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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
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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