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 )