1

MJC 2011 H2 MATHS JC2 MID YEAR EXAMINATION PAPER 1

Qn Solution Q1 Inequalities (i) Method 1

Since 2 1 1x + ≥ , Therefore ( )2ln 1 0x + ≥

Method 2 From graph ( )2ln 1 0x + ≥

(ii)

From (i) since ( )2ln 1 0x + ≥ therefore need to solve ( )2 3

30 and 0

2x

xx x−

> ≠−

( )

2 3

30

2x

x x−

>−

( )( )2

30, 0, 2

2x

xx x

−⇒ > ≠

−

Therefore 3x > or 2x < , 0x ≠ Alternative solution: 0 or 0 2 or 3x x x< < < >

0 2 − + + 3 +

y

x O

2

Qn Solution Q2 Transformation of Graphs (a)

3 61 e

2xy − −=

( )

3 6

3 2

e

e

x

x

y − −

− +

=

=

1Q−

3e xy −=

1P−

3e xy =

3e xy∴ =

(b)

8

O 2 x

y

( )2 4,−8y =

2x =

0y =

1R−

3

Qn Solution Q3 Maclaurin Series

1tany x−=

( )

( ) ( )

( ) ( ) ( ) ( )

( )

2

2

22

2

3 2 22

3 2 2

3 22

3 2

d 1d 1

d1 1dd d1 2 0 (proven)d dd d d d1 2 2 2 0d d d dd d d1 4 2 0d d d

yx x

yxxy yx x

x xy y y yx x x

x x x xy y yx x

x x x

=+

+ =

+ + =

+ + + + =

+ + + =

When x = 0,

2

2

3

3

0d 1dd 0dd 2d

yyxy

xy

x

=

=

=

= −

1 3

3

2tan ...3!

1 ...3

y x x x

x x

− −= = + +

= − +

Required area 12 1

012 3

01

2 4 2

0

tan d

1 d3

12 3 4

1 1 08 19223 or 0.120 (3s.f.)

192

x x

x x x

x x

−=

≈ −

= −

= − −

=

∫

∫

Working is not required, i.e. can use G.C.

directly.

4

Qn Solution Q4 Curve Sketching (a)

(b)

11b

−

y

x O

11=x

axy =

y x= y

x O

11=x

1 10

1110

−

( )7.84,4.68

( )14,2,17.3

11by ax

x= +

−

5

Qn Solution Q5 Techniques of Integration (By parts) (i)

( )( )

1 2

2

2 2

2 2

2

2

d sin 1d

1 1 2 121 1

1 11

2 1 (shown)

x x xx

x x xx x

x xx

x

− + −

= + ⋅ − + −− −

− + −=

−

= −

(ii)

( )

( )

22

2 2

2

2

1 1111 1

shown1

xx

x xx

x

− − +− − + =

− −

=−

( )

( )

( )

22

2 2

2

2 2

2

2

Alternative method:

1 1

1 11 11 1

11 shown1

xxx x

x

x x

xx

− −= −

− −

− −= − −

− −

= − − +−

(iii)

( )

1

2 1 2

2

2 1 2

2

2 1 1 2 1

2 1 2

4 sin d

12 sin 2 d1

12 sin 2 1 d1

2 sin sin 1 2sin

2 1 sin 1

x x x

x x x xx

x x x xx

x x x x x x c

x x x x c

−

−

−

− − −

−

= − ⋅−

= + − −

−

= + + − − +

= − + − +

∫

∫

∫

6

Qn Solution Q6 Application of Differentiation (exc. Tangent/Normal) (a) 2 tan 3 5x y xy+ + =

Using implicit differentiation, 2 d d2 sec 3 3 0

d dy yx y x yx x

+ + + =

( ) ( )2d 3 sec 2 3dy x y x yx

+ = − +

( )( )2

2 3dd 3 sec

x yyx x y

+= −

+

(b)

5 1using similar triangles, 15 3

13

rh

r h

= =

⇒ =

2

2

3

13

1 13 3127

V r h

h h

h

π

π

π

=

=

=

( )

( )( )

2

2

2

1

d 1d 9d d dd d d

9 4

9 42

9 cms

V hhh h Vt V t

h

π

π

π

π−

=

= ×

= × −

= × −

= −

7

Qn Solution 7 Loci & Integration application (Vol)

( ) ( )2 2

2 i 2

2 1 2

z

x y

− − =

− + − =

4 i 3iz z− + = −

( )

4 0 1 3Midpoint of line-segment ,2 2

2,1

+ − + =

=

1 3Gradient of line-segment 4 01

− −=

−= −

Equation of perpendicular bisector is

( )( )

1 1 2

1 shown

y x

y x

− = −

= −

( ) ( )( ) ( )

( )

2 2

2 2

2

2 1 2

2 2 1

2 2 1

x y

x y

x y

− + − =

− = − −

= ± − −

( ) ( )( )22 2 2

0

3

Volume generated 1 2 2 1 d

23.9 units (3 s.f)

y y yπ= + − − − −

=

∫

Re

Im

O 1 2 4

−1

3

1

2

2

3

8

Qn Solution Q8 AP/GP

( ) ( ) ( )( )

2 11

12

Interest owed to Bank A 0.01 0.01 1.1 0.01 1.1 0.01 1.1

1.1 10.01

1.1 1

0.21384 21.4% (3 s.f.)

= + + + + +

−=

− = ≈

( ) ( ) ( )

[ ]

Interest owed to Bank B 0.01 0.01 0.01 0.01 0.02 0.01 0.1112 0.01 0.01 0.112

0.12 0.66

k k k

k

k

= + + + + + + +

= + +

= +

Alternative method:

( ) ( ) ( )

( ) ( )

Interest owed to Bank B 0.01 0.01 0.01 0.01 0.02 0.01 0.1112 2 0.01 11 0.012

0.12 0.66

k k k

k

k

= + + + + + + +

= +

= +

Since interest owed to both banks is the same,

0.12 0.66 0.214

0.14240.142 (3 s.f.)

kk

+ ===

( )( )

( )

Interest owed under special loan 12 0.01 1000120

120 0.214 10001.78

pp

pp

=

=

=

=

9

Let n be the number of months in the loan tenure with Bank A. Interest owed on special loan 0.015n=

( ) ( )1.1 1Interest owed on regular loan 0.01 0.1 1.1 1

1.1 1

nn

−= = −

−

( )( )

0.1 1.1 1 0.015

0.1 1.1 1 0.015 0

n

n

n

n

− <

− − <

Using GC, ( )( )

8,0.1 1.1 1 0.015 0.00564 0

9,0.1 1.1 1 0.015 0.000795 0

n

n

n n

n n

= − − = − <

= − − = >

Therefore max n = 8.

Qn Solution Q9 Mathematical Induction + Method of Difference

(i)

Let Pn denotes the statement ( ) ( )1

2 4 1 23 ! 3 3 !

n

r

rr n

=

+= −

+ +∑ for all 1n ≥ .

When 1n = ,

LHS ( )

( )( )

1

1

2 1 42 4 13 ! 1 3 ! 4r

rr

=

++= = =

+ +∑

RHS ( )

1 2 13 1 3 ! 4

= − =+

LHS.

Therefore 1P is true. Assume Pk

is true for some value of k where 1k ≥

i.e

( ) ( )1

2 4 1 23 ! 3 3 !

k

r

rr k

=

+= −

+ +∑ Need to prove that 1Pk+

is also true

i.e ( ) ( )

1

1

2 4 1 23 ! 3 4 !

k

r

rr k

+

=

+= −

+ +∑

(1)

10

LHS ( ) ( )

( )( )

1

1 1

2 1 42 4 2 43 ! 3 ! 1 3 !

k k

r r

kr rr r k

+

= =

+ ++ += = +

+ + + +∑ ∑

( ) ( )1 2 2 63 3 ! 4 !

kk k

+= − +

+ + [from (1)]

( ) ( )1 2 2 63 3 ! 4 !

kk k

+= − − + +

( ) ( )

( )2 4 2 61

3 4 !k k

k + − +

= − +

( )1 2 RHS3 4 !k

= − = +

Therefore Pk

is true 1Pk+⇒

is also true.

Hence since 1P is true and 1P is true P is truek k+⇒ , by mathematical induction, Pn is true for

all 1n ≥ .

(ii)

( ) ( ) ( )0

2 6 f 1 f4 !

n

r

r r rr

=

++ + − +

∑

( ) ( ) ( )0 0

2 6 f 1 f4 !

n n

r r

r r rr

= =

+= + + − +∑ ∑

Method 1

( ) ( ) ( )0

2 6 6 8 2 4 2 64 ! 4! 5! 3 ! 4 !

n

r

r n nr n n

=

+ + += + + +

+ + +∑

( ) ( )1

2 4 2 63 ! 4 !

n

r

r nr n

=

+ += +

+ +∑

( )

1

1

2 43 !

n

r

rr

+

=

+=

+∑

( ) ( )

1 2 1 23 1 3 ! 3 4 !n n

= − = −+ + +

Method 2

(2)

(3)

11

( ) ( ) ( )0

2 6 6 8 2 4 2 64 ! 4! 5! 3 ! 4 !

n

r

r n nr n n

=

+ + += + + +

+ + +∑

( ) ( )1

2 4 2 63 ! 4 !

n

r

r nr n

=

+ += +

+ +∑

( ) ( )

1 2 2 63 3 ! 4 !

nn n

+= − +

+ +

( ) ( ) ( ) ( )

0 0

f 1 f f 1 f 0n n

r r

r r= =

+ − = − ∑ ∑

( ) ( )f 2 f 1+ −

( ) ( )f 3 f 2+ −

( ) ( )( ) ( )

f f 1

f 1 f

n n

n n

+

+ − −

+ + −

( ) ( )f 1 f 0n= + −

( ) ( )11 e 0nn − + = + −

Sub (3) and (4) into (2)

( ) ( ) ( ) ( ) ( ) ( )1

0

2 6 1 2f 1 f 1 e4 ! 3 4 !

nn

r

r r r nr n

− +

=

++ + − = − + + + +

∑

or

( ) ( ) ( ) ( ) ( ) ( ) ( )1

0

2 6 1 2 2 6f 1 f 1 e4 ! 3 3 ! 4 !

nn

r

r nr r nr n n

− +

=

+ ++ + − = − + + + + + +

∑

(iii) Method 1

As n →∞ , ( ) ( )11 e 0,nn − ++ → ( )

2 04 !n

→+

( ) ( ) ( )0

2 6 1f 1 f4 ! 3

n

r

r r rr

=

+⇒ + + − → + ∑

(3)

12

The limit of convergence is 13

.

Method 2

As n →∞ , ( ) ( )11 e 0,nn − ++ → ( )

2 0,3 !n

→+

( )2 6 0

4 !n

n+

→+

( ) ( ) ( )0

2 6 1f 1 f4 ! 3

n

r

r r rr

=

+⇒ + + − → + ∑

The limit of convergence is 13

.

Qn Solution Q10 Functions (i) ( )

( )[ )1

2

f ff

Let 1 2

1 2

Since 1, 1 2

Hence , f 1 2.

Since D R ,D 2,

f : 1 2, , 2

y x

x y

x x y

x x

x x x x

−

= − +

∴ = ± −

≥ = + −

= + −

= = ∞

∴ + − ∈ ≥

(ii) ( )1f f x x− = , ( )1ff x x− =

-1 -11 1

ff f ff f is not equal to ff because D [1, ) [2, ) D− − = ∞ ≠ ∞ =

(iii) Method 1:

( )1 3h2

xxa

− −=

( ) 3g ln 2 1 12

3 ln 1 1

xxa

xa

− = − + − = − +

Method 2: ( ) ( )Let g ln 1x Ax B= + +

13

( ) ( )( ) ( )gh ln 2 3 1 ln 2 1 1

By comparison,12 2

33 1 1

x A ax B x

aA Aa

A B Ba

= + + + = − +

= ⇒ =

+ = − ⇒ = − −

( ) 3g ln 1 1xxa

− ∴ = − +

(iv) For fh to exist, h fR D⊂ .

[ )fD 2,= ∞ ,

( )hR 3,a= + ∞ Since h fR D , fh exists.⊂ 1 1a+ +

( )fhR 1 1,a∴ = + + ∞

( )f 1 2x x= + −

1

2 fD

( )fgR 1 1,a= + + ∞

f ( )gR 3,a= + ∞

h h

1D ,2

= ∞

y

x O 3a +

14

Qn Solution

Q11 Differential Equations (a)

d dyd d

u xyu y xx x

=

= +

Sub. d 1dy y yx x

−= − into d dy

d du y xx x= +

d 1d

11

d 1d

u yy x yx x

y y xyxy

u ux

− = + −

= + − −= −

= −

1 d 1 dx

1ln 1

ln 1

1 e e

1 e e1 e where e

1 e1 e

C x

C x

x C

x

x

uu

u x C

u x C

u

uu A A

xy AAyx

− −

− −

− −

−

−

=−

− − = +

− = − −

− =

− = ±

− = = ±

− =

−=

∫ ∫

(b)

22

2

2

2

d edd 2ed

4e

t

t

t

xtx Ct

x Ct D

−

−

−

=

= − +

= + +

when 0, 7t x= =

( )027 4e 0

7 43

C DD

D

−= + +

= +=

15

24e 34, 3

t

x CtB D

−∴ = + += =

2C =

x

t

C = 0

C = 1

C = 1−

3x =

7

O

16

Qn 12 Solution (i) ( )3 12

5 423 51 4

2 13 5 ,1 4

y zx

xyz

− +− = = =

⇒ = −⇒ = +⇒ = − +

− = + ∈ −

r

λ

λλλ

λ λ

(ii) Method 1 2 13 5 ,1 4

− = + ∈ −

r λ λ

1 1

5 1 1 5 4 0 is parallel to 4 1

l− = − + − = ⇒ Π −

⋅

2 13 1 2 3 1 6 A point on lies in 1 1

l = + + = ⇒ Π − −

⋅

Therefore l lies in Π . Method 2

2 13 5 1 2 3 5 1 4 61 4 1

−λ + λ = −λ + + λ + − λ = − + λ −

⋅

Therefore l lies in Π .

(iii) Method 1 Line passing through A and perpendicular toΠ ,

7 1: 6 1 ,

1 1ANl

− = + ∈ − −

r α α

Since N lies on ANl , 7

61

ON→

− + = + − −

ααα

for some α ∈

At point of intersection of ANl and Π ,

17

7 16 1 61 1

7 6 1 63 6

2

− + + = − − − ⇒ − + + + + + =⇒ =⇒ =

αααα α α

αα

7 2 5

6 2 81 2 3

ON→

− + − = + = − − −

(shown)

Method 2

Let 600

OB =

be a point on Π

6 7 130 6 60 1 1

AB−

= − = − −

Π

A ( 7,6, 1)− − ×

N (6,0,0)B ×

18

AN

= Projection of AB→

onto the normal = ( ) AB ⋅

n n

13 1 11 16 1 13 31 1 1

113 6 1 1

31

12 1

1

222

= − − −

− − = −

= −

= −

7 2 5

6 2 81 2 3

ON OA AN→ → →

− − = + + = + = − − −

(shown)

Using ratio theorem,

( )1 '2

ON OA OA= +

' 25 7

2 8 63 1

310

5

OA ON OA= −

− − = − − − − = −

19

(iv) Method 1 to find P using A

The smallest possible area of ANP∆ occurs when P is the foot of perpendicular from A to l .

To find OP→

Let P be the foot of the perpendicular from A to l .

2 13 5 , .1 4

− = + ∈ −

r λ λ

Since P lies on l then23 51 4

OP−λ

= + λ − + λ

for some λ∈

2 7 93 5 6 3 51 4 1 4

AP OP OA−λ − −λ

= − = + λ − = − + λ − + λ − λ

9 1, 3 5 . 5 0

4 49 15 25 16 0

42 2447

AP l− −

⊥ ⇒ − + =

− + − + + ==

=

λλ

λλ λ λ

λ

λ

427 104 13 5 417 7

941 47

OP

−

= + = − +

Π

A

P N

l

20

Method 2 to find P using N The smallest possible area of ANP∆ occurs when P is the foot of perpendicular from N to l .

To find OP→

Let P be the foot of the perpendicular from N to l .

2 13 5 , .1 4

− = + ∈ −

r λ λ

Since P lies on l then23 51 4

OP−λ

= + λ − + λ

for some λ∈

2 5 73 5 8 5 51 4 3 2 4

NP OP ON−λ − −λ

= − = + λ − = − + λ − + λ − + λ

7 1, 5 5 . 5 0

2 4 47 25 25 8 16 0

42 2447

NP l− −

⊥ ⇒ − + = +

− + − + + + ==

=

λλλ

λ λ λλ

λ

Π

A

P N

l

21

427 104 13 5 417 7

941 47

OP

−

= + = − +

Method 3 to find PN using cross-product The smallest possible area of ANP∆ occurs when P is the foot of perpendicular from A to l .

To find PN→

Let P be the foot of the perpendicular from A to l .

Let 231

OC→

= −

be a point on 2 1

: 3 5 , .1 4

l−

= + ∈ −

r λ λ

5 2 7

8 3 53 1 2

CN− − = − = − − −

2 2 2

7 11 5 5

1 5 4 2 4

PN CN= ×

− − = × + + −

d

Π

A

P N

lC

22

301 3042 30

30 342

30 15 14714

= −

=

= =

After finding OP

or PN

,

5 10 31 158 41 17 7

3 9 215 147

5 7 28 6 23 1 2

' 12

PN

PN

AN

AA

− − = − = − −

=

− − = − = − − −

=

2

1Smallest possible area of 21 15 14 122 7

15 42 13.9 units7

ANP PN AN∆ =

=

= =