2016 H2 Math Promo Exam Suggested Solution
1
3 x 2
2
1 x2 3 2 3 x 2 x2 3
x 3 4 x 12 x 9 2
2
x2 3
0
0
2 x 3 2
x 3 x 3 +
0
-√3
+
−
−
1.5
√3
𝑥 < −√3 or 𝑥 > √3 or 𝑥 = 1.5
2
Let the radius of the surface level of the sand be r cm and the depth of the sand be h cm at time t seconds. Height of the funnel =
5
132 52 12 cm r
13
Using similar triangles,
r 5 5 r h h 12 12
h
1 25 V r 2h h3 3 432 Differentiating both sides wrt t, dV 25 dh h2 dt 144 dt
dV 7 dt 25 2 dh 7 2.5 144 dt dh 7 144 2.05 (3 s.f.) dt 156.25
When h 2.5 ,
the rate at which the depth of the sand falls in the funnel is 2.05 cm s−1 when the depth of the sand is 2.5 cm.
3(i)
3(ii)
3(iii)
4(a) 2n
4r 3 5n 1
r n
n 1 2 n 2n 4 r 3 r 3 5n 1 r 1 r 1 r n 2n 2 (2n 1) 2 n 12 (n) 2 4 5n 1 2n n 1 4 4 2 n 2 4(2n 1) 2 n 1 5n 1 ( n 1)
n 2 2(2n 1) n 1 2(2n 1) n 1 5n 1 ( n 1) n 2 3n 3 5n 1 5n 1 (n 1) 3n 2 1 5n 1 (n 1)
4(b)(i)
LHS = =
4(b)(ii)
2n
r 2
1 1 2 (n 1) ( n 1) 2 (n 1) 2 (n 1) 2
(n 1)(n 1)
2
4n RHS (n 1) 2 2
r
1 2 2 4 (r 1)
2n
r 2
1 1 2 (r 1) 2 (r 1)
1 1 2 2 1 3 1 1 2 2 4 2 1 1 32 52 1 4 1 1 (2n 3) 2 (2n 1) 2 1 1 2 (2n) 2 (2n 2) 1 1 (2n 1) 2 (2n 1) 2 15 1 1 2 2 4 4 4n (2n 1)
4(b)(iii)
As n ,
Thus ,
r 3
1 1 0 2 4n (2n 1) 2 r 5 2 2 (r 1) 16 9 13 = 144 2
5(i)
Let r cm be the radius of the cylinder. 2 r n 2 x n 2x r 2 V r2h n 2x 2
2
2x
n 2 4 nx 4 x 2 2 x 4 2 1 4 x3 4 nx 2 n 2 x 2
5(ii)
dV 1 12 x 2 8nx n2 dx 2
For stationary value of V , let
dV 0. dx
12 x2 8nx n2 0
6 x n 2 x n 0 n n or x (rejected 6 2 Stationary value of V x
n 2x 0 )
3 2 1 n n n 4 4n n2 2 6 6 6
n3 27
d 2V dx 2
1 24 x 8n 2
When x
n , 6
d 2V
n 2n 24 6 8n 0
dx
2
1 2
By 2nd derivative test, V is maximum when x
n . 6
6(i)
Let y f ( x) 3x 5 4x 3 4 xy 3 y 3 x 5 y
x(4 y 3) 5 3 y x
5 3y 4y 3
f 1 ( x)
6(ii)
Since f f
5 3x , x 4x 3 1
3 \ 4
2 1 , f (x) ff ( x) x
for x , x
3 4
3 \ 4
Rf 2
6(iii)
The graph of f is symmetrical about the line y x .
6(iv)
3 \ 4 Since Rg Df , the function fg exists. Rg [3, ) and Df
Method 1
fg(2) f 2 2(2 ) 3
2
= f 16 3(16) 5 4(16) 3 53 = 61
=
You may use your GC to check the accuracy of your answer.
Method 2
fg( x) f x3 2 x 2 =
3 x3 6 x 2 5 4 x3 8 x 2 3
fg(2)
3(2)3 6(2) 2 5
4(2)3 8(2) 2 3 53 = 61
7(a)
x2 cos x 2 1 sin 2 x 1 2 x x2 1 1 1 2 x 2 1
x2 2 1 1 2 x (2 x) ... 2 7 1 2 x x2 2
7(b)(i)
y
3 6e x
Alternative:
y 2 3 6e x
y
dy 6e x dx dy y 3e x ...... dx Differentiat wrt x 2y
3 6e x
1 dy 1 3 6e x 2 6e x dx 2
3 6e ddyx 3e
2
d 2 y dy y 2 3e x dx dx
x
y
x
dy 3e x dx
When x 0, y 3 6e0 9 3 3
dy dy 3e0 1 dx dx
3
d2 y d2 y 2 2 0 1 3e d x2 dx 2 3
2 3 y 3 x x 2 ... 2! 1 3 x x 2 ... 3 2 2 7(b)(ii) ax nx ax e ln 1 nx 1 ax ... nx ... 2! 2 2 2 2 2 a x n x 1 ax ... nx .. 2 2 n2 x2 anx 2 ... 2 n2 nx an x 2 ... 2 Comparing the terms, nx x n 1 nx
n2 2 1 2 n2 1 an x x an 2 3 2 3 1 1 1 1 5 Sub n 1 , a a 2 3 3 2 6
8(a) (i) 8(a) (ii)
$1.03x Mary’s account: Beginning of year 1st yr x nd 2 yr 1.03𝑥 + 𝑥 1.032 𝑥 + 1.03𝑥 + 𝑥 ⋮
3rd yr ⋮ nth yr
(i)
End of year 1.03x 1.03(1.03𝑥 + 𝑥) = 𝑥(1.03 + 1.032 ) 𝑥(1.03 + 1.032 + 1.033 ) ⋮ 𝑥(1.03 + 1.032 + ⋯ + 1.03𝑛 )
Amount at the end of nth year is 1.03 1.03n 1 103x 1.03n 1 x 1.03 1 3
8(a) (iii)
When 𝑥 = 10000, 103 10000 1.03n 1 190000 3 57 1.03n 1 103 57 n ln(1.03) ln 1 103 n 14.9006 Miss Lee will have at least $190 000 in her account at the end of 2030.
8(b)
Ben’s account: Amt at the beginning Interest received at the of each year end of each year st 1 yr y 0.03𝑦 2nd yr 0.03(2𝑦) 2𝑦 3rd yr 0.03(3𝑦) 3𝑦 ⋮ ⋮ ⋮ nth yr 0.03(𝑛𝑦) 𝑛𝑦
Total interest received 0.03 y 1 2 3 ... n
n 0.03 y 1 n 2 0.015 y n(1 n)
9(i)
9(ii)
x a sin 4 , y 2a cos2 , where 0
x a sin 4 dx 4a cos 4 d
y 2a cos 2 dy 4a cos sin d 2a sin 2
dy 2a sin 2 4a cos sin or dx 4a cos 4 4a cos 4 dy sin 2 cos sin or dx 2 cos 4 cos 4 When , 6 sin dy 3 3 dx 2 2 2 cos 3
3 x a sin a 3 2 3 y 2a cos 2 a 6 2 Equation of tangent at the point P 3 3 3 y a a x 2 2 2
3 3 3 x a a 2 4 2 3 3 y x a 2 4
y
2
.
9(iii)
Equation of normal at point P 3 2 3 y a a x 2 2 3 2 3 y xa a 2 3 2 5 y x a 2 3 At point A, When y = 0, 3 3 0 x a 2 4 3 x a 2 At point B, When y = 0, 2 5 0 x a 2 3 x
5 3 a 4
15 3 3 3 a a a Area of Triangle ABP 2 4 2 2
21 3a 2 units 2 16
10(i)
x2 4 x 5 xr Since x 1 is an asymptote, 1 r 0 r 1 x 1 x 5 x 5 y x 1 x 1 The eqn of the oblique asymptote is y x 5 . y
10(ii)
10(iii) Let
x2 4 x 5 2x x 1
x 2 4 x 5 2 x( x 1) x2 6 x 5 0 There are at least one common point 𝑏2 − 4𝑎𝑐 ≥ 0 36 4(5 ) 0 5 9 4 Since ≠ 0, the range of values of is 4 0 or 0 (Accept 4 and 0 ). 10(iv) & (v)
10(vi) 𝑘 > 6
