Post on 05-Apr-2018
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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 ac x
a
=
2 a m a n = a m + n
3 a m a n = a m - n
4 (a m) n = a nm
5 log a mn = log am + log a n
6 log a nm
= log am - log a n
7 log a mn = n log a m
8 log ab =ab
c
c
loglog
CALCULUS
1 y = uv ,dxdu
vdxdv
udxdy
+=
2vu
y = , 2vdxdv
udxdu
v
dydx
= ,
3dxdu
dudy
dxdy
=
3 A point dividing a segment of a line
( x,y) = ,21
+
+
nmmxnx
+
+
nmmyny 21
4 Area of triangle
= )()(21
312312133221 1 y x y x y x y x y x y x ++++
1 Distance = 2212
21 )()( y y x x +
2 Midpoint
( x , y ) =
+
2
21 x x ,
+
2
21 y y
GEOMETRY
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3
STATISTIC
71
11
w I w
I
=
1 Arc length, s = r
2 Area of sector , L = 212
r
TRIGONOMETRY
3 C
c
B
b
A
a
sinsinsin==
4 a2 = b2 + c2 - 2bc cos A
5 Area of triangle = C ab sin2
1
1 x = N
x
2 x = f fx
3 = N
x x 2)(=
2_2
x N
x
4 =
f
x x f 2)(=
22
x f
fx
5 m = C f
F N L
m
+ 2
1
6 10
100Q
I Q
=
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Answer all questions.
1. Diagram 1 shows the linear function f .
(a) State the value of k .
(b) Using function notation, write a relation between set A and set B.[ 2 marks ]
Answer : (a) ..
(b) ......
2.
The following information above refers to the functions f and g .
(a) State the value of h.
(b) Find the value of )3(1 fg .
[ 4 marks ]
Answer : (a) .
(b) .............................. 4
2
2
1
For examiners
use only
h x x
x x f
+ ,
34
:
x xg 21:
1
3
5
7
3
5
k
9
x f ( x)
Set A Set B
Diagram 1
f
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3. Given the function 12: x x f and 16: + x x fg . Find
(a) the function g( x)
(b) the value of x when 4)( = xgf .
[ 4 marks ]
Answer : (a).........
(b).....................................
4. Given that the roots of the quadratic equation k x x = 512
are 3 and p .Find the value of k and of p .
[4 marks ]
Answer : k =.........
p =....................
5. The quadratic equation 0122 2 =+ p x x has two distinct roots. Find the range of values of p .
[2 marks]
Answer : .............
For examiners
use only
4
3
4
4
2
5
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x
5
Diagram 2
)4,1( +k
y
6. Diagram 2 shows the graph of the function p x y += 2)3( where p is a constant and)4,1( +k is a minimum point.
Find
a) the value of k .
b) the value of p.[ 2 marks ]
Answer : (a) ................................
(b) .................................
___________________________________________________________________________
7 . Find the range of values of x for which )52(3 x x .
[3 marks ]
Answer : ..................................
8. Solve the equation 455 12 =++ x x .
[3 marks ]
Answer : ..
2
6
For examiners
use only
3
7
3
8
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9. Given x p =3log and y p =2log , express 12log 8 in terms of x and y.
[3 marks ]
Answer : ..................................
10. Solve the equation 0)1(log)23(log 33 =+ x x
[4 marks ]
Answer : ...................................
11 . The points )2,(,),( r Pt t A and )2,9( B are on a straight line. P divides AB internally in the ratio of 3 : 4 . Find the value of t and of r .
[3 marks ]
Answer : t = ..............................
r = ..............................
3
9
For examiners
use only
4
10
3
11
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12. Diagram 3 shows a straight line PQ with equation 01234 =+ y x .
Find
(a) the value of h and of k
(b) the equation of PQ in intercept form.[3 marks]
Answer : (a) .........
(b) .. .......................................
13 . Find the equation of the straight line that passes through a point )1,3(P and is
perpendicular to the straight line 145
=+ y x
.
[4 marks ]
Answer: ............
3
12
For examiners
use only
4
13
P (k ,0)
Q(0,h)
x
y
O
Diagram 3
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14. The coordinates of point P and Q are )1,3( and )10,6( respectively. The point X movessuch that XP : XQ = 2 : 3 . Find the equation of the locus of X .
[3 marks ]
Answer: ...........................
___________________________________________________________________________
15. Table 1 shows the distribution of the weight of 40 pupils in form 4 Alpha.
Table 1
(a) Find the range of the weight.
(b) Without drawing an ogive, calculate the median of the distribution of weight.
[4 marks ]
Answer : (a) .
(b) ......
Weight (kg) Number of pupils31 35 736 40 441 45 846 50 751 55 656 60 461 65 4
4
15
For examiners
use only
3
14
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16 . The mean of ten numbers is m . The sum of squares of the number is k and thestandard deviation is 4. Express k in terms of m .
[2 marks ]
Answer : .
17. Diagram 4 shows a circle with centre O .
The length of the minor arc AB is 3.9275 cm and the angle of the major sector AOB is
315o
. Using 142.3=
, find
(a) the value of , in radians,(Give your answer correct to four significant figures.)
(b) the length, in cm, of the radius of the circle.
[3 marks ]
Answer : ......
___________________________________________________________________________
2
16
3
17
For examiners
use only
O
A
B
Diagram 4
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18 . Diagram 5 shows a sector OTV of a circle, centre O .
Find the perimeter of the shaded region.
[3 marks ]
Answer : ...
___________________________________________________________________________ 19 . Diagram 6 shows a sector of a circle OPQ with centre O and OPR is a right angle
triangle.
Find the area, in cm 2, of the shaded region.
[4 marks ]
Answer:
For examiners
use only
3
18
O R Q1 cm
5 cm
P
4
19
O
T
V 50
7 cm
Diagram 5
Diagram 6
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20 . Evaluate the following limits,
(a)1
1lim
1 + x x
(b)39
lim2
3
x x
x
[ 3 marks ]
Answer: (a) .............
(b) .................................
___________________________________________________________________________
21 . The straight line 12 += x y is the tangent to the curve x x y 42 = at the point P .Find the x-coordinate of the point P .
. [2 marks ]
Answer : ..
22 . Given that xq
px y = 2 and ,4
2 x x
dxdy
+= where p and q are constants , find the
value of p and of q .
[3 marks ]
Answer : ..
2
21
3
22
For examiners
use only
3
20
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23. Given the curve 221 x
x y += .
(a) Find the coordinates of the turning point.
(b) Hence, determine whether it is a maximum or a minimum point.
[4 marks ]
Answer: (a)............
(b)....................................
___________________________________________________________________________
24. A cylinder has a fixed height of 10 cm and a radius of 5 cm. If the radius decreases by0.05 cm, find (in terms of )
(a) the approximate change in the volume of the cylinder,
(b) the final volume of the cylinder.[4 marks ]
Answer : (a)
(b)
For examiners
use only
4
24
4
23
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25. Given x
x y
2)2( = , find the value of 2
2
dx yd
when 1= x .
[4 marks]
Answer: ............
END OF QUESTION PAPER
4
25
For examiners
use only
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PEPERIKSAAN AKHIR TAHUN TINGKATAN 4 2008MARK SCHEME KERTAS 1
No. Solution and mark scheme Sub marks Full marks 1 (a) k = 7
(b) 2: + x x f or 2)( += x x f
1
1
2
2 (a) h = - 3
(b) - 2
B2 :31)1(4
+
B1: 2
1)(
1 x xg
=
1
3
4
3 (a) 13)( += x xg
B1 : 161)(2 += x xg
(b) 1= x
B1 : 41)12(3 =+ x
2
2
4
4 p = 2, k = 7 (both)
B3 : p + 3 = 5 and 3 p = k 1
B2 : p + 3 = 5 or 3 p = k 1
B1 : 0152 =+ k x x
4 4
5
21
> p
B1 : 0)1)(2(4)2( 2 >+ p
2 2
6 k = -4
p = 4
1
1
2
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3
7
21
,3 x x (both)
B2 : 0)3)(12( + x x
B1 : 0352 2 x x
3 3
8 x = -1
B2 : 155 = x
B1 : 45555 12 = x x
3 3
9
y y x
32+
B2 :2log3
2log2log3log
p
p p p++
B1 : 32log
)223(log
p
p
3 3
10
23
= x
B3 : 123 =+ x x
B2 : 03123
=
+
x x
B1 : 0123
log 3 =+
x x
4 4
11 r = -1
B2 : t = 5
B1 :7
462
t += or
7274
=t
r
3 3
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4
12 a) h = 4
k = 3
b) 143
=+y x
1
1
1
3
134
1945
+= x y or equivalent
B3 : )3(45
)1( += x y
B2 :45
2=m
B1 :54
1=m
4 4
14 04546210255 22 =+++ y x y x
B2 : 22 )1()3(3 ++ y x = 22 )10()6(2 + y x
B1 : 3 XP = 2 XQ
3 3
15 (a) 30
(b) 46.21
B2 : m = 45.5 + 57
192
40
(all values correct)
B1 : 45.5 ,19 , 7 , 5 (at least two are correct)
1
3
4
16 210160 mk +=
B1 : 210
16 mk
=
2 2
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5
17 a) 0.7855
b) r = 5
B1 : 3.9275 = r (0.7855)
1
2
3
18 Perimeter = 12.0263
B2 : length PQ = 2 r sin 25 0 = 5.9167
B1 : arc PQ = r = 7 (0.8728) = 6.1096
3 3
19 Area = 2.045
B3 : )3)(4(21
)6436.0)(5(21 2
B2 : = 0.6436 rad
B1 : tan =43
4 4
20(a)
21
(b) 6
B1 :3
)3)(3(lim
3
+
x x x
x
1
2
3
21 1= x
B1 : 42 = xdxdy
2 2
2221
= p and 4=q
B2 :21
= p or 4=q
B1 : 22 xq
pxdxdy
+=
3 3
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6
23(a)
23
,1
B1 : 01 3 = x
(b) ,0322
>=dx
yd minimum point
B1 : 422
3 = xdx
yd
2
2
4
24 (a) 5
B1 : )5(20 =dr dV
or 100
(b) 245
B1 : 5)5(10 2 =newV
2
2
4
25 8
B3 : 322
8 = xdx
yd or 4
22 )2)(4()2( x
x x x x
B2 : 241 = xdxdy
or 22)2()2(2
x x x x
B1 : 144 += x x y
4 4
END OF MARK SCHEME
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3472/2AdditionalMathematicsKertas 22 jamOKT 2008
SEKTOR SEKOLAH BERASRAMA PENUHBAHAGIAN PENGURUSAN
SEKOLAH BERASRAMA PENUH / KLUSTERKEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN AKHIR TAHUNTINGKATAN 4 2008
ADDITIONAL MATHEMATICS
Kertas 2
Dua jam tiga puluh minit
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
1. This question paper consists of three sections : Section A , Section B and Section C.
2. Answer all question 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 diagram 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 to 3.
8. A booklet of four-figure mathematical tables is provided.
9. You may use a non-programmable scientific calculator.
Kertas soalan ini mengandungi 13 halaman bercetak
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The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used.
ALGEBRA
1
aacbb
x2
42 =
5 nmmn aaa logloglog +=
2 nmnm aaa += 6 nmnm
aaa logloglog =
3 nmnm aaa = 7 mnm an
a loglog = 4 ( ) mnnm aa = 8
a x
xb
ba log
loglog =
KALKULUS ( CALCULUS )
STATISTIK ( STATISTICS )
1 uv y = ,dxdu
vdxdv
udxdy
+=
2vu
y = , 2vdxdv
udxdu
v
dx
dy
=
3dxdu
dudy
dxdy
=
1 N
x x
=
2
= f
fx x
3( ) 2
22
x
N
x
N
x x==
=
4( ) 2
22
x f
fx
f
x x f =
=
6 C f
F N L M
m
+= 2
1
5 1000
1 =Q
Q I
7 =
i
ii
W I W I
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GEOMETRI ( GEOMETRY )
TRIGONOMETRI ( TRIGONOMETRY )
1. Panjang lengkok, js = Arc length , r s =
2. Luas Sektor, 21 2 j L =
Area of sector , 21 2r A =
3.
4.
5.
C c
Bb
Aa
sinsinsin==
Abccba kos2222 += Abccba cos2222 +=
12
Luas segitiga ( )sin
Area of triangleab C =
1. Jarak ( Distance )
( ) ( )221
2
21 y y x x +
2. Titik tengah ( Midpoint )
++
=2
,2
),( 2121 y y x x
y x
3. Titik yang membahagi suatutembereng garis
( A point dividing a segment of a line )
+
+
+
+=
nm
myny
nm
mxnx y x 2121 ,),(
4. Luas segitiga ( Area of triangle ) =
( ) ( )31231213322121
y x y x y x y x y x y x ++++
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SECTION A
[40 marks ]
Answer all questions in this section
1. Solve the simultaneous equations
425 =+ y x
16232 =+ y x x [5 marks ]
2. Express f ( x) = 1 6 x + 2x 2 in the form f ( x) = m( x + n)2 + k , where m, n and k areconstants.
(a) State the values of m, n and k .[3 marks ]
(b) Find the maximum or minimum point.
[1 marks ]
(c) Sketch the graph of f ( x) = 1 6 x + 2 x2.
[2 marks ]
3. (a) The straight line y = 1 tx is a tangent to the curve y2
3 y + 3 x x2
= 0.Find the possible value of t .
[3 marks ]
(b) Given p and q are the roots of the quadratic equations 2 x2 + 7 x = m 5,where pq = 3 and m is a constant. Calculate the values of m, p and q.
[4 marks ]
4 (a) Given that 29 . 27 1 x y = , find the value of x when 4 y =
[2 marks ](b) Solve the equation 3 23 .6 18 x x x=
[2 marks ]
(c) Solve the equation 2 4log ( 2) 2 2 log (4 ) x x = +
[3 marks ]
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5 (a) A set of N numbers have a mean of 8 and standard deviation of 2.121. Given thatthe sum of the numbers, x , is 64. Find
(i) the value of N
(ii) the sum of the squares of the numbers.
[4 marks ]
(b) If each of the numbers is divided by h and is added by k uniformly , the new mean
and standard deviation of the set are 5 and 1.0605 respectively.
Find the value of h and k .
[4 marks ]
6. The gradient of the tangent to the curve 2qx px y = , where p and q are constants, atthe point (1 , 4) is 2.Find
(a) the value of p and q.[4 marks ]
(b) the equation of normal to the curve at the point ( 2, 4)[3 marks ]
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SECTION B
Answer four questions in this section
7 Diagram 1 shows function g maps x to y and function h maps z to y.
Given g ( x) = m x x
,12
1 and h ( z) = 1 + 4 z.
(a) State the value of m.[1 marks ]
(b) Find
(i) gh -1( x) (ii) h g -1(1).
[6 marks ] (c) Find the value of
(i) a(ii) b
[3 marks ]
Diagram 1
x y z g h
a
b
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8. Diagram 2 shows a quadrilateral PQRS with vertices ( 2,5) R
and ( 1,1)S
.
Given the equation of PQ is 1474 = x y . Find ,
(a) the equation of QR[4 marks ]
(b) the coordinates of Q [2 marks ]
(c) the coordinates of P[1 marks ]
(d) the area of quadrilateral PQRS[3 marks ]
R( 2, 5)
S( 1, 1)
P
Q
O
Diagram 2
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9
Table 1 shows the total time spent on watching television by 120 students for a period of 3 weeks. Calculate ,
(a) the mean,[2 marks ]
(b) the standard deviation,[3 marks ]
(c) the third quartile, [5 marks ]
of the distribution
Total Time (hours) Number of students
5 14 12
15 24 17
25 34 26
35 44 31
45 54 16
55 64 10
65 74 8
Table 1
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10. Diagram 3 shows a circle ABCF with radius 6 cm and centre O .
Given that oODB 30= , EBD is the tangent to the circle and OD = OE = 12 cm.Find ,
(a) the length of BD
[3 marks ] (b) the area of shaded region[3 marks ]
(c) the perimeter of the whole diagram[4 marks ]
O
A
D
E
B
C
F
Diagram 3
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11Diagram 4 shows a rectangle PQRS and a parallelogram ABCD .
(a) If L 2cm is the area of the parallelogram,
(i) Show that 22 16 L x x= +
(ii) Find the value of x when L is maximum.
(iii) Find the maximum area of the parallelogram ABCD .[7 marks ]
(b) Given that the rate of change of x is -10.1 cms , find the rate of change of L ,in -1cms , when x is 2 cm.
[3 marks ]
B P Q
A
S D
C
R
x cm
x cm
x cm
x cm
10 cm
6 cm
Diagram 4
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SECTION C Answer two questions in this section
12. Diagram 5 shows two triangles ABC and ACD . BCD is a straight line.
Find ,
(a) ADC [3 marks ] (b) the length of CD
[3 marks ] (c) the area of triangle ABD
[4 marks ]
13
Diagram 6 shows two triangles ABE and BCF , where ABC is a straight line.Given that AE = 5 cm, BE = 7 cm, BC = 8 cm, CF = 9 cm, 050= BAE , 0104 EBF = and .100 0= BCF Calculate
(a) AEB ,[3 marks ]
(b) the length of BF. [3 marks ]
(c) if point E joint with F, find the area of the quadrilateral ACFE .[4 marks ]
A
DC B
55o
8 cm8 cm9 cm
Diagram 5
50o 104 0 100
o
A B C
EF
Diagram 6
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14 Table 2 shows the prices and the price indices of five types of food, H, I, J, K and Lrepresented the cost of food. Diagram 7 shows a percentage according to the foodspyramid.
Types of foodPrice (RM) for the year
Price index for the year 2008based on the year 2006 2006 2008
H 2.20 2.75 125 I m 2.20 110 J 5.00 7.50 150K 3.00 2.70 n
L 2.00 2.80 140
(a) Find the value of m and of n. [3 marks ]
(b) Calculate the composite index for the cost of food in the year 2008 based on theyear 2006.
[3 marks ] (c) The price of each food increases by 30% from the year 2008 to the year 2009.
Given that the cost of food in the year 2006 is RM80, calculate the correspondingcost in the year 2009.
[4 marks ]
L
I 10 %
K 20 %
J 25 %
H 40 %
Table 2
Diagram 7
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3472/2 2008 Hak Cipta SBP [ Lihat sebelahSULIT
15
Table 3 shows the price indices and percentage of usage of four items, J, K, L and M, which are the main ingredients in the production of a brand of cake.
(a) Calculate(i) the price of item M in the year 2003 if its price in the year 2005
was RM2.50.
(ii) the price index of item J for the year 2005 based on the year 2001 if its price index for the year 2003 based on the year 2001 is 108.
[5 marks ]
(b) The composite index of the cost of cake production for the year 2005based on the year 2003 is 113. Calculate ,
(i) the value of t
(ii) the price of a cake in the year 2003 if its corresponding price in the year2005 was RM 25,
[ 5 marks ]
END OF THE QUESTIONS
Item Price Index for the year 2005Based on the year 2003Percentage of usage (%)
J 118 22
K t 12
L 108 31
M 113 35
Table 3
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SKEMA PERMARKAHAN MATEMATIK TAMBAHAN KERTAS 2
PEPERIKSAAN DIAGNOSTIK TINGKATAN 4, 2008
Number Solution and mark scheme Sub Marks Full Marks1
4 5 4 2@2 5 x y y x
= =
2
2
4 53 2 16
2
4 2 4 2@ 3 2 16
5 5
x x x
y y y
+ =
+ =
( )( ) ( )( )10 2 0@ 27 3 0 x x y y + = + = 10, 2 and 27,3 x y= =
P1
K1
K1N1, N1
55
2 (a)2
3 72
2 2 x
m = 2 , n = 32
and k = 72
K1
N 0, 1, 23
6
(b) 72
3( , )
2 N1 1
(c)
ShapeMinimum point and y-intercept
P1P1 2
y
( 32
, 72
)
o x
1
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Number Solution and mark scheme Sub Marks Full Marks3 (a)
(1 tx)2 3(1 tx) + 3 x x2 = 0(t 2 1) x2 + ( t + 3) x 2 = 0
(t + 3) 2 4( t 2 1)(2) = 0(3t + 1) (3 t + 1) = 0
.31
=t
K1
K1
N1 3
7
(b)SOR or POR pq or p + q
POR : pq = 35
32
1.
.
m
m
=
=
7
:2
32
23
22
.
SOR p q
q or
p or
+ =
=
=
K1
N1
N1
N1 4
4 (a)
3
0343)3()3(
0322
=
=+=
x
y x
y x
K1
N1 2
7
(b)3 218 18
3 2
1
x x
x x
x
=
=
=
K1
N1 2
(c) 4log)4(log22)2(log
2
22 x x +=
Change base
)4(log4log)2(log 222 x x +=
( 2) 4(4 )
3.6
x x
x
=
=
K1
K1N1 3
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Number Solution and mark scheme Sub Marks Full Marks5 (a)
i)64
8
8 N
N
=
=
ii)
=
=
99.547
88
121.2
2
22
x
x
K1N1
K1
N1 4
8
(b)New mean = 5
8=+ k
h
or
New Stnd. Deviation = 0605.1121.2 =h
h = 2
85
21
k
k
+ =
=
K1
N1
K1
N1 4
6 (a)
)1(22
2
q p
qx pdxdy
=
= 1
Substitute ( 1, 4 ) in 2qx px y = 4 = p q 2
2 1 ; q = 2 and p = 6
K1
K1
K1
N1 4
7
(b) x = 2, )2(46=
dxdy
21
2
2
1
=
=
m
m
321
)2(21
4
+=
=
x y
x y
K1
K1
N1 3
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Number Solution and mark scheme Sub Marks Full Marks7 (a)
.21
=m P1 1
10
(b)(i)
41
)(1
= z
zh
14
12
1)(1
=
x xgh
.3,3
2
= x
x
(ii) 1 = g( x)
1
1
11
2 11
(1) (1) 1 4(1)
(1) 5
x x
hg h
hg
=
=
= = +
=
K1
K1
N1
K1
K1
N1 6
(c) (i)
1( )41
12( ) 1
42
a g
a
a
=
=
=
(ii)( )
1 4 2
34
h b a
b
b
=
+ =
=
K1
N1
N1 3
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Number Solution and mark scheme Sub Marks Full Marks8 (a) 4 RS M =
14QR
M =
( )15 24
y x = + or other suitable method
4 22 y x= + or equivalent
P1
P1
K1
N1 4
10
(b) Solve simultaneous equation22 7 14 x x+ =
(6,7)Q
K1
N1 2
(c)(2,0)P N1 1
(d) Area of quadrilateral PQRS 2 6 2 1 210 7 5 1 02
1[2(7) 6(5) ( 2)(1) ( 1)(0)] [6(0) ( 2)7 ( 1)5 2(1)]
229.5
=
= + + + + + +
=
K1
K1
N1 3
9 (a)
12(9.5) 17(19.5) 26(29.5) 31(39.5) 16(49.5) 10(59.5) 8(69.5)120
x + + + + + +=
= 36.5
K1
N1 2
10
(b)2
222
)5.36(120
)5.69(8...)5.19(17)5.9(12
+++=
= 266= 16.31
K1
K1N1 3
(c) L = 44.5 ; F = 86
3
3(120) 86
444.5 1016
Q
= +
K1 lower boundry 44.5
K1 using formula= 44.5 + 2.5 = 47
P1, P1
K1
K1N1 5
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Number Solution and mark scheme Sub Marks Full Marks10 (a)
060 6012
DB DOB or Sin = =
0tan 606
10.3923
DB
DB
=
=
P1
K1
N1 3
10
(b) 2 10.3923
20.7846
DE =
=
Area of shaded region
( )( ) ( )( )21 120.7846 6 2.0944 62 224.6546
=
=
N1
K1
N1 3
(c) 0240 or 4.1888 rad Major AOC =
( )6 4.188825.13228
AFC S =
=
Perimeter
25.1328 6 6 20.7846
57.9174
AFC S AD CE DE = + + +
= + + +
=
P1
N1K1N1 4
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Number Solution and mark scheme Sub Marks Full Marks
11 (a)(i)
= )6)(10(
2
12
2
12)6(10 2 x x x L
= x x 162 2 +
(ii)
164 += xdxdL
,0=dxdL
0164 =+ x
x = 4
(iii)2
max 2(4) 16(4)
32
L = +
=
K1
N1
K1
K1
N1
K1
N1 7
10
(b)0.1
dxdt
=
( 4(2) 16) 0.1dLdt
= +
= 0.8
P1
K1
N1 3
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Number Solution and mark scheme Sub Marks Full Marks
12 (a)55
89Sin BCASin
=
0 0
112 51' or 67 9 ' BCA ACD = =
067 9' 67.15 ADC or =
K1
N1
N1 3
10
(b) 0
0
180 2(67 9 ')
45 42' 45.7
CAD
or
=
=
2 2 2 08 8 2(8)(8)cos 45 42 '
6.213
CD
CD
= +
=
K1
K1N1 3
(c) 0 0 0
0
180 55 112 51'
12 9 '
CAB =
=
( )( ) ( )( )0 01 1Area 9 8 sin12 9' 8 8 sin 45 42 '2 230.48
= +
=
N1
K1, K1
N1 4
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Number Solution and mark scheme Sub Marks Full Marks
13 (a)5 7
50SinB Sin=
33.17 33 10' ABE or =
180 50 33.17 AEB =
= 96.83 or 96 50
K1
K1
N1 3
10
(b) 42.83 42 50' CBF or = 9
sin100 sin 42.83 BF
=
BF = 13.04 cm
OR equivalent
P1
K1
N1 3
(c)Area AEB =
15 7 sin 96.83
2 or equivalent
= 17.38
Area BCF = 1
8 9 sin1002
or equivalent
= 35.45
Area BEF = 1
7 13.04sin1042
= 44.28
Area of quadrilateral ACFE = 100.08
K1
K1
K1
N1 4
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Number Solution and mark scheme Sub Marks Full Marks
14 (a)2.20
100 110
2.002.70
1003.00
90
m
m
n
n
=
=
=
=
K1
N1
N1 3
10
(b) (125 40) (110 10) (150 25) (90 20) (140 5)100
12350100
123.5
IW
W
+ + + + =
=
=
K1
K1
N1 3
(c) 08
0906
08
130 123.5100 100 123.5
100 100 80
160.55 98.80
Q I OR
Q RM
= =
= =
0909
09
160.5580 100 130
100 98.80128.44. 128.44
QQ RM
RM Q RM
= =
= =
K1
K1
K1
N1 4
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END OF MARK SCHEME
Number Solution and marking scheme Sub Marks Full Marks
15 (a) (i)
11310050.2
03
=P
21.203 RM P = (ii)
10810001
03 =PP
11810003
05 =PP
44.127100100108
100118
=
K1
N1
K1, K1
N1 5
10
(b)
(i) 113100
)35(113)31(108)12()22(118=
+++=
t I
t = 116.75
(ii) 11310025
03
=P
12.2203 RM P =
K1, K1N1
K1
N1 5
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