1

1 Based on the given ordered pairs {(2, 1), (4, 3), (6, 5), (6, 7)}, an arrow diagram can be drawn as shown below.

2

4

6

1

3

5

7

9

P

Q

(a) The image of 2 is 1.(b) The object of 7 is 6.

2 (a) Let g−1(7) = y Thus, g(y) = 7 4y – 1 = 7 4y = 8 y = 2 ∴ g−1(7) = 2

(b) hg(x) = h(4x – 1) = (4x – 1)2 – 3(4x – 1) + 5 = 16x 2 – 8x + 1 – 12x + 3 + 5 = 16x 2 – 20x + 9

3

9

28

49

3

5

7

11

Set P Set Q

(a) The range is {3, 7}.

‘3’ and ‘7’ are linked to object(s) but ‘5’ and ‘11’ are not linked to any object. Therefore, the range is {3, 7}.

(b) The above relation is a many-to-one relation.

Element ‘7’ in the codomain is linked to two elements, i.e. ’28’ and ‘49’ in the domain. Therefore, it is a many-to-one relation.

4 h : x → 2x + m h(x) = 2x + m

Let h−1(x) = y h(y) = x 2y + m = x

y = x – m2

∴ h−1(x) = x – m2

= x2

– m2

But it is given that h–1(x) = 3kx + 32

.

Hence, by comparison,

3k = 12

⇒ k = 16

and

– m2

= 32

⇒ m = –3.

5 (a) hg(x) = 6x – 2 h[g(x)] = 6x – 2 3g(x) + 1 = 6x – 2 3g(x) = 6x – 3 g(x) = 2x – 1

It is given that h(x) = 3x + 1.Hence, h[g(x)] = 3g(x) + 1.

Form 4: Chapter 1 (Functions)SPM Practice

Fully Worked Solutions

Paper 1

2

(b) gh(x) = g(3x + 1) = 2(3x + 1) – 1 = 6x + 1

When gh(x) = 13

,

6x + 1 = 13

18x + 3 = 1 18x = –2

x = – 19

6 (a) From the given arrow diagram, f(–2) = –5. Hence, f –1(–5) = –2.

(b) gf(–2) = 3

This is a composite function gf(x) which maps x directly onto z.

7 (a) Let w–1(x) = z w(z) = x

63 – 2z

= x

Change the subject of the formula to z.

6 = x(3 – 2z) 6 = 3x – 2xz 2xz = 3x – 6

z = 3x – 62x

∴ w−1(x) = 3x – 62x

, x ≠ 0

(b) w−1h – 52

= w−12– 52 + 3

= w−1(–2)

= 3(–2) – 62(–2)

= 3

8 Let n−1(x) = y n(y) = x 4y – 1 = x 4y = x + 1

y = x + 14

n−1(x) = x + 14

mn−1(x) = m x + 14

=

3

8 x + 14 – 5

= 32x – 3

, x ≠ 32

9 (a) The relation between set P and set Q is a many-to-one relation.

(b) The relation can be represented by f(x) = x4.

Function notation

10 m (2) = 7

2 – hh

= 7

2 – h = 7h 8h = 2

h = 14

11 (a) k = 9

(b) The relation can be represented by f(x) = x – 1.

Function notation

12 f(x) = 9 |x – 4 | = 9 x – 4 = ±9

x – 4 = 9 or x – 4 = –9 x = 13 or x = –5

3

13 g2 (x) = gg(x) = g(px + q) = p(px + q) + q = p2x + pq + q

But it is given that g2 (x) = 49x – 32.

Hence, by comparison, p2 = 49 p = 7 (p > 0)

pq + q = –32 7q + q = –32 8q = –32 q = –4

14 (a) f(x) = 0 |2x – 3 | = 0 2x – 3 = 0

x = 32

k = 32

(b) f(4) = |2(4) – 3 | = |5 | = 5

The range of f(x) is 0 f (x) 5.

15 (a) Let g–1 (6) = y g(y) = 6 3y + 2 = 6 3y = 4

y = 43

∴ g–1 (6) = 43

(b) hg(x) = h(3x + 2) = (3x + 2)2 – 2(3x + 2) + 5 = 9x2 + 12x + 4 – 6x – 4 + 5 = 9x2 + 6x + 5

16 (a) f(6) = 6 + 4 = 10

(b) gf(6) = 24 g(10) = 24 10t – 6 = 24 10t = 30 t = 3

17 (a) The object of 2 is 6.

(b) The range is {1, 2, 3}.

18 (a) fg(3) = 4(3) + 8 = 20

(b) fg(x) = 4x + 8 2g(x) + 6 = 4x + 8 2g(x) = 4x + 2 g(x) = 2x + 1

19 (a) f(5) = 5 + 3 = 8

(b) Let f –1(x) = y f(y) = x y + 3 = x y = x – 3 ∴ f –1(x) = x – 3

2f –1(k) = f(5) 2(k – 3) = 8 k – 3 = 4 k = 7

4

1 (a) f : x → 2x – 3 f(x) = 2x – 3

Let f –1(x) = y f(y) = x 2y – 3 = x

y = x + 32

∴ f –1(x) = x + 32

f –1g(x)

= f –1 x2

+ 2

=

x2

+ 2 + 3

2

=

x + 4 + 622

= x + 104

∴ f –1 g : x → x + 104

(b) hg : x → 2x + 4 hg(x) = 2x + 4

h x2

+ 2 = 2x + 4

Let x2

+ 2 = u

x2

= u – 2

x = 2u – 4

h(u) = 2(2u – 4) + 4 = 4u – 8 + 4 = 4u – 4

∴ h : x → 4x – 4

Paper 2