Amk Prelim 2009 Am1

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ADDITIONAL MATHEMATICS 4038/01

Monday 14 September 2009 2 hours

Name of Setter: Mdm Karen Teng

Additional materials: Answer paper

Graph paper

READ THESE INSTRUCTIONS FIRST

Write your name, class and index number on all the work you hand in.Write in dark blue or black pen.You may use a soft pencil for any diagrams or graphs.Do not use staples, paper clips, highlighters, glue or correction fluid.

Answer all the questions.Write your answers on the separate Answer Paper provided.Give non-exact numerical answers correct to 3 significant figures, or 1 decimal placein the case of angles in degrees, unless a different level of accuracy is specified inthe question.The use of an electronic calculator is expected, where appropriate.You are reminded of the need for clear presentation in your answers.

At the end of the examination, fasten all your work securely together.The number of marks is given in brackets [ ] at the end of each question or partquestion.The total number of marks for this paper is 80.

This question paper consists of 5 printed pages.[Turn over

ANG MO KIO SECONDARY SCHOOLPRELIMINARY EXAMINATION 2009

SECONDARY FOUR EXPRESS/FIVE NORMAL ACADEMIC

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Mathematical Formulae

1. ALGEBRA

Quadratic Equation

For the equation ,02 cbxax

a

acbbx

2

42

Binomial expansion

,......21

)(221 nrrnnnnn bba

r

nba

nba

naba

where n is a positive integer and!

)1(...)1(

)!(!

!

r

rnnn

rnr

n

r

n

2. TRIGONOMETRY

Identities

AAec

AA

AA

22

22

22

cot1cos

tan1sec

1cossin

BA

BABA

BABABA

BABABA

tantan1

tantan)tan(

sinsincoscos)cos(

sincoscossin)sin(

AAA

AAAAA

AAA

2

2222

tan1tan22tan

sin211cos2sincos2cos

cossin22sin

)(2

1sin)(

2

1sin2coscos

)(2

1cos)(

2

1cos2coscos

)(2

1sin)(

2

1cos2sinsin

)(2

1cos)(

2

1sin2sinsin

BABABA

BABABA

BABABA

BABABA

Formulae for ABC

Cab

Abccba

C

c

B

b

A

a

sin2

1

cos2

sinsinsin222

[Turn over

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AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01

1 (a) Solve the simultaneous equations.

48 2 yx 0)132ln( yx [4]

(b) Solve112 363 xx . [4]

2 Find the value ofkfor which the following simultaneous equations has no solution. [3]

43

22

ykyx

kyx

3 Differentiate the following with respect tox

(i) 14ln x , [2]

(ii)1

x

e x. [2]

4 It is given that10

3cossin xx , 25:39cos:2sin yx andx andy are acute angles.

(i) If4

x , show that

13

5cos y . [2]

Hence find the exact value of

(ii) ),2tan( yx [2]

(iii) .sin x [2]

5 (i) Express2

)1)(32(

27

xx

xin partial fractions. [3]

(ii) Hence evaluate .)1)(32(2

275

2 2dx

xx

x

[3]

6 Given that 7563 23 xxxy

(i) finddx

dy, [1]

(ii) find the value ofc for which cyx is a tangent to the curve, [3]

(iii) show thaty decreases asx increases. [2]

[Turn over

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7 The diagram (not drawn to scale) shows a trapezium OPQR in which PQ is parallel to

OR and 90ORQ . The coordinates ofP andR are )3,4( and (4, 2) respectively and

O is the origin.

(i) Find the coordinates ofQ. [3]

(ii) PQ meets they-axis at T. Show that triangle ORTis isosceles. [3]

(iii) The point S is such that ORPS forms a parallelogram, find the coordinates ofS. [3]

(iv) Find the area of the trapezium OPQR. [2]

8 (a) Solve, for x , the equation 5.03

2cos

x , leaving your answers

in terms of . [4]

(b) Solve, for 1800 , the equation .sin5cos4sec2 [5]

9 Find the range of values ofkfor which the line xky 3 meets the curve

xy

2

3at least once. [4]

10 Find, in ascending powers ofx, the first three terms in the expansion of .)1( 6px

Given that the first two non-zero terms in the expansion of )1()1( 6 qxpx are

1 and2

3

7x , find the possible values ofp and q. [5]

[Turn over

O

R (4, 2)

Q

P )3,4(

x

y

T

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11 Variablesx andy are related by the equation xaby , where a and b are constants. The

table below shows measured values ofx andy.

x 1 2 3 4 5

y 470 190 80 30 12

(i) Plot lgy againstx and obtain a straight line graph. [3]

(ii) Use your graph to estimate the value ofa and ofb. [4]

(iii) On the same graph, draw the line representing the equation lgy x = 2

and hence find the value ofx for which 210 xxab . [2]

12

A piece of wire of length 680 m is bent to form an enclosure consisting of a trapezium

PQRS and a quadrant PST. Given in the figure yPQ m, 45SRQ and xST m.

(i) Show that the areaA m2 of the enclosure is given by 2

2

12340 xxA

. [4]

Given thatx can vary,

(ii) find the stationary value ofA, [4]

(iii) determine whether this stationary value is a maximum or a minimum. [1]

END OF PAPER

PQ

R S T

45

y m

x m

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2009 Prelim Exam Additional Mathematics Paper 1 (4 Express / 5 Normal)

1[8]

(a)

)1........(263

22

48

2)2(3

2

yx

yx

yx

)2.......(464,2

232

1132

0)132ln(

0

yx

yx

eyx

yx

(2)-(1), x = 2

Fr (1), 26)2(3 y

3

2 y

M1, M1

A1

A1

(b)

33633

363

12

112

xx

xx

Let xy 3 , 0633

1 2 yy

63.13lg

6lg1

6333

63

0)6)(3(

01892

xorx

or

yory

yy

yy

xx

M1

M1

A2

2

[3]

4

2

13

2

y

x

k

k

No solutions implies

1

13

2

k

kdoes not exist

4.05

2

0322

0)3()1(2

013

2

det

k

kk

kk

k

k

OR

1

32

2

kk

linesforgradientSame

M1

M1

A1

3[4]

(a))14ln(

2

114ln xxy

142

14

4

2

1

x

xdx

dy

M1

A1

(b)

22

2

)1()1(

)1(

))(1(

1

xe

xor

x

xe

x

eex

dx

dy

x

ey

x

x

xx

x

M1

A1

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4[6]

(i)

5

32sin

5

3cossin2

10

3cossin

x

xx

xx

Given25

39

cos

2sin

y

x

)(13

5

5

3

39

25

2sin39

25cos

shown

xy

M1

M1

Majority could get tothe ratio for sin 2x but

could not show theirworking clearly while

trying to find the rationfor cosy

No method mark for 2nd

part if pupil jumpedfrom ratio 39:25 to ratio

13

5:

5

3without showing

any working

(ii)

(iii)

)2(

2

2

4

quadndxx

4

32tan

5

42cos

x

x

quadstacutey 1

5

12cos

13

12sin

y

y

yx

yxyx

tan2tan1

tan2tan)2tan(

56

33

5

12

4

31

5

12

4

3

)1(10

3sin

10

9sin

5

4sin21

5

42cos

2

2

quadstinxx

x

x

x

M1

A1

M1

A1

Pupils who could apply

method correctly get thequadrant wrong for

angle 2x, hence resultedin wrong 'sign' for the

ratios

4

3 5

1213

5

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5[6]

(i)

22 )1(132)1)(32(

27

x

C

x

B

x

A

xx

x

)32()1)(32()1(27 2 xCxxBxAx

1

)32(2)1(7,1

C

Cxlet

2

12

32)

2

3(7,

2

32

A

Axlet

1

332,0

B

CBAxlet

22 )1(

1

1

1

32

2

)1)(32(

27

xxxxx

x

M1

M1

A1

Quite well done

(at least 2 correct values

found)

(ii)

)3(710.0

3

13ln1ln6

16ln7ln2

1

1

1)1ln()32ln(

2

1

)1(

1

1

1

32

2

2

1

)1)(32(2

27

5

2

5

2 2

5

2 2

sf

xxx

dxxxx

dxxx

x

M1

M1

A1

Many could not handle

2)1(

1

xafter

integrating the 1st

2terms in ln, e.g.

2

2)1ln(

)1(

1

x

x

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6[6]

(i) 7563 23 xxxy

51292 xx

dx

dy B1

(ii) cxycyx

gradient of tangent = 1

9

1

6

9

457

3

25

3

26

3

23

3

2

0)23)(23(

04129

15129

23

2

2

yxc

y

x

xx

xx

xxdx

dy

M1

M1

A1

Many pupils let 0dx

dy,

they end up with the eqn

05129 2 xx which gives imaginary

root.

Coincidently it is the

same value3

2x in

their calcuator screenbut with the small

symbol "xy" whichrepresent imaginary

value!!

(iii)

13

29

9

5

3

2

3

2

3

49

9

5

3

49

5129

2

222

2

2

x

xx

xx

xxdx

dy

for all real values ofx, 0dx

dy

hencey decreases asx increases

M1

A1

Very badly done.Majority have no idea

what to do

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7[11]

(i) Gradient ofPQ = gradient ofOR= 0.5

Eqn of PQ: )4(2

13 xy

52

1 xy --------(1)

Gradient of QR = 2 Eqn of QR: )4(22 xy

102 xy ------(2)

(1)=(2)

)6,2(

610)2(2

2

52

5

52

1102

Q

y

x

x

xx

M1

M1

A1

Well done!

(ii) In eqn (1), let 0x ,y = 5, unitsOT 5

525

)52()04(22

RT

RT

Since OT=RT= 5 units

ORT is isosceles.

M1

M1

A1

OK

(iii) Let S (a, b)Midpoint ofRS = Midpoint ofOP

18

32&44

2

3,

2

4

2

2,

2

4

ba

ba

ba

Hence coordinates of )1,8(S

M1

M1

A1

Many equate gradient ofOP = gradient of RS

instead of their midpoint

(iv) Area of trapezium OPQR

225

502

1

62442421

02630

04240

2

1

units

M1

A1

OK

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8[9]

(a)

5.0)3

2cos(

x

3

5

32

3

7

x

x

Basic angle =3

,,3

2,

3,0

3

5,

3

7,

3

5,

3,

332

x

x

M1

M1

A2

Very badly done, only a couplecould list down all the ans in

radian in terms of accurately

(b) sin5cos4sec2

5

42tan

5

22

2cos

2sin

2cos22sin2

5

2sin2

522cos22

2sin

2

5

2

12cos42

cossin5cos42

sin5cos4cos

2

2

Basic angle = 38.66o

)1(3.1093.19

66.21866.382

pldecor

or

M1

M1

M1

M1A1

Common mistakes from step 3

onwards:

2sin5cos42cos

2)sin5cos4(cos

or

9

[4]

023)6(3

3236

2

33

2

2

kxkx

kxkxx

xkx

For line to meet curve at least once implies one or

more real roots, therefore discriminant 0

012

0)12(

012

024363612

0)23)(3(4)6(

2

2

2

kork

kk

kk

kkk

kk

M1

M1

M1A1

Many pupils let D > 0. If end

up with the correct solving

method up to k < 12 or k > 0,

all 3 method marks awarded

able to identify a, b & c fordiscriminant & reduce it to get

k(k+ 2)

-12 0

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10[5]

...1561

...)(2

6

1

61)1(

22

26

xppx

pxpxpx

...15661

)1..)(1561(

)1()1(

222

22

6

xppqxpxqx

qxxppx

qxpx

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

7156

)1........(606

2

ppq

pqpq

Subt (2) into (1)

2),1(

3

1

9

1

371536

3

715)6(6

2

22

2

qfrom

p

p

pp

ppp

A1

M1

M1

A1

A1

Many pupils let coefficient ofx = 1 instead of 0, i.e.

16 pq

Many end up with only 1 set of

ans, i.e. 2&3

1

qp

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11[9]

(i)

axby

abyx

lg)lg(lg

Draw straight line with appropriate scale B3

Quite a number could notdefine the scale

appropriately, hence missingout the lg yintercept

Commonly seen:

baxy

abyx

lglglg

gradient = lg a

intercept = lg b

(ii)

51.2

)36.045.0(4.0lg

2.30

8.108.3

b

toacceptb

gradient

)3(120026.120210

)2.33(08.3lg

08.3

sf

a

toaccepta

M1

A1

M1

A1

(iii) Draw lgy =x + 2

2lg

10lg)2(lg

102

xy

xy

aby xx

From the intersection point,

x = 0.8

B1

B1

Many left blank

lgy =x+2

lgy = (lg b)x + lg a

lgy

(0,3.08)

(0,2)

x

i) Gra h B2

(iii) B1

(3.2, 1.8)

0.8

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12[9]

(i)

xQR

QR

x

QR

x

2

2

1

45sin

xRM

RM

x

RM

x

1

45tan

Given perimeter = 680 m

xxxy

xxxy

yx

yxx

42

2340

2226802

6804

222

2

22

22222

22

22

2

12340

2

1

2

2340

4422340

21

442

2340

2

1

42

1,

xx

xxx

xxxxxx

xxxxxx

xxyxAArea

M1

M1

M1

A1

Badly done or left blank

(ii) 012340 xdx

dA

sfmx

x

3141

8326.14012

340

2

2

9002354.23941

12

340

2

12

12

340340

mA

A

M1

M1

M1

A1

Many could get the method

mark but not the accuracy

mark, check their calculator

skill!

(iii)0)12(

2

2

dx

Ad

Hence A is maximum.B1

left bracket out!

122

2

dx

Ad

PQ

R S T

45

x

xx

y

yM

Amk Prelim 2009 Am1

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