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8/14/2019 Answer Zone C P2 Add Maths 2009
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PEPERIKSAAN PERCUBAAN BERSAMA
JAWATANKUASA KURIKULUM ZON C KUCHING
Model answer
No. Solution Marks
1 x + 2 y = 1
x = 1 2 y ( substitute intox2
+ 2y2
= 5 x y)Therefore,
( 1 2 y ) 2 + 2 y 2 = 5 ( 1 2 y ) y
1 4 y + 4 y 2 + 2 y 2 = 5 y + 2 y 2
1 4 y + 6 y 2 - 5 + y - 2 y 2 = 0
4 y 2 - 3 y 4 = 0
)4(2
)4)(4(4)3()3( 2 =y
y = 1.443 y = -0.693
x = 1 2 ( 1.443 ) x = 1 2 ( -0.693 ) = - 1.886 = 2.386
52(a) 6x -y = k
y = 6x k
The gradient of the tangent, m = 62(3 2) 3y x=
29 12 1 y x x= +
The gradient of the tangent,
tangent 18 12dy
m xdx
= =
18 12 6x =
x = 1
2(3(1) 2) 3y =
y = -2
Hence, coordinatesP= (1, -2) 5
(b) At coordinatesP= (1, -2)
6(1) (-2) = kk= 8
1
1
3 (a) Length of first arc = r = 2
Length of second arc = 5
Length of third arc = 8
This forms an AP with a = 2 and d = 3
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2
[2 ( 1) ]2
[4 ( 1)(3 )] 772
( 3 ) 154
3 154 0
(3 22)( 7) 0
227
3
( )
n
nS a n d
nn
n n
n n
n n
n or n
ignored
= +
+ =
+ =
+ =+ =
= =
(b) Radius:
2, 5, 8, a = 2, d = 3
T7 = 2 + 6(3) = 20Radius of the seventh semicircle = 20 cm
Area of the seventh semicircle =21 ( )(20)
2
= 200 cm2
7
4(a) LHS =
x
xx2sin1
sin1sin1
++
=x2cos
2
= x2sec22
(b) (i)xy 2sin=
The shape of sinx
Two period for the range .3600 x
Modulus of the graphThe amplitude is 1
(ii) Find the equation =y2
x
Draw the straight line =y2
xin the graph
Number of solutions = 86
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5(a)
18.0
20.0
16.0
14.0
12.0
10.0
8.0
6.0
0
2.0
4.0
89.569.549.529.59.5
57. 5
Histogram
Correct scale P1
Shape of Histogram P1
Correct value N1
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5 (a) Histogram
modal mark = 57.5
(b)
5(19.5) 11(39.5) 18(59.5) 8(79.5)
42
2239
53.3142
mean mark+ + +
=
= = 6
6 a) i) BD = BE + ED
= yx 103 +
ii) AD = AB + BD
1(10 ) ( 3 10 )
5
12 3
y x y
y x
= + +
=
b) CD = (m-1) AD
( 1)(12 3 )12( 1) 3( 1)m y x
m y m x= =
CD = nEB + BD
yxn
yxxn
10)33(
)103()3(
+=
++=
Comparing the coefficient of y , 12m 12 = 10, m =11
6Comparing the coefficient ofx , 3n 3 = 3 3m.
3n = 6 3(11
6
)
n =1
6
7
7 (a)
(2x-1) 1 3 5 7 9
logy 0.46 0.8
6
1.2
4
1.6
0
1.97
(b) Graph
(c )2 1xy pk =
logy = (log k) (2x-1)+ logp
From the graph,
(i) log k= gradient of the graph
=1.97 0.3
9 0
= 0.1856
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k=1.533
(ii) logp = 0.3
P = 1.995
10
8 (a) Intersection of curve and straight line;2( 3) 9x x = +
x = 0 or 7
Hence, n = 7
(b) Area of trapezium OABC =1 175
(9 16)(7)2 2
+ =
Area under curve AB =7
2
0( 3) x dx
=7
2
0( 6 9)x x dx +
=73
2
0
3 93
xx x +
=
32(7) 3(7) 9(7) 0
3
+
= 30.33
Area of shaded region, P
= Area of trapezium OABC area under curve AB
=175
2-30.33
= 2157 unit6
(c ) Generated volume =3
2
0y dx
=23
2
0 ( 3)x dx
=
35
0
( 3)
5
x
=
5 5(3 3) (0 3)
5 5
=
3348 unit5
10
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9
10
10 (a) MAB
MCE
Equation of DA
3
(b)
Equation of BC is
Equation of perpendicular bisector of AB is
4
(c ) Area of ABCD
= [ ]1
(8 42 3) (6 7 4)2
+ + +
=25 unit2 3
11 (a) (i) Let X be the student from Science Stream
P(X= number of student selected) = = 0.6
a)
b)
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c)
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For this binomial distribution, n = 6 , ,
[6C0 +6C1
5]
]
= 0.95904
(ii) For this binomial distribution, n = 280 , ,
Standard deviation,
= 8.198
(b) (i) Let X be the diameter of a golf ball.
Given and
0.3085
Percentage = 30.85%.
(ii)
= 0.3721
n = 107510
12(a) t
dt
dva 1015 ==
215,0 == msat
(b)3
5
2
15 32 tts =
,0=s 03
5
2
15 32=
tt
0)29(52
= tt
st2
9=
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1K1
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(c)
3,0
0)3(5
0515
0
2
==
=
=
=
tt
tt
tt
dt
ds
At ,3=t )3(1015=
a= -15 (< 0) maximum
3
)3(5
2
)3(15 32=s
= 22.5 m
10
13 (a) (i)
(a) (ii)
Price index ofPfor 2005 based on the year 2001
(b) (i) Given the composite index of the drink for 2005 based on the year
2003 was 123
x = 122.5
(ii)
Price of a box of drink in 2003 if its corresponding price in the year 2005
= RM 10.1610
14 (a) Using the cosine rule,
BD = 7 + (18.2) - 2 (7)(18.2) cos 52
2
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2001 2003 2005
-- 100 125
100 120 ?
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BD = 14.95 cm
(b) Using the sine rule,
sin < ABD = sin 52
7 14.946
< ABD = 21 40'
2
(c )sin < B'DC = sin 110
6.3 14.946
< BDC = 2320'
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maximum point occur at point (100, 400) and point (400, 100)
By testing the point (100,400):20(100) + 40(400) = k
k= RM18000
By testing point(400, 100):
20(400) + 40(100) = kk = RM12000
So, maximum sales is RM18000.
(ii) Whenx = 200
100 y 300
The fee collection at (200, 300) is maximum=20(200) + 40(300)
=4000+12000
=RM16000
log y
7b
0 2 4 6 8 10 (2x
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
0.3
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8/14/2019 Answer Zone C P2 Add Maths 2009
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y = 100
x +y = 500
y - x = 300
(100,400)
(400,100)
R
0 100 200 300 400
100
200
300
400
500
y15b
0100200 4003002001000
Label axes correctly K1
Draw scale correctly N1
Plot all the points correctly N1
Draw the best line fit N1
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