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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