Page 20 - 1202 Bank Soalan Matematik Tambahan Tingkatan 4

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BAB 1 Katakan y = 2 – x g –1 1 2 2 = –4
4
Kertas 1 x = 2 – y p
4 1 = –4
Bahagian A x = 8 – 4y q + 2
1. (a) 2(6) + 1 = 13 q(y) = 6(8 – 4y) – 3
2(8) + 1 = 17 = 45 – 24y p = –4q – 2 .........2
Maka, fungsi f (p) = 2p + 1. Maka, q(x) = 45 – 24x.  = 2: –2q – 6 = –4q – 2
(b) 2(p) + 1 = q 2x – 5 2q = 4
q = 2p + 1 9. (a) f (x) = x + 2 q = 2
2. (a) y Gantikan q = 2 ke dalam ,
Katakan y = 2x – 5 p = –2(2) – 6
6 x + 2 = –10
y(x + 2) = 2x – 5 10
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–1
4 yx + 2y = 2x – 5 (ii) g (x) = – 2 + x , x ≠ –2
yx – 2x = –2y – 5
10
2 (y – 2)x = –2y – 5 Katakan y = – 2 + x
–2y – 5 2y + xy = –10
x =
0 x y – 2 xy = –10 – 2y
2 4 6 8 10
(b) Banyak kepada banyak Maka, f (y) = –2y – 5 x = –10 – 2y
–1
(c) 2, 4 dan 8 y – 2 y
3. (a) Oleh kerana –6 < y < 4, maka f (x) = –2x – 5 , x ≠ 2 Maka, g(x) = –10 – 2x , x ≠ 0
–1
a = –6 dan b = 4. x – 2 (iii) g(x) = 8 x
(b) –3 2x – 5
(c) –2 < x < 5 (b) f (x) = x + 2 –10 – 2x = 8
x
4. (a) f (x) f (k) = 3 –10 – 2x = 8x
–1
(b) g f (x) atau f g(x) 10x = –10
–1
–1
5. (a) f (p) = –6 2k – 5 = 3 x = –1
k + 2
4p – 5p = –6 2k – 5 = 3k + 6 12. (a)
p = 6 k = –11 x –1 0 1 1 2
2
(b) (i) f (–2) = (–2) – 5(–2) Maka, nilai k ialah –11. 2
= 4 + 10 10. (a) f (x – 2) = 3 – 2(x – 2) f (x) 3 1 0 1 3
= 14 = 3 – 2x + 4 y
(ii) x – 5x = 6 = 7 – 2x
2
x – 5x – 6 = 0 (b) Katakan y = 3 – 2x 3
2
(x – 6)(x + 1) = 0 2x = 3 – y
x = 6 atau x = –1 2 Bab 1
6. ts(x) = t[s(x)] x = 3 – y 1
= t(6 – 4x) 2
= p(6 – 4x) – 3 Maka, f (x) = 3 – x . 0 x
–1
= 6p – 4px – 3 2 –1 1 2
–1
Bandingkan dengan ts(x) = q – 4px: 2f (k) = f (k – 2) (b) (i) (a) ff (x) = f (2x)
3 – k
Maka, q = 6p – 3. 2 1 2 2 = 7 – 2k = 4x
7. (a) gf (–2) = 5 = 2 x
2
g(2(–2) – 6) = 5 3 – k = 7 – 2k (b) fff (x) = ff (2x)
g(–10) = 5 k = 4 = f (4x)
h + 10k = 5 = 8x
h = 5 – 10k Bahagian B = 2 x
3
(b) (i) k = 1 11. (a) (i) –2 (ii) f (x) = 2 x di mana n = 1, 2,
n
n
3x 3,….
1
(ii) = x (ii) f (x) = |3 – x| 5
x – 1 (iii) f (x) = 16
3x = x – x 2 2 (x) = 16
5
2
x – 4x = 0 32x = 16
2
1
x(x – 4) = 0 f (1) = |3 – (1)| 1
x = 0, x = 4 2 x = 2
3x = 5 13. (a) (i) Jika f memetakan set L kepada
(iii) h(x) = x – 1 2 M, maka unsur dalam set M
3x diwakili oleh f (x). Maka,
Katakan y = x – 1 | 1 fungsi yang memetakan M
2
yx – y = 3x (iii) 3 – x| = 1 kepada N mestilah g(x) kerana
yx – 3x = y 1 1 dari L kepada N ialah gf(x).
x(y – 3) = y 3 – 2 x = –1 dan 3 – x = 1 Oleh itu, g(x) = x – 1 .
2
y 1 1 2
x = y – 3 – 2 x = –4 – x = –2 (ii) gf (x) = x – x + 2
2
2
x x = 8 x = 4 f (x) – 1
Maka, h (x) = , x ≠ 3. = x – x + 2
2
–1
x – 3 Maka, domain ialah 4 < x < 8. 2
2
p f (x) – 1 = 2x – 2x + 4
f (x) = 2x – 2x + 5
–1
1 2 2 1 2 2
2
8. (a) qp – 1 = 6 – 1 – 3 (b) (i) Diberi g (x) = q + x , x ≠ –q (b) (i) Katakan y = 3x – 4
–1
= –6 g (3) = –2 3x = y + 4
(b) Diberi qp(x) = 6x – 3 p = –2 y + 4
q + 3 x = 3
1
q 2 – x 4 2 = 6x – 3 p = –2q – 6 ............. Maka, g (x) = x + 4 .
–1
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Jaw(A)_1202 BS MateTamb Tg4.indd 99 01/08/2022 3:55 PM