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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
]
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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 ]
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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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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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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 --------
Bendemeer Secondary SchoolSec 4E / 2011 Preliminary Two Exam / A. Maths Paper 2 / 4038/02
6 cm
4 cm
O P
Q
S
R
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-------- End of Summary --------
Bendemeer Secondary SchoolSec 4E / 2011 Preliminary Two Exam / A. Maths Paper 2 / 4038/02
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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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 ]
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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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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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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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(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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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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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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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 ]
Bendemeer Secondary SchoolSec 4E / 2011 Preliminary Two Exam / A. Maths Paper 2 / 4038/02
T
S
R
QP
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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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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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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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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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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 ]
Bendemeer Secondary SchoolSec 4E / 2011 Preliminary Two Exam / A. Maths Paper 2 / 4038/02
6 cm
4 cm
x
O P
Q
S
R
-
8/22/2019 2011 BDMS 4E Prelims 2 AM Paper 2.doc
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8
18tan =
= 0.66 ( 1 decimal place )
( )+=+ 0.66sin388cos18sin8 xxx
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[ A1 ]
11(iii) Given that ( ) xx cos18sin82 =
xx cos9sin8 =
8
9tan =x
= 4.48x (1 decimal place)
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[ M1 ]
[ A1 ]
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