2016 HCI Prelim Paper 1 Solutions

Qn Solution

1 * Let Pn be statement sin for all nU nx n .

When 1n , 1LHS sinU x , RHS sin x

1P is true.

* Assume Pk is true for some k , i.e. sinkU kx .

Want to prove that Pk + 1 is true, i.e. 1 sin 1kU k x .

LHS

1

2 12cos sin

2 2

k

k

U

k x xU

2 1 1

sin 2cos sin2 2

kkx x x

sin sin 1 sinkx k x kx

sin 1 RHSk x

*Since P1 is true, Pk is true implies Pk + 1 is true, by MI Pn is true for all n .

2 2

21

4( 1) 1x

2

(2 1)(2 3)0

4( 1) 1

x x

x

Since 24( 1) 1 0x for all x,

(2 1)(2 3) 0x x

3 1

2 2x

31

221

21 d

4( 1) 1x

x

1 31

2 212 212

2 21 d 1 d

4( 1) 1 4( 1) 1x x

x x

311

1 1 221

12

tan (2 2) tan (2 2)x x x x

1 1 11 3 1tan 1 1 1 tan 3 tan 1

2 2 2

31

6 2

3

(a)

3 3 3 2

2 3 2 4 3

: 2 4 3 ,

: ,

PQ

OB

OP a AB a b a b a

PQ OQ OP a b a a b

l r a a b

l r b

At point of intersection, 2 4 3a a b b

Comparing coefficients of a and b , 1 3

, 2 2

position vector of the point of intersectio2

n3

b

(b) 2 1 1

4 2 2 2

2 1 1

a b n

Let F be the foot of perpendicular.

Method 1

0 1

: 3 2 ,

4 1

FCl r s s

,

1

: 2 0

1

OAB r

0 1 1

3 2 2 0

4 1 1

s

6 4 1 4 1 0s

1

3s

0 1 0 1 11 1 1

3 2 9 2 73 3 3

4 1 12 1 13

OF

1 7 13, ,

3 3 3F

Method 2

ˆ ˆ

1 1

2 20

1 13

1 14

2 2

1 1

1

2

16 4

6 6

11

23

1

FC OC n n

0 1 0 1 11 1 1

3 2 9 2 73 3 3

4 1 12 1 13

OF OC CF

1 7 13, ,

3 3 3F

4

Method 1

2 2

2 1

2 1

n

n n n

2 2

2 2 2 2

2 1 2 1

2 1 2 1

n n n n

n n n n n n

2 2

2 2

2 1 2 1

2 1

n n n n

n n n

2 22 1n n n

Method 2

2 2 2 22 1 2 1n n n n n n

2 22 1n n n

2 1n

2 2

2 2

2 12 1

2 1

nn n n

n n n

2 21

2 1

2 1

N

n

n

n n n

2 2

1

2 1N

n

n n n

2 2

3 0

8 3

...

2 1N N N

2 2N N

(a) Replace n by 1n ,

22

1

2 21

2

2

2 1

1 2

2 1

2 1

1 2 1

1

N

n

N

n

n

n n n

n

n n n

N N

N

(b) Notice that 2 2n n n and

2

22 1 1 2 2 0n n n .

2 1 1n n

2 22 1 2 1n n n n

2 2

1 1

2 12 1 nn n n

2

2 21 1

2 1 2 12

2 1 2 1

N N

n n

n nN N

n n n n

5 2 i 5 5w

52 i 5

2 iw

2 i 1w circle centre 2, 1 , radius 1

(i)

1 1

2 2

1 1sin sin

52 1

min arg 2 2.2143 2.21 3sfw

Re

Im

O

Re

Im

w

α

α

O

(ii)

min 2 2 1z w

(iii)

4

Re

Im

w

z

–1

Re

Im

w

z

O

O

6

(i) 0 3

0 , 1.5

2 2

OD OE

3 0 3 2

1.5 0 1.5 1.5 1

2 2 0 0

DE

0 2

: 0 1 ,

2 0

DEl r

(ii) 0 3 3

0 0 0

2 0 2

AD

2 3 2 2

1.5 1 0 1.5 4 4

0 2 3 3

DE AD n

2 0 2

: 4 0 4 6

3 2 3

ADE r

2 4 3 6x y z

(iii) 0 2

0 , 4

1 3

OABC ADEn n

1

1

1

angle between planes

0 2

0 4

1 3cos

0 2

0 4

1 3

3cos

4 16 9

3cos

29

56.1 1dp

1 3Angle 180 2 cos

29

67.7 (1 d.p.)

7

(i)

2 2

2 2 2 2

d ( 2) ( 2)(2 1) 4

d ( 2) ( 2)

y kx x x kx kx kx

x kx x kx x

When 0x , 2

d 00

d ( 2)

y

x

and

21

2y

Hence required equation of tangent is 1y .

(ii) For axial intercepts, when 0y , 2x .

when 0x , 1y .

For vertical asymptotes, 2 2 0kx x

1 1 8

2

kx

k

For turning points, d

0d

y

x

2 4 0

( 4) 0

kx kx

kx x

0 or 4x x

(iii) At 2x ,

2 2

d 4 8 4 1

d (4 ) 16 4

y k k k

x k k k

gradient of normal 4k

Hence required equation of normal is 0 4 ( 2)y k x

4 8y kx k

O x

y

A(0,1)

B(2,0)

(iv)

4 8y kx k

y = 1

When 1y , 8 1

1 4 84

kkx k x

k

required area 1 8 1

2 (1)2 4

k

k

16 1

81

281

2 (since 1)8

15

8

k

k

k

k

8 (i)

Area2

02 sin dx x x

2

0 02 cos 2 cos dx x x x x

2

0 02 2( sin sin d )x x x x

2

02 2 cos x

2 22( 4) units

(ii)

y = 2x + 3 y = 2x 5

( 2, 0 )

( 2, 2 )

(iii)

Coordinates of the points of intersections of the 2 curves are ( 1.5374, 2.3623) and

( 2.7626, 2.8238) .

Volume of solid generated

1.5374 1.53742 2 2 2

2.7626 2.7626( sin ) d ( 1 1 4( 2) ) dx x x x x

= 26.8 3units

9

(i)

Let A cm2 be the surface area of the cylindrical container.

Let r cm and h cm be the radius and height of the cylindrical container respectively.

Volume 2r h k

2

kh

r

2

2

2

2

2

2

2

A rh r

kr r

r

kr

r

Hence 2

d 22

d

A kr

r r

When d

0d

A

r ,

O

y

x

Can also express r in terms of

h and find A in terms of h,

then let d

0d

A

h to obtain h

and subsequently r .

2

3

3

22 0

kr

rk

r

kr

13

322

k

k k kh

r

Hence 3 3: : 1:1 (shown)k k

h r

2

2 3

d 42 0

d

A k

r r since p > 0 and k > 0

Hence A is a minimum when 3k

r

(ii) From (i), : 1:1h r

Hence 2 2 22 2 3A rh r r r r r

For new design, : 5 : 2h r

Hence new 2 2 252 2 6

2A rh r r r r r

required ratio is 2 26 :3 2 :1r r

(b) Method 1

Let V cm3 be the volume of the cylindrical container.

2 2 35 5( )2 2

V r h r r r

2 2 252 2 6

2A rh r r r r r

2d 15

d 2

Vr

r

d12

d

Ar

r

2

d d d d

d d d d

212 80

15

128

A A r V

t r V t

rr

r

When 50h , 2

(50) 205

r

Can also find d

d

V

h and

d

d

A

h

and use

d d d d

d d d d

A A h V

t h V t

Hence 2d 128

6.4 cm /sd 20

A

t

Method 2

26A r (reject since 0)6 6

A Ar r r

Hence 2 2 35 5

2 2V r h r r r

32

3 12 2

3

5

2 6

5

2(6)

A

A

12

3 12 2

d 15

d 4(6)

V A

A

When 50h , 2

(50) 205

r

26 (20) 2400A

Hence d d d

d d d

A A V

t V t

3 12 2

12

3 12 2

12

2

4(6)80

15

4(6)80

15(2400 )

6.4 cm /s

A

10

(a)

(i)

sinsin 6

x y

sin

5sin

6

x

y

sin

5 5sin cos sin cos

6 6

x

y

sin

1 3cos sin

2 2

x

y

2sin

cos 3 sin

(shown)

(a)

(ii)

2sin

cos 3 sin

x

y

3

2

2 ...3!

1 3 ...2

x

y

13 2

2 1 33! 2

x

y

2

3

22

1 1 32

23! 1 2

32! 2

x

y

3 222 1 3 3

3! 2

x

y

2 3202 2 3

3

x

y

(b) (i)

Using sine rule,

sinsin 6 3

1

6

x

1sin 3x

(b)

(ii)

Method 1

sin 3x

dcos 3

dx

---- (1)

22

2

d dcos sin 0

d dx x

--- (2)

33 2 2

3 2 2

d d d d d dcos sin 2sin cos

d d d d d d

0 3

x x x x x x

When 0x ,

0 , d

3dx

,

2

2

d0

dx

,

3

3

d27

dx

3 327 93 ... 3 ...

3! 2x x x x

Method 2

1sin 3x

1

2 2

2

d 33 1 9

d 1 9x

x x

3 32

2 22 22

d 13 1 9 18 27 1 9

d 2x x x x

x

5 33

2 22 23

d 8118 1 9 27 1 9

d 2x x x x

x

5 3

2 2 22 2729 1 9 27 1 9x x x

When 0x ,

0 , d

3dx

,

2

2

d0

dx

,

3

3

d27

dx

3 327 93 ... 3 ...

3! 2x x x x

11

(i)

Since f g

10, e 0,e

2R D

, f gR D and gf exists.

gf ln 2,R

e

y

x

x = 0

y

x y = 0

(ii)

Since a horizontal line y = 1 cuts the graph of y = f(x) twice, f is not a one-to-one function and

f -1 does not exist.

(iii) 0b

Let fy x

21

2

1e

2

ln 2 1

1 ln 2

xy

y x

x y

Since 0, 1 ln 2x x y

1 1f : 1 ln 2 , , 0 e

2x x x x

(iv) Step 1

Step 2

1 ln 1 ln2

1 ln 1 ln2 2

xy x y

x xy y

0 e 0 e2

0 2e

xx

x

h : 1 ln , , 0 2e2

xx x x

y

x y = 0

y = 1