2011 bdms 4e prelims 2 am paper 2.doc

8/22/2019 2011 BDMS 4E Prelims 2 AM Paper 2.doc

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Bendemeer Secondary School AM 2011 Prelims Paper 2

Mathematical Formulae

1. ALGEBRA

Quadratic Equation

For the equation 02 =++ cbxax ,

.2

42

a

acbbx

=

Binomial expansion

nrrnnnnn bbar

nba

nba

naba ......

21

221 ,

where n is a positive integer and

!

)1)...(1(

!!

!

r

rnnn

rrn

n

r

n

2. TRIGONOMETRY

Identities


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2

( ) ( )

( ) ( )

( ) ( )

( ) ( )BABABA

BABABA

BABABA

BABABA

A

AA

AAAAA

AAA

BABABA

BABABA

BABABA

AAec

AA

AA

+=

+=+

+=

+=+

=

===

=

=

==

+=

+=

=+

2

1sin

2

1sin2coscos

2

1cos2

1cos2coscos

2

1sin

2

1cos2sinsin

2

1cos

2

1sin2sinsin

tan1

tan22tan

sin211cos2sincos2cos

cossin22sin

tantan1tantan)tan(

sinsincoscos)cos(

sincoscossin)sin(

cot1cos

.tan1sec

.1cossin

2

2222

22

22

22

Formulae forABC

Cab

Abccba

C

c

B

b

A

a

sin

2

1

.cos2

.sinsinsin

222

=

+=

==

Answer all the questions. Electronic calculator may be used in this paper.

The diagrams in this paper are NOT drawn to scale.

1 (a) Solve the simultaneous equations

3cos6sin5 =+ qp

1cos3sin = qp

expressing your answers in the smallest possible positive values of p and q in degrees.

[ 6 ]

(b) (i) Using the same axes, on the same diagram, sketch the curves xy 2cos= and

xy sin= for the interval 20 x , labelling each curve clearly. [ 4 ]

(ii) State the number of solutions in the interval 20 x of the equation

xx sin2cos = . [ 1

]

Bendemeer Secondary SchoolSec 4E / 2011 Preliminary Two Exam / A. Maths Paper 2 / 4038/02


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3

2 Water is poured into a container at a rate of16 cm3s-1. The volume, V cm3, of the water in the

container, when the depth of water is x cm, is given by 316

3xV = . Find

(i) in terms of , the volume of the container when 2=x , [ 2 ]

(ii) the rate of increase in the depth of water when 8=x , [ 4 ](iii) the depth of water when the rate of increase in the depth is 2.0 cms-1. [ 3

]

3 Without using a calculator, solve, for a and b , the simultaneous equations

132312=+

+ ba

( ) 2324311

=

ba

[ 6 ]

4 (i) Write down and simplify, in terms of n , the first four terms of the expansion ( )nx32+ in

the ascending powers of x . [ 3 ]

(ii) From part (i), the ratio of the coefficient of 2x to that of 3x is 9:1 , find the value of

n where n is positive integer. [ 2

]

(iii) With this value of n , write down and simplify the coefficient of the term 7x . [ 3 ]



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4

5 The curve xey 38= intersects the coordinate axes at the points A and B .

(i) Given that the line AB passes through the point with coordinates

c,2

1ln ,

find the value of c . [ 5 ](ii) In order to solve the equation 3 9ln xx = , a graph of a suitable straight line is drawn

on the same set of axes as the graph ofxey 38= . Find the equation of this straight line.

[ 3 ]

6

In the diagram, PR is a chord to the circle PRS . PTQ and ST are tangents to the circle at

P and S respectively such that QSR is a straight line. Prove that

(i) QSTRPS = , [ 3

]

(ii) triangles PQR and SQP are similar, [ 3 ]

(iii) PRSSTQ = 2 . [ 4 ]


T

S

QP


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5

7

The figure shows part of the curve 28 xxy = and part of the line xy = .

Find the(i) coordinates of A , [ 2 ]

(ii) the area of the shaded region, giving your answers as a fraction. [ 4 ]

8 A particleP travelling in a straight line passes a fixed point O with a velocity of 4 ms-1.

Its acceleration, a ms-2, is given by the equation ta 24 = , where t is the time in seconds

after passing O .

(i) Calculate the acceleration of the particle when 2=t . [ 2 ]

(ii) Show that the equation of velocity, v ms-1, is given by 44 2 = ttv . [ 3 ]

(iii) Find the time, t , when P is instantaneously at rest. [ 3

]

(iv) Find the displacement ofP when its velocity is 4 ms-1 again. [ 4 ]



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6

9 Solutions to this question by accurate drawing will not be accepted.

The diagram shows a trapezium ABCD in which BC is parallel to AD , and BD is

perpendicular to AB . The coordinates ofB , C , D and E are the points ( )2,2 ,

( )6,k where 0


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7

11

The diagram shows an isosceles triangle POQ with 4==OQOP cm and an isosceles

triangle ROS with 6== OSOR cm. = 90POS and = xPOQ .

The sum of the areas of the triangles ROS and POQ is S cm2.

(i) Show that xxS cos18sin8 += , [ 4 ]

(ii) Express S in the form ( )+xR sin , and hence find the values of R and . [ 3

]

Given that x can vary, find

(iii) the value ofx for which the area of triangle ROS is twice the area

of the triangle POQ . [ 3 ]

-------- End of Paper --------


6 cm

4 cm

O P

Q

S

R


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emeer Secondary School Bendemeer Secondary School Bendemeer Secondary School Bendemeer Secondary School Bendemeer Secondary School Bendemeer Secondary School









emeer Secondary School Bendemeer Secondary School Bendemeer Secondary School Bendemeer Secondary School Bendemeer Secondary School Bendemeer Secondary Schoolemeer Secondary School Bendemeer Secondary School Bendemeer Secondary School Bendemeer Secondary School Bendemeer Secondary School Bendemeer Secondary School


8

-------- End of Summary --------


1(a) = 150,30 7(i) ( )7,7A

1(b) = 6.45p and = 5.95q 7(ii)6

128 units2 (answer given in fraction)

2(i) See Solution for Proof. 8(i) 0=a ms-2

2(ii)4

1200

+=

x cm / 168.0 cm (3 sig. fig) 8(ii) See Solution for Proof.

2(iii) 02

2


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emeer Secondary School Bendemeer Secondary School Bendemeer Secondary School Bendemeer Secondary School Bendemeer Secondary School Bendemeer Secondary Schoolemeer Secondary School Bendemeer Secondary School Bendemeer Secondary School Bendemeer Secondary School Bendemeer Secondary School Bendemeer Secondary School




9

1 (a) Solve the simultaneous equations

3cos6sin5 =+ qp

1cos3sin = qp

expressing your answers in the smallest possible positive values of p and q in degrees.

[ 6 ]

(b) (i) Using the same axes, on the same diagram, sketch the curves xy 2cos= and

xy sin= for the interval 20 x , labelling each curve clearly. [ 4 ]

(ii) State the number of solutions in the interval 20 x of the equation

xx sin2cos = . [ 1

]

1(a) 3cos6sin5=+

qp .. (1)1cos3sin = qp .. (2)

From Eqn (2), we have 1cos3sin += qp

Substitute into 1cos3sin += qp into Eqn (1), we have

( ) 3cos61cos35 =++ qq

2cos21 =q

21

2cos =q

Basic angle,

= 21

2cos 1

= 535.84

= 535.84180q

= 5.95q (1 decimal place)

Substitute 21

2

cos=

q into Eqn (1), we have

[ M1 ]

[ M1 ]

[ A1 ]


BENDEMEER SECONDARY SCHOOL2011 Preliminary Two ExaminationSecondary Four ExpressAdditional Mathematics Paper 2 Solutions 4038/02


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10

321

26sin5 =

+p

7

5sin =p

Basic angle,

=

7

5

sin

1

= 584.45

= 6.45p (1 decimal place)

= 6.45p and = 5.95q

[ M1 ]

[ M1 ]

[ A1 ]

1(b)

(i)

Correct interval/period (in radian measure) B1

Correct shape xy 2cos= graph B1

Correct shape xy sin= - B1

Correct range (maximum and minimum points) B1

[ B4 ]

1(b)

(ii)

Number of points of intersections = 4 [ A1 ]



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11

2 Water is poured into a container at a rate of16 cm3s-1. The volume, V cm3, of the water in the

container, when the depth of water is x cm, is given by 316

3xV = . Find

(i) in terms of , the volume of the container when 2=x , [ 2 ](ii) the rate of increase in the depth of water when 8=x , [ 4 ]

(iii) the depth of water when the rate of increase in the depth is 2.0 cms-1. [ 3

]

2(i)When 2=x , ( )32

16

3=V

2

3=

/

5.1 cm

3

[ M1 ]

[ A1 ]

2(ii)16=

dt

dVcm3s-1

3

16

3xV =

2

16

9x

dx

dV=

When 8=x , ( )2

816

9

=dxdV

36=

dt

dV

dV

dx

dt

dx=

1636

1=

9

4= / 141.0 cms-1 ( 3 sig. fig.)

[ M1 ]

[ M1 ]

[ M1 ]

[ A1 ]

2(iii)Given that 2.0=

dt

dxcms-1

2.09

16

2

2

=x

( )2.09

162

2

=x cm

73.6=x cm

[ M1 ]

[ M1 ]

[ A1 ]



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12

3 Without using a calculator, solve, for a and b , the simultaneous equations

132312=+

+ ba

( ) 23243 11 = ba [ 6 ]

3 1323 12 =+ + ba .. (1)

( ) 23243 11 = ba .. (2)

From Eqn (2): ( ) 23243 11 = ba

( )

232

22

3

3 2 =ba

223

632 =a

b

From Eqn (1): 1323 12 =+ + ba

( ) 13229

3=+

ba

Substitute2

23

6

32 =

ab into Eqn (1), we have

132

23

6

32

9

3=

+

aa

13233

3

9

3=+

aa

813 =a

4= a

Substitute 4=a into2

23

6

32 =

ab , we have

2

23

6

32

4

=b

22 =b

1= b

Hence, 4=a and 1=b .

[ M1 ]

[ M1 ]

[ M1 ]

[ M1 ]

[ A1 ]

[ A1 ]

4 (i) Write down and simplify, in terms of n , the first four terms of the expansion ( )nx32 + in

the ascending powers of x . [ 3 ]Bendemeer Secondary SchoolSec 4E / 2011 Preliminary Two Exam / A. Maths Paper 2 / 4038/02


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13

(ii) From part (i), the ratio of the coefficient of 2x to that of 3x is 9:1 , find the value of

n where n is positive integer. [ 2

]

(iii) With this value of n , write down and simplify the coefficient of the term 7x . [ 3 ]

4(i) ( ) ( ) ( ) ( ) ( ) 22211

1 3232232 xCxCxnnnnnn ++=+

( ) ( ) ...32 333 ++ xC nn

( )( ) ( )29

4

2

2

13

2

22 x

nnxn

nnn

+

+=

( )( ) ( ) ...27

8

2

6

21 3 +

+ x

nnnn

( ) ( )( )

...16

2219

8

219

2

232

32

+

+

+

+= xnnn

xnn

xn nnnn

[ M1 ]

[ M1 ]

[ A1 ]

4(ii) Given that the coefficient of 2x to that of 3x is 9:1 , we have

( ) ( )( )

16

2219

8

2199

nn nnnnn =

( )

16

2

8

9 = n

20=n

[ M1 ]

[ A1 ]

4(iii) By applying the formulae,rrn

rn

r baCT

+ =1Power of 7x in ( ) ( ) rrrr xCT 32

20201

+ =

7=r

the term of 7x , ( ) ( )71372017 32 xCT =+( )( ) 72187819277520 x=

Coefficient of 7x 121039.1 = ( 3 sig. fig.)

.

[ M1 ]

[ M1 ]

[ A1 ]



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14

5 The curve xey 38= intersects the coordinate axes at the points A and B .

(i) Given that the line AB passes through the point with coordinates

c,2

1ln ,

find the value of c . [ 5 ](ii) In order to solve the equation 3 9ln xx = , a graph of a suitable straight line is drawn

on the same set of axes as the graph ofxey 38= . Find the equation of this straight line.

[ 3 ]

5(i) xey 38=

When 0=x , 7=y

When 0=y , 83

=x

e

2=xe

2ln=x

( )7,0A and ( )0,2lnB

Gradient02ln

70

=

2ln

7

=

Equation of the line AB : ( ) ( )02ln

77 = xy

72ln

7+= xy

At the point

c,2

1ln , we have

7

2

1ln

2ln

7+

=c

14=c

[ M1 ]

[ M1 ]

[ M1 ]

[ M1 ]

[ A1 ]



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15

5(ii) For 3 9ln xx =

3 9 xex =

xe x = 93

x

ex3

9 = xex 381 =

The equation of the straight line to be drawn is 1=xy

[ M1 ]

[ M1 ]

[ A1 ]



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16

6

In the diagram, PR is a chord to the circle PRS . PTQ and ST are tangents to the circle at

P and S respectively such that QSR is a straight line. Prove that

(i) QSTRPS = , [ 3

]

(ii) triangles PQR and SQP are similar, [ 3 ]

(iii) PRSSTQ = 2 . [ 4 ]

6(i) PRSPST = (Alternate Segment Theorem)

=++ 180PSRPRSRPS (sum of )=++ 180PSRPSTQST ( adj.s of a straight line)

Since PRSPST =

QSTRPS =

[ M1 ]

[ M1 ]

[ A1 ]

6(ii) SQPPQR = (common )SPQPRQ = (Alternate Segment Theorem)

Based on the Angle-Angle-Angle Property, PQR is similar to

SQP .

[ M1 ]

[ M1 ]

[ A1 ]

6(iii) PRSSPT = (Alternate Segment Theorem)

PSTSPT = (Tangents from an external point)

PSTSPTSTQ += (Sum of external angle of )

Hence, PRSSTQ = 2

[ M1 ]

[ M1 ]

[ M1 ]

[ A1 ]


T

S

R

QP


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17

7

The figure shows part of the curve 28 xxy = and part of the line xy = .

Find the(i) coordinates of A , [ 2 ]

(ii) the area of the shaded region, giving your answers as a fraction. [ 4 ]

7(i) xy = .. (1)2

8 xxy = .. (2)

Equating both equations, we have

28 xxx =

072 =xx

( ) 07 =xx

0=x or 7=x

Coordinates of ( )7,7A

[ M1 ]

[ A1 ]

7(ii) Area of the shaded region = Area of OAB + Area under the

curve

Area of 772

1=OAB

2

49= or2

124=

Area under the curve

=

8

7

28 dxxx

[ M1 ]

[ M1 ]



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18

8

7

32

34

= xx

3

23=

Area of shaded region3

23

2

124 +=

6

128= units2

[ M1 ]

[ A1 ]



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19

8 A particleP travelling in a straight line passes a fixed point O with a velocity of 4 ms-1.

Its acceleration, a ms-2, is given by the equation ta 24 = , where t is the time in seconds

after passing O .

(i) Calculate the acceleration of the particle when 2=t . [ 2 ](ii) Show that the equation of velocity, v ms-1, is given by 44 2 = ttv . [ 3 ]

(iii) Find the time, t , when P is instantaneously at rest. [ 3

]

(iv) Find the displacement ofP when its velocity is 4 ms-1 again. [ 4 ]

8(i) When 2=t , ( )224 =a

0=a ms-2

[ M1 ]

[ A1 ]

8(ii) ta 24 =

= dttv 24

ctt += 24

At the fixed point O , 0=t and 4=v ms-1

4=c

Hence, 44 2 = ttv (shown)

[ M1 ]

[ M1 ]

[ A1 ]

8(iii) When the particle is instantaneously at rest, 0=v

044 2 = tt

0442

=+ tt

( )( ) 022 = tt

2=t s

[ M1 ]

[ M1 ]

[ A1 ]

8 (iv) When 4=v .

4442

= tt

04 2 =tt

( ) 04 = tt (since 0>t )

4=t s

= dtvd

= dttt 44 2

ct

t

t += 432

32

[ M1 ]



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20

When 0=t , 0=d 0= c

tt

td 43

23

2 =

When 4=t , ( )( )

( )443

442

32 =d

3

15=d m /

= 3.5 m

[ M1 ]

[ M1 ]

[ A1 ]



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21

9 Solutions to this question by accurate drawing will not be accepted.

The diagram shows a trapezium ABCD in which BC is parallel to AD , and BD is

perpendicular to AB . The coordinates ofB , C , D and E are the points ( )2,2 ,

( )6,k where 0


22/25

22

Equation of :AB ( ) ( )( )22

12 = xy

32

1= xy / 062 =++ xy

Equation of :AD ( ) ( )( )920 = xy

182 += xy / 182 =+ xy

Therefore, the coordinates of A :

18232

1+= xx

14=x

Substitute 14=x into 182 += xy , we have

( ) 18142 +=y

10=y

Hence, the coordinates of ( )10,14 A .

[ A1 ]

[ A1 ]

[ M1 ]

[ A1 ]

9(iii) Gradient of 2=FE ,

Equation of the perpendicular bisector: ( )920 = xy

182 = xy

Coordinates ofF , 32

1182 = xx

6=x

Substitute 6=x into ( ) 1862 =y ,

6=y

Therefore, the coordinates of ( )6,6 F .

[ M1 ]

[ M1 ]

[ A1 ]

[ A1 ]



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23

10 (i) Express583

32 +

xx

xin partial fractions. [ 4 ]

(ii) Hence, evaluate +3

2

2 583

26dx

xx

x. [ 4 ]

10(i)

153583

3

2 +

=

+

x

B

x

A

xx

x

( ) ( )5313 += xBxAx

When 1=x , ( )( )513013 += B

1=B

When 0=x , ( ) ( )( )50311003 = A

2=A

1

1

53

2

583

32

=+

xxxx

x

[ M1 ]

[ M1 ]

[ M1 ]

[ A1 ]

10(ii)

+

=+

3

22

3

22 583

32

583

26dx

xx

xdx

xx

x

=3

21

1

53

22 dx

xx

( ) ( )3

2

1ln53ln3

22

= xx

( )( ) ( )( ) [ ]013ln533ln3

22

=

462.0= (3 sig. fig.)

[ M1 ]

[ M1 ]

[ M1 ]

[ A1 ]



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24

11

The diagram shows an isosceles triangle POQ with 4==OQOP cm and an isosceles

triangle ROS with 6== OSOR cm. = 90POS and = xPOQ .

The sum of the areas of the triangles ROS and POQ is S cm2.

(i) Show that xxS cos18sin8 += , [ 4 ]

(ii) Express S in the form ( )+xR sin , and hence find the values of R and . [ 3

]

Given that x can vary, find

(iii) the value ofx for which the area of triangle ROS is twice the area

of the triangle POQ . [ 3 ]

11(i)Area of xPOQ sin44

2

1=

xsin8=

Area of ( )xROS = 90sin662

1

xcos18= ( )( )xx cos90sin =

xxS cos18sin8 += (shown)

[ M1 ]

[ M1 ]

[ M1 ]

[ A1 ]

11(ii) ( )+=+ xRxx sincos18sin8

38818822=+=R [ M1 ]


6 cm

4 cm

x

O P

Q

S

R


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25

8

18tan =

= 0.66 ( 1 decimal place )

( )+=+ 0.66sin388cos18sin8 xxx

[ M1 ]

[ A1 ]

11(iii) Given that ( ) xx cos18sin82 =

xx cos9sin8 =

8

9tan =x

= 4.48x (1 decimal place)

[ M1 ]

[ M1 ]

[ A1 ]

------- End of Solution --------

2011 bdms 4e prelims 2 am paper 2.doc

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