final form 4 2011 ( soalan dan skema), peperiksaan akhir tahun tingkatan 4 2011
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Final Form 4 2011 (Soalan dan Skema), Peperiksaan Akhir Tahun Tingkatan 4 2011TRANSCRIPT

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SEKOLAH MENENGAH SAINS TELUK INTAN
PEPERIKSAAN DIAGNOSTIK TINGKATAN 4 2011
ADDITIONAL MATHEMATICS
Form Four
Paper 2
Two hours and thirty minutes
DO NOT OPEN THIS QUESTION PAPER UNTIL INSTRUCTED TO DO SO
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 For Section C, answer one question from 12 or 13 and one question from 14 or 15.
4 Give only one answer/solution to each question.
5 Show your working. It may help you to get marks.
6 The diagrams in the questions provided are not drawn to scale unless stated.
7 The marks allocated for each question and sub-part of a question are shown in brackets.
8 A list of formulae is provided on pages 3 and 4.
9 You may use a four-figure mathematical table.
10 You may use a non-programmable scientific calculator.
This question paper consists of 13 printed pages.
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Form Four
Additional Mathematics
Paper 2
October 2011
2 ½ hours

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The following formulae may be helpful in answering the questions. The symbols given are
the ones commonly used.
ALGEBRA
1 a
acbbx
2
42 5 nmmn aaa logloglog
2 am x a
n = a
m + n 6 nm
n
maaa logloglog
3 am
an = a
m – n 7 log a m
n = n log a m
4 ( am
)n = a
m n 8
a
bb
c
ca
log
loglog
GEOMETRY
1 Distance = 2
21
2
21 yyxx
3 A point dividing a segment of a line
nm
myny
nm
mxnxyx 2121 ,,
2 Mid point
2,
2, 2121 yyxxyx
4 Area of a triangle =
3123121332212
1yxyxyxyxyxyx
STATISTICS
1 N
xx
5 Cf
FN
Lmm
2
1
2 f
fxx
3 2
22
xN
x
N
xx
6 1
0
100Q
IQ
4 2
22
xf
fx
f
xxf
7 i
ii
W
IWI

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TRIGONOMETRY
1 Arc length, s =r
3 sin sin sin
a b c
A B C
2 Area of a sector, 21
2A r 4
2 2 2 2 cosa b c bc A
5 Area of triangle = 1
sin2
ab C
CALCULUS
1 y = uv , dx
duv
dx
dvu
dx
dy
2 2
,v
dx
dvu
dx
duv
dx
dy
v
uy
3 dx
du
du
dy
dx
dy

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Section A
[40 marks]
Answer all questions in this section.
1 Solve the simultaneous equations 2x – y – 3 = 0 and 2x² - 10x + y + 9 = 0.
[5 marks]
2 (a) Express f(x) = x2 5x + 2m + 2 in the form of f(x) = p(x + q)
2 + r.
[2 marks]
(b) The graph f(x) = x2 5x + 2m + 2 has a minimum point at (k , -
).
Find the values of k and of m.
Hence, sketch the graph for 0 x 3 and state the corresponding range for
f(x).
[6 marks]
3 It is given that and are roots of the equation x2 7x + k = 0 and + 3 and + 3
are roots of the equation x2 - mx + 25 = 0.
Find the values of k and m. [6 marks]
4 If and y = 103 as x = 2 and y = 736 as x = 3, find the values of k and n.
[6 marks]
5 A set of data consists of fifteen numbers. The sum of the numbers x, is 450 while the
sum of the squares 2x , is 13 635.
(a) Find the mean and the variance of the numbers.
[4 marks]
(b) When each of the numbers in the set of data is multiplied with m, followed by
subtracting n from it, the new mean and standard deviation are 83 and 9
respectively. Find the values of m and n.
[4 marks]
6 Given that
and
py
4
Find
(a) the rate of change of p, when x increases at a rate of 0.5 unit per second,
[2 marks]
(b) dy
dx in terms of x,
[3 marks]
(c) the approximate change in y when x changes from 5 to 4.85 .
[2 marks]

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Section B
[40 marks]
Answer four questions from this section.
7 Diagram 1 represents part of the mapping of x onto y by the function f(x) = ax + b,
where a and b are constants and part of the mapping of y onto z by the function g.
f(x) g(y)
x y z
4
1
-1 c
-2
DIAGRAM 1
Find
(a) the values of a and b,
[5 marks]
(b) f -1
(x),
[2 marks]
(c) g(x), given that 2
13 x
[2 marks]
(d) the value of c.
[1 mark]

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8 Solution to this question by scale drawing will not be accepted.
A line passes through A(-3 , 2) and B(2 , 5). The perpendicular bisector of line AB
meets the x-axis at P and the y-axis at Q.
Find
(a) the equation of the perpendicular bisector of AB,
[4 marks]
(b) the area of triangle OPQ,
[3 marks]
(c) the equation of locus R such that APR is always 90.
[3 marks]
9 Table 1 shows the frequency distribution of the marks of a group of pupils in a test.
Marks 1-20 21-40 41-60 61-80 81-100
Number of pupils 7 10 13 6 4
TABLE 1
(a) Draw a histogram for the above data and estimate the modal age.
[3 marks]
(b) Calculate the standard deviation of the distribution.
[4 marks]
(c) Without using an ogive, calculate the median age.
[3 marks]

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10 Diagram 2 shows a circle centre O with radius p cm. The area of rectangle OCTQ is
108 cm². The length CT is three times the breadth OC.
DIAGRAM 2
Using = 3142, find
(a) the length of CT,
[2 marks]
(b) angle COT in radians,
[2 marks]
(c) the area of sector OAB,
[2 marks]
(d) the perimeter of the shaded region.
[4 marks]

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11 (a) The gradient of the curve at the point (1, - 9 ) is 27. Find the
values of a and b.
[5 marks]
(b) Diagram 3 shows a closed cone with diameter of (18 – 2d) cm and height of
(6d – 21) cm. The vertex of the cone is vertically above the center of its base.
(6d – 21) cm
(18 - 2d) cm
DIAGRAM 3
Express the volume of the cone, V in terms of and d.
Hence, calculate the maximum volume of the cone, in term of .
[5 marks]

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Section C
[20 marks]
Answer two questions from this section, in which one question from 12 or 13 and one
question from 14 or 15.
12 Diagram 4 shows triangle ABC with ∠BAC = 34, AB = 7 cm, AC = 9 cm and
BC = t cm.
DIAGRAM 4
(a) Find the value of t.
[2 marks]
(b) Calculate the value of BCA.
[3 marks]
(c) Find the area of triangle ABC.
[2 marks]
(d) If AC is extended to D such that the area of triangle ABD is twice the area of
triangle ABC, find the length of BD.
[3 marks]

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13 In the Diagram 4, ABC and EDC are straight lines.
DIAGRAM 4
Given that AE = 10 cm, BD = 7 cm, BC = 5 cm, CD = 6 cm and DE = 2 cm.
Calculate
(a) BCD,
[2 marks]
(b) AEC,
[3 marks]
(c) the length of AC,
[2 marks]
(d) the area of triangle BDE.
[3 marks]

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14 A particular type of cake is made by using four ingredients K, L, M and N.
Table 2 shows the price indices for the four ingredients and the percentages of usage
of the ingredients in making the cake.
TABLE 2
(a) Calculate
(i) the price per kg of M in the year 2001 if its price per kg in the year
2003 was RM 3.25,
(ii) the price index of K in the year 2003 based on the year 1999 if its price
index in the year 2001 based on the year 1999 was 109.
[5 marks]
(b) The composite index number for the cost of making the cake in the year 2003
based on the year 2001 was 112.5.
Calculate
(i) the value of x,
(ii) the cost of making a cake in the year 2001 if the corresponding cost in
the year 2003 was RM 6.30.
[5 marks]

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15 Table 3 shows the price indices of five components A, B, C, D and E that are needed
to make an electronic equipment for the year 2003 based on the year 2001. The
changes in the price indices of the components from the year 2003 to the year 2004
and their weightages are shown.
Components Price index in 2003
based on 2001
Change in price from
2003 to 2004 Weightage
A 140 Increase 10% 3
B 130 No change 2
C 125 No change 6
D 110 Increase 20% 5
E x Decrease 5% 4
TABLE 3
(a) If the price of component B was RM8 in the year 2001, calculate its price in
2003.
[2 marks]
(b) Given that the prices of component E in the year 2001 and 2003 were RM5.50
and RM6.60 respectively, find the value of x.
[2 marks]
(c) Calculate the composite index for the cost of making the electronic equipment
in the year 2004 based on the year 2003.
[3 marks]
(d) (i) If the cost of making the electronic equipment in the year 2003 is
RM2986, calculate the cost of the electronic equipment in the year
2004.
(ii) The cost of making the electronic equipment is expected to increase by
30% from the year 2004 to the year 2009. Find the expected composite
index for the year 2009 based on the year 2003.
[3 marks]
END OF THE QUESTION PAPER

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Form Four
Additional Mathematics
Paper 2
October 2011
SEKOLAH MENENGAH SAINS TELUK INTAN
PEPERIKSAAN DIAGNOSTIK TINGKATAN 4 2011
ADDITIONAL MATHEMATICS
Form Four
Paper 2
Marking Scheme
.
This marking scheme consists of 11 printed pages.

CONFIDENTAL 14 3472/2
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No. Solution Marks
1 y = 2x – 3
2x² - 10x + (2x – 3) + 9 = 0
(x – 1)(x – 3) = 0
x = 1 , 3
y = -1 , 3
K1
K1
K1
N1
N1
5
2 (a)
(b)
k =
,
= - ¼ m = 2
y
6
x
3
-
Range :
K1
N1
K1
N1, N1
N1
Shape of
graph
N1
(0 , 6), (
,-
)
and (3,0)
N1
8
3 7 or k
( +3)( +3) = m or 259)(3
6 m 16)7(3
k = = -5
-5 = m – 6
m = 1
K1
K1
K1, K1
N1, N1
6

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4 or
n = 5
96)2( 5 k
k = 3
P1
K1
K1
N1
K1
N1
6
5 (a)
(b)
x =
= 30
22 3015
13635
= 9
and
m = 3 , n = 7
K1
N1
K1
N1
K1, K1
N1, N1
8
6 (a)
(b)
(c)
= - 1.5
342
pdx
dy
2
)10(3
12
x
= -0.008
K1
N1
K1
K1
N1
K1
N1
7

CONFIDENTAL 16 3472/2
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7 (a)
(b)
(c)
4)1()1( baf
2)1()1( baf
a = 3 dan b = 1
Let yxf )(1 xyf )(
3
1
yx
3
1)(1 x
xf
2
2313
xxg
3
113
yxxy
2
13
13
)(
y
yg
2)(
xxg
2)4( g
K1
K1
N1, N1
K1
N1
K1
N1
N1
N1
10

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8 (a)
(b)
(c)
Midpoint of AB
=
=
=
=
15
2APm dan
2
15PRm
10
5
8
15
2
x
y
5
8
2
15 xy
K1
K1
K1
N1
K1
K1
N1
K1
K1
N1
10

CONFIDENTAL 18 3472/2
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9 (a)
(b)
(c)
Uniform scale
All bars are correctly drawn
Mode = 46.5
23.98
40.5
45.12
K1
K1
N1
K1
K1
K1
N1
P1
K1
N1
10

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10 (a)
(b)
(c)
(d)
CT = 18 cm
= 71.57
= 1.249 rad
= 5.791 cm²
= 7.494 cm
= 18.9737 cm
BT = 18.9737 – 6
= 12.9737 cm
38.47 cm
K1
N1
K1
N1
K1
N1
K1
K1
K1
N1
10

CONFIDENTAL 20 3472/2
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11 (a)
(b)
cm³
K1
K1
K1
N1
N1
K1
K1
K1
K1
N1
10
12 (a)
(b)
(c)
(d)
K1
N1
K1, P1
N1
K1
N1
K1
K1
N1
10

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13 (a)
(b)
(c)
(d)
= 19.5957 cm²
= 14.6968 cm²
K1
N1
K1
K1
N1
K1
N1
K1
K1
N1
10
14
(a)(i)
(ii)
(b)(i)
(ii)
K1
N1
K1, K1
N1
K1, K1
N1
K1
N1
10

CONFIDENTAL 22 3472/2
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15 (a)
(b)
(c)
(d)(i)
(ii)
= 105.5
= 137.15
K1
N1
K1
N1
K1, K1
N1
N1
K1
N1
10
END OF MARKING SCHEME