Page 70 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Answers

2. 1 + x (c) y
1 – 3x y = – 3x – 3 + 4 y = 3x + 3 + 4
1 1
— – —
2
= (1 + x) (1 – 3x) 2
6
1
1
3 1 1 21 – 1 2 4 5
2
2
= 1 + (x) + 2! (x) + … (–7, 4) (–4, 4) (–1, 4) 4 3 (2, 4) (5, 4)
2
2
1
3 1 1 – 2 1 21 – – 1 2 4 2 1
2
1 – (–3x) + 2! (–3x) + … –7 –6 –5–4 –3 –2 –1 0 1 2 3 4 5 x
2
2
1
3
1
= 3 1 + x – x + …43 1 + x + 27 x + … 4
2
2
2 8 2 8
3
1
3
1
2
2
2
= 1 + x + 27 x + … + x + x + … – x + … 3 –2 5 1 0 0
2 8 2 4 8 4. 1 1 0 2 u 0 1 0 2
= 1 + 2x + 4x + … 2 –1 4 0 0 1
2
1
1
Expansion is valid for : 5 x : – , x , 6
3 3 1 0 2 u 0 1 0
1 + 1 R ↔ R 2 1 3 –2 5 1 0 0 2
1
1
By taking x = , 1 + x = 9 → 2 –1 4 0 0 1
1
9 1 – 3x 1 – 31 2
9 R = –3R + R 2 1 0 2 u 0 1 0
2
1
= 5 → 1 0 –2 –1 1 –3 0 2
3 R = –2R + R 3 0 –1 0 0 –2 1
1
3
5 ≈ 1 + 21 2 + 41 2 2
1
1
1
0
3 9 9 R ↔ R 3 1 1 0 –1 2 0 u 0 0 –2 0 1 2
2
1
1
5 × 3 ≈ 1 + 21 2 + 41 2 2 → 0 –2 –1 1 –3 0
3 × 3 9 9
1
1
15 ≈ 1 + 21 2 + 41 2 2 1 0 2 u 0 1 0
3 9 9 R = –R 1 0 1 0 0 2 –1 2
2
2
1
1
2
15 ≈ 33 1 + 21 2 + 41 2 4 → 0 –2 –1 1 –3 0
9 9
≈ 103 1 0 2 u 0 1 0
27 R = 2R + R 1 0 1 0 0 2 –1 2
3
3
2
Alternative solution: → 0 0 –1 1 1 –2
5 ≈ 1 + 21 2 + 41 2 2 1 0 2 0 1 0
1
1
3 9 9 R = –R 1 0 1 0 u 0 2 –1 2
3
3
1
1
5 × 5 ≈ 1 + 21 2 + 41 2 2 → 0 0 1 –1 –1 2
3 × 5 9 9
5 2 1 0 0 u 2 3 –4
1
1
0
2
15 ≈ 1 + 21 2 + 41 2 R = –2R + R 1 1 0 0 1 0 0 1 –1 –1 –1 2
1
3
9
9
→
2
15 ≈ 5 2
1
1
1 + 21 2 + 41 2 3 –2 5 2 3 –4
9
9
0
2
=
0
≈ 405 The inverse matrix of 1 1 2 –1 2 4 2 1 –1 –1 –1 2
2
103
3. (a) 27x – 9y + 54x + 72y – 360 = 0
2
2
27(x + 2x) – 9(y – 8y) – 360 = 0 3 –2 5 x 45
2
2
y = 15
2
2
27[(x + 1) – 1] – 9[(y – 4) – 16] – 360 = 0 1 1 0 2 21 2 1 2
2
27(x + 1) – 9(y – 4) – 27 + 144 – 360 = 0 2 –1 4 z 32
2
2
27(x + 1) – 9(y – 4) = 243 2 3 –4 3 –2 5 x 2 3 –4 45
2
y = 0
(x + 1) 2 – (y – 4) 2 = 1 1 0 2 –1 21 1 0 2 21 2 1 2 –1 21 2
15
9 27 –1 –1 2 2 –1 4 z –1 –1 2 32
2
(b) a = 3, c = a + b = 9 + 27 = 6 x 90 + 45 – 128
2
y =
Centre = (–1, 4) 1 2 1 0 + 30 – 32 2
Vertices = (–4, 4) and (2, 4) z –45 – 15 + 64
Foci = (–7, 4) and (5, 4) x 7
1 2 1 2
y = –2
Equation of asymptotes: z 4
(y – 4) = ± 3 3 (x + 1)
3 Hence, x = 7, y = –2, z = 4.
y = – 3x – 3 + 4 and y = 3x + 3 + 4
289
Answers STPM Math T S1.indd 289 3/28/18 4:25 PM