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(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
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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