sbp f4 p2 final exam 2007
TRANSCRIPT
SULIT 3472/2
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SEKTOR SEKOLAH BERASRAMA PENUHKEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN AKHIR TAHUN TINGKATAN 4, 2007
ADDITIONAL MATHEMATICS
Form FourPaper 2
Two hours and thirty minutes
DO NOT OPEN THIS QUESTION PAPER UNTIL INSTRUCTED TO DO SO
3472/2Form FourAdditional MathematicsPaper 2Oct 20072 ½ hours
INFORMATION FOR CANDIDATES
1 This question paper consists of three sections : Section A, Section B and Section C.
2 Answer all questions in Section A, four questions from Section B and two questions from
Section C.
3 Give only one answer/solution to each question.
4 Show your working. It may help you to get marks.
5 The diagrams in the questions provided are not drawn to scale unless stated.
6 The marks allocated for each question and sub-part of a question are shown in brackets.
7 A list of formulae is provided on pages 2 and 3.
8 You may use a four-figure mathematical table.
9 You may use a non-programmable scientific calculator.
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This question paper consists of 9 printed pages
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The following formulae may be helpful in answering the questions. The symbols given are the onescommonly used.
ALGEBRA
12 4
2
b b acx
a
2 am an = a m + n
3 am an = a m - n
4 (am) n = a nm
5 loga mn = log am + loga n
6 logan
m= log am - loga n
7 log a mn = n log a m
8 logab =a
b
c
c
log
log
9 Tn = a + (n-1)d
10 Sn = ])1(2[2
dnan
11 Tn = ar n-1
12 Sn =r
ra
r
ra nn
1
)1(
1
)1(, (r 1)
13r
aS
1, r <1
CALCULUS
1 y = uv ,dx
duv
dx
dvu
dx
dy
2v
uy ,
2vdx
dvu
dx
duv
dy
dx
,
3dx
du
du
dy
dx
dy
4. Area under a curve
= b
a
y dx or
= b
a
x dy
5 Volume generated
= b
a
y 2 dx or
= b
a
x 2 dy
5 A point dividing a segment of a line
( x,y) = ,21
nm
mxnx
nm
myny 21
6 Area of triangle
= )()(2
1312312133221 1
yxyxyxyxyxyx
1 Distance = 221
221 )()( yyxx
2 Midpoint
(x , y) =
221 xx
,
221 yy
3 22 yxr
42 2
ˆxi yj
rx y
GEOMETRY
SULIT 3472/2
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STATISTICS
TRIGONOMETRY
9 sin (A B) = sinA cosB cosA sinB
10 cos (A B) = cosA cosB sinA sinB
11 tan (A B) =BA
BA
tantan1
tantan
12C
c
B
b
A
a
sinsinsin
13 a2 = b2 + c2 - 2bc cos A
14 Area of triangle = Cabsin2
1
6 1
0
100Q
IQ
14 z =
x
3
1 Arc length, s = r
2 Area of sector , L = 21
2r
3 sin 2A + cos 2A = 1
4 sec2A = 1 + tan2A
5 cosec2 A = 1 + cot2 A
6 sin 2A = 2 sinA cosA
7 cos 2A = cos2A – sin2 A= 2 cos2A - 1= 1 - 2 sin2A
8 tan 2A =A
A2tan1
tan2
472/2
1 x =N
x
2 x =
f
fx
3 =N
xx 2)(= 2
2
xN
x
4 =
f
xxf 2)(= 2
2
xf
fx
5 m = Cf
FNL
m
2
1
71
11
w
IwI
8)!(
!
rn
nPr
n
9!)!(
!
rrn
nCr
n
10 P(AB)=P(A)+P(B)-P(AB)
11 p (X=r) = rnrr
n qpC , p + q = 1
12 Mean µ = np
13 npq
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Section A
[40 marks]
Answer all questions in this section.
1 Solve the simultaneous equations 3x + y = 5 and x2 – 2y + 5 = 0.Give your answer correct to three decimal places.
[5 marks]
2 The function f(x) = 2x2 + 6x – 5 has a minimum point at ( h , k ).
(a) By using the method of completing the square, express f(x) in the formf(x) = a( x + p )2 + q , where a, p and q are constants.
[3 marks](b) State the value of h and of k.
[2 marks]
(c) Find the range of values of t if 2x2 + 6x – 5 = t , where t is a constant,has no roots.
[2 marks]3 Solution to this question by scale drawing will not be accepted.
Diagram 1 shows the straight lines AN and AL.
Given that the equation of the straight line AN is 042 xy and LAN=900.
(a) Find the equation of the straight line AL.[4 marks]
(b) The straight line LA is extended until it intersects the y-axis at point B such thatLA : AB = 1 : 2 . Find the coordinates of point L.
[3 marks]
042 xy
DIAGRAM 1
y
x
O
A
L
N
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4 Table 1 shows the scores obtained by a group of 50 students in a competition.
Score Number of students
10 8
20 x
30 9
40 6
50 y
60 6
TABLE 1
(a) Calculate the value of x and of y, given the mean of the scores is 31.
[4 marks]
(b) Hence, determine the variance of the students’ score.[3 marks]
5 Diagram 2 shows the triangle AOB and the sector of a circle , AOC with thecenter O.
DIAGRAM 2
It is given that the radius of the sector is 12 cm and AOC 0.52 rad.[ Use = 3.142 ].
Find(a) the length of the arc AC ,
[2 marks](b) the area of the shaded region.
[5 marks]
B
C
OA12
0.52 rad
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6 Given that a curve y = f(x) with the gradient function,13
x
x
dx
dy.
(a) Find2
2
dx
yd.
[3 marks]
(b) The straight line 2y = kx – 7 is normal to the curve y = f(x) at x = 1, findthe value of k .
[4 marks]
Section B
[40 marks]
Answer four questions from this section.
7 Given that the function nxxg 3: and3
22:1 mxxg , where n and m are
constants.
Find
(a) the value of m and of n, [3 marks]
(b) g -1 (8) , [2 marks]
(c) g 2 (-2) , [2 marks]
(d) the value of t if g(2t) = gg -1( 3 – 2t ). [3 marks]
8 (a) Solve the equation 3 x + 1 = 7[ 2 marks]
(b) Given that x = a m and y = a n , show that loga
2
3
ay
x= nm 21
3
1 .
[3 marks]
(c) The relation between the variable x and y is given by the equation
(3 y ) ( 9 2x ) = 1 and )2(log1log 33 xy .
Find the value of x and of y which satisfy both the equations.[5 marks]
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9 Diagram 3 shows a trapezium OABC.
DIAGRAM 3
The equation of the straight line AB is 123 xy and the equation of the straight line
BC is 1253 xy .
(a) Find(i) the coordinates of B ,
(ii) the area of the trapezium OABC,
(iii) the equation of the straight line that passes through point C and perpendicularto the straight line AB.
[7 marks]
(b) The straight line AC is extended to point D ( 5 , k ) such that AC = m CD.Find the value of k and of m.
[3 marks]
10 It is given that 523
2 23 xxy is an equation of a curve.
Find
(a)dx
dy, [2 marks]
(b) the turning point of the curve,[4 marks]
(c) the rate of change of y at ( 1, 4) when the rate of change of x is 3 units per second,
[2 marks](d) the approximate change in y, if x decreases from 1.0 to 0.95 .
[2 marks]
x
y
B
C
O
(3,1)
A(0,4)
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11 Table 2 shows the frequency distribution of the marks obtained by 40 pupils in atest.
Marks Frequency1 – 10 611 – 20 921 – 30 1431 – 40 741 – 50 4
TABLE 2
(a) Use the graph paper to answer this question.
Using a scale of 2 cm to 10 marks on the horizontal axis and 2 cm to 2 pupils onthe vertical axis, draw a histogram to represent the frequency distribution of themarks. Find the mode mark.
[4 marks](b) Without using an ogive, calculate the median mark.
[3 marks](c) Determine the standard deviation of the marks of the pupils .
[3 marks]
Section C
[20 marks]Answer two questions from this section.
12. Table 3 shows the prices and weightages of four types of items A, B , C and D.
Price(RM) Price(RM)Item
Year 2003 Year 2005Price Index Weightage
A 7.00 8.40 w 100B 13.50 x 130 80C y 13.00 115 70D 11.00 12.10 110 z
TABLE 3(a) Calculate the value of w, x and y.
[4 marks]
(b) The composite index of these items for the year 2005 based on the year 2003is 120.Calculate the value of z.
[3 marks]
(c) The total cost of all these items is expected to increase by 20% from the year2005 to the year 2007. Find the expected composite index for the year 2007based on the year 2003.
[3 marks]
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13. Diagram 4 shows two triangles ABC and CDE. The two triangles are joined at C suchthat AE and BD are straight lines. The CED is an obtuse angle.
DIAGRAM 4
(a) Calculate
(i) ACB ,
(ii) DEC .[5 marks]
(b) The straight line CE is extended to F such that DE = DF.Find the area of triangle CDF.
[5 marks]
END OF THE QUESTION PAPER
A
4 cm 7 cm
B 5 cm C D
E
9 cm
6.5 cm
PEPERIKSAAN DIAGNOSTIK ADDITIONAL MATHEMATICS - 1 -SBP 2007 Paper 2
SBP 2007
SULIT3472/2AdditionalMathematicsPaper 2Okt2007
SEKOLAH BERASRAMA PENUHKEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN AKHIR TAHUNTINGKATAN 4
2007
ADDITIONAL MATHEMATICS
PAPER 2
MARKING SCHEME
This marking scheme consists of 8 printed pages
PEPERIKSAAN DIAGNOSTIK ADDITIONAL MATHEMATICS - 2 -SBP 2007 Paper 2
SBP 2007
NO SOLUTION AND MARK SCHEME MARKSSUB
MARKS TOTAL
y = 5 – 3x or2
5
3
5 2
xyor
yx
P1
05)35(22 xx or
52
53
0523
5
2
2
xxor
yy
K1
)1(2
)5)(1(466 2 x or
)1(2
)70)(1(42828 2 y
K1
x = 0.742 , - 6.742 or y = 2.774 , 25.226 N1
1
y = 2.774 , 25.226 or x = 0.742 , - 6.742 N1
2
532)( 2 xxxf
K1
5)2
3()
2
3(2)( 22
xxf
K1
2 (a)
23 19
22 2
x
N1
3
(b)2
19,
2
3 kh
N1 , N1 2
)5)(2(462 t < 0
t <2
19
K1
N1 2
A = (4 , 0)
12,2
12121 mmusemm
P1
K1
y-0 = -2 (x - 4)K1
3 (a)
y = -2x + 8N1
4
(c)
5
7
PEPERIKSAAN DIAGNOSTIK ADDITIONAL MATHEMATICS - 3 -SBP 2007 Paper 2
SBP 2007
NO SOLUTION AND MARK SCHEME MARKSSUB
MARKS TOTAL
B ( 0,8) P1
3
)1(820
3
)1(024
yor
x K1
3(b)
L (6,-4)N1 3
3150
)6(6050)6(40)9(3020)8(10
yx K1
506698 yxK1
6052
4222
yx
yx(solve the simultaneous equations)
K1
4(a)
y =6 , x = 15N1 4
6110013050)( 22 xorxxf N1
Using
f
xxf 22 )(
0r2
2
xfx
fx
=50
130500r 961
50
61100
K1
(b)
= 261N1
3
s = 0.52 x 12 K15(a)
= 6.24 N1 2
tan 0.52 radian =12
AB
AB = 6.8697
K1
12869762
1OABofAreaL1 . K1
).()(torsec 520122
1OACofAreaL 2
2
K1
Area of the shaded region = L1 - L2
= 41.2182 - 37.44K1
(b)
= 3.7782 N1 5
7
7
7
PEPERIKSAAN DIAGNOSTIK ADDITIONAL MATHEMATICS - 4 -SBP 2007 Paper 2
SBP 2007
NO SOLUTION AND MARK SCHEME MARKSSUB
MARKS TOTAL
31 dx
dvand
dx
du P1
2)13(
)3)(()1)(13(
x
xx K1
6 (a)
2)13(
1
x N13
2
1,
221 m
km
P1,P1
12
1
2
k
K1
(b)
k = -4N1
4
3
3
nyx
ynx
K1
3
2
32
3
1
norh
K1
7(a)
6
1h and n = 2
N1
3
3
2)8(
3
1)8(1 g
K1(b)
= 2 N1
2)2(3)2( g or
2)23(3)2(2 xg
K1(c)
g ( -4) = -10 or 10)2(2 g N14
1(2 ) 3(2 ) 2 (3 2 ) 3 2g t t or gg t t K1
3 (2t ) + 2 = 3 – 2t K1
(d)
t =8
1 N1
3
7log3log)1( x K18(a)
x = 0.7712 N1 2
7
10
PEPERIKSAAN DIAGNOSTIK ADDITIONAL MATHEMATICS - 5 -SBP 2007 Paper 2
SBP 2007
NO SOLUTION AND MARK SCHEME MARKSSUB
MARKS TOTAL
23
1
logloglog yax aaa
(Use any law of logarithm)K1
=
23
1
logloglog naa
ma aaa
K1
(b)
= nm
213
N1
3)2(22 39 xx OR 031 OR 3log1 3 P1
0)2(2 xyK1
)2(3 xy OR 32
x
y K1
042)63( xx (solve simultaneously)K1
(c)
x = 2, y = 0 N1 5
Solve simultaneously:1253
123
xy
xyK19(a)(i)
B (6,6) N1
624182
1
K1a(ii)
= 18N1
m= -3 K1
)3(31 xyK1
(iii)
103 xyN1 7
3
1
510
m
mOR
1
1
14
m
mkK1
(b)
1,2
3 km ,
N1N1 3
10 (a) xxdx
dy42 2 K1N1
2
042 2 xx K1(b)
0)2( xxK1
10
10
PEPERIKSAAN DIAGNOSTIK ADDITIONAL MATHEMATICS - 6 -SBP 2007 Paper 2
SBP 2007
NO SOLUTION AND MARK SCHEME MARKSSUB
MARKS TOTAL
(0,-5) and
3
23,0 N1 ,
N14
Using
dx
dy
dt
dx
dt
dy= )42(3 2 xx
K1(c)
= - 6 N12
Using xdx
dyy (-2)(-0.05)
K1(d)
=0.1 N1 2
Histogram (Refer to Graph)All frequencies and label for x-axis correct for atleast 3 bars
N1
Correct HistogramN1
Correct methodK1
11(a)
Mode= 24.5N1
4
L = 20.5 or fm = 14 or F = 15 N1
1014
152
20
5.20
mK1
(b)
= 16.93 N13
40
960x =24
K1
40
5510 or 2)24(
40
28550
K1
( c )
= 11.74N1
3
(i) Using 1002003
2005 Q
QI
K1
w = 120N1
x = 17.55 N1
12(a)
y = 11.30 N1
4
100(*120) + 130(180) + 115(70) + 110(z) K1
z
z
250
)(110)70(115)80(130)120(*100120
K1
(b)
z = 45 N1 3
( c) 120 P1
10
10
PEPERIKSAAN DIAGNOSTIK ADDITIONAL MATHEMATICS - 7 -SBP 2007 Paper 2
SBP 2007
NO SOLUTION AND MARK SCHEME MARKSSUB
MARKS TOTAL
120100
120I
K1
= 144 N13
)5)(7(2
457cos
222 ACB
K113(a)(i)
ACB = 34.050 N1
5.6
05.34sin
9
sin
CED K1
50.830 N1a(ii)
CED = 129.170 N1
5
)(2180
180
180
DEFEDFOR
CEDDEFOR
CEDACBCDE
K1
CDE = 16.780 ,DEF = 50.830 andDEF = 78.340 N1
L1 = Area of ∆ CDE = CDEsin5.692
1
OR
L2 = Area of ∆ EDF = EDFsin5.65.62
1
K1
Area of ∆ CDF = L1 + L2 K1
= 29.13 N1
5
ALTERNATIVE
)(2180
180
180
DEFEDFOR
CEDDEFOR
CEDACBCDE
K1
CDE + EDF K1
CDF = 95.120
N1
Area of ∆ EDF = CDFsin5.692
1 K1
(b)
= 29.13 N1 5
10
10
PEPERIKSAAN DIAGNOSTIK ADDITIONAL MATHEMATICS - 8 -SBP 2007 Paper 2
SBP 2007
marks
2
4
6
8
12
14
Fre
qu
ency
0.5 10 .5 20.5 30.5 40.5 50.5
10