ADD MATHS Final F4 Module 3 _ OKT 2015 _ P2 _ Skema.pdf
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Transcript of ADD MATHS Final F4 Module 3 _ OKT 2015 _ P2 _ Skema.pdf
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7/24/2019 ADD MATHS Final F4 Module 3 _ OKT 2015 _ P2 _ Skema.pdf
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Nama Pelajar : Tingkatan 5 : .
3472/2
Additional
Mathematics
Otober 2015
MODUL PENINGKATAN PRESTASI TINGKATAN 4
TAHUN 2015
ADDITIONAL MATHEMATICS
Paper 2
( MODUL 3 )
MARKING SCHEME
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SULIT 3472/2
MARKING SCHEME
ADDITIONAL MATHEMATICS PAPER 2 2015
N0. SOLUTION MARKS
1 mn 26 or 26 nm
22632 2 mm
02062 2 mm
01032 mm 025 mm
5m and 2m (both)2n and 16n (both)
P1
K1 Eliminate m/n
K1 Solve quadratic equation
N1
N1
5
2
(a)
(i)
(i)
(ii)
(b)
5)2(2)2( f
= 1
952
952
x
x
7x dan 2x
72 x
3
1)(
1 xgxgh
= 5
3
1x
=15
1x
K1
N1
K1
N1
N1
K1
K1
N1
8
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N0. SOLUTION MARKS
3
(a)
(b)
22 0x hx k
S.O.R :
2 3
2
h
2h
P.O.R. :
2 32
k
12k
22 2 12x x m
22 2 12 0x x m
2
2 4 2 12 0m
100 8 0m
12 5m
K1 for finding S.O.R or P.O.R
N1
N1
K1 ( 2 4 0b ac )
K1
N1
6
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N0. SOLUTION MARKS
4
(a)
(b)
Quadratic curve a < 0
Maximum point ( 2 , 8 )
Axis of symmetry x = 2
The value of x= -2 , y=0 or x=8 , y= -10
(i) 0 8x
(ii) 2 different roots
P1
P1P1
P1
N1
N1
6
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5
N0. SOLUTION MARKS
5
(a)
(i)
(ii)
(b)
3
2
5125
25 5
a ba b
aa
=a
ba
2
33
5
55
= ba 355
3xy
5log x a ,
5log y b
44
5 5 5 5log log log 125 log125
xx y
y
4 3a b
2 1 212 12x
2 1 2x
1
2x
K1
K1
N1
K1 Use log log loga a am m nn
K1 Use mnm an
a loglog
N1
K1
K1
N1
9
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N0. SOLUTION MARKS
6
(a)
(b)
Joseph: 25 3 23 4 24 6 27 1 100 4x
2 2 2 2 2
25 3 23 4 24 6 27 1 2527 22x
Mean100 4
25 14
Standard deviation = 21.254
22.2527
= 340.1
Firdaus: 26 1 24 6 25 8 23 9 100 4x
2 2 2 2 226 1 24 6 25 8 23 9 2523 22x
Mean100 4
25 14
Standard deviation =21.25
4
22.2523
0 8916
Firdauss performance is more consistent.
Prestasi Firdaus lebih konsisten.
The standard deviation is smaller.
Sisihan piawainya adalah lebih kecil.
K1 (usingN
xx
)
K1 (using 22
xN
x
)
N1
N1
N1
N1
6
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N0. SOLUTION MARKS
7
(a)
(b)
Correct axes & uniform scale
Draw 5 bar correctly
Show the method to find mode
Mode = 2.25
Midpoint = 1.2, 1.7, 2.2, 2.7, 3.2
Mean=
1.2 5 1.7 8 2.2 20 2.7 10 3.2 7
50
2.26
222222 )2.3(7)7.2(10)2.2(20)7.1(8)2.1(5 fx
22
xf
fx
=2)26.2(
50
7.271
= 5713.0
K1
K1
K1
N1
K1
K1
N1
K1
K1
N1
10
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8
12
14
16
10
8
2
4
6
Weight
Number of
Durian
18
00.95 1.45 1.95 2.45 2.95 3.45 3.95
20
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N0. SOLUTION MARKS
8
(a)
(b)
(c)
(i)
(ii)
(d)
2 2 2 2( 6) (6 12) 6 (12 4)
14
h
h
2 3(6) 2 3(12)14 6
5 5
(26, 3)
x yor
m =4
3
4( 3) ( 26)
3y x
4 113/ / 3 4 113
3 3y x y x
89(6, )
3
Area =
1 896( 3) 26(12) 6(4) 6(4) 26( ) 6( 3)
2 3
2541
3
K1
N1
K1
N1
P1
K1
N1
N1
K1
N1
10
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N0. SOLUTION MARKS
9
(a)
(b)
(c)
18060
= 1.047
5.237 300rad or
42.31
237.56
RQS or95.20
237.54
PSS
Perimeter = 31.42 + 20.95 +2 +2
= 56.37
21 6 5.2372
94.27
Area
214 5.237
2
41.896
Area
Area = 94.2741.896
= 52.374
K1
N1
P1
K1 Use rs
K1
N1
K1 Use21
2A r
N1 For
94.27 or 41.896
K1
N1
10
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N0. SOLUTION MARKS
10
(a)
(b)
6 4 2
6 61
dyx
dx
xx
1
3 4 1
(1 , 1 )
x
y
H
1 1
1 2
1 1
2 2
y
x
y x
2
2
2 3
(6 )
(6 )
V r h
r r
r r
2(12 3 ) 0
(3 )(4 ) 0
0( ) 4
dVr r
dr
r r
r ignored r
2 26(4) (4)
32
V
K1
N1
K1
N1
K1
N1
K1
N1
K1
N1
10
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N0. SOLUTION MARKS
11
(a)
(i)
(ii)
(iii)
(b)
(i)
(ii)
6.8
5.40sin
1.12
sin
PQR
04.66PQR
PSRkos )8.4)(4.9(28.44.91.12 222
83.112PSR
46.73sin1.126.82
1AreaPQR
Or
83.112sin8.44.92
1AreaPSR
79.2088.49 AreaPQRS
= 70.67
113.96
K1
N1
K1
N1
See 73.46
P1
Find Area of
PQR or PSR
K1
K1
N1
N1
N1
10
Q
PR
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N0. SOLUTION MARKS
12
(a)
(b)
(c)
(d)
10020.2
110 x
00.2x
118
10000.5
90.5
y
y
20.16
13510000.12
z
z
1210
120(18) 110(10) 118(45) 135(27)
100I
= 122.15
10040
15.122 12 Q
86.4812 Q
100
15.122115
10
15 I
=140.5
K1 Use
10010
12
Q
QI
N1
N1
N1
K1
N1
K1
N1
K1
N1
10
END OF MARKING SCHEME