Math IB SL Answer Sheet - [PDF Document] (2024)

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AnswersChapter 1Skills check

1 a yA

,.F,..,

v

,.,••

-s - 11.0-..

-a4

A c

• X

E 8

b A O, 2), B l , 0), C - 1, 0),D O, 0), E 2 , 1), F -2, -2),G(3, -1 , H -1, 1)

2 a 34 b 82

c 16 d _ _ 2

3 a 4

4 a yA

........-

A- J: ·

60

b -2

11 j)

J

c 10

.

v ,.R - b r.O 'I

X

I " vJ

b yA

.v

A

'"' ''

''I :

-

'02 1c y

6

A

I

' f,-,1 J\ 17

b h 1\. 1.. 0 ,/1'. v-

X

fL.

Answers

5 a x2 + 9x = 20b x2 - x + 3

c x2 + x 20

Investigation handshakes

a 6b

c

Numberof people

y

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xercise l C

1 Horizontal asymptote: y = 0

2 Horizontal asymptote: y =0 ,Vertical asymptote: x = 0

3 Horizontal asymptote: 0,Vertical asymptote: x = - 1

4 Horizontal asymptote: y = 2,Vertical asymptote: x = - 2

5 Horizo ntal asymptote: y = 2,Vertica l asymptote: x = 1

6 Horizo nt a l asymptote: y = 0,Vertica l asymptot e: x = -3

xercise 1

1 Function , domain {2 3, 4, 5,6, 7, 8, 9, 1 0 } range {1 , 3,6,10 , 15 , 21 , 28 , 36 , 45 }.

2 a domain {x: -4 < x < 4},range {y: 0 $ y $ 4}

b domain {x: - 1 < x < 5},range{y : 0 $ y $ 4}

C domain {x ; - oo X oo},range {y: 0 $ y < oo}

d domain{x:-oo < x < - 2< x < oo},range{y: - oo <y $ 3 4 $y < 8}

e domain{x:-5 < x < 5},range{y: -3 < y < 4}

f domain {x:-oo < x < oo}range {y: - 1 < y < I }

g domain {x: -2 < x < 2},range{y: -2 < y < 2}

h domain {x: - oo < x < oo},range {y: - oo < x <oo}

•domain xe R , x t 1, rangey e y;t: 0

3 a xe IR y e lRY.

8

4

by

16-

14-

•

- -·12- L

-10- _

8- -1\ 14-

- \ v... /-4 -2 0 2

cy

2

15

10

a-j -2 0

d X E JR, y Ey

5

-2 -1 0

- 10

-15

ey

10 - .. .-- .. .....v

8

6

4

2

--

.

t

v

·-

v/ v - -

v-t- - t

4

....

0 20 40 60 80 100 X

X

X

f X 4 , y > 0y

10

8

2

100 80 60 40 20 0 X

g xeJ Rx;t; O, y e lR y;t: Oy

,-1-

.

4

I 8 4 8-4

8

h X E JR, y > 0

y87-6 f-

543

-2 -1 0 1 2

i x e JR x ; t : - 2 , y e lR y:;t;O,y

0-r

l 8- 1-t- 6-

- 4- -\ -

-8 2 6 X-4 -t-6-

_Q

' 0

Answers

X

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j X E JR X: 2, y E JR y :# 1

y

1 B

_Q

'-- 0 ,.G. 10 20-..

4cv

)o-

X

k XE x:t: - 3 ,y E y:t: -6y

4/ v

/ vb 11 0 / vP'

'./ .....8

l X E JR, 0 < y $ 2y

;; \1'•• \

v •-f- ' • V ....

/

X

--

--5 -4 -3 -2 - 1 0 1 2 3 4 5 6 X

xercise l

• •• • •• 11 a I 5 II -5 Ill - 1 -2• - 2 a - 2V v

' 1b • •• •••I 21 II - 9 Ill 1 -2• 0 3aV v• 7 .. - 3 • •• 1c III - Ill -

4 4 810V v - a4

d.

19•• -1 • •• 6II Ill

.5 2a + 5V v

2_ _51••

11• ••e I II Ill

4• 2 a2 + 2V v

2 a a2 - 4 b a 2 + l Oa + 21c a2 - 2a - 3 d a 4 - 4a 2

e 2 1 - lOa+ a2

3 a 2 b 11 c 2

4 a 1- -9

b x = 6, denominato r = 0and h (x) undefined .

Answers

5 a 125b The volume of a cube of

side 5.

6 • 1.. 5a I - - II

9 4... lIll - - iv 0.

2

b. - 4

..- 11 ... - 67II Ill

. -697 - 6 997V v•

- 6 9997I

c T he value of g(x) is gettingincreasingly smaller asxapproaches 2.

d 2e asymptote at x = 2.

yr , \

18r-....

0 A 0 \ 1 ) 1 &-;v

ir.

7 a - 9 m s - 1c 91 m s - 1

b 7 m s - 1

d 3s

8 a / ( 2 + 2h) - / ( 2 + h)h

b / (3 + 2h) - / (3 + h)h

xercise I F

' x

1 a 12 b 3 c - 15d 3x + 3 e 13 f 16

-17 h 3x + 1 • 18I•

38 k 3x2 + 6Jl 9x2 + 2 m 12 n 180 r+ 3 p x2 + x + 3

2 a 3 b 0 c - 12d -1 e - 5 f 48

g 3 - 4 x + x2

h - x

+ x2

3 a x 2 + 4x + 4 b 254 a 5x 2 + 5 b 5x 2 + 15 a x 2 - 8x + 19 bx 2 - 1

c 2.5

6 (ros) (x) = x 2 - 4 , x E JR, y > - 4

xerciselG

1 b , c

2 a y

l /0 // v

Jt....

4/'C

v , I..... ,(j -R ' X/ I IA

//

, A'

b y

l o/

/.....I /

//

'I // /-

/ /

R v . ........, v

l

4

X

v/

/

lo/ I , ..../

c y

i' /r-.... /........ \ /......... /........ \ //

/ \ .........../ .........-R kt ,(j \ .

........ X- . - \A/ \

/l > \

, i\

d y........ /

l > ......... /

//-

/ \, // \/

/ I"-./

/ /

-R - IL ,(jI - XI / I4 v - --//

// 8

/

e y....v

.A - /_-r/

/ .- ', ;

v ;;/ I,I l x4 -B - ,.:::[ 0 2 ' 4

... '--- A

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f y

-3 ; .2 1....-

Exercise l1 a i -2 an d 1

ii _ _ an d - 32

••• •Ill X IV X

b They are inverses of eachother.

2 a x+ 1 b3

c 4(x - 5) d (x + 3) 3

1ex+2

3xg l x

3 a 1 x

f Jx 3

hx+2Sx

b X C 1X

4 a 1 b -5 c . l20

S l +2xx-r6 a-c

y

8-

4- j .

/ r / -- I/ . ,.-- •- - I

,...-L

I _,J .- ' 0 2 4 X

-4- I I

d / x): X E IR y > 0/ - 1 x): x > 0, y E R

7 g- 1(x) = r . Th e range of g(x)is x 0 so the domain of

8

g - 1(x) is x 0. The domain off x ) is x E R so g - 1(x) # fx)

f x ) = mx + c thenf x)=l . .x -.£.m m

tn x , ,= not -1 so notperpendicular.

Invest igation functions Exercise

1 Changi ng the constant term 1 atranslates y = x alongthey-axis.

y

e x--1--1-J

2 Changing the x-coefficienta lt ers the gradient of theline.

y

X

V: l I I dJ3 y = Ix + hI is a translation of

- h a long the x-axisy

Y l x 3 7

X

-2

-3

b

c

d

-8 -6

I

y

8 ....._- ...

6

-4

y

6

4

-6

y6

- 4 t -4

4 2 4

-4-6-8

6 8

Answers

X

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g y

-15

2 g(x) = x) + 2h(x) = x ) - 4

1q(x) = 2 f x)

3 q(x) = f x + 4 ) - 2s x) = f x + 4)t(x) = f x - 2)

10 15 X

4 a D omain -1 < x < 7, range- 4 < y < 6

y86

b D omain - 3 ::; x ; 1, rangeO < y < 5

y

4

b

c

d

e

g

-4 f -

g

f -4 -

g

f -

y

64

g f 2

0-8 -4 -2

-4

y

4

2

-4

y

4

-4

2

2 4 X

4 X

4 X

4 6 8 10 X

-6 -4 -2 0 2 x 6 a Reflection in x-axis.-2

5 a y

4

f

g -4 2 4 X

-4

Answers

b Horizontal trans lation 3units .

c Ve rt ical st retch SF2,reflection x-axis, vertical

translation of 5 units.7 a b

yc...

A I....

/r )L _L2-

X)v 1 II

4 - i -2 - l1 0,

:2 3 X

Review exercise non-GDC

1 a 4a - 13 b 2 ; x

2 a 2x2 - l 5x + 28

b - 2x2 + 9

3 a 2x-173

b 2 3

4 f -1

(x) = - 5 x - 5y

' - 'c.

5 a

6 a

b

I

4

....(1 -

'l

L

A

x - 53

-- )

/

/

I

' n

6W

y

4 rr _ 1 )' '2 \

/

-- /

/

:2

2 \ \1

I + \

yA

•-3

1...

_ 1 /

-... l \o1

b . i l - 2

/

X

-

I/ ......

,..'_... . / r / f

X

,111 v X-I-,. I1

-

'>' '

7 a D omain x E IR, y > 0

b Domain x E JR. x * 3,Rangey E JR, y * 0

8 a f x )=2- l - 3x - 9 +2

1 ( x - 5 lb f x ) = - - 3- 3 1- 14 )

9 a Inverse function graph isthe reflection in y = x.

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b yA•

1 :2/

13 0 P_ '> In ,. .... X

lA v\0 -, -;: • J

I A

•

10 a - 2 b -13

c J - c x)=vx 2 3

l l a yI

10,J

A

1\

21

\ v-B -Q -1 0 , . J X

b P is ( 4, 1)

12 a (jog) x) = 3x + 6

b j -l x) = x g- 1(x) = x - 23

j l J2) = 1:= 4g - 1(12) = 12 - 2 = 10

j l (12) + g-l (12) = 4 + 101(12) + g 1 (12) = 14

13 a3(2x- l )

(hog)(x) = (2x -1 ) -2

_ 6 x - 3-

2x - 3

bl

x= -2

eview exercise GD

3 Domain x E JR, x :;t: - 2, rangey E JR, x:;t:. 0

4 a

y

1210

86

42

0-2 - t2-4-6-8

y

8

2

1 2 X

-3 -2 1 -20 1 2 X

-4

-6-8

b x-intercept -1.5,y-intercept 3.

5 a y

6

4

-6 -4 -2 0 2 4 6 X

1 Domain: x > -2 , range: y > 0 b 0

2 Domain: x E JR, range: y > -4

y

16

1412

10864

2

- 4 -3 -2 - 0 4 X

c Domain x E JR x :;t: lR,rangey > 0

6 a x = - 2 , y = 2

b y6

2

8

4f

-12 -8 -4 0 4 8 12 X-4

c (2.5, 0), (0, -2.5)

7 a y6

4

3 X

-4

-6

b X = +Ji

8 a 1x 3

b y...

j V

4, .I'

v11/ h 0

Jv

c 1.67

9 y·8

X -1

,v

j = - ' \-- - - - - 1- -p - f:l -B - 0

I f \V

l O a f - (x)= x 23

'' X

717

-- ,.X

b (g - 1o/ x) = 3 x - 2) + 3= 3x + 1

C ( / - 0 = x - + 2 = 31

x - 1 =3x+13

x - 1 = 3(3x + 1)x - 1 = 9 x + 3

8 x = - 41x = - -2

Answers

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d yI ' > '....

-- 4; _ - -.. - --

_

6 - 4 - \A

e x = 3, y = 3

Chapter 2Skills check

1 a a= 6

b X =+JSc n = -11

2 a 2k k - 5)

I

I1I

I \II

,_ .;. --:x - 3IIII

b 7a 2a 2 + a 7)

c 2x + 3) x+2y)

d Sa - b) a - 2)

e n + l)(n + 3)

f (2x - 3)(x + 1)

g m + 6) m- 6)

h Sx + 9y) 5x- 9y)

Exercise 2A1 a 1 2

b -8 , 7

c 5, 6

d -5 5'

e - 8 6'

f -3

2 4 1a3 2

b 4-2 - 5

c 5-1 - 2

d I9

- 2 2

e 2- 4 - - 3

f3 4

- 2 3

Answ rs

- - .

--

X

Exercise 28

1 a -5, 4

b - 23

3c2

d - 2 252

e - 9 , 41f 14

2 - 3 or 4

23 - or 3

5

Investigation perfectsquare trinomials

1 -5

2 -3

3 -7

4 4

5 9

6 10

Exercise 2C

1 - 4 ± M

2 5±J372

3 3± 2J2

4 - 7±- 652

5 l + J7

6 -l + J U

2

Exercise 2

1 - 3 + 2J3

2 1+ J 2

3

4- 3+-fi9

4

35 22

6 - 2+3J610

Exercise 2E

1 - 9 + J I938

2 4- 2 - 3

3 1- 15

4 3 ± J 55 no soluti on

6- 5+2.Jl0

3

7 3±. f 0

4

89 ±-Jl i3

4

9- 9 + J l 2 9

x

4

10 3+m4

Exercise 2F

1 18,32

2 24m, 11 tn

3 10

4 18 em, 21 em

5 2. 99 seconds

Investigation roots of

quadratic equations

ba 42

1c5

2 a -7 2 b'

3+ J89c10

4± F o3

3 a No solution

b No solution

c No solution

Exercise2G

1 a 3 7; two different real roots

b 8; two different real roots

c - 79; no real roots

d 0; two equal real roots

e - 23; no real roots

f - 800; no real roots

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2 a p < 4 b p < 3.125

c i I>4v 2 d2iP >-3

3 a k = 25 b k = 1.125

c k=± J lS d k = 0, - 0.754 a m > 9 b 2 < m < 2

c 16m> d m > 123

5 O < q < I

Investigation graphs o fquadratic functions

a Discriminant, 6. = 29

y

b 6 . = 12y

c 6. = 2 4

d 6. = 7 1y

X

X

y

X

X

e 6. = 0y

X

f 6. = 0

y

X

g 6. = 33

y

X

h 6. = 37y

X

f b2 - 4ac > 0, graph cuts

x-axis twice; if b2 - 4ac = 0,graph is tangential to x-axis; if

b2 4ac < 0, graph do es not

intersect x-axis.

Exercise H

1 a x = - 4; 0, 5)

b x=3 ; (0 , -3 )

C X - 1; 0, 6)

d X S. (0 9)3

2 a (7,-2) ;(0,47)

b ( - 5, I) ; 0, 26)

c (1 ,6 ) ;(0,10)

d ( -2 , -7) ; (0 ,5)

3 a f x) = x + 5) 2 - 31y

0 0, -6) X

(-5,-31)

b f x) = x - 2.5) 2 - 4.25

y

,2)

0 X

2.5, -4.25)

C f x) = 3 x - 1)2 + 4y

0, 7)

1, 4)

0 X

d J x) = -2 x 2) 2 + 5

0, -3

2, 5)

X

Answers

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Exercise 2

1 a -3 ,0) ; 7 , 0); 0,-21)

2

b (4,0); (5,0); 0,40)

c -2 , 0); -1 , 0); 0 , -6)

d -6 ,0) ; 2 ,0) ; 0 , -60)

a y = x - 8) x + 1)Y

( 1 0) (8,0}

X

(-0, - 8}

b y = x - 3)(x - 5)

0, 15)

(3,0} (5,0}0 X

c y = 2 x + l ) x - 2.5)y

(0,5)

(-1, 0} (2.5,0)0 X

d y = 5 x + 2 x - ;

Y

( 2, 0) :.o)0 X

(0, -8

Answers

3 a y = x + 3) 2 - 25;

y = x + 8) x - 2)

Y

( 8 , 0) (2, 0)

0 X

0, -16)

(-3, -2 5)

b y = - x + 2) 2 + 25;

y = - x + 7) x- 3)

y(-2,25)

(0,21}

( 7 0) (3,0)

0 X

c y = - 0 .5 x - 3.5) 2 + 3.125;

y = - 0 x - 1) x - 6)

Y

3.5, 3.125)

0 X

(0, -3

d y = 4 x - 2.25) 2 - 12.25;y = 4 x - 0.5) x- 4)

Y

(0,8)

(0.5,0}0

4 a i 0

b x= 3

(4,0}

X

(2.25, - 12.25}

ii 6

c (3, - 1 8)

5 a f o g) x) = x 2) 2 + 3

b (2, 3)

c h x) = x 2 - 14x + 50

d 50

Exercise 2J

1 y = x 2 - 4x + 52 y = x 2 - 4 x 123 y = 3 x 2 - 6x + 5

y = 2.x 2 -32 2

5 y = 2x 2 + 7x + 46 y = -0 .4x 2 + 8x7 y = - x 2 + 4x + 218 y =12x 2 - 12x + 3

Exercise K

1 a 14.5 metres

b 1.42 seconds

2 14 em, 18 em

3 a 1 0 x

c 50 cm 2

4 12.1 em

5 17m, 46 m

6 7, 9, 11

l + J572

8 28.125m 2

9 60 km, 70 h- 2

10 6 hours

Review exerci se non GDC

1 a -6, 2

b 8

7c - - 13

d 3, 4

e - 1±Ji3

f 7±J136

2 a 4

b - 4, 1c x =-1 .5

d -1 . 5

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3 a 5 1 1 11 92 ba -b - 2 5 32 32

4 -3 , - 6 3•

0.21••

0.33a I II

1 b 2525 a A = {3, 6, 9, 12, 15}

b -2 4 a 0.27 B = {1, 2, 3, 5, 6, 10, 15}

c 12 b No - the frequencies are4 7 8 1113 14

very different

5J3 A

c 4506 a f x ) = 2 x + 3 2 - 13 2 6 9 12 3 15 12 5 105 a b 0

b (1 - 5 11

y =I .x - x - 12c 5

7 1126 0.2 • 1 •• 2

Review exercise GD c I II -5 51 137 a b1 0 .907, 2.57 -a 240

b -4.35, 0.345 6 Ac - 2.58, 0.581 Exercise 8

d -1.82, 0.220 133 24

bl br

2 a 2 0 m

b 3 1 5 m 6 4 10

c 3.06 s 550

d 4.07 s 15 c3 21, 68 4

4 a= 0.4, b = 3, c = 2 7a 0.33 b 0.24

5 60 km h- 1 2 c 0.3Fr m

Chapter7

85 Exercise C

Skills check 151a

5250

1 4 b 1_i_a -7 35 8 b

53

c 2 d 22 25100

15 27 3 299A G c

19 3 500e f -27 7 27 11 2 a -

2 a 0.625 b 0.7 5

c 0.42 d 0.16 b 3-1 5e 15 f 4.84 Six have both activities. 1c -g0.0096

6 112

a b -25 25 3 17-

Exercise A 4 20

1G p

1 b 1 4 9a - 4 a b2 4 13 2611 7 9

1 3c d - 2 14 4 c d -3 5

13 2e -

8 Five play neither. 5 a 0.5 b 0.5

nswer s

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660

7 a 14

8 a 0.6

c 0.9

Exercise 30

1

2

3

a Nc Ne Ng Nyes57

89

4 a 212

c 13

60Exercise 3E

b 34

b 0.4

b y

d y

f N

b 4760

1 HH H , HHT, HTH, HTT, THH,THT, TTH, TTT

a b 32 8

c . :.4

2BLUE

1 2 3 4

1 (1, 1) (1, 2) (1 , 3) (1, 4)RED 2 (2, 1) (2, 2) (2 3) (2,4)

3 (3 , 1) (3, 2) (3 3) (3, 4)

4 (4, 1) (4, 2) (4, 3) (4, 4)

3 b 3a - -8 81 d 9c -4 16

3

Box 1

1 2 3

2 (2, 1) (2, 2) (2, 3)

Box 2 3 (3, 1) (3, 2) (3, 3)

4 (4 , 1) (4, 2) (4, 3)

5 (5, 1) (5, 2) (5, 3)

1 b1

a - -6 33 d 5c4 12

e 2-3

Answers

4 First draw

Second

draw

a 16

13c -

18

5e -9

5 a 16

c 29

1

2

3

4

5

Exercise 3F

1

2

125

2169

3 64125

4 0.6375

0 1

(0, 0) (0, 1)

(1, 0) (1, 1)

(2 0) (2, 1)

(3 0) (3, 1)

(4, 0) (4, 1)

(5 , 0) (5, 1)

b 2336

d 1336

b 19

2

(0, 2)

(1, 2)

(2 2)

(3 2)

(4, 2)

(5, 2)

5 a P(B)=0 .2 ;P(BnC)=0 .16b Not independent

6

7

512

1

59049

8 1256

9 a 0.4b P(E) x P(F) = P E F )

c P E F ) ;e 0

d 0.6410 _

27

2712 a 0.27

c 0.07

13 0.18, 0.28

14 a 11296

b 0.63

b 1216

3 4 5

(0, 3) (0, 4) (0, 5)

(1, 3) (1, 4) (1, 5)

(2 , 3) (2 4) (2 5)

(3, 3) (3 4) (3 5)

(4, 3) (4 , 4) (4, 5)

(5 3) (5 , 4) (5 5)

15 Rolling a six on four throwsof one dice

16 a 0.729 b 0.271

Exercise 3G

1 12 take both subjects

a8

27

4c5

2 a 0.2

c 215

3 3948

4 1a -3

c 3- >

56-95

6 1-6

7 a 0

c 0.63

8 67.3

9 3447

10 a 1-1

c13

11 0.3

123

b 2327

b3

b2-5

d1-2

b 0

b 43s

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ChapterSkills check

1 a 1 b 81128 256

c 1 X 10- 9

2a 5 b 3 c 4

3 y

4

Y= x2y= x - 2) 2

-4 -2 0 2 4 6 X

Inve stigation - folding paper

Number Number Thickness As thick

of folds of layers km) as a

0 1 1 X 10 -7Piece of

paper

1 2 2 X 10 - 7

2 4 4 X 10· 7Cred itcard

3 8 8 X 10 - 7

4 16 1.6 X 10 6

5 32 3 .2 X 1Q-6

6 64 6.4 X 10 -s

7 128 1.28 X 10 · 5 Textbook

8 256 2.56 X 1 0-5

9 512 5. 12 X 1 0- 5

3 a 13 fol ds

b 15 fo lds

4 113 000 OOO km

Exercise 4A

1 a x

b 6p6q2

c 1 3 3x y3

d x4y6

2 a x3

b a4

ac

4

d 2x y3

3 a x l b 27 t 6

c 3x6y4 d y 6

Answers

Exercise 4 8

1 a 3

d 4

2 a 1-8

d 116

Exercise 4C

1 a

d

2 a

8a 3

d

3c

o .

a•

b

Exercise 4

1 a x= 5

b

e

b

e

b

e

c x = 3 -1

e x = 3

25

a x= -2

c - 3X - -5

3 x= -6

Exercise E

1 a x = 3 b

1

5 c 16

4-9

I 1- c4 3

25 - ) 916 16

2 c q3x l

- 4-p .3

4

b x= - 2

d3

x

2

b x = - 4

d4

x=

5

x = 2 1c x = -4

I 3-d X= 2 e X= 3 3 f x = -42 a x = 8 b x= 625

1c x= d x = 64

2561

X= 32 f x = -16

13

1 bx= x =3125 216

27X= 512 d x= -64

Investigation - graphs ofexponential functions 1

yy = wx y = sx

0, 1)

0 X

Investigation - graphs ofexponential functions 2

(1 )Xy= -3

(0, 1)

Investigation - compoundinterest

Half-2.25

yearly

Quarterly (1• 1 2.44 1 4062512

Mo nth ly

(1• 12.61 303529022 ..

Week ly r1-r2 2. 692 596 954 44 ..31)5

Daily (1•-1l 2. 714567 48202 .365Hourly

1 . , •1•- ]760 2. 718126 690 63 ..

Every[l- 516 ) 2.7182 79215 4 ..min ute

Eve ry . 3 1536000

ll•1 2.71 8282 4 7254 ..

second 31536000

Exercise 4F

1 Curves of

a y10

8

6g x) = 2x + 3

2 f X) = 2x

-3 -2 -1 0 1 2 3 X

b y5

4

g x) = 3 x 3 f x) = 3x

-3 -2 -1 0 1 2 3 X

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3 a 2 b 3 c 2d 3 e 1

Exercise 4N

1 a p + q b 3p c q - p

2

dq

2e 2q - P

2

6 x - 3y - 6z3 a 1 log x b 2 - 2Iogx

1 1 1c - + - l o a x d - 1- l ogx2 2 b 2

4 y = 3a - 4 5 -3 - 2log 3x

Exercise 4

1 a 2.81 b -1.21 c 0 .325

d 0.514 e 12.4

2 2:2

3 a YX

d 2x+ y

4 a2

1

b Xy

e x+yy

logxY= log4

c 2yX

f y - xX

- 1 0 1 2 3 4 X- 1

-2

b

-4 -3 -2

5 a 2b

y2

1

bb -

2

c - 2b d

Exercise 4P

b

4

logx2

log5

3 4 X

1 a 2.32 b 3.56 c -1.76d 0.425 e 0.229 f -3.64g 1.79 h -11 .0

2 a 6.78 b 2.36c -3 .88 d 0.263e 0.526 f 2.04

g - 99 9

Answers

Exercise4Q

1 a 1.16 b 1.41 c -0.314

d 0 .0570 e 11.1

ln5002 a

ln64

1n3

3 0 b Xln3a x=

Exercise4R

I1 a x = b x = 1

5

ln2

c x= d x = J 27

e x = 1.62

Exercise45

1 a x = 8 3 b x = 14

C X 9532

2 a x = 9 b x= 6c no solutions

3 A = x 2x + 7 = 2x2 + 7xX 0.5

4 x = 4

5X

16Exercise 4T

1 a 450 x 1.032

b 10 years

2 a i 121 ii 195

b 9.6 days (10 days)

3 49.4 hours

4 a v8

20v = 9 + 29e - 0.063t

-3 -2 - 1 0 1 2 3 t

b 38ms- 1

d 10. 7 m s - 1

5 a = b = 3

e 17

Review exercise GD

1 3.52

2 a 0.548 b 0 .954

c - 1 .183 a 5 b 2

c 3.60 d 1,4e I 00, 1

1004 a f x) > 0, range of g(x) is all

real numbers

b They are 1- 1 functions ;

f c ) I l n . I 2 xx =2 x ,g - x )=e C f o g ) x ) = x 3;

g o f ) x ) = 3 x

d x = J 3

5 a 218 393 insects

b 8.66 days

Review exercise non·GD

1 0

2log( )

loge:3 4.5

5 a x= 7

C X 1,4

n6 a m

c 2m

b x = 2

d x = 67

b m - n

d m + nn

7 Shift one unit to the right,1

stretch factor - parallel to3

x-axis, shift 2 units up.

8 a

b- 1f 1 x)= - 1ogx

3

c

9 a = 2 b = 4

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Chapter 5Skills check1 a -8x+ 20 b 12x -1 8

c -.x3 7 x

2

d x 4 + 6x 3 + 9x 2

e x 3 + 5x 2 - 24xy

- - 2 6- x•O 3•)(

y = 4....

2-y•O

T ,.-3 -Q - 1 0 1 2 4 X

-2

y - - 3 -4

3 A is a h orizo ntal shift of 4units to the right. Function Aisy = x - 4) 3B is a vertical shift of 2 units

down . Function B is y = .x3- 2Investigation graphingproductpairs

X 24 12 8 3 6 4 2 1y 1 2 3 8 4 6 2 24

y

24

18xy= 24

12

6

-2 O 2 4 6 X-6

- 8

-2

As y gets bigger, x gets smalJerand vice versa.The graph getscloser and closerto the axes as x- andy -values.mcrease.

Exercise SAl1 a2

d - 1

2g

3

2 a

d

2

13

I

3x

b . .3

e 32

h7

b 1

e

X

1

4y

c 13

f7

1c -

f

y

92x

g 5 h 3d • tIa 2 d

x lJ

x l

3 a 16 x - = 1 b 3 4- x - = 1

45

6 4 3

c3d 2c

a 4 b X•

0 .5..

a I II 0.05...

0.005•

Ill IV 0.0005b y gets sm a ller, nearer to

zero.

c24 .

y so 1t can never beX

zero.

d • 0.5..

I II 0.05

iii 0.005.

0.0005V

e x gets sma ller neare r to zero.f 24 .x = - so 1t can neverbe

y

zero.

Investigation graphs ofreciprocal functions1 a y

2

6

4

2

2 4 X

The nu m erator indicates thescale factor of the stretchparallelto they-axis.

y

-6 -4 -2 0-2

-4

-6

Changing the sign of thenumerator reflects the graphsof theorigi na l functions inthe x-axis.

3 aX 0.25 0.4 0.5 1 2 4 8 10 16

f x} 16 10 8 4 2 1 0.5 0.4 0.25

I

b T he values of xandf(x) are the samenumbers but inreverseorder.

c d e

'-14 /- - ' 1- -

//

i7l/

... \.............

i 1 0 1 2 1 1i>Xf The function reflects onto

itself.

g The function is its own.mverse.

Exercise 8

1 y10-I

18- 5II 6- y • X4- \ I2' .........

10 :-:8 -I, 0 2 6 xp - 10__. -2-

II -.. 8\

:_j_ - 10

Y

8-y •- I· 6- X -1-

4- \ i-2- -.....

•• -2 0 4 lOX0 -:S- u 2 6-2 I.._ •A

T

( llCAnswers

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d y=O x .=-1Domain x E IR x t - 1

Range y E IR y t 0

e y = 2 x = -1Domain x E R x t - 1

Range y E lR y t 2

f y= - 2 x = 1

Domain x E x = - 1

Range y E IR , y t - 2

g y = 2 X = 3Domain x E IR x t 3Range y E IR , y t 2

h y = -2 X = -3Domain x E R x t - 3

Range y E IR y t - 2

ay

86

4

2

4y= -

x

2 4 6 8 X

Domain x E IR x 0RangeyE R y t 0

b y8

36y = X - 3

4

2

-8 -6 -4 - 2 2 4 6 8 x-2- 4

-6

Domain x E IR x :t; 3

Range y E R y 0

c

- 10 -8 -

-4y = - 8x 5

y

4

-4 -2 0-4

-8

- 12

-14

Domain x E R x 5Rangey E IR y = - 8

2 X

d y

8 16 y - 7 3

4

2

-2 0 2 4 6 8 10 12 X-2

Domain x E R x :t; 7

Range y E R y 3

e y

-12 -8 -4 x-4

6 -8y= - 6x 2

Domain x E R x 2Range v E JR v :t: - 6

f y8

6

4

-6 -4 -2 0

5y= - 4X

2 4 6 X

Domain XE R x = 0

Range y E IR y:t:

4g 1y - - 2x 12

y

1

-5 - 4 - - 2 -1 Q 1 X- 1

- 3

- 4

Domain x E R x = -3Range y E IR y :t: - 2

h y6

43

2y

X

- 6 -4 - 2 4 6 X

-4

- 6

Domain x E R x 0Range y E IR y :t; 0

3

•I y

10

86

42

-8 -6 -4 -2 °2-4-6

4 6 8 X

= 4 5Y 3x - 6

Domain xE IR x t 2

Range y E IR y 5

t

2.52

1.51

0.5

-20 - 10 0 10 20 30 40 50 c

b 3.9°C

4 a y86

4

- 10 -8 -6 - -2 O 2 4 6 8 10 X-2-4

-6

The linear function is a line ofsymmetry for the rational

function. The l inear function

crosses the x-axis at the sameplace as the verticalasymptote.

1y =X 1

y

0 2 4 6 8 10 X

The linear function is a line ofsymmetry for the rational

function. The linear function

crosses the x-a xi s at the same

place as the vertical asymptote ofthe rational function.

n sw rs

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Investigation - graphingrational functions 2

a y432 Xy = x 31

-10 -8 -6 -4 - 2 4 6 8 10 X1-2-3

y4

321

x ly= x 3

2 4 6 8 10 X-10 -8 -6 -4 --1-2-3

y86

4

2

4

-6

y8

642

2 4 6 8 lOX

x - 1Y = x 3

-10 -8 -6 - 4 -2 2 4 6 8 10 X

b

Rational Verticalfunction asymptote

XX= -3Y = x 3

x 1X = - 3y = X 3

2 xY = x 3 X= - 3

2x-1Y = X= -3

x 3

-4

-6

Horizontalasymptote

y = 1

y = 1

y = 2

y = 2

Domain Range

X E JR ., y E I {,X - 3 y;;; 1

X E JR, y e R,X ;t; - 3 y :1

X EJR y E R,X t; - 3 y :2

x e lR, y E I(,X ;t; - 3 y;o2

c The horizontal asymptoteis the quotient ofthex-coefficients.

Answers

d The domain excludes thex-value of the verticalasymptote.

Exercise 5

1 a y = 1, x = 3Domain x E IR x :t. 3

3

Range y E IR, y 12 1

b y = 3 x = 3

D . 1omam x e R x :t. -2 3

Range y e IR ,y :t. -3

3 5C y = - ,X = - -

4 4

Domain x E R ,x - 24

3

Range y e R , y -417 1

d y s x - 4

a

a

b

Domain x E IR x _ 4

17Range y e IR, y -

8••• b i • d iill c IV

y

6x + 2

4 y = x 3

2

- 10 -8 -6 -4 -2 0-2

2 4 X

-4-6

Domain x E R , x :t. - 3

Range y E lR, y 1

y

XY = 4x 3

-6 -4 -2 2 4 6 8 X

Domain x E IR, x4

1Range y E y =1 -

4

c

d

e

f

y4

321

-15- 10 -5 0- 1-2

-3

x - 7y = 3 x - 8

10 15 20 X

Domain x E x :;t:3

1Range y E R , y -

3

y

86

4

9x 1y=3 x - 2

-20-15-10 - 50 5 10 15 20 X-2-4

Domain x E IR,x ::;t:3

Range y E lR , y 3

y

4

2

-3x 10y= 4x- 12

-6

Domain x E R , x ::;t:3

3Range y E R , y ::;t:.- -. 4

y4

35x 2

y =4x

2

X

-8 -6 -4 -2 2 4 6 8 X-1

-2

-3

-4

Domain x E IR,x 0

5Range y E R , y ::;t: -

4

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b

c

d

4 a

b

x = O , y = - 3

Domain x E R, x :t 0Range y E JR, y ::t -3X= - 6 y = - 2Domain xE JR, x ::t - 6

Ra n ge y E JR, y ::t - 2

x = l , y = 5

Domain x E R, x t 1Range y E JR, y :t 5

2

2

1

1

c =

c40 -

00 -

60 -

20 -

80 -

40 -

300

s

-300c - s

\ -

.

0 5 10 15 20 25 30 5

c The domain and rangeare limited to JR+ and thedomain to z

5 a. 2I y = - = 2

ljj X= - 2

iii ( -2 , 2)

b] 1

(0, - 2 ) (2, 0)

cy

1:)( )

I I

u

= k 1c X l LJ A

/ • I I.../

8 -6I

2 p 2I

64 - 4I

x- -2I I

Review exe rcise GD

1 a yf IX)

l 1 2· - i

8 - 2 0 \ 8 X- - - 2-

f 1 -4- rr--- t----- ·-- ·-- -- r:--1---r- - - d- - t - -6 6-1--... f X = - 5f \ )(

-I u-io-

Domain x E JR., x t 0

Range y E JR.,y t 5

Answers

j

8'

b

c

d

X

e

y( )

t::..

A\ _ f x ,

- -r- --r--8 -6 -tl. '·2,., ) X-

A

a 2u 1==.

( ) l X I

Dom ain X E JR, X ::f. 0

Range y E JR, y :t 3

y0

af x

u r.t-

II

f xA I - 5I•,., ill...

/ :\4 0 2 4: Vl

IA

I

c

0 __l -

Domain x E R, x t 5

Range y E JR, y t 0

y....

f x) rI

110.

-

0I

p 1: 1.... II

A If x) = 8 1

X 7 Ia 1\.v

I 'II1 \ I... v \ I..... I......

Domain x E JR, x ::f. 7

Range y E JR., y t - 8

y.l2

j.6

f( )iII \ -l > .I - ......I

e.. -6 -g 0 21 2I • ft.

\ I f( ) ..u

I It+

:D I

Domain x E R, x t - 3Range y E JR, y ::f. 0

,12 X

X

f ycv

A

II I ,.,

/...

..v ·ru -8 -6 - 4 Q_.? 2. t::. ,...

f(x)- x 4- 2 IL G I

I0Domain x E JR, x :t - 4

Range y E JR., y t - 2

2 a Using the equation

S ddistance

pee = .orne

X

5600distance= 5600, s

t

b s km h-1)

1200 f

1000

800 5600s= -

600 t

400

200

0 4 8 12 16 20

t hours)

c 560kmh - 1

3 a

m minutes)

300

250

200

150

100

22.2s + 14.28m =s

50

0 20 40 60 80 100120 s

b • 165 min..

57.9 minI...

36.5 minll

c m = 22.2

d The number of minutesthat can be spent in directsunlightwithout skindamage on a day whens = 1.

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4 a cX 106

x10 6750 OO m100 -m

x1Q6 v-x106

/

2

1

0 20 40 60 80 100

b i 187 500 Thai bahtii 750 000 Thai baht

m ( )

iii 6750 000 Thai baht.

c No. When m = 100, thefunction is undefined.

5 ay f ) 2 + 1X =

2x - 5A

:.., Io.J I --...... I

I I 2'

J

-I

I X = 2.p2 -1 0 I

,

'I

X

5b •I x = 2•Y = 2

•• 2.25I•• • 1.8ll

Chapter 6Skills check

1 a - 6 b - 3, 5

2 a k =1 5 - 3m4

c 5

bp

3 a 108 b - 12.22

4 a 5 b 163

c - -32

Investiga tion saving money

a Week Weekly Totalnumber savings •savrngs

1 20 20

2 25 45

3 30 75

4 35 110

5 40 150

6 45 195

7 50 345

8 55 300

b Sa vings in 1Oth week: $65;Savings in 17th week: $100

c Total saved in 1st year (52weeks): $7670

d 1 000 saved afte r 17 weeks. Exercise 6C

e M=20+5 n - l ) o r M = 1 5 + 5 n 1 d =0 .9

f T = n (35 + Sn) or T = Sn (7 + n) 2 d = - 3 u = 64' 12 2 35.5

Exercise 6A

1 a 19,23,27 b 16,32, 64

c 18,24,31 d 80 , - 160,320e 9 11 13

14 j 17 20

f 6. 01234 , 6.012345,6.012 345 6

2 a 10, 30,90, 270b 3, 7, 15, 31

c3 1 1 2

3 a u = 2 and u = u + 2I n +l nb u = 1 and u + 1= 3u11 n

uc u1 = 64 and u ,, 1 = ;:

d u 1 = 7 a n d u n 1=un+54 a 3, 9,27, 81.

b -3 -9 - 15 -2 1' ' '

c 1,2 ,4 , 8

d 1,4,27, 2565 a u = 2n b u = 3n-l

n n

c u = 2 7- ll d u = 5n + 2n n

n

f u = nxu = n + 1 11

6 a 610b u 1 = 1, u2 = 1, and

u = u + un+ l n n-1

Exercise 68

1 a i u 5 = 45

b i u 15 =235

••11 u = 3n

ii u =15n + 10nc i u 15 = 106

ii u = 5n + 31nd i u = - 8215

ii u = 113 - 13nne i u 15 = 14

ii u = 0.6n + 5nf i u 5 = x + 14a

••11 u = x +an - a

2 a 51d 15

17

b 169

e 27

n

c 37f 10

4 8

Exercise 6

1 11 a r = u =2 7 4

b r = - 3 u = - 2 916' 7

c r = 10, u 7 = 1 000 000

d r= 0.4, u7 = 0 .1 024

e r= 3x u = 1458x 6' 7

f r = .. u = ab 7a ' 7

Exercise 6E

1 r = 0.4, u 1 = 125

2 r = 2, u = 4 .53 a n = 12 b n = 9c n = 7 d n = 33

4 r = +4, u2 = +36

5 p = +276 x = 8

Exercise 6F8

1 a L nn=l

6 6

c I , (29-2n) d I,240(0.5 '-1 )n= JI ' J l

I() I 8

e I , an f I , (3n +1)1=5

I I;

I , 3 L h I na u=-1 n=J

g

2 a 4 + 7 + 1 0 + 1 3 + 1 6 + 19+ 22 + 25

b 4 + 16 + 64 + 256 + 1024

c 40 + 80 + 160 + 320 + 640

d x5

+ x6

+ x7

+ x8

+ x9

+x 1o + x l l

3 a 315 b 363c 140 d 315

Exercise 6 6

1 234

2 108

3 594

4 40 x + 152

Answers

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5 a n = 246 2292

b 1776

Exercise 6H

1 3

2 a3 a

3n 2 - 2n b 17

1.7 5n 2 - 31. 75n b 21

4 a5 a

1600 b 12600n = 24 b S

24= 1776

6 d = 2.5, S 20 = 575

Exercise 6

1 a 132 86 0 b 1228.5c 42. 65 625d 4095x + 4095

2 a 435 848 050b = 11 81 9 .58

c - 1048 575d 1og(a 1048575

3 a ib i

•C I

d i

9

6g

11

Exercise 6J

1 a 6 b 5

II

II

II

c 19 d 659048

2 r = 3 S =3 r= 3

15

4 a 1 5 b 215 20596 3

76684

3685.5

1.626 5375

885.73

Investigation converging•senes

1 i 1a r -22b r =5

- 1

c r = 4.Inspect va l ues on GDC

2 a T he va lues are approachi ng4 as n 7 oo

Exercise K

1 I rl < 1-2 a S 4 = 213.3,S 7 z 2 15.9,

an d S = 216

b S4 = 1476, S7 = 1975.712,an d S = 2500

c S 4 = 88 .88, S 7 = 88.888 88,and Soo = 88 .8

-d S4 = 10.83,S 7 z l2 .71 ,

and S = 13.5-

3 13.44 192

5 16 or 48

6 150

7 4118

Exercise 6L

1 - 20

2 a 26.25 em3 a 3984.62

c 4035.364 425 18

6 2327 ::=:19 6 years8 a 1,8 ,21

c 6n - 59 a 4, 12, 28

c 4(2''- 1)10 z 86 months

11 About 16. 30

Exercise6M

1 10

2 283 354 84

5 156 120

b 119b 4025. 81

b 1, 7,13

b 4, 8, 16

Investigation patterns inpolynomials

1 a + b

2c T he va lues are approachi ng

19 2 as n 7 oo3

b T he va lues are approachi ng125 as n 7 oo

a2 + 2ab + b2

a 3 + 3a 2 b + 3ab 2 + b3

. ( 1 )so3 Results like 1-2

. are

beyond the limit of the display.

Answers

4

5

a4 + 4a 3 b + 6a 2b2 + 4ab 3 + b4

a 5 + 5a 4 b + 10a 3b2 + 10a 2b3 +5ab 4 + b5

6 a 6 + 6a 5b + 15a 4 b2 + 20a 3 b3 +15a 2 b4 + 6ab 5 + b6

The co efficients are from Pascal'striang le.

a+ b f = a7 + 7a 6 b + 21a 5 b2 +35a 4 b3 + 35a 3b4 + 21 a2b5

+ 7ab 6 b7

Exercise 6N

1 y 5 + 15y 4 + 90y 3 + 270y 2

+ 40 5y + 243

2 16b 4 - 32b 3 + 24b 2 - 8b + 1

3 729a 6 2916a 5 + 48 6 0a 4

+432 0a 3 576a 64

5 X 8 + 8x 7 y + 28 x 6 y 2 + 56x 5 y 3

+ 70x4

Y4

+ 56x3

y5

+ 28 x2

y6

8x y 1+ ys

6 8Ja• - 2 16a 3 b 216a 2 b2

- 96ab 3 +16b 4

7 243 cs + 810c4

+ 1080c3

+ 720c2

d d 2 d 3

240c 32+ d d s

Exercise6

1 336x 5

2 - 1280y 4

3 4860a 2 b4

4 - 5125 26 +4

7 179208 48 6 0

9 810 7

Review exercise non GDC

1 a 42 a 1-

43 a 44 a 305 120

6 a 1-4

7 +4

b 283

b 1

b 5b 262

b 32003

c 25c 256

3

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8 720.x3

9 a 17 b 323

Review exercise GDC

1 a 3 b 522 a 96 b 323 a u = 7 d = 2

Lb 720

4 a 2 b 115 186 u = 5 r = -3

1

7 - 945x•16

81-4

9 a 5.47 million

Chapter 7Skills check

b

1 a 3x(3x 3 - 5x 2 + 1)b (2x - 3) 2x + 3)c (x - 3) x - 2)

d (2x + 1 )(x - 5)

2 a x 3 + 6x 2 + 12x + 8b 81x 4 - 108x 3 + 54 x 2

- l2x + 1

2056

c 8x 3 + 36 x 2y + 54xy 2 + 27 y 3

3 a x-6 b 4x - 3I 5- d -c 5x 2 X ;3--e 7x 2

Investigation - creating asequence

Portion of the paper

Round you have at the endnumber of the round Fraction

Decimal 3 sf)

1 1- 0.333

3

24- 0.4449

313

0.48127

440- 0.49481

5121

0.498243

6364

0.499729

1 The portion gets closer to .

2 The portion gets closer and

closer to I , yet never reaches . ..2 2

Exercise 7A

1 Divergent

2 Conve rgent; 3.5

3 C on vergent; 0

4 Convergent; 0. 75

5 Divergent

Exercise 78

1 10

2 1

3 1

4 Does not exist

5 4

6 Does not exist

Investigation - secant andtangent lines

1 y

2

X

Point LineGradient

p

A

8

cD

EF

3 04

Coordinates

0 , 1)

- 1.5, 3.25)

-1 , 2)

-0 .5 , 1 .25)

(0.5, 1.25)

1, 2)(1.5, 3.25)

y

or slope

- -

AP - 1.5

BP - 1

P - 0 . 5

DP 0.5

EP 1FP 1.5

X

Exercise 7C

1[3( x +h)+4]- (3x +4) =

3h

2[2(x +h) 2 - I ] - 2x 2 - 1

= 4 x + 2hh

3[ (x +h) 2 +2(x +h)+3

J(x 2 +2x + 3

h= 2 x + h + 2

Exercise 70

1 2; m = 2

2 6x + 2 · m = - 16

3 2x - J· m = 1

ln v .estigation - the derivativeo f f(x)

=1 f ( x ) = x 2f (x) =lim (x + h) z - x z

IH O h

= l im(2x+h)h -+ 0

=2x

f ( x ) = X 3

( ) 1. ( x + h) ' - x3

X = 1mh - . 0 h

= lim(3x 2 + 3xh + h 2 )h - t O

f ( x ) = X 4

f ' (x ) = lim x + h) 4 - x•,, .. o h

= l im(4x 3 + 6x 2 h+ 4xh 2 + h 3 )h -+ 0

2 To find the derivative o ff (x ) = x", multiply x bytheexponent n and subtract onefrom the exponent to get the

new exponent. If f (x) = x",thenf ' x) = x n - l

3 Prediction: f x) = 5x 4

f ( x ) = x 5

f ( x ) = lim (x + h i - xsh-+0 h

= lim 5x 4 + 1Ox 3h + 1Ox 2h2h-+0

+5xh 3 + h 4 )

Answers

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Exercise 7E

1 5x 4

2 8x 7

43 ·x s

1 12 or fx33 23x 3 . x

4

5 1 1- or - ; : : =2Fxl

2x 2

6

Exercise F

161 - -x92 0

4 5n x 4

5 2x - 8

6l 4

7 3-2x 3

83

-8 i 3

5 310 - , + - ,- -

6x 0 4x 4

11 12 x 3 - 4x

12 4x + 32 2

13 . _ + .3x 3 3x 3

14 6x 2 - 12x

15 3x 2 + 4x - 3

Exercise 7G

1 y + 3 = 2(x - 3);I

y + 3 = - - x - 3)2

y1y+3= - -- x-3)2 1

Answers

f(x) = x2 4x

2 a y - 4 = -4 (x + 3)

b 6 = l ( x - 1)

1c y - 5 = - ( x - 3 )

315

d y - 9= - - (x -1 )4

13 a y - 3 = - - (x -2 )

71b y+5= - (x + I)6

c y - 25 = (x - 2)20

d y + 2=-2_ (x - I)26

4 x = l · x = - 1

5 5

Investigation the derivativesof ex and In x

1 Co nj ecture: f'(x) = e

2 Co nj ecture: f ' (x ) =. .

Exercise H

14

X

12 e + - 1

2x 2

X

4 8x + 315 2e + -X

6 5e-" + 4

7 y - 5 = 12(x- In 3)

8 y-9=. . . (x +3)6

19 y - 1= - (x-e)e

10 y-7=- . . . (x -2)9

11 2e 3 ; 40.25

12 - ; 0.20824

X

Investigation the derivativeof the product of twofunctions

11

f (x) = 11x 10

u'(x) = 4x 3 ; v' (x) = 7x 6

u' (x) · v'(x) = 28 x 9

5

6

7

8

No

f ' (x ) = x 4 · 7x 6 +x 7 · 4x 3 = l l x 10

f ' (x ) = u(x) · v'(x) +v(x)·u'(x)

f(x) = (3x + l)(x 2 - 1)

= 3x 3 + x 2 - 3x - 1

f (x) = 9x 2 + 2 x - 3

f (x ) = (3x + 1)(x 2 - 1)f x) = (3x+ 1)(2x) + (x 2 -1) 3)

= 6x 2 + 2x + 3x 2 - 3

= 9x 2 + 2x - 3

This supports the co nj ecture.

Exercise 7

1x - 4 2

2 l 0x 4 + 4x 3 +9x 2 + 2 x + I

31- 1nx

x

ex4 - e l n x

X

65 (x + 4) 2

ex

9 -1

Exercise 7J

1 2x 2 -5

3

2 4 x 3

3 4xe- + 2 x 2e

2xe - 4e4

5 32 16

X - 9

ex

2x7

x2 + Y

8 3 + 31nx19 1 - -

x 2

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10

11X+ I

(x - 1)1

12 10x 4 + 12x 3 - 3x 2 - 18x - 15

I13 y = - - (x -1)

e14 y X ]

15 - 9n + 3.5

16 4n r 2

17 718 4

Investigation findingthe derivative of acomposite function

1 a I (x) = 2 - x) 3= 8 - 12x + 6x 2 - x 3

l ( x ) = 1 2 + 1 2 x - 3x 2b l ( x ) = 3 ( 2 - x)l · ( -1 )

2 a I (x) = (2x + 1 2

= 4x 2 + 4x + 1

l ( x ) = 8x + 4b I (x) = 2(2xl)· 2

3 a l ( x ) = (3x 2 + 1)2= 9x 4 + 6x 2 + 1

I x) = 36 x 3 + 12xb I x) = 2(3x 2 + 1) · (6x)

4 The derivative o f a compositefunction is the derivative oftheoutside function with

respect to tbe inside functionmultiplied by the derivativeof theinside function.

5 f(x) = x4 + x2)3= x 12 + 3x O + 3xs + x6

f x)

=2x 11 + 30 x 9 + 24x 7 + 6x 5

l ( x ) = 3(x 4 + x 2) 2 · (4x 3 +2x)

=3( x 8 + 2x 6 + x 4) ( 4x 3 + 2x)= (4x + 10 x 9 + 8x 7 + 2x 5)=12x 11 + 30 x 9 + 24 x 7 +6x 5

Exercise 7K

1 x 5 · 3x 4 + 2x·I5(3x 4 + 2x) 4 (12x 3 + 2)

2 4x 3• 2x 2 + 3x + I·' .12(2x 2 + 3x + 1 )2 (4x + 3)

3

4

5

6

7

8

9

10

lnx; 3x ' ;X

2x+3;2

,•-

3(2x +3)1

3(1nx)2

x 1 ; In x;X

. 63 • 9x + 2· - - - - - : -

' I I-(9x + 2 3

e · 4 x ' · 12x ,e' '

Exercise L1 8x 2 (2x - 3) 3 + 2x(2x- 3) 4

or 6x(2x - l )( 2x - 3) 3

2

3

4

5

6

7

8

9

10

e'

-8x(x 2 + 3)2

-x 1 x+l+ ' or -3- -(2x + I ) (2x + 1)2 (2x + 1 2

6x 2

2x l

I

xlnx

- 2(e' - e ) - 2e ' (eh - I )or ,1(e' +e ' )2 (e · +I)

-2x+3

I 1

x s (x 2 +3 ) 2 +4.x3(x2 + 3) 25x 5 + 12x.1

or - - - - - - ,1,..-

(x 2 +3 ) 2

'11 a (2x-2)e ' 2 'b 2

c y - I = 2(x - 2)I-12 e

13 h (x) =6

. Since( l - 2x)

6 > 0 and(l - 2x) 4 > 0 forall x where his defined,thegradient of h is alwayspositive.

14 a 6b 8

Exercise 7M

31

2

3

4

5

6

7

8

x

3e J , (6n + 5)

8

-3

1

equals 0

d y = e · - e •dxdl_::...Y = e• +e rdx 1

d)_::... = e· - e 'dx 3

d4_::...Y = e' +e · ·dx 4

When n is odd

dY = e ' - e ' and when

dx .

n 1s even

d_::... = e ' +e ' .dx

9 dy -1

10

-

= -

•y {- l )"n

dx

-1 8

- · ·

Answers

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Exercise 7N

1 ab

c

2 ab

c

d

1 .4m;2 1 m

9 .8ms -

9.8ms - ; Om s - ; - 9.8ms - 1;The ball is moving upwardat 1 sat rest at 2 s anddownward at 3 s.

4000 litres; 1778 litres

- l l l l i t res/min; D uringthe time interval 0 to20 minutes,water is beingpumped out of the tank at

an average rate of111 litres per minute .

- 89 l i tres/min; at20 minutes, water is beingpumped out of thetank atan average rate of 89 litresper minute.

V (t) is negative for0 < t < 40 minutes, whichmeans wateris flowingout of the tank during thist ime interval. Thereforetheamount of water in the

tank is never increasing

from t = 0 minutes tot = 40 minutes .

3 a 112 bacteria/dayb P' (t) = 25e 0 ·251

c 305 bacteria/ day; on day10 the number of bacteriaareincreasing at a rate of305 bacteria/ da y .

4 a 20.25 dollars/unit;20.05 dollars/unit

b C (n) = O. ln + 10

c 20 dollars/unit ; t costs20 dollars per un it to

produce units after thelOOth unit.

Exercise 70

1 a Ocm· 9cms - 1

b 1 sand 3 s

ct = 3

t t = 1t = 0• •0 4

Answers

s

2 a 4ft

b s(2) = - 16(2) 2 + 40(2)+ 4 = - 64 + 80 + 4 = 20ft•

- 16t 2 + 40t + 4 = 20I1

II t = - 2s2

d.I

ds- = - 3 2 t 4 0

dt..40fts- 1II

. 5Ill - s

4

iv 29ft

3 a v(t) = s (t)

- e ( l) - t(e )- (e )2

-e 1 - t)

-e 2 t

v(t) = - tet

b l second

Investigation velocityacceleration and speed

1 a Let acceleration be 2 m s- 2 •

Time Velocity Speeds) m s- 1 ) m s -1 )

0 10 10

1 12 12

2 14 143 16 16

4 18 18

d Let acceleration be 2 m s- 2 •

Time Velocity

(s) m s- 1 )

0 - 10

1 -8

2 -6

3 - 4

4 - 2

2 a Spee d ing upb Slowing downc Spee ding upd Slowing down

3 a Speeding upb Slowing down

Exercise 7P

Speedm s- 1 )

10

8

6

4

2

1 a v ( t ) = 8 t 3 - l2t , t 2 : .0

a (t) = 24 2 - 12, t > 0b 84cms - 2 ; Velocity is

increasing 84 em s- 1 at

t ime 2 seconds .

c v(t) = 0 when t = 0 and1.22 s; a(t) = 0 whent = 0. 707 s;speeding upfor 0 < t< 0.707 s a n dt > 1.22; slowing downfor0.707 < t< 1.22

2 a v(t) = - 3 t 2 + 2 4 t - 36,0 < t< 8a(t) = - 6 t + 24,0 < t< 8

b s(O) = 2 0 m ;b Let acceleration be - 2 m s - 2 . v(O) = - 36ms- 1;

Time Velocity

s) m s -1 )

0 10

1 8

2 6

3 4

4 2

Speedm s -1 )

10

8

6

4

2

a(O) = 24ms - 1;

c t = 2, 6 s; moving left on0 ; t ; 2 and 6 ; t ; 8,moving right2 ; t:::; 6

d t = 4 s; speeding up on2 < t < 4 and 6 < t <8,slowing down on 0 < t < 2and 4 ; t :::; 6

c Let acceleration be - 2 m s - 2 . 3 a v(t) = - 9.8 t + 4.9

Time Velocity

s) m s- 1 )

0 -10

1 -12

2 -14

3 - 16

4 - 1 8

Speedm s- 1 )

10

12

14

16

18

a(t) = .8b 2.01 s

c O.Ss; 11.2m

d v(0.3) = 1.96 > 0 anda(0.3) = -9.8 < 0. Since thesignsof v(0.3) and a(0.3)are different the particle isslowing down at0.3 seconds.

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•a••II

b i

..II

1 1v(t) t

2 t+11 second

1 1a t) = - + -----,_-

2 ( t+1) 2

Since > 0 and2

1 > 0t+IY

1 1a t) = 2 + ( t+1) z > 0

for t :: 0 and so velocityis never decreasing.

Exercise 7Q

Decreasing (-oo, oo

2 Increasing (-oo, 2); decreasing(2, 00

3 Increasing ( -1 , 1) ; decreasing

(-oo, -1 ) and (1, ooDecreasing (-oo, 0); increasing(0, 00

5 Increasing (- 1, 0) and (1, oo ;decreasing (-oo, -1) and (0,1)

6 Decreasing (-oo, 3) and (3, oo

7 Decreasing (0, oo

8 Increasing (- 3, oo ; decreasing( -oo, -3)

9 Increasing - oo, - J 3 ) and

( 3 oo ; decreasing (-J3, - 1 ,( -1 , 1) and (1,J3)1 Increasing(-oo, -2) and

(4, oo ); decreasing (-2, 4)

Exercise 7R

relative minimum (1, -5)

2 relative minimum (2, -21);relative maximum ( -2 , 11)

3 no relative extrema

relative minimum ( - 1, - 1and (1, -1); relativemaximum (0,0)

(3 2187)5 relative minimum -4

, -256

6 relative minimum (0, 0);

relative maximum 2, )7 no relative extrema

8 relative minimum (1, 0);relative maximum ( - 3, - 8)

Exercise 7

concave up ( -oo, oo

2 concave up (0, 2); concavedown (- oo, 0) and (2, oo ;inflexionpoints (0, 0)and (2, 16)

3 concave up (2, oo ); concave

down (-oo, 2); inflexionpoint (2, 8)

concave up ( -oo, oo

5 concave up (-2, oo ; concavedown (-oo, -2); inflexion

point (- -: )6 concave up -oo - J3 and

3

7 a

J3 oo ; concave down3

J3J3- - - ; inflexion points3 , 3

J33- and3 , 4 J333 4

- 48xf ( x ) = (x z + 12) z

f (x)

_ x 2 + 12) 2 ( - 48) - (- 48x)[2 x 2 + 12)(2x)]

(x 2 + 12 t

_ x 2 +12) 2 (- 48 ) + 192x 2 (x 2 +12)- x

2 +I2t

Exercise 7T

y

4

2

3

4

( 4, 0)

-3 -2 -1 O 1 2 3 4 X-4

-2

y

8

6

-8)

y

-10

x= 4IIII

0, -2) 4 I

_-6 -4 -2 6 8 10 12 X

-2

y

1086

4

2

-4-6-8

0 3 4 5 x

_ 48(x 2 + 12)[- (x 2 + 12) + 4x 2 ] 5 y-(x 2 + 12)

_ 48(x 2 + 12)(3x 2 - 12)(x 2 + 12) 4

_ 144(x 2 +12)(x 2 - 4 )

x 2 + 12) 4

_ 144(x 2 - 4)

(x 2 +12) 3

b i relative maximum(0, 2)

inflection points

8 concave up (-oo, -2) and(4, oo ; concave down ( -2 ,4);inflection points at x = - 2 , 4

6

10864

2(0,0)

-4 -3 -2-4-6-8

(-1, 0

2 3 4 X

- ;Answers

X

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Exercise U

1 y y = f (x)I

y = f'(x)

; f\I I J 1 I I X

-3 -2 1 11 1 2

- - 1 X

y = f (x)

3 vY= f x)

6 8 X

y = f (x)

Exercise V

1 relative minimum (3, -75)

2 relative min imum (1, 0) and( - 1, 0); relative maximum 0 ,1)

3 relative minimum (3, -27)

relative minimum (

5 relative minimum (1, 0)

6 relative maximum (0, 1)

Exercise W

1 A - neither; B - relative andabsolute minimum;C - absolutemaximum

2 A -neither; B - r e lativeminimum; C - relative andabsolutemaximum;D - absolute minimum

absolute maximum 8;abso lute minimum - 8

Answers

4 absol ute maximum 16;absol ute minimum - 9

5 absol ute maximum 2;

abso lute minimum _2

Exercise X

79 11 - a n d -4 4

2 100 and 50

3 X = 50 ft; y = 200 ft3

Exercise Y

1 40cm by 40cm by 20cm

2 3 items

3 224 a

3 - 3hr=

5

b V(h) =reo5 3hJ h) orV(h) =

9r ( lO Oh - 20h 2 + h 3 )

25

c dV = 9P(100-40h + 3h 2 ) ·dh 25 '

d2V = 9p - 40+6h)dh 2 25

10d r = 4cm; h= - cm3

5 a p(x) = 4 2 x 2

b d2

pdx I d 2 3- X -x2 x z

c 0.630 th ousand unitsor 630 units

Review exercise non GDC

1 a 12x 2 + 6x - 2

12c - -x5d 10x 4 - 4x 3 - 3x 2 + 2x - 1

e

h

•I

•J

11

(x + 7) 2

2x + 3

1- 2ln x

4 1x - -

3 3

k ex 3x 2 + 6x + 1)

l 6e

3m.J2x - 5

n 2xe 2x(x + 1)

01

X

2 a x 3 + 3x 2h + 3xh2 + h3

b

f x)

1. [2(x+h/ - 6(x+h)] - 2x 3 - 6x)=

h-O h

1. 2x 3 +6x 2h+6xh 2 +2h 3 - 6x - 6h - 2x 3 6x=

h

=lim 6x2h+6xh

2+2 h

3-6h)

h

1. h(6x2

+6 xh - 2h2

- 6)= 1m ___ : ___________ :h

=lim (6x 2 + 6xh - 2h 2 - 6)h-o

c p = - 1 ;q= 1d f (x) = 12xe (O,oo)

3 y - 4 = - __ :_(x- 1)12

2../3 9- 2../3) -2../3 9+2../33 ' 9 3 9

5 a / ( 2 ) > / 9 ) > / ' ( 2 ) ;b f (2)> 0; since thegraph

of f is concave up,/(2) = 0a n d / ' (2) < 0 since thegraphof fis decreasing

6 a.

4x 3 - 12x 2I

c

••12x 2 -24xI

• (0,0), (4,0)• •

(3, - 27)I...

(0,0),(2, - 16)llY

20

15

0-4 -3 -2 -1 5 2 3

101520 2 ,-16)

25

4,0)

X

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Exercise SF

1 a 9 5 em b 6 7. 5d 92.5 e 35

c 57.5

y

. f" r • .j ·;

'

0 20 40 60 80 100 120 X

2 a 14d 82y

M

b 79e 7

1* II

c 75

Q3[ 1 a t*

71 75 79 82 85 X

3 a 19d 27y

1

...l

b 21e 15

iI

J ...l10 20

4 a 5 b 8d 10 e 3

S a m b i

Exercise 86

1 a 75cm

c 12

'" X' 3JI

...l

c 7

c 11

b (77.5 - 72) em= 5.5 em

c The middle 50% of da tahas a spread of 5.5 em.

2 y

3

4035

> .

30Q):::l0 25

';; 20>·.;::;ro 15:::l

E 10:::l

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c 2.47d The standard deviation

remains the same. This

is because the standarddeviation only measures the

spread of the numbers , andthat remains const a nt if thesamenumber is added to eachitem in the list.

e The mean is doubled.f 4.94g T he variance wi ll be

multiplied by 4 b ec au se thevariance is the stan d arddeviation squared .

Review exercise non-GDC

1 a 3 b 5 c 5d 9

2 a 4.2 b 4 c 43 Mea n = 27.5 yrs, st a n da r d

deviation = 0.4 yrs.

Type A

4 a 52 b 14 c 8Type B

a 525 a 426

b 8 c 3b 72 c 62

6 ay

140,_

c::

1000

8060

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Exercise 9C

1 f x) = + 4 r + 83

1 42 y - x 5 + - x 4 + 9

5 53 s(t) = f - f 6

4 115ncm 3

5 a -5ms- 2

b2

Exercise 901

2

3

4

5

6

7

8

9

10

2lnx + C x > 0

3 e + c. .In t + C, t > 04

_ _x2 +C2

i x 3 + 6x 2 + 9x + C32

- x3

+ 3x2 + 51n x + C, x > 01

. .u 3 + C31 _A __1 3 2- .x -- x- + - x - x+ C4 21- e + x) +C2

2 2 _- x z + - x 2 +5 3

Exercise 9E

1 . .c2x + 5) 3 + c6

2 - - 1 - 3 x + 5) 4 + C12

3

4

5

6

7

8

9

I- x -3

e 2 +C

_ _ln(5x + 4) + C x > _i5 5

3 7ln(7 - 2x) + C x > -2 , 2e x +l + C3

- 4 x - 3) 8 + C162 _

- (7x+2) 2 +C21

e4x + 4 In(3x- 5) + C4 3 '

5x> -

3

101

+ c12(4x - 5) 2

11 a 12(4x + 5) 2

b - 1 (4x + 5) 4 + C16

12 s = _ _ e-3t + 3f2 + . 23 3

Answers

Exercise 9F

1 . .(2x2 + 5)3 + c3

2 ln(x-1 + 2x) + C, x3 + 2x > 0

3

4

5

6

7

8

e + c1

- - - C

x2

+ 3x + 1e Fx + C

I- 2x3 + 5) 5 + c304- x 2 + x) 4 +C3

9 . .(0 - x2)4 + c2

10 - ln(x3 - 4x) + C x3 - 4x > 0

11 / (x) = ln(4x 2 + 1) + 4

2 f (x) = e + 4e

Investiga tion area and thedefinite integral

1 a i 0.5 ii 1; 1.25; 2; 3. 25• • •

ll 3. 75

b i 0.5 ii 1.25; 2; 3_25; 5...ll 5. 75

c 4.67; 3.75 < 4.67 < 5.75; thearea of the shadedregion

1 22 2 (3)(6) = 9; 2x + 2)dx = 9;

- I

they are equala

3 f x)dxb

4 a . .(2.5 + 1)(3) = 5.25;2

5

( -. . x+ 3) dx = 5.251 2

b _ _1t(42) z 25 .1 ;2

4.

-J16- x 2 dx :::::25.1-4

Exercise 9G

16

(. . x + 1) dx = 16 ·2 .- 2

_ _(8)(4) = 162

2 (x3 - 4x)dx = 4; no area-2

formula3

3 3dx = 12; (4)(3) = 12- I

4

3

-J9 - x 2 dx :::::7.07;0

1 24 n(3 ) ::::::7.07

53 1

- d x z 1.10; no areaI X

formula6

6 ( . .x+2)dx = 18;0 3

_ _(2 + 4)(6) = 182

Exercise9H

1 12

2 143 -4

4 - 8

5 12

6 07 11

8 -39 20

10 12

11 a 412 a 4

b 12

b i a= 3· b = 7 ii

9 4 Exercise 9

12

3

4

5

6

7

8

9

1

10

31

2

36

5

4(e3 - 1)

1

16

316

a 24

10 12

Exercise 9J

1

2

3

4

ln 31 1

b 323

k = 3

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8

9

0.3841

-1.952

' 2.68

y

3

2 - x - x-2) - e-"dx

Y.

8

4

2

-4 - 2 2 4 6 8 10 X-2

- 4

9.275 1 ) X+ 2x+6 - dx

1.725 2 x - 1

;:::9.68

10 ay

43

21

-2

b i

2 3 4 5 6 x

0.Jx- x )dx

..11 2.67 or -

k 3

2-Fx - x)dxor•

C I

Exercise LJ

1 x3 - 2x 2) - 2x2 - 3x) )dx +0

3

2x 2 - 3x) - x 3 - 2x 2)) dxI

3.08

Answers

1 Exercise M2 x - 1) 3 - x - 1))dx +

5

12 0

x - 1) - x - 1) 3)dx= 0.5 V n- 4 2) 5) 2513

3

4

l

- I .J3J

1.131

0( xe-x)- x3 - x))dx

1.18- 0.707l

- .0 + IO.x-2 - 9) --3

0.7071

2

3

4

n- 6 - 2x) 2 dx

0 1V

3n 6 2) 3) 113

2

:rr .J4-x 2 Ydx 33 .5;-2

4V 3 n 2

3) 33.5

4

:rr .J I 6 x 2 Ydx 134;0

- 9x2))dx + x-4- 9x2)V= :rr 4

3) ) 1340.7071

- - x-4 + 1 Ox-2- 9 )) dx +3

((-.0 + 10x 2 - 9) -0.7071

x-4- 9x2))dx 11 0

5 a i 4, 4)

ii f (x) = _ _ x2

m = / (4) = 2y - 4 = 2 x - 4)y = 2x - 4

b i 1.236, - 1.528)

Exercise N2

1 n( x3)2 dx = 127 1rI 7

1 281r2 n x2 + 1)2 dx =

150

381.1r

••II 1.236 )

±x2 x2) dx +3 3x-x2fdx= -

0 10

4 4 :rr .. ..)2 dx = 31Z4

1.236 4

I X 4

2.55 5 aln4 . 2

:rr e 4 dx0

Investigation: Volume ofrevolution

1

Interval

O x 1

1 :s:x:s:2

2 < x < 3

3 < x < 4

4 x:s:5 I5 < x < 6 I

2 7 1. 5; greater6

adius

f(1) = 0.5

f 2) = 1

f 3) = 1.5

f(4) = 2

f 5) = 2.5

f 6) = 3

3 n 0.5x) 2dx 56.50

b 2

Height Volume

1 0 = 1 n(0.5) 2 1) 0 . 785 4

2 - 1 = 1 n( 1)2 1) 3. 142

3 2 = 1 n( 1.5) 2 1) 7. 06 9

4 - 3 = 1 n(2 )2 1) 1 2.5 75 4 = 1 l 7r(2.5) 2 1) 19.636 5 = 1 n3) 2 1) 28.27

6 a , {J;)'dx4 Volume= n- 3) 2 6) 56.5 b e

3

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Exercise9

1 a v t) = 2t - 6

bt= 3 C t=4 • t = 0

.. I I I I I I I I I I .,. S(t)-1 0 1 2 3 4 5 6 7 8

4

c 2 t - 6 ) d t = 8 m ;

0 4

12t - 6 Idt = 10 m0

2 a v t) = ? - 6t + 8b

t=4 st = 0 t = 2•0

3 36

t - 6•

12s(t)

c ? - 6t+ 1 2 m ;0

6

IP - 6 t+ 14.7 m• 0

3 a v t) = 3 t - 2?b t 0 t = 2 t = 4

...-- . - - - - . - - - . . - + s(t)-8 0 84

c 3 t - 2?dt = 1 6m ;0

4

l3 t - 2? Idt = 16 m0

4 a12 1

v t)dt = - (6)(6)

c

b

2 2

- . _(4 + 2)(2) = 12 m212 1

lv(t) ld t = - (6)(6)2 2

1+

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

51

v t)dt = - (2)(2)0 21

+ s2 (3)(6) = 11 m

1Iv t) Idt = - (2)(2)2

I+ - (3)(6) = 1 1 m

212 1 1

v t)dt = 2(2)(2) +2

(6)(6)

1- 2 (4 + 2)(2) = 14m

12 1Iv t) Idt = - (2)(2) + . _(6)(6)

0 2 2

+ . _(4 + 2)(2) = 2 6 m2

5 a 2ms- 21

b s t) = -P - 9t + 123

8

c I? - 9 Idt11

9 m26 a 2ms- 2

b 2 < t < 1c 28

Exercise 9P

1

2

10I

18.4e 20 dt 239 billions0

of barrels1.5

1375P - P)dt 15500

spectators3 36.5 +

8

,te < O .Or OI3r

0240 cm 3

20

4 4000 +0

-133 1 - ) t1780 gallons

Review exercise non GDC1 a .0 - 4.x2 + 6x + C

3 2.

b- xJ + C7

1c - - +Cx 3

d 5 3 11- x - - n x +C x>O18 2 '1e - e 4 " +C4

1f - x 3 + l) s + C151 3

g l n 2x +3)+C, x > - -2 2

h . _ ln X ? + C, X > 02 .•I

.J

1- 3x 2 +1) 2 + C2

21n e + 3) + C3

k 2x - 5)2 + C

l

2 a

1 ( }- e 2x·+ C2

4

b 16c 8

d e6 - e3

e - 20

f InS2

2

3 a xl - l)dx1

b 4-

3 2 1c x2 - l ) d x - x2 - l)dx

I - J2

d Jr xl - 1)2dxI

4 f x) = - 2x+42

5 a 5b 28

6 s t) = 2e 2' + 2t + 67 13

Review exercise GD

1 107

2 a a t) = 4t - 11

b a= 1.5, b = 4

c 7.83ma y = 3xb (2, 6)c

-2

2

Y

864

2

-4

d 3x - x3- 2))dx = 6. 75- I

Chapter 1

Skills check1 a 32 b 27 c 343

d 181

e128 256

f 0.000000001 or 1 x I0 -9

X

2 a n = 4 b n = 5 c n = 3

d n = 4 e n = 3 f n = 3

nsw r s

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Exercise lOA

1 a Positive, strongb Negative, weakc Negative, strongd Positive, weake No correlation

a i Positive •• Linear...

Strongb i Negative

••

LinearI• • •

Strong..

LinearI.

c 1 Po s itive...

Moderate

d i No association••

No n-linear...

Zeroe i Positive •• Linear

• ••Weak

f•I Negative

Non-linear...

Strong

3 a Increases b Decreases4 a

Y Rainfall inTennessee60

I

·I-

01999 2001 2003 2005 2007 2009 X

b Strong, negativec As the year increases the

rainfall decreases.5 a

y

10 0

80

25 60c::Q)

b5 40

2 0

Scores

•-••

,

20 40 60 80 100 XMathematics

b Strong, positive, linear

Answe rs

Investigation - leaning tow ro f Pisa continued)

a,}j Scatterplot of lean vs year

750 •725 •

c:: • •C O 700) • ••

675 ••650 •• •

•

3 a 4, 6.67)

b

Q)j)

C O

.)c::

14

12

10

8

6

42

/

/

/

//

.

v/ ban oi t

I

87 .5 X5.0 77.5 80.0 82.5 85.0v/

0 xyear

b Strong, positive

c The lean is increasing. Thedanger with extrapolation

is that it assumes that thecurrent trend will continueand thisis not always the

case.

Exercise lOB

1 a 96.7, 44.1)

b

yRelationship between leaf length

70

60

EE 40

3020

10

II

v1/I'

and width

M

_ A 4

•• v

• // •

0 40 80 120 160 XLength mm)

a i 175 em

b

y

190185180

E175

+-

170Q)

:r: 165

160155

..II 66 kg

I

e_@

/

L v/ •

//

;>

60 65 70

Weight kg)

/

/

75X

2 4Hours

c Strong, positive

6 8

d An increase in the numberof hours spent studyingmathematicsproduces an

increase in the grade .

Exercise lOC

1 a (x ,y)=(75 ,7 .03)

Y

-o 12.3 •Q)

j)C OQ)

j)

-og),.sc::Q)

2.3 ----=-. ...70

Temperature

b y =-

0. 9 x +79

c 7

80x

a £220000

b 75.4

c and d Note the values

140

120100

80

60

40

20

of m and b in the equationy = mx +bareapproximate.Y

y -x 300

0160 200 240

e Approximately 70 houses

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Exercise 1

The slope is - 0.3. As astudent plays one moreday of sport theydo 18minutes less homework.They-i ntercept is 40,which meansthatthe average studentwho does no hours ofsport does 40 hoursofhomework.

The slope is 6. For every timea person has been convictedof acrime they know 6 morecriminals.

They-intercept is 0.5,which means that peoplewho have notbeen

convicted of a crimeknow 0.5 criminals onaverage.

3 The slope is 2.4. For everypack of cigarettes smoked perweekthere are 2.4 more sickdays per year.

They-intercept is 7, whichmeans that the average personthat doesnot smoke has 7 sickdays per year.

4 The slope is 100. 100 morecustomers come to his shopevery year.

They-intercept is - 5,which means that -5people visited his shopinyear zero; they-interceptis not suitable forinterpretation.

5 The slope is 0.8. Every 1mark increase in mathematic sresultsin a 0.8 increase in

.science.

They-intercept is - 10 which isnot suitable for interpretationas a zero in mathematicswould mean a - 10 in

.science.

Exercise lOE

a

c::0

·_.:;

-::( )uc::0

u

14

12

10

8

6

4 /v

. /

/v

/ 'v

/

0 1 2 3 4 5 6Time hours)

b y = 1.84x 1.99c 8.43 3 sf)a

Y

30

8 200

15

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6 0.994. Strong, positivecorrelation.

Review exercise non-GDC

1 .. b11 v

c d.

111 1

a and bY.

60

50-40

•; ::::

03 30 •::::l

u

20

10

0 200 400 600 800Di stance km)

c 321itres3 a and c

}

13.6

13.2

- 12.8C/)

0g 12.4_)

Q)

. .. 12.0Q)

•

11.6 1 1.

11.2

••

• •

•

•0.85020 30 40

Age yea rs)

b Mean age = 34 years,mean time = 12 seconds

c Approximately 11.6 s

Review exercise GDC

1 a

•

•0 X

b As the time increases,the numbe r of push-upsdecreases.

An swers

3

4

N

c y = 1 29x + 9

d r = 0 .929. Thereis a strong, negativecorrel ation.

a w = -22.4 + 55.5h

b 66 .4kg

a r=0 785b y = 30.7 + 0.688xc 99 .5

This should be reasonablyaccurate since the product-momentcorrelation

coefficient shows fairly strongcorrelation.

aY.

50

40 •• ••t ) 30 ••

20 ••

10

0 20 40 60Test 1

b Positive, strong

c high

d y =0 50x

+ 0.48e 20.48

5 a c and fy

333

3E

c

c::

2

22

8642

086

42 v

I

L/vv

[7I

. //

80

/

0 1 2 3 4 5 6 7 8 XLoad kg)

b 4,30) d•I r= 0.986

ii (very) strong positivecorrelation

e y = 1 83x + 22.7g 30 .9cm

6

h Not possible to find ananswer as the value liestoo far outsidethe givendata set.

ay

40

35C/)

E 30Q)

. 0e 25Cl.

.o 20>ro

15co

10

5

0 1Agreeab leness

b Behavior problemsdecrease .

c -0.797

d Strong, negativecorrelation

e Fewerf y = - 10. 2x + 51.0

g 5.1

7 a y=10.7x+121

6 X

b i Ev e ry coat on averagecosts 10.65 toproduce.

••When the factory

does not produce any

clothes it has to pay

costs of 121.

c 870

d 4

ChapterSkills check

1 a x = 90b x= 50C X 68

70d x=

3

e x = 6.09 (3sf)

f x = 14.7 (3sf)

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Exercise 11A

1 b = 16, A 36.9°, B= 53. F2

3

A

B = 50°, a = 31.0, c= 48.3

A 35°, a = 2.58, b = 3.69A A

4 a = 36, A = 36.9°, B = 53.1°A

5 B = 55°, b = 15.7, c= 19 .2

6 c = 12. 9,A

41.2°,i

=48.8°7 X= 5, A= 22 .6°, B = 67.4°Exercise 118

1 a b= l2J3 A 30°,A

B = 60°

b B= 45° a = 9 c = 9 2\fL.C A = 3 0 ° a = 2 . 2 5

b = 9 J34

d a = 2J3 A = 30°,A

B = 60°

e b = sJ2 A= 45°,

2 x = 8J2 , y = 81 3 - 8, z = 16

3 X= 2 J3 + 2 AC = 4.J3 + 2J3 3

4 x = 1 AB = 3 2 or x = 3\j L.AB = 11 . J2

5 w = 9.8em, x = 13.9em,y = 6.5em, z = 15.4em

Exercise 11C

1 a 1 0 J 2 e m

b BAC = 70.5°A B C = 38.9°

2 a AE = 29.1, B E = 34.4

b AED = 74 .1 °,A

EBA = 54.5° ,AEB = 51.5°

3 7 5 8 m

4 71.5° and 108.5°

5 4.78 km, N2l . l 0 W

6 70 .7 m

7 44 .8 km, 243.5°

8 135. 7m, 202.2cm

9 91.2 m

10 40 .7 m

11 4.01 s

12 a 20.6°

c 35.1°

Exercise 11

d 50 .0°

1 a (0.940, 0.342)

b (0 . 956, 0.292)c (0.5, 0.866)

d (0.276, 0.961)

e (0, I)2 a 66°

c 45°

3 a 0.470

c 0.203

b 81°

d 14°

b 0.308

d 0.25

Investigation Obtuse angles

1y

-0 .766,0.643) (0. 766, 0.643)

0 X

2y

( 0.906, 0.423)155°

(0.906, 0.423)

X

3y

(-0.375, 0.927) (0.375, 0.927)

Exercise 11E

1 a B (0.866, 0.5),c ( -0 .866, 0.5)

b B (0.545, 0.839),c ( -0 .545, 0.839)

c 8(0 .7 07 ,0 .707),c ( -0 .707 0. 707)

d B (0.974, 0.225),c (- 0. 974, 0.225)

e B (0.087, 0.996),c (- 0. 087, 0.996)

2 a 70 .6°

b 17.3°

c 25.4°d 39.7°

3 a 0.2588, 165°b 0.5878, 144°c 0.9877, 99°

d 0.8988, 116°4 a 60.6°, 1 I 9.4°

b 25.8°, 154.2°

c 30.3°, 149.7°d 30° 150°

Exercise 11F

1 a 1.50c -0.910

b -1.92

d 1

2 a y = 1.09x, e= 48°b y = 1.87x,

e= 62°

c y = -2 .80x, e= 11 0°d y = - 1. 21x e= 129°e y = - 0 .75x e =143°f y = 2.36x e= 113°

Exercise 11GA

1 a C = 5 0 ° a = 17.7em,c = 18.5A

b B = 68°, a = 1.69em,b = 2.44emA A

c B = 40.9°, C =84.1°,

c = 5.46emt

d A = 40° a = 149)

c = 190A

e c = 110°, a= 2.80, b = 4.212 26 .9 em

3 3.37 km, 2.24 km

4 15.8 m

Investigation Ambiguoustriangles

A A

1 C = 62°, C2

= 118°. Theangles are supplementary.

A A

2 B = 86° B = 30°1 2

b = 5.65 em, b2 = 2.83 em

Exercise 11HA A

1 a C = 61.0°, B = 89.0°,

h1

= 8.0emA A

C 2 = 119.0°, B 2 = 31.0°,

b2 = 4.1 em

An swe rs

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8 gob C, = 1.1°, A 1 = 5 . ,a = 19.0em

A - 0C2 =108 .9° A 2 =21 .1

a 2 = 8.0em

A 9 soC B 1 =68 .5° A 1 = 1. ,

a = 7.3emA

B2 = 111 .5°, A2 = 48.5°,a

2 = 5.5emA A

d C = 30.5°, B = 107.5°,

b = 47.0em

e Triangle does not existf B = 77.8°, c, = 32.2°,

c = 14.2emA A

B2 = 102.2°, C 2 = 7.8°,

c2 = 3.6 em

g i = 26 . 7°, c = 108.3°,c = 29.5 etn

h c, = 67. 1° A = 56.9°,a

= 45.5 em

C1

= l l .9° A 2 = 11.1°,

a2 = 10.4em

a BE= 8m CE= 6mDE= 15m

b EAB=53 .1°A

BCE= 53.1°,BCD= 126.9°,

A

ABD = 98.8°,CBD = 25.1°

c Given side BD = 1 7 min :::ABD and angle

A •

D = 28 .1 o and sideAB = 10, th en there are 2possibletriangles, fittingth is data namely DBA andDBC.

3 b 5.80 km c 24 .9 km

d 143.5°

Exercise

1 a a= 65 . 7m B = 36.0°,

c = 80.0°A

b A = 28.9°, B = 52.8°,{; = 98.4°

Answ rs

A

A = 44.4°, B = 107.8°,A

c = 27.8°d b = 7.48m A= 43.5°,

A

c = 105 .5°e c = 92.8m A= 49.4°,

s = 60.6°f

A

A= 48.6°, B = 56.4°,

c = 75.0°12. 1 km

3 4.07 em 6.48 em

4 18.8 km

5 043.5° or 136. 5°

6 a 45°

b 71.8°c 63.8°

Exercise 11J

1 a 26.7 em 2

b 40. 8em 2

c 152cm 2

d 34 .1 cm 2

e 901 cm 2

f 435 cm 2

47 .8°

3 22.7 em

4 a 76.7°

b 81.4em 2

5 x=2 5 em

6 5.31 mm 18.5 mm

Exercise 11K

1 9.52 em

39 em

3 5 radians

43000 em

2, 220 em

5 22.95 em 2 , 21.3 em

6 8 = 1 .7 r=16

7 7.96 cm 2

Exercise

1 a 57r12

4 7rb

3

4 7rc

9

dl1.7r

6

a 0.977 rad

b I .87 rad

c 5.65 radd 4.01 rad

3 a 150°

b 300°c 270°

d 225°4 a 85.9°

b 20.6°c 136°

d 206°

Exercise 11M

J1 a -

2

Ib - -

2

cJ33

d J32

a 0.892b 0.949

c - 1.12d 0.667

3 a 9.76 em 2

b 5.45 emc 50.5 em 2

4 10.9 m 2

5 a 17.1 em 2 b 12.1 em 2

c 2.63 rad d 15 .8 em

Review exer cise non GDC

1 7J2 ema 30°

3 25

4 10 em 2

5 a 25 em

b 8·f3 em

b 125 em 2

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Review exercise GD

1 72.7 m

2 a (0.848, 0. 530)

b 72.9°

c ( 0 .600, 0.800)3 a 54 .7° b 10. 9cm

4 a 18.0 m b 34.3°

5 a 121 b 8.60 em

6 54.1km

7 a 3 1.9 b 1 3. 9 em

c 119 d 2 7.4 em

8 a 2 1. 6 em b 14.5 em

c 11 .16 em d 4 7. 3 em

hapter 12

Skills check

1 a (3, 0,0)b 3,4,0)

c 3,0,2)

d 3,4,2)

e 1.5,4,2)2 6. 71

3 a 20cm

b 101°

Exercise 12A

1 a x = - 2i + 3j

b y = 7j

c z = i + j - k__.. .. 2

2 a A B =3

- 1...b C D = 6

- 1

0...

c EF = 0

3 a =

b =

1

- 3= 3 i - 5j

- 5

- 24 = - 2i + 4j

3c = 8 = 3i + 8j

0d =

6= 6j

-3e =

6= - 3 i 6j

4 a 5b J1 = 3.16

c J 9 = 5.39d 5.3

e J 9 = 5.39

5 a J38 =6.16

b .J26=5.1 0

c 3

d 7

e F = 1.41

Exercise 128

1 a c = 3b1

d = a2

e = -5 b

f = -2 a

b They are per pendicular .

2 a , b, e

3 a - 247

bs 8

4 t = - 25, s = -55 a OG = j + k

....b BD = - i - j + k

....c A D = - i + k

_____ ... ... 1d OM

2i + j + k

,..6 a OG = 4j +3 k...

b BD = - 5i - 4j + 3k...c AD = - 5i + 3k

_____,...,..

d OM = i + 4j + 3k2

Exercise 12C_ ... -5 ... 5

1 PQ = QP =1 I

__,...,... - 42 a AB =

4

____.,.. 4b BA =

4

_.,.. -7c AC =

3

____.,.. 3d CB =

- 7

3 a 2i - 3j + 5kb - i + 5 j 6k

c - i + 5j - 6kd i - 5j + 6k

5...

4 L M - 4-3

5 us = 2i + 8j - 3k6 x O , y = 7 , z = 9

Exercise 1 2

- 3 3

5 AC = -5AB =-4 4

6...BC = - 10 . Any two of

8

these are scalar mult iples ofeach other

3

16

- 3... ... ....orBC= 2 soAB = B C

- 8

-3 - 6... ...

3 P P 2 = - 1 p lp3 = - 2

0 0

- 3

; 4 2_ 43

54 X = 3; AB : BC = 1 : 2

swers

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Exercise 2E

5...

1 AB = 0 ; J 9 = 5.39-2

____..,.. ...

2 IAB I = Jl29 IA C I = J42lo r;;::;;

IBC I = -v129. Two sidesequal length therefore

isosceles. Angle C A B = 46 .8°

3 t = +6

4 x= ±JS

5 a= +2

6 a 15b 10

c 13

Exercise 2F

1

2

3

4

5

6

7

8

(3.2

(4)2

5 + 5 = 1

J = 1. _(4i - 3j5

- 11

= = I -5J42

4

I.( 2i + 2 k )3

1

J5

_2_(2i j )J5

1

7---= -3J14

2

9 acosB

sinB

bcos a

sm a

Exercise 2G

1 a Si + jb 2i + 3j

c 2i + 4j

d 8i + 4j

e - i - 3jf 2i

Answers

2 a

b

c

d

e

- 22

l

8

- 1.5

- 3

-5

15

3- 34

3 a 8i j 3k

b - i + 2j + 3k

c i - 2j - 3kd 8i - 6j - 1 0k

4 19

4 X= -5.5 y = 3

z =-610

- 16

5 X - 4 .5 , y = 10.5

6 s = 4 . 5 t = 9 u = 9

Exercise 2H

4 a i b - a..11 b - a

iii 2b - 2a•IV b - 2a

v 2b - 3ab A B is parallel to and ha lf

the length of FC

c FD and A C are parallel__. .. ...

5 d M X = MP

Investigation cosine rule

Exercise 2

1 a - 18

c 20e - 13

b 5

d -13

2 a -9 b 20

3

c 20 d -5 8

e 13a Pe r p endicul ar

b Neither

c Parallel

d Neither

4

5

67

e Pe r pe nd i cu la r

f Parallelg Parallel

- 152

d = 1

3

45°

a 94.8 °

b 161.6°

c 136.4°- 1 1

8 a AB = A C =5 -2b - I I

- 11c J26JS

9 a 79.0°

b 90°

c l l 8 .1°

10 a A B = JU; A C = J 61

b cosBA C = 0 2617 26

c 10 .5

11 54.7°____..,.. .

12 a OA ·O = 0 thereforeperpen d ic u lar

b J6i13 A = 2.5

14 A = +915 p = ±3

Exercise 12J

1- 1

a r =2

b- 1

r=0

3

c r= 1- 2

3+ t

2

5+ t

- 2

3

+ t - 28

d r = 2j - k + t(3i - j + k)

24 - 1

a E.g. r = + t5 - 7

4b E .g. r =

- 2

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Chapter 3Skills check

1 a f i2

b J3

f id -

f j2 2

2 af j b - 12

c 1

3 a - 1.48d - 0 .5b +2

4 a -0.182, 2.40 b +1.14

Investigation - Sine cosineand tangent o the unit circle

1 sin90° = 1 cos90° = 0 tan90°does not exist

2 sin180° = 0 cos180° = 1tan180° = 0

3 sin270° = -1 cos270° = 0tan270° does not exist

4 sin360° = 0 cos360° = 1tan360° = 0

5 sin -90° ) = -1, cos(- 90°) = 0,tan( -90°) does not exist

6 sin( -180°) = 0, cos( -1 8 0°) = -1 ,tan -180°) = 0

7 sinO = 0, cosO = I , tanO = 0

8 •1C 1C 0 1C

sm2 = 1,

cos2 = , ta n 2

does not exist9 sin1t = 0, cos1t = - 1, tan1t = 0

. 3rc 3rc 3rc10 s m - = -1, c o s - = 0, tan-

2 2 2does not exist

. 3rc 37r11 sm = 1 cos- - = 02 231Z d .

ta n - 2 oes not exist12 sin41t = 0, cos41t = I,

tan41t = 0

Exercise 3A1 a

b

nswers

c

d

e

f

g

h

2 a

b

c

d

-270°

5rc...--r-... 3

1l

2

e

f

g

h

1l

3

21l

For questions 3 to 8 , there are manyother possible correctanswers.

3 a 120°, - 240°, - 300°

b 340° - 20° -160°

c 255° 285°, - 105°

d 65°, -245°, -295°

4 a - 35°, ±325°

b - 130°, +230°

c -295°, +65°

d 240°, ±120°5 a 230°, - 130°, - 310°

b 280°, - 80°, - 260°

c 40°, - 140°, - 320°

d 155°, 335°, -2 05°

6 arc 4rc 57r

3 3 3

b 7rc _ rc _ 37r4 , 4 4

c 31t - 4.1, 4 I - 2 1t , 1t - 4 .1

d 1t + 3, 21t - 3, 3 - 1t

7 a - 1Z + 111Z6 - 6

b -1, ± 1 - 21t )

c -2.5, ±(2.5 - 21t )

d 31Z + 7rc5 , - 5

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8 a51l 31l 71l- - - - -4 4 4

b 1.3 + 7r, 1.3 - 7r, 1.3 - 27r

c121l 21l 9rc

- - - -7 7 7

d 27r- 5 7 r 5 - 5 - 7r

Exercise 138

1 a 0.940c - 0 .342

2 a

c

12

1

b 0.342d - 0 .940

J3b - -2

d J32

3 a 0.8 b 0.6 c 0.6

d - 0.8

g - 0 .8

4 a .b

db

4e -3

h 43

b a

e a

g a h b

Exercise 13C

f

c b

f b

1 a -300°, -240°, 60°, 120°b ±120°, ±240°

c -315°, -135°, 45°, 22SO

4

3

d - 360°, - 180°,0°,180°,360°

e +45°, + 135°,+225°,+3 15°f +30°,+150°,±210°,±330°

2 a _11Jr rc _c _ rc6 6 6 6

b 0, n +2n

+ 1l 11 rcc _- , +-6 6

d - .1C 31l2 2.rc 21l 4rc 5.rc

e +- + - + - + -,_ , - , _3 3 3 3

71l 31l 1l 51Zf - - - 4 4 43 a 0°, 360°, 720°

b -135°, -45°,225°, 315°,585°, 675°

c -225° -45° 135° 315°,495° 675°

d ±60°, ± 120°, 240°, 300°,420° 480° 600° 660°

4 a2

5.rc .rcb - - ·6 ) 6

.1C 3.rcC +- .+-- -4 4

.rc 5.rcd + - , + -

6 6

Exercise 13

1 a +15°,+ 165°b -165° -105° 15° 75°c 90°d +180°

2 a 5 1C .1C 7 Z' ll.rc- - - 2 12 12,.1211.rc 7 rc .rc .rc 5 rc3rcb - -

2 ) 12 4 ) 12 12 4.1Cc +-2

2.rc 4 rc3 a 3

3.rc 7rcc

4

Exercise 13E1 a sJU

2 a

3 a

c

4 a

c

5 a

c

6 a

c

7 a

184- 5

9Jl1

5

7-18j63

32

j6331

3-52425

7- -25

336-625

a

2ab

Exercise 13F

d 3.rc+-4

b 7 1C 3.rc 11.rc6 ) 2 6

d ..2

7b c - sJil7

b

18

1

9c 4 J5

b sJU18

d 5Ji.T.7

b

d

b

d

31

32

31163

512

4

57

25

b7

527d -

625

b

db 2 2- a

1 a 30° 90° 150°b 22.5°, 112.5°c 135°d 45° 135°

2 a -150° , -120° ,30° ,60°

b 90°

c + 150°, +30°

d -90°, 30°, 150°3 a 0 1r

b .1C 7.rc8 8

02.rcc -

3

4 a.1C 5.rc

8 8

c 0, 7r

6 k 6

7 b = 8

Exercise 13G

b .1C2rc 3rc

d 4 4

1 - 346° - 194 14° 166°

2

3

+27°, 333°

244° 296°

4 55°, 235°, 415°5 - 5.33, - 4.10, 0.955, 2.19

6 +1.71, 4.58

7 - 0 .739

8 - 0 .637, 1.41

Investigation: graphing tan x

1

Angle Tangent

measure x) valuedegrees) tan x)

0 0

- 30,+30 11

-J3'J3- 45,+45 - 1 , 1

-60, +60 - ../3 ) J3120 - /3135 - 1

1150 -

J3180 0

2101

-J3

225 1

240 .J3300 - ../3315 - 1

3301

- J3360 0

An swe rs

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3 ta n + 90° and tan + 270°are undefined. The limitof thetangent as the ang leapp roaches ± 90° or ± 270° isinfini te .Asymptotes are oftenshown on graphs for valuesth at do notexist.

Exercise3H

- 297° - 117° 63° 243°' ' '

-107° 73° 253°' '

3 124° 304°'

4 38°, 142°, 398°, 502°5 - 5.88, - 2.74, 0.405, 3.55

6 -1.88, 1.26

7 4.558 - 4.66, 1.20, 2.28, 4.77

Investigation: transformationso f sin x and cos x

Y

X

y

X

0 X

3

0 X

4 Y

0 X

Answe rs

5

Exercise 3

' I'

v"-,

' \ '..I•-2n -n

3 y

r-

J J'1.17' 0

".4

4

\

\11T [\.

\

5

v \

y

//

J

I/

I I\n.," . I - .rr \./

6 y' 11T 0-

\. ,..;\ 7

A

7

11T - r;r\I Iv

1.. /

X

y

0 TI"i -

4- 7 '/ -

y

A........

r\. /i' ' ./ v

J) TT X

y

1\v \0 \ I

I• \ v

y

11v \

I r \0 v \

[\, /

1\ J\ v

y

0 .T/ \

/ I[ .., I

/l l"'

rr x?_

/

.

II

2-

\..

2

\\

8 y

I i J vI I. rr - 0 ) 'TT Xl 4

-2

9 y =cos x -

2

; ) or

y =sin

1 y = sinx +l

y = tan ( x - : )

12 y = cos ( x - : ) - 1.5

Exercise 3J

I \.I \

' ftr- ' \

j \1 '

' ' -

J

3

-I

v- rr

4

1 1

1 \n- -Jfrr ir

y

1 \ 'I \.

I \() "\d \.y

A

I i\2

I

0 . 7-

"'-1 J

yA

,.,..

/15 I

I

A

y

I \ 1 1

A/

)

'

I/

)

1

71,.

\ 1 : - \ ;-. \ f r\ IJ: r•

5 y·"1 \ / 1

1 T T 0 T -21 -

J \ i/ \ J \

frr X

'rr X

'rr x

\

') rr X

prX

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6 y

,) ' I I '1\ /\--3-'-

- - - c- 1-2 1 - 1 .o

-

.

7 yr - r

v - \2 T 0 \..;. -

- I ',_

8

9 y=7.5sinx

10 y =cos(0.25x)

11 y=tan(0.2Sx)

-....:

12 y = - 3cos 0.5x) or

y = sin(x J-1 5

Exercise 13K

'

' i- ?

I

1 For questions 1 to 4, answers may vary.

1 y= 3.Ssin(x- 2; -I s ,y = 3.5cos( x +

5: )-1.5

2 ))- 2,

3 y =2 in(2x) +1,

y = 2cos( ))+1

4 ;)).

y x+: ))

X

5 y

6

X

72-

1\ \

1•

I c X' Iv v v v

- ,_2

8 y76-

Vi\ 5-4-

. I- - -· -- 1- - - -

-: 7l -1 -lQ )r 2f- 3rx

Exercise 13L

1 a ,b

f'lot2 f'lot),y1a4.8cos

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3 a, b

F"l.;.t2 F l.;.n,y1a0.8cos

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8

9

10

11

2sin (2x)

cos 2 (2x)

- 8.1Z COS .7rX)(1r x )

[cos( sin x)] cos x

3x 1a

b - 4cos 3 x sin x

12 a 3cos 3x - 4)

b - 9sin (3x - 4)

Exercise 148

1y - 1 = t(x-;-} -1 = -l(x-:)

2y - 2 = 4(x- ;} y - 2 = - ;)

3 - 2I4 a - - b - 2sin (2x)2

C y + i --13:.. 51Z3 ' 3

Exercise 4C

1

2

3

4

5

6

7

8

9

10

11

- 12sin (

1( l+cosx)

e sin2l cos 2t

2ex sin xt

- + tantcos t

3e 3... cos 4x - 4e 3... sin 4xI

cos2 2xJ tan 2x

cosx .lnxsmxX

.SLO X

or - tanxcosx

2 I Xa b - cos -

c

X 2 2

I 2 X 2 . X- In 3x c o s - + - s m -2 2 X 2

12 a= 1 b = 2'

Exercise 14

1 Relative minimum: -2)·3 ' '

relative maximum: ( f 2)2 relative minimums :

( I).e;, - 3 ; relativemaximums : ( :..).).(5 6 2 6 ' 2

3 d .creasmg: - < x < JZ;2

increasing 0 < x < :..; concave2

down: 0 < x < n; relative

maximum: (f·1)f(x}

1

1l 1l 37l- - -4 2 4

4 decreasing:1Z 1Z J1Z .

O 0relative minimum atx = 4.91

7 a j ' (x) = - r Sin X 2x COS Xb minimum : - 11.6;

maximum : 7.09

8 ad 1 fJ) 2 . 0 4sinBcosB=- Sill - r = = = = =

.J25 - 4sin 2 B

2. O 2sin2B

o r - sm r = = = = =

. J z s -4s in2

)

b

1r 3Jr 5Jr 7 f(x}0 < X < - < X < - - < X < ; r· (5.05 , 2.16}

f X)

1

8 8 8 8 2

relative maximum: ( ; 1}

relative minimums:

1

0-1

1l

451T M . 2n 64 2 4

n o) o ; -2 d'(O) - -2sin8 - -4sm6cos6(1.23, -2.16} h5- 4Sif128I I 4 )x points:

( T l ) (37T 1) (57T I) (71Z 1)8 1 2 I 8 2 J 8 1 2 J 8 l2

IT IT 37l 1l 57l 37l 71l IT X4 8 2 8 4 8

-3

•c 1 The blade is closestthe center of thewheel when d(8) hasarelative minimum orat an endpoint. Thereis a relative minimumwhen d(e) changes fromnegative to positive at8 =n. Testing theendpointsand critical

Answe rs

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numbers we findd O) = 7, d 2rr) = 7and d rr) = 3. So theclosestdistance is 3metres and it occurswhen the angle ofrotation isrr.

ii The distance isch a n ging fastestwhen d ((J) has arelativeminimum ormaximum. This occurswhen e s 1.23 radiansor 5.05radians.

Exercise 4E

1 2sin x 3cos x + C

2 x 3 + 3s inC