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VJC H2 Maths Preliminary Examination P1 2017 Solutions

Q Solution Comments

1

2

2

2

2

6 131

4

6 13 40

4

30

2 2

x

x

x x

x

x

x x

2 2 or 3x x

2i Let $ , $ and $ be the unit cost of electricity,

water and gas, respectively.

514 + 18.8 + 134 = 155.54

309 11.3 89 94.99

639(1.2) 21.7 108 208.40

Using G.C,

0.2137, 1.1749, 0.1761.

E W G

E W G

E W G

E W G

E W G

2ii Let be the water usage for August 2017

(0.2137)(1.2)(555) 1.1749 0.1761(128) 184.84

17

w

w

w

3i

3ii

2 2 3

+ - + +

(3,0 )

(8, -1)

(7, 0)

(2, -1)

x

(6,3)

O

(0, 3)

y

(1,0 )

(10,1)

(9, 0)

(6,1 )

(4, 0) x

(10,3)

O

(0, 4)

y

y = f(x)

11 f

2y x

Q Solution Comments

4

Intersect at least once: : 1 or 1 m m m

5i

2

d 5 55

d 4

T x

x x

2

12 2

12 2

12 2

1

2

5 55 d

4

5 52 4 d 5 d

2

45 55

2

5 5 4 5 1

xT x

x

x x x x

xx C

x x C

5ii When t = 30,

d0

d

T

x : 2

2

5 55 5 4

4

xx x

x

2 25 4 1x x x

Since 0, 1x x

Substitute 1 and 30x T into equation (1)

1

230 5 5 1 4 5 10C C

1

2 25 5 4 5 10T x x

5iii

When x =0, 32.361T . When x =2, 31.623T Longest time taken by Alvin is 32.4 mins.

y

7y x

1y x

O

y

x

(–3, –4)

(–3, –6)

(–3, –2)

T

Q Solution Comments

6i 2

1

1

= e 1

ln ( 1) 2

h ( ) ln ( 1) 2

Domain of h range of h ( 1 , )

xy

x y

x x

6ii

6iii Using G.C, 1h ( ) and h( )y x y x intersects at

0.94753x and 3.50524x

Set of values of x = : 0.948 3.51x x .

7i

ii Since 3x is the vertical asymptote, 3c

Given that 1y x is an oblique asymptote,

f 13

Ax x

x

21 3 2 33 3

x x A x x A

x x

By comparing coefficient of x with 2

3

x ax b

x

: 2a

1x

Q Solution Comments

Since 8

0,3

is on the curve,

20 2 0 8

0 3 3

8

b

b

8i 2 2

1 1 1

1 1

3 3 3

r r

r rr r r

r r

13 3

12 13

5

4

8ii

2 22

24 2 4

22

2

222

11

2

2

2

2

2

2

2(replace with 2)

3 3

3

1

3 3

2 3 2 33 1

2 32 3

7 4 4 3 6 3

6 2 3

7 1

6 2 3

n r n

r rr r

n

rr

n

rr

n

n

n

r rr r

r

r

n n

n n n

n n

7, 6, 1p q a

9a 0

sin d = 0.5a

x x x

0 0

cos cos d 0.5aa

x x x x

0

cos 0 sin 0.5a

a a x

cos sin 0.5a a a --- (1)

Using GC, a = 1.20249 =1.20 (3 s.f.)

sin cosy a a a

0.5y

Q Solution Comments

9b

Area =

4

0

d3

xx

x

Let 3u x

d 1 d

2 3d d2

u xu

x ux

When 0, 3

When 4, 1

x u

x u

4 1

0 3

3 d 2 3 d

3

x ux u u

ux

3 2

1

2 3 d

uu

u

=

3 2

1

9 62 d

u uu

u

3

1

92 6 du u

u

32

1

2 9ln 62

uu u

9 1

2 9ln3 18 2 62 2

18ln3 16

10 Since 1 3i is a root,

3 2

2 1 i 3 1 i 3 1 i 3 4 0

2 8 2 2 3i 1 i 3 4 0

20 2 2 3 3 i 0

p q

p q

p q p q

Compare real and imaginary parts:

2 20 1

2 3 3 0 2

p q

p q

5, 10p q

10i Since 1 3i is a root, and all coefficients are real

1 3i is also a root.

3 2

2

2 5 10 4 1 3i 1 3i 2

2 4 2

z z z z z z a

z z z a

Q Solution Comments

By observation: 1a

2 3

1, 1 3i

2z z

10ii 1 1 3 2 z 2 2 2 w

1

1

arg arg 1 3i

3tan

1

3

z

arg arg 2 i 2

4

w

i3

1 2e

z , i

42e

w

10iii

Quadrilateral OPRQ is a rhombus

10iv 1 14arg arg * 4arg arg

4

3 4

19

12

z w z w

4119

arg * 212

5

12

z w

11 1 1

2 0 ,

3 1

r

Q Solution Comments

11i 1 1

0 . 1

1 1cos

2 3

35.3

o

90 35.3

54.7

11ii Intersection of light beam with reflective surface:

1 1 1

2 0 . 1 4

3 1 1

6 2 4

1

Coordinates of point of intersection = 0, 2, 2 .

11iii Let F be the foot of perpendicular from device to normal

line and A be the point 1, 2, 3 :

ˆ ˆ

1 0 1 1

2 2 . 1 1

3 2 1 1

3 3

12

13

1

BF BA n n

Using Ratio Theorem,

1, 2, 3A

0, 2, 2B

F'A

Q Solution Comments

'

2

1 1 14 1

' 1 0 43 3

1 1 1

BA BABF

BA

Equation of reflected light path:

0 1

2 4 ,

2 1

r

12i

ii Volume of the vase =

2

2d

k

x y

2

2

2

e d

e

e e

ky

ky

k

y

iii Volume of water ,

2e d

yyV y

2

2

ln 2

2 2

e e

e e

e

y

x

x

Given d

2d

V

t ,

d d d

d d d

1 2

2

1 =

x V x

t t V

x

x

O

1,0 1,0

0x

O

Q Solution Comments

Hence the rate at which the radius of the water surface is

increasing is 1

xcm per second.

iv For the insect,

d0.03

d

x

t .

t seconds later, the location of the insect is at

0.03 ex t

For the movement of the water,

d 1

d

x

t x

2

d 1 d

2

x x t

xt C

When t = 0, x = 1

2

2

2 2

C

xt

When the insect first comes into contact with water,

2

0.03 e

2 2

tt

2

0.03 e 2t t

2 2

0.03 et

t

Using GC, t = 13.858

13.8580.03 e = 3.1340x

2

ln 3.1340 2.28y

Hence coordinates of the point = 3.13, 2.28

y = 2

0.03 et y = 2t

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VJC H2 Maths Prelim P2 2017 Solutions/Mark Scheme

Q Solution

Section A: Pure Mathematics [40 marks]

1

2

2 2

2

d d d

d d d

1 2 1 1

1 1

2

y y x

x t t

t t t t t

t t

t t

1i At point

2

,1 1

p p

p p

, t p

Equation of tangent at point 2

,1 1

p p

p p

,

22

2 3 2

2

2

21 1

22

1 1 1

12

1

2

p py p p x

p p

p p py p p x

p p p

p py p p x

p

y p p x p

1ii Tangents pass through 2,5

2

2

5 2 2

4 5 0

5 or 1

p p p

p p

p p

Equations of tangents are

3 1 and 15 25y x y x

Required acute angle between the 2 tangents

1 1tan 15 tan 3

0.255 rad or 14.6

2 OC a b

2,5

Q Solution

2i Length of projection of OC

onto OA

2

ˆ

ˆ ˆ ˆ

2cos

3

11 4

2

3

a b a

a a b a a a b a a a

a b a

2ii

0 1

1 0 2

Sub (2) into (1): 1 0

a b d

a b d

b a a d

AB DA

Since AB // DA and A is a common point, A , B and D are collinear

Given 4 , 5 4d a b

Area of triangle OBD

1

2

15 4

2

15 4

2

50

2

5

2

b d

b a b

b a b b

b a b b

a b

5

2k

3i (8 1)

7a a

ar

ad

rd

22 (13 1)

12ar a d

ar ad

2

2

127

12 12 7 7

ar a ar a

r r

2 12 5 07r r

Q Solution

3ii From the GC,

5

7r or 1r .

Since 0d , the terms of the geometric series are distinct

we conclude that 1r . Hence, 5

7r .

As 5

17

r , the geometric series is convergent.

3iii 0.001

51

70.001

5 5 51 1 1

7 7 7

51 1 0.001

5 571 1

7 7

n

n

n

S S S

aa a

a a

5

0.001 07

5ln ln 0.001

7

n

a

n

ln 0.001

5

7ln

n

20.53n

Smallest value of n is 21.

3iv 5

27

7 7 49

a aar a

d a

The sum of the first 2017 terms of the arithmetic series

20172 (2017 1)

2

42 9a a

566777

7a

4ai 2

1f 1

3x x nx

Comparing with

2f '' 0f f 0 f ' 02!

x x x

f 0 1

Q Solution

1

f ' 03

4aii 2 3

dGiven e sin 2

d

xymy y xx

When 0,x

2 3 011 1 e sin 0

3

11

3

3

m

m

m

Differentiate (2) w.r.t. x: 22

2 2

2

d d d3 6 3 e sin e cos

d d d

x xy y yy y x xx x x

When 0,x

2

2 21 13 1 2 6 3 1 1

3 3

26

9

1

9

n

n

n

4bi Method 1

2

ext angle of a triangle2 6 3

ACD

Using Sine Rule in ACD

2 2sin sin

3 3

32

sin3

AD AC

AD

Q Solution

32

sin cos cos sin3 3

32

3 1cos sin2 2

3

3 cos sin

Method 2

In right-angled ABC , 1

3

tan6

AB

.

angle sum of a triangle6 2 3

ADB

Using Sine Rule in ABD

sin sin6 3

3 sin6

sin3

AD AB

AD

32

sin cos cos sin3 3

32

3 1cos sin2 2

3

3 cos sin

4bii When is a sufficiently small angle,

Q Solution

2

1

2

12

22 2

2 2

2

3

3 12

33 3

2

123

1 21

2 2! 23 3

12 33

1 51

63

AD

1 5,

63a b

Section B: Statistics [60 marks]

5

5i

P Peter made first move | Peter won the game

P Peter made first move and Peter won the game

P Peter won the game

0.5 0.5

0.5 0.2 0.5 0.5

5

7

5ii P John wins 0.5 0.3 0.5 0.4 0.35

1st Move

John

Peter

0.5

0.5

John

Peter

Draw

John

Peter

Draw

Q Solution

P John wins in exactly 1 game

3!0.35 0.65 0.65

2!

35490.443625 or

8000

0.444 to 3 s.f.

Alternative

Let X be the number of games won by John out of 3 games.

~ (3,0.35)X B

P John wins in exactly 1 game

P 1

35490.443625 or

8000

0.444 to 3 s.f.

X

6i

6iia From GC, r = 0.93639 = 0.936 (3 s.f)

6iib From GC, r = 0.98775 = 0.988 (3 s.f)

6iii Since

1) the points on the scatter diagram seem to lie close to an increasing curve with decreasing gradient (or

close to a curve in which y increases by decreasing

amounts as x increases), and

2) the product moment correlation coefficient between ln x and y of 0.988 is closer to 1 than the product

moment correlation coefficient between x and y of

0.936,

hence lny c d x is the better model.

6iv From (iii), we should use the regression line of y on ln x.

From GC, the equation of the regression line of y on ln x is

90

100

110

120

130

140

150

160

170

0 20 40 60 80 100

Q Solution

20.8496 31.539ln

20.8 31.5ln (3 s.f)

y x

y x

When y = 144, 144 20.8496 31.539ln x 49.635 50 (nearest gram)x

6v Since x = 110 is outside the range of data values

(15 90x ), hence the estimated value of y may not

be reliable.

7i

4 4

4 4

9 9

4 4

P(no odd digits) P(all even digits)

or

1

126

C P

C P

7ii

5 4

1 3

9

4

10P 1

63

C CX

C

x 0 1 2 3 4

P X x 1

126

10

63

10

21

20

63

5

126

7iii

4

0

E P

10 10 20 51 2 3 4

63 21 63 126

20

9

x

X x X x

22

4 2

2

0

2

Var E E

20P

9

10 10 20 5 201 4 9 16

63 21 63 126 9

50

81

x

X X X

x X x

Q Solution

7iv

1 2 1 2

1 2

1 2

1 2

P 3 P 3 3

1 2P 0 & 4

2P 0 & 3

2P 1 & 4

1 5 1 201 2 2

126 126 126 63

10 52

63 126

7793or 0.982

7938

X X X X

X X

X X

X X

8 Let X kg and Y kg be the mass of a randomly chosen D25

durian and Musang Queen durian respectively.

2 2N(1.5,0.02 ), N(1.8,0.035 )X Y

8ii Let 1 2 3 1 29 18T X X X Y Y

1 2 3 1 2E E 9 18

9 3 1.5 18 2 1.8

105.3

T X X X Y Y

1 2 3 1 2

2 2 2 2

Var Var 9 18

9 3 0.02 18 2 0.035

0.891

T X X X Y Y

N 105.3,0.891T

P 107 0.035852

0.0359 (3 s.f)

T

8ii The masses of all the durians are independent of each other.

8iii Let 1 2 n

X X XX

n

20.02N 1.5,X

n

Given P 0.1X m

1.5P 0.1

0.02

1.5P 0.9

0.02

mZ

n

mZ

n

Q Solution

From the GC, P 1.28155 0.9Z

1.51.28155

0.02

1.5 0.025631

when 1.51

1.51 1.5 0.025631

6.5695

m

n

m n

m

n

n

Largest value of n is 6

9i Let 240y x

unbiased estimate of population mean

240

240

120240

50

242.4

x

y

y

n

Unbiased estimate of population variance

2

2

2

2

1

1

1 12011200

49 50

222.69 223 (3 s.f)

s

yy

n n

9ii Let be the population mean of X.

0

1

H : 240

H : 240

Level of significance: 10%

Test Statistic : since n = 50 is sufficiently large,

By Central Limit Theorem,

X is approximately normal.

When 0H is true,

240

N(0,1) approximately

50

XZ

S

Q Solution

Computation :

242.4

222.69 14.923

value 0.128 (3 s.f)

x

s

p

Conclusion : Since p-value = 0.128 > 0.10, H0 is not rejected

at the 10% significance level. So there is insufficient

evidence that the population mean waiting time is more than

240 seconds.

9iii No assumption is needed. Since the sample size is large, by

Central Limit Theorem, the distribution of the sample mean

( X ) is approximately normal.

9iv There is a probability of 0.10 that the test will conclude the

population mean waiting time is more than 240 seconds

when it is actually 240 seconds.

9v 0

1

H :

H :

k

k

Level of significance: 10%

For 0H to be rejected,

1.6449 or 1.6449

1.6449 or 1.6449

50 50

242.4 242.41.6449 or 1.6449

14.923 14.92350 50

242.4 3.4714 or 242.4 3.4714

z z

x k x k

s s

k k

k k

245.87 or 238.93

: 239 (3 s.f) or 246 (3 s.f)

k k

k k k

10i The assumptions are

(1) The probability that a customer turn up for the flight is

100

p for all the 154 customers.

(2) Customers turn up independently of each other.

10ii Customers may be travelling in a group or as a family.

Therefore, customers may not turn up independently of the

others in their group.

Q Solution

10iii B 154,

100

pX

Given P 153 0.05X

153 154

153 154

P 153 P 154 0.05

154 1541 0.05

153 154100 100 100

154 1 0.05100 100 100

X X

p p p

p p p

From the GC, 96.9568 97.0 to 3 s.f.p

10iv

a B 154,0.94X

P 141 148 P 148 P 140

0.825

X X X

10iv

b P 150 0.98443 0.984 to 3 s.f.X

10v Let Y be the number of days (out of 7) in which every

customer who turns up gets a seat on the flight

B 7,0.98443Y

P 5 1 P 5

0.995

Y Y

SA2 Victoria Junior College