spm addmath2 ans (kedah)
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8/14/2019 SPM AddMath2 Ans (Kedah)
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3472/2 Additional Mathematics Paper 2 [Lihat sebelah
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1. 3
2
yx
+= or
2y x= 3
2
23 3 102 2
y yy y
+ + + =
0
0
or ( ) ( )22 2 3 2 3 1 x x x x + =
2 2 29 6 6 2 4 40 0 y y y y y+ + + =
or
2 2 22 3 4 12 9 10 x x x x x + + + =
y = 3.215 , 3.215
or
x = 3.107 ,
0.107
x = 3.107 / 3.108 , 0.107 / 0.108
or
y = 3.214 / 3.215 , 3.214 / 3.215
Answer must correct to 3 decimal places. 5
Make x or y asthe subject
P1
Eliminate
x or y
Solve quadratic
equation23 31 0y =
2
31
3y =
or23 9 1 0x x =
or
using formula
or
completing the
square
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2(a)
(b)
16 ,8 , 4 ,......
16a = 1
2r=
7
7
116 1
2
11
2
S
=
=3
314
or 31.75
64 ,16 , 4 ,......
64a = 1
4r=
64
11
4
S
=
= 1853
or 85.33
3
3 6
3(a) 2( ) f x x px= + + q
=
2 2
2
2 2
p p x px q
+ + +
=
2 2
2 4
p p
x q
+ +
32
p = p =6
365
4q + = q = 4
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K1
User
raS
n
n
=
1
)1(K1
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K1
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P1
P1
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1
aS
r =
Use
2
2 )2b
(
b
use x + bx = ( x + ) or
32
b
a =use axis of symmetry
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Alternative solution
( )
223 5 x px q x+ + =
=2 6 9 5x x +
=2
6 4x x +
6p =
q = 43
3(b)2 6 4 31x x + 0
0
0
2 6 27x x
( 9)( 3)x x +
3 9x 2 5
4(a)
tanx +cos
1 sin
x
x+
=sin
cos
x
x+
cos
1 sin
x
x+
=2sin (1 sin ) cos
cos (1 sin )
x x x
x x
+ +
+
=2 2
sin sin coscos (1 sin ) x x x
x x+ +
+
=sin 1
cos (1 sin )
x
x x
+
+
=1
cosx
= secx3
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Use ( ) 31 0f x andfactorization
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Usesin
tancos
xx
x=
Use identity2 2
sin cos 1x x+ =
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Comparing coefficient of x
or constant term
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4(b)(i)
Negative sine shape correct.
Amplitude = 3 [ Maximum = 3 and Minimum = 3 ]
Two full cycle in 0 x 2
3
4(b)(ii)
53sin 2 2
xx
=
or
52
xy
=
Draw the straight line
5
2
x
y =
Number of solutions is 3 .
3 9
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y
3
3
2
2
3
2 x
52
xy =
O
y 3sin2x=
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5 (a)
(b)
1220
x=
240x =
The mean
240 5 + 8 + 10 + 11 + 14 288
25 25X
+= =
= 11.52
2
212 3
20
x =
2 3060x =
The standard deviation
( )2 2 2 2 2
23060 5 8 10 11 1411.52
25
+ + + + +=
= ( )23566
11.5225
= 9.9296
= 3.151
7 7
K1
xx N=
Use
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K1
Use formula
22x
xN
=
N1
N1
For the new
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Xand2
x
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6(a)
(b)
(i) PR PO OR= +uuur uuur uuur
= 6 15a b +% %
(ii) OQ OP PQ= +uuur uuur uuur
=3
65
a O+uuur
%R
r
5%
5%
k
k
= 6 9a b+% %
(i) OS hOQ=uuur uuu
= (6 9 )h a b+% %
(ii) OS OP PS= +uuur uuur uuur
= 6a k PR+uuur
%
= ( )6 6 1a k a b+ +% %
=(6 9 )h a b+% %
( )6 6 1a k a b+ +% %
6 6 6h = 9 15h k=
1h = 5
3h k=
5 13
k k=
3
8k=
5 3
3 8h
=
=
5
8
3
5 8
K1
N1Use
PR PO OR= +uuur uuur uuur
or
OQ OP PQ= +uuur r uuuruuu
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K1 Equate coefficient
of a or b % %and
Eliminate
h or k
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7(a)
(b)(i)
(ii)
kn
xn
y 4log1
log1
log +=
log x 0.18 0.30 0.40 0.60 0.74
log y 0.48 0.54 0.59 0.69 0.76
6
410
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P1
All points plotted
correctly
3472/2 Additional Mathematics Paper 2 [Lihat sebelah
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Correct axes and
scale
Line of best-fit
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intercept
= kn
4log1
n = 2 k= 1.51
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K1
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gradient
=n
1
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0.6
0.7
Graph of log10y against log10x
0.8
log 10y
0.5
0.4
0.1
0.2
0.3
0.9
0
0.39
0.4 0.5 0.6 0.7
log 10 x
0.1 0.2 0.8 0.90.3
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8.
(a)
(b)
(c)
Solving simultaneous equation
P( 2, 2) Q(4, 8)
Use dxyy )( 12
+ dxx
x )2
4(2
Integrate dxyy )( 12
64
2
32 xx
x+
18
Note : If use area of trapezium and ydx , give the marksaccordingly.
Integrate dxx
2
2
)2
(
=
20
5x
5
8
3
4
310
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3472/2 Additional Mathematics Paper 2 [Lihat sebelah
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N1
K1Use correct
limit into
4
2K1
64
2
32x
xx
+
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K1
Use correct
limit into2
0
20
5
x
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9(a)
(b)
(c)
(d)
Equation ofAD :
y 6 = 2 (x 2 )
y =2x + 10 or equivalent
y =2x + 10
and
x 2y = 0
D(4, 2)
p = 3 or C(8, 4)
Substitute (8, 4) intoy = 3x + q
q = 20
OADOABCArea = 4
Using formula
0
0
6
2
2
4
0
0
2
1= OAD
40
Alternative solution :
B(10, 10)
Using formula
0
0
6
2
10
10
4
8
0
0
2
1=OABCArea
40
2
2
3
3
10
Use m = 2 and find
equation of straight lineK1
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Solving
simultaneous
equations
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Find area of
triangle
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K1
P1
Find area of
parallelogram
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10(a)
(b)
(c )
=
10
6sin 1 or equivalent
= 2POQ
radPOQ 287.1=
Alternative solution :
122
= 102
+ 102
2(10)(10) cos POQ
+=
)10)(10(2
121010cos
2221POQ
radPOQ 287.1=
Using (2 1.287)
Major arc PQ
= 10 ( 2 1.287 )
= 49.96 cm
Lsector= 287.1)10(2
1 2
Ltriangle = 287.1sin)10(2
1 2
= 16.35 cm2
= 3.9149 cm
3
3
4 10
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K1 Use cosine
rule
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Use formula
s = rK1
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K1
Using formula
Lsector= 21 r
2K1
K1
K1Lsector - L
Using formula
L = absin C 21
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11
(a)(i)
(ii)
b)(i)
(ii)
p =5
3, p + q = 1
P( X = 0 )
=5C0(
5
3)0(
5
2)
5
= 0.01024
Using P( X 4 )= P( X = 4 ) + P( X = 5 )
=5C4(
5
3)4(
5
2)
1+ (
5
3)5
= 0.337
P ( 30 X 60 )
= P (10
3530 Z 60 3510 )
Use P( 0.5 Z 2.5 )
= 1 P( Z 0.5 ) P( Z 2.5 )
= 1 0.30854 0.00621
= 0.68525
Number of pupils = P( X 60 ) 483
3
3
2
3
2 10
3472/2 Additional Mathematics Paper 2 [Lihat sebelah
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K1
N1
Use P(X = r) = n Crprqnr,
p + q = 1
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K1
use
Z =
X
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12.
(a)
(b)
(c)
(d)
Subst. t= 0 intodv
dt
a= 15 6t15 ms
-2
1 134 4
ms75 /18ms
Integrate
Use s = 0
15
2/
17 / 7.5
2
S4 S3
115
2
Note : If use , give the marks accordingly.4
3vdt
2
2
3
3
10
K1
N1
0dv
dt= K1and subst. t in v = 15t 3t
2Use
5[t= ]
2
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K1
K1
K12 315
2dt t t s v = =
Subst. t = 3 or t = 4 in
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2 315
2s t t=
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13.
(a)
(b)
(i)
(ii)
(iii)
Use I = 2007
2005
100P
P
Value of m : 25, m, 80, 30 or equivalent
120 25+130m+135 80+139 30
Use i i
i
I WIW
=
132.1 =12025+130m+13580+139 30
135+m
m =65
100150.00x
132.1
RM 113.55
/I . ( . x .132 1 132 1 0 3= + )08 05
171.73
3
3
2
2
10
K1
N2, 1, 0x = 48.6y = 135z = 80
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14.
(a)
(b)
(c)
x + y 80 or equivalent
y 4x or equivalent
x + 4y 120 or equivalent
x=30
(16,64)
x + 4y = 120
x + y = 80
y = 4x
100
90
80
70
60
50
40
30
20
10
10080604020 9070503010
x
y
At least one straight line is drawn correctlyfrom inequalities involving x and y
All the three straight lines are drawn correctly
Region is correctly shaded
(i) minimum = 23
(ii) (16,64)
Subst. point in the range
in 20x + 40y
RM2880
3
3
4 10
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15.
(a)
(b)
(c)
(d)
120 9.3 6 sin
2
BCD=
45o48 / 45.8
o
0.74545 4555Use cosine rule in BCD
BD2
= 9.32
+ 62 29.36 cos 4548
6.685
Use sine rule in BCDo
sin CBD sin '
. .
= 45 489 3 6 685
94
o
10
4444
qqq11111aaaaaaaaaaaaaaaaaa 14s4
5.555555
5555 z5555555555555555
Sum of area:
20 cm2
+ ABD 555 102
58.82 cm2
4
2
4 10
Use area = ab sin c in BCD K1
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K1 Obtain ADB by using180o 85o50 BAD or equivalent
Use area ADB = 6.685 13 sinADB
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