Math AI HL

A nswers 543

y by
8
b 6
4
8 2
6
4 −6 −4 −2 2 4 6 x
2 −2
−8 −6 −4 −2 2 4 6 8 x −4
−2
−4 cy
−6
−8

27 a Vertical stretch with scale factor 3, vertical 8
6
   4
2
b Vertical stretch scale factor 5, vertical
translation 1 unit −8 −6 −4 −2−2
−4
28 a Horizontal stretch scale factor 2, vertical −6 2 4 6 8 10 x
−8
  

b Horizontal stretch scale factor 2, vertical
translation 3 units

29 a Horizontal translation 2 units, vertical

translation 5 units

b     
  
30 a factor 1 , ref lection in 32 y = 12x2 + 56x + 60
b Horizontal stretch scale factor 3 ref lection in
y-axis 1 33 y = 3ex−2
stretch scale 2 , 34 a (x − 5)2 − 14
Horizontal
y-axis b Horizontal translation 5 units, vertical

31 a y   
35 a 5(x + 3)2
6 b      

4 scale factor 5

36 a y = 9x2 b3

2 37 a y = 16x3 b 1
38 y = 2x3 + 5x2 2
2 4 6 8x
−4 −2 39 a Translation 1 right followed by a horizontal
stretch factor 0.5
−2

−4 b A horizontal stretch factor 0.5 followed by
a translation 1 right (or a translation 2 right
−6
      
40 a = 9, b = 6, c = −10

544 Answers

41 A horizontal stretch factor 0.25 followed by a 24 a 5 years b 43.2 years
translation 2 left or a translation 8 left followed by b 0.140
a horizontal stretch factor 0.25. 25 a A = 70, B = 20

c 6.06 minutes

Exercise 5D 26 a 200, 178 b 178

c 0.173

1 a y = 10e−0.347t b y = 7e−0.0693t d 12.6 minutes

2 a y = 10 + 2e−0.139t b y = 15 + 5e−0.0990t 27 a 5 years b 0.139

3 a y = 1.5 − 0.5e−6.93t b y = 120 − 20e−13.9t c 9.6 g

4 a y = 3 + 2 ln x b y = 10 + 5ln x 28 a PA = 20 sin ⎛π t⎞ + 30

⎝12 ⎠

5 a y = −0.419 + 4.93ln x b y = −2.21+ 4.48ln x b PP = 10 sin ⎛π (t − 4)⎞⎠ + 15

6 a y = 11.5 − 4.98ln x b y = 6 − 2.89ln x c t = 14 ⎝12

7 a y = 3 + 5sin(1.57(x − 1)) dy

b y = 10sin(2.09(x − 2)) 80

8 a y = 2 + 6sin(0.628(x + 2)) PA + PP

b y = −4 + 10sin(1.05(x + 1)) 60

9 a y = 1.5 + 1.5sin(0.785(x − 4)) PA
b y = −1 + 3sin(3.14(x − 0.5))
40
10 a y = 1+ 500 b y = 1 800
4e−0.980x 19e−3.46x PP
+
20
11 a y = 1+ 2000 b y = 70
3e−0.275x 1 + 1.8e−0.292x

12 a y = 1+ 200 b y = 1+ 200 t
4e−1.24x e−0.549x 12 18 24
13 a −2 b4 6

14 a 1.5 b 0.5 71.5 MW

15 a 6 b −2 29 a y

16 a 3 b5 3

17 a 1 ln 1 b ln4
2 2
1 2
18 a e b 2 e4

19 a 3 b 10 1 y2
y1
20 a 0.0462 b 22.9 minutes x
0 Cc e.g. The temperature of the ice remains ° .
21 a p = 0.396, q = 2.73 b 4.55 −1

c 0.16  A  313 −2

22 a 25 b5 −3

c 30 seconds

23 P = P0eln32t

A nswers 545

by (8.19, 2.65) d e.g. The number of tigers is modelled
3 (1. 90, 2. 65) as continuous, which is only a valid
2 y3 approximation when the number is large;
for example, the ability to reproduce may be
severely impacted when N < 2.

1 y2 20 80e− 3 ln 2 t
y1 q
35 a +
−1
x q ln 3
ln 28
b − 3

−2 Chapter 5 Mixed Practice

− 3 (5.05, − 2.65) (11.3, − 2.65) 1a y

c y3 = 2.65sin(x − 0.333)
1
30 b C = 4 e5k c k = 0.327, C = 1.28

d 219 e 11

31 a 300; initial population is 0, no rabbit death or 3 y = ex
breeding y = 3ex

b 610 1 x
y=0
32 a L = 2000, C = −0.2 by
b k ≈ 0.8
cN x = −1

3000

(1, 2200)
2000

2x

1000
(2.5, −2)

1 2 3t cy
x = −6
33 a N(t) = ⎧ N0eαt for 0  t  0.5 t  1 −2 15 x
⎪ (12, −2)
⎨ N e e0 0.5α −β(t−0.5) for 0.5 <

⎪⎩

b α>β

34 a N = 200e−(ln22)t

b N <1

c 15.3 years

546 A nswers

f 3 12 a −1(x) = x + b −4 c i 2.5 y

3a 5 ii

4 0.462

5a 3 b1

6 −3 x

7 a x>2 b 2 + e 3 , g−1(x) > 2 2x

8 3 x − 1

3

9 a 2x3 + 3 b −1.14

10 a i −1 ii 0

b −3  x  3

c y d i 2.5 ii 0
5
e1

22 29
4
23 a 0.577
y = f(x) y xb Reflection in the line =
−5 5 x c 32
y = f−1(x)
1 3, 3 2 3, 0
y=x 24 a i 2x + x ≠ − 2 ii + x ≠
−5 x

b (−1, 1)

11 a 1 1 − 2, x ≠ 1 25 2x3 − 18x2 + 50x − 42
2 526 a (x + )2 +
− b      
x

4x +1 translation by 5 units
12 3− x 3
y
27 a = x + 2

13 b 9.39 b 1 by
14 a 3 −4
x = −2
e 215 a x + , y > 2 b x−2
1− −
16 a −0.5 b 2 x
17 a 28
b 9 − x2 23
y xc i Reflection in the line =
iii y  1 y=0 x

ii x − 3 4x + 11
x −1
18 a i 22 ii iii 9x + 4

b f(x) can be 1, which is not in the domain of g.

c ii y ≠ 1

19 a 9 b5
x+2
21 a 3

A nswers c 547
y
28 b Translation 5 units to the right and 2 units up

c x = 5, y = 2
29 a y

1 x y=2 3

4x

3
7

x=0 x = 7
by 2

x=2 34 a x = −5, y = 3 b x ≠ −5, f(x) ≠ 3

35 a Translation 3 units to the left

by

3x x = −3

y = ln(x2 + 6 x + 9)

cy 2ln 3 x
ln3 y = ln(x + 3)

31 e5 x 36 a p = −4, q = 6 b y = −x2 + 12x − 31

37 a 3 b7 c y=7
b b=2
38 a a = 800 c 40
ln e 339
⎛ 2⎞

⎝⎜ x − ⎠⎟

x=0 40 a x = 5, y = 2 b α = 2, β = 1

a 3, b 430 == c Horizontal translation 5 to the right, vertical
131 Vertical stretch scale factor , translation 5 units translation 2 up
3to the left
d 5x − 9 , x ≠ 0
x−2
32 y = ln(e6 (x + 2)2) e Reflection in y = x

33 a x = − 7 , y = 2 b ⎛3 , 0⎞, ⎛0, − 3⎞ 41 Horizontal stretch factor 2
2 ⎝4 ⎠ 7⎠
⎝ 42 Vertical stretch factor 1 / ln10

548 A nswers

43 y 7 a 8−i b 16 + 2i
x=4 8 a 8 + 6i b 7 − 24i
9 a 4 − 2i b −3 + 3i
log 16 10 a 4 + 3i b 1−i
35 1 1 2 23
11 a 2 + 2 i b − 13 − 13 i
44 y
x 12 a a = 5, b = 7 b a = −3, b = 9

13 a a = − 3 , b = 31 b a = 1 , b = −8
2 2 2
4 i14 a z = − − b z = 1+ 2i
2 5 1
15 a z = 3 + 7i b z = − 3 + 3 i

16 a x = ±3i b x = ±6i
17 a 2x = ± 2i b 5x = ± 3i
18 a x = 1 ± 2i b x = 2 ± 3i
2 1 5
19 a 1x = − ± 2 i b x = 3 ± 3 i

20 a Im

−1 1 x 
z

45 2 − f(x)     Re
 z 
z∗


Chapter 6 Prior Knowledge

1 1.58, 0.423 b Im
π 5π
2 a 6 b 4

3 8.06, 0.519 z

4 a 10e4x b 8e14x c 32e10x

Exercise 6A b −1    Re
 z
1ai b5
z ∗ 
2 a 4i b −8i

3 a −9 b6

4 a −20 b −4 + 2i
5 a 11 + 4i b −6 + 10i
6 a 1− 2i

A nswers 549
Im
21 a Im b

2 iz 6 z+w

2    10 Re 4w

2 z

 z   4 Re



 2z

b Im 23 a Im

2 24 Re 6

8 6 4 2 iz 4
2
zw 
z 4
6 4  4 Re
8 z w 

2z 

22 a Im b Im
10
6 z+w
8
4 w zw z6
z

2

2

4 2 24 Re 2  6 Re
2
    w
2


24 a x = − 3 ± 4 i
5 5

550 Answers

b Im 29 a a + 2(1 − a) i b 2a − a2i

i 30 a = 0, b = 4 or a = 12, b = 0
−0.6 + 0.8i 31 z = −1 + 2i

32 3 – 5i

33 2 – 3i, –2 + 3i

−1 1 Re 34 k2 − 1 + 2k 1i
k2 + 1
−0.6 − 0.8i k2 +
−i
35 a = ±3

37 z = 2 − 5i, w = 4 + i
3 1
38 z = 2 + 2 i, w= 2i

39 a = 8, b = 1
a = −1, b = 10
5 i25 z* = + 3
26 z = 2 + 3i 40 a = 4, − 2
27 a 3 + 4i, 2 + 11i
Im Exercise 6B
b 12
10
8 z3 1 a 5cis32π b 7cis32π
6
4 z2 2 a 6cis 5π b 4 2cis 7π
2 z 3 4
24
3 a 2 2cis54π b 2cis 4π
4 a z = −10i 3
5 a z = 2 + 2 3i b z = 8i
6 a 2z = − 6 − 2 6i b z =1+ i
7 a z = 4 3 − 4i b 1z = − + 3i
8 a 12cis81π5 b z = 2 − 2i
−6 −4 −2 6 Re 9 a 3cis3158π π
−2 5cis 8
b

28 Im b 1 cis 4π
3 7

10 a    b   

z3 11 a 0.643 – 0.766i b 0.5 + 0.866i

12 a 15e−0.1i b 2e2i

i z2 13 a 2e34πi b 4e− πi
12

z 14 a 1 3 i b 1 1 i
2+ 2 2+ 2

15 a i b 1 3
2− − 2
1 Re

A nswers 29 a 3.03, 0.093 551

16 Im 30 a 3.50e0.317 b 3.03sin(θ + 0.093)
b 350cos(0.317 + t)
iz
31 a V V2
10
z2
z 8

Re t

17 a 8 b 3π −8 V1
4 − 10
3π
c 8, 2 d −8i

18 cis1 b 14.5cos(30t + 5.56)

19 a π b 7π 32 260 + 36.5 sin ⎛2π t + 0.521⎞
3 12 ⎝365 ⎠

20 a i b ⎛1 + 23⎠⎞(1 + i) 33 b 0.2 b ln2 + iπ
⎝2
π⎞ 34 a 2eiπ
z=2 2cis⎛− 4⎠ b −128 + 128i
21 a ⎝ 35 a ei2π iπ
2
π⎛ ⎞ b
6⎝ ⎠
22 a 2 cis b 8i c iπ
Could be 2 + 2 kπi , where k ∈

23 a w = 2cis⎛− 3π⎞ b 64i Exercise 6C
⎝ 4⎠

24 −162 2 − 162 2i 1a Im
25 n = 24

26 n = 9 13π z+w
= 24 w
27 z = 4, arg z

w = 2, arg w = 5π
24
π z
28 a i 1 ii 4 Re

iii 2 iv π z∗
3
b 2cis1π2

c 6+ 2 6− 2i
2 + 2

d 6+ 2
4

552 A nswers
b
Im 5 3 3− 2 + i 3 + 2 3
z+w 2 2
6 (7,−3), (2,−5)
w
z 7 a 60 b (6 − 3, 3 + 2 3)
Im
8a

z∗ z2 z

Re

z3
Re

2a Im

zw

w b Rotation through 1.85 radians about the origin

z 9 a 4−i

Re B⎛ 3− 4 , 4 3 + 1 ⎞ , C⎛− 4 + 3, −4 3 − 1⎞
⎝ 2 2 ⎠ ⎝ 2⎠
b 2

w 10 b ⎛3 − 4 3 , 2 − 3⎞
b Im ⎝2
2 ⎠

Chapter 6 Mixed Practice

1 Im

5i

w z2

w Re Re
5
z
z3
zw

3 a 17, 0.245; 34, 0.540 b 2, 0.295 (16.9°) z1

4 b Rotation 1.76 radians (101°) anticlockwise 2 a 1±i
about the origin

A nswers 553

b Im 14 z = 4 + 3i
1+i π
3 a 3 ± 3i 15 1, 6
1i
b Im 16 ± 3
17 2i
2 ±

Re 18 5 + 12i
19 b = −2, c = 5
20 3 + 4i
b 0.387cos(3t + 1.22)
21 a 0.387, 1.22

22 a 17, 2.90 b eiπ6
c (−3.96, −1.13)

23 2 ± i 5

3 + i 3 24 x2 + y 2 − 2x = 3
25 a (1 − 3) + (1 + 3) i

b z= 2 ⎛cos π + i sin π⎞ , w = 2 ⎛cos π + i sin π⎞ ,
4 4⎠ 3 3⎠
⎝ ⎝

zw =2 2 , arg (zw) = 7π
12

Re c 6+ 2
4
26 a 2, π6
b −128 3

27 3 + 4i

3  i 3 28 1.30 radians, 2.28 b π
12
29 a 2 2 − 3

4 3 + 1 i Chapter 7 Prior Knowledge
5 5
5 −0.5 − i
i ⎛ 106 91.8 67.2 69.4 ⎞
−3 ⎜ 245 220 163 153 ⎟
6 1 203 149

7 z = −2 − 2i ⎜ 222 136 ⎟
⎜ ⎟
⎜⎝ 130 112 81.9 83.2 ⎠⎟
8a z= 2, arg z π b 8i
= 4

9 16 2 cis34π 1 2 1 2 Exercise 7A
3 3 3 3
10 a p = 3, q = 0.5 b p = − i , q = + i 1 a Connected

1 3 i 3 1 i b Complete, connected, simple
2+ 2 2 2
11 a b − + 2 a Connected, simple, tree

12 a − 1 1i b None

a2 + 1 a2 + 3 a No b Yes

3i13 x = − 4 a Yes b No

554 A nswers

A ⎛0 0 1 1⎞ A ⎛0 1 1 1⎞ 8a A
B
5a B ⎜ 0 0 1 0 ⎟ b B ⎜ 1 0 0 0 ⎟
C ⎜ 1 1 0 1 ⎟ C ⎜ 1 0 0 1 ⎟ C
⎜ ⎟ D ⎜ 1 0 1 0 ⎟ A
D ⎜⎝ 1 0 1 0 ⎠⎟ ⎝⎜ ⎠⎟
B
A ⎛2 2 0 1⎞ A ⎛2 0 0 2⎞⎜ E
B ⎜ 2 0 1 1 ⎟ B ⎜ 0 0 1 0 ⎟
6a C ⎜ 0 1 2 2 ⎟ b C ⎜ 0 1 0 1 ⎟
⎜ ⎟ ⎟
D ⎜⎝ 1 1 2 0 ⎠⎟ D ⎜⎝ 2 1 0 2 ⎠⎟

7a A B

D

b

E

DC

bA B
DC

9 a i B, D ii B, E

iii 3 iv 2

b i A, C ii B, D, E

iii 2 iv 3

10 a i B, D ii B, D, E

iii 3 iv 5

b i C, D, E ii A, B, C

iii 8 iv 3

11 a i 2 ii 3
D C bi 2
ii 2

12 a i 4 ii 6

bi 2 ii 3

A nswers 555

13 a A ⎛ 0 1 1 0 1 1 1 ⎞ 9 a ACBDEA (there are alternative solutions)
B ⎜ 1 0 1 1 1 0 0 ⎟ b ABDCEA
C ⎜ 1 1 0 1 1 0 0 ⎟
⎜ ⎟ 10 a ABFECDA
D ⎜ 0 1 1 0 0 0 0⎟ b ADEBCFA (there are alternative solutions)
E ⎜ 1 1 1 0 0 1 1 ⎟
F ⎜ 1 0 0 0 1 0 1 ⎟ 11 a Four vertices have odd degree
⎜ ⎟
G ⎝⎜ 1 0 0 0 1 1 0 ⎠⎟ b SMNPQNRQMRS (there are alternative
solutions)

b 4; C is friends with 4 out of the other 6 people. 12 There are two vertices of odd degree:

c Everyone is friends with the other six people. UWXVYZWUVZ (there are alternative solutions)

d8 13 a B and C Y

14 a There is a loop ABDEA . bX

b So that power can be supplied to each device.

c Any one of AB, BD, DE or EA .

15 a A

EB

W Z

DC c 16

b It is not possible to get from E to any other 14 a e.g. Hamiltonian cycle ABEDCA
vertex.
b 80
c EA or ED
Exercise 7C

Exercise 7B 1a A B CDE
b A – 34 20 18 –
1 a AEDB, ADCB b CBAE, CDAE B 34 – 20 – 25
2 a ABCDA, ABDEA b CDABC C 20 20 – 12 –
D 18 – 12 – 15
3a6 b7 E – 25 – 15 –

4 a 14 b 19 A B CDE

5a0 b1 A– 9– – 9

6a1 b1 B9– 6 12 –

7 a Eulerian; e.g. ADBECABCDEA C– 6 – – 14

b Semi-Eulerian; e.g. DCBAECA D – 12 – – 8

8 a Semi-Eulerian; e.g. FEDCABCFD E 9 – 14 8 –

b Neither

556 A nswers

2a A B CD 4a A

A–6 57 10 13

B 8 – 10 –

C– – – –

D7 – 6–

b ABCD D8 B

A––88 12 12
B 5– 6 –
C 10 – – 6 C
D– – 7 – bA

3a A

76

D 8B 9 8
7 B
12
58 D 10

18

C C
bA

25 17 5 a ⎛ 0 1/2 0 1/4 1/3 ⎞
12 ⎜ 1/3 0 0 1/ 4 0 ⎟
⎜ 0 0 0 1/ 4 1/3 ⎟
⎜ 1 / 3 1/ 2 1/ 2 0 1/3 ⎟
⎜ 1 / 3 0 1/ 2 1/ 4 0 ⎟
D 20 B ⎜⎝⎜ ⎠⎟⎟

15 b ⎛ 0 1/4 0 1/4 1/3 ⎞

15 ⎜ 1/3 0 1/ 2 1/ 4 1/3 ⎟
⎜ 0 1/ 4 0 1/ 4 0 ⎟
⎜ 1 / 3 1/ 4 1/ 2 0 1/3 ⎟
⎜ 1 / 3 1/ 4 0 1/ 4 0 ⎟
C ⎝⎜⎜ ⎠⎟⎟

A nswers 557

6 a ⎛ 0 1/2 0 0 ⎞ 16 a D (0.387)

⎜ 1/ 2 0 1 0 ⎟ b ⎛ 0 0 0 0.3 0 ⎞
⎜ 0 0 0 1 ⎟ ⎜ 0.45 0 0 0.3 0 ⎟
⎜ ⎟ ⎜ 0.45 0.45 0 0.3 0 ⎟
⎝⎜ 1 / 2 1 / 2 0 0 ⎟⎠ ⎜ ⎟

b ⎛ 0 0 0 1/2 ⎞⎜ ⎜ 0 0.45 0.9 0 0 ⎟
⎜ 1/3 0 0 0 ⎟ ⎝⎜⎜ 0.1 0.1 0.1 0.1 1 ⎟⎠⎟
⎜ 1/3 1 0 1/ 2 ⎟
⎟ c No, relative probabilities stay the same.

⎝⎜ 1 / 3 0 1 0 ⎠⎟ Exercise 7D

7a 0.143 b 0.125 1a A
8a 0.167 b 0.400
9a B (0.352) b C (0.421) 2 B
10 a A and B (0.286) b B and D (0.333) G
11 a 1500 km (via A) b $350 (via E )
12 a
B 24

3 5 F C
A C 2 2
D
5 54 E 4
A B
Weight = 16
E F b
4 3
4
E

D 3 1
C
b ABD, 8 minutes D 4
3
13 a 12 km (SCBDT)
b SCBT or SCDT (13km) F

14 a ⎛ 0 1 / 4 0 0 0 ⎞ Weight = 15
⎜ 1 0 1/ 2 1/3 1/ 2 ⎟
⎜ 0 1/ 4 0 1/3 0 ⎟
⎜ ⎟
⎜ 0 1/4 1/2 0 1/2 ⎟
⎜ 0 1/ 4 0 1/3 0 ⎟
⎝ ⎠

b 0.145 c B (33.3%)

15 A(0.293), E (0.220), C (0.195), B and D (0.146)

558 A nswers

2a A bW
SV

EB
T PU

D X Z
CY B
Weight = 41
(can replace BE by CE) Weight = 92
(can replace PS by WS)
bA
4a A
E
BE

D K C D A C
L B
Weight = 28 M Weight = 24
3a C
b
R

E

S
Q

P N D

Weight = 47 Weight = 52

(can replace RP by RS)

A nswers 559

5a A bA

G BF B

F C
D
E A EC
B
Weight = 120 D
b C
D Weight = 19
G 7 a AB, BD, DF, AC, CE
B
F b AB, AD, DF, FC, CE
8 a AB, AE, ED, EC (or AC)
E A
or AE, ED, AB, AC (or EC)
Weight = 155 b AB, BC, CD, DE

6a or AE, AB, ED, BC
or other combinations
F 9 a AB, BG, BH, HE, HD, DC, GF
b AG, GB, BC, GD, DE, AF (or EF)
or AG, GD, DE, GB, BC, AF (or EF)
10 a AD, DB, BC, DE
b AD, AB, BC, EA
11 a ED, EF, FG, GC, CA, DB
or ED, DG, EF, GC, CA, DB
b EF, ED, EG, DB, BA, AC
12 a CF, FE, EB, BA, DA (or DC)
or CF, FE, CB, BA, DC (or DA)
b CF, FD, EC, EB, BA

E C

D

(or DA instead of DC)
Weight = 21

560 A nswers
13 a
A 3 a AB, EG (78) b BC, CD , AF , FE
E (93)
5
B 4 a AB, CF, FH (137) b BE, EF , DG (107)

5 a O, C b9
c 81
d OB , BC

6 a F and D (FABFEDCBD); 6800 m

32 b Vertices D and F have odd degrees.

c 8050 m

D2 C d FE and ED
3
7 e.g. CBAJIHGFEDBDFHJCFDC, 8.7 km
F
NB SB8 a 99 minutes ( , I1, , I1, I2, SB, I1, NB)

b 130 minutes

b 15 km c e.g. The time to walk from one bridge to another

14 a BD, CD, CE, AB, AF (or EF) 9 a AD and EH

b Add CF first, then apply the usual algorithm. b 109 minutes

15 a Kruskal adds one edge at a time, Prim adds c 100 minutes
one vertex at a time.
d e.g. ABCDACFEDEHFGH
bA b FG, DH
10 a 208 m

c 208 m d 230 m

10 11 a C and F have odd degrees.
F
B b CEF (15)

98 c DEFDBACBECFECD (180)

d i 165 ii C and F

10 7 Exercise 7F
E C
1 a 36 ( ADCBA) b 85 ( ACBDA)
2 a 63 ( ACDEBA ) b 98 ( ADBECA )
3 a 98 ( AECBDA ) b 110 ( ADCBEA )

D 4 a 32 b 77
b 81
5 a 46 b 105

c 44 m 6 a 95

d i Each workstation is connected to the power 7a AB CDE
supply.

ii There are no unnecessary connections. A–4 10 9 5

16 a 49 b 58 B 4– 659

17 0 < x < 7 or 21 < x < 28 C 10 6 – 11 15

D 95 11 – 4

Exercise 7E E 59 15 4 –

1 a DE, EF (70) b DF , FG (45)
2 a AF , FE (52) b AF , FE (34)

A nswers 561

b ABCDE c PQR S TU

A – 10 9 20 8 P – 25 40 24 26 14
B 10 – 12 18 6 Q 25 – 20 21 23 11
C 9 12 – 12 11 R 40 20 – 16 28 26
D 20 18 12 – 12
E 8 6 11 12 – S 24 21 16 – 12 10
T 26 23 28 12 – 12
8a A B CD U 14 11 26 10 12 –

A – 8 10 7 d 119 km e 79 km

B 8– 79 14 a 40 b 40

C 10 7 – 6 c The lower and upper bounds are equal.

D 79 6– d ABCDEFA
15 a i BDE (7)
b A B CD
9a A – 30 28 12 ii BAC (13)
B 30 – 12 18 iii CED (9)
C 28 12 – 17 bA
D 12 18 17 –
83
A B CD
A–476 E 7B
B4–38 10 10
C7 3 – 5
D6 8 5 – 2 7 5 13

b ABCD D 9C
c 27 (CEDBAC)
A – 20 38 35
d 26
B 20 – 18 15 e The lower bound is unattainable; the paths

C 38 18 – 35 CD and CE cannot be linked to either end of a
route, nor can CE twice. CD and CA is the next
D 35 15 35 – best option, and can form a route.

10 a ABCDA, ABDCA, ACBDA f 27 minutes: CEDBAC
b ABDCA (45)
Chapter 7 Mixed Practice
11 a 32
b 31 1 a The second one, all vertices have even degree.
c It is at most 31km long.
b e.g. ABCDEBDA
12 a The cycle ACBEDA has length 29.
c C or E
b 28
c 24 2 a RS
d It is 24, 25, 26, 27 or 28 minutes. b e.g. PQRSUV
c e.g. PQRSVUP
13 a PUSR is 40 km.
b PUQ, 25 km

562 A nswers

3a A B C D E 9 C(0.293), E(0.244), D (0.195), B(0.171), A (0.098)

A00011 10 276
B00111
C0 1 0 1 0 11 a 1
D1 1 1 0 0
E11000 23 4
76 5
b 20
c B and D
4 a A tree of minimum weight which includes

every vertex of the graph.
b BH, NF, HN, HA, BE, CN; weight = 48
c BH, NF, HN, HA, BE, CN
5 a Kruskal’s adds one edges at a time, Prim’s

adds one vertex at a time.

bA B C D

IF 8 0
0
J HGE b Vertices 1 and 8 have odd degrees. 0
c 0 1/2 1/4 1/4 0 0 0 0
Length = 117 76 1/3
1/3 0 0 0 0 1/4 0 1/3
6 a i 11 10 9 1 8 1/3 0 0 1/4 1/4 0 1/2 1/3
1/3 0 1/4 0 1/4 1/4 0 0
0 0 1/4 1/4 0 1/4 0
0 1/2 0 1/4 1/4 0 0
0 0 1/4 0 0 0 0
0 0 0 0 1/4 1/4 1/2

12 13 2345 d 0.215 e 11.5%

12 a A and E have odd degrees.

ii deg 2: 2, 5, 6, 8, 9, 10, 11, 12, 13; b 250 m c A and E
deg 4: 1, 3, 4, 7
13 b i e.g PQRSTSRTQP
iii All vertices have even degree.
ii 34
b i G: 15, H : 22
c i To find a Hamiltonian cycle of least weight.
ii G: exactly two vertices have odd degrees
(E and F). ii There is no cycle containing vertex P.

iii Both graphs have vertices of odd degree. 14 a A B C

7 a Hamiltonian cycle F E ii 12 m D

b 111 (BAFCDEB )

c 106

d 106  T  111

8 a n−1

b i AB, BC, AD, DE

c i All vertices have even degree.

ii e.g. ABCDEACEBDA, 46

A nswers 563

b The number of walks of length 2 from a vertex 6 x = −10.25, y = 7.25, z = 10.75
to itself
1, 0.1 54 , 117 a λ =
c8 , v = ⎛ ⎞ ⎛− ⎞
⎜⎝
d ABEDC, AFEBC, AFEDC ⎠⎟ ⎜⎝ ⎠⎟

15 a 63 b ⎛5 −1 ⎞⎛ 1 0 5⎞⎛ 1 −1
4⎝⎜ 1 0⎟⎠⎜⎝ 0.1 ⎟⎠⎝⎜ 4
b 63 −⎞

c ADCBEA is a solution to the travelling 1 ⎟⎠
salesman problem.
Exercise 8A

16 a i There are vertices of odd degree. 1 a E(Y) = 21, Var(Y) = 20
b E(Y) = 31, Var(Y) = 80
ii No, there are more than two vertices of odd 5
degree. 2 a E (Y ) = −4, Var (Y ) = 9

b i All degrees have doubled, so all are even b E(Y) = 4.9, Var(Y) = 0.05

ii 306 km 3 a E(Y) = −25, Var(Y) = 45
b E(Y) = −3, Var(Y) = 5
17 $92, repeat AE , EB , DG, GC

18 a A B CDE F 4 a E(Y) = 24, Var(Y) = 20

A – 8 8 16 18 18 b E(Y) = 21, Var(Y) = 45

B 8 – 7 15 17 17 5 a E(Y) = 1, Var(Y) = 0.2

C 8 7 – 8 10 10 b E(Y) = 5, Var(Y) = 1.25

D 16 15 8 – 6 8 6 a E(Z) = 30, Var(Z) = 35

E 18 17 10 6 – 9 b E(Z) = −37, Var(Z) = 107

F 18 17 10 8 9 – 7 a E(Z   Z) = 48.5

b 52 c 56 E(Z) 7 , Var(Z) 7
d 52  T  56 6 3
e CBACDEFC b = =

19 a CD , AB, BC, BF, CE 8 a E(Z) = 28.5, Var(Z) = 45

b i 231 b E(Z) = −74, Var(Z) = 224

ii 239 9 a E(Z) = 4.7, Var(Z) = 1.16

iii 239 b E (Z) = 3, Var (Z) = 11
16
c 239  L  253 10 a E(Z) = 1.5, Var(Z) = 6

20 a 33 b E(Z) = 0, Var(Z) = 6

b There is a cycle of length 33 (ABCDEA). 11 a E(Z) = 17, Var(Z) = 14

Chapter 8 Prior Knowledge b E(Z) = 17, Var(Z) = 14
12 a E(X) = 5, Var(X) = 0.171
b E(X) = 6, Var(X) = 0.208
1 a 0.4 b 2.5
b 0.833
2 0.345 13 a E(X) = −4.7, Var(X) = 0.04

3 0.174 b E(X) = −15.1, Var(X) = 0.0467

4 a 0.12 14 a E(X) = 12, Var(X) = 0.9

⎛ 1319 ⎞ b E(X) = 8, Var(X) = 0.0208
5 ⎜ 1082 ⎟ 15 a E(X) = 3, Var(X) = 0.15
⎜ 2541 ⎟ b E(X) = 3.6, Var(X) = 0.315
⎝⎜ ⎠⎟

564 A nswers

16 a 4.2 b 13.6 15 0.0352
b 13.8
c 44.64 b 34.25 16 0.0336
8 17 0.0820
17 a 15
18 a 0.4 kg, 0.223 kg2
18 a 25
8 b 0.198
c 0.0209
c 85.9 19 a 0.196
b 0.0211
19 a 15 b 17
20 0.272 m
20 6.6 cm, 0.6 cm
21 0.0281
21 198.8 g, 4.16 g 22 a 2500g, 1.79g

22 102 g, 1.92 g b 0.995

23 E(B) = 3.1, Var(B) = 0.3 c CLT shows that the distribution of the mass of
24 E(V) = −6, Var(V) = 2.56 a ream is approximately normal.

25 1076 kg, 30.2 kg 23 a 0.321

26 0, 1.41 b The distribution of the times for each company
were unknown.
27 0, 35.4
24 a 0.208
28 68.3 g b 0.196

29 65 minutes, 8.06 minutes 25 μ = 7.33mm, σ = 0.525mm

30 a 19 b 10 26 a 0.0228
b i 0.868
Exercise 8B ii 0.315
iii 0.868
1 a W ~ N (21, 16) b W ~ N (42, 67) c 0.691
2 a W ~ N (7, 111) b W ~ N (−1, 7) d 0.645
3 a W ~ N(7, 1.48) b W ~ N (−5.5, 6.91)
4 a W ~ N (0, 8) b W ~ N (15, 9) 27 a 0.0173
5 a W ~ N (10, 0.5) b W ~ N (5, 0.3)
6 a X ~ N (30, 1.25) b X ~ N (30, 0.5) b The mean of normal variables is normal.
28 44
7 a Cannot say b Cannot say
Exercise 8C
50
1 a No, X ~ B(60, 0.801)
8 a ∑Xi ~ N (1500, 5000) b No, X ~ B(5, 0.195)

i=1 2 a Yes, X ~ Po(3)

40 Conditions well met.

b ∑Xi ~ N (1200, 4000) b Yes, X ~ Po(4)

i=1 But occurrences of fish unlikely to be
independent of each other as live in shoals.
9 a 0.543 b 0.996
3 a No – this is a continuous distribution.
10 a 0.331 b 0.137 b No – this is a continuous distribution.

11 a N (91.3,16.3) b 0.0156

12 a 0.3 s, 0.721 s b 0.339

c 0.166

13 a 65 cm, 0.005 cm2 b 0.00235

14 a 0.252 b 0.0175

A nswers 565

4 a No – not Poisson as not a number of events in a Exercise 8D
given period; not binomial as n not fixed.

b No – not Poisson as not a number of events in a 1 a ⎛ 0.25 0.7 ⎞ b ⎛ 0.45 0.9 ⎞
⎜⎝ 0.75 0.3 ⎟⎠ ⎝⎜ 0.55 0.1⎠⎟
given period; not binomial as n not fixed.

5 a 0.132 b 0.108 ⎛0.82 0.08 0.22⎞
6 a 0.0789 b 0.666
7 a 0.774 b 0.399 2 a ⎜ 0.06 0.73 0.11⎟
⎜ 0.12 0.19 0.67 ⎟
8 a 0.843 b 0.990 ⎝⎜ ⎟⎠

9 a 0.085 b 0.601 ⎛0.64 0 0.45⎞
10 a 0.507 b 0.458
b ⎜ 0.02 0.57 0 ⎟
11 a 0.138 b 0.590 ⎜ 0.34 0.43 0.55 ⎟
12 a 0.249 b 0.213 ⎝⎜ ⎠⎟

13 a 3.1 b 3.1 ⎛ 0.71 0 0.24 0.45⎞
⎜ 0.29 0.88 0 0⎟
14 a 2.30 b 5.3 3 a 0⎜ 0.05 0.57 0.07 ⎟
15 a X ~ Po(36) b X ~ Po(24) ⎟
⎜
0⎝⎜ 0.07 0.19 0.48⎠⎟
16 a X ~ Po(2.4) b X ~ Po(3)
17 a X ~ Po(54) b X ~ Po(9.6)
18 a X ~ Po(6.5) b X ~ Po(8.9) ⎛0.35 0.15 0.1 0.2 ⎞
⎜ 0.1 0.4 0.05 0.2 ⎟
19 a X ~ Po(14) b X ~ Po(7) b ⎜ 0.25 0.2 0.5 0.05 ⎟
⎜ ⎟
20 0.0116 ⎜⎝ 0.3 0.25 0.35 0.55⎟⎠
21 a 0.0595 b 0.0548
22 a 0.298 b 0.973 ⎛ 0.14 0.23 ⎞ ⎛ 0.5 1⎞
4 a ⎜⎝ 0.86 0.77 ⎟⎠ b ⎝⎜ 0.5 0 ⎟⎠

23 a 0.868

b The rate of arrival of messages is unlikely ⎛ 0.3 0 0.4⎞

to be constant – there will probably be more 5 a ⎜ 0.1 1 0.4⎟

at some times of the day than others. Within ⎜ 0.6 0 0.2 ⎟
⎝⎜ ⎟⎠
each distribution messages are not likely to

be independent as they may occur as part ⎛0.22 0.5 0.12⎞

of a conversation. The two distributions are b ⎜ 0.33 0.29 0.8 ⎟
0.21
also probably not independent of each other, ⎜ 0.45 0.08 ⎟
⎝⎜ ⎟⎠
as times when more emails arrive might be

similar to times when more texts might arrive. ⎛ 0 0.55 0.05 0.3⎞

24 a 0.916 b 0.0656 6a ⎜ 0.3 0 0.7 0.1⎟
25 a 0.430 0.25 0
⎜ 0.35 0.6 ⎟
b No – likely to be more outbreaks at certain ⎜ ⎟
⎝⎜ 0.35 0.2 0.25 0 ⎠⎟
times of the year so not constant rate, and

outbreaks not independent since contagious. ⎛0.84 0 0.31 0.34⎞

26 a i 0.161 ii 0.0514 ⎜ 0 0 0.13 0.22⎟
0 ⎟
b 0.0132 b 0.321 b⎜ 0.5 0.56 0.44 ⎟

27 a 0.0537 ⎜
⎝⎜ 0.16 0.5 0 0 ⎠⎟
28 a 0.475
29 a 0.273 b 1 c 0.00413 7 a 0.666 b 0.590
30 a 0.747 8 a 0.316 b 0.378
b 0.143 c 201 9 a 0.112
b 78 b 0.0426

566 A nswers

10 a ⎛ 0.610 ⎞ b ⎛ 0.503⎞ ⎛0.8 0.2 0.1⎞ ⎛ 433⎞
⎝⎜ 0.390⎠⎟ ⎝⎜ 0.497⎠⎟ ⎜ 0.1 0.7 00..63 ⎟⎟⎟⎠ ⎜ 357 ⎟
21 a ⎜⎝⎜ 0.1 0.1 b ⎜⎝⎜ 210 ⎠⎟⎟

⎛ 0.277 ⎞ ⎛ 0.264 ⎞
⎜ 0.325 ⎟ ⎜ 00..336751⎟⎟⎟⎠
11 a ⎜⎜⎝ 0.398 ⎠⎟⎟ b ⎜⎜⎝ ⎛ 450 ⎞

c ⎜ 325000⎟⎟⎠⎟
⎜⎝⎜
⎛ 0.260 ⎞ ⎛ 0.153⎞
⎜ 0.254 ⎟ ⎜ 00..243478⎟⎟⎟ 22 a 0.1 0.4 0.6
12 a ⎜ 0.267 ⎟ b ⎜ E L
⎜ ⎟ ⎜
⎝⎜ 0.220⎟⎠ ⎜⎝ 0.162 ⎟⎠ 0.1

13 a ⎛ 0.591⎞ b ⎛ 0.517 ⎞ 5.0 3.0
⎝⎜ 0.409⎟⎠ ⎜⎝ 0.483⎠⎟

⎛ 0.209 ⎞ ⎛ 0.333 ⎞ 0.6 0.4
⎜ 00..527193 ⎟⎟⎟⎠ ⎜ 00..333333⎟⎟⎟⎠
14 a ⎝⎜⎜ b ⎜⎜⎝

⎛ 0.253⎞ ⎛ 0.153⎞ H

⎜ 00..225691⎟⎟⎟ ⎜ 0.444 ⎟ b 0.319 c 0.5
⎜ ⎜ 0.239 ⎟
15 a ⎜ b ⎜ ⎟ ⎛0.85 0.1 0.25⎞

⎜⎝ 0.227 ⎟⎠ ⎝⎜ 0.164⎟⎠ 23 a ⎜ 0.05 0.75 00.2.55⎟⎟⎠⎟
⎜⎜⎝ 0.1 0.15
⎛13 22⎞ ⎛61 118⎞
16 a ⎜⎝9 22⎟⎠ b ⎜⎝57 118⎟⎠ b 0.278

c Bull 51.5%, Bear 29.4%, Stagnant 19.1%

⎛49 235⎞ ⎛1 3⎞ ⎛1 0.5 0 0⎞
⎜ 235⎟⎟ ⎜ 1 3⎟⎟
17 a ⎜ 136 b ⎜ ⎜ 0 0 0.5 0 ⎟ 2
⎜ 0 0.5 0 0 ⎟ 3
⎝⎜ 10 47 ⎠⎟ ⎜⎝1 3⎠⎟ 24 a ⎜ ⎟ b

⎛249 985⎞ ⎛ 9 59 ⎞ ⎜⎝0 0 0.5 1⎟⎠

18 a ⎜ 51 197 ⎟ b ⎜ 49 ⎟ 25 a ⎛ 0.7 0.4 ⎞ b 0.561
⎜ ⎟ ⎜ ⎟ ⎝⎜ 0.3 0.6⎟⎠
⎜⎝⎜⎜ 225274 998855⎟⎠⎟⎟ ⎝⎜⎜⎜ 12297157371⎟⎟⎟⎠
⎧ 0.7a + 0.4b = a a 4 , b 3
⎛ 0.875 0.875⎞ c ⎪⎨ 0.3a + 0.6b = b d = 7 = 7
⎝⎜ 0.125 0.125⎠⎟ ⎩⎪ a + b = 1
19 a 0.35 b

c 12.5% 26 a 0.125

20 a ⎛ 0.95 0.8⎞ b 0.938 b ⎧ 0.5d + 0.25h = d = h
⎜⎝ 0.05 0.2 ⎟⎠ ⎪⎨⎪ 0.5d + 0.5h + 0.5r
⎪ 0.25h + 0.5r = r
c ⎛ 0.941⎞ ⎩⎪ d+ h+r =1
⎝⎜ 0.059⎠⎟
c d = 0.25, h = 0.5, r = 0.25

A nswers 567

27 a ⎛ 0.9 0.15 ⎞ 8 a Independent events; constant rate of success
⎜⎝ 0.1 0.85 ⎠⎟
b 0.731

b 1, 3 9 a The average rate must be constant. However,
4
λ = we might expect it to vary over different

23 , 11v1 times of the day and with different weather

= ⎛ ⎞ v2 ⎛ ⎞ conditions. Birds must arrive independently,
⎜⎝ ⎠⎟ = ⎜⎝ − ⎟⎠
but we know they may come in flocks.

⎛ 3 −11⎞⎠⎟, ⎛ 1 0⎞ b 0.156 ⎛ 24.6%⎞
⎜⎝ 2 0 3⎟
c P = D = ⎜ 4⎟ 10 a 0.04 b ⎜ 42.6%⎟
⎜
⎝ ⎠ ⎜ 3 2.8% ⎟
⎝⎜ ⎟⎠
d 0.6 + 0.05(0.75n)
e 0.6 ⎛16.2% ⎞
⎜ 33.7%⎟
⎛ 0.8 0.6 ⎞ c ⎜ 50.1% ⎟
⎝⎜ 0.2 0.4 ⎟⎠ ⎝⎜ ⎟⎠
28 a

b λ = 1, 1 11 a i 2μ, 2σ 2 ii 3μ, 9σ 2
5
iii μ, 3σ 2 ,iv μ σ2
13 , 11v1 12 0.00537
⎛ ⎞ v2 ⎛− ⎞ 13 a 0.110 n
⎝⎜ ⎟⎠ ⎝⎜ ⎠⎟
= =

1 0⎞ b 0, 168 c 0.350
0 1⎟
P ⎛ 3 −11⎞⎠⎟, D ⎛ 5⎟ 14 0.882
⎜⎝ 1
c = = ⎜ ⎠ 15 a X ~ Po(45), calls arrive independently and at a
⎜
⎝ constant rate.

d 1 − 1 1n b 0.204 c 0.0207
4 4
1 ⎛⎞ 16 a 0.547 b3
4
5⎝ ⎠

e 17 a 0.362 b 0.279
c 0.659 d 0.0464
e 0.247 f 0.641
Chapter 8 Mixed Practice 18 a 0.189 b 0.372
c 0.208

1 a 3.2 b −7.6 c 14.0 19 a R F S
2 a 0.1
b 4.2 c 0.6 d 4.28 ⎛0.6 0.4 0.5⎞ R
⎜ 0.1 0.4 0.2⎟ F
3a x 1 23 4 ⎜ 0.3 0.2 0.3 ⎟ S
⎝⎜ ⎟⎠
P(X = x) 1 54 11
13 26 13 26 b 0.526

b 40 c 22.9 ⎧ 0.6r + 0.4 f + 0.5s = r
13
ii 0.52 kg2 ⎪ 0.1r 0.4 f + 0.2s f
4 a i 6.4 kg c i ⎪ + =

b Masses of individual components are ⎨ 0.3r + 0.2 f + 0.3s = s
⎪
r f s 1
independent. ⎪ + + =
⎩
5 a 20 m, 1.57 m2 b 0.0553
6 0.123 ⎛ 38 71⎞
⎜ 13 71 ⎟
7 a 45, 6.71 ii ⎜ ⎟

b Number of tweets on any given day ⎜ 20 71 ⎟
independent from other days. ⎝ ⎠

20 a 0.785 b 0.122 c 0.00976

568 Answers

21 i 0.549 ii 0.0231 Exercise 9A
iii 0.964 iv 0.169
22 a i 0.122 ii 0.248
8 1 a Not valid, reliable b Not valid, reliable
n
bi E(X) = 8, Var (X) = 2 a Valid, reliable b Valid, reliable

ii E(X) ≠ Var(X) for n > 1 3 a Not valid, not reliable

c i 0.846 b Not valid, not reliable

ii k = 0.736 4 a 20.8, 17.4 b 24.5, 171.5

23 85; assume independent lengths 5 a 0.833, 67.4 b 1.83, 221

24 8.20 g 6 a 0.1, 0.072 b 1.3, 4.46

25 a 0.431 b 0.274 7 13.1
b 0.131
26 a 0.261 b 42 c 1.72 85
c 58
27 a 0.112 9 a 4.5 b 3.5

28 a ⎛ 0.1 0.6⎞ c The sampling is not representative of the whole
⎝⎜ 0.9 0.4 ⎟⎠ snake population.

b λ = 1, − 1 d She could repeat the experiment and use a
2 t-test to see if the population means were very

different.

23 , 11v1=⎛ ⎞ v2 ⎛− ⎞ 10 a It is not a feasible result – someone is likely not
⎝⎜ ⎠⎟ = ⎝⎜ ⎟⎠ taking the questionnaire seriously.

2 −11⎟⎠⎞ , ⎛ 1 0⎞ b 2, 1.5
3 0 1⎟
c P = ⎛ D = ⎜ 2− ⎟ c e.g. Criminal activities is not clearly defined.
⎝⎜ ⎜ Affecting could be very indirect. In the last
⎝ ⎠ year might mean last calendar year or most
2 3 1n recent 12 months.
d 5 + 5 ⎛− ⎞
⎝
2⎠
2 d The sample was self selecting, and based on
e 5 those attending a police station, so biased and

therefore not valid.

Chapter 9 Prior Knowledge e Overestimate, since people attending a police
station are more likely to be victims of crime.
1 x = 5.8, σ = 2.32
103 > 12.8 H0. f Use a larger sample size.
2 so reject Data do not come from
B(5, 0.7). 11 a Reliability is when similar conclusions are be
reached on each occasion the test is conducted
3 p = 0.0248 < 0.1 so reject H0. Data are not
in similar circumstances.
independent.
b Parallel forms
4 a r = 0.920: strong positive correlation
b y = 1.20x + 0.862 c Give the same test again some time later (test-
p = 0.192 > 0.1 H0. retest) and check that the results are similar.

5 so do not reject Insufficient 12 a Validity is the extent to which you are
6 measuring what you really want to measure.
evidence that μ1 < μ2 .

0.227 b e.g. Success in business might not be just
7 0.834 measured by salary. People’s own opinions are
8 0.268
not the same as actual success.

c e.g. Politeness might not then be a confounding
variable.

d Criterion validity

A nswers 569

Exercise 9B 13 a H0: Po(3.5) is a good model, H1: Po(3.5) is not
a good model.
1 a χ2 = 3.10, ν = 2, p = 0.212, plausibly
independent b 0 and 1
b χ2 = 0.287, ν = 2, p = 0.866, plausibly
independent c5

2 a χ 2 = 6.10, ν = 2, p = 0.0473, not independent d 0.737; Po(3.5) is a good model.
b χ2 = 1.86, ν = 2, p = 0.394, plausibly
14 a 4.46, 5.20, 5.57, 5.57, 5.2
independent
b Sufficient evidence that the model is not
3 a χ 2 = 10, ν = 3, p = 0.0186, not independent appropriate (p = 0.0811).
b χ2 = 4.02, ν = 6, p = 0.674, plausibly
15 Insufficient evidence of bias (p = 0.459)
independent
6, 0;16 ν = p ≈ sufficient evidence that city and
4 a p = 0.0258; not Poisson more of transport are dependent.
b p = 0.200, plausibly Poisson
17 a i p = 0.0235, not plausible
5 a p = 0.263, plausibly Poisson ii p = 0.0575, plausible
b p = 1.16 × 10−8, not Poisson iii p = 0.00404, not plausible

6 a p = 0.426, plausibly Poisson b If there were theoretical reasons to believe that
b p = 0.191, plausibly Poisson the mean is 25 before the data is observed.

7 a p = 0.477, plausibly normal 18 a H0: B(3, 0.6) is a good model.
b p = 7.38 × 10−6 , not normal H1: B(3, 0.6) is not a good model.

8 a p = 0.219, plausibly normal b 12.8, 57.6, 86.4, 43.2, ν = 3
b p = 0.805, plausibly normal c p = 1.04 × 10−7 < 0.05 so reject H0. There is

9 a p = 0.000214, not normal evidence, at the 5% significance level, that
b p = 0.0167, not normal
B(3, 0.6) is not a good model.
10 p = 0.436, accept Mendel’s suggestion. d Mean = 2.14, p ≈ 0.713
11 a H0: N (12, 2.52) is a good model, e Binomial is a good model ( p = 0.249 > 0.05).
19 a 2.01
H1: N (12, 2.52) is not a good model.
b H0: a Poisson model is appropriate.
b 4.60, 9.18, 12.4, 9.18, 4.60 H1: a Poisson model is not appropriate.

c Some expected frequencies are smaller c 13.40, 26.93, 27.07, 18.13, 9.11, 5.36, ν = 4
than 5.
d No reason to conclude a Poisson model is
d2
inappropriate ( p = 0.060 > 0.05).
e p = 0.578; no evidence that N (12, 2.52) is not 20 a 1.22
a good model.
b 5.44, 11.59, 15.95, 11.59, 5.44. Expected
12 a H0: Diet choices are independent of age; 20value for t < needed combining with
H1: Diet choices are dependent on age. 20 < t < 21.5 and likewise for t > 26 with
24.5 < t < 26.
b The expected frequency of the 17–18
vegetarian group is less than 5. c3

c Sufficient evidence that diet choices depend on d No reason to conclude a Normal distribution
age (p = 0.0485). with mean 23 mins is inappropriate

d i e.g. Ask for the response which is the best ( p = 0.103 > 0.05).
description of their usual behaviour or have 21 a 28.410, 40.265, 5.264

   b2

ii Having too many categories with a fixed c H0: a normal model is appropriate.
sample size would make the chi-squared H1: a normal model is not appropriate.
test invalid as some expected frequencies
would drop below 5. d No reason to conclude that a normal model is

inappropriate ( p = 0.194 > 0.1).

570 Answers

22 a 2.67 13 a ABssgb::iorylSSssSS==rreess−−=0=1..18267.6.371t55t22++1193.1.2tt++b1235.1M.1odel A
14 a
b 1.57
b Different number of data points in each
c i Mean and variance are very different,
suggesting not a Poisson distribution. model

ii χ2 = 6.67, p = 0.0830 > 0.05; possibly c Rg2irls = 0.915
Poisson
Rb2oys = 0.844
iii χ2 = 60.7, p = 2.08 × 10−12 < 0.05; not
Poisson So, girls’ model is better fit.

d The first method does not have any measure of 15 a T = 0.2t2 − 7.13t + 99.8
uncertainty. It is subjective. The third method b 0.994
takes into account the fact that no observations
of 5 or more were made, so it provides a more c No – this quadratic model predicts that after
authentic view of the observed data so is
more valid. t ≈ 17.8 minutes the temperature will increase

Exercise 9C again (without bound). Clearly the temperature
model needs to predict the temperature tending
1 a y = 1.70x2 − 4.97x + 4.26 to room temperature.

b y = −3.72x2 + 12.6x + 6.14 16 The extra parameter in a cubic model compared
2 a y = −5.89x3 + 17.7x2 − 6.86x + 4.53 to a quadratic model makes the comparison
b y = 0.497x3 − 3.15x2 + 7.42x + 0.868 invalid.

3 a y = 1.68e0.217x b y = 72.7e−1.07x Exercise 9D
4 a y = 55.1x−0.706 b y = 1.37x4.39
5 a y = −3.80sin(2.18x + 0.785) + 3.56
Hint: The answers in this section are all given using

b y = 2.83sin(0.754x − 0.824) + 2.14 interval notation, but you can use inequality notation

6 a R2 = 0.938 b R2 = 0.819 or words if you prefer.

7 a R2 = 0.851 b R2 = 0.712 1 a (6.06, 9.94) b (2.76, 5.35)
8 a R2 = 0.650 b R2 = 0.899 2 a (135, 205) b (377,567)
9 0.773 3 a (5.90,8.20) b (5.63, 8.47)
10 a T = −0.00160x3 + 0.179x2 + 1.57x − 0.671 4 a (11.4, 13.2) b (12.2, 16.2)
5 a (165, 207) b (209, 245)
b i 242 years 6 a (8.04, 8.36) b (7.48, 7.92)
7 a (16.6, 21.2) b (15.5, 22.3)
ii Need to extrapolate so treat with caution. 8 a (11.5, 26.3) b (29.1, 43.9)
9 a (156, 190) b (325, 385)
11 a h = 1.9sin(0.52t − 0.15) + 6.2

b i 8.1 m

ii 4.3 m 10 a (7.07, 8.93) b (3.55, 4.56)

12 a 0.861, 0.911 11 a (175, 193) b (455, 489)

b Model B a better fit 12 a (6.87, 7.23) b (6.52, 7.58)

c D = 89.5p −0.785 13 (3.66, 4.62)
d 10.7 ≈ 11
14 (62.1, 65.5)
e Demand tends to zero as price increases so
potentially suitable, but extrapolation means 15 (780, 832)

any predictions should be treated cautiously.

A nswers 571

16 a (23.8, 29.8) Exercise 9E

b z -interval because the population standard 1 a p = 0.0852, insufficient evidence to reject H0
deviation is known. b p = 0.0196, sufficient evidence to reject H0

17 a (8.33, 8.87) 2 a p = 0.0138, sufficient evidence to reject H0

b Population distribution is normal. b p = 0.0227, insufficient evidence to reject H0

18 a 15.5 (°C)2 b (16.8, 20.4) 3 a p = 0.438, insufficient evidence to reject H0

19 a Population of concentrations is normally b p = 0.207, insufficient evidence to reject H0
distributed. 4 a p = 0.00338, sufficient evidence to reject H0
b p = 0.0125, sufficient evidence to reject H0
b (20.1, 23.9)

c No; the confidence interval does not include 5 a x < 78.1 b x < 91.4
20.
6 a X > 85.5 b X > 753
20 a (2.85, 3.55) ; t-interval as the population 7 a X < 55.1, X > 64.9 b X < 117, X > 123
variance unknown. Population distribution is
normal. 8 a 0.0288 b 0.0190

b There is no evidence against his belief, as 9 a 0.303 b 0.0707
the confidence interval extends below
10 a 0.00455 b 0.238
3 minutes.

21 a 12 hours2 11 a 0.0702 b 0.0271

b (25.2, 27.8); assuming the population 12 a Sufficient evidence to reject H0 (p = 0.000919)
distribution is normal. b Insufficient evidence to reject H0 (p = 0.221)

c The claim is not supported, as the confidence 13 a Sufficient evidence to reject H0 (p = 0.00691)
interval does not include 28. b Sufficient evidence to reject H0 (p = 0.00106)

22 a 4.72 14 a Insufficient evidence to reject H0 (p = 0.155)
b (147, 151) b Insufficient evidence to reject H0 (p = 0.199)

c No; the confidence interval is entirely below 15 The suspicion is justified (p = 0.00329).
153.
16 a 13.3
d No; 150 is contained within the confidence
interval. b Population variance has been estimated from
the sample.
23 a (166, 172)
c Insufficient evidence for belief (p = 0.179)
b Used midpoints instead of actual values.
17 a H0: μ = 7.8, H1: μ < 7.8
24 a (16.7, 18.9) 0.8 ⎞
N ⎛7.8, 10⎠ c X < 7.38
b Yes – the entire interval is above 16.5. b
⎝
c Yes – the new interval is (16.4, 19.2) and
contains 16.5. 18 a The population variance is known.

25 a (3.13, 4.27) b Insufficient evidence (p = 0.366)

b Yes, because the sample mean is 19 a We have a pair of values for each tree.
approximately normal because of the central
b p = 0.0352, evidence that claim is justified.
limit theorem.
20 a 0.2, 0.3, 0, 0.1, 0.4, 0.1, 0.1, 0.3, 0.2, 0
26 a          
b (0.507, 12.8) b 0.127

c Yes – the confidence interval suggests that the c H0: μD = 0, H1: μD > 0, belief supported
average increase is greater than zero.
(p = 0.00151).

572 A nswers

21 Yes ( p = 0.00921) 7 a X  37 b X  34
22 X < 59.1 or X > 60.9 8 a X  22 b X  21

23 a 144, 122.4 9 a 0.155, do not reject H0

b Sufficient evidence to support Celine’s belief b 0.287, do not reject H0

( p = 0.00949 < 0.1) 10 a 0.0986, reject H0 b 0.0430, reject H0

c Homework times distributed normally, with 11 a 0.156, do not reject H0
equal variances.
b 0.201, do not reject H0
24 a Evidence to support claim (p = 0.0188)
12 a 0.0185, reject H0 b 0.00987, reject H0
b No; large sample, so CLT means that the
distribution of sample mean is approximately 13 a X 6 b X 12
14 a X  4 b X5
normal. 15 a X 19 b X  29

25 a Weights before and after follow normal 16 a X  3 b X4
distributions with equal variances.

b 0.359 17 a Reject H0 (p = 0.0169)
b Do not reject H0 (p = 0.0649)
c 6, 16, 0, 3, 6, 1
18 a Reject H0 (p = 0.0631)
d Neither; the change in weight needs to be b Reject H0 (p = 0.0189)
normally distributed.
19 a Reject H0 (p = 0.0482)
e Fewer assumptions; eliminates differences b Reject H0 (p = 0.00163)
between individual mice.
20 a Reject H0 (p = 0.00251)
f 0.0368 b Do not reject H0 (p = 0.0362)

g Paired test; there is evidence that the weight is 21 Not supported (p = 0.0659)

increased after taking the drug.

26 a 89 b N (0, 89) 22 Justified (p = 0.0761)

c Insufficient evidence that the mean is positive 23 Insufficient evidence (p = 0.0707)

( p = 0.174). 24 a Po(47)

d There is insufficient evidence to support b Sufficient evidence (p = 0.0420)
Johannes’s belief.

27 a H0: μD = 0, H1: μD < 0 25 a Po(2.4)

b D ~ N (0, 1.4792), D ~ N (0, 0.1233) b Insufficient evidence of increase (p = 0.0959)
c D < −0.702
26 a 0.900

Exercise 9F b H0: ρ = 0, H1: ρ > 0

c 0.0185

1 a 0.110, do not reject H0 d Sufficient evidence of positive correlation

b 0.169, do not reject H0 27 Sufficient evidence of positive correlation
(p = 0.0447)
2 a 0.0321, reject H0
28 a H0: ρ = 0, H1: ρ < 0
b 0.0478, reject H0
b Sufficient evidence of negative correlation
3 a 0.0221, do not reject H0 (p = 0.0415)

b 0.00635, reject H0 29 a H0: ρ = 0.38, H1: ρ > 0.38
b X 88
4 a 0.0393, reject H0

b 0.0603, do not reject H0 c Insufficient evidence that the support has
 5 a X 23 increased.
b X 22
6 a X  10 b X  10

A nswers 573

30 Insufficient evidence of correlation (p = 0.0884) 16 a Type II – they are judging the dice to be fair

31 a 0.820 (not rejecting H0) when in fact they are biased
(H0 is false).
b Insufficient evidence of correlation (p = 0.0239) 1
6
c Sufficient evidence of positive correlation b H0: p = c 0.0649
(p = 0.0120)
1
32 a H0: λ = 1, H1: λ < 1, where λ is the mean H1: p < 6

number of busses in a ten-minute interval. 17 a 0.05 b 0.114

b X2 c X  22 2018 a H0: λ = b 0.0343
33 a H0: λ = 6.3, H1: λ > 6.3 b X 42 20H1: λ > b 0.756

34 a At least 5% b At least 6.5% 19 a 5%

35 a H0: λ = 1.7, H1: λ < 1.7 c Increase the sample size.

b P(X = 0) = 0.183 > 0.05; H0 will never be 20 a i Claiming the proportion that germinate is
rejected.
lower than 80% when it isn’t.
c 19 d X  10 e3
ii Claiming the proportion that germinate
36 a H0: p = 0.01, H1: p > 0.01, where X ~ B(10, p) ;
isn’t lower than 80% when it is.
sufficient evidence of increase (p = 0.00427) b 0.0867
c 0.786
b Sufficient evidence of increase (p = 0.00468)

c e.g. 0.45% d Test more seeds, increase the limit of less than

Exercise 9G 14 to reject the distributor’s claim.

1 a 0.05 b 0.1 21 a H0: λ = 2, H1: λ > 2 c 56.8%

b 4.87%

2 a 0.1 b 0.01 22 a No change in chance of Type I error.

3 a 0.0119 b 0.0331 b Reduces 2the chance of Type II error.
23 a H0: p = 3
4 a 0.136 b 0.0477 b 0.619

5 a 0.0181 b 0.0592 H1: p > 2
6 a 0.0548 b 0.0781 3
7 a 0.238 b 0.202 24 a H0: λ = 18.4 b 0.183

8 a 0.0909 b 0.0527 H1: λ < 18.4
9 a 0.829 b 0.581
10 a 0.538 b 0.589 25 a 17 b 3
56 14
11 a 0.304 b 0.485
12 a 0.618 b 0.633 Chapter 9 Mixed Practice
13 a 0.731 b 0.735
14 a 0.450 b 0.285 1 a 25.3 cm2 b 121 < μ < 126

H15 a 0: defendant is innocent. 2 a 7 and 8
H1: defendant is guilty.
b Insufficient evidence of dependence
(p = 0.173)

b Finding an innocent person guilty. c Not valid; the test is looking for evidence of
dependence, not independence.
c Finding a guilty person innocent.
3 a Qingqing’s model: 0.983 > 0.870

b 7.36

574 A nswers

4 a p = 0.0918; evidence of a decrease d R2 = 0.979, which suggests the quadratic

b It is still 3.7 minutes. The training may have model is a good fit.
decreased differences between individual
16 a Model BA6.::0S1SS,Srwreessh==ic5h6..1s0u71ggests
students. b
Model
c Repeat the test with a different sample. Model B is a
5.17 <
5 a 73.2 g, 134 g2 b (66.5, 79.9) better fit.

6 (85.8, 90.6) 17 a (21 200, 27 800)

7 a H0: μ1 = μ2, H1: μ1 ≠ μ2 b No, cannot reject H0.
b 0.197
c The population of wages is normal.
c Insufficient evidence that the means are
different 18 a (31.3, 53.5) b (25.4, 31.4)

8 a H0: ρ = 0, H1: ρ < 0 c Confidence intervals barely overlap, so not
b Both are normal. very reliable.

c 0.0312 d Increase sample size.

d Significant evidence of negative correlation 19 a H0: μ = 2.7; H1: μ ≠ 2.7

e e.g. The season has a greater effect on b x < 2.53 or x > 2.87
temperature.
c Evidence that the average height is different.

9 a 0.119 d Yes; X is approximately normal by CLT.

b H0: λ = 17, H1: λ > 17 20 a 7.09

c Insufficient evidence of increase (p = 0.119) b His0n:oPtoaisssuointabisleamsuoidtaebll.e model, H1: Poisson

10 a e.g. Students may not answer honestly. c 6.95, 7.90, 11.2, 13.2, 13.4, 11.9, 9.35, 6.63, 9.45

b The expected frequency in the bottom left cell d7
is less than 5.
e Suff i2c.i8e9nt×e1v0i−d4e)n. cSeugtogeresjtescPtoHis0so(nχ 2 = 27.3,
c They are not adjacent values. p=
is not a
d p = 0.0740
suitable model.
11 H0: p = 0.5, H1: p > 0.5
P(X 19) = 0.100 > 0.05 21 a HX0: p2=6 0.72, H1: p > 0.72
b
Insufficient evidence c 0.0495

12 Insufficient evidence of linear correlation 22 a 24, 32, 26, 23 b 21.5 g to 31.0 g
(p = 0.0578)
c The differences in weights are normally
13 a 9.761 to 9.825
distributed.

b If a large number of intervals are formed in 23 a X 2 b 0.0824 c 0.857
this way, 99% of them will contain μ.
24 a H0: a normal distribution is a suitable model.
14 a H0: ρ = 0, H1: ρ > 0 H1: a normal distribution is not a suitable
b 0.853 model.

c 0.00173; sufficient evidence of positive b x = 15.3, sn−1 = 4.99
correlation (e.g. at 1% SL)
c 0.61, 3.69, 8.65, 30.11, 20.96, 11.08, 13.51,
d y = 1.78x + 40.5 2.24, 0.15

e 74.3 d Some expected frequencies are < 5; ν = 2.

15 a H0: ρ0.=0507,4H1: ρ > 0 e Evidence that normal is not a good model

p= (χ2 = 11.2, p = 0.00370).

25 a e.g. Students may not answer honestly.

b Only shows insufficient evidence of a linear b H0: p= 0.6, H1: p < 0.6
relationship. c 23
X
c P = −0.202n2 + 1.72n + 1.79
d 0.283 e 0.989

A nswers 575

26 a H0: p = =0.01,.0H618:4p>>00.0.15 c Insufficient evidence of improvement (p = 0.312)

p-value 34 a X  243

HDproobnaobtirleitjyecttrain0s; insufficient evidence that b Reto (0.0461 vs 0.0437)
are late is greater than 0.1.
35 a H0: μ = 1.2, H1: μ < 1.2
b 0.0170 c 0.854 d 0.146 b i 3.45% (accept between 3.45% and 5.13%)

27 b H0: X ∼ B(6, 0.4), H1: X does not follow B(6, 0.4) ii 0.257
χ2 = 6.06, p = 0.194; claim consistent with data
36 a 0.0835 b 0.105
28 a H0: μ = 2.5, H1: μ ≠ 2.5
b x < 2.45 or x > 2.55 37 a X  4.71 b 0.361 c 0.109

c 0.228 Chapter 10 Prior
Knowledge
29 a i x < 2.863
f 21 ′′(x) =
ii Rejecting H0 when it is true. 2 y − 1 = 3(x − 1)
iii Accepting H0 when it is false. 3 x = ln5 ≈ 1.61
iv 0.05 4 (2, −16), (−2, 16)

v 0.0877

b i t-test

ii Do not reject H0, p = 0.0509. Exercise 10A

iii 2.719 < μ < 3.001 1 a f′(x) = 2 1 b f′(x) = 3 1
30 a p = −0.000964 y3 + 0.111y2 − 3.43y + 151 3 x−3 4 x−4

b p = −0.000140m3 + 0.0628m2 − 6.72m + 332 3 7
c Using R2, Nathan’s model accounts for 76.7%
f 212 a ′(x) = − x−2 f 43b ′(x) = − x−3
of the variability in p whereas Marc’s only 1 f 23
3 a f′(x) = 9 x 2
accounts for 68.0% . b ′(x) = x−4

d Form a model that depends on both age and

weight. f 65 f 47

31 a T = 4.21Q2 − 44.9Q + 656 4 a ′(x) = − x−3 b ′(x) = − x−5

b R2 = 0.973 ii $5600 f1 b f′(x) = 3 3
c i $630 2 x−4
5 a ′(x) = x−2
10d Interpolation for Q = with good fit of model f6
40so reliable. Extrapolation for Q = so treat f 44
b ′(x) = x−5
6 a ′(x) = x−3

with caution. 47 35

e T = 0.257Q2 + 31.7Q + 192 3 437 a − x− 3 − x− 3 23b −x−2 + x−2
3x14 3 2x43 2
f Different number of data points in each model 8 a − 1 x−4 b + 1 x− 3
20 21
g RRJI22ruilniaan = 0.973 41 51
= 0.935
23 29 a − x− 3 − x−3 45 6b − x−4 + x−4
So, Irina’s model is better fit.
10 a f′(x) = 3cos x b f′(x) = −4sin x
h R2 is better for the cubic model (0.993 cf. 0.973)

but the extra parameter makes the comparison 1
2
invalid. 11 a f′(x) = − sin x − 5cos x

32 a HH10:: μ = 6.4 b 10% 3
μ ≠ 6.4 4
b f′(x) = cos x + 2 sin x

33 a Test B; Test A also measures the difference 2 1
cos2 cos2
between the two groups. 12 a f′(x) = b f′(x) = 1 −

b 0.494 x x

576 Answers

13 a dy 3 b dy 4 7 a f′(x) = 3sin(2 − 3x)
dx = x dx = − x
b f′(x) = − 1 sin ⎛1 x + 4⎞
dy 2 dy 1 2 ⎝2 ⎠
14 a dx = x b dx = − x
8 a 3e3x 1 e x
15 a y ′ = 5ex b y ′ = −6ex 3 e9 a − x2 −x3 b 2
2

y ex y 3ex b 8xe4x2
2 4
16 a ′ = − b ′ = 10 a 1 b 1
x x
17 0.5 2x 8x
11 a x2 + 1 b − 3 − 4x2
18 ; falcon is descending.

19 (0, 1) 12 a y′ = 6sin3x cos3x
b y′ = −8cos4x sin4x
20 5 3x π 33
2 4 2 13 a y′ = −2sin 2xecos2x
21 y = − + + b y′ = 5cos5xesin5x

22 y = x + 3 − 3ln3 1 1 1 2
2x 2 3x 3
23 (9, 3) 14 a y ′ = (ln 3x)− b y ′ = (ln 2 x)−

24 a y = 2x − 5 15 −e−x
2
1 23
25 y − tan k = cos2 k (x − k) 16 4x 9
17 y= 5 5
+

26 a 1 million b 9 hours 18 (3, 0)

c 1.5 million per hour d No upper limit π2x π
4 2
27 x > 0.5 19 y = −

28 0.5 π2 π2 y x 5 ln3
4 4 3 3
29 a b 20 = − +

30 (4, 2) 21 a 2 sin x cos x b π , π, 3π
25 x = 3 2 2
⎛13π
31 ⎝6 , 0⎞ 26 b No, also true for e.g. y = 3e2x
⎠

32 0.5

Exercise 10B Exercise 10C

1 a dy 12(3x 2)3 b dy 10(2x 7)4 1 a dy = x cos x + sin x
dx dx dx
= + = − 1 1
2 2
dy 3 1 dy 4 1 b dy x cos x 1 x − sin x
dx 2 4 dx 3 3 dx 2
x (x2 3)− x (4 x 2 )− = +

2 a = + b = − −

3 a dy 1 (2x 3 x)− 1 (6x 1) 2 a dy = −x2 sin x + 2x cos x
dx 2 2 dx
= + + dy
dx
b dy 1 (5x x 3 )− 2 (5 3x2) b = x− −1 sin x − x −2 cos x
dx 3 3
= − −

4 a f′(x) = 2cos2x f′(x) 1 cos 1 x 3 a dy x − 1 e x 1 x − 3 e x
3 3 dx 2 2 2
b = = −

5 a f′(x) = −π sin πx b f′(x) = −5sin5x b dy = x3ex + 3x2ex
dx
6 a f′(x) = 3cos(3x + 1)

b f′(x) = −4cos(1 − 4x)

A nswers 577

4 a dy x − 1 2 x − 1 ln x 22 a = 3, b = 4
dx 3 3 3 23 a = 0, b = −2
= +

b dy = x3 + 4x3 ln x 24 a 12en ; derivative > 0 for all n
dx (en + 3)2
11 1)12
5a f′(x) = x (2 x + (2x + b i 1000 rabbits
+ )−2

f′(x) 9 x2 (3x 4)− 1 2 x (3x 4)23 ii 750 rabbits per year
2 2
b = − + − c4

6 a f′(x) = 3x2 cos3x + 2xsin3x dP

b f′(x) = 2x 3 cos 2x + 3 1 sin 2x
4 4 x−4

7 a f′(x) = −4x sin(4x − 1) + cos(4x − 1) 4

b f′(x) = 5− x−2 sin5x − 2x−3 cos5x

f 4 e 12 e8 a 1 1
′(x) = x2 4x+5 + x−2 4x+5

f e 2 eb ′(x) = −x−2 1−x − x−3 1−x

9 a f′(x) = 2 x −1 − x−2 ln(2x − 3) 1
2x − 3

b f′(x) = − 5 x3 x + 3x2 ln (5 − x) n

−

10 a y′ = x2 + 4x b y ′ = −12 25 b y = x
(x + 2)2 (x − 4)2

y′ 1 x+2 y x −5 Exercise 10D
2 + 1)23 (x − 3)3
11 a = (x b ′ = 1 a  b4

ex (3x − 5) ex (2x − 7) 2 a 12 3 b 7e
(3x + 1)3 (2x − 1)4 3 a 4
12 a y′ = b y′ = − per second b   

4 4x 4 a −2 per hour b − 3 per hour
π π2 2
13 y = − 5  

2 1 6  
y π x 2 15
14 = − 7 16

15 7e6 8 20 cm2 s−1

16 ex ((x + 1) ln x + 1) 9 22.6 mm2 per day

17 exsinx (sin x + x cos x) 10 1.92 cm s–1

18 (ln2, 8 − 11ln2) 11 a 10 cm2 s–1 b 0.8 cm s–1
5
⎛1, 1⎞ 12 9π cm s–1
e⎠
19 ⎝ 13 7.65cm

20 (e, e) 14 2.68 m s–1

21 k 15 Increasing
(k + x2)1.5
16 0.2 ≈ 0.0894 m s−1
5

578 A nswers

Exercise 10E 17 e.g. y

1 a 10 b 12 x

2 a 62 b8

3 a 4e−4 b 18e6
1
4a −4 b1

5 a i Increasing, concave-down

ii Concave-up

iii Concave-down

b i Decreasing, concave-up

ii Increasing

iii Decreasing, concave-down
4
6a x < 3 b x > −3 18 a k = 3

7 a All x b No values b It is a point of inflection.

8 a Max: (1, 11); min: (2, 10) 19  
b Max: (−2, 20); min: (2, −12)
9 a Min: (−3, −84); max: (0, −3); min: (3, −84) 20 b = −2, c = 3
21 b = −4, c = 1
⎛1 5⎞ 22 (1, −1), local min
b Min: ⎝2 , 8⎠ 1
− (a + x)2
⎛1 , 4⎞ b Min: (0.25, 1) 25
⎝9 3⎠
10 a Max:

11 a Min: (ln5, 5 − 5ln5) 26 a ⎛1 , − 1 ⎞ b e− 3
⎝e e2 ⎠ 2

b Max: (− ln 2, 1 (−1 − ln 2) 27 a x 5 b ⎛5 , 124⎞
2 3 ⎝3 27 ⎠
< −

12 a Max: ⎛3 , ln ⎛27⎞ − 3⎞ 1 − ln x ⎛e, 1⎞
⎝2 8⎝ ⎠ e⎠
⎠ 28 a x2 b ⎝

b Min: (4, 2 − ln 4) d Local max

13 2 1 , 229 p⎟⎠⎞ 1 , 2Min:⎛ p⎠⎞⎟
Max: ⎛ −
14 a = 1, b = −1 ⎝⎜ p
2 e15 (x + ) x ⎜⎝− p
16  x
π , e30 ⎛ ⎞, local max ⎛3π , 1 ⎞, local min
y 2⎝ ⎝2 e
⎠ ⎠

31 Max: ⎛π , 3 − π⎞ Min: ⎛5π , − 3 − 5π⎞
⎝3 2 6⎠ ⎝3 2 6⎠
x
32 Min: (0, 0) and max:(−2, −4)

33 a 4000 b $39 000

A nswers 579

34 4a3 cm3 10 
27
1 , 1e35 11 b 10 c 600 cm3
⎛ − p ⎠⎞⎟; max if p < 0 , min if p > 0
12 y = x b (6 + 3 2)cm
⎜⎝− p 13 2e2x (2x2 + 4x + 1) b (4, −94)
b (2, 2e−2)
36 t = 1, dV 2 14 a 4.5 cm2
dt =
15 6e5
37 250
16 a 12
⎧f(x) > a(1 − aln a) (if a > 0)
17 ex ((x + 1)sin x + x cos x)
f x 038⎪ (if a = 0) 18 a = 1, b = 8, c = 1.5
⎨ ( )> (if a < 0)
f x⎪ 19 a x < 2
( ) ∈
⎩

39 (0, 3) local max ⎛ − π ⎞
4
(ln 2, ln16) local min π , e20 ⎜ ⎟
4 2⎜⎝
40 1.53 2 m ⎠⎟

41 a  b b > 12 21 0 5 2b ⎛ + ⎞ million m3
e⎝ ⎠
42 a 6 22 a 5 million m3

43 5x2 + 10x + 2 y c 2 hours after the storm

44 a 23 0.577 mg1−1

24 ds = 25 (constant)
dt
R 1, 125 b ⎛−
P Q Px 9⎝ ⎞ c 1 < k < 1
R ⎠ 9

27 a 100 b 34.3
51
c 0 < f(x) < 100
e5

28 b No

 29 a −2 x 2

f 0d No; ′(x) ≠ at these points 30 a 0.5 b ±2

31 0.243ms−1

Chapter 10 Mixed Practice Chapter 11 Prior Knowledge

1 b = −4, c = 7 1 y = 1 x3 + 2x + 4
2 2xex 3
2
3 (x + 2)3 2 48.4

xcos x − sin x 3 dy = 2e2x + cos x
dx
4 x2 1
f 34 )− 2
5 1 + ln x ′(x) = x (x2 +

6 0.25 Exercise 11A

7 y = x −1 3 x 5 c 4 x 7 c
x 5 3 7 4
8 y 1− 2 1 a + b +
=
1 1
9 259.2cm3 s−1 2 a 2x 2 c b −3 x − 3 c

+ +

580 A nswers

3 a 4x 5 + c b 4x 5 + c 30 − 3 cos (2 x) − 2 sin (3x) + c
2 4 2 3
1
4 a 12 x 3 + c b 5x 3 + c 31 1 (x2 + 1)6 + c
5 6

5 a 3 x 4 + c b 5 x 6 + c 1 cos (3x2) + c
8 3 18 5 6

3 1 32 −
4 2
6 a 8x + c b 14x + c 33 3 x2 + 2 + c

7 a y = 3sin x + c b y = − sin x + c 13
b y = − 12 cos x + c
8 a y = 2cos x + c 34 − 2(2x + 3)(+ln2x)2
35 f(x) = +3
9 a y = 5ex + c b y = − 4 ex + c 1
3
1 36 2 ln3 + 5
10 a y = 2 ln x + c b y = 2 ln x + c kx3
3
1 (2 x 1)23 c 3 (1 2x)43 c 37 − cos4 x + +x+c
3 8
11 a + + b − − + 1
3
2 5x)21 2 7)43 38 ex + e−3 x +c
5 3
12 a − (3 − + c b (2x − + c 39 13(x2 + 1)1.5 + c
40 a 1 − 2sin2 x
13 a − 1 sin (2 − 3x) + c b 1 sin (4 x + 3) + c 41 − ln cos x + c 1 x 1 sin 2 x c
3 4 2 4
⎛1 b − +
⎝2 1
14 a −2 cos x − 5⎞ + c b 2 cos(5 − 2 x) + c
⎠

15 a 1 e5 x + 2 + c b − 1 e1− 3 x + c 42 −cos x + 1 cos3 x +c
5 3 3

16 a 1 ln 4 x − 5 + c b − 1 ln 3 − 2x + c 43 ln ln x + c
4 2
2)52 Exercise 11B
17 a 1 (x2 + 4)4 + c b 2 (x 4 − + c
4 5
1 1 1 a 14 b 12
18 a 4 sin4 x c b 3 cos3 x x 3
+ + 33 1
5 −6
19 a ln x3 + 2x + c b ln x4 − 5x + c 2 a b

20 a 2 (x3 + 4)23 + c b − 3 (1 − x2)43 + c 3 a 26 b −60
9 8 3
1
21 a −2e−x2 + c b 3ex3 + c 4 a 2 b1

22 a 1 x3 + +c b 1 ln x4 − 3 + c 5a2 b5
3 4
6 a ln5 ln 3
23 y = 2x1.5 − 42 b 2

24 2sin x + 3cos x + 2 7 a 0.774 b −0.152
25 y = 2(ex − e) − 5ln x
3 8 a 1.12 b 3.02
x2 2 ln x +c 4 4
26 + 9 a 3 b 3

27 V = 1 t2 + 12 cost + 3 10 a 27 b 5
2 2 2 2
28 x = 5t − 2et + 7
1 11 a 1.63 b 9.77
14 (5 2 x)7 +c 12 a 2.83 b 1.83
29 − −

A nswers 581

13 112 Exercise 11C
9
26 45
14 3 1 a 3 b 4

15 3.29 2 a 1.83 b 0.848

16 25 3 a 5.10 b 8.77

17 a (−3, 0), (3, 0) b 36 4 a 2.48 b 0.527

18 a 8 , − 5 b 37 5 a 1370 b 230
3 12 12
6 a 91.7 b 512
19 11.0
7 a 11.8 b 33.0
52
20 81 8 a 3.14 b 7.07

21 −5ln3 9 a 101 b 133
, 223 a ⎛π
4 2⎝ ⎞ b 2− 2 10 a 12.6 b 45.7
⎠ 11 a 3.59 b 0.771
12 a 93.2 b 48.7
24 a y

2 y = 2 − 2sin x( 2) 13 a ln5 b 4 π
5
14 3

1 15 a = ±2

16 a   b 18π
32 512π
17 a 3 b 15

√2π x 18 a 2.43 b 7.75
19 b 5.25
y = cos x( 2) c i 44.1 ii 30.1
20 2π
b 0.0701 b 0.115
7 21 3646
3 b1
26 22 π2
2
3 (1 e−2 23 a y
27 2 − )

28 1 ln ⎛11⎞
2 5⎝ ⎠

29 ln 2⎛ ⎞ y =x
3⎝ ⎠

30 a = 6

31 a = 2 or 8
−3
1⎞
32 a A (0, 1), B ⎛1, e⎠

⎝

33 27

34 17.5 0 x

35 a ln x + 1 b 127

36 −2 c 153

582 Answers

24 a A (0, 2), B(2, 0) b 4.39 c 32.7 7 a 1 ln 3+ 4x +c b − ln 2 − x + c
20 4
25 3 π 1 e(x3) 1
3 5
32 π 8 a + c b sin (x5) + c
5
26 1
6
27 a a − 1 9 a ln 1 + 3x 2 +c b ln 1 + sin x + c

28 a hx + ry = rh 10 a sin2 x + c b − ln cos x + c
x y r29 a 2 + 2 = 2 2
3 2 1)52 2 1)23
30 a (0, 0) and (1, 1) b 10 π 11 5 (x + − 3 (x + + c

31 π2 − 2π 8π 12 arcsin(x) + c
4 3
32 a (0, 0) and (2, 4) b
Chapter 11 Mixed Practice
33 1 : 3 π
b 2 (e2 + 8e − 1) 41 a3 − a
34 a y = ln(x − 2)

35 a y 2 y = 1 x +2
2
3−1
3 2

4 3 ln x + 1 + c
4 2x

1 y = cos x 5 19.0

  6 π2 ≈ 2.47
4
1 x 7
3
7

8 6 ln ⎛2 x− 5⎞

⎝ 3 ⎠

9 2x 3 4 x − 1 c
2 2
+ +

10 y = sin3 x + 2

c 3π2 11 a y

Exercise 11D

1 a 1 e2x + c b 1 sin (5x) + c 1 y = cos x
2 5
1 1
2 a − 5 ln 2 − 5x b 4 e1+ 4 x + c

3 a ln 1 + ex + b 1 ln 1+ x3 +
3
4 a 13(x2 + 2)23 + c 2 x)23 x
5 a −2cos x + c b 3 (sin + c 2

b sin(ln x) + c b A (0, 1) , B ⎛π , 0⎞ c y = − 2 x +1
⎝2 ⎠ π

6 a − 1 cos (4 x) + c b 1 e3x + c d 1 − π
4 3 4

A nswers 583

12 4 Exercise 12A
3
13 a A (0, 1), B(2, 5), C(7, 0) b 17.2 1 a v = 3t2 + 6t b v = 4t3 − 5
a = 6t + 6 a = 12t2
14 a ln11 b 1.52 c 5.33 1 t
2 a v = −2sin 2t 4 2
15 π (e4 − 1) a = −4cos2t b v = − sin

16 π π 3 a v = 3e3t 1 t
4 a = 9e3t 4 2
18 a π tan a b ≈ 0.786 a = − sin

19 1 ln17 b v = −4e−4t
2
5π a = 16e−4t
20 3
v 1 v 1
21 a y = 0.000545x3 − 0.0582x2 + 1.69x + 10 4 a = b =
t t
1 1
b 74.4 l a = − t2 a = − t2

22 6 b a < x < b and x > c 5 a 3s5 − 3s2 b 2s3 + 3s2 + s
1
x d23 a 0 < < 6 a eb − −2s
c 15 − s3
7 a 45 b 124
24 a hx + (a − b) y = ah 4
5
25 a y 8a4 b

9 a 33 b3
5 b ln136
r2 y = r2 x2 10 a ln 3
 r y = r2 x2
11 a 1.21 b 0.935
12 a 1.27 b 0.984
13 a 0.672 b 2.53
14 a 1.54 b 0.795
15 a 0.657 b 1.435
r r x b 4 m s c 
16 a 16 m

4 17 2 − 2e−1.5 ≈ 1.55m
3
b r = 18 a  b 176 ≈ 58.7 m
3
26 a a − 1 b a ln a − a + 1 19 23 m s
20 a 256 m s
Chapter 12 b 
Prior Knowledge
c 4s d 819.2 m

21 e2 + ln16 − 7 ≈ 3.16

3e1 a − −3t 1 cos⎛2t π⎞ 22 a 0 s, 6 s b 4s c 4s
2 ⎝ 6⎠
b − 23 a sint + t cost b 0 m s

2 1 c 4 + π m s−1 d 2 m s
4 2
− t2

3 1.5t2 1 cos(3t) c e 0.556 s, 1.57 s, 5.10 s f 13.3 m
3
+ + 26 a  b 0.5 s, 61.25 m

45 c 4s d 62.5 m

5 60.3° e e.g. No obstructions to path of ball

584 A nswers

27 a 18 m s b 3s 10 a 60.3° b 3.61ms−2
d 108 m
c  f 36 m 11 a 0.249ms−1 43ee tb ⎛− −⎞ c 0.249m
e  ⎜
⎝− − t ⎟
28 10 m s ⎠

29 8 m s 12 a 53.9° b 11.9ms−1 c 3.03ms−2
13 a 16.8ms−1 b 3.61ms−2
30 0.5 m 14 a 9.8m s−2 downwards b 20m s−1
b 44.6 m s c 11.3 m s c 256 + (12 − 9.8t)2; 16ms−1
31 a 10 m s

32 a 0 m b 50 m
0
33 a 6.06 m s b 7 94.915 a
ii ⎛t ⎞ b 12.9 m

c 71 ≈ 2.96 m s−1 ⎜ t− t 2 ⎟
24 ⎝ ⎠

v A t2 +u s A t3 + ut 16 a i 26.6° above the horizontal
34 a i 6 ii 25.6° below the horizontal
= 2 =

35 6 s b 1.02 seconds

36 0.938 m ⎛ 20t ⎞ 5.10m
− 4.9t2 ⎟⎠
v37 = 1 s4 + s2 + k c ⎜⎝10t ;
2
⎛ 12t ⎞
Exercise 12B 17 a ⎝⎜ 9t − 4.9t2 ⎠⎟ b 22.0m

a 46 , r 21 a ⎛ t2 + t⎞ 18 a 5m
⎛⎞ = ⎜
⎝ t − t3 ⎟
= ⎜⎝ − t ⎟⎠ ⎠ ⎛ ⎛t⎞⎞

a 162 , r 32 3b⎛ t⎞ ⎛ t3 − t ⎞ 19 a 6sin⎜ − ⎝ ⎠⎟ b 6ms−1
= ⎜⎝ =⎜ 2v = ⎜
⎠⎟ t3 + t ⎟ ⎛t⎞ ⎟
⎝ ⎠ ⎜ ⎟

4et 4et 6cos 2⎜⎝ ⎝ ⎠ ⎠⎟
3et 3et
2 a a = ⎛ ,⎞ r = ⎛ ⎞ d 3ms−2
⎜ ⎜ ⎟ 20 a 0.9ms−1
⎝ ⎟ ⎝ ⎠
⎠

a 35ee , r 35eeb
⎛ t⎞ ⎛ t⎞ 0.3 sin 3t
−0.3cos3t
=⎜ t ⎟ =⎜ t ⎟ b ⎛ ⎞⎠⎟; r = 0.3 m
⎝− ⎠ ⎝− ⎠ ⎜⎝

a 4188ee , r 32ee3 a⎛ 4t ⎞ ⎛ 4t ⎞ ⎛ −2.7sin3t ⎞⎟⎠ ;
=⎜ ⎝⎜ 2.7cos3t
−3t ⎟ 3= ⎜ − t ⎟ c k = −9
⎝− ⎠ ⎝− ⎠

a 505ee , r 25eeb⎛− −t ⎞ ⎛ − −t ⎞ d Towards the origin ⎛ 0.5 ⎞
=⎜ 0⎜⎝ ⎠⎟
5t ⎟ =⎜ 5t ⎟ 21 a 0.5m b
⎝ ⎠ ⎝ ⎠

4 a a = ⎛ −6 sin 2t ,⎞ r = ⎛ 1.5 sin 2t ⎞ c 1.57 seconds d v ⎛ −2 sin 4t ⎞
⎝⎜ 8 cos 2t ⎜⎝ −2cos2t ⎠⎟ ⎝⎜ 2 cos 4t ⎟⎠
⎟⎠ =

b a = ⎛ −25cos5t ,⎞ r = ⎛ cos5t ⎞ 22 a ⎛ 4t − 0.75t2 ⎞ b 20.2 m
⎜⎝ −50sin5t ⎝⎜ 2 sin 5t ⎠⎟
⎟⎠ ⎜ 0.1t3 + 5t ⎟
⎝ ⎠
5 a 14.2 ms−1 b 57.0ms−1
6 a 8.24ms−1 b 19.4ms−1 v 34 , r 1.2523 a ⎛ t2 ⎞
⎛ t⎞ =⎜
7 a 59.5ms−1 b 18100ms−1 = ⎜⎝ − t ⎠⎟ ⎝− t2 ⎟
8 a 3.98ms−1 b 8.26ms−1 ⎠

9 a ⎛1.5cos3t⎞ b 4.20ms−1 b 4 seconds
4⎜⎝ ⎟⎠
c         
straight line

A nswers 585

y 5 10 15 20 3 a v(t)
5
−5 x

− 10 23π π 43π t
− 15 π3 5

− 20

Hint: You can find the equation of the line −5

expressing y in terms of x, where x = 1.5t2 and b 1.05s, 2.09s, 3.14s, 4.19s
y t= −2 2 . c 16.3m

v 13 2 , r 3 0.524 a
⎛ − t⎞ ⎛ t − t2 ⎞ 704 a
= ⎜⎝ =⎜
+ t ⎟⎠ ⎝ t+ t2 ⎟ ⎛⎞ b 7cm c 21cm s−1
⎠ ⎜⎝ ⎠⎟

b 49.4° 5 a 21.3ms−1 ⎛ 16 ⎠⎟⎞;
c         ⎜⎝ −5.6 down
   b

y c ⎛ 16t ⎞ d 33.1m
⎜ 14t 4.9t 2 ⎟
⎝ − ⎠

10 6 a 2cms−1 31b⎛⎞
⎜⎝ − ⎠⎟

7 a −31.7cms−2 b 7.41cm

8 a v(t)

5 25

5 5x 20 maximum
15

25 4.7 ms−2 10
5
27 18.2 ⎛ 20(t − 1) ⎞

28 a ⎛ 12t ⎞ b ⎜ b(t 1) 4.9 (t 1)2 ⎟
⎜⎝30t − 4.9t2⎠⎟ ⎜⎝ ⎠⎟
− − −

c 36.9 d 53.6m 5 10 15 20 25 t

29 a 12002 + (1390 − 98t)2 9
b i d = ∫(15 t − 3t)dt
b 14.2 seconds

0
ii 149m
Chapter 12 Mixed Practice 9 a −9 m s−2
b 18.5 m
1 v = −10.3ms−1; a = −3.10ms−2
2 a −0.147 ms−2 b 1.96m 4 1.45e10 a⎛ t2 ⎞
⎜ b 6.51 m
⎝ −0.5t ⎟
− ⎠

586 Answers

11 a r = ⎛ −3cost ⎞ c 4m b 0  t < 0.785 and t > 2.55
⎝⎜ 4 sin t ⎠⎟
c 0  t < 0.785, 2 < t < 2.55 and t > 3
d 73.7°
d xA 1 t4 5 t3 3t2
2⎛ ⎞ = 4 − 3 +
⎝⎜ 17 − 9.8t ⎠⎟
12 a e vB = −20e−2t f t = 4.41

⎛ 2t ⎟⎞ ; 3.47 20 a They meet when t = 3.
− 4.9t2
b ⎜⎝17t ⎠ seconds b 37.7°

c (2t)2 + (17t − 4.9t2)2 21 a v = 8 − 9.8t, s = 8t − 4.9t2

b 1.17 seconds

d 15.2 m 22 a v = ⎛ 8 ,⎞ r = ⎛ 5t 8t ⎞
− 9.8t ⎜⎝ − 4.9t2 ⎟⎠
⎛⎞ ⎛ −8sin 2t ⎞ 5⎝⎜ ⎠⎟
⎜⎝ 20cos4t ⎠⎟
0413 a ⎝⎜ ⎠⎟ b b 0.260 seconds

c 21.5 m s d e.g. When t = 0.785 23 a r = ⎛ 8 sin 2t ⎞ b 2.17seconds,
⎝⎜ −8 cos 2t ⎟⎠ 10.9cm
e t = 3.14
24 a i −10m s−2 ii −100ms−1
14 a y

y = 2sin x + 1 b dt 1 v 2 98ed = − − −5(t−10)
dv = −10 − 5v
98
e s = −2t + 5 e−5(t−10) + 500.4

1 f t = 250

2 x Chapter 13
Prior Knowledge

b 2.20ms−1 c 6.28m d 9.02m 1 1 e2x + c
1 e15 −s(s − ) −2s 2
16 4s
17 17.6ms−1 22

3 x2

4 1035

18 a Displacement = A, acceleration = B

b t=3 Exercise 13A

v t19 a ( ) db kb da k a
dt dt
1 a = b =

(0.785, 2.11) 2 a dr k b dh k
dt = r2 dt = h

3 a dF = −k F b dv k
dt dt = − v
dR kR (N − R)
4 a dI = kI (N − I) b dt = t
dt
0 23 t x3
(2.55, −0.631) 5 a y = + c b y =8 x+c
3

6 a y = e2t + c b y = 3lnt + c
2

A nswers 587

7 a s = − cos3x + c b s = 2 tant + c 1a y
3
1 3
8 a F = m − 3ln m + c b F = m + 3m + c 2
1
9 a y = Ae2x b y = Ae− y 0
−1
10 a y = Aex − 1 b y = Ae−x + 1

11 a y = − c 3 b y= 2 x
c − x4
12 a y = Ax x+ 3 b y = x2 + c

13 y = − x 1 c −2

+

14 y = 3 3sin x + c −3−3 −2 −1 0 1 2 3

15 y = ln(x2 + c) by
16 y = Ax 3

17 y = − 1 c
x2 +
18 a y = Aetan x b y = 4etan x 2

19 2 y2 = 3x3 + 18 1
9
20 y = 3 2 (x2 + 2) 0x

21 a y = 1 + Aex2+4x t −1
5
23 a k = 1 b m = 25e−
5
−2
c 3.47 seconds

24 a 1 b 24 000 −3−3 −2 −1 0 1 2 3
4 b 40.8 m

25 b V = 6000t + 90000 2a y
26 a v = 100 − 100e−0.1t
27 y = −2 2 − e−2x 3

28 y = − ln(c − ex) 2
1
29 y = 2 ln(4ex − 3)

y30 = 102 − 2cos x 1

31 b π 0x
±2
dr
32 a dt = −0.0328 b 15 minutes −1

Exercise 13B −2
−3−3 −2 −1 0 1 2 3
The slope fields shown in these answers are produced
using technology. You would only be expected to
sketch three or four trajectories to indicate behaviour.

588 Answers

by 4 a 2.49 b 1.4
3
2 5 a 2.98 b 0.655
1
0 6 a 1.36 b 1.46
−1
−2 7 a 4.56 b 3.82

8 a x = 0.224, y = −0.186
b x = −1.50, y = −3.16
9 a x = 0.271, y = −0.697
x b x = −1.33, y = −0.458

10 a x = 1.41, y = 0.969
b x = 0.667, y = 0.133

11 a i 0.9 ii 3.8

−3−3 −2 −1 0 1 2 3 b i 0.8 ii 3.6

c y = x2, b ii is furthest away.

3a y 12 a y

3 3

22

11

0 x0 x

−1 −1

−2 −2

−3−3 −2 −1 0 1 2 3 −3−2 −1 0 1 2 3 4

by c Solution is constant (or in equilibrium).
3 13 a 31.4

b Use a smaller step length.

14 157 rabbits, 37 foxes

2 15 x(0.25) ≈ 8.12, y(0.25) = 9.46

1

0x

−1

−2

−3−3 −2 −1 0 1 2 3

A nswers 589

16 a, b y 19 a 1.46 m b 3 seconds

3 20 a 0.825 x2 1
2 2
2 (− 2, 2) b y = ln− ⎝⎜⎛− + + e−0.3⎞⎟⎠
1
c i 11.0% ii Take smaller h.

21 a i 0.615 ii Use smaller step

0 x 1 e13x3; length.
123 2
−1 b y = f(1) = 0.698
x
−2 cy
5
−3−3 −2 −1 0 x y = e21 13 x3 12
c 1.629 1234
17 y 34 x

2 The tangent is always below the curve.

1 22 183 m
0
−1 23 177 m y
−2
−3 24 a

−4 5
−5
−60 4
18 a y

2

1 3

2

0

−1 1

−20 1 2 0−2 −1 0 1 2 3 x

b 1.1 b 3.8

590 A nswers

25 a y 4a y

5 4

4 3

2

3 1

2 0x

−1

1 −2

0−1 0 1 2 3 4x −3

b 3.1 −4−4 −3 −2 −1 0 1 2 3 4

Exercise 13C by
4
1 a ⎛ x ⎟⎠⎞ = A e2t ⎛ 2 ⎟⎞⎠ + Be3t ⎛1⎞ 3
⎜⎝ y ⎜⎝ 1 ⎜⎝ 1⎟⎠ 2
1
b ⎛ x ⎞ A e2t ⎛ 3 ⎞⎟⎠ + Be4t ⎛ 1⎞ 0x
⎜⎝ y ⎟⎠ = ⎜⎝ −5 ⎜⎝ −1 ⎟⎠ −1
−2
2 a ⎛ x ⎞ A e−t ⎛ 2 ⎞⎠⎟+ Be−11t ⎛ 4⎞ −3
⎜⎝ y ⎟⎠ = ⎝⎜ 1 ⎜⎝ −3 ⎟⎠ −4−4 −3 −2 −1 0 1 2 3 4

b ⎛ x ⎠⎟⎞ = A e−t ⎛ 1 ⎠⎟⎞ + Be−2t ⎛ 4 ⎞
⎜⎝ y ⎜⎝ 1 ⎜⎝ 3 ⎠⎟

3 a ⎛ x ⎞ A e3t ⎛ 5 ⎞⎟⎠ + Be−t ⎛1⎞
⎜⎝ y ⎠⎟= ⎜⎝ 1 ⎜⎝ 1⎟⎠
y
⎛ x ⎞ ⎛ −5 ⎠⎟⎞ + ⎛ −3 ⎞ 5a
b ⎝⎜ y ⎟⎠ = A e2t ⎜⎝ 1 Be−5t ⎝⎜ 2 ⎠⎟
4

3

2

1

0x

−1

−2

−3

−4− 4 −3 −2 −1 0 1 2 3 4

A nswers 591

by 7a y
4
3 4
2
1 3
0x
−1 2
−2
−3 1
−4−4 −3 −2 −1 0 1 2 3 4
0x

−1

−2

−3

−4−4 −3 −2 −1 0 1 2 3 4

6a y by
4
4 3
2
3 1
0x
2 −1
−2
1 −3
−4−4 −3 −2 −1 0 1 2 3 4
0x

−1

−2

−3

−4−4 −3 −2 −1 0 1 2 3 4

by 8a y
4
3 4
2
1 3
0x
−1 2
−2
−3 1
−4−4 −3 −2 −1 0 1 2 3 4
0x

−1

−2

−3

−4−4 −3 −2 −1 0 1 2 3 4

592 Answers

by 10 a i
4
3 x
50

2 40
1

0 x 30

−1 20
−2

−3 10

−4−4 −3 −2 −1 0 1 2 3 4 0 0 0.5 1 1.5 2 2.5 t

9a y ii y

4

3 20

2
1 16

0 x 12
−1

−2 8

−3 4
−4−4 −3 −2 −1 0 1 2 3 4
00 10 20 30 40 50 x
by
4 bix
3 200
2

1 150
0x

−1 100
−2

−3 50
−4−4 −3 −2 −1 0 1 2 3 4

00 0.5 1 1.5 2 2.5 t