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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
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[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