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Exercise 2
1(a) 2e (ln )xy x
2
2
2
d 2e (ln ) ( 1) e
d
2e ln
x x
x
y xx
x x
xx
Alternative solution: 2e (ln ) e (2ln )x xy x x
d 2e (2ln ) ( 1) e
d
2e 2ln
x x
x
yx
x x
xx
1(b) 2 1cos 2 1 siny x x x
2 1
2
2 1
d 12sin 2 1 sin 2
d 1
2sin 2 1 2 sin
yx x x x
x x
x x x x
2(i) d3
d
y
t ,
d2 4
d
xt
t
d 3
d 2 4
y
x t
For gradient to be parallel to the y-axis, 2 4 0t ,
2t
(ii) Gradient of tangent at P =
3 1
2(1) 4 2
Gradient of normal at P = −2
At t = 1, x = 5, y = 2
Equation of tangent at P
12 ( 5)
2y x
1 1
2 2y x
Equation of normal at P
2 2( 5)
2 12
y x
y x
Area of triangle formed = 1 1 125
(12 )(5)2 2 4
(shown)
3(i) Let h be the height of the cylinder
2 22 10h r
2
2 2 2
2 2 2
π
π (2 10 )
2π 10
V r h
r r
r r
(ii) 1
2 2 2 2 2 2
3
2 2
d 14π ( 10 ) 2π ( )(10 ) ( 2 )
d 2
400π 6π
10
Vr r r r r
r
r r
r
For maximum V, d
0d
V
r
3
2 2
d 400π 6π0
d 10
V r r
r r
Since 0r , 2400π 6π 0r
200
3r
Using first derivative test,
r 200
3
200
3
200
3
Sign ofd
d
V
r + 0
/ \
Hence V is maximum when 200
3r .
Maximum V
2
2200 200 4000π 4000 3ππ 2 10
3 3 93 3
4(i) ln(1 sin )y x
Differentiating wrt x:
d cos d
1 sin cosd 1 sin d
y x yx x
x x x
Differentiating wrt x:
2
2
d d1 sin cos sin
d d
y yx x x
x x
Differentiating wrt x:
3 2 2
3 2 2
d d d d1 sin cos cos sin cos
d d d d
y y y yx x x x x
x x x x
3 2
3 2
d d d1 sin 2cos sin cos
d d d
y y yx x x x
x x x
(ii) When 0x , 0y ,
d1
d
y
x ,
2
2
d1
d
y
x ,
3
3
d1
d
y
x
2 3
...2 6
x xy x
(iii)
2 3
ln 1 ...2 3
x xx x
1
ln ln 1 ln 1 sin1 sin
xx x
x
2 3 2 3 31ln ... ... 2
1 sin 2 3 2 6 2
x x x x x xx x x
x
5(a) 2
2 1 d
1 4
xx
x
2 2
1 8 1 d d
4 1 4 1 4
xx x
x x
2 11 1ln 1 4 tan (2 )
4 2x x c
(b) Let tanx
2dsec
d
x
When 0x , tan 0 0
When 1x , π
tan 14
1
2 20
1d
(1 )x
x
= π
242 20
1sec d
(1 tan )
= π
420
1d
sec
= π
24
0cos d
= π
4
0
1(1 cos 2 ) d
2
=
π
4
0
1 sin 2
2 2
= 1 π 1 1
π 22 4 2 8
6(i)
1~
1 2
: 4 4 ,
3 1
l r
2~
3 2
: 1 1 ,
2 1
l r
Let be the acute angle between 1l and 2l .
1
2 22 2 2 2
2 2
4 1
1 1cos
2 4 1 2 1 1
1 1
cos126
= 84.9
C
A
M l2
(ii)
Let
3
1
2
OC
.
1 3 2
4 1 3
3 2 1
CA OA OC
Shortest distance = 2 2 2
2
1
1
2 1 1
CA
2 2 2
2 2
3 1
1 1
2 1 1
4
0
8
6
2 24 ( 8)
6
2
30 or = 3.653
(3s.f.) units
(iii) Let B be the reflection of A in the line 2l .
Since M lies on line 2l ,
Let
3 2
1
2
OM
for some .
AM OM OA
3 2 1 2 2
1 4 3
2 3 1
Since 2AM l
2 2 2
3 1 0
1 1
4 4 3 1 0
1
3
71
23
7
OM
By Ratio Theorem, 1
2OM OA OB
2OB OM OA
7 12
2 43
7 3
111 1
8 or 11 8 53 3
5
OB OB
i j k
7(a)
d dcosec cosec cosec cot .
d d
y uy u x x u x x
x x
3
3
dcot 4 cosec
d
dcosec cosec cot cosec cot 4 cosec
d
yy x x x
x
ux u x x u x x x x
x
3
3
dcosec 4 cosec
d
d4 .
d
ux x x
x
ux
x
Alternatively
d dcosec sin sin cos
d d
u yy u x u y x x y x
x x
3
3
d dcosec cot
d d
dcosec 4 cosec
d
d4 .
d
u yx y x
x x
ux x x
x
ux
x
Hence, 4 ,u Cx where C is an arbitrary constant.
4sin .y x x C
Therefore, 4 cosecy x xC is the general solution, where C is an arbitrary
constant.
(b) 1
1
1
2
1tan
tan
1tan
1
dd 1d
d
d , where is an arbitrary constant.
zz t
t
z z z t
z
z
z
Cz
C
Hence, 1 21ln 1
2tan zt z z C is the general solution, where C is an
arbitrary constant.
1 21, 0 1 0 tan 01
ln 1 0 1.2
t z C C
Therefore 1 21ln 1 1
2tan zt z z is the particular solution.
8(i) 2
f ( ) 4 3x x for , 4x x
Horizontal line test
Every horizontal line , ( , ) 3y b b cuts the graph of f ( )y x exactly once.
(ii) Let 2
4 3y x
2
4 3x y
4 3x y
4 3x y
Since 4x , 4 3x y
1f ( ) 4 3x x , 3x -1fD (3, )
(iii) 1g( ) 4
3x
x
, 3x
Rg = (4, )
Df = (4, )
Since Rg Df , fg exists.
(iv) 1f ( ) f ( )x x
2
4 3x x
2 8 19x x x
2 9 19 0x x
9 81 4(1)(19)
2x
9 5
2x
Since 4x , 9 5
2x
9(i)
fy x
(ii)
f 12
xy
(iii)
1
f ( )y
x
x
x = 0
y
B (2, 0)
y = 1
B’ (2, 0)
x
x = 2
y
B’ (2, 0) A’ (−4, 1)
y = 1
x = 6
(2, 0)
A’ (−1, 1)
y = 1
(0, 0) x
y
x = 2
(iv) f 'y x
10(i) 23 36 7 , 6 3
8 8r d r d
2 23 16 3 2
8 8r d d r
2
2
2
3 16 7 2
8 8
3 76 14
8 8
114 6 0
2
1 1 or (reject since 0)
2 14
r r
r r
r r
r r r
612
11
2
S
(ii) Let nS be the sum of first n terms of the geometric progression.
99
100nS S
12
12
6 199
121001
n
x
x = 2 x = 0
y
B’ (2, 0)
A’ (−1, 0)
y = 0
12
12
12
1 0.99
0.01
ln(0.01)6.6439
ln( )
n
n
n
Alternative method:
Using GC,
12
n
6 0.01563 > 0.01
7 0.00781 < 0.01
Thus least value of n is 7.
(iii)
Let nS be the sum of first n terms of the arithmetic progression
21 1 3
22 8 8
d
3 3(2) ( 1)
2 8 8n
nS n
3 3( 1)
2 4 8
nn
2
2013
3 3( 1) 2013
2 4 8
3 3( 1) 2013
8 16
6 3 ( 1) 32208
3 3 32208 0
nS
nn
n n n
n n n
n n
Using GC,
104.12 103.12n
0 103.12n since 0n
Largest value of n is 103.
11(a)(i)
(a)(ii) Solving for points of intersection,
2
2
2 2
3
2
1
(1 ) 2
( 2) 0
Using GC, since 0, 1
xx
x
x x x
x x x
x x
2
21 1 2
20 0
3
2Required volume π d π d
1
1.16 units
xx x x
x
(b)
At point S, 1, 3, 4t x y
At point T, 3 27 63
, ,2 4 8
t x y
Area of the region
= 4
63
8
1 27 633 4 d
2 4 8x y
= 1
2 23
2
3705( 3 )(3 3) d
64t t t
= 29.3 units2
R
O
2
2
1
xy
x
y 2y x
x
12(i) 4 10 2
5 8 3 1
7 1 2
AB
Equation of 1l is
4 2
5 1 ,
7 2
r
Since 1l intersects
1p at C,
4 2 2
5 2 13
7 2 1
2 4 2 2 5 7 2 13
3
4 2
5 3 1
7 2
OC
2
2
13
(ii) 4 2 6
5 2 3
7 13 6
CB
Shortest distance from B to 1p = 2 2
6 21
3 22 2 1
6 1
= 4 units
(iii)
A vector parallel to 2l =
2 2
1 2
2 1
5
6
2
(iv) 2 2 2
1 2 1 28
2 13 2
r
Cartesian Equation of 2p = 2 2 28x y z
(v) Let be the angle between 1p and
2p .
2 2
2 1
1 2cos
2 2
2 1
1 2
4
9
63.6 (to nearest 0.1 )