Page 20 - 1202 Bank Soalan Matematik Tingkatan 4

2
(c) Apabila f(x) = x + 5x + 4 dipantulkan Panjang sisi segi tiga = 5 – 1 Kertas 2
pada paksi-x, fungsi kuadratik = 4 cm
2
menjadi f(x) = –x – 5x – 4 di mana 19. Luas segi empat sama – Luas segi tiga Bahagian A
nilai a berubah kepada −a. = 40 1. (a) 4 2 (b) 6 2
14. Harga = RM300 2y(2y) – [ 1 (y + 2)(2y) = 40 (c) 8 1 (d) 2 5
]
(8x + 14)(5x) = 300 2 2. (a) 3 × 7 = 21 (b) 4 × 5 = 500
1
3
40x + 70x – 300 = 0 4y – y – 2y – 40 = 0 (c) 7 × 9 = 63 (d) 2 × 3 = 162
2
1
4
2
2
(x – 2)(4x + 15) = 0 3y – 2y – 40 = 0 0 1
2
15 3. (a) 32 = 2 × 4 + 3 × 4
4
x = 2, x = – (3y + 10)(y – 4) = 0
4 = 2 + 12
(Tidak diterima) y = – 10 (Tidak diterima) atau y = 4 = 14
3 (b) 11101
2
1
0
3
2
Bilangan buku yang dibeli oleh Sofea Perimeter gabungan objek = 1 × 2 + 0 × 2 + 1 × 2 + 1 × 2 +
= 8(2) + 14 = (4 + 6) + (4 + 2) + [3 × 2(4)] 1 × 2 4
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= 30 buku = 10 + 6 + 8 + 8 + 8 = 1 + 4 + 8 + 16
= 40 cm
15. Isi padu kotak = 4 500 1 = 29
2
1
0
(x + 5)(x)(30) = 4 500 20. (a) L = (y)(y + 4) (c) 413 = 3 × 5 + 1 × 5 + 4 × 5
5
2
2
30x + 150x – 4 500 = 0 1 = 3 + 5 + 100
2
2
x + 5x – 150 = 0 L = (y + 4y) = 108
2
1
0
(x – 10)(x + 15) = 0 L = y + 2y (d) 624 = 4 × 7 + 2 × 7 + 6 × 7 2
1
7
2
x = 10, x = –15 2 = 4 + 14 + 294
(Tidak diterima) (b) Luas segi tiga = 48 = 312
Maka, nilai x ialah 10 cm. 1 y + 2y = 48 4. (a) 534 kepada asas dua
2
2 10
16. Luas kotak = 432 1 2
(2x + 4)(3x + 3) = 432 2 y + 2y – 48 = 0 2 534 Baki
2
2
6x + 12x + 6x + 12 = 432 y + 4y – 96 = 0 2 267 0
2
6x + 18x – 420 = 0 (y + 12)(y – 8) = 0 2 133 1
2
x + 3x – 70 = 0 y = 8, y = –12
(x + 10)(x – 7) = 0 (Tidak diterima) 2 66 1
x = 7, x = –10 Maka, y = 8 cm. 2 33 0
(Tidak diterima)
Gantikan x = 7, = 21 + 3 (c) Luas poligon = 6 × 48 2 2 2 2 16 8 1 0

BAB 1 – BAB 2 Jadi, diameter 1 bola = 24 ÷ 4 21. (a) L = (35 + y)(65 + y) 2 534 = 1000010110

= 288 cm
Maka, luas poligon ialah 288 cm

Diameter 4 bola = 3(7) + 3
dan nama poligon ialah heksagon.

4
2
0
= 24 cm
2
L = 2 275 + 65y + 35y + y
2
0

= 6 cm
L = y + 100y + 2 275
2
0
2
1
(b)
Luas jubin = 6 175
Gantikan x = 7 ke dalam 2x + 4,
1
0
y + 100y + 2 275 = 6 175

2
Diameter 3 bola = 2(7) + 4
y + 100y – 3 900 = 0
2

= 14 + 4
2
10
= 18 cm
10
y = 30, y = –130 (Tidak diterima)
y = 30 cm.
Maka,
8
Baki
534
Jadi, diameter 1 bola = 18 ÷ 3 (y – 30)(y + 130) = 0 (b) 534 kepada asas lapan
= 6 cm
(c) Bahagian terkecil diwakili kawasan
Bahagian C DEFI. 8 66 6
17. (a) L = (x + 6)(x + 3) Luas DEFI = 0.3 m × 0.3 m 8 8 2
2
L = x + 6x + 3x + 18 = 0.09 m 2 8 1 0
2
L = x + 9x + 18 Bilangan jubin yang diperlukan
(b) Luas kek = 270 = 1.08 ÷ 0.09 0 1
2
x + 9x + 18 = 270 = 12 jubin 534 = 1026
10 8
x + 9x + 18 – 270 = 0 22. (a) Luas A – Luas B = 12 cm 2
2
x + 9x – 252 = 0 2x(x + 3) – x(x + 5) = 12 (c) 534 kepada asas lima
2
10
(x – 12)(x + 21) = 0 2x + 6x – x – 5x – 12 = 0 5 534 Baki
2
2
x = 12 cm x + x – 12 = 0
2
(c) Bilangan kek (x – 3)(x + 4) = 0 5 106 4
2
= 270 cm ÷ 9 cm 2 x = 3, x = – 4 (Tidak diterima) 5 21 1
= 30 kek Maka, x = 3 cm.
Maka, bilangan kek cukup untuk (b) Perimeter susunan yang baru 5 4 1
diberikan kepada rakan Safi. = 6 + 6 + 6 + 8 + 3 + 8 + 3 0 4
18. (a) Luas segi tiga, L = 40 cm 534 = 4114
1 10 5
= (x + 3)(x – 1)
2 BAB 2 (d) 534 kepada asas empat
10
x + 2x – 3
=
2 Kertas 1 4 534 Baki
(b) Luas segi tiga = 16 1. A 2. D 3. D 4. A 5. C 4 133 2
2
x + 2x – 3 = 32 6. C 7. A 8. B 9. B 10. D
x + 2x – 35 = 0 11. C 12. D 13. B 14. D 15. C 4 33 1
2
(x – 5)(x + 7) = 0 16. A 17. C 18. C 19. B 20. A 4 8 1
x = 5, x = –7 21. C 22. B 23. B 24. D 25. A
(Tidak diterima) 26. C 27. D 28. A 29. B 30. D 4 2 0
Apabila x = 5, 31. C 32. C 33. A 34. B 35. D 0 2
Tinggi segi tiga = 5 + 3 36. D 37. B 38. C 39. D 40. C 534 = 20112
= 8 cm 41. B 42. B 10 4
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