ANSWERSwashburnmath.weebly.com/uploads/1/0/7/5/107524597/... · 2018-02-22 · ANSWERS 695 EXERCISE...

ANSWERS 695

EXERCISE 1A

1 a, d, e 2 a, b, c, e, g 3 No, for example x = 1

4 y = §p9¡ x2

EXERCISE 1B

1 a 2 b 8 c ¡1 d ¡13 e 1

2 a 2 b 2 c ¡16 d ¡68 e 17

4

3 a ¡3 b 3 c 3 d ¡3 e 15

2

4 a 7¡ 3a b 7 + 3a c ¡3a¡ 2 d 10¡ 3b

e 1¡ 3x f 7¡ 3x¡ 3h

5 a 2x2 + 19x+ 43 b 2x2 ¡ 11x+ 13

c 2x2 ¡ 3x¡ 1 d 2x4 + 3x2 ¡ 1

e 2x4 ¡ x2 ¡ 2 f 2x2 + 4xh+ 2h2 + 3x+ 3h¡ 1

6 a i ¡ 7

2ii ¡ 3

4iii ¡ 4

9

b x = 4 c2x+ 7

x¡ 2d x = 9

5

7 f is the function which converts x into f(x) whereas f(x) is the

value of the function at any value of x.

8 a 6210 euros, value after 4 years

b t = 4:5, the time for the photocopier to reach a value of

5780 euros.

c 9650 euros

9 10 f(x) = ¡2x+ 5

11 a = 3, b = ¡2

12 a = 3, b = ¡1, c = ¡4,

T (x) = 3x2 ¡ x¡ 4

EXERCISE 1C

1 a Domain = fx j x > ¡1g, Range = fy j y 6 3gb Domain = fx j ¡1 < x 6 5g, Range = fy j 1 < y 6 3gc Domain = fx j x 6= 2g, Range = fy j y 6= ¡1gd Domain = fx j x 2 R g, Range = fy j 0 < y 6 2ge Domain = fx j x 2 R g, Range = fy j y > ¡1gf Domain = fx j x 2 R g, Range = fy j y 6 25

4g

g Domain = fx j x > ¡4g, Range = fy j y > ¡3gh Domain = fx j x 2 R g, Range = fy j y > ¡2gi Domain = fx j x 6= §2g, Range = fy j y 6 ¡1 or y > 0g

2 a x < ¡6 b x = 0 c x > 3

2

3 a Domain = fx j x 2 R g, Range = fy j y 2 R gb Domain = fx j x 2 R g, Range = f3gc Domain = fx j x 2 R g, Range = fy j y > 2gd Domain = fx j x 6 ¡2, x > 2g, Range = fy j y > 0ge Domain = fx j x 6= 2g, Range = fy j y 6= 0gf Domain = fx j x 6 2g, Range = fy j y > 0gg Domain = fx j x > 5

2g, Range = fy j y > 0g

h Domain = fx j x 6= 5g, Range = fy j y 6= 2g4 a Domain = fx j x > 0g, Range = fy j y > 0g

b Domain = fx j x 6= 0g, Range = fy j y > 0g

c Domain = fx j x 6 4g, Range = fy j y > 0gd Domain = fx j x 2 R g, Range = fy j y > ¡2 1

4g

e Domain = fx j x 2 R g, Range = fy j y 6 2 1

12g

f Domain = fx j x 6= 0g, Range = fy j y 6 ¡2 or y > 2gg Domain = fx j x 6= 2g, Range = fy j y 6= 1gh Domain = fx j x 2 R g, Range = fy j y 2 R gi Domain = fx j x 6= ¡1 or 2g,

Range = fy j y 6 1

3or y > 3g

j Domain = fx j x 6= 0g, Range = fy j y > 2gk Domain = fx j x 6= 0g, Range = fy j y 6 ¡2 or y > 2gl Domain = fx j x 2 R g, Range = fy j y > ¡8g

EXERCISE 1D

1 a 5¡ 2x b ¡2x¡ 2 c 11

2 f(g(x)) = (2¡x)2, Domain fx j x 2 R g, Range fy j y > 0gg(f(x)) = 2¡ x2, Domain fx j x 2 R g, Range fy j y 6 2g

3 a x2 ¡ 6x+ 10 b 2¡ x2 c x = § 1p2

4 a Let x = 0, ) b = d and so

ax+ b = cx+ b

) ax = cx for all x

Let x = 1, ) a = c

b (f ± g)(x) = [2a]x+ [2b+ 3] = 1x+ 0 for all x

) 2a = 1 and 2b+ 3 = 0

) a = 1

2and b = ¡ 3

2

c Yes, f(g ± f)(x) = [2a]x+ [3a+ b]gEXERCISE 1E

1 a b

c d

e f

g h

i j

k l

2 a b

c d

e f

g h

ANSWERS

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IB SL 2nd ed

magentacyan yellow black

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Y:\HAESE\IB_SL-2ed\IB_SL-2ed_an\695IB_SL-2_an.CDR Monday, 16 March 2009 4:06:35 PM PETER

696 ANSWERS

i j

k l

m n

o

3 a b

c d

e f

g h

i

4 a b

c d

e f

g h

i j

k l

EXERCISE 1F

1 2

EXERCISE 1G

1 a i vertical asymptote x = 2, horizontal asymptote y = 0

ii as x ! 2¡, y ! ¡1 as x ! 1, y ! 0+

as x ! 2+, y ! 1 as x ! ¡1, y ! 0¡

iii no x-intercept, y-intercept ¡ 3

2

iv

b i vertical asymptote x = ¡1, horizontal asymptote y = 2

ii as x ! ¡1¡, y ! 1 as x ! 1, y ! 2¡

as x ! ¡1+, y ! ¡1 as x ! ¡1, y ! 2+

iii x-intercept 1

2, y-intercept ¡1

iv

c i vertical asymptote x = 2, horizontal asymptote y = 1

ii as x ! 2¡, y ! ¡1 as x ! 1, y ! 1+

as x ! 2+, y ! 1 as x ! ¡1, y ! 1¡

iii x-intercept ¡3, y-intercept ¡ 3

2

iv

d i vertical asymptote x = ¡2, horizontal asymptote y = 3

ii as x ! ¡2¡, y ! 1 as x ! 1, y ! 3¡

as x ! ¡2+, y ! ¡1 as x ! ¡1, y ! 3+

iii x-intercept 1

3, y-intercept ¡ 1

2

iv

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All have vert. asympt. x = 0and horiz. asympt. y = 0.

They are positive for x > 0and negative for x < 0.

All have vert. asympt. x = 0and horiz. asympt. y = 0.

They are positive for x < 0and negative for x > 0.

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Y:\HAESE\IB_SL-2ed\IB_SL-2ed_an\696IB_SL-2_an.CDR Wednesday, 18 March 2009 9:40:54 AM PETER

ANSWERS 697

EXERCISE 1H

1 a i

ii, iii f¡1(x) =x¡ 1

3

b i

ii, iii f¡1(x) = 4x¡ 2

2 a i f¡1(x) =x¡ 5

2ii

b i f¡1(x) = ¡2x+ 3

2

ii

c i f¡1(x)

= x¡ 3

ii

3 a b

c d

e f

4 a fx j ¡2 6 x 6 0gb fy j 0 6 y 6 5gc fx j 0 6 x 6 5gd fy j ¡2 6 y 6 0g

6

5 fy j ¡2 6 y < 3g

REVIEW SET 1A

1 a 0 b ¡15 c ¡ 5

42 a = ¡6, b = 13

3 a x2 ¡ x¡ 2 b x4 ¡ 7x2 + 10

4 a i Range = fy j y > ¡5g, Domain = fx j x 2 R gii iii is a function

b i Range = fy j y = 1 or ¡3g, Domain = fx j x 2 R gii iii is a function

5 a b

6 a = 1, b = ¡1

7 a b

8 a f¡1(x) =x¡ 2

4b f¡1(x) =

3¡ 4x

5

9 (f¡1 ± h¡1)(x) = (h ± f)¡1(x) = x¡ 2

REVIEW SET 1B

1 a Domain = fx j x 2 R g, Range = fy j y > ¡4gb Domain = fx j x 6= 0, 2g, Range = fy j y 6 ¡1 or y > 0g

2 a 2x2 + 1 b 4x2 ¡ 12x+ 11

3 a b

4 a x = 0 b

c Domain = fx j x 6= 0g, Range = fy j y > 0g

y

x

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7 a f : x 7! 1

x, x 6= 0 satisfies both the vertical and horizontal

line tests and so has an inverse function.

b f¡1(x) =1

xand f(x) =

1

x) f = f¡1

) f is a self-inverse function

8 a y =3x¡ 8

x¡ 3is symmetrical about y = x,

) f is a self-inverse function.

b f¡1(x) =3x¡ 8

x¡ 3and f(x) =

3x¡ 8

x¡ 3

) f = f¡1 ) f is a self-inverse function

9 a f¡1(x) = 2x+ 2

b i (f ± f¡1)(x) = x ii (f¡1 ± f)(x) = x

10 a 10 b x = 3

11 a i 25 ii 16 b x = 1

12 (f¡1 ± g¡1)(x) =x+ 3

8and (g ± f)¡1(x) =

x+ 3

8

13 a Is not b Is c Is d Is e Is

14 b i is the only one

x-int ¡1, 5, y-int ¡ 25

9

no x-intercepts, y-intercept 1

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Y:\HAESE\IB_SL-2ed\IB_SL-2ed_an\697IB_SL-2_an.CDR Wednesday, 18 March 2009 4:04:07 PM PETER

698 ANSWERS

5 a a = 2, b = ¡1

b Domain = fx j x 6= 2g, Range = fy j y 6= ¡1g6 a vertical asymptote x = 2, horizontal asymptote y = ¡4

b as x ! 2¡, y ! 1 as x ! 1, y ! ¡4¡

as x ! 2+, y ! ¡1 as x ! ¡1, y ! ¡4+

c x-intercept ¡ 1

4, y-intercept 1

2

d

7 a (g ± f)(x) =2

3x+ 1b x = ¡ 1

2

c i vertical asymptote x = ¡ 1

3,

horizontal asymptote y = 0

ii

iii Domain = fx j x 6= ¡ 1

3g, Range = fy j y 6= 0g

8 a b f¡1(x) =x+ 7

2

9 a

b Range = fy j 0 6 y 6 2gc i x ¼ ¡1:83 ii x = ¡3

REVIEW SET 1C

1 a

b

2 a 10¡ 6x b x = 2

3 a 1¡ 2px b

p1¡ 2x

4 a = 1, b = ¡6, c = 5

5 a b

6 a f¡1(x) =7¡ x

4b f¡1(x) =

5x¡ 3

2

7 (f¡1 ± h¡1)(x) = (h ± f)¡1(x) =4x+ 6

158 16

EXERCISE 2A

1 a 4, 13, 22, 31 b 45, 39, 33, 27

c 2, 6, 18, 54 d 96, 48, 24, 12

2 a Starts at 8 and each term is 8 more than the previous term.

Next two terms 40, 48.

b Starts at 2, each term is 3 more than the previous term; 14,

17.

c Starts at 36, each term is 5 less than the previous term; 16,

11.

d Starts at 96, each term is 7 less than the previous term; 68,

61.

e Starts at 1, each term is 4 times the previous term; 256,

1024.

f Starts at 2, each term is 3 times the previous term; 162, 486.

g Starts at 480, each term is half the previous term; 30, 15.

h Starts at 243, each term is 1

3of the previous term; 3, 1.

i Starts at 50 000, each term is 1

5of the previous term; 80,

16.

3 a Each term is the square of the term number; 25, 36, 49.

b Each term is the cube of the term number; 125, 216, 343.

c Each term is n(n + 1) where n is the term number; 30,

42, 56.

4 a 79, 75 b 1280, 5120 c 625, 1296

d 13, 17 e 16, 22 f 14, 18

EXERCISE 2B

1 a 2, 4, 6, 8, 10 b 4, 6, 8, 10, 12

c 1, 3, 5, 7, 9 d ¡1, 1, 3, 5, 7

e 5, 7, 9, 11, 13 f 13, 15, 17, 19, 21

g 4, 7, 10, 13, 16 h 1, 5, 9, 13, 17

2 a 2, 4, 8, 16, 32 b 6, 12, 24, 48, 96

c d ¡2, 4, ¡8, 16, ¡32

3 17, 11, 23, ¡1, 47

EXERCISE 2C

1 a 73 b 65 c 21 1

2

2 a 101 b ¡107 c a+ 14d

3 a u1 = 6, d = 11 b un = 11n¡ 5 c 545

d yes, u30 e no

4 a

5 b u1 = 1, d = 3 c 169 d u151 = 451

6 b u1 = 32, d = ¡ 7

2c ¡227 d n > 68

7 a k = 17 1

2b k = 4 c k = 4 d k = 0

e k = f

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y

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u1 = 87, d = ¡4, b un = 91¡4n c ¡69 d u97

¡2 or 3 k = ¡1 or 3

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ANSWERS 699

8 a un = 6n¡ 1 b un = ¡ 3

2n+ 11

2

c un = ¡5n+ 36 d un = ¡ 3

2n+ 1

2

9 a 6 1

4, 7 1

2, 8 3

4b 3 5

7, 8 3

7, 13 1

7, 17 6

7, 22 4

7, 27 2

7

10 a u1 = 36, d = ¡ 2

3b 100 11 100 006

EXERCISE 2D.1

1 a b = 18, c = 54 b b = 2 1

2, c = 1 1

4

c b = 3, c = ¡1 1

2

2 a 96 b 6250 c 16

3 a 6561 b 19 683

64c 16 d ar8

4 a u1 = 5, r = 2 b un = 5£ 2n¡1, u15 = 81 920

5 a u1 = 12, r = ¡ 1

2

b un = 12£ (¡ 1

2)n¡1, u13 = 3

1024

6 u1 = 8, r = ¡ 3

4, u10 = ¡0:600 677 49

7 u1 = 8, r = 1p2

, un = 272¡n

2

8 a k = §14 b k = 2 c k = ¡2 or 4

9 a un = 3£ 2n¡1 b un = 32£ (¡ 1

2)n¡1

c un = 3£ (§p2)n¡1 d un = 10£ (§p

2)1¡n

10 a u9 = 13 122 b u14 = 2916p3 ¼ 5050:66

c u18 ¼ 0:000 091 55

EXERCISE 2D.2

1 a $3993:00 b $993:00 2 E11 470:393 a U43 923 b U13 923 4 $23 602:32

5 U148 024:43 6 $51 249:06 7 $14 976:01

8 $ 11 477:02 9 E19 712:33 10 U19 522:47

EXERCISE 2D.3

1 a i 1550 ants ii 4820 ants b 12:2 weeks

2 a 278 animals b Year 2044

EXERCISE 2E.1

1 a i Sn = 3 + 11 + 19 + 27 + ::::+ (8n¡ 5) ii 95

b i Sn = 42 + 37 + 32 + ::::+ (47¡ 5n) ii 160

c i Sn = 12 + 6 + 3 + 1 1

2+ ::::+ 12( 1

2)n¡1 ii 23 1

4

d i Sn = 2 + 3 + 4 1

2+ 6 3

4+ ::::+ 2( 3

2)n¡1 ii 26 3

8

e i Sn = 1 + 1

2+ 1

4+ 1

8+ ::::+

1

2n¡1ii 1 15

16

f i Sn = 1 + 8 + 27 + 64 + ::::+ n3 ii 225

2 a 10 b 25 c 168 d 310 320P

n=1

(3n¡ 1) = 610

EXERCISE 2E.2

1 a 820 b 3087:5 c ¡1460 d ¡740

2 a 1749 b 2115 c 1410 1

2

3 a 160 b ¡630 c 135 4 203 5 ¡115:5 6 18

7 a 65 b 1914 c 47 850

8 a 14 025 b 71 071 c 3367

10 a un = 2n¡ 1 c S1 = 1, S2 = 4, S3 = 9, S4 = 16

11 56, 49 12 10, 4, ¡2 or ¡2, 4, 10

13 2, 5, 8, 11, 14 or 14, 11, 8, 5, 2

EXERCISE 2E.3

1 a 23:9766 ¼ 24:0 b ¼189 134

c ¼4:000 d ¼0:5852

2 a Sn =3 +

p3

2

¡(p3)n ¡ 1

¢b Sn = 24(1¡ ( 1

2)n)

c Sn = 1¡ (0:1)n d Sn = 40

3(1¡ (¡ 1

2)n)

3 a u1 = 3 b r = 1

3c u5 = 1

27

4 a 3069 b 4095

1024¼ 3:999 c ¡134 217 732

5 c $26 361:59

6 a 1

2, 3

4, 7

8, 15

16, 31

32b Sn =

2n ¡ 1

2n

c 1¡ ( 12)n =

2n ¡ 1

2nd as n ! 1, Sn ! 1

EXERCISE 2E.4

1 a i u1 = 3

10ii r = 0:1 b S = 1

3

2 a 4

9b 16

99c 104

3334 a 54 b 14:175

5 a 1 b 4 2

76 u1 = 9, r = 2

3

7 u1 = 8, r = 1

5and u1 = 2, r = 4

5

8 b Sn = 1 + 18(1¡ (0:9)n¡1) c 19 seconds

REVIEW SET 2A

1 a arithmetic b arithmetic and geometric

c geometric d neither e arithmetic

2 k = ¡ 11

23 un = 33¡ 5n, Sn = n

2(61¡ 5n)

4 k = § 2p3

35 un = 1

6£ 2n¡1 or ¡ 1

6£ (¡2)n¡1

6 21, 19, 17, 15, 13, 11

7 a un = 89¡ 3n b un =2n+ 1

n+ 3

c un = 100(0:9)n¡1

8 a 1 + 4 + 9 + 16 + 25 + 36 + 49

b 4

3+ 5

4+ 6

5+ 7

6+ 8

7+ 9

8+ 10

9+ 11

10

9 a 10 4

5b 16 + 8

p2 10 18 metres

11 a un = 3n+ 1

REVIEW SET 2B

1 b u1 = 6, r = 1

2c 0:000 183

2 a 81 b ¡1 1

2c ¡486

3 a 1587 b 47 253

256¼ 47:99 4 u12 = 10 240

5 a E8415:31 b E8488:67 c E8505:75

6 a 42 b un+1 ¡ un = 5 d 1672

7 un =¡3

4

¢2n¡1 a 49 152 b 24 575:25

8 u11 = 8

19 683¼ 0:000 406 9 a 17 b 255 511

512¼ 256:0

10 a 1331

2100¼ 0:634 b 6 8

1511 $13 972:28

12 a 3470 b Year 2014

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700 ANSWERS

REVIEW SET 2C

1 a d = ¡5 b u1 = 63, d = ¡5 c ¡117

d u54 = ¡202

2 a u1 = 3, r = 4 b un = 3£ 4n¡1, u9 = 196608

3 un = 73¡ 6n, u34 = ¡131

4 anP

k=1

(7k ¡ 3) bnP

k=1

¡1

2

¢k+15 a 70 b ¼ 241:2

6 64

18757 12 8 a $18 726:65 b $18 855:74

9 a u1 = 54, r = 2

3and u1 = 150, r = ¡ 2

5

b jrj < 1 in both cases, so the series will converge.

For u1 = 54, r = 2

3, S = 162

For u1 = 150, r = ¡ 2

5, S = 107 1

7

10 a 35:5 km b 1183 km 11 a 0 < x < 1 b 35 5

7

EXERCISE 3A

1 a 21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32, 26 = 64

b 31 = 3, 32 = 9, 33 = 27, 34 = 81, 35 = 243,

36 = 729c 41 = 4, 42 = 16, 43 = 64, 44 = 256, 45 = 1024,

46 = 4096

2 a 51 = 5, 52 = 25, 53 = 125, 54 = 625

b 61 = 6, 62 = 36, 63 = 216, 64 = 1296

c 71 = 7, 72 = 49, 73 = 343, 74 = 2401

EXERCISE 3B

1 a ¡1 b 1 c 1 d ¡1 e 1 f ¡1

g ¡1 h ¡32 i ¡32 j ¡64 k 625 l ¡625

2 a 16 384 b 2401 c ¡3125 d ¡3125

e 262 144 f 262 144 g ¡262144

h 902:436 039 6 i ¡902:436 039 6

j ¡902:436 039 6

3 a 0:1 b 0:1 c 0:027 d 0:027

e 0:012 345 679 f 0:012 345 679 g 1 h 1

4 3 5 7

6 a Yes, N = 2s¡1 b 239 ¼ 5:50£ 1011

c 264 ¡ 1 ¼ 1:84£ 1019

EXERCISE 3C

1 a 511 b d8 c k5 d 1

7e x10 f 316

g p¡4 h n12 i 53t j 7x+2 k 103¡q l c4m

2 a 22 b 2¡2 c 23 d 2¡3 e 25 f 2¡5

g 21 h 2¡1 i 26 j 2¡6 k 27 l 2¡7

3 a 32 b 3¡2 c 33 d 3¡3 e 31 f 3¡1

g 34 h 3¡4 i 30 j 35 k 3¡5

4 a 2a+1 b 2b+2 c 2t+3 d 22x+2 e 2n¡1

f 2c¡2 g 22m h 2n+1 i 21 j 23x¡1

5 a 3p+2 b 33a c 32n+1 d 3d+3 e 33t+2

f 3y¡1 g 31¡y h 32¡3t i 33a¡1 j 33

6 a 4a2 b 27b3 c a4b4 d p3q3 em2

n2

fa3

27g

b4

c4h 1 i

m4

81n4j

x3y3

8

7 a 4a2 b 36b4 c ¡8a3 d ¡27m6n6

e 16a4b16 f¡8a6

b6g

16a6

b2h

9p4

q6

8 aa

b2b

1

a2b2c

4a2

b2d

9b2

a4e

a2

bc2

fa2c2

bg a3 h

b3

a2i

2

ad2j 12am3

9 a a¡n b bn c 3n¡2 d anbm e a¡2n¡2

10 a 1 b 4

7c 6 d 27 e 9

16f 5

2

g 27

125h 151

5

11 a 3¡2 b 2¡4 c 5¡3 d 3£ 5¡1 e 22 £ 3¡3

f 2c¡3£3¡2 g 32k £2¡1 £5¡1 h 2p£3p¡1£5¡2

12 a 53 = 21 + 23 + 25 + 27 + 29

b 73 = 43 + 45 + 47 + 49 + 51 + 53 + 55

c 123 = 133 + 135 + 137 + 139 + 141 + 143 + 145 + 147+149 + 151 + 153 + 155

EXERCISE 3D

1 a 215 b 2

¡ 15 c 2

32 d 2

52 e 2

¡ 13

f 243 g 2

32 h 2

32 i 2

¡ 43 j 2

¡ 32

2 a 313 b 3

¡ 13 c 3

14 d 3

32 e 3

¡ 52

3 a 713 b 3

34 c 2

45 d 2

53 e 7

27

f 7¡ 1

3 g 3¡ 3

4 h 2¡ 4

5 i 2¡ 5

3 j 7¡ 2

7

4 a 2:28 b 1:83 c 0:794 d 0:435 e 1:68

f 1:93 g 0:523

5 a 8 b 32 c 8 d 125 e 4

f 1

2g 1

27h 1

16i 1

81j 1

25

EXERCISE 3E.1

1 a x5 + 2x4 + x2 b 22x + 2x c x+ 1

d 72x + 2(7x) e 2(3x)¡ 1 f x2 + 2x+ 3

g 1 + 5(2¡x) h 5x + 1 i x32 + x

12 + 1

2 a 4x + 22+x + 3 b 9x + 7(3x) + 10

c 25x ¡ 6(5x) + 8 d 4x + 6(2x) + 9

e 9x ¡ 2(3x) + 1 f 16x + 14(4x) + 49

g x¡ 4 h 4x ¡ 9 i x¡ x¡1 j x2 + 4 +4

x2

k 72x ¡ 2 + 7¡2x l 25¡ 10(2¡x) + 4¡x

EXERCISE 3E.2

1 a 5x(5x + 1) b 10(3n) c 7n(1 + 72n)

d 5(5n ¡ 1) e 6(6n+1 ¡ 1) f 16(4n ¡ 1)

2 a (3x+2)(3x¡2) b (2x+5)(2x¡5) c (4+3x)(4¡3x)

d (5+2x)(5¡ 2x) e (3x +2x)(3x ¡ 2x) f (2x +3)2

g (3x + 5)2 h (2x ¡ 7)2 i (5x ¡ 2)2

3 a (2x + 3)(2x + 6) b (2x + 4)(2x ¡ 5)

c (3x + 2)(3x + 7) d (3x + 5)(3x ¡ 1)

e (5x + 2)(5x ¡ 1) f (7x ¡ 4)(7x ¡ 3)

4 a 2n b 10a c 3b d1

5ne 5x f ( 3

4)a

g 5 h 5n

5 a 3m + 1 b 1 + 6n c 4n + 2n d 4x ¡ 1

e 6n f 5n g 4 h 2n ¡ 1 i 1

2

6 a n 2n+1 b ¡3n¡1

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Y:\HAESE\IB_SL-2ed\IB_SL-2ed_an\700IB_SL-2_an.CDR Tuesday, 17 March 2009 3:01:04 PM PETER

ANSWERS 701

EXERCISE 3F

1 a x = 3 b x = 2 c x = 4 d x = 0

e x = ¡1 f x = ¡2 g x = ¡3 h x = 2

i x = ¡3 j x = ¡4 k x = 2 l x = 1

2 a x = 5

3b x = ¡ 3

2c x = ¡ 3

2d x = ¡ 1

2

e x = ¡ 2

3f x = ¡ 5

4g x = 3

2h x = 5

2

i x = 1

8j x = 9

2k x = ¡4 l x = ¡4

m x = 0 n x = 7

2o x = ¡2 p x = ¡6

3 a x = 1

7b has no solutions c x = 2 1

2

4 a x = 3 b x = 3 c x = 2

d x = 2 e x = ¡2 f x = ¡2

5 a x = 1 or 2 b x = 1 c x = 1 or 2

d x = 1 e x = 2 f x = 0

EXERCISE 3G

1 a 1:4 b 1:7 c 2:8 d 0:3 e 2:7 f 0:4

2 a b

c d

3 a b

c d

4 a b

c d

5 a y ¼ 3:67 b y ¼ ¡0:665 c y ¼ 3:38

d y ¼ 2:62

6 a as x ! 1, y ! 1as x ! ¡1, y ! 1 (above) HA is y = 1

b as x ! 1, y ! ¡1as x ! ¡1, y ! 2 (below) HA is y = 2

c as x ! 1, y ! 3 (above)

as x ! ¡1, y ! 1 HA is y = 3

d as x ! 1, y ! 3 (below)

as x ! ¡1, y ! ¡1 HA is y = 3

EXERCISE 3H.1

1 a 100 grams

b i 132 g

ii 200 g

iii 528 g

c

2 a 50

b i 76ii 141

iii 400

c

3 a V0 b 2V0 c 100%d 183% increase, percentage increase at 50oC compared

with 20oC

4 a 12 bears b 146 bears c 248% increase

EXERCISE 3H.2

1 a 250 g b i 112 g ii 50:4 g iii 22:6 g

c d ¼ 346 years

2 a 100oC

b i 81:2oC

ii 75:8oC

iii 33:9oC

c

x

y

@\=\2!

1

-1

1

@\=\-2

@\=\2 -2!

x

y

@\=\2!

1

2

@\ \��!

x

y

@\=\2!

1Qr_@\=\2!-2

x

y

@\=\3!

@\=\12

1

@\=\3 +1!

x

y

@\=\2 +1!@\=\1

2

x

y

@\=\2-2!

@\=\2

1

x

y@\=\3!

@\=\3!-1

1

Qe_

x

y@\=\3!

@\=\-3!

1

-1

x

y

@\=\2-!��

@\=\3

4

x

y

@\=\3-2 !-

@\=\32

x

y

@\=\2!

1

xy �� 2

x

y

@\=\3!

1

@\=\3 !-

Wt (grams)

t (hours)(4' 132"0)(10' 200)

(24' 528)

Wtt\=\100 2�� 0.1

100

Pn

n (years)

Pnn\=\50 2�� 0.3

(2' 76)(5' 141)

(10' 400)

50

Tt (°C)

t (min)

Ttt\=\100 2�� 0.02

(20' 75"8)

(78' 33"9)(15' 81"2)

100

1000

250

200

150

100

50

W t( )

t

( )��,

( )�� , .( )��, .

W t( ) ( )�� t

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702 ANSWERS

3 a 1000 g

b i 812 g

ii 125 g

iii 9:31£ 10¡7 g

c

4 a W0 b 12:9%

EXERCISE 3I

1 e1 ¼ 2:718 281 828 :::: 2

The graph of y = ex lies between y = 2x and y = 3x.

3 One is the other

reflected in

the y-axis.

4 a

5 a ex > 0 for all x

b i 0:000 000 004 12 ii 970 000 000

6 a ¼ 7:39 b ¼ 20:1 c ¼ 2:01 d ¼ 1:65

e ¼ 0:368

7 a e12 b e

32 c e

¡ 12 d e¡2

8 a e0:18t b e0:004t c e¡0:005t d ¼ e¡0:167t

9 a 10:074 b 0:099 261 c 125:09 d 0:007 994 5

e 41:914 f 42:429 g 3540:3 h 0:006 342 4

10

Domain of f , g and h is fx j x 2 R gRange of f is fy j y > 0g, Range of g is fy j y > 0gRange of h is fy j y > 3g

11

Domain of f , g and h is fx j x 2 R gRange of f is fy j y > 0g, Range of g is fy j y < 0gRange of h is fy j y < 10g

12 a i 2 g ii 2:57 g iii 4:23 g iv 40:2 g

b

13 a i 64:6 amps

ii 16:7 amps

b

c 28:8 seconds

14 f¡1(x) = loge x

REVIEW SET 3A

1 a ¡1 b 27 c 2

32 a a6b7 b

2

3xc

y2

5

3 a 2¡4 b 2x+2 c 22x¡3

4 a1

x5b

2

a2b2c

2a

b25 a 33¡2a b 3

52¡ 9

2x

6 a 4 b 1

97 a

m

n2b

1

m3n3c

m2p2

nd

16n2

m2

8 a 9¡ 6(2a) + 22a b x¡ 4 c 2x + 1

9 a x = ¡2 b x = 3

410 a x = 1

3b x = ¡ 4

5

REVIEW SET 3B

1 a 2n+2 b ¡ 6

7c 3 3

8d

4

a2b4

2 a 2:28 b 0:517 c 3:16 3 a 3 b 24 c 3

4

4 a 34 b 30 c 3¡3 d 3¡5

5 a y = 2x has y-intercept 1and horizontal asymptote

y = 0

b y = 2x ¡ 4 has

y-intercept ¡3 and

horizontal asymptote

y = ¡4

6 a 80oC

b i 26:8oC

ii 9:00oC

iii 3:02oC

d ¼ 12:8 min

c

7 a x ¡2 ¡1 0 1 2

y ¡4 8

9¡4 2

3¡4 ¡2 4

b as x ! 1, y ! 1; as x ! ¡1, y ! ¡5 (above)

c d y = ¡5

Wtt\ \��.

Wt (grams)

t (years)

(10' 812)

(100' 125)

1000

x

yy e� x

y 2� x

y 3� x

x

y

y e� xy e� �x

y

x

y x��ƒ

ƒ��

1

1

75

I

I��

t

tetI 15.075)( ��

y

x

42 �� xy

4��y

2� xy

1

-3

y

x

�y� �= ��

53 �� xy

y

x

y � 10

y e�� x

y e�� x

y e�� x

y

x

2�� xey

1

y e�� x

y e�� x

y��

40302010

80

60

40

20

T (°C)

t (minutes)

T�� ( ). t

W

t2

2t

W t e( )��

( )� �� . , .

( )� ��. , .

IB SL 2nd ed


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ANSWERS 703

8

Domain of f , g and h is fx j x 2 R gRange of f is fy j y > 0g Range of g is fy j y > 0gRange of h is fy j y < 3g

9 a x ¡2 ¡1 0 1 2

y ¡1 1 2 2 1

22 3

4

b as x ! 1, y ! 3 (below); as x ! ¡1, y ! ¡1c d y = 3

10 a 1500 g

b i 90:3 g

ii

d 386 years

c

REVIEW SET 3C

1 a 8 b ¡ 4

52 a a21 b p4q6 c

4b

a3

3 a 2¡3 b 27 c 212 4 a1

b3b

1

abc

a

b

5 22x 6 a 50 b 532 c 5¡

14 d 52a+6

7 a 1 + e2x b 22x + 10(2x) + 25 c x¡ 49

8 a x = 5 b x = ¡4 9 a x = 4 b x = ¡2

5

10 a x ¡2 ¡1 0 1 2

y 15:8 6:44 3 1:74 1:27

b as x ! 1, y ! 1 (above); as x ! ¡1, y ! 1c d y = 1

EXERCISE 4A

1 a 104 = 10000 b 10¡1 = 0:1 c 1012 =

p10

d 23 = 8 e 2¡2 = 1

4f 31:5 =

p27

2 a log2 4 = 2 b log2(1

8) = ¡3

c log10(0:01) = ¡2 d log7 49 = 2

e log2 64 = 6 f log3(1

27) = ¡3

3 a 5 b ¡2 c 1

2d 3 e 6 f 7 g 2

h 3 i ¡3 j 1

2k 2 l 1

2m 5 n 1

3

o n p 1

3q ¡1 r 3

2s 0 t 1

4 a ¼ 2:18 b ¼ 1:40 c ¼ 1:87 d ¼ ¡0:0969

5 a x = 8 b x = 2 c x = 3 d x = 14

6 a 2 b 2 c ¡1 d 3

4e ¡ 1

2f 5

2g ¡ 3

2h ¡ 3

4

EXERCISE 4B

1 a 4 b ¡3 c 1 d 0 e 1

2f 1

3g ¡ 1

4h 1 1

2

i 2

3j 1 1

2k 1 1

3l 3 1

2m n n a+2 o 1¡m

p a¡ b

2 The following include calculator keys for the TI-84+ :

a log 10 000 enter , 4 b log 0:001 enter , ¡3

c log 2ndp

10 ) ) enter , 0:5

d log 10 ^ ( 1 ¥ 3 ) ) enter , 0:¹3

e log 100 ^ ( 1 ¥ 3 ) ) enter , 0:¹6

f log 10 £ 2ndp

10 ) ) enter , 1:5

g log 1 ¥ 2ndp

10 ) ) enter , ¡0:5

h log 1 ¥ 10 ^ 0:25 ) enter , ¡0:25

3 a 100:7782 b 101:7782 c 103:7782

d 10¡0:2218 e 10¡2:2218 f 101:1761

g 103:1761 h 100:1761 i 10¡0:8239

a 10¡3:8239

4 a i 0:477 ii 2:477 b log 300 = log(3£ 102)

5 a i 0:699 ii ¡1:301 b log 0:05 = log(5£ 10¡2)

6 a x = 100 b x = 10 c x = 1

d x = 1

10e x = 10

12 f x = 10

¡ 12

g x = 10000 h x = 0:000 01 i x ¼ 6:84

j x ¼ 140 k x ¼ 0:0419 l x ¼ 0:000 631

EXERCISE 4C.1

1 a log 16 b log 4 c log 8 d log 20

e log 2 f log 24 g log 30 h log 0:4

i log 10 j log 200 k log 0:4 l log 1 or 0

m log 0:005 n log 20 o log 28

2 a log 96 b log 72 c log 8 d log¡25

8

¢e log 6 f log 1

2g log 20 h log 25

i 1

3 a 2 b 3

2c 3 d 1

2e ¡2 f ¡ 3

2

5 a p+ q b 2p+ 3q c 2q + r d r + 1

2q ¡ p

e r ¡ 5p f p¡ 2q

6 a x+ z b z + 2y c x+ z ¡ y d 2x+ 1

2y

e 3y ¡ 1

2z f 2z + 1

2y ¡ 3x

7 a 0:86 b 2:15 c 1:075

EXERCISE 4C.2

1

x

y � ��( )x ex

g x e( )�� x��

h( )x e�� x

��

800400

1500W (grams)

t (years)

tW )993.0(1500��

x

y

y��y e�� x

3

x

y

y��y��x

2

5:44 g

a log y = x log 2 b log y ¼ 1:301 + 3 log b

c logM = log a+ 4 log d d logT ¼ 0:699 + 1

2log d

e logR = log b+ 1

2log l f logQ = log a¡ n log b

g log y = log a+ x log b h logF ¼ 1:30¡ 1

2logn

i logL = log a+log b¡log c j logN = 1

2log a¡ 1

2log b

k logS ¼ 2:30 + t log 2 l log y = m log a¡ n log b

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704 ANSWERS

2 a D = 2e b F =5

tc P =

px d M = b2c

e B =m3

n2f N =

13pp

g P = 10x3 h Q =100

x

3 a x = 9 b x = 2 or 4 c x = 25p5

d x = 200 e x = 5 f x = 3

EXERCISE 4D.1

1 a 3 b 0 c 1

3d ¡2

3 x does not exist such that ex = ¡2 or 0

4 a a b a+ 1 c a+ b d ab e a¡ b

5 a e1:7918 b e4:0943 c e8:6995 d e¡0:5108

e e¡5:1160 f e2:7081 g e7:3132 h e0:4055

i e¡1:8971 j e¡8:8049

6 a x ¼ 20:1 b x ¼ 2:72 c x = 1

d x ¼ 0:368 e x ¼ 0:006 74 f x ¼ 2:30

g x ¼ 8:54 h x ¼ 0:0370

EXERCISE 4D.2

1 a ln 45 b ln 5 c ln 4 d ln 24

e ln 1 = 0 f ln 30 g ln 4e h ln¡6

e

¢i ln 20 j ln 4e2 k ln

¡20

e2

¢l ln 1 = 0

2 a ln 972 b ln 200 c ln 1 = 0 d ln 16 e ln 6

f ln¡1

3

¢g ln

¡1

2

¢h ln 2 i ln 16

3 For example, for a, ln 27 = ln 33 = 3 ln 3

5 a D = ex b F =e2

pc P =

px

d M = e3y2 e B =t3

ef N =

13pg

g Q ¼ 8:66x3 h D ¼ 0:518n0:4

EXERCISE 4E

1 a x ¼ 3:32 b x ¼ 2:73 c x ¼ 3:32

d x ¼ 37:9 e x ¼ ¡3:64 f x ¼ ¡7:55

g x ¼ 7:64 h x ¼ 32:0 i x ¼ 1150

2 a t ¼ 6:340 b t ¼ 74:86 c t ¼ 8:384

d t ¼ 132:9 e t ¼ 121:5 f t ¼ 347:4

3 a x ¼ 2:303 b x ¼ 6:908 c x ¼ ¡4:754

d x ¼ 3:219 e x ¼ 15:18 f x ¼ ¡40:85

g x ¼ ¡14:63 h x ¼ 137:2 i x ¼ 4:868

EXERCISE 4F

1 a ¼ 2:26 b ¼ ¡10:3 c ¼ ¡2:46 d ¼ 5:42

2 a x ¼ ¡4:29 b x ¼ 3:87 c x ¼ 0:139

3 a x ¼ 0:683 b x ¼ ¡1:89

4 a x = 16 b x ¼ 1:71

5 x =log 8

log 25or log25 8

EXERCISE 4G

1 a i x > ¡1, y 2 R iii

ii VA is x = ¡1,

x and y-intercepts 0

iv x = ¡ 2

3

v f¡1(x) = 3x ¡ 1

b i x > ¡1, y 2 R iii

ii VA is x = ¡1,x-intercept 2,y-intercept 1

iv x = 8

v f¡1(x) = 31¡x ¡ 1

c i x > 2, y 2 R iii

ii VA is x = 2,

x-intercept 27,

no y-intercept

iv x = 7v f¡1(x) = 52+x + 2

d i x > 2, y 2 R iii

ii VA is x = 2,

x-intercept 7,

no y-intercept

iv x = 27

v f¡1(x) = 51¡x + 2

e i x 2 R , x 6= 0,

y 2 Riii

ii VA is x = 0,

x-intercepts §p2,

no y-intercept

iv x = §2

v if x > 0, f¡1(x) = 21¡x2

if x < 0, f¡1(x) = ¡21¡x2

2 a i f¡1(x)

= ln(x¡ 5)

ii

iii domain of f is

fx j x 2 R g,

range is fy j y > 5gdomain of f¡1 is

fx j x > 5g,

range is fy j y 2 R giv f has a HA y = 5,

f¡1 has a VA x = 5

b i f¡1(x)

= ln(x+ 3)¡ 1

ii

iii domain of f is

fx j x 2 R g,

range is

fy j y > ¡3gdomain of f¡1 is

fx j x > ¡3g,

range is

fy j y 2 R giv f has a HA y = ¡3, f¡1 has a VA x = ¡3

c i f¡1(x)

= ex+4

ii

iii domain of f is

fx j x > 0g,

range of f is

fy j y 2 R gdomain of f¡1 is

fx j x 2 R g,

range is

fy j y > 0giv f has a VA x = 0, f¡1 has a HA y = 0

y

x

y x��

x��

y��

�

5)( �� xexf

1�f

y

x

y x��

x��

y��

3)( 1 �� xexf

1�f

y

x

e4

y x��

1�f

f

e4

��

y

x

y x�� log�( )

��

��

��

y

x

y x�� log�( )

��

��

y

x

y x�� log ( )

��

��

y

x

y x�� log ( )

y

x~`2-~`2

y x�� log� X

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ANSWERS 705

d i f¡1(x)

= 1 + ex¡2

ii

iii domain of f is

fx j x > 1g,

range is

fy j y 2 R gdomain of f¡1 is

fx j x 2 R g,

range is fy j y > 1giv f has a VA x = 1,

f¡1 has a HA y = 1

3 f¡1(x) = 1

2lnx

a (f¡1 ± g)(x) = 1

2ln(2x¡ 1)

b (g ± f)¡1(x) = 1

2ln

³x+ 1

2

´4 a A is y = lnx

as its x-intercept

is 1

b

c y = lnx has

VA x = 0y = ln(x¡ 2)has VA x = 2y = ln(x+ 2)has VA x = ¡2

5 y = ln(x2) = 2 lnx, so she is correct.

This is because the y-values are twice as large for y = ln(x2)as they are for y = lnx.

6 a f¡1 : x 7! ln(x¡ 2)¡ 3

b i x < ¡5:30 ii x < ¡7:61 iii x < ¡9:91

iv x < ¡12:2 Conjecture HA is y = 2

c as x ! ¡1, ex+3 ! 0 and y ! 2 ) HA is y = 2

d VA of f¡1 is x = 2, domain of f¡1 is fx j x > 2gEXERCISE 4H

1 a 3:90 h

b 15:5 h

2 a

b 13:9 h

3 a see graph

alongside

b n ¼ 2:82

4 In 6:17 years, or 6 years 62 days

5 8:65 years, or 8 years 237 days

6 a8:4%

12= 0:7% = 0:007 r = 1 + 0:007 = 1:007

b after 74 months

7 a E12 000 b A6 = E17 919:50

c A3:25 is the value after 2 years 3 months

d 8:64 years e

8 a 17:3 years b 92:2 years c 115 years 9 8:05 s

10 a 50:7 min b 152 min

11 a 25 years b 141 years c 166 years

12 a 10 000 years b 49 800 years

13 166 seconds 14

REVIEW SET 4A

1 a 3 b 8 c ¡2 d 1

2e 0

f 1 g 1

4h ¡1 i 1

3j 1

2

2 a 1

2b ¡ 1

3c a+ b+ 1

3 a ln 144 b ln¡3

2

¢c ln

³25

e

´d ln 3

4 a 3

2b ¡3 c ¡ 3

2

5 a log 144 b log2¡16

9

¢c log4 80

6 a logP = log 3 + x log b b logm = 3 logn¡ 2 log p

7 a 2x b 2 + x c 1¡ x

8 a T =x2

yb K = n

pt

9 a 5 ln 2 b 3 ln 5 c 6 ln 3

10 a 2A+ 2B b A+ 3B c 3A+ 1

2B

d 4B ¡ 2A e 3A¡ 2B

REVIEW SET 4B

1 a ¼ 101:51 b ¼ 10¡2:89 c ¼ 10¡4:05

2 a x = 1

8b x ¼ 82:7 c x ¼ 0:0316

3 a k ¼ 3:25£ 2x b Q = P 3R c A ¼ B5

4004 a x ¼ 1:209 b x ¼ 1:822

6 x ¼ 2:32

7 a x ¼ 148b x ¼ 0:513

8 a x ¼ 5:99

b x ¼ 0:699

c x ¼ 6:80 d x ¼ 1:10 or 1:39

9 a 3 years b 152%

10 a g¡1(x)

= ln

³x+ 5

2

´ b

c domain of g is fx j x 2 R g,

range is fy j y > ¡5gdomain of g¡1 is

fx j x > ¡5g,

range is fy j y 2 R g

REVIEW SET 4C

1 a 3

2b 2

3c a+ b

2 a 5 b 1

2c ¡1

3 a ¼ e3:00 b ¼ e8:01 c ¼ e¡2:59

4 a x = 1000 b x ¼ 4:70 c x ¼ 6:28

5 a ln 24 b ln 3 c ln 4 d ln 125

y

x

x��

y��

y x��

1�f

2)�ln()( �� xxf

y

x

)2ln( �� xy

ln� xy)2ln( �� xy

��

12 000�

An1

Ann

�� ( ).

n

) approximately 2:8 weeks

n4321

12000

10000

8000

6000

4000

2000

AnA en

n�� .

––––3000

t�

Wt = 2500 × 3

t

Wt

x��

y��

1�g

52)( �� exg x

x

y

6:93 h

12:9 seconds

5 a 2500 g d

b 3290 years

c 42:3%

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706 ANSWERS

6 a logM = log a+ n log b b logT = log 5¡ 1

2log l

c logG = 2 log a+ log b¡ log c

7 a x ¼ 5:19 b x ¼ 4:29 c x ¼ ¡0:839

8 a P = TQ1:5 b M =e1:2pN

9 a x > ¡2, y 2 R e

b VA is x = ¡2,

x-intercept is 7,

c

d g¡1(x) = 3x+2 ¡ 2

10 a 13:9 weeks b 41:6 weeks c 138 weeks

EXERCISE 5A

1 a 2x b x+ 2 cx

2d 2x+ 3

2 a 9x2 bx2

4c 3x2 d 2x2 ¡ 4x+ 7

3 a 64x3 b 4x3 c x3 + 3x2 + 3x+ 1

d 2x3 + 6x2 + 6x¡ 1

4 a 4x b 2¡x + 1 c 2x¡2 + 3 d 2x+1 + 3

5 a ¡ 1

xb

2

xc

2 + 3x

xd

2x+ 1

x¡ 1

6 a

b i ¡1 1

2ii 3 iii 2

7 a

b x-ints are ¡1 and 5y-int is ¡5

8 a,b

9 When x = 0,

y = 20 = 1 X

2x > 0 for all x as

the graph is always

above the x-axis. X

10

EXERCISE 5B.1

1 a,b c i If b > 0, the function

is translated vertically

upwards through bunits.

ii If b < 0, the function

is translated vertically

downwards jbj units.

2 a b

c d

3 a

b i If a > 0, the graph is translated a units right.

ii If a < 0, the graph is translated jaj units left.

4 a b

��

��

y��y��

x��x��

��

��

y

x

y x��

y

x-1 5

-5

( ) 922 �� xy

( )��

3

-1\Qw_

y

x

y x��

y

x

1

xy 2�

y

x1

y

x

2

-3 3)( 2 �� xxf

2)( 2 �� xxf

)( 2� xxf

y

x

2)( �� xfy

1)( �� xfy

f )(� xy

11

��

22

y

x

2)( �� xfy

1)( �� xfy

f )(� xy

��

��

y

x

2)( �� xfy

1)( �� xfyf )(� xy

-2-2

11

y x�� ln

y

x-2 3)3( �� xfy)2( �� xfy

2� xy

y

x-21

)1( �� xfy

)2( �� xfy

3� xyy

x2

)1( �� xfy

)2( �� xfy

ln� xy

11-1-1

x��x��

y

x

1)( �� xfy

f )(� xy

2)( �� xfy

��

��

y

xy x�� log�( )

y

x

-5

(1' 0)

(2'-1)

1 Tw_

y x x x�� C X

When y = 0,

lnx = 0

) x = e0 = 1.

��

y-intercept is ¼ ¡1:37

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V:\BOOKS\IB_books\IB_SL-2ed\IB_SL-2ed_an\706IB_SL-2_an.CDR Monday, 30 November 2009 10:42:59 AM PETER

ANSWERS 707

c d

y = f(x¡ a) is a horizontal translation of y = f(x)

through

³a0

´:

5 a

b

c

6 A translation of

³2¡3

´.

a b

7 a i (3, 2) ii (0, 11) iii (5, 6)

b i (¡2, 4) ii (¡5, 25) iii (¡1 1

2, 2 1

4)

EXERCISE 5B.2

1 a b

c d

e

2 a b

c

3 p affects the vertical stretching of the graph of y = f(x) by a

factor of p. If p > 1 the graph moves further away from the

x-axis. If 0 < p < 1 the graph moves closer to the x-axis.

4 a b

y

x

)1( �� xfy

)2( �� xfy

1� xyx 1�

x 2��

x

y

4��y

3�y

3)2( �� xfy

4)1( �� xfy

� ey x

y

x

)(3 xfy �

)(2 xfy �

2xy � �f x�( )

x

y

)(2 xfy �

)(3 xfy �

xf x e�( ) �y �

y

x

)(2 xfy �

)(3 xfy �

3xy � �f x�( )

x

y

)(2 xfy �

)(3 xfy �

xy 1� �f x�( )

x

y

)(2 xfy �

)(3 xfy �

f x x�( ) ln�y �

x

y

)(41 xfy �

( )21 xfy �

2xy � �f x�( )

x

y

)(41 xfy �

)(21 xfy �

3xy � �f x�( )

xf x e�( ) �y �

x

y

)(41 xfy �

)(21 xfy �

y

x

3)2( �� xfy

)(� xfy

y

x

)(� xfy

3)2( �� xfy x

y

& *2

2� xy

2� xy

x

y

& *22� xy

2� xy

x

y

4��y

3�y

3)2( �� xfy

4)1( �� xfy

2�x

1� xy

1��x

y

x

)1( �� xfy)2( �� xfy

2)1( 2 �� xy

x

y

-3

3)2( �� xfy

4)1( �� xfy

2� xy

7

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708 ANSWERS

c

5 a b

c

6 a b

c

7 q affects the horizontal stretching of

y = f(x) by a factor of q.

If q > 1 it moves further from the y-axis.

If 0 < q < 1 it moves closer to the y-axis.

8 a

b

c

9 a i ( 32

, ¡15) ii ( 12

, 6) iii (¡1, 3)

b i (4, 1

3) ii (¡6, 2

3) iii (¡14, 1)

EXERCISE 5B.3

1 a b

c d

e f

2 y = ¡f(x) is the reflection of y = f(x) in the x-axis.

3 a i f(¡x) = ¡2x+ 1 ii f(¡x) = x2 ¡ 2x+ 1

iii f(¡x) = ¡x3

b i ii

x

y

2)2(� xy

2� xy

x

y

2)12( �� xy

2)1( �� xy

Qw_ 1

x

y

2)32( �� xy

2)3( �� xy

�3 -Ew_

x

y

3� xy

� xy

x

y

2)3(� xy

2� xy

x

y3� xey

� xey1

x

y

3�� xy3� xy

x

y

�� ey x

� ey x

��

�

x

y

2�� xy

2� xy

x

y

ln�� xy

ln� xy

�

x

y

3 2�� xy

3 2�� xy

��

�

x

y

2)1(2 �� xy

2)1(2� �y x

�1

x

y

1

12 �� xy12 �� xy

Qw_-Qw_x

y

122 �� xxy

122 �� xxy

�1 1

22

)2( �� xy

2)2( �� xy

�4 �2

y

x

x

y

V(-1'-3)V(-1'-3)

V(2' 1)V(2' 1)

x��

x��

y x�� X

y x�� ( )X

y x�� ( )X

x

y

1)52( 241 �� xy

, 125�V & *

2� xy

yy

xx

22

)3(2 �� xy

22

)3( �� xy

2)3( �� xy

2� xy

V 3, 0( ) V ,( )��

4)3(2 2

2�� xy

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ANSWERS 709

iii

4 y = f(¡x) is the reflection of y = f(x) in the y-axis.

5 a i (3, 0) ii (2, 1) iii (¡3, ¡2)

b i (7, 1) ii (¡5, 0) iii (¡3, 2)

6 a i (¡2, ¡1) ii (0, 3) iii (1, 2)

b i (¡5, ¡4) ii (0, 3) iii (¡2, 3)

7 a A rotation about the origin through 180o.

b (¡3, 7) c (5, 1)

EXERCISE 5B.4

1 a b

c

2 a b

c

3 a A b B c D d C

4

5

6

REVIEW SET 5A

1 a 3 b 8 c 4x2 ¡ 4x d x2 + 2x e 3x2 ¡ 6x¡ 2

2 a ¡15 b 5 c ¡x2 + x+ 5

d 5¡ 1

2x¡ 1

4x2 e ¡x2 ¡ 3x+ 5

3 a b i 2

3ii ¡2 iii 3

4 g(x) = 3x3 ¡ 11x2 + 14x¡ 6

5

6

7 a

b x = ¡2 c A0(¡1, ¡1)

y

x

3xy �3xy ��

x

yy x��ƒ( )

y x��ƒ( )

�

��

x

y y x��ƒ( )

y x��ƒ( )

�

y x��ƒ( )

y x��ƒ( )

y

x

x

y

y x��ƒ( )y x�� ƒ( )x

y

y x��ƒ( )y x�� ƒ( )

��

y

x

)(

)(

)(

1)(

)(

2

21

xhy

xhy

xhy

xhy

xhy

�

��

�

��

��

)(

)1(

)(

2)(

)(

xgy

xgy

xgy

xgy

xgy

��

��

�

y

x��

��

��

y

x

��

)(

)(2

)2(

)2(

)(

21 xfy

xfy

xfy

xfy

xfy

��

��

)(21 xfy �

x

y

y��

x�� x��

y x��ƒ( )y x�� ƒ( )

y

xa b

a c��y x��( ) y x c�� ( )

b c��

c

y

x

��

y x��

y

x 2)(

)2(

)(

)(

)(

��

��

�

xfy

xfy

xfy

xfy

xfy

y

xA ,( )��

y x��( )

x��x��

A' ,( )��

y g x�� ( )

c i y = ¡1:1 ii x = 0:9

IB SL 2nd ed


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710 ANSWERS

8

REVIEW SET 5B

1 a 7 b x2 + 8x+ 14 c 3x2 ¡ 1

2 a

b i 1 and ¡3 ii ¡3 c V(¡1, ¡4)

3

4 a

i true ii false

iii false iv true

5 a

b

c

6 a i y =1

x 1¡ 2

ii

iii For y =1

x, VA is x = 0, HA is y = 0

For y =1

x¡ 1¡ 2, VA is x = 1, HA is y = ¡2

iv For y =1

x, domain is fx jx 6= 0g,

range is fy j y 6= 0gFor y =

1

x¡ 1¡ 2, domain is fx jx 6= 1g,

range is fy j y 6= ¡2g

b i y = 2x¡1 ¡ 2

ii

iii For y = 2x, HA is y = 0, no VA

For y = 2x¡1 ¡ 2, HA is y = ¡2, no VA

iv For y = 2x, domain is fx jx 2 R g,

range is fy j y > 0gFor y = 2x¡1 ¡ 2, domain is fx jx 2 R g,

range is fy j y > ¡2gc i y = log4(x¡ 1)¡ 2

ii

iii For y = log4 x, VA is x = 0, no HA

For y = log4(x¡ 1)¡ 2, VA is x = 1, no HA

iv For y = log4 x, domain is fx jx > 0g,

range is fy 2 R gFor y = log4(x¡ 1)¡ 2, domain is fx jx > 1g,

range is fy 2 R g7 a b x-intercepts ¡5 and ¡1,

y-intercept 5

c (¡3, ¡4)

8

REVIEW SET 5C

1 a ¡1 b2

xc

8

xd

10¡ 3x

x+ 2

2

y

x

�

�xy 2�

yx��

x

��

��

V��

x

y

y 2��

y 2� x

y 2 2� �x�1

1

2

x

y x 1�y xlog� 4

y xlog ( 1) 2� � �4

� ��

5

y

x

g x x( ) ( )X��

y

x��

3)2(2

)2(2

)2(

��

��

xfy

xfy

xfy

)( 2� xxf�y

x

y

2)1(3

)1(3

)1(

��

��

xfy

xfy

xfy

)( 2� xxf

1

�y

g(x) = (x¡ 1)2 + 8

fy j y > 4gfy j y > 8g

y

x

3)()2()(

1)( 2

�

�

��

xf

xf

xf

xxf�y

�y

�y

�y

y

x

)2(

)2(

)(

)(

��

��

xfy

xfy

xfy

xfy

)( 2�� xxf�y

�

¡y

x

x��

y��

xy

1�

21

1 ��

�x

y

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ANSWERS 711

3 a b (2, ¡4) and

(4, 0)

4 g(x) = ¡x2 ¡ 6x¡ 7

5

6 g(x) = x3 + 6x2 + 8x+ 10

7 a i y = 3x+ 8 ii y = 3x+ 8

b f(x+ k) = a(x+ k) + b = ax+ b+ ka = f(x) + ka

EXERCISE 6A.1

1 a x = 0, ¡ 7

4b x = 0, ¡ 1

3c x = 0, 7

3

d x = 0, 11

2e x = 0, 8

3f x = 0, 3

2

g x = 3, 2 h x = 4, ¡2 i x = 3, 7

j x = 3 k x = ¡4, 3 l x = ¡11, 3

2 a x = 2

3b x = ¡ 1

2, 7 c x = ¡ 2

3, 6

d x = 1

3, ¡2 e x = 3

2, 1 f x = ¡ 2

3,

g x = ¡ 2

3, 4 h x = 1

2, ¡ 3

2i x = ¡ 1

4, 3

j x = ¡ 3

4, 5

3k x = 1

7, ¡1 l x = ¡2, 28

15

3 a x = 2, 5 b x = ¡3, 2 c x = 0, ¡ 3

2

d x = 1, 2 e x = 1

2, ¡1 f x = 3

EXERCISE 6A.2

1 a x = ¡5§p2 b no real solns. c x = 4§ 2

p2

d x = 8§p7 e x = ¡3§p

5 f x = 2§p6

g x = ¡1§p10 h x = ¡ 1

2§ 1

2

p3 i x = 1

3§

p7

3

2 a x = 2§p3 b x = ¡3§p

7 c x = 7§p3

d x = 2§p7 e x = ¡3§p

2 f x = 1§p7

g x = ¡3§p11 h x = 4§p

6 i no real solns.

3 a x = ¡1§ 1p2

b x = 5

2§

p19

2c x = ¡2§

p7

3

d x = 1§p

7

3e x = 3

2§p

37

20f x = ¡ 1

2§

p6

2

EXERCISE 6A.3

1 a x = 2§p7 b x = ¡3§p

2 c x = 2§p3

d x = ¡2§p5 e x = 2§p

2 f x = 1

2§ 1

2

p7

g x = ¡ 4

9§

p7

9h x = ¡ 7

4§

p97

4

2 a x = ¡2§ 2p2 b x = ¡ 5

8§

p57

8c x = 5

2§

p13

2

d x = 1

2§ 1

2

p7 e x = 1

2§

p5

2f x = 3

4§

p17

4

EXERCISE 6B

1 a 2 real distinct roots b 2 real distinct roots

c 2 real distinct roots d no real roots

e a repeated root

2 a, c, d, f

3 a ¢ = 16¡ 4m

i m = 4 ii m < 4 iii m > 4

b ¢ = 9¡ 8m

i m = 9

8ii m < 9

8iii m > 9

8

c ¢ = 9¡ 4m

i m = 9

4ii m < 9

4iii m > 9

4

4 a ¢ = k2 + 8k

i k < ¡8 or k > 0 ii k 6 ¡8 or k > 0

iii k = ¡8 or 0 iv ¡8 < k < 0

b ¢ = 4¡ 4k2

i ¡1 < k < 1 ii ¡1 6 k 6 1

iii k = §1 iv k < ¡1 or k > 1

c ¢ = k2 + 4k ¡ 12

i k < ¡6 or k > 2 ii k 6 ¡6 or k > 2

iii k = ¡6 or 2 iv ¡6 < k < 2

d ¢ = k2 ¡ 4k ¡ 12

i k < ¡2 or k > 6 ii k 6 ¡2 or k > 6

iii k = 6 or ¡2 iv ¡2 < k < 6

e ¢ = 9k2 ¡ 14k ¡ 39

i k < ¡ 13

9or k > 3 ii k 6 ¡ 13

9or k > 3

iii k = ¡ 13

9or 3 iv ¡ 13

9< k < 3

f ¢ = ¡3k2 ¡ 4k

i ¡ 4

3< k < 0 ii ¡ 4

36 k 6 0

iii k = ¡ 4

3or 0 iv k < ¡ 4

3or k > 0

EXERCISE 6C.1

1 a y = (x¡ 4)(x+ 2) b y = ¡(x¡ 4)(x+ 2)

c y = 2(x+ 3)(x+ 5) d y = ¡3x(x+ 4)

x

y

( )��,

( )��,

y x��( )

( )��,

( )��, y g x�� ( )

y

x��

��

4

y

x

8

��

y

x

�

y

x��

30

4m

�

Oi_m

�

Or_m

�

��k

�

0

��k

�

1

�

��k

�

2

��k

�

6

-\Ql_E_k

�

3

-\Re_k

�

0

�

y

x 3)()1(

)()(

�

�

�

xf

xf

xf

xf

��

�y

�y

�y

�y�

(5'\\Ow_)

¡2

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y

x

x=1

V(1' 3)4

@ = (! - 1)W + 3 @ = 2(! + 2)W + 1y

x

x=-2

V(-2' 1)

9

y

x

-3

18

y

x��

��

@ = -2(! - 1)W - 3

y

xx=1

V(1'-3)

-5

@ = Qw_ (! - 3)W +2y

xx=3

V(3' 2)

Qs_E_

y

x

x=1

V(1' 4)3\We_

@ = - Qe_ (! - 1)W + 4

y

x

x=-2

V(-2'-3)

-3\Wt_@ = - Aq_p_ (! + 2)W - 3

y

x

V( )��

x� �= 2�

��-2-~`5

-2+~`5

y

x

2

Qw_ 2

45�x

),(89

45 �V

y

x� �

��

23�x

41

23 ),(V

712 ANSWERS

e y = 2(x+ 3)2 f y = ¡ 1

4(x+ 2)2

2 a x = 1 b x = 1 c x = ¡4

d x = ¡2 e x = ¡3 f x = ¡2

3 a C b E c B d F e G f H g A h D

4 a b

c d

e f

5 a G b A c E d B e I

f C g D h F i H

6 a (2, ¡2) b (¡1, ¡4) c (0, 4) d (0, 1)

e (¡2, ¡15) f (¡2, ¡5) g (¡ 3

2, ¡ 11

2) h ( 5

2, ¡ 19

2)

i (1, ¡ 9

2)

7 a §3 b §p3 c ¡5 and ¡2

d 3 and ¡4 e 0 and 4 f ¡4 and ¡2

g ¡1 (touching) h 3 (touching) i 2§p3

j ¡2§p7 k 3§p

11 l ¡4§p5

8 a i x = 1ii (1, 4)

iii no x-intercept,

y-intercept 5iv

b i x = ¡2ii (¡2, ¡5)

iii x-int. ¡2§p5,

y-intercept ¡1iv

c i x = 5

4

ii ( 54

, ¡ 9

8)

iii x-intercepts 1

2, 2,

y-intercept 2iv

d i x = 3

2

ii ( 32

, 1

4)

iii x-intercepts 1, 2,

y-intercept ¡2iv

e i x = 2

3

ii ( 23

, 1

3)

iii x-intercepts 1

3, 1,

y-intercept ¡1iv

f i x = 1

4

ii ( 14

, 9

8)

iii x-intercepts ¡ 1

2, 1,

y-intercept 1iv

g i x = 3ii (3, 9)

iii x-intercepts 0, 6,

y-intercept 0iv

h i x = ¡3ii (¡3, 1)

iii x-int. ¡2, ¡4,

y-intercept ¡8iv

i i x = 4ii (4, 5)

iii x-int. 4§ 2p5,

y-intercept 1

iv

EXERCISE 6C.2

1 a y = (x¡ 1)2 + 2 b y = (x+ 2)2 ¡ 6

c y = (x¡ 2)2 ¡ 4 d y =¡x+ 3

2

¢2 ¡ 9

4

y

x-\Qw_ 11

!=\Qr_

V&Qr_\' Oi_*

y

x

x��

V( )��

1

�� ~`�� ~`

y

x

x��

V( )��

�

y

x

��

Qe_ 1

32�x

31

32 ),(V

y

x

3V(1' 2)

y

x

V&-\Ew_\'-\Or_*

��

y

x-2

V(-2'-6)

y

x

V(2'-4)

�

y

x5

x� �= 1

V ,( )��

y

x

x��

V( )��

��

��

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y

x-2

V&-\Tw_\'-\Ef_E_*

y

x

2

V&Ew_\'-\Qr_*

y

x

5

V(3'-4)

y

x-2

V(-4'-18)

y

x1

V&Tw_\'-5\Qr_*

y

x

5

V(1' 2)

563 �� xxXyy

x

1

V&Ew_\'-\Uw_*

162 �� xxXy

y

x

3

V(2'-5)

@ = 2!W - 8! + 3� � �

y

x

5

V(-1' 3)

@ = 2!W + 4! + 5� � � �

ANSWERS 713

e y =¡x+ 5

2

¢2 ¡ 33

4f y =

¡x¡ 3

2

¢2 ¡ 1

4

g y = (x¡ 3)2 ¡ 4 h y = (x+ 4)2 ¡ 18

i y =¡x¡ 5

2

¢2 ¡ 5 1

4

2 a i y = 2(x+ 1)2 + 3

ii (¡1, 3) iii 5

iv

b i y = 2(x¡ 2)2 ¡ 5

ii (2, ¡5) iii 3

iv

c i y = 2(x¡ 3

2)2 ¡ 7

2

ii ( 32

, ¡ 7

2) iii 1

iv

d i y = 3(x¡ 1)2 + 2

ii (1, 2) iii 5

iv

e i y = ¡(x¡ 2)2 + 6

ii (2, 6) iii 2

iv

f i y = ¡2(x+ 5

4)2 + 49

8

ii (¡ 5

4, 49

8) iii 3

iv

3 a y = (x¡ 2)2 + 3 b y = (x+ 3)2 ¡ 6

c y = ¡(x¡ 2)2 + 9 d y = 2¡x+ 3

2

¢2 ¡ 17

2

e y = ¡2¡x+ 5

2

¢2+ 27

2f y = 3

¡x¡ 3

2

¢2 ¡ 47

4

EXERCISE 6C.3

1 a cuts x-axis twice b touches x-axis

c cuts x-axis twice d cuts x-axis twice

e cuts x-axis twice f touches x-axis

2 a a = 1 which is > 0 and ¢ = ¡15 which is < 0

b a = ¡1 which is < 0 and ¢ = ¡8 which is < 0c a = 2 which is > 0 and ¢ = ¡40 which is < 0

d a = ¡2 which is < 0 and ¢ = ¡23 which is < 0

3 a = 3 which is > 0 and ¢ = k2 + 12 which is always > 0

fas k2 0 for all kg4 ¡4 < k < 4

EXERCISE 6D

1 a y = 2(x¡ 1)(x¡ 2) b y = 2(x¡ 2)2

c y = (x¡ 1)(x¡ 3) d y = ¡(x¡ 3)(x+ 1)

e y = ¡3(x¡ 1)2 f y = ¡2(x+ 2)(x¡ 3)

2 a y = 3

2(x¡ 2)(x¡ 4) b y = ¡ 1

2(x+ 4)(x¡ 2)

c y = ¡ 4

3(x+ 3)2

3 a y = 3x2 ¡ 18x+ 15 b y = ¡4x2 + 6x+ 4

c y = ¡x2 + 6x¡ 9 d y = 4x2 + 16x+ 16

e y = 3

2x2 ¡ 6x+ 9

2f y = ¡ 1

3x2 + 2

3x+ 5

4 a y = ¡(x¡ 2)2 + 4 b y = 2(x¡ 2)2 ¡ 1

c y = ¡2(x¡ 3)2 + 8 d y = 2

3(x¡ 4)2 ¡ 6

e y = ¡2(x¡ 2)2 + 3 f y = 2(x¡ 1

2)2 ¡ 3

2

EXERCISE 6E

1 a (1, 7) and (2, 8) b (4, 5) and (¡3, ¡9)

c (3, 0) (touching) d graphs do not meet

2 a (0:59, 5:59) and (3:41, 8:41) b (3, ¡4) touching

c graphs do not meet d (¡2:56, ¡18:81) and (1:56, 1:81)

3 a (2, 4), (¡1, 1) b (1, 0), (¡2, ¡3)

c (1, 4) d (1, 4), (¡4, ¡1)

5 c = ¡9

6 m = 0 or ¡8

7 ¡1 or 11

8 a c < ¡9b example: c = ¡10

EXERCISE 6F

1 7 and ¡5 or ¡7 and 5 2 5 or 1

53 14

4 18 and 20 or ¡18 and ¡20 5 15 and 17 or ¡15 and ¡17

6 15 sides 7 3:48 cm

8 b 6 cm by 6 cm by 7 cm 9 11:2 cm square 10 no

12 221 ha 13 2:03 m 14 52:1 km h¡1

15 554 km h¡1 16 61:8 km h¡1 17 32

18 No, tunnel is only 3:79 m wide 4:8 m above ground level.

19 a y = ¡ 1

100x2 + 70

b supports are 21 m, 34 m, 45 m, 54 m, 61 m, 66 m, 69 m

y

x

��

��

y x��

y x x�� X

y

x

2

V(2' 6)

y x x�� X

y

x

3

V&-\Tr_\\' Rk_O_*

y x x�� X

>

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714 ANSWERS

EXERCISE 6G

1 a min. ¡1, when x = 1 b max. 8, when x = ¡1

c max. 8 1

3, when x = 1

3d min. ¡1 1

8, when x = ¡ 1

4

e min. 4 15

16when x = 1

8f max. 6 1

8, when x = 7

4

2 40 refrigerators, $4000 4 500 m by 250 m

5 c 100 m by 112:5 m

6 a 41 2

3m by 41 2

3m b 50 m by 31 1

4m 7 b 3 1

8units

8 a y = 6¡ 3

4x b 3 cm by 4 cm 9 125 10 40 11 157

REVIEW SET 6A

1 a ¡2, 1 e

b x = ¡ 1

2

c 4

d (¡ 1

2, 9

2)

2 a x = 15 or ¡4b x = ¡ 5

3or 2 c

3 a x = ¡5§p13

2b x = ¡11§p

145

6

4 x = ¡ 7

2§

p65

2

5 a b

6 a y = 3x2 ¡ 24x+ 48 b y = 2

5x2 + 16

5x+ 37

5

7 a = ¡2 which is < 0 ) a max.

max. = 5 when x = 1

8 (4, 4) and (¡3, 18) 9 k < ¡3 1

8

10 a m = 9

8b m < 9

8c m > 9

811 6

5or 5

6

13 a m = ¡2, n = 4 b k = 7 c (2, 5)

d f(x) has domain fx jx 2 R g, range fy j y > 3gg(x) has domain fx jx 2 R g, range fy j y > 5g

REVIEW SET 6B

1 a y = 2¡x+ 3

2

¢2 ¡ 15

2d

b (¡ 3

2, ¡ 15

2)

c ¡3

2 a x ¼ 0:586 or 3:414 b x ¼ ¡0:186 or 2:686

3 4 x = 4

3, V( 4

3, 12 1

3)

5 a two distinct rational roots

b a repeated root

6 12:9 cm 7 a c > ¡6b example: c = ¡2, (¡1, ¡5) and (3, 7)

8 a x = ¡1 d

b (¡1, ¡3)

c y-intercept ¡1,

x-ints. ¡1§ 1

2

p6

9 13:5 cm by 13:5 cm 10 touch at (¡2, 9)

11 a min. = 5 2

3when x = ¡ 2

3

b max. = 5 1

8when x = ¡ 5

4

12 b A = x

³600¡ 8x

9

ć 37 1

2m by 33 1

3m d 1250 m2

13 a k = ¡12 or 12 b (0, 4)

c horizontal translation of 4

3units

REVIEW SET 6C

1 a x = 2 d

b (2, ¡4)c ¡2

2 a x = 5

2§

p37

2

b x = 7

4§

p73

4

3 a x = 7

2§

p37

2

b no real roots

4 a y = 20

9(x¡ 2)2 ¡ 20

b y = ¡ 2

7(x¡ 1)(x¡ 7)

c y = 2

9(x+ 3)2

5 a graph cuts

x-axis twice

b graph cuts

x-axis twice

6 a neither b positive definite

7 a y = 3(x¡ 3)(x+ 3) b y = ¡6(x¡ 2)2 + 25

8 17 cm 9 1

210 k < 1

11 y = ¡4x2 + 4x+ 24 12 m = ¡5 or 19

13 a i A(¡m, 0), B(¡n, 0) ii x =

b i positive ii negative

EXERCISE 7A

1 a p3 + 3p2q + 3pq2 + q3 b x3 + 3x2 + 3x+ 1

c x3 ¡ 9x2 + 27x¡ 27 d 8 + 12x+ 6x2 + x3

e 27x3 ¡ 27x2 + 9x¡ 1 f 8x3 + 60x2 + 150x+ 125

g 27x3 ¡ 9x2 + x¡ 1

27h 8x3 + 12x+

6

x+

1

x3

2 a 1 + 4x+ 6x2 + 4x3 + x4

b p4 ¡ 4p3q + 6p2q2 ¡ 4pq3 + q4

c x4 ¡ 8x3 + 24x2 ¡ 32x+ 16

d 81¡ 108x+ 54x2 ¡ 12x3 + x4

e 1 + 8x+ 24x2 + 32x3 + 16x4

f 16x4 + 96x3 + 216x2 + 216x+ 81

g x4 + 4x2 + 6+ 4

x2 +1

x4

h 16x4 ¡ 32x2 + 24¡ 8

x2 +1

x4

y

x

-2

(2'-4)

x��

@=\Qw_\(!-2)X-4

y

x-2

&-\Qw_\' Ow_\*

1

4

x=-\Qw_@=-2(!+2)(!-1)

y

x

-3

&-\Ew_\'-\Qs_T_*

@=2!X+6!-3

y

x2

V( )��

@=-!X+2!

(-4' 6) y

x

-2

x��

@=-\Qw_\(!+4)X+6

@=2!X+4!-1

y

x

(-1'-3)

-1

-1-\Qw_\~`6 -1+\Qw_\~`6

y

x

(2'-4)

4

@=(!-2)X-4x��

x = 0 or 4

¡m¡ n

2

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ANSWERS 715

3 a x5 + 10x4 + 40x3 + 80x2 + 80x+ 32

b x5 ¡ 10x4y + 40x3y2 ¡ 80x2y3 + 80xy4 ¡ 32y5

c 1 + 10x+ 40x2 + 80x3 + 80x4 + 32x5

d x5 ¡ 5x3 + 10x¡ 10

x+

5

x3¡ 1

x5

4 a 1 6 15 20 15 6 1

b i x6 + 12x5 + 60x4 + 160x3 + 240x2 + 192x+ 64

ii 64x6 ¡ 192x5 + 240x4 ¡ 160x3 + 60x2 ¡ 12x+ 1

iii x6 + 6x4 + 15x2 + 20 +15

x2+

6

x4+

1

x6

5 a 7 + 5p2 b 161 + 72

p5 c 232¡ 164

p2

6 a 64 + 192x+ 240x2 + 160x3 + 60x4 + 12x5 + x6

b 65:944 160 601 201

7 2x5 + 11x4 + 24x3 + 26x2 + 14x+ 3

8 a 270 b 4320

EXERCISE 7B

1 a 111+¡11

1

¢(2x)+

¡11

2

¢(2x)2+ ::::+

¡11

10

¢(2x)10+(2x)11

b (3x)15 +¡15

1

¢(3x)14

¡2

x

¢+¡15

2

¢(3x)13

¡2

x

¢2+ ::::

::::+¡15

14

¢(3x)

¡2

x

¢14+¡2

x

¢15c (2x)20 +

¡20

1

¢(2x)19

¡¡ 3

x

¢+¡20

2

¢(2x)18

¡¡ 3

x

¢2+ ::::

::::+¡20

19

¢(2x)

¡¡ 3

x

¢19+¡¡ 3

x

¢202 a T6 =

¡15

5

¢(2x)1055 b T4 =

¡9

3

¢(x2)6y3

c T10 =¡17

9

¢x8¡¡ 2

x

¢9d T9 =

¡21

8

¢(2x2)13

¡¡ 1

x

¢83 a

¡10

5

¢3525 b

¡6

3

¢23(¡3)3 c

¡6

3

¢23(¡3)3

d¡12

4

¢28(¡1)4

4 a¡15

5

¢25 b

¡9

3

¢(¡3)3

5 a

1 11 2 1

1 3 3 11 4 6 4 1

1 5 10 10 5 1

b sum

2481632

c The sum of the

numbers in row nof Pascal’s triangle

is 2n.d After the first part

let x = 1.

6 a¡8

6

¢= 28 b 2

¡9

3

¢36 ¡

¡9

4

¢35 = 91 854

7 T3 =¡6

2

¢(¡2)2x8y8

8 a 84x3 b n = 6 and k = ¡2 9 a = 2

REVIEW SET 7

1 a x3 ¡ 6x2y + 12xy2 ¡ 8y3

b 81x4 + 216x3 + 216x2 + 96x+ 16

2 20 000 3 60

4 (a+b)6 = a6+6a5b+15a4b2+20a3b3+15a2b4+6ab5+b6

a x6 ¡ 18x5 + 135x4 ¡ 540x3 + 1215x2 ¡ 1458x+ 729

b 1 +6

x+

15

x2+

20

x3+

15

x4+

6

x5+

1

x6

5 362 + 209p3 6 64:964 808

7¡12

6

¢£ 26 £ (¡3)6 8 8

¡6

2

¢¡ 6¡6

1

¢= 84

9 k = 180 10 c = 3

11 a 7 b¡6

4

¢£ 32 = 135

EXERCISE 8A

1 a ¼2

c b ¼3

c c ¼6

c d ¼10

c e ¼20

c

f 3¼4

cg 5¼

4

ch 3¼

2

ci 2¼c j 4¼c

k 7¼4

cl 3¼c m ¼

5

c n 4¼9

co 23¼

18

c

2 a 0:641c b 2:39c c 5:55c d 3:83c e 6:92c

3 a 36o b 108o c 135o d 10o e 20o

f 140o g 18o h 27o i 150o j 22:5o

4 a 114:59o b 87:66o c 49:68o d 182:14o

e 301:78o

5 a Degrees 0 45 90 135 180 225 270 315 360

Radians 0 ¼4

¼2

3¼4

¼ 5¼4

3¼2

7¼4

2¼

b Deg. 0 30 60 90 120 150 180 210 240 270 300 330 360

Rad. 0 ¼6

¼3

¼2

2¼3

5¼6

¼ 7¼6

4¼3

3¼2

5¼3

11¼6

2¼

EXERCISE 8B

EXERCISE 8C.1

1 a b c

2 a i A(cos 26o, sin 26o), B(cos 146o, sin 146o),

C(cos 199o, sin 199o)

ii A(0:899, 0:438), B(¡0:829, 0:559),

C(¡0:946, ¡0:326)

b i A(cos 123o, sin 123o), B(cos 251o, sin 251o),

C(cos(¡35o), sin(¡35o))

ii A(¡0:545, 0:839), B(¡0:326, ¡0:946),

C(0:819, ¡0:574)

3 µ (degrees) 0o 90o 180o 270o 360o 450o

µ (radians) 0 ¼2

¼ 3¼2

2¼ 5¼2

sine 0 1 0 ¡1 0 1

cosine 1 0 ¡1 0 1 0

tangent 0 undef 0 undef 0 undef

4 a i 1p2¼ 0:707 ii

p3

2¼ 0:866

b µ (degrees) 30o 45o 60o 135o 150o 240o 315o

µ (radians) ¼6

¼4

¼3

3¼4

5¼6

4¼3

7¼4

sine 1

2

1p2

p3

2

1p2

1

2¡

p3

2¡ 1p

2

cosinep3

2

1p2

1

2¡ 1p

2¡

p3

2¡ 1

2

1p2

tangent 1p3

1p3 ¡1 ¡ 1p

3

p3 ¡1

1 a i 49:5 cm ii 223 cm2 b i 23:0 cm ii 56:8 cm2

2 a 3:14 m b 9:30 m2 3 a 5:91 cm b 18:9 cm

4 a 0:686c b 0:6c

5 a 0:75c, 24 cm2 b 1:68c, 21 cm2 c 2:32c, 126:8 cm2

6 10 cm, 25 cm2

8 a 11:7 cm b 11:7 c 37:7 cm d 3:23c

9 a ® ¼ 18:43 b µ ¼ 143:1 c 387 m2

10 25:9 cm 11 b 2 h 24 min 12 227 m2

1

1��

y

x

11

y

x

��

��

2

2

��

y

x��

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716 ANSWERS

5 a i 0:985 ii 0:985 iii 0:866 iv 0:866

v 0:5 vi 0:5 vii 0:707 viii 0:707b sin(180o ¡ µ) = sin µ

d i 135o ii 129o iii 2¼3

iv 5¼6

6 a i 0:342 ii ¡0:342 iii 0:5 iv ¡0:5

v 0:906 vi ¡0:906 vii 0:174 viii ¡0:174b cos(180o ¡ µ) = ¡ cos µ

d i 140o ii 161o iii 4¼5

iv 3¼5

7 a ¼ 0:6820 b ¼ 0:8572 c ¼ ¡0:7986

d ¼ 0:9135 e ¼ 0:9063 f ¼ ¡0:6691

8 a

QuadrantDegree

measure

Radian

measurecos µ sin µ tan µ

1 0o < µ < 90o 0 < µ < ¼2

+ve +ve +ve

2 90o < µ < 180o ¼2< µ < ¼ ¡ve +ve ¡ve

3 180o < µ < 270o ¼ < µ < 3¼2

¡ve ¡ve +ve

4 270o < µ < 360o 3¼2

< µ < 2¼ +ve ¡ve ¡ve

b i 1 and 4 ii 2 and 3 iii 3 iv 2

9 a AbOQ = 180o ¡ µ

b [OQ] is a reflection of [OP] in the y-axis and so Q has

coordinates (¡ cos µ, sin µ).

c cos(180o ¡ µ) = ¡ cos µ, sin(180o ¡ µ) = sin µ

EXERCISE 8C.2

1 a cos µ = §p3

2b cos µ = § 2

p2

3c cos µ = §1

d cos µ = 0

2 a sin µ = §3

5b sin µ = §

p7

4c sin µ = 0

d sin µ = §1

3 a sin µ =p5

3b cos µ = ¡

p21

5c cos µ = 4

5

d sin µ = ¡ 12

13

4 a ¡ 1

2p2

b ¡2p6 c 1p

2d ¡

p7

3

5 a sinx = 2p13

, cosx = 3p13

b sinx = 4

5, cosx = ¡ 3

5

c sinx = ¡p

5

14, cosx = ¡ 3p

14

d sinx = ¡ 12

13, cosx = 5

13

EXERCISE 8C.3

1 a b c d e

sin µ 1p2

¡ 1p2

¡ 1p2

0 ¡ 1p2

cos µ 1p2

¡ 1p2

1p2

¡1 ¡ 1p2

tan µ 1 1 ¡1 0 1

2 a b c d e

sin¯ 1

2

p3

2¡ 1

2¡

p3

2¡ 1

2

cos¯p3

2¡ 1

2¡

p3

2

1

2

p3

2

tan¯ 1p3

¡p3 1p

3¡p

3 ¡ 1p3

3 a 3

4b 1

4c 3 d 1

4e ¡ 1

4f 1

gp2 h 1

2i 1

2j 2 k ¡1 l ¡p

3

4 a 30o, 150o b 60o, 120o c 45o, 315o

d 120o, 240o e 135o, 225o f 240o, 300o

5 a ¼4

, 5¼4

b 3¼4

, 7¼4

c ¼3

, 4¼3

d 0, ¼, 2¼ e ¼6

, 7¼6

f 2¼3

, 5¼3

6 a ¼6

, 11¼6

, 13¼6

, 23¼6

b 7¼6

, 11¼6

, 19¼6

, 23¼6

c 3¼2

, 7¼2

7 a µ = ¼3

, 5¼3

b µ = ¼3

, 2¼3

c µ = ¼

d µ = ¼2

e µ = 3¼4

, 5¼4

f µ = ¼2

, 3¼2

g µ = 0, ¼, 2¼ h µ = ¼4

, 3¼4

, 5¼4

, 7¼4

i µ = 5¼6

, 11¼6

j µ = ¼3

, 2¼3

, 4¼3

, 5¼3

EXERCISE 8D

1 a y =p3x b y = x c y = ¡ 1p

3x

2 a y =p3x+ 2 b y = ¡p

3x c y = 1p3x¡ 2

REVIEW SET 8A

1 a 2¼3

b 5¼4

c 5¼6

d 3¼

2 a ¼3

b 15o c 84o

3 a 0:358 b ¡0:035 c 0:259 d ¡0:731

4 a 1, 0 b ¡1, 0

6 a sin¡2¼3

¢=

p3

2, cos

¡2¼3

¢= ¡ 1

2

b sin¡8¼3

¢=

p3

2, cos

¡8¼3

¢= ¡ 1

27 1p

15

8 a ¡0:743 b ¡0:743 c 0:743 d ¡0:743

9 §p7

410 a

p3

2b 0 c 1

211 a ¡ 3p

13b 2p

13

REVIEW SET 8B

1 a (0:766, ¡0:643) b (¡0:956, 0:292)

2 a 1:239c b 2:175c c ¡2:478c

3 a 171:89o b 83:65o c 24:92o d ¡302:01o

4 111 cm2

5 M(cos 73o, sin 73o) ¼ (0:292, 0:956)

N(cos 190o, sin 190o) ¼ (¡0:985, ¡0:174)

P(cos 307o, sin 307o) ¼ (0:602, ¡0:799)

6 ¼ 103o

7 a 150o, 210o b 45o, 135o c 120o, 300o

8 a µ = ¼ b µ = ¼3

, 2¼3

, 4¼3

, 5¼3

9 a 133o b 14¼15

c 174o

10 perimeter = 34:1 cm, area = 66:5 cm2

11 r 8:79 cm, area 81:0 cm2

REVIEW SET 8C

1 a 72o b 225o c 140o d 330o

2 3 a 0, ¡1

b 0, ¡1

4 a 0:961 b ¡0:961 c ¡0:961 d ¡0:961

5 a i µ = 60o ii µ = ¼3

b ¼3

units 6 3

8 ap7

4b ¡

p7

3c ¡

p7

49 a 2 1

2b 1 1

2c ¡ 1

2

10 a 0 b sin µ 11 a y = ¡ 1p3x b k = ¡2

p3

y

x

¼ ¼

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ANSWERS 717

EXERCISE 9A

1 a 28:9 cm2 b 384 km2 c 28:3 cm2 2 x 19:0

3 18:9 cm2 4 137 cm2 5 374 cm2 6 7:49 cm

7 11:9 m 8 a 48:6o or 131:4o b 42:1o or 137:9o

9 1

4is not covered

10 a 36:2 cm2 b 62:8 cm2 c 40:4 mm2 11 4:69 cm2

EXERCISE 9B

1 a 28:8 cm b 3:38 km c 14:2 m

2 bA ¼ 52:0o, bB ¼ 59:3o, bC ¼ 68:7o 3 112o

4 a 40:3o b 107o 5 a cos µ = 0:65 b x ¼ 3:81

6 a x = 3 +p22 b x =

¡3 +p73

2c x = 5p

3

7 a x ¼ 10:8 b x ¼ 9:21 c

EXERCISE 9C.1

1 a x ¼ 28:4 b x ¼ 13:4 c x ¼ 3:79

2 a a ¼ 21:3 cm b b ¼ 76:9 cm c c ¼ 5:09 cm

EXERCISE 9C.2

EXERCISE 9D

1 17:7 m 2 207 m 3 23:9o 4 77:5 m

5 a i 5:63 km ii 115o b i Esko ii 3:68 min

c 295o

6 9:38o 7 69:1 m 8 a 38:0 m b 94:0 m 9 55:1o

10 AC ¼ 11:7 km, BC ¼ 8:49 km

11 a 74:9 km2 b 7490 hectares

12 9:12 km 13 85:0 mm 14 10:1 km 15 29:2 m

16 37:6 km

REVIEW SET 9A

1 14 km2

2 If the unknown is an angle, use the cosine rule to avoid the am-

biguous case.

3 a x = 3 or 5 b Kady can draw

2 triangles:

4 5 42 km

REVIEW SET 9B

1 a x ¼ 34:1 b x ¼ 18:9

2 AC ¼ 12:6 cm, bA ¼ 48:6o, bC ¼ 57:4o 3 113 cm2

4 7:32 m 5 204 m 6

REVIEW SET 9C

1 a x ¼ 41:5 b x ¼ 15:4 2 x ¼ 47:5

3 EbDG ¼ 74:4o 4 a 10 600 m2 b 1:06 ha

5 179 km, bearing 352o

6 a The information

given could give

two triangles:

b ¼ 2:23 m3

EXERCISE 10A

1 a

Data exhibits periodic behaviour.

b

Not enough information to say data is periodic.

It may in fact be quadratic.

c


It may in fact be quadratic.

d


2 a

b The data is periodic. i y = 32 (approx.) ii ¼ 64 cm

iii ¼ 200 cm iv ¼ 32 cm

c A curve can be fitted to the data.

3 a periodic b periodic c periodic

d not periodic e periodic f periodic

44°A C

B'

B

6 m

8 m

3 cm

8 cm

7 cm

5 cm

60°

��

�

� � � � ��

y

x

�

�

�

�

� � � �

y

x

�

��

�

� � � �

y

x

�

��

� � � � ��

y

x

��

��

��

��

�

��

� ��

height aboveground(cm)

distance travelled

x ¼ 1:41 or 7:78

1 bC 62:1o or bC 117:9o

2 a bA ¼ 49:5o b bB ¼ 72:0o or 108o c bC ¼ 44:3o

3 No,sin 85o

11:46= sin 27o

9:84 AbBC = 66o, BD ¼ 4:55 cm

5 x ¼ 17:7, y ¼ 33:1

6 a

c

7 Area ¼ 25:1 cm2 8 x = 8 + 11

2

p2

12

13

560 m, bearing 079:7o

¼

¼ ¼

or 132:5

91:3o b 91:3o

.... cosine rule as it avoids the ambiguous case.

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718 ANSWERS

EXERCISE 10B.1

1 a

b

c

d

2 a

b

c

3 a ¼2

b ¼2

c 6¼ d 10¼3

4 a b = 2

5b b = 3 c b = 1

6d b = ¼

2

e b = ¼50

EXERCISE 10B.2

1 a

b

c

d

e

f

2 a 2¼5

b 8¼ c ¼ 3 a 2

3b 20 c 1

50d ¼

25

4 a vert. translation ¡1 b horiz. translation ¼4

right

c vert. stretch, factor 2 d horiz. stretch, factor 1

4

e vert. stretch, factor 1

2f horiz. stretch, factor 4

g reflection in the x-axis h translation

³¡2¡3

í vert. stretch, factor 2, followed by a horiz. stretch, factor 1

3

j translation

µ¼3

2

¶EXERCISE 10C

��

��

��

y

x

y x�� sin

�

��

��

��

�� x

y y x�� sin

��

�

��

��

y

x

y x�� sin

�

�

��

��

��

y

x

y x�� \Ew_\sin

��

��

��

y

x

y x�� \Ew_\sin

�

��

��

��

�� x

y y x�� sin ( )

�

�

�

��

��

��

y

x��

y x�� sin

�

�

�

��

y

x��

y x�� sin

�

�

��

��

y

x��

y xsin� & \r_\*�

y xsin� �& \y_\* + 1��

�

��

y

x��

Qw_

�

��

��

��

�� x

y)sin(2

xy �

�

�

��

y

x��

y x�� sin ( )

��

�

��

��

y

x��

y x�� sin ( )��

1 a

2 a

3

4 a

T ¼ 6:5 sin ¼6(t¡ 4:5) + 20:5

T ¼ 4:5 sin ¼6(t¡ 10:5) + 11:5

T ¼ 9:5 sin ¼6(t¡ 10:5)¡ 9:5

H ¼ 7 sin 0:507(t¡ 3:1)

IB SL 2nd ed


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ANSWERS 719

b

5

EXERCISE 10D

1 a y = cosx+ 2

b y = cosx¡ 1

c y = cos¡x¡ ¼

4

¢

d y = cos¡x+ ¼

6

¢

e y = 2

3cosx

f y = 3

2cosx

g y = ¡ cosx

h y = cos¡x¡ ¼

6

¢+ 1

i y = cos¡x+ ¼

4

¢¡ 1

j y = cos 2x

k y = cos¡x2

¢

l y = 3cos 2x

2 a 2¼3

b 6¼ c 100

3 jaj = amplitude, b =2¼

period, c = horizontal translation,

d = vertical translation

4 a y = 2cos 2x b y = cos¡x2

¢+ 2

c y = ¡5 cos¡¼3x¢

EXERCISE 10E.1

EXERCISE 10E.2

1 a i y = tan(x¡ ¼2)

ii y = ¡ tanx

H

t

3.13.1 9.39.3

��

�

15.515.5

�

��

��

y

x

�

��

��

y

x

��

y

x

��

�

��

��

y

x

We_We_

��

��

y

x

Ew_Ew_

��

y

x

�

��

��

y

x

��

y

x

�

��

��

y

x

��

��

y

x

��

��

y

x��

��

��

y

x

��

O N

T

45°

1

1

�

�

��

��

��

y

x

�

�

��

��

��

y

x

H = 10 sin( ¼50

(t¡ 25)) + 12

1 a 0 b ¼ 0:268 c ¼ 0:364

d ¼ 0:466 e ¼ 0:700 f 1

g ¼ 1:19 h ¼ 1:43

2 triangle TON is isosceles, ON = TN

IB SL 2nd ed


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720 ANSWERS

iii y = tan 3x

2 a translation through

³10

´b reflection in x-axis

c horizontal stretch, factor q = 2

3 a ¼ b ¼2

c ¼n

EXERCISE 10F

1 a 1 b undefined c 1

2 a ¼ b 6¼ c ¼

3 a b = 1 b b = 3 c b = 2 d b = ¼2

4 a

b

c

d

e

f

5 a b c d e f

maximum value 1 3 undef. 4 3 ¡2

minimum value ¡1 ¡3 undef. 2 ¡1 ¡4

6 a vertical stretch, factor 1

2b horizontal stretch, factor 4

c reflection in the x-axis

d vertical translation down 2 units

e horizontally translate ¼4

units to the left

f reflection in the y-axis

7 m = 2, n = ¡3 8 p = 1

2, q = 1

EXERCISE 10G.1

1 a x ¼ 0:3, 2:8, 6:6, 9:1, 12:9 b x ¼ 5:9, 9:8, 12:2

2 a x ¼ 1:2, 5:1, 7:4 b x ¼ 4:4, 8:2, 10:7

3 a x ¼ 0:4, 1:2, 3:5, 4:3, 6:7, 7:5, 9:8, 10:6, 13:0, 13:7

b x ¼ 1:7, 3:0, 4:9, 6:1, 8:0, 9:3, 11:1, 12:4, 14:3, 15:6

4 a i ii

b i x ¼ 1:1, 4:2, 7:4 ii x ¼ 2:2, 5:3

EXERCISE 10G.2

1 a x ¼ 1:08, 4:35 b x ¼ 0:666, 2:48

c x ¼ 0:171, 4:92 d x ¼ 1:31, 2:03, 2:85

2 x ¼ ¡0:951, 0:234, 5:98

EXERCISE 10G.3

1 a x = ¼6

, 13¼6

, 25¼6

b x = ¡¼3

, 5¼3

c x = ¡ 7¼2

, ¡ 5¼2

, ¡ 3¼2

, ¡¼2

, ¼2

, 3¼2

, 5¼2

, 7¼2

d x = ¼3

, 5¼6

, 4¼3

, 11¼6

, 7¼3

, 17¼6

, 10¼3

, 23¼6

3 X = ¼3+ k¼

a x = ¼2

, 3¼2

b x = ¼12

, ¼3

, 7¼12

, 5¼6

, 13¼12

, 4¼3

, 19¼12

, 11¼6

c x = ¼3

, 2¼3

, 4¼3

, 5¼3

4 a x = 0o, 90o, 180o b x = ¼4

, 5¼4

, 9¼4

5 a

b x = ¼4

or 5¼4

c x = ¼4

or 5¼4

�

�

��

��

y

x

��

��

323

83

735

34

32

3

617

25

613

611

23

67

65

26

�

��

y

x

-\We_

y x�� We_\cos

�

2

�2 2

3�

�

�

��

y

x

y x�� sin

� 2�2 2

3� �

�

�

��

��

y

x

� 2�

)tan(2�� xy

�

�

��

��

y

x

��2 2

3�2�

y x�� tan

�

�

��

��

y

x

��2 2

3�

y x�� cos

2�

�

��

y

x��2 2

3��4 4

5�2�

1)sin(4

�� xyx

y x�� sin y x�� cosy

��2 2

3�2�

¼ 1:6 ¼ ¡1:1

2 a x = 2¼3

, 4¼3

, 8¼3

, 10¼3

, 14¼3

b x = ¡330o, ¡210o, 30o, 150o

c x = 5¼6

, 7¼6

, 17¼6

d x = ¡ 5¼3

, ¡¼, ¼3

, ¼

e x = ¡ 13¼6

, ¡ 3¼2

, ¡¼6

, ¼2

, 11¼6

, 5¼2

f x = 0, 3¼2

, 2¼

g x = ¼2

, 3¼2

, 5¼2

h x = 0, ¼4

, ¼2

, 3¼4

, ¼

i x = ¡ 8¼9

, ¡ 4¼9

, ¡ 2¼9

, 2¼9

, 4¼9

, 8¼9

j x = 0, ¼6

, ¼, 7¼6

, 2¼

k 2 Z,

IB SL 2nd ed


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ANSWERS 721

EXERCISE 10H

2 a 20 m b at t = 3

4minute c 3 minutes

d

4 a H(t) = 3 cos(¼t2) + 4 b t ¼ 1:46 s

5 a i true ii true b 116:8 cents L¡1

c

d 98:6 cents L¡1 on the 1st and 15th day

EXERCISE 10I.1

1 a 2 sin µ b 3 cos µ c 2 sin µ d sin µ

e ¡2 cos µ f ¡3 cos µ

2 a 3 b ¡2 c ¡1 d 3 cos2 µ

e 4 sin2 µ f cos µ g ¡ sin2 µ h ¡ cos2 µ

i ¡2 sin2 µ j 1 k sin µ l sin µ

3 a 2 tanx b ¡3 tanx c sinx d cosx

e 5 sinx f2

cosx

4 a 1 + 2 sin µ + sin2 µ b sin2 ®¡ 4 sin®+ 4

c tan2 ®¡ 2 tan®+ 1 d

e 1¡ 2 sin¯ cos¯ f ¡4 + 4 cos®¡ cos2 ®

EXERCISE 10I.2

1 a (1¡ sin µ)(1 + sin µ)

b (sin®+ cos®)(sin®¡ cos®)

c (tan®+ 1)(tan®¡ 1) d sin¯(2 sin¯ ¡ 1)

e cosÁ(2 + 3 cosÁ) f 3 sin µ(sin µ ¡ 2)

g (tan µ + 3)(tan µ + 2) h (2 cos µ + 1)(cos µ + 3)

i (3 cos®+ 1)(2 cos®¡ 1)

2 a 1 + sin® b tan¯ ¡ 1 c cosÁ¡ sinÁ

d cosÁ+ sinÁ e1

sin®¡ cos®f

cos µ

2

EXERCISE 10J

1 a 24

25b ¡ 7

252 a ¡ 7

9b 1

9

3 a cos® = ¡p5

3, sin 2® = 4

p5

9

b sin¯ = ¡p21

5, sin 2¯ = ¡4

p21

25

4 a 1

3b 2

p2

35 3

2

6 a sin 2® b 2 sin 2® c 1

2sin 2® d cos 2¯

e ¡ cos 2Á f cos 2N g ¡ cos 2M h cos 2®

i ¡ cos 2® j sin 4A k sin 6® l cos 8µ

m ¡ cos 6¯ n cos 10® o ¡ cos 6D p cos 4A

q cos® r ¡2 cos 6P

8 a x = 0, 2¼3

, ¼, 4¼3

, 2¼ b x = ¼2

, 3¼2

c x = 0, ¼, 2¼

EXERCISE 10K

1 a x = 0, ¼, 7¼6

, 11¼6

, 2¼ b x = ¼3

, ¼2

, 3¼2

, 5¼3

c x = ¼3

, ¼, 5¼3

d x = 7¼6

, 3¼2

, 11¼6

e no solutions

2 a x = 0, 2¼3

, 4¼3

, 2¼ b x = ¼3

, 5¼3

c x = ¼2

, 7¼6

, 11¼6

d x = 0, ¼6

, ¼2

, 5¼6

, ¼, 7¼6

, 3¼2

, 11¼6

, 2¼ e x = ¼4

f x = ¼6

, 5¼6

REVIEW SET 10A

1

2 a minimum = 0, maximum = 2

b minimum = ¡2, maximum = 2

3 a x = 7¼6

, 11¼6

, 19¼6

, 23¼6

b x = ¡ 7¼4

, ¡ 5¼4

, ¼4

, 3¼4

4 a 4¼9

, 5¼9

, 10¼9

, 11¼9

, 16¼9

, 17¼9

b 3¼4

, 7¼4

, 11¼4

5 x = 0, 3¼2

, 2¼, 7¼2

, 4¼

6 a 1¡ cos µ b1

sin®+ cos®c

¡ cos®

2

7 cos® = ¡p7

4, sin 2® = 3

p7

89 c x =

10 a 5000 b 3000, 7000 c 0:5 < t < 2:5 and 6:5 < t 6 8

REVIEW SET 10B

1

2 a 6¼ b ¼4

3 a b x ¼ 5:42

4 a

b

321

40

30

20

10

H t( )

t

(3, 20)

�

�

��

��

y

x��

��

y x�� sin

�

��

y

x��

y x�� sin

�

��

��

��

y

x

��

y x�� cos

y x�� cos

�

��

��

��

y

x

��

y x�� cos

y cos� &! - \r_\*�

6 a x = 3¼4

or 7¼4

b x = ¼12

, 5¼12

, 3¼4

, 13¼12

, 17¼12

, 7¼4

c x = ¼6

, 2¼3

, 7¼6

, 5¼3

1 a i 7500 grasshoppers ii 10 300 grasshoppers

b 10 500 grasshoppers, when t = 4 weeks

c i at t = 1 1

3wks and 6 2

3wks ii at t = 9 1

3wks

d 2:51 6 t 6 5:49

3 a 400 water buffalo

b i 577 water buffalo ii 400 water buffalo

c 650, which is the maximum population.

d 150, after 3 years e t ¼ 0:26 years

on the 5th, 11th, 19th and 25th days

1 + 2 sin® cos®

x ¼ 0:392, 2:75, 6:68

16

3

IB SL 2nd ed


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722 ANSWERS

c

d

5 a x ¼ 1:12, 5:17, 7:40 b

6 a 120

169b 119

169

7 a i x ¼ 1:33, 4:47, 7:61 ii x ¼ 5:30

iii x ¼ 2:83, 5:97, 9:11

b i x = ¡¼2

, ¼2

ii x = ¡ 2¼3

, ¡¼6

, ¼3

, 5¼6

iii x = ¡ 2¼3

, ¡¼3

, ¼3

, 2¼3

c x ¼ 0:612, 3:754, 6:895

8 a

REVIEW SET 10C

1 a x = 3¼2

b x = ¼6

, 5¼6

, 7¼6

, 11¼6

2 a

b 1 6 k 6 3

3 a y = ¡4 cos 2x b y = cos ¼4x+ 2

4 a x = ¼2

, 3¼2

, 5¼2

, 7¼2

b x = ¡¼, ¡¼3

, ¼, 5¼3

5 a cos µ b ¡ sin µ c 5 cos2 µ d ¡ cos µ

6 a 4 sin2 ®¡ 4 sin®+ 1 b 1¡ sin 2®

8 sin µ = 2p13

, cos µ = ¡ 3p13

9 a 28 milligrams per m3 b 8:00 am Monday

EXERCISE 11A

1 a 1£ 4 b 2£ 1 c 2£ 2 d 3£ 3

2 a¡2 1 6 1

¢b

0@ 1:952:350:150:95

1A c total cost of

groceries

3

0@ 1000 1500 12501500 1000 1000800 2300 13001200 1200 1200

1A 4

0@ 40 50 55 4025 65 44 3035 40 40 3535 40 35 50

1AEXERCISE 11B.1

1 a

³9 13 3

´b

³6 8¡1 1

ć

³3 4¡6 ¡1

´d

³0 0

¡11 ¡3

´

2 a

Ã20 1 ¡88 10 ¡21 ¡5 18

!b

Ã¡14 9 ¡1412 ¡6 14¡5 3 ¡4

!

c

Ã14 ¡9 14¡12 6 ¡145 ¡3 4

!3 a Friday SaturdayÃ

859252

! Ã10213749

! b Ã187229101

!

4 a i

0BB@1:7227:850:922:533:56

1CCA ii

0BB@1:7928:751:332:253:51

1CCA c

0BB@0:070:900:41¡0:28¡0:05

1CCAb subtract cost price from selling price

5 a

0@L R

fr 23 19

st 17 29

mi 31 24

1A b

0@L R

fr 18 25

st 7 13

mi 36 19

1A c

0@L R

fr 41 44

st 24 42

mi 67 43

1A6 a x = ¡2, y = ¡2 b x = 0, y = 0

7 a A + B =

³1 35 2

´, B + A =

³1 35 2

´8 a (A + B) + C =

³6 3¡1 6

´, A + (B + C) =

³6 3¡1 6

ÉXERCISE 11B.2

1 a

³12 2448 12

´b

³2 48 2

ć

µ1

21

2 1

2

¶d

³ ¡3 ¡6¡12 ¡3

´2 a

³3 5 62 8 7

´b

³1 1 40 4 1

ć

³5 8 113 14 11

´d

³5 7 142 16 9

´3 a

0@ 122412060

1A b

0@ 363015

1A c

0@ 9189045

1A

4 a

0B@A B C D

35 46 46 69

58 46 35 86

46 46 58 58

12 23 23 17

1CA b

0B@A B C D

26 34 34 51

43 34 26 64

34 34 43 43

9 17 17 13

1CA5 a

Ã7527102

!Ã DVD

Ã VHS

Ã games

Ã13643129

!Ã DVD

Ã VHS

Ã games

b

Ã21170231

!Ã DVD

Ã VHS

Ã gamesc total weekly average hirings

6 12F

EXERCISE 11B.3

1 a 3A b O c ¡C d O e 2A + 2B

f ¡A ¡ B g ¡2A + C h 4A ¡ B

i 3B

�

��

y

x��

��y xcos�

y x3cos 2�

�

�

��

y

x��

y xcos�

y 2cos� &! - \e_\* + 3�

�

�

��

y

x

��

y sin� &! - \e_\* + 2�

��

x ¼ 0:184, 4:62

T ¼ 7:05 sin(¼6(t¡ 10:5)) + 24:75

IB SL 2nd ed


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ANSWERS 723

2 a X = A ¡ B b X = C ¡ B c X = 2C ¡ 4B

d X = 1

2A e X = 1

3B f X = A ¡ B

g X = 2C h X = 1

2B ¡ A i X = 1

4(A ¡ C)

3 a X =

³3 69 18

´b X =

µ1

2¡ 1

4

3

4

5

4

¶c X =

³¡1 ¡61 ¡ 1

2

ÉXERCISE 11B.4

1 a (11) b (22) c (16) 2¡w x y z

¢0BB@1

4

1

4

1

4

1

4

1CCA3 a P =

¡27 35 39

¢, Q =

Ã432

!

b total cost =¡27 35 39

¢Ã 432

!= $291

4 a P =¡10 6 3 1

¢, N =

0@ 3242

1Ab total points =

¡10 6 3 1

¢0@ 3242

1A = 56 points

EXERCISE 11B.5

1 Number of columns in A does not equal number of rows in B.

2 a m = n b 2£ 3 c B has 3 columns, A has 2 rows

3 a i does not exist ii¡28 29

¢b i

¡8¢

ii

Ã2 0 38 0 124 0 6

!

4 a¡3 5 3

¢b

Ã¡211

!5 a

b

6 a R =

Ã1 11 22 3

!b P =

³7 3 196 2 22

ć

³48 7052 76

´d My costs at store A are E48, my

friend’s costs at store B are E76.e store A

EXERCISE 11B.6

1 a

Ã16 18 1513 21 1610 22 24

!b

Ã10 6 ¡79 3 04 ¡4 ¡10

!

c

Ã22 0 132 176 19844 154 88 110 0176 44 88 88 132

!d

0@ 11513646106

1A

2 a¡3 3 2

¢b

Ã125 150 14044 40 4075 80 65

!c¡657 730 670

¢d¡369 420 385

¢e

³657 730 670369 420 385

´3 $224 660

4 a¡125 195 225

¢£Ã

15 12 13 11 14 16 84 3 6 2 0 4 73 1 4 4 3 2 0

!

¡¡85 120 130

¢£Ã

15 12 13 11 14 16 84 3 6 2 0 4 73 1 4 4 3 2 0

!= $7125

b¡125 195 225

¢£Ã

15 12 13 11 14 16 84 3 6 2 0 4 73 1 4 4 3 2 0

!

¡¡85 120 130

¢£Ã

20 20 20 20 20 20 2015 15 15 15 15 15 155 5 5 5 5 5 5

!= ¡$9030, which is a loss of $9030

c¡¡

125 195 225¢¡¡85 120 130

¢¢£Ã

15 12 13 11 14 16 84 3 6 2 0 4 73 1 4 4 3 2 0

!EXERCISE 11B.7

1 AB =

³¡1 1¡1 7

´, BA =

³0 23 6

´, AB 6= BA

2 AO = OA = O

5 a

³7 00 7

´b

³97 ¡59118 38

´6 a A2 does not exist b when A is a square matrix

EXERCISE 11B.8

1 a A2 + A b B2 + 2B c A3 ¡ 2A2 + A

d A3 + A2 ¡ 2A e AC + AD + BC + BD

f A2 + AB + BA + B2 g A2 ¡ AB + BA ¡ B2

h A2 + 2A + I i 9I ¡ 6B + B2

2 a A3 = 3A ¡ 2I, A4 = 4A ¡ 3I

b B3 = 3B ¡ 2I, B4 = 6I ¡ 5B, B5 = 11B ¡ 10I

c C3 = 13C ¡ 12I, C5 = 121C ¡ 120I

3 a i I + 2A ii 2I ¡ 2A iii 10A + 6I

b A2 + A + 2I

c i ¡3A ii ¡2A iii A

4 a AB =

³0 00 0

´b A2 =

µ1

2

1

2

1

2

1

2

¶c false as A(A ¡ I) = O does not imply that

A = O or A¡ I = O

d

³0 00 0

´,

³1 00 1

´,

Ãa b

a¡ a2

b1¡ a

!, b 6= 0

5 For example, A =

³0 10 0

´, gives A2 =

³0 00 0

´6 a a = 3, b = ¡4 b a = 1, b = 8

7 p = ¡2, q = 1 a A3 = 5A ¡ 2I b A4 = ¡12A + 5I

C =

³12:509:50

´, N =

³2375 51562502 3612

´³78 669:5065 589

íncome from day 1income from day 2

c $144 258:50

4 b yes, I =

³1 00 1

´

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724 ANSWERS

EXERCISE 11C.1

1 a

³3 00 3

´= 3I,

³1 ¡2

¡ 2

3

5

3

´b

³10 00 10

´= 10I,

³0:2 0:4¡0:1 0:3

´2 a ¡2 b ¡1 c 0 d 1

3 a 26 b 6 c ¡1 d a2 + a

4 a ¡3 b 9 c ¡12

5 Hint: Let A =

³a bc d

´6 a jAj = ad¡ bc, jBj = wz ¡ xy

b AB =

³aw + by ax+ bzcw + dy cx+ dz

´,

jABj = (ad¡ bc)(wz¡xy)

7 a i ¡2 ii ¡8 iii ¡2 iv ¡9 v 2

8 a 1

14

³5 ¡41 2

´b

³1 01 ¡1

ć does not exist

d

³1 00 1

é does not exist f ¡ 1

15

³7 ¡2¡4 ¡1

´g 1

10

³2 ¡41 3

´h

³¡3 ¡12 1

ÉXERCISE 11C.2

1 a

³x+ 2y3x+ 4y

´b

³2a+ 3ba¡ 4b

´2 a

³3 ¡12 3

´³xy

´=

³86

´b

³4 ¡33 2

´³xy

´=

³11¡5

ć

³3 ¡12 7

´³ab

´=

³6¡4

´3 a x = 32

7, y = 22

7b x = ¡ 37

23, y = ¡ 75

23

c x = 17

13, y = ¡ 37

13d x = 59

13, y = ¡ 25

13

e x = ¡40, y = ¡24 f x = 1

34, y = 55

34

4 b i X =

³¡1 32 4

íi X =

µ13

7

3

7

¡ 2

7¡ 8

7

¶5 a i k = ¡3 ii

1

2k + 6

³2 ¡16 k

´, k 6= ¡3

b i k = 0 ii1

3k

³k 10 3

´, k 6= 0

c i k = ¡2 or 1

ii1

(k + 2)(k ¡ 1)

³k ¡2¡1 k + 1

´, k 6= ¡2 or 1

6 a i

³2 ¡34 ¡1

´³xy

´=

³811

´, jAj = 10

ii Yes, x = 2:5, y = ¡1

b i

³2 k4 ¡1

´³xy

´=

³811

´, jAj = ¡2¡ 4k

ii k 6= ¡ 1

2, x =

8 + 11k

2 + 4k, y =

5

1 + 2k

iii k = ¡ 1

2, no solutions

EXERCISE 11C.3

1 X =

³1

4

3

4

1 0

´2 b

³1 00 1

´,

³¡1 00 ¡1

´,

³0 11 0

´,

³0 ¡1¡1 0

´3 a A¡1 =

³0 ¡11

2

1

2

´, (A¡1)¡1 =

³1 2¡1 0

´b (A¡1)¡1(A¡1) = (A¡1)(A¡1)¡1 = I

c (A¡1)¡1 = A

4 a i

µ1

3

1

3

2

3¡ 1

3

¶ii

µ3

2

1

2

1 0

¶iii

µ5

6

1

3

1

3

1

3

¶iv

µ5

6

1

6

2

3

1

3

¶v

µ5

6

1

6

2

3

1

3

¶vi

µ5

6

1

3

1

3

1

3

¶c (AB)¡1 = B¡1A¡1 and (BA)¡1 = A¡1B¡1

d (AB)(B¡1A¡1) = (B¡1A¡1)(AB) = I

AB and B¡1A¡1 are inverses

5 (kA)

³1

kA¡1

´=

³1

kA¡1

´(kA) = I

kA and1

kA¡1 are inverses

6 a X = ABZ b Z = B¡1A¡1X

7 A2 = 2A ¡ I, A¡1 = 2I ¡ A

8 a A¡1 = 4I ¡ A b A¡1 = 5I + A c A¡1 = 3

2A ¡ 2I

10 If A¡1 exists, so jAj 6= 0.

EXERCISE 11D.1

1 a 41 b ¡8 c 0 d 6 e ¡6 f ¡12

2 a x = 1 or 5b When x = 1 or 5, the matrix does not have an inverse.

3 a abc b 0 c 3abc¡ a3 ¡ b3 ¡ c3 4 k 6= ¡3

5 for all values of k except 1

2or ¡9

6 a k = 1 or 4 b k = 5

2or 2

EXERCISE 11D.2

1

Ã2 0 00 2 00 0 2

!= 2I,

0@¡ 11

2

9

2

15

2

¡ 1

2

1

2

1

2

4 ¡3 ¡5

1A2 a

0@ 5

4

3

4¡ 7

4

¡ 1

4¡ 3

4

3

4

¡ 3

4¡ 1

4

5

4

1A b

Ã¡5:5 4:5 7:5¡0:5 0:5 0:54 ¡3 ¡5

!

3 a

Ã0:050 ¡0:011 ¡0:0660:000 0:014 0:028¡0:030 0:039 0:030

!

b

EXERCISE 11E

1 a

Ã1 ¡1 ¡11 1 39 ¡1 ¡3

! Ãxyz

!=

Ã27¡1

!

b

Ã2 1 ¡10 1 21 ¡1 1

! Ãxyz

!=

Ã3613

!

Ã1:596 ¡0:996 ¡0:169¡3:224 1:925 0:6292:000 ¡1:086 ¡0:396

!

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ANSWERS 725

c

Ã1 1 ¡11 ¡1 12 1 ¡3

! Ãabc

!=

Ã76¡2

!2 AB = I, a = 2, b = ¡1, c = 3

3 MN = 4I, u = ¡1, v = 3, w = 5

4 a x = 2:3, y = 1:3, z = ¡4:5

b x = ¡ 1

3, y = ¡ 95

21, z = 2

21

c x = 2, y = 4, z = ¡1

5 a x = 2, y = ¡1, z = 5 b x = 4, y = ¡2, z = 1

c x = 4, y = ¡3, z = 2 d x = 4, y = 6, z = ¡7

e x = 3, y = 11, z = ¡7

f x ¼ 0:33, y ¼ 7:65, z ¼ 4:16

6 a x represents the cost per football in dollars,

y represents the cost per baseball in dollars,

z represents the cost per basketball in dollars

b 12 basketballs

7 a 2x+ 3y + 8z = 352

x+ 5y + 4z = 274

x+ 2y + 11z = 351

b x = 42, y = 28, z = 23

c E1 201 000

8 a Cashews $12, Macadamias $15, Brazil nuts $10

b $11:80 per kg

9 a 5p+ 5q + 6r = 405

15p+ 20q + 6r = 1050

15p+ 20q + 36r = 1800

b p = 24,

q = 27,

r = 25

10 a a = 50 000, b = 100 000, c = 240 000 b yes

c 2009, ¼ $284 000, 2011, ¼ $377 000

REVIEW SET 11A

1 a

³4 2¡2 3

´b

³9 60 ¡3

ć

³¡2 04 ¡8

´d

³2 22 ¡5

é

³¡5 ¡4¡2 6

´f

³7 64 ¡11

´g

³¡1 82 ¡4

´h

³3 2¡6 ¡8

í

µ1

3

2

3

0 ¡1

¶j

³9 40 1

´k

³¡3 ¡106 8

´l

µ1

3

2

3

1

6

1

12

¶2 a a = 0, b = 5, c = 1, d = ¡4

b a = 2, b = ¡1, c = 3, d = 8

3 a Y = B ¡ A b Y = 1

2(D ¡ C) c Y = A¡1 B

d Y = CB¡1 e Y = A¡1(C ¡ B) f Y = B¡1 A

4 a = 3 5

³1 00 1

´6 a

¡10¢

b

Ã4 3 28 6 40 0 0

!c¡15 18 21

¢d CA does not exist e

Ã575

!7 b 2A ¡ I 8 AB = BA = I, A¡1 = B 9 k 6= ¡3 or 1

10 X =

³1 3 2¡1 1 3

´11 m = 6 or ¡3

13 M =

³0 ¡25 1

´

14 a a = 1, b = ¡1,Ã2 1 11 1 12 2 1

! ¡1

=

Ã1 ¡1 0¡1 0 10 2 ¡1

!b x = ¡5, y = 4, z = 7

REVIEW SET 11B

1 x = 1, y = ¡1, z = 2

2 a x = 0, y = ¡ 1

2b x = 12

7, y = 13

7

c X =

³¡1 8¡2 6

´d X =

µ¡ 1

2

3

2

¶e X =

µ14

3

1

3

¶f X =

µ1

2

3

2

3

2¡ 1

2

¶

3 a A¡1 =

0@ 1

4¡ 1

52

5

26

1

4¡ 5

52¡ 1

26

1

4

7

52¡ 9

26

1A b

4 x = ¡1, 2 or ¡4

5 a

³10 ¡12¡10 4

´b

³2 6 ¡3¡4 ¡2 11

ć not possible d

³2:9 ¡0:3¡0:3 2:1

´6 a i jBj 6= 0 ii AB = BA b k 6= 3, ¡2 or 2

7 x = ¡2, y = 3, z = ¡1

8 a a = ¡3, b = 18, c = 48 ) s(t) = ¡3t2 + 18t+ 48

b 48 m c 8 seconds

9 X =³0 ¡21 1

´10 a d = 80 b a = 2, b = 8, c = 10

12 a 3x+ 2y + 5z = 267

2x+ 3y + z = 145

x+ 5y + 4z = 230

b Opera

Play

Concert

E32E18E27

c E200

REVIEW SET 11C

1 a

Ã4 80 26 4

!b

Ã1 20 1

2

3

21

!c¡11 12

¢d BA does not exist.

2 a

Ã4 22 43 4

!b

Ã2 ¡20 4¡1 ¡2

!c

0@¡ 3

23

1

2¡4

2 7

2

1A3 k 6= 3

44 k = ¡6

5 a

µ7

2¡4

¡5

23

¶b does not exist c

µ1 5

3

¡2 ¡ 11

3

¶6 a

³¡9 6 63 ¡3 0

´b

³¡10 ¡65 3

ć

Ã¡2 0 410 ¡7 ¡6¡1 0 2

!d not possible

e

Ã0 227 ¡120 11

!7 x = 5

8 a A( 53

A ¡ 2I) = I b A¡1 = 5

3A ¡ 2I

x = 2, y = 1, z = 3

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726 ANSWERS

9 a

³¡1 ¡260 ¡27

´b X =

³1 13¡3 17

´10 A3 = 27A + 10I, A4 = 145A + 54I,

A5 = 779A + 290I, A6 = 4185A + 1558I

11 $56:30

12 a AB =

³c d¡ 2ac ad¡ b

´b A¡1 =

1

b¡ 2a

³b ¡2¡a 1

ć c = ¡1, d = 0

13 a A3 = ¡I, A4 = ¡A, A5 = ¡A + I, A6 = I, A7 = A,

A8 = A ¡ I

b A6n+3 = (A6)nA3 = ¡I, A6n+5 = ¡A + I

c A¡1 = ¡A + I

EXERCISE 12A.1

1 a b c

d

2 a

b

3 a b

c d

EXERCISE 12A.2

1 a p, q, s, t b p, q, r, t c p and r, q and t d q, t

e p and q, p and t

2 a true b true c false d false e true f false

EXERCISE 12B.1

1 a b

c d

e f

2 a¡!AC b

¡!BD c

¡!AD d

¡!AD

3 a i ii

b yes

EXERCISE 12B.2

1 a b

c d

2 a b

25 m s��

100 m s� ��

30 N

135°

35 m70°

50 m s� ��

10°

30 NN

Scale: 1 cm 10 N

45°

N

10 m s� ��Scale: 1 cm

146°

40 m s� ��

Scale: 1 cm 10 km

25 km32°

Scale: 1 cm km h� �� 30

150 km h� ��

8°

p

q

p q p q

p q

pq

p q

p

qp q

pq

p q p

q

p q

p q

p qpq

q p

p

�qp q�

p�q

p q�

p

�q

p q� �qp

p q�

�r

p

�q

p q r� �

�r

p q

p q r �

75 m s� ��45°

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�r Qw_\r-\Ew_\s2s

�s

r

r

2r s�

s

s

s

r

r

� �r s

s

s

Qw_\r

Qw_\r s�

sr

r s�

Qw_\& *r s�

s

s

ANSWERS 727

c

3 a¡!AB b

¡!AB c 0 d

¡!AD e 0 f

¡!AD

4 a t = r + s b r = ¡s ¡ t

c r = ¡p ¡ q ¡ s d r = q ¡ p + s

e p = t + s + r ¡ q f p = ¡u + t + s ¡ r ¡ q

5 a i r + s ii ¡t ¡ s iii r + s + t

b i p + q ii q + r iii p + q + r

EXERCISE 12B.3

1 a 24:6 km h¡1 b 9:93o east of south

2 a 82:5 m b 23:3o west of north c 48:4 seconds

EXERCISE 12B.4

1 a b c d

e f

g

h

2 a b c

d e

3 a

b a parallelogram

4 a¡!AB = b ¡ a

EXERCISE 12C.1

1 a b c d

2 a

³73

´b

³¡60

ć

³2¡5

´d

³06

é

³¡63

´f

³¡5¡5

ÉXERCISE 12C.2

1 a

³¡26

´b

³¡26

ć

³¡1¡1

´d

³¡1¡1

é

³¡5¡3

´f

³¡5¡3

´g

³¡64

´h

³¡41

´2 a

³¡37

´b

³¡4¡3

ć

³¡8¡1

´d

³¡69

é

³0¡5

´f

³6¡9

´3 a

³¡54

´b

³12

ć

³6¡5

´4 a

³24

´b

³¡25

ć

³3¡3

´d

³1¡5

é

³6¡5

´f

³13

ÉXERCISE 12C.3

1 a

³ ¡3¡15

´b

³¡12

ć

³014

´d

³5¡3

é

µ5

2

11

2

¶f

³¡77

´g

³511

´h

µ317

3

¶2 a

³8¡1

´b

³8¡1

ć

³8¡1

ÉXERCISE 12C.4

1 ap13 units b

p17 units c 5

p2 units d

p10 units

ep29 units

2 ap10 units b 2

p10 units c 2

p10 units d 3

p10 units

e 3p10 units f 2

p5 units g 8

p5 units h 8

p5 units

ip5 units j

p5 units

r

�p �q

r q p� �

pq

pp q= 2

q

pq

p p q= 3�

qp

p q= Qe_q

i

iiiii

P

N

M

X

Z

Y

4

3

�5

2

�3

�12

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Y:\HAESE\IB_SL-2ed\IB_SL-2ed_an\727IB_SL-2_an.CDR Tuesday, 17 March 2009 10:02:22 AM PETER

728 ANSWERS

4 a¡!AB =

³16

´, AB=

p37 units

b¡!BA =

³¡1¡6

´, BA=

p37 units

c¡!BC =

³¡4¡1

´, BC=

p17 units

d¡!DC =

³37

´, DC=

p58 units

e¡!CA =

³3¡5

´, CA=

p34 units

f¡!DA =

³62

´, DA=2

p10 units

EXERCISE 12D

1 a b

c d

2 a ip14 units ii

b ip14 units ii

¡1, ¡ 1

2, 3

2

¢c i

p21 units ii

¡1, ¡ 1

2, 0¢

d ip14 units ii

¡1, 1

2, ¡ 3

2

¢4 a isosceles b right angled c right angled

d straight line 5 (0, 3, 5), r =p3 units

6 a (0, y, 0) b (0, 2, 0) and (0, ¡4, 0)

EXERCISE 12E

1 a b¡!OT =

Ã3¡14

!c OT =

p26 units

2 a¡!AB =

Ã4¡1¡3

!,¡!BA =

Ã¡413

!b AB =

p26 units BA =

p26 units

3¡!OA =

Ã310

!,¡!OB =

Ã¡112

!,¡!AB =

Ã¡402

!

4 a¡!NM =

Ã5¡4¡1

!b

¡!MN =

Ã¡541

!c MN =

p42 units

5 a¡!OA =

Ã¡125

!, OA =

p30 units

b¡!AC =

Ã¡2¡1¡5

!, AC =

p30 units

c¡!CB =

Ã5¡13

!, CB =

p35 units

6 ap13 units b

p14 units c 3 units

7 a a = 5, b = 6, c = ¡6 b a = 4, b = 2, c = 1

8 a a = 1

3, b = 2, c = 1 b a = 1, b = 2

c a = 1, b = ¡1, c = 2

9 a r = 2, s = 4, t = ¡7 b r = ¡4, s = 0, t = 3

10 a¡!AB =

Ã2¡5¡1

!,¡!DC =

Ã2¡5¡1

!b ABCD is a parallelogram

11 a S(¡2, 8, ¡3)

EXERCISE 12F

1 a x = 1

2q b x = 2n c x = ¡ 1

3p

d x = 1

2(r ¡ q) e x = 1

5(4s ¡ t) f x = 3(4m ¡ n)

2 a y =

µ¡13

2

¶b y =

³24

ć y =

µ3

2

¡ 1

2

¶d y =

µ5

4

3

4

¶4 a B(¡1, 10) b B(¡2, ¡9) c B(7, 4)

5 a M(1, 4) b¡!CA =

³75

´,¡!CM =

³53

´,¡!CB =

³31

´6 a x =

Ã4¡6¡5

!b x =

0@ 1

¡ 2

3

5

3

1A c x =

0@ 3

2

¡15

2

1A7

¡!AB =

Ã34¡2

!, AB =

p29 units

9 C(5, 1, ¡8), D(8, ¡1, ¡13), E(11, ¡3, ¡18)

10 a parallelogram b parallelogram

c not parallelogram

11 a D(9, ¡1) b R(3, 1, 6) c X(2, ¡1, 0)

12 a¡!BD = 1

2a b

¡!AB = b ¡ a c

¡!BA = ¡b + a

d¡!OD = b + 1

2a e

¡!AD = b ¡ 1

2a f

¡!DA = 1

2a ¡ b

13 a

Ã¡15¡1

!b

Ã¡34¡2

!c

Ã¡36¡5

!

14 a

Ã31¡2

!b

Ã1¡34

!c

Ã14¡9

!

d

Ã2¡410

!e

Ã32¡5

!f

0@ ¡13

2

¡ 7

2

1A

OP = 3 units

P , ,( )��

Z

X

Y

OP =p5 units

P , ,( )�� Z

X

Y

2

�1

OP =p26 units

Z

X

YP , ,( )��

13

4

pOP = 14 units

Z

X

Y

P , ,( )��

�1

�2

33

¡¡ 1

2, 1

2, 2¢

Z

X

Y

T , ,( )��

3

4

�1�1

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ANSWERS 729

g

Ã1¡47

!h

Ã42¡2

!15 a

p11 units b

p14 units c

p38 units

dp3 units e

0@p11

¡3p11

2p11

1A f

0B@¡ 1p11

1p11

3p11

1CA16 a r = 2, s = ¡5 b r = 4, s = ¡1

EXERCISE 12G

1 r = 3, s = ¡9 2 a = ¡6, b = ¡4

4 a¡!AB k ¡!

CD, AB = 3CD

b¡!RS k ¡!

KL, RS = 1

2KL opposite direction

c A, B and C are collinear and AB = 2BC

5 a¡!PR =

Ã¡1¡33

!,¡!QS =

Ã¡2¡66

!b PR = 1

2QS

EXERCISE 12H

1 a unit vector b unit vector c not a unit vector

d unit vector e not a unit vector

2 a 2i ¡ j b ¡3i ¡ 4j c ¡3i d 7j e 1p2

i ¡ 1p2

j

3 a

³35

´b

³5¡4

ć

³¡40

´d

³03

é

µ p3

2

¡ 1

2

¶

4 a

Ã1¡11

!p3 units

b

Ã3¡11

!p11 units

c

Ã10¡5

!p26 units

d

0@ 01

2

1

2

1A1p2

units

5 a k = §1 b k = §1 c k = 0

d k = §p11

4e k = § 2

3

6 a 5 units bp6 units c 3 units

d ¼ 6:12 units

7 a 1p5(i + 2j) b 1p

13(2i ¡ 3k) c 1p

33(¡2i ¡5j ¡2k)

8 a 3p5

³2¡1

´b ¡ 2p

17

³¡1¡4

ć 6p

18

Ã¡141

!

d ¡ 5

3

Ã¡1¡2¡2

!EXERCISE 12I

1 a 7 b 22 c 29 d 66 e 52 f 3 g 5 h 1

2 a 2 b 2 c 14 d 14 e 4 f 4

3 a ¡1 b 94:1o 4 a 1 b 1 c 05 a 5 b ¡9

7 a t = 6 b t = ¡8 c t = 0 or 2 d t = ¡ 3

2

8 a t = ¡ 3

2b t = ¡ 6

7c t = ¡1§p

5

2d impossible

9 Show a ² b = b ² c = a ² c = 0 10 b t = ¡5

6

11 a BbAC is a right angle b not right angled

c BbAC is a right angle d AbCB is a right angle

12¡!AB ² ¡!

AC = 0, ) BbAC is a right angle

13 b

c 0, the diagonals of a rhombus are perpendicular

14 a 78:7o b 63:4o c 63:4o d 71:6o

15 a k

³¡25

´, k 6= 0 b k

³¡21

´, k 6= 0

c k

³13

´, k 6= 0 d k

³34

´, k 6= 0

e k

³01

´, k 6= 0

16 AbBC ¼ 62:5o, the exterior angle 117:5o

17 a 54:7o b 60o c 35:3o

18 a 30:3o b 54:2o 19 a M( 32

, 5

2, 3

2) b 51:5o

20 a t = 0 or ¡3 b r = ¡2, s = 5, t = ¡4

21 a 74:5o b 72:5o

REVIEW SET 12A

1 a

b

2 a¡!AC b

¡!AD

3 a q = p + r b l = k ¡ j + n ¡ m

4 a

³43

´b

³3¡5

ć

³0¡4

´

5

³14

´6 a p + q b 3

2p + 1

2q 7 m = 5, n = ¡ 1

2

8

Ã8¡87

!9 a ¡13 b ¡36

11 a x =

Ã¡115

¡10

!b x =

Ã211

!12 k = 6

13 k

³54

´, k 6= 0 14 a i p + q ii 1

2p + 1

2q

15 a ² b = ¡4, b ² c = 10, a ² c = ¡10 16 a = ¡2, b = 0

17 a q + r b r + q, DB = AC, [DB] k [AC]

18 a t = ¡4 b¡!LM =

Ã5¡3¡4

!,¡!KM =

Ã¡2¡2¡1

!So,

¡!LM ² ¡!

KM = 0 ) bM = 90o

Scale: 1 cm 10 m

45 mN

60°

Scale: 1 cm m s� �� 10

60 m s� ��

8°

3 a

0@ 2

3

¡ 1

3

¡ 2

3

1A or

0@¡ 2

3

1

3

2

3

1A b

0@¡ 4

3

¡ 2

3

4

3

1A or

0@ 4

3

2

3

¡ 4

3

1A

j¡!AB j =p14 units, j¡!BC j =p

14 units, ABCD is a rhombus

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730 ANSWERS

REVIEW SET 12B

1 a b

2 4:84 km, 208o

3 AB = AC =p53 units and BC =

p46 units ) ¢ is isosceles

4 ap17 units b

p13 units c

p10 units d

p109 units

5 80:3o 6 r = 4, s = 7

7 a

Ã¡613

!b

p46 units c (¡1, 3 1

2, 1

2)

8 a ¡1 b

Ã4¡17

!c 60o

9 c = 50

310 a

Ã7

¡12¡7

!b

0@ 1

¡ 5

3

¡ 2

3

1A 11 64:0o

12 (0, 0, 1) and (0, 0, 9) 13 t = 2

3or ¡3 14 72:3o

15 a¡!AC = ¡p + r,

¡!BC = ¡q + r 16 a 8 b 62:2o

17 a r = ¡2, s = 15

2b § 4p

14(3i ¡ 2j + k) 18 16:1o

REVIEW SET 12C

1 a¡!PQ b

¡!PR

2 a

Ã3¡311

!b

Ã7¡3¡26

!c

p74 units

3 a AB = 1

2CD, [AB] k [CD] b C is the midpoint of [AB].

4 a¡!PQ =

Ã¡3123

!b

p162 units c

p61 units

5 a r + q b ¡p + r + q c r + 1

2q d ¡ 1

2p + 1

2r

6 a

³¡4¡2

´b

³ ¡1¡13

ć

³¡48

´7 a X =

³¡11

3

´b X =

³1

¡10

´8 v ² w = §6

11 a k = § 7p33

b k = § 1p3

12 40:7o

14 bK ¼ 64:4o, bL ¼ 56:9o, bM ¼ 58:7o

15¡!OT =

³48

ór

³2¡2

´16 a k = § 1

2b ¡ 5p

14

Ã32¡1

!17 a 10 b 61:6o

18 r = 3, s = ¡ 5

2, t = 1

419 sin µ = 2p

5

EXERCISE 13A.1

3 a When t = 1, x = 3, y = ¡2 ) yes b k = ¡5

4 a (1, 2) b

cp29 cm s¡1

5 x = 1¡ t, y = 5 + 3t or 3x+ y = 8

EXERCISE 13A.2

1 a

Ãxyz

!=

Ã13¡7

!+ t

Ã213

!t 2 R

b

Ãxyz

!=

Ã012

!+ t

Ã11¡2

!, t 2 R

c

Ãxyz

!=

Ã¡221

!+ t

Ã100

!, t 2 R

2 a x = 5¡ t, y = 2 + 2t, z = ¡1 + 6t, t 2 Rb x = 2t, y = 2¡ t, z = ¡1 + 3t, t 2 Rc x = 3, y = 2, z = ¡1 + t, t 2 R

3 a

Ãxyz

!=

Ã121

!+ t

Ã¡211

!, t 2 R

b

Ãxyz

!=

Ã013

!+ t

Ã30¡4

!, t 2 R

c

Ãxyz

!=

Ã125

!+ t

Ã0¡30

!, t 2 R

d

Ãxyz

!=

Ã01¡1

!+ t

Ã5¡24

!, t 2 R

4 a (¡ 1

2, 9

2, 0) b (0, 4, 1) c (4, 0, 9)

5 (0, 7, 3) and ( 203

, ¡ 19

3, ¡ 11

3)

EXERCISE 13A.3

1 75:5o 2 75:7o

3

³5¡2

´²³

410

´= 0, so the vectors are perpendicular.

4 a 28:6o b x = ¡ 48

7

EXERCISE 13B.1

1 a i (¡4, 3) ii

³125

íii 13 m s¡1

b i (0, ¡6) ii

³3¡4

íii 5 m s¡1

c i (¡2, ¡7) ii

³¡6¡4

íii

p52 m s¡1

d i (5, ¡5) ii

³84

íii

p80 m s¡1

x

x y

y

�x�x

yy x��

y

x( )��,

(3 ),��

(5 ),��

(7 ),��

9 t = 2§p2 10 bK ¼ 123:7o, bL ¼ 11:3o, bM = 45o

1 a i

³xy

´=

³3¡4

´+ t

³14

´, t 2 R ii 4x¡ y = 16

b i

³xy

´=

³52

´+ t

³28

´, t 2 R ii 4x¡ y = 18

c i

³xy

´=

³¡60

´+ t

³37

´, t 2 R ii 7x¡3y = ¡42

d i

³xy

´=

³¡111

´+ t

³¡21

´, t 2 R ii x+2y = 21

2 x = ¡1 + 2t, y = 4¡ t, t 2 R

Points are: (¡1, 4), (1, 3), (5, 1), (¡3, 5), (¡9, 8)

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ANSWERS 731

2 a

³120¡90

´b

³123:5

ć

µ20

p5

10p5

¶3

Ã¡1230¡84

!EXERCISE 13B.2

1 a

³¡3 + 2t¡2 + 4t

´d

2 a A is at (4, 5), B is at (1, ¡8)

b For A it is

³1¡2

´. For B it is

³21

´.

c For A, speed isp5 km h¡1. For B, speed is

p5 km h¡1.

d

³1¡2

´²³21

´= 0

3 a

³x1y1

´=

³¡54

´+ t

³3¡1

´) x1(t) = ¡5 + 3t,

y1(t) = 4¡ t

b speed =p10 km min¡1

c a minutes later, (t¡ a) min have elapsed.

)

³x2

y2

´=

³157

´+ (t¡ a)

³¡4¡3

´) x2(t) = 15¡ 4(t¡ a); y2(t) = 7¡ 3(t¡ a)

d Torpedo is fired at 1:35:28 pm and the explosion occurs at

1:37:42 pm.

4 a

Ã¡31

¡0:5

!b ¼ 19:2 km h¡1

c

Ãxyz

!=

Ã693

!+ t

Ã¡31

¡0:5

!, t 2 R d 1 hour

EXERCISE 13B.3

1 a 6i ¡ 6j b

³6¡ 6t¡6 + 8t

ć when t = 3

4hours

d t = 0:84 and position is (0:96, 0:72)

2 a

³¡120¡40

´b

³xy

´=

³200100

´+ t

³¡120¡40

ć

³8060

´d

¯̄̄³8060

´¯̄̄= 100 km

e at 1:45 pm and dmin ¼ 31:6 km

3 a A(18, 0) and B(0, 12) b R is at

³x,

36¡ 2x

3

ć

¡!PR =

µx¡ 436¡2x

3

¶and

¡!AB =

³¡1812

´d¡108

13, 84

13

¢and distance ¼ 7:77 km

4 a A(3, ¡4) and B(4, 3)

b For A

³¡12

´, for B

³¡3¡2

ć 97:1o

d at t = 1:5 hours

5 a (2, ¡1, 4) bp27 units

6 a (2, 1

2, 5

2) b

p3

2units

EXERCISE 13B.4

1 a b A(2, 4),

B(8, 0),

C(4, 6)

c BC = BA

=p52 units

) isosceles ¢

2 a b A(¡4, 6),

B(17, 15),

C(22, 25),

D(1, 16)

3 a A(2, 3), B(8, 6), C(5, 0) b AB = BC =p45 units

4 a P(10, 4), Q(3, ¡1), R(20, ¡10)

b¡!PQ =

³¡7¡5

´,¡!PR =

³10¡14

´,¡!PQ ² ¡!

PR = 0

c QbPR = 90o d 74 units2

5 a A is at (2, 5), B(18, 9), C(14, 25), D(¡2, 21)

b¡!AC =

³1220

ánd

¡!DB =

³20¡12

ć Diagonals are perpendicular and equal in length, and as their

midpoints are both (8, 15), ABCD is a square.

EXERCISE 13C

1 a They intersect at (1, 2, 3), angle ¼ 10:9o.

b Lines are skew, angle ¼ 62:7o.

c They are parallel, ) angle = 0o.

d They are skew, angle ¼ 11:4o.

e They intersect at (¡4, 7, ¡7), angle ¼ 40:2o.

f They are parallel, ) angle = 0o.

REVIEW SET 13A

1 a

³xy

´=

³¡63

´+ t

³4¡3

´b x = ¡6 + 4t, y = 3¡ 3t, t 2 R

2 m = 10 3 x = 3 + 2t, y = ¡3 + 5t or 5x¡ 2y = 21

4 a¡!PQ =

Ã14¡3

! ¯̄¡!PQ¯̄=

p26 units,

¡!QR =

Ã¡4¡14

!b x = 2 + t, y = 4t, z = 1¡ 3t, t 2 R

5 1p74

i + 8p74

j + 3p74

k or ¡ 1p74

i ¡ 8p74

j ¡ 3p74

k

6 a A(5, 2), B(6, 5), C(8, 3)

b¯̄¡!AB¯̄=

p10 units,

¯̄¡!BC¯̄=

p8 units,

¯̄¡!AC¯̄=

p10 units

c isosceles

7 a b x = 7, y = 3 + 1

3t,

z = ¡4 + 1

3t, t 2 R

c (7, 3 3

4, ¡3 1

4)

8 (4, 1, ¡3) and (1, ¡5, 0)

��

�

�

��

y

x

t��

5

5 10

y

x

line 1

line 2

line 3

A

C

B

10

20

10 20

C(22, 25)

B(17, 15)

A( 4, 6)�

D(1, 16)

y

x

A

B

C

O

�1

�2

a

a

b

b

a b��OABC is a

rhombus.

So, its

diagonals

bisect its

angles.

b

³28

ć i t = 1:5 s

ii t = 0:5 s

f 2:30 pm

ip544 units ii

p544 units iii 0

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732 ANSWERS

REVIEW SET 13B

1

³xy

´=

³08

´+ t

³54

´, t 2 R

2 a¡!PQ =

Ã5¡2¡4

!b 41:8o

3 a i ¡6i + 10j ii ¡5i ¡ 15j

iii (¡6¡ 5t)i + (10¡ 15t)j

b t = 0:48 h

c shortest dist. ¼ 8:85 km, so will miss reef.

4 a i

³xy

´=

³2¡3

´+ t

³4¡1

´, t 2 R

ii x+ 4y = ¡10

b i

³xy

´=

³¡16

´+ t

³6¡8

´, t 2 R

ii 4x+ 3y = 14

5 a (17, ¡9, 0) b (0, 7

3, 17

3) c ( 7

2, 0, 9

2) 6 8:13o

7 a X23, x1 = 2 + t, y1 = 4¡ 3t, t > 0

b Y18, x2 = 13¡ t, y2 = 3¡ 2a+ at, t > 2

c interception occurred at 2:22:30 pm

d bearing ¼ 193o, ¼ 4:54 units per minute

8 a intersecting at (4, 3, 1) angle ¼ 44:5o

b skew, angle ¼ 71:2o

REVIEW SET 13C

1 2p10(3i ¡ j)

2 a

Ãxyz

!=

Ã32¡1

!+ t

Ã¡405

!, t 2 R

b (¡5, 2, 9) or (11, 2, ¡11)

3 a (¡4, 3) b (28, 27) c 10 m s¡1 d

³86

´4 a (KL) is parallel to (MN) as

³5¡2

ís parallel to

³¡52

´b (KL) is perpendicular to (NK) as

³5¡2

´²³

410

´= 0

and (NK) is perpendicular to (MN) as

³410

´²³

5¡2

´= 0

c K(7, 17), L(22, 11), M(33, ¡5), N(3, 7) d 261 units2

5 30:5o

6 a¯̄¡!AB¯̄=

p22 units

b

Ãxyz

!=

Ã3¡11

!+ t

Ã¡33¡2

!, t 2 R

7 26:4o

8 a Road A:

³xy

´=

³¡92

´+ t

³4¡3

´, t 2 R

Road B:

³xy

´=

³6

¡18

´+ s

³512

´, s 2 R

b Road B, 13 km

9 a¡!AB ² ¡!

AC =

Ã¡2¡16

!²Ã

522

!= 0

b x = 4¡ 2t, y = 2¡ t, z = ¡1 + 6t, t 2 Rc x = 4 + 5s, y = 2 + 2s, z = ¡1 + 2s, s 2 R

EXERCISE 14A

1 a Heights can take any value from 170 cm to 205 cm, e.g.,

181:37 cm.

b

c The modal class is 185 6 H < 190 cm, as this occurred the

most frequently.

d slightly positively skewed

2 a

b Stem Leaf

0 3 6 8 8 8 81 0 0 0 0 0 2 2 2 4 4 4 4 5 5 5 5 6 6 6 6 7 8 8 8 8 92 0 0 0 1 2 4 5 5 5 6 7 7 83 1 2 2 2 3 4 5 7 8

4 0 2 5 5 5 6 1 j 2 means 12 minutes

c positively skewed

d The modal travelling time was between 10 and 20 minutes.

3 a column graph b frequency histogram

4 a b 20c 58:3%

d i 1218

ii 512

EXERCISE 14B.1

1 a i 5:61 ii 6 iii 6

b i 16:3 ii 17 iii 18

c i 24:8 ii 24:9 iii 23:5

2 a A : 6:46 B : 6:85 b A : 7 B : 7c The data sets are the same except for the last value, and the

last value of A is less than the last value of B, so the mean

of A is less than the mean of B.

d The middle value of the data sets is the same, so the median

is the same.

3 a mean: $29 300, median: $23 500, mode: $23 000

b The mode is the lowest value, so does not take the higher

values into account.

c No, since the data is positively skewed, the median is not in

the centre.

4 a mean: 3:19, median: 0, mode: 0

b The data is very positively skewed so the median is not in

the centre.

�

��

�

��

Heights of basketball players

height (cm)

freq

uen

cy

frequency

no. of matches

47

48

49

50 51

52 53

54

550

5

10

15frequency

length(cm)

0

5

10

15

120

130

140

150

160

170

0

10

20

30

40

50

300 325 350 375 400 425 450

Seedling height

freq

uen

cy

mm

Continuous numerical, but has been rounded to become

discrete numerical data.

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ANSWERS 733

c The mode is the lowest value so does not take the higher

values into account.

d yes, 21 and 42 e no

5 a 44 b 44 c 40:2 d increase mean to 40:3

6 $185 604 7 3144 km 8 116 9 17:25 goals per game

10 a mean = $163 770, median = $147 200 (differ by $16 570)

b i mean selling price ii median selling price

EXERCISE 14B.2

1 a 1 b 1 c 1:43

2 a i 2:96 ii 2 iii 2 b

c positively skewed

d The mean takes into

account the larger

numbers of phone

calls.e the mean

c The sample of only 30 is not large enough. The company

could have won its case by arguing that a larger sample would

have found an average of 50 matches per box.

4 a i 2:61 ii 2 iii 2

b This school has more children per family than the average

Australian family.

c positive

d The mean is larger than the median and the mode.

5 a i 69:1 ii 67 iii 73 b i 5:86 ii 5:8 iii 6:7

6 a x = 5 b 75%

7 a i 5:63 ii 6 iii 6 b i ii 7 iii 7c the mean d yes

8 a ¼ 70:9 g b ¼ 210 g c 139 g 9 10:1 cm

10 a mean for A ¼ 50:7, mean for B ¼ 49:9

b No, as to the nearest match, A is 51 and B is 50.

EXERCISE 14B.3

1 31:72 a 70 b ¼ 411 000 litres, ¼ 411 kL c ¼ 5870 L

3 a 125 people b ¼ 119 marks c 3

25d 137 marks

EXERCISE 14C.1

1 a i 6 ii Q1 = 4, Q3 = 7 iii 7 iv 3

b i 17:5 ii Q1 = 15, Q3 = 19 iii 14 iv 4

c i 24:9 ii Q1 = 23:5, Q3 = 26:1 iii 7:7 iv 2:6

2 a median = 2:45, Q1 = 1:45, Q3 = 3:8

b range = 5:2, IQR = 2:35

c i ..... greater than 2:45 min ii ...... less than 3:8 min

iii The minimum waiting time was 0 minutes and the

maximum waiting time was 5:2 minutes. The waiting

times were spread over 5:2 minutes.

3 a 3 b 42 c 20 d 13 e 29 f 39 g 16

4 a i 124 cm ii Q1 = 116 cm, Q3 = 130 cm

b i ...... 124 cm tall ii ...... 130 cm tall

c i 29 cm ii 14 cm d ...... over 14 cm

5 a i 7 ii 6 iii 5 iv 7 v 2

b i 10 ii 7 iii 6 iv 8 v 2

EXERCISE 14C.2

2 a ...... was 98, ...... was 25b .... greater than or equal to 70 c .... at least 85 marks

d ...... between 55 and 85 ...... e 73 f 30 g ¼ 67

3 a i min = 3, Q1 = 5, median = 6, Q3 = 8, max = 10

ii

iii range = 7 iv IQR = 3

b i min = 0, Q1 = 4, median = 7, Q3 = 8, max = 9

ii

iii range = 9 iv IQR = 4

c i min = 117, Q1 = 127, med. = 132, Q3 = 145:5,

max = 151

ii

iii range = 34 iv IQR = 18:5

4 a Statistic Year 9 Year 12

min value 1 6Q1 5 10

median 7:5 14Q3 10 16

max value 12 17:5

b i Year 9: 11,

Year 12: 11:5ii Year 9: 5,

Year 12: 6c i true

ii true

5 a median = 6, Q1 = 5, Q3 = 8 b 3c

6 a Min = 33, Q1 = 35, Q2 = 36, Q3 = 37, Max = 40b i 7 ii 2 c

EXERCISE 14D

1 a b i 40 ii 40

2 a Length (x cm) Frequency C. frequency

24 6 x < 27 1 127 6 x < 30 2 330 6 x < 33 5 833 6 x < 36 10 1836 6 x < 39 9 2739 6 x < 42 2 2942 6 x < 45 1 30

b

Phone calls in a day

3

00 1 2 3 4 5 6 7 8 9 1011

6

9

12

15

freq

uen

cy

number ofphone calls

mean (2.96)mode, median (2)

3 4 5 6 7 8 9 10

3210 4 5 6 7 8 9

115 120 125 130 135 140 145 150 155

2 3 4 5 6 7 8 9 10 11 12 13

30 31 32 33 34 35 36 37 38 39 40 41

�

��

�

��

�

��

�

��

cumulative frequency

length (cm)

median

11 x = 15 12 a = 5 13 37 14 14:8 15 6 and 1216 9 and 7

3 a i 49 ii 49 iii 49:0 b no

6:81

11 a i E31 500 ii E28 000 iii E33 300 b The mean.

1 a i 35 ii 78 iii 13 iv 53 v 26

b i 65 ii 27

25.2 cm

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734 ANSWERS

c median ¼ 35 cm

d actual median = 34:5, ) a good approx.

3 a 9 b ¼ 28:3% c 7 cm d IQR ¼ 2:4 cm

e 10 cm, which means that 90% of the seedlings have a

height of 10 cm or less.

4

a ¼ 61 b ¼ 87 students c ¼ 76 students

d 24 (or 25) students e 76 marks

5 a 26 years b 36% c i 0:527 ii 0:0267

6 a 27 min b 29 min c 31:3 min

d IQR ¼ 4:3 min e 28 min 10 s

7 a 2270 h b ¼ 69% c 62 or 63

EXERCISE 14E

1 a x ¼ 4:87, Min = 1, Q1 = 3, Q2 = 5, Q3 = 7,

Max = 9

b

c

d x ¼ 5:24, Min = 2, Q1 = 4, Q2 = 5, Q3 = 6:5,

Max = 9

2 a discrete c

d There are no outliers for Shane.

Brett has outliers of 7 and 8 which must not be removed.

e Shane’s distribution is reasonably symmetrical. Brett’s

distribution is positively skewed.

f Shane has a higher mean (¼ 2:89 wickets) compared with

Brett (¼ 2:67 wickets). Shane has a higher median (3wickets) compared with Brett (2:5 wickets). Shane’s modal

number of wickets is 3 (14 times) compared with Brett, who

g Shane’s range is 6 wickets, compared with Brett’s range of

8 wickets. Shane’s IQR is 2 wickets, compared with Brett’s

IQR of 3 wickets. Brett’s wicket taking shows greater spread

or variability.

h

i Generally, Shane takes more wickets than Brett and is a more

consistent bowler.

3 a continuous

c For the ‘new type’ of globes, 191 hours could be considered

an outlier. However, it could be a genuine piece of data, so

we will include it in the analysis.

Old type New type

Mean 107 134

Median 110:5 132Range 56 84

IQR 19 18:5

d The mean and

median are ¼ 25%and ¼ 19% higher

for the ‘new type’ of

globe compared with

the ‘old type’.The range is higher

for the ‘new type’ of globe (but has been affected by the

191 hours).The IQR for each type of globe is almost the same.

e

f For the ‘old type’ of globe, the data is bunched to the right

of the median, hence the distribution is negatively skewed.

For the ‘new type’ of globe, the data is bunched to the left

of the median, hence the distribution is positively skewed.

g The manufacturer’s claim, that the ‘new type’ of globe has a

20% longer life than the ‘old type’ seems to be backed up by

the 25% higher mean life and 19:5% higher median life.

EXERCISE 14F.1

1 a Sample A

A B

b x 8 8

c s 2 1:06

2 a x s

Andrew 25 4:97

Brad 30:5 12:6

b Andrew

3 a Rockets: range = 11, x = 5:7;

Bullets: range = 11, x = 5:7

b We suspect the Rockets, they have two zeros.

c Rockets: s = 3:9 greater variability

Bullets: s ¼ 3:29d standard deviation

4 a We suspect variability in standard deviation since the factors

may change every day.

b i sample mean ii sample standard deviation

c less variability

5 a x = 69, s ¼ 6:05 b x = 79, s ¼ 6:05c The distribution has simply shifted by 10 kg. The mean

increases by 10 kg and the standard deviation remains

the same.

6 a x = 1:01 kg; s = 0:17 b x = 2:02 kg; s = 0:34c Doubling the values doubles the mean and the standard

deviation.

7 p = 6, q = 9 8 a = 8, b = 6

1 2 3 4 5 6 7 8 9 10

012345

1 2 3 4 5 6 7 8 9score

frequency

0 1 2 3 4 5 6 7 8 9 10

set 1

set 2

�

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�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�wickets per innings

wickets per innings

freq

uen

cyfr

equen

cy

Shane

Brett

‘old’

‘new’

60 80 100 120 140 160 180 200

lifespan (hours)

0 1 2 3 4 5 6 7 8 9 10

Shane

Brett

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��


score

median

has two modal values of 2 and 3 (7 times each).

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ANSWERS 735

9 a b

c the extreme value greatly increases the standard deviation

EXERCISE 14F.2

1 a sn ¼ 6:77 kg ) ¾ ¼ 6:77 kg b ¹ ¼ 93:8 kg

2 a x ¼ 77:5 g, sn ¼ 7:44 g b ¹ ¼ 77:5 g, ¾ ¼ 7:44 g

EXERCISE 14F.3

1 a x ¼ 1:72 children, s ¼ 1:67 children

b ¹ ¼ 1:72 children, ¾ ¼ 1:67 children

2 a x ¼ 14:5 years, s ¼ 1:75 years

b ¹ ¼ 14:5 years, ¾ ¼ 1:75 years

3 a x ¼ 37:3 toothpicks, s ¼ 1:45 toothpicks

b ¹ ¼ 37:3 toothpicks, ¾ ¼ 1:45 toothpicks

4 a x ¼ 48:3 cm, s ¼ 2:66 cm

b ¹ ¼ 48:3 cm, ¾ ¼ 2:66 cm

5 a x ¼ $390:30, s ¼ $15:87

b ¹ ¼ $390:30, ¾ ¼ $15:87

EXERCISE 14G

1 a 16% b 84% c 97:4% d 0:15% 2 3 times

3 a 5 b 32 c 136 4 a 458 babies b 444 babies

REVIEW SET 14A

1 a Diameter of bacteria colonies

0 4 8 91 3 5 5 72 1 1 5 6 8 83 0 1 2 3 4 5 5 6 6 7 7 9

4 0 1 2 7 9 0 j 4 means 0:4 cm

b i 3:15 cm ii 4:5 cm

c The distribution is slightly negatively skewed.

2 a = 8, b = 6 or a = 6, b = 8

3 a Girls Boys

shape pos. skewed approx. symm.

centre (median) 36:3 s 34:9 s

spread (range) 7:7 s 4:9 s

b The girls’ distribution is positively skewed and boys’ distribu-

tion is approximately symmetrical. The median swim times

for boys is 1:4 seconds lower than for girls but the range

of the girls’ swim times is 2:8 seconds higher than for boys.

The analysis supports the conjecture that boys generally swim

faster than girls with less spread of times.

4 a 2:5% b 95% c 68%

5

6 a 58:5 s b 6 s 7 a 2:5% b 84% c 81:5%8 a 88 students b m ¼ 24

REVIEW SET 14B

1 a highest = 97:5 m, lowest = 64:6 m

b use groups 60 6 d < 65, 65 6 d < 70, ......

c Distances thrown by Thabiso

Distance (m) Tally Freq. (f )

60 6 d < 65 j 1

65 6 d < 70 jjj 3

70 6 d < 75 jjjj©© 5

75 6 d < 80 jj 2

80 6 d < 85 jjjj©© jjj 8

85 6 d < 90 jjjj©© j 6

90 6 d < 95 jjj 3

95 6 d < 100 jj 2

Total 30

d

e i ¼ 81:1 m ii ¼ 83:1 m

2 a

b ¼ 25:9 c ¼ 12:0 d x ¼ 26:0, s ¼ 8:31

3 a i 101:5 ii 98 iii 105:5 b 7:5c x = 100:2, s ¼ 7:59

4 a x ¼ 33:6 L, s ¼ 7:63 L b ¹ ¼ 33:6 L, ¾ ¼ 7:63 L

5 a x ¼ 49:6, s ¼ 1:60b Does not justify claim. Need a larger sample.

6 range = 19, lower quartile = 119, upper quartile = 130,

s ¼ 6:38

7 a = 8, b = 4

REVIEW SET 14C

1

2 a 68% b 95% c 81:5% d 13:5%

3 ¼ 414 customers

4 a A B

Min 11 11:2Q1 11:6 12

Median 12 12:6Q3 12:6 13:2

Max 13 13:8

b A B

i Range 2 2:6

ii IQR 1 1:2

0

10

20

30

40

50

60

0

Cumulative frequency

Score

9.95 19.95 29.95 39.95 49.95

median

0

10

20

30

40Frequency

margin

(points)

0 10 20 30 40 50

11 12.5 15 16.5 18

distance (m)

0

3

6

9f

Frequency histogram displaying thedistance Thabiso throws a baseball

60 65 70 75 80 85 90 95 100

0:809 0:150

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736 ANSWERS

c i We know the members of squad A generally ran faster

because their median time is lower.

ii We know the times in squad B are more varied because

their range and IQR is higher.

5 a x = E103:51, s ¼ E19:40 b ¹ = E103:51, ¾ ¼ E19:40

6 a mean is 18:8, standard deviation is 2:6 b 13:6 to 24:0

7 a 120 students b 65 marks c 54 and 75

d 21 marks e 73% of them f 81 marks

EXERCISE 15A

1 a 0:78 b 0:22 2 a 0:487 b 0:051 c 0:731

3 a 43 days b i ¼ 0:047 ii ¼ 0:186 iii 0:465

4 a ¼ 0:089 b ¼ 0:126

EXERCISE 15B

1 a fA, B, C, Dg b fBB, BG, GB, GGgc fABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD,

BADC, BCAD, BCDA, BDAC, BDCA, CABD, CADB,

CBAD, CBDA, CDAB, CDBA, DABC, DACB, DBAC,

DBCA, DCAB, DCBAgd fGGG, GGB, GBG, BGG, GBB, BGB, BBG, BBBg

2 a b

c d

3 a b

c d

EXERCISE 15C.1

1 a 1

5b 1

3c 7

15d 4

5e 1

5f 8

15

2 a 4 b i 2

3ii 1

3

3 a 1

4b 1

9c 4

9d 1

36e 1

18f 1

6

g 1

12h 1

3

4 a 1

7b 2

7c 124

1461d 237

1461fremember leap yearsg

5 fAKN, ANK, KAN, KNA, NAK, NKAga 1

3b 1

3c 1

3d 2

3

6 a fGGG, GGB, GBG, BGG, GBB, BGB, BBG, BBBgb i 1

8ii 1

8iii 1

8iv 3

8v 1

2vi 7

8

7 a fABCD, ABDC, ACBD, ACDB, ADBC, ADCB,

BACD, BADC, BCAD, BCDA, BDAC, BDCA,

CABD, CADB, CBAD, CBDA, CDAB, CDBA,

DABC, DACB, DBAC, DBCA, DCAB, DCBAgb i 1

2ii 1

2iii 1

2iv 1

2

EXERCISE 15C.2

1

a 1

4b 1

4c 1

2d 3

4

2 a

b 10 c i 1

10ii 1

5

iii 3

5iv 3

5

3 a 1

36b 1

18c 5

9d 11

36e 5

18

f 25

36g 1

6h 5

18i 2

9j 13

18

EXERCISE 15D

1 a 0:476 b 0:241 c 0:483 d 0:578 e 0:415

2 a 7510 b i 0:325 ii 0:653 iii 0:243

3 a 0:428 b 0:240 c 0:758 d 0:257 e 0:480

EXERCISE 15E.1

1 a 6

7b 36

49c 216

3432 a 1

8b 1

8

3 a 0:0096 b 0:8096 4 a 1

16b 15

16

5 a 0:56 b 0:06 c 0:14 d 0:24

6 a 8

125b 12

125c 27

125

EXERCISE 15E.2

EXERCISE 15F

1 a b 1

4

c 1

16

d 5

8

e 3

4

coin

dieH

T

1 2 3 4 5 6

die 1

die 2

1

1

2

2

3

3

4

4

5

5

6

6

spinner 1

spinner 2

A

1

B

2

C

3

D

4

die

spinner

1

A

2

B

3

C

4

D

5 6

5-cent 10-centH

H

H

T

T

T

coin spinner

A

A

B

B

H

T

C

C

spinner 1 spinner 2

X

X

X

Y1

Y2

Y3

Z

Z

Z

draw 1 draw 2

PBP

B

W

W

B

R

Y

B

R

Y

B

R

Y

B

R

Y

Qw_

Qr_

Qr_

Qw_Qr_

Qw_

Qr_Qr_

Qr_

Qw_Qr_

Qr_

1st spin 2nd spin

5-cent

10-centH

H

T

T

coin

H

T

21 3 4 5

spinner

PBW

PBW

1 a 14

55b 1

552 a 7

15b 7

30c 7

15

3 a 3

100b 3

100£ 2

99¼ 0:0006

c 3

100£ 2

99£ 1

98¼ 0:000 006 d 97

100£ 96

99£ 95

98¼ 0:912

4 a 4

7b 2

7

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ANSWERS 737

2 P(win) = 7

50

3 0:032 4 17

405 9

386 a 11

30b 19

30

EXERCISE 15G

1 a 20

49b 10

212 a 3

10b 1

10c 3

5

3 a 2

9b 5

94 a 1

3b 2

15c 4

15d 4

15

These are all possibilities, so their probabilities must sum to 1.

5 a 1

5b 3

5c 4

56 19

45

7 a 2

100£ 1

99¼ 0:0002 b 98

100£ 97

99¼ 0:9602

c 1¡ 98

100£ 97

99¼ 0:0398

8 7

339 7 to start with

EXERCISE 15H

1 a (p+ q)4 = p4 + 4p3q + 6p2q2 + 4pq3 + q4

b 4( 12)3( 1

2) = 1

4

2 a (p+ q)5 = p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5

b i 5( 12)4( 1

2) = 5

32ii 10( 1

2)2( 1

2)3 = 5

16

iii¡1

2

¢4 ¡ 1

2

¢= 1

32

3 a¡2

3+ 1

3

¢4=¡2

3

¢4+ 4¡2

3

¢3 ¡ 1

3

¢+ 6¡2

3

¢2 ¡ 1

3

¢2+4¡2

3

¢¡1

3

¢3+¡1

3

¢4b i

¡2

3

¢4= 16

81ii 6

¡2

3

¢2 ¡ 1

3

¢2= 8

27iii 8

9

4 a¡3

4+ 1

4

¢5=¡3

4

¢5+ 5¡3

4

¢4 ¡ 1

4

¢1+ 10

¡3

4

¢3 ¡ 1

4

¢2+10

¡3

4

¢2 ¡ 1

4

¢3+ 5¡3

4

¢¡1

4

¢4+¡1

4

¢5b i 10

¡3

4

¢3 ¡ 1

4

¢2= 135

512ii 53

512iii 47

128

5 a ¼ 0:154 b ¼ 0:973 6 a ¼ 0:0305 b ¼ 0:265

7 ¼ 0:000 864 8 ¼ 0:0341 9 4 dice

EXERCISE 15I.1

1 a A = f1, 2, 3, 6g, B = f2, 4, 6, 8, 10gb i n(A) = 4 ii A[B =f1, 2, 3, 4, 6, 8, 10g

iii A\B =f2, 6g2 a b

c d

e f

3 a 29 b 17 c 26 d 5

4 a 65 b 9 c 4 d 52

5 a 19

40b 1

2c 4

5d 5

8e 13

40f 7

20

6 a 19

25b 13

25c 6

25d 7

19

7 a 7

15b 1

15c 2

15d 6

7

8 a b

c d

9 a b

c d

e f

EXERCISE 15I.2

1 For each of these draw two diagrams, shade the first with the LHS

set and the second with the RHS set.

2 a A = f7, 14, 21, 28, 35, ......, 98gB = f5, 10, 15, 20, 25, ......, 95g

i n(A) = 14 ii n(B) = 19 iii 2 iv 31

3 a ib+ c

a+ b+ c+ dii

b

a+ b+ c+ d

iiia+ b+ c

a+ b+ c+ div

a+ b+ c

a+ b+ c+ d

b P(A or B) = P(A) + P(B) ¡ P(A and B)

win

lose

win

lose

rain

no rain

Qt_

Rt_

Qw_

Aw_p_

Qw_

Qw_Op_

A BU

A BU

A BU

A BU

A BU

A BU

A B

'AU

is shaded.

A B

' BA � is shaded.U

A B

CU

A B

'B'A � is shaded.U

A B

'BA � is shaded.U

A B

CU

A B

CU

A B

CU

A B

CU

A B

CU

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738 ANSWERS

EXERCISE 15J

1 a 22 study both

b i 9

25ii 11

20

2 a 3

8b 7

20c 1

5d 15

23

3 a 14

25b 4

5c 1

5d 5

23e 9

144 5

6

5 a 13

20b 7

20c 11

50d 7

25e 4

7f 1

4

6 a 3

5b 2

37 a 0:46 b 14

238 70

163

9 a 0:45 b 0:75 c 0:65

10 a 0:0484 b 0:3926 11 2

3

EXERCISE 15K

1 P(R \ S) = 0:2 and P(R) £ P(S) = 0:2) are independent events

2 a 7

30b 7

12c 7

10No, as P(A \B) 6= P(A) £ P(B)

3 a 0:35 b 0:85 c 0:15 d 0:15 e 0:5

4 14

155 a 91

216b 26

6 Hint: Show P(A0 \B0) = P(A0) P(B0)using a Venn diagram and P(A \B)

7 0:9

8 a i 13

20ii 7

10b No, as P(C \D) 6= P(C) P(D)

REVIEW SET 15A

1 ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC,

BCAD, BCDA, BDAC, BDCA, CABD, CADB, CBAD, CBDA,

CDAB, CDBA, DABC, DACB, DBAC, DBCA, DCAB, DCBA

a 1

2b 1

3

2 a 3

8b 1

8c 5

83 a 2

5b 13

15c 4

15

4 a 0 b 0:45 c 0:8

5 a Two events are independent if the occurrence of each event

does not influence the occurrence of the other. For A and Bindependent, P(A) £ P(B) = P(A and B)

b Two events A and B are disjoint if they have no common

outcomes. P(A or B) = P(A) + P(B)

6 a 2

9

b 5

12

7 a 1

4b 37

40c 2

58 5

9

REVIEW SET 15B

1 P(N wins)

2 a 4

500£ 3

499£ 2

498¼ 0:000 000 193

b 1¡ 496

500£ 495

499£ 494

498¼ 0:023 86

3 a ¼ 0:259 b ¼ 0:703

4 a 0:09

b 0:52

5 1¡ 0:9£ 0:8£ 0:7 = 0:496

6 a¡3

5+ 2

5

¢4=¡3

5

¢4+ 4¡3

5

¢3 ¡ 2

5

¢+ 6¡3

5

¢2 ¡ 2

5

¢2+4¡3

5

¢¡2

5

¢3+¡2

5

¢4b i ii 328

625

7 a Female Male Total

smoker 20 40 60

non-smoker 70 70 140

total 90 110 200

b i 7

20ii 1

2c ¼ 0:121

REVIEW SET 15C

1 BBBB, BBBG, BBGB, BGBB, GBBB, BBGG, BGBG, BGGB,

GGBB, GBBG, GBGB, BGGG, GBGG, GGBG, GGGB, GGGG.

P(2 children of each sex) = 3

8

2 a 5

33b 19

66c 5

11d 16

33

3 a 3

25b 24

25c 11

12

4 a 5

8b 1

4

5

6 a 31

70b 21

31

7 a

b i ¼ 0:0205 ii ¼ 0:205

EXERCISE 16A

1 a 7 b 7 c 11 d 16 e 0 f 3

2g 5 h ¡2

2 a ¡3 b 6 c ¡8 d 1

2e 1

2f 5

3 a 0 b 3 c ¡ 2

3d ¡1 e 1 f 1

EXERCISE 16B

1 a vertical asymptote x = ¡3, horizontal asymptote y = 3as x ! ¡3¡, f(x) ! 1as x ! ¡3+, f(x) ! ¡1

as x ! 1, f(x) ! 3¡

as x ! ¡1, f(x) ! 3+

b horizontal asymptote y = 1: as x ! 1, y ! 1¡

as x ! ¡1, y ! 1¡

c horizontal asymptote y = 0: as x ! 1, f(x) ! 0+

as x ! ¡1, f(x) ! 0¡

M P

0

18 1022

U

N (0.4)

N (0.4)

N (0.4)

N (0.4)

N (0.4)

R (0.6)

R (0.6)

R (0.6)

R (0.6)R (0.6)

W (0.95)

W (0.95)W (0.95)

W (0.05)'

W (0.05)'

W (0.05)'

W (0.36)

W (0.36)R (0.25)

W (0.64)'

W (0.64)'

R (0.75)'

die 1

die 2

1

1

2

2

3

3

4

4

5

5

6

6

= 44

125

= 0:352

¡4

5+ 1

5

¢5=¡4

5

¢5+ 5¡4

5

¢4 ¡ 1

5

¢+ 10

¡4

5

¢3 ¡ 1

5

¢2+10

¡4

5

¢2 ¡ 1

5

¢3+ 5¡4

5

¢¡1

5

¢4+¡1

5

¢5

216

625

0:9975

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ANSWERS 739

d vertical asymptotes x = ¡2, x = 1,

horizontal asymptote y = 0:

as x ! ¡2¡, f(x) ! ¡1as x ! ¡2+, f(x) ! 1as x ! 1¡, f(x) ! 1as x ! 1+, f(x) ! ¡1

as x ! 1, f(x) ! 0+

as x ! ¡1, f(x) ! 0¡

EXERCISE 16D.1

1 a i 0:6 units2 ii 0:4 units2 b 0:5 units2

2 a 0:737 units2 b 0:653 units2

EXERCISE 16D.2

1 n AL AU

10 2:185 0 2:485 025 2:273 6 2:393 650 2:303 4 2:363 4100 2:318 4 2:348 4500 2:330 3 2:336 3

converges to 7

3

2 a i n AL AU

5 0:160 00 0:360 0010 0:202 50 0:302 5050 0:240 10 0:260 10100 0:245 03 0:255 03500 0:249 00 0:251 001000 0:249 50 0:250 5010 000 0:249 95 0:250 05

ii n AL AU

5 0:400 00 0:600 0010 0:450 00 0:550 0050 0:490 00 0:510 00100 0:495 00 0:505 00500 0:499 00 0:501 001000 0:499 50 0:500 5010 000 0:499 95 0:500 05

iii n AL AU

5 0:549 74 0:749 7410 0:610 51 0:710 5150 0:656 10 0:676 10100 0:661 46 0:671 46500 0:665 65 0:667 651000 0:666 16 0:667 1610 000 0:666 62 0:666 72

iv n AL AU

5 0:618 67 0:818 6710 0:687 40 0:787 4050 0:738 51 0:758 51100 0:744 41 0:754 41500 0:748 93 0:750 931000 0:749 47 0:750 4710 000 0:749 95 0:750 05

b i 1

4ii 1

2iii 2

3iv 3

4c area =

1

a+ 1

3 a n Rational bounds for ¼

10 2:9045 < ¼ < 3:304550 3:0983 < ¼ < 3:1783100 3:1204 < ¼ < 3:1604200 3:1312 < ¼ < 3:15121000 3:1396 < ¼ < 3:143610 000 3:1414 < ¼ < 3:1418

b n = 10000

EXERCISE 16D.3

1 a b

n AL AU

5 0:5497 0:749710 0:6105 0:710550 0:6561 0:6761100 0:6615 0:6715500 0:6656 0:6676

cR 1

0

px dx ¼ 0:67

2 a

b n AL AU

50 3:2016 3:2816100 3:2214 3:2614500 3:2373 3:2453

cR 2

0

p1 + x3 dx ¼ 3:24

3 a 18 b 4:5 c 2¼

REVIEW SET 16

1 a ¡4 b 1

4c 8 d ¡ 1

2

2 a horizontal asymptote y = ¡3

as x ! 1, y ! 1 as x ! ¡1, y ! ¡3+

b no asymptotes

c vertical asymptote x = 0, horizontal asymptote y = 0

as x ! 0+, f(x) ! ¡1 as x ! 1, f(x) ! 0+

d vertical asymptote x = 3

2as x ! 3

2

+, y ! ¡1

3 a

lower rectangles upper rectangles

b n AL AU

5 2:9349 3:334950 3:1215 3:1615100 3:1316 3:1516500 3:1396 3:1436

1

1

0.8

0.6

0.4

0.2x

y xy �

x1

4

y

21

4

xy

��

x

10.80.60.40.2

4

3

2

1

y

x

10.80.60.40.2

4

3

2

1

y

x

1.510.5

3

2

1

y

2

31 xy ��

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740 ANSWERS

c

Z 1

0

4

1 + x2dx ¼ 3:1416

4 a

b AU ¼ 0:977 units2, AL ¼ 0:761 units2

c AU ¼ 0:8733 units2, AL ¼ 0:8560 units2

5 a 2¼ b 4

6 a A = 17

4, B = 25

4bR 2

0(4¡ x2) dx ¼ 21

4

EXERCISE 17A.1

1 a 3 b 0 2 a 4 b ¡1 3 f(2) = 3, f 0(2) = 1

EXERCISE 17A.2

1 a 1 b 0 c 3x2 d 4x3 2 f 0(x) = nxn¡1

3 a 2 b ¡1 c 2x¡ 3

d 4x+ 1 e ¡2x+ 5 f 3x2 ¡ 4x

EXERCISE 17B

1 a 3 b ¡12 c 9 d 10 2 a 12 b 108

EXERCISE 17C

1 a 3x2 b 6x2 c 14x

d3px

e1

3px2

f 2x+ 1

g ¡4x h 2x+ 3 i 2x3 ¡ 12x

j6

x2k ¡ 2

x2+

6

x3l 2x¡ 5

x2

m 2x+3

x2n ¡ 1

2xpx

o 8x¡ 4

p 3x2 + 12x+ 12

2 a 7:5x2 ¡ 2:8 b 2¼x c ¡ 2

5x3

d 100 e 10 f 12¼x2

3 a 6 b3px

2c 2x¡ 10

d 2¡ 9x2 e 2x¡ 1 f ¡ 2

x3+

3px

g 4 +1

4x2h 6x2 ¡ 6x¡ 5

4 a 4 b ¡ 16

729c ¡7 d 13

4e 1

8f ¡11

5 b = 3, c = ¡4

6 a2px+ 1 b

1

33px2

c1

xpx

d 2¡ 1

2px

e ¡ 2

xpx

f 6x¡ 3

2

px

g¡25

2x3px

h 2 +9

2x2px

7 ady

dx= 4 +

3

x2,

dy

dxis the gradient function of y = 4x¡ 3

x

bdS

dt= 4t+4 ms¡1,

dS

dtis the instantaneous rate of change

in position at the time t, or the velocity function.

cdC

dx= 3 + 0:004x $ per toaster,

dC

dxis the instantaneous

rate of change in cost as the number of toasters changes.

EXERCISE 17D.1

1 a g(f(x)) = (2x+ 7)2 b g(f(x)) = 2x2 + 7

c g(f(x)) =p3¡ 4x d g(f(x)) = 3¡ 4

px

e g(f(x)) =2

x2 + 3f g(f(x)) =

4

x2+ 3

2 a g(x) = x3, f(x) = 3x+ 10

b g(x) =1

x, f(x) = 2x+ 4

c g(x) =px, f(x) = x2 ¡ 3x

d g(x) =10

x3, f(x) = 3x¡ x2

EXERCISE 17D.2

1 a u¡2, u = 2x¡ 1 b u12 , u = x2 ¡ 3x

c 2u¡ 12 , u = 2¡ x2 d u

13 , u = x3 ¡ x2

e 4u¡3, u = 3¡ x f 10u¡1, u = x2 ¡ 3

2 a 8(4x¡5) b 2(5¡2x)¡2 c 1

2(3x¡x2)¡

12 £(3¡2x)

d ¡12(1¡ 3x)3 e ¡18(5¡ x)2

f 1

3(2x3 ¡ x2)¡

23 £ (6x2 ¡ 2x) g ¡60(5x¡ 4)¡3

h ¡4(3x¡x2)¡2 £ (3¡ 2x) i 6(x2 ¡ 2

x)2 £ (2x+

2

x2)

3 a ¡ 1p3

b ¡18 c ¡8 d ¡4 e ¡ 3

32f 0

4 ady

dx= 3x2,

dx

dy= 1

3y¡

23 Hint: Substitute y = x3

bdy

dx£ dx

dy=

dy

dy= 1

EXERCISE 17E

1 a 2x(2x¡ 1) + 2x2 b 4(2x+ 1)3 + 24x(2x+ 1)2

c 2x(3¡ x)12 ¡ 1

2x2(3¡ x)¡

12

d 1

2x¡ 1

2 (x¡ 3)2 + 2px(x¡ 3)

e 10x(3x2 ¡ 1)2 + 60x3(3x2 ¡ 1)

f 1

2x¡ 1

2 (x¡ x2)3 + 3px(x¡ x2)2(1¡ 2x)

2 a ¡48 b 4061

4c 13

3d 11

23 x = 3 or 3

5

EXERCISE 17F

1 a b2x(2x+ 1)¡ 2x2

(2x+ 1)2

c(x2 ¡ 3)¡ 2x2

(x2 ¡ 3)2d

1

2x¡ 1

2 (1¡ 2x) + 2px

(1¡ 2x)2

e2x(3x¡ x2)¡ (x2 ¡ 3)(3¡ 2x)

(3x¡ x2)2

f(1¡ 3x)

12 + 3

2x(1¡ 3x)¡

12

1¡ 3x

2 a 1 b 1 c ¡ 7

324d ¡ 28

27

3 b i never f dy

dxis undefined at x = ¡1g

ii x 6 0 and x = 1

4 b i x = ¡2§p11 ii x = ¡2

x21.510.5

1

y

xey ��

7

(2¡ x)2

x

from which the gradient at any point can be found.

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ANSWERS 741

c

EXERCISE 17G

1 a y = ¡7x+ 11 b 4y = x+ 8 c y = ¡2x¡ 2

d y = ¡2x+ 6 e f

2 a 6y = ¡x+ 57 b 7y = ¡x+ 26 c 3y = x+ 11

d x+ 6y = 43

3 a y = 21 and y = ¡6 b ( 12

, 2p2)

c k = ¡5 d y = ¡3x+ 1

4 a a = ¡4, b = 7 b a = 2, b = 4

5 a 3y = x+ 5 b 9y = x+ 4

c 16y = x¡ 3 d y = ¡4

6 a y = 2x¡ 7

4b y = ¡27x¡ 242

3

c 57y = ¡4x+ 1042 d 2y = x+ 1

7 a = 4, b = 3

8 a (¡4, ¡64) b (4, ¡31)

c does not meet the curve again

9 a y = (2a¡ 1)x¡ a2 + 9; y = 5x, contact at (3, 15)

y = ¡7x, contact at (¡3, 21)

b y = 0, y = 27x+ 54

c y = 0, y = ¡p14x+ 4

p14

10 a b 16x+ a3y = 24a

c A( 32a, 0), B

³0,

24

a2

´d

EXERCISE 17H

1 a 6 b 12x¡ 6 c3

2x52

d12¡ 6x

x4e 24¡ 48x f

20

(2x¡ 1)3

2 a ¡6x b 2¡ 30

x4c ¡ 9

4x¡ 5

2

d8

x3e 6(x2 ¡ 3x)(5x2 ¡ 15x+ 9)

f 2 +2

(1¡ x)3

3 a x = 1 b x = 0, §p6

4 x ¡1 0 1

f(x) ¡ 0 +

f 0(x) + ¡ +

f 00(x) ¡ 0 +

REVIEW SET 17A

1 a ¡17 b ¡17 c ¡6 2 y = 4x+ 2

3 a 6x¡ 4x3 b 1 +1

x24 2x+ 2 5 x = 1

6 a = 5

2, b = ¡ 3

2

7 a f 0(x) = 8x(x2 + 3)3

b g0(x) =1

2x(x+ 5)¡

12 ¡ 2(x+ 5)

12

x3

8

10 x = ¡ 1

2, 3

211 a = 64 12 P(0, 7:5), Q(3, 0)

REVIEW SET 17B

1 ady

dx= 5 + 3x¡2 b

dy

dx= 4(3x2 + x)3(6x+ 1)

cdy

dx= 2x(1¡ x2)3 ¡ 6x(1¡ x2)2(x2 + 1)

2 y = 7, y = ¡25 3 3267

152units2 4 (¡2, ¡25)

5 a = 1

26 (¡2, 19) and (1, ¡2)

7 a ¡2(5¡ 4x)¡12 b ¡4(5¡ 4x)¡

32 8 5y = x¡ 11

10 g(x) = ¡2x2 + 6x+ 3

11 a b y = ¡ 4

k2x+

8

k

c A(2k, 0) B

³0,

8

k

´d Area = 8 units2

e k = 2

REVIEW SET 17C

1 ady

dx= 3x2(1¡ x2)

12 ¡ x4(1¡ x2)¡

12

bdy

dx=

(2x¡ 3)(x+ 1)12 ¡ 1

2(x2 ¡ 3x)(x+ 1)¡

12

x+ 1

2 y = 16x¡ 127

2

3 a ¡ 2

xpx¡ 3 b 4

³x¡ 1

x

´3 ³1 +

1

x2

ć 1

2(x2 ¡ 3x)¡

12 (2x¡ 3)

4 f(3) = 2, f 0(3) = ¡1

5 a f 0(x) =3(x+ 3)2

px¡ 1

2x¡

12 (x+ 3)3

x

b f 0(x) = 4x3px2 + 3 + x5(x2 + 3)¡

12

6 a = ¡1, b = 2

7 a = 2 and the tangent is y = 3x ¡ 1 which meets the curve

again at (¡4, ¡13)

8 BC = 8p10

3Hint: The normal is y = ¡3x+ 8.

9 ad2y

dx2= 36x2 ¡ 4

x3b

d2y

dx2= 6x+ 3

4x¡ 5

2

10 a = 9, b = 2, f 00(¡2) = ¡18 11 4y = 3x+ 5

EXERCISE 18A

y

x

28)(x

xf �

4321

4

3

2

1

xy 4�

x

y

dy

dxis zero when the tangent to the function is horizontal

(gradient 0), at its turning points or points of horizontal

inflection.dy

dxis undefined at vertical asymptotes.

y = ¡5x¡ 9 y = ¡5x¡ 1

a 23

4b ¡ 1

8p2

9 a = ¡14, b = 21

1 a $118 000 bdP

dt= 4t¡ 12 $1000 per year

cdP

dtis the rate of change in profit with time

d i t 6 3 years ii t > 3 years

e minimum profit is $100 000 when t = 3

Area =18

jaj units2,

area ! 0 as jaj ! 1

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742 ANSWERS

fdP

dt

¯̄̄t=4

= 4 Profit is increasing at $4000 per

year after 4 years.

dP

dt

¯̄̄t=10

= 28 Profit is increasing at $28 000 per

year after 10 years.

dP

dt

¯̄̄t=25

= 88 Profit is increasing at $88 000 per

year after 25 years.

2 a 19 000 m3 per minute b 18 000 m3 per minute

3 a 1:2 m

b s0(t) = 28:1¡ 9:8t represents the instantaneous velocity of

the ball

c t = 2:87 s. The ball has stopped and reached its maximum

height.

d 41:5 m

e i 28:1 m s¡1 ii 8:5 m s¡1 iii 20:9 m s¡1

s0(t) > 0 when the ball is travelling upwards.

s0(t) 6 0 when the ball is travelling downwards.

f 5:78 s

gd2s

dt2is the rate of change of

ds

dt, or the instantaneous

acceleration.

4 b 69:6 m s¡1

EXERCISE 18B

1 a i Q = 100 ii Q = 50 iii Q = 0

b i decr. 1 unit per year ii decr. 1p2

units per year

2 a 0:5 m

b t = 4: 9:17 m, t = 8: 12:5 m, t = 12: 14:3 m

c t = 0: 3:9 m year¡1, t = 5: 0:975 m year¡1,

t = 10: 0:433 m year¡1

d asdH

dt=

97:5

(t+ 5)2> 0, for all t > 0, the tree is always

growing, anddH

dt! 0 as t increases

3 a i E4500 ii E4000

b i decr. of E210:22 per km h¡1

ii incr. of E11:31 per km h¡1

cdC

dv= 0 at

4 a The near part of the lake is 2 km from the sea, the furthest

part is 3 km.

bdy

dx= 3

10x2 ¡ x+ 3

5.

dy

dx

¯̄̄x= 1

2

= 0:175, height of hill is increasing as gradient

is positive.

dy

dx

¯̄̄x=1 1

2

= ¡0:225, height of hill is decreasing as gradient

is negative.

) top of the hill is between x = 1

2and x = 1 1

2:

c 2:55 km from the sea, 63:1 m deep

5 adV

dt= ¡1250

³1¡ t

80

´b at t = 0 when the tap

was first opened

cd2V

dt2=

125

8

6 a

b c

7 a

b

c

d

EXERCISE 18C.1

1

2 a ¡14 cm s¡1 b (¡8¡ 2h) cm s¡1

c ¡8 cm s¡1 = s0(2) ) velocity = ¡8 cm s¡1 at t = 2

d ¡4t = s0(t) = v(t)

3 a 2

3cm s¡2 b

³2p

1 + h+ 1

ćm s¡2

c 1 cm s¡2 = v0(1)

d1pt

cm s¡2 = v0(t), the instantaneous accn. at time t

4 a velocity at t = 4 b acceleration at t = 4

EXERCISE 18C.2

1 a v(t) = 2t¡ 4, a(t) = 2

b The object is initially 3 cm to the right of the origin and is

moving to the left at 4 cm s¡1. It is accelerating at 2 cm s¡2

to the right.

c The object is instantaneously stationary, 1 cm to the left of

the origin and is accelerating to the right at 2 cm s¡2.

d At t = 2, s(2) = 1 cm to the left of the origin.

e f 0 6 t 6 2

2 a v(t) = 98¡ 9:8t, a(t) = ¡9:8

b s(0) = 0 m above the ground, v(0) = 98 m s¡1 skyward

c

d 490 m e 20 seconds

3 a v(t) = 12¡ 6t2, a(t) = ¡12t

b s(0) = ¡1, v(0) = 12, a(0) = 0Particle started 1 cm to the left of the origin and was travelling

to the right at a constant speed of 12 cm s¡1.

t

dVdt 80

�1250

s(t):t

1 3�

0

v(t):t

2�

0

a(t):t

0

v(t):t

10

�

0

s(t):t

0

a(t):t

�

0

0�1 3s

v = 3p500 000 ¼ 79:4 km h¡1

This shows that the rate of change of V is constantly

increasing, so the outflow is decreasing at a constant rate.

WhendP

dt= 0, the population is not changing over time,

so it is stable.4000 fish 8000 fish

C0(x) = 0:0009x2 + 0:04x+ 4 dollars per pair

C0(220) = $56:36 per pair. This estimates the additional

cost of making one more pair of jeans if 220 pairs are

currently being made.

$56:58 This is the actual increase in cost to make an extra

pair of jeans (221 rather than 220).

C00(x) = 0:0018x+ 0:04.

C00(x) = 0 when x = ¡22:2: This is where the rate of

change is a minimum, however it is out of the bounds of the

model (you cannot make < 0 jeans!).

a 7 m s¡1 b (h+ 5) m s¡1 c 5 m s¡1 = s0(1)d average velocity = (2t+ h+ 3) m s¡1,

limh!0

(2t+ h+ 3) = s0(t) ! 2t+ 3 as h ! 0

t = 5 Stone is 367:5 m above the ground and moving

skyward at 49 m s¡1. Its speed is decreasing.

t = 12 Stone is 470:4 m above the ground and moving

groundward at 19:6 m s¡1. Its speed is increasing.

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ANSWERS 743

c t =p2, s(

p2) = 8

p2¡ 1 d i t >

p2 ii never

4 a v(t) = 3t2 ¡ 18t+ 24 a(t) = 6t¡ 18

b x(2) = 20, x(4) = 16

c i 0 6 t 6 2 and 3 6 t 6 4 ii 0 6 t 6 3 d 28 m

5 Hint: s0(t) = v(t) and s00(t) = a(t) = g

Show that a = 1

2g, b = v(0), c = 0.

EXERCISE 18D.1

1 a i x > 0 ii never

b i never ii ¡2 < x 6 3

c i x 6 2 ii x > 2

d i all real x ii never

e i 1 6 x 6 5 ii x 6 1, x > 5

f i 2 6 x < 4, x > 4 ii x < 0, 0 < x 6 2

2 a increasing for x > 0, decreasing for x 6 0

b decreasing for all x

c increasing for x > ¡ 3

4, decreasing for x 6 ¡ 3

4

d increasing for x > 0, never decreasing

e decreasing for x > 0, never increasing

f incr. for x 6 0 and x > 4, decr. for 0 6 x 6 4

g increasing for ¡p

2

36 x 6

p2

3,

decreasing for x 6 ¡p

2

3, x >

p2

3

h decr. for x 6 ¡ 1

2, x > 3, incr. for ¡ 1

26 x 6 3

i increasing for x > 0, decreasing for x 6 0

j increasing for x > ¡ 3

2+

p5

2and x 6 ¡ 3

2¡

p5

2

decreasing for ¡ 3

2¡

p5

26 x 6 ¡ 3

2+

p5

2

k increasing for x 6 2¡p3, x > 2 +

p3

decreasing for 2¡p3 6 x 6 2 +

p3

l increasing for x > 1, decreasing for 0 6 x 6 1

m increasing for ¡1 6 x 6 1, x > 2decreasing for x 6 ¡1, 1 6 x 6 2

n increasing for 1¡p2 6 x 6 1, x > 1 +

p2

decreasing for x 6 1¡p2, 1 6 x 6 1 +

p2

3 a

b increasing for ¡1 6 x 6 1decreasing for x 6 ¡1, x > 1

4 a

b increasing for ¡1 6 x < 1decreasing for x 6 ¡1, x > 1

5 a

b increasing for ¡1 6 x < 1, 1 < x 6 3decreasing for x 6 ¡1, x > 3

6 a

b increasing for x > 2, decreasing for x < 1, 1 < x 6 2

EXERCISE 18D.2

1 a A - local max B - horiz. inflection C - local min.

b

c i x 6 ¡2, x > 3 ii ¡2 6 x 6 3

d

e For b we have intervals where the function is increasing (+)

or decreasing (¡). For d we have intervals where the function

is above (+) and below (¡) the x-axis.

2 a b

c d

e f

g h

i j

3 x = ¡ b

2a, local min if a > 0, local max if a < 0

4 a = 9

5 a a = ¡12, b = ¡13

b (¡2, 3) local max. (2, ¡29) local min

6 P (x) = ¡9x3 ¡ 9x2 + 9x+ 2

7 a greatest value is 63 when x = 5,least value is ¡18 when x = 2

b greatest value is 4 when x = 3 and x = 0,least value is ¡16 when x = ¡2.

t2 4

�

0

t3

�

0

160 20x

x�1 1

� �

x�1 1

� �

x�1 1 3

� �

x�2 0 3

� �

x�4 0 5

��

( )��,

x

ƒ( )x

~`2�~`2

local min.

( )��,

x

ƒ( )x �

�1

horizontal

inflection

( )��,

( )��,

2

x

ƒ( )x

local min.

local max.

�2

~`2�~`2

( )��, ( )��,

( )��,

x

ƒ( )x

local min. local min.

localmax.

( )��,

1

x

ƒ( )x

horizontal

inflection

x

ƒ( )x �

local min.

1

( Qr_ Qr_ )� � � �,

( )��,

( )��,

x

ƒ( )x

�3

�3

local min.

horizontal

inflection

( )��,

x

ƒ( )x �

local max.

( )��,( )��,

( )��,

2 x

ƒ( )x �


localmax.

�2

2

x

ƒ( )x �

(no stationary points)

increasing for x >p3 and x 6 ¡p

3

decreasing for ¡p3 6 x < ¡1, ¡1 < x < 1,

1 < x 6p3

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744 ANSWERS

8 Maximum hourly cost is $680:95 when 150 hinges are made per

hour. Minimum hourly cost is $529:80 when 104 hinges are

made per hour.

EXERCISE 18E.1

1 a f(x) = 3¡ 5

x+ 1H.A. y = 3, V.A. x = ¡1

b f(x) = 1

2¡ 7

2(2x¡ 1)H.A. y = 1

2, V.A. x = 1

2

c f(x) = ¡2 +2

x¡ 1H.A. y = ¡2, V.A. x = 1.

2 a i H.A. y = ¡3,

V.A. x = 4iv

ii f 0(x) = (4¡ x)¡2

iii x-int. is 11

3,

y-int. is ¡ 11

4.

b i H.A. y = 1,

V.A. x = ¡2iv

ii f 0(x) =2

(x+ 2)2

iii x-int. is 0,

y-int. is 0.

c i H.A. y = 4,

V.A. x = 2iv

ii f 0(x) =¡11

(x¡ 2)2

iii x-int. is ¡ 3

4,

y-int. is ¡ 3

2.

d i H.A. y = ¡1,

V.A. x = ¡2iv

ii f 0(x) =¡3

(x+ 2)2

iii x-int. is 1,

y-int. is 1

2.

EXERCISE 18E.2

1 a H.A. y = 0, V.A. x = 2 and x = ¡2

b H.A. y = 0, V.A. x = ¡2

c H.A. y = 0, no V.A

2 a i H.A. y = 0

ii f 0(x) =¡4(x+ 1)(x¡ 1)

(x2 + 1)2

(1, 2) is a local max. (¡1, ¡2) is a local min.

iii x-intercept is 0, y-intercept is 0

iv

b i H.A. y = 0, V.A.s x = 5 and x = ¡1

ii f 0(x) =¡4(x2 + 5)

(x¡ 5)2(x+ 1)2, no stationary points


iv

c i H.A. y = 0, V.A. x = 1

ii f 0(x) =¡4(x+ 1)

(x¡ 1)3, (¡1, ¡1) is a local minimum


iv

d i H.A. y = 0, V.A. x = ¡2

ii f 0(x) =¡3(x¡ 4)

(x+ 2)3, (4, 1

4) is a local maximum

iii x-intercept is 1, y-intercept is ¡ 3

4

iv

EXERCISE 18E.3

1 a V.A. x = 1 and x = ¡1, H.A. y = 2b no V.A.s, H.A. y = ¡1 c V.A. x = ¡2, H.A. y = 3

2 a i H.A. y = 1, V.A.s x = 3 and x = ¡2

ii¡1

2, 1

25

¢is a local maximum

iii x-intercepts are 0 and 1, y-intercept is 0

iv

x�� x��

y��

y

x

maxlocal

1

62

2

��

��

xx

xxy

&\Qw_\' wA_t_\*

y

xQd_Q_-\Qf_Q_

3��y

4�x

41

3)(�

��x

xf

y

x

1�y

2��x

2)(

��

xx

xf

y

x

4�y

2�x

234)(

��

�xx

xf

-\Er_

-\Ew_

y

x

1��y

2��x2

1)(��

�x

xxf

1

Qw_

y

x

local min ( )��

local max ( )��

y��

1

42 �

�x

xy

x4

x��

x�

� �

x��

� �

y

x

y��

x�� x��

54

42 ��

�xx

xy

2)1(

4)(

��

x

xxf

x��

y

x

local min ( )��

y

x

x��

2)2(

33)(

�

��

x

xxf

local max ( \ Qr_)��

-\Er_ 1

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ANSWERS 745

b i H.A. y = 1, no V.A.

ii (0, ¡1) is a local minimum

iii x-intercepts are 1 and ¡1, y-intercept is ¡1

iv

c i H.A. y = 1, V.A.s x = ¡4 and x = ¡1

ii (2, ¡ 1

9) is a local min., (¡2, ¡9) is a local max.


iv

d i H.A. y = 1, V.A. x = ¡1

ii (2, ¡ 1

3) is a local minimum


iv

EXERCISE 18F.1

1 a no inflection b horizontal inflection at (0, 2)

c non-horizontal inflection at (2, 3)

d horizontal inflection at (¡2, ¡3)

e horizontal inflection at (0, 2)

non-horizontal inflection at (¡ 4

3, 310

27) f no inflection

2 a i local minimum at (0, 0) v

ii no points of inflection

iii decreasing for x 6 0,

increasing for x > 0

iv function is concave up

for all x

b i horizontal inflection at (0, 0) v

ii horizontal inflection at (0, 0)

iii increasing for all

real xiv concave down for x 6 0,

concave up for x > 0

c i f 0(x) 6= 0, no stationary points v

ii no points of inflection

iii increasing for x > 0,

never decreasing

iv concave down for x > 0,

never concave up

d i local max. at (¡2, 29) v

local min at (4, ¡79)

ii non-horizontalinflection at (1, ¡25)

iii increasing forx 6 ¡2, x > 4decreasing for¡2 6 x 6 4

iv concave down forx 6 1,concave up for x > 1

e i horiz. inflect. at (0, ¡2) v

local min. at (¡1, ¡3)

ii horiz. inflect. at (0, ¡2)non-horizontal inflectionat (¡ 2

3, ¡ 70

27)

iii increasing for x > ¡1,decreasing for x 6 ¡1

iv concave down for¡ 2

36 x 6 0

concave up for

x 6 ¡ 2

3, x > 0

f i local min. at v(1, 0)

ii no points ofinflection

iii increasing forx > 1,decreasing forx 6 1

iv concave up forall x

g i local minimum at (¡p2, ¡1) and (

p2, ¡1),

local maximum at (0, 3)

ii non-horizontal inflection at (p

2

3, 7

9)

non-horizontal inflection at (¡p

2

3, 7

9)

iii increasing for ¡p2 6 x 6 0, x >

p2

decreasing for x 6 ¡p2, 0 6 x 6

p2

iv concave down for ¡p

2

36 x 6

p2

3

concave up for x 6 ¡p

2

3, x >

p2

3

v

h i no stationary points v

ii no inflections

iii increasing for x > 0,

never decreasing

iv concave down for x > 0,

never concave up

( )��, x

ƒ( )x

local min.

( )��, x

ƒ( )x �

horizontal

inflection

x

ƒ( )x �

( )�� ,

( )��,

( )��,

x

ƒ( )x �

local min

local max.

non-horizontal

inflection

�80

��

non-horizontal

inflection

( )��,

( )��,

x

ƒ( )x �

local min.horizontal

inflection

1�1�2

&-\We_\'- Uw_Pu_\*

x

y

1

1local min (1, 0)

y x�� ( )

3

x

ƒ( )x �

Ql_Y_

y��

��

local min ( )��

y

x

1

12

2

�

��

x

xy

localmin

( \Qo_\)��

1

1 x

y��

x�� x��

y

local max ��

4

45

452

2

��

��

xx

xxy

y ��

1 5

5

x�� local min ( \Qe_)��

2

2

)1(

56

�

��

x

xxy

x

y

( )��,

( ~` )� �2, 1 (~` )2, 1�

x

ƒ( )x �


local max.

non-

horizontal

inflection

non-

horizontal

inflection³¡p

23

, 79

´ ³p23

, 79

´

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746 ANSWERS

EXERCISE 18F.2

1 a

b

c

2 a

b

EXERCISE 18G

1 50 fittings 2 250 items 3 10 blankets 4 25 km h¡1

5 b c Lmin ¼ 28:28 m,

x ¼ 7:07 md

6 a 2x cm b V = 200 = 2x£ x£ h

c Hint: Show h =100

x2and substitute into the surface area

equation.

d e SAmin ¼ 213 cm2,

x ¼ 4:22 cmf

7 a recall that Vcylinder = ¼r2h and that 1 L = 1000 cm3

b recall that SAcylinder = 2¼r2 + 2¼rh

c d A ¼ 554 cm2,

r ¼ 5:42 cme

8 b 6 cm £ 6 cm

9 a 0 6 x 6 63:7 c x ¼ 63:7, l = 0

10 a Hint: Show that AC = µ360

£ 2¼ £ 10

b Hint: Show that 2¼r = AC

c Hint: Use the result from b and Pythagoras’ theorem.

d V = 1

3¼¡

µ36

¢2q100¡

¡µ36

¢2e f µ ¼ 294o

11 a For x < 0 or x > 6, X is not

c x ¼ 2:67 km This is the distance from A to X which

minimises the time taken to get from B to C.

12 3:33 km 13 r ¼ 31:7 cm, h ¼ 31:7 cm

14 4 m from the 40 cp globe

15 a D(x) =p

x2 + (24¡ x)2

bd[D(x)]2

dx= 4x¡ 48

c Smallest D(x) ¼ 17:0 Largest D(x) = 24, which

is not an acceptable solution

as can be seen in the diagram.

16 a Hint: Use the cosine rule.

b 3553 km2 c 5:36 pm

17 a QR =

³2 + x

x

´m

c Hint: All solutions < 0 can be discarded as x > 0.

d 416 cm

y

x

max

non-stationary

inflection

stationary

inflection

y x�� '( )

��

y x��( )

y

x��

min

non-stationary

inflection

max

y x�� '( )

y x��( )

y

x

min

��

� '( )x

� ''( )x

�( )x

y

x

�( )x

� '( )x

� ''( )x

min

max

��

y

x

�( )x

� '( )x� ''( )x

min

maxnon-stationary

inflection

�

y ( )m

x (m)

14.14 m

7.07 m

y (cm )X

x (cm)

450

10 8.43 cm 4.22 cm

5.62 cm

A (cm )X

r (cm)

1500

1510.84 cm

5.42 cm

V (cm )C

� ( )°

500

360

12�

0 24

x

12 m

17 m

24 m

t4.605

�

0

on [AC].

(circular)

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ANSWERS 747

REVIEW SET 18A

1 a v(t) = (6t2 ¡ 18t+ 12) cm s¡1 a(t) = (12t¡18) cm s¡2

v(t): a(t):

b s(0) = 5 cm to left of origin

v(0) = 12 cm s¡1 towards origin

a(0) = ¡18 cm s¡2 (reducing speed)

c At t = 2, particle is 1 cm to the left of the origin,

is stationary and is accelerating towards the origin.

d t = 1, s = 0 and t = 2, s = ¡1

e

f Speed is increasing for 1 6 t 6 1 1

2and t > 2.

2 b k = 9

3 6 cm from each end

4 a x = ¡3 b x-int. 2

3, y-int. ¡ 2

3

c f 0(x) =11

(x+ 3)2

d no stationary points

5

6 a

b local min. (0, 1), local max. (4, 1

9)

c x-intercept 2, y-intercept 1

d

REVIEW SET 18B

1 bd[A(x)]2

dx= 5000x¡ 4x3

Area is a maximum when x ¼ 35:4, A = 1250 m2.

2 a a = ¡6

b local max. (¡p2, 4

p2), local min. (

p2, ¡4

p2)

c

3 a v(t) = 3¡ 1

2pt

a(t) =1

4tpt

b x(0) = 0, v(0) is undefined, a(0) is undefined

c Particle is 24 cm to the right of the origin and is travelling

to the right at 2:83 cm s¡1. Its speed is increasing.

d Changes direction at t = 1

36, 0:083 cm to the left of the

origin.

e Particle’s speed is decreasing for 0 t 6 1

36.

4 a i $535 ii $1385:79

b i ¡$0:27 per km h¡1 ii $2:33 per km h¡1

c 51:3 km h¡1

5 a y-int. at y = ¡1. x-int. at x = 1, x = ¡1:

b x2 + 1 > 0 for all real x (i.e., denominator is never 0)

c local minimum at (0, ¡1)

e

6

7 b A = 200x¡ 2x2 ¡ 1

2¼x2 c

REVIEW SET 18C

1 a local maximum at (¡2, 51), local minimum at (3, ¡74)

non-horizontal inflection at ( 12

, ¡11:5)

t1 2

�

0

t1\Qw_

�

0

0�1�5s

stationary

inflection

non-stationary

inflectionmax

y x��( )

y

x

y x��'( )

x

��

x

y

y��

x�� x��

min ,� ��( )max ,� � �( \ Qo_)

2

22 ��

��xx

xy

�

t

0

t�

0

s

y

x

( ~` \' ~` \)� � ��

(~` \' ~` \)� ��

� �� ( ) Cx x x

v(t): a(t):t�

0Ae_y_

t

0

non-horizontalinflection at

non-horizontalinflection at

local min

,( )��

�21

31 ,& *��

21

31 ,& *

x

y

y

x

y x��'( )

y x��( )

28.0 m

56.0 m

e p < 0, 0 < p < 1

9, p > 1

7 a v(t) = 2 +4

t2

a(t) = ¡ 8

t3

b

c The particle never changes direction

d

e i never ii never

8 x =k

2

³1¡ 1p

3

´

V.A.s x = ¡2, x = 1, H.A. y = 0

The particle is 2 cm to the left of O, moving right at 6 cm s¡1,

and slowing down.

<

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748 ANSWERS

b increasing forx 6 ¡2, x > 3decreasing for¡2 6 x 6 3

d

c concave downfor x 6 1

2,

concave up

for x > 1

2

2 a y-intercept at y = 0, x-intercept at x = 0 and x = 2

b local maximum at¡2

3, 32

27

¢, local minimum at (2, 0),

non-horizontal inflection at¡4

3, 16

27

¢c

3

4 a v(t) = 15 +120

(t¡ 1)3cm s¡1, a(t) =

¡360

(t¡ 1)4cm s¡2

b At t = 3, particle is 30 cm to the right of the origin, moving

to the right at 30 cm s¡1 and decelerating at 22:5 cm s¡2.

c 0 6 t < 1

5 a 2 m b H(3) = 4 m, H(6) = 4 2

3m, H(9) = 5 m

c H0(0) = 4

3m year¡1, H0(3) = 1

3m year¡1

H0(6) = 4

27m year¡1, H0(9) = 1

12m year¡1

d H0(t) =12

(t+ 3)2> 0 for all t > 0

) the height of the tree is always increasing.e

6 a Hint: Use Pythagoras to find h as a function of x and then

substitute into the equation for the volume of a cylinder.

b radius ¼ 4:08 cm, height ¼ 5:77 cm

7

EXERCISE 19A

1 a 4e4x b ex c ¡2e¡2x d 1

2e

x2

e ¡e¡x2 f 2e¡x g 2e

x2 + 3e¡x

hex ¡ e¡x

2i ¡2xe¡x2

j e1x £ ¡1

x2

k 20e2x l 40e¡2x m 2e2x+1

n 1

4e

x4 o ¡4xe1¡2x2

p ¡0:02e¡0:02x

2 a ex + xex b 3x2e¡x ¡ x3e¡x cxex ¡ ex

x2

d1¡ x

exe 2xe3x + 3x2e3x f

xex ¡ 1

2ex

xpx

g 1

2x¡ 1

2 e¡x ¡ x12 e¡x h

ex + 2 + 2e¡x

(e¡x + 1)2

3 a 4ex(ex + 2)3 b¡e¡x

(1¡ e¡x)2

ce2xp

e2x + 10d

6e3x

(1¡ e3x)3

e ¡ e¡x

2

¡1¡ e¡x

¢¡ 32 f

1¡ 2e¡x + xe¡x

p1¡ 2e¡x

5 Hint: Finddy

dxand

d2y

dx2and substitute into the equation.

6 k = ¡9

7 a local maximum at (1, e¡1)

b local max. at (¡2, 4e¡2), local min. at (0, 0)

c local minimum at (1, e) d local maximum at (¡1, e)

EXERCISE 19B

1 a 2 b 1

2c ¡1 d ¡ 1

2e 3 f 9 g 1

5h 1

4

2 a eln 2 b eln 10 c eln a d ex lna

3 a x = ln 2 b no real solutions c no real solutions

d x = ln 2 e x = 0 f x = ln 2 or ln 3 g x = 0

h x = ln 4 i x = ln

³3+

p5

2

ór ln

³3¡p

5

2

´4 a (ln 3, 3) b (ln 2, 5) c (0, 2) and (ln 5, ¡2)

5 a A¡ln 3

2, 0¢

, B(0, ¡2) b f 0(x) = 2e2x > 0 for all x

c f 00(x) = 4e2x > 0 for all x

d

e as x ! ¡1,

6 a f(x): x-int. at x = ln 3, y-int. at y = ¡2

g(x): x-int. at x = ln¡5

3

¢, y-int. at y = ¡2

b f(x): as x ! 1, f(x) ! 1as x ! ¡1, f(x) ! ¡3 (above)

g(x): as x ! 1, g(x) ! 3 (below)as x ! ¡1, g(x) ! ¡1

c intersect at (0, ¡2) and (ln 5, 2)

d

local min

,( )��

local max

,&\We_\ Ew_Wu_\*axis intercept

at ,( )��

non-horizontalinflection

&\Re_\' Qw_Yu_\*

y

x

y

x

min

NSPI

max

y x��( )

y x��'( )

x

y

local min ,( )��

local max ( 2, 51)�non-horizontal

inflection

,(\Qw_\ \Qw_\)��

7

x108642

2

H y��

*3

21&6

��

tH

y�

x

32 �� xey

A

B

y��

y

xln 3

ln Te_

(0, 2)�

( )ln ,� �

y 3��

y 3�

y g x( )�

y f x( )�

a y =1

x2, x > 0 c base is 1:26 m square, height 0:630 m

e2x ! 0

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ANSWERS 749

d

EXERCISE 19C

1 a1

xb

2

2x+ 1c

1¡ 2x

x¡ x2

d ¡ 2

xe 2x lnx+ x f

1¡ lnx

2x2

g ex lnx+ex

xh

2 lnx

xi

1

2xplnx

je¡x

x¡ e¡x lnx k

ln(2x)

2px

+1px

llnx¡ 2px(lnx)2

m4

1¡ xn ln(x2 + 1) +

2x2

x2 + 1

2 a ln 5 b3

xc

4x3 + 1

x4 + xd

1

x¡ 2

e6

2x+ 1[ln(2x+ 1)]2 f

1¡ ln(4x)

x2

g ¡ 1

xh

1

x lnxi

¡1

x(lnx)2

3 a¡1

1¡ 2xb

¡2

2x+ 3c 1 +

1

2x

d1

x¡ 1

2(2¡ x)e

1

x+ 3¡ 1

x¡ 1f

2

x+

1

3¡ x

g9

3x¡ 4h

1

x+

2x

x2 + 1i

2x+ 2

x2 + 2x¡ 1

x¡ 5

4 ady

dx= 2x ln 2

5 a x =e3 + 1

2¼ 10:5 b no, ) there is no y-int.

c gradient = 2 d x > 1

2

e f 00(x) =¡4

(2x¡ 1)2< 0 for all x > 1

2, so f(x) is

concave down

f

6 a x > 0

7 Hint: Show that as x ! 0, f(x) ! ¡1,

and as x ! 1, f(x) ! 0.

8 Hint: Show that f(x) > 1 for all x > 0.

EXERCISE 19D

1 y = ¡1

ex+

2

e2 3y = ¡x+ 3 ln 3¡ 1

3 A is¡2

3, 0¢

, B is (0, ¡2e) 4 y = ¡ 2

e2x+

2

e4¡ 1

5 y = eax+ea(1¡a) so y = ex is the tangent to y = ex from

the origin

6 a x > 0

b f 0(x) > 0 for all x > 0, so f(x) is always increasing.

Its gradient is always positive. f 00(x) < 0 for all x > 0,

so f(x) is concave down for all x > 0.

c

normal has equation

f(x) = ¡ex+ 1 + e2

7 ¼ 63:43o

8 a k = 1

50ln 2 ¼ 0:0139

b i 20 grams ii 14:3 grams iii 1:95 grams

c 9 days and 6 minutes (216 hours)

d i ¡0:0693 g h¡1 ii ¡2:64£ 10¡7 g h¡1

e Hint: You should finddW

dt= ¡ 1

50ln 2£ 20e¡

150

ln 2t

9 a b 100oC

d i decreasing by 11:7oC min¡1

ii decreasing by 3:42oC min¡1

iii

10 a 43:9 cm b 10:4 years

c i growing by 5:45 cm per year

ii growing by 1:88 cm per year

11 a A = 0 b k =ln 2

3(¼ 0:231)

c 0:728

12 a f(x) does not have any x or y-intercepts

b as x ! 1, f(x) ! 1as x ! ¡1, f(x) ! 0 (below)

c local minimum at (1, e)

d e ey = ¡2x¡ 3

13 a v(t) = 100¡ 40e¡t5 cm s¡1, a(t) = 8e¡

t5 cm s¡2

b s(0) = 200 cm on positive side of origin

v(0) = 60 cm s¡1, a(0) = 8 cm s¡2

c as t ! 1, v(t) ! 100 cm s¡1 (below)

d e after 3:47 s

�2 &\Qw_\\ *ln 3, 0�


x

y

x

f x( )

1

( , 1)e

t (s)

v t( ) (cm s )��

60

y��

7 a P( 12ln 3, 0),

Q(0, ¡2)

bdy

dx= ex + 3e¡x

> 0 for all x

c yx

x

is concave down below

the -axis and concave up

above the -axis

x

x � Qw_��( )x

213e

� �� ( ) ( )x xln

k = 1

15ln¡19

3

¢¼ 0:123

decreasing by 0:998oC min¡1

litres of alcohol produced per hour

x

f x( )local min

(1, )e

vertical asymptote0x �

xe

xfx

�)(

horizontal asymptote0y �

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750 ANSWERS

14 a at 4:41 months old b

15 a There is a local maximum at

³0, 1p

2¼

´.

f(x) is incr. for all x 6 0 and decr. for all x > 0.

b Inflections at

³¡1, 1p

2e¼

ánd

³1, 1p

2e¼

ć as x ! 1, f(x) ! 0 (positive)

as x ! ¡1, f(x) ! 0 (positive)

d

19 a Hint: They must have the same y-coordinate at x = b and

the same gradient.

c a = 1

2ed y = e¡

12 x¡ 1

2

20 after 13:8 weeks

21 a t-intercept 0, y-intercept 0

b local maximum at

³1

b,A

be

ć non-stationary inflection at

³2

b,2A

be2

´d

e After 40 minutes

REVIEW SET 19A

1 a 3x2ex3+2 b

1

x+ 3¡ 2

x

2 y =e

2x+

1

e¡ e

23 a P(ln 2, 4)

4 a y-intercept at y = ¡1, no x-intercept

b f(x) is defined for all x 6= 1

c f 0(x) 6 0 for x < 1 and 1 < x 6 2and f 0(x) > 0 for x > 2.

f 00(x) > 0 for x > 1, f 00(x) < 0 for x < 1.The function is decreasing for all defined values of x 6 2,

and increasing for all x > 2. The curve is concave down

for x < 1 and concave up for x > 1.

d e tangent is y = e2

5 Tangent is y = ln 3, so it never cuts the x-axis.

6 p = 1, q = ¡8 7 a x = ln¡2

3

¢or 0 b x = e2

REVIEW SET 19B

1 a 60 cm b

c i 16 cm per year ii 1:95 cm per year

2 a v(t) = ¡8e¡t10 ¡ 40 m s¡1

a(t) = 4

5e¡

t10 m s¡2 ft > 0g

b s(0) = 80 m,

v(0) = ¡48 m s¡1,

a(0) = 0:8 m s¡2

d

c as t ! 1,

v(t) ! ¡40 m s¡1 (below)

e t ¼ 6:93 seconds

3 100 or 101 shirts, $938:63 profit

4 a $20 000 b $146:53 per year

5 197 or 198 clocks per day

6 a v(t) = 25¡ 10

tcm min¡1, a(t) =

10

t2cm min¡2

b

c As t ! 1, v(t) ! 25 cm min¡1 from below

d

REVIEW SET 19C

1 a3x2 ¡ 3

x3 ¡ 3xb

ex(x¡ 2)

x3

2 (0, ln 4¡ 1) 3 a x = ln 3 b x = ln 4 or ln 3

4 a local minimum at (0, 1) b As x ! 1, f(x) ! 1c f 00(x) = ex, thus f(x) is concave up for all x.

d

5 a f 0(x) =ex

ex + 3b f 0(x) =

2(x¡ 1)

x(x+ 2)

6 a x > 0 b Sign diag of f 0(x) Sign diag of f 00(x)

f(x) is increasing for all x > 0 and is concave

downwards for all x > 0.

t (years)

A t( )

minimum(e , 0.632)�1

(5, 5ln 5 1)

t

v t( ) m s��

�48��

y e� 2

x 1�

1��

xex

y

x

y

x

��( )x



local max

y

t

point of inflection

2,

22be

A

b

µ ¶

b2

b1

local max*,1&

beA

b

21�1�2

4

2

x

yy e x� �x

11

1284

25

15

5

v t( )

t

x

0

x�

0

³1, 1p

¼́2e

³1, 1p

¼́2e¡

³0, 1p

¼

´2

1p2¼

¡ 12

x2

e�� ( )x

i 4:24 years ii 201 years

e t = 2 min

16 20 kettles 17 C 1p2

, e(¡12) 18 266 or 267 torches

s(e) = 25e¡ 10 cm, v(e) = 25¡e

cm min¡1,

a(e) =102

cm min¡2

x

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ANSWERS 751

10 a b

c d

e

EXERCISE 20B

1 ¼ 109:5o 2 c µ = 30o

3 1 hour 34 min 53 s when µ ¼ 36:9o 4 c 4p2 m

5 9:87 m

REVIEW SET 20

3 a f 0(x) = 3 cosx+ 8 sin(2x),f 00(x) = ¡3 sinx+ 16 cos(2x)

b f 0(x) = 1

2x¡ 1

2 cos(4x)¡ 4x12 sin(4x),

f 00(x) = ¡ 1

4x¡ 3

2 cos(4x)¡ 4x¡ 12 sin(4x)

¡16x12 cos(4x)

4 a x(0) = 3 cm, x0(0) = 2 cm s¡1, x00(0) = 0 cm s¡2

b t = ¼4

s and 3¼4

s c 4 cm

5 a for 0 6 x 6 ¼2

and 3¼2

6 x 6 2¼

b increasing for 3¼2

6 x 6 2¼, decreasing for 0 6 x 6 ¼2

c

6 a v(0) = 0 cm s¡1, v( 12) = ¡¼ cm s¡1, v(1) = 0 cm s¡1,

v¡3

2

¢= ¼ cm s¡1 v(2) = 0 cm s¡1

b 0 6 t 6 1, 2 6 t 6 3, 4 6 t 6 5, etc.So, for 2n 6 t 6 2n+ 1, n 2 f0, 1, 2, 3, ....g

7 b 1p2

m above the floor

8 a 2x+ 3y = 2¼3

+ 2p3 b

9 a f(x) = ¡5 sin 4x b x = ¼8

, 3¼8

, 5¼8

, 7¼8

EXERCISE 21A

1 a ix2

2ii

x3

3iii

x6

6iv ¡ 1

xv ¡ 1

3x3

vi 3

4x

43 vii 2

px

b the antiderivative of xn isxn+1

n+ 1:

2 a i 1

2e2x ii 1

5e5x iii 2e

12x iv 100e0:01x

v 1

¼e¼x vi 3e

x3

b the antiderivative of ekx is1

kekx

( )��,

x

y

( )� ��,¼

¼ 2¼

2

�2

x

y

local min.

stationary inflection5¼ �3 3~`6 2( ),

local max.

¼ 3 3~`6 2( ),

3¼2( ), 0

y x x�� sin cos( )

�!�

��!!�

� ��x

y

xxf cos)( �1

¼ 2¼

1

�1

x

y

y x= sin

max.local)��,(2�

min.local)1,(2

3 ��

�

��

x

y

min.min.

y x= cos 2

¼¼ 2¼2¼

( )��,max.

( )��,max.

(2 )��,max.

)1,(2

�� )1,(2

�3�

1

x

y

max.max.y x= sin2

( )��,min.

( )��,min.

(2 )��,min.

)1,(2� )1,(

23�

� ��x

max

min

xey sin�

),(2

e�

),( 12 e

3�

�

c d normal is

x+ 2y = 3

7 A

³1

2,1

e

ÉXERCISE 20A

1 a 2 cos(2x) b cosx¡ sinx

c ¡3 sin(3x)¡ cosx d cos(x+ 1) e 2 sin(3¡ 2x)

f5

cos2(5x)g 1

2cos(x

2) + 3 sinx

h3¼

cos2(¼x)i 4 cosx+ 2 sin(2x)

2 a 2x¡ sinx b1

cos2 x¡ 3 cosx

c ex cosx¡ ex sinx d ¡e¡x sinx+ e¡x cosx

ecosx

sinxf 2e2x tanx+

e2x

cos2 x

g 3 cos(3x) h ¡ 1

2sin¡x2

¢i

6

cos2(2x)

j cosx¡ x sinx kx cosx¡ sinx

x2l tanx+

x

cos2 x

3 a 2x cos(x2) b ¡ 1

2pxsin(

px) c ¡ sinx

2pcosx

d 2 sinx cosx e ¡3 sinx cos2 x

f ¡ sinx sin(2x) + 2 cosx cos(2x) g sinx sin(cosx)

h ¡12 sin(4x) cos2(4x) i ¡ cosx

sin2 x

j2 sin(2x)

cos2(2x)k ¡8 cos(2x)

sin3(2x)l

¡12

cos2(x2) tan4(x

2)

4 b f 00(x) = 3 sinx cos 2x+ 6cosx sin 2x

6 a y = x b y = x c 2x¡ y = ¼3¡

p3

2d x = ¼

4

7 a rising b rising at 2:73 m per hour

8 a ¡34 000¼ units per second b V 0(t) = 0

9 b i 0 ii 1 iii ¼ 1:11

11 a x(0) = ¡1 cm v(0) = 0 cm s¡1 a(0) = 2 cm s¡2

b At t = ¼4

seconds, the particle is (p2¡ 1) cm left of the

origin, moving right atp2 cm s¡1, with increasing speed.

c changes direction when t = ¼, x(¼) = 3 cm

d increasing for 0 6 t 6 ¼2

and ¼ 6 t 6 3¼2

1 a 5 cos(5x) ln(x) +sin(5x)

xb cosx cos(2x)¡ 2 sinx sin(2x)

c ¡2e¡2x tanx+e¡2x

cos2 xd 10¡ 10 cos(10x)

e tanx f 5 cos(5x) ln(2x) +sin(5x)

x

p2y ¡ 4x = 1¡ 2¼

IB SL 2nd ed


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752 ANSWERS

3 ad

dx(x3 + x2) = 3x2 + 2x

) antiderivative of 6x2 + 4x = 2x3 + 2x2

bd

dx(e3x+1) = 3e3x+1

) antiderivative of e3x+1 = 1

3e3x+1

cd

dx(x

px) = 3

2

px

) antiderivative ofpx = 2

3xpx

dd

dx(2x+ 1)4 = 8(2x+ 1)3

) antiderivative of (2x+ 1)3 = 1

8(2x+ 1)4

EXERCISE 21B

2 a 1

4units2 b 3 3

4units2 c 24 2

3units2 d 4

p2

3units2

3 a 4:06 units2 b 2:41 units2 c 2:58 units2

4 c iR 1

0(¡x2)dx = ¡ 1

3, the area between y = ¡x2

and the x-axis from x = 0 to x = 1 is 1

3units2:

iiR 1

0(x2 ¡ x)dx = ¡ 1

6, the area between y = x2 ¡ x

and the x-axis from x = 0 to x = 1 is 1

6units2.

iiiR 0

¡23xdx = ¡6, the area between y = 3x

and the x-axis from x = ¡2 to x = 0 is 6 units2

d ¡¼

EXERCISE 21C.1

1dy

dx= 7x6;

Rx6 dx = 1

7x7 + c

2dy

dx= 3x2 + 2x;

R(3x2 + 2x) dx = x3 + x2 + c

3dy

dx= 2e2x+1;

Re2x+1 dx = 1

2e2x+1 + c

4dy

dx= 8(2x+ 1)3;

R(2x+ 1)3 dx = 1

8(2x+ 1)4 + c

5dy

dx= 3

2

px;R p

x dx = 2

3xpx+ c

6dy

dx= ¡ 1

2xpx

;

Z1

xpxdx = ¡ 2p

x+ c

7dy

dx= ¡2 sin 2x;

Rsin 2xdx = ¡ 1

2cos 2x+ c

8dy

dx= ¡5 cos(1¡5x);

Rcos(1¡5x)dx = ¡ 1

5sin(1¡5x)+c

10dy

dx=

¡2p1¡ 4x

;

Z1p

1¡ 4xdx = ¡1

2

p1¡ 4x+ c

11 2 ln(5¡ 3x+ x2) + c fsince 5¡ 3x+ x2 is > 0g

EXERCISE 21C.2

1 ax5

5¡ x3

3¡ x2

2+ 2x+ c b 2

3x

32 + ex + c

c 3ex ¡ lnx+ c, x > 0 d 2

5x

52 ¡ 2 lnx+ c, x > 0

e ¡2x¡12 + 4 lnx+ c, x > 0

f 1

8x4 ¡ 1

5x5 + 3

4x

43 + c g 1

3x3 + 3 lnx+ c, x > 0

h 1

2lnx+ 1

3x3 ¡ ex + c, x > 0

i 5ex + 1

12x4 ¡ 4 lnx+ c, x > 0

2 a ¡3 cosx¡ 2x+ c b 2x2 ¡ 2 sinx+ c

c ¡ cosx¡ 2 sinx+ ex + c d 2

7x3

px+ 10 cosx+ c

e 1

9x3 ¡ 1

6x2 + sinx+ c f cosx+ 4

3xpx+ c

3 a 1

3x3 + 3

2x2 ¡ 2x+ c b 2

3x

32 ¡ 2x

12 + c

c 2ex +1

x+ c d ¡2x¡

12 ¡ 8x

12 + c

e 4

3x3 + 2x2 + x+ c f 1

2x2 + x¡ 3 lnx+ c, x > 0

g 4

3x

32 ¡ 2x

12 + c h 2x

12 + 8x¡

12 ¡ 20

3x¡

32 + c

i 1

4x4 + x3 + 3

2x2 + x+ c

4 a 2

3x

32 + 1

2sinx+ c b 2et + 4 cos t+ c

c 3 sin t¡ ln t+ c, t > 0

5 a y = 6x+ c b y = 4

3x3 + c

c y = 10

3xpx¡ 1

3x3 + c d y = ¡ 1

x+ c

e y = 2ex ¡ 5x+ c f y = x4 + x3 + c

6 a y = x¡ 2x2 + 4

3x3 + c b y = 2

3x

32 ¡ 4

px+ c

c y = x+ 2 lnx+5

x+ c, x > 0

7 a f(x) = 1

4x4 ¡ 10

3xpx+ 3x+ c

b f(x) = 4

3x

32 ¡ 12

5x

52 + c

c f(x) = 3ex ¡ 4 lnx+ c, x > 0

8 a f(x) = x2 ¡ x+ 3 b f(x) = x3 + x2 ¡ 7

c f(x) = ex + 2px¡ 1¡ e d f(x) = 1

2x2 ¡ 4

px+ 11

2

9 a f(x) =x3

3¡ 4 sinx+ 3

b f(x) = 2 sinx+ 3cosx¡ 2p2

10 a f(x) = 1

3x3 + 1

2x2 + x+ 1

3

b f(x) = 4x52 + 4x

32 ¡ 4x+ 5

c f(x) = ¡ cosx¡ x+ 4 d f(x) = 1

3x3 ¡ 16

3x+ 5

EXERCISE 21D

1 a 1

8(2x+ 5)4 + c b

1

2(3¡ 2x)+ c c

¡2

3(2x¡ 1)3+ c

d 1

32(4x¡ 3)8 + c e 2

9(3x¡ 4)

32 + c f ¡4

p1¡ 5x+ c

g ¡ 3

5(1¡ x)5 + c h ¡2

p3¡ 4x+ c

2 a ¡ 1

3cos(3x) + c b ¡ 1

2sin( 4x) + x+ c

c 6 sin¡x2

¢+ c d ¡ 3

2cos(2x) + e¡x + c

e ¡ cos¡2x+ ¼

6

¢+ c f 3 sin

¡¼4¡ x¢+ c

g 1

2sin(2x)¡ 1

2cos(2x) + c

h ¡ 2

3cos(3x) + 5

4sin(4x) + c

i 1

16sin(8x) + 3 cosx+ c

3 a y = 1

3(2x¡ 7)

32 + 2 b (¡8, ¡19)

4 a 1

2x+ 1

4sin(2x) + c b 1

2x¡ 1

4sin(2x) + c

c 3

2x+ 1

8sin(4x) + c d 5

2x+ 1

12sin(6x) + c

e 1

4x+ 1

32sin(8x) + c f 3

2x+ 2 sinx+ 1

4sin(2x) + c

5 a 1

2(2x¡ 1)3 + c b 1

5x5 ¡ 1

2x4 + 1

3x3 + c

c ¡ 1

12(1¡ 3x)4 + c d x¡ 2

3x3 + 1

5x5 + c

¡

e ¼ 3:48 units2 f 2 units2 g ¼ 3:96 units2

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ANSWERS 753

e ¡ 8

3(5¡ x)

32 + c f 1

7x7 + 3

5x5 + x3 + x+ c

6 a 2ex + 5

2e2x + c b 3

5e5x¡2 + c

c ¡ 1

3e7¡3x + c d 1

2ln(2x¡ 1) + c, x > 1

2

e ¡ 5

3ln(1¡ 3x) + c, x < 1

3

f ¡e¡x ¡ 2 ln(2x+ 1) + c, x > ¡ 1

2

g 1

2e2x + 2x¡ 1

2e¡2x + c h ¡ 1

2e¡2x ¡ 4e¡x + 4x+ c

i 1

2x2 + 5 ln(1¡ x) + c, x < 1

7 a y = x¡ 2ex + 1

2e2x + c

b y = x¡ x2 + 3 ln(x+ 2) + c, x > ¡2

c y = ¡ 1

2e¡2x + 2 ln(2x¡ 1) + c, x > 1

2

8 Both are correct. Recall that:d

dx(ln(Ax)) =

d

dx(lnA+ lnx) =

1

x, A, x > 0

9 p = ¡ 1

4, f(x) = 1

2cos( 1

2x) + 1

2

11 a f(x) = ¡e¡2x + 4

b f(x) = x2 + 2 ln(1¡ x) + 2¡ 2 ln 2, x < 1

c f(x) = 2

3x

32 ¡ 1

8e¡4x + 1

8e¡4 ¡ 2

3

12 x¡ 1

2cos(2x) + c 13 1

4sin 2x+ 2 sinx+ 3

2x+ c

EXERCISE 21E.1

1 a 1

4b 2

3c e¡ 1 (¼ 1:72) d 1

2

e 1 1

2f 6 2

3g ln 3 (¼ 1:10) h 1

2

EXERCISE 21E.2

1 aR 4

1

px dx ¼ 4:67,

R 4

1(¡p

x) dx ¼ ¡4:67

bR 1

0x7 dx = 1

8,R 1

0(¡x7)dx = ¡ 1

8

2 a 1

3b 7

3c 8

3d 1 3 a ¡4 b 6:25 c 2:25

4 a 1

3b 2

3c 1 5 a 6:5 b ¡9 c 0 d ¡2:5

6 a 2¼ b ¡4 c ¼2

d 5¼2

¡ 4

7 aR 7

2f(x) dx b

R 9

1g(x) dx 8 a ¡5 b 4

9 a 4 b 0 c ¡8 d k = ¡ 7

410 0

REVIEW SET 21A

1 a 8px+ c b ¡ 3

2ln(1¡ 2x) + c, x < 1

2

c ¡ 1

4cos(4x¡ 5) + c d ¡ 1

3e4¡3x + c

2 a 12 4

9b

p2

3dy

dx=

xpx2 ¡ 4

;R

ispx2 ¡ 4 + c

4 b = ¼4

, 3¼4

5 a 2x¡ 2 sinx+ c b9x

2¡ 4 sinx+ 1

4sin(2x) + c

6d

dx(3x2 + x)3 = 3(3x2 + x)2(6x+ 1)R

(3x2 + x)2(6x+ 1) dx = 1

3(3x2 + x)3 + c

7 a 6 b 3 8 a = lnp2 9 f

¡¼2

¢= 3¡ ¼

2

10 ¼12

¡ 1

4

REVIEW SET 21B

1 a y = 1

5x5 ¡ 2

3x3 + x+ c b y = 400x+ 40e¡

x2 + c

2 a 3:528 b 2:963 32(lnx)

x, 1

2(lnx)2 + c

4 f(x) = 3x3 + 5x2 + 6x¡ 1 5 a 1:236 17 b 1:952 49

6 a f(x) = 1

4x4 + 1

3x3 ¡ 10

3x+ 3 b 3x+ 26y = 84

7 a e3x + 6e2x + 12ex + 8 b 1

3e3 + 3e2 + 12e¡ 7 1

3

REVIEW SET 21C

1 a ¡2e¡x ¡ lnx+ 3x+ c, x > 0

b 1

2x2 ¡ 2x+ lnx+ c, x > 0

c 9x+ 3e2x¡1 + 1

4e4x¡2 + c

2 f(x) = 1

3x3 ¡ 3

2x2 + 2x+ 2 1

63 2

3(p5¡p

2)

4 ¼6+

p3

45 e¡¼

6 if n 6= ¡1,1

2(n+ 1)(2x+ 3)n+1 + c

if n = ¡1, 1

2ln(2x+ 3) + c, x > ¡ 3

2

7 a = 1

3, f 0(x) = 2

px+ 1

3px

is never 0 aspx > 0 for all x

) f 0(x) > 0 for all x

8 a = 0 or §3

EXERCISE 22A.1

1 a 30 units2 b 9

2units2 c 27

2units2 d 2 units2

2 a 1

3units2 b 2 units2 c 63 3

4units2 d e¡ 1 units2

e 20 5

6units2 f 18 units2 g ln 4 units2 h ln 3 units2

i 4 1

2units2 j 2e¡ 2

eunits2

3 2

3units2

4 a 2:55 units2 b 0:699 units2 c 1:06 units2

EXERCISE 22A.2

1 a 4 1

2units2 b 1+e¡2 units2 c 1 5

27units2 d 2 units2

e 2 1

4units2 f ¼

2¡ 1 units2 g ¼

2units2

2 10 2

3units2

3 a, b

c 1 1

3units2

4

5 a, b

c enclosed area = 3 ln 2¡ 2 (¼ 0:0794) units2

6 1

2units2

y = 1�

y = 2

x

y1�� xey

22 �� xey

(0, 0) (ln 2, 1)�

(1, 2)�

(3, 0)(0, 3)�

x

y 32 �� xxy

3�� xy

i ¼ 1:52 j 2 k e¡ 1 (¼ 1:72) l 1

3

m ¼8+ 1

4n ¼

4

2 a ¼ 1:30 b ¼ 1:49 c ¼ ¡0:189

1

3units2

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754 ANSWERS

7

enclosed area = 1

12units2

8 a b 9¼4

units2 (¼ 7:07 units2)

9 a 40 1

2units2 b 8 units2 c 8 units2

10 a C1 is y = sin 2x, C2 is y = sinx

b A(¼3

,p3

2) c 2 1

2units2

11 aR 5

3f(x) dx = ¡ (area between x = 3 and x = 5)

bR 3

1f(x) dx¡

R 5

3f(x) dx+

R 7

5f(x) dx

12 a C1 is y = cos2 x, C2 is y = cos(2x)

b A(0, 1) B(¼4

, 0) C(¼2

, 0) D( 3¼4

, 0) E(¼, 1)

13 a 2:88 units2 b 4:97 units2 14 k ¼ 1:7377

15 b ¼ 1:3104 16 a =p3

EXERCISE 22B.1

1 110 m

2 a i travelling forwards

ii travelling backwards (opposite direction)

b 16 km c 8 km from starting point (on positive side)

3 a b 9:75 km

EXERCISE 22B.2

1 a 1

2cm b 0 cm 2 a 5 1

6cm b 1 1

2cm left

3 a 41 units b 34 units 4 b 2 units

5 a 40 m s¡1 f

b 47:8 m s¡1

c 1:39 seconds

d as t ! 1, v(t) ! 50

e a(t) = 5e¡0:5t and asex > 0 for all x,a(t) > 0 for all t.

g 134:5 m

6 900 m

7 a Show that v(t) = 100¡ 80e¡120

t m s¡1 and as t ! 1,

v(t) ! 100 m s¡1

b 370:4 m

EXERCISE 22C

1 E4250

2 a P (x) = 15x¡ 0:015x2 ¡ 650 dollars

b maximum profit is $3100, when 500 plates are made

c 46 6 x 6 954 plates (you can’t produce part of a plate)

3 14 400 calories 4 76:3o C

EXERCISE 22D.1

1 a 36¼ units3 b 8¼ units3 c 127¼7

units3

d 255¼4

units3 e 992¼5

units3 f 250¼3

units3

g ¼2

units3 h 40¼3

units3

2 a 18:6 units3 b 30:2 units3

3 a 186¼ units3 b 146¼5

units3 c ¼2(e8 ¡ 1) units3

4 a 63¼ units3 b ¼ 198 cm3

5 a a cone of base radius r and height h

b y = ¡¡

rh

¢x+ r c V = 1

3¼r2h

6 a a sphere of radius r 7 a ¼2

4units3 b ¼2

8units3

8 a

b ¼¡¼4+ 1

2

¢units3

9 a b 2¼2 units3

EXERCISE 22D.2

1 a A(¡1, 3), B(1, 3) b 136¼15

units3

2 a A(2, e) b ¼(e2 + 1) units3

3 a A(1, 1) b 11¼6

units3

4 162¼5

units3 5 a A(5, 1) b 9¼2

units3

REVIEW SET 22A

1 a 2 + ¼ b ¡2 c ¼

2 A =R b

a[f(x)¡ g(x)]dx+

R c

b[g(x)¡ f(x)] dx

+R d

c[f(x)¡ g(x)] dx

3 no,R 3

1f(x) dx = ¡ (area from x = 1 to x = 3)

4 k = 3p16

5 Hint: Show that the areas represented by the integrals can be

arranged to form a 1 £ e unit rectangle.

6 4:5 units2

7 a v(t):

y

x3

3

�3

�3

922 �� yx

10

2 4 6 8 10 12 14 16 18 20

20

30

40velocity (km h )��

t (mins)

4050

v t( ) (m s )��

t(sec)

1050)( 5.0�� tetv

t (seconds)2 4

�

0

y

x

2� xy

y x�� X

(\Qw_\ ),��

y

x�!�

�!�

1

y x x�� sin cos

y

4

x

y x�� sin( )

�!�

& ' ~`2\*�!�

& ' 1 \*�!�

IB SL 2nd ed


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Y:\HAESE\IB_SL-2ed\IB_SL-2ed_an\754IB_SL-2_an.CDR Thursday, 19 March 2009 9:43:44 AM PETER

ANSWERS 755

b The particle moves in the positive direction initially, then at

t = 2, 6 2

3m from its starting point, it changes direction. It

changes direction again at t = 4, 5 1

3m from its starting

point, and at t = 5, it is 6 2

3m from its starting point.

c 6 2

3m d 9 1

3m

8 (3¡ ln 4) units2 9 128¼5

units3

REVIEW SET 22B

1 29:6 cm

2 a local maximum at (1, 1

2), local minimum at (¡1, ¡ 1

2)

b as x ! 1, f(x) ! 0 (above)

as x ! ¡1, f(x) ! 0 (below)

c d 0:805 units2

3 2:35 m

4 a v(0) = 25 m s¡1, v(3) = 4 m s¡1

b as t ! 1, v(t) ! 0

c

REVIEW SET 22C

1 269 cm

2 a

b¡1¡ ¼

4

¢units2

3 a = ln 3, b = ln 5 4

³¼2

2¡ 2

únits2

5¡2¡ ¼

2

¢units2 6 k = 1 1

3

7 a 312¼ units3 b 402¼ units3 c ¼2

2units3

d ¼¡3¼¡8

4

¢units3

8 ¼2

units3 10 a 128¼3

units3

EXERCISE 23A

1 a continuous b discrete c continuous d continuous

e discrete f discrete g continuous h continuous

2 a i height of water in the rain gauge ii 0 6 x 6 200 mm

iii continuous

b i stopping distance ii 0 6 x 6 50 m iii continuous

c i number of switches until failure ii any integer > 1iii discrete

3 a 0 6 X 6 4

b YYYY YYYN YYNN NNNY NNNN

YYNY YNYN NNYN

YNYY YNNY NYNN

NYYY NNYY YNNN

NYNY

NYYN

(X = 4) (X = 3) (X = 2) (X = 1) (X = 0)

c i X = 2 ii X = 2, 3 or 4

4 a X = 0, 1, 2, 3

b HHH HHT TTH TTT

HTH THT

THH HTT

(X = 3) (X = 2) (X = 1) (X = 0)

c P(X = 3) = 1

8

P(X = 2) = 3

8

P(X = 1) = 3

8

P(X = 0) = 1

8

d

EXERCISE 23B

1 a k = 0:2 b k = 1

7

2 a P (2) = 0:1088

b a = 0:5488, the probability that Jason does not hit a home

run in a game.

c P (1) +P (2) +P (3) +P (4) +P (5) = 0:4512 and is the

probability that Jason will hit one or more home runs in a

game.

d

3 aP

P (xi) > 1 b P (5) < 0 which is not possible

4 a The random variable represents the number of hits that Sally

has in each game.

b k = 0:23c i P(X > 2) = 0:79 ii P(1 6 X 6 3) = 0:83

5 a6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

roll 1 4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

1 2 3 4 5 6

roll 2

y x�� sin

y x�� sin�

x

y

�

�

2 310

P( )X x��

x

Wi_

Ri_

2 3 4 51

Probability

0.2

0.4

x

x

ƒ( )x �

)(1 2

��

xfx

x( Qw_)��, local max.

( Qw_)�� , local min.

v t( ) m/s

t (sec)

25

100)( �tv( )Xt�

d 3 seconds e a(t) =¡200

(t+ 2)3, t > 0 f k = 1

5

5 a 0 and ¡0:7292 b 0:2009 units2

6 $408 7 m = ¼3

8 a a ¼ 0:8767 b ¼ 0:1357 units2

9 a ¼

³3¼32

¡ 1

8p2

únits3 b ¼ 124 units3

( is the number of heads)X

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756 ANSWERS

b P (0) = 0, P (1) = 0, P (2) = 1

36, P (3) = 2

36,

P (4) = 3

36, P (5) = 4

36, P (6) = 5

36, P (7) = 6

36,

P (8) = 5

36, P (9) = 4

36, P (10) = 3

36, P (11) = 2

36,

P (12) = 1

36

c

6 a k = 1

12b k = 12

25

7 a P (0) = 0:1975k; P (1) = 0:0988k; P (2) = 0:0494k;

P (3) = 0:0247k; P (4) = 0:0123k

8 a P (0) = 0:665 b P(X > 1) = 0:335

9 a x 0 1 2

P(X = x) 3

28

15

28

10

28

b x 0 1 2 3

P(X = x) 1

56

15

56

30

56

10

56

10 a Die 21 2 3 4 5 6

1 2 3 4 5 6 72 3 4 5 6 7 8

Die 1 3 4 5 6 7 8 94 5 6 7 8 9 105 6 7 8 9 10 116 7 8 9 10 11 12

b 1

6

d 15

26

c d 2 3 4 5 6 7 8 9 10 11 12

P(D = d) 1

36

2

36

3

36

4

36

5

36

6

36

5

36

4

36

3

36

2

36

1

36

11 a Die 21 2 3 4 5 6

1 0 1 2 3 4 52 1 0 1 2 3 4

Die 1 3 2 1 0 1 2 34 3 2 1 0 1 25 4 3 2 1 0 16 5 4 3 2 1 0

b

c 1

6d 2

5

EXERCISE 23C

1 102 days 2 a 1

8b 25 3 30 times

4 $1:50 5 15 days 6 27

7 a i 0:55 ii 0:29 iii 0:16

b i 4125 ii 2175 iii 1200

8 a E3:50 b ¡E0:50, no c i k = 3:5 ii k > 3:5

9 a i 1

6ii 1

3iii 1

2

b i $1:33 ii $0:50 iii $3:50

c lose 50 cents d lose $50

10 a $2:75 b $3:75 11 a k = 0:03 b ¹ = 0:74

12 ¹ = 2:5

EXERCISE 23D

1 a

b The binomial distribution applies, as this is equivalent to

tossing one coin 100 times.

c The binomial distribution applies as we can draw out a red

or a blue marble with the same chances each time.

d The binomial distribution does not apply as the result of each

draw is dependent upon the results of previous draws.

e The binomial distribution does not apply, assuming that ten

bolts are drawn without replacement. We do not have a

repetition of independent trials.

2 a ¼ 0:268 b ¼ 0:800 c ¼ 0:200

3 a ¼ 0:476 b ¼ 0:840 c ¼ 0:160 d ¼ 0:996

4 a ¼ 0:231 b ¼ 0:723 c 1:25 apples

5 a ¼ 0:0280 b ¼ 0:002 46 c ¼ 0:131 d ¼ 0:710

6 a ¼ 0:998 b ¼ 0:807 c 105 students

7 a i ¼ 0:290 ii ¼ 0:885 b 18:8

REVIEW SET 23A

1 a a = 5

9b 4

92 4:8 defectives

3 a k = 0:05 b ¹ = 1:7

4 a i 1

10ii 3

5iii 3

10b 1 1

5

5 a $7 b No, she would lose $1 per game in the long run.

REVIEW SET 23B

1 a k = 8

5b 0:975 c 2:55

2 a 0:302 b 0:298 c 0:561

3 6:43 surgeries

4 a 0:849 b 2:56£ 10¡6 c 0:991

d 0:000 246

5 a 42 donations b 0:334

6 a i 0:100 ii 0:114 b 3:41 games

REVIEW SET 23C

1 a k = 12

11b k = 1

2

2 a

b ¹ = 2

3 480 4 a 0:259 b 0:337 c 0:922

5 a i 2

5ii 1

10iii 1

10b $2:70

1 2 3 4 5 6 7 8 9 10 11 12

probability

Se_y_Fe_y_

He_y_

sum

N 0 1 2 3 4 5

P(N = n) 6

36

10

36

8

36

6

36

4

36

2

36

13 a m 1 2 3 4 5 6

P(M = m) 1

36

3

36

5

36

7

36

9

36

11

36

b ¹ 4:47

14 a P(X 6 3) = 1

12, P(4 6 X 6 6) = 1

3,

P(7 6 X 6 9) = 5

12, P(X > 10) = 1

6

c a = 5 d organisers would lose $1:17 per game

e $2807

The binomial distribution applies, as tossing a coin has two

possible outcomes (H or T) and each toss is independent of

every other toss.

x 0 1 2 3 4

P(X = x) 0:0625 0:25 0:375 0:25 0:0625

b k = 81

31(¼ 2:61), P(X > 2) = 0:226

¼

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ANSWERS 757

EXERCISE 24A

1 b 2

3

2 a a = ¡ 3

32b c 5

32d 2

3 a k = ¡ 4

375b 3 1

3c 0:64 4 a = 5

32, k = 2

5 a k = 3

16b 2:4 hours c 0:369 d a = 3

p16

EXERCISE 24B.1

1

2 a,b The mean volume (or diameter) is likely to occur most often

with variations around the mean occurring symmetrically as

a result of random variations in the production process.

EXERCISE 24B.2

1 a 0:341 b 0:383 c 0:106

2 a 0:341 b 0:264 c 0:212

d 0:945 e 0:579 f 0:383

3 a a ¼ 21:4 b a ¼ 21:8 c a ¼ 2:82

EXERCISE 24C.1

1 a 0:885 b 0:195 c 0:302 d 0:947 e 0:431

2 a 0:201 b 0:524 c 0:809 d 0:249 e 0:249

3 a 0:383 b 0:950

4 a a = 1:645 b a = ¡1:282

5 a Physics 0:463, Chemistry 0:431, Maths 0:990,

German 0:521, Biology 0:820

b Maths, Biology, German, Physics, Chemistry

6 65:6%

EXERCISE 24C.2

1 a 0:159 b 0:309 c 0:335

2 a 0:348 b 0:324 c 0:685

3 a 0:585 b 0:805 c 0:528

EXERCISE 24D

1 a

k ¼ 0:878

b

k ¼ 0:202

c

k ¼ ¡0:954

2 a

k ¼ ¡0:295

b

k ¼ 1:17

c

k ¼ ¡1:09

3 b i k ¼ 0:303 ii k ¼ 1:04

4 a k ¼ 79:1 b k ¼ 31:3

EXERCISE 24E

1 0:378 2 a 90:4% b 4:78% 3 83

4 a 0:003 33 b 61:5% c 23 eels

5 ¹ ¼ 23:6, ¾ ¼ 24:3

6 a ¹ = 52:4, ¾ = 21:6 b 54:4% 7 112

8 0:193 m 9 a ¹ = 2:00, ¾ = 0:0305 b 0:736

REVIEW SET 24A

1 a i 2:28% ii 84% b 0:341

2 a a = 6:3 grams b b 32:3 grams

3 a a = ¡ 3

10b

c 1:2

d 13

20

4 k ¼ 0:524 5 29:5 m 6 a 0:136 b 0:341

REVIEW SET 24B

1 a i 81:9% ii 84:1% b 0:477 c x ¼ 61:9

2 ¹ ¼ 31:2

3 a ¹ = 29, ¾ ¼ 10:7 b i 0:713 ii 0:250

4 a 1438 students b 71 marks

5 a 0:260 b 29:3 weeks

6 a ¹ = 61:218, ¾ ¼ 22:559 b ¼ 0:244

REVIEW SET 24C

1 a 0:364 b 0:356 c k ¼ 18:2

2 ¾ ¼ 0:501 mL 3 0:207 4 ¹ ¼ 80:0 cm 5 0:0708

6 0:403

EXERCISE 25A

4

x��

x

�( )x

A

B

Cx

y

604020

0.2

0.15

0.1

0.05

z0

z0

z0

0.81 0.58 0.17

k k k

z0

z0

z0

0.384 0.878 0.1384

k k k

2x

y)3(

103 �� xxy

3 a 0:683 b 0:477

4 a 84:1% b 2:28% c i 2:15% ii 95:4%

d i 97:7% ii 2:28%

5 a 0:954 b 2:83

6 a ¹ = 176 g, ¾ = 24 g b 81:9%

7 a i 34:1% ii 47:7%

b i 0:136 ii 0:159 iii 0:0228 iv 0:841

c k = 178

8 a ¹ = 155 cm ¾ = 12 cm b 0:84

9 a ¼ 41 days b ¼ 254 days c ¼ 213 days

1 a r = 3 b 2£ 319 2 Hint: u1 = ln 2 = d

3 a b2x b 2 ln b+ x c x¤ =2 ln b

b2 ¡ 14 a (b, 2) b y-intercept is 2¡ 2b2, x-intercepts are b§ 1

c i b = ¡2 ii b < ¡2 iii b =1§p

17

45 a x3 ¡ 6x2 + 12x¡ 8 b 29

6 a 1 b 3 c fx j x 6 1

2, x 2 R g d fy j y > 0, y 2 R g

7 a a b ¡b c a dap

1¡ a2

8 a x = 0, ¼, 2¼ b x = ¼3

, 5¼3

¼

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758 ANSWERS

9 a Constant a b c d e h

Sign > 0 < 0 > 0 < 0 > 0 = 0

Constant ¢ of f(x) ¢ of g(x)

Sign < 0 > 0

10 a x = ¡5 b a = ¡ 1

2

11 a A = 1

3(B¡1¡I) b A =

µ¡2 ¡1 1

31

30

¶c 4

9

12 a i p = 1 iip

6p2 + 60

13 a¡!BA =

Ã311

!¡!BC =

Ã11¡3

!b both are

p11 units

c a rhombus d 1

11e

p120

11f 2

p30

14 a g b i m¡ a iij + k ¡ c¡ d

2

15 a 35, 6:4 b 19:5, 3:2 c 57:5, 9:6

17 a b

18 a 4 sin(2x)

b

c x = ¡¼, ¡¼2

, 0, ¼2

, ¼

d M is at (¡¼, 0), (0, 0) or (¼, 0)

19 a P(A [B) = x+ 0:57 b x = 0:16

20

21 b¡1

2, 1

2

¢c i x > 0 ii x < 1

2

22 a v(t) = k ¡ 8e2t m s¡1 b k = 72

23 a x = 3 b x =p7 c x =

5¡ ln 8

2d x = 3

24 a 1 m s¡1, the initial velocity

b 0, uniform (constant) velocity

c 4, 4 m displacement on 1 6 t 6 3

25 a ¡8 b k = 1

226 b ¼

4¡ 1

227 a = 0:3, b = 0:2

28 a r =1

e2b e¡199 c

e3

e2 ¡ 1

29 a x = 3 b x = 2

30 a,b

31 a 70%

b i m ¼ 27:5 ii n ¼ 35 iii p ¼ 42:5 iv q = 100

32 a p = 10p3 b x+

p3y = 40

33 a v(t) = t¡ 3

2sin¡2t+ ¼

2

¢+6 1

2cm s¡1 b

¼ + 26

4

34 b 1

3ln¡7

2

¢35 a ¡ 2p

21b ¡ 4

p21

25

36 a 5

16

p2 b 40 + 20

p2

37 a

Ã¡226

!b

p11 units s¡1

c x = 3¡ 2t, y = 1 + 2t, z = ¡2 + 6t, t > 0

38 a PQ =

³ ¡2 3b+ 38¡ 4a ¡3a¡ 4b

´b a = 2, b = ¡1, k = ¡2 c P¡1 = ¡ 1

2Q

39 a,c

y

x��

��

A ,( )��

B ,( )��

stationary

inflectionnon-stationary

inflection

C ,( )��local min

x#

k

#$

0.7

¼��.

y

x

y g' x�� ( )

��2

�2

��

y��

y��

b fy j 1 6 y 6 egd fx j 1 6 x 6 eg,

fy j 0 6 y 6 4ge g¡1(x) = 4 lnx

y

( )��, e

1

1

y x��

g x x��( )�� ln

4)(x

exgy ��

x

( )e,��

y

x

�'( )x

y

x

�"( )x

units2

) 5 solutions

c i 0:3

ii 0:2iii 0:541

d 0:1

40 c SR = 5p3 cm

d perimeter = 15 + 5p3 cm, area = 25

2

p3 cm2

41 a i ¡ 1

2x+ 3 ii x+ 2y = 20 iii A(12, 4)

b iR 6

2(¡ 1

4x2 + 3x+ 4)dx ii 46 2

3units2

iii ¼R 6

2(¡ 1

4x2 + 3x+ 4)2dx

42 a i r = ¡3 ii ¡4£ 313

b i x = 4 or ¡1ii S = 8 when x = 4; when x = ¡1, S does not exist

c i ¡55 ii ¡2300

43 a i

Ã¡1¡3¡7

!ii 1p

59

Ã137

!b no c a = 1

5

d¡!OM = 1

2(5i + j ¡ 9k)

e r1 = 1

2

Ã51¡9

!+ t

Ã32¡1

!, t 2 R

f i

ii m = ¡45 1

2iii P(¡30 1

2, ¡21 1

2, 6 1

2)

cm s¡1

Ã32¡1

!6= k

Ã2¡31

!for some k 2 R

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y

x1 3

-1\Qw_

Er_

y��

x��

region

B

Y

B

Y

C

D

Wy_

Ry_

Rt_

Qt_

Wt_

Et_

1 2 3 4

1

2

3

4

die 2

die 1

60000

20000

P t( )

t

y P t�� ( )

y

x

��

( )��,

y x��( )

ANSWERS 759

44 a i

³4 30 1

íi

³8 70 1

íii

³16 150 1

ć

³1024 10230 1

´d i

³2 10 1

´+

³4 30 1

´+ ::::::+

³2n 2n ¡ 10 1

íii p = 2 + 4 + 8 + 16 + ::::::+ 2n

fuse sum of a geometric seriesgq = 2n+1 ¡ 2¡ n

iv

³215 ¡ 2 215 ¡ 16

0 14

´45 a ii x = 6 iii 16

b i y = 12¡ x ii y2 = x2 + 64¡ 16x cos µ

e 8p5 units2 when x = y = 6 f isosceles

46 a f¡1(x) = x+3

4, g¡1(x) = x¡ 2

b 4x¡ 11

c x = 47

15

d i,iv

ii A = 4, B = ¡11 iii 12¡ 11 ln 4

47 a The probabilities do not add to 1.

b a+ b = 0:3, 0 6 a 6 0:3, 0 6 b 6 0:3

c i 0:16 ii 0:84

48 a b 2

5

c 7

15

d 1

2

e $7:40

49 a

b X = 2, 3, 4, 5, 6, 7, 8

c i 3

16ii 5

8iii 3

10d d = 8 1

3

50 a 0 cm s¡2,¡3¼2

¡ 1¢

cm s¡2

b v(t) = 3

2t2 + cos t+ 2 cm s¡1

c

³¼3

16+ ¼ + 1

ćm, which is positive as ¼ > 3.

d the integral is the displacement in the first ¼2

seconds.

51 a a = 7, b = ¼8

, c = 1, d = 10

b i A0(7, 28) ii y = 14 sin ¼8(t¡ 3) + 14

c a vertical stretch (x-axis invariant) of factor 1

2, followed by

a translation of

³¡23

´:

52 a (2x + 4)(2x ¡ 5) b x = log2 5

c i x =1

pii x =

1

3p+ 1

53 b 2a¡ b when x = 3¼4

, 7¼4:

c Max TP’s: (0, a), (¼, a), (2¼, a)

Min TP’s: (¼2

, b¡ a), ( 3¼2

, b¡ a)

54 c S(x) d1

[C(x)]2

55 a P 0(t) =30 000 e¡

t4

(1 + 2e¡t4 )2

and use the fact that e¡t4 is never

negative.

b P (t) is increasing for all t > 0

c P 00(t) =7500e¡

t4 (2e¡

t4 ¡ 1)

(1 + 2e¡t4 )3

d 3750 per year when t = 4 ln 2 years

e P (t) ! 60 000 and P (0) = 20 000

f

56 b 1

2

p7 units when P is at

³¡

p3p2

, 3

2

ór at

³p3p2

, 3

2

´:

57 a fy j ¡1 6 y 6 1g b 2 solutions c ¡3 sinx cos2 x

d ¼ units3

58 a 25 sin® cm2 b

³25¼

2¡ 25 sin®

ćm2

c Amax = 25¼2

cm2 when ® = 0 or ¼

Amin = 25(¼2¡ 1) cm2 when ® = ¼

2.

59 a i h = 4 ii k = 18 iii a = ¡2 b 18 2

3units2

60 a i x = 1 ii x = 5p7 b x = 0 or 1

61 a ii µ = ¼3

b cosx = 1¡p3

2

62 u1 = 2, un = 3n2 ¡ 3n+ 3, n > 1

63 x = ¡ 3¼2

or ¼2

64 a ¡e2 b e2 ¡ 3

65 (0, ¡1, ¡1)

66 a y = 2x¡ 3 b

d 3

2< x < 2

67 x = 2, y = 1

8or x = 64, y = 4 69 a = 1

2

70 P(A [B) = 1 or P(A \B) = 0

71 µ = ¡ 11¼12

, ¡ 7¼12

, ¼12

, 5¼12

72 a fx j x < 0 or x > 2g b1

x+

1

x¡ 2c 4x¡ 3y = 12¡ 3 ln 3

73 a 24

49b 16

2574 a ¼ 0:34 b ¾ ¼ 5

75 a = 3

576

8

x77 9b = 2a2

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Y:\HAESE\IB_SL-2ed\IB_SL-2ed_an\759IB_SL-2_an.CDR Thursday, 19 March 2009 10:08:29 AM PETER

x��

�

�

�

non-stationary

inflection

(2' To_\)

min (1' Qw_\)

y��

x��

y

23

),(41

6�

�2� 2��

y x��sinXy

x

A CB D Ex

k

y

x

� ��( )x e�x�

y

x

y��

x��y x��

¼��.

¼��.

y x��( )

y x��( )

�

�

��

��

y x��( )

�� x

y

y x�� ln

760 ANSWERS

78 a (0, 4), a translation of

³21

´b (0, 6), a translation of

³20

´followed by a vertical stretch

of factor 2 (with x-axis invariant)

c

d (¡2, 1

3) e (3, ¡2), a reflection in y = x

79 y = 4 sin¡¼2x¢¡ 1 80 a x = 0 b x = 0:2 or 0:3

81 a = ¡2, b = 3, A11 =

³¡2 ¡13 2

´82 a VA is x = ¡1, HA is y = 1

b f 0(x) =2(x¡ 1)

(x+ 1)3; local min (1, 1

2)

c f 00(x) =¡4(x¡ 2)

(x+ 1)4; inflection

¡2, 5

9

¢d

83 a3x

x¡ 2b

2x+ 1

x¡ 1

84 a i 13

21ii 11

21b 2

385 x =

2

a2 ¡ 1

86 a a5 ¡ 5a4b+ 10a3b2 ¡ 10a2b3 + 5ab4 ¡ b5 b 1

c 32x5 + 80x3 + 80x+40

x+

10

x3+

1

x5

87 a a2 ¡ 2 b a3 ¡ 3a

88 a i A(4, 0), B(¡4, 0) ii C(0, 2), D(0, ¡2)

b y =

r4¡ x2

4c area = 4

Z 4

0

r4¡ x2

4dx

d volume = 64¼3

units3

89 a x 0 ¼4

¼2

3¼4

¼ 5¼4

3¼2

7¼4

2¼

f(0) 0 1

21 1

20 1

21 1

20

b

c when x = ¼6

, y = 1

4X d fy j 0 6 y 6 1g

e ¼2

units2 f x¡ y = ¼4¡ 1

2

90 a f(0) = 2 b

c k = ln 2

d 2¡ 2pe

91 a f 0(x) = 1¡x¡2, x = 1 b A(1, 2) c .... is at least 2

d i no solutions ii one solution iii two solutions

92 a r =

Ã20¡3

!+ t

Ã1¡12

!, t 2 R

b x = 2 + t, y = ¡t, z = ¡3 + 2t, t 2 R

c it represents any point on the line d

Ãt+ 3¡t¡ 32t¡ 8

!e 6t¡ 10 f t = 5

3g¡11

3, ¡ 5

3, 1

3

¢93 a i A is the minimum value of X ii B is Q1

iii C is the median iv D is Q3

v E is the maximum value of X

b i the range ii the IQR c i 0:5 ii 0:75

d

EXERCISE 25B

1 n = 30

2 a,c

b x ¼ 4:82 d x+ 5y = 15

3 ¡ 84

125

4 a m = ¡2, n = 4 b k = 7 c vertex is (2, 5)

d Domain of f is fx j x 2 R gRange of f is fy j y > 3gDomain of g is fx j x 2 R gRange of g is fy j y > 5g

5 a 1950 b 10 500

6 a x = §1 b x ¼ §0:671 c x = ¡0:2

d x = ¡ 1

2or 2

5

7 a r ¼ 35:4 cm b ¼ 1530 cm2 c 59:4 cm

8 a,c

b x ¼ §1:68

d i A ¼R 1:245

0:0501(lnx¡ x sinx+ 3cosx)dx

ii A ¼ 1:37 units2

( 12

, 3), a horizontal stretch of factor 1

2, followed by a

translation of

µ3

2

0

¶

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50

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150

20 40 60

37\Qw_

75

Qz Qx Qct

CF

x

y y x��( )

2�

W

G

W

G

W

G

Qq_Pt_

qG_t_

qL_r_

qG_r_Qq_Pr_

qF_r_

1st 2nd

y

x

x��

y��

-\Qw_

21)(

��

xxg

y

x

y x��( )

�� '

�'

'��

y g x�� ( )

ANSWERS 761

9 a A¡1 =

0@ ¡2 ¡3 ¡2

¡ 1

2¡ 1

2¡ 1

2

¡ 7

10¡ 9

10¡ 1

2

1A b x = ¡10

y = ¡2:5

z = ¡2:5

10 a m = 4, n = ¼4

, p = 1, r = 8

b i ¼ 5:17 ii t = 2 1

3

11 a ¼ 30:9o

b ii No, as equations have inconsistent solutions

c (¡8, 2 1

2, ¡ 1

2) d a = ¡5 1

2

12 a x-intercepts are ¡1 and ¯, y-intercept is ¡¯

b,c

13 a Time f Cumulative freq.

0 < t 6 10 15 15

10 < t 6 20 10 25

20 < t 6 30 25 50

30 < t 6 40 50 100

40 < t 6 50 30 130

50 < t 6 60 20 150

b i ¼ 35 min ii ¼ 19 iii ¼ 0:5

14 ¼ 0:0548

15 a ¼ 0:470

b No, it is¡5

3

¢(0:86)3(0:14)2 where

¡5

3

¢= 10:

16 1

2

17 a z-score for 100 m ¼ ¡1:86z-score for 200 m ¼ ¡1:70

b the 100 m

18 a 0, ¼ 1:46

b

c i ¼ 1:64 ii y ¼ 1:64x¡ 0:820 d P(0:903, 0:671)

19 a i ¼ 0:672 ii ¼ 0:705

b Method is ok. Although not strictly binomial, the binomial

distribution is very close in this case.

20 a ¡12e1¡4x b ¡ 3

4e1¡4x + c c ¼ 2:04

21 a A(1, 0), B(¼, 0) b C(2:128, 0:641)

c (1:101, 0:086) d a non-stationary inflection

22 a 186 months b 371 months

23 a +80x4 + 80x2 + 32

b 1

11x11 + 10

9x9 + 40

7x7 + 16x5 + 80

3x3 + 32x+ c

24 a f(x) = ¡4(x¡ 1)2 + 4

b i A ¼R 1:89

0:106[f(x)¡ g(x)]dx ii ¼ 4:77 units2

25 b P(B) = 0:6, P(A) = 0:2

26 a x = 70:5 b 76 kg c s ¼ 15:1

d about 1:92 standard deviations above the mean.

27 a ¼ 0:0355 b ¼ 0:974

28 a i ¼ 0:544 ii ¼ 0:456 b i (0:97)n ii n = 12

29 a i ii 11

21

b n = 6

30 ¹ ¼ 679 kg, ¾ ¼ 173 kg 31 a ¼ 2000 m b 350o

32 a z = 0:1 b y = 0:1 c x = 0:7

33 a m = ¡1, n = 2

b i g(x) = ¡ 1

x+ 2iv

ii VA is x = ¡2,HA is y = 0

iii ¡ 1

2

34 a i (¡1, 3) ii (39, 23) b ¼ 44:7 m

d no

35 k ¼ ¡0:969

36 a A(0, 1), B

³2,

1

e4

´b ¼ 0:882 units2

37 a 0:8 b i ¼ 0:0881 ii ¼ 0:967

38 a ¹ ¼ 40:4 b ¼ 0:0117 c a ¼ 55:8

39 a

³¡8 00 ¡8

´b show jAj 6= 0

d a = ¡ 1

8, b = 5

8

40 a

b Range is fy j ¡2:41 6 y 6 0:91g c b = ¼2

d A ¼ 0:785 e ¼ 1:721 units3

� ��( ). , .

( )��. , .

x

y

b

�

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die 2

die 11

1

2

2

3

3

4

4

5

5

6

6

U

G T

a

.��b

.��c

.��

0.42

y

x

),(22

1 e

),(22

1 e��

y

x

��

�

2521 �� xexy

0.718

762 ANSWERS

41 b 3x+ e3¼2 y = 1 + 3¼

2c ¼ 0:0847 units2

42 b x = 0, x = ¼2

are VAs c 2 when x = ¼4

d ¼ 2:046

43 a i ¼ 0:0362 ii ¼ 0:610 iii ¼ 0:566 b k ¼ 74:4c a ¼ 81:0, b ¼ 101:6 d i ¼ 0:506 ii ¼ 0:168

44 a

Ã23¡5

!,p38 units b D(4, 3, 2) c F(7, ¡4, ¡2)

d 33, 11

3p38

e 3p221 units2

45 a t = 1

4b 2:675 c the mean of the Y distribution

46 a e¡ 2 ¼ 0:718

b

c ¼ ¡0:134 d

47 a a ¼ 17:2, b ¼ 30:0 b i ¼ 0:514 ii ¼ 0:538

48 a 2:59 days b i ¼ 0:279 ii ¼ 0:799

49 a A(¡3, 4, ¡2) b Yes, at (4, ¡3, 5) c ¼ 75:0o

50

51 a ¡1 b x ¼ 1:857 at P, ¼ 4:536 at Q

c f 0(x) = x2e¡x(3¡ x), B(3, 0:344)

d 3¡p3 at A and 3 +

p3 at C e ¼ 0:595 units2

52 a

b 11

36c i ¼ 0:227 ii ¼ 0:635

53 b a = ¡3 1

2, b = 5 c D(10, ¡11, 11)

54 a DB ¼ 4:09 m, BC ¼ 9:86 m

b AbBE ¼ 68:2o, DbBC ¼ 57:5o c ¼ 17:0 m2

d ¼ 10:9 m

55 a a = ¡1, b = 2 b y-intercept is ¡2 1

2

c ¡1¡p21

2and ¡1+

p21

2d D

¡¡ 1

2, ¡ 2 1

3

¢e i A ¡

Z k

p21¡1

2

³¡1 +

3

x2 + x¡ 2

´dx

ii ¼ 0:558 units2

56 a i ii 0:12

iii are independent

b iii b ¼ 0:104, a ¼ 0:124 iv ¼ 0:228

57 b 9a+ 3b+ c = 14, ¡4a+ 2b¡ c = 1

c

Ã1 1 19 3 1¡4 2 ¡1

! Ãabc

!=

Ã¡4141

!d a = 2, b = 1, c = ¡7 e p = 2, q = 3

58 b i ¼ 0:927c ii ¼ 0:644c

c i ¼ 2:16 cm2 ii ¼ 29:3 cm2

59 a x = 3 b x = ln 2

ln 3(or log3 2)

60 a ¼ 1:48 units b ¼ 3:82 units

61 a 2p2 ¡ p4 b p ¼ 0:541

62 a 3

5b 5 cm or 2:2 cm c AB = 5 cm is not possible

63 ¼ 6:40 cm 64 ¼ 0:114 65 ¼ 0:842

66 a a = 13, b = 12, c = ¼30

, d = 15 b ¼ 24:9 m

67 AB = I, a = 2, b = ¡1, c = 3

68 a x = ¼2

b f 00(x) = esin2 x(2¡ 4 sin4 x), sin2 x = 1p

2

c ¼ (0:999, e1p2 ), ¼ (2:14, e

1p2 )

69 31 1

7or 46 6

7

70 a f 0(x) = e1¡2x2

(1¡4x2), f 00(x) = e1¡2x2

(16x3¡12x)

b local min at

³¡ 1

2,pe

2

´, local max at

³1

2,pe

2

ć x = 0 or §

p3

2

d as x ! 1, f(x) ! 0 (from above)

as x ! ¡1, f(x) ! 0 (from below)e

71 a x ¼ 16:0 b s ¼ 2:48

72 c µ ¼ 1:02, 2:59, 4:16, 5:73

73 a no solutions exist b x ¼ 3:82

74 a k = 2 b ¹ = 3:2 c 47

50

75 a f 0(x) = 6 cos3 x¡ 5 cosx

c local max. at (0:421, 0:272), (2:72, 0:272),

local min. at¡¼2

, ¡ 1¢

d

2�

( )��. , . ( )��. , .

�x

y

(\w_\' )��

3

a A(1, 0), B(2, 0), C(0, 2) b y = 0 is a HA

d local max. at x ¼ ¡0:618, local min. at x ¼ 1:618

f x ¼ 2:05 g area ¼ 0:959 units2

=

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INDEX 763

INDEX

addition rule

algebraic test

ambiguous case

amplitude

angle between vectors

angular velocity

antiderivative

antidifferentiation

arc

arc length

area of triangle

arithmetic mean

arithmetic sequence

arithmetic series

asymptotic

average acceleration

average speed

average velocity

axis of symmetry

bimodal

binomial

binomial coefficient

binomial expansion

binomial experiment

binomial probability distribution

binomial theorem

box-and-whisker plot

Cartesian equation

census

chain rule

chord

coincident lines

column graph

column matrix

common difference

common ratio

complementary events

completing the square

component form

composite function

compound interest

concave downwards

589

47

225

238

341

570

581

580

200

200

218

73

72

83

60

514

468

513

164

385

190

193

190

639

639

193

397

355

376

491

200

368

378

275

72

76

429

159

322

55

79

533

concave upwards

continuous random variable

continuous variable

coplanar lines

cosine function

cosine rule

critical value

cubic function

cumulative distribution function


cumulative frequency graph

data

definite integral

dependent events

derivative

differentiation

dilation factor

direction vector

discrete numerical variable

discrete random variable

discriminant

disjoint sets

displacement

displacement function

distribution

distributive law

domain

dot product

double angle formulae

empty set

equal vectors

expectation

experimental probability

exponent

exponential

exponential equation

exponential function

first derivative

five-number summary

frequency

frequency histogram

function

general term

geometric mean

geometric sequence

geometric series

geometric test

global maximum

533

630

377

369

249

221

56

144

654

389

401

377

474, 598

436

481

485

149

355

377

630

162

447

314, 615

513

377

288

46

341

267

447

312

636

422

94

554

104

105

501

397

422

378

47

71, 76

76

76

85

47

525

e 113,

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764 INDEX

global minimum

golden ratio

gradient function

graphical test

horizontal asymptote

horizontal inflection

horizontal line test

horizontal translation

identity function

identity matrix

included angle

independent events

index

infinite geometric series

initial conditions

instantaneous acceleration

instantaneous speed

instantaneous velocity

integral

integrating constant

interquartile range

intersecting lines

intersection of sets

interval notation

inverse function

invertible matrix

laws of indices

laws of logarithms

life table

limit

linear function

linear speed

local maximum

local minimum

logarithmic function

lower quartile

lower rectangles

magnitude

mapping

matrix

mean

median

midpoint

modal class

mode

monotone decreasing

monotone increasing

motion diagram

525

181

480

539

61, 529

526

65

244

62

289

218

434, 456

94

87

515

514

468

514

587

590

395

368

447

52

62

298

96

125

421

462

144

570

525

525

144

394

471

311

47

274

382

382

328

378

382

521

521

615

motion graph

multiplicative inverse

mutually exclusive events

natural exponential

natural logarithm

negative definite quadratic

negative matrix

negatively skewed

non-stationary inflection

normal

normal curve

normal distribution

Null Factor law

number of trials

number sequence

optimisation

outcome

outlier

parabola

parallel lines

parallel vectors

parameter

parametric equation

Pascal’s triangle

percentile

period

periodic function

point of discontinuity

point of inflection

population

position vector

positive definite quadratic

positively skewed

power rule

principal axis

probability

probability density function

probability distribution

probability generator

product rule

quadratic formula

quadratic function

quantile

quotient rule

radian

radius

random sample

random variable

513

292

452

128

128

173

281

383

534

497

648

412, 649

157

422

70

182, 538

422

377

164

368

337

377, 409

209

191

400

245

238

463

533

376

311

173

383

589

238

420

646

630

444

493

160

144

659

495

198

200

409

630

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INDEX 765

range

rate of change

rational function

reciprocal function

relation

relative frequency

relative frequency histogram

row matrix

sample

sample space

scalar

scalar multiplication

scalar product

second derivative

second derivative test

sector

segment

self-inverse function

series

sigma notation

sign diagram

sign diagram test

sine curve

sine rule

singular matrix

skew lines

solid of revolution

speed

square matrix

standard deviation

stationary inflection

stationary point

statistic

survey

table of outcomes

tangent

tangent function

tangent ratio

theoretical probability

translation

tree diagram

turning point

two-dimensional grid

union of sets

unique solution

unit circle

unit vector

universal set

46, 394

509

466

60

47

422

379

275

377

426, 447

310

320

341

501

539

200

200

64

82

82

56

539

240

224

298

369

619

517

275

406

533

525

377, 409

377

431

469, 497

251

204

427

244

438

525

430

447

293

203

338

426, 447

upper quartile

upper rectangles

variance

vector

vector equation

vector product

velocity

velocity vector

Venn diagram

vertex

vertical asymptote

vertical line test

vertical translation

volume of revolution

-intercept

-intercept

zero matrix

zero vector

-value

394

471

406

310

354

341

310

354

446

164

61, 529

47

244

620

164

164

281

316

653

x

y

z

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766 NOTES

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NOTES 767

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768 NOTES

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ANSWERSwashburnmath.weebly.com/uploads/1/0/7/5/107524597/... · 2018-02-22 · ANSWERS 695 EXERCISE...

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