  5 j 

c

cos  



2 3

cos  

7

a

   45

9

2

 2 2       2  2  1  5  4  5  1  1  5   2    1

22  12 

2    5

2 5  2

2

5 4  25 

1  5 29 cos   1   cos 1     94.8  145 

29

29 

 2i  5 j 

a

 1  1   5   2 

= −1 −10 = −11  | AB |  12  52  26  | AC |  12  22  5     AB ⋅ AC = | AB || AC |cosθ 11 

6  2 3 cos  1

2 2  52 

21    cos 1    136.4  29    1   2   1 8 a AB =          9   4   5   3   2   1 AC =          2   4   2     1  1  b AB ⋅ AC =       5   2 

a  b  a  b cos 

6



  21  29 cos 

⎜⎜ ⎟⎟ 3 ⎝ ⎠

6

10

5  26 cos  11 130

⎛ −1⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 2 ⎟ ⋅ ⎜ −3 ⎟ = − 2 − 6 + 12 = 4 ⎜ 2⎟ ⎜ 6 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ −1⎞ ⎜ ⎟ 2 2 2 ⎜ 2⎟ = 1 + 2 + 2 = 9 = 3 ⎜ 2⎟ ⎝ ⎠ ⎛ 2⎞ 2 ⎜ ⎟ 2 2 ⎜ −3 ⎟ = 2 + ( −3 ) + 6 = 4 + 9 + 36 = 7 ⎜ 6⎟ ⎝ ⎠ so 4 = 3 × 7 cos  cos  =

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

4 , 21

= 79°

Worked solutions: Chapter 12

9

WORKED SOLUTIONS b

c

12 a ⎛2⎞ ⎛ 4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ ⋅ ⎜ −2 ⎟ = 8 − 6 − 2 = 0 ⇒ perpendicular vectors ⎜ 1 ⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎝ ⎠ θ = 90° ( 2i − 7 j + k ) ⋅ ( i + j − k )

= ( 2 × 1) + ( −7 × 1) + (1 × −1) = −6

( 2i − 7 j + k ) = 22 + 72 + 12 (i + j − k) = 1+ 1+ 1 = 3

b

= 54

= −3i − 2j + 7k  2 2 | AB |   3    2   72  13

 6   cos 1     118.1  162 

b

 17, AC

2

14

a − b = (5 − 1)i + (−3 − 1)j + (7 − λ)k = 4i − 4j + (7 − λ)k 26

Now (a + b) ⋅ (a − b) = (6 × 4) + (−2 × −4) + (λ + 7)(7 − λ) = 24 + 8 + 49 − λ2 = 0

λ2 = 81

1  cos  442   Area ABC = 1 | AB || AC |sin BÂC 2

  1 442 sin  cos 1 1   2  442 

10.5cm 2

 p   2     15 a  b   2     p    p   3       p  2   a  b  2  p    p  3  

 1   2 1  1 and  1   1  

so 1  3 cos   1    cos 1    54.7  3

1  1  1

3

 p  2    2 p   p  3  

λ = ±9

 p  2  p  2   a  b    a  b    2  p    2  p    p  3   p  3    

1   11 The x-axis has unit direction vector 0   0  1  1        1 so  1   0    1  0     

1   0  0  

a + b = (5 + 1)i + (−3 + 1)j + (7 + λ)k = 6i − 2j + (7 + λ)k

 1   1       AB ⋅ AC  4    0  1  0   5          AB ⋅ AC = | AB || AC |cosθ



(2i + λj + k) ⋅ (i − 2j + 3k) = (2 × 1) + (λ × −2) + (3 × 1)

5 = 2λ 

1  17 26 cos 

c

62  7.87

= 2 − 2λ + 3 = 0 for perpendicular vectors.

 2   1  1         AB =  3    1  4  4  4 0        AB =  12  4 2  17 2 1 1  ⎛⎜ ⎞⎟ ⎛⎜ ⎞⎟ ⎛⎜ ⎟⎞ AC = ⎜ −1⎟ − ⎜ −1⎟ = ⎜ 0 ⎟ ⎜ −1⎟ ⎜ 4 ⎟ ⎜ −5 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠  2 AC 12   5  26  so AB

= 4 + 8 −12 = 0   = OA and OB are perpendicular    AB = OB − OA = (1 − 4)i + (2 − 4)j +(3 − (−4))k

so  6 162 cos 

10 a

  OA = 4i + 4j − 4k, OB = i + 2j + 3k   (OA) . (OB ) = (4 × 1) + (4 × 2) + (−4 × 3)



p

2

 4    4  p2   9  p2 

 p 2  9  0 for perpendicular vectors. 9, p   p2  3

Exercise 12J 1

a

⎛ −1⎞ ⎛ 3 ⎞ r = ⎜ ⎟ + t ⎜ ⎟ , t ∈ . ⎝ 2⎠ ⎝2⎠

b

⎛ −1⎞ ⎛ 5 ⎞ r = ⎜ ⎟ + t ⎜ ⎟ , t ∈ . ⎝ 0 ⎠ ⎝ −2 ⎠

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Worked solutions: Chapter 12

10

WORKED SOLUTIONS

2

c

⎛ 3⎞ ⎛ 3⎞ ⎜ ⎟ ⎜ ⎟ r = ⎜ 1⎟ + t ⎜ −2 ⎟ , t ∈ . ⎜ −2 ⎟ ⎜ 8 ⎟ ⎝ ⎠ ⎝ ⎠

d

r = 2 j − k + t ( 3i − j + k ) , t ∈ .

Then a ⋅ p = 0, and line is

b

⎛4⎞  3 Position vectors are ⎜ ⎟ and   ⎝5⎠  2  Line joining the 2 points has direction ⎛ 3 −4 ⎞ ⎛ −1 ⎞ ⎜ ⎟=⎜ ⎟ ⎝ −2 − 5 ⎠ ⎝ −7 ⎠ a

c

⎛ 4 ⎞ ⎛1⎞ Line is r = ⎜ ⎟ + t ⎜ ⎟ , t ∈ . ⎝ 5 ⎠ ⎝ 7 ⎠ 4   5 b Position vectors   and    2   2  Line joining 2 points has direction ⎛ 5 −4 ⎞ ⎛ 1 ⎞ ⎜ ⎟=⎜ ⎟ ⎝ −2 ( +2) ⎠ ⎝ 0 ⎠

d

⎛ 4⎞ ⎛1⎞ Line is r = ⎜ ⎟ + t ⎜ ⎟ , t ∈ . ⎝ −2 ⎠ ⎝0⎠ c

d

3  2   Position vectors  5  and  4  2  5     Line joining 2 points has direction −2 ⎞ ⎛ 1⎞ ⎛3 ⎜ ⎟ ⎜ ⎟ ⎜ 5 − ( −4 ) ⎟ = ⎜ 9 ⎟ ⎜2 −5 ⎟⎠ ⎜⎝ −3 ⎟⎠ ⎝

⎛3⎞ ⎛ 1⎞ ⎜ ⎟ ⎜ ⎟ Line is r = ⎜ 5 ⎟ + t ⎜ 9 ⎟ , t ∈ . ⎜2⎟ ⎜ −3 ⎟ ⎝ ⎠ ⎝ ⎠ 0  1   Position vectors  0  and  1 1  0    

Take p1 = 0, p2 = 4, p3 = 3 for example Then line is r = 5k + t ( 4 j + 3k ) , t ∈ . 4

a

b

a

⎛ p1 ⎞ We need a vector p = ⎜ ⎟ which is ⎝ p2 ⎠ perpendicular to a ⎛3⎞ ⎛ p ⎞ a ⋅ p = 0 ⇒ ⎜ ⎟ ⋅ ⎜ 1 ⎟ = 3 p1 + 2 p2 = 0 ⎝ 2 ⎠ ⎝ p2 ⎠ Take p1 = 2, p2 = −3

We need to know if there is a value of t for which ⎛2⎞ ⎛1⎞ ⎛ 4 ⎞ r = ⎜ ⎟ + t⎜ ⎟ = ⎜ ⎟ ⎝1⎠ ⎝2⎠ ⎝ 5⎠ ⎛1⎞ ⎛ 4 ⎞ ⎛2⎞ Take t = 2 Then ⎜ ⎟ = ⎜ ⎟ + 2 ⎜ ⎟ ⎝2⎠ ⎝ 5⎠ ⎝1⎠ so (4, 5) lies on the line. 5 4 5 Is there t so that ? t 1 3 2 5 + t ( 4 ) = 5 and 1 − 3t = − 2 t = 0 and t = 1 ⇒ no such t.

c

⎛0⎞ ⎛ −1⎞ ⎜ ⎟ ⎜ ⎟ Line is r = ⎜ 0 ⎟ + t ⎜ 1⎟ ⎜1⎟ ⎜ 1⎟ ⎝ ⎠ ⎝ ⎠ 3

⎛4⎞ ⎛1⎞ ⎜ ⎟ line is r = ⎜ 2 ⎟ + t ⎜⎜ 0 ⎟⎟ , t ∈ . ⎜1⎟ ⎜3⎟ ⎝ ⎠ ⎝ ⎠ ⎛ p1 ⎞ ⎜ ⎟ We require p = ⎜ p2 ⎟ so that p ⋅ a = 0 ⎜p ⎟ ⎝ 3⎠ p ⋅ a = p1 − 3 p2 + 4 p3 = 0

Line joining 2 points has direction −1⎞ ⎛ −1⎞ ⎛0 ⎜ ⎟ ⎜ ⎟ ⎜ 0 − ( −1) ⎟ = ⎜ 1⎟ ⎜1 −0 ⎟⎠ ⎜⎝ 1⎟⎠ ⎝

⎛ 2⎞ ⎛ −1⎞ r = ⎜ ⎟ + t ⎜ ⎟ , t ∈ . ⎝ 6⎠ ⎝ −3 ⎠ Using the same technique as in part a, we see ⎛2⎞  5 ⎜ ⎟ is perpendicular   5 ⎝ ⎠  2  ⎛ −1⎞ ⎛2⎞ Line is r = ⎜ ⎟ + t ⎜ ⎟ , t ∈ . ⎝ 0⎠ ⎝5⎠ ⎛1⎞  3 ⎜ ⎟ is perpendicular to   0 ⎜ ⎟  0 ⎜3⎟  1 ⎝ ⎠  

d

so (5, –2) does not lie on the line.  1   1   3        Is there t so that  5   t  0   5 ?  3   2   1        −1 + t = −3 ⇒ t = −2 5 + 0(t) = 5 ⇒ t = anything −3 − 2t = 1 ⇒ t = −2  3   1   1       so  5   5   2  0  i.e. (−3, 5, 1) lies on  1   3   2  line.       Is there t so that (2i + j + k) = (2i – j – 3k) + t(–2j –3k) 1 = − 1 − 2t and 1 = − 3 − 3t 4 = − 3t −2 = 2t

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Worked solutions: Chapter 12

11

WORKED SOLUTIONS t = −1

and t =

−4 3

5 5  4      s   No such s ⇒  3  1   3  lines NOT co-incident.

⇒ no such t.

so (2,1,1) does not lie on line. ⎛ −2 ⎞ ⎛2⎞ ⎜ ⎟, t  ⎟ ⎜ 5 r = 4 +t ⎜ 3⎟ ⎜ ⎟ ⎜ 8⎟ ⎜5 ⎟ ⎝ ⎠ ⎝ ⎠ 10 = 4 + 3t ⇒ 6 = 3t, t = 2 p = 2 − 2 t = 2 − 2( 2 ) = − 2 q = 5 + 8t = 5 + 8(2) = 21

d

−1 . Then 3 lines parallel. (2)

Are lines parallel? ⎛ 1⎞ ⎛1 ⎞ Is there t so that ⎜ ⎟ = t ⎜ ⎟ No such t ⎝ 1⎠ ⎝2⎠ ⇒ NOT parallel.

(2)

⎛0⎞ 6 A vertical line will have direction ⎜ ⎟ ⎝1 ⎠ − 6 0 ⎛ ⎞ ⎛ ⎞ so r = ⎜ ⎟ + t ⎜ ⎟ , t ∈  5 ⎝ ⎠ ⎝1 ⎠ 7 a (1) Are the 2 lines parallel?  2  6  Is there t such that    t    1  3 Take t =

(1)

Are lines perpendicular? Take dot product of direction vectors: ⎛ 1 ⎞ ⎛ 1⎞ ⎜ ⎟ . ⎜ ⎟ = 1 + 2 = 3 ⇒ NOT perpendicular. ⎝ 2 ⎠ ⎝ 1⎠

e

(1)

Are lines parallel? Is there t so that ⎛4⎞ ⎛ 4⎞ ⎜ ⎟ = t ⎜ ⎟ No such t ⇒ ⎝3 ⎠ ⎝ −3 ⎠

⎛ 2 ⎞ −1 ⎛ −6 ⎞ ⎜ ⎟= ⎜ ⎟ so ⎝ −1⎠ 3 ⎝ 3 ⎠

lines not parallel. Are lines perpendicular?

(2)

Take dot product of direction vectors: ⎛ 4⎞ ⎛4⎞ ⎜ ⎟ . ⎜ ⎟ = 16 − 9 = 7 ⇒ ⎝ −3 ⎠ ⎝ 3 ⎠

Are 2 lines co-incident? ⎛ −9 ⎞ Does ⎜ ⎟ lie on r1? ⎝ 10 ⎠ ⎛ 2⎞ ⎛ −9 ⎞ ⎛ 3 ⎞ ⎜ ⎟ = ⎜ ⎟ + s ⎜ ⎟ . Take s = –6. ⎝ −1⎠ ⎝ 10 ⎠ ⎝ 4 ⎠ ⎛ 2⎞ ⎛ −9 ⎞ ⎛ 3 ⎞ ⎜ ⎟ = ⎜ ⎟ −6 ⎜ ⎟ ⎝ −1⎠ ⎝ 10 ⎠ ⎝ 4 ⎠

NOT perpendicular. 1 8

a

(1)

(2)

Are lines parallel? ⎛ 1⎞ ⎛ −4 ⎞ Is there t so that ⎜ ⎟ = t ⎜ ⎟ ⇒ No such ⎝2⎠ ⎝ 2⎠ t, so NOT parallel.

b

(2)

Are lines co-incident? ⎛5⎞ Does ⎜ ⎟ lie on r1? ⎝3⎠

−1

0 ⋅ −2

Are lines perpendicular?

Are the lines parallel? ⎛ 8⎞ ⎛ 4⎞ Is there t so that ⎜ ⎟ = t ⎜ ⎟ ⎝ −6 ⎠ ⎝ −3 ⎠ t = 2 gives ⎛ 8⎞ ⎛ 4⎞ ⎜ ⎟ = 2 ⎜ ⎟ ⇒ lines parallel. ⎝ −6 ⎠ ⎝ −3 ⎠

1

2

cos A

1

3 = 1

2

−1

0

3

−2

1

cos A

−2 − 2 = 22 + (−2)2 (−1)2 + 32 + 12 cos A −4 =

8 11 cos A  −4  A = cos−1   = 115.2°  8 11 

⎛ −4 ⎞ ⎛ 1 ⎞ ⎜ ⎟ . ⎜ ⎟ = − 4 + 4 = 0 ⇒ perpendicular. ⎝ 2⎠ ⎝2⎠ (1)

4 ⋅ 1 = 4 0 1 0

6 = 17 6 cos A ⎛ 6 ⎞ A = cos−1 ⎜ ⎟ = 53.6° ⎝ 17 6 ⎠

Take dot product of direction vectors:

c

2

2 + 4 = 12 + 42 22 + 12 + 12 cos A

⎛ −9 ⎞ so ⎜ ⎟ lies on r1 ⇒ lines co-incident ⎝ 10 ⎠ b

1

2

9

a

⎛ −2 ⎞ ⎜ ⎟ A has position vector ⎜ −3 ⎟ . We require t so ⎜ −4 ⎟ that ⎝ ⎠  2   1 1         3   1  t  2  Taking t = –1, we see:  4   2  6      

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Worked solutions: Chapter 12

12

WORKED SOLUTIONS

b

−2 = − 1 + ( −1)1 −3 = − 1 + ( −1)2 −4 = 2 + ( −1) 6  6   2   4         AB   7    3    4   2   4   2        Taking dot product,

⎛ −4 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ −4 ⎟ . ⎜ 2 ⎟ = − 4 − 8 + 12 = 0 ⎜ 2⎟ ⎜6⎟ ⎝ ⎠ ⎝ ⎠  ⇒ AB perpendicularr toL1

10 a

b

c

 OF = 2i + 5j + 3k  ii AG = −2i + 5j + 3k  i |OF | = 22 52 32 38  ii | AG | = 22 + 52 + 32 = 38   iii OF . AG = (2i + 5j + 3k) . (−2i + 5j + 3k) = 30     OF . AG = |OF | | AG |cosθ 30 = 38 38 cos

e

1 P= 177

b

1

2

(1)

−2 + 2 s = − 3 + 6t (1) ⇒ 8 s = 2 + 9t

(2)

(1) ⇒ s 

30 177

+ 5j − 2k + µ(7i − 8j − 8k)   = OA + µ AB which is the position vector of a point on the line that passes through A with direction vector AB , and hence also passes through B. OP . AB = 0

⎛ 1 + 7μ ⎜ ∴ ⎜ 5 − 8μ ⎜ ⎝ −2 + 8μ

⎞ ⎛ 7⎞ ⎟ ⎟ ⎜ ⎟ . ⎜ −8 ⎟ = 0 ⎟ ⎜ 8⎟ ⎠ ⎠ ⎝

7 + 49μ − 40 + 64μ − 16 + 64μ = 0 49 μ= 177

1 2 8



54 15

4 intersec at     2 

49

= (i

 2  9t 

t 

30 177

Let r be the position vector for the point P.

1 8

(2) ⇒ 2  1  2  9t   3  6t 4 −8 + 2 + 9t = − 12 + 24 t 6 = 15t

|AO||AB| −1 × 7 + (−5) × (−8) + 2 × 8

Then r = (1 + 7µ)i + (5 − 8µ)j + (−2 + 8µ)k

d

4 + 8 s = 6 + 9t

AO . AB

= c

Equating components:

 s

30 cos 7.9 38    AB = OB – OA = 7i – 8j + 8k

=

Equating components of r1 & r2: (1) 4  2 11   (2) 2  4  16  2  (1) ⇒  = − 7 + 2 (2) ⇒ 2 − 4  = 16 + 2 ( −7 + 2 ) 2−4 = 2+4  =0 (1) ⇒  = − 7 so intercept at (4,2)

i

cos OAB =

⎛ 520 ⎞ ⎜ ⎟ ⎜ 493 ⎟ ⎜ 38 ⎟ ⎝ ⎠

Exercise 12K

1

11 a

Use the value of µ from part d to get:

3

5 + 2t = 3 + 2 s

6 15

  

7 10

 48  5  8    48  7    1     10  2   3  5  3   5   (1)

−1 + t = − 2 + s

(2)

2 − t = − 4 + 2s (2) ⇒ s = 1 + t

(3)

(1) ⇒ 5 + 2t = 3 + 2 (1 + t ) 5 + 2t = 5 + 2t

(so (1) & (2) are consistent)

(3) 2 − t = − 4 + 2 (1 + t ) 2 − t = − 2 + 2t 4 = 3t t = (2) ⇒ s 1 

4 3 4 7 3 3

Thus l1 & l2 intersect.

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Worked solutions: Chapter 12

13

WORKED SOLUTIONS  3 2  23    7   1    23 , 1 , 2  r  2    1    1 at   3 3 3  4  3  2  3  2         4

1 + 3t = − 1

(1)

1− t = s (1) ⇒ 3t = − 2

(2)

t 

7

a

3  2  1  a  a  5.

2 3

  5 (2) ⇒ s  1   2    3  3

  j, i.e. at  1, 5  3  5 If lines intersect then there are s & t so that

Intersect at i 

5 3

3 − t =1+ s

(1)

t =4+s 5 + 2t = s

(2)

b

s = −13 in (2) ⇒ t = 4 − 13 = − 9 check in (1): 3 − ( −9 ) ≠ 1 − 13 12 ≠ − 12 so there are no such s & t ⇒ skew a

3 − s = 14 + 3t

18

(1)

(2) −2 + 3s = −20 − 4t 5 − 5 s = 6 − 3t (3) (1) ⇒ s = − 11 − 3t (2) ⇒ 2  3(11  3t )  20  4 t −35 − 9t = − 20 − 4 t −15 = 5t t = −3 (1) ⇒ s = −11 −3(−3) = −2 check in (3): 5 − 5(−2) = 6 − 3(−3) 15 = 15 so lines intersect. Point of intersection = 14i − 20j + 6k − 3(3i − 4j − 3k) when t = −3 = 5i − 8j + 15k b

6  1  b        9  t  2   13  3   2   1       9  2t  13  t  2. 6  2  b  b  8. ⎛6⎞ ⎜ ⎟ OP has position vector ⎜ 9 ⎟ + ⎜3 ⎟ ⎝ ⎠

Take dot product of direction vectors: (−i + 3j − 5k) ⋅ (3i − 4j − 3k) = (−1 × 3) + (3 × −4) + (−5 × −3) = −3 −12 + 15 =0 ⇒ perpendicular.

⎛ 1⎞ ⎜ ⎟ t ⎜ 2 ⎟ for some t ⎜ −2 ⎟ ⎝ ⎠

 8 5  3        AB =  13    7    6  1  5   6         6  t   3       0 (OP ) ⋅ ( AB ) = 0   9  2t  .  6    3  2t   6      3 6 t 6 9 2t 6 3 2t 0

(3)

sub (2) into (3): s = 5 + 2 (4 + s), s = 13 + 2s

6

6  1  5        9  t  2  7  3   2   a        6  t 5  t  1

3t

54

12t

18 12t 27t 54 t

 so OP

c

 |OP |

8

a

0 0 −2

6

1

4

9 −2 3

2 = 5 , P is (4, 5, 7) 2 7

42 + 52 + 72 = 3 10

AB = b − a = ( 3 − 2 ) i + ( −2 − ( −1) ) j + ( −1 − 2 ) k

= i − j − 3k line is  2i  j  2 k     i  j  3k  for    b

2 +  = 7 + 2s −1 −  = s 2 − 3 = 3 + 2 s sub (2) in (1) ⇒ 2

(1) (2) (3) +  = 7 + 2 ( −1 −  )

2 +  = 5 − 2 3 = 3 ⇒  = 1 (2) ⇒ s = −1 − 1 = −2 (3) ⇒ 2 = − 3(1) − 1 = 3 + 2(−2)

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Worked solutions: Chapter 12

14

WORKED SOLUTIONS ∴ lines intersect.

1

(2) ⇒ t = (5 × 3 − 7) = 4

point is ( 2 + 1) i + ( −1 − 1) j + ( 2 − 3 ) k

c

2

 20  When t = 4, LS =   +  −8 

a − c = (2 − 3)i + ( −1 − ( −2)) j + (2 − ( −1)) k = − i + j + 3k

 4 When s = 3, RS =   + 3  −8 

AC = 1 + 1 + 32 = 11 d

3

Take dot product of direction vectors: ( i − j − 3k ) .( 2i + j + 2k ) = 2 − 1 − 6 = −5

a

want to find when a = b .

Exercise 12L Position of ship relative to buoy is

2

a

b

15 10

152

velocity =

10 2

displacement time

122

52

speed = v(t)

d

s( t )

4i

j

t (12i

5 j)

s(3)

4i

j

3(12i

5 j)

(1)

−8 − 2t = − 1 − 5 s (2) ⇒ t 1 (5 s  7) (2) ⇒ −7 + 5 s = 2t  2

(1) ⇒ 20

5 2

(5 s

7)

4

(2)

b

Collide at (3i + 3 j) + 1( 4 i + 3 j) = 7 i + 6 j

a

⎛ 11⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ rx = ⎜ 3 ⎟ + t ⎜ −1⎟ ⎜ −3 ⎟ ⎜ 4 ⎟ ⎝ ⎠ ⎝ ⎠

⎛2⎞ ⎛ 1⎞ ⎜ ⎟ ⎜ ⎟ ry = ⎜ −7 ⎟ + t ⎜ 1 ⎟ ⎜9⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎝ ⎠

Vy = 22 + 12 + 9 2 = 86 ms−1 b

40 i

402 + 162 = 8 29 m ⎛ 5⎞ ⎛ 4 ⎞ ⎛ 12 ⎞ ⎛ 20 ⎞ want ⎜ ⎟ + t ⎜ ⎟ = ⎜ ⎟ + s ⎜ ⎟ ⎝ −2 ⎠ ⎝ −1⎠ ⎝ −5 ⎠ ⎝ −8 ⎠ for collision

20 + 5t = 4 + 12 s

3 + 3t = 3 + 3s (2) ⇒ t = s

Vx = 12 + ( −1)2 + 4 2 = 18 = 3 2 ms−1 Meet if rx = ry at the same time ⎛2⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ 11⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ + t ⎜ −1⎟ = ⎜ −7 ⎟ + s ⎜ 1 ⎟ ⎜9⎟ ⎜ 4 ⎟ ⎜ −2 ⎟ ⎜ −3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

13 ms−1

distance =

e

4

⎛ 20 ⎞ ⎛ 5⎞ s(t ) = ⎜ ⎟ + t ⎜ ⎟ ⎝ −8 ⎠ ⎝ −2 ⎠ ⎛ 20 ⎞ ⎛ 5 ⎞ ⎛ 50 ⎞ = ⎜ ⎟ + 6⎜ ⎟ = ⎜ ⎟m ⎝ −8 ⎠ ⎝ −2 ⎠ ⎝ −20 ⎠

c

(1)

They collide 1 hour after 3 pm, ie at 4 pm.

5 13 km ⎛ 20 ⎞ ⎜ 4 ⎟ ⎛ 5⎞ = ⎜ −8 ⎟ = ⎜ ⎟ ms −1 ⎜⎜ ⎟⎟ ⎝ −2 ⎠ ⎝ 4 ⎠

3 + 4t = 4 + 3s

(1) ⇒ s = t = 1

⎛ 60 ⎞ ⎛ 45 ⎞ ⎛ 15 ⎞ ⎜ ⎟ − ⎜ ⎟ = ⎜ ⎟ ie 10Km North, 15Km East ⎝ 30 ⎠ ⎝ 20 ⎠ ⎝ 10 ⎠ b

Let A’s position be given by a  (3i  3 j)  t (4 i  3 j) Let B’s position be given by b = ( 4 i + 3 j) + s (3i + 3 j)

⎛ −5 ⎞ = cos −1 ⎜ ⎟ = − 120º (nearest degree) ⎝ 3 11 ⎠

a

 12   40    =  −16    −5  

Hence particles will collide.

Then −5 = 11 9 cos 

1

 20   40    =  −16    −8  

3i − 2 j − k ie (3, −2, −1)

16 j

11 + t = 1 + 2 s

(1)

3 − t = −7 + s

(2)

−3 + 4 t = − 2 + 9 s (1) ⇒ t = 2s − 10

(3)

(2) ⇒ 3 − (2 s − 10 ) = − 7 + s 13 − 2 s = − 7 + s 20 = 3 s , s =

20 3

⎛ ⎞ (1) ⇒ t = 2 ⎜ 20 ⎟ − 10 = ⎝3⎠ so ships do not collide.

10 3

≠s

12 s

40 + 25 s − 35 = 8 + 24 s s=3

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 12

15

WORKED SOLUTIONS ⎛ 1⎞ ⎛ 21⎞ ⎛ 11⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ rx (10 ) = ⎜ 3 ⎟ + 10 ⎜ −1⎟ = ⎜ −7 ⎟ ⎜ 4 ⎟ ⎜ 37 ⎟ ⎜ −3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 1 ⎛ 2 ⎞ ⎛ 21⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ry (10 ) = ⎜ −7 ⎟ + 10 ⎜ 1 ⎟ = ⎜ 3 ⎟ ⎜ 9 ⎟ ⎜ 88 ⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

c

⎛ 0⎞ ⎜ ⎟ ry − rx = ⎜ 10 ⎟ , ⎜ 51⎟ ⎝ ⎠ ry − rx 

10 2  512 

2701  51.97m

 5  1   6     3 a  b   1  3    2   3  ( 5)   8       5 1   4      a  b   1  3    4   3  ( 5)   2       6  4      (a  b ).(a  b)   2  .  4   24  8  16  0  8   2      ⇒ a − b and a + b are perpendicular. 4

Review exercise 1

 2   1       AB   3    2    5 3     

 3     1  2  

7 s = 3 + 2t 6 + 3s = 1 + 4t

(1)

−1 + s = 2 − t (3) ⇒ s = 3 − t

(3)

(2)

(1) ⇒ 7 (3 − t ) = 3 + 2t 21 − 7t = 3 + 2t 18 = 9t t =2 (3) ⇒ s = 3 − 2 = 1 check in (2) ⇒ 6 + 3(1) = 1 + 4 (2)

 7   2   9         BC   0    3   3   1  5   6       

⎛ 9⎞ ⎛ −3 ⎞  ⎜ ⎟ ⎟ 1⎜ Now AB = ⎜ 1 ⎟ = − 3 ⎜ −3 ⎟ = ⎜ −6 ⎟ ⎜ 2⎟ ⎝ ⎠ ⎝ ⎠

We need s & t so that

−1 3

 BC

LS = RS = 9 so s and t exist. ⎛ 0⎞ ⎛7 ⎞ ⎛7 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ so P has position vector ⎜ 6 ⎟ + ⎜ 3 ⎟ = ⎜ 9 ⎟ ⎜ −1⎟ ⎜ 1 ⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Since they contain a common point (B ), A, B, C are collinear.

Point (7, 9, 0) 2

The sides of the triangle are given by the vectors AB, AC, and BC

5

a

AB = (2l + 2J) – (5i – j + 6k) = – 3i + 3j – 6k AC = (–3i – 5j + 8k) – (5i – j + 6k)

b

= –8i – 4j + 2k Now AB · AC = (–3i + 3j – 6k) · (–8i – 4j + 2k)

c

  1   2   3  AB        7   4  3   3   2   1  AC           2   4   2     3   1  AB ⋅ AC   .    3  6  9  3   2      ˆ AB ⋅ AC = | AB || AC | cos BAC

ˆ ( 1)2  ( 2)2 cos BAC

= (–3 × –8) + (3 × –4) + (–6 × 2)

 9

= +24 – 12 – 12 =0

ˆ 9  18 5 cos BAC ˆ 9  3 2 5 cos BAC

Since AB · AC = 0, the vectors are perpendicular. Hence, A, B, and C form a right-angled triangle.

32  32

ˆ  cos BAC

3 2 5

6

a

P has position vector  6    2  4  3   

 2   6  8   2         2    2  8    10   1  3  4   1      

P is (−2, 10, 1) © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 12

16

WORKED WORKED SOLUTIONS SOLUTIONS ⎛ −1⎞ ⎛ −−22⎞ ⎛ 00⎞ ⎟⎞ ⎜⎛ −1⎟⎞ ⎜⎛ ⎟⎞ ⎜⎛ b Suppose ⎜10 ⎟ = ⎜−12 ⎟ + t ⎜11 ⎟ for some t b Suppose ⎜ 10⎟ = ⎜ −12⎟ + t⎜ 11⎟ for some t ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎝⎜ 11⎠⎟ ⎝⎜ 77⎠⎟ ⎝⎜−−33⎠⎟ ⎝ ⎠ ⎠ ⎝ ⎠ ⎝ (1) −2 = − t (1) −2 = − t (2) 10 = − 12 + 11t (2) 10 = − 12 + 11t (3) 1 = 7 − 3t (3) 1 = 7 − 3t t =2 (1) ⇒ (1) ⇒ t = 2 10 = − 12 + 11(2) = 10 (2) ⇒ (2) ⇒ 10 = − 12 + 11(2) = 10 (3) ⇒ 1 = 7 − 3(2) = 7 − 6 = 1 (3) ⇒ 1 = 7 − 3(2) = 7 − 6 = 1 so equations are consistent. so equations are consistent. ⎛ −−22⎞ ⎜⎛ ⎟⎞ so t = 2 gives ⎜⎜10 ⎟⎟ ∴ P lies on L so t = 2 gives ⎜ 10 ⎟ ∴ P lies on L22 ⎜ 1⎟ ⎝⎜ 1⎠⎟ ⎝ ⎠ 7 a 7 a

⎛⎛22⎞⎞ ⎛⎛11⎞⎞ ⎟ ⎜ ⎟ ⎜ L2: r = ⎜2 ⎟ + s ⎜3 ⎟ L2: r = ⎜ 2⎟ + s⎜ 3⎟ ⎜⎜⎜2 ⎟⎟⎟ ⎜⎜⎜4 ⎟⎟⎟ ⎝⎝ 4⎠⎠ ⎝⎝ 2⎠⎠

⎛ 11⎞ ⎛ 44⎞ ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ 0 = ⎜⎜7 ⎟⎟. ⎜⎜x ⎟⎟= 4 + 7 x + 3 = 7 + 7 x ⇒ x = − 1 0 = ⎜ 7 ⎟ . ⎜ x ⎟ = 4 + 7x + 3 = 7 + 7x ⇒ x = − 1 ⎜ 3 ⎟ ⎜1 ⎟ ⎝⎜ 3⎠⎟ ⎝⎜ 1 ⎠⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 44⎞ ⎛1 ⎞ ⎛ 7 ⎞ ⎛ 22⎞ ⎛ ⎟⎞ ⎜⎛ 1⎟⎞ ⎜⎛ 7⎟⎞ ⎜⎛ ⎟⎞ ⎜ c t q + = + − 3 7 5 ⎜−1 ⎟ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ c ⎜ −3⎟ + t⎜ 7⎟ = ⎜ 5⎟ + q⎜ −1⎟ ⎜⎜ 1⎟⎟ ⎜⎜−3 ⎟⎟ ⎜⎜3 ⎟⎟ ⎜⎜1 ⎟⎟ ⎝⎜ 1⎠⎟ ⎝⎜ −3⎠⎟ ⎝⎜ 3⎠⎟ ⎝⎜ 1⎠⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2 + t = 7 + 4q (1) 2 + t = 7 + 4q (1) −3 + 7t = 5 − q (2) −3 + 7t = 5 − q (2) (3) −3 + 3t = 1 + q (3) −3 + 3t = 1 + q b b

8 8

Suppose r1 = r2 Suppose r1 = r2 ⎛ 4⎞ ⎛4⎞ ⎛ −12 ⎞ ⎛ −4 ⎞ Then ⎜⎛ −4⎟⎞ +  ⎜⎛ 4⎟⎞ = ⎜⎛ 4⎟⎞ +  ⎜⎛ −12⎟⎞ Then⎝⎜ 3 ⎠⎟ + ⎝⎜17 ⎠⎟ = ⎝⎜9 ⎠⎟ + ⎝⎜ 5 ⎠⎟ ⎝ 17 ⎠ ⎝ 9 ⎠ ⎝ 5⎠ ⎝ 3⎠ −4 + 4  = 4 − 12  (1) −4 + 4  = 4 − 12  (1) (2) 3 + 17 = 9 + 5 (2) 3 + 17 = 9 + 5 4  = 8 − 12  (1) ⇒ (1) ⇒ 4  = 8 − 12   = 2 − 3  = 2 − 3 (2) ⇒ 17 (2 − 3 ) = 6 + 5 (2) ⇒ 17 (2 − 3 ) = 6 + 5 34 − 51 = 6 + 5 34 − 51 = 6 + 5 28 = 56  28 = 56  1 = 1  =2 2 3 1 3= 1 (1) ⇒  = 2 − (1) ⇒  = 2 − 2 = 2 2 2 1 So ships collide after 1 So ships collide after 2 2

hour, ie 12.30 pm. hour, ie 12.30 pm.  2   4   4   2 collide at  4  11 4  23  collide at  3   2 17    3 2 17  223       2  

 3  44 1  44  3 At 12.15, A has position   1   At 12.15, A has position    4     29   33 417 17  429  4  so after 12:15, A’s position given by so after 12:15, A’s position given by ⎛⎛−−33⎞⎞ ⎛ 16 ⎞ rr1 == ⎜⎜29 ⎟⎟ ++ tt⎜⎛ 16⎟⎞ where t is time after 12:15 ⎜⎜⎜ 29⎟⎟⎟ ⎝⎜17 ⎠⎟ where t is time after 12:15 1 17 ⎝ ⎠ ⎝⎝⎜ 44 ⎠⎟⎠ b b

At At12.30, 12.30, A’s A’s position position isis ⎛⎛−−33⎞⎞ ⎛1 ⎞ ⎛⎛16 ⎞⎞ ⎜⎛ 1 ⎟⎞ 1 ⎜ ⎟ 16 rr1 == ⎜29 ⎟ ++ 1 ⎜ ⎟ == ⎜23 ⎟ ⎜⎜ 29 ⎟⎟ 44 ⎝⎜17 ⎠⎟ ⎜⎜ 23⎟⎟ 1 ⎝ 17 ⎠ ⎝ 22 ⎠ ⎝⎝ 44 ⎠⎠ ⎝ ⎠  2 1  2  1  33      Distance is   Distance is 23   23   23   23  0   22   22   0  so soships ships are are 3km 3km apart. apart.

Review Review exercise exercise –– GDC GDC 1 1

2 3 33 ⋅⋅ 2 == 3 55 55 −4 −4

22 cos cos AA −4 −4

2 2 2 2 66 −− 20 20 == 332 ++ 552 222 ++ (−4) (−4)2 cos cos AA −14 −14 = cos A = cos A 34 34 20 20 AA == 122.4 122.4 ≈≈ 122° 122°

2 a 2 a

LQR : QR = OR − OQ 2 = OR − OQ 2 3 −1 ⎛ 2⎛⎜⎞ ⎞⎟⎛ 3⎛⎜⎞ ⎞⎟⎛ −1⎛⎜⎞ ⎞⎟ ⎜ ⎟−1 ⎜ − ⎟−1 ⎜ = ⎟ 0 −1⎜⎟ −⎟⎜ −1⎜⎟ =⎟⎜ 0⎜⎟ ⎟ = ⎜= ⎜ 5⎜⎝⎟ 5 ⎟⎠⎜ 0⎜⎝⎟ 0 ⎟⎠⎜ 5⎜⎝⎟ 5 ⎟⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = OP − OQ QPQP = OP − OQ 3 3 0 ⎛ 3⎛⎜⎞ ⎞⎟⎛ 3⎛⎜⎞ ⎞⎟⎛ 0⎛⎜⎞ ⎞⎟ ⎜ ⎟−2 ⎜ − ⎟−1 ⎜ = ⎟−1 −2⎜⎟ −⎟⎜ −1⎜⎟ =⎟⎜ −1⎜⎟ ⎟ = ⎜= ⎜ 1⎜⎝⎟ 1 ⎟⎠⎜ 0⎜⎝⎟ 0 ⎟⎠⎜ 1⎜⎝⎟ 1 ⎟⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

b b

⎛ −1 ⎞ ⎛ −1 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⋅ ⎜ −1 ⎟ = ⎜ 0 ⎟ ⎜ 5⎟ ⎜ 5⎟ ⎜ 1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 0+0+5=

⎛ 0⎞ ⎜ ⎟ ˆ ⎜ −1 ⎟ cos PQR ⎜ 1⎟ ⎝ ⎠

(−1)2 + 02 + 52

ˆ 02 + (−1)2 + 12 cos PQR 5 26 2

ˆ = cos PQR = 46.1 ≈ 46°

c c

1 1|QR||QP| sin 46 = Area 2 |QR||QP| sin 46 = Area 2 == 11 52 52 sin sin 46 46 2 2

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

2 == 2.60 2.60 units units2

Worked solutions: Chapter 12 Worked solutions: Chapter 12

17 17

WORKED SOLUTIONS 3

a

b

i

OC = 4j

ii

OB = i +

iii

OD = 2i + 4j

i

BC = BO + OC

22 − 12 k = i + 3 k

b

c

= −i − 3k + 4j = −i + 4j − 3k

c

ii

BD = i + 4j − 3k

i

|BC| =

12 + 16 + 3 =

20 = 2 5

ii

|BD| =

12 + 16 + 3 =

20 = 2 5

iii

(−i + 4j −

3k)·(i + 4j − 3k) = −1 + 16 + 3 = 18 ˆ 18 = 2 5 × 2 5 cos DBC 18 20

=

9 10

a

If perpendicular, a ∴ b = 0 ∴ x3 − 2x(x − 2) − 12x = 0 x3 − 2x2 + 4x − 12x = 0 x(x2 − 2x − 8) = 0 x(x − 4)(x + 2) = 0 x = 0, x = 4, or x = −2

1=2+µ

3 + 2λ = 2 − 2µ LHS 3 + 2 ×

11 149

a

1 2

in r1

⎛ 1 + 0 ⎞ ⎛ 1⎞ Position vector = ⎜⎜ −1 + 3 ⎟⎟ = ⎜ 2 ⎟⎟ ⎜ ⎜ 3 + 1 ⎟ ⎜ 4⎟ ⎝ ⎠ ⎝ ⎠ d

⎛ 0⎞ ⎛ 1⎞ ⎛ 0⎞ ⎛ 1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 6 ⎟ ⋅ ⎜ −3 ⎟ = ⎜ 6 ⎟ ⎜ −3 ⎟ cos A ⎜ 2 ⎟ ⎜ −2 ⎟ ⎜ 2 ⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ −18 + −4 =

11 149 cos C

= cos C

PQ = OQ − OP ⎛ 1⎞ ⎛ 1⎞ ⎛ 0⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜ 5 ⎟ − ⎜ −1 ⎟ = ⎜ 6 ⎟ ⎜ 5⎟ ⎜ 3⎟ ⎜ 2⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ 1⎞ ⎛ 0⎞ OP.PQ = ⎜⎜ −1 ⎟⎟ . ⎜⎜ 6 ⎟⎟ = 0 − 6 + 6 = 0 ⎜ 3⎟ ⎜ 2⎟ ⎝ ⎠ ⎝ ⎠ ∴OP is perpendicular to PQ

02 + 62 + 22

12 + (−3)2 + (−2)2 cos A

−22 = 40 14 cos A

−22 40 14

(−1)2 + (−3)2 + 12

82.9° = C 5

=4

consistent values for λ and µ in all 3 equations

= cos A 22

A = 158°

12 + 22 + 122 cos C

5

1 2

1 2

RHS 2 − 2 × −1 = 2 + 2 = 4

a⋅b = |a||b| cos C

5=

∴ µ = −1

−1 + 6λ = −1 − 3µ ∴ 6λ = −3µ = 3 ∴ λ =

Let x = −1. a = −1i + −3j + k, b = i + 2j + 12k −1 + −6 + 12 =

If intersect, r1 = r2 ⎛ 1⎞ ⎛ 0⎞ ⎛ 2⎞ ⎛ 1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ −1 ⎟ + λ ⎜ 6 ⎟ = ⎜ −1 ⎟ + µ ⎜ −3 ⎟ ⎜ 3⎟ ⎜ 2⎟ ⎜ 2⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Let λ =

(xi + (x − 2)j + k) ∴ (x2i − 2x j − 12x k) = 0

b

⎛ 0⎞ ⎜ ⎟ ⎜ 6⎟ ⎜ 2⎟ ⎝ ⎠

∴ lines intersect

ˆ cos DBC

ˆ 25.8° = DBC 4

⎛ 1⎞ r1 = ⎜⎜ −1 ⎟⎟ + λ ⎜ 3⎟ ⎝ ⎠

158 158

22

acute angle between lines is 22° ⎛ 0⎞ ⎜ ⎟ 6 a t = 0 A = ⎜ 0⎟ ⎜ 6⎟ ⎝ ⎠ ⎛ 6⎞ ⎜ ⎟ t = 2 B = ⎜ −2⎟ ⎜ 6⎟ ⎝ ⎠ AB = OB − OA ⎛ 6⎞ ⎛ 0⎞ ⎛ 6⎞ ⎜ ⎟ = ⎜⎜ −2⎟⎟ − ⎜ 0 ⎟ = ⎜⎜ −2⎟⎟ ⎜ 6 ⎟ ⎜⎝ 6 ⎟⎠ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠ b

Position vector = initial position + t (directional vector of AB) ⎛ 3⎞ = initial position + t ⎜⎜ −1⎟⎟ ⎜ 0⎟ ⎝ ⎠

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Worked solutions: Chapter 12

18

WORKED SOLUTIONS c

(36, 18, 0)

d

 −3    ν =  −4  , speed = |ν| =  1  

(−3)2 + (−4)2 + 12

= 9 + 16 + 1 =

26

= 5.10 ms−1 e

⎛ 36 ⎞ ⎜ ⎟ ⎜ 18 ⎟ + t ⎜ 1⎟ ⎝ ⎠

⎛ −3 ⎞ ⎛ 0 ⎞ ⎛ 3⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ −4 ⎟ = ⎜ 0 ⎟ + s ⎜ −1 ⎟ ⎜ 1⎟ ⎜ 6⎟ ⎜ 0⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

36 − 3t = 3s 36 − 18 = 18 ∴ s = 6 18 − 4t = −s 18 − 24 = −s ∴ s = 6 consistent t=6 t = 6 seconds f

c = (36 −18, 18 −24, 0 + 6) = (18, −6, 6)

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Worked solutions: Chapter 12

19

WORKED SOLUTIONS

13

Circular functions

Answers

5

Skills check You should know these values,without using your GDC. a

2 2

b

3

c

3 − 2

d

2 − 2

a

3 2

b

–1

c

–1

d

–0.5

3

a

–1.48

b

±2

4

a

–0.182, 2.40

b

±1.14

1

2

–90° (0, –1)

sin(–90°) = –1, cos(–90°) = 0, tan(–90°) does not exist 6

(–1, 0) –180°

Investigation 1

sin(–180°) = 0, cos(–180°) = –1, tan(–180°) = 0 7

(0, 1) 90°

(1, 0)

sin90° = 1, cos90° = 0, tan90° does not exist 2

(–1, 0)

sin0 = 0, cos0 = 1, tan0 = 0 8

(0, 1) r 2

180°

π

sin180° = 0, cos180° = –1, tan180° = 0 9

3 270°

sin π = 1, cos π = 0, tan 2 does not exist 2

2

r

(–1, 0)

(0, –1)

sin270° = –1, cos270° = 0, tan270° does not exist 4

sinπ = 0, cosπ = –1, tanπ = 0 10 3r 2

360° (1, 0)

(0, –1) 3π

sin360° = 0, cos360° = 1, tan360° = 0

sin 3π = –1, cos 2 = 0, tan 3π does not exist 2

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2

Worked solutions: Chapter 13

1

WORKED SOLUTIONS 11

c

(0, 1)

d

2

e

sin⎛⎜ − 3π ⎞⎟ = 1, cos ⎛⎜ − 3π ⎞⎟ = 0, tan − 3π does not exist ⎝

11r 6

r 2

– 3r

2 ⎠

⎝

2 ⎠

f

2

12 r 3

4r

5r 6

(1, 0)

g

h

sin4π = 0, cos4π = 1, tan4π = 0

2r

3

Exercise 13A 1

a

b

For questions 3–8, there are many other possible correct answers.

75°

110°

3

a 60° O

c

d

250°

120°, –240°, –300°

330°

b

200°

e

f

100°

340°, –20°, –160°

270°

c g

h 75° 40° 180°

255°, 285°, –105° d 2

a

b 115° 5r 3 r 6

65°, –245°, –295° © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 13

2

WORKED SOLUTIONS 4

d

a

25°

35°

155°, 335°, –205°

–35°, ±325° 6

b

a r 3

130°

2π 4π 5π ,− ,− 3 3 3

–130°, ±230° c

b 5r 4

295°

7π π 3π ,− ,− 4 4 4

–295°, ±65° d

c 240°

4:1

240°, ±120° 5

3π – 4.1, 4.1 – 2π, π – 4.1

a

d 50° 3

230°, –130°, –310° b

π + 3, 2π – 3, 3 – π 7

a r 6

100°

280°, – 80°, – 260° c

−

π 6

,±

11π 6

b 220°

40°, –140°, –320°

1

–1, ±(1 – 2π)

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Worked solutions: Chapter 13

3

WORKED SOLUTIONS c

2 2.5

–2.5, ±(2.5 – 2π) d

3

a

sin ⎛⎜ 7π ⎞⎟ = − sin ⎛⎜ π ⎞⎟ = − 1

b

cos ⎛⎜ 5π ⎞⎟ = − cos ⎜⎛ π ⎞⎟ = − 3 2 ⎝ 6 ⎠ ⎝6⎠

c

sin ⎜⎛ − π ⎟⎞ = − sin ⎛⎜ π ⎞⎟ = − 1

d

cos ⎛⎜ − 11π ⎞⎟ = cos ⎛⎜ π ⎞⎟ =

a

sin (180 − A ) = sin ( A ) = 0.8

b c

3r 5

3π 7π ,± 5 5

8

a r 4

4

5π 3π 7π ,− ,− 4 4 4

b 1.3

c

⎝6⎠

⎝ 6⎠

⎝6⎠

6 ⎠

⎝

2

2

⎝6⎠

3 2

cos ( − A ) = cos ( A ) = 0.6

cos ( 360 − A ) = cos ( A ) = 0.6

d

sin (180 + A ) = − sin ( A ) = −0.8

e

tan ( A ) =

f

tan ( − A ) = − tan ( A ) = − 4

g

sin ( 360 − A ) = − sin ( A ) = −0.8

h

tan (180 + A ) = tan ( A ) = 4

sin( A ) 0.8 4 = = cos ( A ) 0.6 3 3

3

= sin θ = a cos θ b

a

tanθ

b

sin (π −θ ) = sinθ = a

c

cos (π + θ ) = − cosθ = −b

d

tan (π + θ ) = tanθ = a

e

sin (π + θ ) = − sinθ = −a

f

cos ( −θ ) = cosθ = b

g

sin ( 2π −θ ) = − sinθ = −a

h

1.3 + π, 1.3 – π, 1.3 – 2π

⎝ 6 ⎠

b

cos (θ − π ) = − cosθ = −b

Exercise 13C 1

a

5r 7

60° O

12π 2π 9π ,− ,− 7 7 7

d

x = –300°, –240°, 60°, 120° b 120°

5

2π – 5, π – 5, –5 – π

Exercise 13B 1

a

sin110 = sin 70 = 0.940

b

cos ( −70 ) = cos 70 = 0.342

c

cos 250 = − cos 70 = −0.342

d

sin 290 = − sin 70 = −0.940

x = ±120°, ±240° c 45°

x = –315°, –135°, 45°, 225°

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Worked solutions: Chapter 13

4

WORKED SOLUTIONS d

d

r 2

x = –360°, –180°, 0°, 180°, 360° e

cos x = ±

θ=

1 2

e

π 3π − , 2 2

tan 2 θ = 3 tanθ = ± 3

45°

r 3

x = ±45°, ±135°, ±225°, ±315° f

tan x = ±

π

θ = ±3,±

1 3

f

2π 4π 5π ,± ,± 3 3 3

tan = 1

r 4

30°

x = ±30°, ±150°, ±210°, ±330° 2

a

θ=− 3

7π 3π π 5π ,− , , 4 4 4 4

a

r 6

θ=−

11π 7π π 5π ,− , , 6 6 6 6

θ = 0°, 360°, 720° b

b

45°

θ = −135°, −45°, 225°, 315°, 585°, 675°

θ = 0, ±π, ±2π c

c

tan θ = −1

r 6

45°

θ=

π 11π ± ,± 6 6

θ = −225°, −45°, 135°, 315°, 495°, 675°

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Worked solutions: Chapter 13

5

WORKED SOLUTIONS d

3 tan2 θ = 9

Exercise 13D

3 tan θ = 3 tanq = ± 3

1

2

a

30° 60°

2x = ±30, ±330 x = ±15°, ±165°

θ = ±60°, ±120°, 240°, 300°, 420°, 480°,

600°, 660° 4

b

a

6sin(2x) = 3 sin ( 2 x ) = 1 2

30°

x= b

π 2

2sin x = −1 sin x = −1

2x = −330, −210, 30, 150

2

x = −165°, −105°, 15°, 75° c

r 6

sin ⎛⎜ x ⎞⎟ = cos ⎛⎜ x ⎞⎟ ⎝2⎠

⎝2⎠

tan ⎛⎜ x ⎞⎟ = 1 ⎝2⎠

x = − 5π , − π 6

c

6

45°

5 1 = 10 2 =± 1 2

sin 2 x = sin x

x 2

r 4

x = 90° d

x = d

tan 2 ⎛⎜ x ⎟⎞ = 3 ⎝3⎠

tan ⎛⎜ x ⎞⎟ = ± 3

± p , ± 3p 4 4

3 4cos2 x = 3 cos2 x = 4

= 45

⎝3⎠

cos x =

±

3 2

60°

r 6

x 3

x = ± p , ± 5p 6

= ± 60

x = ±180°

6

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Worked solutions: Chapter 13

6

WORKED SOLUTIONS 2

a

b

(2sin x + 1)(sin x +1) = 0 sin x = − 1 or sin x = −1 2

r 6

r 6

x =

7p 3p 11p , , 6 2 6

2q = − 5p , − p , 7p , 11p

q =

6 6 6 6 5p p 7p 11p − ,− , , 12 12 12 12

c

b

(tan x + 1)(tan x + 1) = 0 tan x = −1

r 4

r 4

3q = − 11p − 7p , − 3p , p , 5p , 9p 4

4

4 4 4

4

d

q = − 11p , − 7p , − p , p 12

c

5p 3p , , 4 12 12 4

12

cos ⎛⎜ q ⎞⎟ = ± ⎝2⎠

1 2

x =

sin2 x − 6 sin x + 5 = 0 (sin x − 1)(sin x − 5) = 0 The second factor gives sin x = 5, which has no solution. sin x −1 = 0

r 4

q 2

q = d

x =

= ±p

1

a

= ±1

2

=

q = ± 3p

11 6

⎛ 11 ⎞ ⎟ 6 ⎝ ⎠ ⎝ 6 ⎟⎠

2

⎝6⎠

sin ( 2q ) cos ( 2q )

a

sin2 x + cos2 x = 1

2

sin 2 x +

⎛ 2⎞ ⎜− ⎟ ⎝ 3⎠

sin x =

5 3

2

=

⎛ 5 11 ⎞ ⎜⎜ ⎟⎟ ⎝ 18 ⎠ ⎛ 7 ⎞ ⎜− ⎟ ⎝ 18 ⎠

x =

2p 4p , 3 3

36

= −5

2

⎝ 3⎠

⎛ 5 ⎞⎛ 2 ⎞ ⎟⎟ ⎜ − ⎟ ⎝ 3 ⎠⎝ 3 ⎠

b

c

7 18

11 7

= 1 → sin 2 x = 1 − ⎜⎛ − 2 ⎞⎟ =

sin ( 2x ) = 2 sin x cos x = 2 ⎜⎜ 2r 3

5 11 18

2

tan ( 2q ) =

The second factor gives cos x = 3, which has no solution. cos x = − 1

=

cos ( 2 ) = 1 − 2 sin2  = 1 − 2 ⎛⎜ 5 ⎞⎟ = 1 − 50 = −

c

4

11 36

⎝6⎠

sin ( 2 ) = 2 sin cos = 2 ⎛⎜ 5 ⎞⎟ ⎜⎜

±p 2

(2cos x + 1)(cos x −3) = 0

2

+ cos2 q = 1 → cos2 q = 1 − ⎛⎜ 5 ⎞⎟ =

cosq =

b

a

sin 2 θ + cos 2 θ = 1 ⎛5⎞ ⎜ ⎟ ⎝6⎠

2q 3

3

p 2

Exercise 13E

4 ±p 2

sin ⎛⎜ 2q ⎞⎟ ⎝ 3 ⎠

3p 7p , 4 4

= −4

5 9

5 9

2

cos(2x) = 2cos 2 x −1 = 2 ⎛⎜ − 2 ⎞⎟ −1 = 8 −1 = − 1 tan = ( 2x )

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

sin ( 2 x ) = cos ( 2 x )

⎝ 3⎠ ⎛ 4 5⎞ ⎜⎜ − ⎟ 9 ⎟⎠ ⎝ = ⎛ 1⎞ ⎜− ⎟ ⎝ 9⎠

9

9

4 5

Worked solutions: Chapter 13

7

WORKED SOLUTIONS 3

a

sin2 θ + cos2 θ = 1 sin q + 2

⎛5⎞ ⎜ ⎟ ⎝6⎠

sin q cosq

tanq =

c

4

1 − ⎛⎜ 5 ⎞⎟ ⎝6⎠

= 1 → sin q = 2

=

sin ( 2q ) cos ( 2q )

a

sin 2 x + cos 2 x = 1

=

⎛ 5 11 ⎞ ⎜⎜ ⎟⎟ ⎝ 18 ⎠ ⎛7 ⎞ ⎜ ⎟ ⎝ 18 ⎠

5

=

sin x =

=

7 18

c

sin ( 2x ) = 2 sin x cos x = 2 ⎜

63 64

d

63 ⎞ ⎟ 8 ⎟⎠

= 1 − 2 ⎛⎜ − 1 ⎞⎟ ⎝ 8⎠

⎛ 63 ⎞ ⎜⎜ ⎟⎟ ⎝ 32 ⎠ ⎛ 31 ⎞ ⎜ ⎟ ⎝ 32 ⎠

2

=

= 1− 1 32

63 32

=

31 32

1

a

63 31

=

⎛ ⎞ 2 ⎜⎜ 63 ⎟⎟ ⎛⎜ 31 ⎞⎟ 32 ⎝ ⎠ ⎝ 32 ⎠

=

31 63 512

5

sinq =

3 5

b

cosq =

4 5

c

sin ( 2 ) = 2 sin cos = 2 ⎛⎜ 3 ⎞⎟ ⎛⎜ 4 ⎞⎟ =

d

cos ( 2 ) = 2 cos2  −1 = 2 ⎛⎜ 4 ⎞⎟ −1 =

a

sin 2 (2x) + cos 2 (2x) = 1

⎝ 5 ⎠⎝ 5 ⎠

24 25

⎝5⎠

⎛

⎞ ⎟ 2 2 ⎝ a +b ⎠

cos ( 2x ) = cos2 x − sin2 x = ⎜

b

2

−

⎛ ⎞ a ⎜ 2 ⎟ 2 ⎝ a +b ⎠

2

b 2 − a2 a2 + b 2

2sin x cos x = cos x 2sin x cos x − cos x = 0 cos x (2sin x − 1) = 0 cos x = 0 or 2sin x −1 = 0 1 2

cos x = 0

or sin x =

x = 90°

or x = 30°, 150°

tan(2x) = 1 2x = 45, 225 x = 22.5°, 112.5°

c

sin x + cos x = 0 sinx = −cos x tan x = −1 x = 135°

d

cosx = ±

32 −1 25

2

=

a

1 2

2

→ cos2 ( 2x ) = 1 − ⎛⎜ 24 ⎞⎟ = ⎝ 25 ⎠

cos ( 2x ) = −

49 625

=

⎛ 24 ⎞ ⎜ ⎟ ⎝ 25 ⎠ ⎛ 7 ⎞ ⎜− ⎟ ⎝ 25 ⎠

=

24 − 7

3 2

b

sin x − sin2 x = cos2 x sin x = sin2 x + cos2 x = 1 x = 90°

c

cos2 x − sin 2 x = cos2x =

7 25

sin ( 2 x ) cos ( 2 x )

sin2 x =

2x = −300, −240, 60, 120 x = −150°, −120°, 30°, 60°

7 25

+ cos2 ( 2x ) = 1

tan ( 2x ) =

⎞⎛ ⎞ a b ⎟⎜ 2 ⎟ 2 2 2 + + a b a b ⎝ ⎠⎝ ⎠ 2ab a2 + b 2

x = 45°, 135°

2

b

⎛

b

i

2

a2 + b 2

Exercise 13F

a

⎛ 24 ⎞ ⎜ ⎟ ⎝ 25 ⎠

b

cos x =

4

6

a a2 + b 2

b

sin ( 4 x ) = 2 sin ( 2x ) cos ( 2x )

3

625

√a2 + b2

=

= 1 − 2 sin2 x

=

⎝ 25 ⎠

63 8 ⎛

tan ( 2x ) =

2

=

⎝ 8 ⎠⎝

d

50 −1 36

⎝ 8⎠

sin ( 2 x ) cos ( 2 x )

336 625

a

5 11 18

2

cos ( 2x )

cos ( 4 x ) = 1 − 2 sin 2 ( 2 x ) = 1 − 2 ⎛⎜ 24 ⎞⎟ = − 527

−

b

5 11 7

=

sin ( 2x ) = 2 sin x cos x = 2 ⎛⎜ − 1 ⎞⎟ ⎜⎜ −

c

=

+ cos2 x = 1 → cos2 x = 1 − ⎜⎛ − 1 ⎞⎟ =

cos x = −

d

⎝ 25 ⎠ ⎝ 25 ⎠

X ⎛ ⎞ 2 ⎜⎜ 11 ⎟⎟ ⎛⎜ 5 ⎞⎟ 6 ⎝ ⎠⎝ 6 ⎠ ⎝6⎠

tan ( 2q ) =

sin ( 4 x ) = 2 sin ( 2x ) cos ( 2x ) = 2 ⎛⎜ 24 ⎞⎟ ⎛⎜ − 7 ⎞⎟ =

a

2

d

b

=

c

11 5

=

cos ( 2 ) = 2 cos2  −1 = 2 ⎛⎜ 5 ⎞⎟ −1 =

2

11 36

7

⎛ 11 ⎞ ⎜⎜ ⎟⎟ ⎝ 6 ⎠ ⎛5⎞ ⎜ ⎟ ⎝6⎠

sin ( 2 ) = 2 sin cos =

⎛ 1⎞ ⎜− ⎟ ⎝ 8⎠

2

11 6

sinq =

b

2

1 2

1 2

2x = ±60, ±300 x = ±30°, ±150°

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Worked solutions: Chapter 13

8

WORKED SOLUTIONS d

1 − 2 sin2 x = sin x 2 sin2x + sinx −1 = 0 (2 sin x −1)(sin x + 1) = 0) sin x = 1 or sin x = −1

b

2

x = 30°, 150° 3

a

b

sin x cos x

or

c

= sin x

2

c

x = d

2

2 3

or

x =0

2

cos4x − sin4x = (cos2x − sin2x)(cos2x + sin2x) cos2x × 1 = cos2x

cos4x = cos2(2x) = 1 − 2sin2(2x) = 1 − 2(sin(2x))2 = 1 − 2(2sin x cos x)2 = 1 − 2(4sin2 x cos2 x) = 1 − 8sin2 x cos2 x ⇒b=8

Exercise 13G 1

2  5 , 3 3  5 , 6 6

x =

−346°, −194°, 14°, 166° 2

sin2 2x + 2 sin 2x cos 2x + cos2 2x = 2 2 sin 2x cos2x + 1 = 2 sin 4x = 1 4 x = p , 5p x =

b

+ cosq

2 sin (2x) cos(2x) = sin(2x) 2 sin (2x) cos(2x) − sin(2x) = 0 (sin(2x))(2 cos(2x) −1) = 0 sin(2x) = 0 or 2cos(2x) −1 = 0 sin(2x) = 0 or cos(2 x ) = 1 x = 0,  ,  or

a

sin 2q cosq

7

2x = 0,  , 2 or 2x =

4

=

+ cosq

2sin3x cos3x = sin 2(3x) = sin 6x ⇒k=6

p 7p , 4 4 p 7p , 8 8

2 cos2 x −1 = cos x 2 cos2 x − cos x −1 = 0 (2 cos x + 1)(cos x −1) = 0 cos x = − 1 or cos x = 1

1 cosq

sinq cosq

6

2x = x =

= sinq

1 = sin2 θ + cos2 θ

x = − 90°

sinx = sinx cosx sinx cosx − sinx = sinx (cosx −1) = 0 sin x = 0 or cosx = 1 x = 0, π cos 2x = 1

1 cosq

2 2 p 5p , 8 8

sin x −1 = 1 − sin2 x sin2 x + sin x −2 = 0 (sin x − 1)(sin x + 2) = 0 sin x = 1 or sin x = −2 which is invalid x =p

±27°, 333° 3

2

c

d

cos 2 x = 2 cos2 x −1 cos 2 x = 1 cos x = ±1 x = 0, π sin 2 x = 1 2

sin x = x = 5

±

244°, 296°

1 2

p 3p , 4 4

working may vary a sin2 x + 2 sin x cos x + cos2 x = 1 + sin(2x) LHS: 1 + 2 sin x cos x = 1 + sin(2x) RHS

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Worked solutions: Chapter 13

9

WORKED SOLUTIONS Investigation: graphing tan x

4

1

Angle measure (x) (degrees)

Tangent value (tan x)

0

0

-30, +30

55°, 235°, 415°

-45, +45 -60, +60

5

−

1 3

- 3 , - 3

135

-1 -

3

3

1 3

0

180

1

210

6

1

-1, 1

120

150

−5.33, −4.10, 0.955, 2.19

,

3

225

1

240

3

300 315

- 3 -1

330

−

360

1 3

0

3 tan ± 90° and tan ± 270° are undefined. The limit of the

tangent as the angle approaches ± 90° or ± 270° is infinite. Asymptotes are often shown on graphs for values that do not exist.

±1.71, 4.58 7

Exercise 13H 1

−0.739 8

−297°, −117°, 63°, 243° 2

−0.637, 1.41 −107°, 73°, 253° 3

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Worked solutions: Chapter 13

10

WORKED SOLUTIONS 124° , 304° 4

Investigation – transformations of sinx and cosx

38°, 142°, 398°, 502° 5

1

−5.88, −2.74, 0.405, 3.55 6

−1.88, 1.26

2

3

7

4.55

4

8

5

−4.66, 1.20, 2.28, 4.77

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Worked solutions: Chapter 13

11

WORKED SOLUTIONS

Exercise 13I

10

Sine curve shifted upwards by 1 unit. y = sin x + 1

1

11

 π⎞ ⎛ Tangent curve shifted right by 4 . y = tan ⎜ x − 4 ⎟

y 1 –2r

0

–r

12

2r x

r

Cosine curve shifted right

Exercise 13J

–6

2

1

y 4 3 2 1 –r

y –0.5 –2r

0

–r

4

2r x

r

2

y

0

r 2

2 –2

r

3r 2

–r

2r x

–2r

0

r

–r

0

–4

4

y 1

y 1 0

–r

r

–2r – 3r –r – r 0 2 2 –1

2r x

5

–1

6

r

2r x

–2r

–r

3r 2r x 2

0

r

2r x

r

2r x

–1

–4

–2

7

y

6

0

–r

r

y

2r x

3

–1

1

–2 –2r

–3

–r

0 –1 –3

y

7

8

y

4

3

–2r –r 0 –4

r

1

2r x

for questions 9 − 12, answers may vary.

9

r

1

–2r –r 0 –2

8

r 2

y 2

y

–2r

2r x

–2

2r x

–1

–2r

r

y 4 2

1

5

2r x

–4

3

y

–r

0 –2

–4

–2r

r

2 –2r

4

2r x

y 4

2 –r – r

r

–0.5 0

3

2

and downwards by

4⎠

⎝

–4

–2r – 3r

⎠

1.5= units. y cos ⎛⎜ x − π ⎞⎟ −1.5

–2

–2r

⎝

 by 4

Cosine curve shifted to the right by y = cos ⎛⎜ x − 2π ⎞⎟ ⎝

2 . 3

–2r – 3r –r – r –10 2 2 –3

r 2

r

3r 2r x 2

3 ⎠

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 13

12

WORKED SOLUTIONS 8

1 –2r

3

2r x

r

–1

5

2 1 3

x 

10

Cosine graph, period of functions is 8π. y = cos(0.25x)

Vertical shift =

Tangent graph, period of functions is 4π. y = tan(0.25x)

12

Cosine graph, reflected in x-axis, amplitude is 3, period is 4π. y = −3 cos(0.25x)

Exercise 13K Want to write as y = asin(b(x + c)) + d and y = pcos(q (x + r)) + s

3

 3 

–3r –2r –r –10

6

 4

Vertical shift = y –r –r 0 2 –1

= 5.

So b = q =

7

–2

–r

 3, x 

2 . 3 2

y

0

r

x

–2

8

s

Amplitude = 2, period:

2 x

 1 , x = 4π. 2

Vertical shift = 4, horizontal shift = 0. y 6

2 –3r –2r –r

0

r

2r 3r x

Exercise 13L

= 2. 31 2

1

=1

period: 2 − 0 = 2 amplitude: 11.8 − 2.2 = 4.8 2

 

5  ( 5) 2

a

 4

 y 2cos  2  x  So y = 2sin(2x) + 1,

= 5.

   4 

vertical shift :

1 b

11.8 + 2.2 2

=7

horizontal shift: 0 (first maximum) 2 y = 4.8cos ⎛⎜ x ⎞⎟ + 7

Period = 2π − (−π) = 3π. 2 3

2 x

4

Horizontal shift: c = 0, r 

So b = q =

x

r

–1

 4 

Amplitude: a = p =

= −2, x = −π.

 = −1, horizontal shift =  4 .

02 2

Amplitude = 1.5, period:

2

Vertical shift: d = s =

4

r 2

2 x

Vertical shift = 0, horizontal shift =   .

2



2r 3r x

–2

  3  = π. 2

r

Amplitude = 1, period:

Amplitude: a = p = 3 − (−1) = 2. Period =

6

–1

5 12   7    4 3 3  3  So b = q = 2  1 4 2 3  1 Vertical shift = = −2 = d = 2   c 4 ,r  Horizontal shift: 3 3 1 4   y sin   x     2 , So  3  2



= 2, horizontal shift =  .

2

Period =

 y

3

4

Amplitude: a = p = 1

 

= 1,

6

Amplitude: a = p =

cos  1  x 2

5 1 2

2 x

5    4 

y

11

2

3

= 6π.

Sine graph, amplitude of functions is 7.5. y = 7.5 sin x

2  ( 5) = 3.5 2 2 Period = 2π. So b = q = 2 = 1. 2  ( 5) Vertical shift = 2 = −1.5 = d Horizontal shift: c = 2 , r  5 3 6 2   So y 3.5sin  x    1.5, 3   y = 3.5cos x + 5 1.5 6

2 

Amplitude = 3, period:

9

1

5 4

,r 

So y 5sin  2  x       , y 5cos  2  x 

0

–r

 2

Horizontal shift: c 

y

⎝ 2

⎠

2 3

Vertical shift: d = s = 0 © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 13

13

WORKED SOLUTIONS c

d

d

3

a

period: 20 − 4 = 16 amplitude: 2.1 − 0.5 = 0.8 vertical

2 2 .1 + 0 .5 shift: 2

= 1 .3

horizontal shift: 4 (first maximum) b

y = 0.8cos ⎛⎜ 2 ( x − 4 ) ⎞⎟ + 1.3 ⎝ 16

⎠

c

2

a

period: 55 − 25 = 30 21.9 − 9.3 = 6 .3 2 shift: 21.9 + 9.3 = 15.6 2

amplitude: vertical

horizontal shift: 25 (first maximum) b c

y = 6.3cos ⎛⎜ 2 ( x − 25) ⎞⎟ + 15.6 ⎝ 30

⎠

d

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 13

14

WORKED SOLUTIONS

Exercise 13M 1

a

period:

2π 0.5236

approximately 12 hours b

d(0) = 5.6 sin(0.5236(0 − 2.5)) + 14.9 ≈ 9.49 m

c

d(14) = 5.6 sin(0.5236(14 − 2.5)) + 14.9 ≈ 13.5 m

d

about 90 days 3

a b

after 10 minutes, the wheel will be at the maximum height, 46 m period: 20 min amplitude: 46 − 1 = 22.5 vertical

2 shift: 46 + 1 2

= 23.5

horizontal shift: 5 min

c

ht

22.5sin

2 20

t 5

23.5

h3

22.5sin

2 20

3 5

23.5

10.3 m

d

first maximum at about 05:30 2

a b

T (32) = 17.5 cos(0.0172(32 − 187)) + 12.5 ≈ − 3.06 °C high temp: 12.5 + 17.5 = 30 °C

4

a

4.8 minutes period: 12 37 − 5 = 2 shift: 37 + 5 2

amplitude:

16

vertical

= 21

horizontal shift: 1 (first minimum) or 7 (first maximum) g ( x ) = −16 cos ⎛⎜ 2p ( x −1) ⎟⎞ + 21 day 187 (about 6 July) c

⎝ 12

b

g(x ) =

−16 cos ⎛⎜ 2p ⎝ 12

⎠

( 4 −1) ⎟⎠⎞ + 21 = 21 gallons

c

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 13

15

WORKED SOLUTIONS

✗

2

3

21 2

tan ( x ) =

c

sin ( 2x ) = 2 sin x cos x = 2 ⎜⎜

=

⎛ 21 ⎞ ⎛ 2 ⎞ ⎟⎟ ⎜ ⎟ ⎝ 5 ⎠⎝ 5 ⎠

=

4 21 25

y –3 –2 –10 1 2 3 4 5 x –2

Review exercise 1

=

⎛ 21 ⎞ ⎜⎜ ⎟⎟ ⎝ 5 ⎠ ⎛2⎞ ⎜ ⎟ ⎝5⎠

b

7

early May and late August

sin ( x ) cos ( x )

–4 –6

a

cos110 = − cos70 = −0.342

b

cos250 = − cos70 = −0.342

Review exercise

c

cos(−290) = cos70 = 0.342

1

a

sin 140 = sin 40 = 0.643

b

sin 320 = − sin 40 = −0.643

c

sin(−140) = − sin 40 = −0.643

a

cos x = − 1

a

2

x = ±120°, ±240° b

tan x =

48.6°, 131.4°

1 3

b

x = −330°, −150°, 30°, 210° c

2 sin2 x − sin x = 1 2 sin2 x − sin x −1 = 0 (2 sin x +1)(sin x −1) = 0 sin x = − 1 or

sin x = 1

2

x = −150°, −30°, 210°, 330°, or x = −270°, 90° 4

±129°, 231° c

sin 2x + sin x = 0 2 sin x cos x + sin x = 0 sin x (2 cos x +1) = 0

5

sin x = 0

or

cos x =

x = 0,

or

x =

a

i

−

2 3

amplitude: a =

1 2

11 − 1 2

=5

−70.3°, 109.7°, 289.7°

horizontal shift: c = 4 vertical shift: d = ii

6

b=

2p , period

11 + 1 2

2

=6

and the period is 8. b =

b

4 0

  du  sin x

3

  cos( x )  C

5

 sin x dx   cos x

cos 1 x dx

x 2dx 1 3 x 3

4

 cos x

(2cos x 3sin x )dx 2 cos xdx 3 sin xdx 2(sin x ) 3( cos x ) C 2sin x − 3cos x + C

2

3

sin x dx , 10  

3 4



2  2  cos  x  dx 3  0

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

 0  

 1.30 and

3 3 4

 1.30

Worked solutions: Chapter 14

5

WORKED SOLUTIONS 4

∫

ln

ln

 3

= ln  4 = ln  3

when x when x ln

∫

5

ex cos ( ex ) dx ⇒ u = ex ; du = ex ;

f ( x ) = a sin(bx )

a

dx

 4

 3

 ln 4

u= 4 u= 3

then then

ex cos ( ex ) dx =

∫

x = ln

 3

 x = ln 4

The sine function has a vertical stretch by a factor of 2 ⇒ a = 2.

and

Since the period of f is 4π we have

⎛ du ⎞ ⎜ ⎟ cos ( u ) dx ⎝ dx ⎠

=

∫

u=

u=

 3

 4

2π = b

cos udu

y

= ⎡⎣ sin u ⎤⎦ = sin ⎛⎜  ⎞⎟ − sin ⎛⎜  ⎞⎟ = 3 − 2 2 ⎝3⎠ ⎝4⎠

∫

ln

 3

 4

ex cos ( ex ) dx ≈ 0.159 and

3− 2 2

1 2

(r, 2)

2

 3  4

ln

4π ⇒ b =

1

≈ 0.159

0 –1

r 2

r

3r 2r 5r 3r 7r 4r x 2 2 2

–2

Exercise 14G 1

2

  0

y = x sin x and 2x – 6 y 3 2 1

6

y = xsin x

2

y = cos x + sin2 x 2

x

1 2 3

1

A

0 –1

2

a

i

y cos x sin2 x

y cos x 2sin x cos x y cos x (1 2sin x ) y cos x (c d sin x ) c 1 and d 2

ii

cos x (1 + 2sin x ) = 0 ⇒ cos x = 0 or sin x = −1

∫ ( x sin x − 2x + 6 ) dx ≈ 12.1 3.041

0

y = x 2 − 2 and y = x + cosx y 3 2 1

y = x2 – 2

π 7π

⇒ x =,

y = x + cos x

b

1 2 3

1.891

k

  cos xdx 0

(x

2

− 2)

c

⎤ ≈ 6.31 ⎥⎦

 1 and 0 ≤ k ≤ π 2

2

k 1    sin x  0 2

1  2 1 1  sin k  sin0   sin k  2 2

1

6

a

y 8 6 f(x) = tan √x 4 2 0 –2

1

2

x=2 2

3

b

2

  0

 tan

x  dx  38.3 2

0

∫

7 6

 2

=2

⎡ cos x + sin(2 x ) ⎤ dx ⎣ ⎦

⎡ cos x + sin(2 x ) ⎤ ⎣ ⎦

2

= 4.25

dx ≈ 9.12

s (t )  e t sin t v (t )  s (t )  e t  (cos t )  (sin t )  e t   e t (cos t  sin t )

b

v(t) = e t (cos t sin t ) a(t ) v (t ) e t ( sin t cos t ) (cos t sin t ) e t 2e t cos t

a

s (t ) = 1 − 2 sin t v (t ) = s′(t ) = −2 cos t ⇒ v (0) = −2 cos(0) = −2 ms−1

b

−2 cost = 0 for 0 < t < π ⇒ t =π s

c

s ⎛⎜ π ⎞⎟ =1 − 2sin ⎛⎜ π ⎞⎟ =1 − 2 =−1m ⎝2⎠ ⎝2⎠

2

0

∫

 2

⎡ cos x + sin(2 x ) ⎤ dx ⎣ ⎦

a

x

 tan x dx  3.97

0

Exercise 14H

 k 

4



 2

2−

ii

⌠ ⎡ ( x + cos x ) − ⎮ ⌡−1.135 ⎢⎣   cos xdx 0

∫

i

x

x 2 − 2 = x + cos x ⇒ x ≈ −1.135,1.891

k

2

2 6

–3 –2 –1–10 –2

3

4 x

B

–2

x sin x = 2 x − 6 ⇒ x = 3.1 2



y y = 2x – 6

–3 –2 –1–10 –2 –3 –4 –5 –6

2sin  1 x  dx  8

2

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 14

6

WORKED SOLUTIONS 3

a

v (t )  esin t cos t during the interval

i

b

v (t ) = e2sin t −1

i

0 ≤ t ≤ 2π ⇒ cos t = 0 ⇒ t = π , 3π 2

sign of v

ii

+

–

t0

v

2

6

+ 3r 2

r 2

5 2r

4

which is on the interval b

π 2

2

< t < 3π

1

2

v (t )  esin t cos t v( t )  esin t  ( sin t )  (cos t )  esin t  (cos t )

0 –1

  sin tesin t  esin t cos2 t c

esin t cos tdt

u sin t ; du cos t

s t

esin t cos tdt

eu

dt

du dt

a

c

3

4

5

6

7

8

9

10 12 13 t

iii

No, the particle does not return to the origin. Looking at the area between the curve and the t-axis, there is more area above the axis than below indicating that the particle moves to the right a greater distance than to the left, so it never returns to the origin.

dt

v (t ) = 4 sin t + 3 cos t , t ≥ 0 displacement after 4 seconds =

2

e2sin t 1 5 t 1.11 s, 2.03 s, 7.39 s, 8.31 s.

eudu eu C esin x C e sin0 C 4 C 4 1 3 s (t ) e sin t 3 4

1

ii

s (0) 4 s t

v(t) = e2sint − 1

3

The particle moves left when v(t) < 0

∫

12 0

e2sin t −1 dt ≈ 24.1 m

4

∫ ( 4 sin t + 3 cost ) dt 0

5

b

∫ ( 4 sin t + 3 cos t ) dt ≈ 4.34 m

a

i

4

✗

0

v (t )   (t  1)sin  t

  2

ii

2

v(1.5) = −2.26 m

c

– 3.54

4

a

tan t 1 ⎞ = e 2 2 ⎟ cos t cos t ⎝ ⎠ 1

d

  0

 (t  1)sin  t

  dt 2 2

⎡

−

⎢⎣ 2

1

⎤

x cos x 2

⎥⎦

sin x 2

e

f ( x ) = x 2 cos x f ′( x )= ( x 2 )(− sin x ) + (cos x )(2 x ) = − x 2 sin x + 2 x cos x

f

y = ln(tan x ) 1 ⎞⎛ 1 ⎞ ⎟⎜ ⎟ 2 ⎝ tan x ⎠ ⎝ cos x ⎠

y ′ = ⎛⎜

 7.37 m

v ( t ) e2sin t 1 for 0  t  12 Use a GDC to evaluate a(1) a(1)  v (1) 5.82 ms2

f ( x ) = sin x 2 = ( sin ( x 2 ) ) 2 f ʹ( x ) = ⎢ 1 ( sin ( x 2 ) ) 2 ⎥ ⎡⎣⎢ ( cos( x 2 ) ) (2x ) ⎤⎦⎥ =

2r

The particle changes direction when velocity changes sign at 2.51 s and 3.54 s.

s (t ) = etan t s′(t ) = etan t ⎛⎜

2

+

3

2

2

2.51

y = sin3 x = ( sin x )

y ′ = 3 ( sin x ) (cos x ) = 3 sin 2 x cos x

t   (t  1)sin  t 2.51,3.54   0 for 0  t  4

–

f ( x ) = cos (1 − 2x )

f ′( x ) = ⎡⎣ − sin (1 − 2x ) ⎤⎦ (−2) = 2 sin (1 − 2x ) b

t0

6

a

Use a GDC to evaluate a (1.5) a (1.5)  v (1.5)  2.52 ms2

sign of v

c

1

where t  0

Since velocity and acceleration are both negative at 1.5 seconds, the particle is speeding up. b

Review exercise

g

1 ⎞ 1 ⎟= 2 ⎝ sin x ⎠ ⎝ cos x ⎠ sin x cos x

or ⎛⎜ cos x ⎞⎟ ⎜⎛

f ( x ) = (ln x )(sin x ) f ′( x ) = (ln x )(cos x ) + (sin x ) ⎛⎜ 1 ⎞⎟ ⎝x⎠

= (ln x )(cos x ) + sin x x

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 14

7

WORKED SOLUTIONS h

y = 2 sin x cos x or y = sin 2x y ʹ = ( 2 sin x ) (− sin x ) + (cos x )(2 cos x )

g

2

2

a

2

3 x dx 4  1 x 4   ( cos x )  C   4 x  sin  4 

c

 x )dx  cos(3

1 2

du dx

2

 sin(4 x  1)dx

1 4

2

 x cos(2x )dx  u  2x

2

  sin(2 t  1)  dt   cos (2t  1)  u 2

1 2

4 x  1  du   4  dx 

; du  dx

x

2

   1 u

2

1 2

(2 t  1)



 u ln x ; du dx

=6∫ u −2du =6 ( −u −1 ) + C = 3

a

∫

 3 −

 3

−6 2 + sin x

+C  3

sin x dx = [ − cos x ]

−

 3

⎛ ⎞ ⎛ ⎞ = − cos ⎜  ⎟ + cos ⎜ −  ⎟ ⎝3⎠ ⎝ 3⎠ =−1 + 1 =0 b

∫

 0

2

2

(1 + sin x ) dx = [ x − cos x ]0 

= [ − cos( )] − [ 0 − cos(0 )] = ( + 1) − (0 − 1) = 2 + 

c

∫

 0



( sin x + cos 2 x ) dx = ⎡⎢ − cos x + 12 sin 2 x ⎤⎥ ⎦0 ⎣

⎤ ⎡ ⎤ ⎡ = ⎢ − cos( ) + 1 sin(2 ) ⎥ − ⎢ − cos(0 ) + 1 sin(0 ) ⎥ 2 2 ⎦ ⎦ ⎣ ⎣ = (1 + 0 ) − ( −1 + 0 ) = 2

1 x

1  sin(ln x )   x dx   sin(ln x )  x  dx          sin(u )  du  dx  dx  

C

1 ⎛ du ⎞ ⎟ dx 2 ⎜ ⌡ u ⎝ dx ⎠

C

   sin(ln x )   f  x dx  sin(ln x )  1x  dx     

C

= 6⌠ ⎮

 1  du     2  dt   dt   

1 2 cos(2 t  1)

1 u e 2

1 cos x ) dx ⇒ 2 ( ⌡ (2 + sin x ) u= 2 + sin x ; du = cos x dx 1 ⌠ 6 cos x dx = 6⌠ cos x ) dx ⎮ ⎮ 2 2 ( ⌡ (2 + sin x ) ⌡ (2 + sin x )

 sin(2t  1)  dt

    1  u 2du   1  1 u 1   C 2  1 2 

dx

⌡ (2 + sin x )

2

   2sin(2t  1)   1  du  2  dt   sin(2t  1)

du dx

⌠ 6 cos x = dx 6⌠ h ⎮ ⎮ 2

 1  sin(2t  1)  dt    cos (2t  1) cos(2t  1);

du dt

eu d u

1 esin x 2 2

1 1  cos udu  sin u  C 4 4 1 sin(2x 2 )  C 4

 sin(2t  1) dt      cos (2t  1)  cos

1 2

eu

  cos(4 x  1)   C

2

x cos x 2 esin x ( x cos x 2 )dx

1 sin(3x )  C 3

 1  du  2  x cos(2x )dx   4  dx  cos(u )dx

e

cos x 2 (2 x )

xesin x cos x 2dx

  1 cos(4 x  1)  C 4 d

sin x 2 ;

du dx

 x 4  cos x  C

b

e sin x ( x cos x 2 )dx

u

= −2 sin x + 2 cos x or y ʹ = ( cos 2x ) (2) = 2 cos 2 x 2

2

xesin x cos x 2dx

d

∫

 2

3 2

5 sin x cos xdx ⇒ u = sin x ;

0

du dx

= cos x ;

  cos u  C  sin(u )du 

hen x =  when x = 0 then u = 0 and wh

  cos(ln x )  C

then u = 1

2

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 14

8

WORKED SOLUTIONS

∫

 2

∫ =5 ∫

3 2

x=

5 sin x cos xdx = 5

0

 2

x =0

a

3 2

⎛ ⎞ u ⎜ du ⎟ dx ⎝ dx ⎠

u =1

The sine graph has been stretched by a scale factor of 2 and shifted up 2. So p = 2 and q = 2.

3 2

u du

b

u =0

∫

Area =

1

⎡ 5⎤ = 5⎢2 u2 ⎥ ⎣ 5 ⎦0

3

⎡ ⎛ ⎞ ⎛ ⎞⎤ = ⎢ −2 cos ⎜ 3 ⎟ + 2 ⎜ 3 ⎟ ⎥ ⎝ 2 ⎠ ⎝ 2 ⎠⎦ ⎣ − ⎡⎣ −2 cos ( 0 ) + 2 ( 0 ) ⎤⎦ = 3 + 2

y = cos ( 3 x − 6 )

Review exercise

y (2) = cos(0 ) = 1 y′ =

− 1 sin 3

1

(3x − 6 )

a

y = 2cos 2x + cosx + 1, x = 0, x = 2 and the x-axis y

mtan = y′(2) =

− 1 sin(0 ) = 0 3

4 3 2 1

Therefore at (2, 1) the tangent line is horizontal, so the normal line is the vertical line through (2, 1) or the line x = 2 5

( 2 sin( x ) + 2 ) dx

0

= [ −2 cos x + 2 x ]02

5 5 ⎤ ⎡ = 5 ⎢ 2 (1) 2 − 2 (0 ) 2 ⎥ 5 ⎦ ⎣5 =2

4

3 2

Tangent line parallel to y = 1 x + 3 ⇒ m = 4

⎛ ⎞ = y sin ⎜ x ⎟ , 0 ≤ x ≤= π ⇒ y′ ⎝2⎠

⎛ 1 cos ⎜ x 2 ⎝2

⎛ 1 cos ⎜ x 2 ⎝2

⎞ 1 ⎛x⎞ 1 ⎟ = 4 ⇒ cos ⎜ 2 ⎟ = 2 ⇒ ⎠ ⎝ ⎠ ⎛ ⎛ 2π ⎞ ⎞ ⎛π ⎞ 3 = sin ⎜ 1 = ⎜ 3 ⎟ ⎟ sin ⎜3⎟ 2 2 ⎝ ⎠ ⎝ ⎝ ⎠⎠

The point is

x 2

0

1 4

2

⎞ ⎟, 0 ≤ x ≤ π ⎠

b

3

∫ ( 2 cos 2

Area =

2

0

4

x

x + cos x + 1) dx ≈ 4.53

y = 2sin x and y = 0.5x 2 sin x = 0.5 x ⇒ x = 0, 2.366

= π ⇒ x = 2π 3

1

y

3

2 1

(2.366, 1.183)

0

⎛ 2π 3⎞ , ⎜⎜ ⎟ 2 ⎟⎠ ⎝ 3

1

2

Area =

3

4

  2.366

x



2sin x  0.5 x dx  1.36

0

6

f (0) 2 f ( x ) x sin x f (x)

2

a

y = sinx and the x-axis for 0 ≤ x ≤ π y

x sin x dx 1 1 2

1 (0)2 2

cos x C

cos(0) C 1 2

f (x ) 7

x

2

x2

2

C

cos x 1

1

0

3r 4

r 2

Volume =

∫



 0

5r x 4

r

( sin x )

2

dx ≈ 4.93

 , x 0 and x 2 b y e cos x

f ( x ) = p sin( x ) + q , p, q ∈ `

y

y

3

4

2 1

2

0

0

r 4

r 2

r

3r 2

2r x

r 4

Volume = 

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

3r 4

r 2

∫

2 0

r

(e )

5r 4

cos x 2

3r 2

7r 2

2r

x

dx ≈ 45.0

Worked solutions: Chapter 14

9

WORKED SOLUTIONS 3

The area under the curve y = cosx between x = 0 and x = k, where 0 < k < π , is 0.942.

iii

2



k 0

cos xdx  0.942   sin x 0  0.942  k

sin(k )  sin(0)  0.942  sin k  0.942

k ≈ 1.23 4

a

i

⎛ ⎞ ⎛π ⎞ s′ ⎜ π = ⎟ 0 and s ″ ⎜ 5 ⎟ ≈ 18.4 > 0 5 ⎝ ⎠ ⎝ ⎠ Therefore by the second derivative test s has a relative minimum at t = π . 5

b

Total distance =

 s (t ) 2ecos(5t )  4 s (t ) 2e

cos(5 t )

(  sin 5t )(5)



2 0

10 sin(5t )ecos (5 t ) dt

 14.2 m

 10 sin(5t )ecos(5 t ) ii

s (t )   10 sin(5t ) ecos(5t ) ( sin(5t ))(5)  ecos(5 t )   10(cos(5t ))(5).

 50ecos(5t ) sin 2 (5t )  50ecos(5 t ) cos(5t ) s (t ) 50e cos(5t )  sin 2 (5t )  cos(5t )  

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 14

10

WORKED SOLUTIONS

15

Probability distributions

Answers

P (2 sixes) =

Skills check

P (1 six) =

1

a

x 

= b

x 

= 2

3

 fx f



P (1 six) =

(3  3)  (4  5)  (5  7)  (6  9)  (7  6)  (8  2) 357962

6   2



b

8   5

8!  8  7  6  5!3! 3  2  1 = 56

c

9   6

(0.3)3 (0.7)6 = 0.267

a

5.5 x

b

x  2.5 1.2

c

9x 0.2

6!  6  5 2!4! 2

∴x=

= 1.71875

∴ x − 2.5 = 0.48 ∴ x = 2.98

= 1.6 ∴ 9 − x = 0.32 ∴ x = 9 − 0.32 = 8.68

Exercise 15A 1

2

a

discrete

b

continuous

c b

× ×

5 6 1 6

5 6

×

1 6

=

1 36

5

= 36 5 = 36 5 6

=

25 36

0

1

2

25 36

10 36

1 36

1

1 1

2 1

3 1

4 1

5 1

6 1

2

1

2

2

2

2

2

3

1

2

3

3

3

3

4

1

2

3

4

4

4

5

1

2

3

4

5

5

6

1

2

3

4

5

6

c

= 3.2 ∴ 5.5 = 3.2x = 0.4

n P (N = n)

= 15

5.5 3.2

×

P (0 sixes) =

176 = 5.5 32 (10  3)  (12  10)  (15  15)  (17  9)  (20  2) 3  10  15  9  2 568 = 14.6 (3 sf) 39

a

1 6 5 6

1 6

n

1 2 3 4 5 6

P (N = n)

11 9 7 5 3 1 36 36 36 36 36 36

1

2

3

4

5

6

1

1

2

3

4

5

6

discrete

2

2

4

6

8

10 12

continuous

3

3

8

9

12 15 18

d

2 3

3 4

4 5

5 6

6 7

4

4

10 12 16 20 24

1

1 2

5

5

12 15 20 25 30

2

3

4

5

6

7

8

6

6

14 18 24 30 36

3

4

5

6

7

8

9

4

5

6

7

8

9

10

5

6

7

8

9

10 11

6

7

8

9

10 11 12

a

p P (P = p) p P (P = p)

s 2 3 4 5 6 7 8 9 10 11 12 P(S = s) 1 2 3 4 5 6 5 4 3 2 1 36 36 36 36 36 36 36 36 36

b

1 6 1 6

5 6

6

6 P(2 sixes) =

5 6

6' P(1 six)

1 6

6 P(1 six)

1 1 1 × = 6 6 36

5 1 5 = × = 6 6 36

=

5 5 1 × = 6 6 36

6' P(0 sixes) =

1 5 5 × = 6 6 36

6' 5 6

36

36

3

a

1

2

3

4

5

6

8

9 10

1 36

2 36

2 36

3 36

2 36

4 36

2 36

1 36

2 36

12 15 16 18 20 24 25 30 36 4 36

2 36

1 36

2 36

2 36

2 36

1 36

2 36

1 36

The faces are numbered 1, 2, 2, 3, 3, 3 1

2

2

3

3

3

1

2

3

3

4

4

4

2

3

4

4

5

5

5

2

3

4

4

5

5

5

3

4

5

5

6

6

6

3

4

5

5

6

6

6

3

4

5

5

6

6

6

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 15

1

WORKED SOLUTIONS t P (T = t)

4

2

3

4

5

6

1 36

4 36

12 36

12 36

9 36

12 36

9 + 36 =

a

no. on dice

1

2

3

4

5

6

s

2

1

6

2

10

3

b

P (S > 2) =

a

1 3

b

+

1 3

1

2

3

6

10

1 6

2 6

1 6

1 6

1 6

3 6

=

1 3

∴ c=

= 31 × 61 + 31 ×

+1×1= 6 3 6 18

5 = 31  16  31  32  18

1 P(C = 6) = P(A = 3 and B = 3) = 31  16  18

c P (C = c)

1 6

P(1 < X < 4) = P(X = 2) + P(X = 3) = 31 + 16 =

2

3

4

5

6

1 18

5 18

6 18

5 18

1 18

Investigation – dice scores

P(Y = y) = cy 3 y = 1, 2, 3 2 8c

1 2

3 27c

1

1 0

2 1

3 2

4 3

5 4

6 5

1 36c = 1 ∴ c = 36

2

1

0

1

2

3

4

3

2

1

0

1

2

3

2k + 4k + 6k + k = 1

4

3

2

1

0

1

2

10k + 3k − 1 = 0

5

4

3

2

1

0

1

(5k − 1) (2k + 1) = 0

6

5

4

3

2

1

0

2

1

2

k = 15 (k cannot be negative) P(X = x) = k x P (X = x)

9

2 3

P(C = 5) = P(A = 2 and B = 3) + P(A = 3 and B = 2)

2

8

1 = 5 = 92 + 18 18

P(C = 4) = P(A = 1 and B = 3) + P(A = 2 and B = 2) + P(A = 3 and B = 1)

1 2

1 c

1 6

1 P(C = 2) = P(A = 1 and B = 1) = 31  16  18

b

+c+c=1

y P (Y = y)

7

7 = 12

P (T > 4) =

2c =

6

= 31 × 32 + 31 ×

b

s P (S = s)

5

21 36

P(C = 3) = P(A = 1 and B = 2) + P(A = 2 and B = 1)

10 a

1 k 9

 1 3

x 1

x = 1, 2, 3, 4

1

2

3

4

k

1 3k

1 k 9

1k 27

1 k= 27

k+

1 k 3

40 k 27

=1 ∴ k=

a

x P (X = x)

+

d P (D = d )

+

2

1 3

27 40

0

1

2

3

4

5

a

a

a

b

b

b

3a + 3b = 1 (1)

P(X ≥ 2) = 3P(X < 2)

a + 3b = 6a substitute (2) and (1)

b

b=

P(5, 3) =

5 24

P(5, 4) =

25 576

2

3

4

5

10 36

8 36

6 36

4 36

2 36

Mean =

0

1

2

3

4

5

6

10

8

6

4

2

(0  6)  (1  10)  (2  8)  (3  6)  (4  4)  (5  2) 36

70 36

35 = 18

d Expected frequency

0

1

2

3

4

5

150 9

250 9

200 9

150 9

100 9

50 9

Mean =

3b = 5a (2) a = 18

1

6 36

d Expected frequency

= 4

0

0  1509   1 2509    2  2009   3  1509    4  1009   5  509 

3a + 5a = 1

100

5 24

=

5  25  24 576

P(sum > 7) =

P(3, 5) =

P(4, 5) = 125 576

25 576

25 576

P(5, 5) =

25 576

35 18

5

The means are the same

6

35 18

Exercise 15B

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 15

2

WORKED SOLUTIONS 1

P (R = 1) =

2 10

 51

P (R = 2) =

8 10

 92 

P (R = 3) =

8 10

 79  82 

P (R = 4) =

8 10

 79  86  72 

P (R = 5) =

8 10

 79  86  75  62 

E(Z ) = 5 ∴

P (R = 6) =

17 2  16   3  16   5  16   7x  11y  3

8 10

 79  86  75  46  52 

P (R = 7) =

8 10

 79  86  75  46  35  42 

7x + 11y = 4 (2)

8 7654322  P (R = 8) = 10 9 8 7 6 5 4 3

x P (X = x)

E (X) = = 2

3 6

1

4

9

16

24

36

1 6

1 6

1 6

1 6

1 6

1 6

8

a

1  16    4  16   9  16   16  16   25  16   36  16  91 6

 15.2 (3 sf) 1 2

+x+y=1 ∴x+y=

(1)

2

3

Solving (1) and (2), x = 38 , 3

x P (X = x)

y = 18

P (R = 9) =

1

2

3

5

8

13

1 6

1 6

1 6

1 6

1 6

1 6

1 6

16 3

E(X ) = (1 + 2 + 3 + 5 + 8 + 13) = 4

x P (X )

E(X ) = = 5

a

1 36



204 36

1

2

3

4

5

6

7

8

1 36

2 36

3 36

4 36

5 36

6 36

7 36

8 36

4 36



9 36

 16  36

x

1

2

a



3

4

5

7

8 9

x

2

3

0.2

1−k

k − 0.2

0.2 ≤ k ≤ 1.2

10 90



5 45 8 90

4 45



6 90

 4 90

3 45 2 45



1 45

1

2

3

4

5

6

7

8

9

9 45

8 45

7 45

6 45

5 45

4 45

3 45

2 45

1 45

1 45

(9 + 16 + 21 + 24 + 25 + 24 + 21

P (R = 3) =

c

P (R = n) = 0.8n−1 × 0.2

d

1

10 a

4 25

16 (108 ) × 102 = 125 2

P (Z = 0) + 0.2 + 0.05 + 0.001 + 0.0001 = 1 P (Z = 0) = 1 − 0.2511 = 0.7489 E (Z ) = 0 + 0.4 + 1 + 0.2 + 0.1

A ticket costs $2, but you only expect to win $1.70. Therefore you expect to lose $0.30

Investigation – the binomial quiz You would expect to get 2.5 questions right P (3 right) =

2

4

a

0.3

b

∴ a + 0.6 + 4b = 2.8 a + 4b = 2.2

1

(0.5)3 (0.5)2 = 0.3125

 

X~ B 4, 12

( ) ( 12 )

⎛4⎞ 1 ⎝ ⎠ 2

1

3

=

1 4

a

P (X = 1) = ⎜ 1 ⎟

b

P (X < 1) = P (X = 0) =

c

1 +1= 5 P (X ≤ 1) = P (X = 0 or 1) = 16 4 16

d

1 15 = 16 P (X ≥ 1) = 1 − P (X = 0) = 1 − 16

(2)

solving (1) and (2), a = 0.2, b = 0.5

5   3

Exercise 15C

1

(1)

∴ P (x = 1) = 0.2

2 90

b

and 0 ≤ k − 0.2 ≤ 1

= 0.2 + 2 − 2k + 3k − 0.6 = k + 1.6

mean = 2.8

6 45



 79  86  75  46  35  42  31  22

8 × 2 = 16 = P (R = 2) = 10 10 100

c

Mean = 0.2 + 2(1 − k) + 3(k − 0.2)

a + b = 0.7

12 90

$1.70 is the expected winnings on a ticket

∴ 0.2 ≤ k ≤ 1

x P (x = x)

7 45



= 1.7

1

0≤1−k≤1

14 90

a

b

1≥k≥0

7

9

64 36

6

Mean =

8 45



+ 16 + 9) = 3 32

1 25

−1 ≤ − k ≤ 0

c

b

From symmetry, E (X) = 5 P (X = x)

b

49 36

2k 3k 4k 5k 4k 3k 2k k

25k = 1 ∴ k =

6

36   36

r P (R = r)

= 5 32

P (X = x) k

b

25 36



5 31

or

8 10

16 90

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

 12   161 4

Worked solutions: Chapter 15

3

WORKED SOLUTIONS using a GDC: a 0.329 b 0.351 c 0.680 d 0.649 3 using a GDC: 2

a b c d

P(X = 5) = 0.0389 P(X < 5) = 0.952 P(X > 5) = 0.00870 P(X ≥ 1) = 0.932

2

X ~ B (n, 0.01) P (x = 0) > 0.5 n 0.99 > 0.5 n log 0.99 > log 0.5 log 0.5 n < log 0.99 = 68.9, so the largest sample size is 68

3

X ~ B (n, 0.2) P (X ≥ 1) > 0.75 1 − P (X = 0) > 0.75 1 − 0.8n > 0.75 0.25 > 0.8n

Exercise 15 D

0.8n < 0.25

(Using a GDC where possible) 1 X ~ B (4, 0.25)

n log 0.8 < log 0.25 n>

x 0 1 2 3 4 P (X= x) 0.316 0.422 0.211 0.0469 0.00391

Most likely outcome is 1 red face with probability 0.422 2

3

4

5

n > 6.21 ∴ least value of n = 7 4

X ~ B (n, 0.3) P (x ≥ 1) > 0.95

X ~ B (8, 0.55)

1 − P (x = 0) > 0.95

a

P (X= 5) = 0.257

1 − 0.7n > 0.95

b

P (at least 5 times) = P (X < 3) = 0.260

0.05 > 0.7n

X ~ B (16, 0.01)

0.7n < 0.05

a

P (X = 0) = 0.851

n log 0.7 < log 0.05

b

P (13 not faulty) = P (X = 3) = 0.000491

n>

c

P (X ≥ 2) = 0.0109

n > 8.399

X ~ B (10, 0.25) a

P (X = 5) = 0.0584

b

P (at least 3 free) = P (X ≤ 7) = 0.9996

5

X ~ B (n, 0.5) P (X ≥ 1) > 0.99 1 − P (X = 0) ≥ 0.99

X ~ B (5, 0.4)

1 − 0.5n ≥ 0.99

7

P (X > 1) = 0.224

b

P (X = 1) = 0.399

0.5n ≤ 0.01 n log 0.5 ≤ log 0.01 n≥

X ~ B (15, 0.05) a

b

P (X = 3) = 0.0307

ii

P (X = 0) = 0.463

iii

P (X ≥ 2) = 0.171

Exercise 15F

i

(0.46329...)2 = 0.215

1

ii

(0.17095....)2 = 0.0292 0.46329..... × 0.17095.... × 2 = 0.158

Exercise 15 E X ~ B (n, 0.6) P (X < 1) = 0.0256 P (X = 0) = 0.0256 (0.4)n = 0.0256 n log 0.4 = log 0.0256 n=4

log 0.01 log 0.5

n > 6.64 so the coin must be tossed 7 times.

i

iii

1

0.01 ≥ 0.5n

X ~ B (6, 0.15) a

log 0.05 log 0.7

∴ least number of attempts is 9

P (X ≤ 3) = 0.913 6

log 0.25 log 0.8

2

a

X ~ B (40, 0.5)

b

X ~ B 40,

c

X ~ B (40, 0.25)



X ~ B (n, p)

1 6



E (X) = 40 × 0.5 = 20 E (X) = 40 ×

1 6

= 6 32

E (X) = 40 × 0.25 = 10

mean = 10 p = 0.4 np = 10

n × 0.4 = 10 ∴ n = 25 3

a

X ~ B (15, 0.25)

b

mean = 15 × 0.25 = 3.75

c

P (X ≥ 10) = 0.000795

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 15

4

WORKED SOLUTIONS 4

a

total number of girls = (1 × 34) + (2 × 40) + (3 × 13) = 158

7

X ~ B(n, p) E(X) = 7.8

total number of children = 100 × 3 = 300 153 300

∴ P(girl) = b

p = 0.3

= 0.51

a

E(X ) = np = 7.8

X ~ B (3, 0.51) P (x = 2) = 0.382

0.3n = 7.8

expected number of families = 0.382 × 100 = 38.2

n=

n = 26 b

Exercise 15G 1

( )

= 26 × 0.3 × 0.7

1 4

= 5.46

=0

(

1 4

Variance = 0 × 2

1 4

× 1−

)=0

B(12, 0.6) Variance = 12 × 0.6 × 0.4 = 2.88 Standard deviation = 2.88 =1.70 (3 sf)

(

X ~ B 40, 12

)

Mean = 40 ×

5

6

X ~ B(n, p) E(X) = 9.6 E(X) = np = 9.6 Var(X) = npq = np(1 – p) = 1.92 9.6(1 – p) = 1.92 9.6 – 9.6p = 1.92

1 2

= 20

Variance = 40 × 4

8

Var(X) = 1.92

Mean = 12 × 0.6 = 7.2

3

Var(X ) = npq = np(1 – p)

X ~ B 0, 41

Mean = 0 ×

7.8 0.3

1 2

×

9.6p = 7.68 1 2

= 10

p = 0.8 9.6 0.8

Standard deviation = 10 = 3.16 (3 sf)

n=

X ~ B 10, 16

n = 12

(

)

1 6

=

5 3

1 6

×

a

E(X ) = 10 ×

b

Var(X ) = 10 ×

c

P(X < μ) = P X < 35 = 0.485 (3 sf ) (using binomial CDF on the GDC)

(

(

X ~ B 22, 15

5 6

=

)

25 18

)

22 5

a

E(X) = 22 ×

1 5

=

b

Var(X) = 22 ×

1 5

c

P(X < 4) = 0.332 (3 sf ) using binomial CDF on the GDC)

×

4 5

=

88 25

X ~ B(n, p) E(X ) = 4.5, Var (X) = 3.15 E(X ) = np = 4.5 Var(X ) = npq = np(1 – p) = 3.15 4.5(1 – p) = 3.15 1–p=

3.15 4.5

= 0.7 p = 1 – 0.7 = 0.3 np = 4.5 n=

4.5 p

=

4.5 0.3

= 15

P(X = 6) = 0.0155 (using binomial PDF on the GDC)

Exercise 15H Using GDC: 1 a 0.683 b 0.954 c 0.997 2 a P (1 < Z < 2) + P (−2 < Z < −1) = 0.272 b P (0.5 < Z < 1.5) + P (−1.5 < Z < −0.5) = 0.483 3 a P (Z > 1) = 0.159 b P (Z > 2.4) = 0.00820 4 a P (Z < −1) = 0.159 b P (Z < −1.75) = 0.0401 5 a 0.742 b 0.236 c 0.0359 d 0.977 e 0.390 6 a 0.306 b 0.595 c 0.285

P(X ≥ 3) = 1 – P(X < 3) = 1 – 0.126828 = 0.873 (3 sf ) © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

Worked solutions: Chapter 15

5

WORKED SOLUTIONS 7

a b

P (|Z| < 0.4) = P (−0.4 < Z < 0.4) = 0.311 P (|Z| > 1.24) = 1 − P (|Z| < 1.24) = 1 − P (−1.24 < Z < 1.24) = 0.215 3

Exercise 15I

0.997

c

0.494

2

∴ P (Z < a) = 0.9452 4

4

5

2

a

P (Z < a) = 0.95

∴ a = 1.64

b

P (Z < a) = 0.8

∴ a = 0.842

X ~ N (5.5, 0.22) P (X > a) = 0.235 P (x < a) = 0.765 ∴ a = 5.64 2 M ~ N (420, 102) a P (M < a) = 0.25 ∴ a = 413 b P (M < b) = 0.9 ∴ b = 433 1

X ~ N (100, 20 ) 2

a

P(X < 130) = 0.933

b

P(X > 90) = 0.691

c

P(80 < X < 125) = 0.736

X ~ N (4, 0.252)

X ~ N (14, 42) a

P (X > 20) = 0.0668

b

P (X > 10) = 0.159 = 15.9%

3

X ~ N (502, 1.62) a P (x < 500) = 0.106 b P (500 < x < 505) = 0.864 or 86.4% c P (x < b) = 0.975 b = 505.1 a = 498.9 a = 499 b = 505

4

X ~ N (550, 252) a P (520 < X < 570) = 0.673 b P (X > a) = 0.1 ∴ P (X < a) = 0.9 ∴ a = 582

5

a

X ~ N (55, 152), P(x > d) = 0.05, d = 79.7 P (x < f ) = 0.90, f = 35.8

b

X ~ N (551.3, 152)

Exercise 15M

P (X > 550) = 0.535 = 53.6%

1

X ~ N (500, 202) a

P (X < 475) = 0.106

b

(0.1056 …) = 0.00118

a

P (Z < a) = 0.922, a = 1.42

b

P (Z > a) = 0.342 ∴ P (Z < a) = 0.658, a = 0.407

c

P (Z > a) = 0.005 ∴ P (Z < a) = 0.995, a = 2.58

a

P (1 < Z < a) = 0.12 P (Z < 1) = 0.8413 ∴ P (Z < a) = 0.9613 ∴ a = 1.77 P (a < Z < 1.6) = 0.787 P (Z > 1.6) = 0.0548 ∴ P (Z < a) = 1 − (0.787 + 0.0548) = 0.1582 ∴ a = −1.00

b

X ~ N (30, σ2) Z= ∴

3

Exercise 15K 1

∴ a = 1.60

Exercise 15L

P (3.5 < X < 4.5) = 0.9545 number of acceptable bolts = 0.9545 × 500 = 477 3

∴ a = 0.385

P (|Z| > a) = 0.1096

b

Exercise 15J 1

a

∴ P (Z < a) = 0.65

Using GDC: 1 a 0.655 b 0.841 c 0.186 d 0.5 2 a 0.672 b 0.748 c 0.345 3 a 0.994 b

P (a < Z < −0.3) = 0.182 P (Z > −0.3) = 0.6179 ∴ P (Z < a) = 1 − (0.182 + 0.6179) = 0.2001 ∴ a = −0.841 P (−a < Z < a) = 0.3

c

2

40  30



10

 10 

= 1.2004 ∴ σ = 8.33



X ~ N ( μ, 42) 20.5   4 20.5   = 4

Z= ∴

P (X > 40) = 0.115

P (X < 20.5) = 0.9 

∴ P  Z 

20.5     4 

= 0.9

1.28155

∴ μ = 15.4 3

X ~ N (μ, σ 2) P (X > 58.39) = 0.0217 P (X < 41.82) = 0.0287 Z= 

P  Z

59.39  



  58.39    

∴ P  Z  

P  Z 

Z=

∴ � = 49.9

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute



= 0.0217

58.39      

41.82      

41.82  

= 0.9783 ∴

= 0.0287 ∴

58.39  



41.82  



= 2.0198

= −1.9003

σ = 4.23 Worked solutions: Chapter 15

6

WORKED SOLUTIONS 4

X ~ N( �, σ 2) P (X < 89) = 0.90 P (X < 94) = 0.95 89  

Z= 

P  Z  

P  Z 

= 0.90 ∴

89  

94     

= 0.95 ∴

94  

145  130



6

9





11

= 1.64485

P





X ~ N (�, 20 )



P (X < 500) = 0.01

= 0.01 ∴

x + 4 = ± 19

X ~ N (0.85, σ2) Z=

a

1.1  



P Z



 0.25 

✗

= −2.326



1

X ~ N (�, 72)

3 k

= 0.01



2 k

+ 0.1 + 2.1 = 1

= 0.5 ∴ k = 6





E (X ) = (−2 × 0.3) + 1  16 + (1 × 0.1) + (2 × 0.1)

= 0.6433

−7 15

=

P  Z  

∴

68    7 

68   7

= 0.025 ∴ P  Z  

= 1.95996

X ~ N (2.9, σ2) 3  2.9 5

Z=



P Z ∴

2

P (X > 68) = 0.025

a

68   7

Z=

10 a

490     

P (x > 1) = 0.350 = 35.0%

b

9

1 k

0.3 +

a

∴ σ = 0.389 kg 8



= 0.99 ∴ P  Z 

= −2.32635 (2)



b 0.25

= 0.05

Review exercise



∴

495     

Solving (1) and (2) simultaneously gives � = 507.1, σ = 7.34



 = 0.74

490     

490  

P (X < 1.1) = 0.74

=



= 0.95 ∴ P  Z 

= −1.64485 (1)



∴

∴ � = 546.5 or 547 g 7

495  



Z  50020  200    20 



 495   

 Z 

P  Z 

∴P Z  9 = 0.88  ∴ σ = 7.66 cm

2

490  



∴

= 1.175

P  Z 

X ~ N ( �, σ 2) P (X > 495) = 0.95 P (X > 490) = 0.99 495  

P (X > 145) = 0.12

P Z  9 = 0.12 ∴

= 1.28155

 9







σ = 13.8

X ~ N (136, σ2) Z=

yes, this is consistent with the normal distribution



89     

∴ � = 71.4 5

94  

Z=



P (X > 117) = 0.605 = 60.5%

b

 0.1 

0.1



 = 0.35



Px= P

Z



−7 ± 65 2

154     Z      154  

∴



1 4



X ~ N ( �, σ 2) P (X < 10.8) = 0.3 P (X > 154) = 0.2 Z

154  

4

108  

= 0.2 ∴ P



 Z 

= −0.5244  154   

= 0.8

= 0.8416

∴ � = 125.7

σ = 33.67

� = 126

σ = 33.7

5

P (X = x)

5c

8c

9c

8c

5c

+

1 4

+

1 8

1 4

+ x = 1 ∴ x = 38

 38  81  81  38  41 = 13 64 2

2

4

4

1

2

2

4

4

2

4

4

8

8

3

6

6

12 12

4

8

8

16 16

a



= 0.3 ∴

4

P (6) = P (2, 4) + P (3, 3) + P (4, 2)



0.1 ∴ P Z   = 0.65

∴ σ = 0.260 m

108  

3

from symmetry, E (X) = 3

b 3



2

1 ∴ c = 35

∴ � = 54.3 cm

0.1

1

35c = 1

= 0.975

P (X > 3) = 0.35

= 0.3853



68    7 

x

possible values of P are 2, 4, 6, 8, 12, 16 b

x

2

4

6

8

P (X = x)

1 8

2 8

1 8

2 8

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

12 16 1 8

1 8

Worked solutions: Chapter 15

7

WORKED SOLUTIONS c

E (P) =

2 8

 88 

6 8

 16  12  8 8

4

6 8

X ~ B (10, 0.2)

a

= 7.5 x

d

£10

£5

1 4

3 4

1 4

+5×

P (X = x)

E (X ) = 10 ×

 

X ~ B 5, 31

3 4

= 6.25

7

a

P (X > 5) = 0.00637

5 3

1 3

= 3) = 0.201 the probabilities continue to decrease after this

∴ most likely number is 2

3

P (X ≥ 1) > 0.95

X ~ B (n, 0.2)

c

1 − P (X = 0) > 0.95 0.05 > P (X = 0)

2

2 3

P (X = 0) < 0.05

1 4   10  27 9

X ~ B (2, 0.1)

ii

P (X

P (X = 3) =

6

P (X = 4) = 0.0881

P (X = 0) = 0.107 P (X = 1) = 0.268 P (X = 2) = 0.302

b

After 10 weeks, expected total = 6.25 × 10 = £62.50 5

i

(0.8)n < 0.05

40 243

n log 0.8 < log 0.05

E (X) = nP = 2 × 0.1 = 0.2

n>

log 0.05 log 0.8

n > 13.4 ∴ need 14 points in this sample 5

P (| Z | ≤ a) = 0.85 P (−a ≤ Z ≤ a) = 0.85 ∴ P (Z ≤ a) = 0.925 ∴ a = 1.44

0.954 6

65

75

Z=

x

a

1  0.954 2

7

1

 

a

X ~ B 3,

b

x P (X = x)

c

i

ii 2

3

P P (X ≥ 1) =

−5

1

8 27

19 27

E (X) = −5 ×

8 27

+1×

19 27

30  



30     

= 0.15 ∴

50  



or −$0.78

8

a

30  



= −1.03643

50     ∴ P  Z    = 0.9

= 1.28155

� = 38.9 hours = −$



50     P  Z    = 0.1

7 9

= 1.0364 ∴ σ = 8.68

50  

Z=



 Z 

∴ 19 27

9



X ~ N(�, σ2) P (X < 30) = 0.15 P (X > 50) = 0.1 Z=

Review exercise 1 3



 9

P (X > 65) = 0.755

b

= 0.023





From symmetry, a = 85 P (X > a) =

80  71

P Z  9 = 0.85 ∴

P (X < 65) = P (X > a) b

X ~ N (71, σ 2) p (x < 80) = 0.85

a

X ~ N(�, 2)

σ = 8.63 hours P (x > 35) = 0.2

35   2

∴ lose $0.78 (or $ 79 )

Z=

7 9

35    35      P  Z  2  = 0.2 ∴ P  Z  2  = 0.8

× 9 = 7 ∴ lose $7

X ~ B (8, 0.3) a

P (X = 3) = 0.254

b

P (X ≥ 3) = 0.448

 

1 X ~ B 6, 6

∴

= 0.8416

∴ � = 33.3

P (X = 3) = 0.05358

Y ~ B (5, 0.05358)

35   2

P (Y = 2) = 0.0243

b

X ~ P (5, 0.2)

c

P (X ≥ 2) = 0.263

© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute

P (X = 0) = 0.328

Worked solutions: Chapter 15

8