1

Marking Scheme for Additional Mathematics Trial 1 (2009)

Paper 2

Question 1

2 2

2 2

2 2

2

Given the following equationslet 2 24 be (eq 1)and 8 be (eq 2)By substitution,Substitute (eq 2) into (eq 1)

we have, 8 2 24Expanding and solve for 64 16 2 242 18 64 24 02

x y yx y

y y yy

y y y yy yy

2

2

18 40 0Simplifying the quadratics by dividing by 2

9 20 0Using simple factorization,

4 5 04 or 5

Now for each value of we will get thecorresponding value of .For 4 and using (eq 2)

8

y

y y

y yy y

yx

yx

4

4(4, 4)For 5 and using the same method,

8 53

(3,5)Therefore, the possible pair of solutions are(3,5) and (4,4)

x

yxx

Question 2

20

20

20Given that 2 (20 1) 1102

Simplifying the information above we have,10 2 (19) 110

*(actually in mathematics we cannot have twoequations sign on the same line)2 19 11 ............ let th

S a d

S a d

a d

8

is be (eq 1)Second information,

(8 1) 27 2 ............... let this be (eq 2)

T a da d

2 19 11 ............ let this be (eq 1)7 2 ............... let this be (eq 2)

Notice that we can solve for and using the simultaneous equation method.For this, we use elimination method but,of

a da d

a d

15 6

course you can use substitution.But first, (eq 2) 22 14 4 ............ let this be (eq 3)(eq 1 eq 3)5 15

3Substitute 3 into (eq 2)

7(3) 22 2123

(b)15 6( 46 14(3)) ( 46 5(3))2 2

1

a d

dd

daaa

S S

5 ( 46 42) 3( 46 15)2

15 ( 4) 3( 31)230 93

63

Question 3

score

Upper limit Lower limit frequency (f) x fx fx2

10 19 5 14.5 72.5 1051.25 20 29 4 24.5 98 2401 30 39 8 34.5 276 9522 40 49 15 44.5 667.5 29703.75 50 59 43 54.5 2343.5 127720.75 60 69 20 64.5 1290 83205 70 79 5 74.5 372.5 27751.25 100 5120 281355

5120100

51.2

fxx

N

x

x

22

2281355 51.2100

192.1113.8604

fxx

N

1

1

1

1

1

5

1

1 1

1

1 6 1 1

1

1

1

1 7 1

Question 4

3

3

2

2

2

4 4

2 3 42 3 4

1 2 3

2 3

2 2 44

42

2( 1)

14 )( 2)( 1) 2

1 1 2 2

21 2 31 1 2

2 3

dx x dxx

xc

cx

a iix x x xdx dx

x x

dx x x x dxx x x

x x x c

cx x x

Question 5 Please note: For part (a) 1 Mark is given to students who draw a table and plotted at least 6 points. 1 Mark is given when the students label the axis correctly (either in degrees or in radians) 1 Mark is given for the shape of the graph correct. 1 Mark is given for labeling the graph.

4 2

4 2

4 2

4 2

5 3

5 3

( )

Given that the general gradient of an

equation is 15 12 at a certain point .

Thus, 15 12

15 12

15 12

15 125 3

3 4At the point (1, 4)4 3 4

5The equati

bdydx

x x xdy x xdx

dy x x dx

dy x x

x xy c

y x x c

cc

5 3on is 3 4 5y x x For part (b) 1 Mark is given for showing the correct equation of the straight line. 1 Mark is given for sketching the straight line correctly. 1 Mark is given for stating the correct number of solutions in this case the answer is 4 solutions.

1

1

1

1

1

1

1

7

7

2cos( ) 1y x

12xy

●

A B

F

C D E

b

3a

Question 6 From the diagram above, we were given several

clues, , , and that

. We were also told that is parallel

to

(since it is par

3

al

43

DAD b

DE k DB AE

B

CAB a AB

C

AE qBC

lel, this relation is truewhere is a const

(3

an

4343

t)

( )

( )

( )

( )

( )

......... l t

43

et

)

q

AE q

AE

BD

BA AD

BA AD

a

q

AE q

AE q

AE

a

q a b

AE qa qb

ab

DC

AB

his be eq. 1

Now we also know that,

which means; ( )

( )

3

3 ( 1) ....... let this be eq.Comparing eq. 1 and eq

2

3.

3

2

a

DE k DB

DA AE k DB

b AE k

AE ka kb b

AE ka k b

q k

b

and 1 (equating the together)1 3

4 114

From parallel to and

, where is a consta

Next we have to find the value o .

t

f

n .

k q qk k

k

k

EF AB BF mBC

EF pAB p

BF mBC

BE E

m

F

( ) from the above part.

But (refer to diagram above)

m a b

BE BA AE

1(from the first part 4

1 3sub. into 3 we have )4 4

3 3 4 4

3 334 4

9 34 4

9 34 4

9 3( ) ( ) ...... let this be eq. 14 4

But,

EA a b

BE a a b

BE

k

k q

a b

EF a b ma mb

EF m a m b

E

q

F

k

or 3

by comparing this with eq. 19 3( ) 3 and ( ) 04 43 3 ( , not needed)4 2

pAB EF pa

m p m

m p

( )3 2

Given that and 5 7

3 25 7

512

The modulus of is the magnitude of OA

25 144 169

13 units

b

AB OB

AB AO OB

AB

OA

OA

OA

OA

SECTION B Question 7

0

4

01

1 202

4

401

2

4

4Evaluate 1 2

(1 2 )4 1 2 4 1 22

4 (1 2 ) 4 1 3

4 1 3 8

dxx

xx dx

x

1

1

1

1

1

1

1

1 8

1

1

2

2

2

2

3 2

3

33

3

First find the roots of the curve 9When 0, then 9 0

93. Here the roots are ( 3,0) and (3,0)

The area bounded by equation 9 and -axis,

9

93

27 9 27 9 18 18

36

y xy x

xx

y x x A

A x dx

xA x

A

A

2

9

0

922

0

3

unit)

Volume generated about the -axis,

Volume of cone 9

1 (12 )(12) 93 2

81576 81 02

576 40.5535.51682.541 unit

iiy

V y dy

yV y

V

VVV

Question 8

Given that the 100 and 5)

110 100( 110) ( )5

( 2) 1 ( 2)From the normal distribution table1 0.0228 0.9772

xz

a

P X P Z

P Z P Z

6 55

)The probability the a bar of chocolate with massless that 100gm is 0.9772Thus the probability of a box of chocolates which hasless than 5 chocolates < than 100gm1 ( ( 5) ( ( 6))1 ( (0.9772) (0.0

b

P X P XC

6 6 06

4 1 31

228) (0.9772) (0.0228) )1 (6(0.89108)(0.0228) 0.870764 1)1 0.1218997 0.8707640.0073363

)(0.0073363) (0.9926637)

0.028704

C

iiC

Question 9 2

2 3

1

py qxx

y p qx x

From the graph, Please refer to suggested graph. p = 0.5 q = 5.01 Note: 1 mark will be given if the student writes the complete expression of the linear function. 2 marks will be given if the tables for values

2

yx

and 3

1x

are shown.

1 mark will be given for the correct axis shown on the graph. 1 mark will be given for correct plotting of all the points. 1 mark will be given for the best line drawn 1 mark will be given for construction on the line to show the working to get the gradient hence the value for p. 1 mark will be given for the accuracy of the value of p 1 mark will be given for the line to touch the vertical axis 1 mark will be given for the accuracy for the value of q.

x 1 1.5 2 2.5 3 3.5 y 5.5 11.6 20.3 31.5 45.2 61.4

2

yx

5.5 5.16 5.08 5.04 5.02 5.01

3

1x

1 0.29 0.125 0.064 0.037 0.023

1

1

1

1

1

1

1

1 10

1

1

1

1 1

1

1

1 1

1 10

Question 10

1

o

)Consider the triangle where 12, 6

6tan tan12

tan 0.5

tan (0.5)26.5650.4637

)The angle and angle are equalSo, using the property of

180 26.

aAOM

OA OM

PAO MAO

MAO

MAOMAOMAO radians

bPAO APO

AOP

o

2

o

2

2

565 26.565126.872.2146

)1Area of major sector is 2

233.13 or 4.0694 1 12 (4.0694)2292.997

)Length of arc, 12(2.2146)

26.5752

Length of chord 2 sin2

AOPAOP radians

c

r

radians

A

A cmd

ss cm

r

AP

o126.872(12)sin( )2

24 0.89442721.4663

Perimeter of the segment 26.5752 21.4663 48.0415

APAP cm

cmcm

Question 11

1

Given that 1.5 rate of water entering the cone

Rate of water leaving the cone 2

We are asked to find where is the depth of water

at time . *The net rate would be (1.5 2) 0.5

dVdt

dVdt

dh hdt

t cms

Based on similar triangles,

2

2

3

3

2

2

2

1

420420 5

1Volume of cone is 3

13 5

13 25

753

75

25

25 0.5 and when 12

25(288)

0.02762

that is to say the wat

rh

h hr

V r h

hV h

hV

hV

dV hdhdV hdhdh dh dVdt dV dtdh h cmdt hdhdtdh cmsdt

er level is falling atthe rate of 0.0276 every second.cm

4

20 r

h 1

1

1

1

1

1

1

1

1

1 10

1

1

1

1

1

1

1

1

1

1

10

SECTION C Question 12

240 4)

Intial velocity of the ball, 0

40 8

40 /)

To find the range of time the object movesupwards from 0 would be to find whenthe time stops instantaneously; 0

40 840 8 08 40

s t ta

tds v tdtv m sb

tv

v tt

tt

2

5Therefore, the range of time would be 0 5)

The maximum height reached would be when theobject is instantaneously at rest; from ) 5

40 4 when 5200 4(25)200 100100

Maximum distance

st

c

b ts t t tsss m

2

2

2

2

from the ground is 100 96196

)Time taken to travel that distance

40 4 If the object moves 96 m passes O96 40 4

4 40 96 010 24 02 12 0

12It took 12 to reach the ground after passi

mm

d

s t tt t

t tt tt t

t ss

1

ng 40 8 when 1240 96

56

Ov t t svv ms

Question 14

2

1

o o

2 2 2

Given that the area of 18.6 )

1 5.2 7.9 sin 18.62

18.6 2sin5.2 7.937.2sin41.08

sin 0.90555

sin 0.90555

64.897 or 64 53')

Using cosine rule.5.2 7.9 2 5.2 7.9 cos64

PQS

PQS cma

A QPS

QPS

QPS

QPS

QPS

QPSb

QS

o

2 o

2

2

o o o o

.897

27.04 62.41 82.16cos64.89789.45 34.8654.59

54.597.389

)Angle 180 85 58 37Using sine rule7.389sin 58 sin 37

7.389 sin 37sin 58

7.389 0.60180.8480

5.2437 )

Area of

QSQSQS

QSQSc

SQR

SR

SR

SR

SR cmd

Q

2

1 7.389 5.2437 sin852

1 38.7457 0.99619219.29904

QSR

QSR

SR

A

A cm

1

1

1

1

1

1

1

1

1

1 10

1

1

1

1

1

1

1 1

1

1

10

Question 15

9898

96

98

98

98

9696

98

)

100

1101.30 100

110 1.30100

1.43 Index for the year 1996 based on 1998

100

1.30 100 90.911.43

aP IPP

P

P RM

P IP

Question 13

2 1000 ........ inequality 1

2 3 1200 ........ inequal

)8 4 4000

4 6 2400

3 3ity 2

200 ......... inequality 600

3

x y

x y

x y

ax y

x y

x y

c) From the graph, if the 200 units of B are produced, then the maximum of product A would be 300 units. The Profit line 5 8 400x y 1 Mark is given if there is any indication of rough estimation of the profit line (like the above, an example) Drawing the profit line on the graph and translating it up parallel to the vertices that shows the maximum profit gets 1 mark. The maximum profit would be the production of 200 of B and 300 of A

20022002

2000

2002

2002

15 )

110( ) (4) 120(5) 1009

110 4 600 900 10010 4 3004 300 10

75 2.5)

100

1.3 100

130

bIw

Iwx yI

xx y x

x yy x

y xcP IP

x Ix

I

13 b)

5 8x y k

Profit line

1

1

1

1

1

1

1

1

1

1

10

1

1

1

1 1

1

1

1 Area shaded correctly

1 All the straight line graphs are correctly drawn

Indicate the profit point

1

10