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Nama Pelajar : ………………………………… Tingkatan 5 : …………………….
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Additional
Mathematics
August 2016
PROGRAM PENINGKATAN PRESTASI AKADEMIK
SPM 2016
ADDITIONAL MATHEMATICS
Paper 2
( MODULE 1 )
.
MARKING SCHEME
2
MARKING SCHEME
ADDITIONAL MATHEMATICS TRIAL EXAMINATION AUGUST 2016
MODULE 1 ( PAPER 2 )
N0. SOLUTION MARKS
1 3 2x y or
3
2
xy
3 52
3 2y y
24 17 9 0y y
2( 17) ( 17) 4(4)(9)
2(4)y
3.63y and 0.620y (both)
4.26x and 1.76x
P1
K1 Eliminate x/y
K1 Solve quadratic equation
N1
N1
5
2
(a)
(b)
(i)
(ii)
30x
2 21456030
15
= 70.667
450 232
17
n
47n
2214560 2(32)
3217
9.61
N1
K1
N1
K1
N1
K1
N1
7
3
3
(a)
(b)
13 4 12x x x
1 484
3
xx
x
16x
24 x
1 2x x
1x
3log9 5
x
3log lg9 lg5x
3
lg5log
lg9x
0.7325
0.73253x
2.2361@ 5
K1 (48
3
x
x or 16x
)
N1
K1
K1
N1
5
4
(a)
P1 for cosine curve
P1 for amplitude 0 and 4
P1 for 3
2 cycle 0 to 2
P1 for 3
42
y kos x
4
(b)
2 1 3
3 1
2 2
y k
y k
3 14
2 2k
3k
K1
K1
N1
7
5
(a)
(b)
(c)
15 1 (14)(2)T
29
20 1 (19)(2)T
39
10
3 392
nS or equivalent
210
2
2(1) ( 1)4 4352
4 2 870 0
5 2 29 0
nn
n n
n n
15n 29
2n
Bilangan baris 2 15
30
K1
N1
K1
K1
N1
K1
K1
N1
8
5
6
(a)
(b)
y = 3x 2 + 1
1
h
y
0 x
Shape of quadratic function
Minimum = 1
Complete graph with axis of symmetry x = 0
2 1
3
yx
1
1 9
3 6
h ydy
1
2
21 9
3 6
1 91
2 2 2
2
h
hh
yy
2 2 8 0
4 2 0
h h
h h
4 , 2h h
Panjang Jeli
4 1
3
P1
P1
P1
K1
K1 integrate and sub. the limit
correctly
K1
N1
N1
8
6
N0. SOLUTION MARKS
7
(a)
(b)
(c)
(i)
(ii)
(iii)
x+2
2 3 4 5 6 7
10log y 0.30 0.53 0.78 1.02 1.26 1.50
10log y
10 10 10log ( 2log logy x b a
10log 4.5 = 0.65
3.5x from graf
10log a=*y-intercept
0.6310a
10log b = *gradient
1.754b
N1 6 correct
values of 10log y
K1 Plot 10log y
vs 2x
Correct axes &
uniform scale
N1 6 points plotted
correctly
N1 Line of best-fit
P1
N1
K1 for y-intercept
N1
K1 finding gradient
N1
10
0 -0.2
x+2
7
N0. SOLUTION MARKS
8
(a)
(b)
1.4033 rad 0.3349 rad
6
6
2
BA
2 2 2
2 2 2
6 2 6 2 2 6 cos
24 cos 2 6 6 4
1cos
6
80 4059 @ 1 4033
A
A
A
A rad
2 6 12
2 2 2 1 4033 6.953
Big circle
Small circle s rad
12 6 953
44 652
length
211 2 2 2 1 4033 6.953
2A rad
2 21 12 6 0 3349 6 sin 0 3349 0 1121
2 2A rad rad
6 953 2 0 1121
6 729
Area
K1
N1
K1 Use s r or 2 r
K1
N1
K1 Use formula 21
2A r
K1 in rad
K1
K1
N1
10
8
N0. SOLUTION MARKS
9
(a)
(b)
(i)
(ii)
(c)
2BDm
2 6y x
0 6 6 01
6 4 2 62Area
1
0 12 36 36 24 02
160
2
230 unit
3 2 2 30p q 2 6q p
2 2 10p q
2 2 6 2 10p p
4p , 2q
4, 2B
6 unitOD
PD OD
2 2
0 6 6x y
2 2 12 36 36x y y
2 2 12 0x y y
P1
N1
K1
N1
K1 ( 30ABC )
K1 for substitude q
N1
K1
K1
N1
10
9
N0. SOLUTION MARKS
10
(a)
(i)
(ii)
(b)
(i)
(ii)
Mean, 5.25np
7 5.25p
0.75p
5 5 6 7P X P X P X P X
5 2 6 1 7 07 7 7
5 6 70.75 0.25 0.75 0.25 0.75 0.25C C C
0.31146 0.31146 0.13348
0.7564
1.5 1.3
1.50.5
P X P Z
0.4P Z
0.3446
Percentage = 34.46%
0.9 1.5P X
0.9 1.3 1.5 1.3
0.5 0.5P Z
0.8 0.4P Z
1 0.8 0.4P Z P Z
1 0.2119 0.3446
0.4435
P1
N1
P1
K1 use n r n r
rC p q
N1
K1 Use Z =
X
N1
K1
K1
N1
10
10
N0. SOLUTION MARKS
11
(a)
(i)
(ii)
(b)
(c)
9 x
9 4
19 4
5
PC PH HC
x y
PG x y
13 9 4
5
124 4
5
EG EP PG
x x y
x y
12 2
ER EH HR
x y
212 2
5
12 2
: 2 : 5
5
x yEG
x yER
EG ER
h
12
21
32
2
3
tinggi
tinggi
N1
K1
K1
K1
N1
K1
K1
N1
K1
N1
10
11
N0. SOLUTION MARKS
12
(a)
(b)
(c)
(d)
dva
dt
, a ht k 2
( ) ( ) ( )h k or h k23 3 0 2 3 9
h and k 3 9
tS t
23 9
2
.4 5t s
6(4.5) 9
18
a
( ) ( )( ) ( )S or S
2 23 3
5 3
9 5 9 35 3
2 2
Total distance = 1 1
13 2 122 2
= 1
392
K1
K1
N1
K1
(for integration)
N1
K1
N1
K1
K1
(for summation)
N1
10
12
N0. SOLUTION MARKS
13
(a)
(b)
(c)
(d)
( )( ) cos
.
.
oAC
AC cm
AE cm
2 2 218 26 2 18 26 35
15 27
10 57
sin sin
.
x 35
8 4 7
x = 77.50o
.
.
O OCED
0
180 77 50
102 50
Area = ( )( )sin 0118 26 35
2
= 134.22
.
.
h
h cm
126 134 22
2
10 32
K1
N1
N1
K1
K1
N1
K1
(for using
area= ½absinc)
N1
K1
N1
10
13
N0. SOLUTION MARKS
14
(a)
(b)
(c)
(i)
(ii)
0 35100
0 25
140
p
6 50100 130
5
q
q
4 2 10
4
4
x y
x y
y x
112 5 4 140 130 135 2125
10
720 140 130 4 1250
1240 10 1250
10 10
1
3
x y
x x
x
x
x
AND
y
16
10
125 110
100
137 5
I
137 5 0 60
100
RM0 825 / RM0 83
K1
N1
N1
K1
K1
N1
K1
N1
K1
N1
10
14
N0. SOLUTION MARKS
15
(a)
(b)
(c)
(i)
(ii)
7
50 100 1000
2 20
1
2
x y
x y
x y
y x
R
7
7 80 20
6( 8 , 6 )
x + y = 7x + 2y = 20
x = 2y
x
y
10
At least one straight line is drawn correctly from
inequalities involving x and y.
All the three straight lines are drawn correctly
Region is correctly shaded
8,6
8 6 14
max
15 25
8 , 6
15 8 25 6
RM270
P x y
P
N1
N1
N1
K1
N1
N1
N1
N1
K1
N1
10
END OF MARKING SCHEME