2009 Prelim ESpring S4E AM P2

8/8/2019 2009 Prelim ESpring S4E AM P2

1/15

Name : _____________________________________________________ ( )

S 4E___

Additional MathematicsPaper 2

Friday 28th August 2009

Additional materials:6 Writing papers

INSTRUCTIONS TO CANDIDATES

Write your Name 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.

Answerall the questions.

Write your answers on the separate Answer Paper provided.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimalplace in the case of angles, unless a different level of accuracy is specified in thequestion.

The use of a scientific 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 100.

2 hours 30 mins9:35 am 12:05 pm

This question paper consists of 6 printed pages including the cover page.

4038/2

East Spring Secondary SchoolTowards Excellence and Success

Preliminary Examinations 2009Sec 4 Express


2/15

East Spring Secondary SchoolMathematics DepartmentDo It Right. Always

Mathematical Formulae

1. ALGEBRA

Quadratic Equation

For the equation 02 =++ cbxax

a

acbbx

2

42 =

Binomial expansion

nrrnnnnn bbar

nba

nba

naba ++

++

+

+=+ ......

21)( 221 ,

where n is a positive integer and!)!(

!

rrn

n

r

n

=

( ) ( )!

1...1

r

rnnn +=

2. TRIGONOMETRY

Identities

1cossin 22 =+ AAAA 22 tan1sec +=AA 22 cot1cosec +=

( ) BABABA sincoscossinsin =( ) BABABA sinsincoscoscos =

( )BA

BABA

tantan1

tantantan

=

AAA cossin22sin = AAAAA 2222 sin211cos2sincos2cos ===

A

AA

2tan1

tan22tan

=

( ) ( )BABABA +=+2

1cos

2

1sin2sinsin

( ) ( )BABABA +=2

1sin

2

1cos2sinsin

( ) ( )BABABA +=+2

1cos

2

1cos2coscos

( ) ( )BABABA += 21

sin2

1

sin2coscos

Formulae for ABC

C

c

B

b

A

a

sinsinsin==

Abccba cos2222 +=

Abc sin2

1=

49229412.docTowards Excellence and Success page 2/6


3/15


1. Given that

=

12

32M , find M 1 and hence solve the simultaneous equations

01032 =+ yx 062 =+ xy [6]

2. (a) Differentiate with respect tox

(i) 47 2 x [2]

(ii) ( )12sin 3 +x [2]

(b) Show that xecxxx

x

dx

d5cos55cot

5tan

2=

. [5]

3. (a) Express )32()3(

152

+

+

xx

x

in the form 323 ++ x

B

x

A

, whereA andB are

constants. [4]

(b) Hence, or otherwise, evaluate xxx

xd

)32()3(

152

+

+ . [3]

4. The roots of the quadratic equation 0423 2 =+ xx are and .

(a) State the value of + and of. [2](b) Find the quadratic equation in x whose roots are +2 and 2+ . [6]

5. (a) Prove the identityBA

BA

BA

BA

tantan

tantan

)sin(

)sin(

+

+

. [4]

(b) Find all the angles between 0 and 2 which satisfy the equation

xxx 2sin3sinsin =+ . [5]

6. (a) Given that ( ) ( ) xxx 212 1553 = + , show that5

915 =x .

Hence or otherwise, find the value ofx. [5]

(b) Solve the equation 01)23(log6)23(log2 82 =++ xx . [4]



4/15


7.

In the circle,Pis the centre andAD is the diameter.DCis a tangent to the circle and BCD

is a right angle.AB produced andDCproduced meet at T.

(a) Show with reasons that ABD and DCB are similar.

Name three other triangles which are similar to ABD. [5]

(b) Stating the reasons clearly, show that

(i) CBDADB =2 ,

(ii) TBTADT =2 ,

(iii) CBDATBTATB =2 . [6]


P

D C

B

A

T


5/15


8. A curve has the equation xxy 2coscos2 = , where2

0 < x .

(a) Find expressions fordx

dyand

2

2

dx

yd. [2]

(b) Given that sin 2x can be expressed as 2 sinx cosx, find thex-coordinate of the

stationary point of the curve and determine the nature of this stationary point. [4]

(c) Evaluate 2

3

dxy . [3]

9. A curve has the equation 4)3( 2 ++= xkxy where k is an integer.

(a) Given that the curve has a stationary value at 3=x , find the value of k . [4]

(b) Find the equations of the tangent and normal to the curve at the point where the curve

intersects they-axis.

Hence, find the area of the triangle bounded by these 2 lines and thex-axis. [6]

10. The diagram shows a vertical section through a tent in whichPQ = 2 m, QR = 3 m,== QRSQPT

(in degrees) where < 90 , andRTis horizontal. The diagram issymmetrical about the verticalPT.

(a) IfH represents the length ofPT , show that sin3cos2 +=H . [1]

(b) Express H in the form )cos( R . [3]

(c) Find the maximum value ofH and the corresponding value of. [3]

(d) Find the value of for which 5.3=H m. [3]


P

Q

RS

U

T

2 m

3 m


6/15


11. The figure shows the curve y = (x 2)2 + 3 and the straight liney +x = 7 intersecting at the

pointsPand Q. The straight line cuts thex-axis atR.

Find

(a) the coordinates ofP, Q andR, [6]

(b) the area of the region bounded by the curve, the straight line and the axes. [6]

END OF PAPER


y = (x 2)2 + 3

y +x = 7

P

y

Rx

O

Q


7/15

Additional Mathematics Paper 2 (4038/02)

Secondary 4 Express

Marking Scheme

Question Solution Marks

1.

=

12

32M 8)6(2det == M

=

=

4

1

4

18

3

8

1

22

31

8

11M

01032 =+ yx 1032 = yx062 =+ xy 62 =+ yx

=

=

=

=

4

1

32

8

8

1

6

10

22

31

8

1

6

10

12

32

y

x

y

x

1=x , 4=y

M1

A1

M1

M1

M1

A1

2. (a)(i)

(a)(ii)

( ) ( )47

714

472

147

22

2

=

=

x

xx

xx

dx

d

( )[ ] ( ) ( ) ( ) ( )12cos12sin612cos212sin312sin 223 ++=++=+ xxxxxdx

d

M1, A1

M1, A1

49229412.doc

7/5

For correct placing of the inverse

matrix.

For correct solution of matrices


8/15


2. (b) ( ) ( )

( )

xecxx

xxx

x

x

xx

x

x

xx

x

x

x

xxx

x

xxx

x

x

dx

d

5cos55cot

5sin

155cot

5sin

5cos

5cos

15

5tan

1

5tan

5sec5

5tan

5tan

5tan

5sec55tan

5tan

)5sec5(1)5(tan

5tan

2

2

2

2

2

2

2

2

2

2

2

2

=

=

=

=

=

=

M1

M1

M1

A1

3. (a) Let)32()3(

152

+

+

xx

x

=32

B

3

A

+

+ xx,

A(2x 3) + B( 3 +x) = 2x + 15

Letx = 3, A = 1;

Let x = 1.5, B = 4.

)32()3(

152

+

+

xx

x=

32

4

3

1

+

+

xx

B1

B1

B1

A1

3. (b)

xxx

xd

)32()3(

152

+

+ = dxxx)( 32

4

3

1

+

+ [ECF from (a)]

=ln(3 + x) + 2ln(2x 3)

or lnx

x

+

3

)32( 2

B1

A2

4. (a)

0423 2 =+ xx

3

2

3

2=

=+

3

4=

A1

A1


A1For correct differentiation


9/15


(b)

( ) ( )

2

3

23

)(3

3322

=

=

+=+=+++

( )( )

[ ]

9

20

3

4

3

22

)(2

52)(2

5)(2

24222

2

2

2

22

22

=

+

=

++=

++=

++=

+++=++

09

2022 =+ xx

020189 2 =+ xx

M1

A1

M1

M1

A1

A1

5. (a) LHS =)sin(

)sin(

BA

BA

+

( ) ( )

BA

BA

BABA

BA

BA

BA

BA

BA

BA

BA

BA

BA

BABA

BA

BABA

BABA

BABA

tantan

tantan

tantantantan

coscos

sincos

coscos

cossin

coscos

sincos

coscos

cossin

coscos

sincoscossin

coscos

sincoscossin

sincoscossin

sincoscossin

+=

+=

+=

+=

+=

= RHS

M1

M1

M1

A1



10/15


5. (b)

0)1cos2(2sin

02sincos2sin22sincos2sin2

2sin2

3cos

2

3sin2

2sin3sinsin

===

=

+

=+

xx

xxxxxx

xxxxx

xxx

02sin =x or2

1cos =x

3,2,2 =x3

5,

3

=x

2

3,,

2

=x

3

5,

2

3,,

2

,

3

=x

M1

M1

M1

A1

A1

6. (a) ( ) ( )

x

x

xx

x

xx

x

xxx

15

15

5

9

53

15

5

9

155

533

1553

2

2

22

212

=

=

=

= +

5

915 =x

8.1lg15lg

5

9lg15lg

=

=

x

x

217.015lg

8.1lg==x (3 sig figs)

M1

A1

M1

M1, A1



11/15


6. (b)

947.03

22

223

4

1)23(lo g

1)23(lo g4

01)23(lo g2)23(lo g2

013

)23(lo g6)23(lo g2

018lo g

)23(lo g6)23(lo g2

01)23(lo g6)23(log2

25.0

25.0

2

2

22

22

2

22

82

=+

=

=

=

==++

=+

+

=+

+

=++

x

x

x

x

xx

xx

xx

xx

M1

M1

M1

A1

7.

(a) DCBABD == 90 (rt.

in semi

; given)BDCDAB = ( in alt segment;DCis tangent to atD)

CBDBDA = (alt ;AD//BC)

ABD and DCB are similar by AAA property (shown)

ABD is also similar to DBT, BCTand ADT.

( 2m for all correct, 1m for 2 correct )

M3

A2


P

D C

B

A

T


12/15


(b) (i)

(ii)

(iii)

Since ABD and DCB are similar,CB

BD

DB

AD=

CBDADB =2 (shown)

TBTADT =2 (Tangent-Secant Theorem) (shown)

2DT 22 DBTB += (Pythagoras Theorem)

2TB 22 DBDT = ( using answers from i and ii)

CBDATBTA = (from (ii) & (i)) (shown)

M1

A1

A1

M1

M1

A1

8. (a)dxdy = 2 sinx + 2 sin 2x,

2

2

dx

yd= 2 cosx + 4 cos 2x.

A1

A1

(b)dx

dy= 2 sinx + 4 sinx cosx = 0 at stationary point.

sinx(1 + 2 cosx) = 0,

sinx = 0 or cosx = 0.5

x=0(rejected)

=x ,

2

2

dx

yd= 2 cos

+ 4 cos 2

= ve,

hence

=x is a maximum point.

B1

B1B1

A1

(c) 2

3

dxy = 2

3

d)2coscos2( xxx

= 2 sinx 2

1sin 2x ]

2/

3/

= (2 0) ( 3 34

1)

= 2 34

3

B1

B1

A1



13/15


9. (a)

4)3( 2 ++= xkxy

dx

dy

42

34)6(

2

+

+++=

x

kxxx

42

4815

42

3)4(12

2

2

+++=

++++

=

x

kxx

x

kxxx

When 3=x , = 0dx

dy0

432

)3(48)3(15 2 =+

++ k

9=k

M1

M1

M1

A1

(b) 4)93(2 ++= xxy &

42

94815 2

+++=

x

xx

dx

dy

When the curve intersects they-axis, 0=x

18= y &4

9

42

9==

dx

dy

Equation of the tangent at (0, 18) is 184

9 += xy (1)

& equation of the normal at (0, 18) is 1894 += xy (2)

Subst 0=y : (1): 8=x

(2):2

81=x

Area of the triangle bounded by these 2 lines and thex-axis

)18(

2

818

2

1

+=

2

1436= units2

M1

A1

A1

M1

M1

A1


Having both values

ofx correct.

Ecf for value of 9

from part (a).


14/15


10. (a)

In PQU, cos22

cos == PUPU

In QRS, UTQSQS

=== sin33

sin

sin3cos2 +=+== UTPUPTH (shown)A1

(b) Let )cos(sin3cos2 =+= RH

1332 22 =+=R

2

3tan = = 310.56

)3.56cos(13 = H

M1

M1

A1

(c) Max value ofH= 13 m

Note that


15/15


11. (a)

(x 2)2 + 3 = 7 x

x2 3x = 0

x = 0 or 3

P = (0, 7), Q = (3, 4) and R = (7, 0).

B1

B1B1

A3

(b) the area below PQ = +3

0

2 3])2([ dxx

=3

0

3 3)2(3

1

+ xx

= 9 3

1

(3

8

) = 12

The area under QR =2

1(4)(4) = 8

Total area = 12 + 8 = 20 sq units

B1

B2

B1

B1

A1

End of Paper

Qn Answers

1

2ai

2aii

2b

3a

b

4ai

4aii

4b

5a

5b

5c

6a

Qn

6bi

6bii

6biii

7ai

7aii

8ai

8aii

8b

9ai

aii


Documents

2009 Prelim ESpring S4E AM P2