Maths Quest 12. Mathematical Methods: VCE Units 3 & 4 - Solutions Manual 9780730323129, 0730323129

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mathsquest12 mathematical methods

solutions manual vce units 3 and 4

mathsquest12 mathematical methods

solutions manual vce units 3 and 4

author margaret swale Contributing authors Sue Michell | Catherine Smith

First published 2016 by John Wiley & Sons Australia, Ltd 42 McDougall Street, Milton, Qld 4064 Typeset in 9/11 pt Times LT Std © John Wiley & Sons Australia, Ltd 2016 The moral rights of the authors have been asserted. ISBN  978 0 7303 2312 9 Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. Trademarks Jacaranda, the JacPLUS logo, the learnON, assessON and studyON logos, Wiley and the Wiley logo, and any related trade dress are trademarks or registered trademarks of John Wiley & Sons Inc. and/or its affiliates in the United States, Australia and in other countries, and may not be used without written permission. All other trademarks are the property of their respective owners. Cover and internal design images: © John Wiley and Sons, Australia Ltd; © MPFphotography/Shutterstock Illustrated by diacriTech and Wiley Composition Services Typeset in India by diacriTech Printed in China by Printplus Limited 10 9 8 7 6 5 4 3 2 1

Table of contents About eBookPlus vi Topic 1 — Solving equations  1 Exercise 1.2 — Polynomials  1 Exercise 1.3 — Trigonometric symmetry properties  5 Exercise 1.4 — Trigonometric equations and general solutions 8 Exercise 1.5 — L  iteral and simultaneous equations  14 Topic 2 — Functions and graphs  19 Exercise 2.2 — Polynomial functions  19 Exercise 2.3 — Other algebraic functions  25 Exercise 2.4 — Combinations of functions  33 Exercise 2.5 — Non-algebraic functions  38 Exercise 2.6 — Modelling and applications  49 Topic 3 — Composite functions, transformations and ­inverses  53 Exercise 3.2 — C  omposite functions and ­functional equations 53 Exercise 3.3 — Transformations  56 Exercise 3.4 — Transformations using matrices  59 Exercise 3.5 — Inverse graphs and relations  62 Exercise 3.6 — Inverse functions  64 Topic 4 — Logarithmic functions  69 Exercise 4.2 — Logarithm laws and equations  69 Exercise 4.3 — Logarithmic scales  72 Exercise 4.4 — Indicial equations  74 Exercise 4.5 — Logarithmic graphs  78 Exercise 4.6 — Applications  83 Topic 5 — Differentiation  87 Exercise 5.2 — Review of differentiation  87 Exercise 5.3 — Differentiation of exponential functions  94 Exercise 5.4 — Applications of exponential functions  97 Exercise 5.5 — D  ifferentiation of trigonometric functions 100 Exercise 5.6 — Applications of trigonometric functions  103 Topic 6 — Further differentiation and applications  107 Exercise 6.2 — The chain rule  107 Exercise 6.3 — The product rule  111 Exercise 6.4 — The quotient rule  116 Exercise 6.5 — Curve sketching  121 Exercise 6.6 — Maximum and minimum problems  133 Exercise 6.7 — Rates of change  137

Topic 7 — Antidifferentiation  143 Exercise 7.2 — Antidifferentiation  143 Exercise 7.3 — Antiderivatives of exponential and trigonometric functions  145 Exercise 7.4 — Families of curves and applications  147 Topic 8 — Integration  153 Exercise 8.2 — The fundamental theorem of integral calculus 153 Exercise 8.3 — Areas under curves  159 Exercise 8.4 — Applications  162 Topic 9 — Logarithmic functions using calculus  171 Exercise 9.2 — The derivative of f(x) = loge(x) 171 1 Exercise 9.3 — The antiderivative of f(x) =  177 x Exercise 9.4 — Applications  182 Topic 10 — Discrete random variables  189 Exercise 10.2 — Discrete random variables  189 Exercise 10.3 — Measures of centre and spread  196 Exercise 10.4 — Applications  204 Topic 11 — The binomial distribution  211 Exercise 11.2 — Bernoulli trials  211 Exercise 11.3 — The binomial distribution  212 Exercise 11.4 — Applications  216 Topic 12 — Continuous probability distributions  219 Exercise 12.2 — C  ontinuous random variables and probability functions 219 Exercise 12.3 — The continuous probability density function 223 Exercise 12.4 — Measures of centre and spread  229 Exercise 12.5 — Linear transformations  237 Topic 13 — The normal distribution  243 Exercise 13.2 — The normal distribution  243 Exercise 13.3 — C  alculating probabilities and the standard normal distribution  245 Exercise 13.4 — The inverse normal distribution  246 Exercise 13.5 — Mixed probability application problems  247 Topic 14 — Statistical inference  253 Exercise 14.2 — P  opulation parameters and sample statistics 253 Exercise 14.3 — The distribution of p 253 Exercise 14.4 — Confidence intervals  255

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TOPIC 1 Solving equations • EXERCISE 1.2   |  1

Topic 1 — Solving equations b n 2 p2 − 4 m 2 − 4 m − 1

Exercise 1.2 — Polynomials 1 a 15u − u − 2 = (5u − 2)(3u + 1)

(

2

)

2 2 b 6d − 28d + 16 = 2 3d − 14 d + 8 = 2(3d − 2)(d − 4)

c 3 j 2 + 12 j − 6

( (

=3 j +4j−2

)

= 3 j 2 + 4 j + (2)2 − (2)2 − 2

(

= 3 ( j + 2) − 6 2

(

)

= ( np ) − ( 2m + 1)2 = ( np − ( 2m + 1))( np + ( 2m + 1)) = ( np − 2m − 1)( np + 2m + 1)

)

7 Let P ( x ) = x 3 − 2 x 2 − 21x − 18 P (−1) = ( −1)3 − 2 ( −1)2 − 21( −1) − 18 P (−1) = −1 − 2 + 21 − 18 P (−1) = 0 Thus ( x + 1) is a factor.

( 6) ) 6 )( j + 2 + 6 ) 2

= 3 ( j + 2) − 2

(

= 3 j+2− 2

2 a f − 12 f − 28 = ( f − 14)( f + 2)

(

x 3 − 2 x 2 − 21x − 18 = ( x + 1) x 2 − 3 x − 18

b g 2 + 3g − 4 = ( g + 4)( g − 1)

8 Let P ( x ) = x 4 − 5 x 3 − 32 x 2 + 180 x − 144

3 a 125a − 27b = ( 5a ) − ( 3b ) 3

3

3

( (

= ( 5a − 3b ) ( 5a )2 + ( 5a )( 3b ) + ( 3b )2 = ( 5a − 3b ) 25a 2 + 15ab + 9b 2 3

2

2

b 2c + 6c d + 6cd + d

3

(

= 2 c3 + 3c 2 d + 3cd 2 + d 3 = 2 ( c + d )3

(

)

c 40 p3 − 5 = 5 8 p3 − 1

(

= 5 ( 2 p ) − 13 3

(

)

(

)

2

)

2

)

2

4 a 27 z − 54 z + 36 z − 8 = ( 3z )3 − 3 ( 3z )2 ( 2 ) + 3 ( 3z )( 2 )2 + 23 = ( 3z − 2 )3 3 3

b m n + 64 = ( mn )3 + 4 3

( (

) )

= ( mn + 4 ) ( mn )2 − 4 mn + 4 2 = ( mn + 4 ) m 2 n 2 − 4 mn + 16 5 a 9 x 2 − xy − 3 x + y = 9 x 2 − 3 x − xy + y = 3 x ( x − 1) − y( x − 1) = ( x − 1)(3 x − y) b 3 y3 + 3 y 2 z 2 − 2 zy − 2 z 3 = 3 y2 y + z 2 − 2z( y + z 2 ) 2

(

)

= ( y + z ) 3 y2 − 2z

(

)

Q(2) = 23 − 4 ( 2 )2 − 36 ( 2 ) + 144 ≠ 0

= 5 ( 2 p − 1) 4 p + 2 p + 1

(

)

)

P (1) = (1)4 − 5 (1)3 − 32 (1)2 + 180 (1) − 144 P (1) = 1 − 5 − 32 + 180 − 144 P (1) = 181 − 181 P (1) = 0 Thus ( x − 1) is a factor. x 4 − 5 x 3 − 32 x 2 + 180 x − 144 = ( x − 1) x 3 − 4 x 2 − 36 x + 144 Let Q( x ) = x 3 − 4 x 2 − 36 x + 144

= 5 ( 2 p − 1) ( 2 p ) + 2 p + 1 3

)

= ( x + 1)( x − 6 )( x + 3)

c b 2 − 1 = (b − 1)(b + 1) 3

= ( np )

2 2

2

( ) 2 − (( 2m ) + 2 ( 2m ) + 1)

= ( np ) − 4 m 2 + 4 m + 1

2

)

6 a 9a 2 − 16b 2 − 12a + 4 = 9a 2 − 12a + 4 − 16b 2 = ( 3a )2 − 2 ( 3a )( 2 ) + 22 − ( 4 b )2 = ( 3a − 2 )2 − ( 4 b )2 = ( 3a − 2 − 4 b )( 3a − 2 + 4 b )

Q(4) = 4 3 − 4 ( 4 )2 − 36 ( 4 ) + 144 = 64 − 64 − 144 + 144 =0 Thus ( x − 4 ) is a factor.

(

)

x 3 − 4 x 2 − 36 x + 144 = ( x − 4) x 2 − 36 = ( x − 4)( x − 6)( x + 6) So x 4 − 5 x 3 − 32 x 2 + 180 x − 144 = ( x − 1)( x − 4 )( x − 6 )( x + 6 ) 9 a x 4 − 8 x 3 + 17 x 2 + 2 x − 24 = 0 ( x − 4 )( x − 3)( x − 2 ) ( x + 1) = 0 x − 4 = 0, x − 3 = 0 x−2=0 x = 4, x = 3, x = 2, b a 4 + 2a 2 − 8 = 0 Let x = a 2 x 2 + 2x − 8 = 0 ( x − 2)( x + 4) = 0 x = 2, −4 Substitute x = a 2 a 2 = 2, a 2 = −4 (no solution) ∴a = ± 2 10 a

2 x 3 − x 2 − 10 x + 5 = 0 x 2 (2 x − 1) − 5(2 x − 1) = 0

(

(2 x − 1)( x 2 − 5) = 0

)(

)

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

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

x +1 = 0 x = −1

2 | TOPIC 1 Solving equations • EXERCISE 1.2 2a 2 − 5a = 9

b

b 4 z 2 + 28 z + 49 = 0

2

( 2 z )2 + 2 ( 2 z ) ( 7 ) + 7 2 = 0 ( 2 z + 7 )2 = 0

2a − 5a − 9 = 0 5 ± (−5)2 − 4 × 2 × −9 2×2 5 ± 97 = 4 3 11 Ax + ( B − 1) x 2 + ( B + C ) x + D ≡ 3 x 3 − x 2 + 2 x − 7 A = 3, B − 1 = −1, B + C = 2 and D = −7 B=0 0+C = 2 C=2 3 2 3 12 x + 9 x − 2 x + 1 ≡ x + ( dx + e )2 + 4 x 3 + 9 x 2 − 2 x + 1 ≡ x 3 + d 2 x 2 + 2dex + e 2 + 4 a=

c

5m 2 − 10 m + 3 = 0

(

d

)

= ( r − 7 ) 7r 2 + 1 3

x2 − 4x + 3 = 0 ( x − 3)( x − 1) = 0 x − 3 = 0 or x − 1 = 0 x=3 x =1

2

b 36 v + 6 v + 30 v + 5 = 6 v 2 ( 6 v + 1) + 5 ( 6 v + 1)

(

= ( 6 v + 1) 6 v 2 + 5 3

)

2

c 2m + 3m − 98m − 147 = m 2 ( 2m + 3) − 49 ( 2m + 3)

(

= ( 2m + 3) m − 49 2

2

24 p − 48 p + 18 = 0 4p 2 − 8 p + 3 = 0 ( 2 p − 1)( 2 p − 3) = 0 2 p − 1 = 0 or 2 p − 3 = 0 1 3 p= p= 2 2

= ( 2m + 3) ( m − 7 )( m + 7 ) d 2 z 3 − z 2 + 2 z − 1 = z 2 ( 2 z − 1) + ( 2 z − 1)

(

)

2

e 4 x − 28 x + 49 − 25 y 2 = ( 2 x − 7 )2 − ( 5 y ) = ( 2 x − 7 − 5 y )( 2 x − 7 + 5 y ) 2

f 16a 2 − 4 b 2 − 12b − 9 = ( 4 a )2 − 4 b 2 + 12b + 9

(

)

= ( 4 a )2 − ( 2 b + 3 )2 = ( 4 a − ( 2 b + 3)) ( 4 a + 2b + 3) = ( 4 a − 2b − 3)( 4 a + 2b + 3) g v 2 − 4 − w 2 + 4 w = v 2 − w2 − 4 w + 4

(

)

= v 2 − ( w − 2 )2 = ( v − ( w − 2 ))( v + ( w − 2 )) = ( v − w + 2 )( v + w − 2 ) h 4 p2 − 1 + 4 pq + q 2 = 4 p2 + 4 pq + q 2 − 1 = (2 p + q) − 1 = ( 2 p + q − 1)( 2 p + q + 1) 2

14 a

81y 2 = 1 81y 2 − 1 = 0

(9 y ) − 12 = 0 (9 y − 1)[9 y + 1] = 0 9 y − 1 = 0 or 9 y + 1 = 0 1 1 y= y=− 9 9 2

48 p = 24 p2 + 18

e

)

= ( 2 z − 1) z 2 + 1

10 ± (−10)2 − 4 × 5 × 3 2×5 10 ± 40 = 10 10 ± 2 10 = 10 5 ± 10 = 5 x 2 − 4 x = −3

m=

d 2 = 9, 2 ( ±3) e = −2 d = ±3 ± 6e = −2 1 e=± 3 13 a 7r 3 − 49r 2 + r − 7 = 7r 2 ( r − 7 ) + ( r − 7 )

2z + 7 = 0 2 z = −7 7 z=− 2 5m 2 + 3 = 10 m

f 39 k = 4 k 2 + 77 2 4 k − 39 k + 77 = 0 ( 4 k − 11)( k − 7 ) = 0 4 k − 11 = 0 or k − 7 = 0 11 k= k=7 4 2 m + 3m = 4 g m 2 + 3m − 4 = 0 ( m + 4 )( m − 1) = 0 m + 4 = 0 or m − 1 = 0 m = −4 m =1 4 n 2 = 8 − 5n

h

4 n 2 + 5n − 8 = 0 −5 ± (5)2 − 4 × 4 × −8 2×4 −5 ± 153 = 8 −5 ± 3 17 = 8 15 a 2 x 3 + 7 x 2 + 2 x − 3 = 0 n=

Let P ( x ) = 2 x 3 + 7 x 2 + 2 x − 3 P(−1) = 2(−1)3 + 7(−1)2 + 2(−1) − 3 = −2 + 7 − 2 − 3 =0 Thus ( x + 1) is a factor.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 1 Solving equations • EXERCISE 1.2   |  3 2x3 + 7x 2 + 2x − 3 = 0

(

If b 3 + 5b 2 + 2b − 8 = 0 (b − 1)(b + 2)(b + 4) = 0 b − 1 = 0 or b + 2 = 0 or b + 4 = 0 b =1 b = −2 b = −4

)

2

( x + 1) 2 x + 5 x − 3 = 0

( x + 1)( 2 x − 1)( x + 3) = 0 x + 1 = 0 or 2 x − 1 = 0 or x + 3 = 0 1 x = −1 x= x = −3 2 l 4 − 17l 2 + 16 = 0 b

b −2m 3 + 9m 2 − m − 12 = 0 2m 3 − 9m 2 + m + 12 = 0 Let P (m) = 2m 3 − 9m 2 + m + 12

(l 2 − 1)(l 2 − 16) = 0

P(−1) = 2(−1)3 − 9(−1)2 − 1 + 12 = −2 − 9 − 1 + 12 = −12 + 12 =0 Thus m + 1 is a factor. 2m 3 − 9m 2 + m + 12 = (m + 1) 2m 2 − 11m + 12 = (m + 1)(2m − 3)(m − 4)

( l − 1)( l + 1)( l − 4 )( l + 4 ) = 0 l − 1 = 0 or l + 1 = 0 or l − 4 = 0 or l + 4 = 0 l =1 l = −1 l=4 l = −4

(

c c3 + 3c 2 − 4c − 12 = 0 Let P (c) = c 4 + c3 − 10c 2 − 4c + 24 P(2) = (2)4 + (2)3 − 10(2)2 − 4(2) + 24 = 16 + 8 − 40 − 8 + 24 = 48 − 48 =0 Thus (c − 2) is a factor.

(

c 4 + c3 − 10c 2 − 4c + 24 = (c − 2) c3 + 3c 2 − 4c − 12 Let Q(c) = c3 + 3c 2 − 4c − 12 Q(2) = 23 + 3(2)2 − 4(2) − 12 = 8 + 12 − 8 − 12 =0 Thus (c − 2) is a factor.

(

)

)

c3 + 3c 2 − 4c − 12 = (c − 2) c 2 + 5c + 6 = (c − 2)(c + 2)(c + 3) 4 3 Therefore c + c − 10c 2 − 4c + 24 = (c − 2)2 (c + 2)(c + 3) (c − 2)2 (c + 2)(c + 3) = 0 c − 2 = 0 or c + 2 = 0 or c + 3 = 0 c=2 c = −2 c = −3 4

3

2

d p − 5 p + 5 p + 5 p − 6 = 0 Let P ( p) = p4 − 5 p3 + 5 p2 + 5 p − 6 P(1) = 14 − 5(1)3 + 5(1)2 + 5(1) − 6 = 1− 5 + 5 =0 Thus ( p − 1) is a factor.

(

p4 − 5 p3 + 5 p2 + 5 p − 6 = ( p − 1) p3 + 4 p2 + p + 8 Let Q( p) = p3 − 4 p2 + p + 6 Q(2) = 23 − 4(2)2 + 2 + 6 = 8 − 16 + 8 =0 Thus ( p − 2) is a factor.

(

)

(

(

If 2 x 3 − x 2 − 6 x + 3 = 0

(

)

)

b 3 + 5b 2 + 2b − 8 = (b − 1) b 2 + 6b + 8 = (b − 1)(b + 2)(b + 4)

)

)

(2 x − 1) x 2 − 3 = 0 2

2 x − 1 = 0 or x − 3 = 0 1 x= x2 = 3 2 x=± 3 17 a Let P (t ) = 3t 3 + 22t 2 + 37t + 10 P(−5) = 3(−5)3 + 22(−5)2 + 37(−5) + 10 = −375 + 550 − 185 + 10 = −560 + 560 =0 Thus t + 5 is a factor. 3t 3 + 22t 2 + 37t + 10 = (t + 5) 3t 2 + 7t + 2 = (t + 5)(t + 2)(3t + 1) If 3t 3 + 22t 2 + 37t + 10 = 0 (t + 5)(t + 2)(3t + 1) = 0 t + 5 = 0 or t + 2 = 0 or 3t + 1 = 0 1 t = −5 t = −2 t=− 3 b Let P (d ) = 3d 3 − 16d 2 + 12d + 16

(

p3 − 4 p2 + p + 6 = ( p − 2) p2 − 2 p − 3 = ( p − 2)( p + 1)( p − 3) Therefore p4 − 5 p3 + 5 p2 + 5 p − 6 = ( p − 1)( p − 2)( p + 1)( p − 3) ( p − 1)( p − 2)( p + 1)( p − 3) = 0 p − 1 = 0 or p − 2 = 0 or p + 1 = 0 or p − 3 = 0 p =1 p=2 p = −1 p=3 16 a Let P (b) = b 3 + 5b 2 + 2b − 8 P(1) = 13 + 5(1)2 + 2(1) − 8 = 8 − 8 = 0 Thus b − 1 is a factor.

If 2m 3 − 9m 2 + m + 12 = 0 (m + 1)(2m − 3)(m − 4) = 0 m + 1 = 0 or 2m − 3 = 0 or m − 4 = 0 3 m = −1 m= m=4 2 c Let P ( x ) = 2 x 3 − x 2 − 6 x + 3 3 2  1  1  1  1 P   = 2  −   − 6  + 3  2  2  2  2 1 1 = − −3+3 4 4 =0 Thus ( 2 x − 1) is a factor. 2 x 3 − x 2 − 6 x + 3 = (2 x − 1) x 2 − 3

)

)

P(2) = 3(2)3 − 16(2)2 + 12(2) + 16 = 24 − 64 + 24 + 16 = 64 − 64 =0 Thus d − 2 is a factor. 3d 3 − 16d 2 + 12d + 16 = (d − 2) 3d 2 − 10 d − 8 = (d − 2)(d − 4)(3d + 2)

(

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

)

4 | TOPIC 1 Solving equations • EXERCISE 1.2 If 3d 3 − 16d 2 + 12d + 16 = 0 (d − 2)(d − 4)(3d + 2) = 0 d − 2 = 0 or d − 4 = 0 or 3d + 2 = 0 d=2

d=4

d=−

a 4 − 10 a 2 + 9 = 0

18 a

2 3

( a2 − 1)( a2 − 9) = 0

(a − 1)(a + 1)(a − 3)(a + 3) = 0 a − 1 = 0 or a + 1 = 0 or a − 3 = 0 or a + 3 = 0 a =1 a = −1 a=3 a = −3 4 2 b 4 k − 101k + 25 = 0

( 4 k 2 − 1)( k 2 − 25) = 0

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

2 k − 1 = 0 or 2 k − 1 = 0 or k − 5 = 0 or k + 5 = 0 1 1 k= k=− k=5 k = −5 2 2 9 z 4 − 145z 2 + 16 = 0

c

(9z 2 − 1)( z 2 − 16) = 0

(3z − 1)(3z + 1)( z − 4 )( z + 4 ) = 0

3z − 1 = 0 or 3z + 1 = 0 or z − 4 = 0 or z + 4 = 0 1 1 z= z=− z=4 z = −4 3 3

(

) − 47 ( x 2 − 2 x ) − 48 = 0 Let A = ( x 2 − 2 x )

d x 2 − 2 x

2

A2 − 47 A − 48 = 0 ( A − 48)( A + 1) = 0

( x 2 − 2 x − 48)( x 2 − 2 x + 1) = 0

( x − 8)( x + 6)( x − 1)2 = 0 x − 8 = 0 or x + 6 = 0 or x − 1 = 0 x =8 x = −6 x =1 19 a 5z 3 − 3z 2 + 4 z − 1 ≡ az 3 + bz 2 + cz + d a = 5, b = −3, c = 4 and d = −1 b x 3 − 6 x 2 + 9 x − 1 ≡ x ( x + a)2 − b x 3 − 6 x 2 + 9 x − 1 ≡ x ( x 2 + 2ax + a 2 ) − b x 3 − 6 x 2 + 9 x − 1 ≡ x 3 + 2ax 2 + a 2 x − b 2a = −6 b = 1 a = −3 20 2 x 3 − 5 x 2 + 5 x − 5 ≡ a( x − 1)3 + b( x − 1)2 + c( x − 1) + d

(

) (

)

≡ a x 3 − 3 x 2 + 3 x − 1 + b x 2 − 2 x + 1 + cx − c + d 3

2

2

≡ ax − 3ax + 3ax − a + bx − 2bx + b + cx − c + d ≡ ax 3 + ( −3a + b ) x 2 + ( 3a − 2b + c ) x + ( − a + b − c + d ) Equating coefficients a = 2 − 3a + b = −5 3a − 2b + c = 5 − a + b − c + d = −5 − 3(2) + b = −5 3(2) − 2(1) + c = 5 − 2 + 1 − 1 + d = −5 − 6 + b = −5 − 2 + d = −5 6−2+c = 5 b =1 d = −3 4+c=5 c =1 Thus 2 x 3 − 5 x 2 + 5 x − 5 ≡ 2( x − 1)3 + ( x − 1)2 + ( x − 1) − 3 21 a  kx 2 − 3 x + k = 0 has no solutions if the discriminant is less than zero. ∆ 1.5} b kx 2 + 4 x − k + 2 = 0 ∆ = 16 − 4 × k × (− k + 2) = 16 + 4 k 2 − 8 k = 4( k 2 − 2 k + 4) = 4( k 2 − 2 k + 12 − 12 + 4) = 4 ( k + 1)2 + 3 

= 4( k + 1)2 + 12 As ( k + 1)2 > 0, ∴ 4( k + 1)2 > 0

and 4( k + 1)2 + 12 > 0 ∆ is always greater than zero, therefore the equation will always have a solution for all values of k. 22 P ( x ) = ax 3 + bx 2 − 4 x − 3 where ( x + 3) and ( x − 1) are factors. a(−3)3 + b(−3)2 − 4(−3) − 3 = 0 −27a + 9b = −9 3a − b = 1.....................(1) a(1)3 + b(1)2 − 4(1) − 3 = 0 a + b = 7....................(2) (1) + (2) 4a = 8 a=2 Substitute a = 2 into (2) 2 + b = 7 so b = 5. 23 P ( x ) = x 4 + ax 3 + bx 2 + cx + 24 where ( x + 2), ( x − 3) and ( x + 4) are factors. (−2)4 + a(−2)3 + b(−2)2 + c(−2) + 24 = 0 16 − 8a + 4 b − 2c + 24 = 0 −8a + 4 b − 2c = −40 4 a − 2b + c = 20.........................(1) (3)4 + a(3)3 + b(3)2 + c(3) + 24 = 0 81 + 27a + 9b + 3c + 24 = 0 27a + 9b + 3c = −105 9a + 3b + c = −35.......................(2) (−4)4 + a(−4)3 + b(−4)2 + c(−4) + 24 = 0 256 − 64 a + 16b − 4c + 24 = 0 −64 a + 16b − 4c = −280 16a − 4 b + c = 70.........................(3) Solve using CAS: a = 2, b = −13 and c = −14 5 − 2m  24 ( m − 1) x 2 +  x + 2m = 0  2  This has two solutions if the discriminant is greater than zero. b 2 − 4 ac > 0 2

 5 − 2m    − 4 ( m − 1) 2m > 0 2  Using CAS: −7m 2 + 3m +

25 >0 4

 3 − 2 46 3 + 2 46  m ∈ , \ {1} 14 14   Note: m ≠ 1 as the coefficient of x 2 would be zero if m = 1, therefore no parabola would exist.

Exercise 1.3 — Trigonometric symmetry properties 1 a sin ( 2π − θ ) = − sin (θ ) = −0.4695 b cos (π − α ) = − cos (α ) = −0.5592 c tan ( − β ) = − tan ( β ) = −0.2680

d sin (π + θ ) = − sin (θ ) = −0.4695 e cos ( 2π − α ) = cos (α ) = 0.5592 f tan (π + β ) = tan ( β ) = 0.2680

2 a sin 2 (θ ) + cos 2 (θ ) = 1 cos 2 (θ ) = 1 − sin 2 (θ ) cos (θ ) = ± 1 − sin 2 (θ ) cos (θ ) = ± 1 − ( 0.4695)2 cos (θ ) = ±0.8829 As we are dealing with the fourth quadrant cos (θ ) = 0.8829. Thus cos ( −θ ) = 0.8829 b tan (θ ) =

sin (θ ) 0.4695 = = 0.5318 cos (θ ) 0.8829

tan (180° − θ ) = − tan (θ ) = −0.5318 c sin 2 (α ) + cos 2 (α ) = 1 sin 2 (α ) = 1 − cos 2 (α ) sin (α ) = ± 1 − cos 2 (α ) sin (α ) = ± 1 − ( 0.5592 )2 sin (α ) = ±0.8290 As we are dealing with the first quadrant sin (α ) = 0.8290. d tan ( 360° − α ) = − tan (α ) sin (α ) tan ( 360° − α ) = − cos (α ) 0.8290 tan ( 360° − α ) = − 0.5592 tan ( 360° − α ) = −1.4825 π 3 a sin  + θ  = cos (θ ) = 0.8829 2  3π  b cos  − θ  = − sin (θ )  2  sin 2 (θ ) + cos 2 (θ ) = 1 sin 2 (θ ) = 1 − cos 2 (θ ) sin(θ ) = 1 − cos 2 (θ ) sin(θ ) = 1 − ( 0.8829 )2 sin(θ ) = 0.4696  3π  cos  − θ  = − sin (θ ) = −0.4696  2  π sin  − θ    π 2 c tan  − θ  = 2  π  cos −θ 2  cos (θ ) = sin (θ ) 0.8829 = 0.4695 = 1.8803

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

6 | TOPIC 1 Solving equations • EXERCISE 1.3 d

 3π  sin  + α  = − cos (α )  2  sin 2 (α ) + cos 2 (α ) = 1 cos 2 (α ) = 1 − sin 2 (α ) cos 2 (α ) = 1 − sin 2 (α ) cos (α ) = 1 − ( 0.1736 )2 cos (α ) = 0.9848  3π  sin  + α  = − cos (α ) = −0.9848  2 

π e sin  − α  = cos (α ) = 0.9848 2   3π  + α  = − cot (α ) f tan   2  3π − sin  + θ   2  3π cos  + θ   2  − cos (θ ) = sin (θ ) −0.9848 = 0.1736 = −5.6729

 3π  tan  +θ =  2 

4 sin 2 (θ ) + cos 2 (θ ) = 1 cos 2 (θ ) = 1 − sin 2 (θ ) cos (θ ) = 1 − sin 2 (θ ) cos (θ ) = ± 1 − ( 0.8290 )2 cos (θ ) = ±0.5592 Since we are dealing with the first quadrant cos (θ ) = 0.5592. sin 2 ( β ) + cos 2 ( β ) = 1

sin 2 ( β ) = 1 − cos2 ( β ) sin ( β ) = 1 − cos 2 ( β ) sin ( β ) = ± 1 − ( 0.7547 )2

sin ( β ) = ±0.6561

Since we are dealing with the first quadrant sin ( β ) = 0.6561. a sin ( 90° − θ ) = cos (θ ) = 0.5592

b cos ( 270° + θ ) = sin (θ ) = 0.8290 sin ( 90° + θ ) c tan ( 90° + θ ) = − cos ( 90° + θ ) cos (θ ) = − sin (θ ) −0.5592 = 0.8290 = −0.6746 d sin ( 270° − β ) = − cos ( β ) = −0.7547 sin ( 90° − β ) cos ( 90° − β ) cos ( β ) = sin ( β ) 0.7547 = 0.6561 = 1.1503 f cos ( 270° − β ) = − sin ( β ) = −0.6561 e tan ( 90° − β ) =

3π π π 5 a tan   = tan  π −  = − tan   = −1   4  4  4 π 3  5π   π b cos   = cos  π −  = − cos   = −   6  6  6 2 1 π π c sin  −  = − sin   = −  4  4 2 1 7π  π π   d cos   = cos  2π +  = cos   =  3    3 2 3 π π e tan  −  = − tan   = − 3  3  3 1 π π  11π  f sin  = sin  2π −  = − sin   = −   6  6  6 2 1 5 π π π 6 a tan   = tan  π −  = − tan   = −   6  6  6 3 1 14π  π π b cos  = cos  5π −  = − cos   = −   3  3  3 2 5π 5π π π c tan  −  = − tan   = − tan  π +  = − tan   = −1   4  4   4  4 1 π π  3π   3π  d cos  −  = cos   = cos  π −  = − cos   = −   4  4   4  4 2

π π 2π 2π 3 e sin  −  = − sin   = − sin  π −  = − sin   = −   3  3   3  3 2 1 π π  17π   5π  = sin   = sin  π −  = sin   = f sin       6   6  6 6 2 7 a sin (π − θ ) = sin (θ ) b cos ( 6π − θ ) = cos (θ ) c tan (π + θ ) = tan (θ ) d cos ( −θ ) = cos (θ ) e sin (180° + θ ) = − sin (θ ) f tan ( 720° − θ ) = − tan (θ ) π 8 a cos  − α  = sin (α ) 2  1 b tan ( 90° + α ) = − tan (α ) c sin ( 270° − α ) = − cos (α ) 11π 1  d tan  −α =  2  tan (α )  3π  + α  = sin (α ) e cos   2  f sin ( 90° − α ) = cos (α ) 9 a sin ( 2π − θ ) = − sin (θ ) = −0.9511 b sin (π − θ ) = sin (θ ) = 0.9511

π c cos  − θ  = sin (θ ) = 0.9511 2  sin (θ ) d tan (θ ) = cos (θ ) sin 2 (θ ) + cos 2 (θ ) = 1 cos 2 (θ ) = 1 − sin 2 (θ ) cos (θ ) = 1 − sin 2 (θ ) cos (θ ) = 1 − ( 0.9511)2 cos (θ ) = 0.3089 sin (θ ) 0.9511 tan (θ ) = = = 3.0792 cos (θ ) 0.3089

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 1 Solving equations • EXERCISE 1.3   |  7 e cos ( 3π + θ ) = − cos (θ ) = −0.3089 f tan ( 2π − θ ) = − tan (θ ) = −3.0792 10 a cos (180° − α ) = − cos (α ) = −0.8572 b cos ( −α ) = cos (α ) = 0.8572 3π  c sin  + α  = − cos (α ) = −0.8572  2  tan (180° − α ) = − tan (α )

d

sin (α ) + cos 2 (α ) = 1 2

sin 2 (α ) = 1 − cos 2 (α ) sin (α ) = 1 − cos (α ) 2

sin (α ) = 1 − ( 0.8572 )2 sin (α ) = 0.5150 sin (α ) 0.515 tan (α ) = = = 0.6008 cos (α ) 0.8572 tan (180° − α ) = − tan (α ) = −0.6008 e cos ( 360° − α ) = cos (α ) = 0.8572

π sin  + α  2  π cos  + α  2  cos(α ) = − sin(α ) 0.8572 = −0.5150 = −1.6645

π f tan  + α  = 2 

11 For the right angle triangle of ( 3, 4,5) in the first quadrant 4 3 4 sin ( β ) = , cos ( β ) = and tan ( β ) = 5 5 3 3 a cos ( β ) = − 5 4 b tan ( β ) = − 3 2 2  4  3 2 c sin ( β ) + cos 2 ( β ) =   +  −   5  5 16 9 2 2 + sin ( β ) + cos ( β ) = 25 25 sin 2 ( β ) + cos 2 ( β ) = 1 2 2  3  4 d cos ( β ) − sin ( β ) =  −  −    5  5 16 7 9 − =− cos 2 ( β ) − sin 2 ( β ) = 25 25 25 5 12 a sin (θ ) = 13 5 b tan (θ ) = 12 12 c cos (θ ) = 13 12 d sin ( 90° − θ ) = cos (θ ) = 13 5 e cos ( 90° − θ ) = sin (θ ) = 13 1 12 f tan ( 90° − θ ) = = = 2.4 tan (θ ) 5 2

2

3 7π π π 13 a sin   = sin  2π +  = sin   =  3    3 3 2 1 7π π π b cos   = cos  2π +  = cos   =  3    3 2 3 1 3 5π π π c tan   = tan  π −  = − tan   = − =−      6  6 6 3 3 d sin (150° ) = sin (180° − 30° ) = sin ( 30° ) =

1 2

3 7π π π e cos   = cos  π +  = − cos   = −  6    6 6 2 1 3  7π   π π =− f tan  −  = tan  −  = − tan   = −  6   6  6 3 3 g cos  

π  =0 2

h tan ( 270° ) = undefined i sin ( −4π ) = 0 j tan (π ) = 0 k cos ( −6π ) = 1  3π  l sin   = −1  2  7π π π  2π  14 a cos   + cos   = cos  π +  + cos  π −   6     3  6 3 π π   = − cos   − cos    6  3 =−

3 1 − =− 2 2

(

)

3 +1 2

7π 5π π π b 2 sin   + 4 sin   = 2 sin  2π −  + 4 sin  π −   4     6  4 6 π π = −2 sin   + 4 sin    4  6 2 1 = −2 × +4× =− 2 +2 2 2 5 π 5 π π π        c 3 tan   − tan   = 3 tan  π +  − tan  2π −     4   3  4 3 π π = 3 tan   + tan    4  3 = 3+ 3=2 3

π π  8π   9π    d sin 2   + sin   = sin 2  3π −  + sin  2π +     3   4  3 4 π π = sin 2   + sin    4  3 2

 3 1 = +  2 2   =

3+ 2 2 4

5π 5π e 2 cos 2  −  − 1 = 2 cos 2   − 1  4   4  2  1  = 2 − −1   2 = 1−1 =0

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

8 | TOPIC 1 Solving equations • EXERCISE 1.4 17π  π cos ( −7π ) tan  4π +  cos ( −π ) tan   4   4 f = 11 11π  π  sin  − − sin   6   6  π tan   × −1  4 = π − sin    6 1 = (1 × −1) ÷ − 2 =2 15 a sin 2 ( x ) + cos 2 ( x ) = 1 sin 2 ( x ) cos 2 ( x ) 1 + = cos 2 ( x ) cos 2 ( x ) cos 2 ( x ) 1 tan 2 ( x ) + 1 = as required cos 2 ( x ) sin( x ) = 0.6157

b 2

(0.6157) + cos2 ( x ) = 1 0.3791 + cos 2 ( x ) = 1 cos 2 ( x ) = 1 − 0.3791 cos( x ) = 1 − 0.3791 cos( x ) = 0.788 1 2 tan ( x ) + 1 = (0.788)2 tan 2 ( x ) = 1.6105 − 1 tan( x ) = 0.6165 = 0.7814 16 Let height of tree be h metres and length of shadow be s metres.

π h sin   =  3  30 3 30 × =h 2 15 3 = h

π s cos   =  3  30 1 30 × = s 2 15 = s

Height of tree is 15 3 metres and the length of the shadow is 15 metres.

πt 17 a v = 12 + 3sin    3 Initially t = 0 v = 12 + 3sin ( 0 ) = 12 cm/s b When t = 5  5π  v = 12 + 3sin    3  π  v = 12 + 3sin  2π −   3 π v = 12 − 3sin    3 3 3 cm/s v = 12 − 2 c When t = 12  12π  v = 12 + 3sin   3  v = 12 + 3sin ( 4π ) = 12 cm/s

b At 2 pm t = 8  8π  h(8) = 0.5 cos   + 1.0  12   4π  h(8) = 0.5 cos  + 1.0  3  π h(8) = 0.5 cos  π −  + 1.0  3 π  h(8) = 0.5 cos   + 1.0  3 1 3 1 h(8) = × − + 1 = 0.75 m or m 2 2 4 c At 10 pm t = 16  16π  h(16) = 0.5 cos  + 1.0  12   4π  h(16) = 0.5 cos  + 1.0  3  π h(16) = 0.5 cos  π +  + 1.0  3 π h(16) = −0.5 cos   + 1.0  3 3 1 1 h(16) = − × + 1 = 0.75 m or m 2 2 4

Exercise 1.4 — Trigonometric equations and general solutions 0 ≤ θ ≤ 2π 1 a 2 cos (θ ) + 3 = 0 cos (θ ) = −

3 2

3 suggests 60°. Since cos is negative 2

π π ,π+ 6 6 5π 7π , θ= 6 6 θ=π−

b tan ( x ) + 3 = 0

0 ≤ x ≤ 720°

tan ( x ) = − 3 3 suggests 60°. Since tan is negative

x = 180° − 60°, 360° − 60°, 540° − 60° , 720° − 60° x = 120°, 300°, 480°, 660° −π ≤θ ≤ π c 2 cos(θ ) = 1 1 cos(θ ) = 2 π 1 suggests . Since cos is positive 3 2

πt 18 h(t ) = 0.5 cos   + 1.0  12  a At 6 am t = 0 h(0) = 0.5 cos ( 0 ) + 1.0 = 1.5 m or

3 m 2

θ=−

π π , 3 3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 1 Solving equations • EXERCISE 1.4   |  9 2 a sin (θ ) + 0.5768 = 0 sin (θ ) = −0.5768 0.5768 suggests 36.2258°. Since sin is negative

θ = 180° + 36.2258°, 360° − 36.2258° θ = 215.23°, 324.77° Or solve on CAS b sin( x ) = 1 3π π x=− , for − 2π ≤ x ≤ 2π . 2 2 3 a 2 cos ( 3θ ) − 2 = 0 cos ( 3θ ) =

2 2

0 ≤ θ ≤ 2π 0 ≤ 3θ ≤ 6π

π 2 suggests . Since cos is positive 4 2

π π π π π π , 2π − , 2π + , 4π − , 4π + , 6π − 4 4 4 4 4 4 π 7π 9π 15π 17π 23π 3θ = , , , , , 4 4 4 4 4 4 π 7π 3π 5π 17π 23π , , , , θ= , 12 12 4 4 12 12 b 2sin ( 2 x + π ) + 3 = 0 −π ≤ x ≤ π 3θ =

sin ( 2 x + π ) = −

3 2

− π ≤ 2 x + π ≤ 3π

π 3 suggests . Since sin is negative 3 2

π π π π , − , π + , 2π − 3 3 3 3 2π π 4π 5π 2x + π = − , − , , 3 3 3 3 2π 4π 5π π 2x = − − π, − − π, − π, −π 3 3 3 3 5π 4π π 2π 2x = − , − , , 3 3 3 3 5π 2π π π x=− , − , , 6 3 6 3 π 0 ≤ θ ≤ 2π 4 2 cos  3θ −  + 3 = 0  2 2 x + π = −π +

π 3 cos  3θ −  = − 0 ≤ 3θ ≤ 6π  2 2 3 π π π π cos  3θ −  = − − ≤ 3θ − ≤ 6π −  2 2 2 2 2 3 π suggests , Since cos is negative 2 6

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

10 | TOPIC 1 Solving equations • EXERCISE 1.4

π π π π π π π = π − , π + , 3π − , 3π + , 5π − , 5π + 2 6 6 6 6 6 6 π 5π 7π 17π 19π 29π 31π 3θ − = , , , , , 2 6 6 6 6 6 6 5π 3π 7π 3π 17π 3π 19π 3π 29π 3π 31π 3π + + + + + + 3θ = , , , , , 6 6 6 6 6 6 6 6 6 6 6 6 4π 5π 10π 11π 16π 17π 3θ = , , , , , 3 3 3 3 3 3 4π 5π 10π 11π 16π 17π , , , , , θ= 9 9 9 9 9 9 3θ −

5 cos 2 (θ ) − sin (θ ) cos (θ ) = 0

0 ≤ θ ≤ 2π

cos (θ ) ( cos (θ ) − sin (θ )) = 0 cos (θ ) = 0 or cos (θ ) − sin (θ ) = 0 π 3π θ= , cos (θ ) = sin (θ ) 2 2 cos (θ ) sin (θ ) = cos (θ ) cos (θ ) 1 = tan (θ ) π 1 suggests . Since tan is positive 4

π π ,π+ 4 4 π 5π θ= , 4 4 θ=

Therefore θ = 6

π π 5π 3π , , , 4 2 4 2

2 cos 2 (θ ) + 3 cos (θ ) = −1 2 cos (θ ) + 3cos (θ ) + 1 = 0 2

( 2 cos (θ ) + 1)( cos (θ ) + 1) = 0

0 ≤ θ ≤ 2π 0 ≤ θ ≤ 2π

2 cos (θ ) + 1 = 0

or cos (θ ) + 1 = 0 1 cos (θ ) = −1 so θ = π cos (θ ) = − 2 π 1 suggests . Since cos is negative 3 2

π π ,π+ 3 3 2π 4π , θ= 3 3 2π 4π Therefore θ = , π, 3 3 θ =π −

7 2 sin (θ ) − 3 = 0 sin (θ ) =

3 2

π 3 suggests . 3 2 π π θ = 2nπ + and (2n + 1)π − , n ∈ Z 3 3 6nπ + π 3(2n + 1)π − π = , 3 3 6nπ + π 6nπ + 2π , , n ∈Z = 3 3 Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 1 Solving equations • EXERCISE 1.4   |  11 8

3 tan ( 2θ ) + 1 = 0 1 3 1 π Basic angle for − is − (quadrant 4) 3 6 tan ( 2θ ) = −

π 2θ = nπ +  −  , n ∈ Z  6 6nπ − π = 6 6nπ − π = , n ∈Z 12

π , 3 π 3θ = , 3 π θ= , 9 3θ =

π π π π π , 2π + , 3π + , 4π + , 5π + 3 3 3 3 3 4π 7π 10π 13π 16π , , , , 3 3 3 3 3 4π 7π 10π 13π 16π , , , , 9 9 9 9 9

π+

π d tan  θ −  + 1 = 0  2

For θ ∈[ −π , π ] solutions are: 13π 12 7π n = −1; θ = − 12 π n = 0; θ = − 12 5π n = 1; θ = 12 11π n = 2; θ = 12 7π π 5π 11π ∴θ = − ,− , , 12 12 12 12

0 ≤ θ ≤ 2π

π π π π tan  θ −  = −1 − ≤ θ − ≤ 2π −  2 2 2 2 π 1 suggests . Since tan is negative 4

n = −2; θ = −

9 a

π π π =− , π− 2 4 4 π π 3π θ− =− , 2 4 4 π π 3π π + θ=− + , 4 2 4 2 π 5π θ= , 4 4 θ−

2 sin (θ ) = −1

0 ≤ θ ≤ 2π 1 sin (θ ) = − 2 1 π suggests . Since sin is negative 2 4

π π , 2π − 4 4 5π 7π , θ= 4 4

θ =π +

x = 180° − 60°, 180° + 60° x = 120°, 240°

0 ≤ θ ≤ 2π b 2 cos (θ ) = 1 1 cos (θ ) = 2 π 1 suggests . Since cos is positive 3 2

b 2sin ( 2 x ) + 2 = 0 sin ( 2 x ) = −

0° ≤ x ≤ 360° 2 2

0° ≤ 2 x ≤ 720°

2 suggests 45°. Since sin is negative 2

2 x = 180° + 45°, 360° − 45°, 540° + 45°, 720° − 45° 2 x = 225°, 315°, 585°, 675° x = 112.5°, 157.5°, 292.5°, 337.5°

π π , 2π − 3 3 π 5π θ= , 3 3

θ=

c tan ( 3θ ) − 3 = 0

0° ≤ x ≤ 360° 10 a 2 cos ( x ) + 1 = 0 2 cos ( x ) = −1 1 cos ( x ) = − 2 1 suggests 60°. Since cos is negative 2

0 ≤ θ ≤ 2π

tan ( 3θ ) = 3 0 ≤ 3θ ≤ 6π π 3 suggests . Since tan is positive 3

11 a 3sin (θ ) − 2 = 0 0 ≤ θ ≤ 2π 2 sin (θ ) = 3 2 suggests 0.7297c . Since sin is positive 3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

12 | TOPIC 1 Solving equations • EXERCISE 1.4

θ = 0.7297, π − 0.7297 θ = 0.73, 2.41 or solve on CAS

π 13 a 2sin  2 x +  = 2  4

b 7 cos ( x ) − 2 = 0 0° ≤ x ≤ 360° 2 cos ( x ) = 7 2 suggests 73.3985°. Since cos is positive 7

x = 73.3985°, 360° − 73.3985° x = 73.40°, 286.60° or solve on CAS 12 a 2sin ( 2θ ) + 3 = 0

−π ≤θ ≤π

3 sin ( 2θ ) = − − 2π ≤ θ ≤ 2π 2 π 3 suggests . Since sin is negative 3 2

π 2 sin  2 x +  =  4 2 π 2 sin  2 x +  =  4 2

b 2 cos ( 3θ ) = 1

−π ≤θ ≤π 1 cos ( 3θ ) = − 3π ≤ θ ≤ 3π 2 1 π suggests . Since cos is positive 4 2

π π π π π π 3θ = −2π − , − 2π + , − , , 2π − , 2π + 4 4 4 4 4 4 9π π π 7π 9π 7π 3θ = − ,− ,− , , , 4 4 4 4 4 4 7π π π 7π 3π 3π ,− , , , θ=− , − 4 12 12 12 12 4 −π ≤ θ ≤ π c tan ( 2θ ) + 1 = 0 tan ( 2θ ) = −1 − 2π ≤ θ ≤ 2π π 1 suggests . Since tan is negative 4

π π π π , − , π − , 2π − 4 4 4 4 5π π 3π 7π 2θ = − ,− , , 4 4 4 4 π 3π 7π 5π , θ=− , − , 8 8 8 8

2θ = −π −

− 2π +

π π π ≤ 2 x + ≤ 2π + 4 4 4

π π π π π π = −2π + , − π − , , π − , 2π + 4 4 4 4 4 4 7π π 5π π 3π 9π 2x + = − ,− , , , 4 4 4 4 4 4 π 3π 2 x = −2π , − , 0, , 2π 2 2 π 3π , 0, , π x = −π , − 4 4

2x +

cos ( x + π ) =

π π π π , − , π + , 2π − 3 3 3 3 2π π 4π 5π 2θ = − ,− , , 3 3 3 3 π π 2π 5π , θ=− , − , 3 6 3 6

− 2π ≤ 2 x ≤ 2π

2 π suggests . Since sin is positive 4 2

b 2 cos ( x + π ) = 3

2θ = −π +

−π ≤ x ≤ π

3 2

−π ≤ x ≤π 0 ≤ x + π ≤ 2π

π 3 suggests . Since cos is positive 6 2

π π , 2π − 6 6 π 11π x +π = , 6 6 π 11π x = −π, −π 6 6 5π 5π , x=− 6 6 x +π =

c tan ( x − π ) = −1 −π ≤ x ≤π tan ( x − π ) = −1 − 2π ≤ x − π ≤ 0 π 1 suggests . Since tan is negative 4

π π ,− 4 4 π 5π x −π = − ,− 4 4 π 5π x=− +π, − +π 4 4 π 3π x=− , 4 4 14 a tan 2 (θ ) − 1 = 0 0 ≤ θ ≤ 2π ( tan (θ ) − 1)( tan (θ ) + 1) = 0 tan (θ ) − 1 = 0 or tan (θ ) + 1 = 0 tan (θ ) = 1 tan (θ ) = −1 3π 7π π 5π θ= , θ= , 4 4 4 4 x − π = −π −

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 1 Solving equations • EXERCISE 1.4   |  13

π 3π 5π 7π , , , 4 4 4 4 2 b 4 sin (θ ) − 2 + 2 3 sin (θ ) + 3 = 0 θ=

(

)

( 2sin (θ ) − 3 )( 2sin (θ ) − 1) = 0

0 ≤ θ ≤ 2π

2sin (θ ) − 3 = 0 or 2sin (θ ) − 1 = 0 sin (θ ) =

3 2

2sin (θ ) =

π π 3 1 suggests and suggests 3 6 2 2 Since sin is positive

π , 3 π θ= , 3 π θ= , 6 θ=

π−

π 3

1 2

π π ,π− 6 6 π 5π θ= , 6 6 θ=

2π 3 π 2π 5π , , 3 3 6

15 a sin (α ) − cos 2 (α ) sin (α ) = 0

(

)

−π ≤α ≤π

sin (α ) 1 − cos (α ) = 0 2

sin (α ) (1 − cos (α ))(1 + cos (α )) = 0 sin (α ) = 0 or 1 − cos (α ) = 0 or 1 + cos (α ) = 0 α = −π , 0, π cos (α ) = 1 cos (α ) = −1 α =0 α = −π , π Thus α = −π , 0, π . b sin ( 2α ) = 3 cos ( 2α ) −π ≤ α ≤ π sin ( 2α ) cos ( 2α ) = 3 − 2π ≤ 2α ≤ 2π cos ( 2α ) cos ( 2α ) tan ( 2α ) = 3 π 3 suggests . Since tan is positive 3

2α = −2π + 5π , 3 5π α =− , 6

2α = −

π π π π −π + , , π + 3 3 3 3 2π π 4π − , , 3 3 3 π π 2π − , , 3 6 3

c sin 2 (α ) = cos 2 (α ) −π ≤α ≤π sin 2 (α ) − cos 2 (α ) = 0 (sin (α ) − cos (α ))(sin (α ) + cos (α )) = 0 sin (α ) − cos (α ) = 0 or sin (α ) + cos (α ) = 0 sin (α ) = cos (α ) sin (α ) = − cos (α ) tan (α ) = 1 tan (α ) = −1 π 3π 3π π π 3π 3π π α =− , α =− ,− , , α = − ,       4 4 4 4 4 4 4 4 d 4 cos 2 (α ) − 1 = 0 −π ≤α ≤π

( 2sin (α ))2 − 12 = 0 ( 2sin (α ) − 1)( 2sin (α ) + 1) = 0

π 1 suggests . Since sin is both positive and negative, 3 2 all four quadrants. π π π π α = −π + , − , , π − 3 3 3 3 π π 2π 2π ,− , , α =− 3 3 3 3 16 a 2 cos ( x ) + 1 = 0 1 cos ( x ) = − 2 1 2π Basic angle for − is (quadrant 2) 3 2 General solution: 2π x = 2 nπ ± , n ∈Z 3 6nπ ± 2π = , n ∈Z 3 b 2 sin ( x ) − 2 = 0 sin ( x ) =

2 2

π 2 suggests . 4 2 π π θ = 2nπ + and (2n + 1)π − , n ∈ Z 4 4 8nπ + π 4(2n + 1)π − π = , 4 4 8nπ + π 8nπ + 3π = , , n ∈Z 4 4 17 2 sin ( 2 x ) + 1 = 0 sin ( 2 x ) = −

1 2

π 1 is − (quadrant 4) 6 2 π π  2θ = 2nπ +  −  and (2n + 1)π −  −  , n ∈ Z  6  6 12nπ − π 6(2n + 1)π + π 2θ = , 6 6 12nπ − π 12nπ + 7π = , , n ∈Z 6 6 12nπ − π 12nπ + 7π , , n ∈Z θ= 12 12 π 7π n = 0: θ = − , 12 12 11π 19π n = 1: θ = , 12 12 23π 31π , n = 2: θ = 12 12 7π 11π 19π 23π ∴θ = , , , 12 12 12 12 π π 18 3 sin  x +  = cos  x +    2 2 π  3 tan  x +  = 1  2 π 1  tan  x +  =  2 3 π is the base angle 6 Basic angle for −

2sin (α ) − 1 = 0 or 2 sin (α ) + 1 = 0 1 1 sin (α ) = − sin (α ) = 2 2 Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

14 | TOPIC 1 Solving equations • EXERCISE 1.5 General solution: π π x + = nπ + , n ∈ Z 2 6 π x = nπ − 3 3nπ − π , n ∈Z = 3 4π n = −1: x = − 3 π n = 0: x=− 3 2π n = 1: x= 3 π 2π ∴x = − , 3 3 19 sin ( 3θ ) = cos ( 2θ ) Using CAS:

θ = 0.314, 1.571, 2.827, 4.084, 5.341 1 20 2 sin ( 2 x ) − 1 = − x + 1 2 Using CAS: x = 0.526, 1.179

Exercise 1.5 — Literal and simultaneous equations 1 a my − nx = 4 x + kx my − kz = nx + 4 x my − kz = x ( n + 4 ) my − kz x= n+4 b

2p m 3c − = x x−c x 2 p ( x − c ) − mx = 3c( x − c) 2 px − mx − 3cx = 2 pc − 3c 2 x ( 2 p − m − 3c ) = 2 pc − 3c 2 x=

2 pc − 3c 2 2 p − m − 3c

x − my =2 px + y x − my = 2 ( px + y ) x − my = 2 px + 2 y x − 2 px = my + 2 y x (1 − 2 p ) = y ( m + 2 ) x (1 − 2 p ) y= m+2 3 x + y = 2 k ...................(1) mx + ny = d................(2) From (1) y = 2 k − x ..........(3) Substitute (3) into (2) mx + n ( 2 k − x ) = d mx + 2nk − nx = d mx − nx = d − 2nk x ( m − n ) = d − 2nk d − 2nk x= m−n d − 2nk Substitute x = into (3) m−n 2

d − 2nk m−n 2 k ( m − n ) − d + 2nk y= m−n 2 km − d y= m−n 4 a nx − my = k ...................(1) nx + my = 2d ................(2) (1) + (2) 2nx = k + 2d k + 2d x= 2n k + 2d into (1) Substitute x = 2n  k + 2d  n − my = k  2n  1 ( k + 2d ) − my = k 2 1 k + d − k = my 2 1 d − k = my 2 2d − k y= 2m b nx + my = m..................(1) mx + ny = n ..................(2) (1) × m and (2) × n mnx + m 2 y = m 2 ...........(3) mnx + n 2 y = n .............(4) (3) - (4) y = 2k −

m2 y − n2 y = m2 − n2

(

)

y m2 − n2 = m2 − n2 m2 − n2 m2 − n2 y =1 Substitute y = 1 into (1) nx + m(1) = m nx = 0 x=0 5 2 x + ky = 4....................(1) ( k − 3) x + 2 y = 0.......... (2) There is a unique solution for all values of k except when the gradients are the same. From (1) ky = −2 x + 4 2 2 4 y = − x + so m = − k k k From (2) 2 y = − ( k − 3) x ( k − 3) ( k − 3) y=− x so m = − 2 2 Equating gradients, we have 2 ( k − 3) − =− k 2 2(2) = k ( k − 3) y=

0 = k 2 − 3k − 4 0 = ( k − 4)( k + 1) 0 = k − 4 or 0 = k + 1 k=4 k = −1 If k = −1 or 4 the equations will have the same gradient so for all other values of k there will be a unique solutions. i.e. k ∈ R \ {−1, 4}

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 1 Solving equations • EXERCISE 1.5   |  15 6 mx − 2 y = 4.....................(1) x + (m − 3) y = m.............(2) There will be infinitely many solutions if the equations are identical. From (1) mx − 4 = 2 y m x−2= y 2 From (2) x + (m − 3) y = m x − m = − (m − 3) y x − m = (3 − m) y 1 y= x−m 3− m For equations to be identical then m = 2. To check we should look at the two gradients. m 2 For (1) Gradient = = = 1 2 2 1 1 For (2) Gradient = = =1 3− m 3− 2 Gradients are also the same, therefore m = 2 for an infinite number of solutions. 7 2m − 4 n − p = 1..................(1) 4 m + n + p = 5...................(2) 3m + 3n − 2 p = 22............. (3) (1) × 2 and (2) × 2 4 m − 8n − 2 p = 2...............(4) 8m + 2n + 2 p = 10..............(5) (3) − (4) − m + 11n = 20....................(6) (3) + (5) 11m + 5n = 32....................(7) (6) × 11 −11m + 121n = 220............(8) (7) + (8) 126n = 252 n=2 Substitute n = 2 into (6) − m + 11(2) = 20 22 − 20 = m m=2 Substitute m = 2 and n = 2 into (2) 4(2) + 2 + p = 5 10 + p = 5 p = −5 8 2d − e − f = −2....................(1) 3d + 2e − f = 5....................(2) d + 3e + 2 f = 11.................. (3) (1) × 2 and (2) × 2 4 d − 2e − 2 f = −4...............(4) 6d + 4e − 2 f = 10................(5) (3) + (4) 5d + e = 7................(6) (3) + (5) 7d + 7e = 21. d + e = 3................(7) (6) − (7) 4d = 4 d =1 Substitute d = 1 into (7)

1+ e = 3 e=2 Substitute d = 1 and e = 2 into (1) 2(1) − 2 − f = −2 f =2 9 x + 2 y + 2 z = 1....................(1) 2 x − 2 y + z = 2...................(2) a (1) + (2) 3 x + 3z = 3 b Let z = λ 3 x + 3λ = 3 x +λ =1 x = 1− λ Substitute z = λ and x = 1 − λ into (2) 2 (1 − λ ) − 2 y + λ = 2 2 − 2λ − 2 y + λ = 2 −λ = 2 y λ y=− 2 c This solution describes the line along which the two planes are intersecting. 10 x + 2 y + 4 z = 2....................(1) x − y − 3z = 4......................(2) (2) − (1) −3 y − 7 z = 2 3 y + 7 z = −2 Let z = λ 3 y + 7λ = −2 3 y = −7λ − 2 −7λ − 2 y= 3 7λ + 2 y=− 3 Substitute z = λ and y = −

7λ + 2 into (2) 3

7λ + 2 − 3λ = 4 3 7λ + 2 x = 3λ + 4 − 3 9λ + 12 − 7λ − 2 x= 3 2λ + 10 x= 3 2(λ + 5) x= 3 kx + dy 11 a = −2 k x + 3y kx + dy = −2 k ( x + 3 y ) kx + dy = −2 kx − 6 ky kx + 2 kx = −6 ky − dy 3kx = − y ( 6 k + d ) y (6k + d ) x=− 3k mx + ny b = x+q p mx − px = pq − ny x ( m − p ) = pq − ny pq − ny x= m− p x+

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

16 | TOPIC 1 Solving equations • EXERCISE 1.5 m 3k −k= +m x x m − kx = 3k + mx m − 3k = mx + kx m − 3k = x ( m + k ) m − 3k x= m+k k 2d d = m+x m−x k 2d = m+x m−x k ( m − x ) = 2d ( m + x ) km − kx = 2dm + 2dx km − 2dm = 2dx + kx m ( k − 2d ) = x ( 2d + k ) m ( k − 2d ) x= k + 2d 12 a 2mx + ny = 3k ......................(1) mx + ny = − d .....................(2) (1) − (2) 2mx − mx = 3k + d mx = 3k + d 3k + d x= m 3k + d Substitute x = into (2) m  3k + d  m + ny = − d  m  3k + d + ny = − d ny = −2d − 3k 2d + 3k y=− n x y + = 2.................(1) b 2a b 2x 4 y + = 8...............(2) b a (1) × 2ab ⇒ xb + 2ay = 4 ab...........(3) (2) × ab ⇒ 2ax + 4 by = 8ab............(4) c

(3) × 2a ⇒ 2abx + 4 a 2 y = 8a 2 b.......(5) (4) × b ⇒ 2abx + 4 b 2 y = 8ab 2 .........(6) (5) − (6) 2

2

2

4 a y − 4 b y = 8a b − 8ab

2

a 2 y − b 2 y = 2a 2 b − 2ab 2 y(a 2 − b 2 ) = 2ab(a − b) 2ab(a − b) y= (a − b)(a + b) 2ab = a+b Substitute into (3)  2ab  xb + 2a  = 4 ab  a + b 

4a2 b = 4 ab a+b 2 xb(a + b) + 4 a b = 4 ab(a + b) xb +

x (a + b) + 4 a 2 = 4 a(a + b) x (a + b) = 4 a 2 + 4 ab − 4 a 2 4 ab x= a+b

13 x + my = 3....................(1) 4 mx + y = 0..................(2) From (1)  my = − x + 3 1 1 3 y = − x + where gradient1 = − m m m From (2)  y = −4 mx where gradient2 = −4 m There is no solution when the lines have the same gradient and are parallel. Gradient1 = Gradient2 1 − = −4 m m 1 = 4m 2 1 = m2 4 1 m=± 2 1 No solution if m = ± . 2 14 x + 3ky = 2.....................(1) ( k − 1) x − 1 = −6 y..........(2) From (1)  3ky = − x + 2 1 1 2 where gradient m1 = − y=− x+ 3k 3k 3k From (2)  1 − ( k − 1) x = 6 y 1 ( k − 1) k −1 x = y where gradient m2 = − − 6 6 6 The lines have a unique solution for all vales of k except for when the gradients are the same. m1 = m2 k −1 1 − =− 3k 6 6 = 3k 2 − 3k 0 = 3k 2 − 3k − 6 0 = k2 − k − 2 0 = ( k − 2 )( k + 1) k − 2 = 0 or k + 1 = 0 k=2 k = −1 Lines have a unique solution when k ∈ R \ {−1, 2}. 15 a −2 x + my = 1....................(1) ( m + 3) x − 2 y = −2m.......(2) From (1)  my = 2 x + 1 2 1 2 y = x + where gradient1 = m m m From (2)  ( m + 3) x + 2m = 2 y m+3 ( m + 3) x + m = y where gradient2 = 2 2 The lines have a unique solution for all vales of k except for when the gradients are the same. Gradient1 = Gradient2 2 m+3 = m 2 4 = m 2 + 3m 0 = m 2 + 3m − 4 0 = ( m + 4 )( m − 1) m + 4 = 0 or m − 1 = 0 m = −4 m =1 Lines have a unique solution when m ∈ R \ {−4,1}.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 1 Solving equations • EXERCISE 1.5   |  17 b The lines have a no solution when the gradients are the same because the lines are parallel. If m = −4 −2 x − 4 y = 1.........(1) − x − 2 y = 8.........(2) (2) × 2 ⇒ −2 x − 4 y = 16 Therefore, if m = −4, the gradients are equal, but the y-intercepts are not, therefore the lines are parallel c The lines have infinitely many solutions when both the equations are identical. This is when their gradients are the same and so too are the y-intercepts. If m = 1 −2 x + y = 1............(1) 4 x − 2 y = −2.........(2) (1) × −2 ⇒ 4 x − 2 y = −2 The lines are identical, so there are infinitely many solutions when m = 1. 16 CAS can be used for all parts or the equations can be solved by hand. a 2 x + y − z = 12.....................(1) − x − 3 y + z = −13................(2) −4 x + 3 y − z = −2................(3) (2) + (3) −5 x = −15 x=3 (1) ⇒ 6 + y − z = 12 y − z = 6................(4) (2) ⇒ −3 − 3 y + z = −13 − 3 y + z = −10.......(5) (4) + (5) −2 y = −4 y=2 Substitute into (4) 2− z = 6 z = −4 b m + n − p = 6........................(1) 3m + 5n − 2 p = 13...............(2) 5m + 4 n − 7 p = 34...............(3) (1) × 3 3m + 3n − 3 p = 18................(4) (2) − (4) 2n + p = −5............... ( 5) (1) × 5 5m + 5n − 5 p = 30...............(6) (6) − (3) n + 2 p = −4...............(7) (7) × 2 2n + 4 p = −8...............(8) (8) − (5) 3 p = −3 p = −1 Substitute p = −1 into (5) 2n − 1 = −5 2n = −4 n = −2 Substitute p = −1 and n = −2 into (1) m − 2 +1= 6 m=7

c u + 2 v − 4 w = 23.........................(1) 3u + 4 v − 2 w = 37......................(2) 3u + v − 2 w = 19........................(3) (1) × 3 3u + 6 v − 12 w = 69....................(4) (4) − (2) 2 v − 10 w = 32 v − 5w = 16....................(5) (4) − (3) 5v − 10 w = 50 v − 2 w = 10....................(6) (6) − (5) 3w = −6 w = −2 Substitute w = −2 into (6) v − 2(−2) = 10 v + 4 = 10 v=6 Substitute w = −2 and v = 6 into (1) u + 2(6) − 4(−2) = 23 u + 12 + 8 = 23 u=3 d a + b + c = 4......................(1) 2a − b + 2c = 17............... (2) − a − 3b + c = 3.................(3) (1) + (3) − 2a + 2c = 7................(4) (3) × 2 −2a = 6b + 2c = 6.............(5) (4) × −2 4 b − 4c = −14........(6) (6) + (7) − 3b = 9 b = −3 Substitute b = −3 into (4) −2(−3) + 2c = 7 6 + 2c = 7 2c = 1 1 c= 2 1 Substitute b = −3 and c = into (1) 2 1 a−3+ = 4 2 5 8 a− = 2 2 13 a= 2 17 Let a be the smallest angle, b be the largest angle and c be the third angle. b = a + 20 ....................... (1) a + b = c + 60 ....................... (2) a + b + c = 180 ...........................(3) Substitute (1) into (2) and (3) a + a + 20 = c + 60 2a − c = 40..................... (4) a + a + 20 + c = 180 2a + c = 160....................(5) (5) + (4) 4 a = 200 a = 50°

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

18 | TOPIC 1 Solving equations • EXERCISE 1.5 Substitute a = 50° into (1) b = 50 + 20 = 70° Substitute a = 50° and b = 70° into (3) 50 + 70 + c = 180 120 + c = 180 c = 60° The largest angle is 70°, the smallest angle is 50° and the third angle is 60°. 18 x + y − 2 z = 5....................(1) x − 2 y + 4 z = 1..................(2) (1) − (2) 3 y − 6z = 4 Let z = λ 3 y = 6λ + 4 2 ( 3λ + 2 ) y= 3 2 ( 3λ + 2 ) into (1) Substitute y = 3 2 ( 3λ + 2 ) − 2λ = 5 x+ 3 3 x + 6λ + 4 − 6λ = 15 3 x = 11 11 x= 3 19 −2 x + y + z = −2....................(1) x − 3z = 0..................... (2) Let z = λ From (2) x − 3λ = 0 x = 3λ Substitute z = λ and x = 3λ into (1) −2 ( 3λ ) + y + λ = −2 y − 5λ = −2 y = 5λ − 2 20 3 x + 2 y = −1..................(1) mx + 4 y = n...................(2) From (1)  2 y = −3 x − 1 3 3 1 y = − x − where gradient1 = − 2 2 2

From (2)  4 y = − mx + n m n m y = − x + where gradient2 = − 4 4 4 If gradient1 = gradient2 3 m − =− 2 4 12 = 2m m=6 a The lines have a unique solution for all values of k except for when the gradients are the same. Therefore m ∈ R \ {6} and n ∈ R. b The lines have infinitely many solutions when both the equations are identical. This is when the gradients are the same and so too are their “c” values. If the gradients are the same then m = 6 and if the “c” values are the same then 1 n − = 2 4 −4 = 2n n = −2 Therefore, for an infinite number of solutions, m = 6, n = −2 c The lines have no solution when the gradients are the same but the y-intercepts are different (lines are parallel). Therefore, m = 6 and n ∈ R \ {−2}. 21 2 x − y + az = 4...............................(1) (a + 2) x + y − z = 2....................(2) 6 x + (a + 1) y − 2 z = 4....................(3) Solve using CAS: 2 ( a + 2) 4 ( a + 2) 4 x= , y= and z = a (a + 4) a (a + 4) a 22 w − 2 x + 3 y − z = 10 2w + x + y + z = 4 − w + x + 2 y − z = −3 3w − 2 x + y = 11 Solve using CAS: w = 1, x = −3, y = 2 and z = 3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 2 Functions and graphs • EXERCISE 2.2   |  19

Topic 2 — Functions and graphs Exercise 2.2 — Polynomial functions 1 a  A vertical line would cut the graph no more than once so the graph is of a function. A horizontal line could cut the graph up to two times. The function has a many-to-one correspondence. b Domain [−4, 2) and range [0,16]. c f : [−4, 2) → R, f ( x ) = x 2 d Image of −2 3 is f −2 3 .

(

) (

)

2

(

)

f −2 3 = −2 3 = 12 The image is 12. 2 f ( x ) = x 2 − 4 2  2 a The image of is f   3 3 2  2  2 f   =  −4  3  3 4 = −4 9 32 =− 9 b f (2a)

 36 48  The point closest to the origin is  , −   . 25 25 4 A(−1, −3) and B(5, −7) a Gradient of AB: −7 + 3 m= 5+1 2 =− 3 Equation of AB: 2 y + 3 = − ( x + 1) 3 ∴ 3 y + 9 = −2 x − 2 ∴ 2 x + 3 y = −11

= ( 2 a )2 − 4 = 4a2 − 4 c The implied domain is R. 3 L = {( x ⋅ y) : 3 x − 4 y = 12}. a Let y = 0 ∴ 3 x = 12 ∴x = 4 Let x = 0 ∴ −4 y = 12 ∴ y = −3 Line goes through (4, 0) and (0, −3). y 3x – 4y = 12 0

(4, 0)

4 3 x − 4  − x  = 12  3  ∴ 9 x + 16 x = 36 36 ∴x = 25 4 36 ∴y = − × 3 25 48 ∴y = − 25

x

(–3, 0)

 −1 + 5 −3 − 7  , b Midpoint of AB is   = (2, −5). 2  2 3 Gradient of line perpendicular to AB is . 2 Equation of the perpendicular bisector is 3 y + 5 = ( x − 2) 2 3x ∴y = −3−5 2 3x ∴y = −8 2 c m = tan(θ ) 3 ∴ tan(θ ) = 2 ∴θ = tan −1 (1.5) ∴θ ≈ 56.3° 5 a y = 2(3 x − 2)2 − 8

b Rearranging the equation, 3 x − 4 y = 12 ∴ 3 x − 12 = 4 y 3 ∴y = x −3 4 3 The gradient is . 4 c Let the point on the line closest to the origin be P. Then the line OP is perpendicular to the given line. 4 The gradient of OP is m = − and OP contains the origin. 3 4 The equation of OP is y = − x. 3 4 P is the point of intersection of y = − x and 3 x − 4 y = 12. 3 At intersection,

2 Turning point: When 3 x − 2 = 0, x = . 3 Therefore the graph has a minimum turning point at 2   , −8  . 3 y intercept: Let x = 0 ∴ y = 2(−2)2 − 8 ∴y = 0 (0, 0) 2 Axis of symmetry x = , so the other x intercept must be 3 4   , 0   . 3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

20 | TOPIC 2 Functions and graphs • EXERCISE 2.2 y y = 2(3x – 2)2 – 8

(0, 0) 0

(–43 , 0)

∴ 0 = −4( x + 2)3 + 16 ∴ ( x + 2)3 = 4 ∴x = 3 4 − 2

( 3 4 − 2,0 ) x

y

(–23 , –8)

y = –4(x + 2)3 + 16 (–2, 16)

Domain R, range [−8, ∞). 1 b x intercept at x = − ⇒ (2 x + 1) is a factor. 2 x intercept at x = 4 ⇒ ( x − 4) is a factor. Let the equation be y = a(2 x + 1)( x − 4) Substitute the point (0, 2) ∴ 2 = a(1)(−4) 1 ∴a = − 2 1 The equation is y = − (2 x + 1)( x − 4). 2 6 f : R + ∪ {0} → R, f ( x ) = 4 x 2 − 8 x + 7. a The discriminant determines the number of x intercepts. ∆ = b 2 − 4 ac, a = 4, b = −8, c = 7 ∆ = 64 − 4 × 4 × 7 = 64 − 112 6. There will two intersections. Check: A point below the axis such as (1, −5) for example, would still lie below the x axis if it was vertically translated up one unit so the graph must cross the axis to reach this point.

∴9 − 4x = 0 9 ∴x = 4 9   ,0 ⇒ 4 

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

22 | TOPIC 2 Functions and graphs • EXERCISE 2.2 y

16 a − x 2 + 2 x − 5 = − x 2 − 2x + 5

( ) 2 = − ( ( x − 2 x + 1) − 1 + 5) = − ( ( x − 1)2 + 4 )

(0, 9)

= − ( x − 1)2 − 4 b    y = − x 2 + 2 x − 5 ∴ y = − ( x − 1)2 − 4 Turning point is (1, −4). c Turning point is a maximum so the range is (−∞, −4]. There are no x intercepts. The y intercept is (0, −5).

y = 9 – 4x

(–49 , 0) x

0

y

Range is R. b g : (−3,5] → R, g( x ) =

3x 5

x

0

g(0) = 0 ⇒ (0, 0) 9  9 g(−3) = − ⇒  −3, −  is open endpoint 5  5

y = –x2 + 2x – 5 (0, –5)

g(5) = 3 ⇒ ( 5,3) is closed endpoint.

(1, –4)

y 3x y = –– 5 (5, 3)

(–3 , – –59 )

(0, 0) 0

d x

y (0, 3)

y=x+3

x

(–3, 0) 0

9 Range is  − ,3  .  5  15 A(5, −3), B(7,8) and C (−2, p) 9 a The line 9 x + 7 y = 24 has a gradient of − . 7 This is the gradient of AC since it is parallel to this line. p+3 m AC = −2 − 5 9 p+3 ∴− = 7 −7 ∴9 = p + 3 ∴p=6 7 b A line perpendicular to AC would have a gradient of . 9 The line through B(7,8) perpendicular to AC has equation: 7 y − 8 = ( x − 7) 9 ∴ 9 y − 72 = 7 x − 49 ∴ 9 y − 7 x = 23 c Let the point where 9 y − 7 x = 23 meets AC be Q. The length of PQ is the shortest distance from B to AC. To find Q, solve the pair of simultaneous equations 9 x + 7 y = 24....(1) 9 y − 7 x = 23....(2)  11 75  Q is the point  ,   . 26 26 11 2 75 2 The length of BQ is  7 −  +  8 −  ≈ 8.3 units.   26  26 

y = –x2 + 2x – 5 (0, –5)

(1, –4)

The line y = x + 3 passes through the points (−3, 0) and (0,3). As the diagram shows, this line will never intersect the concave down parabola. e The graphs of y = x + k and y = − x 2 + 2 x − 5 will intersect when x + k = − x 2 + 2x − 5 ∴ x2 − x + k + 5 = 0 For one intersection, ∆ = 0. ∆ = b 2 − 4 ac, a = 1, b = −1, c = k + 5 = 1 − 4( k + 5) = −4 k − 19 ∴ −4 k − 19 = 0 19 ∴k = − 4 19 For exactly one intersection, k = − . 4 1 7 a Let the equation be y = a( x − h)2 + k Turning point is (−6,12) ∴ y = a( x + 6)2 + 12 Substitute the point (4, −3)

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 2 Functions and graphs • EXERCISE 2.2   |  23 ∴−3 = a(10)2 + 12 ∴100 a = −15 3 ∴a = − 20

3 ( x + 6)2 + 12. 20 1 b The points (−7, 0) and  −2 ,0  are the two x intercepts.  2  1 The equation has linear factors ( x + 7) and  x + 2  or  2 (2 x + 5). Let the equation be y = a( x + 7)(2 x + 5) Substitute the point (0, −20) ∴ −20 = a(7)(5) 4 ∴a = − 7 4 The equation is y = − ( x + 7)(2 x + 5). 7 c As the points (−8,11) and (8,11) have the same y co-ordinate, the turning point lies between them and the axis of symmetry lies midway between the, −8 + 8 Axis of symmetry: x = ⇒ x = 0. 2 The minimum value of a quadratic function is the y value of its minimum turning point. Therefore, the turning point is (0, −5). Let the equation be y = ax 2 − 5 Substitute the point (8,11) ∴11 = a(64) − 5 ∴ 64 a = 16 1 ∴a = 4 1 The equation is y = x 2 − 5. 4 1 8 a y = x 3 − x 2 − 6 x ∴ y = x ( x 2 − x − 6) ∴ y = x ( x − 3)( x + 2) The factors show the graph cuts the x axis at (0,0), (3,0) and (−2, 0). The leading term has a positive coefficient. The equation is y = −

y

(–2, 0)

(0, 0)

(3, 0)

0

x intercept: Let y = 0 1 ∴ 0 = 1 − ( x + 1)3 8 ∴ ( x + 1)3 = 8 ∴ x +1 = 2 ∴x =1 (1, 0) End points: 1 x = −3, y = 1 − (−3 + 1)3 8 1 = 1 − (−2)3 8 1 = 1 − × −8 8 =2 ∴ ( −3, 2 ) 1 x = 2, y = 1 − (2 + 1)3 8 1 = 1 − (3)3 8 27 = 1− 8 3 3  = −2 ∴  2, −2   8 8 y

1 (x + 1)3 y=1–– 8

(–3, 2)

(0, –78 )

(1, 0) x

0

(2, –2 –38 ) c y = 12( x + 1)2 − 3( x + 1)3 ∴ y = 3( x + 1)2 (4 − ( x + 1)) ∴ y = 3( x + 1)2 (3 − x ) The factors show the graph touches the x axis at (−1, 0) and cuts at (3, 0). y intercept: Let x = 0 ∴ y = 3(1)2 (3) = 9 (0,9) y

x

y = 12(x + 1)2– 3(x + 1)3

(0, 9) y = x3 – x2 – 6x

1 b y = 1 − ( x + 1)3 8 Stationary point of inflection (−1,1). y intercept: Let x = 0 1 ∴ y = 1 − (1)3 8 7 ∴y = 8  0, 7    8

(–1, 0) 0

(3, 0)

x

19 The x intercept at x = −4 ⇒ ( x + 4) is a factor. 5 The x intercept at x = ⇒ (4 x − 5) is a repeated factor of 4 multiplicity two. Let the equation be y = a( x + 4)(4 x − 5)2 Substitute the point (0,10)

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

24 | TOPIC 2 Functions and graphs • EXERCISE 2.2 ∴10 = a(4)(−5)2 ∴10 = 100 a 1 ∴a = 10 1 The equation is y = ( x + 4)(4 x − 5)2. 10 2 0 a f ( x ) = −2 x 3 + 9 x 2 − 24 x + 17 f (1) = −2 + 9 − 24 + 17 = 0 ∴ ( x − 1) is a factor. By inspection, −2 x 3 + 9 x 2 − 24 x + 17 = ( x − 1)(−2 x 2 + 7 x − 17). Consider the discriminant of the quadratic factor −2 x 2 + 7 x − 17. ∆ = 49 − 4(−2)(−17) = 49 − 136 1. Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 2 Functions and graphs • EXERCISE 2.3   |  25 y y = x6 (–1, 1)

(1, 1) x

0 (0, 0)

(–1, –1)

y = x7

The points of intersection of the two graphs are (0, 0) and (1,1). ii {x : x 6 + x 7 ≥ 0} x 6 − x 7 ≥ 0 when x 6 ≥ x 7. The graph of y = x 6 lies above that of y = x 7 for x < 0 and 0 < x < 1 and the two graphs intersect at x = 0 and x = 1. Hence, {x : x 6 − x 7 ≥ 0} = {x : x ≤ 1} b y = 16 − ( x + 2)4 Even degree so maximum turning point at (−2,16). y intercept: Let x = 0 ∴ y = 0 ⇒ (0, 0). Axis of symmetry x = −2 ⇒ (−4, 0) is the other x intercept. y = 16 − ( x + 2)5 Odd degree so stationary point of inflection at (−2,16). y intercept: Let x = 0 ∴ y = 16 − 32 = −16 ⇒ (0, −16). x intercept: Let y = 0 ∴ 0 = 16 − ( x + 2)5 ∴ x = −2 + 16 5

( −2 +

(

)

16 − ( x + 2)4 = 16 − ( x + 2)5 ∴ ( x + 2)4 = ( x + 2)5 ∴ ( x + 2)4 [1 − ( x + 2) ] = 0 ∴ ( x + 2)4 (−1 − x ) = 0 ∴ x = −2 or x = −1 Points of intersection are (−2,16) and (−1,15). y



(–2, 16) y = 16 – (x + 2)5 (–1, 15) (0, 0) x y = 16 – (x + 2)4

a + k. x−h Asymptotes are x = −3 and y = 1. a +1 ∴ y= x+3 Graph has an x intercept at x = −9. Substitute the point (−9, 0) a ∴0 = +1 −6 ∴a = 6 6 The equation is y = + 1. x+3 5x − 2 b i y = x −1 If x = 1 the denominator would be zero and the function undefined. Its maximal domain is R \ {1}. 5 x − 2 5( x − 1) + 3 ii = x −1 x −1 3 = 5+ x −1 3 has asymptotes x = 1, y = 5. y = 5+ x −1 y intercept: Let x = 0, then y = 2. (0, 2) x intercept: Let y = 0 5x − 2 =0 ∴ x −1 ∴ 5x − 2 = 0 2 ∴x = 5 2   , 0 5 

1 a Let the equation be y =

16, 0 Points of intersection of the two graphs occur when

(–4, 0)

)

Exercise 2.3 — Other algebraic functions

∴ ( x + 2)5 = 16 5

The equation is of the form y = a( x + 3)2 ( x + 1)( x − 2)3. Since it is a monic polynomial, a = 1. Therefore the equation is y = ( x + 3)2 ( x + 1)( x − 2)3, degree 6. ii An additional cut at x = 10 ⇒ ( x − 10) is also a factor. The graph would now have the behaviour that as x → ∞, y → −∞, showing it to be an odd degree with a negative coefficient of its leading term. The degree is 7 and a possible equation is y = ( x + 3)2 ( x + 1)( x − 2)3 (10 − x ). 23 Draw the graphs using CAS. At the intersection of the two graphs x 4 − 2 = 2 − x 3. And therefore, x 4 + x 3 − 4 = 0. The roots of the equation are the x co-ordinates of the points of intersection of the two graphs. Use CAS to obtain these. To two decimal places the roots are x = −1.75 and x = 1.22. 24 Sketch each graph and use the tools to obtain the points required. a y = ( x 2 + x + 1)( x 2 − 4) Minimum turning points (−1.31, −3.21) and (1.20, −9.32), maximum turning point (−0.636, −2.76). b y = 1 − 4 x − x 2 − x 3 No turning points or stationary point of inflection. 1 c y = ( x − 2)5 ( x + 3) + 80 4 Minimum turning point (−2.17, −242), stationary point of inflection (2, 20).

(0, –16)

c i The x intercepts determine the factors. Graph touches x axis at x = −3 ⇒ ( x + 3)2 is a factor. Graph cuts x axis at x = −1 ⇒ ( x + 1) is a factor. Graph saddle cuts x axis at x = 2 ⇒ ( x − 2)3 is a factor. These factors imply the degree is 2 + 1 + 3 = 6. The shape suggests the long term behaviour of an even degree polynomial function with positive leading term.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

26 | TOPIC 2 Functions and graphs • EXERCISE 2.3 y

5x— –2 y=— x–1

y=5

(0, 2) (0.4, 0) 0

x x=1

Range is R \ {5}. 4 2 y = 1 − 2x 1 Vertical asymptote: 1 − 2 x = 0 ⇒ x = is the vertical 2 asymptote. Horizontal asymptote is y = 0. There is no x intercept. y intercept: Let x = 0 ∴ y = 4 (0, 4) y (0, 4)

a −1  1 2   2 ∴ 0 = 4a − 1 1 ∴a = 4 1 The equation is y = 2 − 1. 4x 5 a y = − x + 9 + 2 i For the function to have real values, x + 9 ≥ 0. This means x ≥ −9 so the maximal domain is [−9, ∞). ii endpoint: (−9, 2) y intercept: Let x = 0 ∴y = − 9 + 2 ∴ y = −1 (−1, 0) x intercept: Let y = 0 ∴0 = − x + 9 + 2 ∴0 =

x = 0.5

∴ x+9 = 2 ∴x + 9 = 4 ∴ x = −5 (−5, 0)

y y = – x+ 9+2

(–9, 2) 0

y=0

(–5, 0)

x

0

4 — y=— 1 – 2x (1, – 4)

Domain R \ 3 y =

{}

1 and range R \ {0}. 2

8 −2 ( x + 2)2

Asymptotes: x = −2, y = −2 y intercept: Let x = 0 8 ∴y = −2 (2)2 ∴y = 0 The origin is the intercept on both the axes. By symmetry about the vertical axis there is another x intercept at (−4, 0). y 8 –2 y=— (x + 2)2 (0, 0)

(–4, 0) 0

x y = –2

x = –2

x (0, –1)

Range is (−∞, 2]. b Let the equation be y = a 3 ( x − h) + k. Point of inflection is (1,3) ∴ y = a3 x −1 + 3 Substitute the point (0,1) ∴1 = a 3 −1 + 3 ∴1 = − a + 3 ∴a = 2 The equation is y = 2 3 x − 1 + 3. x intercept: Let y = 0 ∴23 x −1 + 3 = 0 3 ∴ 3 x −1 = − 2  3 3 ∴ x − 1 = −   2 27 ∴ x = 1− 8 19 ∴x = − 8  19  The x intercept is  − , 0  . 8 y

Domain R \ {−2} and range (−2, ∞). a 4 Let the equation be y = +k ( x − h) 2 Asymptotes are x = 0 and y = −1 a ∴ y = 2 −1 x 1  Substitute the point  , 0  2

(1, 3)

y = 2 3 x – 1+ 3

(0, 1)

(

)

0

19 – —, 0 8

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

x

TOPIC 2 Functions and graphs • EXERCISE 2.3   |  27 y

6 a y = 3 4 x − 9 − 6 9 Domain requires 4 x − 9 ≥ 0 ⇒ x ≥ . 4 9 Maximal domain is  ,∞  . 4  9 Endpoint: When 4 x − 9 = 0, x = 4 9  The endpoint is  , −6 . 4 There is no y intercept. x intercept: Let y = 0 ∴3 4x − 9 − 6 = 0 ∴ 4x − 9 = 2 ∴ 4x − 9 = 4 ∴ 4 x = 13 13 ∴x = 4  13   ,0  4

y=x

(1, 1)

3 –

y = x4

(0, 0) 0

x

4

b y = x 3 Since y = 3 x 4 the cube root of x 4 is required. The x 4 polynomial lies in first and second quadrants. The cube root of both sections can be formed so the graph has two branches and domain R. As 4 > 3, the polynomial shape dominates. The graph contains the points (0, 0), (1,1), (−1,1) and lies below y = x for 0 < x < 1 and above y = x for x > 1. y 4 –

y=x

y = x3

y

(1, 1)

(–1, 1) y = 3 4x – 9 – 6 0

x (0, 0)

1

) ) 13 —, 0 4

0

)

x

)

9 , –6 — 4

Range is [−6, ∞). 1

b y = (10 − 3 x ) 3 . This is the cube root function y = 3 10 − 3 x . When 10 − 3 x = 0, x =  10   , 0   . 3

10 , so the point of inflection is 3

8 a y = x 5 y = 5 x ⇒ 5th root of x The line y = x lies in first and third quadrants. The fifth root of both sections can be formed so the graph has two branches and domain R. For the first quadrant, the graph contains the points (0, 0) and (1,1) and lies above y = x for 0 < x < 1 and below y = x for x > 1. By symmetry for the third quadrant, the graph contains the points (0, 0) and (−1, −1) and lies below y = x for −1 < x < 0 and above y = x for x < −1. y

1 –

y = x5 (0, 0)

The point of inflection lies on the x axis. y intercept: Let x = 0

(

∴ y = 3 10 ⇒ 0, 3 10 y

x

1– 3

(0, 10 ( 0

— , 0( (10 3

(1, 1)

(–1, –1)

)

1

y = (10 – 3x) 1– 3

y=x

x

b y = x 8 y = 8 x ⇒ 8th root of x. As the even root of the third quadrant section of the y = x line cannot be taken, the graph has one first quadrant branch with domain R + ∪ {0}. The graph contains the points (0, 0) and (1,1) and lies above y = x for 0 < x < 1 and below y = x for x > 1. y=x

y

3

7 a y = x 4 y = 4 x 3 ⇒ 4 th root of x 3 As the even root of the third quadrant section of the x 3 polynomial cannot be taken, the graph has one first quadrant branch with domain R + ∪ {0}. As 4 > 3, the root shape dominates. The graph contains the points (0, 0) and (1,1) and lies above y = x for 0 < x < 1 and below y = x for x > 1.

1 –

y = x8 (0, 0)

(1, 1) 0

x

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

28 | TOPIC 2 Functions and graphs • EXERCISE 2.3 x−6 x+9 If x + 9 = 0 then x = −9 and the function would be undefined. Domain is R \ {−9}. b y = 1 − 2 x The domain requires 1 − 2 x ≥ 0 1 ∴x ≤ 2 1 Domain  −∞,   2 −2 c ( x + 3)2 Denominator would be zero if x = −3, so the domain is R \ {−3}. 1 d 2 x +3 Since the denominator is the sum of two positive terms, it can never be zero, so the domain is R. 4 1 0 a y = + 5 x Asymptotes: x = 0, y = 5 No y intercept x intercept: Let y = 0 4 ∴ +5= 0 x ∴ 4 = −5 x 4 ∴x = − 5  4  − , 0   5 

y

9 a y =

y

(–12 , 0)

x (0, –1)

x = –1

Domain R \ {−1} and range R \ {2}. 4x + 3 c y = 2x + 1 2(2 x + 1) + 1 ∴y = 2x + 1 1 ∴y = 2+ 2x + 1 1 Asymptotes: 2 x + 1 = 0 ⇒ x = − , y = 2 2 y intercept: Let x = 0 3 ∴y = = 3 1 (0,3) x intercept: Let y = 0 4x + 3 =0 2x +1 ∴4x + 3 = 0

∴

∴x = −

3 4

y

y=5

4x + 3 y=– 2x + 1

x

0

y=2

0

 3  − , 0   4 

4+5 y =– x

(– –45 , 0)

3 y=2– – x+1

(0, 3)

0

x=0

Domain R \ {0} and range R \ {5}. 3 b y = 2 − x +1 Asymptotes: x = −1, y = 2 y intercept: Let x = 0 3 ∴ y = 2 − = −1 1 (0, −1) x intercept: Let y = 0 3 =0 ∴2 − x +1 3 ∴2 = x +1 ∴ 2x + 2 = 3 1 ∴x = 2 1   , 0 2 

1 x = –– 2

y=2 x

(– –34 , 0)

{ }

Domain R \ −

1 and range R \ {2}. 2

d xy + 2 y + 5 = 0 ∴ y( x + 2) = −5 −5 ∴y = x+2 Asymptotes: x = −2, y = 0 No x intercept. y intercept: Let x = 0 −5 ∴y = = −2.5 2 (0, −2.5)

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 2 Functions and graphs • EXERCISE 2.3   |  29 y xy + 2y + 5 = 0 y=0

0

x

(0, –2.5) x = –2

Domain R \ {−2} and range R \ {0}. 10 e y = −5 5− x Asymptotes: 5 − x = 0 ⇒ x = 5, y = −5 y intercept: Let x = 0 10 ∴y = − 5 = −3 5 (0, −3) x intercept: Let y = 0 10 ∴ −5= 0 5− x ∴10 = 5(5 − x ) ∴5 − x = 2 ∴x = 3 (3, 0 ) y (3, 0) 0 y = –5

x =5

10 – 5 y=– 5–x

(0, –3)

Domain R \ {5} and range R \ {−5}. a 1 1 a Let the equation be y = + k. x−h Asymptotes x = −3 and y = 6 a +6 ∴y = x+3 Substitute the point (−4,8) a +6 ∴8 = −1 ∴ a = −2 −2 The equation is y = + 6. x+3 a b Let the equation be y = + k. x−h 3 Asymptotes x = −2 and y = − 2 3 a − ∴y = x+2 2 Substitute the point (−3, −2) a 3 ∴−2 = − −1 2 1 ∴a = 2 1 3 The equation is y = − . 2( x + 2) 2 2 2 + 1 or y = +1 (3 − x )2 ( x − 3)2 Asymptotes: x = 3, y = 1 y intercept: Let x = 0

12 a y =

2 11 +1= 9 9  11   0,  9 No x intercepts as graph lies above its horizontal asymptote.  11  The point  6,  is symmetric about the vertical 9  11  asymptote to  0,  . 9

∴y =

x

y 2 y= –+1 (3 – x)2

(0, –119 )

(6, –119 ) y=1 0

x x=3

Domain R \ {3} and range (1, ∞) −3 b y = −2 4( x − 1)2 Asymptotes: x = 1, y = −2 y intercept: Let x = 0 11 −3 ∴y = −2=− 4 4 11    0, −  4 No x intercepts as graph lies below its horizontal asymptote. 11   The point  2, −  is symmetric about the vertical 4 11   0, asymptote to  −  . 4 y y=

0

–3 –2 4(x – 1)2

x=1

x y = –2

(0, – 11–4)

(2, – 11–4)

Domain R \ {1} and range (−∞, −2). 1 c y = −1 (2 x + 3)2 3 Asymptotes: 2 x + 3 = 0 ⇒ x = − , y = −1 2 y intercept: Let x = 0 1 8 ∴ y = −1= − 9 9 8   0, −  9

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

30 | TOPIC 2 Functions and graphs • EXERCISE 2.3 a +2 ( x − 4)2 Substitute the point (5, −1) a ∴ −1 = + 2 1 ∴ a = −3 −3 The equation is y = + 2. ( x − 4)2

x intercepts: Let y = 0 1 ∴ −1 = 0 (2 x + 3)2 ∴1 = (2 x + 3)2 ∴ 2 x + 3 = ±1 ∴ 2 x = −4 or − 2 ∴ x = −2 or x = −1 (−2, 0),(−1, 0)

∴y =

1 y= – –1 (2x + 3)2

(–2, 0)

(–1, 0) x

0 y = –1

(0, – –98 )

3 x = –– 2

{ }

Domain R \ − d

a +k ( x − h) 2 As the range is (−4, ∞), the horizontal asymptote is y = −4. a ∴y = −4 ( x − h) 2

b Let the equation of the graph be y =

y

3 and range (−1, ∞). 2

25 x 2 − 1 5x 2 25 x 2 1 − 2 ∴y = 2 5x 5x 1 ∴y = 5− 2 5x Asymptotes: x = 0, y = 5 No y intercept x intercepts: Let y = 0 y=

Given f ( −1) = 8 and f (2) = 8, the points (−1,8) and (2,8) lie on the graph. As these points have the same y coordinate, they must be symmetrically placed around the vertical asymptote. −1 + 2 1 Therefore, the vertical asymptote is x = = 2 2 a The equation becomes y = 4 − 2  x − 1   2 4a b This can be written as y = − 4 or y = −4 ( 2 x − 1)2 ( 2 x − 1)2 where b = 4 a. Substitute the point (2,8) b ∴8 = − 4 9 ∴ b = 108 108 The equation of the graph is y = − 4. ( 2 x − 1)2 1 so the function is The domain is R \ 2 1 108 f :R\ → R, f ( x ) = − 4. 2 (2 x − 1)2

{}

{}

14 a i ( y − 2)2 = 4( x − 3)

25 x 2 − 1 ∴ =0 5x 2 2 ∴ 25 x − 1 = 0 1 ∴ x2 = 25 1 ∴x = ± 5  ± 1 ,0   5 

∴ y − 2 = ± 4( x − 3) ∴ y = 2 ± 2 ( x − 3) The upper branch is the function with rule y = 2 ( x − 3) + 2. Its domain is [3, ∞). Its endpoint is (3, 2) so its range is [2, ∞). The other function is the lower branch y = −2 ( x − 3) + 2 with the same domain [3, ∞) but range (−∞, 2]. ii y 2 + 2 y + 2 x = 5 Completing the square

y y=5 25x2 – 1 y= – 5x2 (0.2, 0) x

(–0.2, 0) 0

x=0

Domain R \ {0} and range (−∞,5). a 13 a Let the equation be y = +k ( x − h) 2 Asymptotes are x = 4 and y = 2

( y2 + 2 y + 1) − 1 + 2 x = 5

∴ ( y + 1)2 = 6 − 2 x ∴ y + 1 = ± 6 − 2x ∴ y = ± 6 − 2x − 1 The upper branch is the function with rule y = −2( x − 3) − 1 and endpoint (3, −1). Its domain requires 6 − 2 x ≥ 0 ⇒ x ≤ 3. The domain is (−∞,3] and range [−1, ∞). The lower branch is the function with rule y = − −2( x − 3) − 1. Its domain is (−∞,3] and range (−∞, −1]. b i y = 1 − 3 x Domain: 3 x ≥ 0 ⇒ x ≥ 0. Domain is [0, ∞) Endpoint: (0,1) which is also the y intercept.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 2 Functions and graphs • EXERCISE 2.3   |  31 x intercept: Let y = 0

iv y = − 3 − 12 − 3 x Domain: 12 − 3 x ≥ 0 ⇒ x ≤ 4, (−∞, 4]. Endpoint: (4, − 3) y intercept: Let x = 0

∴1 − 3 x = 0 ∴ 3x = 1 ∴ 3x = 1 1 ∴x = 3  1 ,0  3 

∴ y = − 3 − 12 ∴y = − 3 − 2 3 ∴ y = −3 3

( 0, −3 3 )

y (0, 1)

(

No x intercept as range is −∞, − 3 . y

y = 1 – 3x

x

0

x

0

(4, – 3 )

( , 0) 1 – 3

(0, –3 3 ) Range is (−∞,1]. ii y = 2 − x + 4 Domain: − x ≥ 0 ⇒ x ≤ 0 Domain is (−∞, 0]. Endpoint: (0, 4) which is also the y intercept. x intercept: Let y = 0 ∴2 −x + 4 = 0 ∴ − x = −2 This is not possible so there is no x intercept (also possible to anticipate this as a > 0). Let x = −1 then y = 6 so point on the graph is (−1,6). y

y = 2 –x + 4

y = – 3 – 12 – 3x

15 a f : [5, ∞) → R, f ( x ) = a x + b + c The endpoint of the graph is (5, −2) so f ( x ) = a x − 5 − 2 The point (6, 0) is on the graph so f (6) = 0. ∴0 = a 6 − 5 − 2 ∴0 = a − 2 ∴a = 2 Hence, f ( x ) = 2 x − 5 − 2 with a = 2, b = −5, c = −2. b f : (−∞, 2] → R, f ( x ) = ax + b + c i Let y = ax + b + c Endpoint is (2, −2) so a(2) + b = 0 and c = −2 ∴ b = −2a and c = −2 y = ax − 2a − 2 Substitute (0, 0)

(–1, 6) (0, 4)

∴ 0 = −2a − 2 0

x

Range is [4, ∞). iii y = 2 4 + 2 x + 3 Domain: 4 + 2 x ≥ 0 ⇒ x ≥ −2, [−2, ∞) Endpoint: (−2,3) y intercept: Let x = 0 ∴y = 2 4 +3= 7 (0, 7) There will not be an x intercept. The range is [3, ∞). y

∴ −2a = 2 ∴ −2a = 4 ∴ a = −2 Since b = −2a, b = 4 f ( x ) = −2 x + 4 − 2 with a = −2, b = 4, c = −2. ii Reflecting the graph in the x axis would make the endpoint (2, 2) and the range (−∞, 2]. The graph would still pass through the origin. The equation of the reflected graph would be y = − −2 x + 4 + 2. 16 a {( x , y) : y = 3 x + 2 − 1} y = 3 x + 2 −1 Point of inflection (−2, −1) y intercept: Let x = 0 ∴ y = 3 2 −1

( 0, 3 2 − 1)

(0, 7)

x intercept: Let y = 0

y = 2 4 + 2x + 3

∴0 = 3 x + 2 −1 ∴3 x +2 =1

(–2, 3) 0

x

∴ x + 2 = 13 ∴x + 2 =1 ∴ x = −1 (−1,0)

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

32 | TOPIC 2 Functions and graphs • EXERCISE 2.3 y

y = 3x + 2 – 1

(0,

)

32–1

(–1, 0) x

0 (–2, –1)

1− 3 x + 8 2 Let y = f ( x ) 1 ∴ y = 1− 3 x + 8 2 1 1 = − 3 x +8 2 2 The implied domain is R and the range is R. 1  Point of inflection  −8,  2

b f ( x ) =

(

)

y intercept: Let x = 0

e Let the equation be y = a 3 x − h + k The tangent is vertical at the point of inflection so (−1, −2) is the point of inflection. ∴ y = a3 x +1 − 2 Substitute the point (−9,5) ∴ 5 = a 3 −8 − 2 ∴ 5 = −2a − 2 ∴ 2a = −7 7 ∴a = − 2 The equation is y = −

1 (1 − 3 8) 2 1 = (1 − 2) 2 1 =− 2 1   0, −  2

∴y =

73 x + 1 −2 2

f ( y + 2)3 = 64 x − 128 Take the cube root of each side ∴ y + 2 = 3 64 x − 128 ∴ y = 3 64( x − 2) − 2 ∴ y = 4 3 ( x − 2) − 2 The point of inflection is (2, −2). 1

x intercept: Let y = 0 1 ∴0 = 1− 3 x + 8 2 ∴3 x +8 =1 ∴x +8 =1 ∴ x = −7 (−7,0)

(

(–8, –12)

d Let the equation be y = a 3 x − h + k Point of inflection is (0, −2) ∴ y = a3 x − 2 Substitute the point (1, 0) ∴0 = a3 1 − 2 ∴0 = a − 2 ∴a = 2 The equation is y = 2 3 x − 2.

y = x 3 is y = 3 x and its graph could be formed by drawing 17 a  the line y = x and constructing its cube root. The root shape dominates since 3 > 1. b The two graphs both contain the points (1,1),(0, 0) and (−1, −1). The line lies in quadrants 1 and 3. Since cube roots

)

1

of negative numbers can be taken, the graph of y = x 3 will exist in both quadrants 1 and 3. y

y

)

(

y = –12 1– 3 x + 8

y=x

1 –

y = x3

(–7, 0)

(–1, –1)

x

0

(

(1, 1) x

0

)

1 0, – – 2

c g : [−3,6] → R, g( x ) = 3 − x + 5 Endpoints: g(−3) = 3 8 = 2, so (−3, 2) is an endpoint. g(6) = 3 −1 = −1, so (6, −1) is an endpoint. Point of inflection: (5, 0) which is also the x intercept.

(

y intercept: g(0) = 3 5 ⇒ 0, 3 5

)

5

y = g(x) = 3 –x + 5

(0, 3 5) (5, 0) x

0

(6, –1)

Domain [−3,6] and range [−1, 2].

1

18 a y = x 2

y

(–3, 2)

1

c x 3 − x > 0 when x 3 > x. From the diagram this occurs for 0 < x < 1 and if x < −1 The solution set is {x : x < −1} ∪ {x : 0 < x < 1}.

∴ y = x5 The graph of y = x 5 lies in the first and third quadrants. However, where x 5 < 0, the square root of these values cannot be formed. 5

Therefore, the graph of y = x 2 lies only in the first quadrant and has domain R + ∪ {0} and range R + ∪ {0}. As 5 > 2, the polynomial shape dominates the function y = x5 . The graph intersects the line y = x at (0, 0) and (1,1).

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 2 Functions and graphs • EXERCISE 2.4   |  33 5 – 2

y

y=x

y=x

(0, 0)

(1, 1) x

0

5

b

y = x3 ∴ y = 3 x5 The cube root of both the negative and positive sections of y = x 5 can be formed. 5 Therefore, the graph of y = x 3 lies in both the first and third quadrants and has domain R and range R. The graph intersects the line y = x at (0, 0), (1,1) and (−1, −1). As 5 > 3, the polynomial shape dominates the function y = 3 x5 . y 5 –

y = x3 y=x

(0, 0)

1 1 1 and y = . , y= 2 x2 − 4 x +4 ( x − 4)2 Draw each of the graphs on your device and determine where their asymptotes lie. Vertical asymptotes will exist for any value of x which makes the denominator zero, so these could be found by reasoning. Maximal domains may also be obtained by reasoning.

19 y =

Domain

Range

Asymptotes

y=

1 x −4

R \ {±2}

1 R \  − ,0   4 

x = ±2, y = 0

y=

1 x +4

R

 0, 1   4 

y=0

y=

1 ( x − 4)2

R \ {4}

(0, ∞)

x = 4, y = 0

2

2

1 is a truncus. ( x − 4)2 20 y = (2 − x )( x + 3) For the graph to exist, (2 − x )( x + 3) ≥ 0 Solve (2 − x )( x + 3) = 0 ⇒ x = 2, −3. Sketch the graph to solve the inequality. The graph of y =

y

(1, 1)

y = (2 – x)(x + 3)

x (–1, –1)

c

y=

(–3, 0)

3 x5

(2, 0) x

∴ y = 5 x3 The graph lies in both the first and third quadrants and has domain R and range R. The graph intersects the line y = x at (0, 0), (1,1) and (−1, −1). As 3 < 5, the root shape dominates the function y = 5 x 3 .

x ∈[ −3, 2 ]

y

y=x (–1, –1)

(1, 1) x

0

d

3 –

y = x5

(0, 0)

y = x 0.25 1

∴y = x4 ∴y = 4 x The line y = x lies in the first and third quadrants. The fourth root of its negative sections cannot be formed. Therefore the graph of y = x 0.25 lies only in the first quadrants and has domain R + ∪ {0} and range R + ∪ {0}. The graph intersects the line y = x at (0, 0) and (1,1). As 1 < 4, the root shape dominates the function y = 4 x1 . y

y=x y = x0.25 (0, 0)

(1, 1) 0

Exercise 2.4 — Combinations of functions − 3 x , x < −1  1 a f ( x ) =  x 3 , −1≤ x ≤1 2 − x , x >1  f (−8): Use the rule f ( x ) = − 3 x f (−8) = − 3 (−8) =2 f (−1): Use the rule f ( x ) = x 3 f (−1) = (−1)3 = −1 f (2): Use the rule f ( x ) = 2 − x f (2) = 0 b f ( x ) = − 3 x If x = −1, f (−1) = 1. Point (−1,1) is open for the cube root function. The point (−8, 2) lies on this branch. f (x) = x3 Stationary point of inflection at the origin. The points (−1, −1) and (1,1) are closed points for this cubic. f (x) = 2 − x The point (1,1) is an open point for the line and the point (2, 0) lies on the line.

x

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

34 | TOPIC 2 Functions and graphs • EXERCISE 2.4 y 4 3 2 1

y (–1, 1) (1, 1) (2, 0)

(0, 0)

x

0

x1  x − 2, a The branch to the left of x = 1 has the rule f ( x ) = 1 − x , so f (1) = 0. The branch to the right of x = 1 has the rule f ( x ) = x − 2, so f (1) → −1 (open circle). These branches do not join so the hybrid function is not continuous at x = 1. b The cubic function’s point of inflection is not in its restricted domain. The point (−3, 2) is an open point. The square root function has closed domain endpoints (−3, 2) and (1, 0). The line has open endpoint (1, −1) and contains point (2, 0).

x

(1, –1)

Many-to-one correspondence c f ( x ) = 4 The graph shows only the linear branch has a point with y = 4. Let x − 2 = 4 ∴x = 6 10 The left branch is a parabola on domain section where x < 0. Let its equation be y = a( x + 3)( x + 1) Substitute point (0, 4) ∴ 4 = a(3)(1) 4 ∴a = 3 4 Parabola branch has equation y = ( x + 3)( x + 1). 3 The middle branch is y = 4 for domain section x ∈[0, 2]. The line through points (3, 2) and (4, 0) has gradient m = −2. Its equation is y = −2( x − 4). The rule for the hybrid function could be expressed as 4 x 4 ii one solution if k = 4 or k ≤ 0 iii two solutions if 0 < k < 4 c {x : f ( x ) = 1} There will be two solutions, one on the cube root branch and one on the hyperbola branch. Let 3 x + 2 = 1

Domain: d fg = d f ∩ d g = R

5  Graph is a concave down parabola with x intercepts  , 0  2  and (1, 0); y intercept (0, −10). Turning point: x =

y = 2(5 – 2x)(x – 1)

y

a y = ( f + g ) ( x ) Rule: y = 5 − 2x + 2x − 2 ∴y = 3 Domain d f +g = d f ∩ dg ∴ d f +g = R

(1, 0)

(–74 , –49 )

x

0

( –52 , 0)

(0, –10)

Graph of y = ( f + g ) ( x ) is the horizontal line through (0,3). Its range is {3}. y

y=3 (0, 3)

13 f ( x ) = x 2 − 1 and g( x ) = x + 1 a i ( g − f ) (3) = g(3) − f (3) = 4 −8 = −6 ii ( gf ) (8) = g(8) f (8)

x

= 9 × 63 = 189 b d f = R, d g = [−1, ∞) d f + g = d f ∩ dg = [−1, ∞) c i Graph of f + g Add the ordinates together

b y = ( f − g ) ( x ) Rule: y = 5 − 2 x − ( 2 x − 2) ∴ y = 7 − 4x Domain: d f −g = d f ∩ dg ∴ d f −g = R Graph of y = ( f − g ) ( x ) is a straight line through (0, 7) and  7 , 0  . Its range is R.   4 



y = f(x)

y (0, 7)

+1 7 = 2 4

7 7 y = 2  5 −   − 1  2  4  3 3 = 2× × 2 4 9 = 4  7 9  is the maximum turning point.  ,  4 4 9 Range is  −∞,   4 

∴ 3 x = −1 ∴ x = −1 32 Let =1 x ∴ x = 32 The solution set is {−1,32}. 12 f ( x ) = 5 − 2 x, d f = R and g( x ) = 2 x − 2, d g = R.

0

5 2

(–1, 0)

y 6 5 4 3 2 1

–4 –3 –2 –1 0 –1 –2

y = 7 – 4x

y = f(x) + g(x) y = g(x) 1 2 3 4

x

ii Graph of g − f Subtract ordinates

) –74 , 0) x

0

y = f(x)

c y = ( fg ) ( x ) Rule: y = (5 − 2 x )(2 x − 2) ∴ y = 2(5 − 2 x )( x − 1)

(–1, 0)

y 6 5 4 3 2 1

–4 –3 –2 –1 0 –1 –2

y = g(x) – f(x) y = g(x) 1 2 3 4

x

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 2 Functions and graphs • EXERCISE 2.4   |  37 b The graphs of y = f ( x ) and y = ( f ( x ) ) will intersect at places where y = 0 or y = 1. For the function f ( x ) = x 3 − 2 x, let f ( x ) = 0. ∴ x ( x 2 − 2) = 0

iii Graph of fg Multiply ordinates together y 6 5 4 3 2 1

y = f(x) (–1, 0)

2

y = f(x)g(x)

∴ x = 0, x = ± 2 Let f ( x ) = 1 ∴ x3 − 2x = 1

(0, 1) y = g(x) (1, 0) x 1 2 3 4

–4 –3 –2 –1 0 –1 –2

∴ x3 − 2x − 1 = 0 ∴ ( x + 1)( x 2 − x − 1) = 0 ∴ x = −1 or x 2 − x − 1 = 0 Consider x 2 − x − 1 = 0

14 y = x + − x Draw the line y1 = x and the square root function y2 = − x . The common domain is (−∞, 0]. y 6 5 4 y = –x 3 2 y=x (–1, 0) 1 x –5 –4 –3 –2 –1 0 1 2 3 4 5 –1 y = x + –x –2 (0, 0) y 6 5 4 3 2 1

15 a

y = f(x) + g(x) –3

b

(

17

y = g(x)

–4

1

2

–2

y 50 40

x 4 y = f(x)

30

16 a g ( x ) = (2 x − 1)3. There is a stationary point of inflection at 1   , 0  . 2 y = ( g( x ) ) Square ordinates 2

20 10 –5

(–4, 0) –4 –3 –2

0

–1 –10

y

–20

2 y = (g(x))2

–30

1

0

–1

(0, 0) (2, 0) 1 2

3

4

5x

–40 (0, 1)

–50 (0.5, 0)

–1

x

(1, –2)

The minimum turning point of the parabolic branch needs to be obtained to form the range. The parabola  7 25  y = (2 x − 1)( x − 3) has minimum turning point at  , −  . 4 8 25 The range of the hybrid function is  − , ∞ .   8 18

0

(3, 0)

(–47 , – ––258 )

3 x

2

2

–2

(1, 1) 0

4

y = f(x) + g(x)

y

(–1, 1)

y y = g(x)

)

y = f(x)

–1 –10 –2 –3 –4 –5 –6

–2

1± 1+ 4 2 1± 5 = 2 The two graphs intersect at 1± 5  (0, 0), ± 2, 0 ,(−1,1),  ,1 .  2  x=

1 y = g(x) (0, –1)

2

x

Where the polynomial graph cuts the x axis, the cube root graph has vertical points of inflection; where the polynomial touches the x axis, the cube root graph also touches the x axis but at a sharp point. Wherever the polynomial graph has the values y = 0, y = −1 and y = 1, the cube root graph must have the same value. There are 9 points of intersection.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

38 | TOPIC 2 Functions and graphs • EXERCISE 2.5 y

Exercise 2.5 — Non-algebraic functions 1 f ( x ) = −10 x a f (2) = −10 2 = −100 b The graph of y = 10 x contains (0,1) and (1,10). The graph of y = −10 x contains the points (0, −1) and (1, −10). The graph of y = 10 − x contains the points (0,1) and (−1,10).

x

0 y = –3 (0, –5)

y

(–1, 10)

(1, 10)

y = 10–x

y = 10x

y = –2ex – 3

Domain R, range (−∞, −3). b y = 4e −3 x − 4 Asymptote: y = −4 y intercept: Let x = 0

y=0

(0, 1)

x

0 y = –10x

(1, –10)

1 x c y = 10 − x can be expressed as y =   or y = 0.1x.  10 

2 Domain is R for both f ( x ) = 2 x and g( x ) = 2− x ⇒ d f ∩ d g = R Domain of f − g is R. y = ( f − g)( x ) = f ( x ) − g( x ) = 2 x − 2− x To sketch the difference function, sketch it as y = f ( x ) + ( − g( x ) ) and add y cordinates. x −1 1 0 1 f ( x) 1 2 2 1 − g( x ) − −2 −1 2 f ( x ) + ( − g( x ) )

−1 1 2

0

1 12

As x → ∞, the graph of f dominates and as x → −∞, the graph of − g dominates. y y = f(x) y = (f – g)(x)

(0, 1) 0 (0, 0) (0, –1)

∴ y = 4e 0 − 4 ∴y = 0 (0,0) The origin is also the x intercept. 1 Point: Let x = − 3 ∴ y = 4e − 4 > 0 y

y = 4e–3x – 4

(0, 0) x

y = –4

Domain R and range (−4, ∞). c y = 5e x − 2 Asymptote: y = 0 There is no x intercept. y intercept: Let x = 0 ∴ y = 5e −2 (0,5e −2 ) Point: Let x = 2 ∴ y = 5e 0 ∴y = 5 (2,5)

y y = 5ex – 2

y=0 x y = –g(x)

(0, 5e–2) 0

The range of the graph is R. 3 a y = −2e x − 3 Asymptote: y = −3 y intercept: Let x = 0 ∴ y = −2e 0 − 3 ∴ y = −5 (0, −5) There will not be an x intercept.

(2, 5)

y=0

x

Domain R and range R +. 4 a y = 2e1− 3 x − 4 Asymptote: y = −4 y intercept: Let x = 0 ∴ y = 2e1 − 4 (0, 2e − 4) This point lies above the asymptote so there will be an x intercept. Approximately, 2e − 4 = 1.4. x intercept: Let y = 0

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 2 Functions and graphs • EXERCISE 2.5   |  39 b y = a × 10 kx Substitute point (4, −20) ∴−20 = a × 10 4 k .....(1) Substitute point (8, −200) ∴−200 = a × 108 k .....(2) Divide equation (2) by equation (1)

∴ 2e1− 3 x − 4 = 0 ∴ 2e1− 3 x = 4 ∴ e1− 3 x = 2 Convert to logarithm form ∴1 − 3 x = loge (2) ∴ 3 x = 1 − loge (2) 1 ∴ x = (1 − loge (2)) 3 1  The x intercept is  (1 − loge (2)), 0  which is 3  approximately (−0.1, 0). 1 Point: Let x = 3 ∴ y = 2e 0 − 4 ∴ y = −2 1   , −2 3

−200 a × 108 k = −20 a × 10 4 k ∴10 = 10 4 k ∴1 = 4 k 1 ∴k = 4 1 Substitute k = in equation (1) 4 ∴ −20 = a × 101 ∴ a = −2 x

∴

The equation is y = −2 × 10 4 . 6 y = a × e kx Substitute point (2,36) ∴ 36 = a × e 2 k .....(1) Substitute point (3,108) ∴108 = a × e3 k .....(2) Divide equation (2) by equation (1)

y

(0, 2e – 4)

(–13(1 – log (2)), 0)

0

e

x

∴ y = 2e1 – 3x – 4 y = –4

108 a × e3 k = 36 a × e 2 k ∴ 3 = ek ∴ k = loge (3)

Substitute e k = 3 in equation (1) 36 = a × e 2 k

b y = 3 × 2 x − 24 Asymptote: y = −24 y intercept: Let x = 0

∴ 36 = a × (e x )2

0

∴ y = 3 × 2 − 24 = −21 (0, −21) x intercept: Let y = 0

∴ 36 = a × (3)2 ∴ 36 = 9a ∴a = 4 Answer is a = 4, k = loge (3) 7 a y = 2 cos(4 x ) − 3, 0 ≤ x ≤ 2π 2π π Period = 4 2 Amplitude 2 Mean position y = −3 Range [−3 − 2, −3 + 2] = [−5, −1] No x intercepts

∴ 3 × 2 x − 24 = 0 ∴ 2x = 8 ∴x = 3 (3,0) y

y

(0, –1)

y = 3 × 2 – 24 x

0

(3, 0) x

0 (0, –21)

y = –24

Domain R and range (−24, ∞). 5 a y = ae x + b From the graph, the asymptote is y = 2 so b = 2 The equation becomes y = ae x + 2. The graph passes through the origin. Substitute (0, 0) ∴ 0 = ae 0 + 2 ∴0 = a + 2 ∴ a = −2 The equation is y = −2e x + 2 with a = −2, b = 2

–5

π – 4

π – 2

3π — 4

π

5π — 4

3π — 2

7π — 4

2π x (2π, –1)

y = 2 cos(4x) – 3

b Let the equation be y = a sin(nx ) + k. The period of the graph is 2. 2π ∴ =2 n ∴n = π The mean position is 5. y = 5 ⇒ k = 5. The range is [−3,13] which means the amplitude is 8. As the graph has an inverted sine shape, a = −8 The equation is y = −8 sin(π x ) + 5.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

40 | TOPIC 2 Functions and graphs • EXERCISE 2.5 3x 8 f : [ 0, 2π ] → R, f ( x ) = 1 − 2 sin    2  3x  y = f ( x ) = 1 − 2 sin    2 3 4π Period 2π ÷ = 2 3 Amplitude 2, inverted graph Mean position y = 1 Range [−1,3] x intercepts: Let y = 0

3π   −6 sin  3 x −  = 0  4  Solving the equation:  3π  sin 3 x −  =0  4  3π ∴ 3x − = 0, π , 2π ,3π 4 3π 7π 11π 15π ∴ 3x = , , , 4 4 4 4 π 7π 11π 5π ∴x = , , , 4 12 12 4

 3x  0 = 1 − 2 sin   , 0 ≤ x ≤ 2π 2  3x  1 3x ∴ sin   = , 0 ≤ ≤ 3π 2 2 2 3x π π π π ∴ = , π − , 2π + ,3π − 2 6 6 6 6 3 x π 5π 13π 17π ∴ = , , , 2 6 6 6 6 π 5π 13π 17π ∴ 3x = , , , 3 3 3 3 π 5π 13π 17π ∴x = , , , 9 9 9 9 y intercept: Let x = 0 y = 1 − 2 sin(0) = 1 (0,1)

(0, 3 2)

(2π, 1)

(0, 1) 5π – 9

13π –– 9

x 17π –– 2π 9

3π 3π   9 a f :  0,  → R, f ( x ) = −6 sin  3 x −   4   2  3 π   y = f ( x ) = −6 sin  3 x −   4  π   ∴ y = −6 sin  3  x −     4  π Horizontal translation units to the right. 4 2π Period 3 Amplitude 6, graph is inverted Mean position y = 0 so range is [−6,6]. Endpoints:  3π  f (0) = −6 sin  −   4  = −6 ×

− 2 2

=3 2  3π   15π  f   = −6 sin   2   4  − 2 = −6 × 2 =3 2

3π  Endpoints are 0,3 2 and  ,3 2  .  2 

(

π – 4

7π — 12

–6

( )

0 π –1 – 9

— , 3 2) (3π 2

0

3x f(x) = 1 – 2 sin — 2

y 3 1

y 6

)

π x intercepts: Either translate those of y = −6 sin(3 x ) units 4 to the right or solve

11π — 12

x 3π — 2

5π — 4

(

)

— f(x) = –6 sin 3x – 3π 4

b y = cos(2 x ) − 3 cos( x ) for x ∈[0, 2π ]. y = y1 + y2 where y1 = cos(2 x ) and y2 = −3 cos( x ) y1 = cos(2 x ) has period π , amplitude 1, range [−1,1]. y2 = −3 cos( x ) has period 2π , amplitude 3, inverted graph, range [−3,3]. y 4 3 2 1 0 –1 (0, –2) –2 –3 –4

(π, 4)

y = cos(2x) – 3 cos(x) y1 = cos(2x)

π – 4

π – 2

3π – 4

π

5π – 4

3π – 4

7π – 4

2π

x

(2π, –2)

y2 = –3 cos(x)

10 To sketch the graph of y = ( sin( x ) ) = sin 2 ( x ) for x ∈[−π , π ], remember that (−1)2 = 1, 0 2 = 0,12 = 1. The squared graph will not lie below the x axis. 2

y = sin2(x)

y 1

(–π, 0) –π

(0, 0) π –– 2 –1 y = sin(x)

π – 2

(π, 0) x π

x 11 a y = 3 tan   for x ∈[−π , π ]  2 1 Period π ÷ = 2π 2 x π First positive asymptote: = ⇒ x = π 2 2 For x ∈[−π , π ], there is only one other asymptote at x = π − 2π = −π . x intercept midway between the two asymptotes is x = 0. π To illustrate the dilation effect: Let x = then 3 3 3 π y = 3 tan   = = 3.  6 3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 2 Functions and graphs • EXERCISE 2.5   |  41 y 6 3 π –– 2

–π

π – 2

0 –3

π

–6

x = –π

x

x=π

b y = − tan ( 2 x − π ) for x ∈[−π , π ]. π   y = − tan  2  x −     2 π Period 2 π Horizontal shift units to the right and the graph is inverted. 2 π An asymptote occurs when 2 x − π = 2 3π ∴ 2x = 2 3π ∴x = 4 3π π π π π π Other asymptotes at x = − = and x = − = − 4 2 4 4 2 4 3π π π and x = − − = − . 4 2 4 π π 3π 3π The asymptotes are x = − ,x = − ,x = ,x = . 4 4 4 4 x intercepts lie midway between the asymptotes as the mean position is y = 0. π π They are at x = − , x = 0, x = and at the endpoints 2 2 x = −π , x = π . y

3π x = – –– 4 –π

π x =– 4

π x = –– 4

π 3π – – ––– 2 4

π –– 4

0

3π x = –– 4

π – 4

π – 2

1 3 −π  = a tan  × 2 6  2 1 π ∴ − = a tan  −   4 2 1 ∴ − = a × −1 2 1 ∴a = 2 1  3x  The equation is y = tan    .  2 2 7 7 13 y = 3 tan(2π x ) − 3 for − ≤ x ≤ . 8 8 1 π Period = 2π 2 Mean position x = − 3 1 π An asymptote occurs at 2π x = ⇒ x = . 2 4 1 1 3 1 1 1 Other asymptotes at x = + = , at x = − = − , 4 2 4 4 2 4 1 1 3 at x = − − = − 4 2 4 1 1 3 3 The asymptotes are x = − , x = − , x = , x = 4 4 4 4 x intercepts: Let y = 0 ∴−

()

x y = 3 tan – 2

3π –– 4

y = –tan(2x – π)

π

x

3 tan(2π x ) − 3 = 0 3 7π 7π , − ≤ 2π x ≤ 3 4 4 π π π ∴ 2π x = , π + or − π + 6 6 6 π 7π 5π ∴ 2π x = , or − 6 6 6 1 7 5 ∴ x = , ,− 12 12 12 7 Endpoints: Let x = − 8 7π y = 3 tan  −   4  =3 7 Let x = , 8 7π y = 3 tan    4  = −3 7 7 Endpoints are  − ,3 and  , −3 .  8  8  ∴ tan(2π x ) =

y

12 The graph of y = a tan(nx ) π 2π The distance between the asymptotes at x = ± is so 3 3 this is the period. π 2π ∴ = n 3 ∴ 3 = 2n 3 ∴n = 2  3x  The equation becomes y = a tan    . 2 π 1  Substitute the point  − , −   6 2

y = 3 tan(2πx) – 3

(– –78 , 3 3 ) 0 –1

3 –– 4

3 x = –– 4

1 –– 2

1 –– 4

1 x = –– 4

1 – 4 (0, – 3)

1 – 2

1 x=– 4

3 x=– 4

 

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

3 – 4

1

(

x 7, – –3 3 8

)

42 | TOPIC 2 Functions and graphs • EXERCISE 2.5

π 14 y = 1 − tan  x +  for 0 ≤ x ≤ 2π .  6 π Period π , horizontal translation units to the left, graph is 6 inverted. Mean position y = 1. π π π An asymptote when x + = ⇒ x = . 6 2 3 4π 2π π π and x = − π = − but Other asymptote at x = + π = 3 3 3 3 this one is not in the domain given. π 4π The asymptotes are x = , x = 3 3 x intercepts: Let y = 0  π 1 − tan  x +  = 0, 0 ≤ x ≤ 2π  6  π π π ∴ tan  x +  = 1, ≤ x ≤ 2π +  6 6 6 π π π ∴x + = ,π + 6 4 4 π π 5π ∴x + = , 6 4 4 π 13π ∴x = , 12 12 Endpoints: Let x = 0 π y = 1 − tan    6

(

–5

4 0, – 5

0

y = 3 × 4–x

(0, 3)

−

y = −5 × 30 = −5 ( 0, −5) Point: Let x = −2

)

y = −5 × 31 = −15 ( −2, −15) y

(2π, 1– –33 ) 13π –– 12

π x=– 3

4 4 × 10 0 = 5 5  4  0,  5 Point: Let x = 1 4 y = × 101 = 8 5 (1,8)

4π –– 3

2π

x

y=0

0

x

(0, –5)

4π x = –– 3

4 × 10 x 5 Asymptote: y = 0 y intercept: Let x = 0 y=

x

c y = −5 × 3 2 Asymptote: y = 0 y intercept: Let x = 0

(

15 a y =

x

y =0

As x → ∞, y → 0 +.

π y = 1 – tan x + – 6

π – 3

y=0 x

As x → ∞, y → ∞. b y = 3 × 4 − x Asymptote: y = 0 y intercept: Let x = 0

0

)

0 π – 12

( (

(–1, 12)

  3 3 Endpoints are  0,1 − . and  2π ,1 − 3  3   

3 0, 1– – 3

4 × 10 x y=– 5

y

3 3

y 5

(1, 8)

y = 3 × 40 = 3 ( 0,3) Point: Let x = −1 y = 3 × 41 = 12 ( −1,12 )

3 = 1− 3 Let x = 2π π y = 1 − tan  2π +   6 π  = 1 − tan    6 = 1−

y

y = –5 × 3

x –– 2

(–2, –15)

As x → ∞, y → 0 −. −x  2 d y = −    3 Asymptote: y = 0 y intercept: Let x = 0 0

 2 y = −   = −1  3 ( 0, −1) Point: Let x = 1

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 2 Functions and graphs • EXERCISE 2.5   |  43  2 y = −   3 (1, −1.5)

−1

=−

y

3 2

y

y = –12 e–4x + 3

(0, 3.5) y=0

0

y=3

x

(0, –1) (1, –1.5)

()

2 –x y =– – 3

x

Range is (3, ∞). d y = 4 − e 2 x Asymptote: y = 4 y intercept: Let x = 0

As x → ∞, y → −∞ 16 a y = e x − 3 Asymptote: y = −3 y intercept: Let x = 0 y = e 0 − 3 = −2 ( 0, −2 ) x intercept: Let y = 0 ex − 3 = 0

y = 4 − e0 = 3 ( 0,3) x intercept: Let y = 0 4 − e2 x = 0 ∴ e2 x = 4 ∴ 2 x = loge (4) 1 ∴ x = loge (4) 2 1  log (4), 0  or ( loge (2), 0 )  2 e 

∴ex = 3 ∴ x = loge (3)

( loge (3), 0 )

y y = ex – 3

y

0

x

(0, –2)

(loge (3), 0) y = –3

0

(loge (2), 0) x y = 4 – e2x

Range (−3, ∞). b y = −2e 2 x − 1 Asymptote: y = −1 y intercept: Let x = 0

e

y = −2e 0 − 1 = −3 ( 0, −3) No x intercept y

x

0

y=4

(0, 3)

y = –1 (0, –3) y = –2e2x – 1

Range is (−∞, 4). y = 4e 2 x − 6 + 2 ∴ y = 4e 2( x − 3) + 2 Asymptote: y = 2 y intercept: Let x = 0 y = 4e −6 + 2 ≈ 2.01 (0, 4e −6 + 2) No x intercept Point: Let x = 3 ∴ y = 4e 0 + 2 = 6 (3,6) y y = 4e2x–6 + 2

Range is (−∞, −1). 1 c y = e −4 x + 3 2 Asymptote: y = 3 y intercept: Let x = 0 1 y = e 0 + 3 = 3.5 2 (0,3.5) No x intercept

(0, 4e– 6 + 2) 0

(3, 6)

y=2 x

Range is (2, ∞).

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

44 | TOPIC 2 Functions and graphs • EXERCISE 2.5 −

x +1

Substitute the point (3,9) 9 = 23− b + c ....(2) Subtract equation (1) from equation (2)

f y = 1 − e 2 Asymptote: y = 1 y intercept: Let x = 0

∴14 = 23− b − 2− b

y = 1 − e −0.5 ≈ 0.39

∴14 = 23 × 2− b − 2− b

( 0,1 − e ) −0.5

∴14 = 8 × 2− b − 2− b

x intercept: Let y = 0

x +1 − 1− e 2

∴e

−

x +1 2

x +1 − ∴e 2

∴14 = 7 × 2− b

=0 =1 = e0

x +1 =0 2 ∴x +1 = 0 ∴ x = −1

∴−

( −1, 0 )

y (0, 1 – e–0.5) y=1 (–1, 0)

x

0

+1 – x––– 2

y =1 – e

Range is (−∞,1). 17 a f ( x ) = ae x + b Asymptote is y = 11 so b = 11. Equation becomes f ( x ) = ae x + 11. The graph passes through the origin so f (0) = 0. ∴ ae 0 + 11 = 0 ∴ a + 11 = 0 ∴ a = −11 The rule for the function is f ( x ) = −11e x + 11 with a = −11, b = 11. The domain of the graph is R so as a mapping the function is written f : R → R, f ( x ) = −11e x + 11. b y = Ae nx + k The asymptote is y = 4 so k = 4 and the equation becomes y = ae nx + 4. Substitute the point (0,5) ∴ 5 = ae 0 + 4 ∴5 = a + 4 ∴a = 1 The equation becomes y = e nx + 4. Substitute the point −1, 4 + e 2 ∴ 4 + e2 = e− n + 4

(

∴ e2 = e− n ∴ 2 = −n ∴ n = −2 The equation is y = e −2 x + 4 c i y = 2 x − b + c Substitute the point (0, −5) −5 = 2− b + c ....(1)

)

∴ 2 = 2− b ∴ b = −1 Substitute b = −1 in equation (1) ∴ −5 = 2 + c ∴ c = −7 Hence, b = −1, c = −7 ii The equation of the graph is y = 2 x +1 − 7. Its asymptote is y = −7 and the given points lie above this asymptote. Therefore, the range of the graph is (−7, ∞). d i y = Ae x −2 + B The long term behaviour x → −∞, y → −2 means there is an asymptote at y = −2. Therefore, B = −2 and the equation becomes y = Ae x − 2 − 2. Substitute the point (2,10) ∴10 = Ae 0 − 2 ∴ A = 12 Answer: A = 12, B = −2 ii The equation is y = 12e x − 2 − 2   6  Substitute the point  a, 2  − 1   e   6 ∴ 2  − 1 = 12e a − 2 − 2 e 

∴12e −1 − 2 = 12e a − 2 − 2 ∴ e −1 = e a − 2 ∴ a − 2 = −1 ∴a = 1 18 a y = 6 sin(8 x ) 2π π Period = , amplitude 6. 8 4 Mean position y = 0 so range is [−6,6]. x b y = 2 − 3 cos    4 1 Period 2π ÷ = 8π , amplitude 3 4 Mean position y = 2 so range is [−1,5]. c y = − sin ( 3 x − 6 ) 2π Period , amplitude 1. 3 Mean position y = 0 so range is [−1,1]. d y = 3 ( 5 + 2 cos ( 6π x )) y = 15 + 6 cos(6π x ) 2π 1 Period = , amplitude 6. 6π 3 Mean position y = 15 so range is [9, 21]. 19 a y = −7 cos(4 x ), 0 ≤ x ≤ π 2π π Period = , amplitude 7, graph is inverted, mean 4 2 position y = 0, range [−7, 7].

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 2 Functions and graphs • EXERCISE 2.5   |  45 y

y

y = 2 – 4 sin(3x)

6

7 y = –7cos(4x)

4 2

(2π, 2)

(0, 2) 0

π – 8

π – 4

π – 6

π – 2

3π –– 8

3π –– 4

3π –– 6

π

(π, –7)

(0, –7)

–7

0 π — 18 –2

x

b y = 5 − sin( x ), 0 ≤ x ≤ 2π Period 2π , amplitude 1, graph is inverted, mean position y = 5, range [4,6].

13π — 18

17π — 25π — 18 18

x

29π — 18

π e y = 2 sin  x +  , 0 ≤ x ≤ 2π  4 π Period 2π , amplitude 2, horizontal shift to left, mean 4 position y = 0, range [−2, 2]. π y = 2 sin x + – 4

(

y

)

2

y 6 (0, 5) 4

5π — 18

(2π, 2)

(0, 2) y = 5 – sin(x) 0

(2π, 5)

π – 4

3π — 4

5π — 4

7π — 4

2π

x

–2 0

π

π – 2

3π –– 2

2π

x

1 cos(2 x ) + 3, − π ≤ x ≤ 2π 2 2π 1 Period = π , amplitude , mean position y = 3, range 2 2 [2.5,3.5].

c y =

y

)–π, –27(

–π

1 3– 2 1 – 22

π 0 –– 2 –1

1 cos(2x) + 3 y =– 2

)2π, –27(

π – 2

π to the left. 4 3π π π f y = −4 cos  3 x −  + 4, − ≤ x ≤  2 2 2 π    y = −4 cos  3  x −   + 4   6  2π Period , amplitude 4, graph is inverted, horizontal 3 π translation to the right, mean position y = 4, range [0,8]. 6 x intercepts: Let y = 0 π π 3π ∴ −4 cos  3 x −  + 4 = 0, − ≤ x ≤  2 2 2 π π  ∴ cos  3 x −  = 1, − 2π ≤ 3 x − ≤ 2π  2 2 π ∴ 3 x − = −2π , −π ,0, π ,2π 2 3π π π 3π 5π ∴ 3x = − ,− , , , 2 2 2 2 2 π π π π 5π ∴ x = − ,− , , , 2 6 6 2 6 π Endpoints: Let x = − 2 3 π π   y = −4 cos  − − +4  2 2  = −4 cos ( −2π ) + 4 =0 3π Let x = 2  9π π  y = −4 cos  − +4  2 2  = −4 cos ( 4π ) + 4 =0 Points on y = 2 sin( x ) are moved

π

3π –– 2

2π

x

d y = 2 − 4 sin(3 x ), 0 ≤ x ≤ 2π 2π Period , amplitude 4, graph is inverted, mean position 3 y = 2, range [−2,6]. x intercepts: Let y = 0 ∴ 0 = 2 − 4 sin(3 x ), 0 ≤ x ≤ 2π 1 ∴ sin(3 x ) = , 0 ≤ 3 x ≤ 6π 2 π π π π π π ∴ 3 x = , π − ,2π + ,3π − ,4π + ,5π − 6 6 6 6 6 6 π 5π 13π 17π 25π 29π , , , ∴ 3x = , , 6 6 6 6 6 6 π 5π 13π 17π 25π 29π ∴x = , , , , , 18 18 18 18 18 18

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

46 | TOPIC 2 Functions and graphs • EXERCISE 2.5

π  3π  , 0 . Endpoints are  − , 0  and   2  2  y

(

(– –π2 , 0)

0

)

π y = –4 cos 3x – – + 4 2

8 (0, 4)

π – 6

π – 2

x – , 0) ( 3π 2

5π — 6

20 a i 2 sin(2 x ) + 3 = 0 for x ∈[0, 2π ]



3 , 2 x ∈[0,4π ] 2 π π π π ∴ 2 x = π + ,2π − ,3π + ,4π − 3 3 3 3 4π 5π 10π 11π ∴ 2x = , , , 3 3 3 3 2π 5π 5π 11π ∴x = , , , 3 6 3 6 ∴ sin(2 x ) = −

ii Graph of y = sin(2 x ) for x ∈[0, 2π ] Period π , amplitude 1, range [−1,1] y 1

y = sin(2x) (2π, 0)

(0, 0) 0 –1

π – 4

π – 2

3π — 4

π

5π — 4

3π — 2

7π — 4

2π

x

iii  x : sin(2 x ) < − 3 , 0 ≤ x ≤ 2π  2   3 on the graph of y = sin( x ). Draw the line y = − 2 3 At their intersections, sin( x ) = − and therefore 2 2 sin( x ) + 3 = 0, the solutions to which were found in 2π 5π 5π 11π . part ai as x = , , , 3 6 3 6 2π 5π The sine curve lies below the line for and 0 ∴ r = 8h − h 2 b

c

1 V = π r 2h 3 1 ∴V = π (8h − h 2 )h 3 1 2 ∴V = π h (8 − h) 3 h > 0 and 8 − h > 0 so the restriction on h is 0 < h < 8. This would be seen on the graph to be the domain interval where V > 0. V

)5 –31 , 79.4) 1 πh2 (8 – h) V= – 3

N N = 38 (5, 30)

+ 20 – N = 38t t+2

(0, 10) 0

t

c The horizontal asymptote shows that as t → ∞, N → 38. The population of quolls will never exceed 38.

t

(8, 0) (0, 0)

d Using CAS the greatest volume is 79 cm3.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

h

TOPIC 3 Composite functions, transformations and ­inverses • EXERCISE 3.2   |  53

Topic 3 — Composite functions, transformations and ­inverses Exercise 3.2 — Composite functions and ­functional equations 1 f ( x ) = ( x − 1)( x + 3) and g( x ) = x 2 Dom

Ran

f (x)

R

[−4, ∞ )

g( x )

R

[0, ∞ )

For f ( g( x ) ), the range of g( x ) is [ 0, ∞ ), which is a subset of the domain of f ( x ), which is R so f ( g( x ) ) exists. f ( g( x ) ) = x 2 − 1 x 2 + 3 = ( x − 1)( x + 1) x 2 + 3 where dom = R

(

)(

)

(

)

For g ( f ( x ) ), the range of f ( x ) is [ −4, ∞ ), which is a subset of the domain of g( x ), which is R so g ( f ( x ) ) exists. g ( f ( x ) ) = (( x − 1)( x + 3)) = ( x − 1)2 ( x + 3)2 where dom = R 2

2 f ( x ) = 2 x − 1 and g( x ) =

1 x−2

Dom

Ran

f (x)

R

R

g( x )

R\{2|

R\{0}

For f ( g( x ) ), the range of g( x ) is R\{0}, which is a subset of the domain of f ( x ), which is R so f ( g( x ) ) exists. 2 f ( g( x ) ) = − 1 where dom = R \ {2} x−2

For g ( f ( x ) ), the range of f ( x ) is R, which is not a subset of the domain of g( x ), which is R\{2} so g ( f ( x ) ) does not exist 3 f ( x ) = x + 3 and g( x ) = 2 x − 5

If the domain of f ( x ) is restricted to R\{−2,2} to produce the function h( x ), then g ( h( x ) ) exists because the range of h( x ) will be R \{4}, which equals the domain of g( x ). \ h (x) = x2, x ∈ R \ {−2, 2} 1 c g ( h( x ) ) = 2 where x ∈ R \ {−2, 2} x −4 3 5 f ( x ) = − x f ( x ) − f ( y) LHS: f ( xy) −3 3 + x y = 3 − xy xy 3x − 3 y ×− = xy 3 = − ( x − y) = y− x RHS: y − x = LHS f ( x ) − f ( y) Therefore = y− x f ( xy) 2x 2y 6 f ( x ) = e and f ( y) = e f ( x + y) = e 2( x + y ) = e 2 x × e 2 y = f ( x ) f ( y) f (x) f ( x − y) = e 2( x − y ) = e 2 x ÷ e 2 y = f ( y) 1 2 7 f ( x ) = x + 1, g( x ) = x and h( x ) = x Dom Ran

f (x)

R

[1, ∞ )

Dom

Ran

g( x )

[0, ∞ )

[0, ∞ )

f (x)

[−3, ∞ )

[0, ∞ )

h( x )

R\{0}

R\{0}

g( x )

R

R

a f ( g( x ) ) is not defined because the range of g( x ), which is R is not contained in the domain of f ( x ), which is [ −3, ∞ ). b We want ran g = [ −3, ∞ ) Solve 2 x − 5 > −3 If the domain of g( x ) is restricted to [1, ∞ ) to produce the function h( x ), then f ( h( x ) ) exists because the range of h( x ) will be [ −3, ∞ ), which equals the domain of f ( x ). \ h (x) = 2x - 5, x ∈ [1, ∞) c f ( h( x ) ) = 2 x − 5 + 3 = 2 x − 2 where x ∈[1, ∞ ) 1 4 f ( x ) = x 2 and g( x ) = x−4 Dom

Ran

f (x)

R

[0, ∞ )

g( x )

R\{4}

R\{0}

a g ( f ( x ) ) is not defined because the range of f ( x ), which is [0, ∞ ) is not contained in the domain of g( x ), which is R\{4}. b We want ran f ≠ 4 Solve x 2 ≠ 4

a f ( g( x ) ) exists because the range of g( x ) which is [ 0, ∞ ), is a subset of the domain of f ( x ), which is R. Dom = [ 0, ∞ ). b g ( f ( x ) ) exists because the range of f ( x ), which is [1, ∞ ), is a subset of the domain of g( x ),which is [ 0, ∞ ) Dom = R. c h ( g( x ) ) does not exist because the range of g( x ), which is [0, ∞ ), is not a subset of the domain of h( x ), which is R\{0}. d h ( f ( x ) ) does exist because the range of f ( x ) , which is [1, ∞ ), is a subset of the domain of h( x ), which is R\{0}. Dom = R. 1 8 f ( x ) = x 2, g( x ) = x and h( x ) = − x Dom

Ran

f (x)

R

[0, ∞ )

g( x )

[0, ∞ )

[0, ∞ )

h( x )

R\{0}

R\{0}

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

54 | TOPIC 3 Composite functions, transformations and ­inverses • EXERCISE 3.2 a f ( g( x ) ) exists because the range of g( x ), which is [ 0, ∞ ) is a subset of the domain of f ( x ), which is R. 2 The rule is f ( g( x ) ) = f x = x = x where Dom = [ 0, ∞ ). b g ( f ( x ) ) exists because the range of f ( x ), which is [ 0, ∞ ) is equal to the domain of g( x ), which is [ 0, ∞ ).

12 f : ( −∞, 2 ] → R, f ( x ) = 2 − x and g : R → R, g( x ) = −

( ) ( )

( )=

The rule is g ( f ( x ) ) = f x

2

2

x = | x | where Dom = R.

c h ( f ( x ) ) does not exist because the range of f ( x ), which is [ 0, ∞ ), is not a subset of the domain of h( x ), which is R \ {0}. d g ( h( x ) ) does not exist because the range of h( x ), which is, R\{0} is not a subset of the domain of g( x ), which is [ 0, ∞ ).

9 f : R → R, f ( x ) = x 2 + 1 where Ran = [1, ∞ ) g : [ −2, ∞ ) → R, g( x ) = x + 2 where Ran = [ 0, ∞ ) f ( g( x ) ) exists because the range of g( x ), which is [ 0, ∞ ), is a subset of the domain of f ( x ), which is R. f ( g( x ) ) = f

(

) (

x+2 =

x+2

)

2

+1= x + 3

where Dom = [ −2, ∞ ) and Ran = [ 0, ∞ ). 1 10 f : ( 0, ∞ ) → R, f ( x ) = where Ran = ( 0, ∞ ) x 1 g : R \ {0} → R, f ( x ) = 2 where Ran = ( 0, ∞ ) x a g ( f ( x ) ) exists because the range of f ( x ), which is ( 0, ∞ ), is a subset of the domain of g( x ), which is R \ {0}.  1 b g ( f ( x ) ) = g   = x 2 where Dom = ( 0, ∞ ) and Ran = ( 0, ∞ ).  x y

c

y = g( f(x)) x

(0, 0)

Dom

Ran

f (x)

( −∞, 2 ]

[0, ∞ )

g( x )

R \ {1}

R \ {2}

a g ( f ( x ) ) is not defined because the range of f ( x ), which is [ 0, ∞ ) is not contained in the domain of g( x ), which is R \ {1}. b For g ( f1 ( x ) ) to have a domain for its existence then the domain must be ( −∞, 2 ] \ {1}. f1 ( x ) = 2 − x , x ∈( −∞, 2 ] \ {1} 1 + 2 where x ∈( −∞, 2 ] \ {1} c g ( f1 ( x ) ) = − 2 − x −1 x 13 f ( x ) = 5 a To show that f ( x + y) = f ( x ) × f ( y). LHS: f ( x + y) = 5x + y RHS: f ( x ) × f ( y) = 5x × 5 y = 5x + y = LHS Therefore f ( x + y) = f ( x ) × f ( y) f (x) b To show that f ( x − y) = . f ( y) LHS: f ( x − y) = 5x − y f (x) RHS: f ( y) 5x = y 5 = 5x − y = LHS Therefore f ( x − y) =

11 g( x ) =

1

( x − 3 )2

− 2 and f ( x ) = x . Ran

g( x )

R\{3)

( −2, ∞ )

f (x)

[0, ∞ )

[0, ∞ )

a f ( g( x ) ) is not defined because the range of g( x ), which is ( −2, ∞ ) is not contained in the domain of f ( x ), which is [0, ∞ ). b For f ( g( x ) ) to have a domain for its existence 1

( x − 3 )2

−2=0

1

( x − 3 )2

f (x) f ( y)

14 f ( x ) = x 3 , f ( y) = y3 and f ( xy) = ( xy )

3

LHS: f ( xy) = ( xy ) RHS: f ( x ) × f ( y) = x 3 × y3

3

Dom

=2

( x − 3 )2 =

1 2

1 2 1 +3 x=± 2

x−3= ±

1     1 + 3 ∪  + 3, ∞  i.e. x ∈ −∞, −   2 2  

1 +2 x −1

= ( xy ) = LHS Therefore f ( xy) = f ( x ) × f ( y) 15 a h( x ) = 3 x + 1 h( x + y) = 3( x + y) + 1 = 3x + 3 y + 1 = (3 x + 1) + (3 y + 1) − 1 = h ( x ) + h ( y) + c Therefore c = −1 1 b h( x ) = 3 x LHS: h( x ) + ( y) 1 1 = 3+ 3 x y y3 + x 3 = 3 3 x y 3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 3 Composite functions, transformations and ­inverses • EXERCISE 3.2   |  55

(

)

RHS: x 3 + y3 h ( xy ) 1 = x 3 + y3 × 3 3 x y x 3 + y3 = 3 3 x y = LHS Therefore h( x ) + h( y) = x 3 + y3 h( xy)

(

)

(

)

16 a f ( x ) = x 2 A: LHS: f ( xy) = ( xy)2 RHS: f ( x ) f ( y) = x 2 × y 2 = ( xy)2 LHS = RHS B: 2  x  x LHS: f   =    y  y 2

f (x) x 2  x  = = f ( y) y 2  y  LHS = RHS C: LHS: f (− x ) = (− x )2 = x 2 RHS: − f ( x ) = − x 2 LHS ≠ RHS D: LHS: f ( x + y) = ( x + y)2 RHS: f ( x ) f ( y) = x 2 y 2 LHS ≠ RHS b f ( x ) = x A: LHS: f ( xy) = xy RHS: f ( x ) f ( y) = x × y = xy LHS = RHS B:  x x LHS: f   =  y y x x f (x) RHS: = = f ( y) y y LHS = RHS C: LHS: f (− x ) = − x RHS: − f ( x ) = − x LHS ≠ RHS D: LHS: f ( x + y) = x + y RHS: f ( x ) f ( y) = x y = xy LHS ≠ RHS 1 c f ( x ) = x A: 1 LHS: f ( xy) = xy 1 1 1 RHS: f ( x ) f ( y) = × = x y xy LHS = RHS B:  x 1 y LHS: f   = =  y x x y RHS:

1 f (x) x y = = RHS: f ( y) 1 x y LHS = RHS C:

1 −x 1 RHS: − f ( x ) = − x LHS = RHS D:

LHS: f (− x ) =

1 x+y 1 1 1 RHS: f ( x ) f ( y) = × = x y xy LHS ≠ RHS d f ( x ) = e x A: LHS: f ( xy) = e xy RHS: f ( x ) f ( y) = e x e y = e x + y LHS ≠ RHS B: x  x LHS: f   = e y  y f (x) ex = = ex− y RHS: f ( y) e y LHS ≠ RHS C: LHS: f (− x ) = e − x RHS: − f ( x ) = − e x LHS ≠ RHS D: LHS: f ( x + y) = e x + y RHS: f ( x ) f ( y) = e x e y = e x + y LHS = RHS 17 f : [ 4, ∞ ) → R, f ( x ) = x − 4 and g : R → R, g( x ) = x 2 − 2 a Dom Ran LHS: f ( x + y) =

f (x)

[4, ∞ )

[0, ∞ )

g( x )

R

[−2, ∞ )

g ( f ( x ) ) exists because the range of f ( x ), which is [ 0, ∞ ) is the subset of the domain of g( x ) which is R.

(

) (

b g ( f ( x ) ) = g x − 4 = domain is [ 4, ∞ ). y c

x−4

)

2

− 2 = x − 6 where the

y = g( f(x)) (6, 0) x (4, –2)

d f ( g( x ) ) does not exist because the range of g( x ), which is [ −2, ∞ ) is not the subset of the domain of f ( x ) which is [4, ∞ ).

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

56 | TOPIC 3 Composite functions, transformations and ­inverses • EXERCISE 3.3

(

e If the domain of g( x ) is restricted to −∞, − 6  ∪ [6, ∞ ) then a new function is formed which is g1 : −∞, − 6  ∪ [6, ∞ ) → R, g1 ( x ) = x 2 − 2 so that f ( g( x ) ) exists.

(

f f1 ( g( x ) ) = x 2 − 2 − 4 = x 2 − 6 where x ∈ −∞, − 6  ∪ [6, ∞ ).

(

d i y = x 3 has been translated up one unit or in the positive in y axis direction to produce y = x 3 + 1 y 3

f (x)

Ran ( −∞, −1 + k ]

g( x )

( −∞, 2 ]

[ k, ∞ )

2

ii ( −2, −8 ) → ( −2, −7 ) 2 a = y sin( x ) has been dilated by a factor of four parallel to the y axis or from the x axis to produce y = 4 sin( x ) y 6

y 2

4

1

2 y = x3

0

1

– π–2

x

2

3 x

–3

1 a i y = x 3 has been dilated by a factor of three parallel to the y axis or from the x axis to produce y = 3 x 3.

–1

2

–2

Exercise 3.3 — Transformations

y = 3x3

0 (0, 0)1

–1

–1

Ran f ⊆ Dom g, so [1, ∞ ) ⊆ [ k , ∞ ), therefore k ≥ 1. Ran g ⊆ Dom f , so ( −∞, −1 + k ] ⊆ ( −∞, 2 ], therefore k ≤ 3. Therefore, overall, k ∈[1,3]

–2

–2

–3

y = x3

1

y = x3 + 1

18 f : [1, ∞ ) → R, f ( x ) = − x + k and g : ( −∞,2] → R, g( x ) = x + k Dom [1, ∞ )

2

(0, 1)

–1

0

–2

y = 4 sin(x) y = sin(x) π

π – 2

2π x

3π — 2

–4 –2

ii ( −2, −8 ) → ( −2, −24 ) b i y = x 3 has been translated two units to the left or in the negative x direction to produce y = ( x + 2 )3 y = (x + 2)3

y 3

–3

–2

0.5 1

2

3 x

–1 –2

c i y = x 3 has been reflected in the x axis two to produce y = − x3 y 3

–1

y = x3

1 0 –1 –2 –3

π

3π — 2

2π x

y = sin(x)

π units to the left or in the 2 π negative x direction to produce y = sin  x +   2

c y = sin( x ) has been translated

–) y = sin (x + π 2

y 1

2

–2

π – 2

–1.5

ii ( −2, −8 ) → ( −4, −8 )

–3

– π–2 0 –0.5 –1

–3

y = –x3

y = sin(2x)

1

1 (0, 0) –1 0

b y = sin( x ) has been dilated by a factor of one half parallel to the x axis or from the y axis to produce y = sin(2 x ) y 1.5

y = x3

2

(–2, 0)

–6

1

2

0.5 3 x

0 – π–2 –0.5 –1

π – 2

π

3π — 2

2π

x

y = sin(x)

ii ( −2, −8 ) → ( −2,8 ) Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 3 Composite functions, transformations and ­inverses • EXERCISE 3.3   |  57 d y = sin( x ) has been translated up two units or in the positive y direction to produce y = sin( x ) + 2 y

– π–2

4

y = sin(x) + 2

2

y = sin(x)

0

π

π – 2

–2

1 cos( x ) 2 1 y = cos( x ) has been dilated by a factor of parallel to the 2 y axis or from the x axis.

7 a y = cos( x ) → y =

y 1 3π — 2

2π

x

0.5 0

π 3 y = sin( x ) → y = −2 sin  2 x −  + 1 2  π    y = sin( x ) → y = −2 sin  2  x −   + 1 4   y = sin( x ) has been • Reflected in the x axis • Dilated by a factor of 2 parallel to the y axis or from the x axis 1 • Dilated by a factor of parallel to the x axis or from 2 the y axis π • Translated units to the right or in the positive 4 x direction • Translated 1 unit up or in the positive y direction  x +1  2 

1  4 y = e x → y = e 3 y = e x has been

π – 2

(2π, 0.5)

π

3π –– 2

2π

x

–1

b y = cos( x ) → y = cos(2 x ) y = cos( x ) has been dilated by a factor of x axis or from the y axis.

1 parallel to the 2

y 1 (0, 1)

(2π, 1)

0.5 0 –0.5

π – 4

π – 2

3π –– 4

π

5π –– 4

3π –– 2

7π –– 4

2π

x

–1

−2

1 • Dilated by a factor of parallel to the y axis or from 3 the x axis • Dilated by a factor of 2 parallel to the x axis or from the y axis • Translated 1 units to the left or in the negative x direction • Translated 2 units down or in the negative y direction

5 g( x ) = x is reflected in the y axis → ( − x ) = x 2

2

is translated 4 units to the right → ( x − 4 )

2

c y = cos( x ) → y = − cos( x ) y = cos( x ) has been reflected in the x axis. y 1 0.5 0 –0.5 –1

2 2

x is dilated by a factor of 2 from the y axis →  − 4   2 2  x − 4 − 3 is translated 3 units down →   2 2 1 x 1 is dilated by a factor of from the x axis →  − 4  − 1   3 2 3 2

1  x − 8 ∴ f (x) =   −1 3 2  1 3 is dilated by a factor of 3 parallel to the x axis → x x 3 is translated up 2 units → + 2 x 3 is reflected in the y axis → − + 2 x 3 is translated 1 unit to the left → − +2 x +1 3 −2 is reflected in the x axis → x +1 3 ∴ f (x) = −2 x +1

6 h( x ) =

–0.5

(0, 0.5)

π – 2

π

(0, –1)

3π –– 2

2π

x

(2π, 1)

d y = cos( x ) → y = cos( x ) − 1 y = cos( x ) has been translated down 1 unit or I unit in the negative y direction. y 0.5 (2π, 0) x

(0, 0) 0 –0.5

π – 2

π

3π –– 2

2π

–1 –1.5 –2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

58 | TOPIC 3 Composite functions, transformations and ­inverses • EXERCISE 3.3 1 → y = f ( x − 2) x 1 y = has been translated by a factor of 2 parallel to the x x axis or 2 units in the positive x direction.

8 a y =

y

xy=0

0

x=2

1 → y = − f (x) x 1 y = has been reflected in the x axis. x

b y =

y

y=0 0

x

1 2 2 2 9 a y = x → y = ( x + 3) − 3 3

1 parallel to the 3 y axis or from the x axis, translated 3 units to the left or 2 3 units in the negative x direction and translated units 3 2 down or units in the negative y direction. 3 b y = x 3 → y = −2 (1 − x )3 + 1 y = x 2 has been dilated by a factor of

y = x 3 has been reflected in the x and y axes, dilated by a factor of 2 parallel to the y axis or from the x axis direction, translated 1 unit to the right or 1 unit in the positive x direction and translated 1 unit up or 1 unit in the positive y direction. 1 3 3 − 1 or y = −1 c y = → y = x 2 ( x + 3) ( 2 x + 6) 1 y = has been dilated by a factor of 3 parallel to the x 1 y axis or from the x axis direction, dilated by a factor of 2 parallel to the x axis or from the y axis direction, translated 3 unit to the left or 3 unit in the negative x direction and translated 1 unit down 1 unit in the negative y direction. 4  4  2  10 a ( −2, 4 ) →  −2,  →  −5,  →  −5,   3  3  3 b (1,1) → ( −1, −1) → ( −1, −2 ) → ( 0, −2 ) → ( 0, −1)

x=0

1 → y = 3 f (x) x 1 y = has been dilated by a factor of 3 parallel to the in the x y axis or from the x axis.

c y =

y

y=0 x

0

x=0

1 d y = → y = f (2 x ) x 1 1 y = has been dilated by a factor of parallel to the x axis x 2 or from the y axis y

y=0 x

0

x=0

 1 c  2,  → ( 2,1) → (1,1) → ( −2,1) → ( −2, 0 )  2

π    11 a y = cos( x ) → y = 2 cos  2  x −   + 3 2   y = cos( x ) has been dilated by a factor of 2 parallel to the 1 y axis or from the x axis, dilated by a factor of parallel to 2 π the x axis or from the y axis, translated units to the right 2 π or units in the positive x direction and translated 3 units 2 up or 3 units in the positive y direction. b y = tan( x ) → y = − tan ( −2 x ) + 1 y = tan( x ) has been reflected in both axes, dilated by a 1 factor of parallel to the x axis or from the y axis and 2 translated 1 unit up or 1 unit in the positive y direction.

π   c y = sin( x ) → y = sin ( 3 x − π ) − 1 or y = sin 3  x −   − 1 3   1 y = sin( x ) has been dilated by a factor of parallel to the x 3 π π axis or from the y axis, translated units to the right or 3 3 units in the positive x direction and translated 1 down up or 1 unit in the negative y direction.  x − 6 x  12 h( x ) = 3 x → 3 − x → 3 − ( x − 3) → 3 −  − 3 = 3 −  2   2   x − 6 Therefore f ( x ) = 3 −   2 

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 3 Composite functions, transformations and ­inverses • EXERCISE 3.4   |  59     1 1 1 1 3 → −3→ − − 3 → −3  − 3 = − +9 x2  ( x + 2 )2   ( x + 2 )2  ( x + 2 )2 ( x + 2 )2 3 →− +9 ( 2 − x )2 3 +9 Therefore f ( x ) = − ( 2 − x )2

13 h( x ) =

2 2 14 f ( x ) = 2 x 2 − 3 → −2 x 2 + 3 → −2 ( 3 x ) + 3 = −18 x 2 + 3 → −18 ( x − 1) + 1 2 Therefore f ( x ) = −18 ( x − 1) + 1 1 1 1 1 2 → → −3→ −3→ −6 15 h( x ) = x+2 2x + 2 2( x + 3) 2(− x + 3) 2(3 − x ) 1 Therefore f ( x ) = −6 (3 − x ) 2x − 5 16 y =   x −1 2( x − 1) − 3 = x −1 2( x − 1) 3 . = − x −1 x −1 3 = 2− x −1 1 y = has been reflected in the y axis or the x axis, dilated by a factor of 3 parallel to the y axis or from the x axis, translated 1 unit x to the right or in the positive x direction and translated 2 units up or in the positive y direction. Dom = R \ {1} and Ran = R \ {2}.

y

y=2 x

0

x=1

17 y = 3 −

5− x →y= x 2

5− x 1 has been reflected in both axes, translated 5 units to the left or in the negative x direction, dilated by a factor of 2 2 parallel to the x axis or from the y axis and translated down 3 units or in the negative y direction. y = 3−

2 2 18 y = −2 ( 3 x − 1) + 5 → y = ( x − 2 ) − 1

1 parallel to the y axis or from the x axis, dilated by a 2 factor of 3 parallel to the x axis or from the y axis, translated 3 units to the left or in the negative x direction and translated 6 units down or in the negative y direction. y = −2 ( 3 x − 1)2 + 5 has been reflected in the x axis, dilated by a factor of

Exercise 3.4 — Transformations using matrices  x      x   x ′   1 0   x   1 T   y  =  = 2 = 2        y′   0 1   y   y      1 x ′ = x or 2 x ′ = x and y ′ = y 2 y = cos ( x ) → y ′ = cos ( 2 x ′ ) The equation of the transformed graph is y = cos ( 2 x ).

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

60 | TOPIC 3 Composite functions, transformations and ­inverses • EXERCISE 3.4   x   x ′   1 0   x   x  2 T   y  =  =  =     y ′   0 −1   y   − y  x ′ = x and y ′ = − y or − y ′ = y 1 1 1 or y ′ = − y = 2 → − y′ = x ( x ′ )2 ( x ′ )2 The equation of the transformed graph is y = −

1 . x2

3 y = x 4 is dilated by a factor of 2 from the x axis → y = 2 x 4 is translated one unit in the negative x direction → y = 2 ( x + 1)4 is 4 ­translated one unit in the negative y direction → y = 2 ( x + 1) − 1.

π 1 parallel to the x axis → y = − cos ( 2 x ) is translated 2 2 π  π      units in the positive x direction → y = − cos  2  x −   is translated 3 units in the negative y direction → y = − cos  2  x −   − 3.     2 2

4 y = cos ( x ) is reflected in the x axis → y = − cos ( x ) is dilated by a factor of

  x   x ′    1 0    1 0   −2   2  5 a T   y  =    = +    y ′    0 2    0 −1   5   0    x   x ′ T   y  =     y′   x   x ′ T   y  =     y′   x   x ′ T   y  =     y′ (−2,5) → (0, −10)

   1 0   −2   2   =   0 2   −5  +  0       −2  =  −10    0  =  −10  

  2  +   0   

  x   x ′    −1 0    2 0   −2 b T   y  =    =    y ′    0 1    0 1   5  T    T  

x   x ′ y  =  y ′ x   x ′ y  =  y ′

  0   +  −2 

   −1 0   −4   0   =   0 1   5  +  −2    

   =     x   x ′   T   y  =  =    y ′   (−2,5) → (4,3)    x   x ′   c T   y  =  =     y′       x   x ′   T   y  =  =  y ′          x   x ′   T   y  =  =     y′       x   x ′   T   y  =  =    y′    5   (−2,5) → ,6 3   1 6  0  

0 1 2

 1 7  4   0

  1 0  1 0   →   0 3   4 1   0

4 + 0  5   −2  4  3   1 0    −1 3    0 0 1    1 0    −1 3    0 0 1    1 0  5  3   6  0 1  5  3  6 

0    −2  +  −3 1    5   1

 

0   −5  1   6 

  −1 0    −1 0  +  −2  →  1  +  −2    0 1   0   0   0  2     0   3 

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 3 Composite functions, transformations and ­inverses • EXERCISE 3.4   |  61 1 parallel to the y axis or from the 2 x axis, translated 3 units to the left or 3 units in the negative x direction and translated 2 units down or 2 units in the ­negative y direction. y x ′ + 3 = x and y ′ = − 2 ⇒ 2 y ′ + 4 = y 2 2 y = 2 ( x − 1)2 → 2 y ′ + 4 = 2 ( x ′ + 3 − 1)

8 Dilated by a factor of

2 y′ + 4 = 2 ( x ′ + 2)

2

2 y′ = 2 ( x ′ + 2) − 4 2

y′ = ( x ′ + 2) − 2 2

Transformed rule is y = ( x + 2 )2 − 2.

π π to the right or 4 4 in the positive x direction and translation of 2 units down or 2 units in the negative y direction.

9 Reflection in the x axis, translation of

π and y′ = − y − 2 4 π x = x′ − y = − y′ − 2 4 π y = cos( x ) → − y′ − 2 = cos  x ′ −   4 π  y′ + 2 = − cos  x ′ −   4 π y′ = − cos  x ′ −  − 2  4

x′ = x +

12 x ′ = x − 3 and y ′ = −2 ( y + 1) y′ x = x′ + 3 − = y +1 2 y′ − −1= y 2 1 y′ 1 y = 2 → − −1= 2 x ( x ′ + 3 )2 y′ 1 − = +1 2 ( x ′ + 3 )2 2 y′ = − −2 ( x ′ + 3 )2 2 −2 Therefore y = − ( x + 3 )2     x   x ′   2 0   x 1 = 13 T   y  =       y′   0 2   y   Check: 1 x ′ = 2 x − 4 or x = ( x ′ + 4 ) and y ′ = 2 2 1 y = x 2 → 2 ( y ′ + 1) = ( x ′ + 4 ) 2

{

 ( x ′ + 4)  2 ( y ′ + 1) =   2 

x ′ = − x + 1 and y ′ = 3 y y′ x = 1 − x′ =y 3 y′ y = x + 1 − 2 → = 1 − x′ + 1 − 2 3 y′ = 3 − x ′ − 6 Transformed rule is y = 3 − x − 6. 11 Dilated by a factor of 2 parallel to the x axis or from the y axis, reflected in the x axis and translated 4 units to the left or 4 units in the negative x direction. x ′ = 2 x − 4 and y′ = − y 1 x = ( x ′ + 4 ) − y′ = y 2 3 1 y = x 3 → − y′ =  ( x ′ + 4 )  2  1 3 − y′ = ( x ′ + 4 ) 8 1 3 y′ = − ( x ′ + 4 ) 8 1 Transformed rule is y = − ( x + 4 )3. 8

1 y − 1 or y = 2 ( y ′ + 1) 2

2

1  ( x′ + 4)  y′ + 1 =  2  2 

2

2

π Transformed rule is y = − cos  x −  − 2  4 10 Reflected in the y axis, dilated by a factor of 3 parallel to the y or from the x axis and translated 1 unit to the right or in the positive x direction.

}

  −4   +  −1  

y′ =

1  ( x′ + 4)  − 1 as required 2  2 

 1   1    x   x ′   0  x   = + 14 T   y  =  3 3        y ′   0 −2   y   4      Check: 1 1 1 x ′ = x + so 3 x ′ − 1 = x and y ′ = −2 y + 4 or − ( y ′ − 4 ) = y 3 3 2 1 1 1 y = → − ( y′ − 4 ) = 2 3x ′ − 1 x 2 y′ − 4 = − 3x ′ − 1 2 y′ = − + 4 as required 3x ′ − 1 15 y =

2 1 −3→ y = x +1 x

  x   x ′   1 0   x   b  T   y  =  =  +    y ′   0 a   y   c  x′ = x + b → x = x′ − b y′ − c y ′ = ay + c → y = a Substituting in the new points: 2 1 y′ − c = −3= x a x′ − b + 1 2a 1 − 3a = y′ − c = x′ − b + 1 x 2a 1 y′ = − 3a + c = x′ − b + 1 x

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

62 | TOPIC 3 Composite functions, transformations and ­inverses • EXERCISE 3.5 b y = (1 − x )( x + 5) is a many-to-one mapping and is a ­function. Inverse is a one-to-many mapping and is a relation. c Function: Dom = R and Ran = ( −∞,9 ] Inverse: Dom = ( −∞,9 ] and Ran = R 2 a & b

Now equate the components: 2a = 1 1 a= 2 −b + 1 = 0 b =1 −3a + c = 0 c = 3a 3 = 2

y Inverse y=x y= x

16 y = − ( 3 x − 1)2 + 2 → y = x 2

(0, 0)

  x   x ′   a 0    x   c  T   y  =  =   +    y ′   0 b    y   d  x ′ = a ( x + c ) and y ′ = b( y + d ) x′ y′ x = −c y= −d a b

x

c y = x is a one-to-one mapping so is a function. y = x 2, x ≥ 0 is a one-to-one mapping so is a function.

Substituting in the new points:

3 a  x-intercepts occur at x = 1, x = −5. The TP is halfway between them, so the x-value of the TP is x = −2. To obtain the maximum domain, we restrict the parabola about the TP. Therefore, a = −2. y b

2

y′   x′   − d = −  3  − c − 1 + 2 = x 2    a b 2

y′  3x ′  = − − 3c − 1 + 2 + d = x 2  a  b 2  3x ′  y′ = − b  − 3c − 1 + b(2 + d ) = x 2  a 

(–2, 9)

y = (1 – x)(x + 5)

Now equate the components: 3 =1 a a=3 −3c − 1 = 0 1 c=− 3 −b = 1 b = −1 b(2 + d ) = 0 −1(2 + d ) = 0 2+d = 0 d = −2

y=x (–5, 0) x

0 (9, –2) (0, –5)

Inverse

c For y = (1 − x )( x + 5), Dom = ( −∞, −2 ] and Ran = ( −∞,9 ]

For x = (1 − y )( y + 5), Dom = ( −∞,9 ] and Ran = ( −∞, −2 ]

Exercise 3.5 — Inverse graphs and relations 1 a y = (1 − x )( x + 5) and inverse is x = (1 − y )( y + 5) y

4 If the domain for y = − ( x − 3)2 is restricted to [3, ∞ ) then the inverse will be a function also. y

Inverse

(–2, 9) y = (1 – x)(x + 5)

(0, 3)

y=x

y=x (3, 0) x

0

(0, 5)

(0, 1) (5, 0)

(–5, 0) 0

x

(1, 0) (9, –2)

(0, –5)

Inverse

y = –(x – 3)2

5 a f ( x ) = cos( x ) is a many-to-one function. b g( x ) = 1 − x 3 is a one-to-one function. c h( x ) = 4 − x 2 is a many-to-one function. 1 is a one-to-one function. d k ( x ) = 2 + x−3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 3 Composite functions, transformations and ­inverses • EXERCISE 3.5   |  63 6 A y = x 2 − 1 is a many-to-one function, so inverse in a one-to-many relation. B x 2 + y 2 = 1 is a many-to-many relation, so inverse in a many-to-many relation. 1 is a one-to-one function, so inverse in a one-toC y = x −1 one function.

9 x = ( y − 2 ) has a turning point at (0,2) and cuts the x axis at (0,4) so inverse will have to have a turning point at (2,0) and cut the y axis at (4,0), so Option A is the answer. 2

10 a & b

y=x

D y = 1 − x 2 is a many-to-one function, so inverse in a one-to-many relation. E y = 10 is a many-to-one function, so inverse in a one-tomany relation. Therefore C is the answer. y 7 a

0 –6 –5 –4 –3 –2 –1 –1 –2 Inverse –3 –4 –5 –6 y 4 3 2 1

0 –4 –3 –2 –1 –1 –2 –3 –4

d

Inverse

y 4 3 2 1

0 –4 –3 –2 –1 –1 –2 –3 –4

0 (0, –1)

x

(0, –3) 1x – 1 y = –– 3 Inverse

1 2 3 4 5 6 x

11 a & b y y= x

y 6 5 4 3 2 1

Inverse

(–3, 0)

Inverse

0 –6 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 –6

c

(–1, 0)

y=x

6 5 4 3 2 1

b

y

y=x

y = (x + 4)(x – 2) Inverse (0, 2) (–8, 0) (–4, 0) 0 (–9, –1) (0, –4)

1 2 3 4 5 6 x

y=x

1 2 3 4 x

(–1, –9)

(0, –8)

c P  arabola is a many-to-one mapping and the inverse is a one-to-many relation. d The inverse is not a function as functions must be one-toone or many-to-one mappings. e Function: Dom = R and Ran = [ −9, ∞ ). Inverse: Dom = [ −9, ∞ ) and Ran = R. f Largest domain for an inverse function is either x ∈( −∞, −1] or x ∈[ −1, ∞ ). 12 a

y y = x3

y=x

x

(2, 0)

y=x Inverse

(0, 0)

x

1 2 3 4 x

8 A y = x 2 a many-to-one function and y = ± x a one-to-many relation. B y = x 2, x ∈( −∞,0 ] a one-to-one function and y = x a one-to-one function but not a pair. C Two one-to-one functions. D y = x 2 + 1, x ∈[ 0, ∞ ) a one-to-one function and y = − x 2 − 1, x ∈[ 0, ∞ ) a one-to-one function but not a pair.

b Graph: One-to-one function. Inverse: One-to-one function. c The inverse is a function because it is a one-to-one ­mapping. d Graph: Dom = R and Ran = R. Inverse: Dom = R and Ran = R.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

64 | TOPIC 3 Composite functions, transformations and ­inverses • EXERCISE 3.6 y

13 a & b

b (2.828, 2.828), (0, 0), (−2.828, −2.828)

y=x

y

16 a

y=x

1 y=– x2

(1, 1)

1 y=3+– x

x

0

y=3

Inverse

Inverse x

0

c Graph: many-to-one function. Inverse: one-to-many relation. d Restricted domain to produce an inverse function is x ∈( −∞, 0 ). e y

x=3

b (3.532, 3.532)

Exercise 3.6 — Inverse functions 1 y = x 3 is a one-to-one function with Dom = R and Ran = R Inverse: swap x and y x = y3

y=x

y= 3 x This is a one-to-one with Dom = R and Ran = R 1 2 y = 2 is a many-to-one function with Dom = R\{0} and x Ran = ( 0, ∞ ) Inverse: swap x and y 1 x= 2 y 1 = y2 x 1 ± =y x This is a one-to-many relation and thus is not a function. Dom = ( 0, ∞ ) and Ran = R\{0}.

1 y = –2 x x

0 Inverse

Graph: Dom = ( −∞, 0 ) and Ran = ( 0, ∞ ). Inverse: Dom = ( 0, ∞ ) and Ran = ( −∞, 0 ). 14 y = 2 x 2 − 12 x + 13 has a turning point where: 13   y = 2  x 2 − 6x +   2 13   2 y = 2  x − 6 x + (3)2 − (3)2 +   2

3 f : ( −∞, 2 ) → R, f ( x ) = − where Ran = ( −∞, 0 ) Inverse: swap x and y

y = 2 ( x − 3) − 18 + 13 2

y = 2 ( x − 3 )2 − 5 TP = (3, 5) so largest possible domain for the inverse to be a function is ( −∞,3] so a = 3. y 15 a y=x y = 9x – x3

Inverse

0

x

x=− −x = 1 x 1 ± − x 1 − − x 1 2− − x −

1

( x − 2 )2

is a one-to-one function

1

( y − 2 )2 1

( y − 2 )2

= ( y − 2)

2

= y−2 = y − 2 since x ∈( −∞, 0 ) =y

This is a one-to-one function x where Ran = ( −∞, 2 ). f −1: ( −∞,0 ) → R, f −1 ( x ) = 2 − −

1 x

4 f : [3, ∞ ) → R, f ( x ) = x − 3 is a one-to-one function where Ran = [ 0, ∞ ). Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 3 Composite functions, transformations and ­inverses • EXERCISE 3.6   |  65 Inverse: swap x and y x = y−3 x2 = y − 3 y = x 2 + 3 where x ∈[ 0, ∞ ) and y ∈[3, ∞ ) f −1 : [ 0, ∞ ) → R, f −1 ( x ) = x 2 + 3 5 a f ( x ) = ( x + 1)2 where Dom = R and Ran = [ 0, ∞ ). This is a many-to-one function. Inverse: swap x and y x = ( y + 1)

2

± x = y +1

y = (x + 1)2

y= x

y = 5± x which is a one-to-many relation where Dom = [ 0, ∞ ) and Ran = R. c f : [ −4, 0 ] → R, f ( x ) = 16 − x 2 is a one-to-one function where Dom = [ −4, 0 ] and Ran = [ 0, 4 ] Inverse: swap x and y x = 16 − y 2 x ∈[ 0, 4 ] x 2 = 16 − y 2 y 2 = 16 − x 2

Inverse

(0, –1)

2

± x = y−5

The inverse is not a function as f ( x ) is not a one-to-one function. b If the domain for the function is restricted to [ −1, ∞ ) then it is a one-to-one function so the inverse is also a function. Thus b = −1. c y

(1, 0)

b y = ( x − 5)2 is a many-to-one function where Dom = R and Ran = [ 0, ∞ ). Inverse: swap x and y x = ( y − 5)

± x − 1 = y where Dom = [ 0, ∞ ) and Ran = R.

(0, 1) (–1, 0) 0

Inverse: swap x and y 1 x = ( y − 3) 3 3x + 3 = y y = 3 ( x + 1) which is a one-to-one function where Dom = R and Ran = R.

y = ± 16 − x 2 y = − 16 − x 2 as y ∈[ −4, 0 ]

x

∴ f −1 ( x ) = − 16 − x 2 which is a one-to-one function. Dom = [ 0, 4 ] and Ran = [ −4, 0 ].

d f

−1

: [ 0, ∞ ) → R, f

−1

d y = ( x − 1)3 is a one-to-one function where Dom = R and Ran = R. Inverse: swap x and y

(x) = x − 1

e The graphs do not intersect. 6 f ( x ) = 2 x + 2 is a one-to-one mapping with Dom = [ −2, ∞ ) and Ran = [ 0, ∞ ). Inverse: swap x and y x = 2 y+2 x = y+2 2 x2 = y+2 4 x2 y= −2 4 This is a one-to-one mapping with Dom = [ 0, ∞ ) and Ran = [ −2, ∞ ). The two functions intersect where 2 x+2 = x 4( x + 2) = x 2 4x + 8 = x2 x2 − 4x − 8 = 0 4 ± (−4)2 − 4 × −8 × 1 2 4 ± 48 = 2 4±4 3 = 2 = 2+2 3

x=

(

When x = 2 + 2 3, y = 2 + 2 3 ∴ POI = 2 + 2 3, 2 + 2 3 1 y ( x − 3) is one-to-one function where Dom = R and 7 a = 3 Ran = R.

)

x = ( y − 1)

3

3

x = y −1

y = 3 x +1 which is a one-to-one function where Dom = R and Ran = R. e y = x is a one-to-one function. Dom = [ 0, ∞ ) and Ran = [ 0, ∞ ). Inverse: swap x and y x= y x 2 = y where x ∈[ 0, ∞ ) which is a one-to-one function. Dom = [ 0, ∞ ) and Ran = [ 0, ∞ ). 1 f y = + 2 is many-to-one function. Dom = R \ {1} and ( x − 1)2 Ran = ( 2, ∞ ). Inverse: swap x and y 1 x= +2 ( y − 1)2 1 x−2= ( y − 1)2 1 2 = ( y − 1) x−2 1 = y −1 ± x−2 1 y =1± x−2 Which is a one-to-many relation with Dom = ( 2, ∞ ) and Ran = R \ {1}.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

66 | TOPIC 3 Composite functions, transformations and ­inverses • EXERCISE 3.6 1 , x ≠ −2 x+2 Inverse: swap x and y 1 x= y+2 1 = y+2 x 1 y= −2 x 1 f −1 ( x ) = − 2, x ≠ 0 x 1 1 1 a f f ( x ) = = = x as required. 1 1 −2+2 x x 1 b f −1 ( f ( x ) ) = − 2 = x + 2 − 2 = x as required. 1 x+2 9 k ( x ) = x 3 − 1 Inverse: Let y = k ( x ), swap x and y x = y3 − 1 8 f ( x ) =

(

)

x + 1 = y3

4x − 7 x−2 4( x − 2) + 1 = x−2 4( x − 2) 1 = + x−2 x−2 1 =4+ x−2

11 f ( x ) =

1 + 4 where Dom = R\{2} and Ran = R\{4} x−2 Inverse: swap x and y 1 +4 x= y−2 1 x−4= y−2 1 = y−2 x−4 1 y= +2 x−4 1 f −1 ( x ) = + 2 where Dom = R\{4} and Ran = R\{2} x−4 f (x) =

y

y = 3 x +1 k −1 ( x ) = 3 x + 1

(

1

) (

)

3

a k k ( x ) = x + 1 − 1 = x +1−1 = x as required. 3

y=4 y=2

(0, 7–2) (0, 7–4)

c f : [ −3,3] → R, f ( x ) = 9 − x 2 is a many-to-one function. The inverse will be a one-to-many relation. d f : [ −2, ∞ ) → R, f ( x ) = x + 2 is a one-to-one function with Dom = [ −2, ∞ ) and Ran = [ 0, ∞ ). Inverse: swap x and y x = y + 2, x ∈[ 0, ∞ ) y = x 2 − 2 where y ∈[ −2, ∞ ) Thus f −1 : [ 0, ∞ ) → R, f −1 ( x ) = x 2 − 2 where y ∈[ −2, ∞ ).

x=4

12 a f ( x ) = ( x + 2 )2. If the maximal domain is ( −∞, −2 ] then the inverse will be a function. Inverse: swap x and y x = ( y + 2)

2

x ∈[ 0, ∞ )

y = − x − 2 where y ∈( −∞, −2 ] Thus f : [ 0, ∞ ) → R, f −1 ( x ) = − x − 2 where Ran = ( −∞, −2 ] −1

b f ( x ) = − 25 − x 2 . If the maximal domain is ( 0,5] then the inverse will be a function. Inverse: swap x and y

y=x

(0, 2 )

x = − 25 − y 2 x ∈( −5, 0 ]

(2, 2)

x 2 = 25 − y 2

y = x2 – 2

(0, –2)

x=2

− x = y+2

x2 = y + 2

0

x

(7–4 , 0)

10 a f : R → R, f ( x ) = x 4 is a many-to-one function. The inverse will be a one-to-many relation. b f : R → R, f ( x ) = 2 x 2 − 7 x + 3 is a many-to-one function. The inverse will be a one-to-many relation.

(–2, 0)

(7–2 , 0)

0

= 3 x3 = x as required.

y = x+2

(3 + 2, 3 + 2)

(3 – 2, 3 – 2)

y = f –1(x)

b k −1 ( k ( x ) ) = 3 x 3 − 1 + 1

y

y=x y = f(x)

( 2, 0)

x

y = 25 − x 2 where y ∈( 0,5] Thus f −1 : ( −5, 0 ] → R, f −1 ( x ) = 25 − x 2 where Ran = ( 0,5] 13 a f ( x ) = x 2 − 10 x + 25 = ( x − 5)2 which is a many-to-one function. For the inverse to be a function the domain must be restricted to [5, ∞ ) so that a = 5.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 3 Composite functions, transformations and ­inverses • EXERCISE 3.6   |  67 b Inverse: swap x and y 2 x = ( y − 5) where x ∈[ 0, ∞ )

d

x = y−5 y = x + 5 where y ∈[5, ∞ ) f −1 : [ 0, ∞ ) → R, f −1 ( x ) = x + 5 where Ran = [5, ∞ ). x 14 f : [ −2, 4 ) → R, f ( x ) = 1 − 3  1 5 a Dom = [ −2, 4 ) and Ran =  − , .  3 3 b Inverse: swap x and y y x = 1− 3 y = 1− x 3 y = 3 (1 − x ) −1  1 5  f :  − ,  → R, f −1 ( x ) = 3 (1 − x ) where Ran = [ −2, 4 )  3 3 y c

y = f(x)

(– –13 , 4) (–2, –35 )

y=x

+ — + — 13, –3 13 ——— ) (–3——— 2 2 (0, –31)

(1, 0) x

( )

y = f –1(x)

16 a f : ( −∞, a ] → R, f ( x ) = x 2 − 2 x − 1 To find TP: y = x 2 − 2 x − 1 y = x 2 −2 x + 1 − 2 y = ( x − 1)2 − 2 TP is at (1, −2) so largest possible value of a is 1. Thus f : ( −∞,1] → R, f ( x ) = x 2 − 2 x − 1 b Inverse: swap x and y x = ( y − 1) − 2

x + 2 = ( y − 1)

(3, 0) 0

)

5 – , –2 3

2

± x + 2 = y −1

x

(4, – –13 )

y = − x + 2 + 1 as dom f = ( −∞,1] f −1 : [ −2, ∞ ) → R, f −1 ( x ) = − x + 2 + 1 y = f(x)

x d x = 1 − 3 3x = 3 − x 4x = 3 3 x= 4  3 3 ∴ POI =  ,   4 4

y

y= x

(– 2 + 1, 0) (–2, 1) (–1, 0) 0

(0, – 2 + 1) (0, –1)

15 a f : D → R, f ( x ) = 1 − 3 x 1 The maximal domain is D =  −∞, .  3 b Inverse: swap x and y x = 1 − 3 y where x ∈[ 0, ∞ ) 2

x = 1 − 3y 3y = 1 − x 2 1 1  y = 1 − x 2 where Ran =  −∞,   3 3 1 1  −1 −1 2 f : [ 0, ∞ ) → R, f ( x ) = 1 − x where Ran =  −∞, .  3 3 c x = 1 − 3 x

(

(0, 1)

0 1 –, 0 3

y = f –1(x)

(0, 1)

(

y= x

2

(0, 3)

(1, 0)

y

y = f(x)

)

(

x 2 = 1 − 3x 0 = x 2 + 3x − 1 −3 ± (3)2 − 4 × 1 × −1 x= 2 −3 ± 13 = 2 −3 + 13 = , dom f = [ 0, ∞ ) 2  −3 + 13 −3 + 13  ∴ POI =  ,  2 2 

)

x

y = f –1(x)

(1, –2)

c x = x 2 − 2 x − 1 0 = x 2 − 3x − 1 3 ± (−3)2 − 4 × 1 × −1 2 3 ± 13 = 2 3 − 13 , dom f = ( −∞,1] = 2

x=

 3 − 13 3 − 13  Therefore POI =  , 2   2 17 f : [1, ∞ ) → R, f ( x ) = x − 1 where Ran = [ 0, ∞ ) a Inverse: swap x and y x = y −1

x ≥ 0, y ≥ 1

y = x2 + 1 f −1 : [ 0, ∞ ) → R, f −1 ( x ) = x 2 +1 where Ran = [1, ∞ )

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

68 | TOPIC 3 Composite functions, transformations and ­inverses • EXERCISE 3.6 b f −1 ( f ( x ) ) = f −1

(

(

x −1

)

2

)

y 2

18

= x −1 +1 = x −1+1 = x where Dom = [1, ∞ ) y

1 –2

–1

0

()

x2 y = 1– – 4

1

2 x

–1 y= x

Two inverse functions are: y2 4 2 y x2 = 1− 4 2 y x2 + =1 4 2 y = 1 − x2 4 y = ± 1 − x2 2 y = ±2 1 − x 2 Thus, we have f −1 : [ 0,1] → R, f −1 ( x ) = 2 1 − x 2 where Ran = [ 0, 2 ] or x = 1−

1

0

(1, 1) x

1

x+2    x + 2  −1  c f −1  − f  − 1  = f −   3   3    x + 2−3 = f −1  −  3   x −1 = f −1  − 3  

f −1 : [ 0,1] → R, f −1 ( x ) = −2 1 − x 2 where Ran = [ −2, 0 ] y

2

 x −1 +1 = − 3   x −1 = +1 3 x −1+ 3 = 3 x+2 = 3

y = f –1(x) (0, 1) 0

y

(0, 2) y = f(x)

y = f(x)

(0, 1)

(1, 0)

(–2, 0) x (1, 0) (2, 0)

y=x

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

y=x

x

0

y = f –1(x) (0, –2)

TOPIC 4 Logarithmic functions • EXERCISE 4.2   |  69

Topic 4 — Logarithmic functions Exercise 4.2 — Logarithm laws and equations 1 a log7 (49) + log2 (32) − log5 (125) 2

5

3

= log7 (7) + log2 (2) − log5 (5) = 2 log7 (7) + 5log2 (2) − 3log5 (5) = 2+5−3 =4 b 5log11 (6) − 5log11 (66) = 5 ( log11 (6) − log11 (66) ) 6   = 5  log11     66    1   = 5  log11     11   

= 5log11 (11)−1 = −5 c log4 ( 25) log 4 ( 625) log4 ( 5)2 log4 ( 5)4 2 log4 ( 5) = 4 log4 ( 5) 1 = 2 =

2

3 ( log2 ( x )) − 5 log2 ( x ) − 2 = 0 (3 log2 ( x ) + 1)(log2 ( x ) − 2) = 0 3 log2 ( x ) + 1 = 0 or log2 ( x ) − 2 = 0 1 log2 ( x ) = − log2 ( x ) = 2 3 22 = x −1 x=2 3 x=4 2

d log5 (4 x ) + log5 ( x − 3) = log5 (7) log5 (4 x ( x − 3)) = log5 (7) 4 x ( x − 3) = 7 4 x 2 − 12 x − 7 = 0 (2 x − 7)(2 x + 1) = 0 7 1 x = ,− 2 2 1 x = − isn’t a valid solution as x > 3 2 7 Therefore x = 2 4 a log3 ( x ) = 5 35 = x x = 243 b log3 ( x − 2) − log3 (5 − x ) = 2

 1  d log2  7  128   = log2  2−7 

c 3 ( log2 ( x ) ) − 2 = 5 log2 ( x )

1



( ) 7 

= log2 (2)−1 = −1 2 a 7 log4 ( x ) − 9 log4 ( x ) + 2 log4 ( x ) = 0 b log7 (2 x − 1) + log7 (2 x − 1)2 = log7 (2 x − 1) + 2 log7 (2 x − 1) = 3log7 (2 x − 1) c log10 ( x − 1)3 − 2 log10 ( x − 1) = 3log10 ( x − 1) − 2 log10 ( x − 1) = log10 ( x − 1) 3

3 a log5 (125) = log5 (5) = 3 log5 (5) =3 log4 ( x − 1) + 2 = log4 ( x + 4) b log4 ( x − 1) + 2 log4 4 = log4 ( x + 4) log4 ( x − 1) + log4 4 2 = log4 ( x + 4) log4 (16( x − 1)) = log4 ( x + 4) 16( x − 1) = x + 4 16 x − 16 = x + 4 15 x = 20 4 x= 3

 x − 2 log3  =2  5 − x  x−2 32 = 5− x x−2 9= 5− x 9 (5 − x ) = x − 2 45 − 9 x = x − 2 47 = 10 x 47 x= 10 loge (12) = 1.2770 5 a  i    log7 (12) = loge (7) 1 loge    4  1 = −1.2619 ii log3   =  4 loge (3) b z = log3 ( x ) 3z = x i 2 x = 2 × 3z ii log x (27) =

log3 (27) log3 ( x )

log3 (3)3 log3 ( x ) 3 log3 (3) = log3 ( x ) 3 = z =

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

70 | TOPIC 4 Logarithmic functions • EXERCISE 4.2 log10 (9) log10 (5) log10 (12) b log 1 (12) = 1 2 log10    2

6 a log5 (9) =

6 = 216 log6 (216) = 3

b

28 = 256 log2 (256) = 8

c

34 = 81 log3 (81) = 4

e

= log4 (2)−1 −

10 −4 = 0.0001 log10 (0.0001) = −4 5−3 = 0.008 log5 (0.008) = −3 1

7 =7 log7 (7) = 1 8 a log3 (81) = x 3 x = 81 f

(

= log4 (2)−6

3

7 a

d

 1  c log4  6  64 

= log4 (4)−2 = −2 log4 (4) = −2 log5 (32) e 3 log5 (16) log2 (2)5 3 log2 (2)4 5 log2 (2) = 12 log2 (2) 5 = 12 6 log2 3 x =

1 216 6 x = 6−3 x = −3 c log x (121) = 2 6x =

f

( )

( )

log2 x 5

1

x 2 = 121 x 2 = 112 x = 11 d log2 (− x ) = 7 27 = − x 128 = − x x = −128 9 a log2 (256) + log2 (64) − log2 (128)  256 × 64  = log2   128  = log2 2 × 26 7

1

= log4 (4) 2 1 = − log4 (4) 2 1 =− 2  16  d log4   256   1 = log4    16 

3 x = 34 x=4 1   b log6  =x  216 

(

1

)6

)

= log2 (2) = 7 log2 (2) =7 b 5 log7 (49) − 5 log7 (343)

= 5 ( log7 (49) − log7 (343) )  49  = 5 log7   343   1 = 5 log7    7 = 5 log7 (7)−1 = −5 log7 (7) = −5

6 log2 ( x ) 3 = log2 ( x )5 2 log2 ( x ) = 5log2 ( x ) 2 = 5 10 a log3 ( x − 4) + log3 ( x − 4)2 = log3 ( x − 4) + 2 log3 ( x − 4) = 3log3 ( x − 4) 3 b log7 (2 x + 3) − 2 log7 (2 x + 3) = 3log7 (2 x + 3) − 2 log7 (2 x + 3) = log7 (2 x + 3)

c log5 ( x )2 + log5 ( x )3 − 5log5 ( x ) = 2 log5 ( x ) + 3log5 ( x ) − 5log5 ( x ) =0 3 2 log d 4 (5 x + 1) + log 4 (5 x + 1) − log 4 (5 x + 1) = log4 (5 x + 1) + 3 log4 (5 x + 1) − 2 log4 (5 x + 1) = 2 log4 (5 x + 1) 11 a log3 (7) = 1.7712  1  b log2   = −6.9189 121  12 If n − log5 ( x ) then 5n = x a 5 x = 5 × 5n = 5n +1

( )

( ) 2n = log5 ( 5 × 5 )

b log5 5 x 2 = log5 5 × (5n )2

= log5 ( 5)2 n +1 = (2n + 1) log5 ( 5) = 2n + 1 Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 4 Logarithmic functions • EXERCISE 4.2   |  71 h log10 (2 x ) − log10 ( x − 1) = 1

c log x (625) log5 (625) = log5 ( x ) =

 2x  log10  =1  x − 1 

( )

2x x −1 10( x − 1) = 2 x 10 x − 10 = 2 x 10 x − 2 x = 10 8 x = 10 5 x= 4 i log3 ( x ) + 2 log3 (4) − log3 (2) = log3 (10)

log5 54

10 =

n 4 n 13 a loge ( 2 x − 1) = −3 e −3 = 2 x − 1 =

e −3 + 1 = 2 x 1 x = e −3 + 1 2  1 b loge   = 3  x loge ( x )−1 = 3 − loge ( x ) = 3 loge ( x ) = −3

(

)

log3 ( x ) + log3 (4)2 − log3 (2) = log3 (10) log3 (16 x ) − log3 (2) = log3 (10)  16 x  log3  = log3 (10)  2  8 x = 10 5 x= 4

x = e −3 c log3 ( 4 x − 1) = 3

j

33 = 4 x − 1 27 + 1 = 4 x 28 = 4 x x=7 d log10 ( x ) − log10 (3) = log10 (5) x log10   = log10 (5)  3 x =5 3 x = 15 e 3 log10 ( x ) + 2 = 5 log10 ( x ) 2 = 5 log10 ( x ) − 3 log10 ( x ) 2 = 2 log10 ( x ) 1 = log10 ( x ) x = 10

( )

f log10 x 2 − log10 ( x + 2 ) = log10 ( x + 3)  x2  log10  = log10 ( x + 3)  x + 2 

Let a = log10 ( x ) 2a 2 − 5a + 3 = 0 (2a − 3)(a − 1) = 0 Substitute back for a = log10 ( x )

( 2 log10 ( x ) − 3)( log10 ( x ) − 1) = 0

2 log10 ( x ) − 3 = 0 or log10 ( x ) − 1 = 0 2 log10 ( x ) = 3 log10 ( x ) = 1 3 log10 ( x ) = 101 = x 2 3

x = 10 2

x = 10

k ( log3 ( x ) ) = log3 ( x ) + 2 2

(log3 ( x ))

2

− log3 ( x ) − 2 = 0

(log3 ( x ) − 2)(log3 ( x ) + 1) = 0

log3 ( x ) − 2 = 0 or log3 ( x ) + 1 = 0 log3 ( x ) = 2 log3 ( x ) = −1

x2 = x+3 x+2 x 2 = ( x + 3)( x + 2)

3−1 = x 1 x =9 x= 3 l log6 ( x − 3) + log6 ( x + 2) = 1 log6 ( x − 3)( x + 2) = 1 6 = ( x − 3)( x + 2) 32 = x

x 2 = x 2 + 5x + 6 0 = 5x + 6 6 x=− 5 g 2 log5 ( x ) − log5 ( 2 x − 3) = log5 x − 2 log5 ( x )2 − log5 ( 2 x − 3) = log5 ( x − 2 )  x2  log5  = log5 ( x − 2 )  2 x − 3  x2 = x−2 2x − 3 x 2 = ( x − 2)(2 x − 3) 2

( log10 ( x )) ( log10 ( x )2 ) − 5 log10 ( x ) + 3 = 0 ( log10 ( x ))( 2 log10 ( x )) − 5 log10 ( x ) + 3 = 0 2 2 ( log10 ( x ) ) − 5 log10 ( x ) + 3 = 0

2

x = 2x − 7x + 6 0 = x 2 − 7x + 6 0 = ( x − 1)( x − 6) x =1 x=6

6 = x2 − x − 6 0 = x 2 − x − 12 0 = ( x − 4)( x + 3) x − 4 = 0 or x + 3 = 0 x=4 x = −3 But x > 3, ∴ x = 4 14 a log19 ( y) = 2 log10 2 − 3 log10 ( x ) log19 ( y) = log10 22 − log10 ( x )3 4 log19 ( y) = log10  3  x  4 y= 3 x

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

72 | TOPIC 4 Logarithmic functions • EXERCISE 4.3 b log4 ( y) = −2 + 2 log4 ( x ) log4 ( y) = 2 log4 ( x ) − 2 log4 (4) log4 ( y) = log4 ( x )2 − log4 (4 2 )  x2  log4 ( y) = log4    16 

8 log2 (4) × log2 (2) = [ log2 ( x ) ]

2 2

16 log2 (2) × log2 (2) = [ log2 ( x ) ]

2

16 = [ log2 ( x ) ] log2 ( x ) = ±4

2

x = 24 , 2−4 1 = 16, 16

3

( )2

log9 ( 3 xy ) = log9 32

8 log x (4) = log2 ( x ) 8 log2 (4) log2 ( x ) = log2 ( x ) log2 (2)

8 log2 (22 ) × log2 (2) = [ log2 ( x ) ]

2

x 16 c log9 ( 3 xy ) = 1.5 3 log9 ( 3 xy ) = log9 9 2 y=

16

log9 ( 3 xy ) = log9 33 3 xy = 27 xy = 9 9 y= x  2x  log8   + 2 = log8 ( 2 ) d  y  2x  log8   + 2 log8 (8 ) = log8 ( 2 )  y  2x  log8   + log8 (8 )2 = log8 ( 2 )  y  128 x  = log8 ( 2 ) log8   y  128 x =2 y y = 64 x 15 a 3 logm ( x ) = 3 logm (27) 3 logm ( x ) = 3 logm (m) + logm (3)3 3 logm ( x ) = 3 logm (m) + 3 logm (3) logm ( x ) = logm (m) + logm (3) logm ( x ) = logm (3m) x = 3m b If x = log10 (m) and y = log10 (n) then 10 x = m and 10 y = n  y 2   100 n 2   100 10  log10  5  = log10  1 5 m n x y  10 10 2 

( ) ( )( )

 2  10 × 10 2 y  = log10  y  5x   10 × 10 2  3y  2  10 × 10 2   = log10  5x   10 

 2+ 3 y −5x  = log10  10 2    3y   = 2+ − 5 x  log10 (10)   2 3y = 2+ − 5x 2

17 a e 2 x − 3 = loge ( 2 x + 1) Solve using CAS x = −0.463, 0.675 b x 2 − 1 = loge ( x ) Solve using CAS x = 0.451, 1

18 ( 3 log3 ( x ) )( 5 log3 ( x ) ) = 11log3 ( x ) − 2 Solve using CAS x = 1.5518, 1.4422

Exercise 4.3 — Logarithmic scales  I  1 L = 10 log10  −12   10  When L = 130 dB,  I  130 = 10 log10  −12   10  13 = log10 I × 1012

(

)

13 = log10 ( I ) + log10 (10 )12 13 = log10 ( I ) + 12 log10 (10 ) 13 = log10 ( I ) + 12 13 − 12 = log10 ( I ) 1 = log10 ( I ) I = 10 Intensity is 10 watt/m2. E 2 M = 0.67 log10   K If M = 5.5 and E = 1013 then  1013  5.5 = 0.67 log10    K   1013  8.2090 = log10    K  1013 108.2090 = K 1013 K = 8.2090 10 K = 10 4.7910 = 61 801.640 Thus K  61 808

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 4 Logarithmic functions • EXERCISE 4.3   |  73 E 3 M = 0.67 log10   K When M = 6.3,

E  6.3 = 0.67 log10  6.3   K   E6.3  6.3 = log10    K  0.67 E  9.403 = log10  6.3   K  E6.3 9.403 = 10 K 252 911 074 K = E6.3

When M = 6.4,

E  6.4 = 0.67 log10  6.4   K   E6.4  6.4 = log10    K  0.67 E  9.5522 = log10  6.4   K  E6.4 9.5522 = 10 K 3 566 471 895K = E6.4 E6.4 : E6.3 = 3 566 471 895K ; 252 911 074 K = 1.4101: 1 6.4 earthquake is 1.41 times bigger than the 6.3 earthquake.  E 4 M = 0.67 log10   K When M = 9 and E = 1017  1017  9 = 0.67 log10   K 

A 500 watt amplifier is 146.9897 − 133.0103 = 13.98 dB louder than the 20 watt amplifier.  I  6 L = 10 log10  −12   10  When I = 10 4 ,  10 4  L = 10 log10  −12   10 

(

L = 10 log10 10 4 × 1012

)

L = 10 log10 (10 )16 L = 160 log10 (10 ) = 160 dB Loudness is 160 dB 7 pH = − log10  H +  When H + = 0.001, pH = − log10 [ 0.001]

pH = − log10 (10)−3 pH = 3 log10 (10) = 3 Lemon juice has a pH of 3 which is acidic. 8 a pH = − log10  H +  When pH = 0,

0 = − log10  H +  0 = log10  H + 

10 0 =  H +  1 moles/litre =  H +  b When pH = 4,

4 = − log10  H + 

13.4328 = log10 (10 )17 − log10 ( K ) log10 ( K ) = 17 log10 (10) − 13.4328 log10 ( K ) = 17 − 13.4328 log10 ( K ) = 3.5672 10 3.5672 = K K = 3691.17

−4 = log10  H + 

10 −4 =  H +  0.0001 moles/litre =  H +  c pH = − log10  H +  When pH = 8,

8 = − log10  H + 

 I  I 5 L = 10 log10   = 10 log10  −12   10   I0  If I = 20;  20  L = 10 log10  −12   10 

−8 = log10  H + 

10 −8 =  H + 

10 −8 moles/litre =  H + 

( ) 13 L = 10 log10 ( 2 × 10 )

d pH = − log10  H + 

L = 10 log10 20 × 1012

When pH = 12,

(

L = 10 log10 ( 2) + 10 log10 1013

L = 10 log10 ( 2) + (13 × 10) log10 (10 ) L = 10 log10 ( 2) + 130 L = 133.0103 dB If I = 500;  500  L = 10 log10  −12   10 

( L = 10 log10 ( 5 × 1014 )

L = 10 log10 5 × 10 2 × 1012

12 = − log10  H + 

)

)

L = 10 log10 (5) + 10 log10 (10)14 L = 10 log10 (5) + (10 × 14) log10 (10) L = 10 log10 (5) + 140 L = 146.9897 dB

−12 = log10  H + 

10 −12

10 −12 =  H +  moles/litre =  H + 

9 a pH = − log10  H +   H +  = 0.0000158 moles/litre pH = − log10 ( 0.0000158 ) pH = 4.8 My hair conditioner has a pH of 4.8 which is acidic. b pH = − log10  H +   H +  = 0.00000275 moles/litre

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

74 | TOPIC 4 Logarithmic functions • EXERCISE 4.4 pH = − log10 ( 0.00000275) pH = 5.56 My shampoo has a pH of 5.56 which is acidic. 10 a N (t ) = 0.5 N 0 0.5 N 0 = N 0 e − mt 1 = e − mt 2  1 loge   = − mt  2 loge (2)−1 = − mt − loge (2) = − mt loge (2) = mt loge (2) t= as required m b N (t ) = 0.3 N 0 When t = 5750 years, loge (2) 5750 = m 5750 m = loge (2) loge (2) = 0.000121 m= 5750 −0.000121t 0.3 N 0 = N 0 e 0.3 = e −0.000121t loge (0.3) = −0.000121t loge (0.3) =t −0.000121 t = 9987.55 The skeleton is 9988 years old. b  11 m2 − m1 = 2.5 log10  1   b2  Sirius: m1 = −1.5 and b1 = −30.3 Venus: m2 = −4.4 and b2 = ?  −30.3  −4.4 − (−1.5) = 2.5 log10   b2   −30.3  −2.9 = 2.5 log10   b2   −30.3  −2.9 = log10  2.5  b2   −30.3  −1.16 = log10   b2  −30.3 10 −1.16 = b2 −30.3 b2 = −1.16 10 −30.3 b2 = 0.0692 = −437.9683 Brightness of Venus is −437.97. f  12 n = 1200 log10  2   f1  f1 = 256, f2 = 512  512  n = 1200 log10   256  n = 361 cents

 I  13 L = 10 log10  −12   10 

(

)

0.22 Rifle: I = 2.5 × 1013 I 0 = 2.5 × 1013 × 10 −12 = 2.5 × 10  2.5 × 10  L = 10 log10   10 −12 

(

L = 10 log10 ( 2.5 × 10 ) − log10 (10)−12

)

L = 10 ( log10 ( 2.5) + log10 (10) + 12 log10 (10) )

L = 10 ( log10 ( 2.5) + 13) L = 133.98 The loudness of the gunshot is about 133.98 dB so ear protection should be worn. E 14 M = 0.67 log10   K San Francisco: M SF = 8.3 E  8.3 = 0.67 log10  SF   K   ESF  12.3881 = log10   K  E 1012.3881 = SF K South America: M SA = 4 ESF  4E  M SA = 0.67 log10  SF   K  E 12.3881 Substitute 10 = SF K

(

)

M SA = 0.67 log10 4 × 1012.3881 = 8.7 Magnitude of the South American earthquake was 8.7.

Exercise 4.4 — Indicial equations 1 a 32 x +1 × 272 − x = 81 32 x +1 × (33 )2 − x = 34 32 x +1 × 36 − 3 x = 34 37 − x = 34 Equating indices 7− x = 4 x=3 2 x −1 −5= 0 b 10 10 2 x −1 = 5 log10 (5) = 2 x − 1 log10 (5) + 1 = 2 x 1 1 x = log10 (5) + 2 2 c 4 x − 16 4 x + 3 = 0

(

)(

x

)

x

4 − 16 = 0 or 4 + 3 = 0 4 x = 16

4 x = −3

4 x = 42 x=2

No solution

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 4 Logarithmic functions • EXERCISE 4.4   |  75 d

( ) ( ) 2 2 (10 x ) − 7 (10 x ) + 3 = 0 ( 2 (10 )x − 1)((10 )x − 3) = 0 2 10 2 x − 7 10 x + 3 = 0

2 (10 ) x − 1 = 0 or (10 ) x − 3 = 0 1 10 x = 10 x = 3 2  1 x = log10   x = log10 (3)  2 1 x +3 =0 2 a 2 − 64 1 2x +3 = 64 2 x + 3 = 2−6 Equating indices

d

( ) ( ) 2 6 ( 9 x ) − 19 ( 9 x ) + 10 = 0 (3 (9x ) − 2)( 2 (9x ) − 5) = 0 3 ( 9 x ) − 2 = 0 or 2 ( 9 x ) − 5 = 0 3 (9 x ) = 2 2 (9 x ) = 5 (9 x ) = 23 (9 x ) = 52 6 92 x − 19 9 x + 10 = 0

2  x = log9    3

4 a 16 × 22 x + 3 = 8−2 x 24 × 22 x + 3 = 23( −2 x ) 22 x + 3+ 4 = 2−6 x 2 x + 7 = −6 x 8 x = −7 7 x=− 8

x + 3 = −6 x = −9 b 22 x − 9 = 0 22 x = 9 log2 (9) = 2 x 1 x = log2 (9) 2 c 3e 2 x − 5e x − 2 = 0

( )2 − 5e x − 2 = 0 (3e x + 1)(e x − 2) = 0 3 ex

3e x + 1 = 0 or e x − 2 = 0 3e x = −1 No solution

ex = 2 x = loge (2)

d e 2 x − 5e x = 0 x

(

b 2 × 3 x +1 = 4 3 x +1 = 2 log3 (2) = x + 1 x = log3 (2) − 1

( )

2 5 x − 12 = −

c

( )2 − 12 (5x ) + 10 = 0 ( 5 x )2 − 6 ( 5 x ) + 5 = 0 (5x − 1)(5x − 5) = 0

2 5x

e e −5 = 0 e x = 0 or e x − 5 = 0 x

e =5 x = loge (5)

No solution 3 a

2 x −1

7 =5 log7 (5) = 2 x − 1 log7 (5) + 1 = 2 x 1 1 x = log7 (5) + 2 2

(

)(

)

b 3 x − 9 3 x − 1 = 0 3x − 9 = 0

or 3 x − 1 = 0

3x = 9

3x = 1

x

2

3 x = 30 x=0

3 =3 x=2 c

25 x − 5 x − 6 = 0

( 52 ) x − 5 x − 6 = 0 (5x )2 − 5x − 6 = 0 (5x − 3)(5x + 2) = 0 x

5x = 1

5x = 5

5 x = 50 x=0

5 x = 51 x =1

d          4 x +1 = 31− x loge ( 4 ) x +1 = loge ( 3)1− x ( x + 1) loge ( 4 ) = (1 − x ) loge ( 3) x loge ( 4 ) + loge ( 4 ) = loge ( 3) − x loge ( 3) x loge ( 4 ) + x loge ( 3) = loge ( 3) − loge ( 4 )  3 x ( loge ( 4 ) + loge ( 3)) = loge    4 3 loge    4 x= loge ( 4 ) + loge ( 3) 3 loge    4 x= loge (12 )

(

)

5 a 2 2 x −1 − 3 + 4 = 0

(

2 2

x

5 − 3 = 0 or 5 + 2 = 0 5x = 3 log5 (3) = x

10 5x

5 x − 1 = 0 or 5 x − 5 = 0

)

x

5  x = log9    2

5 x = −2 No solution

x −1

2

)

− 3 = −4

x −1

− 3 = −2

2 x −1 = 1 2 x −1 = 20 x −1= 0 x =1

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

76 | TOPIC 4 Logarithmic functions • EXERCISE 4.4

(

)

b 2 51− 2 x − 3 = 7

(

e

)

2 51− 2 x = 10 51− 2 x = 5

ex = 3 x = loge (3)

2

2

ex = 2 x 2 = loge (2) x = ± loge (2) e 2 x = e x + 12

8 a

e 2 x − e x − 12 = 0

(e x )2 − (e x ) − 12 = 0 (e x − 4 )(e x + 3) = 0 ex − 4 = 0

or e x + 3 = 0

x

e x = −3 loge (−3) = x No solution

e =4 loge (4) = x 2 loge (2) = x

e x = 12 − 32e − x

b

e x − 12 + 32e − x = 0

(e x )2 − 12 (e x ) + 32 = 0 (e x − 4 )(e x − 8) = 0 ex − 4 = 0

ex − 8 = 0

or

x

2

2

2

×3

x 2 − 4 x +3

6

x 2 − 4 x +3

0

x2 − 4x + 3 = 0 ( x − 1)( x − 3) = 0 x − 1 = 0 or x − 3 = 0 x =1 x=3 7 a ex −2 − 2 = 7 ex −2 = 9 loge (9) = x − 2 loge (9) + 2 = x loge (3)2 + 2 = x x = 2 loge (3) + 2 x

b e 4 − 1 = 3

ex = 8 loge (8) = x

loge (2)2 = x 2 loge (2) = x

loge (23 ) = x 3 loge (2) = x

e 2 x − 2e x − 4 = 0

=1 =6

e =4 loge (4) = x

c e 2 x − 4 = 2e x

23− 4 x × 3−4 x + 3 × 2 x × 3 x = 1

2 ± (−2)2 − 4(1)(−4) 2 2 20 ± ex = 2 2 2 5 ± ex = 2 ex = 1 ± 5 ex =

(

x = loge 1 ± 5

)

(

)

Therefore x = loge 1 + 5 as 1 − 5 −5 ex 2x x e − 12e + 5 = 0

d e x − 12 =

12 ± 144 − 4(1)(5) 2 12 ± 144 − 20 = 2 12 ± 124 = 2 12 ± 2 31 = 2 = 6 ± 31

ex =

+1= 3 x e4

ex − 3 = 0

or

d e x + 2 = 4

23− 4 x × 3−4 x + 3 × ( 2 × 3) x = 1

x e4

)

No solution

2

2

(

ex = 0

23− 4 x × 3−4 x + 3 × 6 x = 1

x 2 − 4 x +3

2x

ex ex − 3 = 0

51− 2 x = 51 1 − 2x = 1 0 = 2x x=0 1 6 a x −1 − 1 1− 1 + x −1 1 = x −1 − 1 1− 1 1+ x 1 −1 =x − 1 1− x +1 x 1 −1 =x − x 1− x +1 1 = x −1 − x +1− x x +1 1 = x −1 − 1 x +1 1 = − ( x + 1) x 1 = − x −1 x b

e 2 x = 3e x − 3e x = 0

c

ex

=2 x loge (2) = 4 x = 4 loge (2)

ex ex ex

(

x = loge 6 ± 31

)

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

0

TOPIC 4 Logarithmic functions • EXERCISE 4.4   |  77 9 y = m (10 )nx

c

x = 2, y = 20; 20 = m (10 )

2n

........................(1)

When x = 4, y = 200; 200 = m (10 )4 n ....................(2) (2) ÷ (1) 200 m (10 )4 n = 20 m (10 )2 n 10 = 10 2 n log10 (10) = 2n 1 = 2n 1 n= 2 1 Substitute n = into (1) 2 20 = m (10 )2( 2 ) 20 = 10 m m=2

5n

= 4 m or 3

= 4m

m=

35n 4

8mx × 4 2 n = 16 2

3mx

 5 + 57  1 , loge  m 4  

m ∈ R \ {0}

13 y = ae − kx

3.033 ae −2 k = 1.1157 ae −6 k 2.7185 = e 4 k loge ( 2.7185) = 4 k

3.033 = ae −2(0.25)

12 a e m − kx = 2n m − kx = loge (2n) − kx = loge (2n) − m loge (2n) − m x= −k m − loge (2n) , k ∈ R \ {0}, n ∈ R + = k b

x=

1 loge ( 2.7185) = k 4 k = 0.25 Substitute k = 0.25 into (1)

2

1 4 × 35n

5 ± (−5)2 − 4(2)(−4) 2(2) 5 ± 25 + 32 e mx = 4 5 ± 57 mx e = 4 5 + 57 mx mx e = , e >0 4  5 + 57  mx = loge  4   e mx =

(1) ÷ (2)

11 ( log3 (4 m) ) = 25n 2 log3 (4 m) = ±5n m=

− 5e mx − 4 = 0

When x = 6, y = 1.1157 so 1.1157 = ae −6 k .................(2)

2 x < 0.3 log2 (0.3) < x −1.737 > x x < −1.737 x b ( 0.4 ) < 2 log0.4 (2) < x −0.756 < x x > −0.756

10 a

3

2e mx = 5 + 4e − mx

When x = 2, y = 3.033 so 3.033 = ae −2 k ....................(1)

1

−5n

( )

2 e

When

mx 2

× 22(2 n ) = 24

23mx + 4 n = 24 3mx + 4 n = 4 3mx = 4 − 4 n 4 − 4n x= , m ∈ R \ {0} 3m

3.033 = ae −0.5 3.033 =a e −0.5 a=5 14 A = Pert When t − 5, A = $12 840.25

12 840.25 = Pe5r .....................(1)

When t − 7, A = $14 190.66

14 190.66 = Pe 7r ....................(2)

14 190.66 Pe 7r = 12 840.25 Pe 7r

(2) ÷ (1)

1.1052 = e 2r loge (1.1052) = 2r 1 loge (1.1052) = r 2 0.05 = r r = 5% Substitute r − 0.05 into (1) 12 840.25 = Pe5(0.05) 12 840.25 =P e 0.25 P = $10 000

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

78 | TOPIC 4 Logarithmic functions • EXERCISE 4.5

Exercise 4.5 — Logarithmic graphs 1 a Graph cuts y axis when x = 0, y = loge (4) = 1.386 Domain = (−4, ∞) and Range = R y

y = loge (x + 4) (–3, 0)

d Graph cuts the x axis where y = 0, − loge ( x − 4) = 0 loge ( x − 4) = 0 e0 = x − 4 1+ 4 = x 5= x Domain = (4, ∞) and Range = R y

(0, loge (4))

y = –loge (x – 4) x

0

(5, 0)

x (6, –loge (2))

0

x = –4

b Graph cuts x axis when y = 0, loge ( x ) + 2 = 0 loge ( x ) = −2

x=4

2 a y = log3 ( x + 2) − 3 Graph cuts the x axis where y = 0, log3 ( x + 2) − 3 = 0 log3 ( x + 2) = 3

e −2 = x 0.1353 = x When x = 2, y = loge (2) + 2 = 2.69 Domain = (0, ∞) and Range = R y y = loge (x) + 2

33 = x + 2 27 = x + 2 25 = x Domain = (−2, ∞) and Range = R y

(1, 2) 0

(e–2,

0)

y = log3 (x + 2) –3 x (25, 0) x

0 (0, log3 (2) –3) x=0

c Graph cuts x axis when y = 0, 4 loge ( x ) = 0 loge ( x ) = 0 e0 = x 1= x When x = 2, y = 4 loge (2) Domain = (0, ∞) and Range = R y

y = 4 loge (x)

x = –2

b y = 3 log5 (2 − x ) Graph cuts the x axis where y = 0, 3 log3 (2 − x ) = 0 log3 (2 − x ) = 0 30 = 2 − x x = 2 −1 x =1 Domain = (−∞,1) and Range = R y y = 3log5 (2 – x)

(2, 4 loge (2)) x

0

(1, 0)

(0, 3log5 (2))

(1, 0) x

0

x=0 x=2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 4 Logarithmic functions • EXERCISE 4.5   |  79 c y = 2 log10 ( x + 1) Graph cuts the x axis where y = 0, 2 log10 ( x + 1) = 0 log10 ( x + 1) = 0 10 0 = x + 1 0=x Domain = (−1, ∞) and Range = R y

5 a Graph cuts x axis when y = 0. loge ( x ) + 3 = 0 loge ( x ) = −3 e −3 = x 0.05  x When x = 1, y = loge 1 + 3 = 3 y y = loge (x) + 3

y = 2log10 (x + 1)

(0, 0)

(1, 3)

(1, 2log10 (2)) x 0

(e–3, 0)

x

x=0

x = –1

−x d y = log2    2  Graph cuts the x axis where y = 0, −x log2   = 0  2  −x 20 = 2 −2 = x Domain = (−∞,0) and Range = R

b Graph cuts x axis when y = 0. loge ( x ) − 5 = 0 loge ( x ) = 5 e5 = x 148.4  x When x = 200, y = loge (200) − 5 = 0.298 y

y

y = loge (x) – 5

( )

x y = log2 – –2

0

(–2, 0) (–1, –1)

(e5, 0)

x

x

0

x=0

3 y = loge ( x − m) + n Vertical asymptote is x = 2 so m = 2. y = loge ( x − 2) + n When x = 4.71828, y = 3 3 = loge (4.71828 − 2) + n 3 = loge (2.71828) + n n = 3 − loge (2.71828) n=2 y = loge ( x − 2) + 2 4 y = p loge ( x − q) When x = 0, y = 0 0 = p loge (− q)........................(1) When x = 1, y = −0.35 −0.35 = p loge (1 − q)............(2) From (1) 0 = loge (−q) e 0 = −q q = −1 Substitute q = −1 into (2) −0.35 = p loge (1 − (−1)) −0.35 = p loge (2) −0.35 =p loge (2) −7 p= 20 loge (2)

x=0

c Graph cuts x axis when y = 0. loge ( x ) + 0.5 = 0 loge ( x ) = −0.5 e −0.5 = x 0.6  x When x = 1, y = loge (1) + 0.5 = 0.5 y

y = loge (x) + 0.5

(1, 0.5)

0

(e , 0) –1 2

x=0

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

x

80 | TOPIC 4 Logarithmic functions • EXERCISE 4.5 7 a Graph cuts x axis when y = 0. 1 loge ( x ) = 0 4 loge ( x ) = 0

6 a Graph cuts x axis when y = 0. loge ( x − 4) = 0 e0 = x − 4 1 x −4 5= x When x = 10, y = loge (10 − 4) = loge (6)  1.8 y

e0 = x 1= x Graph does not cut the y. y

y = loge (x – 4) (10, loge (6))

y = –1 loge (x) 4

(1, 0)

x

0

(5, 0) x

0

x=0

b Graph cuts x axis when y = 0. 3 loge ( x ) = 0 loge ( x ) = 0 e0 = x 1= x Graph does not cut the y.

x=4

b Graph cuts x axis when y = 0. loge ( x + 2) = 0

y

e0 = x + 2 1 x +2 −1 = x When x = 0, y = loge (0 + 2) = loge (2)  0.7

y = 3loge (x)

(1, 0)

y

x

0

y = loge (x + 2) (0, loge (2)) (–1, 0) x

0

x=0

c Graph cuts x axis when y = 0. 6 loge ( x ) = 0 loge ( x ) = 0 e0 = x 1= x Graph does not cut the y.

x = –2

c Graph cuts x axis when y = 0. loge ( x + 0.5) = 0

y

0

e = x + 0.5 1  x + 0.5 0.5 = x When x = 0, y = loge (0 + 0.5) = loge (0.5)  −0.7

y = 6 loge (x) (1, 0) x

0

y y = loge (x + 0.5)

(0.5, 0) x

0 (0, –loge (0.5))

x = – 0.5

x=0

8 a Graph cuts x axis when y = 0. loge (3 x ) = 0 e0 = 3x 1 = 3x 1 =x 3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 4 Logarithmic functions • EXERCISE 4.5   |  81 y

Graph does not cut the y. y y = loge (3x)

y = 1 – 2 loge (x – 1) (e0.5 + 1, 0)

x

( –13 , 0)

0

0

x

x=1

b Graph cuts x axis when y = 0. loge (2 x + 4) = 0

x=0

b Graph cuts x axis when y = 0. x loge   = 0  4 x e0 = 4 x 1= 4 4=x Graph does not cut the y.

e0 = 2 x + 4 1 − 4 = 2x 3 − =x 2 Graph cuts the y axis where x = 0. loge (2(0) + 4) = y loge (4) = y 1.3862 = y

y

y

()

x y = loge – 4 0

(4, 0)

y = loge (2x + 4) x x

0 (–1.5, 0) x=0

x = –2

c Graph cuts x axis when y = 0. loge (4 x ) = 0 e0 = 4 x 1 = 4x 1 =x 4 Graph does not cut the y. y y = loge (4x)

0

x

(0.25, 0)

c Graph cuts x axis when y = 0. 1 x loge   + 1 = 0  2 4 x 1 loge   = −1  4 2 x loge   = −2  4 x e −2 = 4 4e −2 = x 0.5413 = x Graph does not cut the y axis. y

()

y = 1–2 loge –4x + 1 (4e–2,

x=0

9 a Graph cuts x axis when y = 0. 1 − 2 loge ( x − 1) = 0 2 loge ( x − 1) = 1 1 loge ( x − 1) = 2

0)

0

x=0

1

e2 = x −1 1

e2 +1 = x 2.6487 = x Graph does not cut the y.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

x

82 | TOPIC 4 Logarithmic functions • EXERCISE 4.5 10 a f ( x ) = 2 loge (3 x + 3) Domain = ( −1, ∞ ) and Range = R Inverse: swap x and y x = 2 loge (3 y + 3) x = loge ( 3 y + 3) 2

b

y

y = loge (2(x – 1) + 2) (3.68, 3.68)

(1.23, 1.23)

x

y = 1–2 e x – 2 + 1

e 2 = 3y + 3 x

(0,

e 2 − 3 = 3y 1 x y = e2 −1 3 1 x f −1 ( x ) = e 2 − 1 3 Domain = R and Range = ( −1, ∞ ) b f ( x ) = loge ( 2 ( x − 1)) + 2 Domain = (1, ∞ ) and Range = R Inverse: swap x and y

1– –2 e 2

)

+1

y=1

0

(e

1– –2 2

x=1

c

x = loge ( 2 ( y − 1)) + 2

x − 2 = loge ( 2 ( y − 1))

e = 2( y − 1) 1 x −2 e = x −1 2 1 x = ex −2 + 1 2 1 f −1 ( x ) = e x − 2 + 1 2 Domain = R and Range = (1, ∞ ) c f ( x ) = 2 loge (1 − x ) − 2 Domain = ( −∞,1) and Range = R Inverse: swap x and y x = 2 loge (1 − y ) − 2 x = 2 loge (1 − y ) − 2 x + 2 = 2 loge (1 − y ) 1 ( x + 2) = loge (1 − y ) 2 1 ( x + 2)

y = 2 loge (1 – x) – 2 (1 – e, 0)

(–0.81, –0.81)

12 a y = a loge ( bx ) When x = 1, y = loge (2), loge (2) = a loge (b)...................(1) When x = 2, y = 0, 0 = a loge (2b)..........................(2) (2) − (1) 0 − loge (2) = a loge ( 2b ) − a loge ( b ) − loge (2) = a ( loge ( 2b ) − loge ( b ))  2b  − loge (2) = a loge   b − loge (2) = a loge ( 2) loge ( 2) − =a loge ( 2) −1 = a Substitute a = −1 into (1) loge (2) = − loge (b)

1 ( x + 2)

f −1 ( x ) = 1 − e 2

Domain = R and Range = ( −∞,1) 11 a

y=x

loge (2) = loge (b)−1  1 loge (2) = loge    b 1 2= b 1 b= 2

y = 2 loge(3x + 3) (6.12, 6.12) (0, 2 loge (3))

x –

1 e2 – 1 y=– 3

b When x = 3,

(– 2–3, 0)

x = –1

x

(–0.77, –0.77)

x (0, 1 – e) (0, –2) 1– (x + 2) y = 1 – e2 x=1

1 ( x + 2)

(2 loge (3), 0)

y=x y=1

(–2, 0)

y = 1− e2

(0 , – 2–3)

y

0

= 1− y

0

x

)

+ 1, 0

x −2

e2

y=x

y = –1

 3 w = − loge   = −0.4055  2

13 y = a loge ( x − h ) + k Graph asymptotes to x = −1 so h = −1 and y = a loge ( x + 1) + k Graph cuts the y axis at y = −2 (0, −2) ⇒ −2 = a loge (1) + k k = −2 ∴ y = a loge ( x + 1) − 2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 4 Logarithmic functions • EXERCISE 4.6   |  83 Graph cuts the x axis at x = 1 (1,0) ⇒ 0 = a loge ( 2) − 2 2 = a loge ( 2) 2 a= loge ( 2) 2 loge ( x + 1) − 2 Thus y = loge (2) 14 y = m log2 (nx ) When x = −2, y = 3 so 3 = m log2 (−2n).....................(1) When 1 1 n x = − , y = so = m log2  −  ...................(2)  2 2 2 1  n (1) − (2) 3 − = m log2 (−2n) − m log2 −   2 2  5  n  = m  log2 (−2n) − log2 −   2   2  5 n   = m  log2 −2n ÷ −    2 2  5 = m log2 (4) 2 5 = m log2 (2)2 2 5 = 2m 2 5 m= 4 5 5 Substitute m = into (1) 3 = log2 (−2n ) 4 4 12 = log2 ( −2n ) 5 2 −

12 5

= −2n

12 5

2 =n 2 n = −2

Investments equal in value when e 0.05t 5500 = e 0.045t 4200 55 e 0.005t = 42  55  loge   = 0.005t  42   55  loge   ÷ 0.005 = t  42  t = 53.9327 It takes 54 years for the amounts to be equal. 2 a A = Pert A = 3P , t = 15 3P = Pe15r 3 = e15r loge (3) = 15r loge (3) =r 15 r = 0.0732 Interest rate of investment is 7.32%. b A = Pert P = $2000, r = 4.5% = 0.045, A = $9000 9000 = 2000e 0.045t 9000 = e 0.045t 2000  9 loge   = 0.045t  2  9 loge   ÷ 0.045 = t  2 t = 33.42 It takes 33 years and 5 months for the investment to grow to $9000.  T−R 3 t = −10 loge   37 − R  T = 25oC, R = 20 o C

7 5 7

Thus m = 1.25 and n = −2 5 as required. 15 a x − 2 = loge ( x ) Solve on CAS x = 0.159 or 3.146 b 1 − 2 x = loge ( x − 1) Solve on CAS x = 1.2315 16 a x 2 − 2 < loge ( x ) Solve on CAS x ∈( 0.138,1.564 )

 25 − 20  t = −10 loge   37 − 20   5 t = −10 loge   = 12.2378  17  Time of death is 9 am – 12.2378 hours = 8.7622 or 8.46 pm 3 the day before. The person died 1 hours after the telephone 4 call. 4 n(t ) = loge (t + e 2 ),

t≥0

( )

2 a Initially t = 0, n(0) = loge e = 2 loge (e) = 2 Initially there were 2 parts per million.

(

)

2 b When t = 12, n(12) = loge 12 + e = 2.9647 After 12 hours there are 2.96 parts per million. c When n(t ) = 4,

b x 2 − 2 ≤ loge ( x ) Solve on CAS

(

4 = loge t + e 2

x ∈[ 0.136,1.315]

)

e4 = t + e2

Exercise 4.6 — Applications 1 A = Pert Western bank: P = $4200, r = 5% = 0.05 so A = 4200e 0.05t

e4 − e2 = t t = 47.2 It takes 47.2 hours before the four parts in a million of fungal bloom exists.

Common bank: P = $5500, r = 4.5% = 0.045 so A = 5500e 0.045t Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

84 | TOPIC 4 Logarithmic functions • EXERCISE 4.6 5 A = Pert

5 = 0.05, When t = 10, P = $1000 and r = 100 A = 1000e 0.05(10) A = $1648.72

kt 6 P (t ) = 200 + 1000 0 Initially t = 0 so P(0) = 200 + 1000 = 1001 When t = 8 and P = 3 × 1001 = 3003,

3003 = 2008 k + 1000 2003 = 2008 k loge (2003) = loge (200)8 k loge (2003) = 8 k loge (200) loge (2003) = 8k loge (200) loge (2003) =k 8 loge (200) k = 0.1793 7 P (t ) =

(

It takes 1.87 millennia to lose a third of the basic words. 10 a M = a − loge ( t + b ) When t = 0, M = 7.8948, 7.8948 = a − loge ( b ) ..............(1)

)

1 3 , 1 − e − kt and when t = 3 and P = 1500 4

(

1 3 = 1 − e −3 k 1500 4 4 = 1 − e −3 k 4500 4 e −3 k = 1 − 4500 e −3 k = 0.999 loge (0.999) = −3k

2 b W = W0 since one third of the basic words have been lost 3 2 W0 = W0 (0.805)t 3 2 = (0.805)t 3  2 loge   = loge (0.805)t  3  2 loge   = t loge (0.805)  3 2   loge   ÷ loge (0.805) = t  3 t = 1.87

)

1 − loge (0.999) = k 3 k = 0.0003 8 Q = Q0 e −0.000124 t a When Q0 = 100 and t = 1000, Q = 100e −0.000124(1000) Q = e −0.124 Q = 88.3 milligrams 1 b When Q = Q0 = 50, 2 50 = 100e −0.000124 t 0.5 = e −0.000124 t loge (0.5) = −0.000124t loge (0.5) =t −0.000124 t = 5589.897 It takes 5590 years for the amount of carbon-14 in the fossil to be halved. 9 W = W0 (0.805)t a When t = 10, W = W0 (0.805)10 = 0.11428W0 0.114W0 are the words remaining after 10 millennia or 88.57% of the words have been lost.

When t = 80, M = 7.3070, 7.3070 = a − loge (80 + b ) ........(2) (1) – (2) 7.8948 − 7.3070 = a − loge ( b ) − ( a − loge (80 + b )) 0.5878 = a − loge ( b ) − a + loge (80 + b ) 0.5878 = loge (80 + b ) − loge ( b )  (80 + b )  0.5878 = loge    b  (80 + b ) e 0.5878 = b 1.8b = 80 + b 0.8b = 80 b = 100 Substitute b = 100 into (1): 7.8948 = a − loge (100 ) 7.8948 + loge (100 ) = a 12.5 = a M = 12.5 − loge ( t + 100 ) Thus a = 12.5 and b = 100. b When t = 90, M = 12.5 − loge ( 90 + 100 ) M = 12.5 − loge (190 ) = 7.253 g 11 a P = a loge (t ) + c When t = 1, P = 10 000, 10 000 = a loge (1) 10 000 = c P = a loge (t ) + 10 000 When t = 4, P = 6000, 6000 = a loge (4) + 10 000 −4000 = a loge (4) −4000 =a loge (4) a = −2885.4

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 4 Logarithmic functions • EXERCISE 4.6   |  85 b P = −2885.4 loge (t ) + 10 000 P = 10 000 − 2885.4 loge (t ) When t = 8, P = 10 000 − 2885.4 loge (8) = 4000 There are 4000 after 8 weeks. c When P = 1000, 1000 = 10 000 − 2885.4 loge (t ) 2885.4 loge (t ) = 9000 9000 loge (t ) = 2885.4 loge (t ) = 3.1192 e3.1192 = t t = 22.6 After 22.6 weeks there will be less than 1000 trout. 12 a C = A loge ( kt ) When t = 2, C = 0.1, 0.1 = A loge (2 k )...............(1) When t = 30, C = 4, 4 = A loge (30 k )...............(2) (2) ÷ (1) A loge (30 k ) 4 = A loge (2 k ) 0.1 loge (30 k ) = 40 loge (2 k )

loge (30) + loge ( k ) = 40 ( loge (2) + loge ( k ) ) loge (30) + loge ( k ) = 40 loge (2) + 40 loge ( k ) loge (30) − 40 loge (2) = 40 loge ( k ) − loge ( k ) loge (30) − 40 loge (2) = 39 loge ( k ) −24.3247 = 39 loge ( k ) −24.3247 = loge ( k ) 39 −0.6237 = loge ( k ) e −0.6237 = k k = 0.536 Substitute k = 0.536 into (1): 0.1 = A loge (2 × 0.536) 0.1 = 0.0695 A A = 1.440 C = 1.438 loge (0.536t ) b When t = 15, C = 1.438 loge (0.536 × 15) = 3.002 M Concentration after 15 minutes is 3.002 M. c When C = 10 M, 10 = 1.438 loge (0.536t ) 6.9541 = loge (0.536t ) e6.9541 = 0.536t 1047.4385 = 0.536t t = 1934 After 1934 seconds or 32 minutes and 14 seconds the concentration is 10 M. 13 F (t ) = 10 + 2 loge (t + 2) a When t = 0, F (0) = 10 + 2 loge (2) = 11.3863 b When t = 4, F (0) = 10 + 2 loge (4 + 2) = 10 + 2 loge (6) = 13.5835

c When F = 15, 15 = 10 + 2 loge (t + 2) 5 = 2 loge (t + 2) 5 = loge (t + 2) 2 5

e2 = t + 2 5

e2 − 2 = t t = 10.18 After 10.18 weeks Andrew’s level of fitness is 10. 14 Q = Q0 e −0.000124 t When Q = 20% of Q0 = 0.2Q0 0.2Q0 = Q0e−0.000124 t 0.2 = e−0.000124 t loge (0.2) = −0.000124t loge (0.2) =t −0.000124 t = 12 979 Age of painting is 12  979 years. x  15 R( x ) = 800 loge  2 +  and C ( x ) = 300 + 2 x  100  a P ( x ) = R( x ) − C ( x ) x  P ( x ) = 800 loge  2 +  − 300 − 2 x  100  b When P ( x ) = 0, x  800 loge  2 + − 300 − 2 x = 0  100  x  800 loge  2 + = 300 + 2 x  100  x = 329.9728 x  330 330 units are needed to break even. 16 a V = ke mt When t = 0, V = 10 000; 10 000 = ke 0 10 000 = k V = 10 000e mt When t = 12, V = 13 500; 13 500 = 10 000e12 m 1.35 = e12 m loge (1.35) = 12m 1 loge (1.35) = m 12 0.025 = m V = 10 000e 0.025t b When t = 18, V = 10 000e 0.025(18) = $15 685.58 c Profit = P P = 1.375 × 10 000e 0.025t − 10 000 P = 13 750e 0.025t − 10 000 d When t = 24, P = 13 750e 0.025(24) − 10 000 = $15 059.38

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 5 Differentiation • EXERCISE 5.2   |  87

Topic 5 — Differentiation Exercise 5.2 — Review of differentiation 1 a f ( x ) = ( 2 − x )3 + 1 f ( x ) = 2 3 − 3 ( 2 )2 x + 3 ( 2 ) x 2 − x 3 + 1 f ( x ) = 8 − 12 x + 6 x 2 − x 3 + 1 f ( x + h) = ( 2 − ( x + h) ) + 1 3

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

3

(

) (

)

f ( x + h) = 8 − 12 ( x + h ) + 6 x 2 + 2 xh + h 2 − x 3 + 3 x 2 h + 3 xh 2 + h 3 + 1 2

2

3

2

2

3

f ( x + h) = 8 − 12 x − 12h + 6 x + 12 xh + 6h − x − 3 x h − 3 xh − h + 1 f ( x + h) − f ( x ) h 8 − 12 x − 12h + 6 x 2 + 12 xh + 6h 2 − x 3 − 3 x 2 h − 3 xh 2 − h 3 + 1 − 8 + 12 x − 6 x 2 + x 3 − 1 f ′( x ) = lim h→ 0 h −12h + 12 xh + 6h 2 − 3 x 2 h − 3 xh 2 − h 3 f ′( x ) = lim h→ 0 h f ′( x ) = lim −12 + 12 x + 6h − 3 x 2 − 3 xh − h 2 , h ≠ 0 f ′( x ) = lim

h→ 0

h→ 0

(

)

f ′( x ) = −12 + 12 x − 3 x

2

b When x = 1, f ′(1) = −12 + 12 (1) − 3 (1)2 = −3 y

2 a, b

y = x2(4 – x) (– 0.83, 3.31) (–2, 0) 0

(1, 3) (2, 0)

(4, 0)

x

y = (x + 2)(2 – x)

(4.83, – 19.31)

(−0.83, 3.31), (1, 3) and (4.83, −19.31) c

f ( x ) = ( x + 2)(2 − x ) = 4 − x 2 f ( x + h) = (( x + h) + 2)( 2 − ( x + h)) 2

f ( x + h) = 4 − ( x + h ) = 4 − x 2 − 2 xh − h 2 f ( x + h) − f ( x ) f ′( x ) = lim h →0 h 4 − x 2 − 2 xh − h 2 − 4 + x 2 f ′( x ) = lim h →0 h −2 xh − h 2 f ′( x ) = lim h →0 h f ′( x ) = lim (−2 x − h ) , h ≠ 0 h →0

f ′( x ) = −2 x When x = 1, f ′(1) = −2(1) = −2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

88 | TOPIC 5 Differentiation • EXERCISE 5.2 f ( x ) = x 2 (4 − x ) = 4 x 2 − x 3 f ( x + h) = 4( x + h)2 − ( x + h )3

(

) (

f ( x + h) = 4 x 2 + 2 xh + h 2 − x 3 + 3 x 2 h + 3 xh 2 + h 3 2

2

3

2

2

f ( x + h) = 4 x + 8 xh + 4 h − x − 3 x h − 3 xh − h

)

3

f ( x + h) − f ( x ) h 4 x 2 + 8 xh + 4 h 2 − x 3 − 3 x 2 h − 3 xh 2 − h 3 − 4 x 2 + x 3 f ′( x ) = lim h→ 0 h 8 xh + 4 h 2 − 3 x 2 h − 3 xh 2 − h 3 f ′( x ) = lim h→ 0 h f ′( x ) = lim 8 x + 4 h − 3 x 2 − 3 xh − h 2 , h ≠ 0 f ′( x ) = lim

h→ 0

h→ 0

(

)

f ′( x ) = 8 x − 3 x

2

When x = 1, f ′(1) = 8(1) − 3(1)2 = 5 1 1 1 3 3 1 −2 3 a f ( x ) = 4 x + 2 + = 4 x + x + 2 3 2 3x 2 2 f ′( x ) = 12 x 2 − x −3 = 12 x 2 − 3 3 3x 5

2 x − x4 2 −2 1 = x − x 5 5 5x 3 7 − 1 1 1 f ′( x ) = − x 2 − = − 7 − 5 5 x2

b f ( x ) =

4 a f ( x ) = ( x + 3)( x 2 + 1) = x 3 + 3 x 2 + x + 3 f ′( x ) = 3 x 2 + 6 x + 1 b f ( x ) =

4− x x3

f ′( x ) = −6 x

−

5 2

= 4x

−

3 2

− x −1

+ x −2 = −

6 5 x2

+

1 x2

1 + 2 x = − x −2 + 2 x x2 2 f ′( x ) = 2 x −3 + 2 = 3 + 2 x 2  1 f ′−  = +2  2  1 3 −   2 = −16 + 2 = −14 2x − 4 = 2 − 4 x −1 b f ( x ) = x 4 f ′( x ) = 4 x −2 = 2 x 4 If f ′( x ) = 1 then 2 = 1 x 4 = x2 ±2 = x 2(−2) − 4 When x = −2, f ( x ) = =4 −2 2(2) − 4 When x = 2, f ( x ) = =0 −2 Therefore gradient = 1 at (2, 0) and (−2, 4)

5 a

  f ( x ) = −

6 y = ( x − a)( x 2 − 1) = x 3 − ax 2 − x + a dy = 3 x 2 − 2ax − 1 dx dy x = − 2, = 3 ( −2 )2 − 2(−2)a − 1 = 12 + 4 a − 1 = 11 + 4 a dx

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 5 Differentiation • EXERCISE 5.2   |  89 7 a Turning point at (−2, −9) so f ′( x ) cuts the x axis at x = −2. For x < −2, f ′( x ) is negative and for x > −2, f ′( x ) is positive. y

(–2, 0) x

0

y = fʹ(x)

b Domain of the gradient function shown is x ∈( −∞, 2 ) \ {−2}. 8 Stationary point of inflection at x = −2 so turning point at x = −2. Parabola lies above the x axis for all other values of x. y

y = fʹ(x)

(–2, 0)

x

0

(

)(

)

9 a y = x ( x − 2)2 ( x − 4) = x 2 − 4 x x 2 − 4 x + 4 = x 4 − 8 x 3 + 20 x 2 − 16 x dy = 4 x 3 − 24 x 2 + 40 x − 16 dx dy = 4(3)3 − 24(3)2 + 40(3) − 16 = −4 x = 3, dx Equation of tangent which passes through ( x1 , y1 ) ≡ (3, −3) which has mT = −4 y − y1 = mT ( x − x1 ) y + 3 = −4( x − 3) y + 3 = −4 x + 12 y = −4 x + 9 1 b Equation of the line perpendicular to the tangent where mP = . 4 y − y1 = mP ( x − x1 ) 1 y + 3 = ( x − 3) 4 1 3 y+3= x− 4 4 15 1 y= x− 4 4 10 Tangent and perpendicular intersect where: y = −2 x + 5....................(1) 1 5 y = x + ....................(2) 2 2 (1) = (2) 1 5 −2 x + 5 = x + 2 2 −4 x + 10 = x + 5 5 = 5x x =1 Substitute x = 1 into (1):    y = −2(1) + 5 = 3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

90 | TOPIC 5 Differentiation • EXERCISE 5.2 General rule for the parabola is y = ax 2 + bx + c and general dy rule for the derivative is = 2ax + b. dx When x = 0, y = 4 so c = 4 or y = ax 2 + bx + 4. When x = 1, y = 3 so 3 = a + b + 4 or − 1 = a + b. dy = 0 so 0 = b. When x = 0, dx If b = 0 then for − 1 = a + b, a = −1. Hence y = 4 − x 2 is the equation of the parabola.

b i Domain = R y ii y = f '(x) x

0

y

c i Domain = ( −∞, 4 ) \ {−2} y ii

(0, 5)

(0, 4)

(1, 3) y = f ʹ(x)

1x+– 5 y=– 2 2

y = –2x + 5

(–2, 1)

(4, 1) x

0 0

(–2, 0)

(–2, –1)

x

(2, 0)

y = 4 – x2

3 1 3 1 − + 4 = x −5 − x −1 + 4 4 2 4 x5 2x 15 1 dy 15 −6 1 −2 =− x + x =− 6 + 2 dx 4 2 4x 2x

11 a y =

d i Domain = R \{0} ii

y

10 x − 2 x 3 + 1 = 10 x −3 − 2 x −1 + x −4 x4 4 30 2 f ′( x ) = −30 x −4 + 2 x −2 − 4 x −5 = − 4 + 2 − 5 x x x

b f ( x ) =

c y = x − dy 1 = dx 2

1 2 x

1 − x 2

+

1

y = f ʹ(x)

1

1 −2 x 2 1 1 = + 3 2 x 4x 2

(2, 0)

= x2 − 1 4

3 − x 2

0

(3 − x )3 27 − 27 x + 9 x 2 − x 3 27

27 9 1 = = x −1 − + x − x 2 2x 2x 2 2 2 2 27 9 27 −2 9 f ′( x ) = − x + − x = − 2 − x + 2 2 2 2x

d f ( x ) =

x

13 a Gradient of secant =

12 a i Domain = R ii

y

y = fʹ(x)

(– –21 , 0)

(3, 0) 0

x

f ( x + h) − h f ( x + h) − h f ( x + h) − h f ( x + h) − h f ( x + h) − h

f ( x + h) − f ( x ) h

( x + h)2 + 5( x + h) + 6 − x 2 − 5 x − 6 h f ( x ) x 2 + 2 xh + h 2 + 5 x + 5h + 6 − x 2 − 5 x − 6 = h f ( x ) 2 xh + h 2 + 5h = h f (x) = 2 x + h + 5 prov h ≠ 0 f (x)

f (x)

=

= 2(−1) + 1 + 5 = 4 where x = −1 and h = 1

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 5 Differentiation • EXERCISE 5.2   |  91 b Gradient of secant = f ( x + h) − h f ( x + h) − h f ( x + h) − h f ( x + h) − h f ( x + h) − h f ( x + h) − h f ( x + h) − h

f (x) f ( x)

= =

( x + h)3 ( x + h − 3) − x 3 ( x − 3) h x 3 + 3 x 2 h + 3 xh 2 + h3 ( x + h − 3) − x 4 + 3 x 3

(

)

h x 4 + x 3h − 3 x 3 + 3 x 3h + 3 x 2 h 2 − 9 x 2 h + 3 x 2 h 2 + 3 xh3 − 9 xh 2 + xh3 + h 4 − 3h3 − x 4 + 3 x 3 = h f ( x ) 4 x 3h + 6 x 2 h 2 − 9 x 2 h + 4 xh 3 − 9 xh 2 − 3h 3 + h 4 = h f (x) 3 2 2 = 4 x + 6 x h − 9 x +4 xh 2 − 9 xh − 3h 2 + h 3 , h ≠ 0 f ( x)

f (x) f (x)

= 4(2)3 + 6(2)2 (2) − 9(2)2 + 4(2)(2)2 − 9(2)(2) − 3(2)2 + (2)3 where x = 2 and h = 2 = 36

c Gradient of secant = f ( x + h) − h f ( x + h) − h f ( x + h) − h f ( x + h) − h f ( x + h) − h 14 a

f ( x + h) − f ( x ) h

f (x) f (x) f (x)

= =

f ( x + h) − f ( x ) h

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

(

)

h − x 3 − 3 x 2 h − 3 xh 2 − h 3 + x 3 = h

f (x)

= −3 x 2 − 3 xh − h 2 , h ≠ 0

f (x)

= −3 ( 3)2 − 3 ( −3)( 3) − ( 3)2

= −27 + 27 − 9 = −9 where x = −3 and h = 3    f ( x ) = 12 − x

f ( x + h) = 12 − ( x + h) = 12 − x − h f ( x + h) − f ( x ) f ′( x ) = lim h→ 0 h 12 − x − h − 12 + x f ′( x ) = lim h→ 0 h f ′( x ) = lim (−1), h ≠ 0 h→ 0

f ′( x ) = −1 b

  f ( x ) = 3 x 2 − 2 x − 21 f ( x + h) = 3( x + h)2 − 2( x + h) − 21 = 3 x 2 + 3 xh + 3h 2 − 2 x − 2h − 21 f ( x + h) − f ( x ) f ′( x ) = lim h→ 0 h 3 x 2 + 6 xh + 3h 2 − 2 x − 2h − 21 − 3 x 2 + 2 x + 21 f ′( x ) = lim h→ 0 h 6 xh + 3h 2 − 2h f ′( x ) = lim h→ 0 h f ′( x ) = lim ( 6 x + 3h − 2 ) , h ≠ 0 h→ 0

f ′( x ) = 6 x − 2 15 a f ( x ) = x 2 − 3 f ′( x ) = 2 x f ′(2) = 2(2) = 4 b f ( x ) = (3 − x )( x − 4) = − x 2 + 7 x − 12 f ′( x ) = −2 x + 7 f ′(1) = −2(1) + 7 = 5 c     f ( x ) = ( x − 2 )3 = x 3 − 3( x )2 (2) + 3( x )(2)2 − (2)3 = x 3 − 6 x 2 + 12 x − 8 f ′( x ) = 3 x 2 − 12 x + 12   f ′(4) = 3 ( 4 )2 − 12(4) + 12 = 12

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

92 | TOPIC 5 Differentiation • EXERCISE 5.2 1

d f ( x ) = x −

3 + 2 x = x 2 − 3 x −1 + 2 x x

1

1 −2 1 3 x + 3 x −2 + 2 = + +2 2 2 x x2 1 3 1 3 4 3 32 39 f ′(4) = + 2 +2= + +2= + + = 4 16 16 16 16 16 2 4 4

f ′( x ) =

16 a

f ( x ) = ( x + 1)( x + 3) = x 2 + 4 x + 3 f ′( x ) = 2 x + 4 f ′(−5) = 2(−5) + 4 = −6 When x = −5, y = (−5 + 1)(−5 + 3) = 8 Equation of tangent which passes through the point ( x1, y1 ) ≡ (−5,8) where

mT = −6 is y − y1 = mT ( x − x1 ) y − 8 = −6 ( x + 5) y − 8 = −6 x − 30 y = −6 x − 22 b f ( x ) = 8 − x 3 f ′( x ) = −3 x 2 f ′(a) = −3a 2 When x = a, y = 8 − a 3 Equation of tangent which passes through the point ( x1, y1 ) ≡ (a,8 − a3 ) where mT = 3a 2 is y − y1 = mT ( x − x1 )

y − (8 − a ) = −3a ( x − a ) 3

2

y − 8 + a 3 = −3a 2 x + 3a 3 y = −3a 2 x + 2a 3 + 8 1

c f ( x ) = 2 x − 5 = 2 x 2 − 5 f ′( x ) = x

−

1 2

=

1 x

1 3 When x = 3, y = 2 3 − 5 Equation of tangent which passes through the point 1 is x1, y1 ≡ (3, 2 3 − 5) where mT = 3 y − y1 = mT ( x − x1 ) 1 y − (2 3 − 5) = ( x − 3) 3 1 3 y−2 3+5= x− 3 3 1 3 y= x− +2 3−5 3 3 3 y= x+ 3−5 3 2 d f ( x ) = − − 4 x = −2 x −1 − 4 x x 2 f ′( x ) = 2 x −2 − 4 = 2 − 4 x 2 7 f ′(−2) = −4=− 2 (−2)2 f ′(3) =

(

)

When x = −2, y = −

Equation of tangent which passes through the point ( x1, y1 ) ≡ (−2,9) where mT = − 72 is y − y1 = mT ( x − x1 ) 7 y − 9 = − ( x + 2) 2 7 y−9= − x−7 2 7 y=− x+2 2 17 a Equation of perpendicular line which passes through the point ( x1, y1 ) ≡ (−5,8) 1 where mP = is 6

y − y1 = mP ( x − x1 ) 1 y − 8 = ( x + 5) 6 1 5 y−8= x + 6 6 1 5 48 y= x+ + 6 6 6 53 1 y= x+ 6 6 b Equation of perpendicular line which passes through the point x1, y1 ≡ (a,8 − a 3 )

(

)

1 is 3a 2 y − y1 = mP ( x − x1 ) 1 y − (8 − a 3 ) = − 2 ( x − a ) 3a 1 1 y − 8 + a3 = − 2 x + 3a 3a 1 1 y=− 2 x+ + 8 − a3 3a 3a c Equation of perpendicular line which passes through the point x1, y1 ≡ (3, 2 3 − 5) where mP = − 3 is where mP = −

(

)

y − y1 = mP ( x − x1 )

y − (2 3 − 5) = − 3 ( x − 3) y − 2 3 + 5 = − 3x + 3 3 y = − 3x + 5 3 − 5 d Equation of perpendicular line which passes through the point x1, y1 ≡ (−2,9)

(

)

2 is 7 y − y1 = mP ( x − x1 ) 2 y − 9 = ( x + 2) 7 2 4 y−9= x+ 7 7 2 4 63 y= x+ + 7 7 7 67 2 y= x+ 7 7

where mP = −

2 − 4(−2) = 9 (−2)

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 5 Differentiation • EXERCISE 5.2   |  93 18 a Any line parallel to y = 3 x + 4 has gradient of 3. f ( x ) = − ( x − 2 )2 + 3 = − x 2 + 4 x − 4 + 3 = − x 2 + 4 x − 1 f ′( x ) = −2 x + 4 3 = −2 x + 4 1 =x 2 1 1 2 1 1 3 When x = , y = −   + 4   − 1 = − + 1 =  2  2 2 4 4 1 3 Equation of tangent which passes through the point x1, y1 ≡  ,  where  2 4 mT = 3 is y − y1 = mT ( x − x1 ) 3 1  y − = 3 x −   4 2 3 3 y − = 3x − 4 2 6 3 y = 3x − + 4 4 3 y = 3x − 4 1 b A line perpendicular to 2 y − 2 = −4 x or y = −2 x + 1 will have a gradient of . 2 2 −2 f ( x ) = − 2 + 1 = −2 x + 1 x 4 f ′( x ) = 4 x −3 = 3 x 1 4 = 2 x3 x3 2= 4 8 = x3 2= x 2 1 1 When x = 2, y = − 2 + 1 = − + 1 = 2 2 2  1 Equation of perpendicular line which passes through the point x1, y1 ≡  2,  2 1 with mP = is 2 y − y1 = mP ( x − x1 ) 1 1 y − = ( x − 2) 2 2 1 1 y − = x −1 2 2 1 1 y= x− 2 2 19 Tangent to parabola at x = 4 is y = − x + 6. When x = 4, y = −4 + 6 = 2 So parabola y = ax 2 + bx + c passes through the points (4, 2), (0, −10) and (2, 0). At (0, −10), −10 = a(0)2 + b(0) + c so c = −10. At (2, 0), 0 = a(2)2 + 2b − 10 or 4 a + 2b = 10 ⇒ 2a + b = 5..........(1) At (4, 2), 2 = a(4)2 + 4 b − 10 or 16a + 4 b = 12 ⇒ 4 a + b = 3............(2) (2) − (1)  2a = −2 ⇒ a = −1 Substitute a = −1 into (1)   2(−1) + b = 5 so b = 7 Equation of parabola is y = − x 2 + 7 x − 10. 20 Tangent to a cubic function at x = 2 has the rule y = 11x − 16. If x = 2 then y = 11(2) − 16 = 6. The cubic passes through the points  (0, 0), (−1, 0) and (2, 6). General rule for the cubic is y = ax 3 + bx 2 + cx + d d=0 ∴ y = ax 3 + bx 2 + cx

(

)

(

)

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

94 | TOPIC 5 Differentiation • EXERCISE 5.3 y − y1 = mT ( x − x1 ) 27 15  3 = x−  y+ 8 4 2 27 15 45 = x− y+ 8 4 8 15 45 27 − y= x− 4 8 8 15 y= x−9 4 Equation of perpendicular line which passes through  3 27  the point ( x1 , y1 ) ≡  , −  which has a gradient of 2 8 4 mP = − is 15 y − y1 = mP ( x − x1 ) 27 4 3 =− x+  y+ 8 15  2 27 4 2 =− x+ y+ 8 15 5 4 2 27 y=− x+ − 15 5 8 119 4 y=− x− 15 40

(−1, 0) ⇒ 0 = − a + b − c...........(1) (2, 6) ⇒ 6 = 8a + 4 b + 2c............(2) dy = 3ax 2 + 2bx + c dx m = 11, x = 2 11 = 12a + 4 b + c...............(3) Using CAS to solve: a = 1, b = 0, c = −1 ∴ y = x3 − x 1 21 Line perpendicular to y = 2 x = 2 x 2 has the equation 1 y = −2 x + m. Thus the gradient of the curve is . 2 1 1 − dy 1 2 2 =x = If y = 2 x then . dx x 1 1 If = then x = 2 ⇒ x = 4 x 2 When x = 4, y = 2 4 = 4. Perpendicular line passes through the point (4, 4). Thus 4 = −2(4) + m 4 = −8 + m m = 12 y 22 a y = x(x – 2)(x + 3) (–3, 0)

(2, 0)

(0, 0)

(3, 0) x

Exercise 5.3 — Differentiation of exponential functions 1 a y = e

1 − x 3 1



dy 1 − x =− e 3 dx 3

b y = 3 x 4 − e −2 x (0, –18) y = (2 – x)(x + 3)(x – 3)

3 27 b Point of intersection is  , −  . 2 8 3

2

c y = x ( x − 2)( x + 3) = x + x − 6 x dy = 3x 2 + 2 x − 6 dx When 2 3 dy  3  3 x= , = 3  + 2  − 6  2  2 2 dx 27 = −3 4 27 12 = − 4 4 15 = 4 Equation of tangent which passes through the point 15  3 27  ( x1 , y1 ) ≡  , −  which has a gradient of mT = is 2 8 4

2

2 2 dy = 12 x 3 + e −2 x ( −4 x ) = 12 x 3 + 4 xe −2 x dx

4e x − e − x + 2 3e3 x 4 1 2 y = e −2 x − e −4 x + e −3 x 3 3 3 dy 8 −2 x 4 −4 x = − e + e − 2e −3 x 3 3 dx c y =

(

d y = e 2 x − 3

)

2

y = e 4 x − 6e 2 x + 9 dy = 4e 4 x − 12e 2 x dx 1 2 f ( x ) = e3 x + e − x 2 3 f ′( x ) = e3 x − e − x 2 1 3 f ′(0) = e 0 − e 0 = 2 2 3 y = e 2 x dy = 2e 2 x dx dy = 2e 2(0) = 2 dx x = 0 When x = 0, y = e 0 = 1

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 5 Differentiation • EXERCISE 5.3   |  95 Equation of tangent which passes through ( x1 , y1 ) ≡ (0,1) with mT = 2 is y − y1 = mT ( x − x1 ) y − 1 = 2x y = 2x + 1

4 y = e −3 x + 4 dy = −3e −3 x dx dy = −3e −3(0) = −3 dx x = 0 When x = 0, y = e 0 + 4 = 5 Equation of tangent which passes through ( x1 , y1 ) ≡ (0,5) with mT = −3 is

y − y1 = mT ( x − x1 ) y − 5 = −3 x y = −3 x + 5 1 Equation of perpendicular line which passes through ( x1 , y1 ) ≡ (0,5) with mP = is 3 y − y1 = mP ( x − x1 ) 1 y−5= x 3 1 y= x+5 3 d 5 a 5e −4 x = −20e −4 x dx

(

b

)

1 1 d  −2x 1 3 1 − x + x  = − e 2 + x2 e dx  3  2

c d  4e3 x − 1 e6  dx  2 d

d  e5 x − e − x + 2  d 3 x −3 x −2 x = 3e3 x + 3e −3 x − 4e −2 x  = dx e − e + 2e dx  e2 x

(

(

 e x 2 − e −3 x e d  dx  e− x f

3e6 x  − 3e −3 x + 2  = 12e3 x − + 9e −3 x + 2  2 x

x

((

)  =

(

)

d 2e 2 x − e − x = 4e 2 x + e − x dx



))

)(

)

(

)

d d x e2 x + 3 e− x − 1 = e − e 2 x + 3e − x = 3 = e x − 2e 2 x − 3e − x dx dx

6 a y = 2e − x dy = −2e − x dx dy = −2e 0 = −2 dxx = 0 4 b y = 2 x = 4e −2 x e dy = −8e −2 x dx  1

−2  dy = −8e  2  dx x = 1 2

= −8e −1 8 =− e

1 3x c y = e 2 dy 3 3 x = e dx 2  1

dy 3 3  3 = e  3 = e 1 dx x = 2 2 3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

96 | TOPIC 5 Differentiation • EXERCISE 5.3 d y = 2 x − e x dy = 2 − ex dx dy = 2 − e0 = 2 − 1 = 1 dxx = 0

Equation of perpendicular line which passes through (3, e 2 3 with mP = − is e 3 y - y1 = mP (x - x1)

7 y = e −2 x dy mT = = −2e −2 x dx 1 −2 −  1 When x = − , mT = −2e  2  = −2e 2 1 −2 −   2

1 =e When x = − , y = e 2 Equation of tangent with mT = −2e which passes through the point 1 ( x1 , y1 ) =  − , e is given by 2 y − y1 = mT ( x − x1 ) 1  y − e = −2e  x +   2 y = −2ex − e + e y = −2ex 8 y = e −3 x − 2 dy mT = = −3e −3 x dx 1 When x = 0, mT = −3e −3(0) = −3 and mP = 3 When x = 0, y = e −3( 0 ) − 2 = 1 − 2 = −1 Equation of tangent with mT = −3 which passes through the point ( x1 , y1 ) = ( 0, −1) is given by y − y1 = mT ( x − x1 ) y + 1 = −3 ( x − 0 ) y = −3 x − 1

3

)

+ 1 with

3

e is 2 3 y − y1 = mT ( x − x1 )

mT =

y−e

3

−2 3 e

3

( x − 3) 6 3

x+

e

3

+e

3

+1

f ( x ) = e −2 x + 3 − 2e

10 a

f ′( x ) = −2e −2 x + 3 f ′(−2) = −2e −2( −2) + 3 = −2e 7 b −2e −2 x + 3 = −2 e −2 x + 3 = 1 e −2 x + 3 = e 0 Equating indices −2 x + 3 = 0 −2 x = −3 3 x= 2 e3 x + 2 = e 2 x + 2e − x 11 a f ( x ) = ex f ′( x ) = 2e 2 x − 2e − x 2 f ′(1) = 2e 2 − 2e −1 = 2e 2 − e 2e 2 x − 2e− x = 0 e 2 x − e− x = 0 e

−x

(e3x − 1) = 0

Not possible e3 x = 1 e3 x = e 0 x=0 12 y = e x

2

+3x − 4

du = 2x + 3 dx

dy = eu du By the Chain rule: dy dy du = × dx du dx dy = ( 2 x + 3) e u dx 2 dy = ( 2 x + 3) e x + 3 x − 4 dx 2 dy = ( 2(1) + 3) e1 + 3(1) − 4 dx x =1 dy = 5e 0 = 5 dx x =1 y = eu so

dy e x = dx 2 x dy e 3 = x = 3, dx 2 3 When x = 3, y = e 3 + 1 Equation of tangent which passes through the point

(

3

e

Let u = x 2 + 3 x − 4 so

+1

( x1 , y1 ) ≡ 3, e

2 3

b e − x = 0 or e3 x − 1 = 0

y − y1 = mP ( x − x1 ) 1 y + 1 = ( x − 0) 3 1 y = x −1 3 x

−1= − y=



1 Equation of perpendicular line with mP = which passes 3 through the point ( x1 , y1 ) = ( 0, −1) is given by

9 y = e

3

y−e

e 3 ( x − 3) 2 3 3e 3 e 3 y= x− +e 2 3 2 3

When x = 1, y = e1 + 3(1) − 4 = e 0 = 1 Equation of tangent which passes through ( x1 , y1 ) ≡ (1,1) where mT = 5 is y − y1 = mT ( x − x1 ) y − 1 = 5( x − 1) y = 5x − 4 2

−1=

3

+1

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

3

+ 1)

TOPIC 5 Differentiation • EXERCISE 5.4   |  97 13 f ( x ) = Ae x + Be −3 x x

f ′( x ) = Ae − 3 Be

−3 x

When f ′( x ) = 0. Ae x − 3 Be −3 x = 0 e −3 x ( Ae 4 x − 3 B) = 0 e −3 x ≠ 0, no real solution 0 = Ae 4 x − 3 B So, 3 B = Ae 4 x 3B e4 x = A 14 a A = A0 e −0.69t When t = 0, A = 2 2 = A0 e −0.69(0) 2 = A0 Thus A = 2e −0.69t dA = −0.69 × 2e −0.69t b dt dA = −1.38e −0.69t dt dA When t = 0, = −1.38e −0.69(0) = −1.38 dt 15 f ( x ) = 32 x − 4 f ′( x ) = 2 loge (3) × 32 x − 4 f ′(2) = 2 loge (3) × 32(2)− 4 f ′(2) = 2 loge (3) 16 a y = 2e −2 x + 1............(1) y = x 3 − 3 x ...............(2) Point of intersection occurs where (1) = (2) 2e −2 x + 1 = x 3 − 3 x x  1.89 y = (1.8854)3 − 3(1.8854) y  1.05 Therefore POI = (1.89, 1.05) dy b mT = = 3x 2 − 3 dx When x = 1.8854, mT = 3(1.8854)2 − 3  7.66.

Exercise 5.4 — Applications of exponential functions 1 a M = Ae −0.00152t When t = 0, M = 20 20 = Ae −0.00152(0) 20 = A Thus M = 20e −0.00152t dM b = −0.00152(20)e −0.00152t dt dM = −0.0304e −0.00152t dt c When t = 30 then dM = −0.0304e −0.00152(30) dt dM = −0.0291 dt Rate of decay is 0.0291 g/year. 2 a I = I 0 e −0.0022 d When d = 315 then I = I 0 e −0.0022(315) I = 0.5I 0

dI = −0.0022 I 0 e −0.0022 d dd When d = 315 then dI = −0.0022 I 0 e −0.0022(315)   dd dI = −0.0022 I 0 × 0.05 dd dI = −0.0011I 0 dd Intensity is decreasing at a rate of 0.0011I 0. 3 a f ( x ) = e 2 x + qe x + 3 When x = 0, f (0) = 0, so e 2(0) + qe 0 + 3 = 0 1+ q + 3 = 0 q+4=0 q = −4 b f ( x ) = e 2 x − 4e x + 3 Graph cuts the x axis where y = 0. b

e 2 x − 4e x + 3 = 0

( e x )2 − 4 ( e x ) + 3 = 0

(e x − 3)(e x − 1) = 0

x

e − 3 = 0 or e x − 1 = 0 ex = 3

ex = 1

e x = e0 x=0 Thus (m, 0) = (loge 3, 0). x = loge 3

c f ′( x ) = 2e 2 x − 4e x d f ′(0) = 2e 2(0) − 4e 0 = 2 − 4 = −2 4 a f ( x ) = e −2 x + ze − x + 2 f (0) = 0 e −2(0) + ze 0 + 2 = 0 1+ z + 2 = 0 z+3= 0 z = −3 b y = f ( x ) = e −2 x − 3e − x + 2 Graph cuts the x axis where y = 0 e −2 x − 3e − x + 2 = 0

( e − x )2 − 3 ( e − x ) + 2 = 0 (e − x − 1)(e − x − 2) = 0

e

−x

− 1 = 0 or e − x − 2 = 0

e− x = 1 −x

e− x = 2 0

=e − x = loge 2 x=0 x = − loge 2 Thus (n, 0) = (− loge (2), 0). e

c f ( x ) = e −2 x − 3e − x + 2 f ′( x ) = −2e −2 x + 3e − x d f ′(0) = −2e −2(0) + 3e 0 = −2 + 3 = 1 5 a A = A0 e − kt When t = 0, A = 120 120 = A0 e − k (0) 120 = A0 Thus A = 120e − kt

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

98 | TOPIC 5 Differentiation • EXERCISE 5.4 Therefore the temperature is decreasing at a rate of 1.531°C/min. Temperature is falling at rate of 1.531°C/min

dA = −120 ke − kt dt dA = − k × 120e − kt dt dA = −k × A dt dA αA dt c A = 120e − kt b

8 a P = P0 e 0.016t When t = 0, P = 8.2 million

90 = 120e −2 k 3 = e −2 k 4  3 −2 k = loge    4 1  3 k = − loge    4 2 1 4   k = loge    3 2 1  4 d A = 120e kt , k = loge    3 2 dA = −120(0.139)e kt dt dA = −16.68e kt dt When t = 5 dA = −16.68e k (5)  −8.408 units/min dt Therefore the gas is decomposing at a rate of 8.408 units/min. e As t → ∞, A → 0. Technically the graph approaches the line A = 0 (asymptotic behavior, so never reaches A = 0 exactly) however, the value of A would be so small, that in effect, after a long period of time, there is no gas left. 6 a L = L0 e 0.599t When t = 0, L = 11 11 = L0 e 0.599(0) 11 = L0 So L = 11e 0.599t dL = 0.599 × 11e 0.599t b dt dL = 6.589e 0.599t dt c When t = 3 then dL = 6.589e 0.599(3) = 39.742 mm/month dt Therefore the temperature is decreasing at a rate of 1.531°C/min. 7 a When t = 0, T = 95 − 20 = 75 75 = T0 e − z (0) 75 = T0 So T = 75e − zt b T = 75e −0.034 t dT = −0.034 × 75e −0.034 t dt dT = −2.55e −0.034 t dt When t = 15 then dT = −2.55e −0.034(15) = −1.531°C/min dt

8.2 = P0 e 0.016(0) 8.2 = P0 Thus P = 8.2e 0.016t When 2015, t = 2015 − 1950 = 65 P = 8.2e 0.016(65) = 23.2 million b 20 = 8.2e 0.016(t ) Solve for t using CAS: t = 55.72 Therefore September, 2005 dP c = 0.016 × 8.2e 0.016t dt dP = 0.1312e 0.016t dt When 2000, t = 2000 − 1950 = 50 dP = 0.1312e 0.016(50) = 0.29 199 dt Change in population is 0.29 million/year. dP = 0.016 × 8.2e 0.016t d dt dP = 0.1312e 0.016t dt 0.1312e 0.016t > 0.4 Solve using CAS: 0.1312e 0.016t > 0.4 t > 69.67 Therefore in the year 2019. 9 a P = P0 e − kh When h = 0.5, P = 66.7 → 66.7 = P0 e −0.5 k When h = 1.5, P = 52.3 → 52.3 = P0 e −1.5 k Solve using CAS: P0 = 75.32 cm of mercury, k = 0.24 So P = 75.32e −0.24 h b P = 75.32e −0.24 h dP = −0.24 × 75.32e −0.24 h dh dP = −18.08e −0.24 h dh When h = 5 dP = −18.08e −0.24(5) = −5.45 cm of mercury/km dh The rate is falling at 5.45 cm of mercury/km 10 a h = 0.295(e x + e − x ) Height of post at x = 0.6, so h = 0.295(e 0.6 + e −0.6 ) h = 0.6994 m Height of chain at lowest point x = 0 h = 0.295(e 0 + e 0 ) = 0.295(2) = 0.59 m Amount of sag is 0.6994−0.59 = 0.1094 metres or 10.94 centimetres. b h = 0.295(e x + e − x ) dh = 0.295(e x − e − x ) dx When x = 0.6

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 5 Differentiation • EXERCISE 5.4   |  99 dh = 0.295(e 0.6 − e −0.6 ) dx dh = 0.3756  0.4 metres dx If θ is the angle the chain makes with the post then tan(θ ) = 0.3756

θ = tan −1 (0.3756) θ = 20.6° 11 a Graph cuts the y axis where x = 0. f (0) = e 0 − 0.5e 2(0) = 1 − 0.5 = 0.5 Therefore (0, 0.5) b f ′( x ) = − e − x + e −2 x c Max TP occurs where f ′( x ) = 0. − e − x + e −2 x = 0

b When t = 7 days and k = 0.387 T = 30e 0.378(7) T = 30e 2.709 T = 450 000 dT = 0.387 × 30e 0.387t c dt dT = 11.61e 0.387t dt When t = 3 then dT = 11.61e 0.387(3) = 37 072.2/day dt d C = C0 e mt When t = 0, C = 450 450 = C0 e m (0) 450 = C0

e − x (e − x − 1) = 0 e − x − 1 = 0 as e − x > 0 for all x e

−x

=1

−x

= e0 x=0 Maximum TP at (0, 0.5). e

d f ′( x ) = − e − x + e −2 x When x = − loge (2) f ′(− loge (2)) = − e

loge (2)

+e

−2( − loge (2))

2

= −2 + e loge (2 ) = −2 + 4 =2 Let θ be angle graph makes with the positive direction of the x axis. tan(θ ) = 2

θ = tan −1 (2) θ = 63.4° e mT = f ′( x ) = − e − x + e −2 x mT = f ′(1) = − e −1 + e −2(1) = −0.2325 f (1) = e −1 − 0.5e −2(1) = 0.3002 Equation of tangent with mT = −0.2325 which passes through the point

( x1 , y1 ) = (1, 0.3002) is given by

y − y1 = mT ( x − x1 ) y − 0.3002 = −0.2325( x − 1) y − 0.3002 = −0.2325 x + 0.2325 y = −0.2325 x + 0.5327 1 = 4.3011 0.2325 Equation of perpendicular line with mP = 4.3011 which passes through the point ( x1 , y1 ) = (1, 0.3002) is given by

f m N =

y − y1 = mP ( x − x1 ) y − 0.3002 = 4.3011( x − 1) y − 0.3002 = 4.3011x − 4.3011 y = 4.3011x − 4.0009 12 a T = T0 e kt When t = 0, T = 30 30 = T0 e k (0) 30 = T0 Thus T = 30e kt

So C = 450e mt e When t = 7 then 90 C = 450 − × 450 100 C = 450 − 405 C = 45 45  000 cane toads remain after 7 days. f

45 = 450e 7 m 0.1 = e 7 m loge (0.1) = 7m 1 loge (0.1) = m 7

1 m = − loge (10) 7 (m = −0.3289)

C = 450e −0.3289t dC = −0.3289 × 450e −0.3289t dt dC = −148.0233e −0.3289t dt When t = 4 then dC = −148.0233e −0.3289(4) = −39.710 156 thousand/day dt Decrease of 39 711 cane toads per day. 2 13 a y = Ae − x When x = 0, y = 5 5 = Ae 0 A=5 2 Thus y = 5e − x . 2 dy = −2 x × 5e − x b dx 2 dy = −10 xe − x dx 2 dy c i When x = −0.5, = −10(−0.5)e − ( −0.5) = 3.89 dx 2 dy ii When x = 1, = −10(1)e − (1) = −3.68 dx x2 − 5 14 a y = 2 2e x dy x 3 − 6 x = 2 dx ex

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

100 | TOPIC 5 Differentiation • EXERCISE 5.5 b When

y − y1 = mT ( x − x1 ) π y − 0 = 1  x −   2 π y=x− 2

dy = 0 then dx x3 − 6x

=0 2 ex x − 6x = 0 3

(

)

x x2 − 6 = 0 2

x = 0 or x − 6 = 0

2

x −

( 6)

2

=0

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

x=− 6

When x = ± 6; y =

(± 6 )

2

−5

± 6 2e( )

2

=

6−5 1 = 2e6 2e6

1   Co-ordinates are  ± 6, 6   2e  c When x = 1.5,

dy (1.5)3 − 6(1.5) = = −0.593 2 dx e(1.5)

Exercise 5.5 — Differentiation of trigonometric functions 1 a b c d e f

d (5 x + 3 cos( x ) + 5sin( x )) = 5 − 3sin( x ) + 5 cos( x ) dx d sin(3 x + 2) − cos(3 x 2 ) = 3 cos(3 x + 2) + 6 x sin(3 x 2 ) dx d 1   sin(9 x ) = 3 cos(9 x ) dx  3 d 5 tan(2 x ) − 2 x 5 = 10 sec 2 (2 x ) − 10 x 4 dx d   x 2  x  8 tan    = 2 sec   4 4 dx d π x π πx d   ( tan(9 x °) =  tan    = sec 2   20  20 20 dx dx 

( (

)

)

sin( x ) cos 2 (2 x ) − sin( x ) 2 sin( x ) sin(2 x ) cos 2 (2 x ) − 1 = , sin( x ) ≠ 0 sin(2 x ) − sin 2 (2 x ) = sin(2 x ) = − sin(2 x ), sin(2 x ) ≠ 0 Thus d  sin( x ) cos 2 (2 x ) − sin( x )  d  = dx ( − sin(2 x ) ) = −2 cos(2 x ) sin( x ) sin(2 x ) dx  3 y = − cos( x ) dy mT = = sin( x ) dx π π When x = ; mT = sin   = 1  2 2 π π When x = ; y = − cos   = 0  2 2 Equation of tangent with mT = 1 which passes through the point π ( x1 , y1 ) =  , 0 is given by 2

4 y = tan(2 x ) dy = 2 sec 2 (2 x ) mT = dx π 2 2 1 When x = − ; mT = = =2÷ =4 2 π 2 8 cos 2  −   1   4  2 π π When x = − ; y = tan  −  = −1  4 8 Equation of tangent with mT = 4 which passes through the point π ( x1 , y1 ) =  − , −1 is given by 8 y − y1 = mT ( x − x1 ) π y + 1 = 4  x +   8 π y = 4x + − 1 2 5 a y = 2 cos(3 x ) dy = −6 sin(3 x ) dx b y = cos ( x ° ) πx  y = cos   180  π πx  dy =− sin    180 180  dx

π c y = 3 cos  − x  2  dy π  = 3sin  − x  2  dx x d y = −4 sin    3 dy 4 x = − cos    3 dx 3 e y = sin (12 x ° ) πx y = sin   15  πx dy π = cos   15  dx 15 π f y = 2 sin  + 3 x  2  dy π  = 6 cos  + 3 x  2  dx 1 2 g y = − tan 5 x 2 dy 1 = − × 10 x × sec 2 5 x 2 dx 2 = −5 x sec 2 5 x 2

( )

( )

( )

h y = tan(20 x ) dy = 20 sec 2 ( 20 x ) dx

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 5 Differentiation • EXERCISE 5.5   |  101 x 6 y = −2 sin   for x ∈[ 0, 2π ]  2 dy x = − cos    2 dx 1 x = − cos    2 2 x x 1 − = cos   for ∈[ 0, π ]  2 2 2 π 1 suggests . Since cos is negative, the second quadrant. 3 2 π x =π − 2 3 x 2π = 2 3 4π x= 3 π π 4π  4π 1   2π  ×  = −2 sin   = −2 sin  π −  = −2 sin   = − 3 When x = , y = −2 sin         3 3 3 2 3 3 4π   Point is  , − 3   .  3  π π 3 3 3 = 7 When x = , y = 3 cos   = 3 ×   6 6 2 2 dy mT = = −3sin( x ) dx π π 1 3 When x = , m T = −3sin   = −3 × = −   6 6 2 2 3 Equation of tangent with mT = − which passes through the point 2 π 3 3 ( x1 , y1 ) =  ,  is given by 6 2  y − y1 = mT ( x − x1 ) 3 3 3 π = −  x −  2 2 6 3 3 3 π =− x+ y− 2 2 4 3 π 3 3 y=− x+ + 2 4 2 π π  8 When x = , y = 2 tan   = 2  4 4 dy mT = = 2 sec 2 ( x ) dx π π 2 1 = 2÷ = 4 When x = , m T = 2sec 2   =  4 4 2 2π cos  4 Equation of tangent with mT = 4 which passes through the point π ( x1 , y1 ) =  , 2 is given by 4 y − y1 = mT ( x − x1 ) π y − 2 = 4  x −   4 y − 2 = 4x − π y = 4x + 2 − π y−

9 y = sin(2 x ) dy = 2 cos(2 x ) dx Let θ be the angle the curve makes with in the positive direction of the x axis. π tan θ = 2 cos  2 ×   2 tan θ = 2 cos(π ) tan θ = −2

θ = tan −1 (−2) θ = 116.6° Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

102 | TOPIC 5 Differentiation • EXERCISE 5.5 10 a y = sin(3 x ) 2π  2π  When x = , y = sin  3 ×  = sin(2π ) = 0  3  3 dy mT = = 3 cos(3 x ) dx When 2π 2π  1 x= , m T = 3cos  3 ×  = 3cos(2π ) = 3 and mP = −  3 3 3 Equation of tangent with mT = 3 which passes through the point 2π ( x1 , y1 ) =  , 0 is given by 3 y − y1 = mT ( x − x1 ) 2π   y − 0 = 3 x −   3  y = 3 x − 2π

1 Equation of perpendicular line with mP = − which passes 3 through the point 2π  , 0  is given by 3  y − y1 = mP ( x − x1 ) 1 2π  y−0= − x−  3 3  2π 1 y=− x+ 3 9 x  b y = cos    2 π When x = π , y = cos   = 0  2 1 dy x = − sin   mT = 2  2 dx 1 1 π When x = π , m T = − sin   = − and m N = 2 2 2  2 1 Equation of tangent with mT = − which passes through 2 the point ( x1 , y1 ) = (π , 0 ) is given by y − y1 = mT ( x − x1 ) 1 y − 0 = − (x − π ) 2 π 1 y=− x+ 2 2 Equation of perpendicular line with mP = 2 which passes through the point ( x1 , y1 ) = (π , 0 ) is given by y − y1 = mP ( x − x1 ) y − 0 = 2 (x − π ) y = 2 x − 2π 11 a f ( x ) = sin( x ) − cos( x ) f (0) = sin(0) − cos(0) = −1 b f (x) = 0 sin( x ) − cos( x ) = 0 sin( x ) = cos( x ) tan( x ) = 1 π 1 suggests . Since tan is positive 1st and 3rd quadrants. 4 π π x = ,π + 4 4 π 5π x= , 4 4 c f ′( x ) = cos( x ) + sin( x )

( x1 , y1 ) = 

f ′( x ) = 0 cos( x ) + sin( x ) = 0 sin( x ) = − cos( x ) tan( x ) = −1 π 1 suggests . Since tan is negative 2nd and 4th quadrants. 4 π π x = π − , 2π − 4 4 3π 7π , x= 4 4 12 a f ( x ) = 3 cos( x ) + sin( x ) d

b

f (0) = 3 cos(0) + sin(0) = 3 f (x) = 0 3 cos( x ) + sin( x ) = 0 sin( x ) = − 3 cos( x )

tan( x ) = − 3 π 3 suggests . Since tan is negative 2nd and 4th quadrants. 3 π π x=− , π− 3 3 π 2π x=− , 3 3 c f ′( x ) = − 3 sin( x ) + cos( x ) d f ′( x ) = 0 − 3 sin( x ) + cos( x ) = 0 cos( x ) = 3 sin( x ) 1 = 3 tan( x ) 1 = tan( x ) 3

1 π suggests . Since tan is positive 1st and 3rd quadrants. 6 3 π π x = −π + , 6 6 5π π x=− , 6 6 sin( x ) cos( x ) + sin 2 ( x ) sin( x )(cos( x ) + sin( x )) 13 a = sin( x ) cos( x ) + cos 2 ( x ) cos( x )(sin( x ) + cos( x )) sin( x ) = prov sin( x ) ≠ − cos( x ) cos( x ) = tan( x ) b

d  sin( x ) cos( x ) + sin 2 ( x )  d = ( tan( x ) ) = sec 2 ( x ) dx  sin( x ) cos( x ) + cos 2 ( x )  dx

14 f ( x ) = sin(2 x ) so f ′( x ) = 2 cos(2 x ) f ( x ) = cos(2 x ) so f ′( x ) = −2 sin(2 x ) When the gradients are equal 2 cos(2 x ) = −2 sin(2 x ) where x ∈[ −π , π ] cos(2 x ) = − sin(2 x ) where 2 x ∈[ −2π , 2π ] cos(2 x ) − sin(2 x ) = cos(2 x ) cos(2 x ) 1 = − tan(2 x ) −1 = tan(2 x ) π 1 suggests . Since tan is negative 2nd and 4th quadrants. 4 π π π π 2 x = −π − , − , π − and 2π − 4 4 4 4 5π π 3π 7π 2x = − , − , and 4 4 4 4 7π 5π π 3π and x =− ,− , 8 8 8 8

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 5 Differentiation • EXERCISE 5.6   |  103

π π ≤x≤ . 2 2

d When t = 7.30 am then 1.5 hours. dH π 2 2π π π  π 3 π = cos  ×  = cos   = × = x = −0.524, 0.524  4  12 2  6 2  12 dt 12 24 Points where the gradient is zero are (−0.524, 0.342) and (0.524, −0.342) π πt 2π e cos   = (−0.524, 0.342) and (0.524, −0.342)  6 12 24 π 16 On CAS, solve f ′( x ) = 0, 0 ≤ x ≤ . πt 2π 12 2 2 cos   = × =   6 24 π 2 x = 0.243, 0.804 2 π Points where the gradient is zero are (0.243,1.232) and (0.804, 0.863) suggests . Since cos is positive then 1st and 4th 4 2 (0.243,1.232) and (0.804, 0.863) quadrants. πt π π = , 2π − Exercise 5.6 — Applications of trigonometric 6 4 4 functions π t π 7π = , 1 6 4 4 1 a A = ab sin(c) 2 π 6 7π 6 t= × , × 1 4 π 4 π A = (6)(7) sin(θ ) 2 3 21 t= , A = 21sin(θ ) as required 2 2 dA The second time when t = 10.5 hours or at 4.30 pm. b = 21cos(θ ) dθ 1 5 a 2 sin(4 x ) + 1 = 1 π π  2 c When θ = ; A = 21cos   = 21 × = 10.5  3 1 3 2 2 sin(4 x ) = − Rate of change is 10.5 cm2/radian 2 1 2 a BD = a sin(θ ) and CD = a cos(θ ) sin(4 x ) = − b Length of sleepers required is 4 = 2a sin(θ ) + 2 + a + 2 + a cos(θ ) x  0.849, 1.508 = 2a sin(θ ) + a cos(θ ) + a + 4 b Max/min values occur when f ′( x ) = 0 dL f ′( x ) = 8 cos(4 x ) c = 2a cos(θ ) − a sin(θ ) dθ 0 = 8 cos(4 x ) d a = 2 ⇒ L = 4 sin(θ ) + 2 cos (θ ) + 6 0 = cos(4 x ) Using CAS: π 3π 0 suggests , dL 2 2 = 0,θ = 1.1c dθ π 3π 4x = , 3 a L (t ) = 2 sin (π t ) + 10 2 2 π 3π When t = 0; L (0) = 2 sin(0) + 10 = 10 x= , 8 8 dL π π π b = 2π cos (π t ) Maximum when x = , f   = 2 sin   + 1 = 2 + 1 = 3 dt  2 8  8 dL Minimum when c When t = 1 second then = 2π cos(π ) = −2π cm/s. dt 3π  3π   3π  x= , f   = 2 sin   + 1 = −2 + 1 = −1 π  8   2  8 4 a Period of function is 2π ÷ = 12 hours 6  π ,3 ,  3π , −1 Coordinates are:  πt   8   8 b Low tide occurs when sin   = −1 so  6 π π c f ′   = 8 cos  4 ×  = −8 H LOWTIDE = 1.5 + 0.5(−1) = 1 m.  4  4 πt  3π π  π  1.5 + 0.5sin   = 1 d Rate of change is positive for  0,  ∪  ,  .  6  8   8 2  πt 100 cos (θ) 6 a 0.5sin   = −0.5  6 πt sin   = −1  6 100 cm π 100 sin (θ) 100 sin (θ) 1 suggests . Since sin is negative 3rd quadrant. 2 πt π 100 cos (θ) =π + θ 6 2 π t 3π θ = 6 2 3π 6 t= × = 9 or 3 pm 100 sin (θ) 100 sin (θ) 2 π 100 cm Low tide = 1 metre at 3 pm dH π 1 πt π πt c = × cos   = cos    6  12  6 dt 6 2 100 cos (θ) 15 On CAS, solve f ′( x ) = 0, −

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

104 | TOPIC 5 Differentiation • EXERCISE 5.6 L = 3 × 100 cos(θ ) + 4 × 100 sin(θ ) + 2 × 100 L = 300 cos(θ ) + 400 sin(θ ) + 200 as required b

dL = −300 sin(θ ) + 400 cos(θ ) dθ

c Maximum length occurs when

8 a

π –θ – 2

dL = 0. dθ

M

−300 sin(θ ) + 400 cos(θ ) = 0 400 cos(θ ) = 300 sin(θ ) 400 = 300 tan(θ ) 4 = tan(θ ) 3  4 tan −1   = θ  3 Lmax Lmax

200 m

O

(

θ

N

)

Distance ÷ Time = Velocity Distance = Velocity × Time Distance ÷ Velocity = Time So d ( PM ) = 400 cos(θ ). 400 cos(θ ) Tobstacles = 2 Tobstacles = 200 cos(θ ) d ( PM ) = 200 × 2θ d ( PM ) = 400θ 400θ = 80θ Thurdles = 5 Ttotal = Tobstacles + Thurdles Ttotal = 200 cos(θ ) + 80θ Ttotal = 40(5 cos(θ ) + 2θ ) as required

Therefore the maximum length is 700 cm when θ = 0.93c. Z

θ – 2 h O r

π –θ – 2 200 m

θ

π –θ π–2 – 2 = π – π + 2θ = 2θ

θ = 0.93c = 300 cos(0.9273) + 400 sin(0.9273) + 200 = 700 cm

7 a

P

θ h–r

T = 40(5cos(θ ) + 2θ ) dT = 40(−5sin(θ ) + 2) dθ 0 = 40(−5sin(θ ) + 2) 0 = −5sin(θ ) + 2 5sin(θ ) = 2 2 sin(θ ) = 5 2 π θ = sin −1   0 < θ <  5 2 θ = 0.4115 c Tmax = 40(5 cos(0.4115) + 2(0.4115)) Tmax = 216.2244 seconds Tmax = 3 mins 36 seconds b

X

M

Y

∠XOY = 2θ because the angle at the centre of the circle is twice the angle at the circumference. 1 ∠XOM = ∠YOM = × 2θ 2 ∠XOM = θ as required XM = r sin(θ ) XM = tan(θ ) h−r r sin(θ ) sin(θ ) = cos(θ ) h−r 1 r = h − r cos(θ ) h−r = cos(θ ) r h − 1 = cos(θ ) r h = cos(θ ) + 1 r h c r = 3 cm, = cos(θ ) + 1 3 h = 3 cos(θ ) + 3 dh = −3sin(θ ) dθ b

d When θ =

−3 3 π dh π . , = −3sin   =  6 2 6 dθ

9 a h = a cos(nt ) + c Amplitude = 50 so a = 50 2π and n = 2π Period = 1 second so 1 = n Vertical translation is 50 so c = 50 Thus h = 50 cos(2π t ) + 50 dh b = −100π sin(2π t ) dt c When t = 0.25 seconds dh = −100π sin(2π × 0.25) = −100π mm/sec dt πt 10 a h = 5 − 3.5 cos    30  When t = 0; h = 5 − 3.5 cos(0) = 5 − 3.5 = 1.5 m

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 5 Differentiation • EXERCISE 5.6   |  105 b hmax = 5 − 3.5(−1) = 8.5 m 2π c period = π = 60 s 30 Therefore 1 rotation takes 60 seconds πt d Solve 5 − 3.5 cos   > 7 for 0 ≤ t ≤ 60.  30  t = 20.808,39.192 Time spent above 7 m = 39.192 - 20.808 = 18.4 seconds e

dh 3.5π πt = sin    30  dt 30 dh 7π πt = sin    30  dt 60

dh = −0.2 m/s dt 7π πt −0.2 = sin    30  60 πt −0.2 × 60 = sin    30  7π πt −0.5456 = sin    30  0.5456 suggests 0.5772. Since sin is negative 3rd and 4th quadrants. πt = π + 0.5772, 2π − 0.5772 30 πt = 3.7188, 5.7060 30 30 30 t = 3.7188 × , 5.7060 × π π t = 35.51 s, 54.49 seconds 7 πx 5 0 ≤ x ≤ 20 y = cos  + 11 a  20  2 2 7 5 ymax = × 1 + = 6 m 2 2 f

dy 7π πx =− sin   20  dx 40 7π π dy c i When x = 5; =− sin   = −0.3888  4 40 dx 7π π dy ii When x = 10; =− sin   = −0.5498  2 40 dx b

d i When y = 0 then 7 πx 5 cos   + = 0  20  2 2 πx 7 cos   + 5 = 0  20  πx 7 cos   = −5  20  5 πx cos   = −  20  7 πx 5 = cos −1  −   7 20 πx = 2.3664 20 2.3664 × 20 x= π x = 15 m

ii When x = 15.0649 then dy 7π  π × 15.0649  =− sin    40 20 dx dy = −0.3786 dx If θ is the required angle, then tan(θ ) = −0.3786

θ = tan −1 (−0.3786) θ = 180° − 20.7343° θ = 159.27° 7x 12 a h( x ) = 10 cos   − 5 x + 90 0 ≤ x ≤ 4.5  2 When x = 0; h(0) = 10 cos ( 0 ) − 5(0) + 90 = 100 cm When 7 × 4.5  x = 4.5; h(4.5) = 10 cos  − 5(4.5) + 90 = 57.5 cm  2  Therefore coordinates are (0, 100) and (4.5, 57.5) 7x b h′( x ) = −35sin   − 5  2 Min value occurs when h′( x ) = 0. 7x −35sin   − 5 = 0  2 7x 1 sin   = −  2 7 7x  1 = sin −1  −   7 2 1 suggests 0.1433. Since sin is negative 3rd quadrant. 7 7x = π + 0.1433 2 7x = 3.2849 2 x  0.94  7 × 0.9386  x = 0.9386, h(0.9386) = 10 cos  − 5(0.9386) + 90   2 h(0.9386) = 75.41 cm Co-ordinates are (0.94, 75.41) c When x = 0.4 then  7 × 0.4  h′(0.4) = −35sin  −5  2  h′(0.4) = −39.5 13 a P = −2 cos(mt ) + n When t = 0, P = 4 4 = −2 cos(0) + n 4+2=n n=6 Period: 3 2π = 2 m 3m = 4π 4π m= 3 4π t  b P = −2 cos  +6  3  dP 8π  4π t  = sin   3  dt 3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

106 | TOPIC 5 Differentiation • EXERCISE 5.6 c When t = 0.375 then 8π π dP 8π  4π × 0.375  8π = sin  sin   = m/min  =  2 3  3 3 3 dt  x 14 a h( x ) = 2.5 − 2.5 cos   − 5 ≤ x ≤ 5  4  5 h(5) = 2.5 − 2.5 cos    4 h(5) = 1.7117 Maximum depth is 1.7 metres. b

b i T (t ) = x (t ) + y(t )

πt πt T (t ) = 1.5sin   + 1.5 + 2 − 2 cos    3  3 πt πt T (t ) = 3.5 + 1.5sin   − 2 cos    3  3 T(t) 6

dh 2.5  x  = sin    4 dx 4 dh x = 0.625sin    4 dx

c When x = 3 then dh  3 = 0.625sin    4 dx dh = 0.426 dx dh d When = 0.58 then dx x 0.58 = 0.625sin    4 −5 x 5 x 0.928 = sin   ≤ ≤  4 4 4 4 0.928 suggests 1,1890. Since sin is positive 1st quadrant because of the domain. x = 1.1890 4 x = 4.756 metres πt 15 a x (t ) = 1.5sin   + 1.5 0 ≤ t ≤ 12  3 πt y(t ) = 2.0 − 2.0 cos   0 ≤ t ≤ 12  3 Solve x (t ) = y(t ) on CAS. First time emissions are equal is at t = 1.9222 or 1 hour and 55 minutes after 6 am. So at 7.55 am the emissions are both 2.86 units.

( )

(12, 1.5)

(0, 1.5) 0

c 16 a

b

c

d

( )

πt πt T(t) = 1.5 sin — + 3.5 – 2.0 cos — 3 3

6

12

t

ii Maximum emission of 6 units at t = 2.3855 or 2 hours and 23 minutes after 6 am which is 8.23 am, and at t = 8.3855 or 8 hours and 23 minutes after 6 am which is 2.23 pm. Minimum emission of 1 unit at t = 5.3855 or 5 hours and 23 minutes after 6 am which is 11.23 am, and again at t = 11.3855 or 11 hours and 23 minutes after 6 am which is 5.23 pm. As emissions range is 1 - 6 units they lie within the required range. πt πt N (t ) = 45sin   − 35 cos   + 68 0 ≤ t ≤ 15  5  3  π (0)   π (0)  N (0) = 45sin  − 35 cos  + 68  5   3  N (0) = 33 Sketch the graph on CAS and find the minimum. Coordinate is (6.4643, 1.2529) Quietest time when t = 6.4643 or 6 hours and 28 minutes after 8 am which is 2.28 pm. When t = 4 (midday)  π (4)   π (4)  N (0) = 45sin  − 35 cos  + 68  5   3  N (0) = 112 At midday there were 112 customers in line. Maximum number of customers in the queue between 3 pm (t = 7) and 7 pm (t = 11) is 86.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 6 Further differentiation and applications • EXERCISE 6.2   |  107

Topic 6 — Further differentiation and applications Exercise 6.2 — The chain rule

(

(

−1

) 2=

dy 1 = (2 x − 7) x 2 − 7 x + 1 dx 2 2

1

)2

1 a y = x 2 − 7 x + 1 = x 2 − 7 x + 1

2x − 7 2 x 2 − 7x + 1

3

b y = (3 x + 2 x − 1) dy = 3(6 x + 2)(3 x 2 + 2 x − 1)2 = 6(3 x + 1)(3 x 2 + 2 x − 1)2 dx 2 2 a y = sin ( x ) = ( sin( x ) ) dy = 2 cos( x ) sin( x ) dx b y = e cos(3 x ) dy = −3sin(3 x )e cos(3 x ) dx 3 3 y = sin 3 ( x ) = ( sin( x ) ) dy = 3 cos( x ) sin 2 ( x ) dx

2

2

When x =

π dy π π 9 1  3 = 3 cos   sin 2   = 3 × ×  , =  3  3 3 dx 8 2  2 

2

4 y = esin ( x ) 2 dy = 2 cos( x ) sin( x )esin ( x ) dx  π

 2 2 

2 π dy π π 2 2    = 2 cos   sin   esin 4 = 2 × × e When x = ,     2 dx 4 4 2 2 1 5 y = = (2 x − 1)−2 (2 x − 1)2 dy 4 = −2(2)(2 x − 1)−3 = − dx (2 x − 1)3 4 dy When x = 1, =− = −4 dx (2 − 1)3 1 When x = 1, y = =1 (2 − 1)3

2

1

= e2 = e

Equation of tangent with mT = −4, which passes through the point ( x1 , y1 ) ≡ (1, 1), is given by y − y1 = mT ( x − x1 ) y − 1 = −4 ( x − 1) y − 1 = −4 x + 4 y = −4 x + 5

6 f ( x ) = ( x − 1)3 and g( x ) = e x a f ( g( x )) = f (e x ) = (e x − 1)3 b h( x ) = (e x − 1)3 and h ′( x ) = 3e x (e x − 1)2 c At (0,0), h ′(0) = 3e 0 (e 0 − 1)2 = 0 Equation of tangent is y = 0.

(

)

2 7 a y = g( x ) = 3 x + 1

−1

du = 2x dx dy 3 so = −3u −2 = − 2 du u

Let u = x 2 + 1 so Let y = 3u −1

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

108 | TOPIC 6 Further differentiation and applications • EXERCISE 6.2 dy dy du = × dx du dx 3 6x dy = − 2 × 2x = − 2 dx u x +1

(

)

dy dy du = × dx du dx 2 dy = 5eu × 6 x = 30 xe3 x −1 dx −2 2  h y = f ( x ) =  x 3 − 2  = x 3 − 2 x −2  x 

2

(

b y = g( x ) = e cos( x ) Let u = cos( x ) so Let y = eu so

du = − sin( x ) dx

dy dy du = × dx du dx dy = eu × − sin( x ) = − sin( x )e cos( x ) dx

(

c y = g( x ) = ( x + 1)2 + 2 = x 2 + 2 x + 3

du 4  = 3 x 2 + 4 x −3 =  3 x 2 + 3   dx x  2 dy Let y = u −2 so = −2u −3 = − 3 du u dy dy du = × dx du dx dy 2  4 = − 3 ×  3x 2 + 3  dx u  x   3x 5 + 4  2 =− × 3  x 3   3 2  x − 2  x 6x5 + 8 =− 3 2  x3  x3 − 2   x 

1

)2

du = 2x + 2 dx 1 1 1 dy 1 − 2 Let y = u 2 so = u = du 2 2 u dy dy du = × dx du dx 1 dy x +1 = × 2 ( x + 1) = 2 dx 2 u x + 2x + 3 Let u = x 2 + 2 x + 3 so

i y = f ( x ) =

1 −2 = ( sin( x ) ) sin 2 ( x ) du Let u = sin( x ) so = cos( x ) dx 2 dy Let y = u −2 so = −2u −3 = − 3 du u dy dy du = × dx du dx 2 2 cos( x ) dy = − 3 × cos( x ) = − dx sin 3 ( x ) u

d y = g( x ) =

Let y = u

(

1

du = 2x − 4 dx 1

dy 1 − 2 1 = u = du 2 2 u

dy dy du = × dx du dx 1 dy = × 2 ( x − 2) = dx 2 u

(

2

x−2 2

)

1 2− x 1 = = ( 2 − x )− 2 2− x 2− x

Let u = 2 − x so

e y = f ( x ) = x 2 − 4 x + 5 = x 2 − 4 x + 5

Let y = u 2 so

x − 4x + 5

1

)2

−1 2

du = −1 dx 3

dy 1 − 1 =− u 2 =− 3 du 2 2u 2

so

dy dy du = × dx du dx 1 1 dy = − 3 × −1 = 3 dx 2(2 − x )2 2u 2 3 j y = f ( x ) = cos3 ( 2 x + 1) = ( cos(2 x + 1) du Let u = cos(2 x + 1) so = −2 sin(2 x + 1) dx dy Let y = u 3 so = 3u 2 du dy dy du = × dx du dx dy = 3u 2 × −2 sin(2 x + 1) = −6 sin(2 x + 1) cos 2 (2 x + 1) dx 8 a f ( x ) = tan(4 x + π ) 4 cos 2 (4 x + π ) 4 4 π = =4 f ′  =  4  cos 2 (2π ) 2 π cos  4 +π   4  f ′( x ) =

f y = f ( x ) = 3 cos x − 1

du = 2x dx dy Let y = 3 cos(u) so = −3sin(u) du dy dy du = × dx du dx dy = −3sin(u) × 2 x = −6 x sin x 2 − 1 dx 2 g y = f ( x ) = 5e3 x −1 du = 6x Let u = 3 x 2 − 1 so dx dy Let y = 5eu so = 5eu du Let u = x 2 − 1 so

(

−2

Let u = x 3 − 2 x −2 so

dy = eu du

Let u = x 2 − 4 x + 5 so

)

b

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

)

2

( 2 − x )3

27 16 1 2 =2÷ = f ′   =  2  8 27 13 − 2   2 c

f ( x ) = e2 x

2

f ′( x ) = 4 xe 2 x

2

f ′(−1) = 4(−1)e 2( −1) = −4e 2 2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 6 Further differentiation and applications • EXERCISE 6.2   |  109 4

( ) = (3x 2 − 2) 3 1 4 f ′( x ) = ( 3 x 2 − 2 ) 3 × 6 x = 8 x 3 3 x 2 − 2 3 4

d f ( x ) = 3 3 x 2 − 2

f ′ (1) = 8(1) 3 3(1)2 − 2 = 8 e

f ( x ) = ( cos(3 x ) − 1)

5

f ′( x ) = 5 × −3sin(3 x ) ( cos(3 x ) − 1) = −15sin(3 x ) ( cos(3 x ) − 1) 4

4

4

π 3π  3π  f ′   = −15sin    cos   − 1 = −15 ( −1) −14 = 15  2  2   2   1 1 9     f ( x ) = 2 so g( x ) = f ( f ( x )) = = x4 2 x  1  2  x

( )

g ′( x ) = 4 x 3 f ( x ) = sin 2 ( 2 x ) = ( sin(2 x ) )

2

10

f ′( x ) = 2 cos ( 2 x ) sin ( 2 x ) , 0 = 2 cos ( 2 x ) sin ( 2 x ) , cos ( 2 x ) = 0 or sin ( 2 x ) = 0 π 3π 2x = , 2 2 π 3π x= , 4 4

0≤ x ≤π 0 ≤ 2 x ≤ 2π 2 x = 0, π , 2π x = 0,

π ,π 2

f ( 0 ) = sin 2 ( 2(0) ) = 0 π  π  f   = sin 2  2    = 1  4   4  π  π  f   = sin 2  2    = 0  2   2 3π  3π  f   = sin 2  2    = 1  4    4  f (π ) = sin 2 ( 2 (π )) = 0

π π 3π  Therefore coordinates are: ( 0, 0 ) ,  , 1 ,  , 0  ,  , 1 , π , 0)  4   2   4  ( 11 z = 4 y 2 − 5 and y = sin(3 x ) z = 4 ( sin(3 x ) ) − 5 dz = 4 ( 2 )( 3cos(3 x ) )( sin(3 x ) ) = 24 cos(3 x )sin(3 x ) dx 12 a  f ( x ) = g [ cos( x ) ] f ′( x ) = − sin( x ) g ′ [ cos( x ) ] 2

( ) f ′( x ) = 6 x 2 g ′ ( 2 x 3 )  f ( x ) = g ( 3e 2 x +1 ) f ′( x ) = 6e 2 x +1 g ′ ( 3e 2 x +1 )

b  f ( x ) = g 2 x 3 c

d f ( x ) = g

(

)

1   2x 2 − x = g  2x 2 − x 2   

(

1 f ′( x ) = ( 4 x − 1) 2 x 2 − x 2 13 a f ( x ) = [ h( x ) ]

−2

(

)

) 2 g′  ( 2 x 2 − x ) 2  = 2 42xx−2 1− x g′ ( −1



1

2x 2 − x

)

so f ′( x ) = −2h ′( x ) [ h( x ) ]

−3

b f ( x ) = sin 2 [ h( x ) ] so f ′( x ) = 2h ′( x ) sin [ h( x ) ] c f ( x ) = 3 2h( x ) + 3 so f ′( x ) =

2 1 ( 2h′( x ))( 2h( x ) + 3)− 3 = 3

d f ( x ) = −2e h ( x ) + 4 so f ′( x ) = −2h ′( x )e h ( x ) + 4

2h′ ( x ) 2

3 ( 2h( x ) + 3) 3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

110 | TOPIC 6 Further differentiation and applications • EXERCISE 6.2 14 a g( x ) = f (h( x )) = f (2 x − 1) = 3 (2 x − 1)2

c

2 3

1 2 4 b g( x ) = ( 2 x − 1) so g′( x ) = (2) ( 2 x − 1)− 3 = 3 3 3 2x − 1 4 4 c mT at x = 1: g ′(1) = 3 = 3 2(1) − 1 3 4 Equation of tangent with mT = which passes through 3 ( x1 , y1 ) = (1,1) is given by

y − y1 = mT ( x − x1 ) 4 y − 1 = ( x − 1) 3 4 4 y = x − +1 3 3 1 4 y= x− 3 3 4 4 =− 3 3 3 2(0) − 1 4 Equation of tangent with mT = − which passes through 3 ( x1 , y1 ) = (0,1) is given by y − y1 = mT ( x − x1 ) 4 y − 1 = − ( x − 0) 3 4 y = − x +1 3 d Tangents intersect where 4 1 y = x − ................(1) 3 3 4 y = − x + 1...............(2) 3 (1) = (2) 4 1 4 x − = − x +1 3 3 3 8 1 x = 1+ 3 3 8 4 x= 3 3 4 3 1 x= × = 3 8 2 4  1 1 2 1 1 1 When x = , y =   − = − = 3  2 3 3 3 3 2 1 1 Tangents intersect at  ,   2 3 mT at x = 0 : g ′(0) =

15 a

d (h( g( x )) = dx d (h( g( x )) = dx d (h( g( x )) = dx d (h( g( x )) = dx

)

(

)

(

−1

)2

x 2 − 6x − 7

d When x = −2, gradient −5 5 −2 − 3 =− = = 3 4 + 13 − 7 (−2)2 − 6(−2) − 7 1 1  1 1 = − x2 16 y = g ( f ( x ) ) = g   = −  x x  12 x   x dy 1 −2 = − x − 2x = − 2 − 2x dx x Perpendicular equation is given by y = − x + a so mP = −1 and mT = 1. dy =1 dx 1 − 2 − 2x = 1 x −1 − 2 x 3 = x 2 0 = 2x3 + x 2 + 1 Let P ( x ) = 2 x 3 + x 2 + 1 P (−1) = 2 ( −1)3 + ( −1)2 + 1 = 0 ( x + 1) is a factor

(

)

2 x 3 + x 2 + 1 = ( x + 1) 2 x 2 − x + 1 Quadratic can’t be factorised, x +1= 0 x = −1 If x = −1, y =

1 − ( −1)2 = −2 and −2 = 1 + a ⇒ a = −3 −1

17 f ( x ) = 2 sin( x ) and h( x ) = e x

( )

( )

a i m( x ) = f (h( x )) = f e x = 2 sin e x

ii n( x ) = h( f ( x )) = h ( 2 sin( x ) ) = e 2sin( x )

( )

b m ′( x ) = 2e x cos e x and n ′( x ) = 2 cos( x )e 2sin( x ) Solve using CAS for 0 ≤ x ≤ 3 m ′( x ) = n ′( x )

( ) cos ( e x ) = cos( x )e 2sin( x )

2e x cos e x = 2 cos( x )e 2sin( x )

h( x ) = x 2 − 16 and g( x ) = x − 3

e

2

h( g( x )) = ( x − 3) − 16

x

x = 1.555, 2.105, 2.372

2

h( g( x )) = x − 6 x + 9 − 16

18 a m(n( x )) = m( x 2 + 4 x − 5) = 3 x

h( g( x )) = x 2 − 6 x − 7 h( g( x )) = ( x − 7)( x + 1)

b

If h( g( x )) = ( x + m)( x + n) then m = −7 and n = 1 b Maximum domain for ( x − 7)( x + 1) ≥ 0

–1

(

d x 2 − 6x − 7 dx 1 d 2 x − 6x − 7 2 dx 1 (2 x − 6) x 2 − 6 x − 7 2 x−3

(

2

+ 4 x −5

)

2 d x 2 + 4 x −5 3 = 1.0986(2 x + 4)3 x + 4 x − 5 dx 2 = 2.1972( x + 2)3 x + 4 x − 5

When x = 1,

7

x

(

)

2 d x 2 + 4 x −5 3 6.5916 = 2.1972(1 + 2)3(1) + 4(1) − 5 dx = 2.1972 × 3 × 30 = 6.5916

{ x : x ≤ −1} ∪ { x : x ≥ 7}

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 6 Further differentiation and applications • EXERCISE 6.3   |  111

Exercise 6.3 — The product rule 1 a f ( x ) = sin(3 x ) cos(3 x ) f ′( x ) = −3sin(3 x ) sin(3 x ) + 3 cos(3 x ) cos(3 x ) f ′( x ) = 3 cos 2 (3 x ) − 3sin 2 (3 x ) b f ( x ) = x 2 e3 x f ′( x ) = 3 x 2 e3 x + 2 xe3 x c f ( x ) = ( x 2 + 3 x − 5)e5 x f ′( x ) = 5( x 2 + 3 x − 5)e5 x + (2 x + 3)e5 x f ′( x ) = (5 x 2 + 17 x − 22)e5 x 2 f ( x ) = 2 x 4 cos(2 x ) f ′( x ) = −4 x 4 sin(2 x ) + 8 x 3 cos(2 x )

π π 3 π π 4 π f ′   = 8   cos  2 ×  − 4   sin  2 ×   2  2   2  2 2 8π 3 (−1) = 8 3 = −π 3    f ( x ) = ( x + 1) sin( x ) f ′( x ) = ( x + 1) cos( x ) + sin( x ) × 1 f ′(0) = sin(0) + cos(0) = 0 +1 =1 4 y = ( x 2 + 1)e3 x dy mT = = 3( x 2 + 1)e3 x + 2 xe3 x dx When x = 0, mT = 3(0 + 1)e3(0) + 2(0)e3(0) = 3 When x = 0, y = (0 + 1)e3(0) = 1 Equation of tangent with mT = 3 which passes through ( x1 , y1 ) = (0,1) is given by y − y1 = mT ( x − x1 ) y − 1 = 3( x − 0) y = 3x + 1 5 Let y = f ( x ) = 2 x 2 (1 − x )3 f ′( x ) = 2 x 2 × −3 (1 − x )2 + (1 − x )3 × 4 x = −6 x 2 (1 − x )2 + 4 x (1 − x )3 = −2 x (1 − x )2 (3 x − 2(1 − x )) = −2 x (1 − x )2 (5 x − 2) If f ′ ( x ) = 0 −2 x (1 − x )2 (5 x − 2) = 0 x = 0 or 1 − x = 0 or 5 x − 2 = 0 2 x = 0,1, 5 f ′(0) = 2 ( 0 )2 (1 − 0 )3 = 0 f ′ (1) = 2 (1)2 (1 − 1)3 = 0 2 2 3  2  2  f ′   = 2  1 −   5  5  5 4 27 = 2× × 25 125 216 = 3125

 2 216  Therefore the coordinates are: (0, 0),(1, 0),  ,  5 3125 

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

112 | TOPIC 6 Further differentiation and applications • EXERCISE 6.3 d y = x cos( x )

−x

6 a f ( x ) = e 2 sin( x ) f ( x ) = 0 for x ∈[ 0,3π ] e

−x 2

sin( x ) = 0

Let u = x and v = cos( x ) so

−x e 2

sin( x ) = 0 since > 0 for all x x = 0, π , 2π , 3π b Max/min values occur when f ′( x ) = 0. −x 1 −x f ′( x ) = e 2 cos( x ) − e 2 sin( x ) 2 −x  1  2 0 = e  cos( x ) − sin( x )   2 −x 2

1 − sin( x ) + cos( x ) = 0 since e > 0 for all x 2 1 cos( x ) = sin( x ) 2 1 1 = tan( x ) 2 2 = tan( x ) x = 1.11, 4.25, 7.39 7 a y = x 2 e5 x Let u = x 2 and v = e5 x so dy dv du =u +v dx dx dx dy 2 5x = 5 x e + 2 xe5 x dx b y = e 2 x +1 tan(2 x )

du dv = 2 x and = 5e5 x dx dx

Let u = e 2 x +1 and v = tan(2 x ) so du dv 2 = 2e 2 x +1 and = dx dx cos 2 (2 x ) dy dv du =u +v dx dx dx dy 2e 2 x +1 = + 2e 2 x +1 tan(2 x ) dx cos 2 (2 x ) 2e 2 x +1 2e 2 x +1 sin(2 x ) dy = + 2 cos(2 x ) dx cos (2 x ) dy 2e 2 x +1 + 2e 2 x +1 sin(2 x ) cos(2 x ) = dx cos 2 (2 x ) 2 x +1 dy 2e (1 + sin(2 x ) cos(2 x )) = dx cos 2 (2 x ) c y = x −2 (2 x + 1)3 −2

3

Let u = x and v = (2 x + 1) du dv so = −2 x −3 and = 3(2)(2 x + 1)2 = 6(2 x + 1)2 dx dx dy dv du =u +v dx dx dx dy = 6 x −2 (2 x + 1)2 − 2 x −3 (2 x + 1)3 dx dy 6(2 x + 1)2 2(2 x + 1)3 = − dx x2 x3 2 dy 6 x (2 x + 1) − 2(2 x + 1)3 = dx x3 2 dy 2(2 x + 1) ( 3 x − (2 x − 1) ) = dx x3 2 dy 2(2 x + 1) ( x − 1) = dx x3

dy dv du =u +v dx dx dx dy = − x sin( x ) + cos( x ) dx

du dv = 1 and = − sin( x ) dx dx

1

e y = 2 x ( 4 − x ) = 2 x 2 ( 4 − x ) 1

Let u = 2 x 2 and v = 4 − x so

dv du = x and = −1 dx dx

dy dv du =u +v dx dx dx dy 4− x = 2 x (−1) + dx x dy −2 x + 4 − x = x dx dy 4 − 3 x = dx x f y = sin ( 2 x − π ) e −3 x Let u = sin ( 2 x − π ) and v = e −3 x so du dv = 2 cos ( 2 x − π ) and = −3e −3 x dx dx dy dv du =u +v dx dx dx dy = −3e −3 x sin ( 2 x − π ) + 2e −3 x cos ( 2 x − π ) dx 2 g y = 3 x −2 e x 2 2 du dv Let u = 3 x −2 and v = e x so = −6 x −3 and = 2 xe x dx dx dy dv du =u +v dx dx dx 2 2 dy = 3 x −2 × 2 xe x + e x × −6 x −3 dx 2 2 6e x dy 6e x = − 3 dx x x 2 x2 dy 6e x − 1 = dx x3

(

)

(

1

)2

h y = e 2 x 4 x 2 − 1 = e 2 x 4 x 2 − 1

(

1

) 2 so

Let u = e 2 x and v = 4 x 2 − 1

(

1

) 2=

du dv = 2e 2 x and = 4x 4x2 − 1 dx dx

−

dy dv du =u +v dx dx dx dy 4 xe 2 x + 2e 2 x 4 x 2 − 1 = dx 4x2 − 1 2x 2x 2 dy 4 xe + 2e 4 x − 1 = dx 4x2 − 1 2x 2 dy 2e 4 x + 2 x − 1 = dx 4x2 − 1

(

(

)

)

i y = x 2 sin 3 (2 x ) = x 2 ( sin(2 x ) )

3

Let u = x 2 and v = ( sin(2 x ) ) so du dv = 2 x and = 6 cos(2 x ) sin 2 (2 x ) dx dx 3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

4x 4x2 − 1

TOPIC 6 Further differentiation and applications • EXERCISE 6.3   |  113 c f ( x ) = (1 − x ) tan 2 ( x )

dy dv du =u +v dx dx dx dy 2 = 6 x cos(2 x ) sin 2 (2 x ) + 2 x sin 3 (2 x ) dx dy = 2 x sin 2 (2 x ) ( 3 x cos(2 x ) + sin(2 x ) ) dx j y = ( x − 1)4 ( 3 − x )−2 Let u = ( x − 1)4 and v = ( 3 − x )−2 so du dv 2 = 4 ( x − 1)3 and = −2 ( 3 − x )−3 × −1 = dx dx ( 3 − x )3 dy dv du =u +v dx dx dx 2 ( x − 1)4 4 ( x − 1)3 = + ( 3 − x )3 ( 3 − x )2 2 ( x − 1)4 + (3 − x )4 ( x − 1)3 = (3 − x )3 3 2 ( x − 1) ( x − 1 + 2(3 − x ) ) = ( 3 − x )3 3 2 ( x − 1) ( 5 − x ) = ( 3 − x )3 2 ( x − 1)3 ( x − 5) = ( x − 3)3 k y = ( 3 x − 2 )2 g( x ) Let u = ( 3 x − 2 )2 and v = g( x ) so du dv = 6 ( 3 x − 2 ) and = g′( x ) dx dx dy dv du =u +v dx dx dx dy 2 = ( 3 x − 2 ) g′( x ) + 6 ( 3 x − 2 ) g( x ) dx dy = ( 3 x − 2 ) (( 3 x − 2 ) g ′ ( x ) + 6 g( x ) ) dx l y = − e5 x g x

( )

Let u = − e5 x and v = g

dv g′( x ) = − 5e5 x and = ( x ) so du dx 2 x dx

( ) ( ))

f ′(−1) = − e −1 + e −1 =0

(

2 b f ( x ) = x x + x

)

4

(

) + (x2 + x) 3 = ( x 2 + x ) ( x 2 + x + 8x 2 + 4 x ) 3 = ( x 2 + x ) (9 x 2 + 5x ) 3 f ′(1) = (12 + 1) ( 9(1)2 + 5(1) )

f ′( x ) = +4 x ( 2 x + 1) x 2 + x

= 112

3

( )

d f ( x ) = x sin 2 (2 x 2 ) = x

4

1 2

(sin(2 x 2 ))

2

sin(2 x 2 ) 2 x 8 x 2 cos(2 x 2 ) sin(2 x 2 ) + sin(2 x 2 ) = 2 x 8π cos(2π ) sin(2π ) + sin(2π ) f′ π = 2 π 8π (1)(0) + (0) = 2 π =0 9 y = f ( x ) = x 4 e −3 x du dv Let u = x 4 and v = e −3 x so = 4 x 3 and = −3e −3 x dx dx dy dv du =u +v dx dx dx dy 4 −3 x = −3 x e + 4 x 3e −3 x dx dy = e −3 x −3 x 4 + 4 x 3 = e −3 x (4 x 3 − 3 x 4 ) dx dy If = e −3 x (ax 3 + bx 4 ) then a = 4 and b = −3. dx f ′( x ) = 4 x x cos(2 x 2 ) sin(2 x 2 ) +

( )

(

dy dv du =u +v dx dx dx dy e5 x g ′( x ) =− − 5e5 x g x dx 2 x e5 x g ′( x ) + 10 xg x dy =− dx 2 x 8 a f ( x ) = xe x f ′( x ) = xe x + e x

(

2 (1 − x ) tan( x ) − tan 2 ( x ) cos 2 ( x ) π π 2 ×  1 −  tan     3    π  2 π 3  f ′  = −  tan    2  3   3  1   2 π 2 × 1 −  × 3  2 3 = − 3 1 4 π  = 8 3  1 −  − 3  3 f ′( x ) =

)

10 a y = f ( x ) = ( x − a)2 g( x ) Let u = ( x − a)2 and v = g( x ) so du dv = 2( x − a) and = g′( x ) dx dx dy dv du =u +v dx dx dx dy = ( x − a)2 g ′( x ) + 2( x − a) g( x ) dx b f ( x ) = g( x ) sin(2 x ) where g( x ) = ax 2 dv du Let u = ax 2 and v = sin(2 x ) so = 2ax and = 2 cos(2 x ) dx dx dy dv du =u +v dx dx dx dy 2 = 2ax cos(2 x ) + 2ax sin(2 x ) dx π 2 π dy = 2a   cos(π ) + 2a   sin(π ) = −3π  2  2 dx x = π 2

−

π2 a + 0 = −3π 2 π 2 a = 6π 6 a= π

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

114 | TOPIC 6 Further differentiation and applications • EXERCISE 6.3 11 y = 2 x tan(2 x ) du dv 2 Let u = 2 x and v = tan(2 x ) so = 2 and = dx dx cos 2 (2 x ) dy dv du =u +v dx dx dx 4x dy = + 2 tan(2 x ) dx cos 2 (2 x ) π When x = , 12 π  4  12  dy π = + 2 tan    6 dx 2π cos  6 1 dy π 1 = × +2× dx 3 3 2π cos  6 2 1 dy π = × + dx 3  3  2 3  2  dy π 4 2 3 = × + 3 dx 3 3 dy 4π 2 3 4π + 6 3 = + = dx 9 3 9 x 12 y = xe du dv Let u = x and v = e x so = 1 and = ex dx dx dy dv du =u +v dx dx dx dy x x = xe + e = e x ( x + 1) dx dy 1 When x = 1, mT = = e1 (1 + 1) = 2e and m N = − dx 2e When x = 1, y = (1)e1 = e Equation of tangent with mT = 2e which passes through the point ( x1 , y1 ) = (1, e) is given by y − y1 = mT ( x − x1 ) y − e = 2e( x − 1) y − e = 2ex − 2e y = 2ex − e 1 Equation of perpendicular with mP = − which passes 2e through the point ( x1 , y1 ) = (1, e) is given by y − y1 = mP ( x − x1 ) 1 y − e = − ( x − 1) 2e 1 1 y−e= − x+ 2e 2e 1 1 y= − x+ +e 2e 2e  1 + 2e 2  1 y= − x+ 2e  2e  13 a y = − cos( x ) tan( x ) sin( x ) y = − cos( x ) × cos( x ) y = − sin( x ), cos( x ) ≠ 0 dy = − cos( x ) dx

b y = − cos( x ) tan( x ) Let u = − cos( x ) and v = tan( x ) so du dv 1 = sin( x ) and = sec 2 ( x ) = dx dx cos 2 ( x ) dy dv du =u +v dx dx dx 1 dy = − cos( x ) × + tan( x ) sin( x ) dx cos 2 ( x ) 1 sin( x ) dy =− + × sin( x ) prov cos( x ) ≠ 0 cos( x ) cos( x ) dx dy sin 2 ( x ) − 1 = dx cos( x ) dy 1 − cos 2 ( x ) − 1 = dx cos( x ) dy cos( x ) cos( x ) =− dx cos( x ) dy = − cos( x ), cos( x ) ≠ 0 dx 14 y = e − x (1 − x ) a Graph cuts the y axis where x = 0, y = e 0 (1 − 0) = 1. Graph cuts the x axis where y = 0 2

e − x (1 − x ) = 0 2

1 − x = 0 as e − x > 0 for all x x =1 Therefore, coordinates are: (0, 1) and (1, 0) 2 2 dy b = − e x − 2 xe x (1 − x ) dx 2 = − e x (1 + 2 x (1 − x )) 2

2

= − e x (1 + 2 x − 2 x 2 )

( (

) )

2 dy = ex 2x 2 − 2x − 1 dx 2 0 = ex 2x 2 − 2x − 1

2

0 = 2 x 2 − 2 x − 1 as e x > 0 for all x 2 ± ( −2 )2 − 4(2)(−1) 2(2) 2 ± 12 = 4 x = −0.366, 1.366 x=

− −0.366 2 ) When x = −0.366, y = e ( (1 + 0.366) = 1.1947 − 1.366 2 ) When x = 1.366, y = e ( (1 − 1.366) = −0.057

Therefore coordinates are: (-0.366, 1.195) and (1.366, -0.057) 2 1 c When x = 1, mT = e −(1) 2 (1)2 − 2 (1) − 1 = − e 1 Equation of tangent with mT = − which passes through e ( x1 , y1 ) ≡ (1, 0) is given by

(

)

y − y1 = mT ( x − x1 ) 1 y − 0 = − ( x − 1) e 1 1 y=− x+ e e 2 −( 0 ) 2 ( 0 )2 − 2 ( 0 ) − 1 = −1 so mP = −1 d When x = 0, mT = e Equation of perpendicular with mP = 1 which passes through ( x1 , y1 ) ≡ (0,1) is given by

(

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

)

TOPIC 6 Further differentiation and applications • EXERCISE 6.3   |  115 y − y1 = mP ( x − x1 ) y −1= x y = x +1 e Tangent and perpendicular intersect where 1 1 x +1= − x + e e x = −0.462 ∴ y = −0.462 + 1 = 0.538 POI = (−0.46, 0.54) CD 15 a i = sin(θ ) 3 CD = 3sin(θ ) AD BD = = cos(θ ) ii 3 3 AD = BD = 3 cos(θ ) 1 b S = 4 × × 6 cos(θ ) × 3sin(θ ) + ( 6 cos(θ ) )2 2 S = 36 cos(θ ) sin(θ ) + 36 cos 2 (θ )

(

)

S = 36 cos 2 (θ ) + cos(θ ) sin(θ ) as required c Let S1 = 36 cos (θ ) = 36 ( cos(θ ) ) dS So 1 = 2 × 36 × − sin(θ ) cos(θ ) = −72 sin(θ ) cos(θ ) dθ Let S2 = 36 cos(θ ) sin(θ ) Let u = 36 cos(θ ) and v = sin(θ ) so du dv = −36 sin(θ ) and = cos(θ ) dθ dθ dS2 dv du =u +v dθ dθ dθ dS2 = 36 cos(θ ) cos(θ ) − 36 sin(θ ) sin(θ ) dθ dS2 = 36(cos 2 (θ ) − sin 2 (θ )) dθ S = S1 + S2 dS dS1 dS2 = + dθ dθ dθ dS = −72 sin(θ ) cos(θ ) + 36(cos 2 (θ ) − sin 2 (θ )) dθ dS = −72 sin(θ ) cos(θ ) + 36 cos 2 (θ ) − 36 sin 2 (θ ) dθ dS = −72 sin(θ ) cos(θ ) + 36 cos 2 (θ ) − 36(1 − cos 2 (θ )) dθ dS = −72 sin(θ ) cos(θ ) + 36 cos 2 (θ ) − 36 + 36 cos 2 (θ ) dθ dS = 72 cos 2 (θ ) − 72 sin(θ ) cos(θ ) − 36 dθ 16 a y = f ( x ) = 3 x 3e −2 x du dv Let u = 3 x 3 and v = e −2 x so = 9 x 2 and = −2e −2 x dx dx dy dv du =u +v dx dx dx dy 3 −2 x = −6 x e + 9 x 2 e −2 x dx dy = 3e −2 x (3 x 2 − 2 x 3 ) dx dy If = ae −2 x (bx 2 + cx 3 ) then a = 3, b = 3 and c = −2 dx 2

2

b Stationary points occur where 3e −2 x (3 x 2 − 2 x 3 ) = 0

dy =0 dx

x 2 (3 − 2 x ) = 0 as e −2 x > 0 for all x x = 0 or 3 − 2 x = 0 3 = 2x 3 =x 2 If x = 0, y = 0  3

3 3  3  −2 2  = 81 e −3 = 81 If x = , y = 3   e  2 2 8 8e3 Stationary point (0,0) is a point of inflection and stationary  3 81  point  , 3  is a maximum turning point.  2 8e  3 c When x = 1, y = 3(1)3 e −2(1) = 3e −2 = 2 e 3 dy When x = 1, mT = = 3e −2(1) (3(1)2 − 2(1)3 ) = 3e −2 = 2 dx e 3 Equation of tangent with mT = 2 which passes through e  3 the point ( x1 , y1 ) =  1, 2  is given by  e  y − y1 = mT ( x − x1 ) 3 3 y − 2 = 2 ( x − 1) e e 3 3 3 y− 2 = 2 x− 2 e e e 3 y= 2 x e

17 a When x = −2, y = (4(−2)2 − 5(−2))e −2 = 26e −2  3.5187 so they have made the correct decision. b Graph cuts the x axis where y = 0. (4 x 2 − 5 x )e x = 0 x (4 x − 5) = 0 as e x > 0 for all x x = 0 or 4 x − 5 = 0 4x = 5 5 x= 4 5  T is the point  , 0  4  c y = (4 x 2 − 5 x )e x Let u = 4 x → 2 − 5 x and v = e x so dy dx dy dx dy dx dy dx

=u

du dv = 8 x − 5 and = ex dx dx

dv du +v dx dx

= (4 x 2 − 5 x )e x + (8 x − 5)e x = (4 x 2 − 5 x + 8 x − 5)e x = (4 x 2 + 3 x − 5)e x

Stationary points occur when

dy = 0. dx

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

116 | TOPIC 6 Further differentiation and applications • EXERCISE 6.4 (4 x 2 + 3 x − 5)e x = 0 2

x

4 x + 3 x − 5 = 0 as e > 0 for all x −3 ± (3)2 − 4(4)(−5) 2(4) −3 ± 9 + 80 x= 8 −3 ± 89 x= 8 x=

Point B: When x =

−3 + 89  0.804, 8

y = (4(0.804)2 − 5(0.804))e 0 /804  −3.205 B has the coordinates (0.804, −3.205). 18 y = 2 x sin( x ) dy = 2 x cos( x ) + loge (2) × 2 x sin( x ) dx When x =

π π π dy π π = 2 2 cos   + loge (2) × 2 2 sin    2.06 ,  2  2 2 dx

Exercise 6.4 — The quotient rule 1 a y =

e2 x ex + 1

Let u = e 2 x and v = e x + 1 du dv So = 2e 2 x and = ex dx dx du dv v −u dy dx dx = dx v2 2x x dy 2e (e + 1) − e 2 x × e x = dx (e x + 1)2 3x dy 2e + 2e 2 x − e3 x = dx (e x + 1)2 3x dy e + 2e 2 x = x dx (e + 1)2 cos(3t ) b y = t3 Let u = cos(3t ) and v = t 3 du dv So = −3sin(3t ) and = 3t 2 dt dt du dv v −u dy dt dt = dt v2 dy −3t 3 sin(3t ) − 3t 2 cos(3t ) = dt t6 2 dy −3t (t sin(3t ) + cos(3t )) = dt t6 dy −3(t sin(3t ) + cos(3t )) ,t≠0 = dt t4 x +1 2 y = 2 x −1

du dv −u dx dx v2 2 ( x − 1) − 2 x ( x + 1) = ( x 2 − 1)2 2 x − 1 − 2x 2 − 2x = ( x 2 − 1)2 2 − ( x + 2 x + 1) = ( x 2 − 1)2 − ( x + 1)2 = 2 ( x − 1)2 − ( x + 1)2 = ( x + 1)2 ( x − 1)2 −1 = ( x − 1)2 sin( x ) 3 y = 2 x e Let u = sin( x ) and v = e 2 x du dv So = cos( x ) and = 2e 2 x dx dx du dv v −u dy = dx 2 dx dx v dy e 2 x cos( x ) − 2e 2 x sin( x ) = dx e4 x dy e 2 x (cos( x ) − 2 sin( x )) = dx e4 x dy cos( x ) − 2 sin( x ) = dx e2 x dy cos(0) − 2 sin(0) When x = 0, = =1 dx e 2(0) 5x 4 y = 2 x +4 Let u = 5 x and v = x 2 + 4 du dv So = 5 and = 2x dx dx du dv v −u dy dx dx = dx v2 2 dy 5( x + 4) − 5 x × 2 x = dx ( x 2 + 4)2 2 dy 5 x + 20 − 10 x 2 = dx ( x 2 + 4)2 dy 20 − 5 x 2 = dx ( x 2 + 4)2 dy 5(4 − x 2 ) = dx ( x 2 + 4)2 dy 5 ( 3) 15 3 = = = When x = 1, dx 12 + 4 2 25 5 dy = dx

v

(

5 a y =

Let u = x + 1 and v = x 2 − 1 du dv So = 1 and = 2x dx dx

)

sin( x ) x 1

Let u = sin( x ) and v = x = x 2 so du dv 1 −12 1 = cos( x ) and = x = dx dx 2 2 x

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 6 Further differentiation and applications • EXERCISE 6.4   |  117

dy dx dy dx dy dx dy dx

=

v

du dv −u dx dx v2

e y =

sin( x )   =  x cos( x ) − ÷  2 x  2 x cos( x ) − sin( x ) 1 = × x 2 x 2 x cos( x ) − sin( x ) = 2x x tan(2 x ) b y = ex

dy dx dy dx

Let u = 3 x − 1 and v = 2 x 2 − 3 so

( x )2

Let u = tan(2 x ) and v = e x so

dy = dx dy = dx

du 2 dv = and = ex dx cos 2 ( 2 x ) dx

du dv −u dx dx = v2  2e x  = − e x tan(2 x ) ÷ e 2 x  cos 2 ( 2 x ) 

dy  2e e sin(2 x )  1 = − × dx  cos 2 ( 2 x ) cos(2 x )  e 2 x

( 5 − x ) 2 − 4 ( 5 − x )2 2 5 − x (5 − x )

5 − x − 20 + 4 x 2 5− x 3 x − 15 = 2 5− x 3 (5 − x ) =− 2 5− x 3 5− x =− 2 sin 2 x 2 =

( )

x

(

)

2

Let u = sin( x 2 ) and v = x so du dv = 4 x cos( x )sin( x ) and =1 dx dx dy = dx dy = dx

du dv = 4 x cos( x )sin( x ) and =1 dx dx

du dv −u dx dx v2 4 x 2 cos( x 2 ) sin( x 2 ) − sin 2 x 2 v

x2

( )

2

dy 6 x 2 − 9 − 12 x 2 + 4 x = 2 dx 2x 2 − 3

(

)

2

dy −6 x + 4 x − 9 = 2 dx 2x 2 − 3

)

x − 4x2 2 x

−1 du dv 1 = 1 − 8 x and =x 2= dx x dx du dv v −u dy dx dx = dx v2

dy  −4 ( 5 − x )2 + ( 5 − x )2  =  ÷ (5 − x ) dx  2 5− x 

d y =

( 2 x 2 − 3)

1

du = −2 ( 5 − x ) = 2 x − 10 Let u = ( 5 − x )2 and v = ( 5 − x ) so dx 1 1 dv 1 − and = − (5 − x ) 2 = − 2 dx 2 5− x du dv v −u dy dx dx = dx v2 dy  −2 ( 5 − x ) 5 − x ( 5 − x )2  ÷ (5 − x ) = + dx  1 2 5 − x 

=

)

Let u = x − 4 x 2 and v = 2 x 2 so

1 2

dy dx dy dx dy dx dy dx dy dx

(

f y = f ( x ) =

x

dy e x ( 2 − sin(2 x ) cos(2 x ) ) 1 = × 2x dx e cos 2 ( 2 x ) dy 2 − sin(2 x ) cos(2 x ) = dx e x cos 2 ( 2 x ) ( 5 − x )2 ( 5 − x )2 = c y = f ( x ) = 5 − x ( 5 − x ) 12

du dv −u dx dx v2 3 2 x 2 − 3 − 4 x ( 3 x − 1)

du dv = 3 and = 4x dx dx

v

(

v

x

3x − 1 2x 2 − 3

dy = dx dy dx dy dx dy dx dy dx

=

2 x (1 − 8 x ) −

( 2 x )2

(

−1 du dv 1 = 1 − 8 x and =x 2= dx dx x

x − 4x2 x

2 x (1 − 8 x ) − x − 4 x 2

)

4x x 2 x − 16 x 2 − x + 4 x 2 = 4x x x − 12 x 2 = 4x x 1 = −3 x 4 x ex g y = cos(2 x + 1) Let u = e x and v = cos(2 x + 1) so dv du = e x and = −2 sin(2 x + 1) dx dx du dv v −u dy dx dx = dx v2 x dy e cos(2 x + 1) + 2e x sin(2 x + 1) = dx cos 2 (2 x + 1) −x e h y = x −1 dv du Let u = e − x and v = x − 1 so = − e − x and =1 dx dx du dv v −u dy dx dx = dx v2 −x dy − e ( x − 1) − e − x = dx ( x − 1)2 dy − e − x x + e − x − e − x = dx ( x − 1)2 dy xe − x =− dx ( x − 1)2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

118 | TOPIC 6 Further differentiation and applications • EXERCISE 6.4 i y =

3 x x+2

1

Let u = 3 x 2 and v = x + 2 so

du 3 dv = and =1 dx 2 x dx

du dv v −u dy dx dx = dx v2 dy  3( x + 2)  = − 3 x  ÷ ( x + 2 )2  dx  2 x dy 3( x + 2) − 6 x = dx 2 x ( x + 2 )2 3x + 6 − 6 x dy = dx 2 x ( x + 2 )2 6 − 3x dy = dx 2 x ( x + 2 )2 cos(3 x ) j y = sin(3 x )

dy = dx dy = dx

du dv −u dx dx v2 2x −2e 1 + e 2 x − 2e 2 x 1 − e 2 x v

(

)

(1 + e )

2x 2

(

)

dy −2e 2 x − 2e 4 x − 2e 2 x + 2e 4 x = 2 dx 1 + e2 x

(

dy −4e 2 x = dx 1 + e 2 x

(

6 a y = f ( x ) =

)

)

2

x+2 sin ( g( x ) )

Let u = x + 2 and v = sin ( g( x ) ) so

du dv = 1 and = g′( x ) cos ( g( x ) ) dt dt

du dv = 1 and = g′( x ) cos ( g( x ) ) dv dt du dt Let u = cos(3 x ) and v = sin(3 x ) so = −3sin(3 x ) and = 3 cos(3 x ) dx dx du dv dv v −u dy = −3sin(3 x ) and = 3 cos(3 x ) dt dt = dx dt v2 du dv dy sin ( g( x ) ) − ( x + 2 ) g′( x ) cos ( g( x ) ) v −u = dy dx dx = dt sin 2 ( g( x ) ) dx v2 g e −2 x dy −3sin 2 (3 x ) − 3 cos 2 (3 x ) = b y = f ( x ) = 2 dx sin (3 x ) ex

(

(

2 2 dy −3 sin (3 x ) + cos (3 x ) = dx sin 2 (3 x ) dy 3 =− 2 dx sin (3 x ) x−2 k y = 2 2x − x − 6

)

(

) du dv = −2e −2 x g ′ ( e −2 x ) and = ex dx dx

Let u = g e −2 x and v = e x so

dy = dx

Let u = x − 2 and v = 2 x 2 − x − 6 so

7 a y =

)

(

(

(

)

2

)

dy dx dy dx dy dx dy dx

)

2x x +1

Let u = 1 − e 2 x and v = 1 + e 2 x so

du dv = 2 and = 2x dx dx

du dv −u dx dx = v2 2 2( x + 1) − 2 x (2 x ) = ( x 2 + 1)2 2 2x + 2 − 4 x 2 = ( x 2 + 1)2 2 − 2x 2 2(1 − x 2 ) = 2 = 2 2 ( x + 1)2 ( x + 1) v

When x = 1,

2x

1− e 1 + e2 x

)

2

Let u = 2 x and v = x 2 + 1 so

2 x − 4x + 4 dy =− dx ( 2 x + 3)2 ( x − 2)2 2( x − 2)2 dy =− dx ( 2 x + 3)2 ( x − 2)2 2 dy , x≠2 =− dx 2 x ( + 3)2 l y =

=

ex

dy 2 x 2 − x − 6 − 4 x 2 + 9 x − 2 = 2 dx 2x 2 − x − 6 dy −2 x 2 + 8 x − 8 = dx 2 x 2 − x − 6 2

du dv −u dx dx v2 x e × −2e −2 x g ′ e −2 x − e x g e −2 x v

( ) ( ( e x )2 −2e −2 x g ′ ( e −2 x ) − g ( e −2 x ) =

du dv = 1 and = 4x − 1 dx dx

du dv v −u dy dx dx = dx v2 2 dy 2 x − x − 6 − ( x − 2)(4 x − 1) = 2 dx 2x 2 − x − 6

(

)

dy 2(1 − 12 ) = =0 dx (12 + 1)2

du dv = −2e 2 x and = 2e 2 x dx dx

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 6 Further differentiation and applications • EXERCISE 6.4   |  119 b y =

sin ( 2 x + π ) cos ( 2 x + π )

8 y =

Let u = sin ( 2 x + π ) and v = cos ( 2 x + π ) so du dv = 2 cos ( 2 x + π ) and = −2 sin ( 2 x + π ) dx dx du dv v −u dy = dx 2 dx dx v dy 2 cos 2 ( 2 x + π ) + 2 sin 2 ( 2 x + π ) = dx cos 2 ( 2 x + π )

(

2 2 dy 2 cos ( 2 x + π ) + sin ( 2 x + π ) = dx cos2 ( 2 x + π ) 2 dy = dx cos 2 ( 2 x + π )

When x =

v

dy = dx

3x + 1

)

d y =

)

3 2

1

3

1

dy 2 ( 4 ) 2 − 9(1) ( 4 ) 2 1 = =− When x = 1, dx 32 ( 4 )3 ex x +2 2

Let u = e x and v = x 2 + 2 so

du dv 3 = 1 and = dx dx 2 3 x + 1

dy dx dy dx dy dx dy dx

2

dv du = e x and = 2x dx dx

du dv −u dx dx = v2 x e ( x 2 + 2) − 2 xe x = ( x 2 + 2)2 x 2 e x + 2e x − 2 xe x = ( x 2 + 2)2 x e ( x 2 − 2 x + 2) = ( x 2 + 2)2 v

dy e 0 (0 2 − 2(0) + 2) 2 1 = = = dx 4 2 (0 2 + 2)2

When x = 0, mT = When x = 0, y =

3(5) − 1 14 7 dy = = = dx 2 3(5) + 1 ( 3(5) + 1) 2(4)(16) 64

e0 1 = 0 +2 2 2

Equation of tangent with mT =  1 ( x1 y1 ) =  0,  is given by  2

Let u = 5 − x 2 and v = e x so

dy = dx

1

dy 2 ( 3 x + 1) − 9 x ( 3 x + 1) 2 = dx ( 3 x + 1)3

5 − x2 ex

dy = dx

3

(

dy 2 ( 3 x + 1) − 3( x + 1) = dx 2 3 x + 1 ( 3 x + 1) dy 6 x + 2 − 3x − 3 = dx 2 3 x + 1 ( 3 x + 1) dy 3x − 1 = dx 2 3 x + 1 ( 3 x + 1) When x = 5,

du dv −u dx dx v2

du dv −u dx dx v2 −2 xe x − e x 5 − x 2

du dv = −2 x and = ex dx dx

(e )

(

)

dy −2 xe x − 5e x + 5e x x 2 = dx e2 x 2 dy 5 x − 2 x − 5 = dx ex When x = 0,

5 dy = − 0 = −5 dx e

1 that passes through 2

y − y1 = mT ( x − x1 ) 1 1 y − = ( x − 0) 2 2 1 1 y= x+ 2 2

v

x 2

du dv 9 = 2 and = 3x + 1 dx dx 2

dy 2 ( 3 x + 1) 2 − 9 x ( 3 x + 1) 2 = 3 2 dx ( 3 x + 1) 2

du dv −u dx dx v2 3( x + 1) 3x + 1 − 2 3x + 1

(

v

dy = dx

9 y =

Let u = x + 1 and v = ( 3 x + 1) so dy = dx

3

π dy 2 2 , = = =2 2 dx cos 2 ( 2π ) 12

1 2

3

( 3 x + 1) 2

Let u = 2 x and v = ( 3 x + 1) 2 so

)

x +1 c y = 3x + 1

2x

10

d  1 + cos( x )  dx  1 − cos( x )  If y =

1 + cos( x ) , let u = 1 + cos( x ) and v = 1 − cos( x ) 1 − cos( x )

dv du = − sin( x ) and = sin( x ) dx dx

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

=

120 | TOPIC 6 Further differentiation and applications • EXERCISE 6.4 du dv v −u dy dx dx = dx v2 dy (1 − cos( x )) × − sin( x ) − (1 + cos( x )) × sin( x ) = dx (1 − cos( x ))2 − sin( x )(1 − cos( x ) + 1 + cos( x ) = (1 − cos( x ))2 −2 sin( x ) = (− (cos( x ) − 1))2 −2 sin( x ) = (cos( x ) − 1)2 11 y = f ( x ) =

(

4 m 2 = 16

m=2

m2 = 4 m=2

e −3 x e2 x + 1

Let u = e −3 x and v = e 2 x + 1 so

If

(

)

( x 2 + 1)

2

dy 10 − 10 x 2 = 2 dx x2 + 1 dy

Minimum TP at  1 , 83   2 24 

b y = ax 2 + bx + c

16 a y =

b

x = −2

When x = −3, f ′ ( −3) = 2 ( −3)2 + 3 ( −3) − 2 = 18 − 9 − 2 = 7 ( + ve )

sin(2 x − 3) 15 a y = f ( x ) = ex

When x = −1.088, y =

Either 2 x − 1 = 0 or x + 2 = 0

Therefore a = −3, b = −6, c = −8 2 a y = x 3 + ax 2 + bx − 11 5 Stationary point when x = 1 and x = . 3 dy 2 = 3 x + 2ax + b dx

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

1 2

122 | TOPIC 6 Further differentiation and applications • EXERCISE 6.5 dy = 3 (1)2 + 2a (1) + b = 0 dx x =1 3 + 2a + b = 0 2a + b = −3...................(1) 2 dy  5  5 = 3   + 2a   + b = 0 5    3 3 dx x = 3

25 10 + a+b=0 3 3 10 a + 3b = −25.............(2) (1) × 3 6a + 3b = −9................(3) (2) − (3) 4 a = −16 a = −4 Substitute a = −4 into (1) 2 ( −4 ) + b = −3 −8 + b = −3 b=5 b When x = 1, y = (1)3 − 4 (1)2 + 5 (1) − 11 = −9 5 5 3 5 2 5 125 100 15 When x = , y =   − 4   + 5   − 11 = − + − 11     3 3 3 3 27 9 3 125 300 135 2700 247 = − + − =− 27 27 27 27 27 dy 2 When x = 0, = 3 ( 0 ) − 8 ( 0 ) + 5 = + ve dx dy When x = 1.5, = 3 (1.5)2 − 8 (1.5) + 5 = − ve dx dy When x = 2, = 3 ( 2 )2 − 8 ( 2 ) + 5 = + ve dx x

5 3

5 247  Minimum TP at  , −  3 27 

3 a y = f ( x ) = 2 x 3 − x 2 = x 2 ( 2 x − 1) Graph cuts the y axis where x = 0, y = 0. Graph cuts the x axis where y = 0 x 2 ( 2 x − 1) = 0 x = 0 or 2 x − 1 = 0 1 x= 2 3 2 1  1  1 When x = , y = 2   −   = 0  2  2 2 dy Stationary points occur where =0 dx dy = 6x 2 − 2x = 0 dx 3x 2 − x = 0 x ( 3 x − 1) = 0 x = 0 or 3 x − 1 = 0 1 x= 3 1 1 3 1 2 2 3 1 When x = , y = 2   −   = − =−    3 3 3 27 27 27 dy When x = −1, = 6 ( −1)2 − 2 ( −1) = + ve dx When x =

2 1 dy  1  1 3 4 , = 6   − 2   = − = − ve  4  4 8 8 4 dx

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 6 Further differentiation and applications • EXERCISE 6.5   |  123 When x = 1, x 0 for all x x=2 x = 2, y = (2 − 1)e −2 = e −2 dy When x = 1, = e −1 (2 − 1) = + ve dx dy When x = 3, = e −3 ( 2 − 3) = − ve dx

When x = −4, y = ( −4 )3 + 3 ( −4 )2 − 24 ( −4 ) + 5 = 85 When x = 2, y = ( 2 )3 + 3 ( 2 )2 − 24 ( 2 ) + 5 = −23 dy When x = −5, = 3(−5)2 + 6(−5) − 24 = 75 − 30 − 24 = + ve dx dy When x = −1, = 3(−1)2 + 6(−1) − 24 = − ve dx dy When x = 3, = 3(3)2 + 6(3) − 24 = 27 + 18 − 24 = + ve dx x

x < −4

x = −4

dy dx Maximum TP at (−4, 85)

−4 < x < 2

x=2

When When When When

x

x

x2

x>2

(

Maximum TP at 2, e −2

Minimum TP at (2, −23)

x2 c y = x +1 dy Stationary points occur where =0 dx 2 dy 2 x ( x + 1) − x = dx ( x + 1)2 2 dy 2 x + 2 x − x 2 = dx ( x + 1)2 2 dy x + 2 x = dx ( x + 1)2 x 2 + 2x 0= ( x + 1)2 0 = x ( x + 2) x = 0 or x + 2 = 0 x = −2 When x = −2, y =

dy =0 dx

9 a

x < −1

x = −1

−1 < x < 3

x=3

dy dx y

y = f(x)

–1 0

b

( −2)2

= −4 −2 + 1 ( 0 )2 x = 0, y = =0 0 +1 dy ( −3)2 + 2 ( −3) x = −3, = = + ve dx ( −3 + 1)2 dy ( −0.5)2 + 2 ( −0.5) x = −0.5, = = − ve dx ( −0.5 + 1)2 dy (1)2 + 2 (1) x = 1, = = + ve dx (1 + 1)2 x < −2 x = −2 −2 < x < 0

x

)

x=0

x

(3, 0)

x < −1

x

−1 < x < 3

x = −1

dy dx y



(–1, 2) 0

x y = f(x)

x>0

dy dx Maximum TP at (−2, −4) Minimum TP at (0, 0)

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

x >3

TOPIC 6 Further differentiation and applications • EXERCISE 6.5   |  127 10 a y = x 3 + bx + c When x = 2, y = −8 −8 = (2)3 + 2b + c −16 = 2b + c..................(1) dy = 3x 2 + b dx Stationary point at ( 2, −8 ). 0 = 3 ( 2 )2 + b 0 = 12 + b −12 = b Substitute b = −12 into (1) −16 = 2(−12) + c −16 = −24 + c 8=c y = x 3 − 12 x + 8 b y = ax 2 + bx + 15 When x = 1.5, y = 6 2  3  3 6 = a   +   b + 15  2  2 9 3 −9 = a + b 4 2 −36 = 9a + 6b −12 = 3a + 2b...................(1) dy = 2ax + b dx Stationary point at (1.5,6 ).

 3 0 = 2a   + b  2 0 = 3a + b........................(2) (1) − (2) −12 = b Substitute b = −12 into (2) 0 = 3a − 12 12 = 3a 4=a y = 4 x 2 − 12 x + 15 c y = x 3 + bx 2 + cx + d When x = −3, y = −10 −10 = (−3)3 + b(−3)2 − 3c + d −10 = −27 + 9b − 3c + d 17 = 9b − 3c + d ....................(1)

dy = 3 x 2 + 2bx + c dx 0 = 3 ( −3)2 + 2 ( −3) b + c −27 = −6b + c......................(3) (1) − (2) 12 = 8b − 4c 3 = 2b − c...........................(4) (3) + (4) −24 = −4 b 6=b Substitute b = 6 into (3) −27 = −6 ( 6 ) + c −27 = −36 + c 9=c Substitute b = 6 and c = 9 into (2) 5= 6+9+d 5 = 15 + d −10 = d y = x 3 + 6 x 2 + 9 x − 10 1 11 a y = f ( x ) = − ( x − 4 )3 + 2 4 Graph cuts the y axis where x = 0, y = 14 + 2 = 16. Graph cuts the x axis where y = 0 ( x − 4 )3 = 8 x−4=2 x=6 dy Stationary points occur where = 0. dx dy 3 = − ( x − 4 )2 dx 4 3 0 = − ( x − 4 )2 4 x−4=0 x=4 1 When x = 4, y = − ( 4 − 4 )3 + 2 = 2 4 3 dy When x = 0, = − ( 0 − 4 )2 = − ve 4 dx 3 dy When x = 5, = − ( 5 − 4 )2 = − ve 4 dx x4

dy dx y (0, 18)

When x = 1, y = 6 6 = (1)3 + b(1)2 + c + d 6 = 1+ b + c + d 5 = b + c + d .......................(2) Stationary point at ( −3, −10 )

(4, 2) (6, 0) 0

x y = f(x)

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

128 | TOPIC 6 Further differentiation and applications • EXERCISE 6.5 x ∈[ −1,1]

b y = g( x ) = 2 x 3 − x 2 ,

dy = 3x 2 − 2 x − 1 dx 0 = ( 3 x + 1)( x − 1) 3 x + 1 = 0 or x − 1 = 0 1 x=− x =1 3 dy When x = −1, = 3 ( −1)2 − 2 ( −1) − 1 = + ve dx dy When x = 0, = 3(0)2 − 2(0) − 1 = − ve dx

Graph cuts the y axis where x = 0, y = 0. Graph cuts the x axis where y = 0 x 2 ( 2 x − 1) = 0 x = 0 or 2 x − 1 = 0 1 x=0 x= 2 dy Stationary points occur where = 0. dx dy = 6 x 2 − 2 x = 2 x ( 3 x − 1) dx 0 = 2 x ( 3 x − 1) x = 0 or 3 x − 1 = 0 1 x= 3 dy When x = −1, = 2(−1) ( 3(−1) − 1) = + ve dx dy When x = 0.1, = 2(0.1) ( 3(0.1) − 1) = − ve dx dy When x = 1, = 2(1) ( 3(1) − 1) = + ve dx x

x 2 This is a positive cubic with a turning point at ( a, 0 ). Stationary points occur where f ′( x ) = 0

3 2

15 y = f ( x ) = xe x

2

dy = 0. dx

)

f ′( x ) = −2 ( a − x ) ( x − 2 ) + ( a − x )2 f ′( x ) = − ( a − x ) ( 2( x − 2) − (a − x ) ) f ′( x ) = − ( a − x ) ( 3 x − 4 − a ) 0 = ( a − x ) (3x − 4 − a ) a − x = 0 or 3 x − 4 − a = 0 a+4 x=a x= 3 When x = a, y = ( a − a )2 ( a − 2 ) = 0 a+4 When x = , 3

x + 1 = 0 as e x > 0 for all x x = −1 Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

132 | TOPIC 6 Further differentiation and applications • EXERCISE 6.5 a + 42  a + 4   y = a− − 2     3   3 2  3a − a − 4   a + 4 − 6  =     3 3 2  2(a − 2)   a − 2  =     3   3  4(a − 2)3 = 27  a + 4 4(a − 2)3  Therefore, stationary points are (a, 0) and  . , 27   3

b

x = a −1

x

x=a

x=

2a + 2 a+4 a+4 a+7 +1= x= x= 3 3 3 3

dy dx  a + 4 4(a − 2)3  Minimum TP at (a, 0) and a maximum TP at  . , 27   3  a + 4 4(a − 2)3  c (3, 4) =  , 27   3 a+4 =3 3 a+4=9 a=5 18 a f ( x ) = ( x − a)( x − b)3 where a < b This is a quartic graph with a stationary point of inflection at x = b since ( x − b ) raised to the power of three. Graph cuts the x axis where f ( x ) = 0. ( x − a)( x − b)3 = 0 x − a = 0 or x − b = 0 x=a x=b Stationary points are (a, 0) and (b, 0). b Stationary points occur where f '( x ) = 0. f ′ ( x ) = ( x − b )3 + 3 ( x − a ) ( x − b ) 2 f ′( x ) = ( x − b)2 ( x − b + 3 x − 3a ) f ′( x ) = ( x − b)2 ( 4 x − 3a − b ) 0 = ( x − b)2 ( 4 x − 3a − b ) x − b = 0 or 4 x − 3a − b = 0 3a + b x=b x= 4 When x = b, f (b) = (b − a)(b − b)3 = 0 This is a point of inflection. 3a + b When x = , 4 3   3a + b   3a + b   3a + b f − a  − b =   4   4   4  3a + b − 4 a   3a + b − 4 b  =     4 4  a − b   3a − 3b  = −  4   4  27 ( a − b )4 =− 256  3a + b −27(a − b)4  , Stationary points are (b, 0),   256  4 c x

x=

3a + b − 4 4

x=

3a + b 4

x=

3a + 5b 8

x=b

x = b +1

dy dx  3a + b −27(a − b)4  , There is a minimum TP at   and a stationary point of inflection at (b, 0) 256  4 Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 6 Further differentiation and applications • EXERCISE 6.6   |  133  3a + b 27 ( a − b )4  d ( 3, −27 ) ≡  ,− 256   4

(

)

Height = 75 − 1 + 2 (15) + 15 = 53.8 cm 2 a V = x (16 − 2 x )(10 − 2 x )

3a + b =3 4 3a + b = 12...................(1) −

c Width = 30 cm

(

)

V = x 160 − 42 x + 4 x 2

27(a − b)4 = −27 256 ( a − b )4 =1 256 ( a − b )4 = 256 a − b = ±4 but a < a − b = −4......................(2)

(1) + (2) 4a = 8 a=2 Substitute a = 2 into (2) so 2 − b = −4 ⇒ b = 6

Exercise 6.6 — Maximum and minimum problems 1 a Area = rectangular area plus triangular area 1 A = 2 xy + × 2 x × x 2 A = 2 xy + x 2

3

2

V = 4 x − 42 x + 160 x b Greatest volume occurs when dV = 12 x 2 − 104 x + 160 = 0 dx 3 x 2 − 26 x + 40 = 0 ( 3 x − 20 )( x − 2 ) = 0 20 x = 2, 3 x = 2, (0 < x < 5)

dV = 0. dx

Therefore, height = 2cm, width = 6 cm and length = 12 cm Vmax = 2(16 − 2(2))(10 − 2(2)) = 2 × 12 × 6 = 144 m 3 3 Point is ( x1 , y1 ) ≡ ( x , y ). Let ( x 2 , y2 ) ≡ ( 0, 0 ).

( x2 − x1 )2 + ( y2 − y1 )2

Minimum distance D =

= ( x − 0 )2 + ( y − 0 )

2

= x 2 + ( 2 x − 5 )2 = 5 x 2 − 20 x + 25 c

dD = 0 gives minimum distance dx

x

x

(

x

By Pythagoras x 2 + x 2 = c 2 2 x 2 = c2 2 x = c, c > 0 Perimeter = 150 = 2 x + 2 y + 2 2 x

(

(

)

75 = y + 1 + 2 x

)

75 − 1 + 2 x = y

(

) ) A = 150 x − ( 2 2 + 2 ) x 2 + x 2 A = 150 x − ( 2 2 + 1) x 2 as required (

Thus A = 2 x 75 − 1 + 2 x + x 2

b Greatest area occurs when

( (

dA = 0. dx

) )

dA = 150 − 2 2 2 + 1 x dx 0 = 150 − 2 2 2 + 1 x

(

)

150 = 2 2 2 + 1 x 75 =x 2 2 +1 x = 19.59 Width = 2 x = 39.2 cm

(

)

Height = 75 − 1 + 2 (19.59 ) + 19.59 = 47.3 cm

)

1

− dD 1 = 5 x 2 − 20 x + 25 2 × (10 x − 20) dx 2 10 x − 20 0= 2 5 x 2 − 20 x + 25 0 = 10 x − 20 10 x = 20 x=2 y = 2×2−5 =1

Distance = 5 x 2 − 20 x + 25 = 5(2)2 − 20(2) + 25 = 5 units −

t

4 P (t ) = 200te 4 + 400, a Initially t = 0

0 ≤ t ≤ 12

P (0) = 200(0)e 0 + 400 = 400 birds b Largest number of birds when P '(t ) = 0. P ′(t ) = 200e

−

t 4

e

− 50te −

t 4

−

t 4

=0

(4 − t ) = 0 −

t

4 − t = 0 as e 4 > 0 for all t t=4 At the end of December the population was at its largest. c P (4) = 200(4)e −1 + 400 = 694 birds

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

134 | TOPIC 6 Further differentiation and applications • EXERCISE 6.6 5 Let one number be m and the other number be n. P is the product of the two numbers. m + n = 32 m = 32 − n...................(1) P = mn..............................(2) Substitute (1) into (2) P = n(32 − n) P = 32n − n 2 Max/min values occur where

dV = 110π − 3π r 2 dr 0 = 110π − 3π r 2 3r 2 = 110 110 r2 = 3 110 r= r>0 3 r = 6.06 cm

dP = 0. dn

dP = 32 − 2n dn 0 = 32 − 2n 2n = 32 n = 16 Substitute n = 16 into (1) m = 32 − 16 = 16 Therefore both numbers are 16.

Substitute r = 6.06 into (1) 110 h= − 6.06 = 12.11 cm 6.055 Vmax = π ( 6.06 )2 (12.11) = 1395.04 cm 3 8 If y = 2 x and the point ( x1 , y1 ) = ( 5, 0 ) the shortest distance is given by D=

(

6 a A = lw...............................(1) 2l + 5w = 550 2l = 550 − 5w 1 l = ( 550 − 5w ) ...............(2) 2 Substitute (2) into (1) 1 A = w ( 550 − 5w ) 2 550 w 5w 2 A= − 2 2 Max/min values occur where

( x − x1 )2 + ( y − y1 )2

D = ( x − 5 )2 + 2 x − 0

)2

D = x 2 − 10 x + 25 + 4 x D = x 2 − 6 x + 25 Min distance occurs when dD 1 2x − 6 = × 2 dx 2 x − 6 x = 25 dD x−3 = dx x 2 − 6 x = 25 x−3 0= 2 x − 6 x = 25 0= x−3 x=3

dA = 0. dw

dA 550 = − 5w dw 2 550 − 5w 0= 2 550 5w = 2 550 w= 10 w = 55 m

dD = 0. dx

When x = 3, y = 2 3 Dmin = 32 − 6(3) = 25 = 4 units y

9

y = 4 – x2

Substitute w = 55 into (2) 1 l = ( 550 − 5(55) ) 2 275 l= 2 l = 137.5 m

x

A = 2x 4 − x

7 SAcylinder = 220π = 2π rh + 2π r 2 Vcylinder = π r 2 h.............(1) 110 = rh + r 2 110 − r 2 = rh 110 − r = h..................(2) r Substitute (2)into (1)  110  V = πr2  − r  r  Max/min values occur when

x

x

Max area occurs when

b Amax = 137.5 × 55 = 7562.5 m 2

V = 110π r − π r 3

0

2

dA = 0. dx

2 x ( −2 x ) dA =− + 2 4 − x2 dx 2 4 − x2

(

)

2 2 dA 2 4 − x − 2 x = dx 4 − x2 0 = 8 − 4x2

4x2 = 8 x2 = 2 x= 2 x>0 When x = 2, Amax = 2 dV = 0. dr

( 2 ) 

4−

( 2)

2

  = 2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

( 2 )( 2 ) = 4 units2 .

TOPIC 6 Further differentiation and applications • EXERCISE 6.6   |  135

π 2 R where A is a constant. 2 Perimeter of pool is given by P = 2l + 2 R + π R = 2l + ( 2 + π ) R π From area equation   A − R 2 = 2lR 2 π 2 A − R = 2lR 2 2 A − π R2 = 2lR 2 2 A − π R2 =l 4R 2 A − π R2 into perimeter equation. Substitute l = 4R

10 a P = 96 = 2 ( 2.5b ) + 2 ( 2a ) 96 = 5b + 4 a 48 = 2.5b + 2a 48 − 2.5b = 2a 24 − 1.25b = a.................(1) A = ab + 2.5ab = 3.5ab..........(2) Substitute (1) into (2) A = 3.5b ( 24 − 1.25b ) A = 84 b − 4.375b 2 Max area occurs where

12 Area of pool is given by A = 2lR +

dA = 0. dx

dA = 84 − 8.75b dx 0 = 84 − 8.75b 8.75b = 84 b = 9.6 Substitute b = 9.6 into (1)

 2 A − π R2  + (2 + π ) R P = 2 4 R  

24 − 1.25 ( 9.6 ) = a 12 = a b Amax = 3.5 ( 9.6 )(12 ) = 403.2 m 2 11 a

4−t3 A(t ) = 1000 − 12te 8

A(0) = 1000 − 12(0)e

, t ∈[ 0,6 ]

4 − 03 8

= $1000

b Least amount of money occurs when A '(t ) = 0. 3 A′(t ) = 12t × t 2 e 8 A′(t ) = A′(t ) = 0=

9 3 t e 2

4−t3 e 8

4−t3 e 8

4−t3 8

4−t3 8

4−t3 − 12e 8

− 12e

4−t3 8

2 A = (π + 4 ) R 2 2A = R2 π +4 2A = R, R > 0 π +4

9 3   t − 12 2

9 3   t − 12 2

9 3 t − 12 = 0 as e 2 9 3 t = 12 2 24 t3 = 9 8 3 t = 3 8 t=3 3 t = 1.387

4−t3 8

Substitute R =

> 0 for all t

A−

2A π into A − R3 = 2lR. π +4 2

 2A  π  2A    = 2l  2 π +4  π + 4  A−

 2A  πA = 2l  π +4  π + 4 

 2A  A(π + 4) − π A = 2l  π +4  π + 4 

A(1.387) = 1000 − 12(1 − 0.387)e

4 −1.38873 8

 2A  2A = l π + 4  π + 4  = $980.34

c The least amount of money occurred 1.387 years after January 1, 2009 which is May 2010. d

2 A − π R2 + 2 ( 2 + π ) R2 2R 2 A − π R 2 + 2π R 2 + 4 R 2 P= 2R 2 A + π R2 + 4 R2 P= 2R A (π + 4 ) P= + R R 2 dP Min value occurs when = 0. dR dP A π +4 =− 2 + dR 2 R −2 A + (π + 4 ) R 2 0= 2 R2 0 = −2 A + (π + 4 ) R 2 P=

4 − 63 A(6) = 1000 − 12(6)e 8

= $1000

2A  2A  =l ÷ π + 4  π + 4  2A =l π +4 If both l and R =

2A , the perimeter is a minimum. π +4

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

136 | TOPIC 6 Further differentiation and applications • EXERCISE 6.6 13 By Pythagoras r + ( h − 12 ) = 12 2

2

Substitute x =

2

(

r 2 = 144 − h 2 − 24 h + 144

)

r 2 = 24 h − h 2 1 Vcone = π r 2 h 3 1 Vcone = π h 24 h − h 2 3 1 Vcone = 8π h 2 − π h 3 3 dV = 16π h − π h 2 dh dV Max volume occurs when = 0. dh 2 16π h − π h = 0 h (16 − h ) = 0 h = 0 or 16 − h = 0 h = 16, (h > 0) 16π Vmax = 24(16) − (16)2 = 2145 cm 3 3

(

(

r 2 − x 2 = y2 1 r 2 − r 2 = y2 2 1 2 r = y2 2 1 y= r, y > 0 2

)

)

The x and y values are the same, thus, the largest rectangle is a square. 15

8 cm

h

(x,

r 2 – x2

10 – h

)

2x

By similar triangles: r : 8 as 10 − h : h

2y x

0

By Pythagoras: r 2 = x 2 + y2

r 10 − h = 8 10 10r = 80 − 8h 8h = 80 − 10r 5 h = 10 − r 4 Vcylinder = π r 2 h

r 2 − x 2 = y2 r 2 − x 2 = y, y > 0 Area of rectangle is given by: A = ( 2 x ) ( 2 y ) = 4 xy A = 4x r2 − x2 dA 8x 2 =− + 4 r2 − x2 dx 2 r2 − x2 2 2 2 dA 8 r − x − 8 x = dx 2 r2 − x2 2 dA 8r − 8 x 2 − 8 x 2 = dx 2 r2 − x2

(

r

10 cm

y

14

1 r into Pythagoras relationship. 2

)

dA 4r 2 − 8 x 2 = dx r2 − x2

Max area occurs when 4r 2 − 8 x 2

=0 r2 − x2 4r 2 − 8 x 2 = 0 4r 2 = 8 x 2 1 2 r = x2 2 1 r = x, x > 0 2

dA = 0. dx

5   Vcylinder = π r 2  10 − r   4  5 Vcylinder = 10π r 2 − π r 3 4 dV 15 = 20π r − r 2 dr 4 Max volume occurs when

dV = 0. dr

15 2 πr = 0 4 15 20r − r 2 = 0 4 15   r  20 − r  = 0  4  15 r = 0 or 20 − r = 0 4 80 16 cm, r > 0 = r= 15 3 20π r −

5  16  20 10 = h = 10 −   = 10 − cm   4 3 3 3 2  16   10  Vmax = π     = 298 cm 3  3  3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 6 Further differentiation and applications • EXERCISE 6.7   |  137 16 Speed =

When h = 5 cm dV π 25π = ( 5 )2 = cm 3 / cm dh 9 9

distance time

Rowing:

5=

AB = tr

x 2 + 16 tr

Walking:

x 2 + 16 5

tr =

8= tw =

8− x tw

2 a

(8 − x )

20 cm

8

x 2 + 16 8 − x + 5 8 2x 1 = − 10 x 2 + 16 8 x 1 = − 2 8 5 x + 16

20 cm h cm

Time for total journey is T = tr + t w = dT dx dT dx

dT Min time occurs when = 0. dx x 1 − =0 2 8 5 x + 16 x 1 = 2 8 5 x + 16 8 x = 5 x 2 + 16

(

64 x 2 = 25 x 2 + 16 2

)

2

64 x = 25 x + 400 64 x 2 − 25 x 2 = 400 39 x 2 = 400 400 = 3.2 km 39 Therefore the rower will row to a point that is 3.2 km to the right of point O. x=

Exercise 6.7 — Rates of change 1 a By similar triangles r : 4 as h :12 r h = 4 12 1 r= h 3 4 cm

r cm 12 cm h cm

1 b V = π r 2 h 3 1  1 2 V = π  h h 3 3  π 3 V= h 27 dV π 2 c = h dh 9

r cm

By Pythagoras: r 2 + h 2 = 20 2 r 2 = 400 − h 2 r = 400 − h 2 , r > 0 1 b V = π r 2 h 3 1 V = π 400 − h 2 h 3 400π h π h 3 V= − 3 3 dV 400π c = − π h2 dh 3 When h = 8 cm dV 400π = − π ( 8 )2 dh 3 dV 400π 192π 208π = − = cm 3 / cm 3 3 3 dh 3 a Initially t = 0 x = 2 ( 0 )2 − 8 ( 0 ) = 0 It is at the origin initially. dx b v = = 4t − 8 dt When t = 0 dx v= = 4 ( 0 ) − 8 = −8 dt Initially it is moving with a velocity of 8 m/s to the left. c When v = 0 When t = 2

(

0 = 4t − 8 8 = 4t 2=t

)

x = 2 ( 2 )2 − 8 ( 2 ) x = −8 m

It is at rest after 2 seconds and is 8 metres to the left of the origin. d When at the origin, x = 0. 2t 2 − 8t = 0 2t ( t − 4 ) = 0 t = 0 or t − 4 = 0 t=4 As expressed in (a) it is initially at the origin then it is there again after 4 seconds. Initially it is at the origin, it then travels 8 metres to the left and at t = 4 it is back at the origin again so a total of 16 metres has been travelled. 16 − 0 e Average speed = = 4 m/s 4−0 v − v0 8 − 8 f Average velocity = 4 = = 0 m/s 4 4

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

138 | TOPIC 6 Further differentiation and applications • EXERCISE 6.7 1 4 a x (t ) = − t 3 + t 2 + 8t + 1 and v (t ) = − t 2 + 2t + 8 3 Initially t = 0 1 x (0) = − ( 0 )3 + ( 0 )2 + 8 ( 0 ) + 1 and v (0) = − ( 0 )2 + 2 ( 0 ) + 8 3 x (0) = 1 metre v (0) = 8 m/s Initially it is 1 metre to the right of the origin travelling at 8 metres per second. b It changes its direction of motion when v = 0. − t 2 + 2t + 8 = 0 ( 4 − t )( 2 + t ) = 0 t = 4, −2 t = 4, t ≥ 0 1 3 64 64 147 2 x (4) = − ( 4 ) + ( 4 )2 + 8 ( 4 ) + 1 = − + 49 = − + = 27 m 3 3 3 3 3 dv c a(t ) = = −2t + 2 dt a(4) = −2(4) + 2 = −6 m/s 2 4 5 a Vsphere = π r 3 3 dV = 4π r 2 dr When r = 10 cm. dV = 4π (10)2 = 400π cm 3 /cm dr b SAcube = 6 x 2 where x is the side length of the cube. d ( SA) = −12 x as the cube is melting dx When x = 6 mm. d ( SA) = −12(6) = −72 mm 2 /mm dx 110 6 N = , t>0 t dN 110 a =− 2 dt t When t = 5 months. dN 110 = − 2 = −4.4 rabbits per month dt (5) Population is decreasing at 4.4 rabbits per month. b When t = 1, N = 110 and when t = 5, N = 22 22 − 110 88 = − = −22 rabbits/month 5−1 4 c As t → ∞, N → 0 so the rabbits will eventually disappear. Average rate of change is

3 7 V = 0.4 (8 − t ) , 0 ≤ t ≤ 8 dV a = −1.2 (8 − t )2 dt When t = 3 minutes dV = −1.2 (8 − 3)2 = −30 litres/min dt Water is leaving the bath at a rate of 30 L/min b When t = 0, V = 0.4 (8 )3 = 204.8 and when t = 3, V = 0.4(5)3 = 50 204.8 − 50 Average rate of change is = −51.6 litres/minute 3− 0 dV c = R( x ) dt R ′( x ) = 0 R ′( x ) = −2.4(8 − t ) × −1 0 = 2.4(8 − t ) t =8 t = 8 corresponds to a minimum, therefore the maximum rate is when t = 0 The rate of water leaving is greatest at the beginning which is when t = 0.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 6 Further differentiation and applications • EXERCISE 6.7   |  139 2 2 t (15 − t ) , 0 ≤ t ≤ 10 3 2 1 a When t = 10, V = (10 )2 (15 − 10 ) = 333 mL. 3 3 dV 2 2 4 b = − t + t (15 − t ) dt 3 3 dV 4 2 = 20t − t 2 − t 2 = 20t − 2t 2 dt 3 3 c When t = 3 seconds. dV = 20 ( 3) − 2 ( 3)2 = 60 − 18 = 42 mL/s dt d  dV  d The flow greatest when   = 0. dx  dt  d  dV    = 20 − 4t dx  dt  0 = 20 − 4t 4t = 20 t=5 dV When t = 5, = 20 ( 5) − 2 ( 5)2 = 50 mL/s dt

8 V =

9 15 cm

h cm

d cm

x cm

1 a V = x 2 h 3 By Pythagoras d 2 = x2 + x2 d 2 = 2x 2 d = 2 x m, x > 0 b By Pythagoras 2

 2  122 = h 2 +  x  2  1 144 = h 2 + x 2 2 1 144 − h 2 = x 2 2 288 − 2h 2 = x 2 as required 1 V = x2h 3 1 V = 288 − 2h 2 h 3 1 V = 288h − 2h 3 3 dV 1 c = (288 − 6h 2 ) dh 3 When h = 3 3 metres.

( (

)

)

x cm

(

)

2 dV 1 = 288 − 6 3 3 dh 3 1 = ( 288 − 6 × 27 ) 3 126 = 3 = 42m 3 / m 1 10 Vcone = π r 2 h 3

(

)

6 cm

r cm

15 cm h cm

By similar triangles: r : 6 as h :15 r h = 6 15 2h r= 5 1  2h  2 Vcone = π   h 3  5 4π 3 V= h 25 dV 4π 2 = h dh 25 15 a When h = 2 dV 4π  15  2 3 =   = 9π cm / cm dh 25  2  b Container is one third full. When full, volume is 180π cm 3 so one third will be 60π cm 3 . 1  2h  2 60π = π   h 3  5 4 3 h 180 = 25 25 = h3 180 × 4 h 3 = 1125 2

h = 5 × 33 2

2 1  dV 4π  3 =  5 × 3 3  = 12 × 3 3 π cm / cm dh 25  

11 x (t ) = 2t 2 − 16t − 18, t ≤ 0 a When t = 2 seconds x (2) = 2 ( 2 )2 − 16 ( 2 ) − 18 = −42 The particle is 42 metres to the left of the origin. dv b = 4t − 16 dt When t = 2 seconds dv = 4 ( 2 ) − 16 = −8 dt t = 2 The speed is 8 m/s.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

140 | TOPIC 6 Further differentiation and applications • EXERCISE 6.7 dv dv = −16 and when t = 2, = −8 dt dt −16 + −8 Average velocity is = −12 m/s 2−0 d When x (t ) = 0 c When t = 0,

2t 2 − 16t − 18 = 0 t 2 − 8t − 9 = 0 ( t − 9 )( t + 1) = 0 t−9= 0 t = 9 as t ≥ 0 dv = 4 ( 9 ) − 16 = 20 m/s dt t = 9 Therefore it reaches O after 9 seconds at 20 m/s 2 12 x = t 3 − 4t 2 , t ≥ 0 3 dx a v = = 2t 2 − 8t dt When t = 0 2 xt = 0 = ( 0 )3 − 4 ( 0 )2 = 0 3 vt = 0 = 2 ( 0 )2 − 8 ( 0 ) = 0 The particle starts from rest at the origin. b When v = 0 2t 2 − 8t = 0 t 2 − 4t = 0 t (t − 4 ) = 0 t = 0 or t − 4 = 0 Initially t=4 2 3 64 1 128 192 − =− = −21 x t = 4 = ( 4 ) − 4 ( 4 )2 = 3 3 3 3 3 1 Velocity is zero after 4 seconds when the particle is 21 metres to the left of the origin. 3 c When x = 0 2 3 t − 4t 2 = 0 3 2  t 2  t − 4 = 0 3  2 t = 0 or t − 4 = 0 3 2 Initially t=4 3 t=6 The particle is at the origin again after 6 seconds. d When t = 6 seconds vt = 6 = 2 ( 6 )2 − 8 ( 6 ) = 72 − 48 = 24 m/s dv = 4t − 8 a= dt at = 6 = 4 ( 6 ) − 8 = 24 − 8 = 16 m/s 2 At the origin the particle’s speed is 24 m/s and the acceleration is 16 m/s2. 13 h = 50t − 4t 2 dh a = 50 − 8t dt When t = 3 seconds dh = 50 − 8(3) = 50 − 24 = 26 m/s dt t = 3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 6 Further differentiation and applications • EXERCISE 6.7   |  141 b When t = 5 seconds dh vt = 5 = = 50 − 8 ( 5) = 10 m/s dt c When v = −12 m/s −12 = 50 − 8t 8t = 62 t = 7.75 After 7.75 seconds the velocity of the ball is 12 m/s and it is travelling downwards. d When v = 0 50 − 8t = 0 8t = 50 t = 6.25 seconds The velocity is zero after 6.25 seconds. e Greatest height is obtained when the velocity is zero. ht = 6.25 = 50 ( 6.25) − 4 ( 6.25)2 = 156.25 metres f When the ball strikes the ground, h = 0. 0 = 50t − 4t 2 0 = 25t − 2t 2 0 = t ( 25 − 2t ) t = 0 or 25 − 2t = 0 Initially 2t = 25 t = 12.5 The ball strikes the ground after 12.5 seconds. vt =12.5 = 50 − 8 (12.5) = −50 m/s The ball hits the ground with a speed of 50 m/s. 2t 14 N (t ) = + 0.5 ( t + 0.5)2 a Initially t = 0 2(0) N (0) = + 0.5 = 0.5 hundred thousand or 50 thousand 0 + ( 0.5)2 2t b N (t ) = + 0.5 ( t + 0.5)2 Let u = 2t and v = ( t + 0.5)2 du dv =2 = 2 ( t + 0.5) = 2t + 1 dt dt du dv v −u dt dt N '(t ) = v2 2 ( t + 0.5)2 − 2t ( 2t + 1) = ( t + 0.5)4 2t 2 + 2t + 0.5 − 4t 2 − 2t = ( t + 0.5)4 2 −2t + 0.5 = ( t + 0.5)4 c Maximum number of viruses occurs when −2t 2 + 0.5 =0 ( t + 0.5)4 −2t 2 + 0.5 = 0

dN = 0. dt

2t 2 = 0.5 t 2 = 0.25 t = 0.5, t ≥ 0 2(0.5) N (1) = + 0.5 = 1.5 hundred thousand after half an hour ( 0.5 + 0.5)2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

142 | TOPIC 6 Further differentiation and applications • EXERCISE 6.7 d When t = 10 dN −2 (10 ) + 0.5 199.5 = =− = −0.01641 dt t =10 10.54 (10 + 0.5)4 After 10 hours the viruses were changing at a rate of -1641 viruses per hour. 150 15 N = 220 − t +1 a When N = 190 150 190 = 220 − t +1 150 220 − 190 = t +1 30 ( t + 1) = 150 t +1= 5 t=4 dN 150 = dt (t + 1)2 150 dN t = 4, = dt (4 + 1)2 150 = 25 =6 Therefore after 4 years, butterflies are growing at a rate of 6 butterflies per year. b Growth rate is 12 butterflies per year. 150 dN = dt ( t + 1)2 150 12 = ( t + 1)2 2 12 ( t + 1) = 150

dy = dt

2

dy = dt

(

dv = 2t dt

)

(

)

2 dy 3 4 − t = 2 dt 4 + t2

(

)

b Maximum concentration of painkiller in the blood occur dy when = 0. dt 3(4 − t 2 ) 0= 2 4 + t2

(

)

2

0 = 3(4 − t ) 0 = 4 − t2 t = 2, −2 t = 2, t > 0 3(2) (4 + 22 ) = 0.75 mg/L Therefore max concentration is 0.75 mg/L after 2 hours 3t c 0.5 = 4 + t2 t = 2, y =

(

)

1 2 + t 2 = 3t 2 t 2 − 6t + 2 = 0

d

dN → 0. dt

(

)

3 4 − 12 dy = 2 dt t =1 4 + 12

(

)

dy 9 = dt t =1 25 dy = 0.36 mg/L/h dt t =1

x

0 (0, 70)

du =3 dt

)

6 ± (6)2 − (4)(1)(2) 2(1) 6 ± 36 − 16 t= 2 6+2 5  5.24 hours, (t > 2) t= 2

( )

e

(

)

2 dy 3 4 − t = 2 dt 4 + t2

(

−0.06 =

Let u = 3t and v = 4 + t 2

2

t=

150 N = 220 – – t+1

(4 + t 2 )

(4 + t 2 ) (

c N

3t

)

dy 12 − 3t 2 = 2 dt 4 + t2

t + 1 = 3.54, t ≥ 0 t = 2.54 years

16 a y =

(

dy 12 + 3t 2 − 6t 2 = 2 dt 4 + t2

( t + 1)2 = 12.5

As t → ∞, N → 220 and

du dv −u dt dt v2 3 4 + t 2 − 3t ( 2t ) v

(

)

3 4 − t2

(

)

)

2 2

4+t t = 2.45 and 6 hours (solved on CAS)

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 7 Antidifferentiation • EXERCISE 7.2   |  143

Topic 7 — Antidifferentiation Exercise 7.2 — Antidifferentiation 3 x − 4 x 2 + 2x3 2 3 4 1 f (x) = x 2 − x3 + x 4 + c 4 3 2

1 a f ′( x ) =

−5

c

3

1  5 a ∫  2 x 2 +  dx  x

1 − x 4 − x −2 5

= 6 x − x4 −

 = ∫  2x 2 

( )

1 5x 2

)

=

1 4 1 3 15 2 x − x − x +c 2 3 2

∫

3x 3 − x dx 2 x 5

6 a

1

1 −2 x +c 4 3 1 = 4 x − + 2 +c x 4x

(

)

( 2 x − 3)

( 3 x − 5)6 3×6

=

dx = ∫ ( 2 x − 3)

((

x

)2 − 2 ( x ) ( x ) + ( x )2 ) dx

1 2 4 2 1 3 x − x + x +c 2 5 3

1 ( 3 x − 5)6 + c 18 −

5 2

dx

( 2 x − 3 )− 2 3 2 1

3

1

1

− 2 2 x + 2x 2 + 6x 2 − 2x 2 + c 5

7 y = (3 x 2 + 2 x − 4)3

3

=

2

5

=

= ∫ 2 x 3 − x 2 + x + 4 dx 1 1 1 = x4 − x3 + x2 + 4x + c 2 3 2

5 2

)

x − x dx

1 1 3  3 − −  = ∫  x 2 + 3 x 2 + 3 x 2 + x 2  dx  

b ∫ ( x + 1)(2 x 2 − 3 x + 4) dx

∫

)

3

= 4 x 2 − 3 x −1 +

1

( )

1   b ∫  x +  dx  x 2 2 3  1 3  1   −1   1   −1   −1   = ∫   x 2  + 3  x 2   x 2  + 3  x 2   x 2  +  x 2   dx             

 −1  1 = ∫  2 x 2 + 3 x −2 − x −3  dx 2  

b

2 3  1  1  2  1   + 3 2 x   +    dx x x x 

5

=

3 1   2 + − dx 2 a ∫   x x 2 2 x 3 

3 a ∫ ( 3 x − 5) dx =

2

3   = ∫  x − 2 x 2 + x 2  dx  

1

5

∫( =∫

3 2 1 2 x − x dx 2 2 7 3 1 3 = x2 − x2 + c 7 3 =∫

( )

+ 3 2x 2

(

= ∫ 2 x 3 − x 2 − 15 x dx

d

3

= ∫ 8 x 6 + 12 x 3 + 6 + x −3 dx 8 1 = x 7 + 3 x 4 − x −2 + c 7 2 1 8 7 4 = x + 3x + 6 x − 2 + c 7 2x

∫ x ( x − 3)(2 x + 5) dx

(

(1 − 2 x )−4

−2 × −4 1 = (1 − 2 x )−4 8 1 = +c 8 (1 − 2 x )4

 −1  2 = ∫  3 x 2 − 4 x 3 + x −3  dx 5   =

2×5 1 = ( 2 x + 3)5 + c 10

b ∫ (1 − 2 x ) dx =

2   3 − 4 x 3 + 3  dx b ∫   x 5x 

1 6x 2

( 2 x + 3)5

4 4 a ∫ ( 2 x + 3) dx =

dy = 3(6 x + 2)(3 x 2 + 2 x − 4)2 dx dy = 6(3 x + 1)(3 x 2 + 2 x − 4)2 dx Therefore

∫ (3x + 1)(3x

2

1 6(3 x + 1)(3 x 2 + 2 x − 4)2 6∫ 1 = (3 x 2 + 2 x − 4)3 6

+ 2 x − 4)2 dx =

2×−

=−

3

3 ( 2 x − 3) 2

+c

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

144 | TOPIC 7 Antidifferentiation • EXERCISE 7.2 1   8 y =  7 x + x −   x

4

1 1  −  y =  7x + x 2 − x 2   

4

d

∫

(

e

3  dy 1 1  1  = 47 + + 7x + x −    dx x 2 x 2 x3   

∫ =

Therefore =

3  1  1 1  + + 7 x x dx 7 + −     ∫  2 x 2 x3   x

f

3  1 1 1  1  = ∫ 47 + + dx 7x + x −    x 4 2 x 2 x3   

=∫

)

3 8x 2

2 2 x +c 5

3

2

9 27 4 x − 9x3 + x 2 − x + c 4 2

1

∫ 4 x 3 dx

∫(



)

1 = x 2 + x − x −1 2 1 1 2 = x +x− 2 x 

∫  2

x−

4   dx x

1

2 2 x − 4 × 2x 2 3 1

4 2 x − 8x 2 3 4 = x x −8 x 3

2

)

+ 13 dx

2

2  d ∫  x 3 − 3  dx  x  2   2  2  = ∫  ( x 3 )2 − 2( x 3 )  3  +  3   dx x  x   

(

)

= ∫ x 6 − 4 + 4 x −6 dx 1 4 = x 7 − 4 x − x −5 7 5 4 1 7 = x − 4x − 5 7 5x

1 = ∫ x −3 dx 4 1 = − x −2 + c 8 1 = − 2 +c 8x 2

)

3

∫ ((3x ) − 3(3x ) (1) + 3(3x )(1) = ∫ ( 27 x 3 − 27 x 2 + 9 x − 1) dx

5

(

= ∫ 6 x 2 + 5 x − 6 dx 5 = 2x3 + x 2 − 6x 2 3 2 x + x +1 dx b ∫ x2 = x + 1 + x −2 dx

=

11 a ∫ (3 x − 1)3 dx =



4 − xdx

3

16 5 x +c 5 16 = x2 x + c 5

c

∫

= 2×

=

b

5

4 2 1 2 x − x +c 5 2

1  1 −  = ∫  2 x 2 − 4 x 2  dx  

dx

=

)

− x dx

12 a ∫ (2 x + 3)(3 x − 2) dx

c

dx

5

=8×

)

3 2 = − (4 − x )2 + c 3

4

∫(

∫(

3 2x 2

1

1  1  7 x + x −  4 x 1 9 f ′( x ) = x 2 − 2 x f ′( x ) = x 2 − x −2 1 f ( x ) = x 3 + x −1 + c 3 1 3 1 f (x) = x + + c 3 x 1 4 3 10 a ∫ x dx = x + c 4 2  2 7 1  2 b ∫  7 x − 3  dx = ∫  7 x 2 − x −3  dx = x 3 + x −2 + c    5 3 5 5x  7 c ∫ (4 x 3 − 7 x 2 + 2 x − 1) dx = x 4 − x 3 + x 2 − x + c 3 3

(

x 2 x − x dx

= ∫ ( 4 − x ) 2 dx

=

2 x

)

= ∫ x − 2 x −2 dx 1 = x 2 + 2 x −1 + c 2 1 2 2 = x + +c 2 x

1 3 3  dy 1 − 1 −  1  = 4  7 + x 2 + x 2   7x + x −  dx 2 2 x  

d

x 4 − 2x dx x3

2

7

∫  x 2 − 3x 5  dx = 7 x 2 −

7

15 5 x +c 7

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 7 Antidifferentiation • EXERCISE 7.3   |  145 e

−3

∫ 2 (1 − 4 x ) dx −3 = 2 ∫ (1 − 4 x ) dx

(

 (1 − 4 x )−2  = 2  −2 × −4  1 = 4 (1 − 4 x )2 2 f ∫ 5 dx

dy 15 x 2 + 8 x = dx 2 5 x 3 + 4 x 2 15 x 2 + 8 x ∫ 5x 3 + 4 x 2 dx 15 x 2 + 8 x dx = 2∫ 2 5x 3 + 4 x 2

( 2 x − 3) 2

−

5

= ∫ 2 ( 2 x − 3) 2 dx −

= 2 5x 3 + 4 x 2

5

= 2 ∫ ( 2 x − 3) 2 dx

19

3 −  2

 1 = 2  − ( 2 x − 3)   3  2 =− 3 1

dy x 3 + 3 x 2 − 3 = = x + 3 − 3 x −2 dx x2 1 y = x 2 + 3 x + 3 x −1 + c 2 1 2 3 y = x + 3x + + c 2 x 14

3

+

1 − x 2

1

2 2 x + 2x 2 + c 3 2 y= x x +2 x +c 3

y =

1

16 y = x 2 + 1 = ( x 2 + 1) 2 1



− dy 1 = (2 x )( x 2 + 1) 2 dx 2 dy x = 2 dx x +1 5x x ∫ x 2 + 1 dx = 5 ∫ x 2 + 1 dx 2

(

= 5 x +1 + c

) 3 dy = 4(10 x + 2) ( 5 x 2 + 2 x − 1) dx 3 dy = 8(5 x + 1) ( 5 x 2 + 2 x − 1) dx 3 3 2 2 ∫ 16(5x + 1) (5x + 2 x − 1) dx = 2 ∫ 8(5x + 1) (5x + 2 x − 1) dx 4 = 2 ( 5 x 2 + 2 x − 1) 2

17 y = 5 x + 2 x − 1

4

(

dx

)

1

∫ 2 ( 3 x + 5) 2 ( 7 x =

3

1 4 2 x − 3× x2 + c 4 3 1 4 y = x − 2x x + c 4

dy 1 = x+ = dx x

x3 + 1

1

20

y=

15

x2

2 3 x +1 2 + c 3 2 3 x +1 + c = 3

dy = x 3 − 3 x = x 3 − 3x 2 dx

1 x2

∫ =

3 ( 2 x − 3) 2

13

1

)2 1 dy = 1 (15 x 2 + 8 x ) ( 5 x 3 + 4 x 2 )− 2 dx 2 3 2 3 2 18 y = 5 x + 4 x = 5 x + 4 x

4 (3 x 405

2

)

+ 4 x − 1 dx

3 + 5) 2 (135 x 2

− 72 x + 35) + c

Exercise 7.3 — Antiderivatives of exponential and trigonometric functions 1 x  x 1 1 a ∫  cos(3 x + 4) − 4 sin    dx = sin(3 x + 4) + 8 cos   + c  2  2 2 6 3  2x  1    2x  1 b ∫  cos   − sin (5 − 2 x ) dx = sin   − cos (5 − 2 x ) + c  3 4   2  3 8 x x  x x  2 a ∫  sin   − 3 cos    dx = −2 cos   − 6 sin   + c  2  2  2   2 b f ′( x ) = 7 cos(2 x ) − sin(3 x ) 7 1 f ( x ) = sin(2 x ) + cos(3 x ) + c 2 3 1 5 1 −4 x 4 −4 x 3 a ∫ x − e dx = x + e + c 5 4 x  1 2x 2 − x  1 2x 4 − b ∫  e − e 2  dx = e + e 2 + c 3 4 3 2 

(

)

x  x x x x 1 4 a ∫  e 2 + sin   +  dx = 3e 2 − 3cos   + x 2 + c     3 3 3 6  

b

−3 x −3 x ∫ ( cos(4 x ) + 3e ) dx = 4 sin(4 x ) −e + c

1

∫ (e − e ) dx 2x 3 2 x 2 −3 x 2x −3 x 2 −3 x 3 = ∫ (( e ) − 3 ( e ) ( e ) + 3 ( e )( e ) − ( e ) ) dx = ∫ ( e6 x − 3e x + 3e −4 x − e −9 x ) dx 5

2x

−3 x 3

3 1 1 = e6 x − 3e x − e −4 x + e −9 x + c 6 4 9

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

146 | TOPIC 7 Antidifferentiation • EXERCISE 7.3 dy = cos(2 x ) − e −3 x dx 1 1 y = sin(2 x ) + e −3 x + c 2 3 dH π2 πt 13 a = 1+ sin    45  9 dt

2

6

 x 1 ∫  e 2 − e x  dx

12

 x   x   1   1 2 = ∫   e 2  − 2  e 2   x  +  x   dx       e   e   2

x   − = ∫  e x − 2e 2 + e −2 x  dx  

 45 π 2  πt H (t ) = t −  × cos    45  9  π πt H (t ) = t − 5π cos    45  b When t = 15, π H = 15 − 5π cos   = 7.146 kilojoules  3

x 2

1 − e −2 x + c 2 4 1 = ex + x − 2x + c 2e e2 = e x + 4e

−

7 y = e cos ( x ) = e( cos( x )) 2 dy = −2 sin( x ) cos( x )e cos ( x ) dx Therefore 2

2

1

∫ sin( x ) cos( x )ecos ( x ) dx = − 2 ∫ −2 sin( x ) cos( x )ecos ( x ) dx 2

2

2 1 = − e cos ( x ) 2

8 y = e( x +1) dy 2 ( x +1)3 = 3(1)( x + 1) e dx 3 dy = 3( x + 1)2 e( x +1) dx Therefore

14 y = 2 xe3 x dy 3x 3x = 2e + 6 xe dx 3x 3x 3x ∫ 2e + 6 xe dx = 2 xe

(

)

3x 3x 3x ∫ 2e dx + 6 ∫ xe dx = 2 xe

6 ∫ xe3 x dx = 2 xe3 x − ∫ 2e3 x dx 2 6 ∫ xe3 x dx = 2 xe3 x − e3 x 3 1 3x 3x 3x 3 ∫ xe dx = xe − e 3 1 3x 1 3x 3x xe dx xe = − e ∫ 3 9

3

∫ 9( x + 1) 9 a

∫ ( 2e

2 ( x +1)

e

3

dx = 3 ∫ 3( x + 1) e

2 ( x +1)

= 3e

3x

3

2x 15 y = e

dx

( x +1)3

+c 2 3x 1 − sin(2 x ) dx = e + cos ( 2 x ) + c 3 2

)

e 2 x + 3e −5 x 3 1  dx = ∫  e x + e −6 x  dx b ∫ 2  2e x 2 1 x 1 −6 x = e − e +c 2 4 1 −x c ∫ 0.5 cos(2 x + 5) − e dx = sin(2 x + 5) + e − x + c 4

( ) 2 d ∫ ( e x − e 2 x ) dx x 2 x 2x 2x 2 = ∫ ( (e ) − 2(e )(e ) + (e ) ) dx = ∫ ( e 2 x − 2e3 x + e 4 x ) dx =

10

∫ ae

1 2 x 2 3x 1 4 x e − e + e +c 2 3 4

bx

dx = −3e3 x + c

a e bx + c = −3e3 x + c b a b = 3 and = −3 3 a = −9 3π x    1 11 ∫  2 + sin  dx  2    4x  1 −2  3π x   dx = ∫  x + sin   2   4 1 2  3π x  = − x −1 − cos   2  4 3π 1 2  3π x  cos  =− −  2  4 x 3π

2

+3 x − 1

2 dy = (4 x + 3)e 2 x +3 x − 1 dx 2 2 2 x 2 +3 x − 1 dx = 2 ∫ (4 x + 3)e 2 x +3 x − 1 dx = 2e 2 x +3 x − 1 ∫ 2(4 x + 3)e

16 y = x cos( x ) dy = cos( x ) − x sin( x ) dx ∫ ( cos( x ) − x sin( x )) dx = x cos( x )

∫ cos( x ) dx − ∫ x sin( x ) dx = x cos( x ) ∫ cos( x ) dx − x cos( x ) = ∫ x sin( x ) dx sin( x ) − x cos( x ) = ∫ x sin( x ) dx

πt 17 x (t ) = 20 + cos    4 dy π = x (t ) − π dt 20 dy π   πt  =  20 + cos    − π dt 20 4 dy π πt   cos   − π =π +  4 dt 20 dy π t π cos   =  4 dt 20 πt 4 π y = × sin   + c π 20  4  πt 1 y = sin   + c 5  4 nx 18 f ′( x ) = a sin(mx ) − be

f ( x ) = cos(2 x ) − 2e −2 x + 3 f ′( x ) = −2sin(2 x ) + 4e −2 x Therefore a = −2, b = −4, m = 2, and n = −2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 7 Antidifferentiation • EXERCISE 7.4   |  147 19

∫e

x

sin( x ) dx = ∫ e x cos( x ) dx

Solve using CAS: π 3π x= , 2 2 e2 x + e x − 1 dx = e x − x + loge (e x + 1) + c 20 ∫ ex + 1

Exercise 7.4 — Families of curves and applications 1 a f ′( x ) = 3 x 2 so f ( x ) = x 3 + c y

(0, 3)

5

(0, 1)

y = x3 + 3

y = x3 – 4 x

(0, 0)

20 3 1 4 t − t 3 4 When t = 20,

V=

y = x3 – 2 (0, –4)

20 1 (20)3 − (20)4 3 4 1 V = 53333 − 40 000 3 1 V = 13333 cm 3 3 dV 6 a = πr2 dr 1 V = π r3 + c 3 V=

3

b    f ( x ) = x + c f (2) = 16 23 + c = 16 8 + c = 16 c=8 f (x) = x3 + 8 2 a f ′( x ) = 2 cos(2 x ) so f ( x ) = −sin(2 x ) + c y

(0, 4)

y = –sin(2x) + 4

(0, 2)

y = –sin(2x) + 2

(0, 0) (0, –2)

dV = 20t 2 − t 3 dt 20 3 1 4 V= t − t +c 3 4 When t = 0, V = 0 so c = 0

y = x3

(0, –2)

y = x3 + 1

4 f ′( x ) = cos(2 x ) − sin(2 x ) 1 1 f ( x ) = sin(2 x ) + cos(2 x ) + c 2 2 f (π ) = 2 1 1 2 = sin(2π ) + cos(2π ) + c 2 2 1 1 2 = (0) + (1) + c 2 2 1 2= +c 2 3 c= 2 1 3 1 f ( x ) = sin(2 x ) + cos(2 x ) + 2 2 2

π

π – 2

3π – 2

y = –sin(2x) x 2π y = –sin(2x) – 2

(0, –5)

y = –sin(2x) – 5

(0, –7)

y = –sin(2x) – 7

b    f ( x ) = − sin(2 x ) + c π f   = 4  2 4 = −sin(π ) + c 4= 0+c c=4 f ( x ) = 4 − sin(2 x ) dy 3 = 2e 2 x + e − x dx y = e2 x − e− x + c

1 When r = 0, V = 0 so c = 0 and V = π r 3. 3 1 64 b When r = r 4, V = π (4)3 = π cm 3 3 3

7 There is a stationary point at x = 1. When x < 1 the parabola has a positive gradient and when x > 1 the parabola has a negative gradient. y Local maximum at x = 1 y = f(x) 0 Positive gradient

1

x

Negative gradient

When x = 0, y = 3 3 = e0 − e0 + c 3 = 1−1+ c c=3 y = e2 x − e− x + 3 Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

148 | TOPIC 7 Antidifferentiation • EXERCISE 7.4 8 Stationary points occur at x = −2 and x = 4. When x < −2 and x > 4 the gradient is positive but when −2 < x < 4 the gradient is negative. Local y maximum at x = –2

y = f(x) 0

–2

4

x

(3t − 1)

Local minimum at x=4

9 a v =

1 dx = ( 3t + 1) 2 dt 1

x = ∫ ( 3t + 1) 2 dt

3

1 ( 3t + 1) 2 + c 3  3    2 3 2 = ( 3t + 1) 2 + c 9 When t = 0, x = 0 =

3

2 (3(0) + 1) 2 + c 9 2 0= +c 9 2 c=− 9 0=

3

2 2 x = ( 3t + 1) 2 − 9 9 2 2 (3t + 1)3 − x= 9 9 1 dx b v = = = (t + 2)−2 dt (t + 2)2 −2

x = ∫ ( t + 2 ) dt 1 = − (t + 2)−1 + c 1 1 =− +c (t + 2) When t = 0, x = 0 1 0=− +c (0 + 2) 1 0= − +c 2 1 c= 2 1 1 x= − 2 (t + 2) dx c v = = (2t + 1)3 dt

When x = 0, t = 0; 1 0 = (2(0) + 1)4 + c 8 1 0= +c 8 1 c=− 8 1 1 x = (2t + 1)4 − 8 8 dx (3t − 1) v= =e 1 0 a   dt x = ∫e dt 1 (3t − 1) = e +c 3 When t = 0, x = 0 1 0 = e(3(0) − 1) + c 3 1 −1 0= e +c 3 1 c=− 3e 1 1 x = e(3t − 1) − 3 3e dx = − sin(2t + 3) dt x = ∫ − sin(2t + 3)dt 1 = cos(2t + 3) + c 2 When t = 0, x = 0 1 0 = cos(2(0) + 3) + c 2 1 0 = cos(3) + c 2 1 c = − cos(3) 2 1 1 x = cos(2t + 3) − cos(3) 2 2 dx 11 v = = sin(2t ) + cos(2t ) dt a When t = 0, b v =

v = sin(0) + cos(0) v = 1 cm/s b x = ∫ ( sin(2t ) + cos(2t ) ) dt 1 1 = − cos(2t ) + sin(2t ) + c 2 2 When t = 0, x = 0 1 1 0 = − cos(0) + sin(0) + c 2 2 1 0= − +c 2 1 c= 2 1 1 1 x = − cos(2t ) + sin(2t ) 2 2 2

3

x = ∫ ( 2t + 1) dt

1 (2t + 1)4 + c 2(4) 1 = (2t + 1)4 + c 8

=

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 7 Antidifferentiation • EXERCISE 7.4   |  149 12 v = 0.25t ( 50 − t ) = 12.5t − 0.25t 2 dv a Greatest velocity occurs when =0 dt dv = 12.5 − 0.5t dt 0 = 12.5 − 0.5t 0.5t = 12.5 t = 25 When t = 25, v = 0.25(25)(50 − 25) = 156.25 m/s

5 = 2e 0 + c 5= 2+c c=3 x

y = 2e 2 + 3 1 6 a f ′( x ) = 5 − 2 x f ( x ) = 5x − x 2 + c When f (1) = 4 4 = 5(1) − (1)2 + c 4= 4+c c=0

b x = ∫ 0.25t (50 − t ) dt x = ∫ 12.5t − 0.25t 2

f ( x ) = 5x − x 2  x b f ′( x ) = sin  2

1 x = 6.25t − t 3 + c 12 When t = 0, x = 0 so c = 0 1 x = 6.25t 2 − t 3 12 2

x f ( x ) = −2 cos   + c  2 When f (π ) = 3

13 a f ′( x ) = 3e −3 x f ( x ) = − e −3 x + c y

(0, 0)

y = –e –3x + 2 y = –e –3x + 1

(0, 1)

y = –e –3x (0, –1) y = –e –3x – 2 (0, –3)

x

π 3 = −2 cos   + c  2 3= 0+c c=3 x f ( x ) = 3 − 2 cos    2 1 c f ′( x ) = = (1 − x )−2 (1 − x )2 1 (1 − x )−1 + c (−1)(−1) 1 f (x) = +c (1 − x ) f (x) =

y = –e –3x – 4 (0, –5)

When f (0) = 4 1 +c (1 − 0) 4 = 1+ c c=3 1 f (x) = +3 (1 − x ) 4=

b f ( x ) = − e −3 x + c When x = 0, y = 1 1 = −e0 + c 1 = −1 + c c=2

y

17 a

f ( x ) = 2 − e −3 x dy 14 = cos(2 x ) + 3e −3 x dx

(

)

y = ∫ cos(2 x ) + 3e dx 1 −3 x = sin(2 x ) − e + c 2 When x = 0, y = 4 −3 x

1 sin(0) − e 0 + c 2 4 = 0 −1+ c c=5 1 y = sin(2 x ) − e −3 x + 5 2 x dy = e2 15 dx 4=

Local maximum at x=2

y = f(x)

–4

0

2

Local minimum at x = –4

x

y = 2e 2 + c When x = 0, y = 5

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

x

150 | TOPIC 7 Antidifferentiation • EXERCISE 7.4 b

12 + 6(0) + c (0 − 1) c = −12 12 x = 6t − − 12 (t − 1)

y

x=−

b When t = 3, 12 − 12 (3 − 1) = 18 − 6 − 12 =0

x = 6(3) − 0

–4

2

x

y = f(x)

dx = 3t 2 + 7t dt 7 x = t3 + t2 + c 2 When t = 0, x = 0; 7 0 = (0)3 + (0)2 + c 2 c=0 7 x = t3 + t2 2 πt dx 1 9 a v = = 3π sin    8 dt

After 3 seconds the particle is at the origin again. dP 2 1 a = 30e 0.3t dt 30 0.3t P= e +c 0.3 P = 100e 0.3t + c When t = 0, P = 50 50 = 100e 0 + c 50 = 100 + c c = −50

18 v =

8 πt × 3 cos   + c  8 π π t x = −24 cos   + c  8

P = 100e 0.3t − 50 b When t = 10, P = 100e3 − 50 = 1959 There are 1959 seals after 10 years. dh π πt 22 = cos    4 dt 2

When t = 0, h = 3 3 = 2 sin ( 0 ) + c c=3 πt h = 2 sin   + 3  4

x=−

When t = 0, x = 0; 0 = −24 cos ( 0 ) + c c = 24 πt x = 24 − 24 cos    8 b x MAX = 24 − 24(−1) = 24 + 24 = 48 x MIN = 24 − 24(1) = 24 − 24 = 0 Maximum displacement is 48 metres. π c When t = 4, x = 24 − 24 cos   = 24  2 After 4 seconds the particle is 24 metres above the stationary position. 12 dx 2 0 a   v= = +6 dt (t − 1)2 dx v= = 12(t − 1)−2 + 6 dt x = −12(t − 1)−1 + 6t + c 12 x=− + 6t + c (t − 1) When t = 0, x = 0;

πt πt 4 π × sin   + c = 2 sin   + c  4 π 2  4

a h =

b Maximum depth = 2(1) + 3 = 5  m Minimum depth = 2(−1) + 3 = 1  m c

h 5 h=4

4

(24, 3)

(0, 3)

1 0

()

πt + 3 h = 2sin – 4 8

16

πt 4 = 2 sin   + 3  4 πt 1 = 2 sin    4 πt 1 = sin    2 4 π 1 indicates . Since sin is positive then 1st and 2nd 6 2 quadrants.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

24 t

TOPIC 7 Antidifferentiation • EXERCISE 7.4   |  151

πt π π π π π π = , π − , 2π + , 3π − , 4π + , 5π − 4 6 6 6 6 6 6 π t π 5π 13π 17π 25π 29π = , , , , , 4 6 6 6 6 6 6 π 4 5π 4 13π 4 17π 4 25π 4 29π 4 × , × , × , × , × t= × , π π π π 6 π 6 π 6 6 6 6 2 10 26 34 50 58 t= , , , , , 3 3 3 3 3 3 2 10 26 34 50 58 h ≥ 4 when h : ≤ t ≤ ∪ h: ≤t ≤ ∪ h: ≤t ≤ 3 3 3 3 3 3 8 8 8 24 This is + + = = 8 hours/day. 3 3 3 3 dN = 400 + 1000 t 2 3 a dt 1 dN = 400 + 1000t 2 dt 3 2000 2 N = 400t + t +c 3 2000 3 N = 400t + t +c 3 When t = 0, N = 40 2000 3 40 = 400(0) + 0 +c 3 c = 40 2000 3 N = 400t + t + 40 3 b When t = 5, 2000 N = 400(5) + (5)3 + 40 3 2000 N = 2000 + 125 + 40 3 N = 9494 families

{

}{

}{

}

dx = 2t cos(t ) dt x = 2t sin(t ) + 2 cos(t ) + c

24 v =

When t = 0, x = 0; 0 = 2(0) sin(0) + 2 cos(0) + c so c = −2 x = 2t sin(t ) + 2 cos(t ) − 2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 8 Integration • EXERCISE 8.2   |  153

Topic 8 — Integration Exercise 8.2 — The fundamental theorem of integral calculus 1 a Left end-point rectangle rule: 2 f (0.5) = 2, f (1) = 1, f (1.5) = , f (2) = 0.5 3 Approximate area 2 = 0.5 × 2 + 0.5 × 1 + 0.5 × + 0.5 × 0.5 3 2   = 0.5  2 + 1 + + 0.5   3 25 2 = units 12 b Right end-point rectangle rule: f (2.5) = 0.4 Approximate area 2 = 0.5 × 1 + 0.5 × + 0.5 × 0.5 + 0.5 × 0.4 3 2   = 0.5  1 + + 0.5 + 0.4    3 77 = units 2 60 3 2 f (1) = −0.01(1) (1 − 5)(1 + 5) = 0.24 f (2) = −0.01(2)3 (2 − 5)(2 + 5) = 1.68

f (3) = −0.01(3)3 (3 − 5)(3 + 5) = 4.32 f (4) = −0.01(4)3 (4 − 5)(4 + 5) = 5.76 Approximate area = 1(0.24 + 1.68 + 4.32 + 5.76) = 12 units 2 1

3 a

∫ (4 x

3

+ 3 x 2 + 2 x + 1) dx

0

1

=  x 4 + x 3 + x 2 + x 

(

)

0

= 14 + 13 + 12 + 1 − 0 =4 π

b

∫ (cos( x ) + sin( x )) dx

−π

π

= [sin( x ) − cos( x ) ]− π = ( sin(π ) − cos(π ) ) − ( sin(−π ) − cos(−π ) ) = ( 0 − (−1) ) − ( 0 − (−1) ) = 1−1 =0

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

154 | TOPIC 8 Integration • EXERCISE 8.2 ( x + 1)3 = x 3 + 3 x 2 + 3 x + 1

4 a 2

3 ∫ ( x + 1) dx =

−3

∫ (x

3

+ x 2 + x + 1) dx

−3

2 1 3 =  x 4 + x 3 + x 2 + x  2 4  −3 3 2 3 1 4  1  3 =  (2) + (2) + (2) + (2) −  (−3)4 + (−3)3 + (−3)2 + (−3) 4  4  2 2 27   81 = ( 4 + 8 + 6 + 2 ) −  − 27 + − 3  4  2  135  = 20 −  − 30   4  80  135 120  = − −  4  4 4  80 15 = − 4 4 65 = 4

1

b

2

∫ (e

x

0

1

+ e− x

)2 dx

(

)

= ∫ e 2 x + 2 + e −2 x dx 0

1 1 1 =  e 2 x + 2 x − e −2 x  2 2 0 1 1 −2(1)   1 0 1 0  2(1) =  e + 2(1) − e  −  e + 2(0) − e  2 2 2 2 1 2 1 −2 1 1 = e +2− e − + 2 2 2 2 = 2 + 0.5e 2 − 0.5e −2 5

5

5

∫ m( x ) dx = 7 and ∫ n( x ) dx = 3 2

2

5

5

2

2

a ∫ 3m( x ) dx = 3 ∫ m( x ) dx = 3(7) = 21 5

b

2

c

5

5

2

2

∫ ( 2m( x ) − 1) dx = 2 ∫ m( x ) − ∫ 1 dx = 2(7) − [ x ]52 = 14 − (5 − 2) = 14 − 3 = 11

2

5

5

2 5

5

2

2

∫ ( m( x ) + 3) dx = − ∫ ( m( x ) + 3) dx = − ∫ m( x ) dx − ∫ 3 dx = −7 − [3 x ]52 = −7 − ( 3(5) − 3(2) ) = −7 − 15 + 6 = −16 5

d

5

5

5

2

2

2

∫ ( 2m( x ) + n( x ) − 3) dx = 2 ∫ m( x ) dx + ∫ n( x ) dx − ∫ 3 dx 2

= 2(7) + 3 − [3 x ]52 = 14 + 3 − ( 3(5) − 3(2) ) = 17 − 9 =8

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 8 Integration • EXERCISE 8.2   |  155 1

∫ (4 x

6

3

)

− 3 x 2 + 1 dx = 0

k

1

 x 4 − x 3 + x  = 0 k

(14 − 13 + 1) − ( k 4 − k 3 + k ) = 0 1− k + k3 − k4 = 0 (1 − k ) + k 3 (1 − k ) = 0

(1 − k )(1 + k 3 ) = 0 (1 − k )(1 + k )(1 − k + k 2 ) = 0 k = ±1 as 1 − k + k 2 cannot be further factorised Verification: 1

∫ (4 x

3

−1

)

1

− 3 x 2 + 1 dx =  x 4 − x 3 + x 

(

−1

) (

)

= 1 − 1 + 1 − (−1)4 − (−1)3 − 1 = 1−1 =0 1

∫ (4 x

3

4

3

)

− 3 x 2 + 1 dx = 0

1

7 a Approximate area = 8 ×1+ 9 ×1+ 8 ×1+ 5 ×1 = 30 units 2 b Approximate area = 9 ×1+ 8 ×1+ 5 ×1 = 22 units 2 8 Approximate area = 1(1.75) + 1(3) + 1(3.75) + 1(4) + 1(3.75) + 1(3) + 1(1.75) = 2(1.75) + 2(3) + 2(3.75) + 4 = 21 units 2 9 a f ( x ) = x (4 − x ) Graph intersects the x axis where y = 0 x (4 − x ) = 0 x = 0 or 4 − x = 0 4=x Thus a = 4. b f (0) = 0 f (1) = 1(4 − 1) = 3; f (2) = 2(4 − 2) = 2.8284; f (3) = 3(4 − 3) = 1.7321; f (4) = 0 Left End-point Rule: Approximate area = 0.5 ( f (0) + f (1) + f (2) + f (3) ) = 0.5 ( 0 + 3 + 2.8284 + 1.7321) = 7.56 units 2 Right End-point Rule: Approximate area = 0.5 ( f (1) + f (2) + f (3) + f (4) ) = 0.5 ( 3 + 2.8284 + 1.7321 + 0 ) = 7.56 units 2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

156 | TOPIC 8 Integration • EXERCISE 8.2 3

10 a

∫ (3x

2

0

)

− 2 x + 3 dx =  x 3 − x 2 + 3 x 

(

3

3 0

)

2

= 3 − 3 + 3(3) − 0 = 27 − 9 + 9 = 27 2 2  2 x 3 + 3x 2  2 = dx b ∫  ∫ 2 x + 3x dx , x ≠ 0  x  1 1

(

)

2 2 3 =  x 3 + x 2  2 1 3  2 3 3 2  2 3 3 2 =  (2) + (2)  −  (1) + (1)  3  3  2 2 16 2 3 = +6− − 3 3 2 28 36 9 = + − 6 6 6 55 = 6 1

1 1 1 − e −2 x dx =  e 2 x + e −2 x  2 2  −1 −1 1 1 1   1  =  e 2(1) + e −2(1)  −  e 2( −1) + e −2( −1)  2    2 2 2 1 2 1 −2 1 −2 1 2 = e + e − e − e 2 2 2 2 = 0 4π 4π x x d ∫ sin   dx =  −3 cos     3  3   2π 

c

∫ (e

)

2x

2π

−1

e

∫

−3

 4π   2π  = −3 cos  + 3 cos    3   3  = 1.5 − 1.5 =0 −1

1

− 2 dx = 2 ∫ (1 − 3 x ) 2 dx 1 − 3x −3 −1

1   1  = 2  2  −  (1 − 3 x ) 2    3   −3 1 1 2  2  = 2  − (1 + 3) 2 + (1 + 9 ) 2   3  3  4 2 10  = 2 − + 3   3 4 = 10 − 2 3

(

π 2

)

π

 1  x   x  2 ∫  cos(2 x ) − sin  2  dx =  2 sin(2 x ) + 2 cos  2  π π −

f

−

3

3

π   1  2π  π  1 =  sin(π ) + 2 cos    −  sin  −  + 2 cos  −    4   2  3   6  2  3 3 1  1 =  (0) + 2  −   − + 2  2   2  2   2   = 2− 5

3 3 4

5

11 Given that ∫ f ( x ) dx = 7.5 and ∫ g( x ) dx = 12.5 0

0

5

5

0

0

a ∫ −2 f ( x ) dx = −2 ∫ f ( x ) dx = −2 × 7.5 = −15

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 8 Integration • EXERCISE 8.2   |  157 0

5

5 5

0

b ∫ g( x ) dx = − ∫ g( x ) dx = −12.5 5

5

0

0

∫ (3 f ( x ) + 2) dx = 3 ∫ f ( x ) dx + ∫ 2 dx

c

0

= 3 × 7.5 + [ 2 x ]50 = 22.5 + (2(5) − 0) = 22.5 + 10 = 32.5 5

5

5

0

0

0

∫ ( g( x ) + f ( x )) dx = ∫ g( x ) dx + ∫

d

5

e

3

5

5

0

3

0

3

∫ x 2 dx = − 1

h

5

0

0

∫ g( x ) dx + ∫ g( x ) dx = ∫ g( x ) dx = 12.5 h

12

5

∫ (8g( x ) − 10 f ( x )) dx = 8 ∫ g( x ) dx − 10 ∫ f ( x ) dx = 8(12.5) − 10(7.5) = 25 0

f

f ( x ) dx = 7.5 + 12.5 = 20

12 5

12 5 1 h 12 3 x −1  = − 1 5 h 3 12   =−  x  5 1 3 12     − 3 = − h 5 3 12 15 =− + h 5 5 3 27 = h 5 h 5 = 3 27 5 5 h= ×3= 27 9

∫ 3x

−2

dx = −

a

13

∫e

−2 x

dx =

0

a  − 1 e −2 x   2  0  1 −2 a   1 0  − e  −− e   2   2  1 1 − 2 2e 2 a 1 1  1 −  2  e2a 

1 1  1 − 8  2 e

1 1  1 − 8  2 e 1 1 = 1 − 8  2 e  1 1 = 1 − 8  2 e  1 1 = 1 − 8  2 e  =

e 2 a = e8 2 a = 8 a=4 1 4 a Graph cuts the x axis when f ( x ) = 0. f ( x ) = x 3 − 8 x 2 + 21x − 14 f (1) = 13 − 8(1)2 + 21(1) − 14 = 0 Thus ( x − 1) is a factor. x 3 − 8 x 2 + 21x − 14 = ( x − 1)( x 2 − 7 x + 14) As x 2 − 7 x + 14 has a discriminant such that ∆ = (−7)2 − (4(1)14) = 49 − 56 = −7 there are no factors. ( x − 1)( x 2 − 7 x + 14) = 0 x −1= 0 x = 1 so a = 1

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

158 | TOPIC 8 Integration • EXERCISE 8.2 5

b

∫ (x

)

3

− 8 x 2 + 21x − 14 dx

1

5 1 8 21 =  x 4 − x 3 + x 2 − 14 x  3 2 4 1 8 21   1 4 8 3 21 2  1 =  (5) − (5) + (5) − 14(5) −  (1)4 − (1)3 + (1)2 − 14(1)  4  4 3 2 3 2 625 1000 525 1 8 21 = − + − 70 − + − + 14 4 3 2 4 3 2 624 992 504 = − + − 56 4 3 2 2 = 156 − 330 + 252 − 56 3 2 = 408 − 386 3 1 = 21 units 2 3

15 a y = x sin( x ) dy = x cos( x ) + sin( x ) dx π 2

π 2

−π

−π

∫ 2 x cos( x ) dx = 2 ∫ x cos( x ) dx

b

π

  2   = 2  x sin( x ) − ∫ sin( x )  −π     π

= 2 [ x sin( x ) + cos( x )]−2π π π   π  = 2  sin  + cos   − 2 (−π sin(π ) + cos(π )) 2  2  2 π  = 2  (1) + 0  − 2 (−π (0) − 1) 2  =π +2 16 a y = e x − 3 x + 2 3 2 dy = 3x 2 − 6 x e x −3x dx 3 2 dy = 3 x 2 − 2 x e x −3x dx 3

2

(

) )

(

1

(

)

(

)

b ∫ 3 x 2 − 2 x e x 0

1

3∫ x 2 − 2x e x 0 1

∫ (x

2

)

− 2x ex

0

−3x 2

dx = e x

3

−3x 2

dx = e1 − 3(1) − e 0

3

−3x 2

dx =

k

(

∫1 ( 2 x − 3) dx = 7 − 3

17

1

3

3

−3x 2

3

(

 0

2

)

)

1 −2 e −1 3 5

k

 x 2 − 3 x  = 7 − 3 5 1

(

)

k 2 − 3k − 12 − 3(1) = 7 − 3 5 2

k − 3k + 2 = 7 − 3 5 Solve using CAS: k = 3 − 5, 5 = 5, k > 1 0

1 + e 2 x − 2 xe 2 x dx = 1.964 18 ∫ (e 2 x + 1)2 −2 (Solved using CAS) Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 8 Integration • EXERCISE 8.3   |  159 Area is

Exercise 8.3 — Areas under curves

0

=−∫

1 Area is 25

25

0

0

−8 8 3

1

∫ 2 x dx = 2 ∫ x 2 dx

0 8

2 = 2  3  0 3 3 2  2 = 2  (25) 2 − (0) 2  3 3 

( )

0

8

3 4  = 2 x3   4  0 4 3  = 2  23 3  − 0 4  3 = 2 × × 16 4 = 24 units 2

3

( )

π

π ∫ ( 2 sin(2 x ) + 3) dx = [− cos(2 x ) + 3x ]0 0

5 Area =

= ( − cos(2π ) + 3π ) − ( − cos(0) + 3(0) ) = 3π units

2

=

−0.5

∫

−0.5

1 dx x2 −2.5

∫

x −2 dx

−2.5

y

3

1

= 2 ∫ x 3 dx

4 2 2 5 3 4 = × 125 3 2 = 166 units 2 3 2

0

= 2 ∫ xdx by symmetry

3 25  x2 

=

8

xdx + ∫ 3 xdx

3

=  − x −1 

(0, 0) x

1

−2.5

1 −0.5 =  −   x  −2.5  1   1  = −  0.5   2.5  = 2 − 0.4

y = 1 – e–x

–1

−0.5

= 1.6 units 2 y

6 a

Area is 0

(

1

)

(

)

= − ∫ 1 − e − x dx + ∫ 1 − e − x dx −1

= −  x + e − x 

(

0 −1

0

) ( − 1)

0

= − (0 + e 0 ) − (−1 + e1 ) + (1 + e −1 ) − (0 + e 0 )

(

= − (1 + 1 − e ) + 1 + e = −2 + e + e −1

−1

) 0

b Area =

3

x

3

1 3  1 −  1 2 dx =  2 x 2  = 2 3 − 2 1 = 2 3 − 2 units 2 = dx x ∫ x ∫      1 1 1 7 a Graph intersects x axis where y = 0. 3

y

3

y= x (0, 0) –8

1

x=0

= e + e −1 − 2 units 2 4

1 y=– x

1

+  x + e − x 

8 x

2 sin( x ) + cos( x ) = 0 2 sin( x ) = − cos( x ) 2 tan( x ) = −1 1 tan( x ) = − 2 1 suggest 0.4636. Since tan is negative 2nd quadrant as 2 0 ≤ x ≤ π. x = π − 0.4636 x = 2.6779 Thus (m, 0) = (2.6779, 0).

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

160 | TOPIC 8 Integration • EXERCISE 8.3 b Area 2.6779

∫ ( 2 sin( x ) + cos( x )) dx

=

0

= [ −2 cos( x ) + sin( x ) ]0 = ( −2 cos(2.6779) + sin(2.6779) ) − ( −2 cos(0) + sin(0) ) = 1.7888 + 0.4473 + 2 2.6779

= 4.2361 units 2 8 a

2

y = ex 2 dy = 2 xe x dx 0

(

b − ∫ 2 xe x −1

2

1

1

) dx + ∫ (2xe ) dx = 2 ∫ (2xe ) dx by symmetry x2

0

x2

0

1

= 2 e x  0 2

(

2

= 2 e(1) − e 0

)

= 2 ( e − 1) units 2 9 a y = f ( x ) = − ( x 2 − 1)( x 2 − 9) Graph cuts the y axis where x = 0, y = − (−1)(−9) = −9. Graph cuts the x axis where y = 0 − ( x 2 − 1)( x 2 − 9) = 0 −( x − 1)( x + 1)( x − 3)( x + 3) = 0 x −1 = 0 x +1 = 0 x − 3 = 0 x + 3 = 0 x = 1, x = −1 x = 3 x = −3 dy TP’s occur where = 0. dx dy = −2 x ( x 2 − 9) − 2 x ( x 2 − 1) dx dy = −2 x 3 + 18 x − 2 x 3 + 2 x dx dy = −4 x 3 + 20 x dx 0 = −4 x ( x 2 − 5) 0 = −4 x ( x − 5)( x + 5) x =0 x− 5 =0 x+ 5 =0 x= 5 When x = 0, y = −9.

x =− 5

((

When x = ± 5, y = − ± 5

)

2

)(( ±

−1

5

)

2

)

− 9 = −4 × −4 = 16

y (–2.2361, 16)

(–3, 0)

(–1, 0)

16

0

(2.2361, 16)

(1, 0)

(3, 0) x

(0, –9) f(x) = –(x2 – 1)(x2 – 9)

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 8 Integration • EXERCISE 8.3   |  161 b By symmetry area is 3  1  2  − ∫ − ( x 2 − 1)( x 2 − 9) dx + ∫ − ( x 2 − 1)( x 2 − 9) dx   0  1

(

)

(

)

3  1  = 2  − ∫ − x 4 + 10 x 2 − 9 dx + ∫ − x 4 + 10 x 2 − 9 dx   0  1

(

)

(

)

1 3   1 10 1 10 = 2  −  − x 5 + x 3 − 9 x  +  − x 5 + x 3 − 9 x   3 3 0  5 1    5 10   1 5 10 3   1 5 10 3   1  = 2  −  − (1) + (1) − 9(1) − 0  +  − (3) + (3) − 9(3) −  − (1)5 + (1)3 − 9(1)    5   5    5 3 3 3 2 = 52.27 units 10 Required area

π  = 4  ∫ ( 2 sin( x ) + 3 cos( x ) ) dx  0 

(

π

= 4 [ −2 cos( x ) + 3sin( x ) ]0

)

= 4 (( −2 cos(π ) + 3sin(π ) ) − ( −2 cos(0) + 3sin(0) )) = 4 (2 + 0 + 2 − 0) = 16 units 2 11 a y = −0.5( x + 2)( x + 1)( x − 2)( x − 3) Graph cuts the x axis where y = 0. −0.5( x + 2)( x + 1)( x − 2)( x − 3) = 0 x + 2 = 0, x + 1 = 0, x − 2 = 0, x − 3 = 0 x = −2, x = −1, x = 2, x =3 Thus a = −2, b = −1, c = 2 and d = 3. b Area is 2

3

= − ∫ ( −0.5( x + 2)( x + 1)( x − 2)( x − 3) ) dx + 2 ∫ ( −0.5( x + 2)( x + 1)( x − 2)( x − 3) ) dx −1 2

2

3

7 7 = − ∫  −0.5 x 4 + x 3 + x 2 − 4 x − 6 dx + 2 ∫  −0.5 x 4 + x 3 + x 2 − 4 x − 6 dx     2 2 2 −1 = 13.05 + 2 × 1.3167 = 15.68 units 2 12 a y = ax ( x − 2) When x = 1, y = 3 3 = a(1)(1 − 2) 3 = −a a = −3 Thus y = −3 x ( x − 2) = −3 x 2 + 6 x b Area of glass is 2

∫ (−3x

2

+ 6 x ) dx

0

=  − x 3 + 3 x 2 

(

2 0

)

= − (2)3 + 3(2)2 − ( (0) − (0) ) = −8 + 12 − 0 = 4 m2 c Two windows = 8 metres2. Cost = $55 × 8 = $440

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

162 | TOPIC 8 Integration • EXERCISE 8.4 1

1 3 3 3 13 a ∫ 3 x 3 dx =  x 4  = (1)4 − 0 = units 2 4 4 4   0 0

b Orange region = 3 × 1 − π 2

14 a

1 3 = 2 units 2 4 4 π 2 0

∫ 2 sin( x ) dx = [−2 cos( x )] 0

π = −2 cos   − ( −2 cos(0) )  2 =0+2 =2 π b Shaded region is 2  2 × − 2 = 2 (π − 2 ) units 2   2 4

15 a

∫(

)

x ( x − 3)2 dx

0

4 1  = ∫  x 2 ( x 2 − 6 x + 9) dx  0 3 1 4 5  = ∫  x 2 − 6 x 2 + 9 x 2  dx  0

3   2 =  4(16) − (4 2 ) 2  − 0 3   2 3 = 64 − × 4 3 1 = 21 units 2 3

2 y = ( x − 3)2 ....................... (1) y = 9 − x ............................(2) (1) = (2) ( x − 3)2 = 9 − x x 2 − 6x + 9 = 9 − x x 2 − 5x = 0 x ( x − 5) = 0 x = 0 or x − 5 = 0 x =5 When x = 0, y = 9 − 0 = 9 When x = 5, y = 9 − 5 = 4 Graphs intersect at (0,9) and (5,4). y

5 3 4  2 2 x + 6x 

 2 7 12 =  x2 − 5  7  0 7 5 3 2  12 =  (4) 2 − (4) 2 + 6(4) 2  − 0 7 5   = 36.5714 − 76.8 + 48 = 7.7714 cm 2 b Area of whole motif is 7.7714 × 2 = 15.5428  15.5 cm 2 c 0.34875 m 2 = 3487.5 cm 2 Scale factor is 3487.5 ÷ 15.5428  224 2

16 a

∫e

−x

2

dx = 1.7642 units 2

−2

3

x 2 + 3x − 4 x 2 + 3x − 4 dx + ∫ dx b − ∫ 2 x +1 x2 + 1 −2 1 = 7.8371 + 2.0959 = 9.933 units

2

Exercise 8.4 — Applications 1 y = 4.................(1) y = x ..............(2) (1) = (2) x =4 x = 16 When x = 16, y = 4, therefore POI = (16, 4) Shaded region is 16

= =

∫ (4 −

0 16 

)

x dx

1  4 − x  ∫  2  dx 0 16

3  2  = 4 x − x 2  3   0

y = (x – 3)2

(5, 4) y=9–x

(3, 0)

x

0

Area is 5

(

)

= ∫ 9 − x − ( x − 3)2 dx 0 5

Solved using CAS. 1

(0, 9)

(

)

= ∫ 9 − x − x 2 + 6 x − 9 dx 0 5

(

)

= ∫ − x 2 + 5 x dx 0

5 1 5 =  − x 3 + x 2  2 0  3 1 5   =  − (5)3 + (5)2  − 0  3  2 125 125 =− + 3 2 250 375 =− + 6 6 125 = 6 5 = 20 units 2 6

3 Point of intersection between the graphs: sin( x ) = − cos( x ) 0 ≤ x ≤ π tan( x ) = −1 π Basic angle = , tan is negative in the 2nd quadrant 4 3π x= 4

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 8 Integration • EXERCISE 8.4   |  163 y

y = sin(x)

1

(0, 0)

(π, 1)

π – 2

3π – 4

π

x

y = – cos(x)

(0, –1)

Area is A=

3π 4

∫ (sin( x ) + cos( x )) dx + 0

3π

π

∫ ( − cos( x ) − sin( x )) dx

3π 4

π

= [ − cos( x ) + sin( x ) ]04 + [ − sin( x ) + cos( x ) ]3π 4

     3π   3π   3π   3π    =  − cos   + sin   − ( − cos(0) + sin(0) ) +  − sin(π ) + cos(π ) −  − sin   + cos      4   4   4   4      2 2 2 2 + +1− 0 + 0 −1+ + 2 2 2 2 = 2 2 units 2 =

4 Area is =

3.92

5

0

3.92

−x −x ∫ ( 4 − x − 4e ) dx + ∫ ( 4e − 4 + x ) dx

5 3.92 1 1 =  4 x − x 2 + 4e −4 x  +  −4e − x − 4 x + x 2  2 2 3.92  0  = 4.6254 units 2

5 Average value = =

b 1 f ( x ) dx ∫ a b−a

1

1

∫ 3e  0

1−0   3

3x

dx

1

1 3 = 3  e3 x  3  0

1  1 = 3  e1 − e 0  3 3  = e −1 b 1 6 Average value = f ( x ) dx b − a ∫a 1 1 = x 2 − 2 x dx 1 − 0.5 ∫0.5 1 1 = 2  x 3 − x 2  3 

(

)

0.5

3 2  1   1  1  1  = 2   (1)3 − (1)2  −    −       3  2  2    3  2 5 = 2 − +   3 24   16 5  = 2 − +  24 24  11 =− 12

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

164 | TOPIC 8 Integration • EXERCISE 8.4 d Distance is

1

7

− dL 4 = = 4t 2 dt t

1 8  = ∫  1 + 3(t + 1) 2  dt  0

Average total increase in length is 36

8

 1 4 t = 8t 2  ∫  6 6 = 8 36 − 8 6 = 48 − 19.6 = 28.4 cm Therefore, the average total increase in length is 28.4 cm dN 8 a & b = 0.8t + 2 dt 36

−

1 2 dt

(

) (

3   = t + 2(t + 1) 2    0 3 3     =  8 + 2(8 + 1) 2  −  0 + 2(0 + 1) 2     

)

dN — dt

( )

t π dx = 3 cos  −   2 4 dt t π x = ∫ 3 cos  −  dt  2 4 t π x = 6 sin  −  + c 2 4

10 a v = (30, 26)

26

3

= 8 + 2 32 2 − 2 = 8 + 54 − 2 = 60 metres

dN — = 0.8t + 2 dt

When t = 0, x = −3 2

π −3 2 = 6 sin  −  + c  4

(0, 2) 0

10

20

 2 −3 2 = 6  − +c  2 

t

30

−3 2 = −3 2 + c c=0 t π Thus x = 6 sin  −   2 4

c Number of bricks 20

∫ ( 0.8t + 2) dt

=

10

=  0.4t 2 + 2t 

(

b When t = 3π ,

20

10

) (

= 0.4(20)2 + 2(20) − 0.4(10)2 + 2(10) = 160 + 40 − 40 − 20 = 140 bricks

 3π π  x = 6 sin  −  2 4   6π π  x = 6 sin  −  4 4   5π  x = 6 sin    4 

)

1

9 v = 1 + 3 t + 1 = 1 + 3 ( t + 1) 2 a Initially t = 0, v = 1 + 3 1 = 4 m/s dv 3 = (1) ( t + 1) dt 2

x = 6×−

3 2 t +1 3 3 i When t = 0, a = = = 1.5 m/s 2 2 1 2 3 3 = = 0.5 m/s 2 ii When t = 8, a = 2 8 +1 6

b a =

c

1 − 2

=

c

2 = −3 2 m 2

v 3

(

0, 3 2 2

v

(4π, 3 2 2 (

(2 4 (

v = 3 cos –t – –π

(

(7π–2 , 0 (

(3π–2 , 0 (

0 (10, 10.95) y = 1 + 3 x+ 1 –3

d Distance is (0, 4)

=

3π 2

t

π

∫ 3 cos  2 − 4 

dt −

0

0

10

t

3π

t

2

3π 2

t π = 2 ∫ 3 cos  −  dt 2 4 0

3π

π

∫ 3 cos  2 − 4  dt 3π

t π 2  = 2 6 sin  −    2 4  0  Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual π    3π π  = 2  6 sin  −  − 6 sin  −  

4π t

3π 2

0 3π 2

t π = 2 ∫ 3 cos  −  dt 2 4 0

TOPIC 8 Integration • EXERCISE 8.4   |  165

3π

t π 2  = 2 6 sin  −    2 4  0  π    3π π  = 2  6 sin  − − 6 sin  −    4   4 4   π π   = 2  6 sin   − 6 sin  −    2  4  

(

)

= 2 6+3 2 = 20.49 m  t π e v = 3cos  −   2 4 dv t π 3 a= = − sin  −  dt 2  2 4 f When t = 3π , 3  3π π  a = − sin  −  2  2 4 3  6π π  a = − sin  −  2  4 4 3  5π  a = − sin   2  4  2 3 2 3 a=− ×− = m/s 2 2 2 4 11 a y = 3 x 3 − x 4 ....................(1) y = 3 − x ..........................(2) (1) = (2) 3x 3 − x 4 = 3 − x 0 = x 4 − 3x 3 − x + 3 0 = x 3 ( x − 3) − ( x − 3) 0 = ( x − 3)( x 3 − 1) 0 = ( x − 3)( x − 1)( x 2 + x + 1) x − 3 = 0 or x − 1 = 0 as x 2 + x + 1 cannot be further factorised x=3 x =1 When x − 1, y = 3 − 1 = 2 When x − 3, y = 3 − 3 = 0 Thus (a, b) = (1, 2) and (c, 0) = (3, 0) b Area is 3

(

)

= ∫ 3 x 3 − x 4 − (3 − x ) dx 1 3

(

)

= ∫ 3 x 3 − x 4 − 3 + x dx 1

3 3 1 1 =  x 4 − x 5 − 3 x + x 2  5 2 1 4 3 1 5 3 4 1 2  =  − x + x + x − 3 x  4 2  5 1 3 1  1 5 3 4 1 2   1  =  − (3) + (3) + (3) − 3(3) −  − (1)5 + (1)4 + (1)2 − 3(1)  5   5  4 2 4 2  243 243 9   1 3 1  = − + + − 9 −  − + + − 3  5 4 2   5 4 2  242 240 =− + +4−6 5 4 968 1200 40 =− + − 20 20 20 192 = 20 = 9.6 units 2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

166 | TOPIC 8 Integration • EXERCISE 8.4 c Average value =

b 1 f ( x ) dx b − a ∫a

(

)

2.5 1 3 x 3 − x 4 dx 2.5 − 1 ∫1 2.5 2 3 1 =  x 4 − x 5  3 4 5 1  2  3  5 4 1  5 5   3 1  =     −    −  (1)4 − (1)5    3   4  2 5  2   4 5

=

2  1875 1250 11  − −   3  64 64 20  = 6.144 = 12 Area is 0.5

x ∫ ( cos( x ) − 0.5e ) dx

=

−1.5

= sin( x ) − 0.5e x 

(

0.5 −1.5

) (

= sin(0.5) − 0.5e 0.5 − sin(−1.5) − 0.5e −1.5 = ( 0.4794 − 0.8244 ) − ( −0.9975 − 0.1116 )

)

= 0.7641 units 2 13 a y 2 = 4 − x ....................(1) y = x − 2......................(2) Substitute (2) into (1)

( x − 2 )2 = 4 − x x2 − 4x + 4 = 4 − x x 2 − 3x = 0 x ( x − 3) = 0 x = 0 or x − 3 = 0 x=3 When x = 0, y = 0 − 2 = −2 When x = 3, y = 3 − 2 = 1 Points of intersection are (0, −2) and (3,1). b Green region is 3

4

2

3

= ∫ ( x − 2) dx + ∫ 4 − x dx 4

3 3 1  2  =  x 2 − 2 x  +  − ( 4 − x ) 2  2 2  3 3 3 3   2  1 2  1 2   2 =  (3) − 2(3) −  (2) − 2(2) +  − ( 4 − 4 ) 2  −  − ( 4 − 3) 2  2  2   3   3  2 9  =  − 6 − ( 2 − 4 ) + 0 + 2  3 3 2 = − +2+ 2 3 7 2 = units 6 c Orange region is 2

(

(

= ∫ x −2− − 4− x 0 2

2

)) dx + ∫ ( − 4

)

4 − x dx

1 1 2  = ∫  x − 2 + (4 − x ) 2  dx − ∫ ( 4 − x ) 2 dx  4 0 2

2

3 3 1   2  2 =  x 2 − 2 x − (4 − x ) 2  −  − (4 − x ) 2  3  2  0  3  4 3 3 3 3 1  1   2  2 2 2 =  (2)2 − 2(2) − (4 − 2) 2  −  (0)2 − 2(0) − (4 − 0) 2  −  − (4 − 2) 2 + (4 − 4) 2  3 3 3 2  2   3  3

3

3

2 2 2 Quest 2 = 2 − 4 − ( 2 ) 2 + Maths (4) + (2) 2 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual 3 3 3

= ∫  x − 2 + (4 − x )  dx − ∫ ( 4 − x ) dx  4 0 2

2

3 3   2  1 TOPIC 8 Integration • EXERCISE 8.4   |  167 2 =  x 2 − 2 x − (4 − x ) 2  −  − (4 − x ) 2  3  2  0  3  4 3 3 3 3 1  1   2  2 2 2 =  (2)2 − 2(2) − (4 − 2) 2  −  (0)2 − 2(0) − (4 − 0) 2  −  − (4 − 2) 2 + (4 − 4) 2  3 3 3 2  2   3  3

3

3

2 2 2 = 2 − 4 − ( 2 ) 2 + (4) 2 + (2) 2 3 3 3 1 = −2 + 5 3 1 = 3 units 2 3

d Area between graphs is = green region + orange region = 1.17 + 3.3  4.5 units 2

14 Area is 3

=

∫ ( −0.5x

2

−3 3

=

∫ ( −0.75x

2

( x − 3)( x + 3) dx

∫ ( −0.75x

2

( x 2 − 9) dx

∫ ( −0.75x

4

+ 6.75 x 2 dx

−3 3

=

−3 3

=

)

( x − 3)( x + 3) − 0.25 x 2 ( x − 3)( x + 3) dx

)

)

)

−3

=  

−0.75 5 6.75 3 3 x + x  5 3  −3

=  −0.15 x 5 + 2.25 x 3 

(

3

−3

) (

= −0.15(3)5 + 2.25(3)3 − −0.15(−3)5 + 2.25(−3)3 = −36.45 + 60.75 − 36.45 + 60.75

)

= 48.6 units2 y

15 a y = 0.5(x + 4)(x – 1)(x – 3)

y = (3 – x)(x + 4) (0, 12)

(0, 6)

(−4, 0)

0 (1, 0)

(3, 0) x

b y = 0.5( x + 4)( x − 1)( x − 3)...........(1) y = (3 − x )( x + 4).........................(2) (1) = (2) 0.5( x + 4)( x − 1)( x − 3) = (3 − x )( x + 4) 0.5( x + 4)( x − 1)( x − 3) − (3 − x )( x + 4) = 0 0.5( x + 4)( x − 1)( x − 3) + ( x − 3)( x + 4) = 0 ( x − 3)( x + 4) ( 0.5( x − 1) + 1) = 0 ( x − 3)( x + 4) ( 0.5 x − 0.5 + 1) = 0 ( x − 3)( x + 4) ( 0.5 x + 0.5) = 0 x=3 x = −4 x = −1 When x = −4, y = (3 + 4)(−4 + 4) = 0 When x = −1, y = (3 + 1)(−1 + 4) = 12 When x = 3, y = (3 − 1)(3 + 4) = 0 Therefore, the co-ordinates are (−4, 0), (−1,12) and (3, 0).

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

168 | TOPIC 8 Integration • EXERCISE 8.4 c Area between curves is = = = = =

−1

3

∫ ( 0.5( x + 4)( x − 1)( x − 3) − (3 − x )( x + 4)) dx + ∫ ((3 − x )( x + 4) − 0.5( x + 4)( x − 1)( x − 3)) dx

−4 −1

−1

3

∫ (( x − 3)( x + 4)( 0.5( x − 1) + 1)) dx + ∫ ((3 − x )( x + 4)(1 + 0.5( x − 1))) dx

−4 −1

∫ (( x

2

3

)

+ x − 12)(0.5 x + 0.5) dx +

−4 −1

−1

∫ ( (− x

2

)

− x + 12)(0.5 x + 0.5) dx

−1

3

3 2 2 3 2 2 ∫ ( 0.5x + 0.5x − 6 x + 0.5x + 0.5x − 6) dx + ∫ ( −0.5x − 0.5x − 0.5x − 0.5x + 6 x + 6) dx

−4 −1

1

∫  2 x

3

+ x2 −

−4

−1

3

11 11   1  x − 6 dx + ∫  − x 3 − x 2 + x + 6 dx   2  2 2 −1

1 3 1 1 11 1 1 11 =  x 4 + x 3 − x 2 − 6 x  +  − x 4 − x 3 + x 2 + 6 x  3 4 3 4 8  −4  8  −1 = 12.375 + 26.7

= 39.04 units 2 v

16 a

()

x y = 0.5 sin –2 + 2

(0, 2)

0

(4π, 2)

2π

4π

t

b Area is =

4π





 x

∫  0.5sin  2  + 2 dx 0

4π

x   =  − cos   + 2 x   2  0 = ( − cos ( 2π ) + 2(4π ) ) − ( − cos ( 0 ) + 2(0) ) = −1 + 8π + 1 = 8π  25 m 2 c Soil required is 0.5 × 25 = 12.5 m3. 17 v = e −0.5t − 0.5 dv a a = = −0.5e −0.5t dt

(

)

b x = ∫ e −0.5t − 0.5 dt = −2e

−0.5t

− 0.5t + c

When x = 0, t = 0 0 = −2e −0.5t (0) − 0.5(0) + c c=2 x = −2e −0.5t − 0.5t + 2 c When t = 4, x = −2e −0.5(4) − 0.5(4) + 2 = −0.2707 metres

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 8 Integration • EXERCISE 8.4   |  169 d Fourth second occurs between t = 3 and t = 4.

b Total cost is 10

Distance 3

(

=∫∫ e 4

−0.5t

=

)

− 0.5 dt

=  −2e −0.5t − 0.5t 

(

= −2e

−0.5(3)

10

) (

− 0.5(3) − −2e

−0.5(4)

− 0.5(4)

(

)

(0, 1)

∫ 0.853e

0.1333t

=

π – 4

π 4

y = sin2(x)

π – 2

(π, 1)

π

3π – 4

∫(

)

cos2 ( x ) − sin 2 ( x ) dx +

0

+

3π 4

∫ (sin π

x

)

2 ( x ) − cos2 ( x ) dx

4 π

∫ ( cos 3π

= 2 units 2 22 Parabola, of the form y = a( x − 2)( x + 2) (0, −3) ⇒ −3 = a(−2)(2) − 3 = −4 a 3 a= 4 Equation of parabola is 3 y = ( x − 2)( x + 2) 4 3 = ( x 2 − 4) 4 3 = x2 − 3 4

17

20 a dC — dt (10, 1750)

y

–2

0

k

2

(

dC = 15t2 + 250 — dt

(0, –3)

x

)

3 k2 – 3 k, – 4 Parabolic canal

(0, 250)

10

)

2 ( x ) − sin 2 ( x ) dx

4

= 0.5 + 1 + 0.5

0

0

)

b Area between curves is

dt

1750

y = cos2(x)

0

= 6.3991e 0.1333t  0 = 6.3991e 0.1333(17) − 6.3991e 0.1333(0) = 6.3991( 9.6417 − 1) = 55.3 million



) (

0.5

0.2 3    = 2  5(5) − (5)  − 0    3   1  = 2  25 − 8   3 1 2 = 33 m 3 1 2 c Stone area = (12 × 7 ) − 33 = 50 m 2. 3 3 2 d Volume of stones is 50 × 3 = 152 m 3. 3 17

y

21 a

0.2 3  b 2 ∫ 5 − 0.2 x 2 dx = 2 5 x − x  3  0 0

19 N =

3

= ( 5000 + 2500 ) − ( 625 = 1250 ) = 7500 − 1875 = $5625

5

)

5

= 5(10) + 250(10) − 5(5)3 + 250(5)

18 a y = a( x − 5)( x + 5) When x = 0, y = 5 5 = a(−5)(5) 5 = −25a 1 − =a 5 a = −0.2 Thus equation of arch is y = 5 − 0.2 x 2.

(

)

+ 250 dt

= 5t 3 + 250t 

= −2e −1.5 − 1.5 + 2e −2 + 2 = 0.3244 metres

5

2

5

3 4

∫ (15t

t

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

170 | TOPIC 8 Integration • EXERCISE 8.4 If the canal were full with water, the cross sectional area would be −2  3  − ∫  x 2 − 3 dx 2 4  2 3  = −2 ∫  x 2 − 3 dx 04  2

1 = −2  x 3 − 3 x  4 0  1   = −2   (2)3 − 3(2) − 0    4  = −2(2 − 6) = 8 m2 When the canal is one-third full, the cross sectional area =

8 2 m. 3

Cross-sectional area for a canal which is one third full is given by k

A = 2∫ 0 k

3 2 3  k − 3 −  x 2 − 3 dx 4  4

8 3  3 = 2 ∫  k 2 − x 2  dx 4  3 4 0 k 3 1 = 2  k 2 x − x 3  4 0 4 3 1   = 2  k 3 − k 3 − 0 4  4  1 3 = 2 k  2 

= k3 8 3 = 1.39 3 y = × (1.39)2 − 3 4 = −1.58 Therefore the depth of water = −1.58 − (−3) = 1.44 m. k=3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.2   |  171

Topic 9 — Logarithmic functions using calculus Exercise 9.2 — The derivative of f(x) = loge (x) 1 a b c

d

e

d   x 7  7 loge    = dx  3 x d 2(3 x 2 + 4 x ) 2 loge x 3 + 2 x 2 − 1 = 3 dx x + 2x 2 − 1 d 1 sin( x ) loge ( x − 2) ) = sin( x ) × + cos( x ) loge ( x − 2) ( dx ( x − 2) sin( x ) = + cos( x ) loge ( x − 2) x−2 loge ( x 2 ) y= (2 x − 1) du 2 x 2 Let u = loge ( x 2 ) so = = dx x 2 x dv Let v = 2 x − 1 so =2 dx du dv v −u dy dx dx = dx v2 2 (2 x − 1) × − 2 loge ( x 2 ) dy x = dx (2 x − 1)2 dy  2(2 x − 1) 1  = − 2 loge ( x 2 ) ×  (2 x − 1)2 dx  x dy 2(2 x − 1) − 2 x loge ( x 2 ) = dx x (2 x − 1)2 3 2e 2 x + e − x d 3e − x (2e3 x + 1) 3(2e3 x + 1) 3 loge e 2 x − e − x = = − x 3x = 2 x − x dx (e3 x − 1) e (e − 1) e −e

d f dx

(

(

(

(

(

))

))

)

( (

)

)

1

d loge (3 − 2 x ) = ( loge (3 − 2 x ) ) 2 dx 1 − −2 1 × ( loge (3 − 2 x ) ) 2 = (3 − 2 x ) 2 −1 = (3 − 2 x ) loge (3 − 2 x ) 1 = (2 x − 3) loge (3 − 2 x ) 2 a  y = −5 loge ( 2 x ) dy 1 = −5 × x dx 5 dy = − , x ∈( 0, ∞ ) dx x 1  b  y = loge   x − 2  = loge ( x − 2 )−1 = − loge ( x − 2 ) dy 1 =− , x ∈( 2, ∞ ) dx x−2  x + 3 = loge ( x + 3) − loge ( x + 1) c  y = loge   x + 1  1 dy 1 = − dx x + 3 x + 1 x + 1 − ( x + 3) = ( x + 3)( x + 1) −2 = ( x + 3)( x + 1) for x ∈( −∞, −3) ∪ ( −1, ∞ ) Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

172 | TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.2

(

d y = loge x 2 − x − 6

)

dy 2x − 1 = 2 , x ∈( −∞, −2 ) ∪ ( 3, ∞ ) dx x − x − 6 3 a Dom = ( 2, ∞ ) and Ran = R b Graph cuts the x axis where y = 0 2 loge ( x − 2) = 0 loge ( x − 2) = 0 e0 = x − 2 1= x − 2 x=3 Thus (a, 0) = (3, 0) so a = 3 c y = 2 loge ( x − 2) dy 2 = dx x − 2 dy 2 When x = 3, mT = = =2 dx (3 − 2) Equation of tangent with mT = 2 which passes through ( x1 , y1 ) = ( 3, 0 ) is given by y − y1 = mT ( x − x1 ) y − 0 = 2 ( x − 3) y = 2x − 6

d Equation of perpendicular line with mP = −

1 which passes through ( x1 , y1 ) = ( 3, 0 ) is given by 2

y − y1 = mP ( x − x1 ) 1 y − 0 = − ( x − 3) 2 3 1 y=− x+ 2 2 4 y = 4 loge (3 x − 1) dy 12 = dx 3 x − 1 If the tangent is parallel to 6 x − y + 2 = 0 or y = 6 x + 2 then the gradient is 6. 12 mT = =6 3x − 1 12 = 6(3 x − 1) 12 = 18 x − 6 18 = 18 x 1= x When x = 1, y = 4 loge (3(1) − 1) = 4 loge (2)

Equation of tangent with mT = 6 which passes through ( x1 , y1 ) = (1, 4 loge (2) ) is given by y − y1 = mT ( x − x1 ) y − 4 loge (2) = 6( x − 1) y − 4 loge (2) = 6 x − 6 y = 6 x + 4 loge (2) − 6

5 y =

1 + loge ( x ) 10 x

Turning point occurs where 1 1 − =0 x 10 x 2 10 x − 1 =0 10 x 2 10 x − 1 = 0 10 x = 1 x = 0.1

dy = 0. dx

1  1 + loge   = 1 + loge (1) − loge (10) = 1 − loge (10)  10  10(0.1) Minimum TP at (0.1, 1 − loge (10)). When x = 0.1, y =

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.2   |  173 6 a f ( x ) = 2 x loge ( x ) Local max/min values occur where f ′( x ) = 0 2x f ′( x ) = + 2 loge ( x ) = 2 + 2 loge ( x ) x 0 = 2 + 2 loge ( x ) −2 = 2 loge ( x ) −1 = loge ( x ) x = e1 When x = 0.1, f ′( x ) = 2 loge (0.1) + 2 = −2.6052 When x = 1, f ′( x ) = 2 loge (1) + 2 = 2 x

x < e −1

x = e −1

x > e −1

f ′( x ) When x = e −1 , y = 2e −1 loge (e −1 ) y = −2e −1 2 y=− e  1 2  is a local minimum TP.  , −  e e loge (2 x ) b y = f ( x ) = x du 1 Let u = loge (2 x ) so = dx x dv Let v = x so =1 dx du dv v −u dy = dx 2 dx dx v 1 x × − loge (2 x ) × 1 dy x = dx x2 dy 1 − loge (2 x ) = dx x2 dy Max/min values occur where =0 dx 1 − loge (2 x ) =0 x2 1 − loge (2 x ) = 0 1 = loge (2 x )

1 x< e 2

x=

1 e 2

x


3 e

3 3 3 3  3 3  When x = , f   = loge  3 ÷  = loge (e) =     e e e e e e  3 3 There is a local maximum at  ,  .  e e d   x 4 7 a  4 loge    = dx  2 x b

d 1  loge dx  2

(

1   d 1 loge ( x − 2 ) 2  x−2  =   dx  2  d 1 1  =  × loge ( x − 2 ) dx  2 2 1 1 = × 4 x−2 1 = 4( x − 2)

)

))

d 3x 2 − 6 x + 7 log x 3 − 3 x 2 + 7 x − 1 = 3 e dx x − 3x 2 + 7 x − 1 sin( x ) d d ( −6 loge (cos( x )) = 6 cos( x ) = 6 tan( x ) dx c

1 x> e 2

(

e

d dx

(

(

)

loge (3 x + 1) =

1 d loge (3 x + 1) ) 2 ( dx

1 1 3 − loge (3 x + 1) ) 2 × ( 2 3x + 1 3 = 2(3 x + 1) loge (3 x + 1)

=

f y =

2 loge (2 x ) e2 x + 1

Let u = 2 loge (2 x ) so

dy dx

3 e

f ′( x )

e1 = 2 x 1 x= e 2 dy 1 − loge (2 × 0.6) When x = 0.6, = = 1.7649 dx (0.6)2 dy 1 − loge (2 × 2) When x = 2, = = − 0.0966 dx (2)2 x

3 c f ( x ) = x loge   x 3  1 f ′( x ) = loge   + x × −  x  x 3 f ′( x ) = loge   − 1 x Max/min values occur where f ′( x ) = 0  3 loge   − 1 = 0  x  3 loge   = 1  x 3 =e x 3 x= e When x = 1, f ′(1) = loge (3) − 1 = 0.0986 When x = 2, f ′(1) = loge (1.5) − 1 = −0.5945

Let v = e 2 x + 1 so

du 2 = dx x

dv = 2e 2 x dx

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

174 | TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.2

dy = dx

v

du dv −u dx dx v2

2 2x (e + 1) − 4e 2 x loge (2 x ) dy x = dx (e 2 x + 1)2 2x dy 2(e + 1) − 4 xe 2 x loge (2 x ) = dx x (e 2 x + 1)2 2x dy 2e + 2 − 4 xe 2 x loge (2 x ) = dx x (e 2 x + 1)2 8 a

b

y = ( x 2 − 3 x + 7) loge (2 x − 1) dy 2 = ( x 2 − 3 x + 7) × + ( 2 x − 3) loge (2 x − 1) dx (2 x − 1) 1  dy 2( x 2 − 3 x + 7) , x ∈  , ∞ = ( 2 x − 3) loge (2 x − 1) + 2  dx (2 x − 1)

( )

y = sin( x ) loge x 2

dy 2x = sin( x ) × 2 + cos( x ) loge x 2 dx x dy 2 sin( x ) = + cos( x ) loge x 2 dx x 2 dy x cos( x ) loge x + 2 sin( x ) = , x ∈ ( 0, ∞) dx x log (3 x ) c y = 3e (x − x) du 1 Let u = log e (3 x ) so = dx x dv Let v = x 3 − x so = 3x 2 − 1 dx du dv −u v dy dx dx = dx v2 1 3 ( x − x ) − (3 x 2 − 1) loge (3 x ) dy x = dx ( x 3 − x )2 2 dy ( x − 1) − (3 x 2 − 1) loge (3 x ) = dx ( x 3 − x )2 2 2 dy x − 3 x loge (3 x ) − 1 + loge (3 x ) = , x ∈( 0, ∞ ) / {1} dx ( x 3 − x )2

( )

( ) ( )

 4− x = loge (4 − x ) − loge ( x + 2) d y = loge   x + 2  −1 1 dy = − dx 4 − x x + 2 − ( x + 2) − (4 − x ) = (4 − x )( x + 2) −6 = (4 − x )( x + 2) where x ∈( −2, 4 ) 9 Use CAS for this question 2 d a 2 log5 ( x ) ) given x = 5 is log5 (e) ( 5 dx d 1 1  log ( x + 1) b  given x = 2 is log3 (e)  3 dx  3 9 d 2 c log6 ( x − 3) given x = 3 is log6 (e) dx 10 a y = loge (2 x − 2) 2 1 dy Gradient of tangent is mT = = = dx 2 x − 2 x − 1

(

)

1 =2 1.5 − 1 Equation of tangent with mT = 2 which passes through ( x1 , y1 ) = (1.5, 0) is given by When x = 1.5, mT =

y − y1 = mT ( x − x1 ) y − 0 = 2( x − 1.5) y = 2x − 3 b y = 3 loge ( x )

Gradient of tangent is mT = When x = e, mT =

3 e

dy 3 = dx x

Equation of tangent with mT =

( x1 , y1 ) = (e,3) is given by y − y1 = mT ( x − x1 )

3 y − 3 = ( x − e) e 3 y−3= x−3 e 3 y= x e 1 c y = loge x 2 = loge ( x ) 2

3 which passes through e

( )

Gradient of tangent mT = When x = e, mT =

1 e

dy 1 = dx x

Equation of tangent with mT =

( x1 , y1 ) = (e,1) is given by y − y1 = mT ( x − x1 )

1 which passes through e

1 y − 1 = ( x − e) e 1 y −1= x −1 e 1 y= x e 11 a y = 2 loge (2 x ) dy 2 = dx x

e e 4 b Gradient of tangent at  , 2 is mT = 2 ÷ = . 2  2 e 4 Equation of tangent with mT = which passes through e e ( x1 , y1 ) =  , e is given by 2 

y − y1 = mT ( x − x1 ) 4 e y − e =  x −  e 2 4 y−e= x−2 e 4 y = x −2+e e 12 y = x is a tangent to y = loge ( x − 1) + b Gradient of tangent is mT = 1 Also gradient of tangent is mT =

1 dy = dx x − 1

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.2   |  175 Thus 1 =1 x −1 1= x −1 x=2 When x = 2, y = 2 2 = loge (2 − 1) + b 2 = loge (1) + b b=2 Thus y = loge ( x − 1) + 2 13 y = −2 x + k is perpendicular to y = loge (2( x − 1)) Gradient of perpendicular line is mP = −2 1 Gradient of tangent is mT = 2 1 dy Also gradient of tangent is mT = = dx x − 1 Thus 1 1 = x −1 2 x −1= 2 x=3 When x = 3, y = loge (2(3 − 1)) = loge (4)  1.3863 1.3863 = −2(3) + k 7.3863 = k k  7.4 Thus y = −2 x + 7.4 14 a The graph cuts the x axis where y = 0. 1 − 2 loge ( x + 3) = 0 x2 1 = 2 loge ( x + 3) x2 By calculator x = −1.841 and − 0.795 Therefore coordinates are (-1.841, 0) and (- 0.795, 0) 1 b y = 2 − 2 loge ( x + 3) x y = x −2 − 2 loge ( x + 3) 2 dy = −2 x −3 − dx x+3 dy 2 2 =− 3 − dx x+3 x dy −2( x + 3) − 2 x 3 = dx x 3 ( x + 3) dy (2 x 3 + 2 x + 6) =− dx x 3 ( x + 3) (2(−1.841)3 + 2(−1.841) + 6) = −1.1989. (−1.841)3 (−1.841 + 3) Equation of tangent with mT = −1.1989 which passes through ( x1 , y1 ) = ( −1.841, 0 ) is given by

At ( −1.841, 0 ) , mT = −

y − y1 = mT ( x − x1 ) y − 0 = −1.1989 ( x + 1.841) y = −1.1989 x − 2.2072

At ( − 0.795,0 ) , mT = −

(2(− 0.795)3 + 2(− 0.795) + 6) = 3.0725. (− 0.795)3 (− 0.795 + 3)

Equation of tangent with mT = 3.0735 which passes through ( x1 , y1 ) = ( −0.795, 0 ) is given by y − y1 = mT ( x − x1 ) y − 0 = 3.0735 ( x + 0.795) y = 3.0735 x + 2.4434

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

176 | TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.2 c Minimum TP occurs when −

(2 x 3 + 2 x + 6) =0 x 3 ( x + 3) 2x3 + 2x + 6 = 0

When

dy =0 dx

x3 + x + 6 = 0 x = −1.2134 since x < 0

1 − 2loge (−1.2134 + 3) = − 0.4814. (−1.2134)2 Minimum TP is (−1.2134, − 0.4814). x = −1.2134, y =

( 2 loge ( x ))2 = 2 loge ( x )

15 a

When x = 0.8364,

( 2 loge ( x )) − 2 loge ( x ) = 0 2 loge ( x ) ( 2 loge ( x ) − 1) = 0 2

loge ( x ) = 0 or 2 loge ( x ) − 1 = 0 e0 = x x =1

2 loge ( x ) = 1 loge ( x ) = 0.5

x = e 0.5 When x = 1, y = 2 loge (1) = 0 When x = e 0.5 , y = 2 loge (e)0.5 = 1 The points of intersection are (1,0) and e 0.5, 1 .

(

b If y = ( 2 loge ( x ) ) 8 loge ( x ) dy  2 = 2   ( 2 loge ( x ) ) =  x dx x dy 8 loge (1) At (1,0) then = =0 1 dx If y = 2 loge ( x ) dy 2 = dx x dy 2 At (1,0) then = =2 dx 1 y c 2

)

{

d 2 loge ( x ) > ( 2 loge ( x ) ) when x :1 < x < e 2

Equation of tangent with mT = − 6 which passes through

( x1 , y1 ) = (1, −1) is given by y − y1 = mT ( x − x1 ) y + 1 = −6( x − 1) y + 1 = −6 x + 6 y = −6 x + 5

1 which passes 6

y − y1 = mP ( x − x1 ) 1 y + 1 = ( x − 1) 6 1 1 y +1= x − 6 6 7 1 y = x − or x − 6 y = 7 6 6 dy d TP occurs where = 0. dx 2 − 8x 2 =0 x3 2 − 8x 2 = 0

1 − 4x2 = 0 (1 − 2 x )(1 + 2 x ) = 0 1 − 2 x = 0 or 1 + 2 x = 0 1 = 2x 2 x = −1 1 1 x= x = − but x > 0 2 2 1 When x = 0.5, y = − − 8 loge (0.5) = −4 + 8 loge (2) (0.5)2 1  Maximum TP at  , − 4 + 8 loge (2) . 2  17 a f : y = −2 loge (2 − x ) − 1 where dom = (−∞, 2) and ran = R f −1 : Interchange x and y.

x

x=0

dy 2 − 8(1)2 = = −6 dx (1)3

through ( x1 , y1 ) = (1, −1) is given by

y = 2 loge (x) (1, 0)

c At (1, −1), mT =

dy 2 − 8(0.8364)2 = = − 6.15 dx (0.8364)3

Equation of perpendicular line with mP =

y = (2 loge (x))2

0

1 − 8 loge ( x ) x2 −2 y = − x − 8 loge ( x ) 8 dy = 2 x −3 − dx x dy 2 8 = − dx x 3 x dy 2 − 8 x 2 = dx x3 dy 2 − 8(0.3407)2 When x = 0.3407, = = 27.09 dx (0.3407)3

b y = −

0.5

}.

16 a Graph cuts the x axis where y = 0. 1 − 2 − 8 loge ( x ) = 0 x 1 − 8 loge ( x ) = 2 x x = 0.3407 or 0.8364 Therefore coordinates are (0.3407, 0) and (0.8364, 0)

x = −2 loge (2 − y) − 1 x + 1 = −2 loge (2 − y) 1 − ( x + 1) = loge (2 − y) 2 e

1 − ( x +1) 2

= 2− y

y = 2−e

1 − ( x +1) 2

f −1 : R → R, f −1 ( x ) = 2 − e Ran = (−∞, 2)

1 − ( x +1) 2

where Dom = R and

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.3   |  177 b If f ( x ) = x then 1  b f ( x ) = ∫  x 3 −  dx  −2 loge (2 − x ) − 1 = x x 4 x = −4.8479 or 1.7467 x = − loge ( x ) + c, x > 0 When x = −4.8479, y = −2 loge (2 + 4.8479) − 1 = −4.8479 4 1 When x = 1.7467, y = −2 loge (2 − 1.7467) − 1 = 1.7467 f (1) = Points of intersection are (−4.8479, −4.8479) and (1.7467,1.7467) 4 1 1 (−4.8479, −4.8479) and (1.7467,1.7467). = − loge (1) + c y 4 4 c c=0 (1.747, 1.747)

(0, 2 – e – 0.5) ( – 2 loge (2) – 1, 0)

(2 – e – 0.5, 0) x

0

y = f(x)

y = f ʹ(x)

(0, – 2 loge (2) – 1)

x4 − loge ( x ), x > 0 4 3 y = 2 loge ( cos(2 x )) dy −2 sin(2 x ) = = − tan(2 x ) dx cos(2 x ) f (x) =

∫ (−2 tan(2 x )) dx = loge (cos(2 x )) −2 ∫ ( tan(2 x )) dx = loge ( cos(2 x )) 1 ∫ ( tan(2 x )) dx = − 2 loge (cos(2 x )) 4 y = ( loge ( x ) )

(–4.848, –4.848)

18 Equation of tangent is y = mT x + c When y = 0, x = 0.3521 0 = 0.3521mT + c −0.3521mT = c Thus y = mT x − 0.3521mT y = mT ( x − 0.3521) 2 dy But mT = = dx 2 x − 1 2 Thus y = ( x − 0.3521) 2x − 1 2 At x = n, y = (n − 0.3521).................(1) 2n − 1 Also at x = n, y = loge (2n − 1).....................(2) (1) = (2) 2 (n − 0.3521) = loge ( 2n − 1) 2n − 1 n = 2 as n is an integer

Exercise 9.3 — The antiderivative of f(x) = 1 x 1 a

2

3

b

2 1

2

∫ 5x dx = 5 ∫ x dx = 5 loge ( x ) + c, x > 0 ∫ 1

3

3 3 4 dx = ∫ dx 4x −1 4 1 4x −1

3 3 =  loge (4 x − 1)  4 1 3  3  =  loge (4(3) − 1) −  loge (4(1) − 1) 4  4  3 3 = loge (11) − loge (3) 4 4 3  11  = loge    3 4 2 x + 2x − 3 2 a ∫ dx = ∫ 1 + 2 x −1 − 3 x −2 dx x2 = x + 2 loge ( x ) + 3 x −1 + c 3 = x + 2 loge ( x ) + + c, x > 0 x

(

)

2

Let u = loge ( x ) so

du 1 = dx x

dy = 2u du dy du dy = × dx dx du dy 1 = × 2u dx x dy 1 = × 2 loge ( x ) dx x dy 2 = loge ( x ) dx x e e 4  2  = x dx log ( ) 2   e ∫ x ∫  x loge ( x ) dx 

Thus y = u 2 so

1

1

e

2 = 2 ( loge ( x ) )   1

(

= 2 ( loge (e) ) − ( loge (1) ) 2

= 2 (1 − 0 ) =2

2

)

1 − 1 cuts the x axis where f ( x ) = 0. x+2 1 Thus −1= 0 x+2 1 =1 x+2 1= x + 2 x = −1 So (a, 0) = (−1, 0), a = −1

5 a f ( x ) =

b Area is 2

1

2   ∫  x + 2 − 1 dx = [ loge ( x + 2) − x ]−1 −1 = ( loge ( 4 ) − 2 ) − ( loge (1) + 1)

= loge ( 4 ) − 3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

178 | TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.3 c

1 1 1 −1= − x + x+2 2 4 1 1 5 =− x+ x+2 2 4 4 = −2 x + 5 x+2 4 = ( x + 2)(−2 x + 5) 4 = −2 x 2 + x + 10 0 = 2x 2 − x − 6 = (2 x + 3)( x − 2) 3 x = − ,2 2 1 3  3  − 1 = 1 ∴  − ,1 x = − ,y= 3  2  2 − +2 2 1 3 3  x = 2, y = −1= − ∴  2, −   2+2 4 4 2

d A =

 1

1  1

 1

5



∫3  − 2 x + 4 −  x + 2 − 1  dx

−

2 2

=

1 

∫3  − 2 x + 4 − x + 2  dx

−

2

2 1 5 =  − x 2 + x − loge ( x + 2)  4  4 − 3 2

 1  −3  2 5  −3  1 5  1 = − (2)2 + (2) − loge (4) −  −   +   − loge         2  4 4 4 2  4 2 5 9 15  1 = −1 + − loge (4) + + + loge    2 2 16 8 63 = − 3 loge ( 2 ) 16 6 a Graphs intersect where 1 1 =− +3 x +1 ( x + 1)2 x + 1 = −1 + 3( x + 1)2 prov x ≠ −1 x + 1 = −1 + 3 x 2 + 6 x + 3 0 = 3x 2 + 5x + 1 −5 ± (5)2 − 4(3)(1) 2(3) −5 ± 13 x= 6 −5 + 13 1 1 6 When x = = = , y= 6 13 + 1 −5 + 13 −5 + 13 + 6 +1 6 6 x=

 −5 + 13

∴ POI = 



6

,

   13 + 1  6

b Required area is 2

1 1    − ( x + 1)2 + 3 − x + 1  dx −0.2324

∫

2

=

1   −( x + 1)−2 + 3 −  dx  + 1 x −0.2324

∫

= ( x + 1)−1 + 3 x − loge ( x + 1)  2

2

−0.2324

 1  = + 3 x − loge ( x + 1)   ( x + 1)  −0.2324 2 = 4.3647 units Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.3   |  179 −2

1 dx x −4

7 A = − ∫ 4

=∫ 2

1 dx x

= [ loge ( x ) ]2 = loge (4) − loge (2) 4

= loge (2) units 2 8 A =

−1

 1



∫  x − 1 + 2 dx

−2 2

 1  = ∫ + 2 dx  − x − 1  1 2

1   = ∫− + 2 dx  x + 1  1

= [ − loge ( x + 1) + 2 x ]1

2

= − loge (3) + 4 − ( − loge (2) + 2 ) = − loge (3) + 4 + log3 (2) − 2  2 = loge   + 2 units 2  3 4 9 a ∫ − dx = −4 loge ( x ) + c, x > 0 x 3 3 4 3 7 b ∫ dx = ∫ dx = loge ( 4 x + 7 ) + c, x > − 4x + 7 4 4x + 7 4 4 x 3 + 2 x 2 + 3x − 1 3   dx = ∫  x + 2 + − x −2  dx 2 ∫   x x 1 1 2 = x + 2 x + 3loge ( x ) + + c, x > 0 2 x 3   d ∫  + cos(4 x ) dx  2− x  1 −1 = −3 ∫ dx + ∫ cos(4 x ) dx = −3 loge (2 − x ) + sin(4 x ) + c, x < 2 (2 − x ) 4 c

4

10 a

4

3 1 ∫ 1 − 2 x dx = 3 ∫ 1 − 2 x 2 2 4 1 = 3  − loge (1 − 2 x )   2 2 1  1  = 3  − loge (1 − 2(4)) + loge (1 − 2(2))  2  2 3 = − ( loge (7) − loge (3) ) 2 7 3 = − loge    2 3 −1

−1

 2   1   dx = 2 ∫   dx b ∫  −3  x + 4  −3  x + 4  = 2 [ loge ( x + 4) ]−3 −1

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

4 2  1  c ∫  e 2 x +  dx =  e 2 x + 2 loge ( x )   x 2 1 1

1  1  =  e8 + 2 loge (4) −  e 2 + 2 loge (1) 2  2  = 1489.56

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

180 | TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.3 dy 5 is equivalent to = dx 2 x + 4 5 1 ∫ 2 x + 4 dx = 5 ∫ 2 x + 4 dx 5 = loge (2 x + 4) + c, x > −2 2 5 3 Thus y = loge (2( x + 2)) + c and when x = − , y = 3. 2 2 5   3  3 = loge  2  −  + 4  + c   2  2 5 3 = loge (1) + c 2 c=3 5 y = loge ( 2 x + 4 ) + 3 2 5 y = loge ( 2( x + 2) ) + 3 2 dy 3 b is equivalent to = dx 2 − 5 x 3 1 ∫ 2 − 5x dx = 3 ∫ 2 − 5x dx 2 3 = − loge ( 2 − 5 x ) + c, x < 5 5 3 1 Thus y = − loge ( 2 − 5 x ) + c and when x = , y = 1. 5 5 3   1  1 = − loge  2 − 5    + c  5   5 3 1 = − loge (1) + c 5 c =1 3 y = − loge ( 2 − 5 x ) + 1 5 12 Let y = f ( x ) = 2 x loge (mx ) 11 a

dy 1 = 2 loge (mx ) + 2 x × dx x dy = 2 loge (mx ) + 2 dx

∫ (2 loge (mx ) + 2) dx = 2 x loge (mx ) 2 ∫ loge (mx ) dx + ∫ 2 dx = 2 x loge (mx ) 2 ∫ loge (mx ) dx = 2 x loge (mx ) − ∫ 2dx 2 ∫ loge (mx ) dx = 2 x loge (mx ) − 2 x ∫ loge (mx ) dx = x loge (mx ) − x + c 13 If y = 3 x loge ( x ) then

3x dy = 3 loge ( x ) + = 3 loge ( x ) + 3 x dx

∫ (3loge ( x ) + 3) dx = 3x loge ( x ) 3 ∫ loge ( x ) dx + ∫ 3 dx = 3 x loge ( x ) 3 ∫ loge ( x ) dx = 3 x loge ( x ) − ∫ 3 dx 3 ∫ loge ( x ) dx = 3 x loge ( x ) − 3 x ∫ loge ( x ) dx = x loge ( x ) − x 2 ∫ loge ( x ) dx = 2 x loge ( x ) − 2 x ∫ 2 loge ( x ) dx = 2 x loge ( x ) − 2 x

(

14 If y = loge 3 x 3 − 4

)

2

dy 9x = dx 3 x 3 − 4 3

3

9x 2 3 ∫ 3x 3 − 4 dx =  loge 3x − 4  2 2

(

3

)

x2 dx =  loge 3 x 3 − 4  3 − 3 4 x 2

(

9∫ 3

x

2

∫ 3x 3 − 4

dx =

2 3

)

(

3

2

)

1  loge 3 x 3 − 4   9

( (

3

2

1 x2 3 3 ∫ 3x 3 − 4 dx = 9 loge 3(3) − 4 − loge 3(2) − 4 2

(

)

))

3

1 x2 ∫ 3x 3 − 4 dx = 9 ( loge (77) − loge (20)) 2 3

x2

1

 77 

∫ 3x 3 − 4 dx = 9 loge  20  2

(

)2

15 If y = loge e x + 1

(

)2

( )(

)

du = 2 e x e x + 1 = 2e 2 x + 2e x dx dy 1 Let y = loge (u) so = du u dy du dy = × dx dx du dy 1 = 2e 2 x + 2e x × dx u dy 1 = 2e 2 x + 2e x × 2 dx ex + 1 Let u = e x + 1

so

( (

) )

(

)

(

)

2e x e x + 1

dy = dx

(e

x

)

+1 x

2

2e dy = x , e x ≠ −1 dx e +1

(

)

Thus 5

∫( 1 5

2∫ 1 5

(

2e x

((

)2 )1

e

)

x

(

)

x

e

1

∫ (e x + 1) dx = 2 ( loge (e 1 5

5

)

2 dx =  loge e x + 1  1  e +1 x

∫ (e x + 1) dx = 2  loge (e 1 5

5

dx =  loge e x + 1  ex + 1

ex

1

ex

1

)

5

x

2 + 1  1

5

+ 1 − loge e 2 + 1

)

2

∫ (e x + 1) dx = 2 (10.0134 − 2.6265) 1 5

ex

∫ (e x + 1) dx = 3.6935 1

16 a y =

10 x 5 + x2

du = 10 dx dv Let v = 5 + x 2 so = 2x dx Let u = 10 x so

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

(

)) 2

TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.3   |  181

dy = dx

du dv −u dx dx v2 10 5 + x 2 − 10 x (2 x )

Let v = x 2 + 1 so

v

(

)

dy dx dy dx dy dx dy dx

dy = dx (5 + x 2 )2 dy 50 + 10 x 2 − 20 x 2 = dx (5 + x 2 )2 dy 50 − 10 x 2 = dx (5 + x 2 )2

Maximum TP occurs when

dy =0 dx

5 = x2 ± 5=x When x = − 5, y =

)

( ) = − 10 5 = − 10 5 = − 2 5+5 10 5 + (− 5 ) 10 − 5

So − 5, − 5 is a minimum TP When x = 5, y =

( 5 ) = 10 5 = 10 5 = 2 5+5 10 5 + ( 5) 10

( 5, 5 ) is a maximum TP d b (loge (5 + x 2 ) = 5 +2 xx 2 dx So

2x

10 x

∫ 5 + x 2 dx = 5 ∫ 5 + x 2 dx = 5 loge (5 + x 2 )

(

)

y = loge 5 + x 2 so 6

6

10 x

2x dy = dx 5 + x 2

2x

∫ 5 + x 2 dx = 5 ∫ 5 + x 2 dx 5 6

10 x

10 x

∫ 5 + x 2 dx = 5 5 6

{

(

6 5

)

10 x

 41

units 2

5

17 a Graph of y =

(

loge 5 + 62 − loge 5 +

∫ 5 + x 2 dx = 5loge  10  5x x2 + 1

5

5

5  1  2  dy  =  5 −  ÷   + 1  4  4   dx  dy  20 − 5  25 = ÷ dx  4  16 dy 15 16 = × dx 4 25 dy 12 = dx 5 12 Equation of tangent with mT = which passes through 5 1 ( x1 , y1 ) ≡  − , −2 is given by 2 y − y1 = mT ( x − x1 ) 12  1 y+2= x+  5 2 12 6 10 y= x+ − 5 5 5 4 12 y = x − or 12 x − 5 y = 4 5 5 dy 2x 2 b If y = loge x + 1 then = dx x 2 + 1 5x 5 2x ∫ x 2 + 1 dx = 2 ∫ x 2 + 1 dx 5 = loge 1 + x 2 2 5x c y = 2 ..................(1) x +1 12 4 y = x − ...............(2) 5 5 (1) = (2) 5x 12 4 = x− 5 x2 + 1 5 25 x = (12 x − 4)( x 2 + 1)

(

5

2 ∫ 5 + x 2 dx = 5  loge (5 + x ) 5 6

2

1 dy At x = − , = 2 dx  1 2  2     − 2  + 1  

50 = 10 x 2

c

du dv −u dx dx = v2 5( x 2 + 1) − 5 x (2 x ) = ( x 2 + 1)2 2 5 x + 5 − 10 x 2 = ( x 2 + 1)2 5 − 5x 2 = 2 ( x + 1)2 v

1 5 − 5  −   2

50 − 10 x 2 =0 (5 + x 2 )2 50 − 10 x 2 = 0

(

dv = 2x dx

1 5  −   2 1 Tangent at x = − , y = 2 2  − 1 +1   2 5 5 y=− ÷ 2 4 5 4 y=− × 2 5 y = −2 dy Gradient of tangent = mT = dx du Let u = 5 x so =5 dx

( 5 ) )} 2

)

(

)

25 x = 12 x 3 − 4 x 2 + 12 x − 4 0 = 12 x 3 − 4 x 2 + 12 x − 4 − 25 x 0 = 12 x 3 − 4 x 2 − 13 x − 4 Let P ( x ) = 12 x 3 − 4 x 2 − 13 x − 4 3 2  1  1  1  1 P  −  = 12  −  − 4  −  − 13  −  − 4  2  2  2  2 12 4 13 8  1 P−  = − − + −  2 8 4 2 2 3 2 13 8  1 P−  = − − + − = 0  2 2 2 2 2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

182 | TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.4 1   x +  or ( 2 x + 1) is a factor 2

5

c

(

12 x 3 − 4 x 2 − 13 x − 4 = (2 x + 1) 6 x 2 − 5 x − 4

)

12 x 3 − 4 x 2 − 13 x − 4 = (2 x + 1)(2 x + 1)(3 x − 4) Thus if (2 x + 1)2 (3 x − 4) = 0 1 4 x = − or 2 3 4 3

Shaded area is

 5x

∫1  x 2 + 1 −

−

12  5x  = ∫ 2  dx −  5 1 x +1 2

5 = 2

4 3

4 3

−

2

2

6

20 a Shaded region is ∫ 5 loge ( x − 3) dx = 6.4792 units 2 4

b y = loge ( x − 3) For inverse, swap x and y x = 5 loge ( y − 3) x = loge ( y − 3) 5 x

4 3

−

2

4 3

e5 = y − 3

∫

x

y = e5 + 3

2

4 3

x dx +

−1 2

4 ∫1 5 dx

−

c Area = Area of rectangle – area under the curve 5loge (3)

 x  e 5 + 3 dx   0   5log (3)  x  e = 30 loge (3) − 5e 5 + 3 x    0

= 6 × 5 loge (3) −

2

4

5 12 4 3 =  loge ( x 2 + 1) − x 2 + x  10 5 − 1 2

2

= 2.5541 − 2.13 + 1.07 − 0.5579 + 0.3 + 0.4 = 1.6295 units 2 1 18 a The curve y = + x 3 − 4 x 1 Tangent at x = 1, y = + 13 − 4 = −2 1 Gradient of tangent 3x 4 − 1 dy 1 = mT = = − x −2 + 3 x 2 = 2 + 3 x 2 = dx x x2 4 3(1) − 1 When x = 1, mT = =2 (1)2

(

= 30 loge (3) − 15 − 15 loge (3) + 5

Exercise 9.4 — Applications 1  1 a y = f ( x ) = 2 loge (4 x ), x ∈  , ∞ 4 Inverse is x = 2 loge (4 y), x ∈[ 0, ∞ ) 1 x = loge (4 y) 2 1

y − y1 = mT ( x − x1 ) y + 2 = 2( x − 1) y + 2 = 2x − 2 y = 2x − 4

x

e2 = 4y 1

1 2x e =y 4

2

1 b Shaded region is ∫  + x 3 − 4 − (2 x − 4) dx x 

Thus f −1 ( x ) =

1

2 loge 8

2

2 1 =  loge ( x ) + x 4 − x 2  4  1 1 4 1     =  loge (2) + (2) − (2)2  −  loge (1) + (1)4 − (1)2      4 4 1 = loge (2) + 4 − 4 − 0 − + 1 4 3 2 = + loge (2) units 4 19 a Dom = (1, ∞ ) and ran = R b Graph cuts the x axis where y = 0 2 loge ( x − 1) = 0 loge ( x − 1) = 0

)

= 6.4792 units 2

Equation of tangent with mT = 2 which passes through

1  = ∫  + x 3 − 2 x  dx x  1

∫

= 30 loge (3) − 5e loge (3) + 3 × 5 loge (3) − (5 + 0 )

( x1 , y1 ) ≡ (1, −2) is given by

e0 = x − 1 1= x −1 x=2 Thus ( a, 0 ) ≡ (2, 0), a = 2

2

Solved using CAS

4 ∫1 x dx + ∫1 5 dx

12  2x  ∫1  x 2 + 1 dx − 5

−

4  dx 5

2

4 3

−

12 x+ 5

∫ 2 loge ( x − 1) dx = 5.0904 units

b

∫ 0

x

1 2 e 4

2 loge 8 1 1 1x   x  e 2  dx = 1 ∫ e 2  dx 4   4 0    

1 2 loge 8 x 1 = 2e 2  4   0 1  (2 log   e 8) 1 = 2e 2 − 2e 0  4   1 = (8 − 1) 2 7 = units 2 2 7 c Required area is 4 loge (8) − units 2. 2 2 a Graph cuts the y axis where x = 0, y = 2 loge (5) Graph cuts the x axis where y = 0

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.4   |  183 2 loge ( x + 5) + 1 = 0 2 loge ( x + 5) = −1 1 loge ( x + 5) = − 2 e

−

(

1 2

= x+5

x=e

At the point e

−0.5

−

1 2

−5

− 5, 0

)

b h : dom = ( −5, ∞ ) and ran = R h −1 : dom = R and ran = ( −5, ∞ ) If y = 2 loge ( x + 5) + 1 Inverse is x = 2 loge ( y + 5) + 1 x − 1 = 2 loge ( y + 5) 1 ( x − 1) = loge ( y + 5) 2 1

e2

( x −1)

= y+5 1

y = e2 Thus h −1 ( x ) = c

( x −1)

−5

1 ( x −1) e2

−5 y

x = –5

y = h(x)

(e

1 –– 2–

(0, 2 loge(5) + 1) 5, 0

)

(2 loge(5) + 1, 0)

0

(0, e

1 –– 2

)

–5

x

y = h–1(x) y = –5

h( x ) = h −1 ( x )

d

1

( x −1)

−5 2 loge ( x + 5) + 1 = e 2 x = −4.9489, 5.7498 1   ( x −1) 2 2 log ( x + 5) + 1 − e + 5 dx = 72.7601 units 2 ∫  e  −4.9489 5.7498

e

3 a Graph cuts the x axis where y = 0 2 x loge (2 x ) = 0 x = 0 or loge (2 x ) = 0 e0 = 2 x 1 = 2x 1 x= 2 1  Cuts x axis at  , 0  . 2 b y = x 2 loge (2 x ) dy x2 = 2 x loge (2 x ) + dx x dy = 2 x loge (2 x ) + x dx

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

184 | TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.4 2

c Shaded region is ∫ 2 x loge (2 x ) dx. 1 2

2

∫1 ( 2 x loge (2 x ) + x ) dx =  x

2

2

loge (2 x )  1

2

2

2

2

∫1 ( 2 x loge (2 x )) dx + ∫1 x dx = ((2) 2

2

2

 1   1 loge (2(2)) −    loge  2       2     2 2

)

2

2

1

∫1 ( 2 x loge (2 x )) dx = 4 loge (4) − 4 loge (1) − ∫1 x dx 2 2

2

2  1 x2  2 log (2 ) 5.5452 = − x x dx ( ) e ∫1  2  1

2

2 2

 1 2 1  12 ∫1 ( 2 x loge (2 x )) dx = 5.5452 −  2 (2) − 2  2   2 2

1

∫1 ( 2 x loge (2 x )) dx = 5.5452 − 2 + 8 2 2

∫1 ( 2 x loge (2 x )) dx = 3.670 units

2

2

4 a Graph intersects the x axis where y = 0 − loge (5 x ) = 0 loge (5 x ) = 0 e0 = 5x 1 = 5x 1 x= 5 1  Cuts x axis at  , 0  . 5 b  y = − x loge (5 x ) + x dy 1 = − x × + loge (5 x ) × −1 + 1 dx x dy = − loge (5 x ) dx 2

c Shaded region is − ∫ ( − loge (5 x ) ) dx = 1 5

2

∫1 ( loge (5x )) dx 5

2

2

2

1 5

1 5

1 5

2 ∫ ( − loge (5x ) − 1) dx = ∫ ( − loge (5x )) dx − ∫ (1) dx = [ − x loge (5x )]1 2

5

∫1 ( loge (5x )) dx + [ x ]1 = − [ − x loge (5x )]1 5

2

2

5

5

2

∫1 ( loge (5x )) dx = − [ − x loge (5x )]1 − [ x ]15 5 2

2

2

5

1







1

∫1 ( loge (5x )) dx = −  −2 loge (10) + 5 loge (1) −  2 − 5  5 2

1

1

∫1 ( loge (5x )) dx = 2 loge (10) − 5 loge (1) − 2 + 5 5 2

9

∫1 ( loge (5x )) dx = 2 loge (10) − 5

units 2

5

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.4   |  185 5 a Graph cuts the y axis where x = 0, y = loge (2), i.e. at ( 0, loge (2) ). Graph cuts the x axis where y = 0 loge (2 − 4 x ) = 0 e0 = 2 − 4 x 1 = 2 − 4x 4x = 1 1 x= 4 1  At the point  , 0  . 4 b Largest possible domain occurs when 2 − 4x > 0 −4 x > −2 1 x< 2 1  Dom =  −∞,  2 y = loge (2 − 4 x ) dy −4 = dx 2 − 4 x dy −2 = dx 1 − 2 x 1 dy Since x < then 2 x < 1 and 2 − 4 x < 0 for all x. Therefore < 0. 2 dx d i y = loge (2 − 4 x ) Inverse: swap x and y x = loge (2 − 4 y) c

ex = 2 − 4 y 4 y = 2 − ex 1 y = 2 − ex 4 1 So h −1 ( x ) = 2 − e x 4 1  ii Dom = R and ran =  −∞,  2

(

)

(

e

)

y x = –1 2

y = h(x)

(0, –41 (

(0, loge(2))

0

y = h – 1(x)

( –41 , 0( 6 a

y = –1 2

x

(loge(2), 0)

y = x loge ( x ) 1 dy = loge ( x ) + x × dx x dy = loge ( x ) + 1 dx

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

186 | TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.4 Consequentially e2

∫ ( loge ( x ) + 1) dx = [ x loge ( x )]1

e2

1

e2

e2

∫ loge ( x ) dx + ∫ 1dx = [ x loge ( x )]1 1

e2

1

e2

(

)

e2

2 2 ∫ loge ( x ) dx = e loge (e ) − (1) loge (1) − ∫ 1dx

1 e2

∫ loge ( x ) dx = 2e

2

loge (e) − [ x ]1e

∫ loge ( x ) dx = 2e

2

− e2 + 1

∫ loge ( x ) dx = e

+1

1 e2 1 e2

2

1

2

1

b y = x ( loge ( x ) ) 1 dy m m −1 = ( loge ( x ) ) + mx ( loge ( x ) ) × dx x dy m m −1 = ( loge ( x ) ) + m ( loge ( x ) ) dx c Consequentially m

e2

∫ (( loge ( x ))

m

+ m ( loge ( x ) )

m −1

∫ ( loge ( x ))

m

∫ ( loge ( x ))

m

∫ ( loge ( x ))

m

∫ ( loge ( x ))

m

1 e2 1 e2 1 e2 1

e2

dx + m ∫ ( loge ( x ) )

m −1

dx + m ∫ ( loge ( x ) )

m −1

dx + m ∫ ( loge ( x ) )

m −1

dx + m ∫ ( loge ( x ) )

m −1

1 e2 1 e2 1 e2

 1

e2

m dx =  x ( loge ( x ) )   1

(

dx = e 2 loge (e 2 )

)

m

− (1) ( loge (1) )

m

dx = e 2 ( 2 loge (e) ) − 0 m

dx = 2m e 2

1

e2

If I m =

2 m e

e

1

e2

) dx =  x (log (x))

∫ ( loge ( x ))

m

dx then

1

I m + mI m −1 = 2m e 2 as required e2

e2

∫ ( loge ( x )) dx if I m = ∫ ( loge ( x ))

d I 3 =

3

1

m

dx

1

Remembering that I1 = e 2 + 1 I 2 + 2 I1 = 2 2 e 2 I 2 = 2 2 e 2 − 2 I1 I 2 = 4e 2 − 2(e 2 + 1) I 2 = 4e 2 − 2e 2 − 2 I 2 = 2e 2 − 2 I 3 + 3 I 2 = 23 e 2 I 3 = 23 e 2 − 3 I 2 I 3 = 8e 2 − 3(2e 2 − 2) I 3 = 8e 2 − 6e 2 + 6 I 3 = 2e 2 + 6 e

So

2

∫ ( loge ( x )) dx = 2e 3

2

+6

1

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.4   |  187 7 a y = 4 x loge ( x + 1) When x = −0.5, y = 4(−0.5) loge (1 − 0.5) = −2 loge (0.5) = −2 loge (2−1 ) = 2 loge (2) dy 4x Gradient of tangent mT = = + 4 loge ( x + 1) dx x + 1 4(−0.5) When x = −0.5, mT = 4 loge (−0.5 + 1) + = 4 loge (0.5) − 4 −0.5 + 1 Equation of line with mT = 4 loge (0.5) − 4 which passes through ( x1 , y1 ) ≡ (−0.5, 2 loge (2)) is given by y − y1 = mT ( x − x1 ) y − 2 loge (2) = (4 loge (0.5) − 4)( x + 0.5) y = (4 loge (0.5) − 4) x + 0.5(4 loge (0.5) − 4) + 2 loge (0.5) y = (4 loge (0.5) − 4) x − 2 loge (0.5) − 2 + 2 loge (0.5) y = (−4 loge (2) − 4) x − 2 b

y = 2( x 2 − 1) loge ( x + 1) dy 2( x 2 − 1) = 4 x loge ( x + 1) + dx ( x + 1) dy 2( x − 1)( x + 1) = 4 x loge ( x + 1) + ( x + 1) dx dy = 4 x loge ( x + 1) + 2 x − 2, x > −1 dx

c

∫ 4 x loge ( x + 1) dx = 2( x

2 2

− 1) loge ( x + 1) + ∫ (2 x − 2) dx

= 2( x − 1) loge ( x + 1) + x 2 − 2 x

0.5

∫

−0.5

4 x loge ( x + 1) dx =  2( x 2 − 1) loge ( x + 1) − x 2 + 2 x 

(

0.5 −0.5

) (

)

= 2 × (0.5 − 1) loge (1.5) − 0.5 + 1 − 2 × ((−0.5)2 − 1) loge (0.5) − (−0.5)2 − 1 2

2

3  3 3 3  1 5 = − loge   + + loge   +  2 4 2  2 4 2 3  1 = loge   + 2  3 2 3 = − loge (3) + 2 units 2 2 2x 2 8 a m( x ) = loge x + 1 so m′( x ) = 2 x +1 2(−2) 4 m′(−2) = = − = − 0.8 5 (−2)2 + 1

(

)

( ) c Area = ∫ loge ( x 2 + 1) dx = 3.4058 × 2 = 6.8117 units 2 −3 3

b Area = ∫ loge x 2 + 1 dx = 3.4058 units 2 0 3

9 a When t = 7, C7 = 25 loge (1 + 0.5(7)) C7 = 25 loge (1 + 3.5) C7 = 25 loge (4.5) C7 = 37.6 mg b C (7, 37.6)

C = 25 loge (1 + 0.5t)

0

7

t

dC 25 × 0.5 12.5 c = = dt 1 + 0.5t 1 + 0.5t 12.5 12.5 dC When t = 3, = = = 5 mg/day dt 1 + 0.5(3) 2.5

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

188 | TOPIC 9 Logarithmic functions using calculus • EXERCISE 9.4 b Graph cuts the y axis when x = 0.

7

d ∫ 25 loge (1 + 0.5t ) dt 0

y=e

7

= 25 ∫ loge (1 + 0.5t ) dt

−

1 2

+3

1   − Point is  0, e 2 + 3   Graph cuts x axis where y = 0. 2 loge ( x − 3) + 1 = 0 2 loge ( x − 3) = −1 1 loge ( x − 3) = − 2

0

= 25 × 6.5367 = 163.4 mg

x 10 POI between y = −4 and y = loge   is x = −0.03663  2 1 1 x A = 2 ∫ 2e x dx − ∫ loge   dx + 0.03663 × 4  2 0 0.03663 = 5.093 km 2

e

= 5.1 km 2 11 A = 4.6 loge (t − 4) a Graph intersects the t axis where A = 0 4.6 loge (t − 4) = 0 loge (t − 4) = 0 e0 = t − 4 1= t − 4 t=5 Thus (a, 0) ≡ (5, 0), a = 5 b When A = 15, 15 = 4.6 loge (t − 4) 15 = loge (t − 4) 4.6 e3.2609 = t − 4 e3.2609 + 4 = t t = 30 It takes the patient 30 minutes to reach level 15. dA 4.6 c = dt t − 4 4.6 4.6 46 23 dA When t = 10, = = = = units/min 6 60 30 dt 10 − 4 d Total change is given by 30

∫ 4.6 loge (t − 4 ) dt 5

= 4.6 × 59.7105 = 274.6683 Change in alertness is 274.6683 units. 12 a Graphs intersect where g( x ) = h( x ). x 2 loge ( x ) = loge ( x + 1) x = 1.5017 Consequently y = loge (1.5017 + 1) = 0.9170. Therefore POI = (1.5017, 0.9170) b Area between curves is given by 1.5017

2 2 ∫ ( loge ( x + 1) − x loge ( x )) dx = 0.7096 units

x=e

Inverse:

x=e

1 2

+3

c

5 1 ( x −1) 2

∫  e 0

5  + 3 dx − ∫ ( 2 loge ( x − 3) + 1) dx  e −0.5 + 3

= 28.5651 − 1.9857 = 26.58 m 2 d Solve ( −5 = 2 loge ( x − 3) + 1) for x x = 3.049787

(

)

5 e −0.5 + 3 −

e −0.5 + 3

∫ ( 2 loge ( x − 3) + 1) dx

3.049787

= 18.90 m 2 e Total area of Australian native garden is 26.5794 + 18.90 = 45.5 m 2 14 a i W  hen x = 2.3942, y = e −2.3942 + 3 = 3.0915 i.e the point (2.4, 3.1) ii When x = −2.8654, y = − loge (−2.8654 + 3) = 2.0054 i.e the point ( −2.9, 2.0 )

v When x = 2.3942, y = − loge (2.3942 + 3) = −1.6853 i.e the point ( 2.4, −1.7 ) vi When x = −1, y = 2e( −1− 3) + 3 = 2.0366 i.e the point ( −1.0, 2.0 ) b i Region I: Area is 2.3942

( x − 3) ∫ ( 2e + 2 + loge ( x + 3)) dx

−1

= 12.1535 units 2 ii Region II: Area is 2.3942

−x ( x − 3) ∫ (e + 3 − 2e − 2) dx

−1 2.3942

+3

1 ( y −1) 2

−

 −1  Point is  e 2 + 3, 0   

0

1 ( y −1) 2

= x−3

iv When x = −1, y = − loge (−1 + 3) = −0.691 i.e the point ( −1.0, −0.7 )

30

= 4.6 ∫ loge ( t − 4 ) dt

13 a y = f ( x ) = e

1 2

iii When x = −1, y = e1 + 3 = 5.7128 i.e the point ( −1.0,5.7 )

5

1 ( x −1) 2

−

+3

x−3=e 1 loge ( x − 3) = ( y − 1) 2 2 loge ( x − 3) = y − 1 y = 2 loge ( x − 3) + 1 Thus f −1 : (3, ∞) → R, f −1 ( x ) = 2 loge ( x − 3) + 1

=

−x ( x − 3) ∫ (e − 2e + 1) dx

−1

= 4.9666 units 2 iii Region III: Area is −1

( x − 3) ∫ ( 2e + 2 + loge ( x + 3)) dx

−2.8654

= 3.5526 units 2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 10 Discrete random variables • EXERCISE 10.2   |  189

Topic 10 — Discrete random variables Exercise 10.2 — Discrete random variables 1 a

R G Y R G Y R G Y R G Y R G Y R G Y R G Y R G Y R G Y

R R

G Y R G

G

Y R G

Y

Y

ξ = {RRR, RRG, RRY, RGR, RGG, RGY, RYR, RYG, RYY, GRR, GRG, GRY, GGR, GGG, GGY, GYR, GYG, GYY, YRR, YRG, YRY, YGR, YGG, YGY, YYR, YYG, YYY} b Y is the number of green balls obtained. Y = {0, 1, 2, 3} 3 3 3 27 Pr(Y = 3) = Pr(GGG) = × × = 10 10 10 1000 Pr(Y = 2) = Pr(RGG + Pr(GRG) + Pr(GGR) + Pr(GGY) + Pr(GYG) + Pr(YGG) 3 3 3 3 3 3 3 3 3 Pr(Y = 2) = × × + × × + × × 10 10 10 10 10 10 10 10 10 3 3 4 3 3 3 4 3 3 + × × + × × + × × 10 10 10 10 10 10 10 10 10 27 36 189 ×3+ ×3= Pr(Y = 2) = 1000 1000 1000 Pr(Y = 1) = Pr(RRG) + Pr(RGR) + Pr(RGY) + Pr(RYG) + Pr(GRR) + Pr(GRY) + Pr(GYR) + Pr(GYY) + Pr(YRG) + Pr(YGR) + Pr(YGY) + Pr(YYG) 3 3 3 3 3 3 3 3 4 3 4 3 Pr(Y = 1) = × × + × × + × × + × × 10 10 10 10 10 10 10 10 10 10 10 10 3 3 3 3 3 4 3 4 3 3 4 4 + × × + × × + × × + × × 10 10 10 10 10 10 10 10 10 10 10 10 4 3 3 4 3 3 4 3 4 4 4 3 + × × + × × + × × + × × 10 10 10 10 10 10 10 10 10 10 10 10 27 36 48 ×3+ ×6+ ×3 Pr(Y = 1) = 1000 1000 1000 441 Pr(Y = 1) = 1000 Pr(Y = 0) = 1 − ( P(Y = 1) + P (Y = 2) + P (Y = 3))  441 189 27  + + Pr(Y = 0) = 1 −    1000 1000 1000  657 343 1000 − = Pr(Y = 0) = 1000 10 000 1000 c 0 1 2 3 y Pr(Y = y) d

343 1000

441 1000

189 1000

27 1000

∑ Pr(Y = y) = 1 and all probabilities are between 0 and 1, therefore this is a discrete probability function.

all y

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

190 | TOPIC 10 Discrete random variables • EXERCISE 10.2 2 Let X be the number of sixes obtained. ξ = {11, 12, 13, 14, 15, 16 21, 22, 23, 24, 25, 26 31, 32, 33, 34, 35, 36 41, 42, 43, 44, 45, 46 51, 52, 53, 54, 55, 56 61, 62, 63, 64, 65, 66} X = 1, 2, 3 1 1 1 × = 6 6 36 = 1) = Pr(61, 62, 63, 64, 65, 16, 26, 36, 46, 56) 1 1 10 = 1) = 10 × × = 6 6 36 = 0) = 1 − ( Pr( X = 1) + Pr( X = 2) )  1 10  36 11 25 − = = 0) = 1 −  +  =  36 36  36 36 36

Pr( X = 2) = Pr(66) = Pr( X Pr( X Pr( X Pr( X

x

0

1

2

Pr( X = x )

25 36

10 5 = 36 18

1 36

3 a i 0 ≤ Pr(Y = y) ≤ 1 for all y and the sum of the probabilities is 1. This is a discrete probability density function. ii 0 ≤ Pr(Y = y) ≤ 1 for all y and the sum of the probabilities is 1. This is a discrete probability density function. b

∑ Pr( X = x ) = 1

all x

5k + 3k − 0.1 + 2 k + k + 0.6 − 3k = 1 8 k + 0.5 = 1 8 k = 0.5 0.5 1 k= = 8 16 4 a i 0 ≤ Pr(Y = y) ≤ 1 for all y and the sum of the probabilities is 0.9. This is not a discrete probability density function. ii Negative probabilities so this is not a discrete probability density function. b

∑ Pr( X = x ) = 1

all x

0.5k 2 + 0.5k + 0.25( k + 1) + 0.5 + 0.5k 2 = 1 k 2 + 0.75k + 0.75 = 1 k 2 + 0.75k + 0.25 = 0 100 k 2 + 75k + 25 = 0 4 k 2 + 3k − 1 = 0 (4 k − 1)( k + 1) = 0 1 k = ,−1 4 1 k = , as k ≥ 0 4 5 a ξ = {11, 12, 13, 14, 15, 16 21, 22, 23, 24, 25, 26 31, 32, 33, 34, 35, 36 41, 42, 43, 44, 45, 46 51, 52, 53, 54, 55, 56 61, 62, 63, 64, 65, 66} b Z = number of even numbers so Z = {0, 1, 2} Pr( Z = 0) = Pr(11) + Pr(13) + Pr(15) + Pr(31) + Pr(33) + Pr(35) + Pr(51) + Pr(53) + Pr(55) Pr( Z = 0) = (0.1 × 0.1) × 9 Pr( Z = 0) = 0.09

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 10 Discrete random variables • EXERCISE 10.2   |  191 Pr( Z = 2) = Pr(22) + Pr(24) + Pr(26) + Pr(42) + Pr(44) + Pr(46) + Pr(62) + Pr(64) + Pr(66) Pr( Z = 2) = (0.2 × 0.2) + (0.2 × 0.25) + (0.2 × 0.25) + (0.25 × 0.2) + (0.25 × 0.25) + (0.25 × 0.25) + (0.25 × 0.2) + (0.25 × 0.25) + (0.25 × 0.25) Pr( Z = 2) = 0.49 Pr( Z = 1) = Pr(12) + Pr(14) + Pr(16) + Pr(21) + Pr(23) + Pr(25) + Pr(32) + Pr(34) + Pr(36)+ Pr(41) + Pr(43) + Pr(45) + Pr(52) + Pr(54) + Pr(56) + Pr(61) + Pr(63) + Pr(65) Pr( Z = 1) = 1 − ( Pr( Z = 2) + Pr( Z = 0) ) = 1 − ( 0.49 + 0.09 ) = 0.42 z

0

1

2

Pr( Z = z )

0.09

0.42

0.49

c Pr( Z = 1) = 0.42 6 a ξ = {SSS, SSA, SAS, SAA, ASS, ASA, AAS, AAA} where S means Simon won and A means Samara won. b X = number of sets Simon wins Pr( X = 0) = Pr( AAA) = 0.63 = 0.216 Pr( X = 1) = Pr( AAS ) + Pr( ASA) + Pr( SAA) = (0.6)2 × 0.4 × 3 = 0.432 Pr( X = 2) = Pr( SSA) + Pr( SAS ) + Pr( ASS ) = (0.6)2 × 0.4 × 3 = 0.288 Pr( x = 3) = Pr( SSS ) = 0.4 3 = 0.064 x

0

1

2

3

Pr(X = x)

0.216

0.432

0.288

0.064

c Pr( X ≤ 2) = 1 − Pr( X = 3) = 1 − 0.064 = 0.936 7 a 0 ≤ Pr( X = x ) ≤ 1 for all x and

∑ Pr( X = x ) = 0.1 + 0.5 + 0.5 + 0.1 = 1.2 ≠ 1

all x

This is not a discrete probability density function. b Negative probabilities so this is not a discrete probability density function. c 0 ≤ Pr( Z = z ) ≤ 1 for all z and

∑ Pr( Z = z) = 0.25 + 0.15 + 0.45 + 0.35 = 1.1 so ∑ Pr( Z = z) ≠ 1. This is not a discrete probability

all z

density function. d 0 ≤ Pr( X = x ) ≤ 1 for all x and

all z

∑ Pr( X = x ) = 0.1 + 0.25 + 0.3 + 0.25 + 0.1 = 1

all x

This is a discrete probability density function. 8 a

∑ Pr( X = x ) = 1

all x

3d + 0.5 − 3d + 2d + 0.4 − 2d + d − 0.05 = 1 d + 0.85 = 1 d = 0.15 b

∑ Pr(Y = y) = 1

all y

0.5k + 1.5k + 2 k + 1.5k + 0.5k = 1 6k = 1 1 k= 6 c

∑ Pr( Z = z) = 1

all z

1 1 1 − a2 + − a2 + − a2 + a = 1 3 3 3 1 + a − 3a 2 = 1 a − 3a 2 = 0 a(1 − 3a) = 0 1 − 3a = 0 as a > 0 1 1 = 3a or a = 3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

192 | TOPIC 10 Discrete random variables • EXERCISE 10.2 1 1 4 (5 − x ), p(1) = (5 − 1) = , 7 7 7 1 2 1 1 p(3) = (5 − 3) = , p(4) = (5 − 4) = 7 7 7 7 Each probability lies between 0 and 1. Sum of probabilities is 1 so this is a discrete probability function.

9 a p( x ) =

x2 − x 40 (−1)2 + 1 2 12 − 1 =0 p(−1) = = , p(1) = 40 40 40 2 2 2 −2 2 3 −3 6 = = p(2) = , p(3) = 40 40 40 40 52 − 5 20 4 2 − 4 12 = = p(4) = , p(5) = 40 40 40 40 Each probability lies between 0 and 1. Sum of probabilities is greater than 1 so this is not a discrete probability function. 1 c p( x ) = x 15 1 1 1 2 1 3 p(1) = 1 = , p(4) = 4 = , p(9) = 9= 15 15 15 15 15 15 1 4 1 5 p(16) = 16 = , p(25) = 25 = 15 15 15 15 Each probability lies between 0 and 1. Sum of probabilities is 1 so this is a discrete probability function. b p( x ) =

1 (15 − 3 x ) a 1 12 1 9 1 6 p(1) = (15 − 31) = , p(2) = (15 − 3(2)) = , p(3) = (15 − 3(3)) = a a a a a a 1 3 1 p(4) = (15 − 3(4)) = , p(5) = (15 − 3(5)) = 0 a a a 12 9 6 3 + + + +0 =1 a a a a 30 =1 a a = 30

10 p( x ) =

11 a F = female and M = male F M F M F M F M F M F M F M F M

F F F

M F

M

M F

F

M

M F

M

M

 FFFF , FFFM , FFMF , FFMM , FMFF , FMFM ,    ξ =  FMMF , FMMM , MFFF , MFFM , MFMF , MFMM ,   MMFF , MMFM , MMMF , MMMM ,    b X is the number of females in the litter X = {0,1, 2,3, 4} 4 4 4 1 4 6  1  1  1 Pr( X = 0) =   = , Pr( X = 1) = 4   = , Pr( X = 2) = 6   =  2     16 16 2 16 2 4 4 4 1  1  1 Pr( X = 3) = 4   = , Pr( X = 4) =   =  2  2 16 16

x

0

1

2

3

4

Pr( X = x )

1 16

4 1 = 16 4

6 3 = 16 8

4 1 = 16 4

1 16

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 10 Discrete random variables • EXERCISE 10.2   |  193 c Pr( X = 4) =

1 16

d Pr( X ≥ 1) = 1 − Pr( X = 0) = 1 −

1 15 = 16 16

e Pr( X ≤ 2) = Pr( X = 0) + Pr( X = 1) + Pr( X = 2) =

1 4 6 11 + + = 16 16 16 16

12 a ξ = {11, 12, 13, 14, 15, 16, 17, 18, 19, 110, 111, 112 21, 22, 23, 24, 25, 26, 27, 28, 29, 210, 211, 212 31, 32, 33, 34, 35, 36, 37, 38, 39, 310, 311, 312 41, 42, 43, 44, 45, 46, 47, 48, 49, 410, 411, 412 51, 52, 53, 54, 55, 56, 57, 58, 59, 510, 511, 512 61, 62, 63, 64, 65, 66, 67, 68, 69, 610, 611, 612 71, 72, 73, 74, 75, 76, 77, 78, 79, 710, 711, 712 81, 82, 83, 84, 85, 86, 87, 88, 89, 810, 811, 812} b X is the number of primes obtained as a result of a toss Pr( X = 0) = Pr(11, 14, 16, 18, 19, 1 10, 1 12, 41, 44, 46, 48, 49, 4 10, 4 12, 61, 64, 66, 68, 69, 6 10, 6 12, 81, 84, 86, 88, 89, 8 10, 8 12) 1 1  = 28 ×  ×   8 12  28 = 96 Pr( X = 1) = Pr(12, 13, 15, 17, 1 11, 21, 24, 26, 28, 29, 2 10, 2 12, 31, 34, 36, 38, 39, 3 10, 3 12, 42, 43, 45, 47, 4 11, 51, 54, 56, 58, 59, 5 10, 5 12, 62, 63, 65, 67, 6 11, 71, 74, 76, 78, 79, 7 10, 7 12, 82, 83, 85, 87, 8 11) 1 1  = 48 ×  ×   8 12  48 = 96 Pr( X = 2) = Pr(22, 23, 25, 27, 2 11, 32, 33, 35, 37, 3 11, 52, 53, 55, 57, 5 11, 72, 73, 75, 77, 7 11) 1 1  = 20 ×  ×   8 12  20 = 96 c Pr(Win) = Pr( X = 2) × Pr( X = 2) × Pr( X = 2)  5 3 =   = 0.009  24  13 a Possible scores are: PP = 20 points, PJ or JP = 15 points, PS or SP = 12 points, JJ = 10 points JS or SJ = 7 points, SS = 4 points 3 3 9 6  3 1 × = , Pr(15) = 2  ×  = ,  13 13  169 13 13 169 1 1 1  3 9  54 Pr(12) = 2  ×  = , Pr(10) = × = ,  13 13  169 13 13 169 81 9 9  1 9  18 Pr(7) = 2  ×  = , Pr(4) = × =  13 13  169 13 13 169

b Pr(20) =

4

x Pr( X = x )

7

81 18 169 169

10

12

15

20

1 169

54 169

6 169

9 169

6 169 54 6 9 69 ii Pr( X ≥ 12) = + + = 169 169 169 169 Pr( X = 15 ∩ X ≥ 12) 6 169 6 2 iii Pr( X = 15 | X ≥ 12) = = × = = Pr( X ≥ 12) 169 69 69 23

c i Pr( X = 15) =

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

194 | TOPIC 10 Discrete random variables • EXERCISE 10.2 14 a

∑ Pr( X = x ) = 1

all x

3m + 2n = 1.....................(1) 3 Pr( X = 0) = Pr( X = 5) 3m = n.....................(2) Substitute (2) into (1) n + 2n = 1 3n = 1 1 n= 3 1 Substitute n = into (2) 3 1 3m = 3 1 m= 9 b i Pr( X ≥ 0) = Pr( X = 0) + Pr( X = 1) + Pr( X = 5) 1 1 1 5 Pr( X ≥ 0) = + + = 9 9 3 9 Pr( X = 1) ∩ Pr( X ≥ 0) 1 5 1 ii Pr( X = 1 | X ≥ 0) = = ÷ = Pr( X ≥ 0) 9 9 5 15 a S = Success and F – Failure

S S

F S

F

F S S

F

F S

F

S F S F S F S F S F S F S F S F S F S F S F S F S F S F S F S F

S F S F S F S F S F S F S F S F

S

F

E = {SSSSS, SSSSF, SSSFS, SSSFF, SSFSS, SSFSF, SSFFS, SSFFF, SFSSS, SFSSF, SFSFS, SFSFF, SFFSS, SFFSF, SFFFS, SFFFF, FSSSS, FSSSF, FSSFS, FSSFF, FSFSS, FSFSF, FSFFS, FSFFF, FFSSS, FFSSF, FFSFS, FFSFF, FFFSS, FFFSF, FFFFS, FFFFF} X = {0, 1, 2, 3, 4, 5} Pr( X = 0) = Pr(5 failures) = 0.4 5 = 0.01 024 Pr( X = 1) = Pr(4 failures) = 5 × 0.4 4 × 0.6 = 0.0768 Pr( X = 2) = Pr(3 failures) = 10 × 0.4 3 × 0.62 = 0.2304 Pr( X = 3) = Pr(2 failures) = 10 × 0.4 2 × 0.63 = 0.3456 Pr( X = 4) = Pr(1 failure) = 5 × 0.4 × 0.64 = 0.2592 Pr( X = 5) = P (0 failures) = 0.65 = 0.0778 x

0

1

2

3

4

5

Pr( X = x ) 0.0102 0.0768 0.2304 0.3456 0.2592 0.0778 b Pr( X = 3) + Pr( X = 4) + Pr( X = 5) = 0.3456 + 0.2592 + 0.0778 = 0.6826 It is a success helping three or more patients.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 10 Discrete random variables • EXERCISE 10.2   |  195 16 a

R Game over R Game over R Game over B R Game over G R Game over Y R Game over R Game over B R Game over G R Game over Y R Game over R Game over B R Game over G R Game over Y

B

G

Y

b Wins $10 with BBB, GGG or YYY c X = {0, 1, 10} 2  1 2  1 1 1 Pr( X = 0) = + 3  ×  + 9  × ×   5 5  5 5 5 5 2 6 9 = + + 5 25 125 50 30 9 98 = + + = 125 125 125 125 3 3  1 Pr( X = 10) = 3   =  5 125 Pr( X = 1) = 1 − ( Pr( X = 0) + Pr( x = 10) ) 3  24 125  98 = − + =  125 125 125  125 x

$0

$1

$10

Pr( X = x )

98 125

24 125

3 125

17 a Pr( H ) =

2 1 and Pr(T ) = and X = the number of heads obtained. 3 3 6

 1 Pr( X = 0) =   = 0.0014  3 5  1  2 Pr( X = 1) = 6C1     = 0.0165  3  3 4 2  1  2 Pr( X = 2) = 6C2     = 0.0823  3  3 3 3  1  2 Pr( X = 3) = 6C3     = 0.2195  3  3 2 4  1  2 Pr( X = 4) = 6C4     = 0.3292  3  3 5  1  2 Pr( X = 5) = 6C5     = 0.2634  3  3 6

 2 Pr( X = 0 = 6) =   = 0.0878  3 x

0

1

2

3

4

5

6

Pr( X = x )

0.0014

0.0165

0.0823

0.2195

0.3292

0.2634

0.0878

b i Pr( X > 2) = Pr( X = 3) + Pr( X = 4) + Pr( X = 5) + Pr( X = 6) = 0.2195 + 0.3292 + 0.2634 + 0.0878 = 0.8999 ii Pr( X > 2 | X < 5) Pr( X = 3) + Pr( X = 4) 1 − ( Pr( X = 5) + Pr( X = 6) ) 0.2195 + 0.3292 = 1 − ( 0.2634 + 0.0878 ) = 0.8457 =

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

196 | TOPIC 10 Discrete random variables • EXERCISE 10.3 18

∑ Pr(Y = y) = 1

all y

0.5k 2 + 0.3 − 0.2 k + 0.1 + 0.5k 2 + 0.3 = 1 k 2 − 0.2 k + 0.7 = 1 k 2 − 0.2 k − 0.3 = 0 k = −0.4568 or k = 0.6568 k can be positive or negative due to the two places of k: 0.5k­2 and 0.3 − 0.2k For both values of k , 0 < 0.5k 2 < 1 and 0 < 0.3 − 0.2 k < 1

Exercise 10.3 — Measures of centre and spread 1 (2 x − 1), x ∈{ 1, 2, 3, 4} 16 1 1 p(1) = ( 2(1) − 1) = 16 16 1 3 p(2) = ( 2(2) − 1) = 16 16 1 5 p(3) = ( 2(3) − 1) = 16 16 7 1 p(4) = ( 2(4) − 1) = 16 16

1 a p( x ) =

y

1

2

3

4

Pr(X = x)

1 16

3 16

5 16

7 16

1 3 5 7 b E(X ) = 1  + 2   + 3   + 4    16   16   16   16  1 6 15 28 50 1 + + + = = 3.125 or 3 E(X ) = 16 16 16 16 16 8 1 1 1 1 2 a Pr(Y ) = Pr(−$2) = , Pr(G ) = Pr($3) = , Pr( R) = Pr($6) = , Pr( P ) = Pr($8) = 4 4 4 4 x

−$2

$3

$6

$8

Pr( X = x )

1 4

1 4

1 4

1 4

1 1 1 1 b E(X ) = −2   + 3   + 6   + 8    4  4  4  4 2 3 6 8 15 E(X ) = − + + + = = $3.75 4 4 4 4 4  1  3  2  3  1  15 3 a E(Y ) = −5   + 0   + 5   + d   + 25   =  10   10   10   10   10  2 5 10 3d 25 15 + = − +0+ + 10 10 10 10 2 3d + 30 15 = 10 2 3d + 30 = 75 3d = 45 d = 15 b i E(2Y + 3) = 2E(Y ) + 3 = 2(7.5) + 3 = 18 ii E(5 − Y ) = 5 − E(Y ) = 5 − 7.5 = −2.5 iii E(−2Y ) = −2E(Y ) = −2(7.5) = −15

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 10 Discrete random variables • EXERCISE 10.3   |  197 1 2 ( z − 4), 2 ≤ z ≤ 5 38 1 1 5 p(2) = (22 − 4) = 0, p(3) = (32 − 4) = , 38 38 38 1 12 1 21 p(4) = (4 2 − 4) = , p(5) = (52 − 4) = 38 38 38 38  5  12   21  E( Z ) = 2(0) + 3   + 4   + 5    38   38   38  15 48 105 E( Z ) = 0 + + + 38 38 38 168 E( Z ) =  4.42 38

4 p( z ) =

5 a E( X ) = 1(0.3) + 2(0.15) + 3(0.4) + 4(0.1) + 5(0.05) E( X )  $2.45 b E( X 2 ) = 12 (0.3) + 22 (0.15) + 32 (0.4) + 4 2 (0.1) + 52 (0.05) E( X 2 ) = 7.35 Var(X ) = E( X 2 ) − [ E( X )]2 Var(X ) = 7.35 − 2.452 = $1.35 SD(X ) = 1.35 = $1.16 6 a k + k + 2 k + 3k + 3k = 1 10 k = 1 1 k= 10  1  1  2  3  3 b E(X ) = −2   + 0   + 2   + 4   + 6    10   10   10   10   10  2 4 12 18 32 E(X ) = − + 0 + + + = = 3.2 10 10 10 10 10 1 1 2 3 3 c E( X 2 ) = (−2)2   + 0 2   + 22   + 4 2   + 62    10   10   10   10   10  4 8 48 108 168 +0+ + + = = 16.8 E( X 2 ) = 10 10 10 10 10 Var(X ) = E( X 2 ) − [ E( X )]2 = 16.8 − 3.22 = 6.56 SD(X ) = 6.56 = 2.56 7 a p( x ) = p(1) =

x2 x = 1, 2, 3, 4. 30 12 22 32 4 2 16 1 4 9 = , p(2) = = , p(3) = = , p(4) = = 30 30 30 30 30 30 30 30

x

1

2

3

4

Pr( X = x )

1 30

4 2 = 30 15

9 3 = 30 10

16 8 = 30 15

1

4

9

16

∑ Pr( X = x ) = 30 + 30 + 30 + 30 = 1

all x

1 4 9 16 b i E(X ) = 1  + 2   + 3   + 4    30   30   30   30  1 8 27 64 100 10 E(X ) = + + + = = 30 30 30 30 30 3  1  4  9  16  ii E( X 2 ) = 12   + 22   + 32   + 4 2    30   30   30   30  1 16 81 256 2 + + + E( X ) = 30 30 30 30 354 118 = E( X 2 ) = 30 10

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

198 | TOPIC 10 Discrete random variables • EXERCISE 10.3

( )

Var(X ) = E X 2 − [ E( X ) ]

2

2

118  10  −  10  3  1062 1000 62 31 − = = = 0.69 Var(X ) = 90 90 90 45

Var(X ) =

c i Var(4 X + 3) = 4 2 Var( X ) = 16(0.69) = 11.02 ii Var(2 − 3 X ) = (−3)2 Var( X ) = 9(0.689) = 6.2 8 a E( Z ) = −7(0.21) + m(0.34) + 23(0.33) + 31(0.12) = 14.94 − 1.47 + 0.34 m + 7.59 + 3.72 = 14.94 0.34 m + 9.84 = 14.94 0.34 m = 5.1 5.1 m= 0.34 m = 15 2 2 2 2 2 b E( Z ) = (−7) (0.21) + 15 (0.34) + 23 (0.33) + 31 (0.12) 2 E( Z ) = 10.29 + 76.5 + 174.57 + 115.32

E( Z 2 ) = 376.68 Var( Z ) = E( Z 2 ) − [ E( Z )]2 Var( Z ) = 376.68 − 14.94 2 Var( Z ) = 153.48 Var(2( Z − 1)) = VAR(2 Z − 2) Var(2( Z − 1)) = 22 VAR( Z ) Var(2( Z − 1)) = 4 × 153.48 Var(2( Z − 1)) = 613.91 Var(3 − Z ) = (−1)2 Var( Z ) Var(3 − Z ) = 153.48 9 a

∑ Pr( X = x ) = 1

all x

 1  1  1  2  2  1  1 E(X ) = −3   + (−2)   + (−1)   + 0   + 1  + 2   + 3   i  9  9  9  9  9  9  9 3 2 1 2 2 3 1 E(X ) = − − − + 0 + + + = 9 9 9 9 9 9 9 1  2  2  1  1    1  1 ii E( X 2 ) = (−3)2   + (−2)2   + (−1)2   + 0 2   + 12   + 22   + 32    9  9  9  9  9  9  9 9 4 1 2 4 9 2 E( X ) = + + + 0 + + + 9 9 9 9 9 9 29 2 E( X ) = 9 2 Var( X ) = E( X 2 ) − [ E( X ) ] 29  1  2 −  9  9 261 1 260 Var( X ) = − = = 3.2099 81 81 81 SD( X ) = 3.2099 = 1.7916 Var( X ) =

b

∑ Pr(Y = y) = 1

all y

i E(Y ) = 1(0.15) + 4(0.2) + 7(0.3) + 10(0.2) + 13(0.15) E(Y ) = 0.15 + 0.8 + 2.1 + 2 + 1.95 = 7 ii E(Y 2 ) = 12 (0.15) + 4 2 (0.2) + 72 (0.3) + 10 2 (0.2) + 132 (0.15) E(Y 2 ) = 0.15 + 3.2 + 14.7 + 20 + 25.35 E(Y 2 ) = 63.4 Var(Y ) = E(Y 2 ) − [ E(Y )]2 Var(Y ) = 63.4 − 72 Var(Y ) = 63.4 − 49 = 14.4 SD(Y ) = 14.4 = 3.7947 Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 10 Discrete random variables • EXERCISE 10.3   |  199 c

∑ Pr( Z = z) = 1

all z

1 1 1 1 1 1 i E( Z ) = 1  + 2   + 3   + 4   + 5   + 6    12   4  3  6  12   12  1 6 12 8 5 6 + + + + + E( Z ) = 12 12 12 12 12 12  38  19 E( Z ) =   =  12  6  1  1  1  1  1  1 2 ii E( Z ) = 12   + 22   + 32   + 4 2   + 52   + 62    12   4  3  6  12   12  1 12 36 32 25 36 2 + + + + + E( Z ) = 12 12 12 12 12 12 142 E( Z 2 ) = 12 2 Var( Z ) = E( Z 2 ) − [ E( Z ) ] 142  19  2 −  12  6  1704 1444 Var( Z ) = − 144 144 65 Var( Z ) = = 1.8056 36 65 SD( Z ) = = 1.3437 36 10 a ∑ Pr(Y = y) = 1 Var( Z ) =

all y

1 − 2c + 3c 2 + 1 − 2c = 1 3c 2 − 4c + 1 = 0 (3c − 1)(c − 1) = 0 3c − 1 = 0 or c − 1 = 0 3c = 1 c =1 1 c= 3 1 ∴ c = as 0 < c < 1 3 b

y Pr(Y = y)

−1 1 3

1 1 9

3 1 9

5 1 9

7 1 3

 1  1  1  1  1 E(Y ) = −1  + 1  + 3   + 5   + 7    3  9  9  9  3 1 1 3 5 7 E(Y ) = − + + + + 3 9 9 9 3 3 1 3 5 21 E(Y ) = − + + + + 9 9 9 9 9 27 =3 E(Y ) = 9  1  1  1  1  1 c E(Y 2 ) = (−1)2   + 12   + 32   + 52   + 72    3  9  9  9  3 3 1 9 25 147 E(Y 2 ) = + + + + 9 9 9 9 9 185 2 E(Y ) = 9 Var(Y ) = E(Y 2 ) − [ E(Y )]2 185 2 −3 Var(Y ) = 9 185 81 − Var(Y ) = 9 9 104 Var(Y ) = 9 Var(Y ) = 11.56 SD(Y ) = 11.56 = 3.40 Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

200 | TOPIC 10 Discrete random variables • EXERCISE 10.3 11 a E(2 X − 1) = 2E( X ) − 1 E(2 X − 1) = 2(4.5) − 1 = 8 b E(5 − X ) = 5 − E( X ) E(5 − X ) = 5 − 4.5 = 0.5 c E(3 X + 1) = 3E( X ) + 1 E(3 X + 1) = 3(4.5) + 1 = 14.5 12 SD( X ) = 2.5 so Var( X ) = 2.52 = 6.25 a Var(6 X ) = 62 Var( X ) = 36 × 6.25 = 225 b Var(2 X + 3) = 22 Var( X ) = 4 × 6.25 = 25 c Var(− X ) = (−1)2 VAR( X ) = 6.25 13 a p( x ) = h(3 − x )( x + 1) p(0) = h(3)(1) = 3h p(1) = h(3 − 1)(1 + 1) = 4 h p(2) = h(3 − 2)(2 + 1) = 3h 3h + 4 h + 3h = 1 10 h = 1 1 h= 10 b

x

0

1

2

Pr( X = x )

3 10

4 10

3 10

3 4 3 E( X ) = 0   + 1  + 2    10   10   10  4 6 10 E( X ) = 0 + + = = 1 10 10 10 4 3 2 2 3  E( X ) = 0   + 12   + 22    10   10   10  4 12 16 2 E( X ) = 0 + + = = 1.6 10 10 10 Var( X ) = E( X 2 ) − [ E( X )]2 16 Var( X ) = − (1)2 10 16 10 6 Var( X ) = − = = 0.6 10 10 10 6 = 0.7746 SD( X ) = 10 14 a

∑ Pr( X = x ) = 1

all x

a + 0.2 + 0.3 + b + 0.1 = 1 a + b + 0.6 = 1 a + b = 0.4.............................(1) E( X ) = 2.5 1(a) + 2(0.2) + 3(0.3) + 4(b) + 5(0.1) = 2.5 a + 0.4 + 0.9 + 4 b + 0.5 = 2.5 a + 4 b + 1.8 = 2.5 a + 4 b = 0.7......(2) (2) = (1) 3b = 0.3 b = 0.1 Substitute b = 0.1 into (1)

a + 0.1 = 0.4 a = 0.3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 10 Discrete random variables • EXERCISE 10.3   |  201 b E( X 2 ) = 12 (0.3) + 22 (0.2) + 32 (0.3) + 4 2 (0.1) + 52 (0.1) E( X 2 ) = 0.3 + 0.8 + 2.7 + 1.6 + 2.5 E( X 2 ) = 7.9 Var( X ) = E( X 2 ) − [ E( X )]2 Var( X ) = 7.9 − 2.52 Var( X ) = 7.9 − 6.25 Var( X ) = 1.65 SD( X ) = 1.65 = 1.2845 15 a Var( X ) = 2a − 2 and E( X ) = a Var( X ) = E( X 2 ) − [ E( X )]2 2a − 2 = E( X 2 ) − −a 2 a 2 + 2a − 2 = E( X 2 ) b E( X 2 ) = 6 a 2 + 2a − 2 = 6 a 2 + 2a − 8 = 0 (a + 4)(a − 2) = 0 a + 4 = 0 or a − 2 = 0 a = −4 a=2 ∴ a = 2, a > 0 Thus E( X ) = a = 2 and Var( X ) = 2a − 2 = 2(2) − 2 = 2 ny y = 1, 2, 3, 4 16 a p(n) =  n(7 − y) y = 5, 6 p(1) = n, p(2) = 2n, p(3) = 3n, p(4) = 4 n, p(5) = 2n, p(6) = n

∑ Pr( X = x ) = 1

all x

b

n + 2n + 3n + 4 n + 2n + n = 1 13n − 1 = 0 1 n= 13 y

1

2

3

4

5

6

Pr(Y = y)

1 13

2 13

3 13

4 13

2 13

1 13

 2  1  1  2  3  4 E(Y ) = 1  + 2   + 3   + 4   + 5   + 6    13   13   13   13   13   13  1 4 9 16 10 6 46 + + + + + = = 3.5385 E(Y ) = 13 13 13 13 13 13 13  1  2  3  4  2  1 E(Y 2 ) = 12   + 22   + 32   + 4 2   + 52   + 62    13   13   13   13   13   13  1 8 27 64 50 36 186 2 + + + + + = E(Y ) = 13 13 13 13 13 13 13 Var(Y ) = E(Y 2 ) − [ E(Y )]2 2 186  46  −  13  13  2418 2116 302 Var(Y ) = − = = 1.7870 169 169 169 SD(Y ) = 1.7870 = 1.3368

Var(Y ) =

17 a E = {11, 12, 13, 14, 15, 16, 17, 18 21, 22, 23, 24, 25, 26, 27, 28 31, 32, 33, 34, 35, 36, 37, 38 41, 42, 43, 44, 45, 46, 47, 48 51, 52, 53, 54, 55, 56, 57, 58 61, 62, 63, 64, 65, 66, 67, 68 71, 72, 73, 74, 75, 76, 77, 78 81, 82, 83, 84, 85, 86, 87, 88} Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

202 | TOPIC 10 Discrete random variables • EXERCISE 10.3 2 1  1 b Pr( Z = 1) = Pr(11) =   =  8 64 2 3  1 Pr( Z = 2) = Pr(12, 21, 22) = 3   =  8 64 2 5  1 Pr( Z = 3) = Pr(13, 23, 31, 32, 33) = 5   =  8 64 2 7  1 Pr( Z = 4) = Pr(14, 24, 34, 41, 42, 43, 44) = 7   =  8 64 2 9  1 Pr( Z = 5) = Pr(15, 25, 35, 45, 51, 52, 53, 54, 55) = 9   =  8 64 Pr( Z = 6) = Pr(16, 26, 36, 46, 56, 61, 62, 63, 64, 65, 66) 2 11  1 Pr( Z = 6) = 11   =  8 64

Pr( Z = 7) = Pr(17, 27, 37, 47, 57, 67, 71, 72, 73, 74, 75, 76, 77) 2

13  1 Pr( Z = 7) = 13   =  8 64 Pr( Z = 8) = Pr(18, 28, 38, 48, 58, 68, 78, 81, 82, 83, 84, 85, 86, 87, 88) 2 15  1 Pr( Z = 8) = 15   =  8 64

z

1

2

3

4

5

6

7

8

Pr( Z − z )

1 64

3 64

5 64

7 64

9 64

11 64

13 64

15 64

1 3 5 9  11   13   15  7 c E( Z ) = 1  + 2   + 3   + 4   + 5   + 6   + 7   + 8    64   64   64   64   64   64   64   64  1 6 15 28 45 66 91 120 E( Z ) = + + + + + + + 64 64 64 64 64 64 64 64 372 = 5.8125 E( Z ) = 64 1 3 5 9  11   13   15  7 E( Z 2 ) = 12   + 22   + 32   + 4 2   + 52   + 62   + 72   + 82    64   64   64   64   64   64   64   64  1 12 45 112 225 396 637 960 2388 + + + + = = 37.3125 E( Z 2 ) = + + + 64 64 64 64 64 64 64 64 64 2 2 Var( Z ) = E( Z ) − [ E( Z )] 2 2388  372  −  64  64  152 832 138 384 14 448 − = = 3.5273 Var( Z ) = 4096 4096 4096 14 448 = 1.8781 SD( Z ) = 4096

Var( Z ) =

18 a Area of whole board is π (4 × 5)2 = 400π B and A = π (4)2 = 16π and Pr( A) =

16π 1 = 400π 25

48π 3 = 400π 25 80π 5 B and C = π (12)2 − 64π = 144π − 64π = 80π and Pr(C ) = = 400π 25 112π 7 2 B and D = π (16) − 144π = 256π − 144π = 112π and Pr( D) = = 400π 25 144π 9 B and E = π (20)2 − 256π = 400π − 256π = 144π and Pr( E ) = = 400π 25 b X is the gain in dollars Pr( E ) = −$1, Pr( D) = $0, Pr(C ) = $1, Pr( B) = $4, Pr( A) = $9 B and B = π (8)2 − 16π = 64π − 16π = 48π and Pr( B) =

x

−$1

$0

$1

$4

$9

Pr( X = x )

9 25

7 25

5 1 = 25 5

3 25

1 25

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 10 Discrete random variables • EXERCISE 10.3   |  203 9 3 1 7 5 c i E( X ) = −1  + 0   + 1  + 4   + 9    25   25   25   25   25  9 5 12 9 E( X ) = − + 0 + + + 25 25 25 25 17 = 0.68 cents E( X ) = 25 9 5 3 1 7 ii E( X 2 ) = (−1)2   + 0 2   + 12   + 4 2   + 92    25   25   25   25   25  9 5 48 81 E( X 2 ) = + 0 + + + 25 25 25 25 143 2 E( X ) = 25 Var( X ) = E( X 2 ) − [ E( X )]2 2 143  17  −  25  25  3575 289 3286 − = = $5.26 Var( X ) − 625 625 625 3286 SD( X ) = = $2.29 625

Var( X ) =

19 a Possible Scores: 6 8 2 4 11       1 − 1, 1 − 3, 1 − 5, 1 − 7, 1 − 10    6 8 10 13 4     3 − 1, 3 − 3, 3 − 5, 3 − 7, 3 − 10   6  8 10 15 12       E = 5 − 1, 5 − 3, 5 − 5, 5 − 7, 5 − 10  8 10 17   12 14     7 − 1, 7 − 3, 7 − 5, 7 − 7, 7 − 10   11  17  20 13 15        10 1, 10 3, 10 5, 10 7, 10 10 − − − − −     \ Possible Scores are 2, 4, 6, 8, 10, 11, 12, 13, 14, 15, 17, 20 b Pr(2) = Pr(11) = 0.2 × 0.2 = 0.04 Pr(4) = Pr(13, 31) = (0.2 × 0.2)2 = 0.08 Pr(6) = Pr(15, 33, 51) = (0.2 × 0.3) + (0.2 × 0.2) + (0.3 × 0.2) = 0.16 Pr(8) = Pr(17, 35, 53, 71) = (0.2 × 0.2) + (0.2 × 0.3) + (0.3 × 0.2) + (0.2 × 0.2) = 0.20 Pr(10) = Pr(37, 55, 73) = (0.2 × 0.2) + (0.3 × 0.3) + (0.2 × 0.2) = 0.17 Pr(11) = Pr(1 10, 10 1) = (0.2 × 0.1) + (0.1 × 0.2) = 0.04 Pr(12) = Pr(57, 75) = (0.3 × 0.2) + (0.2 × 0.3) = 0.12 Pr(13) = Pr(1 10, 10 1) = (0.2 × 0.1) + (0.1 × 0.2) = 0.04 Pr(14) = Pr(77) = 0.2 × 0.2 = 0.04 Pr(15) = Pr(5 10,10 5) = (0.3 × 0.1) + (0.1 × 0.3) = 0.06 Pr(17) = Pr(7 10, 10 7) = (0.2 × 0.1) + (0.1 × 0.2) = 0.04 Pr(20) = Pr(10 10) = 0.1 × 0.1 = 0.01 x

2

Pr( X = x ) 0.04

4

6

8

10

11

12

13

14

15

17

20

0.08

0.16

0.2

0.17

0.04

0.12

0.04

0.04

0.06

0.04

0.01

c E( X ) = 9.4 and SD( X ) = 3.7974 20 a

∑ Pr( X = x ) = 1

all x

0.5k 2 + 0.5k 2 + k + k 2 + 4 k + 2 k + 2 k + k 2 + 7 k 2 = 1 10 k 2 + 9 k − 1 = 0 k = −1 or k = 0.1 ∴ k = 0.1, k > 0 b E( X ) = 1.695 c SD( X ) = 1.167

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

204 | TOPIC 10 Discrete random variables • EXERCISE 10.4

Exercise 10.4 — Applications 1 a E( X ) = 5(0.05) + 10(0.25) + 15(0.4) + 20(0.25) + 25(0.05) E( X ) = 0.25 + 2.5 + 6 + 5 + 1.25 E( X ) = 15 b E( X 2 ) = 52 (0.05) + 10 2 (0.25) + 152 (0.4) + 20 2 (0.25) + 252 (0.05) E( X 2 ) = 1.25 + 25 + 90 + 100 + 31.25 E( X 2 ) = 247.5 Var( X ) = E( X 2 ) − [ E( X )]2 Var( X ) = 247.5 − 152 Var( X ) = 22.5 SD( X ) = 22.5 = 4.7434 c µ − 2σ = 15 − 2(4.7434) = 5.5132 µ + 2σ = 15 + 2(4.7434) = 24.4947 Pr( µ − 2σ ≤ X ≤ µ + 2σ ) = Pr(5.5132 ≤ X ≤ 24.4947) Pr( µ − 2σ ≤ X ≤ µ + 2σ ) = Pr( X = 10) + Pr( x = 15) + Pr( X = 20) Pr( µ − 2σ ≤ X ≤ µ + 2σ ) = 1 − (Pr( X = 5) + Pr( X = 25)) Pr( µ − 2σ ≤ X ≤ µ + 2σ ) = 1 − 0.1 = 0.9 2 E( X ) = 0(0.012) + 1(0.093) + 2(0.243) + 3(0.315) + 4(0.214) + 5(0.1) + 6(0.023) E( X ) = 0 + 0.093 + 0.486 + 0.945 + 0.856 + 0.5 + 0.138 E( X ) = 3.018 E( X 2 ) = 0 2 (0.012) + 12 (0.093) + 22 (0.243) + 32 (0.315) + 4 2 (0.214) + 52 (0.1) + 62 (0.023) E( X 2 ) = 0 + 0.093 + 0.972 + 2.835 + 3.424 + 2.5 + 0.828 E( X 2 ) = 10.652 Var( X ) = E( X 2 ) − [ E( X )]2 Var( X ) = 10.652 − 3.0182 Var( X ) = 1.5437 SD( X ) = 1.5437 = 1.2424 µ − 2σ = 3.018 − 2(1.2424) = 0.5332 µ + 2σ = 3.018 + 2(1.2424) = 5.5028 Pr( µ − 2σ ≤ X ≤ µ + 2σ ) = Pr(0.5332 ≤ X ≤ 5.5028) Pr( µ − 2σ ≤ X ≤ µ + 2σ ) = Pr( X = 1) + Pr( X = 2) + Pr( x = 3) + Pr( X = 4) + Pr( X = 5) Pr( µ − 2σ ≤ X ≤ µ + 2σ ) = 1 − ( Pr( X = 0) + Pr( X = 1) ) Pr( µ − 2σ ≤ X ≤ µ + 2σ ) = 1 − ( 0.012 + 0.023) Pr( µ − 2σ ≤ X ≤ µ + 2σ ) = 0.965 3

x

$100

$250

$500

$750

$1000

Pr( X = x )

0.1

0.2

0.3

0.3

0.1

where X is hundreds of thousands of dollars. a Pr( X ≤ $500) = Pr( X = $100) + Pr( X = $250) + Pr( X = $500) Pr( X ≤ $500) = 0.1 + 0.2 + 0.3 = 0.6 Pr( X ≥ $250) ∩ Pr( X ≤ $750) b Pr( X ≥ $250 X ≤ $750) = Pr( X ≤ $750) Pr( X = $250) + Pr( X = $500) + Pr( X = $750) Pr( X ≥ $250 X ≤ $750) = 1 − Pr( X = $1000) 0.2 + 0.3 + 0.3 Pr( X ≥ $250 X ≤ $750) = 1 − 0.1 0.8 8 Pr( X ≥ $250 X ≤ $750) = = 0.9 9 c E( X ) = 100(0.1) + 250(0.2) + 500(0.3) + 750(0.3) + 1000(0.1) E( X ) = 10 + 50 + 150 + 225 + 100 = $535 \the expected profit is $535 000 d E( X 2 ) = 100 2 (0.1) + 250 2 (0.2) + 500 2 (0.3) + 750 2 (0.3) + 1000 2 (0.1) E( X 2 ) = 1000 + 12500 + 75000 + 168 750 + 100 000 E( X 2 ) = 357 250 Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 10 Discrete random variables • EXERCISE 10.4   |  205 Var( X ) = E( X 2 ) − [ E( X ) ]

2

Var( X ) = 357 250 − (535)2 = 71 025 SD( X ) = 71 025 = 266.51

µ − 2σ = 535 − 2(266.51) = 1.98 µ + 2σ = 535 + 2(266.51) = 1068.02 Pr( µ − 2σ ≤ X ≤ µ + 2σ ) = Pr(1.98 ≤ X ≤ 1068.02) Pr( µ − 2σ ≤ X ≤ µ + 2σ ) = 1 4 a

∑ Pr( Z = z) = 1

all z

3m + 3n = 1....................................(1) Pr( Z < 2) = 3 Pr( Z > 4) Pr( Z = 0) + Pr( Z = 1) = 3 Pr( Z = 5) 2m = 3n................(2) Substitute (2) into (1) 3m + 2m = 1 5m = 1 1 m= 5 1 Substitute m = into (2) 5 1   2   = 3n  5 2 n= 15 3 3 3 2 2 2 b E( Z ) = 0   + 1  + 2   + 3   + 4   + 5    15   15   15   15   15   15  3 6 6 8 10 E( Z ) = 0 + + + + + 15 15 15 15 15 33 11 E( Z ) = = as required 15 5  2  2  3  3  3  2 E( Z 2 ) = 0 2   + 12   + 22   + 32   + 4 2   + 52    15   15   15   15   15   15  3 12 18 32 50 + E( Z 2 ) = 0 + + + + 15 15 15 15 15 115 23 = E( Z 2 ) = 15 3 Var( Z ) = E( Z 2 ) − [ E( Z )]2 2 23  11  −  3 5 23 121 Var( Z ) = − 3 25 575 363 − Var( Z ) = 75 75 212 = 2.8267 Var( Z ) = 75 212 SD( Z ) =  1.6813 75 11 c µ − 2σ = − 2(1.6813) = −1.1626 5 11 µ + 2σ = + 2(1.6813) = 5.5626 5 Pr( µ − 2σ ≤ Z ≤ µ + 2σ ) = Pr(−1.1626 ≤ Z ≤ 5.5626) = 1

Var( Z ) =

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

206 | TOPIC 10 Discrete random variables • EXERCISE 10.4

∑ Pr( Z = z) = 1

5 a E(Y ) = 3.5

all z

1(0.3) + 2(0.2) + d (0.4) + 8(0.1) = 3.5 0.3 + 0.4 + 0.4 d + 0.8 = 3.5 0.4 d + 1.5 = 3.5 0.4 d = 2 2 d= 0.4 d =5 Pr(Y ≥ 2) ∩ Pr(Y ≤ 5) b Pr(Y ≥ 2 Y ≤ 5) = Pr(Y ≤ 5) Pr(Y = 2) + Pr(Y = 5) Pr(Y ≥ 2 Y ≤ 5) = 1 − Pr(Y = 8) 0.2 + 0.4 Pr(Y ≥ 2 Y ≤ 5) = 1 − 0.1 0.6 2 = Pr(Y ≥ 2 Y ≤ 5) = 0.9 3

Substitute (2) into (1) 4 m + 2(2m) = 1 4m + 4m = 1 8m = 1 1 m= 8 1 Substitute m = into (2) 8  1 2  = n  8 1 n= 4 1 1 1 1 1 1 b i E( Z ) = 0   + 1  + 2   + 3   + 4   + 5   8  8  8 8 4 4 1 2 3 8 10 E( Z ) = 0 + + + + + 8 8 8 8 8 24 =3 E( Z ) = 8

c E(Y 2 ) = 12 (0.3) + 22 (0.2) + 52 (0.4) + 82 (0.1) E(Y 2 ) = 0.3 + 0.8 + 10 + 6.4 = 17.5 Var(Y ) = E(Y 2 ) − [ E(Y )]2 Var(Y ) = 17.5 − 3.52 Var(Y ) = 5.25 d SD(Y ) = 5.25 = 2.2913 6 a

∑ Pr( Z = z) = 1

ii E( Z 2 ) = 0 2  1  + 12  1  + 22  1  + 32  1  + 4 2  1  + 52  1  8  8  8  8  4 4 1 4 9 32 50 E( Z 2 ) = 0 + + + + + 8 8 8 8 8 96 2 E( Z ) = = 12 8 Var( Z ) = E( Z 2 ) − [E( Z )]2

all z

0.2 + 0.15 + a + b + 0.05 = 1 a + b + 0.4 = 1 a + b = 0.6.....................................(1) E( Z ) = 4.6 1(0.2) + 3(0.15) + 5a + 7b + 9(0.05) = 4.6 0.2 + 0.45 + 5a + 7b + 0.45 = 4.6 5a + 7b + 1.1 = 4.6 5a + 7b = 3.5................(2) From (1) a = 0.6 − b..........(3) Substitute (3) into (2) 5(0.6 − b) + 7b = 3.5 3 − 5b + 7b = 3.5 2b = 0.5 b = 0.25 Substitute b = 0.25 into (3) a = 0.6 − 0.25 = 0.35 2

2

2

Var( Z ) = 12 − 32 = 3 c SD( Z ) = 3 = 1.732 µ − 2σ = 3 − 2(1.732) = − 0.464 µ + 2σ = 3 + 2(1.732) = 6.464 Pr(µ − 2σ ≤ Z ≤ µ + 2σ ) = Pr(− 0.464 ≤ Z ≤ 6.464) = 1 8

2

2

2

b E( Z ) = 1 (0.2) + 3 (0.15) + 5 (0.35) + 7 (0.25) + 9 (0.05) E( Z 2 ) = 0.2 + 1.35 + 8.75 + 12, 25 + 4.05 E( Z 2 ) = 26.6 Var( Z ) = E( Z 2 ) − [ E( Z )]2 Var( Z ) = 26.6 − 4.62 = 5.44 SD( Z ) = 5.44 = 2.3324 c i E(3 Z + 2) = 3E( Z ) + 3 = 3(4.6) + 3 = 15.8 ii Var(3 Z + 2) = 32 Var( Z ) = 9 × 5.44 = 48.96 7 a

4 m + 2n = 1....................(1) Pr( Z ≤ 3) = Pr( Z ≥ 4) 4 m = 2n 2m = n.....................(2)

z

0

1

2

3

4

5

Pr( Z = z )

m

m

m

m

n

n

M

M’

N

0.216

0.264

0.480

N’

0.234

0.286

0.520

0.450

0.550

1.000

a As M and N are independent Pr( M ∩ N ) = Pr( M ) Pr( N ) = 0.45 × 0.48 = 0.216 b Pr( M ′∩ N ′) = 0.286 c Y is the number of times M and N occur. Y = {0, 1, 2} Pr(Y = 0) = 0.286 Pr(Y = 1) = 0.264 + 0.234 = 0.498 Pr(Y = 2) = 0.216 y

0

1

2

Pr(Y = y)

0.286

0.498

0.216

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 10 Discrete random variables • EXERCISE 10.4   |  207 d i E(Y ) = 0(0.286) + 1(0.498) + 2(0.216) = 0 + 0.498 + 0.432 = 0.93 ii E(Y 2 ) = 0 2 (0.286) + 1(0.498) + 22 (0.216) = 0 + 0.498 + 0.864 = 1.362 Var(Y ) = E(Y 2 ) − [ E(Y )]2 Var(Y ) = 1.362 − 0.932 Var(Y ) = 0.4971 iii SD(Y ) = 0.4971 = 0.7050 1 9 a If p( x ) = (4 − x ), where x = {0, 1, 2} 9 4 3 1 2 p(0) = , p(1) = = , p(2) = . 9 9 3 9 b

∞

∑ p( x ) = 1 so this is a probability density function. x =1

∞ 4 3 2 7 i E( X ) = µ = ∑ xp( x ) = 0   + 1  + 2   =  9  9  9 9 x =1 ∞

ii Var( X ) = σ 2 = ∑ ( x − 2)2 p( x ) = x =1

iii SD( X ) =

11  7  2 99 49 50 −  = − = 9 9 81 81 81

50 = 0.7857 81

c µ − 2σ = 2 − 2(1.03) = − 0.06 µ + 2σ = 2 + 2(1.03) = 4.06 Pr(µ − 2σ ≤ X ≤ µ + 2σ ) = Pr(− 0.06 ≤ X ≤ 4.06) Pr(µ − 2σ ≤ X ≤ µ + 2σ ) = Pr(0) + Pr(1) + Pr(2) Pr(µ − 2σ ≤ X ≤ µ + 2σ ) = 1 10 a

∑ Pr( X = x ) = 1

all x 2

5k − 1 3k − 1 4 k − 1 k + + + =1 4 12 12 12 2 3k + 5k − 1 + 3k − 1 + 4 k − 1 = 12 3k 2 + 12 k − 3 = 12 3k 2 + 12 k − 15 = 0 k 2 + 4k − 5 = 0 ( k + 5)( k − 1) = 0

k = −5, k = 1 k = −5 is not applicable ∴ k = 1 x Pr( X = x )

0

1

2

3

3 1 4 1 2 1 = = = 12 4 12 3 12 6

3 12

4 4 9 17  3  4  2  3 = = 1.4 b E( X ) = 0   + 1  + 2   + 3   = 0 + + +  12   12   12   12  12 12 12 12 c Pr( X < 1.4) = Pr( X = 0) + Pr( X = 1) = 11 a

Money Probability

3 4 7 + = 12 12 12

$1000 $15  000 $50  000 $100  000 $200  000 1 5

1 5

1 5

1 5

1 5

E (Bank offer) is  1  1  1  1  1 = 1000   + 15 000   + 50 000   + 100 000   + 200 000    5  5  5  5  5 = $73 200

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

208 | TOPIC 10 Discrete random variables • EXERCISE 10.4 b

Money

$1000

$15  000

$50  000

$100  000

Probability

1 4

1 4

1 4

1 4

E (Bank offer) is 1 1 1 1 = 1000  + 15 000  + 50 000  + 200 000   4 4 4 4 = $66 500 12 a Autobiography

Probability

Cook Book

Probability

New

$65

0.40

New

$54

0.40

Good used

$30

0.30

Good used

$25

0.25

Worn used

$12

0.30

Worn used

$15

0.35

New Autobiography + New Cook Book  $65 + $54 = $119 0.4 × 0.40 = 0.16 New Autobiography + Good Cook Book   $65 + $25 =    $90 0.4 × 0.25 = 0.10 New Autobiography + Worn Cook Book    $65 + $15 =    $80 0.4 × 0.35 = 0.14 Good Autobiography + New Cook Book   $30 + $54 =    $84 0.3 × 0.40 = 0.12 Good Autobiography + Good Cook Book $30 + $25 =    $55 0.3 × 0.25 = 0.075 Good Autobiography + Worn Cook Book  $30 + $15 =    $45 0.3 × 0.35 = 0.105 Worn Autobiography + New Cook Book   $12 + $54 =    $66 0.3 × 0.40 = 0.12 Worn Autobiography + Good Cook Book $12 + $25 =    $37 0.3 × 0.25 = 0.075 Worn Autobiography + Worn Cook Book  $12 + $15 =    $27 0.3 × 0.35 = 0.105 X = cost of two books. x

$119

$90

$84

$80

$66

Pr( X = x )

0.16

0.10

0.12

0.14

0.12

$55

$45

$37

0.075 0.105 0.075

$27 0.105

b E( X ) = 119(0.16) + 90(0.1) + 84(0.12) + 80(0.14) + 66(0.12) + 55(0.075) + 45(0.105) + 37(0.075) + 27(0.105) = $71.70 13 a Let Y be the net profit per day. y

−$120

$230

$580

$930

Pr(Y = y)

0.3

0.4

0.2

0.1

b E(Y ) = −120(0.3) + 230(0.4) + 580(0.2) + 930(0.1) E(Y ) = $265 c E(Y 2 ) = (−120)2 (0.3) + 230 2 (0.4) + 580 2 (0.2) + 930 2 (0.1) E(Y 2 ) = 179 250 Var(Y ) = E(Y 2 ) − [ E(Y ) ]

2

Var(Y ) = 179 250 − 2652 Var(Y ) = 109 025 SD(Y ) = 109 025 = $330

µ − 2σ = 265 − 2(330) = −$395 µ + 2σ = 265 + 2(330) = $925 Pr( µ − 2σ ≤ Y ≤ µ + 2σ ) = Pr(−$395 ≤ Y ≤ $925) Pr( µ − 2σ ≤ Y ≤ µ + 2σ ) = 1 − Pr(Y = $930) Pr( µ − 2σ ≤ Y ≤ µ + 2σ ) = 1 − 0.1 = 0.9 3 1 14 a Coin: Pr( H ) = and Pr(T ) = 4 4 1 1 1 1 1 1 Die: Pr(1) = , Pr(2) = , Pr(3) = , Pr(4) = , Pr(5) = , Pr(6) = 12 12 4 4 12 4 5 5 5 10 1 1 1 1 10 10 1 1             E = {1H , 2 H , 3 H , 4 H , 5 H , 6 H , 1T , 2T , 3T , 4T , 5T , 6T}

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 10 Discrete random variables • EXERCISE 10.4   |  209 1  1 1  1 1  1 1 3 = Pr(10) = Pr(1T , 2T , 5T ) =  ×  +  ×  +  ×  =  12 4   12 4   12 4  48 16 3  1 3  1 3  1 3 9 Pr(5) = Pr(1H , 2 H , 5 H ) =  ×  +  ×  +  ×  = =  12 4   12 4   12 4  48 16 Pr(1) = Pr(3 H , 3T , 4 H , 4T , 6 H , 6T )  1 3  1 1  1 3  1 1  1 3  1 1 = ×  + ×  + ×  + ×  + ×  + ×   4 4  4 4  4 4  4 4  4 4  4 4  3  1 = 3  + 3   16   16  12 = 16 x

1

5

10

Pr( X = x )

12 3 = 16 4

3 16

1 16

 12   3   1  12 15 10 37 = 2.3 b E( X ) = 1 + 5  + 10   = + + =  16   16   16  16 16 16 16 c E(25 tosses) = 25 × 2.3125 = 57.8 d Let n be the number of tosses 2.3125n = 100 100 n= = 43.243 2.3125 Minimum number of tosses required is 44. 15 a Pr(V ∪ W ) = 0.7725 and Pr(V ∩ W ) = 0.2275. Pr(V ∪ W ) = Pr(W ) + Pr(V ) − Pr(W ∩ V ) 0.7725 = Pr(W ) + Pr(V ) − 0.2275 1.0000 = Pr(W ) + Pr(V )...............................(1) V and W are independent events. Pr(W ∩ V ) = Pr(W ) Pr(V ) 0.2275 = Pr(W ) Pr(V ) 0.2275 = Pr(V )..................................(2) Pr(W ) Substitute (2) into (1) 0.2275 1= + Pr(W ) P (W ) 2 Pr(W ) = 0.2275 + [ Pr(W ) ]

0 = [ Pr(W ) ] − Pr(W ) + 0.2275 Pr(W ) = 0.65 or 0.35 2

But Pr(V ) < Pr(W ) so Pr(W ) = 0.65 and Pr(V ) = 0.35 b

W

W’

V

0.2275

0.1225

0.35

V’

0.4225

0.2275

0.65

0.35

0.65

1.000

Note: Pr(V ′ ∩ W ′) = 0.2275 c

x

0

1

2

Pr( X = x )

0.2275

0.545

0.2275

d i   E( X ) = 0(0.2275) + 1(0.545) + 2(0.2275) = 1 ii  E( X 2 ) = 0 2 (0.2275) + 12 (0.545) + 22 (0.2275) = 0 + 0.545 + 0.91 = 1.455 Var( X ) = E( X 2 ) − [ E( X )]2 Var( X ) = 1.455 − 12 Var( X ) = 0.455 iii SD( X ) = 0.455 = 0.6745 Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

210 | TOPIC 10 Discrete random variables • EXERCISE 10.4 16 a

∑ Pr( Z = z) = 1

i E( Z ) = 2.4286 2 ii Var( Z ) = E( Z 2 ) − [ E( Z )] = 9 − 2.42862 = 3.1019

all z 2

k 5 − 2 k 8 − 3k + + =1 7 7 7 k = 2 or 3 But if k = 3, \k = 2 b

8 − 3(3) 1 = − so this is not applicable. 7 7 1

3

5

2 4 = 7 7

5 − 2(2) 1 = 7 7

8 − 3(2) 2 = 7 7

z Pr( Z = z )

2

iii SD( Z ) = 3.1019 = 1.7613 c µ − 2σ = 2.4286 − 2(1.7613) = −1.094 µ + 2σ = 2.4286 + 2(1.7613) = 5.9512 Pr( µ − 2σ ≤ Z ≤ µ + 2σ ) = Pr(−1.094 ≤ Z ≤ 5.9512) = 1

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 11 The binomial distribution • EXERCISE 11.2   |  211

Topic 11 — The binomial distribution Exercise 11.2 — Bernoulli trials 1 a T  his is not a Bernoulli distribution as a successful outcome is not specified. b This is a Bernoulli distribution as a success is getting a hole in one and a failure is not getting a hole in one. c This is a Bernoulli distribution as a success is withdrawing an ace and a failure is withdrawing any other card. 2 a x 0 1 Pr( X = x )

0.58

Pr(Y = y)

0.32

But p > 1 − p so p = 0.7 b E( X ) = p = 0.7 9 a SD(Y ) = 0.4936

0.42

b E( X ) = 0.42 c i Var( X ) = 0.58 × 0.42 = 0.2436 ii SD( X ) = 0.2436 = 0.4936 3 a T  his is a Bernoulli distribution as the arthritis drug is either successful or not. b This is a Bernoulli distribution as the child is either a girl or not. c This is not a Bernoulli distribution as the probability of success is unknown. d This is a Bernoulli distribution as the next person either subscribes or not. 4 a T  he friend does not replace the ball before I choose a ball, so this cannot be a Bernoulli distribution. b There are 6 outcomes not 2, so this is not a Bernoulli distribution. c The probability of success is unknown so this is not a Bernoulli distribution. 5 a Ε ( Z ) = p = 0.63 b Var( Z ) = pq = 0.63 × 0.37 = 0.2331 c SD( Z ) = 0.2331 = 0.4828 6 a y 0

1 ± 1 − 0.84 2 1 ± 0.4 p= 2 p = 0.3 or 0.7 p=

1

Var(Y ) = 0.49362 = 0.2436 b Var(Y ) = p(1 − p) = 0.2436 p − p2 = 0.2436 0 = p2 − p + 0.2436 1 ± (−1)2 − 4(1)(0.21)436 2(1) 1 ± 1 − 0.9744 p= 2 1 ± 0.16 p= 2 0.84 1.16 p= or 2 2 p = 0.42 or 0.58 But p > 1 − p so p = 0.58 Therefore p =

c E(Y ) = p = 0.58 10 a Pr(breast cancer) = 0.0072 b z 0 Pr( Z = z )

Var( Z ) = pq = 0.0072 × 0.9928 = 0.0071

σ = SD( Z ) = 0.0071 = 0.0845 µ − 2σ = 0.0072 − 2(0.0845) = −0.1618 µ + 2σ = 0.0072 + 2(0.0845) = 0.1762 Pr( µ − 2σ ≤ Z ≤ µ + 2σ ) = Pr(−0.1618 ≤ Z ≤ 0.1762) = Pr( Z = 0) = 0.9928

b i E(Y ) = p = 0.68 ii Var(Y ) = pq = 0.68 × 0.32 = 0.2176

Pr( X = x )

0.11

0.89

b i E( X ) = p = 0.89 ii Var( X ) = pq = 0.89 × 0.11 = 0.0979 iii SD( X ) = 0.0979 = 0.3129 c µ − 2σ = 0.89 − 2(0.3129) = 0.2642 µ + 2σ = 0.89 + 2(0.3129) = 1.5158 Pr( µ − 2σ ≤ X ≤ µ + 2σ ) = Pr(0.2642 ≤ X ≤ 1.5158) = Pr( X = 1) = 0.89 8 a Var( X ) = p(1 − p) = 0.21 p − p2 = 0.21 0 = p2 − p + 0.21 Therefore p =

0.0072

c µ = Ε ( Z ) = 0.0072

0.68

iii SD(Y ) = 0.2176 = 0.4665 c µ − 2σ = 0.68 − 2(0.4665) = −0.253 µ + 2σ = 0.68 + 2(0.4665) = 1.613 Pr( µ − 2σ ≤ Y ≤ µ + 2σ ) = Pr(−0.253 ≤ Y ≤ 1.613) = Pr(Y = 0) + Pr(Y = 1) =1 7 a x 0 1

0.9928

1

11 a y Pr(Y = y)

0

1

0.67

0.33

b µ = E(Y ) = p = 0.33 c Var(Y ) = pq = 0.33 × 0.67 = 0.2211 σ = SD(Y ) = 0.2211 = 0.4702

µ − 2σ = 0.33 − 2(0.4702) = −0.6104 µ + 2σ = 0.33 + 2(0.4702) = 1.2704 Pr( µ − 2σ ≤ Y ≤ µ + 2σ ) = Pr(−0.6104 ≤ Y ≤ 1.2704) = Pr(Y = 0) + Pr(Y = 1) =1 12 a

x

0

1

Pr( X = x )

4 5

1 5

1 ± (−1)2 − 4(1)(0.21) 2(1) Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

212 | TOPIC 11 The binomial distribution • EXERCISE 11.3 b E( X ) = p =

1 5 5

 1 c Pr( X = 5) =   = 0.00032  5 13 a Var( Z ) = p(1 − p) = 0.1075 p − p2 = 0.1075 0 = p2 − p + 0.1075 p = 0.1225 or 0.8775 Since p > 1 − p. p = 0.8775.

b z

Pr( Z = z )

0

1

0.1225

0.8775

c Ε ( Z ) = p = 0.8775 14 a SD( X ) = 0.3316 Var( X ) = 0.33162 = 0.11 b Var( Z ) = p(1 − p) = 0.11 p − p2 = 0.11 0 = p2 − p + 0.11 p = 0.1258 or 0.8742 Since p > 1 − p, p = 0.8742.

Exercise 11.3 — The binomial distribution  3 1 a Y ~ Bi 5,   7  4 5 1024 = 0.0609 Pr(Y = 0) =   =  7  16 807  4 4  3  3840 = 0.2285 Pr(Y = 1) = 5     =  7   7  16 807  4 3  3 2 5760 = 0.3427 Pr(Y = 2) = 10     =  7   7  16 807  4 2  3 3 4320 = 0.2570 Pr(Y = 3) = 10     =  7   7  16 807  4   3 4 1620 = 0.0964 Pr(Y = 4) = 5     =  7   7  16 807  3 5 243 = 0.0145 Pr(Y = 5) =   =  7  16 807 y Pr(Y = y)

0

1

2

3

4

5

0.0609

0.2285

0.3427

0.2570

0.0964

0.0145

b Pr(Y ≤ 3) = Pr( X = 0) + Pr( X = 1) + Pr( X = 2) + Pr( X = 3) = 0.0609 + 0.2285 + 0.3427 + 0.2570 = 0.8891 Pr(Y ≥ 1) ∩ Pr(Y ≤ 3) c Pr(Y ≥ 1 | Y ≤ 3) = Pr(Y ≤ 3) Pr( X = 1) + Pr( X = 2) + Pr( X = 3) = 0.8891 0.2285 + 0.3427 + 0.2570 = 0.8891 0.8282 = 0.8891 = 0.9315 4 3 4 4 d Pr(Miss, Bulls-eye, Miss, Miss) = × × × = 0.0800 7 7 7 7

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 11 The binomial distribution • EXERCISE 11.3   |  213 2 a Y ~ Bi (10, 0.42) Pr( Z = 0) = (0.58)10 = 0.0043 Pr( Z = 1) = 10(0.58)9 (0.42) = 0.0312, Pr( Z = 2) = 45(0.58)8 (0.42)2 = 0.1017 Pr( Z = 3) = 120(0.58)7 (0.42)3 = 0.1963 Pr( Z = 4) = 210(0.58)6 (0.42)4 = 0.2488 Pr( Z = 5) = 252(0.58)5 (0.42)5 = 0.2162 Pr( Z = 6) = 210(0.58)4 (0.42)6 = 0.1304 Pr( Z = 7) = 4120(0.58)3 (0.42)7 = 0.0540 Pr( Z = 8) = 45(0.58)2 (0.42)8 = 0.0147 Pr( Z = 9) = 10(0.58)(0.42)9 = 0.0024 Pr( Z = 10) = (0.42)10 = 0.0002 b z

0

1

Pr( Z = z ) 0.0043

2

3

4

0.0312 0.1017 0.1963 0.2488

5

6

7

8

9

10

0.2162

0.1304

0.0540

0.0147

0.0024

0.0002

Pr( Z ≥ 5) ∩ Pr( Z ≤ 8) Pr( Z ≤ 8) Pr( Z = 5) + Pr( Z = 6) + Pr( Z = 7) + Pr( Z = 8) = 1 − ( Pr( Z = 9) + Pr( Z = 10) ) 0.2162 + 0.1304 + 0.0540 + 0.0147 = 1 − (0.0024 + 0.0002) 0.4153 = 1 − 0.0026 = 0.4164

c Pr( Z ≥ 5 | Z ≤ 8) =

 1 3 a X ~ Bi  25,   6 1 1 E( X ) = np = 25 × = 4  4.1667 6 6 1 5 17 b Var( X ) = npq = 25 × × = 3  3.472 6 6 36 SD( X ) = 3.472 = 1.8634 4 a E( Z ) = np = 32.535 Var( Z ) = npq = np(1 − p) = 9.02195 Re-iterating, we have np = 32.535....................(1) np(1 − p) = 9.02195........(2) (2) ÷ (1) np(1 − p) 9.02195 = np 32.535 1 − p = 0.2773 1 − 0.2773 = p 0.7227 = p b Substitute p = 0.7227 into (1): 0.7227n = 32.535 32.535 n= = 45 0.7227 5 X ~ Bi ( n, 0.2) Pr( X ≥ 1) ≥ 0.85 1 − Pr( X = 0) ≥ 0.85 1 − 0.8n ≥ 0.85 1 − 0.85 ≥ 0.8n n ≥ 8.50 Thus 9 tickets would be required.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

214 | TOPIC 11 The binomial distribution • EXERCISE 11.3 6 X ~ Bi ( n, 0.33) Pr( X ≥ 1) > 0.9 1 − Pr( X = 0) > 0.9 1 − 0.67n > 0.9 1 − 0.9 > 0.67n n > 5.75 They need to play 6 games. 7 X ~ Bi (15, 0.62) a Pr( X = 10) = 10C10 (0.62)10 (0.38)0 = 0.1997 b Pr( X ≥ 10) = 0.4665 Pr( X < 4) c Pr( X < 4 | X ≤ 8) = Pr( X ≤ 8) 0.0011 = 0.3295 = 0.0034 8 X ~ Bi (15, 0.45) a E( X ) = np = 15 × 0.45 = 6.75 b Pr( X = 4) = 0.0780 c Pr( X ≤ 8) = 0.8182 d If T = buys a ticket and N = does not buy a ticket Pr(T , T , N , N ) = (0.45)2 × (0.55)2 = 0.0613 9 X ~ Bi (8, 0.63) a 0 1 2 3 x Pr( X = x ) 0.0004

0.0048

0.0285

0.0971

4

5

6

7

8

0.2067

0.2815

0.2397

0.1166

0.0248

b Pr( X ≤ 7) = 1 − Pr( X = 8) = 1 − 0.0248 = 0.9752 Pr( X ≥ 3) ∩ Pr( X ≤ 7) c Pr( X ≥ 3 | X ≤ 7) = Pr( X ≤ 7) Pr(3 ≤ X ≤ 7) = 0.9752 0.9416 = 0.9752 = 0.9655 d Pr( B′, B, B, B, B, B) = 0.37 × 0.635 = 0.0367 10 a X ~ Bi ( 45, 0.72) i E( X ) = np = 45 × 0.72 = 32.4 ii Var( Z ) = np(1 − p) = 45 × 0.72 × 0.28 = 9.072  1 b Y ~ Bi 100,   5 1 i E(Y ) = np = 100 × = 20 5 1 4 ii Var(Y ) = np(1 − p) = 100 × × = 16 5 5  2 c Z ~ Bi  72,   9 i E( Z ) = np = 72 ×

2 = 16 9

2 7 4 ii Var( Z ) = np(1 − p) = 72 × × = 12  12.4 9 9 9 11 Z ~ Bi ( 7, 0.32) a z Pr( Z = z )

0

1

2

3

4

5

6

7

0.0672

0.2215

0.3127

0.2452

0.1154

0.0326

0.0051

0.0003

b E( Z ) = np = 7 × 0.32 = 2.24 Var( Z ) = np(1 − p) = 7 × 0.32 × 0.68 = 1.5232

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 11 The binomial distribution • EXERCISE 11.3   |  215 c SD( Z ) = 1.5232 = 1.2342 µ − 2σ = 2.24 − 2(1.2342) = −0.2284 µ + 2σ = 2.24 + 2(1.2342) = 4.7084 Pr( µ − 2σ ≤ Z ≤ µ + 2σ ) = Pr(−0.2284 ≤ Y ≤ 4.7084) = 1 − ( Pr( X = 5) + Pr( X = 6) + Pr( X = 7) ) = 1 − ( 0.0326 + 0.0051 + 0.0003) = 0.9623 12 a Let X be the number of females on the executive.  1 X ~ Bi 12,   2 Pr( X ≥ 8) = 0.1938 b Let Y be the number of females on the executive. Y ~ Bi (12, 0.58) Pr(Y ≥ 8) = 0.3825 13 a Let X ~ Bi ( n, p) E( X ) = np = 9.12........................(1) Var( X ) = np(1 − p) = 5.6544.....(2) (2) ÷ (1) np(1 − p) 5.6544 = 9.12 np 1 − p = 0.62 1 − 0.62 = p 0.38 = p b Substitute p = 0.38 into (1): n × 0.38 = 9.12 9.12 n= = 24 0.38 14 a Let X ~ Bi ( n, p) E( X ) = np = 3.8325........................(1) Var( X ) = np(1 − p) = 3.4128.........(2) (2) ÷ (1) np(1 − p) 3.4128 = 3.8325 np 1 − p = 0.8905 1 − 0.8905 = p 0.1095 = p b Substitute p = 0.1095 into (1): n × 0.1095 = 3.8325 3.8325 = 35 n= 0.1095 15 a Let X ~ Bi (16, p) E( X ) = np = 10.16 E( X ) = 16 p = 10.16 10.16 p= = 0.635 16 b Var( X ) = np(1 − p) Var( X ) = 16(0.635)(0.365) Var( X ) = 3.7084 SD( X ) = 3.7084 = 1.9257  1 16 X ~ Bi 10,   7 1 = 1.4286 7 1 6 Var( X ) = np(1 − p) = 10 × × = 1.2245 7 7

a E( X ) = np = 10 ×

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

216 | TOPIC 11 The binomial distribution • EXERCISE 11.4 b SD( X ) = 1.2245 = 1.1066

µ − 2σ = 1.4286 − 2(1.1066) = −0.7846 µ + 2σ = 1.4286 + 2(1.1066) = 3.6410 Pr( µ − 2σ ≤ X ≤ µ + 2σ ) = Pr(−0.7846 ≤ Y ≤ 3.6410) = Pr(0 ≤ Y ≤ 3) = 0.9574 1 7 X ~ Bi ( n, 0.75) Pr( X ≥ 1) ≥ 0.95 1 − Pr( X = 0) ≥ 0.95 1 − 0.25n ≥ 0.95 1 − 0.95 ≥ 0.25n n ≥ 2.16 Thus 3 shots would be required. 18 a X ~ Bi(12, 0.2) Pr( x = 3) = 0.2362 b Y ~ Bi(14, 0.2362) Pr(Y ≥ 6) = 0.0890

Exercise 11.4 — Applications 1 Y ~ Bi (10, 0.3) a Pr(Y ≥ 7) = 0.0106 b E(Y ) = np = 10 × 3 = 3

Var(Y ) = np(1 − p) = 10 × 0.3 × 0.7 = 2.1 SD(Y ) = 2.1 = 1.4491 2 Z ~ Bi (5, 0.01) a Pr( Z ≤ 3) = 0.951 + 0.0480 + 0.0010 + 0 × 3 = 1 b i E( Z ) = np = 5 × 0.01 = 0.05 ii Var( Z ) = np(1 − p) = 5 × 0.01 × 0.99 = 0.0495 SD( Z ) = 0.0495 = 0.2225 c µ − 2σ = 0.05 − 2(0.0495) = −0.049 µ + 2σ = 0.05 + 2(0.0495) = 0.149 Pr( µ − 2σ ≤ Z ≤ µ + 2σ ) = Pr(−0.049 ≤ Z ≤ 0.149) = Pr( Z = 0) = 0.9510 3 X ~ Bi (15, 0.3) a Pr( X ≤ 5) = 0.7216 b i E( X ) = np = 15 × 0.3 = 4.5 Var( X ) = np(1 − p) = 15 × 0.3 × 0.7 = 3.15 ii SD( X ) = 3.15 = 1.7748 4 Y ~ Bi (6, 0.08) a i E(Y ) = np = 6 × 0.08 = 0.48 ii Var(Y ) = np(1 − p) = 6 × 0.08 × 0.92 = 0.4416 SD(Y ) = 0.4416 = 0.6645 b µ − 2σ = 0.48 − 2(0.6645) = −0.849 µ + 2σ = 0.48 + 2(0.6645) = 1.809 Pr( µ − 2σ ≤ Y ≤ µ + 2σ ) = Pr(−0.849 ≤ Y ≤ 1.809) = Pr(Y = 0) + P (Y = 1) = 0.6064 + 0.3164 = 0.9228 c Z ~ Bi (6, 0.01) i E( Z ) = np = 6 × 0.01 = 0.06 ii Var( Z ) = np(1 − p) = 6 × 0.01 × 0.99 = 0.0594 SD( Z ) = 0.0594 = 0.2437

d µ − 2σ = 0.06 − 2(0.2437) = −0.4274 µ + 2σ = 0.06 + 2(0.2437) = 0.5474 Pr( µ − 2σ ≤ Z ≤ µ + 2σ ) = Pr(−0.4274 ≤ Z ≤ 0.5474) = Pr( Z = 0) = 0.9415 e There is a probability of 0.9228 that a maximum of 1 male will be colour blind, whereas there is a probability of 0.9415 that no females will be colourblind. 5 Z ~ Bi (12, 0.85) a Pr( Z ≤ 8) = 0.0922 Pr( Z ≥ 5) ∩ Pr( Z ≤ 8) b Pr( Z ≥ 5 | Z ≤ 8) = Pr( Z ≤ 8) Pr(5 ≤ Z ≤ 8) = 0.0922 0.092213 = 0.0922 = 0.9992 c i E( Z ) = np = 12 × 0.85 = 10.2 ii Var( Z ) = np(1 − p) + 12 × 0.85 × 0.15 = 1.53 SD( Z ) = 1.53 = 1.2369 6 Let Z be the number of offspring with genotype XY.  1 Z ~ Bi  7,   2 6

7  1  1 = 0.0547 Pr( Z = 6) = 7C6     =  2  2 64 7 Let Z be the number of chips that fail the test. Z ~ Bi ( 250, 0.02) Pr( Z = 7) =

250

C7 (0.98)243 (0.02)7 = 0.1051

8 a X ~ Bi (3, p) Pr( X = 0) = (1 − p)3 , Pr( X = 1) = 3(1 − p)2 p, Pr( X = 2) = 3(1 − p) p2 , Pr( X = 3) = p3 0

x Pr( X = x )

(1 − p)

1 3

2 2

3(1 − p) p 3(1 − p) p

3 2

p3

b Pr( X = 0) = Pr( X = 1) (1 − p)3 = 3(1 − p)2 p 3

(1 − p) − 3(1 − p)2 p = 0 (1 − p)2 (1 − p − 3 p) = 0 (1 − p)2 (1 − 4 p) = 0 (1 − p)(1 + p)(1 − 4 p) = 0 1 − p = 0, 1 + p = 0 or 1 − 4 p = 0 p=1 p = −1 1= 4p 1 p= 4 1 ∴ p = because 0 < p < 1 4 1 3 = 4 4 1 3 9 ii Var( X ) = np(1 − p) = 3 × × = 4 4 16 9 3 σ = SD( X ) = = 16 4 9 a Let X be the number of people who suffer from anaemia. X ~ Bi (100, 0.013) Pr( X ≥ 5) = 0.0101 c i µ = Ε ( X ) = np = 3 ×

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 11 The binomial distribution • EXERCISE 11.4   |  217 b Pr( X = 4 | X < 10) =

Pr( X = 4) Pr( X < 10)

Pr( X = 4) = 0.0319 Pr( X < 10) = 0.9999

Pr( X = 4) 0.0319 = = 0.0319 Pr( X < 10) 0.9999 c µ = np = 100 × 0.013 = 1.3 Var( X ) = np(1 − p) = 100 × 0.013 × 0.987 = 1.2831 Pr( X = 4 | X < 10) =

σ = SD( X ) = 1.2831 = 1.1327 µ − 2σ = 1.3 − 2(1.1327) = −0.9654 µ + 2σ = 1.3 + 2(1.1327) = 3.5654 Pr( µ − 2σ ≤ X ≤ µ + 2σ ) = Pr(−0.9654 ≤ X ≤ 3.5654) = Pr(0 ≤ X ≤ 3) = 0.9580 This means there is a 96% chance that a maximum of 3 people per 100 will suffer from anaemia. 10 X ~ Bi(20, 0.2) a Pr( X ≥ 10) = 0.0026 b Pr( X ≥ 10) = 1 × 1 × 1 × 1 × Pr( X ≥ 6) = 0.0489 11 X ~ Bi(6, 0.7) X = kicking 50 m a i Pr(YYYNNN ) = ( 0.7 )3 ( 0.3)3 = 0.0093 ii Pr( X = 3) = 6C3 ( 0.7 )3 ( 0.3)3 = 0.1852 0.7 × Pr( X ≥ 2) iii Pr( X ≥ 3 1st kick > 50 m) = 0.7 0.678454 = 0.7 = 0.9692

b X ~ Bi(n, 0.95) Pr( X ≥ 1) ≥ 0.95 1 − Pr( X = 0) ≥ 0.95 1 − 0.3n ≥ 0.95 1 − 0.95 ≥ 0.3n n ≥ 2.48 Therefore, 3 footballers are needed. 12 X ~ Bi(12, 0.85) a Pr( X ≥ 9) = 0.9078 b Pr(3 M ,9G ) = ( 0.15)3 ( 0.85)9 = 0.0008 Pr( X = 1) × last 9 are goals c Pr( X = 10 last 9 are goals ) = Pr(last 9 are goals) 0.057375 × ( 0.85)9 = ( 0.85)9 = 0.0574 13 X ~ Bi(n, 0.08) Pr( X ≥ 2) > 0.8 1 − ( Pr( X = 0) + Pr( X = 1)) > 0.8 1 − 0.8 > Pr( X = 0) + Pr( X = 1) 0.2 > (0.92)n + n(0.92)n −1 (0.08) n = 36.4179 so at least 37 tickets must be bought. 14 X ~ Bi (10, p) Pr( X ≤ 8) = 1 − Pr( X ≥ 9) = 1 − ( Pr( X = 9) + Pr( X = 10) )

(

= 1 − 10(1 − p) p9 + p10

)

(

If Pr( X ≤ 8) = 0.9 thus solve 0.9 = 1 − 10(1 − p) p9 + p10

(

9

10

0.9 = 1 − 10(1 − p) p + p 10

9

p + 10(1 − p) p − 0.1 = 0 p = 0.6632

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

)

)

TOPIC 12 Continuous probability distributions • EXERCISE 12.2   |  219

Topic 12 — Continuous probability distributions Exercise 12.2 — Continuous random variables and probability functions

2 a

f (x)

loge 3

∫

A=

0

(– –π2, 0) 2

π 2

loge 3 1 1 =  × e 2 x  4 2 0 loge 3 1 =  e 2 x  8 0 1 2 loge 3 1 0 = e − e 8 8 1 loge 9  1  = e −  × 1 8  8 1 loge 9 = e −1 8 1 = (9 − 1) 8 =1

∫π

−

2 π 2

2 π 2

0

π – 2

π – 4

4

3π – 2

x

π 2 1 1 cos( x ) dx =  sin( x )  2 2 − π

2

1

1

1

1

π

1

 π

1

∫π 2 cos( x ) dx = 2 + 2

−

2 π 2

1

∫π 2 cos( x ) dx = 1

−

f (x) = –14 e2x loge 3, –94

(

2

This is a probability density function. b f(x)

)

1 f(x) = – 2x (0.5, 0.71)

(0, ) 1 – 4

0.71 (loge 3, 0) x

(0, 0)

This is a probability function as the area is 1 f (x)

4



f (x) = 0.25 (–2, 0.25)

x

4

0.5 x −0.5 dx =  x 0.5 

∫

0.5 x −0.5 dx = 4 − 0.5

∫

0.5 x −0.5 dx = 2 − 0.7071

∫

0.5 x −0.5 dx = 1.2929

0.25 4

(2, 0) x

(4, 0)

∫

0.25 4

(2, 0.25)

0

(4, 0.25)

0 (0.5, 0)

units2.

0.25 4

(–2, 0)

– –π

−

f (x)

b

2

(–π2 , 0)

∫π 2 cos( x ) dx = 2 sin  2  − 2 sin  − 2 

)

9 – 4

– –π

– – 3π

1 2x e dx 4

(

f (x) = 0.5 cos(x)

(0, 0.5)

 1 2x  e , 0 ≤ x ≤ loge 3 1 a f ( x ) =  4 elsewhere 0,

0.5

0.25

This is not a probability density function. 2

∫ 0.25 dx = [0.25x ]−2 2

−2 2

∫ 0.25 dx = 0.25(2) − 0.25(−2)

−2 2

∫ 0.25 dx = 0.5 + 0.5 = 1

−2

This is a probability density function.

3

3

∫ n( x

3

− 1) dx = 1

1

3 1 n  x 4 − x  = 1 4 1  1 4   1 4  n   (3) − 3 −  (1) − 1  = 1  4   4 81 1   n  − 3 − + 1 = 1  4 4  18n = 1 1 n= 18

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

220 | TOPIC 12 Continuous probability distributions • EXERCISE 12.2 0

3

−2

0

∫ ( − ax ) dx + ∫ ( 2ax ) dx = 1

4

0  − 1 ax 2  +  ax 2 3 = 1  0  2  −2   1 2 2  0 −  − a(−2)   + a(3) − 0 = 1 2 2a + 9a = 1 11a = 1 1 a= 11 10 + 26 36 9 5 a i Pr( X ≤ 2) = = = 100 100 25 16 4 ii Pr( X > 4) = = 100 25 26 + 28 + 20 74 37 b i Pr(1 < X ≤ 4) = = = 100 100 50 P ( X > 1 ∩ X ≤ 4) P (1 < X ≤ 4) = ii Pr( X > 1 | X ≤ 4) = P ( X ≤ 4) P ( X ≤ 4) 37 84 37 100 37 = ÷ = × = 50 100 50 84 42

(

)

6 a Number of batteries is 100. 29 b Pr( X > 45) = 100 82 41 = c Pr(15 < X ≤ 60) = 100 50 3 d Pr( X > 60) = 100 7 a 200 shot-put throws were measured. 200 − 75 125 5 = = 100 200 8 62 31 = ii Pr(1 < X ≤ 2) = 200 100 Pr(0.5 < X < 1) 63 138 63 200 63 21 = ÷ = × = = c Pr( X < 0.5 | X < 1) = Pr( X < 1) 200 200 200 138 138 46 b i Pr( X > 0.5) =

8 a

f (x) (–1, 1)

f (x) = – –1x

1

(–e, –1e ) (–1, 0) 0

(–e, 0)

−1

x

1

∫ − x dx

−e

=  − loge ( x )−1  −1

−1 −e

  1  =  loge     x  −e  = 1 units 2 This is a probability density function.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 12 Continuous probability distributions • EXERCISE 12.2   |  221 b f(x)

1.65

(

𝜋 – 4,

)

1.71

[cx ]1.65 0.25 = 1

(

3𝜋 , –– 4

0.29

1.65c − 0.25c = 1 1.4c = 1 1 c= 1.4 5 c= 7

)

( , 0)

( 0) 𝜋 – 4,

3𝜋 –– 4

𝜋 x

5

3π 4

10

3π 4

= sin( x ) + x  π

3

  3π  3π    π  π  =  sin   + − sin   +   4  4    4  4  3π π = 0.7071 + − 0.7071 − 4 4 π = units 2 2 This is not a probability density function. c f(x)

f (x) =

1 – 2

∫ f (z ) dz = 1

−1

∫ ( cos( x ) + 1) dx π 4

c dx = 1

0.25

f (x) = cos (x) + 1

0

∫

9

Atriangle = 1 1 ×6× z =1 2 3z = 1 1 z= 3 2

0

2

m ∫ (6 − 2 x ) dx = 1 0

2

m 6 x − x 2  = 1 0

m ( 6(2) − (2)2 − 6(0) + 0 2 ) = 1 8m = 1 1 m= 8

sin(x)

( ) 𝜋, – 1 – 2 2

1 – 2

∫ m(6 − 2 x ) dx = 1

11 a

∞

(0, 0)

(𝜋, 0)

𝜋 – 2

x

b

∫ me 0

−2 x

=1

∞

m ∫ e −2 x = 1

π

1 ∫ 2 sin( x ) dx 0

0

1 ∞ m  − 2 x  = 1  2e  0 1  m 0 +  = 1  2 1 m =1 2 m=2

π 1 =  − cos( x )   2 0  1   1  =  − cos(π ) −  − cos(0)  2   2  1 1 = + 2 2 = 1 units 2

loge (3)

This is a probability density function. d f(x)

∫

c

me 2 x dx = 1

0 loge (3)

1 f (x) = ––––– 2 x–1

m

∫

e 2 x dx = 1

0

log (3)

(2, 0.5) 0.5 0 2

∫2 1

(1, 0)

(2, 0)

x

1 dx = 1 units 2 x −1

This is a probability density function.

e 1 m  e 2 x  =1 2 0 1 1 m  e 2 loge (3) − e 0  = 1 2 2  1 loge (9) 1   m e −  =1 2 2 9 1 m  −  = 1  2 2 4m = 1 1 m= 4

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

222 | TOPIC 12 Continuous probability distributions • EXERCISE 12.2 3

∫ (x

12

2

)

+ 2 kx + 1 dx = 1

0

3

 1 x 3 + kx 2 + x  = 1  3  0 1 3  2  (3) + k (3) + 3 − 0 = 1 3 9 + 9k + 3 = 1 9 k = −11 11 k=− 9 a

∫ f ( x ) dx = 1

13

2

a

1  x ∫ 2 loge  2  dx = 1 2 a

 1 ( x log ( x ) − 1.6934 x )  = 1 e  2  2 1 1 ( a loge (a) − 1.6934 a ) − ( 2 loge (2) − 1.6934(2) ) = 1 2 2 a loge (a) − 1.6934 a − 2 loge (2) + 1.6934(2) = 2 a loge (a) − 1.6934 a − 1.3863 + 3.3868 − 2 = 0 a loge (a) − 1.6934 a + 0.0005 = 0 a = 4.2521, 5.4374 Only a = 5.4374 = 2e gives the answer of 1. 14 a

f(x) 1

(–1, 1)

(a, a)

f(x) = –x

f (x) = x

(0, 0) (a, 0)

(–1, 0)

x

a

∫ f ( x ) dx

−1

=

0

a

−1

0

∫ − x dx + ∫ x dx

0 a 1 1 =  − x 2  +  x 2   2  −1  2  0 1 1 1 1 = − (0)2 + (−1)2 + a 2 − (0) 2 2 2 2 2 a2 + 1 = 2

b f(y) (1, 1)

1

f(y) = –1y

(e, ) 1 – e

1 – e

0

(1, 0)

(e, 0) 𝜋 y

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 12 Continuous probability distributions • EXERCISE 12.3   |  223 e

2

∫ f ( y) dy

c Pr( Z > 0.5) =

1

e

1 = ∫ dy y 1

Pr( Z > 0.5) =

= [ loge ( y) ]1 = loge (e) − loge (1) =1 e

a

−1

e

1

a2 + 1 =1 2 a2 + 1 = 2 a2 = 1 a = ±1 a = 1 since a > 0

0.5

1

a

0

a

 x 4  = 1 0 a4 − 0 = 1

∫ n sin(3x ) cos(3x ) dx = 1 0

2

∫ (1 − z ) dz + ∫ (z − 1) dz

2 a ∫ 4 x 3 dx = 1

π 12

15

f ( z ) dz

0.5 1

1 1 1 Pr( Z > 0.5) =  z − z 2  + 2  0.5 2  1 2  1 1  1 2 1  Pr( Z > 0.5) =  1 − (1)  −  −    +    2 2  2  2 2 1  1 1 1 Pr( Z > 0.5) = −  −  + 2  2 8 2 3 Pr( Z > 0.5) = 1 − 8 5 Pr( Z > 0.5) = 8

∫ f ( x ) dx = ∫ f ( y) dy

c

∫

π 12

a4 = 1 a = ±1 a = 1 since a > 0

n ∫ sin(3 x ) cos(3 x ) dx = 1 0

0.083 n = 1 n = 12

b f (x)

(1, 4)

4

a

∫ loge ( x ) dx = 1

16 a

f (x) = 4x3

1

[ x loge ( x ) − x ]1 = 1 a

( a loge (a) − a ) − ( loge (1) − 1) = 1

a loge (a) − a + 1 = 1 a loge (a) − a = 0 a loge (a) = a loge (a) = 1

1

e1 = a a=e

e

c Pr(0.5 ≤ X ≤ 1) =

3

dx

1

Pr(0.5 ≤ X ≤ 1) =  x 4 

1

0.5

14 Pr(0.5 ≤ X ≤ 1) = 1 − 2 1 Pr(0.5 ≤ X ≤ 1) = 1 − 16 15 Pr(0.5 ≤ X ≤ 1) = 16 4

Exercise 12.3 — The continuous probability density function 1 a

∫ 4x

0.5

b As f ( x ) ≥ 0 and ∫ f ( x ) dx = 1, this is a probability density function.

(1, 0) x

(0, 0)

f (z)

y

3 a (0, 1)

(2, 1)

f (z) = z – 1 (0, 0)

b Pr( Z < 0.75) =

(1, 0) 0.75

∫0

(2, 0)

( − z + 1) dx

0.75 1 Pr( Z < 0.75) =  − z 2 + z   2 0  1  3 2 3 Pr( Z < 0.75) =  −   +  − 0  2  4 4 15 Pr( Z < 0.75) = 32

1 sin(x) y =– 2

1 – 2

f (z) = –z + 1 z

(0, 0)

π – 2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

(π, 0)

x

224 | TOPIC 12 Continuous probability distributions • EXERCISE 12.3

3π  1 π b Pr  < X < = 4 4  2 =

3π 4

∫

0

sin( x ) dx

π 4

−1 0

3π

Pr(−1 ≤ X Pr(−1 ≤ X Pr(−1 ≤ X

2 = 2

Pr(−1 ≤ X

π 3π  Pr  < X <  4 4  3π  Pr  X <   4 

Pr(−1 ≤ X

Pr(−1 ≤ X ≤ 1) Pr( X ≤ 1) 0

sin( x ) dx

0

3π

1 2  + 1 2  2 

2 1 + 4 2 π 3π  Pr  < X <  4 3π  π 4   Pr  X > | X < =   3π  4 4  Pr  X <   4  2 2 = 2+2 4 2 2 = 2+2 =2 2−2 f (x)

∫ k dx = 1

18

30 =1 [ kx ]18

( 30 k ) − (18 k ) = 1

12 k = 1 1 k= 12

1 bh 2 1 1 = ×4×2× k 2 1 = 4k 1 k= 4

b A =

1

∫ 12 dx

20

f (x) = k(2 – x)

0

0

30

a

25

(–2, 0)

1

5 Let X be the amount of petrol sold in thousands of litres.

b Pr(20 < X < 25) =

(0, 2k)

1

0 1 1 3 Pr( X ≤ 1) =  x + x 2  + 8  −2 8 2 1 1  3 Pr( X ≤ 1) = 0 −  (−2) + (−2)2  + 2  8 8 1 3 Pr( X ≤ 1) = 1 − + 2 8 1 3 Pr( X ≤ 1) = + 2 8 7 Pr( X ≤ 1) = 8 3 7 3 8 6 Pr( X ≥ −1X ≤ 1) = ÷ = × = 4 8 4 7 7

=

f (x) = k(2 + x)

1

∫ 4 (2 + x ) dx + ∫ 4 (2 − x )dx

−2

1 [− cos( x )]04 2 1   3π  =  − cos   + cos(0)  4   2

4 a

0

0 1 1 1 1 1 ≤ 1) =  x + x 2  +  x − x 2  8  −1  2 8 0 2 1 1 1   1  ≤ 1) = 0 −  (−1) + (−1)2  +  (1) − (1)2  − 0 2    2 8 8 1 1 1 1     ≤ 1) =  −  +  −   2 8  2 8 3 3 ≤ 1) = + 8 8 3 ≤ 1) = 4

Pr( X ≤ 1) =

=

=

1

d Pr( X ≥ −1X ≤ 1) =

3π 4

∫

0

−1

1 π   3π  =  − cos   + cos     4   4  2 1 2 2 =  + 2 2 2 

3π  1  Pr  X < =  4  2

1

1 1  1 1  Pr(−1 ≤ X ≤ 1) = ∫  + x  dx + ∫  − x  dx 2 4  2 4 

1 [− cos( x )]π4 2 4

3π  π  c Pr  X > | X < =  4 4 

1

1

∫ 4 (2 + x ) dx + ∫ 4 (2 − x ) dx

c Pr(−1 ≤ X ≤ 1) =

(2, 0)

x

1 25 Pr(20 < X < 25) =  x   12  20  1   1  Pr(20 < X < 25) =  ( 25) −  (20)  12   12  5 Pr(20 < X < 25) = 12

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 12 Continuous probability distributions • EXERCISE 12.3   |  225 Pr( X ≥ 26 ∩ X ≥ 22) P ( X ≥ 22) Pr( X ≥ 26) Pr( X ≥ 26X ≥ 22) = Pr( X ≥ 22)

e2

c Pr( X ≥ 26X ≥ 22) =

30

Pr( X ≥ 22) =

b

1

e2

=

1

e2

1 1 = ∫ dz 21z 1 e2 = [ loge ( z )]1 2 1 = loge e 2 − loge 12 2 1 = × 2 loge (e) 2 =1

1 30 Pr( X ≥ 22) =  x   12  22 1 1 Pr( X ≥ 22) = (30) − (22) 12 12 8 Pr( X ≥ 22) = 12 2 Pr( X ≥ 22) = 3 30 1 Pr( X ≥ 26) = ∫ dx 12 26

( ( )

c

26

Pr( X ≥ 26) = Pr( X Pr( X Pr( X Pr( X Pr( X Pr( X

≥ ≥ ≥ ≥ ≥ ≥

26) = 22) 26) = 22) 26) = 22)

( ))

e2

As f ( z ) ≥ 0 and function.

1 30 Pr( X ≥ 26) =  x   12 

Pr( X ≥ 26) =

1

∫ 2z dz 1

∫ 12 dx

22

Pr( X ≥ 26) =

∫ f (z ) dz

∫ f (z ) dz = 1, this is a probability density 1

f (u)

1 1 (30) − (26) 12 12 4 12 1 3 1 2 ÷ 3 3 1 3 × 3 2 1 2

f (u) = e4u

(a, e4a)

(0, 1)

(0, 0)

(a, 0)

6 a f (x)

( , –) 1

1

1 f (x) = – 5

5

( , –) 6

1

a

5

d

∫ f (u) du 0

a

= ∫ e 4 u du 0

0

(1, 0)

(6, 0) 5

b Pr(2 ≤ X ≤ 5) =

a 1 =  e 4 u  4 0 1 4a 1 0 = e − e 4 4 1 4a 1 = e − 4 4

x

1

∫ 5 dx 2

1 Pr(2 ≤ X ≤ 5) =  x  5 

5

e2

2

1

7 a f (z)

0.5

0

(1, 0.5)

(1, 0)

a

∫ f (z ) dz = ∫ f (u) du

1 1 Pr(2 ≤ X ≤ 5) = (5) − (2) 5 5 3 Pr(2 ≤ X ≤ 5) = 5

1 f (z) = – 2z

(

1 e2, – 2e2 (e2, 0)

0

1 1 1 = e4a − 4 4 5 1 4a = e 4 4 5 = e4a loge (5) = 4 a

1 loge (5) = a 4

) z

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

u

226 | TOPIC 12 Continuous probability distributions • EXERCISE 12.3 8 a

f (z)

(0, –12 )

0

(– –π2 , 0) π 2

f (z) = –12 cos(z)

(–π2 , 0)

z

1

∫π 2 cos( z) dz

−

2

π

2 1 =  sin( z )  2 − π

2

π π 1 1 = sin   − sin  −  2  2 2  2 1 1 = + 2 2 =1 This is a probability density function. π

1 π π b Pr  − ≤ Z ≤  = ∫ 4π cos( z ) dz   6 4 2 −6 π

=

1 [sin( z )] 4π − 2 6

π  1 π sin   − sin  −    6 2   4  1  2 1 =  + 2  2 2  =

=

2 +1 4

a

∫ f (u) du = 1

9 a

0

a

(

)

1  2  ∫  1 − 4 2u − 3u  du = 1 0 a



1

3

∫  1 − 2 u + 4 u 0

2

 du = 1

a u − 1 u 2 + 1 u3  = 1 4 4  0 

1 2 1 3   a − a + a  − 0 = 1 4 4 1 3 1 2 a − a + a −1= 0 4 4 1 2 a ( a − 1) + ( a − 1) = 0 4 1 ( a − 1)  a 2 + 1 = 0 4  a =1

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 12 Continuous probability distributions • EXERCISE 12.3   |  227 0.75

b Pr(U < 0.75) =

)

1  2   1 − 2u − 3u  du 4

∫

1 3 2   1 − u − u  du 2 4

0 0.75

Pr(U < 0.75) =

(

∫ 0

0.75 1 1 Pr(U < 0.75) = u − u 2 + u 3  4 4 0  1 1   Pr(U < 0.75) =  0.75 − (0.75)2 + (0.75)3  − 0   4 4 183 Pr(U < 0.75) = 256 0.5

c Pr(0.1 < U < 0.5) =

 2  ∫  1 − 4 ( 2u − 3u ) du

0.1 0.5

Pr(0.1 < U < 0.5) =

1



1

3

∫  1 − 2 u − 4 u

0.1

2

 du

0.5 1 1 Pr(0.1 < U < 0.5) = u − u 2 + u3  4 4  0.1  1 1 1 1     Pr(0.1 < U < 0.5) =  0.5 − (0.5)2 + (0.5)3  −  0.1 − (0.1)2 + (0.1)3      4 4 4 4 Pr(0.1 < U < 0.5) = 0.371

d Pr(U = 0.8) = 0 2

10 a Pr( X > 1.2) =

3

∫  8 x

2

 dx

1.2

3 1 3 2 x 8  3  1.2 3 1 1  6 3 =  (2)3 −    83 3  5  3  8 72  =  −  8  3 125  98 = 125 Pr( X > 1) b Pr( X > 1 X > 0.5) = Pr( X > 0.5) =

2

=

∫  8 x

3

2

3

2

1 2

∫  8 x

0.5

 dx  dx

7 8 = 63 64 8 = 9 n

3 c Pr( X ≤ n) = ∫  x 2  dx 8  0 3 3 1 3 n = x 4 8  3  0 3 3  1 3 =  n  4 83  3 1 3 = n 4 8 3 n =6 1

n = 63

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

228 | TOPIC 12 Continuous probability distributions • EXERCISE 12.3

∫ f (z ) dz = 1 0 a

∫e

−

z 3

Pr( Z ≤ α ) = 0.54

d

a

11 a

α

∫e

dz = 1 a

−3e

+ 3e 0 = 1

−3e

a 3

−3e

a 3

a − e 3

−

α

0α 3

− 3e

+3=1 −

dz = 0.54

z  −   −3e 3  = 0.54   0

z  −   −3e 3  = 1   0

−

z 3

0

0

a − −3e 3

−

+ 3e 0 = 0.54 −

α 3

+ 3 = 0.54

− 3e

= −2

−

α 3

−

α 3

= −2.46

= 0.82 α loge (0.82) = − 3 − 3loge (0.82) = α α = 0.60 e

2 = 3

a  2 loge   = −  3 3  2 −3loge   = a  3

12 a f (x) (0, 3)

−1  3 − loge   = a  2

 3 a = 3loge    2 0.7

b Pr(0 < Z < 0.7) =

∫

e

−

z 3

f (x) = 3e–3x

dz

0

Pr(0 < Z < 0.7) =  

Pr(0 < Z < 0.7) = −3e

  0

− −

0.7 3

+ 3e

Pr(0 ≤ X ≤ 1) = ∫ 3e −3 x dx 0

1

Pr(0 ≤ X ≤ 1) =  − e −3 x 

Pr(0.2 < Z < 0.7) Pr( Z > 0.2)

∫e

−

z 3

∞ 2

0.7

Pr( X > 2) = 0.0025

z  −  Pr(0.2 < Z < 0.7) =  −3e 3    0.2 0.7

a

0.2

∫e

−

z 3

2 ∫ loge ( x ) dx = 1

13 a −

0.2

Pr(0.2 < Z < 0.7) = −3e 3 + 3e 3 Pr(0.2 < Z < 0.7) = −2.3757 + 2.8065 Pr(0.2 < Z < 0.7) = 0.4308 Pr( Z > 2) = 1 − Pr( Z ≤ 2) Pr(0 ≤ Z ≤ 0.2) =

dz

0

1

( ) 2 (a loge (a ) − 2a) − (loge (12 ) − 2(1)) = 1 a loge ( a 2 ) − 2a + 2 − 1 = 0 a

 x loge x 2 − 2 x  = 1  1

2a loge ( a ) − 2a + 1 = 0 a = 2.1555

0.2

z  −  Pr(0 ≤ Z ≤ 0.2) =  −3e 3    0 −

0

Pr(0 ≤ X ≤ 1) = − e3 + e 0 Pr(0 ≤ X ≤ 1) = 0.9502 c Pr( X > 2) = ∫ 3e −3 x dx

dz

0.2

−

∫ f ( x ) dx 0 1

0

0.7

0.7

Pr(0.2 < Z < 0.7) =

y=0

1

b Pr(0 ≤ X ≤ 1) =

Pr(0 < Z < 0.7) = −3e 3 + 3 Pr(0 < Z < 0.7) = 0.6243 c Pr( Z < 0.7Z > 0.2) =

x

(0, 0)

z 0.7  −  3 −3e

0.2 3

+ 3e 0 Pr(0 ≤ Z ≤ 0.2) = −3e Pr(0 ≤ Z ≤ 0.2) = 0.1935 Pr(0.2 < Z < 0.7) 0.43085 = = 0.5342 Pr( Z > 2) 1 − 0.1935

2

b Pr(1.25 ≤ X ≤ 2) =

∫

loge ( x 2 ) dx

1.25

Pr(1.25 ≤ X ≤ 2) = 0.7147 0.25

14 a Pr(−0.25 < Z < 0.25) =

1

dz ∫ 2 −0.25 π ( z + 1)

Pr(−0.25 < Z < 0.25) = 0.1560

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 12 Continuous probability distributions • EXERCISE 12.4   |  229 a

∫

b

0

a

1 dx = 0.5 x2 + 1

0

a

 tan −1 ( x )  = 0.5 0 tan −1 (a) − tan −1 (0) = 0.5

a

2 3   y2  = 1  3  0

tan −1 (a) = 0.5

3

Exercise 12.4 — Measures of centre and spread

1

a

∫  z 1

1.3104

−

1.3104

1.3104

2 5  E(Y ) =  y 2   5  0

a

5

m

∫

c

y dy = 0.5

0

m

9 4

2 3   y 2  = 0.5  3  0

1 9 4

2 2 2 2 m − 0 = 0.5 3 3 m = 0.8255

b i E( Z ) = ∫ z f ( z ) dz

3

E( Z ) = ∫ z dz 1

3

2 E( Z ) =  z  3 1

2  9 2 3 1   − 3 4 3 3 2

2   3 2 −   3   2   3 2 27 2 E( Z ) = × − 3 8 3 9 2 E( Z ) = − 4 3 27 8 E( Z ) = − 12 12 19 E( Z ) = 12 m 1 ii ∫ z dz = 0.5 1 1 2

E( Z ) = 0.7305 Var( Z ) = E( Z 2 ) − [ E( Z ) ]

2

e 2

E( Z 2 ) = ∫ 2 z 2 loge (2 z ) dz 1

E( Z 2 ) = 0.8760 Var( Z ) = 0.876 − (0.7305)2 Var( Z ) = 0.3424 SD( Z ) = 0.3424 SD( Z ) = 0.5851 Median: m

0.5 = ∫ 2loge (2 z ) dz 1

0.5 = [2 z loge ( z ) − 0.6137 z ]1

m

0.5 = 2m loge (m) − 0.6137m − (2(1)loge (1) − 0.6137(1)) m = 1.3010, 0.0120 ∴ m = 1.3010 (m >1)

dz = 0.5

1

m

 2 z  = 0.5 1 2 m − 2 1 = 0.5

∞

4 E( X ) = ∫ 3 xe −3 xdx

2 m = 2.5

0

m = 1.25 m = 1.5625 or

E( Z ) = ∫ 2 z loge (2 z ) dz 1

3

2

3

e 2

8 3 4

E( Z ) =

5

2 2 (1.3104) 2 − (0) 2 5 5 E(Y ) = 0.7863 E(Y ) =

2 a=3 3 a= 2 9 = a 4

−

3

y 2 dy

0

 1 2z 2  = 1   1 2 a−2 1=1

∫z

∫

E(Y ) =

 dz = 1 

m

y y dy

0

1  2

E( Z ) =

∫

b E(Y ) =

1 dz = 1 z

∫

1 a

3

2 2 2 2 a − 0 =1 3 3 3 3 a2 = 2 a = 1.3104

 1 a = tan    2 a = 0.5463

a

1

∫ y 2 dy = 1

2 a

25 16

1 E( X ) = 3

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

230 | TOPIC 12 Continuous probability distributions • EXERCISE 12.4 Var( X ) = E( X 2 ) − [ E( X ) ]

2

∞

E( X ) = ∫ 3 x e 2

2 −3 x

m

c

dx

2 9 2  1 2 Var( X ) = −   9  3 2 1 1 Var( X ) = − = 9 9 9 1 SD( X ) = 9 1 SD( X ) = 3 Median:

1

m

1 1 2  2 x  = 0.5 2   0

2 m −2 0 =1 m = 0.5 m = 0.25 ∞

6 a E(T ) = ∫ tf (t ) dt 0 ∞

m

E(T ) = ∫ 2te −2t dt

0

E(T ) = 0.5 min

−3 x ∫ 3e dx = 0.5

0

m

 − e −3 x  = 0.5 o − e −3m + e 0 = 0.5

∞

b E(T 2 ) = ∫ t 2 f (t ) dt 0 ∞

1 − e −3m = 0.5

E(T 2 ) = ∫ 2t 2e −2t dt

0.5 = e −3m loge (0.5) = −3m

0

E(T 2 ) = 0.5

1 − loge (0.5) = m 3 m = 0.2310 1

1

1 2 x

dx = ∫ 0

Var(T ) = E(T 2 ) − [ E(T ) ]

2

Var(T ) = 0.5 − 0.52 Var(T ) = 0.5 − 0.25 Var(T ) = 0.25

1

1 −2 x dx 2

1

1

SD(T ) = 0.25 = 0.5 min

1

1

1 − ∫ 2 x dx = 2 ∫ x 2 dx 0 0

1

1

∫2

1

∫2

1

∫2

1

0 1

m 0 m

x x

∫ 2e

 0

dx =

1 2 1−2 0 2

dx =

1 ×2 2

(

)

m

 − e −2t  = 0.5 0 − e −2 m + e 0 = 0.5 0.5 = e −2 m loge (0.5) = −2m 1 − loge (0.5) = m 2 m = 0.35 min 3

1

b E( X ) = ∫ xf ( x ) dx 0 1

x 2 x

7 a E(Y ) =

9

∫ yf ( y) dy 0 9

3

E(Y ) =

dx

∫ 0

1

y3 dy 3 3

1 E( X ) = ∫ x dx 20 3 12  E( X ) =  x 2  2  3 

dt = 0.5

1 − 0.5 = e −2 m

dx = 1 x As f ( x ) ≥ 0 for all x-values, and the area under the curve = 1, f ( x ) is a probability density function.

0

−2 t

0

0

E( X ) = ∫

∫ f (t ) dt = 0.5

c

1  1 2 x 2 

1 ∫ 2 x dx = 2   0

0 1

dx = 0.5

x

1 −2 x dx = 0.5 2 ∫0

E( X 2 ) =

0 1

1

0 m

0

5 a ∫

∫2

 y4  E(Y ) =    12  0

1

9

(3 9) − (3 0) 4

E(Y ) = 0

12 3 2 3 E( X ) =  1 − 0   23 3 1 2 E( X ) = × 2 3 1 E( X ) = 3

E(Y ) =

12

4 32 3

4

12

( )

12 E(Y ) = 1.5601

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 12 Continuous probability distributions • EXERCISE 12.4   |  231 m

b

8

0.4809   c E( Z 2 ) = ∫  z 2 ×  dz  z  1

∫ f ( y) dy = 0.5 0

m

∫ 0

y2 dy = 0.5 3

8

E( Z 2 ) = ∫ 0.4809z dz 1

m

 y3    = 0.5  9 0

E( Z ) =  0.2405z 2  2

E( Z 2 = 0.2405(8)2 − 0.2405(1)2

m3 03 − = 0.5 9 9 m 3 = 4.5

E( Z 2 ) = 15.1515 Var( Z ) = E( Z 2 ) − [ E( Z ) ]

2

Var( Z ) = 15.1515 − 3.36632 Var( Z ) = 3.8195

m = 3 4.5 m = 1.6510

SD( Z ) = 3.8195 = 1.9571

Q1

c

∫

f ( y) dy = 0.25

Q1

0

Q1

∫ 0

2

y dy = 0.25 3

1

0.4809 [ loge ( z ) ]1 1 = 0.25 loge (Q1 ) − loge (1) = 0.5199 loge (Q1 ) = 0.5199

0.4809 [ loge ( z ) ]1 1 = 0.75 loge (Q3 ) − loge (1) = 1.5596 loge (Q3 ) = 1.5596

Q1 = e 0.5199 Q1 = 1.6817

Q3 = e1.5596 Q3 = 4.7568

Q

π

1 (sin(2 x ) + 1) dx π 0

9 a ∫

f ( y) dy = 0.75

0

=

y2 dy = 0.75 3

1 π

π

∫ (sin(2 x ) + 1) dx 0

π 1 1 =  − cos(2 x ) + x  π 2 0 1  1   1  =   − cos(2π ) + π  −  − cos(0) + 0     2  π  2 1 1 1 = − +π +  π 2 2 =1

Q

 y3  3   = 0.75  9 0 Q33 03 − = 0.75 9 9 Q33 = 6.75 Q3 = 3 6.75 Q3 = 1.8899 d Inter-quartile range is Q3 − Q1 = 1.8899 − 1.3104 = 0.5795 8

a ∫ z dz = 1 1

8 a

1

0.4809 dz = 0.75 z

e Range = 8 – 1 = 7

Q3

0

∫

Inter-quartile range is Q3 − Q1 = 4.7568 − 1.6817 = 3.0751

Q1 = 3 2.25 Q1 = 1.3104

∫

Q3

0.4809 dz = 0.25    z Q

Q

Q3

∫

d

 y3  1   = 0.25  9 0 Q13 0 3 − = 0.25 9 9 Q13 = 2.25

∫

8 1

8

1 a ∫ dz = 1 z 1

a [ loge ( z ) ]1 = 1 8

a ( loge (8) − loge (1) ) = 1 a loge (8) = 1

1 loge (8) a = 0.4809

a= 8

0.4809   b E( Z ) = ∫  z ×  dz  z 

As f ( x ) ≥ 0 for all values of x and the area under the curve is 1, f ( x ) is a probability density function. π

x (sin(2 x ) + 1) dx π 0 E( X ) = 1.0708

b E( X ) = ∫

π

c   i E( X 2 ) = ∫ 0

x2 (sin(2 x ) + 1) dx π

E( X 2 ) = 1.7191 Var( X ) = E( X 2 ) − [ E( X ) ]

2

Var( X ) = 1.7191 − 1.07082 Var( X ) = 0.5725 ii SD( X ) = 0.5725 = 0.7566

1 8

E( Z ) = ∫ 0.4809 dz 1

E( Z ) = [ 0.4809 z ]18 E( Z ) = 0.4809(8) − 0.4809(1) E( Z ) = 3.3663 Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

232 | TOPIC 12 Continuous probability distributions • EXERCISE 12.4 m

1  ∫  π (sin(2 x ) + 1) dx = 0.5

d

a

0

m 1 1 − cos(2 x ) + x  = 0.5  π 2 0   1   1  − cos(2m) + m  −  − cos(0) + 0  = 1.5708 2 2 1 1 − cos(2m) + m + = 1.5708 2 2 1 m − cos(2m) = 1.0708 2 m = 0.9291 2

10

E( X ) = ∫ xf ( x ) dx = 1 0

2

2 ∫ x ( ax − bx ) dx = 1 0 2

∫ ( ax

2

)

− bx 3 dx = 1

0

2

 a x3 − b x4  = 1  3 4  0  a 3 b 4  (2) − (2)  − 0 = 1 3 4 2  a x3 − b x4  = 1  3 4  0 a b  3 4  (2) − (2)  − 0 = 1 3 4 8a − 4b = 1 3 8a − 12b = 3....................(1)

1

a

∫ 3z

0

0

2

 a x2 − b x3  = 1  2 3  0  a 2 b 3  (2) − (2)  − 0 = 1 2 3 8 2a − b = 1 3 6a − 8b = 3....................(2) 8a − 12b = 3..................(1) 6a − 8b = 3....................(2) (1) × 3 24 a − 36b = 9................(3) (2) × 4 24 a − 32b = 12...............(4) (4) − (3) 4b = 3 3 b= 4 3 Substitute b = into (1) 4  3 8a − 12   = 3  4 8a − 9 = 3 8a = 12 3 a= 2

dz = 1 a

 −3z −1  = 1 1 a  3 −  z  = 1 1 3 3 − + =1 a 1 3a − 3 = a 2a = 3 3 a= 2 3 2

b E( Z ) = ∫ z f ( z ) dz 1 3 2

3  E( Z ) = ∫  z × 2  dz   z 1 E( Z ) =

13 2

3

∫ z dz 1

3

E( Z ) = [ 3 loge ( z ) ]12

 3 E( Z ) = 3 loge   − 3 loge (1)  2 E( Z ) = 1.2164 3 2

E( Z 2 ) = ∫ z 2 f ( z ) dz 1 3 2

2

2 ∫ ( ax − bx ) dx = 1

−2

1

2

∫ f ( x ) dx = 1

3

∫ z 2 dz = 1

11 a

3  E( Z 2 ) = ∫  z 2 × 2  dz   z 1 3 2

E( Z 2 ) = ∫ 3 dz 1

3 2 1

E( Z ) = [3z ]  3 E( Z 2 ) = 3   − 3(1)  2 9 6 2 E( Z ) = − 2 2 3 2 E( Z ) = 2 2 Var( Z ) = E( Z 2 ) − [ E( Z ) ] 3 Var( Z ) = − 1.2164 2 2 Var( Z ) = 0.0204 2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 12 Continuous probability distributions • EXERCISE 12.4   |  233 m

c

∫ f (z ) dz = 0.5 1

m

3

∫ z 2 dz = 0.5 1

m

1  3  − z  = 2 1 3 3 1 − + = m 1 2 − 6 + 6m = m 5m = 6 6 m= 5 Q1

∫

f ( z ) dz = 0.25

1

Q1

3

∫ z 2 dz = 0.25 1

Q

 3 1 1  − z  = 4 1 3 3 1 − + = Q1 1 4 −12 + 12Q1 = Q1 11Q1 = 12 12 Q1 = 11 Q3

∫

f ( z ) dz = 0.75

1

Q3

3

∫ z 2 dz = 0.75 1

Q

 3 3 3  − z  = 4 1 3 3 3 − + = Q3 1 4 −12 + 12Q3 = 3Q3 9Q3 = 12 12 Q3 = 9 4 Q3 = 3 Inter-quartile range is 12 a

4 12 44 36 8 − = − = 3 11 33 33 33

y = 4 − x2

(

1

)2 1 − dy 1 = ( −2 x ) ( 4 − x 2 ) 2 dx 2 y = 4 − x2

dy x =− dx 4 − x2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

234 | TOPIC 12 Continuous probability distributions • EXERCISE 12.4 3

b E( X ) =

∫ xf ( x ) dx 0 3

E( X ) =

3x

∫π

4 − x2

0

E( X ) = −

3 π

3



∫  − 0

dx

  dx 4−x  x

2

3 3 E( X ) = −  4 − x 2    π 0 2 3  E( X ) = −  4 − 3 − 4 − 0 2   π 3 E( X ) = − 1− 4 π 3 E( X ) = − × −1 π 3 E( X ) = π

( )

(

)

13 a f(x)

y = f (x)

(0, 2h)

0

(4, 2h)

4

(2, 0)

x

4

∫ f ( x ) dx = 1 0

1  1   × 2 × 2h  +  × 2 × 2h  = 1 2 2 2h + 2h = 1 4h = 1 1 h= 4 4

b E( X ) = ∫ xf ( x ) dx 0 2

4

0

2

1  1   1 1 E( X ) = ∫  − x 2 + x  dx + ∫  x 2 − x  dx  4 4 2  2  2 4 1 1 1 1 E( X ) =  − x 3 + x 2  +  x 3 − x 2  4  0  12 4 2  12 1 2 1 1   1  1   1 3 E( X ) =  − (2) + (2)  − 0 +  (4)3 − (4)2  −  (2)3 − (2)2     12    4 12 4 12 4 2 16 2 E( X ) = − + 1 + − 4 − + 1 3 3 3 E( X ) = 4 − 4 + 2 E( X ) = 2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 12 Continuous probability distributions • EXERCISE 12.4   |  235 4

c E( X 2 ) = ∫ x 2 f ( x ) dx 0 2

4

0

2

1  1   1 1 E( X 2 ) = ∫  − x 3 + x 2  dx + ∫  x 3 − x 2  dx  4 4 2  2  2 4 1 1 1 1 E( X 2 ) =  − x 4 + x 3  +  x 4 − x 3  6  0  16 6 2  16 1 1 1 1 1       1 E( X 2 ) =  − (2)4 + (2)3  − 0 +  (4)4 − (4)3  −  (2)4 − (2)3     16    6 16 6 16 6 4 32 4 −1+ E( X 2 ) = −1 + + 16 − 3 3 3 2 E( X ) = 14 − 8

E( X 2 ) = 6 Var( X ) = E( X 2 ) − [ E( X ) ]

2

Var( X ) = 6 − 22 Var( X ) = 2 14 a f (x)

f (x) = k (b, k) (a, k)

k

0

(a, 0)

(b, 0)

x

b

b

∫ k dx = 1 a

[ kx ]ba = 1

kb − ka = 1 k (b − a) = 1 k=

1 b−a

b

c E( X ) = ∫ xf ( x ) dx a b

E( X ) = ∫ kx dx a

b

k E( X ) =  x 2   2 a k k E( X ) = b 2 − a 2 2 2 k E( X ) = b 2 − a 2 2 1 (b − a)(b + a) E( X ) = × 1 2(b − a) b+a E( X ) = 2

(

)

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

236 | TOPIC 12 Continuous probability distributions • EXERCISE 12.4 b

d E( X 2 ) = ∫ x 2 f ( x ) dx a b

E( X 2 ) = ∫ kx 2 dx a

b

k E( X 2 ) =  x 3   3 a k k 2 E( X ) = b 3 − a 3 3 3 k E( X 2 ) = b 3 − a 3 3 1 (b − a)(b 2 + ba + a 2 ) 2 E( X ) = × 1 3(b − a) 2 2 b ba a + + E( X 2 ) = 3 2 Var( X ) = E( X 2 ) − [ E( X ) ]

(

)

b 2 + ba + a 2  b + a  2 −  2  3 2 2 b + ba + a b 2 + 2ba + a 2 Var( X ) = − 3 4 4 b 2 + 4 ba + 4 a 2 − 3b 2 − 6ba − 3a 2 Var( X ) = 12 b 2 − 2ba + a 2 Var( X ) = 12 (b − a) 2 (a − b) 2 Var( X ) = = 12 12 Var( X ) =

7.9344

∫

15 a

7.9344

f ( y) dy =

2

∫ 2

7.9344

b E(Y ) =

∫ 2

c E(Y 2 ) =

y 0.2 loge   dy = 1  2

y 0.2 y loge   dy = 5.7278  2

7.9344

∫ 2

y 0.2 y 2 loge   dy = 34.9677  2

Var(Y ) = E(Y 2 ) − [ E(Y ) ]

2

Var(Y ) = 34.9677 − 5.72782 Var(Y ) = 2.1600 SD(Y ) = 2.1600 = 1.4697 m

y d ∫ 0.2 loge   dy = 0.5  2 2

(

) 2 2 0.5 = 0.05 ( 2m loge (m) − 2.3863m ) − 0.05 ( 0.05 ( 2(2)2 loge (2) − 2.3863(2)2 )) m

0.5 =  0.05 2 x 2 loge ( x ) − 2.3863 x 2  2

0.5 = 2m 2 loge (m) − 2.3863m 2 − ( 5.5452 − 9.5472 ) 0.05 10 = 2m 2 loge (m) − 2.3863m 2 + 4.0218 0 = 2m 2 loge (m) − 2.3863m 2 − 5.9782 m = 3.9816 e Range is 7.9344 − 2 = 5.9344 a

∫

16 a

1

a

z − 1dz = 1 1

∫ ( z − 1) 2 dz = 1 1

3 a

2 2  3 ( z − 1)  = 1 1

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 12 Continuous probability distributions • EXERCISE 12.5   |  237 a

2

 2 ( z − 1)3  = 1  3  1 2 2 3 (a − 1) − (1 − 1)3 = 1 3 3 3 (a − 1)3 = 2 9 3 (a − 1) = 4 a = 2.3104

b E( X ) =

−2 0

E( X ) =

ii E( Z 2 ) =

∫

∫

0

2

0

2

 x3   x3  E( X ) =  −  +    12  −2  12  0  0 3 (−2)3   23 0 3  E( X ) =  − + + − 12   12 12   12

z 2 z − 1dz = 3.3085

E( X ) = −

1

iii Var( Z ) = E( Z 2 ) − [ E( Z ) ]

2

E( X ) = 0

Var( Z ) = 3.3085 − 1.78632 Var( Z ) = 0.1176

3 3 + 4 4

Var( X ) = E( X 2 ) − [ E( X ) ]

2

E( X 2 ) =

iv SD( Z ) = 0.1176 = 0.3430

Exercise 12.5 — Linear transformations 1 E(Y ) = 4 and Var(Y ) = 3 a E(2Y − 3) = 2E(Y ) − 3 = 2(4) − 3 = 8 − 3 = 5 b Var(2Y − 3) = 22 Var(Y ) = 4 × 3 = 12

E( X 2 ) =

2

∫x

−2 0

2



f ( x ) dx

∫  x

−2 0

2

2

x x × −  dx + ∫  x 2 ×  dx  4 4 0

2  x3   x3  E( X 2 ) = ∫  −  dx + ∫   dx  4  4 −2

0 4 2

4 0

 x  x  E( X 2 ) =  −  +   16   −2  16  0 4  0 (−2)4   24 0 4  E( X 2 ) =  − + − + 16   16 16   16

c Var(Y ) = E(Y 2 ) − [ E(Y ) ]

2

3 = E(Y 2 ) − (4)2 3 = E(Y 2 ) − 16 19 = E(Y 2 ) d E (Y (Y − 1) ) = E(Y − Y ) = E(Y ) − E(Y ) = 19 − 4 = 15 2 E( X ) = 9 and Var( X ) = 2 2

x



x2 x2 dx E( X ) = ∫ − dx + ∫ 4 4 −2 0

z z − 1dz = 1.7863

1 2.3104

2

x



∫  x × − 4  dx + ∫  x × 4  dx

−2 0

2.3104

b i E( Z ) =

∫ xf ( x ) dx

2

a Y = aX + 5 E(Y ) = E ( aX + 5) E(Y ) = aE( X ) + 5 E(Y ) = 9a + 5 Var(Y ) = Var(aX + 5) = a 2 Var( X ) = 2a 2

E( X 2 ) = 1 + 1 E( X 2 ) = 2 Var( X ) = 2 − 0 2 = 2 c E(5 X + 3) = 5E( X ) + 3 = 5(0) + 3 = 3 Var(5 X + 3) = 52 Var( X ) = 25 × 2 = 50

(

)

(

d E (3 X − 2)2 = E 9 X 2 − 12 X + 4

)

2

= 9E( X ) − 12E( X ) + 4 = 9(2) − 12(0) + 4 = 22

E(Y ) = Var(Y ) 9a + 5 = 2a 2

4 a f (x)

0 = 2a 2 − 9a − 5 0 = (2a + 1)(a − 5) a = 5 since a is a positive integer

(π, 1)

1

b E(Y ) = 9a + 5 = 9(5) + 5 = 50

f (x) = –cos(x)

Var(Y ) = 2a 2 = 2(5)2 = 50 2

∫ f ( x ) dx = 1

3 a

−2 2

0

∫ ( − kx ) dx + ∫ ( kx ) dx = 1

−2

− k x 2   2 

0

0

2

k +  x 2  2  0 −2 k k k k   2 2 2 2  − (0) + (−2)  +  (2) − (0)  2 2 2 2 2k + 2k 4k

=1 =1

=1 =1 1 k= 4

(––π2, 0)

0

(π, 0) x

π

∫ ( − cos( x )) dx

π 2

π

= [ − sin( x ) ]π 2

π = − sin(π ) + sin    2 = 0 +1 =1 This is a probability density function.

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

238 | TOPIC 12 Continuous probability distributions • EXERCISE 12.5 π

2

b E( X ) = ∫ xf ( x ) dx

∫ mx (2 − x ) dx = 1

8 a

π 2 π

0

2

∫ (2mx − mx

E( X ) = ∫ − x cos( x ) dx

2

) dx = 1

0

π 2

2  mx 2 − m x 3  = 1 3  0  m  m(2)2 − (2)3  − 0 = 1   3 8 4m − m = 1 3 12m − 8m =1 3 4m = 3 3 m= 4

E( X ) = 2.5708 2 Var( X ) = E( X 2 ) − [ E( X ) ] π

E( X 2 ) = ∫ x 2 f ( x ) dx π 2 π

E( X 2 ) = ∫ − x 2 cos( x ) dx π 2

E( X 2 ) = 6.7506 2 Var( X ) = E( X 2 ) − [ E( X ) ]

2

b E( X ) = ∫ xf ( x ) dx

Var( X ) = 6.7506 − ( 2.5708 )2 Var( X ) = 0.1416 c E ( 3 X + 1) = 3E( X ) + 1 E ( 3 X + 1) = 3(2.5708) + 1 E ( 3 X + 1) = 8.7124

0 2

E( X ) = ∫ 0 2

3 2 x (2 − x ) dx 4

3  3 E( X ) = ∫  x 2 − x 3  dx 2 4 

2

Var(3 X + 1) = 3 Var( X ) Var(3 X + 1) = 9(0.1416) Var(3 X + 1) = 1.2743

0

(

d E ( (2 X − 1)(3 X − 2) ) = E 6 X 2 − 7 X + 2

2 1 3 E( X ) =  x 3 − x 4  16  0 2 1 3 3  E( X ) =  (2) − (2)4  − 0 2  16 E( X ) = 4 − 3 E( X ) = 1

)

E ( (2 X − 1)(3 X − 2) ) = 6E( X ) − 7E( X ) + 2 E ( (2 X − 1)(3 X − 2) ) = 6(6.7606) − 7(2.5708) + 2 E ( (2 X − 1)(3 X − 2) ) = 24.5079 2

2

E( X 2 ) = ∫ x 2 f ( x ) dx

5 E( Z ) = 5 and Var( Z ) = 2 a E(3 Z − 2) = 3E( Z ) − 2 = 3(5) − 2 = 13 b Var(3 Z − 2) = 32 Var( Z ) = 9(2) = 18

( ) 2 = E ( Z 2 ) − 52 2 = E ( Z 2 ) − 25 27 = E ( Z 2 ) 1 1 27 d E  Z 2 − 1 = E ( Z 2 ) − 1 = −1= 8 3  3 3

c Var( Z ) = E Z 2 − [ E( Z ) ]2

0

3 3 5 2 x E( X 2 ) =  x 4 − 20  0 8 3 3  E( X 2 ) =  (2)4 − (2)5  − 0 8  20 E( X 2 ) = 6 − 4.8 E( X 2 ) = 1.2

a E(2 − Y ) = 2 − E(Y ) = 2 − 3.5 = −1.5

Var( X ) = E( X 2 ) − [ E( X ) ]

2

1 1 Y b E   = E(Y ) = × 3.5 = 1.75  2 2 2 2

2

Var( X ) = 1.2 − 1 Var( X ) = 0.2

c E(5 − 2 X ) = 5 − 2E( X ) = 5 − 2(1) = 3

d Var(2 − Y ) = ( −1) Var(Y ) = 1.44

Var(5 − 2 X ) = ( −2 )2 Var( X ) = 4 × 0.2 = 0.8

2

2 Y 1  1 e Var   =   Var(Y ) = × 1.44 = 0.36  2   2 4 7 Let T be the time for the kettle to boil. E(T ) = 1.5 and SD(T ) = 1.1 so Var(T ) = 1.21

E(5T ) = 5E(T ) = 5 × 1.5 = 7.5 minutes Var(5T ) = 52 Var(T ) = 25 × 1.21 = 30.25 SD(5T ) = 30.25 = 5.5 minutes

0 2

3 3 x (2 − x ) dx 4

3  3 E( X ) = ∫  x 3 − x 4  dx 2 4  2

6 E(Y ) = 3.5 and SD(Y ) = 1.2

c Var(Y ) = [SD(Y ) ] = (1.2)2 = 1.44

0 2

E( X ) = ∫ 2

a

9 a

2

∫ z + 1 dz = 1 0

[ 2 loge (z + 1)]0a = 1

2 loge (a + 1) − 2 loge (0 + 1) = 1 loge (a + 1) = 0.5 e 0.5 = a + 1 e 0.5 − 1 = a 0.6487 = a

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 12 Continuous probability distributions • EXERCISE 12.5   |  239 0.6487

∫

b E( Z ) = E( Z 2 ) =

2z dz = 0.2974 z +1

0 0.6487

∫ 0

2z 2 dz = 0.1234 z +1

Var( Z ) = E( Z 2 ) − [ E( Z ) ]

2

Var( Z ) = 0.1234 − 0.2974 2 Var( Z ) = 0.0349 c i E(3 Z + 1) = 3E( Z ) + 1 = 3(0.2974) + 1 = 1.8922 ii Var(3 Z + 1) = 32 Var( Z ) = 9(0.0349) = 0.3141 iii E( Z 2 + 2) = E( Z 2 ) + 2 = 0.1234 + 2 = 2.1234 10 Y = aX + 3 and E( X ) = 5 as well as Var( X ) = 2 E(Y ) = Var(Y ) E(aX + 3) = Var(aX + 3) aE( X ) + 3 = a 2 Var( X ) 5a + 3 = 2a 2 0 = 2a 2 − 5a − 3 0 = (2a + 1)(a − 3) 1 a=− ,3 2 ∴ a = 3 since a is a positive integer Thus E(Y ) = E(3 X + 3) = 3E( X ) + 3 = 3(5) + 3 = 18 and Var(Y ) = Var(3 X + 3) = 32 Var( X ) = 9(2) = 18 11 Z = aY − 3 and E(Y ) = 4 as well as Var(Y ) = 1 E( Z ) = Var( Z ) E ( aY − 3) = Var(aY − 3) aE(Y ) − 3 = a 2 Var(Y ) 4a − 3 = a2 0 = a2 − 4a + 3 0 = (q − 3)(a − 1) a = 1 or 3 Thus E( Z ) = E(Y − 3) = E(Y ) − 3 = 4 − 3 = 1 and E( Z ) = E(3Y − 3) = 3E(Y ) − 3 = 3(4) − 3 = 9 2 2 Also Var( Z ) = Var(Y − 3) = 1 Var(Y ) = 1 and Var( Z ) = Var(3Y − 3) = 3 Var(Y ) = 9(1) = 9 a

12 a

∫ f (z ) dz = 1

1 a

∫ 3z

−

1 2

dz = 1

1

a

 1 6 z 2  = 1  1 a

6 z  = 1 1 6 a −6 1 =1 6 a −6=1 6 a=7 7 a= 6 49 a= 36

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

240 | TOPIC 12 Continuous probability distributions • EXERCISE 12.5

b E( Z ) =

49 36

∫ zf (z ) dz

1 49 36

E( Z ) =

1

∫ 6z 2 dz 1

49

 3  36 E( Z ) =  2 z 2   1 3

3

 49  2 E( Z ) = 2   − 2(1) 2  36  3

 7 2 2 E( Z ) = 2     − 2  6  3

7 E( Z ) = 2   − 2  6 E( Z ) = 1.1759 E( Z 2 ) =

49 36

∫z

2

f ( z ) dz

1

E( Z 2 ) =

49 36

∫

3

3z 2 dz

1

49

 6 5  36 E( Z 2 ) =  z 2   5 1 5

5

6  49  2 6 E( Z ) =   − (1) 2 5  36  5 2

5

6  7 2 2 6 E( Z ) =     − 5  6  5 2

6 7 5 6 E( Z 2 ) =   − 5  6 5 E( Z 2 ) = 1.3937 Var( Z ) = E( Z 2 ) − [ E( Z ) ]

2

Var( Z ) = 1.3937 − 1.17592 Var( Z ) = 0.0109 c i E(4 − 3 Z ) = 4 − 3E( Z ) = 4 − 3(1.1759) = 0.4722 ii Var(4 − 3 Z ) = (−3)2 Var( Z ) = 9 × 0.011 = 0.0978 3π

13 a

x

 x

∫ kπ sin  3  dx = 1 0

1 kπ

3π

 x

∫ x sin  3  dx = 1 0

3π

x x  1  −3 x cos   + 9sin    = 1  3  0  3 kπ  ( −3(3π ) cos (π ) + 9sin (π )) − ( −3(0) cos ( 0 ) + 9sin ( 0 )) = kπ 9π = kπ k=9

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 12 Continuous probability distributions • EXERCISE 12.5   |  241

b E( X ) = E( X ) =

3π

∫ xf ( x ) dx

0 3π

x2

 x

∫ 9π sin  3  dx 0

E( X ) = 5.61 mm c

W = 2X − 1 E(W ) = E ( 2 X − 1) E(W ) = 2E( X ) − 1 E(W ) = 2(5.6051) − 1 E(W ) = 10.21 mm 1.25

∫

14 a

k (2 y + 1) dy = 1

0.9 1.25

∫

k

(2 y + 1) dy = 1

0.9 1.25

k  y 2 + y 

((

2

) (

0.9

2

))

=1

k 1.25 + 1.25 − 0.9 + 0.9 = 1 k ( 2.8125 − 1.71) = 1 1.1025k = 1 400 k= 441 1.25

∫

b E(Y ) =

yf ( y) dy

0.9 1.25

∫ ( 0.907 y(2 y + 1)) dy

E(Y ) =

0.9 1.25

2 ∫ (1.814 y + 0.907 y ) dy

E(Y ) =

0.9

1.25

E(Y ) =  0.6046 y3 + 0.4535 y 2 

(

0.9

) (

3 3   + 0.4535(1.25)2 − 0.6046(0.9) E(Y ) = 0.6046(1.25) + 0.4535(0.9)2 E(Y ) = 1.8895 − 0.8081 E(Y ) = 1.081 kg

c

)

Z = 0.75Y + 0.45 E( Z ) = E ( 0.75Y + 0.45) E( Z ) = 0.75E(Y ) + 0.45 E( Z ) = 0.75 (1.0814 ) + 0.45 E( Z ) = 1.261 kg a

5loge ( z ) dz = 1 z 1

∫

15 a

a

10 z loge ( z ) − 20 z  = 1 1 10 a loge (a) − 20 a − 10 1 loge (1) − 20 1 = 1

(

) (

)

10 a loge (a) − 20 a + 20 = 1 10 a loge (a) − 20 a + 19 = 0 a = 1.7755

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

242 | TOPIC 12 Continuous probability distributions • EXERCISE 12.5 1.7755

b E( Z ) =

∫

zf ( z ) dz

∫

5 z loge ( z ) dz

1 1.7755

E( Z ) =

1

E( Z ) = 1.4921 E( Z 2 ) = 2

E( Z ) =

1.7755

∫

z 2 f ( z ) dz

∫

5z 2 loge ( z ) dz

1 1.7755

3

1

E( Z 2 ) = 2.2625 2 Var( Z ) = E( Z 2 ) − [ E( Z ) ] Var( Z ) = 2.2625 − 1.49212 Var( Z ) = 0.0361 c E(3 − 2 Z ) = 3 − 2E( Z ) = 3 − 2(1.4921) = 0.0158 Var(3 − 2 Z ) = (−2)2 Var( Z ) = 4 × 0.0361 = 0.1444 16 Y = aX + 1 and E( X ) = 2 as well as Var( X ) = 7 E(Y ) = Var(Y ) E ( aX + 1) = Var ( aX + 1) aE( X ) + 1 = a 2 Var( X ) 2a + 1 = a 2 0 = a 2 − 2a − 1 a = 0.5469 E(Y ) = E(0.5469 X + 1) = 0.5469E( X ) + 1 = 0.5469(2) + 1 = 2.0938 Var(Y ) = Var ( 0.5469 X + 1) = 0.54692 Var( X ) = 0.54692 (7) = 2.0938

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 13 The normal distribution • EXERCISE 13.2   |  243

Topic 13 — The normal distribution (

4 Let X = the length of pregnancy for a human X ~ N 275,14 2

Exercise 13.2 — The normal distribution 1 f ( x ) =

1 e 3 2π

1 x−2 −   2 3 

2

=

1 σ 2π

1 x−µ2 −   e 2 σ 

a µ = 2 and σ = 3 b

f (x)

(

1 3 2π

2, —

1 — 3 2π

(

µ + σ = 275 + 14 = 289 µ − σ = 275 − 14 = 261

Pr(261 < X < 289) = 0.68

µ + 2σ = 275 + 2(14) = 303 µ − 2σ = 275 − 2(14) = 247

Pr(247 < X < 303) = 0.95

µ + 3σ = 275 + 3(14) = 317 µ − 3σ = 275 − 3(14) = 233

Pr(233 < X < 317) = 0.997

Pr( X < 233) ∪ Pr( X > 317) = 0.003 Pr( X < 233) = Pr( X > 317) = 0.003 ÷ 2 = 0.0015 So Pr( X < 233) = 0.0015 1  x+2  4 

∞

–7

–1 0

–4 ∞

∫

2 a Input

−∞

1

2

5

8

11

x

1  x+2  4 

2

1 − 2 ( x − 3) e dx  1 on CAS. 2π

∞

1

1 x−µ   σ 

2

−  1 − 2 ( x − 3) 1 e = e 2 π σ π 2 2 −∞ µ = 3 and σ = 1

b f ( x ) =

∫

c f (x)

(3,

−  1 5 a Input ∫ e 2 4 2π −∞

)

−  1 b f ( x ) = e 2 4 2π µ = −2

1

2

3

4

5

6

x

3 a Let X = the results on the Mathematical Methods test

(

X ~ N 72,82

dx = 0.9999  1 using CAS

2

1 x−µ2  σ 

−  1 = e 2 σ 2π

1  10( x −1) 

2

1 x−µ2

−   10 − 2  3  1 e = e 2 σ  σ 2π 3 2π 3 1 10 = so σ = = 0.3 and µ = 1 σ 3 10 10 b Dilation by a factor of parallel to the y-axis or from the 3 3 parallel to the x-axis or x-axis. Dilation by a factor of 10 from the y-axis and a translation 1 unit in the positive x-direction.

6 a f ( x ) =

c 0

2

f (x) 10 (1,— ) 3 2π

10 — 3 2π

)

µ + σ = 72 + 8 = 80 µ − σ = 72 − 8 = 64 µ + 2σ = 72 + 2(8) = 88 µ − 2σ = 72 − 2(8) = 56 So Pr(56 < X < 88) = 0.95 and Pr( X < 56) ∪ Pr( X > 88) = 0.05 Thus Pr( X < 56) = Pr( X > 88) = 0.05 ÷ 2 = 0.025 So Pr( X > 88) = 0.025 b

µ + 3σ = 88 + 8 = 96 µ − 3σ = 58 − 8 = 50 Pr(48 < X < 96) = 0.997 Pr( X < 48) ∪ Pr( X > 96) = 0.003 Pr( X < 48) = Pr( X > 96) = 0.003 ÷ 2 = 0.0015

c

Pr(64 < X Pr( X < 64) ∪ Pr( X Pr( X < 64) = Pr( X Pr( X < 80) = 1 − Pr( X

< 80) = 0.68 > 80) = 0.32 > 80) = 0.32 ÷ 2 = 0.16 > 80) = 1 − 0.16 = 0.84

0

0.1

0.4

0.7

1  x+4

1

1.3

2

1.6

1.9

1 x−µ2

x

−  −    1 1 7 a f ( x ) = e 2  10  = e 2 σ  σ 2π 10 2π µ = −4, σ = 10 1 b Dilation factor from the x-axis, dilation factor 10 from 10 the y-axis, translation 4 units in the negative x-direction.

c i

σ = SD( X ) = 10 Var( X ) = σ 2 = 10 2 = 100

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

)

244 | TOPIC 13 The normal distribution • EXERCISE 13.2

( ) 100 = E ( X 2 ) − (−4)2 100 = E ( X 2 ) − 16 116 = E ( X 2 )

(

ii Var( X ) = E X 2 − [ E( X ) ]

1  x+4  10 

∞

d

−  1  ∫ 10 2π e 2 −∞

2

13 Let X = the number of pears per tree X ~ N 230, 252 a Pr( X < 280) = 1 − Pr( X > 280) = 1 − 0.025 = 0.975

b Pr(180 < X < 280) = Pr( µ − 2σ < X < µ + 2σ ) = 0.95 c Pr( X > 180  X < 280) =

2

dx = 0.9999  1 f ( x ) ≥ 0 for all

values of x and the area under the curve is 1 so this is a probability density function. 1  5( x − 2) 

2

1 x−µ  σ 

−  −   5 1 e 2 2  = e 2 σ 2π 2 2π 2 1 5 = so σ = and µ = 2 σ 2 5

8 a f ( x ) =

b

2

) ) ) ) ) )

(

)

(

2

( ) 1.04 = E ( X 2 )

)

c i E(2 X + 3) = 2E( X ) + 3 = 2(1) + 3 = 5

(

ii E ( ( X + 1)(2 X − 3) ) = E 2 X 2 − X − 3 E ( ( X + 1)(2 X − 3) ) = 2(1.04) − 1 − 3 E ( ( X + 1)(2 X − 3) ) = −1.92

(

16 X ~ N 72.5,8.4 2 2

)

µ + 3σ = 70 + 3(6) = 88

b Pr( X < 55.7) = 0.025 c

1 − 0.997 = 0.0015 = 0.15% get a mark which is 2 greater than 88.

)

a 68% of values lie between 15 − 5 = 10 and 15 + 5 = 20. b 95% of values lie between 15 − 2(5) = 5 and 15 + 2(5) = 25. c 99.7% of values lie between 15 − 3(5) = 0 and 15 + 3(5) = 30. a Pr( X < 31) = 0.16 + 0.68 = 0.84 b Pr(10 < X < 31) = 1 − (0.025 + 0.16) = 1 − 0.105 = 0.815 c Pr( X > 10  X < 31) =

Pr(10 < X < 31) 0.815 = = 0.9702 Pr( X < 31) 0.84

)

a Pr(64.1 < X < 89.3) = 1 − (0.16 + 0.025) = 1 − 0.185 = 0.815

Pr( X > 88) =

(

( )

)

E ( ( X + 1)(2 X − 3) ) = 2E X 2 − E( X ) − 3

(

12 X ~ N 24, 72

2

0.04 = E X 2 − 12

10 Let X = the results on a biology exam X ~ N 70,6

)

1 x−µ2  σ 

2

b Var( X ) = E X 2 − [ E( X ) ] E( X ) = µ = 1

ii Pr( X > 180) = 0.003 ÷ 2 = 0.0015

(

µ − 3σ = 155 1 − 0.997 Pr( X < 155) = = 0.0015 2 0.0015 signifies 3σ Pr( X < h) = 0.0015 Pr( X < 155) = 0.05 So h = 155

( )

9 a Let X = the scores on an IQ test X ~ N 120, 20

11 X ~ N 15,52

)

b 0.025 signifies 2σ Pr( X < k ) = 0.025 Pr( X < 205) = 0.025 So k = 205

1

104 104 = = 2.8 25 5

i µ − σ = 120 − 20 = 100 µ + σ = 120 + 20 = 140 ii µ − 2σ = 120 − 2(20) = 80 µ + 2σ = 120 + 2(20) = 160 iii µ − 3σ = 120 − 3(20) = 60 µ + 3σ = 120 + 2(20) = 180 b i Pr( X < 80) = 0.5 − 0.475 = 0.025

(

a Pr(205 < X < 355) = 1 − (0.025 + 0.16) = 1 − 0.185 = 0.815

−  5 − 2 (5( x −1)) 1 15 a f (x) = e = e 2 σ 2π 2π 1 SD( X ) = σ = 5 1 2 = 0.04 Var( X ) = σ = 25

c i E ( 5 X ) = 5E( X ) = 5(2) = 10 ii E 5 X 2 = 5E( X 2 ) = 5 ×

Pr(180 < X < 280) 0.95 = = 0.9744 Pr( X < 280) 0.975

14 Let X = the rainfall in millimetres X ~ N 305,50 2

c

2 4  2 Var( X ) = σ 2 =   =  5 25 2 Var( X ) = E X 2 − [ E( X ) ] 4 = E X 2 − 22 25 4 = E X2 − 4 25 4 100 + = E X2 25 25 104 = E X2 25 4.16 = E X 2

( ( ( ( ( (

)

d

1 − 0.95 0.05 = = 0.025 2 2 1 − 0.997 0.003 Pr( X < 47.3) = = = 0.0015 2 2 Pr(47.3 < X < 55.7) = 0.025 − 0.0015 = 0.0235 Pr(47.3 < X < 55.7) Pr( X > 47.3  X < 55.7) = Pr( X < 55.7) 0.025 − 0.0015 Pr( X > 47.3  X < 55.7) = 0.025 Pr( X > 47.3  X < 55.7) = 0.94 Pr( X < 55.7) =

Pr( X > m) = 0.16 1 − 0.68 0.32 Pr( X > µ + σ ) = = = 0.16 2 2 So m = 80.9

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 13 The normal distribution • EXERCISE 13.3   |  245

Exercise 13.3 — Calculating probabilities and the standard normal distribution 1 a i Pr( Z < 1.2) = 0.8849 ii Pr(−2.1 < Z < 0.8) = 0.7703 b X ~ N(45,62 ) i Pr( X > 37) = 0.9088 37 − 45 ii z = 6 4 =− 3 2 X ~ N 50,152 Pr(50 < X < 70) = 0.4088

(

c i Pr( X > 32) = Pr( Z > k ) x − µ 32 − 20 k= = = 2.4 σ 5 ii Pr( X > 32) = Pr( Z > k ) = Pr( Z < − k ) x − µ 12 − 20 −k = = = −1.6 so k = 1.6 σ 5 8 Let X = the score for Jing Jing and Y = the score for Rani

(

)

(

X ~ N 72,92        Y ~ N 15, 4 2 x−µ y−µ z= z= σ σ 85 − 72 18 − 15 z= z= 9          4 3 13 z= z= 4 9 z = 0.75 z = 1.4 Jing Jing did better. 9 Let X = the length of Tasmanian salmon

)

3 a Pr( Z ≤ 2) = 0.9772 b Pr( Z ≤ −2) = 0.0228 c Pr(−2 ≤ Pr ≤ 2) = 0.9545 d Pr( Z < −1.95) ∪ Pr( Z > 1.95) By symmetry

(

X ~ N 38, 2.4 2

)

Pr( X > 39.5) = 0.2659 This is not in the top 15% so is not a gourmet fish.

= 2 Pr( Z < −1.95) = 2 × 0.0256 = 0.0512

(

10 Chemistry X ~ N 68,52

(

(

110 − 98   b Pr( X ≥ 110) = Pr  Z ≥   15  12   Pr( X ≥ 110) = Pr  Z ≥   15  4  Pr( X ≥ 110) = Pr  Z ≥   5 Pr( X ≥ 110) = 0.2119

Physics X ~ N 61,82

(

5 Let X = the speed of cars X ~ N 98,62

)

)

x − µ 68 − 61 7 z= = = = 0.875 σ 8 8 Justine did best in physics compared to her peers.

(

11 Let X = the pulse rate in beats per minute X ~ N 80,52 a Pr( X > 85) = 0.1587 b Pr( X ≤ 75) = 0.1587

−2 − 2 5 − 2 c Pr(−2 < X ≤ 5) = Pr  75) =

03 108 = 0.3695 1 − 0.1587

(

a Pr( X > 1.04) = 0.0668 = 6.68%

)

c Pr(90 < X < 110) = 0.8860

b Pr( X < 0.996) = 0.0019 = 0.19%

(

6 Let X = the score on a physics test X ~ N 72,122

)

12 Let X = the weight of a bag of sugar X ~ N 1.025, 0.012

b Pr( X < 90) = 0.0912

)

a Pr( X > 95) = 0.0276 = 2.76% b Pr( X ≥ 55) = 0.9217 = 92.17% 7 a Pr( X < a) = 0.35 and Pr( X < b) = 0.62 i Pr( X > a) = 1 − 0.35 = 0.65 ii Pr(a < X < b) = Pr( X < b) − Pr( X < a) = 0.62 − 0.35 = 0.27 b Pr( X < c) = 0.27 and Pr( X < d ) = 0.56 i Pr(c < X < d ) = Pr( X < d ) − Pr( X < c) = 0.56 − 0.27 = 0.29 ii Pr( X > c  X < d ) =

)

x − µ 72 − 68 4 z= = = = 0.8 σ 5 5 Mathematical Methods X ~ N 69, 72 x − µ 77 − 69 6 z= = = = 0.86 σ 7 7

61 − 65  4 a Pr( X < 61) = Pr  Z <   3  4  Pr( X < 61) = Pr  Z < −   3 Pr( X < 61) = 0.0912

a Pr( X > 110) = 0.0228

)

13 a Pr( Z ≤ −2.125) ∪ Pr( Z ≥ 2.125) By symmetry = 2 Pr( Z ≤ −2.125) = 2 × 0.0168 = 0.0336 b Pr( X < 252.76) = 0.5684 c Pr(−3.175 ≤ Z ≤ 1.995) = 0.9762 d Pr( X < 5.725) = 0.3081 14 Solve using CAS k = 25.2412

Pr(c < X < d ) 0.29 = = 0.5179 Pr( X < d ) 0.56

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

)

246 | TOPIC 13 The normal distribution • EXERCISE 13.4

Exercise 13.4 — The inverse normal distribution 1 a Pr( X ≤ a) = 0.16, µ = 41 and σ = 6.7 a = 34.34 b Pr( X ≤ a) = 0.21, µ = 12.5 and σ = 7.7 a = 14.68 c Pr(15 − a < X < 15 + a) = 0.32, µ = 15 and σ = 4 By symmetry 0.68 Pr( X < 15 − a) = = 0.34 2 15 − a = 13.35 a = 1.65 2 Pr(m ≤ X ≤ n) = 0.92, µ = 27.3 and σ = 8.2 0.08 Pr( X < m) = 2 Pr( X < m) = 0.04 m = 12.9444 Pr( X < n) = 0.96 n = 41.6556

(

)

2 3 X ~ N 112, σ Pr( X < 108.87) = 0.42 108.87 − 112   Pr  Z <  = 0.42  σ 108.87 − 112 = −0.2019 σ 3.13 = −0.2019σ σ = 15.5

(

)

2 4 X ~ N µ , 4.45 Pr( X < 32.142) = 0.11 32.142 − µ   Pr  Z <  = 0.11  4.45  32.142 − µ = −1.2265 4.45 32.142 − µ = −1.2265 × 4.45 32.142 − µ = −5.4579 32.142 + 5.4579 = µ µ = 37.6

5 a Pr( Z < z ) = 0.39 So z = −0.2793 b Pr( Z ≥ z ) = 0.15 or Pr( Z < z ) = 0.85 So z = 1.0364 c Pr(− z < Z < z ) = 0.28 Pr( Z < − z ) = 0.36 So − z = −0.3585 Therefore z = 0.3585

(

6 X ~ N 37.5,8.622

)

a Pr( X < a) = 0.72 So a = 42.52 b Pr( X < a) = 0.68 So a = 41.53 c Pr(37.5 − a ≤ X < 37.5 + a) = 0.88 0.12 Pr( X < 37.5 − a) = = 0.06 2 37.5 − a = 24.10 a = 13.40

(

7 Z ~ N 0,12

)

a Pr( Z < z ) = 0.57 z = 0.1764 b Pr( Z < z ) = 0.63 z = 0.3319 8 X ~ N 43.5,9.72

(

)

a Pr( X < a) = 0.73 a = 49.4443 b Pr( X < a) = 0.24 a = 36.6489

(

9 X ~ N µ ,5.672

)

Pr( X > 20.952) = 0.09 20.952 − µ   Pr  Z >  = 0.09  5.67  20.952 − µ = 1.3408 5.67 20.952 − µ = 1.3408 × 5.67 20.952 − µ = 7.6023 20.952 − 7.6023 = µ 13.3497 = µ µ = 13.35

(

)

(

)

10 X ~ N µ ,3.52

Pr( X < 23.96) = 0.28 23.96 − µ   Pr  Z <  = 0.28  3.5  23.96 − µ = −0.5828 3.5 23.96 − µ = −0.5828 × 3.5 23.96 − µ = −2.038 23.96 + 2.038 = µ µ = 26 11 X ~ N 115, σ 2

Pr( X < 122.42) = 0.76 122.42 − 115   Pr  Z <  = 0.76  σ 122.42 − 115 = 0.7063 σ 7.42 = 0.7063σ 7.42 =σ 0.7063 σ = 10.5

(

12 X ~ N 41, σ 2

)

Pr( X > 55.9636) = 0.11 55.9636 − 41   Pr  Z > = 0.11   σ 55.9636 − 41 = 1.2265 σ 14.9636 = 1.2265σ 14.9636 =σ 1.2265 σ = 12.2

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 13 The normal distribution • EXERCISE 13.5   |  247

(

13 X ~ N µ , σ 2

)

Pr( X < 33.711) = 0.36 33.711 − µ   Pr  Z <  = 0.36  σ 33.711 − µ = −0.3585 σ 33.711 − µ = −0.3585σ 33.711 = µ − 0.3585σ ...................(1) Pr( X < 34.10) = 0.42 34.10 − µ   Pr  Z <  = 0.42  σ 34.10 − µ = −0.2019 σ 34.10 − µ = −0.2019σ 34.10 = µ − 0.2019σ ..................(2) 33.711 = µ − 0.3585σ ...................(1) (1) − (2) 33.711 − 34.10 = −0.3585σ + 0.2019σ − 0.389 = −0.1566σ −0.389 =σ −0.1566 σ = 2.5 Substitute σ = 2.5 into (1) 33.711 = µ − 0.3585(2.5) 33.711 = µ − 0.8963 µ = 34.6

(

)

14 X ~ N µ , σ 2 Pr( X > 18.35) = 0.31 18.35 − µ   Pr  Z >  = 0.31  σ 18.35 − µ = 0.4959 σ 18.35 − µ = 0.4959σ 18.35 = µ + 0.4959σ ....................(1) Pr( X > 15.09) = 0.45 15.09 − µ   Pr  Z >  = 0.45  σ 15.09 − µ = −0.1257 σ 15.09 − µ = −0.1257σ 15.09 = µ − 0.1257σ ....................(2) 18.35 = µ + 0.4959σ ....................(1) (1) − (2) 18.35 − 15.09 = 0.4959σ + 0.1257σ 3.26 = 0.6216σ 3.26 =σ 0.6216 σ = 5.2 Substitute σ = 5.2 into (1) 18.35 = µ + 0.4959(5.2) 18.35 = µ + 2.5787 18.35 − 2.5787 = µ µ = 15.8

(

15 Pr(a < X < b) = 0.52 and X ~ N 42.5,10.32

)

Pr( X < a) = 0.24 Pr( X > b) = 0.24  and  a = 35.2251 b = 49.7749

So Pr(35.2251 < X < 49.7749) = 0.52 Pr(a < X < b) Pr( X > a  X < b) = Pr( X < b) Pr( X < b) = 0.24 + 0.52 = 0.76 0.52 Pr( X > a  X < b) = 0.76 Pr( X > a  X < b) = 0.6842

(

16 X ~ N µ , σ 2

)

Pr( X < 39.9161) = 0.5789 39.9161 − µ   Pr  Z <  = 0.5789  σ 39.9161 − µ = 0.1991 σ 39.9161 − µ = 0.1991σ 39.9161 = µ + 0.1991σ ....................(1) Pr( X > 38.2491) = 0.4799 38.2491 − µ   Pr  Z >  = 0.4799  σ 38.2491 − µ = 0.0504 σ 38.2491 − µ = 0.0504σ 38.2491 = µ + 0.0504σ ...................(2) 39.9161 = µ + 0.1991σ ....................(1) (1) − (2) 39.9161 − 38.2491 = 0.1991σ − 0.0504σ 1.667 = 0.1487σ 1.667 =σ 0.1487 σ = 11.21 Substitute σ = 11.21 into (1) 39.9161 = µ + 0.1991(11.21) 39.9161 = µ + 2.2319 39.9161 − 2.2319 = µ µ = 37.68

Exercise 13.5 — Mixed probability application problems

(

1 a i W ~ N 508,32

)

Pr(W < 500) = 0.0038 ii Pr(W < w) = 0.01 w = 501.0210 b

Pr(W < 500) ≤ 0.01 500 − 508   Pr  Z <  ≤ 0.01  σ 500 − 508 ≤ −2.3263 σ − 8 ≤ −2.3263σ −8 ≤σ −2.3263 3.4389 ≤ σ Or σ ≥ 3.4389 grams

2 a 2

2

∑ Pr( X = x ) = 1

3k + 2 k + 6 k + 2 k + k 2 + 2 k + 3k = 1

10 k 2 + 9 k − 1 = 0 as required

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

248 | TOPIC 13 The normal distribution • EXERCISE 13.5 b 10 k 2 + 9 k − 1 = 0 (10 k − 1)( k + 1) = 0 1 k= as k ≠ −1 10 c Let X = the chocolate surprises containing a ring X ~ Bi (8, 0.25) E( X ) = 8 × 0.25 = 2 d Pr( X = 2) = 0.3115 e Pr( X = 0) ≤ 0.09 n

C0 (0.75)n ().25)0 ≤ 0.09

(0.75)n ≤ 0.09 n loge (0.75) ≤ loge (0.09) loge (0.09) n≥ loge (0.75) n ≥ 8.3701 Sample size has to be 9. f Let W = the weight of the defective chocolate surprises

(

W ~ N 125, σ 2 Pr(W < 100) = 0.082 100 − 125   Pr  Z <  = 0.082  σ 100 − 125 = −1.3917 σ − 25 = −1.3917σ −25 =σ −1.3917 17.9636 = σ σ = 18

)

3 Let X = the error in a speedometer X ~ N 0, 0.762

(

)

Pr(Unacceptable) = Pr( X < −1.5) ∪ Pr( X > 1.5) Pr(Unacceptable) = 2 Pr( X < −1.5) by symmetry Pr(Unacceptable) = 2 × 0.0242 Pr(Unacceptable) = 0.0484 4 Let X = the height of Perth adult males X ~ N 174,82

(

)

a Pr( X ≥ 180) = 0.2266 or 22.66% b Pr( X ≥ x ) = 0.25 x = 179.396 = 179 cm 5 a L  et X = average weight of David’s avocados X ~ N 410, 20 2

(

)

i Pr( X < 360) = 0.0062 Pr(340 < X < 360) Pr( X < 360) 0.005977 Pr( X > 340  X < 360) = 0.0062 Pr( X > 340  X < 360) = 0.9625

ii Pr( X > 340  X < 360) =

b Let Y = average weight of Jane’s avocados Y ~ N µ,σ 2

(

)

Pr(Y < 400) = 0.4207 400 − µ   Pr  Z <  = 0.4207  σ  400 − µ = −0.2001 σ 400 − µ = −0.2001σ 400 = µ − 0.2001σ ....................(1) Pr(Y > 415) = 0.3446 415 − µ   Pr  Z >  = 0.3446  σ  415 − µ = 0.3999 σ 415 = µ + 0.3999σ ...................(2) 400 = µ − 0.2001σ ....................(1) (2) − (1) 415 − 400 = 0.3999σ + 0.2001σ 15 = 0.6σ 15 =σ 0.6 25 = σ Substitute σ = 25 into (1) 400 = µ − 0.2001(25) 400 = µ − 3.0015 µ = 405 6 Let X = the length of metal rods

(

X ~ N 145,1.4 2

)

a Pr( X > 146.5) = 0.1420 0.15 b Pr( X < µ − d ) = 2 Pr( X < 145 − d ) = 0.075 145 − d = 142.9847 145 − 142.9847 = d 2.0153 = d d = 2.0 c Let Y = the number of rods with a size fault Y ~ Bi (12, 0.15) Pr(Y = 2) = 0.2924 d i a + 0.15 + 0.17 = 1 a + 0.32 = 1 a = 1 − 0.32 a = 0.68 ii E(Y ) = 0.68( x − 5) + 0.15(0) + 0.17( x − 8) E(Y ) = 0.68 x − 3.4 + 0.17 x − 1.36 E(Y ) = 0.85 x − 4.76 iii If E(Y ) = 0 then 0.85 x − 4.76 = 0 0.85 x = 4.76 x = 5.6 Selling price of good rods will be 5.6 − 5 = 6 cents 0.68 iv Production of good rods = 0.68 + 0.17 = 0.8 i.e. 80%

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 13 The normal distribution • EXERCISE 13.5   |  249

(

)

(

7 X ~ N 2500, 700 2 and Y ~ N 3000,550 2

)

a Pr( X < 1250) = 0.0371 b Pr(Y < 1500) = 0.0032 c Pr(Both "special") = Pr( X ∩ Y ) Pr(Both "special") = Pr( X ) × Pr(Y ) as they are independent events Pr(Both "special") = 0.0371 × 0.0032 Pr(Both "special") = 0.0001 d i Pr(One "special") = 0.4 × 0.0371 + 0.6 × 0.0032 Pr(One "special") = 0.0167 Pr( X ∩ One "special") Pr(One "special") 0.4 × 0.0371 Pr( X "special" One "special") = 0.0167 0.00744 Pr( X "special" One "special") = 0.0167 Pr( X "special" One "special") = 0.8856

ii Pr( X "special" One "special") =

8 a Let X = the height of plants X ~ N 18,52 Pr( X < 10) = 0.0548 Pr(10 < X < 25) = 0.8644 Pr( X > 25) = 0.0808

(

)

Plant Small Medium Large

Size

Probability

X < 10 10 < X < 25 X > 25

0.0548 0.8644 0.0808

b E(Cost of one plant) = 2(0.0548) + 3.5(0.8644) + 5(0.0808) = $3.54 E(Cost of 150 plants) = 150 × $3.54 = $531 9 Let W = the weight of perch W ~ N 185, 20 2

(

)

a Pr(W > 205) = 0.1587 = 15.87% (Cannery, 60 cents) b Pr(165 < W < 205) = 0.6827 = 68.27% (Market, 45 cents) c Pr(W < 165) = 0.1587 = 15.87% (Jam, 30 cents) E(Profit) = 60(0.1587) + 45(0.6827) + 30(0.1587) = 45.0045 = 45 cents 10 Let X = the diameter of the tennis ball X ~ N 70,1.52

(

)

a Pr( X < 71.5) = 0.8413 b Pr(68.6 < X < 71.4) = 0.6494 c Let Y = the tennis balls in range Y ~ Bi ( 5, 0.3506 ) Pr(Y ≥ 1) = 1 − Pr(Y = 0) Pr(Y ≥ 1) = 1 − (0.64935)5 Pr(Y ≥ 1) = 0.8845 d

(68.6 < X < 71.4) = 0.995 71.4 − 70   68.6 − 70 100) = 0.0078 f E(Cost of one apple) = 0.1797(0.12) + 0.6703(0.15) + 0.15(0.25) = 0.1596 or 16 cents E(Cost of 2500 apples) = 2500 × 0.1596 = $399 g Let Y = Jumbo apples in a bag Y ~ Bi ( 6, 0.15) Pr(Y ≥ 2) = 1 − ( Pr(Y = 0) + Pr(Y = 1) ) Pr(Y ≥ 2) = 1 − ( 0.3771 + 0.3993) Pr(Y ≥ 2) = 1 − 0.7764 Pr(Y ≥ 2) = 0.2236 12 Let X S = the amount of disinfectant in a standard bottle

(

)

(

)

X S ~ N 0.765, 0.0072

Let X L = the amount of disinfectant in a large bottle X L ~ N 1.015, 0.0092

a Pr( X S < 0.75) = 0.0161 b Pr( X L < 1.00) = 0.0478 Let Y = the large bottles with less than 0.95 litres in them Y ~ Bi (12, 0.0478 ) Pr(Y Pr(Y Pr(Y Pr(Y Pr(Y

≥ 4) = 1 − Pr(Y < 4) ≥ 4) = 1 − ( Pr(Y = 0) + Pr(Y = 1) + Pr(Y = 2) + Pr(Y = 3) ) ≥ 4) = 1 − ( 0.5556 + 0.3347 + 0.0924 + 0.0155) ≥ 4) = 1 − 0.9982 ≥ 4) = 0.0019

13 Let L = the length of antenna of a lemon emigrant butterfly

(

L ~ N 22,1.52

)

a Pr( L < 18) = 0.0038 Let Y = the length of antenna of a yellow emigrant butterfly b

Pr(Y < 15.5) = 0.08 15.5 − µ   Pr  Z <  = 0.08  σ  15.5 − µ = −1.4051 σ 15.5 − µ = −1.4051σ 15.5 = µ − 1.4051σ ....................(1) Pr(Y > 22.5) = 0.08 22.5 − µ   Pr  Z >  = 0.08  σ 22.5 − µ = 1.4051 σ 22.5 − µ = 1.4051σ 22.5 = µ + 1.4051σ ...................(2) 15.5 = µ − 1.4051σ ....................(1)

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 13 The normal distribution • EXERCISE 13.5   |  251 (2) − (1) 22.5 − 15.5 = 1.4051σ + 1.4051σ 2 = 2.8102σ 2 =σ 2.8102 σ = 2.5 mm Substitute σ = 2.5 into (1) 15.5 = µ − 1.4051(2.5) 15.5 = µ − 3.5128 15.5 + 3.5128 = µ µ = 19.0 mm c Pr(Yellow) = 0.45 and Pr(Lemon) = 0.55 Let B = the number of yellow emigrants B ~ Bi (12, 0.45) Pr( B = 5) = 0.2225

14 a Let X = the error in seconds of a clock

(

X ~ N µ,σ 2

)

The clock can be up to 3 seconds fast or 3 seconds slow. Pr( X > 3) = 0.025 µ=0 3− 0  Pr  Z >  = 0.025  σ  3 = 1.95996 σ 3 = 1.95996σ 3 =σ 1.95996 σ = 1.5306 b Let Y = the number of rejected clocks Y ~ Bi (12, 0.05) Pr(Y < 2) = Pr(Y ≤ 1) Pr(Y < 2) = 0.8816

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 14 Statistical inference • EXERCISE 14.3   |  253

Topic 14 — Statistical inference Exercise 14.2 — Population parameters and sample statistics 1 Mr Parker teaches on average 120 students per day. This is the population size. N = 120 He asks one class of 30 about the amount of homework they have that night. This is the sample size. n = 30 2 Lois is able to hem 100 shirts per day. This is the population size. N = 100 Each day she checks 5 to make sure that they are suitable. This is the sample size. n = 5 3 Mr Banner tests his joke on this year’s class (15 students). This is the sample size. n = 15 We don’t know how many students Mr Banner will teach. The population size is unknown. 4 Lee-Yin asks 9 friends what they think. This is the sample size. n = 9 We don’t know how many people will eventually eat Lee-Yin’s cake pops. The population size is unknown. 5 a Population parameter b Sample statistic 6 a Population parameter b Sample statistic 523 7 Number of boys: × 75 = 34.29. Therefore 34 boys. 523 + 621 621 Number of girls: × 75 = 40.71. Therefore 41 girls. 523 + 621 8 Number of boarders: 23% of 90 = 20.7. Therefore 21 boarders. The rest of the sample will be day students. 90 − 21 = 69 day students. 9 a We don’t know how many people will eventually eat the pudding. The population size is unknown. b 40 volunteers to taste test your recipe. This is the sample size. n = 40 10 a We don’t know how many people will eventually receive the vaccine. The population size is unknown. b 247 suitable people test the vaccine. This is the sample size. n = 247 11 Sample statistic 12 Population parameter 13 Sample statistics 14 Population parameter 15 a A systematic sample with k = 10 b Yes – assuming that the order of patients is random 16 The sample is not random, therefore the results are not likely to be random 17 It is probably not random. Tony is likely to ask people that he knows, or people that approach him. 18 Number of male staff: 60% of 1500 = 900 Number of full time male staff: 95% of 900 = 855 855 Number to sample: × 80 = 45.6 1500 Number of part-time male staff: 900 – 855 = 45 45 Number to sample: × 80 = 2.4 1500 Number of female staff: 1500 – 900 = 600 Number of full time female staff: 78% of 600 = 460

460 × 80 = 24.96 1500 Number of part time female staff: 600 – 460 = 140 140 Number to sample: × 80 = 7.47 1500 The sample consists of: Full time male staff: 46 Part time male staff: 2 Full time female staff: 25 Part time female staff: 7 19 Use the random number generator on your calculator to produce numbers from 1 to 100. Keep generating numbers until you have 10 different numbers. Answers will vary. 20 First assign every person in your class a number e.g. 1–25. Decide how many students will be in your sample, e.g. 5. Then use the random number generator on your calculator to produce numbers from 1 to 25. Keep generating numbers until you have 5 different numbers. The students that were assigned these numbers are the 5 students in your random sample. Answers will vary. Number to sample:

Exercise 14.3 — The distribution of pˆ 6 15 2 = 5 6 pˆ = 20 3 = 10 N = 1000 n = 50 p = 0.85 Is 10 n ≤ N ? 10 n = 500, therefore 10 n ≤ N . Is np ≥ 10? np = 50 × 0.85, therefore np ≥ 10. = 42.5 Is nq ≥ 10? nq = 50 × 0.15, therefore nq ≥ 10 = 7.5 The sample is not large. For the sample to be large, nq = 10 0.15n = 10 n = 66.7 n = 67 is the smallest sample size that can be considered large. np = 10 As p < q, if np ≤ 10, then nq ≤ 10. 0.05n = 10 n = 200 Is 10 n ≤ N ? 10 n = 2000, therefore n = 200 is a large sample. a µ pˆ = p = 0.5

1 pˆ =

2

3

4

5

p (1 − p ) n 0.5 × 0.5 = 50 = 0.07

b σ pˆ =

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

254 | TOPIC 14 Statistical inference • EXERCISE 14.3 6 If N = 1000, n = 100 and p = 0.8. a µ pˆ = p = 0.8 p (1 − p ) b σ pˆ = n 0.8 × 0.2 = 100 = 0.04 27 7 pˆ = 30 9 = 10 147 ˆ 8 p = 537 12 4 = 9 a p = 21 7 b 0 females chosen out of 4, 1 chosen out of 4, 2 chosen out of 4, 3 chosen out of 4 or 4 chosen out of 4. 1 1 3 Therefore the possible values for pˆ are 0, , , , 1. 4 2 4 c Relative pˆ X Number of samples frequency 0

12

C0 9C4 = 126

0 1 4 1 2

1 2

12

4

C1 9C3 = 1008

0.168

C2 9C2 = 2376

0.397

12

3 4 1

3

0.021

12

C3 9C1 = 1980

0.331

12

C4 9C0 = 495

0.083

TOTAL samples

0

1 4

1 2

3 4

1

Pr( Pˆ = pˆ )

0.021

0.168

0.397

0.331

0.083

3 d Pr( Pˆ > 0.6) = Pr  Pˆ =  + Pr( Pˆ = 1)  4 = 0.331 + 0.083 = 0.414 Pr Pˆ > 0.5 e Pr( Pˆ > 0.5 Pˆ > 0.3) = Pr Pˆ > 0.3

) )

0.414 0.414 + 0.397 = 0.510

=

1 2 3

0 1 5 2 5 3 5

Pr( Pˆ = pˆ ) C0 ( 0.62 )0 ( 0.38 )5 = 0.008

5

C1 ( 0.62 )1 ( 0.38 )4 = 0.064

5

C2 ( 0.62 )2 ( 0.38 )3 = 0.211

5

C3 ( 0.62 )3 ( 0.38 )2 = 0.344

5

4 5

5

5

1

5

C4 ( 0.62 )4 ( 0.38 )1 = 0.281

C5 ( 0.62 )5 ( 0.38 )0 = 0.092

TOTAL samples

5985

Probability distribution table: pˆ

0

1 5

2 5

3 5

4 5

1

Pr( Pˆ = pˆ ) 0.008 0.064 0.211 0.344 0.281 0.092 3 4   c Pr( Pˆ > 0.5) = Pr  Pˆ =  + Pr  Pˆ =  + Pr( Pˆ = 1)   5 5 = 0.344 + 0.281 + 0.092 = 0.717 np = 10 11 As p < q, if np ≤ 10, then nq ≤ 10. 0.01n = 10 n = 1000 Is 10 n ≤ N ? 10 n = 10 000, therefore n = 1000 is a large sample. 12 µ pˆ = p = 0.15

p (1 − p ) n 0.75 × 0.25 = 100 = 0.043 14 µ pˆ = p p = 0.12

σ pˆ =

pˆ

0

4

p (1 − p ) n 0.15 × 0.85 = 150 = 0.029 13 µ pˆ = p = 0.75

Probability distribution table:

1 2 3 4 10 a 0, , , , , 1 5 5 5 5 b X pˆ

pˆ

σ pˆ =

5985

( (

Pr( Pˆ = pˆ )

X

p (1 − p ) n 0.12 × 0.88 0.0285 = n 0.1056 −4 8.1225 × 10 = n 0.1056 n= 8.1225 × 10 −4 = 130 15 µ pˆ = p p = 0.81

σ pˆ =

p (1 − p ) n 0.81 × 0.19 0.0253 = n 0.1539 −4 6.4009 × 10 = n  0.1539   n= 6.4009 × 10 −4 = 240.4 Therefore n = 240

σ pˆ =

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

16

Relative frequency

TOPIC 14 Statistical inference • EXERCISE 14.4   |  255 0.12 0.10 0.08 0.06 0.04 0.02 0

17

0.05

0.10

0.15

0.20

0.25

0.30

1

pˆ

p (1 − p ) n p (1 − p ) 0.015 = 510 p (1 − p ) −4 2.25 × 10 = 510 0.11 475 = p (1 − p )

σ pˆ =

= p − p2 The quadratic p2 − p + 0.11 475 = 0 can be solved using the quadratic formula. p= =

− b ± b 2 − 4 ac 2a 1 ± 1 − 4 × 1 × 0.11 475

2 1 ± 0.541 = 2 p = 0.87 or p = 0.13 As pˆ > 0.5, the population proportion is p = 0.87. 18

p (1 − p ) n p (1 − p ) 0.0255 = 350 p (1 − p ) −4 6.5025 × 10 = 350 0.2 275 875 = p (1 − p )

σ pˆ =

= p − p2 The quadratic p2 − p + 0.2 275 875 = 0 can be solved using the quadratic formula. p= = =

− b ± b 2 − 4 ac 2a 1 ± 1 − 4 × 1 × 0.2 275 875 2 1 ± 0.08 965

2 p = 0.65 or p = 0.35

Exercise 14.4 — Confidence intervals 1  n = 30 pˆ = 0.78 z = 1.96 The 95% confidence interval is pˆ ± z 0.78 ± 1.96

pˆ (1 − pˆ ) n

0.78 × 0.22 30

63%–93% 2  n = 53 pˆ = 0.82 z = 1.96 The 95% confidence interval is pˆ ± z

pˆ (1 − pˆ ) n

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

256 | TOPIC 14 Statistical inference • EXERCISE 14.4 0.82 ± 1.96

7  n = 250 20 pˆ = 250 = 0.08 z = 1.96

0.82 × 0.18 53

72%–92% 3   n = 116 pˆ = 0.86 z = 2.58 ( Pr( Z < z ) = 0.005) The 99% confidence interval is pˆ ± z 0.86 ± 2.58

0.86 × 0.14 116

78%–94% 4  n = 95 pˆ = 0.3 z = 1.64 ( Pr( Z < z ) = 0.05) The 90% confidence interval is pˆ ± z 0.3 ± 1.64

pˆ (1 − pˆ ) n

pˆ (1 − pˆ ) n

0.3 × 0.7 95

22%–38% 5 95% confidence interval z = 1.96 pˆ will be at the centre of the interval, pˆ = 0.4 The confidence interval is pˆ ± z z

pˆ (1 − pˆ ) = 0.05. n

pˆ (1 − pˆ ) n 0.4 × 0.6 1.96 n 0.24 n 0.24 n z

pˆ (1 − pˆ ) . This means that n

= 0.05 = 0.05

= 6.5077 × 10 −4

0.24 6.5077 × 10 −4 = 368.8 A sample of size 369 is needed. 6 90% confidence interval z = 1.64 pˆ will be at the centre of the interval, pˆ = 0.8 n=

pˆ (1 − pˆ ) z = 0.05. n pˆ (1 − pˆ ) n 0.8 × 0.2 1.64 n 0.16 n 0.16 n z

pˆ (1 − pˆ ) . This means that n

0.08 × 0.92 250 4.6%–11.4% of complaints take more than 1 day to resolve. 8  n = 250 230 pˆ = 250 = 0.92 z = 2.58 0.08 × 0.92 0.92 ± 2.58 250 87.6%–96.4% of complaints are resolved within 1 day. 9  n = 250 92 pˆ = 250 = 0.368 z = 1.64 485 0.368 × 0.632 The 90% confidence interval 0.368 ± 1.64 250 31.8%–41.8% of Australians have Type A blood. 10 n = 250, p = 0.65 Since n is large, we can approximate the distribution of Pˆ to that of a normal curve. Therefore µ = p = 0.65 and 0.65 × 0.35 σ= = 0.030 250 Pr Pˆ < 0.6 = 0.0487

)

11 n = 200, p = 0.8 Since n is large, we can approximate the distribution of Pˆ to that of a normal curve. Therefore µ = p = 0.8 and 0.8 × 0.2 σ= = 0.0283 200 Pr 0.8 < Pˆ < 0.9 Pr 0.8 < Pˆ < 0.9 Pˆ > 0.65 = Pr Pˆ > 0.65

(

)

(

= 0.05

(

0.4998 = 0.9999 = 0.4998

z

pˆ (1 − pˆ ) . This means that n

pˆ (1 − pˆ ) = 0.05. n

pˆ (1 − pˆ ) = 0.05 n 0.3 × 0.7 = 0.05 1.96 n 0.21 = 0.0255 n 0.21 = 6.5077 × 10 −4 n 0.21 n= 6.5077 × 10 −4 = 322.7 A sample of size 323 is needed. z

= 0.0305 = 9.285 × 10 −4

0.16 9.285 × 10 −4 = 172.1 A sample of size 173 is needed. n=

)

)

12 z = 1.96 pˆ will be at the centre of the interval, pˆ = 0.3 The confidence interval is pˆ ± z

= 0.05

pˆ (1 − pˆ ) n

0.08 ± 1.96

(

= 0.0255

The confidence interval is pˆ ± z

The 95% confidence interval is pˆ ± z

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual

TOPIC 14 Statistical inference • EXERCISE 14.4   |  257 13 z = 2.58 pˆ will be at the centre of the interval, pˆ = 0.25 The confidence interval is pˆ ± z z

pˆ (1 − pˆ ) . This means that n

pˆ (1 − pˆ ) = 0.05. n

pˆ (1 − pˆ ) = 0.05 n 0.25 × 0.75 = 0.05 2.58 n 0.1875 = 0.0194 n 0.1875 = 3.756 × 10 −4 n 0.1875 n= 3.756 × 10 −4 = 497.62 A sample of size 498 is needed. 14 95% confidence interval means that z = 1.96 The interval is 0%–5% pˆ will be at the centre of the interval, pˆ = 0.025 (2.5%) pˆ (1 − pˆ ) The confidence interval is pˆ ± z . This means that n pˆ (1 − pˆ ) = 0.025. z n z

pˆ (1 − pˆ ) = 0.025 n 0.025 × 0.975 1.96 = 0.025 n 0.024 375 = 0.012 755 n 0.024 375 = 1.6269 × 10 −4 n 0.024 375 n= 1.6269 × 10 −4 = 149.8 A sample of size 150 is needed. 15 99% confidence interval means that z = 2.58 pˆ will be at the centre of the interval, pˆ = 0.94 (94%) pˆ (1 − pˆ ) The confidence interval is pˆ ± z . This means that n pˆ (1 − pˆ ) = 0.04. z n z

pˆ (1 − pˆ ) n 0.94 × 0.06 2.58 n 0.0564 n 0.0564 n z

pˆ (1 − pˆ ) n 0.15 × 0.85 0.03 = 1.96 n n = 544 The sample size needed is 544 people. 17 pˆ will be at the centre of the interval, pˆ = 0.90 M = 1.96

16

The confidence interval is pˆ ± z z

pˆ (1 − pˆ ) . This means that n

pˆ (1 − pˆ ) = 0.05. n

pˆ (1 − pˆ ) = 0.05 n 0.9 × 0.1 = 0.05 z 100 0.03z = 0.05 z = 1.67 Pr ( −1.67 < z < 1.67 ) = 0.9 Benton’s is 90% sure of their claim. 18 pˆ will be at the centre of the interval, pˆ = 0.775 z

The confidence interval is pˆ ± z z

pˆ (1 − pˆ ) . This means that n

pˆ (1 − pˆ ) = 0.025. n

pˆ (1 − pˆ ) = 0.025 n 0.775 × 0.225 = 0.025 z 250 0.026 z = 0.025 z = 0.947 Pr ( −0.947 < z < 0.947 ) = 0.66 Benton’s is 66% sure of their claim. z

= 0.04 = 0.04 = 0.0155 = 2.404 × 10 −4

0.0564 2.404 × 10 −4 = 234.6 A sample of size 235 is needed. n=

Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual