SECTION A
1(a) Given f x is continuous at x 1 , 2(a) 1
x t 1 3t 2
lim f x lim f x f 1 dx t2 6t y t 2 4t 2
x1 x1 dt
dx 1 6t3
lim x 1 lim x2 x a 12 1 a dt t2 dy 2t 1
dt 2t 2
x1 x1 x a 1 a
11 12 1 a 33
1 a
dy 2t 2 2 2(t 2 1)
dt
21 a a 1 1
t2 t2
a2 dy dy dt
dx dt dx
x2 x 2
lim 3
x2 x 2
2(t 2 1) t2
x 2 x 1 6t3 1
1
lim
x2 x 2 t2
lim x 1 3 33 3
x2
2t 2 (t 2 1) 2t3 2t 2
6t3 1 6t3 1
1
(b) lim 2x 5 3
3x 4
x d2y d dy dt
dx2 dt dx dx
lim 3 2x 5
x 3x 4
1 3
3t 2 2t 2
6t 3 2 3
1 6t 2t 18t 2
2x 5 t2
3
xx 6t3 1 2
lim 3 3x 4 6t 1
x
xx
3 2 0 3 2 0.874 3t2 6t 3 1 2t 2 1 36t 4 t 3 3
30 3
t2 t2
6t3 1 3
When t 1,
d 2 y 36 12 1 3611 3
dx2 6 13 25
2(b) 3
y
y ln sin 2x x
1 cos 2x x
dy 1 1 cos 2x2cos 2x sin 2x2sin 2x river
dx sin 2x
1 cos 2x2
2x y 100 y 100 2x
1 cos 2x
Area xy x 100 2x 100x 2x2
dy 1 cos 2x 2 cos 2x 2 cos2 2x 2 sin 2 2x
dx sin 2x
1 cos 2x2
dy
1 2 cos 2x 2 cos2 2x sin2 2x dA 100 4x 0 x 25
dx
dx sin 2x 1 cos 2x
dy 1 21 cos 2x d2A 4 0 max
dx sin 2x dx2
1 cos 2x
dy 2 when x 25, y 100 225 50
dx sin 2x
dy 2cos ec2x
dx
Alternative:
y ln sin 2x ln sin 2x ln 1 cos 2x
1 cos 2x
dy 1 2cos 2x 1 2sin 2x
dx sin 2x 1 cos 2x
dy 2 cos 2x 2 cos2 2x 2sin2 2x
dx
sin 2x1 cos 2x
dy
dx
2 cos 2x 2 cos2 2x sin2 2x 21 cos 2x
sin 2x1 cos 2
sin 2x1 cos 2x x
dy 2 2cos ec2x
dx sin 2x
SECTION B 2(a)
1(a) 4x 6
4x
ln 24x1 ln 8x5 log2 1612x
4x 64 x 0
4x 1ln 2 3x 15ln 2 4 8x
4x 4x
4x ln 2 ln 2 3x ln 2 15ln 2 4 8x
x ln 2 8x 16ln 2 4 5x 12 0 , 12 12 , 4 4,
4x 5 5
x ln 2 8 16ln 2 4
5x 12
x 44ln 2 1 4x
5x 12
ln 2 8 4x
1(b)
z1 2 3i 4 2i Solution set : x : 12 x 4
z2 4 2i 4 2i 5
z3
2(b)
x6 3 x2
8 4i 12i 6i2 x 62 32 x 22
16 4i2
8 8i 61 x2 12x 36 9 x2 4x 4
16 41
14 8i 7 2 i 8x2 24x 0
20 10 5
8x x 3 0
z3 7 2 2 2 65 By using graphical method, the solution set is x : x 0 x 3
10 5 10
3(a) Rf 1,11 4(a) f g x g2 3 4 x 3 , x 5
5
(b) x
9 1 11 1 x 5 g2 x4
m1 2 3 2 m2 5 3 6 g2 x x 1
3, 1 3, 1 Since g x 0, g x x 1
y 2x c y 6x c Dg 1, \ 5
1 23 c c 5 1 63 c c 19 (b)
g x1 g x2 Alternative : Horizontal Line Test
f x 2x 5, 2 x 3 x1 1 x2 1
6x 19, 3 x5
x1 1 x2 1
(c) f 0 20 5 5 x1 x2
g x is 11 g1 x exists
f f 0 f 5 65 19 11 g g1 x x
g1 x 1 x
g 1 x x2 1
Dg1 0, \2
(c) h g1 x a x2 1 b 5x2 12
ax2 a b 5x2 12
By comparing,
a 5, a b 12 5 b 12 b 7
5(a) f x 2x2 2
x2 4
lim 2x2 2 lim 2x2 2
x x2 4 x x2 4
2x2 2 2x2 2
x2 x2 x2 x2
lim x2 4 lim x2 4
x x
x2 x2 x2 x2
2 2 2 2
x2 x2
lim lim
x 4 x 4
1 x2 1 x2
20 2 20 2
1 0 1 0
y 2 is horizontal asymptote.
lim 2x2 2 lim 2x2 2
x2 4 x2 4
x2 x2
x 2 & x 2 are vertical asymptotes.
6(a) y tan x 6(c) 2x2 3y2 k
4 y2 x3 4x 6y dy 0
dy sec2 x 2 y dy 3x2 dx
dx 4 dx dy 2x
dx 3y
d2y 2 sec x sec x tan x dy 3x2
dx2 4 4 4 dx 2 y
d2y 2 sec2 x tan x At 1,1 At 1, 1
dx2 4 4
312 3 12
dy 2 1 3 dy 2 1 3
dx 2 dx 2
d2y 2y dy 2 sec2 x tan x 2 tan x sec 2 x 0
dx2 dx 4 4 4 4
dy 2 1 2 dy 21 2
dx 31 3 dx 31 3
ex
6(b) g x ex 2 The tangents of the two curves at these two points are perpendicular to
each other.
gx
ex 2 ex ex ex 2ex 2
ex 2 2 ex 2
2
ex ex 2 2
e2x 5ex 4 0
ex 4ex 1 0
ex 4, ex 1
x ln 4, x ln1 0
7.
dV 5cm3 / s
dt
tan 30 r r 3 h
h3
V 1 r2h 1 h3
39
dV h2
dh 3
When V 3000 , 1 h3 3000 h 30
9
dh dh dV
dt dV dt
3 5 1 cm / s
60
302