NJC H2Maths 2014 Prelim P1 Soln · PDF fileSuggested Solutions to 2014 SH2 H2 Mathematics Preliminary Examination Paper 1 ... Microsoft Word - NJC_H2Maths_2014_Prelim_P1_Soln 10/9/2014

Suggested Solutions to 2014 SH2 H2 Mathematics Preliminary Examination

Paper 1

1 (i) Since A(–1, 20) lies on the curve,

a(–1)2 + b(–1) + c = 20 ⇒a – b + c = 20 – (1)

Since B(3, 4) lies on the curve,

a(3)2 + b(3) + c = 4 ⇒9a + 3b + c = 4 – (2)

Since B(3, 4) is a stationary point,

2a(3) + b = 0 ⇒ 6a + b = 0 – (3)

a = 1, b = – 6, c = 13

1 (ii) 2 26 13 ( 3) 4= − + = − +y x x x

Hence 2

2

2 2

translate 3 units in the positive -direction

( 3)

translate 4 units in the positive -direction

( 3) 4 6 13

=

↓

= −

↓

= − + = − +

y x

x

y x

y

y x x x

2

( ) ( )2 22 2

2

1 8d d

81 4 1 4

1 1

8 1 4

x xx x

x x

cx

= ⋅+ +

= − + +

⌠⌡

∫

( ) ( )

( ) ( )

( ) ( )

( )

2

2 22 2

2 2

2 2

1

2

4d 4 d

1 4 1 4

1 14 4 d

8 1 4 8 1 4

1d

2 1 4 2 1 4

1tan 2

42 1 4

x xx x x

x x

x xx x

xx

x x

xx c

x

−

= ⋅+ +

= − − − + +

= − ++ +

= − + ++

∫ ∫

∫

∫

3 (i)

( ) ( )

( )

( )

2

2 2

2 22 2

22

2

2

e

e e

,

d (2 )

d

d(1 )

d

(1 )d

d

dAt 0,

d

e

.

e

t

t t t

t

tx y t

t a

x t a t t t a

t t a t a

yt t

t

t t ay

x t a

ayt a

x a

−

− − −

−

= =−

− − − −= =

− −

= − + = −

− −=

− −

−= = = −

−

Since the tangent to the curve C at t = 0 is perpendicular to the line 4y – x = 0,

( )11 4 (Shown)

4a a− = − ⇒ =

3 (ii)

For a = 4, ( )2

2

2

(1 ) 4d e.

d 4

tt ty

x t

− − −=

− −

As 2t → − , ( )22 (3) 4 4d

0.d 4 4

ey

x

−→ =

− −

The gradient of the curve approaches 0 as 2t → − .

3 (iii) Observe that

2

2as 2, , 2 .e e4

ttt x y t

t

−→ − = → ∞ = → −−

Thus, 22ey = − is a horizontal asymptote.

22ey = −

4 (i) Since A, B and P are collinear,

( )1OP λ λ+= −a b����

.

Since M, N and P are collinear,

( ) 12 1

3OP µ µ= + −a b����

.

Comparing, we have

( )

2

11 1

3

λ µ

λ µ

=

− = −

.

Solving, we have 2

5µ = and

4

5λ = .

Thus, ( )1 14

5

4

55OP + == +a b a b����

.

4 (ii) Method 1

Area of OMN1

3

1 12

2 3× ×= =ba a b

Area of APM1 1 1

2 5 1

1

5 0

× − × =

=

ba aa b

Area of the quadrilateral OAPN

1 1 7

3 10 30= × − × ×=a b a b a b

Method 2

Area of OAB1

2= ×a b

Area of BPN1 4 4 4

2 5 5

2

153

× − ×

= =b a b a b

Area of the quadrilateral OAPN

1 4 7

2 15 30= × − × ×=a b a b a b

Method 3

Area of OAP ( )1

1

2

1 14

5 0× + ×= =a ba a b

Area of ONP

( )1 14

3 5

1

2

2

15

= +

×=

×b a

a

b

b

Area of the quadrilateral OAPN

1 2 7

10 15 30= + =× × ×a b a b a b

5 (i) Method 1 2( )

f ( ) , x k

x x kx k

+= ≠

−

2

2

2

2

2( )( ) ( )f ( ) ,

( )

( )(2 2 ),

( 3)

( )( 3 ),

( )

x k x k x kx x k

x k

x k x k x kx k

x

x k x kx k

x k

+ − − +′⇒ = ≠

−

+ − − −= ≠

−

+ −= ≠

−

For f to be increasing, f ( ) 0′ >x

2

( )( 3 )0

( )

( )( 3 ) 0

or 3

x k x k

x k

x k x k

x k x k

+ −>

−

+ − >

< − >

Method 2 2

2 2

2 2 2

2

( )f ( ) ,

2

3 3 4

43

x kx x k

x k

x kx k

x k

x kx kx k k

x k

kx k

x k

+= ≠

−

+ +=

−

− + − +=

−

= + +−

( )

2

2

4f ( ) 1

kx

x k′⇒ = −

−

For f to be increasing, f ( ) 0′ >x

( )

( )

( )

2

2

2

2

22

41 0

41

4

2 or 2

or 3

k

x k

k

x k

k x k

x k k x k k

x k x k

− >−

<−

< −

− < − − >

< − >

5 (ii)

6 (i) y = cos

4x

π

+ 3.5

x = 4

π

cos −1

(y – 3.5)

f −1

(x) = 4

π

cos−1

(x – 3.5),

for x ∈ �, 5 7 2

2 2x

−< ≤

Range of f −1

is or [3, 4) or 3 < x < 4

6 (ii) As range of f −1

is [3, 4) ⊆ domain of g is (2, 4],

gf −1

exists.

6 (iii)

g(3) = 1, g(4) = 1, g(3.5) = 0

Df

−1 → Rf

−1 = [3, 4) → Rgf

−1 = [0, 1]

Range of gf −1

is [0, 1]

7 (i)

7 (ii)

5sin

10

6

θ

πθ

=

=

Thus, ( )arg 6 3i6

zπ

+ − = .

From the right angled triangle,

5 35sin

3 2y

π= =

55cos

3 2x

π= =

Thus the complex number z representing P is 5 5 3

4 3 i2 2

− + +

, i.e.,

3 5 33 i

2 2

+ +

.

×

P

5

y

x

8 (a)

(i)

Total amount of prize fund needed

( ) ( ) ( )

( )

2 19

20

20

4 4 41000 1000 1000 1000

5 5 5

41000 1

5

41

5

45000 1 4942.35 nearest cents

5

= + + + +

− =−

= − =

�

8 (a)(ii) Assume that no two athletes will arrive at the same time.

8 (a) part

after

(ii)

( ) ( )( )

[ ]2

2 15 1 185 1000002

30 185 185 200000

185 155 200000 0

nn

n n

n n

+ − ≤

+ − ≤

− − ≤

By GC, 32.4 33.3n− ≤ ≤ .

Maximum n is 33.

8 (b) Given

1

1

210

5

n

n nS

+

−= − ,

( ) ( )

( )

1

1 1 2

1

2 1

2 210 10

5 5

2 2

5 5

25 2 5 2 .2

5 5

15 2, for 2

5

n n

n n n n n

n n

n n

n n

n n

n

n

T S S

n

+

− − −

+

− −

= − = − − −

= −

= −

= ≥

( )

1 1

1 1 1 1

1

1

210

5

15 26

5

T S+

−= = −

= =

( )15 2, for all

5

n

n nT n +∴ = ∈Z

( ) ( )

( )( )

1

1

1

1

1

15 2 15 2

5 5

15 2 5 2 2.5 2

5 5 5.2 515 2

−

−−

−

−

= ÷

= × = × =

n n

n

n n

n

n n n n

n n nn

T

T

Since 1

n

n

T

T −

gives a constant, the series is a geometric progression. Common ratio is 2/5.

9 (a) ( ) ( ) ( )( )

( )1

2 2 2

2

2

2

1

2

2

22

cos sin sin

3 4sin

5 5

61 36cos 48sin

61 36 1 482

25 48 18

25 48 18

18

25

18

2

5 6 2 5 6 cos

61 60 cos

61 60 cos

485 1

25

1 1

1 48 2 25 1

2 25 5

α θ

θ α θ α

θ θ

θ θ

θθ

θ θ

θ θ

θ θ

θ θ

−

+

= +

= − −

≈ − − −

= + −

= −

−

≈

= −

− ≈ + − +

= −

− +

+

+

+

AD

AD

2

2

2

48

2! 25

24 63

65 1

25

24

25

63

5 1255

θ

θ θ

θ θ

−

= −

−

−

= −

9 (b) ( ) ( )

( ) ( ) ( )

( )( )

( ) ( ) ( )

1

1

sin

sin

2 2 2

32

3/2 2 22

f e f 0 1

f e f f 01 ( ) 1

f f f f 011 ( )

−

−

= ⇒ =

′ ′= = ⇒ =− −

′′ ′ ′′= + ⇒ =−−

nx

nx

x

n nx x n

nx n x

n x nx x x n

n xnx

( )2

2f 12

= + +n

x nx x

Hence n = 2 and 22

2.2

= =b

10 (i) Let Pn be statement “un =

3

2

n

n +, for n

+∈� ”.

LHS of P1 = u1 = 1 and

RHS of P1 = 3(1)

1 2+ = 1 = LHS of P1

Hence, P1 is true.

Assume that Pk is true, i.e. uk = 3

2

k

k + for some .k +∈Z

Consider Pk+1 i.e. uk+1 = ( )

( )3 1

1 2

k

k

+

+ +.

LHS of Pk+1= 1ku + ku= + 6

( 2)( 3)k k+ +

= 3

2

k

k + +

6

( 2)( 3)k k+ +

= 3 ( 3) 6

( 2)( 3) ( 2)( 3)

k k

k k k k

++

+ + + +

=

23 9 6

( 2)( 3)

k k

k k

+ +

+ +=

( )23 3 2

( 2)( 3)

k k

k k

+ +

+ +

= ( )( )3 1 2

( 2)( 3)

k k

k k

+ +

+ + =

( )3 1

( 1) 2

k

k

+

+ += (RHS)

Thus, Pk is true ⇒ Pk+1 is true.

Since P1 is true and Pk is true ⇒ Pk+1 is true, by mathematical induction, Pn is true for

all .n +∈Z

10 (ii) =

1

6

( 2)( 3)

N

n n n= + +∑

= u2 – u1

+ u3 – u2

+ …

+ uN – uN − 1

+ uN + 1 – uN

= uN+1 − u1

=( )3 1

13

N

N

+−

+

=2

3

N

N + or

62

3N−

+

10 (iii)

10

3

( 2)( 3)n n n

∞

= + +∑

= 9

1 1

1 6 6

2 ( 2)( 3) ( 2)( 3)n nn n n n

∞

= =

− + + + +

∑ ∑

= 3

lim 13N N→∞

− +

−9

9 3+= 1 −

9

12 =

1

4

10 (iv) Let j − 3 = n + 2

⇒ j = n + 5

6

6

( 2)( 3)

N

j j j= − −∑

=6

6

( 2)( 3)

j N

j j j

=

= − −∑

= 5

5 6

6

( 5 2)( 5 3)

n N

n n n

+ =

+ = + − + −∑

= 5

1

6

( 2)( 3)

N

n n n

−

= + +∑ =

2 10

2

N

N

−

−

11 (a) Method 1 2

2

2 2

e

e

d de 2 e

d d

x

x

x x

z y

y z

y zz

x x

−

− −

=

=

= −

( )

( )

( )

2

2

2

2 2 2

2

d2 1 e

d

de 2 e 2 e 1 e

d

d1 e

d

x

x x x x

x x

yy x

x

zz z x

x

zx

x

− − −

+

+ = +

− + = +

= +

Method 2 2

2 2 2

e

d d de 2e e 2

d d d

x

x x x

z y

z y yy y

x x x

=

= + = +

( )

( )

( )

2

2

2

2 2

2

d2 1 e

d

de 2 1 e

d

d1 e

d

x

x x x

x x

yy x

x

yy x

x

zx

x

+

+

+ = +

+ = +

= +

( )

( )

( )

2

2

2

2

2

2

2

2

2

2

2 2

2

d1 e

d

d 12 2 e

d 2

12 2 e d

2

1e , where is an arbitrary constant

2

1e e

2

1e e

2

x x

x x

x x

x x

x x x

x x

zx

x

zx

x

z x x

c c

y c

y c

+

+

+

+

+

−

= +

= +

= +

= +

= +

= +

⌠⌡

11 (b)

(i) ( )d20

dk

t

θθ= − −

As d

1dt

θ= − when 70θ = , ( )1 70 20 0.02− = − − ⇒ =k k

( )d0.02 20

dt

θθ= − −

When 40θ = , ( )d0.02 40 20 0.4

dt

θ= − − = −

It is cooling at a rate of 0.4°C per minute.

( )

0.02

0.02

0.02

0.02

d0.02 20

d

1 d0.02

20 d

1 d d 0.02 d

20 d

ln 20 0.02

20 e

20 e

e , where e

20 e

t c

t c

t c

t

t

t

t tt

t c

A A

A

θθ

θθ

θθ

θ

θ

θ

θ

− +

− +

−

−

= − −

= −−

= −−

− = − +

− =

− = ±

= = ±

= +

⌠⌡ ∫

Given 95θ = when 0t = ,

( )95 20 1A= +

75A = 0.0220 75e tθ −= +

11(b) (ii) It is a good model because the equation reflects the decrease of the temperature to a

steady state temperature, which is what would happen in real life.

12 (i)

12 (ii)

2 2

2 2

1 : 1 14 4

x xC y y+ = ⇒ = −

2 22 2

2 : 1 14 4

x xC y y− = ⇒ = −

Volume generated

( )( )2 2

2 82

0 2π 1 8 1 d π 1 d

4 4

3.4701 (to 4 dec places)

π

= − − − −

=

∫ ∫x x

x x

12 (iii) 22 1

4

xy+ =

2

14

xy⇒ = −

Using the substitution 2 cosx θ= , d

2sind

xθ

θ= − .

When 1,3

πθ= =x ; when 2, 0x θ= = .

Area of region ( )2 2

2 0

13

4cos1 d 1 2sin d

4 4π

θθ θ= − = − −∫ ∫

xx

( )

( )

π

23 3

0 0

3

0

2sin d 1 cos 2 d

1sin 2

2

1 2 1sin 0 sin 0

3 2 3 2

3

3 4

π

π

θ θ θ θ

θ θ

π π

π

= = −

= −

= − − −

= −

∫ ∫