2018 GCE A Level H2 Maths 9758 Paper 1 Solutions

1 (i) Given that ln x

yx

, find d

d

y

x in terms of x. [2]

(ii) Hence, or otherwise, find the exact value of e

21

ln d

xx

x , showing your working. [4]

Solution

(i) By Quotient Rule,

2

2

1ln

d

d

11 ln

x xy x

x x

xx

(ii) Method 1 “Hence” Using (i)

e

e e

2 21 11

ln 1 ln d d

x xx x

x x x

e

1

1 ln e

e

1 11

e e

21

e

x

Method 2 “Otherwise” using integration by parts

ee e

21 11

e

1

ln 1 1 1 d ln d

1 1

e

21

e

xx x x

x x x x

x

From part (i),

2

2 2

d ln 11 ln

d

ln 1 ln d d

xx

x x x

x xC x x

x x x

2 Do not use a calculator in answering this question.

A curve has equation 3

yx

and a line has equation 2 7y x . The curve and the line

intersect at the points A and B.

(i) Find the x-coordinates of A and B. [2]

(ii) Find the exact volume generated when the area bounded by the curve and the line is

rotated about the x-axis through 360 . [4]

Solution

(i) 3

yx

------ (1)

2 7y x ---- (2)

Subst. (1) into (2):

3

2 7xx

22 7 3 0x x

2 1 3 0x x

1

2x or 3

(ii) Required Volume

=

3 2

1 2

2

33

1

2

92 7 d

2 7 9

2 3

125

6

x xx

x

x

3

y

x

3 (i) It is given that d

2 6d

yx y

x . Using the substitution

2y ux , show that the

differential equation can be transformed to d

fd

ux

x , where the function f(x) is to

be found.

(ii) Hence, given that y = 2 when x =1, solve the differential equation d

2 6d

yx y

x , to

find y in terms of x.

Solution

(i) Differentiate 2y ux with respect to x,

2d d2

d d

y uux x

x x

Subst. into the DE:

2 2d2 2 6

d

ux ux x ux

x

3

d 6

d

u

x x

3

6f x

x

(ii) 3

d 6

d

u

x x

3

2

6 d

3

u xx

Cx

2 2

2

3

3

yC

x x

y Cx

Since y = 2 when x =1,

2 3 1C C 23y x

Note that u is a variable and not a

constant

4 (i) Find the exact roots of the equation 22 3 2 2x x x . [4]

(ii) On the same axes, sketch the curves with equations 22 3 2y x x and 2y x .

Hence solve exactly the inequality

22 3 2 2x x x . [4]

Solution

(i) Note that 22 3 2y x x has x-intercepts at

2x and 1

2x .

2

2

2

12 3 2 when 2 or

22 3 2

12 3 2 when 2

2

x x x x

x x

x x x

When 1

2 or 2

x x ,

2

2

2 3 2 2

2 2 0

2 121 3

2

x x x

x x

x

When 1

22

x ,

2

2

2 3 2 2

2 2 0

1 0

0 or 1

x x x

x x

x x

x x

(ii)

y

x

y

x 0

2 y = 2 – x

,

or

5 Functions f and g are defined by

f :x a

xx b

for , , 1x x b a ,

g : x x for x .

It is given that ff g .

Find the value of b.

Find 1f x in terms of x and a. [5]

Solution

fx a

xx b

has asymptotes 1y and x b .

Thus 1x is also the asymptote for 1fy x .

Since ff x g x x ,

1f fx x

Thus 1x is also the asymptote for fy x . 1b

1f1

x ax

x

, , 1x x , 1a

Alternative

ff g

x a

x bx a

x b

x x

ax

b

2

2 2

( )

( )

(1 ) (1 )

(1 ) (1 ) ( )

x a a x bx

x a b x b

a x a ab x b x a b

a x a ab b x a b x

Compare coeff of x2: 0 = 1 + b b = –1

1

Since ff g

f f

x x x

x x

1f1

x ax

x

, , 1x x , 1a

6 Vectors a, b and c are such that a 0 and 3 2 a b a c .

(i) Show that 3 2 b c a , where is a constant. [2]

(ii) It is now given that a and c are unit vectors, that the moduus of b is 4 and that the

angle between b and c is 60 . Using a suitable scalar product, find exactly the two

possible values of . [5]

Solution

(i) 3 2 a b a c 0

3 2 a b c 0

Since a 0 , is parallel to 3 2 a b c and hence 3 2 b c a (Shown)

(ii) Using 3 2 3 2 b c b c a a

2 2 229 6 6 4 b b c b c c a

2 2 29 4 12 cos60 4 1 b c

2 124

2 31

7 A curve C has equation 2 2

2 2

4 1

2

x y

x xy

.

(i) Show that 2d 2

.d 2 16

y x y

x xy y

[3]

The points P and Q on C each have x-coordinate 1. The tangents to C at P and Q meet at the

point N.

(ii) Find the exact coordinates of N. [6]

Solution

(i) 2 2

2 2

4 1

2

x y

x xy

2 2 2 22 8x y x xy

Differentiate with respect to x,

2d d4 16 2 2

d d

y yx y x xy y

x x

2d2 16 2

d

yxy y x y

x

2d 2

d 2 16

y x y

x xy y

(Shown)

(ii) When x = 1, 2

2

2

1 4 1

1 2

1

9

y

y

y

1

3y

Let P be 1

1,3

and Q be 1

1,3

.

At P,

21

2 1d 173

1 1d 542 1 16

3 3

y

x

At Q,

21

2 1d 173

1 1d 542 1 16

3 3

y

x

Equation of tangent at P:

1 17

13 54

17 1 ----- (1)

54 54

y x

y x

Equation of tangent at Q:

1 17

13 54

17 1 ------- (2)

54 54

y x

y x

Solving (1) and (2): 1

, 017

x y

Coordinates of N = 1

,017

8 A sequence 1 2 3, , ,...u u u is such that 1 2 ,n nu u An where A is a constant and 1n .

(i) Given that 1 5u and

2 15u , find A and 3u . [2]

It is known that the nth term of this sequence is given by

2 ,n

nu a bn c

where a, b and c are constants.

(ii) Find a, b and c. [4]

(iii) Find 1

n

r

r

u

in terms of n. (You need not simplify your answer.) [4]

Solution

(i) Let n = 1, 2 12 1u u A

15 2 5

5

A

A

Let n = 2, 3 22 2u u A

2 15 5 2

40

(ii) 2n

nu a bn c

1 5 2 5u a b c ----- (1)

2 15 4 2 15u a b c ----- (2)

3 40 8 3 40u a b c ----- (3)

Solve (1), (2) and (3):

15, 5, 5

2a b c

(iii) 15

2 5 52

n

nu n

1 1

1 1

2

152 5 5

2

152 5 5

2

2 2 115 51 5

2 2 1 2

5 1515 2 15

2 2

n nr

r

r r

n nr

r r

n

n

u r

r n

nn n

n n

9 A curve C has parametric equations

2 sin 2x , 22sin ,y

for 0 .

(i) Show that d

cotd

y

x . [4]

(ii) The normal to the curve at the point where meets the x-axis at the point A.

Show that the x-coordinate of A is k , where k is a constant to be found. [4]

(iii) Do not use a calculator in answering this part.

The distance between two points along a curve is the arc length. Scientists and

engineers need to use arc-length in applications such as finding the work done in

moving an object along the path described by a curve or the length of cabling used on

a suspension bridge.

The arc-length between two points on C, where and , is given by the

formula

2 2d d

dd d

x y

.

Find the total length of C. [5]

Solution

(i) d

2 2cos 2d

x

d4sin cos

d

y

2

2

d 4sin cos

d 2 2cos 2

4sin cos

2 2 1 2sin

4sin cos

4sin

cot

y

x

(ii) At , d

cotd

y

x

Equation of normal:

22sin tan 2 sin 2y x

At point A (x-axis), 0y

2

2

2

0 2sin tan 2 tan tan sin 2

tan 2sin 2 tan tan sin 2

2sin2 sin 2

tan

2sin cos 2 sin 2

2

x

x

x

2k

(iii) Total length of C

2 2

0

2 2

0

0

0

2

0

0

0

2 2cos 2 4sin cos d

4 8cos 2 4cos 2 4sin 2 d

4 8cos 2 4d

8 8cos 2 d

8 8 1 2sin d

4 sin d

4 cos

8

10 An electrical circuit comprises a power source of V volts in series with a resistance of R ohms,

a capacitance of C farads and an inductance of L henries. The current in the circuit, t seconds

after turning on the power, is I amps and the charge on the capacitor is q coulombs. The circuit

can be used by scientists to investigate resonance, to model heavily damped motion and to

tune into radio stations on a stereo tuner. It is given that R, C and L are constants, and that I =

0 when t = 0.

A differential equation for the circuit is d

,d

I qL RI V

t C where

d

d

qI

t .

(i) Show that, under certain conditions on V which should be stated,

2

2

d d0.

d d

I I IL R

t t C [2]

It is now given that the differential equation in part (i) holds for the rest of the question.

(ii) Given that 2eRt

LI At

is a solution of the differential equation, where A is a positive

constant, show that 2

4LC

R . [5]

(iii) In a particular circuit, R = 4, L = 3 and C = 0.75. Find the maximum value of I in

terms of A, showing that this value is a maximum. [4]

(iv) Sketch the graph of I against t. [2]

Solution

(i) d

d

I qL RI V

t C

Differentiate w.r.t t, 2

2

d d 1 d d

d d d d

I I q VL R

t t C t t

2

2

d d d

d d d

I I I VL R

t t C t where

d

d

qI

t

When d

0d

V

t ,

2

2

d d0

d d

I I IL R

t t C

(ii) 2eRt

LI At

2 2 2d

e e 1 ed 2 2

Rt Rt Rt

L L LI AR R

A t A tt L L

2

2 22

de 1 e

d 2 2 2

Rt Rt

L LI R AR R

A tt L L L

Since 2eRt

LI At

is a solution of 2

2

d d0

d d

I I IL R

t t C ---- (*),

substitute I, 2

2

d d,

d d

I I

t t into (*):

2 2 2 2

2 2

2

2

e 1 e 1 e e 02 2 2 2

02 2 4 2

4

4 (Shown)

Rt Rt Rt Rt

L L L LR AR R R At

AL t AR tL L L C

AR AR AR t AR t AtAR

L L C

A AR

C L

LC

R

(iii) 2 2

4 4(3)0.75

4

LC

R

Hence 2eRt

LI At

is a solution of the D.E for the case R = 4, L = 3 and C = 0.75.

At max I, d

0d

I

t

2 2d

e e 0d 2

Rt Rt

L LI AR

A tt L

1 02

4 31 0

6 2

Rt

L

t t

When 3

2t ,

4 3

2 3 23 3e

2 2e

A AI

Substitute d

0d

I

t and

3

2e

AI into (*) :

2

2

2

2

d 33 4 0 0

d 0.75 2e

d 20

d 3e

I A

t

I A

t

3

2e

AI is maximum.

Verify if the given R = 4, L = 3 and C = 0.75

satisfies the condition in (ii)

(iv) 2

32e eRt

tLI At At

where A > 0

As 2

3,e 0, 0t

t I

I

t 0

11 Mr Wong is considering investing money in a saving plan. One plan, P, allows him to invest

$100 into the account on the first day of every month. At the end of each month the total in

the account is increased by a% .

(i) It is given that a = 0.2.

(a) Mr Wong invests $100 on 1 January 2016. Write down how much this $100 is

worth at the end of 31 December 2016. [1]

(b) Mr Wong invests $100 on the first day of each of the 12 months of 2016. Find

the total amount in the account at the end of 31 December 2016. [3]

(c) Mr Wong continues to invest $100 on the first day of each month. Find the

month in which the total in the account will first exceed $3000. Explain

whether this occurs on the first or last day of the month. [5]

An alternative plan, Q, also allows him to invest $100 on the first day of every month. Each

$100 invested earns a fixed bonus of $b at the end of every month for which it has been in

the account. This bonus is added to the account. The accumulated bonuses themselves do

not earn any further bonus.

(ii) (a) Find, in terms of b, how much $100 invested on 1 January 2016 will be

worth at the end of 31 December 2016. [1]

(b) Mr Wong invests $100 on the first day of each of the 24 months in 2016 and

2017. Find the value of b such that the total value of all the investments,

including bonuses, is worth $2800 at the end of 31 December 2017. [3]

It is given instead that a = 1 for plan P.

(iii) Find the value of b for plan Q such that both plans give the same total value in the

account at the end of the 60th month. [3]

Solution

(i) (a) Amount worth = 12

$ 1.002 100 = $102.43

(b)

Month Total amount on 1st day Total amount at end of month

Jan 100 1.002 100

Feb 100 + 1.002 100 1.002 100 1.002 100

= 2

1.002 1.002 100

… …. ….

Dec 2 12

1.002 1.002 ...... 1.002 100

12

12

total amount at end 31 Dec 2016

1.002 1.002 1=$ 100

1.002 1

$50100 1.002 1

$1215.714974

$1215.71

(c) Consider 50100 1.002 1 3000n

Using GC,

When n = 29, amount at end of May = $2988.65 < 3000

When n = 30, amount at end of June = $3094.83 > 3000

Thus, the amount first exceed $3000 in the month of June.

Since on 1st June, $100 is added to the account which gives $3088.65 > 3000,

thus this occurs on the first day of June.

(ii) (a) Amount worth = $(100 + 12b)

(b)

Month Total amount on 1st day

of month

Total amount at end of month

Jan 2016 100 100 + b

Feb 2016 2 100 b 2 100 2b b

Mar 2016 3 100 2b b 3 100 2 3b b b

…

Dec 2017 24 100 2 3 ..... 24b b b b

Total amount at end of 31 December 2017 = 24

24 100 242

b b

24

24 100 24 28002

41.33

3

b b

b

(iii) Given a = 1 for plan P,

60

60

100 1.01 1.01 1 6060 100 60

1.01 1 2

10100 1.01 1 6000 30 61

1.23

b b

b

b