Page 21 - 1202 Question Bank Additional Mathematics Form 5

20
2
3

5 2
dh = – —– – — dx (ii) tan 2A 2x – 15x + 25 . 0
2 tan A
(2x – 5)(x – 5) . 0
2
x
= ————
dx
dV
dV
2
5
3
—– = —– × —– 1 – tan A — , x , — and 5 , x , 6
dh dx dh 2 — 2 2
3
 2 2
 – —– – —2
54x
1
2
2

= 120 – —— ————–– = ———— 2 3. (a) BD = x sin α
5
20
3
3
 2 2
x 2 5 1 – — DE = DB sin α
= (x sin α)(sin α)

3
When h = 1, = ——— = x sin α
2
9
3x
20
—– – —– = 1 1 – — (b) EF = DE sin α
4
x
5
3
12
100 – 3x = 5x = – —– = x sin α 2 3
2
3x + 5x – 100 = 0 5 ∴ x sin α, x sin α, x sin α
2
x sin α
2
p
(3x + 20)(x – 5) = 0 15. P(spoilt) = —–, n = 5 r = ———– = sin α
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x sin α
x = 5 10 x sin α
3
1
dV
54(5)
2
2
2

x sin α
—– = 120 – —–––– ————–– (a) p P(X = 0) = 0.1681 r = ———– = sin α
0 q
5
0  10 2  10 2
dh 5  – —– – —2 5 C —– —– = 0.1681 (c) T = HG 5
20
3
5
5
5
2
= x sin α
q
5
a
7 2

= –150 – — —– = 0.7 (d) S = ——–
10
1 – r
∞
x sin α
= 107.14 Therefore, p = 3 q = 7 = ————
dh
dV
 2
—– = 107.14 —– (b) 7 good, 3 spoilt 1 – sin α
dt dt If 3 spoilt bulbs are arranged If x = 8, α = 60°
8 sin 60°
dV
Given —– = 21 cm s together, then number of ways S = ————–
3 –1
dt ∞ 1 – sin 60°
dh
8!
 2
21 = 107.14 —– = —– = 8 ——
8AB 3
7!
dt
2
dh
—– = 0.196 cm s –1 (c) p + q + 0.2 + 2p + q = 1 S = ———–
∞
AB 3
dt 3p + 2q = 0.8 ...........1 1 – —–
14. (a) BC = 10 sin θ P(X < 2) = p + q = 0.55 – 0.2 .....2 2
8AB 3(2 + AB 3)
AB = 10 cos θ 2p + 2q = 0.7 ................3 S = ———————
(i) Area, A 1 – 3: p = 0.1 ∞ (2 – AB 3)(2 + AB 3)
1
= —(10) sin θ (10 cos θ) q = 0.35 – 0.1 = 0.25 S = 24 + 16AB 3
2 ∞
= 50 sin θ cos θ 4. (a) y = ab x – 2
= (25 sin 2θ) cm 2 Paper 2 log y = log a + (x – 2) log b
10
10
10
25AB 3 Section A Y = log y, X = x – 2, m = log b,
10
10
(ii) 25 sin 2θ = ——– c = log a
2 1. x y z 10
AB 3 x + y + z = 16 .................................1 (b)
sin 2θ = —–
2 z – 2 = x + y x – 2 –2 –1 1 3 4 5
x + y – z = –2 .................................2 log y –0.9 –0.6 0 0.6 0.9 1.2
2 y – 5 = x 10
3 log y
y – x = 5 ...................................3 10
2θ 1 + 2: 2x + 2y = 14
1 1.0
2θ = 60°, 120° x + y = 7 ......................4 0.8
θ = 30°, 60° –x + y = 5 .......................................5
4 + 5: 2y = 12
π
π
θ = — rad, — rad y = 6 0.6
6
3
(b) 2x – x – 3 = 0 x = 1 0.4
2
(2x – 3)(x + 1) = 0 1 + 6 + z = 16 0.2
3 z = 9
x = — or x = –1 Therefore, the number is 169. (x – 2)
2
3 –2 –1 0 1 2 3 4 5
tan A = —, tan B = –1 2. (a) (i) V = 5(15 – 2x)(x) –0.2
2
Sum of roots: V = 75x – 10x 2 –0.4
1
dV
tan A + tan B = — (ii) —– = 75 – 20x = 0 –0.6
2
Product of roots: dx x = —– –0.8
75
3
tan A tan B = – — 20
2 15
(i) tan (A + B) x = —– (c) From the graph,
4
tan A + tan B 15 0.9
= —————— When x = —–, m = —– = 0.3
1 – tan A tan B 4 2 3
15
15
 4 2
 4 2
1 V = 75 —– – 10 —– log b = 0.3
10
2 b = 2
5
= ————– V = 140— cm 3
3
2 2

8
1 – – — (b) 90 , 75x – 10x , 140 log a = c = –0.3
10
a = 0.5
2
1
— 10x – 75x + 90 , 0 ∴ a = 0.5, b = 2
2
2
= —— 2x – 15x + 18 , 0
2
5
1
— 5. (a) —(3 – cos 2x)
2 (2x – 3)(x – 6) , 0 2
3
1
1
2
= — — < x , 6 = —(3 – (1 – 2 sin x))
5 2 2
75x – 10x , 125 1
2
2
10x – 75x + 125 . 0 = —(2 + 2 sin x)
2
2
134
10_1202 QB AMath F5.indd 134 10/01/2022 4:49 PM