amk prelim 2009 am1
TRANSCRIPT
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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
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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]
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AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01
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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AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01
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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AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01
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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AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01
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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AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01
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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AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01
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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AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01
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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AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01
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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AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01
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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AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01
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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AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01
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