A-Level H2 Maths 2018 – Paper 1 - Achevas

Mathematics (Higher 2) 9758/01

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A-LEVEL H2 MATHS 2018 – PAPER 1

Question 1

[ Ans: (i) 2

1 ln x

x

− (ii)

21

e− ]

(i) 2 2

1ln

1 lnx x

dy xx

dx x x

− − = =

(ii) 2 2 2

1 ln 1 lndy x x

dx x x x

−= = −

2 2

1 ln xy dx

x x= −

2 2

ln 1

1

1 ln

xdx dx y

x x

y Cx

xC

x x

= −

= − − +

= − − +

211

ln 1 ln

1 ln 1 ln1

1 1

1 11

21

ee x x

dxx x x

e

e e

e e

e

= − −

= − − − − −

= − − +

= −



Question 2

[ Ans: (i) 1

2x = or 3x = (ii)

125

6

units3 ]

(i) 3

yx

= (1)

2 7y x+ = (2)

Sub. (1) into (2)

32 7x

x+ =

22 7 3 0x x− + =

( )( )2 1 3 0x x− − =

1

2x = or 3x =

x -coordinates of A and B are 1

2 and 3 respectively.

(ii) From GC,

Volume

( )2

3 32

1 1

2 2

37 2x dx dx

x

= − −

( )3 32 21 1

2 2

7 2 9x dx x dx −= − −

( )

( )( )

333

1

11

22

7 29

3 2 1

x x

− − = −

− −

( )3

33

11

22

17 2 9

6x

x

= − − +

3 3 11 6 9 2

6 3

= − − + −

125

6

=

7 2y x= − 3

yx

=



Question 3

[ Ans: (i) show; ( ) 3

6f x

x= − (ii)

23y x= − ]

(i) 2y ux=

( ) 2 22 2dy du du

u x x ux xdx dx dx

= + = +

2 6dy

x ydx

= −

2 22 2 6du

x ux x uxdx

+ = −

2 3 22 2 6du

ux x uxdx

+ = −

3 6du

xdx

= −

3

6du

dx x= − (shown)

( ) 3

6f x

x = −

(ii) 3

6du

dx x= −

3

3

2

2

6

6

62

3

u dxx

x dx

xC

Cx

−

−

= −

= −

= − +

−

= +

2

2 2

33

yC y Cx

x x = + = +

When 1x = ,

2y =

3 2 1C C+ = = −

23y x = −



Question 4

[ Ans: (i) 1x = − or 0x = or 1 3x = − (ii) 1 3 1x− − − or 0 1 3x − + ]

(i) 22 3 2 2x x x+ − = −

( )22 3 2 2x x x+ − = − −

22 2 0x x+ =

( )1 0x x + =

1x = − or 0x =

or 22 3 2 2x x x+ − = − 22 4 4 0x x+ − =

2 2 2 0x x+ − =

( )2

1 3 0x+ − =

( )2

1 3x + =

1 3x = −

(ii)

22 3 2 2x x x+ − −

1 3 1x− − − or 0 1 3x − +



Queston 5

[ Ans: 1b = − ; ( )1

1

x af x

x

− +=

− ]

( )

( )

( )

( )

( ) 2

1

1

x aa

x bff xx a

bx b

x a a x b

x a b x b

a x a ab

b x a b

++

+=+

++

+ + +=

+ + +

+ + +=

+ + +

( ) ( )ff x g x=

( )

( ) 2

1

1

a x a abx

b x a b

+ + +=

+ + +

From observation, since there should be no x in the denominator,

1 0 1b b+ = = −

( ) ( )ff x g x=

( )ff x x=

( )( ) ( )1 1f ff x f x− −=

( ) ( )1f x f x−=

( )1

1

x a x af x

x b x

− + += =

+ −



Question 6

[ Ans: (i) show (ii) 2 31 ]

(i) 3 2a b a c =

3 2 0a b a c − =

( )3 2 0a b c − =

a is parallel to 3 2b c− .

3 2b c a− = (shown)

(ii) ( ) ( ) ( ) ( )3 2 3 2b c b c a a − − =

29 6 6 4b b b c c b c c a − − + =

2 2 2 29 12 4b b c c a− + =

2 2 229 12 cos60 4b b c c a− + =

( ) ( )( ) ( ) ( )2 2 221

9 4 12 4 1 4 1 12

− + =

2 124 =

124 2 31 = =



Question 7

[ Ans: (i) 22

2 16

dy x y

dx xy y

−=

+ (ii)

1,0

17N −

]

(i) 2 2

2 2

4 1

2

x y

x xy

−=

+

2 2 2 22 8x y x xy− = + 2 2 28x y xy− =

( ) ( )2 2 28d d

x y xydx dx

− =

22 16 2dy dy

x y xy ydx dx

− = +

( ) 22 16 2dy

xy y x ydx

+ = −

22

2 16

dy x y

dx xy y

−=

+

(ii) When 1x = for curve C ,

( )2 2 21 8 1y y− =

2 1 1

9 3y y= =

Let coordinates of P and Q be 1

1,3

−

and

11,

3

respectively.

At P ,

( )

( )

21

2 1173

1 1 542 1 16

3 3

dy

dx

− −

= = − − + −

Equation of tangent: ( )1 17 17 1

13 54 54 54

y x y x

− − = − − = − −

(1)

At Q ,

( )

( )

21

2 1173

1 1 542 1 16

3 3

dy

dx

− = =

+

Equation of tangent: ( )1 17 17 1

13 54 54 54

y x y x− = − = + (2)

Solving (1) and (2) using GC, 1

17x = − , 0y =

1,0

17N

−



Question 8

[ Ans: (i) 5A = ; 3 40u = (ii) 15

2a = , 5b = − , 5c = − (iii) ( ) ( )

515 2 1 1 5

2

n n n n− − + − ]

(i) Given 1 2n nu u An+ = + , where 1n

For 1n = ,

( )2 12 1u u A= +

( )2 12 15 2 5 5A u u= − = − =

1 2 5n nu u n+ = +

For 2n = ,

( ) ( )3 22 5 2 2 15 10 40u u= + = + =

(ii) Given ( )2n

nu a bn c= + +

1 5u =

( ) ( )12 1 5 2 5a b c a b c+ + = + + = (1)

2 15u =

( ) ( )22 2 15 4 2 15a b c a b c+ + = + + = (2)

3 40u =

( ) ( )32 3 40 8 3 5a b c a b c+ + = + + = (3)

From GC,

15

2a = , 5b = − , 5c = −

(iii) ( )15

2 5 52

n

nu n= − −

1

n

r

r

u=

( )1 1 1 1

15 152 5 5 2 5 5

2 2

n n n nr r

r r r r

r r= = = =

= − − = − −

( )( )

( )( )2 3115

2 2 2 2 5 1 1 52 2

nn n

n +

= + + + − − − +

( )( )

2 2 115 51 5

2 2 1 2

n

n n n − = − + −

−

( ) ( )5

15 2 1 1 52

n n n n= − − + −



Question 9

[ Ans: (i) show (ii) 2k = (iii) 8 units ]

(i) ( ) ( )2 22 2cos2 2 1 cos2 2 2sin 4sindx

d

= − = − = = ; 4sin cos

dy

d

=

2

4sin cos

4sin

coscot

sin (shown)

dydy d

dxdxd

=

=

= =

(ii) When = ,

2 sin 2x = − 22siny =

cotdy

dx=

Equation of normal:

( ) ( )2 12sin 2 sin 2

coty x

− = − − −

At A ,

( ) ( )2 10 2sin 2 sin 2

cotx

− = − − −

( )22sin cot 2 sin 2x = − −

( )2sin cos 2 sin 2

sin 2 2 sin 2

2 (shown)

x

= + −

= + −

=

2k =

(iii) To obtain the total length of the arc, let 0 = , =

Total length 2 2

0

dx dyd

d d

= +

( ) ( )2 22

04sin 4sin cos d

= +

4 2 2

016sin 16sin cos d

= +

( )2 2 2

016sin sin cos d

= +

04sin d

=

0

4 cos

= −

( )( ) ( )4 cos cos0 4 1 1 8 = − + = − − + =

From GC,



Question 10

[ Ans: (i) show; V is constant (ii) show (iii) max. 3

2

AI

e= ; show maximum (iv) sketch ]

(i) dI q

L RI Vdt C

+ + =

( )d dI q d

L RI Vdt dt C dt

+ + =

2

2

1d I dI dq dVL R

dt dt C dt dt+ + =

2

20

d I dI IL R

dt dt C+ + = , for

dqI

dt= and V is a constant

(ii) 2

Rt

LI Ate−

=

2 2

2 2

2

2

1

2 2

Rt Rt

L L

Rt Rt

L L

dI RAte Ae

dt L

RAte Ae

L

R I RI I

L t t L

− −

− −

= − +

= − +

= − + = −

2

2 2

1 1

2

d I R dII

dt t L dt t

= − −

2

2 22

d I L R dI LL I

dt t dt t

= − −

2

2 22

d I dI L R dI L dIL R I R

dt dt t dt t dt

+ = − − +

2

2 22

d I dI L R dI LL R I

dt dt t dt t

+ = + −

2

2 2

1

2 2

d I dI L R R LL R I I

dt dt t t L t

+ = + − −

2 2

2 2 22 2 4

d I dI L R R R LL R I I

dt dt t t t L t

+ = − + − −

2 2

2 4

d I dI RL R I

dt dt L+ = −

2 2

20

4

d I dI RL R I

dt dt L+ + =

2

2

1 4

4

R LC

C L R = = (shown)



(iii) Let 0dI

dt=

10

2

RI

t L

− =

10

2

R

t L− = ( 0I at maximum since 0I )

( )2 32 3

4 2

Lt

R= = =

When 3

2t = ,

( )2 2

20 0

4

d I RL R I

dt L+ + =

2 2 2

2 2 20

4 4

d I R RI I

dt L L= − = − ( 2 0

Rt

LI Ate−

= for 0t )

I is a maximum

Max. ( )4 3

2 3 23 3

2 2

AI A e

e

−

= =

(iv) When 2L

tR

= ,

2

22 2R L

L RL ALI A e

R eR

−

= =



Question 11

[ Ans: (i)(a) $102.43 (b) $1215.71 (c) June 2018 ; beginning of month

(ii)(a) 100 12b+ (b) 4

3b = (iii) 1.23b = ]

(i) (a) Worth of $100 invested on 1 Jan 2016 by 31 Dec 2016

( )12 12

120.2100 1 100 1 100 1.002

100 100

$102.43

a = + = + =

=

(b)

Month Total amount in the account at end of month

1 (Jan) ( )100 1.002

2 (Feb) ( )( )( )

( ) ( )2

100 100 1.002 1.002

100 1.002 100 1.002

+

= +

3 (Mar) ( ) ( )( )( )

( ) ( ) ( )

2

2 3

100 100 1.002 100 1.002 1.002

100 1.002 100 1.002 100 1.002

+ +

= + +

12 (Dec) ( )( )12100 1.002 1.002 1

1.002 1

$1215.71

−

−

=

(c) Total amount in the account at the end of n th month

( )( )

( )

100 1.002 1.002 1

1.002 1

50100 1.002 1

n

n

−

−

= −

Let ( )50100 1.002 1 3000n −

From GC,

Maximum 29n =

Total amount in the account at the beginning of 30 th month

2988.6 100 3088.6 3000= + =

The total amount in the account will first exceed $3000 in June 2018 , and it will

occur in the beginning of the month.



(ii) (a) Worth of $100 invested on 1 Jan 2016 by 31 Dec 2016

100 12b= +

(b)

Month Total amount in the account at end of month

1 (Jan) 100 b+

2 (Feb) ( ) ( )

( ) ( )

100 2 100

2 100 2

b b

b b

+ + +

= + +

3 (Mar) ( ) ( ) ( )

( ) ( )

100 3 100 2 100

3 100 2 3

b b b

b b b

+ + + + +

= + + +

24 ( ) ( )

2424 100 24

2

2400 300

b b

b

+ +

= +

Let 2400 300 2800b+ =

300 400b =

4

3b =

(iii) Under plan P , total amount in the account at the end of 60 th month

( )( )

( )

60

60

100 1.01 1.01 1

1.01 1

10100 1.01 1

−=

−

= −

Under plan Q , total amount in the account at the end of 60 th month

( ) ( )60

60 100 602

6000 1830

b b

b

= + +

= +

Let ( )6010100 1.01 1 6000 1830b− = +

( )6010100 1.01 1 6000

1830

1.23

b− −

=

=

A-Level H2 Maths 2018 – Paper 1 - Achevas

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