Chapter 2 — Relations and functions · Relations and functions MB11 Qld-2 23 Exercise 2A — Relations and graphs 1 The graph that represents the five points is B. The answer is

R e l a t i o n s a n d f u n c t i o n s M B 1 1 Q l d - 2 23

Exercise 2A — Relations and graphs 1 The graph that represents the five

points is B. The answer is B 2 Dependent variable is

y = {1, 2, 3, 4} The answer is A 3 Rule y = 4 − x x ∈ {0, 1, 2, 3} The answer is E 4 Progressively increasing graph is C. The answer is C 5 a Discrete b Continuous c Continuous d Discrete e Discrete f Continuous 6 a Discrete

b Discrete

c Discrete

d Continuous

e Discrete

f Continuous

7 a

b

c Because the variables are

continuous. d Half of the initial temperature is

40°C. It takes approximately 11 minutes.

8 a P = 300 + 40n n 0 1 2 3 4 5 6

P 300 340 380 420 460 500 540

b

c The variables are discrete. 9

t 0 1 2 3 4 5

V 0 30 80 150 250 350

a

Variables are continuous. b i 2.5 seconds V = 110 km/hr ii 4.8 seconds V = 320 km/hr

10 C = 50 + 6n a N 15 16 17 18 19 20 21 22 23 24 25

C 140146152 158164170 176182188194200

c Dots do not join because values are

discrete.

Exercise 2B — Domain and range 1 a [−2, ∞) b (−∞, 5) c (−3, 4] d (−8, 9) e (−∞, −1] f (1, ∞) g (−5, −2] ∪ [3, ∞) h (−3, 1) ∪ (2, 4] 2 a [−6, 2)

b (−9, −3)

c (−∞, 2]

d [5, ∞)

e (1, 10]

f (2, 7)

g (−∞, −2) ∪ [1, 3)

h [−8, 0) ∪ (2, 6]

3 a {x: −4 ≤ x < 2} = [−4, 2) b {x: −3 < x ≤ 1} (−3, 1] c {y: −1 < y < 3 } (−1, 3 ) d y: {− 1

2 < y ≤ 12

}

(− 12 , 1

2]

e {x: x > 3} (3, ∞) f {x: x ≤ −3} (−∞, −3]

Chapter 2 — Relations and functions

M B 1 1 Q l d - 2 24 R e l a t i o n s a n d f u n c t i o n s

g R h (−∞, 0) ∪ (0, ∞) 4 Domain is −1 ≤ x ≤ 7 = [−1, 7] The answer is C 5 {(x, y) = 2x + 5} x ∈[−1, 4] x = −1 y = 3 x = 4 y = 13 Range is 3 ≤ y ≤ 13 [3, 13] The answer is B 6 a {(3, 8), (4, 10), (5, 12), (6, 14),

(7, 16)} i Domain = {3, 4, 5, 6, 7} ii Range = {8, 10, 12, 14 16}

b {(1.1, 2), (1.3, 1.8), (1.5, 1.6), (1.7, 1.4)}

i Domain = {1.1, 1.3, 1.5, 1.7} ii Range = {2, 1.8, 1.6, 1.4}

or = {1.4, 1.6, 1.8, 2} c i Domain = {3, 4, 5, 6} ii Range = {110, 130, 150, 170} d i Domain = {M, Tu, W, Th, Fr} ii Range = {25, 30, 35}

(specified only once) e y = 5x − 2 2 < x < 6 i Domain = {3, 4, 5} ii Range = {13, 18, 23} f y = x2 − 1 x ∈ R i Domain = R ii Range = [−1, ∞)

7 a Domain = R Range = R b Domain = R Range = (0, ∞) or R+

c Domain = [−2, 2] Range = [0, 2] d Domain = [1, ∞) Range = R e Domain = R Range = (0, 4] f Domain = R Range = (−∞, −3] g Domain = R\{0} Range = R\{0} h Domain = R Range = (−∞, 1] i Domain = R Range = R 8 a {(x, y): y = 2 − x2}

Domain = (−∞, ∞) Range = (−∞, 2] b {(x, y): y = x3 + 1, x ∈ [−2, 2]}

Domain = [−2, 2] Range = [−7, 9]

c {(x, y): y = x2 + 3x + 2} If x = 0 y = 2 If y = 0 0 = x2 + 3x + 2 = (x + 2)(x + 1) x + 2 = 0 or x + 1 = 0 x = −2, −1

y = x2 + 3x + 2 = x2 + 3x + 9

4 + 2 − 94

y = (x + 32 )2 − 1

4

TP at ( 32

− , 14− )

Domain = (−∞, ∞) Range = [− 1

4 , ∞)

d {(x, y): y = x2 − 4, x ∈ [−2, 1]} Parabola y = x2 translated 4 units

down

Domain = [−2, 1] Range = [−4, 0] e {(x, y): y = 2x − 5, x ∈ [−1, 4)} If x = 0, y = −5 If y = 0, 0 = 2x − 5 So 2x = 5, x = 5

2

x = −1, y = −7 x = 4, y = 3 Domain = [−1, 4) Range = [−7, 3) f {(x, y): y = 2x2 − x − 6} If x = 0, y = −6 If y = 0, 0 = 2x2 − x − 6 0 = (2x + 3)(x − 2) 2x + 3 = 0, x − 2 = 0 x = 3

2− or x = 2

y = 2x2 − x − 6 y = 2(x2 − 1

2 x − 3)

= 2[x2 − 12 x + 1

16 − 3 − 116 ]

= 2[(x − 18 )2 − 49

16 ]

= 2(x − 18 )2 − 49

8

TP is at ( 18 , 49

8− )

Domain = (−∞, ∞) Range = [ 49

8− , ∞)

or = [−6 18 , ∞)

9 a y = 10 − x Domain = R b y = 3 x

Domain = [0, ∞)

c y = 216 x− −

Domain = [−4, 4] d y = x2 + 3

Domain = R

e y = 1x

Domain = R\{0}

f y = 10 − 7x2

y = −7x2 + 10

Domain = R

Exercise 2C — Types of relations (including functions) 1 a

One x-value produces two y-values: one-to-many

R e l a t i o n s a n d f u n c t i o n s M B 1 1 Q l d - 2 25

b

One y-value produces two x-values: many-to-one

c

One y-value produces two x-values: many-to-one

d

One x-value produces one y-value: one-to-one

e

One x-value produces one y-value: one-to-one

f

One y-value produces many x-values: many-to-one

g

Many y-values produce many x-values: many-to-many

h

One y-value produces two x-values: many-to-one

i

One x-value produces one y-value: one-to-one

j

One y-value produces two x-values: many-to-one

k

Two y-values produce two x-values: many-to-many

l

One y-value produces two or more x-values: many-to-one

2 a

Vertical line intercepts two points: not a function

b

Vertical line intercepts one point: function

c

Vertical line intercepts one point: function

d

Vertical line intercepts one point: function

e

Vertical line intercepts one point: function

f

Vertical line intercepts one point: function

g

Vertical line intercepts two points: not a function

h

Vertical line intercepts one point: function

i

Vertical line intercepts one point: function

j

Vertical line intercepts one point: function

k

Vertical line intercepts one point: function

l

Vertical line intercepts one point: function

3 A

Function

M B 1 1 Q l d - 2 26 R e l a t i o n s a n d f u n c t i o n s

B

Function

C y2 = x, y = ± x Not a function

D y = 8x − 3 Function

E Function

The answer is C

4 a y ≥ x + 1

If x = 0, y = 1 If y = 0, x = −1 The answer is B b many-to-many The answer is D c Domain = R Range = R The answer is B 5 a

Not a function b {(−3, −2) (−1, 1) (0, 1) (1, 3)(2, −2)}

Is a function Domain = {−3, −1, 0, 1, 2} Range = {−2, −1, 1, 3}

c {(3, −1) (4, −1) (5, −1) (6, −1)}

Function Domain = {3, 4, 5, 6} Range = {−1} d {(1, 2) (1, 0) (2, 1) (3, 2) (4, 3)}

Not a function e {(x, y): y = 2, x ∈ R}

Function Domain = R Range = {2} f {(x, y): x = −3, y ∈ J}

Not a function g y = 1 − 2x If x = 0, y = 1 (0, 1) If y = 0, 0 = 1 − 2x 2x = 1, x = 1

2 ( 12 , 0)

Function Domain = R Range = R h y > x + 2 If y = x + 2 If x = 0, y = 2 (0, 2) If y = 0, 0 = x + 2 x = −2 (−2, 0) Point (0, 0) 0 > 0 + 2 0 > 2 False

Not a function

i x2 + y2 = 25

Not a function

j y = 1x + , x ≥ −1

x = −1, y = 0 = 0 x = 0, y = 1 = 1 x = 3, y = 4 = 2 Function Domain [−1, ∞) Range [0, ∞) k y = x3 + x y = x(1 + x2) If x = 0, y = 0 If y = 0, 0 = x(1 + x2) x = 0 only

Function Domain = R Range = R l x = y2 + 1 or y2 = x − 1 y = ± 1x −

Not a function

Exercise 2D — Function notation and special types of functions

1 a f (x) = 3x + 1 f (0) = 3 × 0 + 1 = 0 + 1 = 1 f (2) = 3 × 2 + 1 = 6 + 1 = 7 f (–2) = 3 × −2 + 1 = −6 + 1 = −5 f (5) = 3 × 5 + 1 = 15 + 1 = 16 b g(x) = 4x + g(0) = 0 4 4 2+ = = g(−3) = 3 4 1 1− + = = g(5) = 5 4 9 3+ = = g(−4) = 4 4 0 0− + = =

c g(x) = 4 − 1x

g(1) = 4 − 11 = 4 − 1 = 3

g( 12 ) = 4 −

12

1 = 4 − 2 = 2

g(− 12 ) = 4 −

12

1−

= 4 + 2 = 6

g(− 15 ) = 4 −

15

1−

= 4 + 5 = 9

R e l a t i o n s a n d f u n c t i o n s M B 1 1 Q l d - 2 27

d f(x) = (x + 3)2

f(0) = (0 + 3)2 = 32 = 9 f(−2) = (−2 + 3)2 = 12 = 1 f(1) = (1 + 3)2 = 42 = 16 f(a) = (a + 3)2 = a2 + 6a + 9

e h(x) = 24x

h(2) = 242 = 12

h(4) = 244 = 6

h(−6) = 246− = −4

h(12) = 2412 = 2

2 a f(x) = 3x − 4, f(x) = 5 5 = 3x − 4 3x = 9 x = 3 b g(x) = x2 − 2, g(x) = 7 7 = x2 − 2 x2 = 7 + 2 = 9 x = ± 9 x = ±3

c f(x) = 1x

, f(x) = 3

3 = 1x

3x = 1 x = 1

3

d h(x) = x2 − 5x + 6, h(x) = 0 0 = x2 − 5x + 6 0 = (x − 3)(x − 2) x − 3 = 0 or x − 2 = 0 x = 3 or x = 2 e g(x) = x2 + 3x, g(x) = 4 4 = x2 + 3x 0 = x2 + 3x − 4 0 = (x + 4)(x − 1) x + 4 = 0 or x − 1 = 0 x = −4 or x = 1 f f(x) = 8 x− , f(x) = 3 3 = 8 x− 32 = 8 − x 9 = 8 − x x = 8 − 9 x = −1

3 f(x) = 10 xx

−

a f(2) = 102 − 2 = 5 − 2 = 3

b f(–5) = 105− − (−5)

= −2 + 5 = 3

c f(2x) = 102x

− 2x

= 5x

− 2x

d f(x2) = 210x

− x2

e f(x + 3) = 103x +

− (x + 3)

= 103x +

− x − 3

f f(x − 1) = 101x −

− (x − 1)

= 101x −

− x + 1

4 a {(1, −1), (2, 1), (3, 3), (4, 5)}

one-to-one function b {(−2, 1), (−1, 0), (0, 2), (1, 1)}

function, not one-to-one c {(x, y): y = x2 + 1, x ∈ [0, ∞)}

one-to-one function d {(x, y): y = 3 − 4x} If x = 0, y = 3 (0, 3) If y = 0, 0 = 3 − 4x, 4x = 3 x = 3

4 ( 34 , 0)

one-to-one function e {(x, y): y = 3 − 2x2} If x = 0, y = 3 If y = 0, 0 = 3 − 2x2, 2x2 = 3 x2 = 3

2 , x = ± 32

function, not one-to-one f f(x) = x3 − 1 If x = 0, f(x) = −1 If f(x) = 0, 0 = x3 − 1, x3 = 1 x = 3 1 = 1

one-to-one function

g y = x2, x ≤ 0

one-to-one function

h g(x) = 21 x− If x = 0, g(x) = 1 = 1

If g(x) = 0, 0 = 21 x− 0 = 1 − x2 x2 = 1 x = ± 1 = ±1

function, not one-to-one 5 a

i function ii not a one-to-one function

b

i function ii not a one-to-one function

c

i function ii is a one-to-one function

d

i function ii not a one-to-one function

e

i not a function ii one-to-many

f

M B 1 1 Q l d - 2 28 R e l a t i o n s a n d f u n c t i o n s

i function ii not a one-to-one function

g

i not a function

h

i function ii is a one-to-one function

i

i function ii is a one-to-one function

j

i function ii not a one-to-one function

k

i function ii is a one-to-one function

l

i function ii not a one-to-one function

6 a f(x) = 11

x xx x

− <⎧⎨ ≥⎩

Graph

The answer is B b Range is from −1 to infinity = (−1, ∞) The answer is C

7 a f(x) = 1 0

1 0

xx

x x

⎧ <⎪⎨⎪ + ≥⎩

b Range of f = (−∞, 0) ∪ [1, ∞)

8 a g(x) = 2 1 0

2 0x x

x x⎧ + ≥⎪⎨

− <⎪⎩

For y = x2 + 1 If x = 0, y = 1 (0, 1) If y = 0, 0 = x2 + 1 x2 = −1 No solutions If x = 2, y = 5 (2, 5) For y = 2 − x If x = −1, y = 3 (−1, 3)

b Range [1, ∞)

c i g(−1) (substitute into 2 − x) = 2−(−1) = 3 ii g(0) = 1 (where x2 + 1 meets

y-axis) iii g(1) (substitute into x2 + 1) = 12 + 1 = 2

9 a f(x) = 2

2 2

4 2 22 2

x x

x xx x

− < −⎧⎪

− − ≤ ≤⎨⎪ + >⎩

For y = x − 2 If x = −4, y = −4 If x = −3, y = −5 For y = x2 − 4 If x = 0, y = −4 If y = 0, x2 − 4 = 0 x2 = 4 x = ± 4 x = ±2 For y = x + 2 If x = 2, y = 4 If x = 3, y = 5

b Range: is from −∞ to zero

then 4 and beyond (−∞, 0] ∪ (4, ∞)

c i f(−3) = x − 2 = −3 − 2 = −5 ii f(−2) = (−2)2 − 4 = 0 iii f(1) = (1)2 − 4 = −3

iv f(2) = 22 − 4 = 0 v f(5) = 5 + 2 = 7

10

For x ≤ 0 Equation y = mx + c c = 2, m = 2

2 = 1

y = x + 2 For x > 0 Equation y = mx + c c = 1, m = 2 y = 2x + 1

So f(x) = 2 0

2 1 0x xx x+ ≤⎧

⎨ + >⎩

11 {(x, y): x2 + y2 = 1, x ≥ 0}

Two one-to-one functions Now transform the equation to find

y: y2 = 1 − x2

y = ± 21 x−

where y = 21 x− top half

and y = − 21 x− bottom half

i f : [0, 1] → R, f(x) = 21 x− Range [0, 1]

ii f : [0, 1] → f(x) = − 21 x− Range [−1, 0]

Exercise 2E — Inverse relations and functions 1 Sketch each point, then interchange

x- and y-values. Plot the new points. a

b

R e l a t i o n s a n d f u n c t i o n s M B 1 1 Q l d - 2 29

c

2 Copy each graph. Draw in the line

y = x, then reflect each graph across the line y = x.

a

b

c

d

e

f

3 Use a graphics calculator to sketch

each graph. Draw in the line y = x. Reflect each original graph across the line y = x.

a y = 4x is a straight line passing through the origin.

b y = x2 + 3 is a parabola with turning

point at (0, 3)

c y = 12

x + 1 is a straight line

passing through y = 1

d y = x3 + 4 is a cubic graph with a

point of inflection at (0, 4)

Exercise 2F — Circles 1 a x2 + y2 = r2

r = 3, x2 + y2 = 9 b x2 + y2 = r2

r = 1, x2 + y2 = 1

c x2 + y2 = r2

r = 5, x2 + y2 = 25 d x2 + y2 = r2

r = 10, x2 + y2 = 100 e x2 + y2 = r2

r = 6 , x2 + y2 = 6 f x2 + y2 = r2

r = 2 2 , x2 + y2 = 8 g x2 + y2 = r2

y2 = r2 − x2

y = ± 2 2r x− r = 3 top half only

So y = 2 23 x−

y = 29 x−

h y = ± 2 2r x−

r = 4, y = ± 2 24 x− But we require bottom half only,

so 216y x= − − 2 a Domain = [−3, 3] Range = [−3, 3] b Domain = [−1, 1] Range = [−1, 1] c Domain = [−5, 5] Range = [−5, 5] d Domain = [−10, 10] Range = [−10, 10] e Domain [− 6 , 6 ] Range [− 6 , 6 ] f Domain [−2 2 , 2 2 ] Range [−2 2 , 2 2 ] g Domain [−3, 3] Range [0, 3] h Domain [−4, 4] Range [−4, 0] 3 a x2 + y2 = 4 Centre (0, 0) Radius 4 2=

b x2 + y2 = 16 Centre (0, 0) Radius 16 4=

c x2 + y2 = 49 Centre (0, 0) Radius 49 7=

M B 1 1 Q l d - 2 30 R e l a t i o n s a n d f u n c t i o n s

d x2 + y2 = 7 Centre (0, 0) Radius 7

e x2 + y2 = 12 Centre (0, 0) Radius = 12 = 4 3×

= 2 3

f x2 + y2 = 1

4 Centre (0, 0) Radius = 1

4

= 12

4 a y = ± 281 x− Circle centre (0, 0) Radius = 9

Not a function

b y = 24 x− Centre (0, 0) Radius = 4 2= Top half of circle

Is a function

c y = − 21 x− Centre (0, 0) Radius = 1 = 1 Bottom half of circle

Is a function

d y = 219 x−

Centre (0, 0) Radius = 1

9 = 13

Top half of circle

Is a function

e y = − 214 x−

Centre (0, 0) Radius = 1

4 = 12

Bottom half of circle

Is a function

f y = 25 x− Circle centre (0, 0) Radius = 5 Top half of circle

Is a function

g y = ± 210 x− Centre (0, 0) Radius 10 Full circle

Is not a function h x2 + y2 = 3 − 3 ≤ x ≤ 0 Circle centre (0, 0) Radius 3 (Half circle − 3 to 0)

Is not a function 5 a Circle centre (2, 0) Radius = 2 Equation (x − 2)2 + y2 = 4 The answer is D b Range [−2, 2] The answer is B 6 a (x + 3)2 + (y − 1)2 = 1 Centre (−3, 1) Radius 1 The answer is C b Domain = [−4, −2] The answer is E 7 a x2 + (y + 2)2 = 1 Centre (0, −2) Radius = 1 = 1 Domain [−1, 1] Range [−3, −1]

b x2 + (y − 2)2 = 4 Circle centre (0, 2) Radius = 4 = 2 Domain = [−2, 2] Range = [0, 4]

c (x − 4)2 + y2 = 9 Circle centre (4, 0) Radius = 9 = 3 Domain = [1, 7] Range = [−3, 3]

d (x − 2)2 + (y + 1)2 = 16 Circle centre (2, −1) Radius = 16 = 4 Domain [−2, 6] Range [−5, 3]

e (x + 3)2 + (y + 2)2 = 25 Circle centre (−3, −2) Radius = 25 = 5 Domain [−8, 2] Range [−7, 3]

f (x − 3)2 + (y − 2)2 = 9 Circle centre (3, 2) Radius = 9 = 3 Domain [0, 6] Range [−1, 5]

g (x + 5)2 + (y − 4)2 = 36 Circle centre (−5, 4) Radius = 36 = 6 Domain [−11, 1] Range [−2, 10]

R e l a t i o n s a n d f u n c t i o n s M B 1 1 Q l d - 2 31

h (x − 1

2 )2 + (y + 32 )2 = 9

4

Circle centre 12( , − 3

2)

Radius = 94 = 3

2

Domain [−1, 2] Range [−3, 0]

8 x2 + y2 = 36, y2 = 36 − x2

y = 236 x− Domain = [−6, 6] Range = [0, 6]

236y x= − − Domain = [−6, 6] Range = [−6, 0] 9 x2 + (y − 2)2 = 9 (y − 2)2 = 9 − x2

y − 2 = ± 29 x−

y = 2 ± 29 x−

y = 2 + 29 x− Domain [−3, 3] Range [2, 5]

y = 2 − 29 x− Domain [−3, 3] Range [−1, 2] 10 a 4 = 2 cm, 190 ≈ 13.8 cm

b 13.8 2 11.83 3

− =

≈ 3.93 Travelling at approximately 3.93 cm/s.

Exercise 2G — Functions and modelling 1 Hours of hire Cost Up to 1 $40 Over 1 up to 2 $70 Over 2 up to 4 $110 Over 4 up to 6 $160 a Let cost C in terms of time t be

C(t)

40 0 170 1 2( )

110 2 4160 4 6

ttc ttt

⎧ < ≤⎪⎪ < ≤= ⎨

< ≤⎪⎪ < ≤⎩

b

2 Time of call

(seconds) Cost ($)

Up to 30 0.70 Over 30, up to 60 1.10 Over 60, up to 90 1.50 Over 90, up to 120 1.90 Over 120, up to 150 2.30 Over 150, up to 180 2.70 Over 180, up to 210 3.10 Over 210, up to 240 3.50 Over 240, up to 270 3.90 Over 270, up to 300 4.30

a Let cost C in terms of time t be C(t)

0.70, 0 301.10, 30 601.50, 60 901.90, 90 1202.30, 120 150

( )2.70, 150 1803.10, 180 2103.50, 210 2403.90, 240 2704.30, 270 300

ttttt

C tttttt

< ≤⎧⎪ < ≤⎪⎪ < ≤⎪

< ≤⎪⎪ < ≤⎪= ⎨ < ≤⎪⎪ < ≤⎪

< ≤⎪⎪ < ≤⎪

< ≤⎪⎩

b

3 i Speed = 60 km/h for 1 1

2 hours

ii Stops for 12 hour

iii 80 km/h for 2 hours a d(t) (see graph for 0 to 1.5) d = mt + c c = 0

90 601.5

m = =

d = 60t i For t = 1.5 to t = 2 d = mt + c m = 0 c = 90 d = 90 ii For t = 2 to 4

60 0 1.5

( ) 90 1.5 280 70 2 4

t td t t

t t

≤ ≤⎧⎪= ≤ ≤⎨⎪ − ≤ ≤⎩

b Domain = [0, 4] Range = [0, 250]

c i Distance after 1 hour d = 60t = 60 × 1 = 60 km ii Distance after 3 hours d = 80t − 70 = 80 × 3 − 70 = 170 km

4 5 minutes per $1. Meter accepts a maximum of 120 dollar coins.

a Rule: 560

nB ×=

12nB =

b

c Number of 5-minute operations

450 905

=

1 dollar per 5 minutes = $90 5 a Tax = $3060 + 34 cents for every

dollar over $20 700. T = 3060 + 0.34(x − 20 700) = 3060 + 0.34x − 7038 = 0.34x − 3978 b

c T = 0.34x − 3978 Taxable income x = $32 000 T = 0.34 × 32 000 − 3978 T = 10 880 − 3978 T = $6902 6

a Perimeter P = 2L + 2W = 2(x + 4) + 2(x − 1) = 2x + 8 + 2x − 2 P = 4x + 6 b Domain (1, 6] Range (6, 30]

M B 1 1 Q l d - 2 32 R e l a t i o n s a n d f u n c t i o n s

7 a

A = L × W A = (x + 4)x A = x2 + 4x b

If x = 8 (max. side length is 12) So A = 82 + 4 × 8 = 64 + 32 = 96 Domain = (0, 8] Range = (0, 96] 8 a A = $100 000 + 0.02 × 100 000 = 100 000 × 1.02

In t years time, value P = 100 000 × (1.02)t

b t = 10 P = 100 000 × (1.02)10

P = 121 899.44 P = $121 899

9 N(t) = 96153t

++

a t = 0

N(t) = 96153

+

= 15 + 32 = 47 b t = 13

N(t) = 961513 3

++

= 961516

+

= 15 + 6 = 21 c N(t) = 23

23 = 96153t

++

963t +

= 23 − 15

963t +

= 8

96 = 8(t + 3) 96 = 8t + 24 8t = 96 − 24 8t = 72

t = 728

t = 9 It will be 9 weeks.

d As t increases 963t +

gets

smaller and approaches zero. N(t) → 15, so no.

10 Cost = 600 × 10 = $6000. For every $1 increase, 50 people less

attend.

Now T = an2 + bn + c T = an2 + bn + 6000 n = 12 T = 0 0 = a × 144 + 12b + 6000 −6000 = 144a + 12b −500 = 12a + b (1) n = 10 T = 2000 2000 = 100a + 10b + 6000 −4000 = 100a + 10b −400 = 10a + b (2) (1) − (2) −100 = 2a a = −50 Substitute into (1) −500 = 12 × –50 + b −500 = −600 + b 100 = b a So T = −50n2 + 100n + 6000 or T = 6000 + 100n − 50n2

b

c Maximum takings at 0dTdn

=

dTdn

= 100−100n

100n = 100 n = 1 T = 6000 + 100 − 50 = $6050 So, admission price = $11

Chapter review 1

Rule: y = 2x x ∈ {1, 2, 3, 4} The answer is B

2

Continuous relation The answer is C 3

Time (t) hours 1 2 3 4 5

No. cars (n) 30 75 180 330 500 a

b Graph not continuous

(n ∈ N). The number of cars is a discrete variable.

c 120 4 {(x, y): y = 1 − x2 x ∈ [−3, 3]} If x = 0, y = 1 If y = 0, 0 = 1 − x2

x2 = 1 x = ±1

Domain = [−3, 3] Range = [−8, 1]

5

Interval = [−5, −1) ∪ (1, 4] The answer is E 6 (0, 2) ∪ (2, ∞)

The answer is C 7

Domain = all x-values except 1. The answer is B 8 y = x + 3, x ∈ R+

Range = (3, ∞) The answer is E

R e l a t i o n s a n d f u n c t i o n s M B 1 1 Q l d - 2 33

9 15

yx

=−

Implied domain = (5, ∞) The answer is A 10 f(x) = ( )2 4 x−

Range = [0, ∞) The answer is D 11 The relation is a many-to-one

function. The answer is D 12 A y = x2

B x2 + y2 = 3 Relation

C {(1, 1), (2, 1), (3, 2), (4, 3)}

D y = 5 − x

E {1, 3, 5, 7, 9} (not a set of

ordered pairs.) Answer (a) (c) (d) (e) are relations. The answer is E 13 Not a function The answer is B

14 A 5xy =

B y = 2 − 7x

C x = 5

D y = 10x2 + 3

E y = −8

The answer is C 15 a y = 2x2 − 1

Is a function b 3x + y = 2 If x = 0, y = 2 If y = 0, 3x = 2, x = 2

3

Is a function c x = y2 + 1 y2 = 1 − x y = ± 1 x−

Not a function d x2 + y2 = 10 Circle centre (0, 0) Radius 10 Not a function

e y3 = x y = 3 x

Is a function f y2 − x2 = 1 y2 = 1 + x2

y = ± 21 x+ Is not a function So (a) (b) (e) are functions 16 A one-to-one

B not one-to-one

C {(x, y): y = 4x} one-to-one

D {(x, y): y = 5 − 2x one-to-one

E f(x) = 2 − x3

one-to-one

The answer is B 17 C is a one-to-one function

The answer is C 18 g(x)= x + 2, x ≥ 0

a g(x)2 = 2x + 2 = x + 2 x ≥ 0 b Graph g(x)

Domain = [0, ∞) Range = [2, ∞) 19 f: {x: x = 0, 1, 2} → R f(x) = x − 4

= {(0, −4), (1, −3), (2, −2)} The answer is A 20 g(x) = 6 − x + x2

g(−2) = 6 − (−2) + (−2)2

= 6 + 2 + 4 = 12 The answer is D 21 f(x) = 3x − 5 f(2x + 1) = 3(2x + 1) − 5 = 6x + 3 − 5 = 6x − 2 The answer is E

M B 1 1 Q l d - 2 34 R e l a t i o n s a n d f u n c t i o n s

22 2

1 0

( ) 0 22 2

x x

f x x xx x

+ <⎧⎪

= ≤ ≤⎨⎪ − >⎩

The answer is A

23 a 1yx

=

f: R\{0} → R, 1( )f xx

=

b y = 2 x−

Domain = (−∞, 2] f: (−∞, 2] → R, f(x) = 2 x−

24 2 , 1

( ) 3, 1 32 5, 3

x xf x x

x x

− ≤ −⎧⎪= − < <⎨⎪ − ≥⎩

,

25 Only the graph shown in E could be

a reflection of the given graph across the line y = x.

The answer is E 26

Centre (3, 0), Radius 2 Equation is (x − 3)2 + y2 = 4 The answer is D

27 a y = 21 x− − y2 = 1 − x2 x2 + y2 = 1 semi circle centre (0, 0) Radius 1 (bottom half)

Domain = [−1, 1] Range = [−1, 0]

b (x − 2)2 + (y + 1)2 = 9 Circle centre (2, −1) Radius = 9 3=

Domain = [−1, 5] Range = [−4, 2] 28 (x + 1)2 + (y − 4)2 = 9 Centre (−1, 4), Radius = 3

Domain = [−4, 2] The answer is E 29 Range = [1, 7] The answer is C 30 Centre (4, −2), Radius = 5 (x − h)2 + (y − k)2 = r2

(x − 4)2 + (y + 2)2 = 2( 5) (x − 4)2 + (y + 2)2 = 5 The answer is B 31 a x2 + y2 = 100 Circle centre (0, 0) Radius = 100 = 10

b i First one-to-one function y2 = 100 − x2

y = ± 2100 x−

First function = 2100 x− (top half circle) f: [−10, 10] → R,

f(x) = 2100 x− Domain = [−10, 10] Range = [0, 10] ii Second one-to-one function

Second function = 2100 x− − (bottom half of circle) f: [−10, 10] → R

f(x) = 2100 x− − Domain = [−10, 10] Range = [−10, 0]

32 Hours (h) Charge (c$) 0 < h ≤ 2 50 2 < h ≤ 3 80

3 < h ≤ 6 100 The answer is E

33 x < 1

2 $50

12 ≤ x ≤ 1 $75

1 < x < 2 $100

Modelling and problem solving 1

a Area in terms of x + y A = + A = 10y + (y − x)x A =10y + xy − x2

A =10y + xy − x2 (1) b Perimeter P = y + 10 + x + y − x + x + x + 10 P = 2y + 2x + 20 P = 2(y + x + 10) (2) c P = 72 cm 72 = 2y + 2x + 20 So 2y = 72 − 2x − 20 y = 36 − x − 10 (3) Substitute (3) into (1) A = 360 − 10x − 100 + 36x − x2 −

10x − x A = 260 + 16x − 2x2

d Domain of A(x) x must be greater than 0 to start

with If x = 14 y = 36 − 14 − 10 = 36 − 24 y = 12 So the dimension y − x is negative. If x = 13 y = 36 − 13 − 10 = 13 y − x = 0 So x must be less than 13 So domain is (0, 13) e Graph A = 260 + 16x − 2x2

If x = 0, A = 260 If A = 0, 0 = 260 + 16x − 2x2

R e l a t i o n s a n d f u n c t i o n s M B 1 1 Q l d - 2 35

Turning point A = −2x2 + 16x + 260 A = −2(x2 − 8x − 130) A = −2[x2 − 8x + 16 − 130 − 16] A = −2[(x − 4)2 − 146] A = −2(x − 4)2 + 292 Maximum turning point at (4, 292)

f Maximum area is obtained from

the turning point. A = 292 m2

2

a Domain = x ∈ (−∞, 4]\{−2} b Range = [0, 18] c Rule for x ∈ (−∞, −2) y = 4 d Rule for x ∈ (−2, 0] y = mx + c, c = 0

m = 42−

= −2

y = −2x e Rule for x ∈ [0, 3] In the form y = ax2 At (3, 18) y = ax2

18 = a × 32

18 = 9a a = 18

9 = 2

So y = 2x2 f Rule for x ≥ 3 y = mx + c At (3,18) 18 = 3m + c (1) At (4, 0) 0 = 4m + c (2) (1) − (2) 18 = −m m = −18

Substitute into (2) 0 = 4 × − 18 + c 0 = −72 + c c = 72 Rule = y = −18x + 72

g2

( , 2)4( 2, 0]2( )

[0, 3]2 [3, 4]18 72

xxxf xxxxx

∈ −∞ −⎧⎪ ∈ −⎪ −= ⎨ ∈⎪⎪ ∈− +⎩

3 f: [−2, a]→R, f(x) = (x − 1)2 − 4 a f(x) = (x − 1)2 − 4 f(−2) = (−2 − 1)2 − 4 = 5 f(−1) = (−1 − 1)2 − 4 = 0 f(0) = (0 − 1)2 − 4 = −3 f(1) = (1 − 1)2 − 4 = −4 f(3) = (3 − 1)2 − 4 = 0 b f(a) = 12 = (a − 1)2 − 4 12 + 4 = (a − 1)2

(a − 1)2 = 16 a − 1 = ±4 a = 4 + 1 or −4 + 1 a = 5 or −3 a = 5 (as a must be greater than

−2) c

Turning point at (1, −4) y-intercept at −3 x-intercept when y = 0 from a

x = −1 and 3. d i Domain [−2, 5] ii Range [−4, 12]

4 a

b w = 2 × x, L = x P = 2w + 2L P = 2(2x) + 2x P = 4x + 2x P = 6x

c x cannot be greater than 4, as then the length becomes greater than 8 m (question puts restriction on side length). So, the largest value for x is 4.

d Domain [0, 4] Range [0, 24] Range is calculated by

substituting in the endpoints of the domain into P = 6x.

e P:[0, 4]→R, P(x) = 6x f Area = L × w = x × 2x = 2x2 or A:[0, 4]→R, A(x) = 2x2 g A(x) = 2x2 18 = 2x2 x2 = 9 x = ±3, x ≥ 0 x = 3 So, length = 3 m width = 6 m 5 a A = cost on plan A ($) t = time (min) A = 0.2 + 0.3t b B = cost on plan B ($) B = 0.5 + 0.2t

c i A = 0.2 + 0.3 × 2 = 0.2 + 0.6 = 0.8 = 80 cents ii B = 0.5 + 0.2t B = 0.5 + 0.2 × 2 = 0.5 + 0.4 = 0.9 = 90 cents d i A = 0.2 + 0.3t 5 = 0.2 + 0.3t t = 16 minutes ii B = 0.5 + 0.2t 5 = 0.5 + 0.2t t = 22.5 minutes

e 0.2 + 0.3t = 0.5 + 0.2t 0.3t − 0.2t = 0.5 − 0.2 0.1t = 0.3 t = 3 minutes f A = 0.2 + 0.3(4) = $1.40 B = 0.5 + 0.2(4) = $1.30 For t > 3 minutes, plan B becomes

cheaper. Thomas should choose plan B.