Page 20 - Modul A+1 MM Tingkatan 4

2
BAB 1 y = 2x – 5x + 2
BAB
x = k, y = 0
0 = 2k – 5k + 2
2
Soalan Berformat SPM (2k – 1)(k – 2) = 0
1
Kertas 2/Paper 2 k = — atau/or k = 2
1. x + 3x – 12 = 2(x + 4) 2
2
x + 3x – 12 = 2x + 8 k ialah integer./k is an integer.
2
x + x – 20 = 0 ∴ k = 2
2
(x – 4)(x + 5) = 0
BAB
x = 4 atau/or x = –5 BAB 2
2. 8x(x + 3) = 13 + 19x
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8x + 24x = 13 + 19x Soalan Berformat SPM
2
8x + 5x – 13 = 0
2
(8x + 13)(x – 1) = 0 Kertas 2/Paper 2
13
x = – —– atau/or x = 1 1. (a) Nilai bagi digit 3/Value of digit 3 Baki
2
8 = 3 × 9 5 263 Remainder
3. 2x(2x – 3) = 14 – 5x = 243 5 52 ……3
4x – 6x = 14 – 5x (b) 524 = 5 × 7 + 2 × 7 + 4 × 7 0 5 10 ……2
2
2
1
4x – x – 14 = 0 7 = 245 + 14 + 4
2
……0
(4x + 7)(x – 2) = 0 = 263 5 2 0 …… 2
7
x = – — atau/or x = 2 = 2023 10
4 5 Baki
4. (a) 6x + 7x = 3(8 – x) 2. (a) (i) y = 11021 2 115 Remainder
2
6x + 7x = 24 – 3x 4 3 3 1 2 57 ……1
2
2
6x + 10x – 24 = 0 (ii) y = 3 + 3 + 2 × 3 + 1 2 28 ……1
3x + 5x – 12 = 0 = 81 + 27 + 6 + 1 2 14 ……0
2
= 115
(3x – 4)(x + 3) = 0 = 1110011 2 7 ……0
10
4
x = — atau/or x = –3 2 2 3 ……1
3
(b) y + 6 = 5x … 2 1 ……1
(y + 6)x = 45 … 0 …… 1
5x = 45 (b) (i) 10003 = 1 × 4 + 0 × 4 + 0 × 4 + 0 × 4 + 3 × 4 0
2
4
2
3
1
x = 9 4 = 256 + 0 + 0 + 0 + 3
2
Baki
x = ±3 = 259 8 259 Remainder
x . 0, ∴ x = 3 = 403 10
Diameter/Diameter = 3 m ∴ m = 403 8 8 32 ……3
5. (a) 3(x – 1)(x – 3) = (2x – 5) 2 8 4 ……0
3(x – 4x + 3) = 4x – 20x + 25 (ii) 10003 = 4 0 3 0 …… 4
2
2
3x – 12x + 9 = 4x – 20x + 25 4 8
2
2
x – 8x + 16 = 0 = 100 000 011
2
(x – 4) = 0 2
2
x = 4 3. (a) 1p1 = 1110110 2
9
4
6
5
2
0
1
(b) (i) (ii) y 1 × 9 + p × 9 + 1 × 9 = 1 × 2 + 1 × 2 + 1 × 2 + 0 × 2 3
y
+ 1 × 2 + 1 × 2 + 0 × 2 0
2
1
81 + 9p + 1 = 64 + 32 + 16 + 0 + 4 + 2 + 0
4
9p + 82 = 118
Baki
x 9p = 36 6 118 Remainder
–3 O x p = 4
–2 O 2 6 19 ……4
141 = 118 6 3 ……1
9 = 314 10
6. (a) (i) (x + 3) = 2(x + 7) 6 0 …… 3
2
x + 6x + 9 = 2x + 14 (b) 2110 – 1201 = 202 1 4 0 3
2
x + 4x – 5 = 0 3 3 3 2110 3
2
(x + 5)(x – 1) = 0 1 1 – 1201 3
x = –5 atau/or x = 1 (c) (i) 4203 6 202 3
3x
1
(ii) ——–– = ——– + 513 6
x – 14 x – 4
3x(x – 4) = x – 14 5120 6 ∴w = 5120
3x – 12x = x – 14
2
1
2
3
3x – 13x + 14 = 0 (ii) 5120 = 5 × 6 + 1 × 6 + 2 × 6 + 0 × 6 0
2
6
Baki
(3x – 7)(x – 2) = 0 = 1080 + 36 + 12 + 0 9 1128 Remainder
7
x = — atau/or x = 2 = 1128 10
3 = 1483 9 9 125 ……3
(b) x = 0, y = 2 9 13 ……8
2 = 0 – 0 + a 9 1 ……4
∴ a = 2
0 …… 1
MG-1