ACE YR IGCSE A TOP APPR' TO ADD MATH

5. 3(1 − cos x) − 5 cos x = 5 (b)
2
3 − 3 cos x − 5 cos x − 5 = 0 y
2
3 cos x + 5 cos x + 2 = 0 2
2
(3 cos x +2 )(cos x + 1) = 0
2
cos x = − or cos x = −1
3 –2π –π 0 π 2π x
x = −131.81° or x = −180°
[4] –2
6. (a) 5 sin q − 2 = 0
sin q = 2 –4
5
0 Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
q = 0.412 rad or q = 2.73 rad –6
[4]
1
(b) tan 2q – π 2 = 0.5 –8
4
π [3]
2q – = 0.4636, 3.6052, 6.747, 9.888
4 9. (a) cot x − tan x
2
2
2
2
q = 0.625 rad, 2.20 rad, 3.77rad, = cosec x − 1 − (sec x − 1)
2
2
5.34 rad = cosec x − sec x
[4] 1 – 2 sin x [3]
2
7. cos x = 1 (b) sin x + cos x
2 sin x + cos x – 2 sin x
2
2
2
x = 60° = sin x + cos x
2
2
cosec x + sec x = cos x – sin x
= 1 + 1 sin x + cos x
cos x sin x (cos x + sin x)(cos x – sin x)
2 = sin x + cos x
= 2 +
√3 = cos x – sin x
= 6 + 2√3 or 3.15 [4]
2
2
3 [4] (c) tan x sin x
= (sec x − 1)(1 − cos x)
2
2
= sec x − sec x cos x − 1 + cos x
2
2
2
2
8. (a)
2
= 1 + tan x – 1 1 × cos x – sin x
2
2
2
2
y cos x
= 1 + tan x – 1 – sin x
2
2
= tan x – sin x
2
2
[4]
3
u
u
10. y = 3 sin x
2
[3]
2
11. (a)
y
2
x
π 2π
[3]
x
π 2π
[2]
Answers 191





Answers Add Math.indd 191 14/03/2022 12:29 PM

(b) sin x = cos 2x + 1 (d) sec 1 3a – 18° = –3
2
2 solutions 2
[2] 3a 1
2
cos 1 – 18° = –
12. (a) Amplitude = 5 2 3
360° π 0° < a < 360°
Period = = 90° =
4 2 3a
[3] −18° < 2 − 18° < 522°
2
(b) Maximum point 1 π , 7 3a − 18° = 109.47°, 250.53°, 469.47°
2
2
a = 84.98°, 179.02°, 324.98°
2
2
Minimum point 1 π , –3 and 1 3π , –3 [2] [4]
4
4
2
2
(b) Maximum point 1 π , 8 and 1 5π , 8 13. (a) Amplitude = 3 16. When x = 2
2(2) + (2) − 13(2) + 6 = 0
2
3
6
6
3
2
2
2
Minimum point 1 π , 2 Period = 120° [3] (2x + x − 13x + 6) ÷ (x − 2) = 2x + 5x − 3
2x + x − 13x + 6 = (x − 2)(2x − 1)(x + 3)
2
3
2
(b) Maximum point (30°, 8) and (150°, 8) 2 tan y + tan y – 13 = – 6
2
Minimum point (90°, 2) tan y
3
2
[2] 2 tan y + tan y − 13 tan y + 6 = 0
3
14. y = 1 + 3 sin x (tan y − 2)(tan y + 3)(2 tan y − 1) = 0
1
2
tan y = 2 or tan y = or tan y = −3
3
a = 3, b = , c = 1 2
2 [3] y = 26.57°, 63.43°, 108.43°, 206.57°, 243.43°, 288.43°
[6]
15. (a) −4 sin x + 8 sin x − 5 + 5 sin x = 0 17. (a) 5 sin q − 4 cos q = 5 cos q + 4 sin q
2
2
sin x + 8 sin x − 5 = 0 sin q = 9 cos q
2
2
sin x = –8 ± √8 – 4(1)(–5) sin q = 9
2(1) cos q
= –4 + √21 or –4 – √21 (rejected) tan q = 9
q = 263.66°
sin x = –4 + √21 [3]
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
x = 35.63°, 144.37°
(b) (5 sin q − 4 cos q) + (5 cos q + 4 sin q)
2
2
[4] = 25 sin q − 40 sin q cos q + 16 cos q +
2
2
2
2
(b) sin 2x = – 2 25 cos q + 40 cos q sin q + 16 sin q
3 = 41 sin q + 41 cos q
2
2
0° < x < 360° = 41
0° < 2x < 720° [3]
2x = 221.81°, 318.19°, 581.81°, 678.19° 18. (a) 1 – 3 sin x = – cos x
x = 110.9°, 159.1°, 290.9°, 339.1° sin x sin x
2
[4] 1 – 3 sin x = – cos x
(c) sec y + sec y tan y = 5 cosec y sin x sin x
2
1 + tan y = 5 1 − 3(1 − cos x) = −cos x
2
cos y cos y sin y 1 − 3 + 3 cos x + cos x = 0

3 cos x + cos x − 2 = 0
2
1 + tan y = 5 (3 cos x − 2)(cos x + 1) = 0
cos y sin y 2
5 cos y cos x = or cos x = −1
1 + tan y = 3
sin y x = 48.19°, 311.81°, 180°
1 + tan y = 5 [4]
tan y (b) 2(1 + cot q) = 14 − 2 cot q
2
2
tan y + tan y – 5 = 0 2 + 2 cot q − 14 + 2 cot q = 0
2
2 cot q − 2 cot q − 12 = 0

2
tan y = – 1 + √21 or tan y = –1 – √21 (cot q − 3)(cot q + 2) = 0
2 2 1 1
y = 60.83°, 109.71°, 240.83°, 289.71° tan q = 3 or tan q = –2
[4] q = 18.43°, 153.43°, 198.43°, 333.43°
[4]
Cambridge IGCSE
TM
192 Ace Your Additional Mathematics
Answers Add Math.indd 192 14/03/2022 12:29 PM

19. (a) 1 = 3 tan x = – 1
sin (x – 25°) √3
1
sin (x – 25°) = x = 5π rad or 11π rad
3 6 6
x − 25° = 19.471°, 160.529° [4]
x = 44.47°, 185.53°
[4] 10 Permutations and Combinations
(b) 1 = 1
1
tan 2x + π 2 1. (a) 4 P 4 = 1
3
1
tan 2x + π 2 = 1 5 3 P 5 5 1 [1]
3
P
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
0 < x < 2π (b) 5 P 3 5 = 20 [1]
π
π < 2x + < 13π
3 3 3 2. (a) 4! × 5! = 2880
π 5π 9π 13π 17π [2]
2x + = , , ,
3 4 4 4 4 (b) (6! × P ) + (6! × P ) = 60480
5
4
3
3
x = 11π , 23π , 35π , 47π [3]
24 24 24 24 (c) 6! × 4! = 17280
x = 1.44 rad, 3.01 rad, 4.58 rad, 6.15 rad [2]
[4] 5 5 5 5 5
1 3. ( C × C ) + ( C × C ) + C = 155 [3]
1
2
3
4
2
20. sin q – cos q
cos q sin q 4. (a) 9! = 362880
= 1 – cos q (b) 8! × 4! = 967680 [1]
sin q cos q sin q
[2]
2
= 1 – cos q (c) 10! = 3628800
sin q cos q [2]
2
= sin q 8
sin q cos q 5. (a) P = 6720
5
= sin q (b) P = 840 [1]
7
cos q 4
= tan q 6720 – 840 = 5880
[4] [2]
3
6
(c) ( C × P ) × 2 + ( C × P ) × 2 = 1680
4
6
21. A = 4 1 3 1 3 [4]
B = 1.8
10
C = –3 6. (a) C = 120
7
[3] [2]
9
(b) C = 84
22. (a) 3 [1] 6 [2]
(b) 180° [1]
10
7. (a) C = 120
3
2
23. 2(cosec q − 1) + 3 + 3 cosec q = 0 [2]
9
2
2 cosec q − 2 + 3 + 3 cosec q = 0 (b) C = 36
2
2
2 cosec q + 3 cosec q + 1 = 0 [2]
(2 cosec q +1 )(cosec q + 1) = 0 8. ( C × C ) + C = 95
6
4
6
1
3
4
1 = – 1 or 1 = –1 [3]
sin q 2 sin q 9. 3! × 5! × 3! × 3! = 25920
sin q = –2 (invalid) or sin q = –1 [2]
q = 3π rad
2 10. C = 56
8
[4] 3 [2]
1
24. √3 e cos x = –e sin x 11. (a) 8 × 2! × 8! = 645120
x
2
x
2
x
2
e = 0 (invalid) [2]
sin x = – 1 (b) 10! – 645120 = 2983680
cos x √3 [2]
Answers 193
Answers Add Math.indd 193 14/03/2022 12:29 PM

(c) 8! × 2! × 2 = 161280 T – T = (x + 1)y – xy
2
1
[3] d = xy + y – xy
12. (a) P = 720 d = y
6
6 2
[1] d = d = y
2
1
(b) 3! × 3! = 36 [2]
[2] (b) 24 = xy + 8y
2! × 4! × 3 1
(c) = 24 = (2y)y + 8y
6 P 5 2
6 [3] 2y + 8y − 24 = 0
y = 6 (rejected) or y = 2
5
5
13. ( C × C × C ) × 3 = 1500
5
3 3 4 When y = 2, x = 4
[4] Area of first rectangle = xy = 2x = 8 cm 2
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
47
14. ( C × C ) + ( C × C ) + C = 2.309 × 10 16 [2]
47
53
47
53
23 2 24 1 25
[4] a 120
2. (a) S = =
5
3
15. ( C ) = 243 ∞ 1 – r 9
2 120(1 – r)
[2] a = …… (1)
9
9
16. C = 126 Second term = ar = a – 5
5
[1] 6
5
6
17. (a) C = 15 a = …… (2)
2 6(1 – r)
[2]
6
(b) C = 6 Equation (1) = Equation (2)
5 120(1 – r) 5
[2] =
9 6(1 – r)
5
12
18. C × P = 95 040
3
5
5 5 r = (accepted) and r = (rejected because it
[2] 4 4
leads to a negative first term)
19. C × C × P = 64800
6
6
4
4 2 6
[2] a = 10 (accepted) and a = –10 (rejected
3
3
20. (a) C × C = 840 because all the terms are positive)
8
6
3 4
[2] [4]
6
8
8
6
(b) 840 + ( C × C ) + ( C × C ) = 1016 (b) S – S
2 5 1 6 7 5
[3] 10 3 7
1 –
(c) C – 1016 + 840 = 3256 3 1 1 2 2 70985
14
4
7 S = =
[3] 7 3 6144
1 –
6
6
12
21. (a) C – C – ( C × C ) = 696 4
6
5 5 4 1 5
3
[4] 10 1 1 2 2
1 –
(b) More pianists than violinists S = 3 4 = 3905
2
2
= ( C × C ) + ( C × C ) + ( C × C × C ) 5 1 – 3 384
4
6
2
4
4
1
1
3
2
2
4
2
= 42 4
Same number of pianists and violinists 70985 – 3905 = 2835 = 1.38
6
4
= ( C × C × C ) + ( C × C × C ) 6144 384 2048
6
2
4
2
2
1
2
3
1
1
= 108 [3]
12 C – 42 – 108 = 642 3. Fourth term
5
[7] = ar n – 1
a
2
= 1 2 (r)
3
11 Series 2 2
= 8a
54
1. (a) T = xy r = 8a 2 × 2
3
1
T = (x + 1)y 54 a 2
2
T = (x + 2)y r = 2
3 3
T – T = (x + 2)y – (x + 1)y a 2
2
3
d = xy + 2y – xy – y S = a = 1 2 = 3a 2
2
d = y ∞ 1 – r 2 2
1 1 –
3 [5]
Cambridge IGCSE
TM
194 Ace Your Additional Mathematics
Answers Add Math.indd 194 14/03/2022 12:29 PM

4. nth term = a + (n − 1)d 11. T = ar = 3
2
Second term = 8k + d = k − 8 a = 3
−8 − 7k = d ……(1) r
Third term = 8k + 2d = 4 − k ……(2) 1 2
2
3
Substitute equation (1) into equation (2): r = 27
8k + 2(−8 − 7k) = 4 − k 1 – r 2
2
k − 6k − 20 = 0 27r – 27r + 6 = 0
2
2
k = −2.39 and k = 8.39
2 1
[3] r = ,
5. (a) C (3) + C (3) (x) + C (3) (x) + C (3) (x) 3 3 3
5
2
5
5
3
5
5
2
4
9
0 1 2 3 a = , 9
243 + 405x + 270x + 90x 2 [3]
3
2
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
[2]
3 1 2
2 1 2
1 1 2
4
(b) C (4) + C (4) 4 –x + C (4) 3 –x 2 + C (4) 2 –x 3 12. (a) 8 (2a + (8 − 1)d) = 2 3 5 (2a + (5 − 1)d)
5
5
5
5
5
0 2 2 2 2 2
1024 − 640x + 160x − 20x
2
3
[2] a = 4d
n 3 n [2]
2
6. (a) 1 n + 8 = (2a + (n – 1)d) (b) T = a + (20 − 1)d
2 4 2 20 = 4d + 19d
3 n + 8 = 2a + nd – d = 23d
4
3 n = nd [1]
4 13. S = a(1 – r )
3
3
d = 3 1 – r
4 = 1.925
2a – d = 8
3
2a – = 8 a = 1.925(1 – r)
3
4 (1 – r )
a = 35 a
8 [3] S = 1 – r
∞
(b) S − S = 2.2
20 7
35
20
3
24
= 3 1 1 2 + (20 – 1) – a = 2.2(1 – r)
2
8
4
2
= 2.2(1 – r)
7
35
3
24
2
= 3 1 1 2 + (7 – 1) 1.925(1 – r) r = 0.5
(1 – r )
3
4
2
8
= 1469
8 [3] a = 2.2(1 − 0.5)
= 1.1
7. (a) C (2x) + C (2x) (−4) + C (2x) (−4) [4]
2
3
5
5
5
5
4
0 1 2
32x − 320x + 1280x 10 2 5 1 5
3
5
4
1
[2] 14. C (ax ) – 3x 22 = −8064
5
(b) (−320x )(−1) + (1280x )(3x) = 4160x 1
4
3
4
1
5
[2] 252a – 243 2 = −8064
–1
8. C (3x) + 1 2 3 = −7654.5 a = 6
9
6
3 2x 2 [4] [2]
9. 2 = 64 15. (a) Perimeter = 176 cm, 132 cm, 99 cm
n
n − 1
n = 6 nth term = ar
6 C (2) (ax) = −576x Second term = ar = 176r = 132
5
1 132
192ax = −576x r =
a = −3 116
4
2
6 C (2) (−3x) = 2160x = 3 4
2
2 b = 2160 3 4
[6] Fifth term (perimeter) = 176 1 2
4
10. S = 1 891
∞ –1 = cm
1 – 1 2 16 [3]
3
= 3
4 [2]
Answers 195
Answers Add Math.indd 195 14/03/2022 12:29 PM

(b) S = a 18. (a) C (1) n − 1 (−3x) = n! (1) n − 1 (−3x)
n
∞ 1 – r 1 (n – 1)!1!
176 = –3nx
=
3
1 – 1 2 n C (1) n − 2 (−3x) = n! (1) n − 2 (−3x)
2
2
4
1
(n – 2)!2!
= 704 cm 9n(n – 1)x 2
[2] = 2
2 9801
(c) Area = 1936 cm , 1089 cm , cm 2 9n(n – 1)
2
16 –3n + 2 = 228
Second term = ar = 1936r = 1089 9n − 15n − 456 = 0
2
r = 1089 = 9
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
1936 16 n = –19 (rejected) and n = 8
Sum of the area of the first 10 squares 3 [6]
1 1 2 2
1936 1 – 9 10 (b) C (1) (−3x) = 5670x
4
8
4
4
= 16 4 [2]
1 – 9
16 3
= 4411 cm 19. S = 2(1 – r ) = 42
2
1 – r
3
[3]
2r − 42r + 40 = 0
3
16. (2) C (2) n − 4 (x) = (3) C (2) n − 5 (x) r = −5, 1, 4
n
n
4
5
4 5
n!
n!
(2) 1 (n – 4)!4! 2 (2) n – 4 = (3) 1 (n – 5)!5! 2 (2) n – 5 r = 4 only
4
S = 2(1 – 4 )
n – 4
2
1 – 4
4
1 n – 52 1 n(n – 1)(n – 2)(n – 3) 2 = 170
(2)
24
2
= (3) 1 n(n – 1)(n – 2)(n – 3)(n – 4) 2 2 [3]
120
2 3 –3
2 4 –3
5
5
2 5
5
3
2
x
x
0
1
2
=
(2) 1 2 1 120 2 (n – 4) 20. (a) C (x ) + C (x ) 1 2 + C (x ) 1 2 +
2 2 –3
1 2
3
24
C (x )

5
n = 32 3 x [3]
3 [6] (b) (−270x)(x) + (90x ) 1 2
1
4
2
17. (a) S − S = 13.5 = −270x + 30x 2 3x 2
5 4 = −240x
2
4
5
a(1 – r ) – a(1 – r ) = 13.5 [2]
1 – r 1 – r
a = 13.5 …… (1) 1 1 + T 22
r 4 21. 1 , T , 2
S − S = –20.25 2 2 2
1
6 5 T = ar = r
6
5
a(1 – r ) – a(1 – r ) = –20.25 2 2
1 – r 1 – r T = ar = r
1
2
2
a = –20.25 …… (2) 3 2
r 5 1 1
2
Equation (1) = Equation (2) 1 r = 2 + r
2
13.5 = –20.25 2 2
2
r 4 r 5 2r − r − 1 = 0
–3 8 r = 1 (rejected) and r = − 1
r = and a = 2
2 3
[5]
5
1 1 1 1 2 2
1 – –
8
1 2 S = 2 2
(b) S = 3 5 1
1
∞ –3 1 – – 2
1 – 1 2 11 2
2
= 16 = 32
15 [2] [5]
Cambridge IGCSE
TM
196 Ace Your Additional Mathematics
Answers Add Math.indd 196 14/03/2022 12:29 PM

3
22. (a) 10 1 2 n – 1 [1] a = −780
4
60
(–780 + l)
S =
2
3
(b) 10 1 2 n – 1 . 0 60 = 9840
4
3
log 3 10 + log 3 1 2 n – 1 = 0 l = 1108 [4]
4 4 4
2
log 3 10 + (n – 1) = 0 26. T = a + (2 − 1)d = 5x + 17x + 2
2
4 n = 1 – log 3 10 25x + 50 = 5x + 17x + 2
2
= 9 4 x = −2.4 and x = 4
[2] When x = −2.4,
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
23. r = 0.8 = 2 T = a + 2d
3
0.4 = 25(−2.4) + 2(50)
0.4(1 – 2 ) . 125000 = 40
n
1 – 2
n = log 312501 = 18.253… When x = 4,
2
n = 19 T = a + 2d
3
[3] = 25(4) + 2(50)
= 200
24. (a) T = ar = 8 [4]
2
a = 8
5
r 27. (3 + x)
5
5
5
3 2
4
T = ar = 4(x + 1) = 3 + C (3) (x) + C (3) x
2
2
1
3
2
a = 4(x + 1) = 243 + 405x + 270x
r 2
2 5
8 4(x + 1) (3 + x − x )
=
r r 2 = [3 + (x − x )]
2
5
2 2
2
r = (x + 1) = 243 + 405(x − x ) + 270(x − x )
2 = 243 + 405x − 405x + 270(x − 2x + x )
2
2
3
4
8
a = = −405x + 270x + …
2
2
(x + 1)
2
2 = −135x + … [4]
= 16
x + 1 [4] 28. a + 2d = ar
a + 8d = ar
2
3
a(1 – r ) ar − a = 2d
(b) S = 1 – r = 28 ar – ar = 6d
2
3
ar – ar = 3ar – 3a
2
3
1 16 21 1 x + 1 2 2 ar – 4ar + 3a = 0
1 –
2
x + 1 2 = 28 r = 3 or r = 1 (rejected)
1 – 1 x + 1 2 3a – a = 2d
2
x – 4x + 3x = 0 2a = 2d
2
3
x = 0, 1, 3 a = d
x = 3 only ar – a = 2a
[4] ar = 3a
(c) T = ar r = 3
27
28
= 4(2)
27
= 536870912 a + 2d = 3a
[2] d = 1, a = 1
1 1 2 2
5
6561 1 – 1 8 S = 1(3 – 1)
25. S = 3 5 3 – 1
8
1 – 1 = 121
3
= 9840 [7]
S = 60 (2a + 59(32))
60 2
= 9840
Answers 197
Answers Add Math.indd 197 14/03/2022 12:29 PM

→ → →
12 Vectors 6. AB = AO + OB
7
–10
+
= 1 2 1 2
→ –4 –12
2
2
1. (a) |AB| = √(–4) + (3) –3
–16
= √25 = 1 2
= 5 → –3
[1] AC = 3 1 2
–16
(b) 1 (–4i + 3j)
–9
5 [1] = 1 2
–48
→
→
→
2. |p – q + r| OC = OA + AC
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
4
3
2
u
1
2
u
= 3i – (–2i) – i + 5j – – i + 14 j = 1 2 1 2
–7
–9
+
3
7
4
u
= u 13 i + 81 j = 1 2 –48
–16
14
3
–44
13
81
2
= 1 2 1 2 2 |OC| = √(–16) + (–44)
+
→
14
3
2
2
= 7.23 = 4√137
[3] [4]
→
2 →
1
u
u
3. 2c – a + b 7. BC = (AC)
5
5
→
2 →
→
→
4
1
7
2 1
u 1
u
= 2 – i + j – –3i + 5j 2 + (–6j) BO + OC = (AO + OC)
5
5
3
2
5
2
5
6
3
8
u
u
= – 14 i + i + j – j – j –µ(–2i + j) + l(–3i – 8j) = [−(4i − 5j) + l(−3i − 8j)]
5
5
2
3
5
2
8
6
2µ – 3l = – – l
5
5
u
u
= – 19 i + 21 j 10µ – 15l = –8 – 6l
6
10
2
21
1
= – 19 2 1 2 2 10µ = 9l – 8
+
9l – 8
10
6
µ =
= 3.8 10 16
[3] –µ – 8l = –2 – l
5
4. a(–3i + 5j) + 10j = b(6i) µ = –8l + 16 l + 2
–3a = 6b 5
a = –2b µ = – 24 l + 2
5
5a + 10 = 0 9l – 8 –24l + 10
5(–2b) + 10 = 0 10 = 5
b = 1 9l – 8 = –48l + 20
a = –2
[3] 57l = 28
→ → → l = 28
5. (a) AB = AO + OB 57
= 3i – 7j + 6i – j 24 28
1 2
= 9i – 8j µ = – 5 57 + 2
[2] 34
1 → → = –
(b) AB = CB 95
3 [7]
1 → → →
AB = CO + OB −3 × 5 + 13 × 3
3 8. (a) d =
→ → 1 → 8
OC = OB – AB = 3
3 e × 5 – 20 × 3
1
= 6i – j – (9i – 8j) 8 = 0
3 5e = 60
5
= 3i + j e = 12
3
[2] [4]
Cambridge IGCSE
TM
198 Ace Your Additional Mathematics
Answers Add Math.indd 198 14/03/2022 12:29 PM

→ → → 2 1 5 10 2
2 1
2
(b) OD = OA + AD = a + 1 – a + d + µ – a + d – d
→ 19 → 3 3 3 9 3
= OA + AB 1 1 10 1
1
2
16 = a + d + µ – a – d
→ 19 → → 9 3 9 3
= OA + (AO + OB)
16 1 1 10 1
1
2
→ 19 → 19 → l(–a + d) = a + d + µ – a – d
= OA – OA + OB 9 3 9 3
16 16 1 10
3 → 19 → –l = – 9 µ
9
= – OA + OB
1
16 16 l = 10 µ –
= – 3 (–3i + 12j) + 19 (13i – 20j) 9 9
1
1
16 16 l = – µ
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
= 16i – 26j 3 3
1
1
1
[4] 10 µ – = – µ
→ → → → 9 9 3 3
9. AB = AD + DC + CB 13 µ = 4
= 4a + 3b – a – 3a – b 9 9
= 2b µ = 4
→ → → 13
BD = BC + CD 3
= 3a + b + a – 3b l = 13
= 4a – 2b [9]
→ → → → → →
AE = l(AB + BC) 11. (a) AB = AO + OB
= l(2b + 3a + b) = –a + b
= 3la + 3lb [1]
→ → → 1 →
BE = µBD (b) (i) OE = OD
2
→
→
= µ(BC + CD) 1 → →
= µ(3a + b + a – 3b) = 2 (OA + AD)
= 4µa – 2µb 1 → →
→ → → = 2 (OA + kAB)
AB = AE + EB 1
2b = 3la + 3lb + 2µb – 4µa = 2 [a + k(–a + b)]
3l – 4µ = 0 1
3l + 2µ = 2 = 2 [a – ka + kb]
k
Solve the simultaneous equations = 1 – k a + b
–6µ = –2 2 2 [3]
µ = 1 → → →
3 (ii) OE = OC + CE
→
→
1
1
3l + 2 1 2 = 2 = jOA + CB
3
3
→
→
→
1
l = 4 = jOA + (CO + OB)
9
[8] 3
→
→
→
1
→ → → = jOA + (–jOA + OB)
10. AE = l(AO + OD) 3
= l(–a + d) = 2 ja + b
1
→ → → → 3 3
AE = AB + BC + CE [3]
2 → 1 → → k 1
= OA + BD + µCO (c) =
3 3 2 3
2 → 1 → → → → 2
= OA + (BO + OD) + µ(CD + DO) k =
3 3 3
2 → 1 5 → → 2 → → 2
2 1
2
= OA + 1 – OA + OD + µ BD – OD 1 –
2
3 3 3 3 3 = j
2 → 1 5 → → 2 3
2
= OA + 1 – OA + OD + 1
3 3 3 j = 4
5 →
→
→
2
4
2
µ 3 1 – OA + OD – OD [2]
3
3
Answers 199
Answers Add Math.indd 199 14/03/2022 12:29 PM

→ → 16
12. PS = 2QM (b) 4fx = k x
→
→
= 2(QP + PM) 4
3
2
1
= 2 –2i – 3j + i + j f = k
2
= –i – 4j 12(k – 1) y = 18(1 – f )y
[3] k
→ → 1 4 2
13. 3rp + 5q = 4s(OP + PA) 2(k – 1) = 3k 1 – k
3 →
2
1
3rp + 5q = 4s p + PQ k = 10
5
= 1 3 → → f = 2
3
4
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
2 3rp + 5q = 4s p + (PO + OQ) 5 [4]
5
= √3 3
3
4
3rp + 5q = 4s p + (–p + 5q) (c) AD = √OA + OD
2
2
5
= √(7) – 3 AD = √(4x) + (18y)
2
2
2
4
3
3rp + 5q = 4s p – p + 3q AD = √16x + 324y
5
2
2
2
2
2
= 1 8 + 3 3rp + 5q = 4s 1 2 5 p + 3q AD = √16 + 324(1.5)
2
2
1 8 AD = 27.295
= 2 2 3rp + 5q = sp + 12sq 2
5
= 5q = 12sq CD = (27.295)
5
s = 5 = 10.92
12
8 [3]
3rp = sp → → →
5 15. (a) (i) SP = SQ + QP
1 2
3r = 8 5 = –8y + 4x
5 12 [1]
→
→
→
r = 2 (ii) RP = RQ + QP
9
[7] 1 → →
= SQ + QP
→ → → 4
14. (a) (i) OC = OE + EC = 1 (–8y) + 4x
2 → 1 → 4
= OD + EB = –2y + 4x
3 k [2]
1 →
→
= 2 (18y) + (EO + OB) → →
3 k (b) PT = kPR
1
= 12y + (–12y + 16x) = 2ky – 4kx
k → → →
PT = PS + ST
= 12(k – 1) y + 16 x = 8y – 4x + h(2x – 3y)
k k = (8 – 3h)y + (2h – 4)x
[3]
→ → → 8 – 3h = 2k
(ii) OC = OA + AC 2h – 4 = –4k
1 → → → 2 – h = 2k
= OB + AD – CD
4 h = 3
1 → → → 1
= OB + AD – fAD k = –
4 2 [4]
1 → →
= OB + (1 – f )AD m –6
4 (c) 2 = –4
1 → → → m = 3
= OB + (1 – f )(AO + OD)
4 [1]
= 1 (16x) + (1 – f )(–4x + 18y)
4 16. Velocity vector
2
= 4fx + 18(1 – f )y = 1 −7.5 − 3.3 i + –2 – 1.4 j km/h
[3] 1.5 1.5
= (−7.2i − 2.267j) km/h
[2]
Cambridge IGCSE
TM
200 Ace Your Additional Mathematics
Answers Add Math.indd 200 14/03/2022 12:29 PM

17. Time taken 20. (a) sin 45° = x
= 3.20 p.m. − 1.30 p.m. 8√2
= 1 hour 50 minutes x = 8
11 y
= hours cos 45° =
6 8√2
y = 8
Position vector of Q Velocity vector = −8i + 8j
1 2
= 1 –35 2 + 11 12 [2]
6 –3
80
10
1 2
–13
= 1 74.5 2 (b) 1 –12 2 = a + 1 –8
8 8
–1
= (–13i + 74.5j) km 1 10 2 = a + 1 2
1 Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
[3] –12 1
→ 20 cos 5° a = 1 11 2
–13
18. (a) OP = 1 20 sin 5° 2 = (11i – 13j) km
= 1 19.9 2 [2]
1.74
11
–8
[2] (c) r = 1 –13 2 + 3.5 1 2
8
(b) OQ = √20 + 30 – 2(20)(30) cos[360° – 250° – (180° – 95°)] = 1 –17 2
2
2
= 14.58 km 15
[3] = (–17i + 15j) km
[2]
(c) Bearing of Q from O
–1 30 sin 25°
= 95° + sin 3 14.58 4 13 Differentiation and Integration
= 95° + 119.59°
= 214.59° Differentiation
[3]
→ 14.58 sin(119.59° + 95° − 180°) dy 2
(d) OQ = 1 −14.58 cos(119.59° + 95° − 180°) 2 1. dx = –18(5 – 3x)
dy
= 1 8.28 2 [3] When x = b, dx = 0
–12
–18(5 – 3b) = 0
2
19. (a) Speed b = 5
3
= 8 + (4√5) 3
2
2
5
3
= 12 km/h –5 = 2 5 – 3 1 24 + a
3
[1] a = –5
(b) Time taken [3]
= 1.30 p.m. − 10.45 a.m. dy
–4
= 11 h 2. dx = –9(x – 5)
4
When x = 4,
r = 1 –10 2 + 11 1 8 2 dy = –9(4 – 5)
–4
28
4 4√5
= [12i + (28 + 11√5)j] km dx = –9
[3] 1 = –9(4) + c
=
(c) –10 + 8t 2 1 4 2 c = 37
28 + 4√5t 28 + 7√5 y = –9x + 37
−10 + 8t = 4 [3]
t = 7 dy –2
4 3. dx = 10x – 3x
7
Time = 10.45 a.m. + h dy = dy × dx
4
= 12.30 p.m. dx –2
[3] = [10(3) – 3(3) ] × a
= 89 a
3
[3]


Answers 201







Answers Add Math.indd 201 14/03/2022 12:29 PM

4. (a) 2πr + 2πrh = 500 dx = dx × dy
2
2πrh = 500 – 2πr dt dy dt
2
250 – πr 2 2
h = = × (–21)
πr 3(4) 2
[1]
= – 7
(b) V = πr h 8 [3]
2
= πr × 250 – πr 2
2
πr 9. (a) Surface area, A = 4πr
2
= 250r – πr 3 dA
dr = 8πr
dV = 250 – 3πr dA
2
dr dA = dr × dr
250 – 3πr = 0 = 8π(1.5) × 0.25
2
r = 5.15 cm
= 3π
250 – π(5.15) 2 [3]
h = 4
π(5.15) (b) Volume, V = πr
3
= 10.3 cm dV 3
= 4πr
2
2
V = π(5.15) (10.3) dr dV
2
= 1 = 858.23 cm [5] dV = dr × dr
2 = 4π(1.5) × 0.25
2
= √3 5. Let the side of cube = s = 2.25π
3
V = s [3]
= √(7) – dV dy
2
2
ds = 3s 10. (a) dx = 6x
2
2
= 1 8 + 3 dV = dV × ds When x = 1.2,
2
1 dt ds dt dy = 6(1.2)
= 2 2 = 3(30) × 0.05 dx = 7.2
2
= Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
= 135 cm s
3 –1
y = 3(1.2) + 5
2

[3]
= 9.32
6. dy = 3x + 6x Equation of the tangent at point A,
2
–3
dx
dy = dy × dx y – 9.32 = 7.2(x – 1.2)
dx y = 7.2x + 0.68
= [3(2) + 6(2) ] × 0.3 [4]
2
–3
= 3.825 (b) dy = dy × dx
[3] dx
= 6(0.95) × (–0.25)
7. A = 4πx + 24πx = –1.425
3
–1
dA = 12πx – 24πx [2]
2
–2
dx
2
12πx = 24πx –2 11. xy + 2y = –2x
2
2
x = 2 y(x + 2) = –2x
4
1 y = –2x 2
x = 2 4 x + 2
1 1 –1 dy –4x(x + 2) + 2x 2
3
A = 4π(2 ) + 24π(2 ) dx = (x + 2) 2
4
4
= 84.54 m 2
2
[4] = –2x – 8x
(x + 2) 2
3
8. dy = x dy
2
dx 2 When x = –3, dx = 6
When y = 8, x = 4 When x = –3, y = 18
y – 18 = 6(x + 3)
y = 6x + 36
[5]
Cambridge IGCSE
TM
202 Ace Your Additional Mathematics
Answers Add Math.indd 202 14/03/2022 12:29 PM

2
2
12. (a) dy = –4x(x – 3) – 2x(3 – 2x ) 15. dy = 5 ln x + 5 – 12x
dx (x – 3) 2 dx
2
6x
= When x = e,
2
(x – 3) 2
[3] dy = dy × dx
(b) When x = 2, dx
dy = 12 = (5 ln e + 5 – 12e)(2q)
dx = 20q – 24eq
y = –5 [3]
2
Equation of the tangent at point (2, –5), 16. A = πr
2
y – (–5) = 12(x – 2) 360π = πr
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
y = 12x – 29 r = 6√10
[3] dA = 2πr
dr
13. (a) πr + 2πrh = 24π dr = dr × dA
2
h = 24π – πr 2 dA
2πr = 1 × Kπ
h = 24 – r 2 2π × 6√10
2r = K
1
V = πr 2 24 – r 2 2 12√10 [4]
2r
= 24πr – πr 3 17. V = π(30) h
2
2 [2] dV
dV 3 dh = 900π
(b) = 12π – πr
2
dr 2 dV = dV × dh
3
12π – πr = 0 dt dh dt
2
2 dh
3
12π = πr –1000 = 900π × dt
2
2
r = 8 dh = – 1000
2
900π
dt
r = 2√2 cm
When r = 2√2 cm, = – 10
9π
–1
V = 24π(2√2 ) – π(2√2 ) 3 = –0.354 cm s
2 [4]
= 71.1 cm 2
3
18. V = πr h
3
2
d V = –3πr
dr 2 h = 3V –2
r
When r = 2√2 cm, π
2
r
= –
d V = –26.66 (, 0, maximum value) dh 6V –3
dr 2 [6] dr π
dr
14. (a) CS = 72 dh = dh × dr
x
2
r × dr
Area of PQRS = (4 + x) 1 72 + 5 = – 6V –3
π
x
= 288 + 92 + 5x
x [2] Percentage change
r × dr
(b) dA = –288x + 5 – 6V –3
–2
dx = π
–288x + 5 = 0 3V –2
–2
r
π
x = 288
2
5 = –2 × 0.03 × 100%
x = 7.589 cm = –6%
[3] 6% decrease
[4]
Answers 203


Answers Add Math.indd 203 14/03/2022 12:29 PM

19. (a) dV = 50 y
dt
dV
2
= 0.3h + 1.3 7
dh
When h = 8, 6
dV = dV × dh
dt dh dt 5
dh = 50
dt 0.3(8) + 1.3 4
2
= 2.44 cm s
–1
[3] 3
= 1 (b) 50 = (0.3h + 1.3) × 0.2
2
2
0.3h + 1.3 = 250 2
2
= √3
h = 250 – 1.3
2
= √(7) – 0.3 1
2
h = 28.8 cm
2
2
= 1 8 + 3 [2] –1 0 1 2 x
2
1 20. Let the other side of the triangle = y [8]
= 2 2 xy = 20 dy 1 2 2
x
2
= y = 20 22. (a) dx = 9 1 6 x + 1 2 1 2
6
2 x
(b) dy = 3x 1 1 x + 1
2
2
2
–2
1 = –20x 6
dx [3]
==
dy = dy × dx dy (16x – 6)x – (3x )(8x – 6x + 1)
2
2
3
dt dx dt (b) dx = x 6
= –20(5) × 0.3 16x – 6x – 24x + 18x – 3
–2
2
2
=
2 Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
= –0.24 cm s
–1
x
4
[4] = –8x + 12x – 3
2
21. dy = 36x – 9 x 4 [3]
2
dx
36x – 9 = 0 23. (a) dy = 5(x – 2) – 5x
2
1 1 dx (x – 2) 2
x = or x = –
2 2 = –10
y = 1 or y = 7 (x – 2) 2 [2]
d y = 72x
2
10
5x
dx 2 (b) – (x – 2) 2 dx = x – 2
1
When x = , 4 10 5x 4
1
dx =
2 – × – 2 2 1 2 × –
d y 10 (x – 2) x – 2 10
2
dx 2 = 36 (> 0, minimum point) 4 dx = – 2x
1
When x = – , (x – 2) 2 x – 2 [3]
2
d y = –36 (< 0, maximum point) π
2
2
dx 24. A = r
10
dA = πr
dr 5
dA = dA × dt
dr dt dr
πr 1
= 3π 1 2
5 0.6
r = 25 cm
[3]
Cambridge IGCSE
TM
204 Ace Your Additional Mathematics
Answers Add Math.indd 204 14/03/2022 12:29 PM

4
25. V = πr 2a – 4 = 0
3
3 a 2
dV = 4πr 2 2a = 4
3
dr a = 2
3
dr = dr × dV a = √2
3
dt dV dt [2]
2
= 1 × 18π (b) d y = 2 + 8
4π(0.5) 2 dx 2 x 3
= 18 cm s When x = √2 ,
3
–1
[3] d y 8
2
= 2 +
3
26. (a) 5x h = 500 dx 2 (√2) 3
2
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
h = 500 = 2 + 4
5x 2 = 6
= 100 6 . 0
x 2 ( √2 , 2) is a minimum point.
3
A = 5x + 2(xh) + 2(5xh) [3]
2
4
1
1
2 2
2 2
= 5x + 2x 100 + 10x 100 (c) y = 2x – dx
2
x
2
x
x
200 1000 2x 2 4x –1
2
= 5x + + = – + c
x x 2 –1
4
2
= 5x + 1200 = x + + c
x
2
x [3] When x = √2 , y = 2.
3
(b) dA = 10x – 1200x 3 4
–2
2
dx (√2) + 3 + c = 2
1200x = 10x 3 √2
–2
3
x = 120 (√2) + 4 + c = 2
3
3
x = 4.932 √2 6
c = 2 – 3
A = 5(4.932) + 1200 √2
2
4
4.932 y = x + + 2 – 6
2
= 364.93 cm 2 x 3 √2 [4]
[3]
(c) Minimum cost = $20 dy
[1] 2. (a) When x = 1, dx = 0.
2(1) – k
1
27. (a) dy = 2x(2x – 1) + x (3x )(2x – 1) 1 2 √1 = 0
2
3
2
2
3
dx k = 2
= 2x√2x – 1 + 3x 4 [1]
3
√2x – 1 (b) y = 2x – 2 dx
3
3
= 2x(2x – 1) + 3x 4 √x
√2x – 1 = 2x – 2 dx
3
4
= 7x – 2x √x √x
3
√2x – 1 3 1
A = 7, B = –2 = 2x 2 – 2x 2 + c
[4] 3 1
2 2
2 x(7x – 2) 3
3
4
(b) = x – 4√x + c
2
1 √2x – 1 3
3
3
4
= x √2x – 1 4 2 = √x – 4√x + c
2
3
3
1 3
= 14.49
[2] When x = 1, y = 2.
4 √1 – 4√1 + c = 2
3
Integration 3 14
c = 3
1. (a) At (a, 2), dy = 0 4 14
dx y = √x – 4√x +
3
3 3 [4]
Answers 205
Answers Add Math.indd 205 14/03/2022 12:29 PM

3. y = x ln x – x 6. (a) f'(x) = 10ax – 5x –6
4
2
dy = x + 2x ln x – 1 By comparison,
dx 10a = 5
1
2 1 2 2 1 1 a =
x ln x dx = 2x ln x + x – 1 dx – x – dx 2
0 2 0 2 0 2 2 2 –25b = –5
1
4 3
= 1 3 x ln x – x – 1 x – x 4 b = 1
2
2
2 0 4 2 0 5 [3]
1
4
= 1 [(2) ln 2 – 2] – 3 1 (2) – (2)
2
2
4
2 4 2 (b) f'(x) = 5x – 5x
–6
= ln 4 – 1 When f'(x) = 0,
4
–6
[5] 5x = 5x
x
4
2
4. dy = 2x(2 – x) – x (–1) x –6 = 1
dx (2 – x) 2 x = 1
10
2
= 4x – 2x + x 2 x = ±1
4 – 4x + x 2
–x + 4x When x = 1,
2
=
(x – 2) 2 y = 1 + 1
5
= x(4 – x) = 2 1 5
(2 – x) 2
3 3 When x = –1,
x(4 – x) dx = 3 x 2 4
(2 – x) 2 2 – x y = (–1) + 1
5
0 0 (–1) 5
= 9 – 0 = –2
2 – 3 The turning points are (1, 2) and (–1, –2).
= –9 [4]
[4] 7. (a) When x = a, dy = 0.
5. dy = 4x + 4 dx 3 dx
3
dx a – 8 = 0
2a
2
4x
4
+ 4x + c
= 5 Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
a = 8
3
4
dy = 3 when x = –1. a = 2
dx d y 6x – 4x(x – 8) [2]
4
3
2
(–1) + 4(–1) + c = 3 (b) dx 2 = (2x )
4
2 2
–3 + c = 3 = 6x – 4x + 32x
4
4
c = 6 4x 4
dy = x + 4x + 6 = 2x + 32x
4
4
dx 4x 4
y = x + 4x + 6 dx When x = 2,
4
d y
4
2
5
2
= x + 4x + 6x + d dx 2 = 2(2) + 32(2)
4(2)
4
5 2 3
1
When x = –1, y = . = 2
2 3 . 0
(–1) 5 + 2(–1) + 6(–1) + d = 1 2
2
5 2 (2, 3) is minimum point.
d = 47 [4]
10 (c) y = x – 8 dx
3
x
y = + 2x + 6x + 47 2x 2
2
5 10 x 3 8
When x = 3, = 2x 2 – 2x 2 dx
x
y = 3 5 + 2(3) + 6(3) + 47 = – 4 dx
2
5 10 2 x 2
= 89.3 = x 2 + + c
4
[6] 4 x
Cambridge IGCSE
TM
206 Ace Your Additional Mathematics
Answers Add Math.indd 206 14/03/2022 12:29 PM

At (2, 3), Area = 0 0 2.5
1 2.5
2 2 4 2 0 0 4.375 0
+ + c = 3
4 2 1 175
c = 0 = 2 16
4
y = x 2 + 175
2
4 x [4] = 32 unit [7]
2
8. (a) 2x – = 3
x 3 3
2x – 3x – 2 = 0 10. 4x + – 5x dx
2
x
1
5
x = 2 or x = – (rejected) y = x + 3 ln x – x + c
2
4
2 2
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
5
When x = 2, y = 1. 2 = 1 + 0 – + c
P(2, 1) c = 3.5 2
[3] y = x + 3 ln x – x + 3.5
5
2
4
2
(b) y = 2x – dx 2
x [4]
= x – 2 ln x + c
2
When x = 2, y = 1 11. 3x + 2x – 7 = 1
2
1 = 2 – 2 ln 2 + c 3x + 2x – 8 = 0
2
2
c = –3 + 2 ln 2 (x + 2)(3x – 4) = 0
y = x – 2 ln x + 2 ln 2 – 3
2
[4] x = –2 (rejected) or x = 4
3
9. (a) dy = 2x + 2 – 2x + 5 2
dx (x + 1) 2 y = 3x + 2x – 7 dx
= 7 = x + x – 7x + c
3
2
(x + 1) 2 4 3 4 2 4
+
The function is not applicable when dy = 0. 0 = 1 2 1 2 – 7 1 2 + c
3
3
3
dx
Therefore, the curve has no turning points. c = 140
[3] 27
(b) When y = 0, y = x + x – 7x + 140
3
2
27
2x – 5 = 0 [5]
x + 1 dy
2x = 5 12. (a) dx = sin x + x cos x
5 [2]
x =
2 (b) sin x dx + x cos x dx = x sin x
2
A 1 5 , 0
2
dy = 7 x cos x dx = x sin x – sin x dx
dx 1 5 + 1 2 2 π = x sin x + cos x
2
3
= 4 4 x cos x dx = x sin x + cos x 4 π 4
7 0 0
m = – 7 = 1.2625 – 1
 4 = 0.263
Equation of normal line: [3]
7
y = – x + 35 13. (a) dy = 8x
2 2
4 8 dx (3 – 4x ) [3]
When x = 0, 8 5x 1
7 35 (b) dx =
2 2
y = – (0) + 5 (3 – 4x ) 3 – 4x 2
4 8
= 35 5x dx = 5 + c
8 (3 – 4x ) 8(3 – 4x )
2 2
2
5
5
5
1
2 4
B 0, 35 2 3 8(3 – 4x ) 1 n = – – 24
8
8
= – 5
6 [3]
Answers 207
Answers Add Math.indd 207 14/03/2022 12:29 PM

1
3
3
14. 1 dx = ln (2x – 5) + c (d) –1 6xe = 3xe – e 2x 4 –1
2x
2x
2x – 5 2 –4 2 –4
a
3
4
3
3 1 2 ln (2x – 5) = ln 2 = 3(–1)e 2(–1) – e 2(–1) 4 –
2
4
1 ln (2a – 5) – ln 3 = ln 2 3
1
3
2 2 3(–4)e 2(–4) – e 2(–4) 4
2
1 ln (2a – 5) = ln 2√3 = –0.604
2
2
ln (2a – 5) = 2.485 Area = 0.604 unit
a = 8.5 [2]
[4] 18. Area under the curve
15. When y = 0, = 3 e + e + 3dx
2x
–2x
1
0 = 3 sin 2x + π 2 –3 1 2x 1 –2x 3
2
4
1
2
–3
= 1 sin 2x + π 2 = 0 = 3 2 e – e + 3x
2
1
1
1
2 π = e – e + 9 – 1 1 e – e – 9
2
–6
–6
6
6
= √3 2x + = π 2 2 2 2
2
x = π = 421.426
= √(7) – 4 When x = 3, y = 406.431
2
π
1
4 3 sin 2x + π 2 dx Area of the shaded region = 6 × 406.431 – 421.426
2
2
2
= 1 8 + 3 0 3 2 π π = 2017.16 unit [5]
2
3
1
1 = – cos 2x + 2 24 4 3π
2
= 2 2 3 0 19. Area = 8 sin 2x + cos 2x dx
= = unit 2 0
2 2 [4] 3π
1
1
3
(b) = – cos 2x + sin 2x 4 8
1 1 5 2 2 0
2
== 16. Area = (3 + 5)(2) – x – 7x + 15 dx = 1 + √2 unit
2
2
3
2
x
7x
5
2
3
3
4
= 8 –
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. [4]
+ 15x
–
2
3
3

= 8 – 1 175 – 45 2 20. (a) When y = 0,
5 – e = 0
–x
2
6
e = 5
–x
= 4 units –x ln e = ln 5
2
3
[4] x = –ln 5
17. (a) dy = 3e + 6xe P(–ln 5, 0)
2x
2x
dx
= e (3 + 6x) When x = 0,
2x
y = 5 – e
–0
[3] = 4
(b) e (3 + 6x) = 0 Q(0, 4)
2x
e = 0 (rejected) [2]
2x
3 + 6x = 0 (b) dy = e
–x
x = –0.5 dx
[2] When x = 0,
2x
2x
(c) dy = 3e + 6xe dy = e
–0
dx dx
= 1
(3e + 6xe ) dx = 3xe
2x
2x
2x
m = –1

3e dx + 6xe dx = 3xe Equation of normal line:
2x
2x
2x
y – 4 = –1(x – 0)
6xe dx = 3xe – 3 e dx y – 4 = –x
2x
2x
2x
y = –x + 4
2
6xe dx = 3xe – 3 1 1 e + c When y = 0,
2x
2x
2x
2
–x + 4 = 0
3
6xe dx = 3xe – e + c x = 4
2x
2x
2x
2 R(4, 0)
[4] [5]
Cambridge IGCSE
TM
208 Ace Your Additional Mathematics
Answers Add Math.indd 208 14/03/2022 12:29 PM

(c) Area 3
–2
–3
1 0 23. 2x – 4x + 3 dx
–x
= (4)(4) + 5 – e dx 0
2 –1 –2 3
4
–ln 5 = 3 2x – 4x + 3x
3
= 8 + 5x + e –x 4 0 –1 –2 3 1
2
4
3
–ln 5 = – + 2 + 3x
= 8 + [5(0) – e ] – [5(–ln 5) + e ] x x 2 1
ln 5
–0
2
= 8 + 4.0472 = – + 2 + 3(3) – – + 2 + 3(1)
2
4
4 3
3
= 12.05 unit 3 3 2 1 1 2
2
[5] = 77 – 3
3
21. (a) y = √10x + 3 9 5
2
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
1
dy = (10x + 3) × 10 = 5 9
3
dx 3 [3]
10 0
= 2
2 24. –x – x + 6 dx
3(10x + 3) 3 –2 0
3
4
= – x 3 – x 2 + 6x
When x = 3, 3 2 –2
4
dy = 10 = 0 – 3 8 – + 6(–2) 4
dx 2 3 2
3[10(3) + 3] 3
1
= 0.324 = 0 – – 34 2
3
1
m = – 0.324 = 34

3
= –3.0865 y = –x – x + 6
2
3
When x = 3, y = √10(3) + 3 = 3.2075 dy = –2x – 1
Equation of PQ: dx
y = –3.0865x + 12.467 When x = –2,
[6] dy = –2(–2) – 1
(b) When y = 0 dx = 4 – 1
–3.0865x + 12.467 = 0 = 3
x = 4.039
1 3 1 4 = 3(–2) + c
3
Area = (1.0392)(3.2075) + (10x + 3) dx c = 10
2
0 y = 3x + 10
4 3
4 4
3 (10x + 3) 3 When x = 0,
= 1.6666 + 10 1 2 0 y = 3(0) + 10
3
R(0, 10)
4 4 = 10
= 1.6666 + [10(3) + 3] 3 – [10(0) + 3] 3 Area of the shaded region
4
4
10 1 2 10 1 2 2(4 + 10) 34
3
3
= 9.28 unit 2 = 2 – 3
8
[6] = unit
2
22. d (4x – 8) = 6(4x – 8) × 20x 3
5
5
5
4
6
dx = 120x (4x – 8) [10]
4
5
5
1 1 1 25. p'(x) = 24x – 12 dx
x (4x – 8) dx = 120x (4x – 8) dx
4
5
5
4
5
5
0 120 0 24x 2
1 1 = – 12x + c
3
4
= (4x – 8) 2
5
6
120 0 = 12x – 12x + c
2
= 1 × [(4 – 8) – (0 – 8) ]
6
6
120 p(x) = 12x – 12x + c dx
2
= 1 × –258048
120 = 12x 3 – 12x 2 + cx
= –2150.4 3 2
[4] = 4x – 6x + cx
2
3
Answers 209
Answers Add Math.indd 209 14/03/2022 12:29 PM

p(1) = 9 Kinematics
4(1) – 6(1) + c(1) = 9
2
3
c = 11 1. (a) When t = 3, v = 0.
p(x) = 4x – 6x + 11x 3a(3) – 2b(3) = 0
2
3
2
[4] 27a – 6b = 0 …… (1)
26. y = –x + 16x – 48 s = 3at – 2bt dt
2
2
dy = –2x + 16
dx = 3at 3 – 2bt 2 + c
2
3
–2x + 16 = 0 = at – bt + c
3
2
x = 8
When t = 0, s = 0, so c = 0.
When x = 8, When t = 3, s = –1
2
y = –(8) + 16(8) – 48 a(3) – b(3) = –1
3
2
= 1 = 16 27a – 9b = –1 …… (2)
2 S(8, 16)
Equation (1) – Equation (2)
= √3 When y = 0, 3b = 1
2
–x + 16x – 48 = 0
= √(7) – x = 4 or x = 12 b = 1 27Rights Reserved.
2
3
1
P(4, 0) and Q(12, 0). 27a – 6 1 2 = 0
2
2
= 1 8 + 3 Equation of line PQ, 3 a = 2
2
1 m = 16 – 0 = – 4 27 [5]
= 2 2 0 – 12 3
2
1
Penerbitan Pelangi Sdn Bhd. All
= 4 (b) V = 3 1 2 t – 2 1 2 t
2
2 y = – x + 16 3
(b) 3 2 2 2
2
3
9
1 – 4 x + 16 = –x + 16x – 48 = t – t
3
4
== –4x + 48 = –3x + 48x – 144 dv = t – 2
2
3x – 52x + 192 = 0 dt 9 3
2
x = 16 or x = 12 When dv = 0,
3 4 2 dt
When x = 16 , 9 t = 3
3 3
4 16 t = s
y = – 1 2 + 16 2
3 3 [3]
= 80
9 2. (a) When t = 0,
T 1 16 , 80 2 a = 4 – 2(0)
= 4 m s
–2
9
3
16 [1]
3
–x + 16x – 48 dx
2
8 8 (b) v = 4 – 2t dt
3
= – x 3 3 + 16x 2 – 48x] 4 16 3 = 4t – 2t 2 + c
2
2
2
1
= – 128 2 1 6400 2 = 4t – t + c
– –
3
81
When t = 0, v = 16.
2
= 2944 4(0) – 0 + c = 16
81 c = 16
2
2
Area = (8 × 16) – 3 1 16 + 80 16 1 2944 2 v = 4t – t + 16

9
3
dv
2 4 – 81 dt = 0
= 2048 4 – 2t = 0
81 2t = 4
= 25.28 unit t = 2
2
[10]
Maximum velocity = 4(2) – 2 + 16
2
= 20 m s
–2
[5]
Cambridge IGCSE
TM
210 Ace Your Additional Mathematics
Answers Add Math.indd 210 14/03/2022 12:29 PM

(c) When v = 0, (ii) Total distance travelled
4t – t + 16 = 0 1 3
2
u
2
2
t = 2 + 2√5 = t – 4t + 3 dt + u t – 4t + 3 dt +
t = 2 – 2√5 (rejected) 0 4 1
So, P = 2 + 25 s. t – 4t + 3 dt
2
[2] 3 3 2 1 3 3
4 u3
2
(d) s = 4t – t + 16 dt = 3 t 3 – 4t + 3t + t 3 – 2t + 3t 4 u +
2
2
1
0
4t 2 t 3 3 t 3 – 2t + 3t 4 4
2
= – + 16t + d 3
2 3 4 4 4 3
t
3
= 2t – + 16t + d = 3 + + 3
2
3
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
3
When t = 0, s = 0. = 4 m
[3]
0
3
2(0) – + 16(0) + d = 0 4. (a) a = 6t – 10
2
3 d = 0 When a = 0,
3
t
s = 2t – + 16t 6t = 10
2
5
3 t = s
When t = 5, 3 [2]
5
3
s = 2(5) – + 16(5) (b) dv = 0
2
3
dt
1
= 88 m 6t – 10 = 0
3 [4]
5
t = s
3
3. (a) (i) When t = 0, v = 3.
0 – 4(0) + s = 3 s = 3t – 10t – 32 dt
2
2
s = 3 3t 3 10t 2
[1] = 3 – 2 – 32t + c
(ii) v = t – 4t + 3 = t – 5t – 32t + c
2
2
3
t – 4t + 3 , 0 When t = 0, s = 0.
2
(t – 3)(t – 1) , 0 0 – 5(0) – 32(0) + c = 0
2
3
1 < t < 3 c = 0
[2] 3 2
5
5
5
(iii) a = 2t – 4 s = 1 2 – 5 1 2 – 32 1 2
3
3
3
2t – 4 . 0 1690
2t . 4 = – 27
t . 2
[1] Distance from O = 1690 m
27
(b) (i) [5]
y
4
3
2
1
0 x
1 2 3 4 5
–1
[2]
Answers 211




Answers Add Math.indd 211 14/03/2022 12:29 PM

Penerbitan Pelangi Sdn Bhd. All Rights Reserved.








































































Cambridge IGCSE
TM
212 Ace Your Additional Mathematics






Answers Add Math.indd 212 14/03/2022 12:29 PM

Format 210mm X 297mm Extent= 224 pgs (11.58 mm) (70gsm paper) Status: CRC Date: 14/3

Cambridge IGCSE TM redue70% DA1302



ACE YOUR

ADDITIONAL


MATHEMATICS




Workbook
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.


Cambridge IGCSE Ace Your Additional Mathematics is written to improve
TM
students’ approach to mathematics by providing them with high-quality
educational materials that uphold top academic standards, while allowing them
to acquire valuable techniques to excel in their examinations.


This workbook is written distinctively based on the Cambridge IGCSE 2022 Cambridge IGCSE TM ACE YOUR ADDITIONAL MATHEMATICS
and 2023–24 syllabuses for Additional Mathematics (0606) course. The scope,
sequence and level of the workbook has been constructed to match the
Cambridge IGCSE Year 10 and Year 11 syllabuses.


The workbook provides ample practice exercises for each topic. Answers
and fully-worked solutions are provided for all questions to enhance students’
understanding in how to master the approach to certain questions. It is highly
recommended as it serves as a good aid in evaluating students’ proficiency in
mathematical skills, concepts and processes, helping them achieve excellent
results in examinations.


About the Author TM
Kung Girly received her Bachelor of Laws (Honours) from the University of London and practiced law Cambridge IGCSE
for a few months. She then decided to venture into her true passion in education and teaching.
Throughout her teaching pathway, she has educated many students who sat for the IGCSE exams
with excellent results. She has also prepared and conducted seminars for exam-going students,
providing them the foundation and tips to excel in the examinations. She also provides coaching ACE YOUR
for students participating in international mathematics competitions.

Ms Kung has seven years of experience in teaching IGCSE Additional Mathematics, and currently
serves as a Director at Teras Murni Education Group. She is actively involved in educating students ADDITIONAL
to ensure they meet their academic potential.
MATHEMATICS







Workbook


www.dickenspublishing.co.uk
DA1302
ISBN: 978-1-78187-260-4

Suite G7-G8, Davina House, 137-149 Goswell Road,
London, EC1V 7ET, United Kingdom.
E-mail: [email protected] Kung Girly