ri h2 maths 2013 prelim p1 solutions

8/11/2019 RI H2 Maths 2013 Prelim P1 Solutions

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2/13

H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 2 of 13

Hence,

2 3 sin 2 3 sin

6 6

2 3 sin sin6 6

2 3 sin cos cos sin sin cos cos sin6 6 6 6

1 1OR 2 3 2cos sin

2 6 6 2 6 6

PQ PR

4 3 cos sin6

34 3 sin

2

6sin

6 (since is small sin ) (shown).

2 3 and 6.a b

No. Solution

3(i) By GC,

2

3

4

14

21

30

u

u

u

3(ii)Conjecture:

2

1 5.nu n

LetnP be the statement

21 5nu n for .n

LHS =1 9u =

21 1 5 RHS

Hence1P is true.

AssumekP is true for some k

, i.e. 2

1 5k

u k .

To prove:1kP is true, i.e. to prove

2

1 ( 1) 1 5ku k

1

2

2

2

2 3

( 1) 5 2 3( 1) 2( 1) 1 5

1 1 5 RHS

k ku u k

k k

k k

k

Hence1kP is also true.

SincekP is true 1kP is true, and 1P is also true, by the

principle of Mathematical Induction,nP is true for all ,n

i.e. the conjecture is true.

No. Solution

4(i)2

4 4 2

4 4 2

ln 64

2ln ln 4 6

yy x x

y y x x


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Differentiate with respect tox:

3 3

43

2 2

44

d 2 d 4 4 12

d d

d 4 24 12

d

d 4 ( 3) 2 ( 3)

d 2 12 2 1

y yy x x

x y x

y yx x

x y

y xy x xy x

x yy

4(ii) When 2,y

4 2

2 2

2 2 2

16 6

8 2 0

8 or 2 (reject as 0)

2 2.

x x

x x

x x x

x

Hence,

4 2 2 (8 3) 4 2 2 (8 3)d or

d 2(16) 1 2(16) 1

40 2 40 2 or .

31 31

y

x

No. Solution

5(i)

OC OB BC k b a

5(ii) By the Ratio Theorem,

1 1 1 1

.2 2 2 2

kOY OC OA k

b a a b a

1 1 1 1

2 2 2 2

k kXY XO OY OA

b + b a

XY is parallel to OA.

1 1 1

2 2

XY XY k k

CB kOA k k

Hence the required ratio is 2 21 : 4k k

OA

BC

X Y

D


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No. Solution

6(i) 2 1 2

1 1 1 7

3 1 1

1

4 2 1 3 72

Position vector of the point of intersectionA is

OA

=2 1 31 11 1 3

2 23 1 5

6(ii) 1 2

1 1

1 1 2 2sin

33 6 3 2

.

6(iii)

BA OA OB

=3 2 11 13 1 1 .

2 25 3 1

Shortest distance

11 2 3 2 1

sin 1 .2 3 2 3 6

1

BA

OrShortest distance

1 221

1 11 12 21 11 12 .2 6 6 61

1

BA


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No. Solution

7 2

2

d

d

1d 1 d

1d 1 d

1ln ln 1 , where is an arbitrary constant

ln1

e , where e1

e 1 1 (Shown).

11 e e

e

t C

t

t t

t

NN kN

t

N tN kN

kN t

N kN

N kN t C C

Nt C

kN

NB B

kN

BN

Bk k Ak

B

Alternatively:

2

2

1d 1 d

1

d 1 d1

1ln , where is an arbitrary

1ln

1e , where

1 1e (Shown).

e

t C

t

t

N tN kN

N N t

kN

k t C C N

k t CN

k A A eN

A k NN k A

1When 0, 250 : 250 .

As , 10000 :

1 10000 0.0001 and so 0.0039.

t NA k

t N

k Ak

Therefore1

.0.0001 0.0039e t

N

Now, whenN= 750, 1

0.00011 750

750 e0.0001 0.0039e 0.0039

t

t


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2 2 1

10

0

2 0

0

1 1sin (2 ) sin (2 )

4 4

f ( ) f ( 1)

f (0) f (1)

f (1) f (2)

f (2) f (3)

f ( 1) f ( )

f ( ) f ( 1)

1 1 f (0) f ( 1) sin (2 )

4 4

nr r

n r rr

n

r

n

S x x

r r

n n

n n

n x

2 1

1

2 2 1

1

sin (2 )

1 sin ( ) sin (2 ) (proved)

4

n

n

n

x

x x

8(iii) Since 2 10 sin 2 1n x for all 0n and

11 0

4n as ,n

therefore 2sinnS x as n . Hence, 2sin .S x

No. Solution

9(i) Let $kbe the amount needed for the fund to award the scholarship for 2014.

(1.025) 2000k and thus2000

1951.22 19511.025

k (correct to the nearest dollar)

For additional amount needed :

Mtd 1 :

Additional amount needed

=2

2000$ $1903.63 $1904

(1.025)

Mtd 2 :

Let x be the additional amount needed.

1.025(1.025( ) 2000) 2000

3854.1951

3854.1951 1951.22 1904 (nearest $).

k x

k x

x

Hence 1904x (nearest $).

9(ii) Method 1

Amount needed

=2 10

2000 2000 2000....

1.025 (1.025) (1.025)

=

10

11

20001.02511.025

11.025

= 17504.13

= 17504 dollars (to nearest dollars)


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Method 2Either

10

1010

10

10

0

1.025 11.025 2000 0

1.025 1

1.025 12000

1.025 1 17504 (nearest $)1.025

a

x

x

Or

9

910

9

10

2000

1.025 11.025 2000(1.025) 2000

1.025 1

1.025 12000 2000(1.025)

1.025 117504 (nearest $)

1.025

b

x

x

9

(iii)

Method 1

Amount needed

2 3

2000 2000 2000...

1.025 (1.025) (1.025)

=2000 1

11.0251

1.025

= 80 000 dollars

Method 2

Either

0

1.025 11.025 2000 0

1.025 1

1.025 12000

1.025 1 180000 1 80000

1.025 1.025

n

nn

n

n n

a

x

x

(as1

01.025n

as n )

Or


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1

1

1

2000

1.025 11.025 2000(1.025) 2000

1.025 1

1.025 12000 2000(1.025)

1.025 1

1.025

2000 82000(1.025 1)

1.025

82000 8000080000

1.025 1.025

n

nn

n

n

n

n

n

b

x

x

(as1

01.025n

as n )

Method 3

For scholarship to continue indefinitely, yearly interest earned should be sufficient for thescholarship awarded for the following year. Hence

Minimum amount =2000

800000.025

No. Solution

10

2

2 2

2

2 5 2f

3 2 3 3

3 5 2 3 2 2 3 6 39 45 8 24 18

2 15 27 0

2 3 9 0

3 3 9 or Since .

2 2

xx

x

x x xx x x x

x x

x x

x x x

10(i)

3

9 2

2 3f ln ln 9 or ln

3 2

0 e or e .

x x x

x x

10(ii) 1 2 3f 0.5 9 (no solution) or 0.5

2 3 2

0.5 1.5 or 0.5 1.5

1 or 2.

x x x

x x

x x


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No. Solution

11(i) 2 k

2

2

2

2 2

16 16 0

2

13 5

2

yx x

yx

2 2

3 11

25 50

x y

Ellipse:

11(ii)

2

2

2 2

16 9 25

3 11

25 25

yx x

k

x y

k

Hyperbola : centre 3,1

Oblique Asymptotes : 3 1 y k x k and

1 3 y k x k

55

x

y 3 1 y k x k 1 3 y k x k

( 3,1)

2 2

3 11

25 25

x y

k

O

y

2 2

3 11

25 50

x y

x

5

5 2

3,1

0


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11(iii) Equation of a line of symmetry: 3 or 1x y .

Translation by 3 units in the positive x - direction

Translation by 1 unit in the negative y - direction

No. Solution

12(a) 22

1

23 32

11

23 3

1

ln d

ln 1d

3 3

ln

3 9

8ln 2 8 1

3 9 9

8ln 2 7

3 9

x x x

x x xx

x

x x x

12(b)

2 2

1

9 9d d

3 2 4 ( 1)

1 9sin

2

x xx x x

xc

221

20 0

9 1d 9sin 9 3

2 6 63 2

xx

x x

1 12 2

1122

2 1 1 1 11 2 d2 4 2 4

x x x x

12

1 12 2

2

20

9 1Since d 3 , so .

23 2

1 2 d 2 1 d 3a

x ax x

x x x x

OR

12

12

11 2 d (2)(1)

2x x

ax

12

y

2 2x


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12

1 12 2

12

2

20

2

2

2

74

9 1Since d 3 , so .

23 2

1 2 d 2 1 d 3

3

1 1( ) 2

4 2

70

4

1 1 4( ) 1 8

2 2

1Since ,

2

1 2 22

a

a

x ax x

x x x x

x x

a a

a a

a

a

a

No. Solution

13(i)

d d2 ; 4 10 2 2 5

d d

x yt t t

t t

Thus, 2 2 5d 2 5

.d 2

ty t

x t t

For min. pt.,d 5

0

d 2

yt

x

Hence coordinates of the min. pt. is 14.25,3.5

Coordinates ofAis (8,16) .

13(ii)

13(iii) The equation of tangent at Pis

O

14.25,3.5 x

y

C


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2 2

3 2 2

3 2 3 2

2

2 5 2 10 16 8

2 10 16 (2 5) 8

2 10 16 2 2 16 5 5 40

2 5 5 40

py p p x p

p

py p p p p x p

py p p p px p p x p

py px x p

2

2 5 5 40.py p x p 13(iv) Since the tangent passes through 0, 0 ,

25 40 0

2 2

p

p

Hence the possible coordinates of Pare

16,32 20 2

1 132 20 2 32 20 2 tan arg tan

16 16

0.228 arg 1.311

z

z

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ri h2 maths 2013 prelim p1 solutions

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