PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Answers

7. (a) y (b) y 12. y

1 5
1
4
x
x –1 0 3
–1 0 –1
1
x
–2 0 2 5
8. (a) y (b) y –2
2 –5
1
13.
y
x
0 1 x 4
_
2 0 1 2
3
(c) y (d) y 2
1
4 1 x
3 x 3 0 1 3 3 5
5 0 – – – –
_
– 2 2 2
2 7
– –
3
x
–1 0 –4
14. y
9. 1 , x , 3
8
y
y = |3x – 5| 6
5
3
2
x
y = |x + 1| –4 –2 –1 0 2 4
x
–1 0 1 5 3 –3
–
3
10. x , 4 – 10, 3 – 3 , x , 3 + 3, x . 4 + 10 Exercise 1.4
y 1
y = |x| 1. (a) 1 (b) 2 (c) 13
2
5
2. (a) –7 (b) –2 (c) 11
6 8
2
2
3. (a) x + 5x + 1 (b) x – x – 2
2
2
(c) 5x + 4x – 3 (d) x + 2x – 1
3
2
2
3
y = |x – 7x + 6| (e) 2x + 3x – x – 1
4. (a) A = 1, B = 2 (b) A = 3, B = 4
5
(c) A = 3, B = –1 (d) A = –2, B = , C = 7 3
3
3
3
x (e) A = , B = , C = –2
0 1 7 _ 6 2 2
2
3
2 5. (a) 2x + 3x + 5x + 14
2
3
4
(b) 6x – 13x + 16x – 17x + 3
11.
4
3
2
y (c) 4x + 11x – 10x + 5x – 2
(d) 5x – 7x + 3x + 2x – 3
3
2
5
6 (e) 9x – 4x + 18x – 8
3
2
5
2
5 6. (a) 3x – 5 ; 10 (b) 2x + x + 4 ; 13
2
2
4
(c) x – 3x + 9 ; –30 (d) x – 2x + 3x + 1 ; 0
2
(e) x – x ; 5x – 5
3
7. a = 3, b = –5
8. A = 3, B = 1
1 9. x + x – 1
2
x 10. a = 2, b = 0, c = 3
–4 –2 0 2 4
270
Answers STPM Math T S1.indd 270 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers
Exercise 1.5 Exercise 1.7
1. (a) 20 (b) 35 (c) 6 1. 2
1
3
(d) 2 (e) 1 (f) –6 1 x + 1
8 8 27 2 1
2. –
2. (a) Yes (b) No (c) Yes x + 2 x – 1
(d) Yes (e) Yes (f) No 3. 1 – 1
2
3. (a) (x – 1) (2x + 1) 2(x – 3) 2(x + 3)
(b) (x + 1)(x – 2)(3x + 1) 4. 1 + 1
(c) (x + 3)(x – 3)(x + 8) 2(x + 1) 2(x – 1)
2
(d) x(1 – x + x )(1 + x + x ) 5. 1 + 3
2
2
(e) (x + 2)(2x – 1)(2x – 3) 4(x + 2) 4(x – 2)
(f) (x + 1)(x – 2)(2x + 1)(2x – 1) 6. 1 + 2
8
4. – 3(x + 4) 3(x + 1)
9 7. 1 + 2 – 3
2
6. a = , b = 7 2 + x 3 + x 4 + x
3 3 2 1 1
7. p = 3, q = –6 8. – x + 2 + x – 1 + x + 1
8. a = –5, b = 7 7 7 2
9. a = 3, b = –5 9. 24(x – 3) + 8(x + 1) – 3x
10. 3, –2 10. 3 + 5 – 4
11. k = 6, 0, –6 2(x + 1) 6(x + 3) 3x
12. a = 3, 3(x – 3)(x + 1) ; a = –3, –3(x + 3)(x – 1) 2 11. 1 – 1
2
13. 1, –1 ± 3 1 + x (1 + x) 2 3
2
1
1
1
14. 2; , –1, –3 12. x – 1 – x + 2 – (x + 2) 2
2
15. a = 3, b = –9, x = –1, 2, 2 13. – 1 + 5 + 4 2
16. a = 10, b = 1; 4 2 3(2x – 1) 3 3(x – 2) (x – 2)
1
Exercise 1.6 14. x – – 2x + 1
2
x
1. (a) –1 < x < 2 (b) x , –5, x . 3 15. – 2 + 1 – 1
1
1
(c) –4 , x , 3 (d) x < – , x > 1 x 2(x – 1) 2(x + 1)
2 2 4 1 1
2
(e) –10 < x < 3 (f) x , – , x . 2 16. 1 + x – 1 – x + 1
5
1
1
2. (a) {x : x < –3 or x > 3} (b) {x : x ∈ R, x ≠ –1} 17. 1 – x + 1 – x + 2
2
(c) {x : –3 < x < –2} (d) {x : – < x < 1} 4 1
5
1
1
(e) {x : –1 , x , } (f) {x : x < – , x > 3} 18. 2 – x + 2 + x + 1
2 2
4
1
3. (a) x , –3, –2 , x , 1 (b) x < –1, x = 2 19. x – 2 + x + 1 – (x + 1) 2
1
(c) x > 1, x = –2 (d) x , –2, – , x , 1 20. x + 2 – 1
2
1
4
3
(e) x < – , < x < 4 (f) –2 < x < , x > 5 x – 1 x + 1
5 2 2 x 1
2
4. (a) –3 , x , –2, x . 1 21. x + 1 – x + 2
x
(b) –1 , x , 1 2 22. 1 x – x + x + 1
2
3
(c) x < – , 2 , x < 4 1 1 – x
2 23. x – 1 + x + x + 1
2
1
(d) x , –2, – , x , 1, x . 3
2 24. 1 – x
2
(e) x , –3, x . 4, x = 1 x – 1 x – x + 1
1
(f) x , (7 – 3 5 ), , x , 2, x . (7 + 3 5 ) 25. 1 – 1 + 1 2 + 1 2
1
1
2 2 2 8(x + 1) 8(x – 1) 4(x – 1) 2(x – 1)
5. (a) 1 , x , 3 (b) x < –2, x > 8 Exercise 1.8
1
(c) x , –3, x . (d) –3 < x < 8 1. (a) y (b) y
3
1 7
(e) x > (f) 1 , x ,
2 3
5
(g) –6 < x < 0 (h) x , – , x . 5 2
6
(i) x . – 3 (j) x , –8, x . – 8 1 1
5 3 x x
(k) –4 < x < –1, 1 < x < 4 0 0
(l) x , 0
271
Answers STPM Math T S1.indd 271 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers

(c) y (d) y 5. (a) y (b) y
1
x
0
x
–1 0 1
1 x
0 2
x
0
(c) y (d) y
2. (a) y (b) y
x
1 0
– e 1 e
y = 1
x x
0 0 1
x
0 Exercise 1.9
y = –1
1. (a) 1 (b) 6 (c) 1 7
1
(d) 27 (e) 8 (f) 1 000
(c) y (d) y (g) 8 (h) 6 (i) 4
(j) 6 (k) 3 (l) 2
1 2. (a) 5 2 (b) 3 2 (c) 5 – 2 6
x 2 (d) 66 – 24 6 (e) 18 (f) 2 5
0 y = 1 (g) 7 (h) 13
y = –1
x 1
0 3. (a) 2 5 (b) 6
3
(c) 2 – 2 (d) 5( 3 – 2 )
(e) 1 (2 15 + 3) (f) 1 (1 + 2 )(1 + 3 )
3. (a) y (b) y 17 2
(g) 1 ( 3 + 2 2 )( 5 + 3 ) (h) 1 (4 – 3 3 )( 6 + 2 3 )
2 6
(i) 2 2 (j) 4
x x
0 1 –1 0 1 1
4. (a) 4 (b) 3 (c) – 2
2
(d) 1 (e) – (f) –3
2 3
(c) y x = 1 (g) 3 2 (h) –2
4
5. (a) log 1 2 (b) 2 log 3
3 3 2
3
x (c) 0 (d) log 1 2
0 2 3 4
1
(e) – (f) –log (x – 1)
3
— 1 p(p – q)
2
(g) log |(x – 1) (x – 2) | (h) log 1 2
2
q 2
6. (a) 5 (b) 2 (c) 3
4. (a) y (b) y
Exercise 1.10
1. (a) 3 (b) 2 (c) 3
1 4
x x 4 5
0 1 –1 0 1 (d) –2 (e) 3 (f) 2
_
2
2. (a) –5 (b) 1
(c) –1 (d) 2
3. (a) 0, 3 (b) 0, 2 (c) 1, 2
(c) y (d) –1, 2 (e) 1, 2 (f) ± 1
x = –1 x = 1 2
a
4. (a) 1 + 2a + b (b) 5a – b (c) a + b
x 1
0
– 2 2 5. (a) 1 + 2p (b) p
(c) 1 p (d) – 1
2 p
272


Answers STPM Math T S1.indd 272 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers

6. (a) 22.627 (b) 0.463 (c) 11.212 (f) y
(d) 1.035 (e) 0.499 (f) 3.824
7. (a) 1.683 (b) –3.822 (c) –0.523 1
(d) –0.339 (e) –1.795
8. (a) 4 (b) 3 (c) 9, 1 9
(d) 2, 64 (e) 1 , 64 (f) 5, 25 0 2π x
2
9. (a) x , 3.170 (b) x . –0.183 (c) x . 8.638 (g)
(d) x . 1.853 (e) x . 6.261 y
10. (a) x , 1, x . 1.585 (b) x , 1.099 2
Exercise 1.11
1. (a) (b)
y y 0 2π x
3
1
(h) y
x x
0 180 360 0 180 360
–1 1
–3
x
(c) 0 2π
–1
y
1
3. (a) (b)
x y y
0 30 180 360
–1
2. (a) 0 π _ π _ 3π π x 0 π _ π _ 3π π x
y
4 2 4 4 2 4
1
(c)
x
0 2π y
–1
(b) y x
0 _ π π _ 3π π
1 4 2 4

x
0 2π Exercise 1.12
–1
3
1. (a) 3 , , 5 (b) 8 , – 17 , – 15
5 4 4 17 15 8
(c) y 24 7 25 5 12 12
(c) – , – , – (d) , – ,
2 25 25 24 13 13 5
2. (a) cos q (b) tan q (c) sin q
4
2
x (d) tan q (e) sec q (f) tan q
0 2π 3. (a) sin q (b) cos q + sin q
4
(c) 1 + cos q (d) sec q – sin q
–2
(d) y Exercise 1.13
1 1. (a) 1 ( 6 + 2 ) (b) 1 ( 2 – 6 )
4 4
(c) 3 – 2 (d) 2 – 3
x
0 2π 2. (a) 1 (b) 1 (c) 0 (d) 1
2 2
–1 1 1
3. (a) 3 (b) (c) 6 (d) 2
(e) y 56 33 3 2
4. , –
3 65 65
5. – 29 , – 8 6
x 35 35
0 2π
6. 1, 225°
–3 1
7. 4
2
273


Answers STPM Math T S1.indd 273 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers

9. (a) sin 3x + sin x (b) cos 8q + cos 2q Exercise 1.16
1
3
(c) 3(cos x – cos 5x) (d) 1 (sin 6q – sin 2q) 1. –π , x , – π, π , x , π
2 4 4
π
(e) cos 2A + cos 2B (f) 1 (1 + 2 sin 2q) 2. – , x , 0.212π, 0.788π , x , π
4 2
10. (a) 2 sin 3x cos 2x 3. –π , x , –0.732π, 0.268π , x , π
(b) 2 cos 5A cos 2A 11 7 5 1
(c) –2 cos 4x sin x 4. – 12 π , x , 12 π, – 12 π , x , – 12 π,
(d) 2 sin 3A sin 2A 1 5 7 11
(e) 2 sin 45° cos 15° 12 π , x , 12 π, 12 π , x , 12 π
(f) 2 cos 60° cos 10° 5. –0.885π , x , –0.615π, –0.385π , x , –0.115π,
(g) 2 cos x sin y 0.115π , x , 0.385π, 0.615π , x , 0.885π
(h) –2 sin 2A sin A 1 1
11. (a) cos x (b) 0 6. – π , x , 0, π , x , π
4
4
12. (a) cot 5q (b) –tan 3q 7. –π , x , –0.580π, 0.580π , x , π
(c) cot 2x (d) –tan x
2
π
8. –π , x , – , 0 , x , π
Exercise 1.14 2 2
1. (a) 23°35´, 156°25´ (b) 60°57´, 240°57´ STPM PRACTICE 1
(c) 138°35´, 221°25´ (d) 226°3´, 313°57´ 1
2. (a) –49°28´, 49°28´ (b) 56°6´, 123°54´ 1. f : x ⟼ 1 x – 3 3 — –1
–1
2 ; graph of f is the reflection of the graph
(c) –37°36´, 142°24´ (d) –111°36´, 111°36´ 4
3. (a) 36°52´, 216°52´ (b) 135°, 315° of f about y = x.
–1
(c) 48°35´, 131°25´ (d) 48°11´, 311°49´ 2. fg : x ⟼ –ln x, f : x ⟼ e x
4. (a) 60°, 120°, 240°, 300° (a) graph of fg is the reflection of the graph of f about the
(b) 60°, 90°, 270°, 300° x-axis.
–1
(c) 0°, 78°28´, 180°, 281°32´, 360° (b) graph of f is the reflection of the graph of f about y = x
(d) 270° 3. (a) y
(e) 0°, 30°, 150°, 180°, 360°
(f) 56°19´, 90°, 236°19´, 270° 6
3
π
5. (a) – π, –0.148π, , 0.852π
4 4
(b) –0.920π, –0.080π
π
(c) – , 0, π 2
3 3
3
π
(d) –0.621π, – , 0.379π, π 0 x
4 4 –3 –2 –1 1 2 3
6. (a) 41°25´, 120°, 240°, 318°35´ –2
(b) 60°, 300°
(c) 70°32´, 120°, 240°, 289°28´ Range of f is [–2, 6]
(d) 19°28´, 30°, 150°, 160°32´
–1
(e) 18°26´, 135°, 198°26´, 315° (b) g : x ⟼ x – 4 – 1
(f) 45°, 75°58´, 225°, 255°58´ 4. (a) f (x) = 1 ± x – 1
–1
4
3
7. (a) 2 π, π (b) 0.232π, 0.768π, π (b) 4 < x < 3
3 3 2 3 2
7
3
5
(c) π , π, π, π (d) 0, 0.580π, π, 1.420π, 2π (c) No
4 4 4 4
5. f is a one-to-one function
1
1
1
1
1
Exercise 1.15 [–1, 1], [– π, π], π, , π
1. (a) 135°, 161°34´, 315°, 341°34´ 2 2 6 2 6
1
(b) 38°58´, 162°50, 218°58´, 342°50´ 6. (a) g : x ⟼ x + 1 (b) g : x ⟼ x + 1
2
2
2. (a) 2 cos (q + 45°); max. 2, q = 315°, min. – 2, q = 135° 7. (a) f : x ⟼ (x + 1) 2 (b) f : x ⟼ x – 1
2
(b) 2 cos (q + 30°); max. 2, q = 330°; min. –2, q = 150°
(c) 13 cos (q – 67°23´); max. 13, q = 67°23´; 8. f : No, [–1, 1]
min. –13, q = 247°23´ g : No, [1  2 ]
(d) 5 cos (q – 26°34´); max. 5, q = 26°34´; h : Yes, [–∞, ∞]
min. – 5 , q = 206°34´ — 1 3
–1
–1
9. (a) f (x) = x – 1 (b) g (x) = (x + 3)
3. (a) 74 sin (q – 0.951) (b) 3 5 cos (q – 0.464) y y g
(c) 5 5 cos (q + 0.464) (d) 65 sin (q + 0.519) f –1
f –1 g
4. (a) 26°10´, 110°14´ (b) 41°36´; 244°40´
(c) 180°, 313°36´ (d) 97°58´, 205°54´ 0 x 0 x
6. (a) 10°54´, 231°1´ (b) 257°36´, 349°48´
7. r = 5, s = 6, a = 36°52´
274
Answers STPM Math T S1.indd 274 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers

–1
(c) h (x) = log x (d) k (x) = 10 x 13. (a) (b)
–1
2
y
y y
h k –1
h –1 k y
3
x x
0 0
x
0
10. (a) x
y = f(x) y = h(x) 0 1 3 (2,–1)
f(x) = e + 1 (2,–1) x = 1 x = 3
x
h(x) = x – 4
2
x
2 0 2 14. (a) (b)
1 y
x y
0 –4
(b) (i) Any horizontal lines y = k cuts the graphs f(x) and x
h(x) only at one point, f(x) and h(x) are one-to- x 0
one functions, therefore f and h exist. 0, 0,
–1
–1
1
1
–
a
( ) – 2 ( ) – 2
a
(ii) f : x ↦ ln (x – 1)
–1
–1
Domain f (x): {x : x . 1} x = –a x = a x = –a x = a
h : x ↦ x + 4
–1
–1
Domain h (x): {x : x > –4} (c) y
2
(c) fh(x) = e x – 4 + 1
Domain f h: {x : x > 0}
°
Range f h: {y : y > 1 + e }
–4
°
(d) g(x) = h f(x) = hf(x) x
°
0
11. (a) (b) x = –a x = a
y y
1 15. (a) a = 3, b = 1
1 2 2
x x (b) (x – 4)(3x + x + 6) + 5x – 2
0 2 4 0 2 4 1 2 71
–1 (c) f(x) = 31 x + 2 +
–1 6 12
1
2

(c) (d) (i) Since 1 x + 2 . 0 for ∀x ∈ R, therefore f(x) is
y 6
y always positive for all real values of x.
1
2 (ii) Minimum value of f(x) = 71 when x = –
1 12 6
1
x (d) {x : x , –2, x . 2, x ∈ R}
0 2 4
x –1
2
0 2 4 16. a = 7, b = 1; f(x) = (x – 1)(x + 1) (3x + 4);
4
(e) 1 – , –12 < (–1, 1)
y 3
y y
4
1
x
4 –1 0 1 y = |f(x)|
x – –
3
–2 0 2 x
4 –1 0 1
–1 y = f(x) – –
3
–4
12. (a) (i) 1 (ii) 1
(c) 2 < y < 2 (d) –1, 1 17. (a) y = g(x)
3 (1, 2)
(e)
y y = 2 – (1 – x) 2
1
1 __
In
2
2 0 1+ 2 x
y = 2e – 1 –1
x
x
– 1 0 1
275
Answers STPM Math T S1.indd 275 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers

(b) {y : y < 2} 46. y
(c) Any horizontal lines y = k for –1 , k , 2 will cut the
graph at two points, therefore g is not a one-to-one y = (2 – 3x) y = –(2 – 3x)
function. (0, 2)
y = 2 – x 2
1
18. x , – , x . 1 (1, 1)
2
y
y = |2x – 1| x
– 2 0 2 __ 2
3
{x : x < 0 or x > 1}
1
47. (a) a =
4
1
y = (b) y
|x|
x
0 1 1 _
— f(x) = 1 + e 4x h(x) = In(x – 1)
2 2 4
1
1
1
1
19. x = – , ; x , – , x . 1 0 1 2 x
3 5 3 5
y y = x
y = 4|x | 48. –1.14
49. –1, 1
50. 1.09
51. –2, 1
y = |x–1| 52. 1.31
1 16
55. (a) –2, –9 (b) 6 (c) –2 (d)
x 9
1 0 1 – 56. 3
– –
3 5
57. x = 1 or x = 4
20. A = 2, B = –3, C = 1 58. 3, 81
3
21. 3x + 4x – 2x + 5 60. 101
2
22. 1 + 2 – 3 61. a , 3a
1 – x 2x + 1 (2x + 1) 2 2
23. 1 – 1 2 – 2x + 1 62. 88
2
x – 1 2(x – 1) 2(x + 1) 64. (a) (b)
24. 1 – 1 + 2 y y
x – 3 x + 2 (x + 2) 2
25. 14, x + 1
27. a = 4, b = –20
x
28. 2 – x 0
x
29. k = –7, (2x – 1)(x – 1)(x + 2)(2x + 3)
30. a = 2, b = 2
31. a = 5, b = 10, c = 0; x , –5 or x . 3 (c)
33. a = 1, x = 2, 4; a = –3, x = 2, 4 9 y
36. 3, –1
37. k = 2, a = 10 –π 3 π 0 π – π x
– –
1
38. (a) k = 2; , –1, –3 (b) –1, 1, 4, –1 4 4
2
39. a = 4, b = 3; x = –1, 2; {x : x , –1 or x . 2}
40. k . 1
3
41. (a) – , x , 3 65. y y
4
(b) – 3 , x , 1 or x . 3
x
43. p + q r + 2q pr 0
2
45. (a) –8 , x , 0 (b) –3 < x < – 5
7 x
(c) –1 , x , 0, 3 , x , 4 (d) x , –2, x . 2 0
276



Answers STPM Math T S1.indd 276 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers

66. y 74. (a) 0°, 54°44´, 125°16´, 180°, 234°44´, 305°16´, 360°
(b) 257°35´, 349°47´
75. (a) 0°, 63°26´, 116°34´, 180°, 243°26´, 296°34´, 360°
y = 5 sin x
5 (b) 45°, 60°, 135°, 225°, 300°, 315°
76. 323 , – 36
325 325
y = 1 – 2 sin 2x 77. 18.4°, 45°, 198.4°, 225°
1 π
3
x 78. 4 sin 1 q + 2
0 90 180 7π π
–1 (a) min. –4, q = 6 ; max. 4, q =
6
6.5°, 143.5°, 366.5°, 503.5° (b) y
67. r = 13, a = 22.6°
(a) 13, 337.4°, –13, 157.4° 4
(b) y 2 3 y = 2 sin θ + 12 cos θ
2
13
x
0 __ π __ π __ π __ __ π 7π 4π 3π 5π 11π 2π
__ __ __ __ __
2π 5π
6 3 2 3 6 6 3 2 3 6
x –2
0
–22.6 90 180 270 360
–4
π 5π 3π 11π
π
,
–13 (c) q = , , , ; 5 q : , q , 5π 3π , q , 11π 6
2 6 2 6 2 6 2 6
(c) 85.3°, 229.5° 79. 90°, 306.87°
68.
y Chapter 2 Sequences and Series
y = cos 2x
1 Exercise 2.1
1 __ 1. (a) 25, 30; 5n (b) 16, 19; 3n + 1
2 (c) 14, 17; 3n – 1 (d) 32, 64; 2 n
n – 1
x (e) 3 , – 3 ; (–1) n – 1 3
0 __ π __ π __ π 2π 5π π 7π 4π 3π 5π 11π 2π 625 3 125 5
__ __ __ __ __
__ __
– 1 __ 6 3 2 3 6 6 3 2 3 6 (f) 9 11 ; 2n – 1
,
2 11 13 2n + 1
–1 (g) 625 3125 ; ( ) (h) 1 , 1 ; 1
,
5 n – 1
16 32 2 30 42 n(n + 1)
π π
,
{x : 0 < x , , , x , 2π 5π , x , 7π 4π , x , 5π , (i) 5, –6; (–1) n – 1 n
,
3
3
6
3
6 3
6
11π , x < 2π} 2. (a) 1 , 2 , 3 , 4 (b) 1 , – 1 , 1 , – 1
6 2 3 4 5 1! 2! 3! 4!
2
3
69. (a) 2 (b) m , – 1 (c) 1 2 , – 2x 4x 2 , – 8x 2 (d) – x , x 3 , – x 5 , x 7
2 ,
2 1 3 5 7 1.3 3.5 5.7 7.9
70. 8 (e) cos x , cos 2x cos 3x cos 4x
2 ,
3 ,
71. 15 cos (x – 0.6435) x 2x 3x 4x 4
(a) min. –15, x = 3.7851 rad; max. 15, x = 0.6435 rad 3. (a) (–1) (2n – 1) (b) 1 – (–1) n
n
(b) x = 2.2143, 5.3559 3n + 2 2
(c) y (c) n + 3 . 1 – (–1) n

15 n + 5 2
12 4. (a) 3, 2, 2, 3, 8, 35, … (b) 1, 3, 5, 11, 21, 43, …
10 (c) 1, 3, 7, 15, 31, 63, 127, …
1.485 6.086 5. (a) Convergent, 1 (b) Divergent
x (c) Convergent, 2 (d) Convergent, 3
0 0.6435 2.2143 3.7851 5.3559 2π 6. (a) 0 (b) 0 (c) ∞
(d) 1 (e) 2 (f) 2
5 3
10 3 50
∑ (2r – 1)
–15 y = 12 cos x + 9 sin x 7. (a) ∑ r (b) r = 1
r = 1
6
{x : 1.485 < x < 2.214, 5.356 < x < 6.086} 9 1 ∑ (–1) 1 2 r – 1
r – 1 1
(c) ∑ r (d) r = 1 3
r = 0 2
π
x
72. tan 1 – 2 , 2 + 1, 2 – 3 7 6
4 2 (e) ∑ (14 – 3r) (f) ∑ (3 + 2 )
r
73. (a) 1, 1 (b) 8 r = 1 r = 1
2 15 10 8 r
(g) ∑ (–1) rx r (h) ∑
r
r = 1 r = 1 (r + 1)(r + 2)
277
Answers STPM Math T S1.indd 277 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers

8. (a) 0 + 2 + 6 + 12 + 20 + 30 + 42 + 56 + 72 + 90 4. (a) 1 ; |x| , 1 (b) 2 ; |x| , 2




1
1
(b) 1 + + + 1 + 1 + 1 + 1 + 1 1 – 2x 2 2 + x
4 9 16 25 36 49 64 3 1 3x
(c) 5 + 8 + 11 + 14 +17 (c) 1 + 2x ; |x| , (d) 3 – x ; |x| , 3
2
(d) 3 – 6 + 9 – 12 + 15 – 18 + 21 – 24 + 27 4 4 4
(e) 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 – 1 – 2 5. 3 , , , 2
9 27
(f) 15 + 24 + 35 + 48 + 63 + 80 + 99 1
(g) 25 + 16 + 9 + 4 + 1 6. 3
(h) –6 – 3 + 4 + 15 + 30 7. 13 m
3
Exercise 2.2 8. 1 , 9
5
1. (a) 31, 4n + 3 (b) –17, 25 – 7n 9. 243; 2433 1 – 1 2 4 ; 8
5
n
1
1
(c) 31, 2n – 9 (d) 12 , (2n + 7) 1 1 9 1
3
3
2. (a) 860 (b) 1225 (c) 437 (d) 291 10. 1 + + 16 + 64 ; all values of r, 1 + r 2
4
3. (a) 819 (b) 745 (c) 54 (d) 120
4. 470 Exercise 2.5
5. 9072 1. (a) 4 + 14 + 36 + 76; 130
6. 16 734 (b) 0 – 21 – 44; –65
1
7. 6 , 2600 (c) 17 + 21 + 25 + 29 + 33 + 37; 162
4 45 1
8. 13; 22 (d) 90 + 45 + 30 + 2 + 18; 205 2
9. 6, 11, 16 (e) 3 + 3 + 0 – 3 – 3 + 0 + 3 + 3 ; 3
10. 4, 6 2 2 2 2 2 2
11. 7 (f) 3 – 5 + 9 – 17 + 33 – 65; –42
12. 64 1

2
3
13. (a) 15 (b) –12 (c) 5 (d) log 9 2. (a) 2 n(n + 2n + 1)
27 1

2
Exercise 2.3 (b) 12 n(n + 1)(3n + 11n + 4)
1

2
1. (a) 64, 2 n – 2 (b) 2 , 1621 2 n – 1 (c) 1 n(n + 8n + 17n – 2)
3
3 3 2
3

3
2
1
(c) 25 , 2001 – 2 n – 1 (d) – 81 , – 1 2 n – 3 (d) 1 4 n(n + 14n + 65n + 112)
32 4 16 2 2
3
2. (a) 9, 127 (b) 9, 5 394 3. (a) r(3r + 1); n(n + 1)
1
2

2
4 729 (b) r (r + 4); 12 n(n + 1)(3n + 19n + 8)
1
(c) 7, 198 7 (d) 8, –7 1541 (c) r(r + 2); n(n + 1)(2n + 7)

16 2187 6
3. (a) 124 3124 (b) 2 85 (d) (3r – 2)(3r + 1)(3r + 4);
1
3125 128 12 [(3n – 2)(3n + 1)(3n + 4)(3n + 7) + 56]
1
1
1 1
k
2n
n
(c) 1 2 [1 – (–3) ] (d) a [a – 1] (e) r(r + 1)(r + 2) ; 3 – (n + 1)(n + 2) 4
2
a – 1
2 2
1
4. 3 (f) r(r + 1)(r + 2); n(n + 1)(n + 2)(n + 3)

4 4
5. m = –5, n = 2 1 4. 22 100, 20 825
2 5. 88 200
6. (a) 9 (b) 1 9 (c) 1 × 10 15 6. (a) 1 n (6n + 3n – 1)
2
7. p = 13 2
1
2
8. –53 ; –385 5 (b) 1 n (2n + 1)(6n + 31n + 37)
3 6 3
9. –40; –8 (c) 1 6 n (28n + 57n + 29)
2
n
10. 1 (3 – 1); 8 3
24 (d) n (3n + 1)(9n + 3n + 2)
2
11. 10 4
2
7. 1 n (n + 1) 2
Exercise 2.4 4 n 2n
1. (a) Yes (b) No (c) Yes 8. (a) 2(n + 2) (b) 3(n + 3)
(d) Yes (e) No (f) Yes (c) 11n + 48n + 49n (d) n(3n + 1)
2
3
1
2. (a) 9 (b) 2 (c) 11 (d) 27 18(n + 1)(n + 2)(n + 3) 4(n + 1)(n + 2)
3 9 1 3 2
16 8 968 9. (a) n(48n + 80n – 6n – 47)
3. (a) (b) (c) 3
33 111 1665 (b) 5n + 30n + 37n
2
3
(d) 1117 (e) 9209 36(n + 1)(n + 2)(n + 3)
3330 9990 r 1
10. , 1 –
(r + 1)! (n + 1)!
278
Answers STPM Math T S1.indd 278 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers
11. U = 6r – 1; 9n + 2n 2. (a) 1 – x + 3 x – 5 x ; |x| , 2
1
2
3
2

r
128
4
12. 1 + 2r 2 , n(n + 2) 4 2 32 40 1
r (r + 1) (n + 1) 2 (b) 1 – x – x – 81 x ; |x| , 2
2

3
2
9
3
13. A = 1, B = –3, C = 2 1 1 1 1
3
(c) 2 – x + x – 16 x ; |x| , 2
2

4
8
Exercise 2.6 (d) 1 – x + x – 5 x ; |x| , 1
3
1
3
2

1. (a) 120 (b) 40 320 (c) 3 628 800 2 1 8 5 16
3
3
2

(d) 336 (e) 720 (f) 4896 (e) 1 – x – x + 16 x ; |x| , 1
2
8
(g) 285 (h) 8008 (i) 1260 (f) –1 + 2x – 2x + 2x ; |x| , 1

3
2
(j) 252 33 2 81 3 1
2. (a) 10! (b) 12! (c) 22! (g) 2 + 4x + 4 x + 4 x ; |x| , 3
7! 10! 19! (h) – + x – 13 x + 51 x ; |x| , 1
3
1
2

3
(d) 18! 2! (e) 21! 5! (f) n! 2 4 8 16 2
15! 5! 18! 8! (n – 3)! (i) 1 – x + x ; – (1 + 5 ) , x , ( 5 – 1)
1
1
3
(g) (n + 2)! (h) (2n + 5)! (i) n! 2 2
(n – 2)! (2n + 2)! (n – 1)! 3! 3. 1.1255
(j) (n – 5)! 2! 4. 30.43168
(n – 9)! 5! 5. 1 + 30x + 420x + 3 640x ; 1.03042
2
3
3. (a) 11(9!) (b) 15(7!) 2 3 4 5 6
(c) 291(7!) (d) (n + 2)n! 6. 1 – x + x – x
2
2
1
(e) (n + n – 1)(n – 1)! (f) n(n + n + 2)n! 7. 1 – x – x – 1 x ; 0.9487

2
1
2
3
2
(g) (n + 1) (n – 1)! (h) 31(7!) 2 8 16
2
30(3! 2!) 8. 1.732
(i) 263(7!) (j) (n – 1)! (nr + n + 1) 9. 3.1623 (5 sig.fig.)
210(3! 2!) (r + 1)! 1 1 3 1 5
3
4
2
4. (a) 70 (b) 120 (c) 1140 (d) 1 10. 4 + x + 16 x + x + 64 x ; 0.309
8
4
1
1
(e) 200 (f) 560 (g) 15 (h) 26 11. 1 + x – x + x ; 2.236
3
2
2
45
4
2
(i) 6188 (j) 84 12. 1 – 7x + 28x – 112x ; |x| , 1
2
3

4
4
3
Exercise 2.7 13. 1 – 5x + 30x – 15x ; 0.950 030
3
3
4
2 2
1. (a) p + 4p q + 6p q + 4pq + q 4 STPM PRACTICE 2
2
5
4
6
3
7
3
4
(b) m – 7m n + 21m n – 35m n + 35m n – 21m n + th
5
2
7mn – n 7 1. n(n – 1); 15
6
1
1
(c) 64 + 192k + 240k + 160k + 60k + 12k + k 12 2. (a) 3, 4 , 4 , 4 + 1 , n . 1
2
6
10
4
8
6
2
n(n – 1)
(d) 32a – 80a b + 80a b – 40a b + 10ab – b 5 (b) 2, 4, 18, n(n!), n . 1
5
4
3 2
2 3
4

3
(e) x + 3x + + 1
3
x x 3 3. (a) 250 (b) 510

(f) y – 4y + 7y – 7y + 35 – 7 2 + 7 4 – 1 6 + 1 (c) n(n + 1) (d) (n + 1)! – 1
8
6
4
2
8 4y 16y 16y 256y 8 n(n + 2)
6.
2. (a) 210x 4 (b) 16 128x 2 3(2n + 1)(2n + 3)
4
3 9
(c) 1760 a b (d) –945p q 6 7. b 2
(e) –20 (f) 36 8. 11, 20
x 3 9. Explicit formula: u = 4n – 1
n
5
3. 64x + 160 + 20 7 Recursive formula: u = u n – 1 + 4 for n = 2, 3, 4, … and u = 3
n
1
x
x
4. 14 or u n + 1 = u + 4 for n = 1, 2, 3, 4, … and u = 3
n
1
5. 15 10. (a) w = 8n + 4
n
6. 2 (b) First term = 12; Common difference = 8
7. n = 8, a = – 1 (c) S = u + w – 3

2 11. 11 n n n
2
8. (a) 1 + 7x + 21x + 35x 3 3 3
3 2
(b) 1 + 7x + 14x – 7x 3 12. (a) U = 23 1 + 1 24 , U = 23 1 + 1 2 + 1 2 4 ,
2
5
5
5
3
2
2
9. 1 + 16x + 136x + 784x 3 3 3 2 3 3
10. 1 + 9x + 24x 2 U = 23 1 + 1 2 + 1 2 + 1 2 4
5
5
4
5
11. (a) –16 464 (b) –151 164 (b) lim U = 5
(c) –1760 (d) 26 r → ∞ r
Exercise 2.8 14. (a) (n + 1)(2n + 1) (b) –975
1. (a) 1 – 2x + 3x – 4x 3 (b) 1 + x + x + x 3 15. (a) mn 1
2
2
1
1
3
2
2
n
(c) 1 – 6x + 24x – 80x 3 (d) 1 4 – x + 16 x – x 3 (b) mn – m + 1; S = m(2mn – m + 1)
2
5
8
4
2
(e) 1 – x + 4 x – 8 x 3 16. (a) u = 2 4r + 1
2
r
3 9 27 81
279
Answers STPM Math T S1.indd 279 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers

1
(b) S = 3 1 – 1 1 2 4 ; lim S = S = 1 6 –11 1 –17 –5
n
n 6 16 n → ∞ n ∞ 6 DA = 1 4 2 2 , DB = 1 2 –2 6 2
17. (a) 138 –7 0 –3 7 –7
(b) 1 (10 – 1); 1 (10 n + 1 – 9n – 10) 4. (a) –4 2 (b) –7 –4
n
9 81 1 10 9 2 1 7 12 2
18. (a) 676 (b) 493 (c) 1168
2475 825 1665 (c) 1 9 1 –1 4 2 2 (d) 1 –8 –2 –4 2
8
15
10
6
1
19. (b) u n + 1 = u n (c) S = lim S = 3 2 5. x = 2, y = 4, z = 1, w = 3
2
∞
n
n → ∞
2
20. 5 x + 15 x + 65 x 3
2 4 8 7. a = 1, b = –3
21. (a) 752 50 –50 100 0 –15 –30
0
0
0
1
1
22. 1 + apx + a p(p – 1)x + a p(p – 1)(p – 2)x ; 3 8. (a) 1 10 –10 20 2 (b) 1 0 0 –15 –30 2
2
2
3
2
30
60
6
8
4
3
– , – ; |x| , .

4 3 3 (c) (57) (d) (5)
2 1249
1
1
23. n = 4; 1 – x – x ; 12 –6 36
2 8 200 (e) 1 2 –1 6 2 (f) (17)
24. p = 3, q = 2, n = 8 –2 1 –6
1
6
8
25. 1 + x + x 2 9. (a) 1 38 27 2 (b) 1 11 16 16 2
2
24
26. 1 1 4 –16 –4 1 –1 12
1 2 , 0.98952
6 n
2

27. (a) 1 n(28n – 1) (b) 1 , 4 10. (a) –1 0 31 (b) 1 –1 –4 –12 2
2
24
8
3 999 37 1 –10 –4 104 2 –3 –12 –36
6
6
28. (a) 167 167; 111 445 (b) A = 13 , B = 1; 13 2 2
1
2
4
29. (a) 1 + x – 3 x 2 (b) 20x + 160x + 64; 1364 (c) 1 –11 –11 2 (d) 1 –5 –5 2
4 32 –18 –18 –32 –32
30. – 1 – 2 + 3 2 ; 0, 5, 6; 3r + 1 + (–1) r + 1 11 4 –135 2 8 24
1 + x 1 – x (1 – x)
(e) 1 –22 –8 270 2 (f) 1 –5 –20 –60 2
2 (1 – x
31. (a) S = 2x 3 n – 1 ) 4 33 12 –405 –32 –128 –384
n – 1 1 – x
15 Exercise 3.2
(b) ∑ (2r – 1)3 = 602 654 097 2 2
r
r = 1 1. (a) 9 (b) 2 (c) x + y – 2
14 2. (a) 0 (b) 0 (c) –14
∑ (2r + 1)3 r + 2 = 1 807 962 291
r = 1 3. (a) x(x – 1) (b) 8a(1 – 2a) (c) axy(2 – ay)
4. (a) –3 (b) 3 (c) 3
Chapter 3 Matrices 5
5. (a) –96 (b) 192 (c) 75
Exercise 3.1 6. (a) 18 (b) 2 (c) – 1
3
5
0
3
1
2
1. (a) 1 –6 12 0 6 2 (b) 1 –3 –5 –1 2 7. (a) 0, –9 (b) 0, –2 (c) –3, 0, 2
0
5
1
(c) 1 1 1 3 5 –3 2 (d) 1 –5 –5 3 1 2 Exercise 3.3
3
5
2
11
1
(e) 1 –13 –15 9 1 2 1. (a) 1 0 –1 2 (b) 1 –3 –3 2
2
1
7
(c) 1 –4 –5 2 (d) 1 3 –2 2
7 –5
3
16
24
24
24
16
16
2. (a) 1 33 –6 –39 2 (b) 1 18 5 –31 2 –2 –1
30 18 9 9 –2 28 (e) None (f) 1 3 2 2
1
3
2
2
3. A = 1 –5 8 5 2 , AB = 1 –1 13 8 3 2 (g) None (h) 1 — 0 2
2
3
1
2
2
BC = 1 3 4 –3 –2 2 , BD = 1 –10 –1 2 0 –1 3 1 0 – — 3 –2
2
2
16
1 5 1 3 2 1 5 10 2 2. (a) 1 1 4 2 (b) 1 2 1 2 (c) 1 0 2 2
C = –2 16 –5 , CD = 11 –1
2
–1
–1
–4 2 –3 –1 –7 3. (a) A B (b) BA
(c) B (d) A BA
–1
280
Answers STPM Math T S1.indd 280 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers

4
5. 1 –3 –5 2 10. (a) x = 3, y = 2, z = –1 (b) x = –2, y = 1, z = –1
(c) x = 3, y = 5, z = –2
(d) x = 3, y = 12, z = 9
4
6. 1 –2 –2 2 STPM PRACTICE 3
6
5
5
4
–3
8
–5
43
9
7. 1 –25 14 2 ; m = –5, n = 1; 1 3 5 1 2 2 1 –120 –24 2 2. (a) 1 –1 13 2 (b) 1 2
1
,
6
67
1 8 0 0 2 1 5 –7 1 2 –2 7 5
1
8. 0 8 0 , 8 –6 10 2 3. (a) (b) None
0 0 8 15 –21 –5 1 – 3
5
16
–20
4
8
–2
3
1
1
10. (a) 81 –10 2 1 2 (b) – 231 –2 –3 –2 2 (c) – 2 3 5 3 (d) – 1 1 5 –7 2
4
1
–2
–5
–5
–2
–1 2 11 –3 3
–2
1
0
1
1
1
(c) inverse does not exist (d) 31 –7 –1 0 3 2 4. x = –4, y = 9 4
1
–2
9
1
(e) – 60 1 7 –13 –3 2 5. (a) 1 –1 –4 –7 2
–29
11
–5 –2
1
13 –7 –9 1 – 2 9 1 9 4 9
13
8
5
2
–1
9
11. (a) 1 –3 3 3 17 2 (b) 1 –3 –11 12 2 (b) Q = (Q + 2Q – 8I) = – 1 9 – 4 9 – 7 9
11
0
0 3 13 1 –5 16 1 9 – 5 9 – 2 9
5
7
10
1
1
(c) 61 –8 –3 –3 2 (d) 31 –5 –11 –23 2 6. (b) det B = –4 det A = –4(p – q)(q – r)(p – r)
25
6
13
2 –1 1 –3 –6 6 7. a + d, ad – bc
–2
4
–1
1
1
13 –11
0
–60 –123
111
(e) 181 116 233 –211 2 (f) 61 –2 –4 6 2 9. 1 4 –3 –1 2
–26 –53 49 0 –3 3 –7 4 2
Exercise 3.4 –1 –1 2 –2
2
1. (a) x = 1, y = –2 (b) x = –2, y = 1 10. W = 1 –3 5 –5 ; x = 2, y = 7, z = –10
(c) x = 0, y = –1 (d) p = –1, q = –2 2 –4 3
(e) r = 4, s = 7 (f) u = 2, v = –3 11. (a) a = 3, b ≠ 5
2. (a) (–1, 2) (b) None (b) a = 3, b = 5
(c) (–1, 1) (d) (3, –1) (c) a ≠ 3, b = any real values (real numbers)
1
3. (a) x = , y = k + 1 , k ≠ 0 3 2 1 x 1
k k 12. (a) 1 9 6 4 21 2 1 2
y =
a
2
(b) x = 2 , y = – 1 ; k ≠ ±2 6 4 –2 z a
k + 2 k + 2 3 2 1 1
4. (a) x = 1, y = –1, z = –1 (b) augmented matrix 1 9 6 4 a 2 2 ;
(b) x = –1, y = 2, z = 2 6 4 –2 a
(c) x = –15, y = 8, z = 2 3 2 1 1
5. (a) unique solution, (1, –2, –2) row-echelon form 1 0 0 1 a – 3 2
2
(b) no solution 0 0 0 4a + a – 14
2
(c) infinitely many solutions
Since row 3 has all zero entries, the system of linear
3
5
6. x = , y = , z = 1 equations does not have a unique solution.
4 4 2 (c) a = , –2 ; x = t, y = – t, z = 1 where t ∈ R
7
3
1 6 0 0 2 4 2
7. AB = 0 6 0 (d) 5 a : a ≠ , a ≠ –2, a ∈ R6
7
0 0 6 4
(a) x = 1, y = 1, z = 2 1 3 1 n
(b) x = –1, y = –2, z = 2 1 13. (a) augmented matrix 1 2 3 m 13 4n 2
5
m 2n ;
z 1 2
– —
x
3
1
1
–5 ;
y =
8. A = – 10 1 1 –2 –5 2 1 2 5 row-echelon from 1 1 0 –1 m – 2 n 0 2
8
4
–1
3
– —
5 0 –5 5 2
–2 0 0 m – 11m + 28 n
3
— (b) (i) m = 4, 7 and n ≠ 0
2
(ii)
m ≠ 4, 7 and n ∈ R
0
2 1 2
8 1
1
3
y =
–1
9. B = 2 3 2 7 –4 ; x z 1 2 (c) m = 4, 7 and n = 0 ; x = –16t, y = 5t, z = t
—
–5 –1 4 4 where t ∈ R
7
—
4
281
Answers STPM Math T S1.indd 281 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers

Chapter 4 Complex Numbers (d) i 3, –i 3, 2i, –2i, –2, 2
(e) ± 2, ±i 2, ±2i, ±2
Exercise 4.1 1 i 3 1 i 3
1. (a) 2 + 2i (b) 9 – 5i (f) –2, ±1, 1 ± i 3 , ± 2 , – ± 2
2
2
(c) 2 + 2 2i (d) –3 – 7 2i Exercise 4.3
(e) 4 + 4 5i 1. (a) y (b) y
2. (a) 8 + 2i (b) 1 – 5i
(c) 3 + 3i (d) –5 + 3i P(0, 3)
3. (a) –4 + 3i (b) 4 + 2i P(–2, 1)
(c) –5 – 12i (d) 11 + 3i θ
(e) –1 + 2 2i (f) –3 + i θ x 0 x
1
4. (a) 1 – i (b) 2 + i 0 π
5 5 |z| = 3, q = |z| = 5, q = 2.678 rad
2
(c) i (d) i (c) y (d)
(e) 18 – 1 i (f) 7 – 1 i y
13 13 25 25
5. (a) 2 (b) 5 0 θ x P(5, 7)
3
4

(c) – – i (d) –3 + 4i
5
5
(e) –3 + 4i (f) 1
5 θ x
6. (a) 7 – 7i (b) 7 + 7i P(2, –4) 0
7
(c) 25i (d) 24 – 25 i |z| = 2 5,  |z| = 74,
q = –1.107 rad
q = 0.951 rad
25
(e) 24 + 7 i
25 25 (e) y (f) y
3
7. (a) x = 9, y = –7 (b) x = – , y = 7 2 x P(–4, 3)
2
(c) x = 3, y = –5 (d) x = 2, y = 4 0 θ
(e) x = –5, y = 3
θ
8. (a) ±(2 – i) (b) ±(4 + 3i) x
(c) ±(3 + 2i) (d) ±(3 – i) P(–4, –4) 0
(e) ±(7 + 5i) 3
|z| = 4 2, q = – π |z| = 5, q = 2.498 rad
4
9. – 11 + 3 i
10 10 2. (a) z = 2 3 cos – i sin 4
π
π
10. a = b = 5 4 4
2 (b) z = 23 cos + i sin 4
π
π
11. z = 3 – 2i 6 6
12. z = 3 – 4i (c) z = 5[cos (0.927) + i sin (0.927)]
(d) z = 13[cos (–1.176) + i sin (–1.176)]
Exercise 4.2 (e) z = 7 [cos (–0.714) + i sin (–0.714)]
1. (a) –1 ± 2i (b) 1 (3 ± 7i) (f) z = 43 [cos (0.867) + i sin (0.867)]
2 3. (a) z = 12 + 5i
1
3
(c) – (7 ± 31i) (d) 1 2 (5 ± 3i) (b) |z | = 13, arg z = 0.395 rad,
4
3
3
3
(e) 1 (2 ± 2i) z = 13 [cos (0.395) + i sin (0.395)
3 4. (a) z = –1 + 3i
2. (a) z – 2 2z + 3 = 0 (b) z – 4z + 7 = 0 (b) |z| = 10, arg z = –1.249 rad,
2
2
(c) z – 3z + 1 = 0 (d) 25z – 30z + 11 = 0 z = 10 [cos (–1.249) + i sin (–1.249)
2
2
2
(e) 7z + 8z + 3 = 0 5. 8 6 8
1
5
3. (a) z = 2, 1 ± i 2 (b) z = – , –2 ± i 3 6. 5 + i, 2, 0.927, 2 [cos (0.927) + i sin (0.927)]
2 π 2
5
2
3
(c) z = 3, ± i 2 (d) z = – , ± i 3 7. (a) 2, 4 (b) 2 , 2 2
2
2
2
2
5
1
(e) z = , 2 ± i 8. –0.977, 2.16
3 Exercise 4.4
1
4. z + 4z + 2; z = – , –2 ± i 2
2
4π
4π
2 1. (a) cos 2π + i sin 2π (b) cos 1 – 2 + i sin 1 – 2
1
2
5. (z – 2z + 5)(z + z + 1); 1 ± 2i, – ± i 3 3 3 3 3
2
2 2 (c) cos 5π + i sin 5π (d) cos 3π – i sin 3π
6. (a) ± 2 , ±2 2i 6 6 7 2 2 –3
1
(b) – , 1, ±i 2. (a) (cos q + i sin q) (b) (cos q + i sin q)
2 q q –5 (d) 1 cos + i sin 2 13
q
q
(c) –3, 2i, 2i, –2i, –2i (c) 1 cos + i sin 2 6 6
3
3
282
Answers STPM Math T S1.indd 282 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers
π π 5. ±(1 + 2i), ±(1 – 2i)
3. (a) 2 3 cos + i sin 4 , 4 + 4i
4 4 π π π π 1
π
π
4
4
4
4
(b) 2 3 3 cos + i sin 4 ; –144( 3 – 3i) 6. (a) 2 1 cos + i sin 2 , 2 3 cos 1 – 2 + i sin 1 – 24 ; , π
2
3
3
6
8
1
π
π
(c) cos 1 – 2 + i sin 1 – 2 , – ( 3 + i) (b) 5 + i ; 2, 0.927 rad
5
6 6 2
5
(d) 13 cos (2.4) + i sin (2.4)], 13 [cos (5.88) + i sin (5.88)] 7. ±1.23 rad.
π
π
3
5
1
7
(e) 4 3 cos 1 – 2 + i sin 1 – 24 , 512(1 + 3i) 8. sin π, sin π, sin π, – 2 + 2
3 3 8 8 8 2
(f) 23 [cos (0.485) + i sin (0.485)], 1 1

2
2
— 5 9. (a) a = (3 + 21), b = (–3 + 21)
2
23 [cos (2.425) + i sin (2.425)] (b) (i) –1, – 3 (ii) 5 , π
4. (a) cos 23 q + i sin 23 q (b) cos 8q + i sin 8q 4 12
12 12 1 3 1 3
2
(c) cos 5q + i sin 5q (d) cos 29 q + i sin 29 q 10. 4 cos q – 2; ±i, + 2 i, – 2 i
2
2
12
12
(e) cos 11 q + i sin 11 q π π π π

12 12 11. 2 3 cos + i sin 4 , 2 2 3 cos 1 – 2 + i sin 1 – 24 ;
4
6
4
6
π
π
5
5
5. 1 3 cos + i sin 4 , i 1 2k + 2 π + i sin 1 2k + 2 π4 , k = 0, 1, 2
2 4 4 32 — 1 3 cos 1 3 36 3 36
— 1 π π 2 6
4
6. (a) ±2 3 cos + i sin 4
8 8 13. a = 5, b = –10, c = 1;
(b) ± 5 [cos (–0.464) + i sin (–0.464)] 2 cos , 2 cos π, 2 cos π, 2 cos π
3
4
2
π
(c) ± 13 [cos (–0.588) + i sin (–0.588)] 5 5 π 5 5
π
π
(d) ± 2 3 cos 1 – 2 + i sin 1 – 24 14. (a) |z| = 4 ; arg z = – 6
12
12
— 3 π π (c) 1 (–1 + 3i)
4
(e) ±2 3 cos 1 – 2 + i sin 1 – 24 32
8
8
2π
— 1 4 π π 15. 83 cos 2π + i sin 4 ;
(f) ±3 3 cos + i sin 4 3 3
8 8
— 1 2k 1 2k 1 1.5321 + 1.2856i, –1.8794 + 0.6840i, 0.3473 – 1.9696i
7. (a) 2 3 cos 1 + 2 π + i sin 1 + 2 π4 , k = 0, 1, 2
6
3 12 3 12 16. (a) w = 1 + i ; real part = 1, imaginary part = 1
π
π
1
—
(b) 5 3 cos 1 2k π – 0.3092 + i sin 1 2k – 0.30924 , k = 0, 1, 2 (b) w = 2 3 cos + i sin 4
3
4
4
3
3
— 1 2k 2k (c) 1.0842 + 0.2905i, –0.7937 + 0.7937i, –0.2905 – 1.0842i
(c) 13 3 cos 1 π – 0.3922 + i sin 1 π – 0.39224 , k = 0, 1, 2
3
3 3 π π 5π 5π
— 1 2k 1 2k 1 17. z = 4 2 3 cos1 – 2 + i sin 1 – 24 , w = 2 3 1 cos + i sin 4 ;
(d) 2 3 cos 1 3 – 2 π + i sin 1 3 – 2 π4 , k = 0, 1, 2 8 4 4 6 6
3
18
18
π
π
— 1 2k 2k w 6 = 81 3 cos + i sin 4
1
1
(e) 2 3 cos 1 – 2 π + i sin 1 – 2 π4 , k = 0, 1, 2 z 128 6 6
3
3 12 3 12
— 1 2k 1 2k 1 18. (a) 3 – i, 1 + i 3, – 3 + i, –1 – i 3
(f) 3 3 cos 1 + 2 π + i sin 1 + 2 π4 , k = 0, 1, 2
6
3 12 3 12 19. (a) real part = 32, imaginary part = –32 3
nπ
π
n
8. |z| = 2, arg z = , z = 2 3 cos nπ + i sin 4 π π
n
3 3 3 (b) z = 83 cos 1 – 2 + i sin 1 – 24 = 4 3 – 4i
1
1 2 2 1 6 6
2
3
9. z = 3 cos π + i sin π4 , z = [cos π + i sin π] 5π 5π
4 3 3 8 z = 83 cos + i sin 4 = –4 3 + 4i
6
6
2
π
π
3
3
10. 23 cos + i sin 4 , 2 2 3 cos π + i sin π4
6 6 4 4
7
7

z = 1 1 3 cos 1 2k – 2 π + i sin 1 2k – 2 π4 , k = 0, 1, 2 Chapter 5 Analytic Geometry
— 3 36 3 36
2 6 Exercise 5.1
π
2
2
11. q = , π 1. 3x + 3y – 16y + 16 = 0
3 4 2. y = 3a(2x – 3a)
2
2
2
STPM PRACTICE 4 3. x + y + 4y = 0
4. 16x + 10y – 15 = 0
7
4
1. (a) p = , q = – 5. x + y – 5x = 0
2
2
5 5
2
(b) p = 2 – i, q = 2 + i 6. (a) (y – 3) = x + 1
2
(b) 27y = 4x 3
3. (1, 2), (–1, –2) (c) (x – 4)y = 9
1
4. |z | = 2 , arg z = – π (d) y = (x – 1)(x – 2)
2
1 1 4 (e) x = 12(y – 1)
3
2
|z | = 2 2 , arg z = – π (f) (y – 1) = (x + 1) 3
4
2
2
(g) y(x – 1) = 8
2
|z | = 1, arg z = – π (h) x (y + 1) = 64
2
3 3 3
283
Answers STPM Math T S1.indd 283 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers
Exercise 5.2 (b) y
3
1. (a) (b) ( )
0, –
y y 2
x = 3
x = –2 x
0
3 2
( )
,0
x x 4
0 (2, 0) (–3, 0) 0

Major axis: 3
3
Minor axis: 2
(c) (d) 2
y y (c)
y
3
( )
0, –
9 2 2
x = – – x = –
8 3
x x
–, 0 – –, 0 0
0 9 2 0 x
5
8 3 ( )
–
,0
2

2. (a) (3, 0) (b) (1, –2) Major axis: 5
Minor axis: 3
y y
(d) y
x
0
x 2 2
0 (3, 0) (1, –2) 2
0 (2, 0) x
(c) (–1, 1) (d) (–4, 3)
Major axis: 4 2
y y
Minor axis: 4
(e) y
(–1, 1) (–4, 3)
x
0 x 3 (1, 3)
0 4

2
3. (a) y = 8x (b) y = 6x 0 x
2
2
(c) y = 12x (d) 3y = 20x Major axis: 8
2
9
9
3
4. (a) x = t , y = 3t (b) x = t , y = t Minor axis: 6
2
2
2 4 2
2
4
2
(c) x = 15 t , y = 15 t + 1 (d) x = t + 1, y = t (f)
2
8 4 3 3 y
2
(e) x = 2(2t + 1), y = 8t + 1
5. (a) y = x + 2; y + x = 6 (b) y = x + 3; y + x = 9 0 x
(c) 2y + 2x + 5 = 0; 2y = 2x – 15 2 3
(d) 2y = 2x + 3a; 2y + 2x = 9a
6. y = 3x + 3, 3y + x + 81 = 0, (–9, –24)
7. ty = x + t , tx + y = t + 21 ; (8, –4 2 ) (–, –2)
2
3
3
2
8. 8 2 a, 32a 2 Major axis: 6
5
9. p + q = 1, pq = –2 ; 1 a, a2 Minor axis: 4
2
10. x = –a 2 2
2
12. 2ax = y – y 2. (a) x + y = 1
16 9
Exercise 5.3 (b) x 2 + (y – 1) 2 = 1
1. (a) y 4 9
(0, — ) Major axis: 4 4(x – 2) 2 (y – 1) 2
4
3 8 (c) + = 1
Minor axis: 81 18
3 2 2 2 2
x (d) x + y = 1 , (x – 2) + y = 1
0 (2, 0) 4 3 4 3

284
Answers STPM Math T S1.indd 284 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers

(e) (x + 4) 2 + (y – 1) 2 = 1 , (x – 6) 2 + (y – 1) 2 = 1 (c) y
36 20 36 20
2
2
3. (a) 4x + y = 4
(b) x 2 + y 2 = 1 0 x
16 9
(c) (x – 5) 2 + (y – 2) 2 = 1
4 9
(d) 4(x – 1) + (y + 1) = 4 (d) y
2
2
4. (a) x = 2 + 3 cos q, y = 3 + 2 sin q
(b) x = –2 + 2 cos q, y = 1 + 3 sin q
(c) x = 2 + 2 cos q, y = –1 + sin q 0 x
(d) x = 1 + 10 cos q, y = –1 + 6 sin q
5. (a) 2x + 3 3y = 12
(b) 5y = 3x + 15 2
(c) 3x + 4 3y + 24 = 0 (e) y
6. (a) x + y = 5; y = x – 1
(b) x – 2y = 4; 2x + y = 3
(c) 9y = 2x – 5; 9x + 2y + 20 = 0
y = 1
7. (a) (2, 1) –2 0 x
(b) (3, 2)
(c) (2, 6)
(d) (2, 1) (f) y
, )
9. 1 , –2; 2y = x + 8, (–6, 1); y + 2x = 14; ( 48 2
2 7 7
2
10. 7m + 12m + 5 = 0; outside x
0
11. y = x ± 13 , y + x = ± 13 –1 x = 2
12. y + x = ±5
13. 35°10´
(g) y
2
2
14. (x + 8) + y = 1
15 9
15. 3y + 2x = ± 43
x
0
Exercise 5.4
1. (a)
y
(h)
y
x
(–4, 0) 0 (4, 0) y = 2
x
0
3 3
y = –x y = – –x x = 1
4 4
(b) y
2. (a) (i) 5 , (± 5 , 0)
2
1
(ii) x = ± 4 5 , y = ± x
5 2
x 2 3
(– 10, 0) 0 ( 10, 0) (b) (i) 3 , (±4, 0)
(ii) x = ±3, y = ± 3 x
3
1 1
y = –x y = – –x
2 2

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Answers STPM Math T S1.indd 285 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers

(c) (i) 5 , (±4 5 , 0) 2. (a) (x – 9) 2 + (y + 12) 2 = 1 ; ellipse
2
2
2 a b
16 5 1 (b) The conic is a circle with centre (9, –12) and radius 15.
(ii) x = ± , y = ± x
5 2 y
(d) (i) 6 , (±3 6 , 0) 0
2 18 x
2
(ii) x = ±2 6 , y = ± x
2 (9, –12)
13
(e) (i) 3 , (1 ± 13, 0) –24
2
(ii) x = 1 ± 9 , y = ± (x – 1)
13 3 4. m x´ – my´ + a = 0
2
(f) (i) 14 , (1 ± 70 , 2) 5. (a) (p + q)y = 2x + 8pq ; py = x + 4p 2
3 2
5
(ii) x = 1 ± 3 70 , (y – 2) = ± (x – 1) (b) M(–4p , 0)
14 7 (c) A(4, 0)
6.
3. (a) xy = 9 (b) 4xy = 25 y
(c) xy + 4 = 0 (d) 9xy + 4 = 0
(e) xy = y + 1 (f) xy – x + 16 = 0 2 x + 4y = 2x
2
2
5
5
4. (a) x = 4t, y = 4 t (b) x = t, y = 3t ( )
1
1, –
3
2
1
(c) x = t, y = – 1 (d) x = 8t, y = 8 x
4 4t t – 4 0 2 4
2
1
(e) x = 1 + 2t, y = (f) x = 2 + t, y = 1 ( )
–
1, –
t t 2
2
2
4
5. (a) t y + x = 6t ; ty – t x = 3(1 – t ) –2 x + 4y = 16
3
2
(b) 9y + 4x = 24 ; 12y – 27x + 65 = 0
(c) t y + x = 3t ; 2ty – 2t x = 3(1 – t )
2
4
3
(d) 6y + x = 9 ; 4y – 24x + 105 = 0 7. (2, 1), (–1, 3)
6. (a) 4y + 3x + 8 3 = 0, 4y + 3x – 8 3 = 0 8. (a) (x – 1) 2 + (y – 4) 2 = 1
(b) y + 4x = 8 9 16
(b) C = (1, 4) ; F = (1, 4 – 7 ) ; F = (1, 4 + 7 )
2
1
3
7. 1 – , – 242 , 51 17 (c) Vertices = (1, 0) and (1, 8)
8 8 y
8. QR = 10 ; Q´R´ = 35 (1, 8)
12
9. x + 3y = 9 8 7
1 16 8 F (1, 4 + 7)
10. , –4 ; 1 , – 2 6 2
2 7 7 5
4
11. 1 , 4k2 (–2, 4) 4 C(1, 4) (4, 4)
k
–1 35
12. tan 1 2 or 69°37´ 3
13 2 F (1, 4 – 7)
14. p = 1 1 1
3 –3 –2 –1 0 1 2 3 4 5 x
b
2
15. ± b + c 2 , y = ± x. 2
a a 12. (b) q y + x = 2cq
3
3
STPM PRACTICE 5 13. (a) (–5, 0), (5, 0) (b) y = – x, y = x
4
4
1. (a) (x + 2) 2 + (y – 3) 2 = 1; ellipse 17. (a) (y – 3) 2 – x 2 = 1 ; centre = (0, 3) ; vertices = (0, 0), (0, 6)
16 12 9 9
(b) Centre, C = (–2, 3); Foci, F = (–4, 3) and F = (0, 3) (b) y = x + 3, y = –x + 3
1 2
(c) y (c) y
y = –x + 3 y = x + 3
6 (0, 6)
F (–4, 3) C(–2, 3) F (0, 3) (0, 3)
2
1
x
x –3 0 3
–6 –5 –4 –3 –2 –1 0 1 2
286
Answers STPM Math T S1.indd 286 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers

Chapter 6 Vectors 12. 339 unit 2
Exercise 6.1 13. 29 unit 2
3
1. (a) 4 i – j (b) 5 i + 12 j 14. 6 unit 2
5 5 13 13 2
(c) – 2 i – 1 j 15. –i + j – k
5 5
2. 12i – 5j 16. l = 7 3
6
3
3. (a) 2 i + j – k (b) –4i – 6j + 12k –i + 2j + 8k
7 7 7 17.
69
4. 5i – 2j + k ; 30 units Exercise 6.4
5. –4 i + 1 j + 3 k
26 26 26 1. (a) L : r = 2i – j + 6k + li
1
(b) L : r = –4i + 3k + lj
2
6. 6i + 3j + 7k (c) L : r = 3j – k + lk
3
7. (a) 21 units (b) 5 units (c) 3 units (d) L : r = 2i – 5j + l(–i + 6j)
4
8. (a) 6 untis (b) 7 units (c) 186 units (e) L : r = –2i + j – 3k + l(i + 2j – k)
5
9. (a) No (b) Yes (c) Yes 2. (a) r = 2i + 3k + l(i – j + k)
11. (a) 2i – 2j + 2k, i – 9k (b) r = i + 2j + 2k + l(i – 4j – 2k)
(c) r = 3i + j + 4k + l(–4i + j + k)
(b) 3(i – j + k)
Exercise 6.2 3. x – 1 = y + 3 = z – 2 (=l)
6
–2
5
1. (a) –2i – 8j (b) 16i + 36j 4. r = i – 5j + 2k + l(–3i + 2j + 6k)
(c) 7 (3i + 2j) (d) –35i – 14j 5. r = –i – 2j – 5k + l(7i + 2j – 3k)
4
(e) 6i + 3j 6. (a) r = 2i + j – k + l(3i + j + 6k)
(b) r = 6i – 2j + k + l(i + 5j + k)
2. u – 2v
3. (a) i + j + 5k (b) 1 2 (–i + j + 23k) 7. (a) r = i – 3j + 2k + l(i – j + 5k);
x – 1
z – 2
y + 3
(c) –12i – 7j + 10k 1 = –1 = 5
4. (a) (–1, 3, 0) (b) (–2, 0, –8) (b) r = i + 2j + k + l(i + 2j – 3k);
(c) (–3, 3, –8) (d) (2, –4, 3) x – 1 = y – 2 = z – 1
–3
2
1
5. (a) (5, –6, –1) (b) (0, 2, –3) 8. r = (2i – j + 5k) + l(–3i + 5j – 2k);
(c) (–1, 0, 5) (d) (0, 0, –4) x – 2 y + 1 z – 5
(e) (0, 9, 0) –3 = 5 = –2 (=l)
7. l = 4, l = 1, l = –2 9. p = –1, q = –5
1 2 3
9. 17i – 3j – 10k 10. r = i – 3j + l(2i + j)
398 Exercise 6.5
10. 3i – k 1. r · (i – j – 3k) = –7, x – y – 3z = –7
10 2. r · (i – 2j – k) = –12, x – 2y – z = –12
Exercise 6.3 3. r · (i – 3j – 5k) = 7
1. (a) –23 (b) 5 (c) –9 4. r · (–10i + 10j – 14k) = –32; –10x + 10y – 14z = –32
(d) 6 (e) 18 5. r · (6i + 2j – 3k) = 17
2. (a) 7°8´ (b) 90° (c) 75°38´ 6. r · (i + 2j + 2k) = –6, x + 2y + 2z = –6
(d) 73°24´ (e) 90° 7. r · (2i – 3j + k) = 9, 2x – 3y + z = 9
3. 80°24´ 8. r · (i – 3j + 2k) = 0, x – 3y + 2z = 0
4. (a) –32 (b) 130°46´ Exercise 6.6
5. 60° 1. (a) 71°34´ (b) 0°
6. (a) orthogonal (b) orthogonal 2. 71°14´
(c) orthogonal 3. 70°32´
7. –2 4. 80°24´
→ 5. 61°1´
9. (a) OM = 6i + 6j + 12k ; 6. 79°6´
→
NM = –6i + 2j + 12k 7. 22°12´
(b) 53.0° 8. 71°34´
10. –5i + 7j – 3k 9. 90°
11. (a) –4i + 72j + 28k (b) 4i – 72j – 28k 10. 49°59´ ≈ 50°

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Mathematics Term 1 STPM Answers
Exercise 6.7 21. (b) 3 26 unit 2 (c) 5 i + j – 13 k
1
1. (1, –2, 5) 4 8 4
2. (4, 5, 9) (d) 35 16 201
16 201
3. (a) (–3, 1, 5) (b) (3, –5, 4) 1 9
4. (a) l = 2, l = 1 (b) c = –5 22. 1 2 1 2
–1 + t 14 where t ∈ R
1
2
(c) (2, –3, –3) 0 1
5. (5, –3, 7) 23. (1, 2, 3)
6. (1, –5, 1) 24. 47°7´
7. (2, 14, 6) 25. 8i + 4j – 4k, –8i + 12j + 16k
5
y + z – 13
8. x = 2y + 5 = 2z – 13 or x = 2 = 2 , 26. 2i – 3j + 4k
7
–17
7 – 17 27. (a) 22 units (b) 12i – 15j – 9k
2 2
5
7
r = – j + 13 k + l(i + j – 17 k) 28. (a) 7 390 (b) 1 329 unit 2
2 2 2 2 390 2
.
(c) r (–10i – 15j + 2k) = –30
4
y + z + 8 29. (a) 13x – 2y + 5z = –4 (b) (–1, –22, –7)
9. x 1 = 3 = 3 (c) 14.9°
5 1
3 3
STPM Model Paper (954/1)
10. x – 11 = y + 8 = z
4 –3 1. (a) f : x ↦ x – 6x = (x – 3) – 9
2
2
STPM PRACTICE 6 y
y = f(x)
1. (a) 474 units (b) 46 units y = k
2. (a) –i – 4j – 2k (b) –5i – 7j + 11k x
3. (a) 1 (5i – j – k) (b) 1 (–i – j + 3k) 0 3
27 11
4. (a) 2 (i + j – k) (b) 5 (–3i – j – k) (3, –9)
3 11
Any horizontal line y = k is drawn horizontally with
5. (a) –8 (b) 4 x-axis, will cut the graph at only one point.
–1
6. (a) 83°44´ (b) 127°25´ Therefore f is a one-to-one function and f exists.
7. (a) 78.7° (b) 36.8° (shown)
–1
Let f (x) = y
–1
7
8. (a) p = 1 ; q = –1 (b) 1 2 x = f(y) 2
x = (y – 3) – 9
9. 35° –2 ± x + 9 = y – 3
→
10. (a) PQ = –6i + 6j + 12k (b) 61.9° y = 3 – x + 9 as y < 3
–1
→ f (x) = 3 – x + 9
PR = 6i + 6j + 12k
–1
f : x ↦ 3 – x + 9

11. (b) 58.7° D –1 = [–9, ∞)
1
(c) 152 unit , 25 unit 3 f
2
3 R –1 = (–∞, 3]
f
12. (a) –2i + j + 4k (b) –4i + 4j – 8k (b) y
1
13. ± (–2i + j + 4k)
21 4 g(x) = 4 – e –x
14. 41 unit 2 3
15. (a) 11i – 29j + 5k (b) 1 987 unit 2
2 x
16. (a) 5 (b) 3 0
18 10 R = (–∞, 4)

1 g
17. (a) (3i – 4j + 6k) (b) 8i + (4 + l)j + 8k R = (–∞, 4) ⊄ D = (–∞, 3]
f
g
61
(c) l = –9 Therefore fg does not exist. (shown)
18. (a) a = 10, b = 2 (b) 64.4° We need R = (–∞, 3]
g
–x
19. (b) x + y – 3z = 7 4 – e < 3
x < 0
20. Yes Therefore the maximum domain of g for which fg exists
= (–∞, 0]
288
Answers STPM Math T S1.indd 288 3/28/18 4:25 PM

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

Mathematics Term 1 STPM Answers

5. Let z = 3 + i (Normal vector n = Parallel vector of the line L )
1
1
2
|z| = ( 3) + (1) = 2 Vector equation of L : r = (3i – 6j + 2k) + l(4i – 9j – 7k)
2
1
–1 1 π Parametric equations of L :
Arg z = tan = 1
3 6 x = 3 + 4l, y = –6 – 9l, z = 2 – 7l
π
π
z = 21 cos + i sin 2 2 5 –3
1 2 1 2 1 2
6 6 →
(b) (i) EF = 2 – 1 = 1
π
π
12
12
( 3 + i) = 3 21 cos + i sin 24 ; 3 –3 6
6 6 Vector equation of L :
π
π
2
12
( 3 – i) = 3 21 cos 1 – 2 + i sin 1 – 224 r = (5i + j – 3k) + µ(–3i + j + 6k)
12
6 6
Parametric equations of L : 1
( 3 + i) + ( 3 – i) x = 3 + 4l, y = –6 – 9l, z = 2 – 7l
12
12
π
π
π
π
12
= 3 21 cos + i sin 24 + 3 21 cos 1 – 2 + i sin 1 – 224 12 Parametric equations of L : 2
6 6 6 6 x = 5 – 3µ, y = 1 + µ, z = –3 + 6µ
12
= 2 3 cos 12π + i sin 12π 4 + 2 3 cos 1 – 12π 2 + i sin 1 – 12π 24
12
6 6 6 6 Equating the two lines:
12
12
= 2 [cos 2π + i sin 2π] + 2 [cos(–2π) + i sin (–2π)] 3 + 4l = 5 – 3µ
= 2 [1 + i(0)] + 2 [1 + i(0)] 3µ + 4l = 2 ......................a
12
12
= 2 (shown) –6 – 9l = 1 + µ
13
µ = –7 – 9l .....................b
2 – 7l = –3 + 6µ .....................c
12
3
12
2z – [( 3 + i) + ( 3 – i) ]i = 0
2z – 2 i = 0 Substitute b into a:
13
3
3
12
z = 2 i 3(–7 – 9l) + 4l = 2
z = 2 (0 + i) –21 – 27l + 4l = 2
12
3
23l = –23
π
π
z = 2 1 cos + i sin 2 l = –1
12
3
1 2 2 1
— π π — When l = –1, from b: µ = –7 – 9(–1) = 2,
12 3
3
z = (2 ) 3 cos 1 + 2kπ2 + i sin 1 + 2kπ24
2 2 checking from equation c:
π
π
z = 16 3 cos 1 + 2kπ 2 + i sin 1 + 2kπ 24 , k = 0, 1, 2 l = –1 , LHS = 2 – 7l = 2 – 7(–1) = 9
6 3 6 3 µ = 2 , RHS = –3 + 6µ = –3 + 6(2) = 9 = LHS

π
π
k = 0, z = 161 cos + i sin 2 The equations are consistent with l = – 1 and µ = 2,
6 6 RHS = LHS, therefore the two lines L and L
1
= 161 3 + i 2 intersect. (shown) 1 2
2 2
= 8( 3 + i) Position vector of point of intersection = –i + 3j + 9k
Coordinates of point of intersection = (–1, 3, 9)
5π
k = 1, z = 161 cos 5π + i sin 2 n . n
6 6 (ii) cos q = 1 2
1
2
1
= 161 – 3 + i 2 |n ||n |
.
2 2 (4i – 9j – 7k) (–3i + j + 6k)
=
= 8(– 3 + i) 146 . 46
3π
k = 2, z = 161 cos 3π + i sin 2 = –63
2 2 6 716
= 16(0 – i) q = 140.2°
= –16i
Acute angle = 180° – 140.2° = 39.8°
The roots are 8( 3 + i), 8(– 3 + i) and –16i.
7. (a) p(x) has a factor (x – 3),
2
3
→ 5 1 4 p(3) = a(3) – (1 + 6a)(3) + 2(8a – b)(3) – 27 = 0
5 –
4 = 1
6. (a) AB = 1 2 1 2 1 2 27a – 9 – 54a + 48a – 6b – 27 = 0
–1 –2 1 21a – 6b = 36
1 2 1 2 1 2
→ 4 1 3 7a – 2b = 12
AC = 3 – 4 = –1 (shown)
1 –2 3
i j k (b) From 7a – 2b = 12,

substitute –2b = 12 – 7a into
→
→
n = AB × AC = 4 1 1
3
2
1 p(x) = ax – (1 + 6a)x + (16a – 2b)x – 27,
3 –1 3 p(x) = ax – (1 + 6a)x + (16a + 12 – 7a)x – 27
2
3
= (3 + 1)i – (12 – 3)j + (–4 – 3)k p(x) = ax – (1 + 6a)x + 3(3a + 4)x – 27
3
2
= 4i – 9j – 7k
290
Answers STPM Math T S1.indd 290 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers

Long division: The set values of x: x ∈ (3, ∞).
2
ax – (3a + 1)x + 9
4x – 9
3
x – 3 ax – (1 + 6a)x + 3(3a + 4)x – 27 (ii) 4x – 9 = (x – 3)(2x – 7x + 9)
2
2
p(x)
ax – 3ax 2
3
= A + Bx + C
2
–(3a + 1)x + 3(3a + 4)x x – 3 2x – 7x + 9
2
–(3a + 1)x + 3(3a + 1)x 4x – 9 = A(2x – 7x + 9) + (Bx + C)(x – 3)
2
2
9x – 27 When x = 3, 4(3) – 9 = A[2(3) – 7(3) + 9]
2
9x – 27 3 = 6A
1
Alternative solution: A = 2
ax – (1 + 6a)x + 3(3a + 4)x – 27 2
2
3
= (x – 3)(ax + px + 9) Coefficients of x : 0 = 2A + B
2
1
Comparing coefficients of x , 2 0 = 21 2 + B
2
–(1 + 6a) = p – 3a B = –1
p = 3a – 1 – 6a
= –3a – 1 Constants: –9 = 9A – 3C
1
= –(3a + 1) –9 = 91 2 – 3C
2
The quotient = ax – (3a + 1)x + 9 9 2
C =
2
(c) Remainder Theorem: p(4) = 13 –x + 9
1
2
2
3
p(4) = a(4) – (1 + 6a)(4) + 3(3a + 4)(4) – 27 = 13 Hence, 4x – 9 = 2(x – 3) + 2x – 7x + 9
p(x)
2
64a – 16 – 96a + 36a + 48 – 27 = 13
4a = 8 or 4x – 9 = 1 + 2 9 – 2x .
a = 2 p(x) 2(x – 3) 2(2x – 7x + 9)
2
(d) (i) p(x) = (x – 3)(2x – 7x + 9) = 0
→
(x – 3) = 0 8. Given that ON = 2i, OS = 4k and OR = 3j + 4k
→
→
x = 3 (real root)
(2x – 7x + 9) = 0 (a) Given that SR : NM = 3 : 5
2
→
→
→
2
–(–7) ± (–7) – 4(2)(9) OM = ON + NM
x =
2(2) → → 5 →
OM = ON + SR
7 ± –23 3
= (complex roots) 2 0 0
1 2
3 1 2 1 2
4 OM = 0 + 3 – 0
5
→
Therefore p(x) = 0 has only one real root. (shown) 0 4 4
2
Alternative solution: → 1 2
p(x) = (x – 3)(2x – 7x + 9) = 0 OM = 5 = 2i + 5j
2
(x – 3) = 0 0
x = 3 (real root) Alternative solution:
(2x – 7x + 9) = 0 →
2
2
2
b – 4ac = (–7) – 4(2)(9) OR = 3j + 4k
→
→
= –23 < 0 OS + SR = 4k + 3j
→
(no real root) \ SR = 3j
Therefore p(x) = 0 has only one real root. (shown)
→
→
Since SR parallel to NM.
2x – 7x + 9 SR : NM = 3 : 5,
2
9
7
→
= 21 x – x + 2 therefore NM = 5j
2
2 2
→
→
→
9
7 2
= 231 x – 2 – 49 + 4 OM = ON + NM
4 16 2 = 2i + 5j
7 2
= 21 x – 2 + 23
4 8 2 0 2
1 2 1 2 1 2
→ → → → →
(b) OT = ON + NT = ON + OS = 0 + 0 = 0
2
For (x – 3)(2x – 7x + 9) . 0, 0 4 4
7 2
(x – 3)3 21 x – 2 + 23 4 . 0. 0 2 –2
1 2 1 2 1 2
→
→
→
4 8 TR = OR – OT = 3 – 0 = 3
7 2
7 2
Since 1 x – 2 > 0, 3 21 x – 2 + 23 4 . 0 4 4 0
4
8
4
for x ∈ R.
Therefore (x – 3) . 0.
291
Answers STPM Math T S1.indd 291 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers

1 2 1 2 1 2
1
→ → → 2 2 0 (d) n = 6i + 4j + 5k From the answer in (b)
TM = OM – OT = 5 – 0 = 5 = a parallel vector of the normal vector
0 4 –4 Choose a point of plane RTM, point R(0, 3, 4)
.

1
→ → i j k Equation of the plane: r n = D
TR × TM = –2 3 0  6 0 6
1 2 1 2 1 2
.
.
0 5 –4 r 4 = 3 4
= (–12 – 0)i – (8 – 0)j + (–10 – 0)k 5 4 5
= –12i – 8j – 10k 6
1 2
.
= –2(6i + 4j + 5k) r 4 = 0 + 12 + 20
→
1 →
Area of triangle RTM = TR × TM  5
2 r (6i + 4j + 5k) = 32
.
2
2
= 1 × 2 × 6 + 4 + 5 2 Cartesian equation: 6x + 4y + 5z = 32
2
= 77 (e) Normal vector of the plane RTM:
n = 6i + 4j + 5k From the answer in (b)
1
(c) Let the acute angle between TR and TM = q Normal vector of the plane OTM:
Method 1 : Cross Product i j k
→ → → → → →  
TR × TM  = TR TM  sin q OT × OM = 2 0 4
→ → 2 5 0
TR × TM  = 2 × 77 From the answer in (b) = (0 – 20)i – (0 – 8)j + (10 – 0)k
2 77 = –20i + 8j + 10k
sin q = = –2(10i – 4j – 5k)
2 + 3 2 . 5 + 4 2 n = 10i – 4j – 5k
2
2
2
2 77
= Let the acute angle between the plane RTM and the
13 . 41 plane OTM = a
q = 49.5° n . n 2
1
cos a =
Method 2 : Scalar Product n n  2
1
10
6
→ →
→
→
TR TM = TR TM  cos q 1 2 1 2
.
.
4
–4

–2
0
–5
5
1 2 1 2 = 6 + 4 + 5 2 . 10 + 4 + 5 2
.
5
3

2
2
2
2

cos q = 0 –4 60 – 16 – 25
2 + 3 2 . 5 + 4 2 =
2
2
= 15 77 . 141
13 . 41 = 19
77 . 141
q = 49.5°
a = 79.5°
292
Answers STPM Math T S1.indd 292 3/28/18 4:25 PM