Page 20 - 1202 Question Bank Additional Mathematics Form 4

CHAPTER 1 Let y = 2 – 4 x  = 2: –2q – 6 = –4q – 2
2q = 4
Paper 1 x = 2 – y q = 2
4 Substitute q = 2 into ,
Section A x = 8 – 4y p = –2(2) – 6
1. (a) 2(6) + 1 = 13 q(y) = 6(8 – 4y) – 3 = –10
2(8) + 1 = 17 = 45 – 24y 10
–1
Thus, the function is f (p) = 2p + 1, Thus, q(x) = 45 – 24x. (ii) g (x) = – 2 + x , x ≠ –2
(b) 2(p) + 1 = q 9. (a) f (x) = 2x – 5 10
q = 2p + 1 x + 2 Let y = – 2 + x
2. (a) y Let y = 2x – 5 2y + xy = –10
6 x + 2 xy = –10 – 2y
y(x + 2) = 2x – 5 –10 – 2y
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yx + 2y = 2x – 5 x = y
4
yx – 2x = –2y – 5 –10 – 2x
(y – 2)x = –2y – 5 Thus, g(x) = x , x ≠ 0
2 –2y – 5 (iii) g(x) = 8
x = y – 2 –10 – 2x = 8
0 x –2y – 5 x
2 4 6 8 10 Thus, f (y) = –10 – 2x = 8x
–1
(b) Many-to-many y – 2 10x = –10
(c) 2, 4 and 8 f (x) = –2x – 5 , x ≠ 2 x = –1
–1
3. (a) Since –6 < y < 4, thus a = –6 and x – 2 12. (a) 1
b = 4. (b) f (x) = 2x – 5 x –1 0 2 1 2
(b) –3 f (k) = 3 x + 2 f (x) 3 1 0 1 3
(c) –2 < x < 5
4. (a) f (x) 2k – 5 = 3 y
–1
(b) g f (x) or f g(x) k + 2
–1
–1
5. (a) f (p) = –6 2k – 5 = 3k + 6 3
k = –11
4p – 5p = –6 Thus, the value of k is –11. 2
p = 6 1
2
(b) (i) f (–2) = (–2) – 5(–2) 10. (a) f (x – 2) = 3 – 2(x – 2)

= 3 – 2x + 4
= 4 + 10 = 7 – 2x –1 0 1 2 x
= 14
2
(ii) x – 5x = 6 (b) Let y = 3 – 2x (b) (i) (a) ff (x) = f (2x)
2x = 3 – y

x – 5x – 6 = 0 3 – y = 4x
2
(x – 6)(x + 1) = 0 x = 2 = 2 x
2
x = 6 or x = –1 3 – x (b) fff (x) = ff (2x)
–1
6. ts(x) = t[s(x)] So, f (x) = 2 . = f (4x)
= t(6 – 4x) 2f (k) = f (k – 2) = 8x
–1
= p(6 – 4x) – 3 3 – k = 2 x
3
= 6p – 4px – 3 2 1 2 2 = 7 – 2k (ii) f (x) = 2 x where n = 1, 2,
n
n
Compare with ts(x) = q – 4px: 3 – k = 7 – 2k 3,….
Thus, q = 6p – 3. k = 4 (iii) f (x) = 16
5
7. (a) gf (–2) = 5 2 (x) = 16
5
g(2(–2) – 6) = 5 Section B 32x = 16
g(–10) = 5 11. (a) (i) –2 x = 1
1
h + 10k = 5 (ii) f (x) = |3 – x| 2
2
h = 5 – 10k 13. (a) (i) If f maps set L to M, then the
1
(b) (i) k = 1 f (1) = |3 – (1)| element in set M is represented
3x 2 by f (x). Thus, the function
(ii) = x 5
x – 1 = 2 that maps M to N must be g(x)
3x = x – x because from L to N is gf (x).
2
1
|
x – 4x = 0 (iii) 3 – x| = 1 x – 1
2
x(x – 4) = 0 2 Thus, g(x) = 2 .
1
x = 0, x = 4 3 – 1 x = –1 and 3 – x = 1 (ii) gf (x) = x – x + 2
2
(iii) h(x) = 3x 2 2 f (x) – 1 2
1
x – 1 – 1 x = –4 – x = –2 2 = x – x + 2
3x 2 2
2
Let y = x – 1 x = 8 x = 4 f (x) – 1 = 2x – 2x + 4
2
yx – y = 3x Thus, the domain is 4 < x < 8. f (x) = 2x – 2x + 5
yx – 3x = y –1 p (b) (i) Let y = 3x – 4
x(y – 3) = y (b) (i) Given g (x) = q + x , x ≠ –q 3x = y + 4
y g (3) = –2 x = y + 4
–1
x = 3
y – 3 p x + 4
–1
x q + 3 = –2 Thus, g (x) = .
Hence, h (x) = , x ≠ 3. 3
–1
x – 3 p = –2q – 6 ............. 2 p p
= g g
1
1 2 1 2
3
3
8. (a) qp – 1 2 = 6 – 1 2 – 3 g –1 1 2 = –4 (ii) g 1 2 3 1 24
2
p
3 1 2 4
= –6 p = –4 = g 3 3 – 4
(b) Given qp(x) = 6x – 3 q + 1 = g(p – 4)
1
q 2 – 4 x 2 = 6x – 3 2 p = –4q – 2 .........2 = 3(p – 4) – 4
= 3p – 16
99
12A_1202 QB AMath F4.indd 99 09/05/2022 11:38 AM