answers
CHAPTER 1 Modelling using
linear functions
Exercise 1A — Solving linear equations
1 a 2 b −8 c −3 d 7
e −8 f 6 g 2 h −11
i − j
2 a 7 b −6 c 5 d −11
e −1 f 18 g − h −
3 a 9 b −4 c −7 d 4
e −1 f −5 g 4 h 3
4 a 10 b 6 c −9 d −12
e 14 f −18
5 a 6 b −2 c 5 d −5
e −13 f 4 g 11 h −3
6 a 12 b −5 c 7 d 7
e −9 f
7 a 3 b −4 c 5 d 8
e −9 f −7
Exercise 1B — Rearrangement and
substitution
1 a P = A − L b c
d e f
g
h or
i j k
l m n
o p
q or
r s t
2 a 56 b 30 c 80 d 16.97
e 33.33 f 0.267 g 350 h 7
i 13 100 j 2.498
3 a , 7.746 b , 6.204
c , 59.161 d , 4.167
e or , 17.108
f , 3.976 g , 10.247
h , 10.75 i , 2622
j , 4.706
4 a 42 cm b or c40 mm
5 a 10 N b c 10.77 m/s2 6 a 240 m2
b or c 18 cm
7 a $1123.60
b c 41.4%
8 a 153° b 17.17 cm
9 a 60.25 cm b 6 cm
10 a b c 150 cm
Exercise 1C — Gradient of a straight
line
1 a 2 b 5 c d −
e −4 f −1 g h −
2 a 2 b 5 c −4 d
e −6 f g h 2
i 0 j − k −2 l
3 a b −1 c d
e 1 f −12 g 0 h Undefined
4 a b 2 c d
e f − g −2 h 0
5 a 1 b c d −
e −6 f − g 12 h
6 a 1.192 b 3.078 c 0.176 d −0.577
e −0.577 f 0 g 1 h 57.290
7 a 0.93 b 2.61 c −0.53 d −3.73
8 a D b C c A d B
9 a B b E c D d D
10 e, b, a, c, d
Answers
7 3
--- 13
6
---31 3
--- 7
2
---76 59
---l A
w
----= t d
v ---=
r C
2π
---= R 100A
PT
---= r 3V
πh
---3
=
I3 I4R4+I2R2–I1R2–E R3
---=
α R1 R2 ---–1
θ
---= α R1–R2
R2θ ---=
β E–αθ θ2
---= r kQq
F
---= φ = Et---n
V2 V1N2 N1
---= n pV
RT
---= a 2(s–ut)
t2
---=
v 2Fd+mu2 m
---= r µI2
2πF ---=
U V f1 f2 ---–V
= U V f1–V f2
f2 ---=
γ v2 rT
---= w S–2lh
2(l+h)
---= H S–2πr2
2πr ---=
l = A r 3V
4π
---3
=
v mg–F k
---= a v–u
t ---=
h S
πr ---–r
= h S–πr2
πr ---=
l g T 2π --- 2
= d = l2–4fl
V H–U P
---= c (1–α)K
α2
---=
u H0v Hi ---=
w P
2 ---–l
= w P–2l
2 ---=
a F
m ----=
a 2A h
---= –b a 2A–bh
h ---=
r 100 A
D ----–1
100 A– D
D
---
= =
f uv
u+v
---= u fv
v–f ---=
1 4
--- 1
3 ---16
5
--- 20
9 ---1 2 ---1
3
--- 1
8 ---1
2
--- 5
2 ---1
2
--- 1
4
--- 8
5
---3 4
--- 1
2
--- 7
6 ---1
4
--- 9
2 ---4 3
--- 1
5
--- 11
5 ---3
2
--- 5
8
---1A
➔
answers
11 a b
c
12 13 14
15 a 4 b 31 c −5 d 3
16 a No b Yes c 224 cm
Exercise 1D — Equations of the
form y
=
mx
+
c
1
2 The higher the number, the steeper the graph. Positive values make the graph slope up when moving (or tracing) to the right; negative values make the graph slope down when moving to the right.
3 4
5 The number is where the graph cuts the y-axis (hence the name ‘y-intercept’).
6 a 5 b 6 c −9 d 2
e −8 f 1 g −1 h 5
i 3 j 0 k 0 l 0
7 a 7 b −4 c 1 d −13
e −5 f 2 g −10 h 0
i 0 j 17 k 2 l 0
8 a y = 2x + 7 b y =−3x + 1 c y = 5x − 2
d y = 3 e y = x f y = x − 5
g y = x + h y =− x − i y =−2x + 12
j y = x − 3
9 a 3, 9 b 7, −42 c −4, 12 d −5, −35
e 3, 10 f −6, 24 g −16, −15h −9, −1
i 1, −23 j 4, −18
10 a 4, 5 b 4, −8 c 4, −6 d −3, 1
e −2, 4 f 3, 11 g −7, −9 h 2, 5
i − , −3 j − , −6 k − , l − ,
m , n − , −
11 C 12 E
13 y =−7x + • 14 y = • x − 6
15 3y + 5x = • 16 3y + • x = 17
17 a y = 4x + 2 b y = −3x − 5 c y = x − 2
d y =− x + 5 e y = 2x − 1 f y = −5x
18 a y = 10.7x b 84.7°
19 a a, b b − , c − , − d 2k, −3h
Exercise 1E — Sketching linear graphs
using intercepts
1
2
y
x
y
x
y
x
2 17
--- –17
300
--- 2
25
---1 2 ---2
3 --- 1
3
--- 3
4 --- 1
2 ---5
2
---1 2
--- 2
11
--- 8
3 --- 2
3
--- 3
4 --- 13
4 ---1
6 --- 5
2
--- 5
2 --- 7
2
---a b
c d
e f
g h
a b
c d
e f
g h
5 ---5
6
---a b --- c
b
--- a
b --- c
b
---y
x –3
18
y
x
–21 7
y
x 12
12 — 5
–
y
x
–3
3 – 2
–
y
x 10
2
y
1 1
y
x 30
10 — 3
y
8
–16
y
x 2
3
y
x 4
5
y
x –2
5 – 4
y
x –3
6
y
x
–7 5
y
x –4
1 – 2
y
x –2
2
y
x
11 — 6
answers
3
4
5
Exercise 1F — Simultaneous equations
12
3
4 15 cents and 35 cents
5 22 and 19
6 16 emus, 41 sheep
7 Basketballs $9.45, cricket balls $3.05
8 Limousine $225 (sedan $75)
Exercise 1G — Formula for finding the
equation of a straight line
1 a i 3x − y − 1 = 0 ii y = 3x − 1
b i 2x − y − 4 = 0 ii y = 2x − 4
c i 5x − y − 19 = 0 ii y = 5x − 19
d i 4x − y + 11 = 0 ii y = 4x + 11
e i x + y − 1 = 0 ii y = −x + 1
f i 3x + y + 5 = 0 ii y = −3x − 5
g i x − 2y + 7 = 0 ii y = x +
h i 4x + 3y − 36 = 0 ii y =− x + 12
i i 4x − 5y + 19 = 0 ii y = x +
j i x + 6y − 60 = 0 ii y =− x + 10
k i 8x − 7y + 60 = 0 ii y = x +
l i 3x + 11y − 33 = 0 ii y =− x + 3
2 a i x − 2y − 1 = 0 ii y = x −
b i x − y = 0 ii y = x
c i x + 2y − 12 = 0 ii y = − x + 6
d i 3x + 2y − 2 = 0 ii y =− x + 1
e i 3x + y + 7 = 0 ii y = −3x − 7
f i x + y − 4 = 0 ii y = −x + 4
g i 14x − 3y + 2 = 0 ii y = x +
h i 3x − 4y − 12 = 0 ii y = x − 3
i i 4x + 7y + 42 = 0 ii y = – x − 6
j i x + y − 1 = 0 ii y = −x + 1
8
9 y =− x −
a b
c d
e
a b
c
D 6 E 7 A
a b
c d
e f
g h
x
–6 –7
y
x 10
–4
y
x 4
16 — 3
–
y
x 2
–6
y
x 9
–2
y
x
(1, –1)
y
x (1, 1)
y
x
(1, –2)
i j
a (1, 4) b (−2, 6) c (−4, −15)d (3, −15)
e (−7, −5) f (3, 3) g ( , ) h ( , )
i ( , − )j (13, −3)
a (7, 9) b (−6, 5) c (6, 5) d (10, 1)
e (1, −2) f ( , ) g ( , ) h (− , − )
i ( , ) j ( , )
9 A 10 D
3 A 4 C 5 y = x − 6 6 y = 3x − 23 7 C
a y = x − b y =− x + 3
c y = 2x − 3 d y =− x −
5 9 --- 17
9
--- 12 5 --- 32
5 ---23
14 --- 20
7
---10 3 --- 4
3
--- 1
2 --- 19
2
--- 9
10 --- 3
5 ---59
8 --- 21
8
--- 84 67 --- 99
67
---1 2 --- 7
2 ---4 3 ---4 5 --- 19
5 ---1 6 ---8 7 --- 60
7 ---3 11 ---1 2 --- 1
2
---1 2 ---3 2
---14 3 --- 2
3 ---3 4 ---4 7
---2 5 --- 43
5
--- 1
2 ---3 4 --- 9
2 ---5
2 --- 3
2
---
1D
➔
answers
10
Exercise 1H — Linear modelling
1 a C = 2.5 + 5tb c $40
2 a C = 60 + 8m
b c $100
3 a P = 32 + 0.1n
b c $197
6 a Opus $24, Belecom $20 b After 14 minutes
7 a PinkCabs $28.50, NoTop $26
b After 6.7 km (6 km)
8 After 4 rides
9 6 visits
10 Savus would be cheaper for up to 9 days hire.
Chapter review
1 D 2 A 3 −2 4 6 5 D
6 D 7 A 8 or
9 C 10 D 11 B 12 B
13 a b − c d −
14 − 15 4.331 16 Undefined 17 B
18 A 19 D
20 a 3, −7 b , 10 c , −2
21 y = x − 3 22 B 23 E
24
25 E 26 E
27
28 ( , −5) 29 ( , )
30 21 two dollar and 46 one dollar coins
31 B 32 C
33 y = −x + 4 34 y = x + 35 D
36 a C = 75 + 65t
b c $302.50
37 No, the points are not co-linear. This may be shown by calculating gradients or equations for lines joining different pairs of points.
38 (−3, −4), (−1, 8), (3, 4)
39 a C = 250 + 55j b 13 jumps
c This is open to question.
CHAPTER 2 Relations and
functions
Exercise 2A — Set notation
12
3
4 E
5 6
Exercise 2B — Relations and graphs
5
= + =− +
c y =− +
11 94 12 C = 22n + 280 13 H = 22 + 6t
4 $960 5 Yes ($410 compared to $450).
a b
7 7
x 8 --- 39
4
---1 2 35
25 30 Cost ($)
Time (h)
1 2 76
60 68 Cost ($)
Time (min)
10 20 34
32 33 Payment ($)
Number of leaflets
2 3
---T 4π2R3 GM
---= 2πR R
GM
---3 4
--- 7
11
--- 5
11
--- 7
8 ---7
3
---5 –
3
--- 1
2 ---2
5
---y
x 24
8
y
x –40
5
c d
a b (−5, −5)
a ∅ b {4, 6} c {6}
d {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
e {4, 5, 6, 7, 8} f {2, 8, 10, 12, 14} g {4, 5}
a {2, 3} b {−3, −2, −1} c {−1}
d {−3, −2, −1, 0, 1}
e {−3, −2, −1, 0, 1, 2, 3, 4}
a ∅ b {b, c, d, f, g, h} c {a, e, i}
d {b, c, d, f, g, h} e {o, u}
a C b B
a T b F c T d F
e F f T g T h F
i T j F k F l T
1 B 2 A 3 E 4 C
a D b C c C
d D e D f C
y
x
–9 7
y
x
(1, –6)
5 3
--- –63
10 --- –33
5
---6 –
7 --- 34
7
---1 2 205
75 140 C ($)
answers
67
8 a
b
c The variables are discrete. 9 a
b i Approx. 110 km/h ii Approx. 320 km/h 10 a
b c The variables are
discrete.
Exercise 2C — Domain and range
1
2
3
4
7
8
9 a
Domain = (−∞, ∞), Range = (−∞, 2] b
Domain = [−2, 2], Range = [−7, 9]
a b
c d
e f
a b
c Because the variables are continuous d Approx. 11 minutes
n 0 1 2 3 4 5 6
P($) 300 340 380 420 460 500 540
n 15 16 17 18 19 20 21 22 23 24 25
C($) 140 146 152 158 164 170 176 182 188 194 200
70
60
Day
Cost (¢)
M T W T F S S (D)
y
x
0 2 3
9
1 4
1 (D)
y
x
0 2
–1
–2 1 –1
–3
–4 –2
(D)
y
x
0 2
–2
y = x – 2
(C)
y
x
0 2
7
1 3 4 5 6
2
1 –1
–2 –2
(D)
y
x
0 2
6
1 –1
4 2
–2
(C)
T (°C)
t (minutes) 0
80 70 60 50 40 30 20 10
2 4 6 8
T (°C)
t (minutes) 0
80 70 60 50 40 30 20 10
2 4 6 8 10
P ($)
n 550
500 450 400 350 300 250 200
1
0 2 3 4 5 6
V (km/h)
t (s) 0
350 300 250 200 150 100 50
1 2 3 4 5
a [−2, ∞) b (−∞, 5)
c (−3, 4] d (−8, 9)
e (−∞, −1] f (1, ∞)
g (−5, −2] ∪ [3, ∞) h (−3, 1) ∪ (2, 4]
a b
c d
e f
g h
i j
k l
a [−4, 2) b (−3, 1]
c d
e (3, ∞) f (−∞, −3]
g (−∞, ∞) h [0, ∞)
i (−∞, 1) ∪ (1, ∞) j (−∞, −2) ∪ (−2, ∞) k (−∞, 2) ∪ (3, ∞) l (−∞, −2] ∪ [0, ∞)
a E b D
5 C 6 B
a i {3, 4, 5, 6, 7} ii {8, 10, 12, 14, 16} b i {1.1, 1.3, 1.5, 1.7} ii {1.4, 1.6, 1.8, 2} c i {3, 4, 5, 6} ii {110, 130, 150, 170} d i {M, T, W, Th, F} ii {25, 30, 35}
e i {3, 4, 5} ii {13, 18, 23}
f i R ii [−1, ∞)
a R, R b R, (0, ∞) c [−2, 2], [0, 2] d [1, ∞), R e R, (0, 4] f R, (−∞, −3] g R\{0}, R\{0} h R, (−∞, 1] i R, R
C ($)
n 190
200 180 170 160 150 140 130
5
0 10 15 20 25
2 0
–6 –9 –3 0
2
0 0 5
10 1
0 0 2 7
1 3 0
–2 –8 0 2 6
4 1
0 –10 5
0 2 –2 0 1
1, 3) –
( –1
2
--- 1
2 ---,
y
x 2 0 2
2 –
y = x3 + 1
x ∈ [–2, 2] y
x 0
9
1
–7
–2 2
1H
➔
answers
c Domain = (−∞, ∞), Range = [− , ∞)
d Domain = [−2, 1],
Range = [−4, 0]
e Domain = [−1, 4),
Range = [−7, 3)
f Domain = (−∞, ∞),
Range = [−6 , ∞)
10
Investigation — Interesting relations
1 2
x2+ 2y2 = 9 x3+ y3 = 1
3 4
sin (x2+ y2) = 1 x2 – y2 = 1
5 6
7x2 – 6 xy + 13y2= 16 x4 = x2 – y2
7 8
x2 + y2< 25 x2 + y2> 25
9 10
9 < x2 + y2< 36 x sin x + y sin y < 1
Exercise 2D — Types of relations
(including functions)
1
2 b, c, d, e, f, h, i, j, k, l 3 C
4 5
Exercise 2E — Function notation and
special types of function
1
2
3
4 a, c, d, f, g, h
5 i a, b, c, d, f, h, i, j, k,l ii c, h, i, k 6
7 a b (−∞, 0) ∪ [1, ∞)
a R b [0, ∞) c [−4, 4]
d R e R\{0} f R
y = x2 + 3x + 2
y
x 0 2
–2 –1
1 4
---y = x2 – 4, x ∈ [–2, 1]
y
x 0
–2 –1
–3
–4 1
y = 2x – 5, x ∈ [–1, 4]
2 y
x
0 3 4
–1
–5 3
–7 1
y = 2x2 – x – 6
2 y
x 0
–1 –2
–6 1
1 8
---–4 –4
4 y
x 4
–2 –2
2
2 y
x
y
x 10 –10
–10 10
–10 –10
10
10 y
x
–2 –2
2
2 y
x
–1 –1
1
1 y
x
3
a One-to-many b Many-to-one
c Many-to-one d One-to-one
e One-to-one f Many-to-one
g Many-to-many h Many-to-one
i One-to-one j Many-to-one
k Many-to-many l Many-to-one
a E b D c B
b {−3, −1, 0, 1, 2}, {−2, −1, 1, 3}
c {3, 4, 5, 6}, {−1} e R, {2}
g R, R j [−1, ∞), [0, ∞)
k R, R
a i 1 ii 7 iii –5 iv 16
b i 2 ii 1 iii 3 iv 0
c i 3 ii 2 iii 6 iv 9
d i 9 ii 1 iii 16 iv a2+ 6a + 9
e i 12 ii 6 iii −4 iv 2
a 3 b −3 or 3 c
d 2 or 3 e −4 or 1 f −1
a 3 b 3 c
d e f
a B b C
–10 –10
10
10 y
x
–10 –10
10
10 y
x
–10 –10
10
10 y
x
–10 –10
10
10 y
x
1 3
---5 x ---–2x 10
x2
---–x2 10
x+3
---–x–3 10 x–1 ---–x+1
y
x 0 2 2 f(x) 1
answers
8 a
b [1, ∞)
c i 3 ii 1 iii 2
9 a
b (−∞, 0] ∪ (4, ∞)
c i −5 ii 0 iii −3 iv 0 v 7
10
11 f: [0, 1] → R, f(x) = with range [0, 1] or f: [0, 1] → R, f(x) = – with range [−1, 0].
Investigation — A special relation
2
Exercise 2F — Circles
1 2 3 4 5 6 1x y x2 y2 x2+y2
On the graph of x2+y2= 25?
0 4 3 0 3 4 7 −4 −5 −4 1 3 9 4 0 6 −3 −2 5 −3 5 2 0 8 4 3 7 −3 0 3 5 −4 0 −3 −5 −6 4 −5 0 −4 0 16 9 0 9 16 49 16 25 16 1 9 81 16 0 36 9 4 25 9 25 4 0 64 16 9 49 9 0 9 25 16 0 9 25 36 16 25 0 16 25 20 9 64 25 25 98 25 25 25 26 25 81 25 25 72 25 29 25 25 Yes No No No Yes Yes No Yes Yes Yes No Yes No Yes Yes No Yes No Yes Yes y x 0 2 3 g(x) 2 1 1 –1 –2 2 3 y x 0 –1 –2 –3 –5 –4 5 4 3 2 1 1
f x( ) x+2, 2x+1,
= x≤0
x>0 1–x2
1–x2
–8 –6 –4 –2 –10 –5 10 8 6 4 2 10 5 y x
a x2+ y2= 9 b x2+ y2= 1 c x2+ y2= 25 d x2+ y2= 100 e x2+ y2= 6 f x2+ y2= 8
g h
a Both [−3, 3] b Both [−1, 1]
c Both [−5, 5] d Both [−10, 10]
e Both [− , ] f Both [−2 , 2 ]
g [−3, 3], [0, 3] h [−4, 4], [−4, 0]
a b c d e f a b c d e f g h
a D b B
a C b E
y = 9–x2 y = – 16–x2
6 6 2 2
y x 0 2 2 –2 –2 y x 0 4 4 –4 –4 y x 0 7 7 –7 –7 y x 0 7 – 7 7 – 7 y x 0
–2 3 2 3
–2 3 2 3
y
x 0
–1– 2
1 – 2
–1– 2 1 – 2 y x 0 3 3 –3 –3 y x 0 2 –2 y x 0 1 –1 y x 0 1 – 3
–1–
3 1–3
y
x 0
–1– 2
answers
7 a b
[−1, 1] and [−3, −1] [−2, 2] and [0, 4]
c d
[1, 7] and [−3, 3] [−2, 6] and [–5, 3]
e f
[−8, 2] and [−7, 3] [0, 6] and [−1, 5]
g h
[−11, 1] and [−2, 10] [−1, 2] and [−3, 0]
8 ; Domain [−6, 6] and range [0, 6] or ; Domain [−6, 6] and range [−6, 0]
9 ; Domain [−3, 3] and range [2, 5] or
; Domain [−3, 3] and range [−1, 2]
10 a 2 cm, 13.8 cm b 3.9 cm/s
Exercise 2G — Functions and modelling
1 a b
2 a
b
3 a
b Domain [0, 4]; range [0, 250]
c i 60 km ii 170 km
4 a
b c $90
5 a T = 0.34x − 3978
b
Domain [20 700, 38 000]; range [3060, 8942]
c $6902
6 a P = 4x + 6
b Domain (1, 6]; range (6, 30]
7 a A = x2+ 4x
b Domain (0, 8]; range (0, 96]
8 a P = 100 000(1.02)t
b $121 899
9 a 47
b 21
c 9 weeks
d No, as t increases approaches zero, so N approaches 15.
10 a T = 6000 + 100n − 50n2
b
c $11
Chapter review
1 A 2 D 3 B 4 C
5 a
b The number of cars is a discrete variable.
c 120 y
x 0 1
–2 –1
–3 –1
y
x
0 2
4
2
–2
y
x
0 1 4 7
3
–3
y
x
0 2 6
–2 3
–5 –1
y
x 0 2
–8 –3
3
–7 –2
y
x
0 6
3 5
2
–1
y
0 1 –5 –11
10
4
–2
y
x
0 2
–3 –1
–3– 2
1 – 2
y = 36–x2
y = – 36–x2
y = 2+ 9–x2
y = 2– 9–x2
C t( ) 40, 70, 110, 160,
=
0<t≤1 1<t≤2 2<t≤4 4<t≤6
70
60
Day
Cost (¢)
M T W T F S S
(D)
C d( )
0.40, 0.60, 0.80, 1.70, 2.00,
=
0<d≤50 50<d≤100 100<d≤200 200<d≤700 d>700
Cost ($)
Distance (km) 0
2.00
0.60 0.80 1.70
100 200 700
0.40
d t( )
60t, 90, 80t–70, =
0≤ ≤t 1.5 1.5≤ ≤t 2 2≤ ≤t 4
B n
12 ---=
B (hours)
n 0
10
5
60 120
8942
38 000 20 700
3060 T ($)
x ($)
96 t+3
---T
n 0
6000 5000 4000 3000 2000 1000
1 2 3 4 5 6 7 8 9 101112
No. of cars (
n
)
t (hours) 0
500
400
300
200
100
2 1 3 4 5
answers
6 a
b Domain = [−3, 3]; range = [−8, 1]
7 E 8 C 9 B 10 E 11 A
12 D 13 D
14 a x + 2, x ≥ 0
b Domain = [0, ∞); range = [2, ∞)
15 E 16 B 17 C 18 B 19 C
20 a, b, e
21 A 22 D 23 E 24 A
25 a f: R \ {0} → R, f(x) =
b f: (–∞, 2] → R, f(x) =
26
27 a
Domain = [−1, 1]; range = [−1, 0]
b
Domain = [−1, 5]; range = [−4, 2]
28 D 29 E 30 C 31 B
32 a
b f1: [−10, 10] → R, f(x) = with dom f = [−10, 10], ran f = [0, 10] and
f2: [−10, 10] R, f(x) = with dom f = [−10, 10], ran f = [−10, 0]
33 E
34
35 a A = xy + 10y − x b P = 2x + 2y + 20 or P = 2(x + y + 10)
c A = 260 + 16x − 2x2 d (0, 13)
e f 292 m2
CHAPTER 3 Other graphs and
modelling
Exercise 3A — The parabola (turning
point form)
y
x 0 2 3
–3 1
1 –1
–8 –3–2
y = 1 – x2
1 x
---2–x
2 3 4 5 y
x 0
–1 –2
5 4 3 2 1
1
y
x
0 1
–1
–1
y
x
0 5
2
–4 –1
(2, –1)
y
x
0 10
10
–10 –10
x2 + y2 = 100
100 ( –x2)
100 ( –x2)
–
0 100
25 50 75
1 2
Cost ($)
Number of truck loads
1 a Dilation by the factor of 2 in the y direction
b Dilation by the factor of in the y direction
c Dilation by the factor of 3 in the y direction, reflection in the x-axis
d Translation 6 units down
e Dilation by the factor of 3 in the y direction, translation of 4 units up
f Dilation by the factor of in the y direction, reflection in the x-axis, translation of 1 unit up
g Translation of 2 units to the right
h Reflection in the x-axis, translation of 3 units to the left
i Dilation by the factor of 2 in the y direction, translation of 3 units to the right
j Translation of 2 units to the left, translation of 1 unit down
k Translation of 0.5 unit to the right, translation of 2 units up
l Dilation by the factor of 2 in the y direction, reflection in the x-axis, translation of 3 units to the left, translation of 1 unit up
m Dilation by the factor of 12 in the y direction, translation of 1.5 units to the right, translation of 0.25 units down
n Dilation by the factor of in the y direction, reflection in the x-axis, translation of units to the right, translation of 2 units up
2 D
3 a (ii) b (v) c (i) d (iv) e (iii)
4 a i (0, 0) (ii) ii Domain: R, range: y ≤ 0
b i (0, − ) ii Domain: R, range: y ≥−
c i (0, 2) ii Domain: R, range: y ≤ 2
d i (6, 0) ii Domain: R, range: y ≥ 0
e i (−2, 0) ii Domain: R, range: y ≤ 0
f i (3, 0) ii Domain: R, range: y ≥ 0
g i ( , 0) ii Domain: R, range: y ≥ 0
h i (−3, −6) ii Domain: R, range: y ≥−6
i i (1, 1) ii Domain: R, range: y ≤ 1
j i ( , 0) ii Domain: R, range: y ≥ 0
k i (− , −5) ii Domain: R, range: y ≥−5
l i ( , 4) ii Domain: R, range: y ≤ 4
A (m2)
x (m) 0
292
130 260
2 4 6 8 10 12 14
1 3
---1 2
---9 2
---4 3
---1 2
--- 1
2
---1 2
---4 3 ---1 2 ---1 2
---
2G
➔
answers
5 a b
c d
e f
g h
i j
Exercise 3B — The cubic function of the
form y
=
a
(
x
-
h
)
3+
k
6 a y =− (x − 2)2+ 2 b y = 2(x + 1)2− 2
c y =−3(x − 1)2+ 3
d y = (x + 2)2− 4
7 E
8
9 a y = x2
b y =−x2 c y = (x − 2)2− 1 d y = 3x2− 2
e y =−(x + 3)2
10 a y = (x − 3)2− 4 b y =−2(x + 1)2+ 1 c y = (x + 3)2− 4
d y =− (x − 2)2+ 2
e y = 3(x − 1)2+ 6 f y =−4(x + 2)2+ 8
x
3
y
x
1 – 2 1 – 2
–
– 4 1
–1 1
y
x
12
2
y
x
20
–3 2
y
x
2 1
–15 1–21 21–2
y
x
2.9
0.1 1
–8 1–
2
1
y
x
1 9
–– 3 4
y
x
4
– 3 2 –
3 4
y
x
11 5
2
y
x
1 2
---–5 13
–3 –4.6
–1.4
y
x
1 2
---1 3 ---1 2
---11 a z = 3 or z = 15
b y = 2(x − 3)2− 8 or y = (x − 15)2− 8
12 a 3
b y =− (x + 4)2+ 3 c x =−7, x =−1
13 1. f(x + 2) − 3, −4 ≤ x ≤ 0 2. f(x − 2) − 3, 0 ≤ x ≤ 4 3. f(x + 4), −6 ≤ x ≤−2 4. f(x − 4), 2 ≤ x ≤ 6 5. −f(x + 4) + 6, −6 ≤ x ≤−2 6. −f(x) + 6, −2 ≤ x ≤ 2 7. −f(x − 4) + 6, 2 ≤ x ≤ 6 8. −f(x + 2) + 9, −4 ≤ x ≤ 0 9. −f(x − 2) + 9, 0 ≤ x ≤ 4
1 a Dilation in the y direction by the factor of 7
b Dilation in the y direction by the factor of , reflection in the x-axis
c Translation by 4 units up
d Reflection in the x-axis, translation by 6 units up
e Translation by 1 unit to the right
f Reflection in the x-axis, translation by 3 units to the left
g Dilation in the y direction by the factor of 4, reflection in the y-axis, translation by 2 units to the right
h Dilation in the y direction by the factor of 6, reflection in the x-axis, reflection in the y-axis, translation by 7 units to the right
i Dilation in the y direction by the factor of 3, translation by 3 units to the left, translation by 2 units down
j Dilation in the y direction by the factor of , reflection in the x-axis, translation by 1 unit to the right, translation by 6 units up
k Dilation in the y direction by the factor of 2, translation by units to the left
l Dilation in the y direction by the factor of , reflection in the x-axis, translation by 8 units to the left, translation by 3 units up
2 a i, iv b iii, v c ii
d i, ii, iv e ii, v f iii, iv 3 a (0, 3) b (0, ) c (1, 0)
d (4, 0) e (−2, 4) f (1, −2)
g (2, 1) h (−3, −4) i (−4, 1)
j ( , )
4 E
5 C
6 B
2 25
---1 3
---2 3
---1 2
---5 2
---1 4
---1 2
---1 6 --- 2
---answers
7 a b
c d
e f
g h
i j
k l
Exercise 3C — The hyperbola
8 a y = x3 b
y =−(x + 5)3
c y = (x − 3)3− 1 d y = 2x3+ 3 e y =−(x + 1)3− 1
9 a y =− x3+ 4 b
y = 2(x − 1)3+ 2
c y =−3(x + 1)3+ 1 d y =− (x − 3)3 e y = 4(x + 1)3−
10 E
11 y = 2(x + 1)3− 4 12 a y =− (2 − x)3+ 1
b Positive cubic
y
x
0.8 1
y
x
2.08 –6
y
x
4
–128
y
x
2 4
y
x
1 4
y
x
1
y
x
–0.3 2 1
–5 –2 –0.6
3
y
x
–4 –6 –1 0.4
y
x
y
x
28 1
–2 ––23
y
x
2 245
3 3.6
y
x
3 35
4 5.8
1 2
---1 2
---1 3 ---1
2
---1 2
---13 a (−3, 1) or (−1, 27)
b
1 a Dilation in the y direction by the factor of 2
b Dilation in the y direction by the factor of 3, reflection in the x-axis
c Translation by 6 units to the right
d Dilation in the y direction by the factor of 2, translation by 4 units to the left
e Translation by 7 units up
f Dilation in the y direction by the factor of 2, translation by 5 units down
g Translation by 4 units to the left, translation by 3 units down
h Dilation in the y direction by the factor of 2, translation by 3 units to the right, translation by 6 units up
i Dilation in the y direction by the factor of 4, reflection in the x-axis, translation by 1 unit to the right, translation by 4 units down
2 a v b iii c i
d v, iii e v, ii, iii f i, iii g v, i, iv h ii, iv
3 a i x = 0, y = 0 ii Domain: R\{0}
iii Range: R\{0}
b i x =−6, y = 0 ii Domain: R\{−6}
iii Range: R\{0}
c i x = 2, y = 0 ii Domain: R\{2}
iii Range: R\{0}
d i x = 3, y = 0 ii Domain: R\{3}
iii Range: R\{0}
e i x = 0, y = 4 ii Domain: R\{0}
iii Range: R\{4}
f i x = 0, y =−5 ii Domain: R\{0}
iii Range: R\{−5}
g i x =−6, y =−2 ii Domain: R\{−6}
iii Range: R\{−2}
h i x = 2, y = 1 ii Domain: R\{2}
iii Range: R\{1}
i i x =−n, y =−m ii Domain: R\{−n}
iii Range: R\{−m}
4 a i x = 4, y = 0 ii Domain: R\{4}
iii Range: R\{0}
b i x = 0, y = 2 ii Domain: R\{0}
iii Range: R\{2}
c i x = 3, y = 2 ii Domain: R\{3}
iii Range: R\{2}
d i x =−1, y =−1 ii Domain: R\{−1}
iii Range: R\{−1}
y
x
28 27 1 –1 –4
–3
y = (x +1)3+27
y = (x +3)3+1
3B
➔
answers
6 a b
c d
e f
g h
i j
10 a b
c d
e
Exercise 3D — The square root function
e i x = m, y = n ii Domain: R\{m}iii Range: R\{n}
f i x = b, y = a ii Domain: R\{b}
iii Range: R\{a}
5
7 E
8 C
9 a y = b y =− + 1
c y =− d y =− − 1
e y = + 2 f y = − 1
y
x x
– 3 2
x
– 32
––34x
––x
3 4
x–3
x
–3
x–1 x
–1
–3
– 3 1
y
x
y
x
–2 –1 –1
–– 2 1
y
x
1 5 ––43
–3–43
y
x
–5 ––
5 2
y
x
–1 1 3
–3
y
x
2 2 6 7–21
– 2 1
y
x
2 1 1
– 2 1
y
x
–11 –1 4–
5 2
– 5 2
y
x
–1 – 4 4
– 3 1
– 2 3 – 8 5
y
x
–1 ––
3 1
– 41 –4
3
2 x–2
--- 3
x ---3
x+4
--- 4
x ---2
x–4
--- 6
x+1
---11 Domain: R\{0}, range: R\{3}
1 a Dilated in the y direction by the factor of 2
b Dilated in the y direction by the factor of , reflected in the x-axis
c Dilated in the y direction by the factor of 3, translated 1 unit to the right
d Dilated in the y direction by the factor of 2, reflected in the x-axis, translated 4 units to the left
e Translated 1 unit down
f Dilated in the y direction by the factor of 3, reflected in the x-axis, translated 2 units up
g Translated 4 units to the right, translated 3 units up
h Dilated in the y direction by the factor of 2, reflected in the x-axis, translated 3 units to the left, translated 6 units up
i Dilated in the y direction by the factor of , reflected in the x-axis, reflected in the y-axis, translated 2 units to the right and units up
2 a (0, 0) b (0, 0) c (1, 0)
d (−4, 0) e (0, −1) f (0, 2)
g (4, 3) h (−3, 6) i (2, )
3 E
4 D
y
x
–2
– 2 1
y
x
–1 1
y
x
–2 ––21
y
x
2 3
1 1– 2 1
y
x
–1 1
y
x
3
– 3 1
1 3
---1 2 ---2 3
---answers
8 a b
c d
e f
g h
i
10 a m = 1 b y =2 − 4
Exercise 3E — The absolute value
function
1 a b
c d
e f
g h
i
4 a b
c d
e f
5 a Domain: x ≥−1, range: y ≥ 0
b Domain: x ≥ 3, range: y ≥ 0
c Domain: x ≥ 0, range: y ≥−3
d Domain: x ≥ 0, range: y ≥ 4
e Domain: x ≥ 0, range: y ≤ 5
f Domain: x ≥ 1, range: y ≥ 3
g Domain: x ≥−2, range: y ≥−1
h Domain: x ≥− , range: y ≤ 4
i Domain: x ≥ , range: y ≤ 2
j Domain: x ≤ 3, range: y ≥−7
k Domain: x ≤ 2, range: y ≥ 6
l Domain: x ≤ 2, range: y ≤ 1
6 D
7 D
9 E
11 y =3 + 3
12 a p = 8 b y =−4 + 8
c x = 3 d x ≥−1
e y ≤ 8 f
1 2 ---4 3
---y
x
–2 1.4
y
x
3
y
x
2
4 (6, 1)
y
x
y
x
2 3.7
–3
y
x
–4
– 2 1
–1–21
–3– 4 3
y
x
– 23
y
x
2 4.4
–2
y
x
2 1 0.4
–1
x–1
y
x
9
(4, 3)
4–x
x+1
y
x
4 3 (–1, 8)
2 C
3 a Domain: R, range: y ≥ 0
b Domain: R, range: y ≥ 1
c Domain: R, range: y ≤ 4
d Domain: R, range: y ≥−2
e Domain: R\{−1}, range: y > 1
f Domain: R\{0}, range: y ≥ 0
y
x
y
x
1
1
y
x
– 2 1
3
y
x
6
6 6 –
y
x
4
2 –2
y
x
4 5
1 3 5
y
x
y
x
1 7
–2 –1
y
x
1 2
y
x
y
x
1 –5 –11 –1
–6
y
x
1 7
3
y
x
2
(–1, 1) (1, 1)
y
x
2 –2 – 2 2
2
y
x
0.7 –2.7
–1
–1 –2 –2
3D
➔
answers
g h
i j
k l
Exercise 3F — Addition of ordinates
3 a b
c d
e f
4
5
6 a
b
c
d
7 E
Exercise 3G — Modelling
5 a y = , −2 ≤ x ≤ 2b Yellow: y = 6 − , −2 ≤ x ≤ 2;
green: y = − 6, −2 ≤ x ≤ 2;
blue: y =− , −2 ≤ x ≤ 2
1 a R\{0} b [0, ∞) c [0, ∞)
d [−2, ∞) e R f R\{3}
g R\{−1} h (−∞, 1] i R\{0}
j [−1, 3]
2 C
y
x
–– 3 4
––43 – 3 4
y
x
– 3 1
3 3
6
y
x
––41
–4
– 41
y
x
–1 –1 1
y
x
3 3.6
5
2 –2
y
x
2
35 63 99 –5
(–1, –6)
3 2 ---x
3 2 ---x 3 2 ---x 3 2 ---x
y
x h(x)
f(x)
g(x)
y
x
h(x)
f(x)
g(x)
y
x
h(x)
f(x)
g(x)
y
x h(x)
f(x)
g(x)
h(x)
f(x)
g(x)
y
x
y
x
h(x)
f(x)
g(x)
1 a y = ax3, a = 0.3 b y = ax2, a =−6 c y = a , a = 1.6 d y = , a = 5
e y = ax3, a =−1.5
2 a iii b ii c i
d iv
3 D
y
x
8 7 6 5 4 3 2 1 –4 –5
–6 –3 –2 –10 1 2
f(x)
g(x)
x2 + 5x + 6
y
x y = x3 + x2 – 1
f(x)
g(x) 1 2
–2
–2 2
3 3
y
x
2 2
1
1 2 – x
y = y = x
x +
y = 2 – x
y
x
x y = – y = 2x
x y = 2x –
– 4 1
y
x
y = –x2
x – 3
y =
x – 3 – x2 y =
y
x
5 2
–5 –5 5 – x
y =
x + 5
y =
5 – x x + 5 +
y =
x a
---answers
4 a b
c a = 2, b =−3.2
6 a b
c f =
Chapter review
1 D 2 E
3 a (3, −4) b Domain: R, range: y ≥−4
c
4 C 5 D
6 y = 1 − 3(x − 1)3
7 C 8 E
9 a x =−2, y =−1
b Domain: R\{−2}, range: R\{−1}
c y = − 1
d
10 C 11 A 12 E 13 B 14 C
15 B
16 a
b
17 E 18 E 19 y =
5 y = x3− 12
7 a
b I =
8 y =3 + 4
9 a
b p =2 + 4
c 10.63, 10.93
y
x
1 2 3 4 5 –100
10 20 30 40 50
y
x2
5 10 15 20 25 –100
10 20 30 40 50
1 4
---f
2 4 6 8 10 0
200 400 600 800 1000
λ
f
0.511.522.53 3.5 0
200 400 600 800 1000
λ 1 —
340 λ
---I
d
2
1 3 4
0 50 100 150 200 250
270 d2
---x
$
Month 4
2 6 8 10
0 1 4 6 9 11 10
5 3 2 8 7
Price
m
y
x
–4 14
3 4.4 1.6
20 a i
A(−2, 0), B(0, −2), C (2, −3), D(4, −6)
ii
A(2, 0), B(0, 2), C(−2, 3), D(−4, 6)
iii
A(0, 0), B(2, 2), C(4, 3), D(6, 6) 4
– x+2
---y
x
–1 –3 –2 –6
y
x
(f + g) (x)
f(x)
g(x)
y
x
0 (f + g) (x)
f(x)
g(x)
100 x2
---y
x
A A'
B
B'
2 –2 –2
C(2, 3)
C'(2, –3) D(4, 6)
f(x)
f(x)
D'(4, –6)
y
x
A'
A B
2 –2
D(4, 6) D'(–4, 6)
f(x)
f(–x)
C(2, 3) C'(–2, 3)
y
x
–2
f(x)
f(x–2)
B'(2, 2) C'(4, 3) C(2, 3)
D'(6, 6) D (4, 6)
A'
A 2 B
3F
➔
answers
CHAPTER 4 Triangle
trigonometry
Investigation — Looking at the tangent
ratio
(Answers may vary slightly because of individual measurements.)
1 a 8 mm b 16 mm c 0.50
2 a 13.5 mm b 26 mm c 0.52
3 a 18 mm b 35 mm c 0.51
4 a 23 mm b 44 mm c 0.52
Investigation — Looking at the sine
ratio
(Answers may vary slightly because of individual measurements.)
1 a 8 mm b 17.5 mm c 0.46
2 a 13.5 mm b 29 mm c 0.47
3 a 18 mm b 38.5 mm c 0.47
4 a 23 mm b 49.5 mm c 0.46
Investigation — Looking at the cosine
ratio
(Answers may vary slightly because of individual measurements.)
1 a 16 mm b 17.5 mm c 0.91
2 a 26 mm b 29 mm c 0.90
3 a 35 mm b 38.5 mm c 0.91
4 a 44 mm b 49.5 mm c 0.89
Exercise 4A — Calculating
trigonometric ratios
1 a 1.540 b 17.663 c 40.460
d 0.657
2 a 0.602 b 2.092 c 15.246
d 51.893
3 a 0.707 b 0.247 c 6.568
d 5.896
4 a 0.5 b 0.9659 c 1
d 548.6 e 64 f 1.301
g 5.306 h 1.374 i 15.77
5 a 0.42 b 1.56 c 0.09
d 5.10 e 2.87 f 0.38
g 7.77 h 73.30 i 0.87
6 10°
7 a 44° b 80° c 57°
8 86°40′
9 a 42°57′ b 31°21′ c 16°5′
Exercise 4B — Finding an unknown
side
1 a
b
c
2 148.1 mm
3 5.08 m
4 30.0 cm
5 a 12.1 cm b 55.2 m c 9.43 km
6 a 12.5 m b 89.3 mm c 10.1 m
7 a 5.42 m b 1.35 km c 2.06 km
d 18.4 mm e 3.20 cm f 66.5 m
g 5.40 m h 5.39 km i 0.240 m
j 41.6 km k 82.4 m l 13.2 cm
8 D 9 E
10 6 m 11 4.2 m 12 20 km
13 a b 30.3 cm
iv
A(−2, 3), B(0, 5), C(2, 6), D(4, 9)
v
A(−2, 0), B(0, 4), C(2, 6), D(4, 12)
vi
A(−3, 1), B(−1, −1), C(1, −2), D(3, −5)
b Add multiples of 2, for example, f(x) + 2, f(x) + 4, f(x) + 6, f(x) − 2 etc. and keep the domain fixed at [−3, 7].
21 a y = a(x − h)2+ k b h = 9
c Straight line (negative gradient)
d a =−0.55, k = 275
e y =−0.55(x − 9)2+ 275
f No, the prices started going down.
g $266 000, $261 000
h About 4 months
y
x
f(x)
f(x) + 3 A'(–2, 3)
C' (2, 6) D'(4, 9)
C(2, 3) D(4, 6) B'
–2 2 5
A B
y
x
C'(2, 6)
D'(4, 12)
D(4, 6) C(2, 3) B'4
A'
2 B A
–2
2f(x)
f(x)
y
x
D'(3, –5) C'(1, –2) B'(–1, –1)
A'(–3, 1) A
B C(2, 3) D(4, 6)
f(x)
1 – f(x + 1)
2
hyp opp
adj
θ
hyp
opp adj
α
hyp opp adj
γ
24°
answers
14 a b 1.6 m
15 9.65 m
16 a b 58 m
c 15.5 m
Exercise 4C — Finding angles
1 a 30° b 75° c 81°
2 a 32°48′ b 45°3′ c 35°16′
3 a 53°8′ b 55°35′ c 45°27′
4 a 50° b 32° c 33°
d 21° e 81° f 34°
5 a 39°48′ b 80°59′ c 13°30′
d 79°6′ e 63°1′ f 19°28′
6 A 7 D 8 37°
9 75°31′ 10 8°38′ 11 7° 12 4°35′
Exercise 4D — Applications of
right-angled triangles
5 a 22.33 m b 13.27 m
7 a b 1319.36 m
8 22 m
9
10 11
14
15 201°48′ T
16 17 18 19 20
Investigation — Fly like a bird
a i 572 m ii 715 mb i 143 m ii 4.29 km/h
Exercise 4E — Using the sine rule to
find side lengths
1 a
b
c
2 a 14.8 cm b 1.98 km c 112 mm
3 a 10.0 m b 22.1 cm c 39.6 km
4 9.8 cm
5 26.9 m
6 37.8 m
7 a b 43.2 m c 33 m
8 43.62 m
9 10
11 22.09 km from A and 27.46 km from B.
Investigation — Bearing east and west
12 27.6 km 3 As for 1 4 27 km
Exercise 4F — Using the sine rule to
find angle sizes
1 a 43° b 34° c 27°
d 75° e 37° f 2°
2 B
3 B
4 38°
5 20°
6 84°
7 a 57° b 63°
8 54°
9 a 13.11 km b N20°47′W
Exercise 4G — The cosine rule
7 2218 m
8 9
12
16 1 571 m 2 30 m 3 91 m 4 43.18 m
6 2°44′
a 325° T b 227° T c 058° T d 163° T
a S66°W b S73°E c N39°W d N74°E
12 a C b D 13 1691 m
a 5.39 km b N21°48′W
a 4.36 km b 156°35′ T
a 12.2 km b 348 T or N12°W
a 29.82 km b 38.08 km c 232° T
a 112.76 km b 5 hours 30 minutes
a 82.08 m b 136.03 m c 301°6′ T 60°
1.4 m
15°
60 m
48°
35°
2500 m Helicopter
S1 S2
50 50 3 3 --- m –
a 6.97 m b 4 m
a 8.63 km b 6.48 km/h c 9.90 km
12 D 13 B 14 Yes, she needs 43 m altogether.
1 7.95 2 55.22 3 23.08, 41°53′, 23°7′ 4 28°57′
5 88°15′ 6 A = 61°15′, B = 40°, C = 78°45′
a 12.57 km b S35°1′E
a 35°6′ b 6.73 m2
10 23° 11 89.12 m
a 130 km b S22°12′E
13 28.5 km 14 74.3 km 15 70°49′
a 8.89 m b 76°59′ c x = 10.07 m
17 1.14 km/h 18 E 19 C 20 B a
sin A --- b
sin B --- c
sin C
---= =
x sin X --- y
sin Y --- z
sin Z
---= =
p sin P --- q
sin Q --- r
sin R
---= =
B
M N 20 m 49° 34°
N
12° 5°
W 8.1 E
4A
➔
answers
Chapter review
1 a 0.7193 b 4.2303 c 2.7400
d 8.1955 e 21.9845 f 14.2998
2 a 54° b 51° c 53°
3 a 78°31′ b 26°34′ c 14°54′
4 a 37.9 cm b 3.8 m c 13.6 cm
d 11.7 cm e 14.7 cm f 14.6 m
g 1.5 m h 4.7 cm i 15.6 mm
j 7.5 m k 10.7 m l 5.3 km
5 8.5 m 6 2.5 km 7 63.9 m
8 a 57° b 27° c 68°
9 a 23°4′ b 61°37′ c 59°35′
10 39° 11 24°
12 9.38 m
13 a 12.59 km b S36°10′E
14 2783 m
15 a 1.67 cm b 81.7 mm c 9.81 km
16 12.4 cm
17 a 52° b 21° c 68°
18 809 cm2
19 a 8.64 m b 8.80 m c 11.8 cm
20 84.0 cm
21 985 m
CHAPTER 5 Graphing periodic
functions
Exercise 5A — Period and amplitude of
a periodic function
1 2
3 4 5
6 A
7 a Oestrogen
b 28 days
Investigation — Ferris wheeling
1 2 2 0.5
Exercise 5B — Radian measure
1
2
3 4 E
5 a 0.855 b 1.361 c −2.182
d 3.334 e 4.084 f 5.707
g 2.967 h 3.787
6 a 20° b 84° c 180°
d 55° e 894° f −155°
g 233° h 458°
Exercise 5C — Exact values
1
2
3
4
5
6 7
8
Exercise 5D — Symmetry
12 3 4
a 4 b 1
a T = 4π A = 2
b T = 2π A = 1
c T = 3π A = 1.5
d A = 4
e T = 2π A = 2
f A = 3
g T =π A = 2.5
h A = 0.5
T = 2 A = 2
a Approximately periodic
b 12 months
c February
d August
e 18° T 3π
2 ---=
T 4π 3 ---=
T 2π 3 ---=
1 4
---a b c
d e f
g h 2π i
j k l
m n
a 36° b 120° c 40° d 220°
e 648° f −30° g −45° h 67.5°
E
a b c d
e f g 1 h
a b c d
e 1 f g h
a P b P c P d P
e N f N g N h N
i P j N
a P b P c N d N
e N f P g P h P
i N j N
a P b P c N d N
e P f P g N h N
i N j P
a Quadrant 3 b Quadrant 1
a C b B c A
d B e C
a −1 b −1 c 0 d 0
e 0 f 1 g Undefined h 0
i −1 j 1
a 0.63 b −0.63 c −0.63 d −0.63
a −0.25 b −0.25 c 0.25 d 0.25
a −2.1 b −2.1 c 2.1 d −2.1
a –0.3 b −0.7 c −0.9 d −0.3
e 0.3 f 0.7 g 0.7 h −0.9
i 0.9 π 6
--- π
4
--- π
3 ---π
9
--- 5π
18
--- π 2 ---3π
2
--- 5π
6 ---5π
– 4
--- 7π 3
--- 5π 3 ---4π
15
--- 2π 5
---3 2
--- 1
2
--- 1
3
--- 1
2 ---1
2
--- 3
2
--- 3
1 2
--- 1
2
--- 3 1
2 ---3
2
--- 1
2
--- 1
---answers
5
6
7
8
9
Exercise 5E — Trigonometric graphs
12
3 a i π ii 1
b i 2π ii 2
c i 4π ii 3
d i π ii 4
e
i ii
f i 4π ii
g i 6π ii 5
h i 5π ii 4
i i ii 2
j i π ii 3
4 5 E
6
a b − c d −
e − f − g − h −
i j − k 0 l −1
a − b c −1 d
e f − g − h −
i − j k −1 l −1
a –0.383 b −0.924 c 0.414 d 0.924
e 0.383 f −0.414
a 0.966 b −0.259 c −3.732 d −0.966
e 0.259 f −3.732
a 0.644 b −0.765 c −0.842 d −0.644
a i 4π ii 2 b iπ ii 1
c i 3π ii 1.5 d i ii 4
e i 2π ii 2 f i ii 3
g iπ ii 2.5 h i ii 0.5
a i 2π ii 1 b i 2π ii 3
c iπ ii 2 d i ii 4
e i 6π ii f i ii 2
g i 6π ii 0.4 h i ii 3
i i 8 ii 2.5 j i ii 1
k i 2 ii l i 4 ii
2
---2 3 3
3 2
--- 3
2
--- 3 3
2 ---1
2
--- 1
2
---1 2
--- 1
2
--- 3
2 ---1
2
--- 1
3
--- 3
2
--- 3
2 ---1
3
--- 1
2
---3π 2 ---4π 3 ---2π 3
---2π 3
---1 2
--- π
2 ---2π
5 ---π 3
---1 5
--- 1
4
---y
x
–1 1
0 π
– 2π
– 2
π —
2π 3
y
x 0
2
–2
2π
y
x
–3 3
0 2π 4π
a D b C c A
a y = 1.5 sin b y = 2 cos 2x
c y = 5 sin d y = 4 cos
e y = −sin f y = −3 cos 3x π
y
x 0
4
–4
– 4 π –2
π —
4π 3
y
x
0 2
— 3 π – 3 π 1 – 2 1 – 2
–
2π 3 --- 12
---y
x 0 π 2π 3π 4π
2 – 3 2 – 3
–
2 3
---y
x 0
5
–5
3π 6π
2
π π 3π 4π y
x 0
4
–4
y
x 0
2
–2
– 8π
– 4π –2
π —
8 π 3
π 2
---y
x 0
3
–3
– 2π π
2x 3 ---x 2
--- 2x
3 ---3x
2
---
5A
➔
answers
7
8
9 a f: [0, ] → R, f(x) = 3 sin
b f: [0, 5π] → R, f(x) = cos
c f: [−1, 1] → R, f(x) = 2 sin πx
d f: [−1, 3] → R, f(x) = 1.8 cos
e f: [0, 3] → R, f(x) = −3 sin
f f: [− , 1] → R, f(x) = −2.4 cos
10
Investigation — How high?
Check with your teacher.
Exercise 5F — Applications
1 a i 1 kg ii 6 daysb W = cos + 3
2 a 110 beats/min
b i 50 ii 60 min
c H = 50 sin + 110
3 a 1.6 m b i 1 m ii 0.7 m
4 a 26°C at 2 pm
b i 18°C ii 22°C iii Approx. 11.1°C
5 a i 12 mm ii s
b 10
c −11.41 mm; if the displacement is positive to the right then the string is 11.41 mm to the left (or vice versa)
6 7
