Learning Area: Algebra

Chapter

1 Functions

1 ff(x)) = 2
f( 5 g(t) = – 7 t

Bab
2x2 + 3x = 2 2
2x2 + 3x – 2 = 0 3t – 5 = – 7 t
2

(2x – 1)(x + 2) = 0 2
1 6t – 10 = –7t
2
x = or –2
2 6t2 + 7t – 10 = 0
(6t – 5)(t + 2) = 0
2 g(x) = x
5
4x + 7 x = or –2
=x 6
x–2
4x + 7 = x2 – 2x 6 y
x – 6x – 7 = 0
2
(1, 3)
(x + 1)(x – 7) = 0
2
x = –1 or 7
(–3, 1)

3 A function h is undefined if its

x
donominator = 0 –2 O
x2 – 4 = 0
x2 = 4 7 x –1 0 1 2 3
x = ±2
f(x) 7 3 1 5 9
m On the x-axis, f(x) = 0
4 f(x) =
2x – k
|4x – 3| = 0
f(2) =– 4 3
3 x=
m 4
=– 4
2(2) – k 3 Thus, the graph will touch the x-axis at 3 .
3m = –4(4 – k) 4
y
3m – 4k = –16 .... ① (3, 9)
f(–2) =– 4 (–1, 7)
11
m
=– 4
2(–2) – k 11
3
11m = –4(–4 – k)
11m – 4k = 16 .... ② O 3
x
② – ①: 4
8m = 32
m=4 8 f(x) = 4
From ①: |3x – 7| = 4
3(4) – 4k = –16 3x – 7 = ±4
–4k = –28 When 3x – 7 = 4
k =7 x = 11
7
2
=3
3

Analisis & Tip SPM F4 Add Maths 2021 -Jwp1.indd 1 22/02/2021 4:18 PM
 When 3x – 7 = –4 16 gf(x) = x2 – 3
x=1 g(x + 2) = x2 – 3
Let x + 2 = u
9 gf(3) = g 3 + 3  x=u–2
3–1
= g(3) g(u) = (u – 2)2 – 3
= 2(3) + 5 = u2 – 4u + 4 – 3
= 11 = u2 – 4u + 1
∴ g(x) = x2 – 4x + 1
10 gf(–26) = g(–26)
= g(|–26|) 2
Form 4

17 fg(x) = x
= g(26) 3
= √26 – 1 2
f [g(x)] = x
1

= √25 3
Chapter

=5 2
g(x) – 9 = x
3
11 h2(–3) = hh(–3) 2
g(x) = x+9
= h[(–3 + 1)2] 3
= h(4) 2
∴ g(–3) = (–3) + 9
= (4 + 1)2 3
= 25 =7

12 fg(2) = 4 18 fg(x) = f [g(x)]

f[3(2)2 – 7| = 4 = f (x – 2)
f(5) = 4 = 4 (x – 2) + k
p – 2(5) = 4 = 4x – 8 + k
p = 14
But it is given that
fg(x) = mx + 8.
13 gf(x) = g 2x  
x–2 By comparison,
m = 4 and –8 + k = 8
= 2x + 4
 
x–2 k = 16
2x + 4(x – 2)
=
x–2 19 gf(x) = g x – 1
 
6x – 8 2
= ,x≠2 x
x–2 =3 –1 +4
 
2
14 g2(x) = gg(x) 3(x – 1) + 8
=
= g(x2 – 3) 2
= (x2 – 3)2 – 3 3x + 5
=
= x4 – 6x2 + 6 2

1 20 fg(x) = f (ax + b)
15 fg(x) = = 9 – 2(ax – b)
x–1
1 = 9 – 2ax + 2b
f [g(x)] =
x–1 But it is given that fg(x) = 1 – 6x.
1 By comparison,
g(x) + 2 =
x–1 9 – 2b = 1 and –2a = –6
1 b=4 a=3
g(x) = –2
x–1
1 – 2(x – 1)
g(x) =
x–1
3 – 2x
g(x) = ,x≠1
x–1

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 21 (a) fg(x) = x2 + 6x + 7 When h = 5, 5k + k = 48
f [g(x)] = x2 + 6x + 7 6k = 48
f(x + 3) = x2 + 6x + 7 k =8
Let x + 3 = u When h = –5, –5k + k = 48
x =u–3 –4k = 48
f(u) = (u – 3)2 + 6(u – 3) + 7 k = –12
= u2 – 6u + 9 + 6u – 18 + 7
= u2 – 2 (b) When h = 5 and k = 8,
∴ f(x) = x2 – 2 f(x) = 5x + 8
f(x) = f (3x + 1)
(b) f(2k) = 8k + 30

Bab
Chapter
5x + 8 = 5(3x + 1) + 8
(2k)2 – 2 = 8k + 30 5x + 8 = 15x + 5 + 8
4k2 – 2 = 8k + 30 10x = –5
4k2 – 8k – 32 = 0
x=– 1

1
k2 – 2k – 8 = 0 2

Form 4
(k + 2)(k – 4) = 0
k = –2 or 4 24 (a) gf(x) = 2x + 9
g(2x + 5) = 2x + 9
22 (a) m(t) = 15 Let 2x + 5 = u
4t 2 + 4t = 15 x= u–5
4t 2 + 4t – 15 = 15 2
u – 5
(2t – 3)(2t + 5) = 0 g(u) = 2 2 +9
t = 3 or – 5 =u+4
2 2
g(x) = x + 4
(b) m(n) = m(2 + 3t)
fg(x) = f (x + 4)
= 4(2 + 3t)2 + 4(2 + 3t)
= 2(x + 4) + 5
= 4(4 + 12t + 9t 2) + 8 + 12t
= 2x + 8 + 5
= 16 + 48t + 36t 2 + 8 + 12t
= 2x + 13
= 36t 2 + 60t + 24
fg : x → 2x + 13
(c) n(t) = 17
(b) fg(k2 + 1) = 5k + 27
2 + 3t = 17
2(k2 + 1) + 13 = 5k + 27
3t = 15
2k2 + 2 + 13 = 5k + 27
3t = 5
2k2 – 5k – 12 = 0
m(t) = 4t 2 + 4t (2k + 3)(k – 4) = 0
m(5) = 4(5)2 + 4(5) 3
= 120 k = – or 4
2

23 (a) f(x) = hx + k 25 (a) gf(x) = g(4 – x)

f 2(x) = ff (x) = h(4 – x)2 + k
= f(hx + k) = h(16 – 8x + x2) + k
= h(hx + k) + k = hx2 – 8hx + 16h + k
= h2x + hk + k But it is given that
But it is given that gf(x) = 2x2 – 16x + 26.
f 2(x) = 25x + 48. By comparison, h = 2
By comparison, and
h2 = 25 16h + k = 26
h = 5 or –5 16(2) + k = 26
hk + k = 48 32 + k = 26
k = –6

Analisis & Tip SPM F4 Add Maths 2021 -Jwp1.indd 3 22/02/2021 4:18 PM
 (b) When h = 2 and k = –6, (x – 1) – 4
g(x) = 2x2 – 6. 4
=
g2(–1) = gg(–1) 4
= g[2(–1)2 – 6] x–5
=
= g(–4) 16
= 2(–4)2 – 6
= 26 30 Let g –1(x) = y
g(y) = x
26 mp(x) = x2 + 8 4y = x
m[p(x)] = x2 + 8 x
y=
Form 4

p(x) – 6 = x2 + 8 4
p(x) = x2 + 14 x
∴ g (x) =
–1

pm(x) = p(x – 6) 4
1

= (x – 6)2 + 14 x
 
Chapter

fg (x) = f
–1
= x2 – 12x + 36 + 14 4
= x2 – 12x + 50 x
+1
4
p(m) = 23 = x
x2 – 12x + 50 = 23 –1
4
x2 – 12x + 27 = 0
x+4
(x – 3)(x – 9) = 0 = ,x≠4
x–4
x = 3 or 9

27 Let f –1(3) = y 31 Let m–1(x) = y

f(y) = 3 m(y) = x
2y – 5 = 3 y
=x
2y = 8 4
y = 4x
y =4 ∴ m –1(x) = 4x
∴ f –1(3) = 4
Let n–1(x) = w
28 Let f –1(x) = y n(w) = x
f(y) = x 2w + 5 = x
y x–5
=x w=
y–2 2
y = xy – 2x x–5
∴ n –1(x) =
y – xy = –2x 2
y(1 – x) = –2x
m–1n–1(x) = m–1 x – 5
 
–2x 2
y=
1–x x – 5
2x
=4  2 
y=
x–1 = 2(x – 5)
2x = 2x – 10
∴ f –1(x) = ,x≠1
x–1 m–1n–1 : x → 2x – 10

29 Let g –1(x) = y 32 Let g–1(x) = y

g(y) = x g(y) = x
4y + 1 = x y–1 =x
4y = x – 1 y =x+1
x–1 ∴ g –1(x) = x + 1
y=
4
x–1 g –1f (–1) = g –1[(–1)2 + (–1) + 2]
∴ g (x) =
–1
= g –1(2)
4
x–1 –1 =2+1

4  =3
g(–1)2(x) =
4
4

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 x+3 2
33 Let y = ∴ f –1(x) =
2x x–1
2xy = x + 3
f –1(3k) = k
2xy – x = 3
f(k) = 3k
x(2y – 1) = 3
k+2
3 = 3k
x= k
2y – 1
3k2 – k – 2 = 0
3 1
∴ g(x) = ,x≠ (3k + 2)(k – 1) = 0
2x – 1 2
2
k=– or 1
3

Bab
Chapter
34 Let f –1(x) = y
f(y) = x
1
2y + p 37 (a) Let y =
=x m–x

1
5
ym – yx = 1

Form 4
2y + p = 5x
xy = my – 1
5x – p
y= x = my – 1
2
y
5x + 3
But it is given that f –1(x) = q . mx – 1
f (y) = x
By comparison, p = –3 and q = 2 mx – 1
∴ f (x) =
x
35 Let g –1(x) =y
g(y) =x (b) ff –1(m2 – 2) = g[(1 – m)2]
y m2 – 2 = 1 + (1 – m)2
=x m2 – 2 = 1 + 1 – 2m + m2
k2 – y
y = k 2x – xy 2m = 4
y + xy = k 2x m =2
y(1 + x) = k 2x
k2x 38 (a) f 2(x) = 4x – 9
y = f (m + n) = 4x – 9
1+x
k2x m(mx + n) + n = 4x – 9
∴ g–1(x) = , x ≠ –1 m2x + mn + n = 4x – 9
1+x
9x By comparison,
But it is given that g–1(x) = . m2 = 4
x+1
By comparison, m=2 The question states that m  0.

k2 = 9 mn + n = –9
k = ±3 2n + n = –9
3n = –9
36 Let f –1(x) = y n = –3
f(y) = x
y+2 (b) f (x) = 2x – 3
=x
y Let f –1(x) = y
y + 2 = xy f (y) = x
xy – y = 2 2y – 3 = x
y(x – 1) = 2 x+3
2 y=
x= 2
x–1 x+3
∴ f –1(x) =
2

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 (f –1)2(x) = f –1 f –1(x) 41 (a) Let g –1(x) = y
x+3 +3 g(y) = x
 2  3y + 8 = x
=
2 x–8
y=
x+3+6 3
=
4 x–8
∴ g (x) =
–1
x+9 3
=
4
fg–1(x) = f x – 8
 
3
4 – px x – 8 +4
39 (a) Let y = =9  
Form 4

2 3
2y = 4 – px = 3x – 24 + 4
px = 4 – 2y = 3x – 20
1

4 – 2y
x=
Chapter

p (b) gf(–x) = 23
4 – 2x g[9(–x) + 4] = 23
∴ f (x) = g(–9x + 4) = 23
p
3(–9x + 4) + 8 = 23
(b) f (x2) = 2g(–x) –27x + 20 = 23
4 – 2x2 –27x = 3
= 2[2(–x)2 – 4]
p 3
x=–
4 – 2x2 = 2p(2x2 – 4) 27
4 – 2x2 = 4px2 – 8p 1
=–
Equating the coefficients of x2, 9
4p = –2
42 (a) f(2) = 1
p=– 1
2 a
=1
b–2
40 (a) Let f –1(x) = y a=b–2 ....①
f(y) = x g(–5) = 1
h – ky = x 5 + b(–5)
=1
ky = h – x 3(–5)
h–x 5 – 5b
y= =1
k –15
h–x 1–b
∴ f (x) =
–1
=1
k –3
1 – b = –3
(b) f –1(8) = –1
b=4
h–8
= –1 From ①:
k
a=4–2
h – 8 = –k
a=2
h+k=8 ....①
f(4) = –2 (b) Let g –1(z) = w
h – k(4) = –2 g(w) = z
h – 4k = –2 ....② 5 + b(w)
=z
3w
① – ②: 5k = 10
5 + bw = 3wz
k=2
bw – 3zw = –5
From ①:
w(b – 3z) = –5
h+2 =8
–5
h =6 w=
6 – 3z
–5
∴ g–1(z) =
b – 3z

Analisis & Tip SPM F4 Add Maths 2021 -Jwp1.indd 6 22/02/2021 4:18 PM
 5 (b) fg(x) = f [(x – 1)2 – 3]
∴ g–1(y) =
3y – b = f(x2 – 2x + 1 – 3)
5 4 = f(x2 – 2x – 2)
∴ g–1(y) = ,y≠
3y – 4 3 = p(x2 – 2x – 2) + q
= px2 – 2px – 2p + q
(c) The function that maps x to z is
But it is given that
g–1f(x) = g–1 2   fg(x) = 3(x – 1)2 – 11
4–x
5 = 3(x2 – 2x + 1) – 11
= = 3x2 – 6x + 3 – 11
2
3 
4–x
–4  = 3x2 – 6x – 8

Bab
Chapter
5 By comparison,
= p = 3 and –2p + q = –8
6 –4
4–x –2(3) + q = –8

1
5 q = –2
=

Form 4
6 – 4(4 – x) (c) f(x) = 3x – 2
4–x g(x) = (x – 1)2 – 3
5(4 – x)
= Let f –1(x) = y
–10 + 4x
f(y) = x
20 – 5x 5
= ,x≠ 3y – 2 = x
4x – 10 2 3y = x + 2
x+2
y=
43 (a) Let f –1(x) = y 3
f(y) = x x + 2
∴ f–1(x) =
3y + 5 x 3
=
y–h
gf –1(x) = g x + 2
 
3y + 5 = xy – hx 3
3y – xy = –hx – 5 x + 2–1 2–3
y(3 – x) = –hx – 5 
=
3 
–hx – 5
= x+2–3 –3
 
2
y=
3–x 3
hx + 5
= x–1 –3
 
2
∴ f (x) =
–1
x–3 3
But it is given that (x – 1)2
= –3
2x + 5 9
f –1(x) = (x – 1)2 – 27
x–k =
9
By comparison, h = 2, k = 3
x2 – 2x + 1 – 27
=
(b) f –1(x) = 3x + 1 9
2x + 5 x2 – 2x – 26
= 3x + 1 =
x–3 9
2x + 5 = (3x + 1)(x – 3)
2x + 5 = 3x2 – 9x + x – 3 45 (a) Let f –1(x) = y
3x – 10x – 8 = 0
2
f(y) = x
(3x + 2)(x – 4) = 0 2y – m = x
x = – 2 or 4 y= x+m
3 2
x+m
∴ f (x) =
–1
44 (a) g2(–1) = gg(–1) 2
= g[(–1 –1)2 – 3] But it is given that
= g(1) 7
f –1(x) = nx +
= (1 – 1)2 – 3 2
= –3 2nx + 7
=
2
7

Analisis & Tip SPM F4 Add Maths 2021 -Jwp1.indd 7 22/02/2021 4:18 PM
 1 2x
By comparison, m = 7, n = 47 (a) f (x) =
2 x–4
f 2(x) = ff(x)
(b) (i) f(2) = 2(2) – 7
= –3
(ii) f –1f (2) = 2
=f  x2x– 4 
2 2x 
k–x x–4
46 (a) It is given that f(x) = . =
2x – 4
Denominator ≠ 0
5 + 2x
x – 4
5+2 ≠0 4x
Form 4

5 x–4
x ≠– =
2 2x – 4(x – 4)
1

It is given that x ≠ h. x–4

Chapter

5 4x
Hence, h = – =
2 2x – 4x + 16
1 4x
= –2 =
2 16 – 2x
2x
(b) (i) It is given that 1 is mapped > 8–x
itself under f.
Therefore, f(1) = 1 Let f –1(x) = y
k–1 f(y) = x
=1 2y
5 + 2(1) =x
k–1 =7 y–4
k =8 2y = xy – 4x
xy – 2y = 4x
8–x
(ii) When k = 8, f(x) = . y(x – 2) = 4x
5 + 2x
4x
For self-mapping, y=
x–2
f(x) = x
4x
8–x f –1(x) = ,x≠2
=x x–2
5 + 2x
8 − x = 5x + 2x2 (b) f 2 (x) = f –1(x)
2x + 6x −8 = 0
2
2x 4x
=
x2 + 3x − 4 = 0 8–x x–2
(x + 4)(x −1) = 0 2x(x – 2) = 4x(8 – x)
x = −4 or 1 2(x – 2) = 4(8 – x)
Hence, another value of x that is 2x – 4 = 32 – 4x
mapped onto itself (other then 1) is 6x = 36
–4. x =6
(iii) Let f –1(2) = y
f(y) = 2 48 (a) f (x) = 2x – 4
8–y Let f –1(x) = y
=2 f(y) = x
5 + 2y
8 − y = 10 + 4y 2y – 4 = x
5y = –2 x+4
y=
2 2
y =– x+4
5 ∴ f (x) =
–1
2
2
f (2) = –
–1
5

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 (b) y y=x
(b) g–1(5p) = p
g(p) = 5p
4 p2 + 6 = 5p
y = f (x) =
–1 x+4 p2 – 5p + 6 = 0
2 2 (p – 2)(p – 3) = 0
x
p = 2 or 3
–4 –2 2 4
O (c) fg(x) = gf (x)
–2 9x2 + 24x + 22 = x2 + 6
y = f(x) = 2x – 4 8x2 + 24x + 16 = 0
–4
x2 + 3x + 2 = 0

Bab
Chapter
(x + 2)(x + 1) = 0
x = –2 or –1
The graph of f –1 is the reflection of the

1
graph of f in the straight line y = x.
50 (a) h(x) = x

Form 4
(c) (i) The domain of f(x) is 0  x  4.
12
The range of f(x) is –4  f(x)  4. =x
x–4
(ii) The domain of f –1(x) is –4  x  4. 12 = x(x – 4)
The range of f –1(x) is 0  f –1(x)  4. x2 – 4x – 12 = 0
Conclusion: (x – 6)(x + 2) = 0
The domain of f –1(x) is the range of f(x). x = 6 or –2
The range of f –1(x) is the domain of f(x).
(b) (i) Let f –1(x) = y
f (y) = x
2 – 2y = x
49 (a) gf : x → 9x2 + 24x + 22
2y = 2 – x
gf (x) = 9x2 + 24x + 22
2–x
g[f(x)] = 9x2 + 24x + 22 y=
2
Let f (x) = u 2–x
∴ f (x) =
–1
3x + 4 = u 2
u–4
x= gf –1 : x → x2 – 4x + 5
3
g[f –1 (x)] = x2 – 4x + 5
g(u) = 9 u – 4 + 24 u – 4 + 22
   
2

3 3 g 2 – x = x2 – 4x + 5
 
(u – 4)2 2
g(u) = 9 × + 8(u – 4) + 22
9 Let 2 – x = u
 
g(u) = u2 – 8u + 16 + 8u – 32 + 22 2
g(u) = u2 + 6 2 – x = 2u
g(x) = x2 + 6 x = 2 – 2u
g(u) = (2 – 2u)2 – 4(2 – 2u) + 5
g(u) = 4 – 8u + 4u2 – 8 + 8u + 5
g(u) = 1 + 4u2
∴ g(x) = 4x2 + 1

Analisis & Tip SPM F4 Add Maths 2021 -Jwp1.indd 9 22/02/2021 4:18 PM