peperiksaan percubaan spm tahun 2018 · 2 marking scheme additional mathematics trial examination...
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Nama Pelajar : ………………………………… Tingkatan 5 : …………………….
3472/2
Additional
Mathematics
August 2018
PEPERIKSAAN PERCUBAAN SPM
TAHUN 2018
JABATAN PENDIDIKAN NEGERI KEDAH
MAJLIS PENGETUA SEKOLAH MALAYSIA ( KEDAH )
ADDITIONAL MATHEMATICS
Paper 2
( MODULE 1 )
.
MARKING SCHEME
2
MARKING SCHEME
ADDITIONAL MATHEMATICS TRIAL EXAMINATION AUGUST 2018
MODULE 1 ( PAPER 2 )
N0. SOLUTION MARKS
1
(90 45) (90 2 )( ) 3300x y
When
= 15
(abaikan)
= 50
50
15
AD
AB
P1
P1
P1
K1 Eliminate x or y
K1 Solve quadratic equation
N1
N1 Both sides of the pond
correct
7
3
N0. SOLUTION MARKS
2
(a)
(b)
2013
219 5 10 23 55
2015
2
10
30
p
m
p
p
N
Mode = 33 5
P1 3
19 5 , 13 , 5 , 10
Any of thesevalue
K1 formula of m
N1 both p & N
K1 Plot / Correct axes &
uniform scale
K1 Correct histogram
( 5 bar )
K1 Find mode
N1
7
0 4.5 14.5 24.5 34.5 44.5 Marks
Students
2
4
5
9
10
mode = 33.5
4
3
(a)
(b)
1, 4, 10a d n
10
102 1 9 4
2S
= 190
Yellow tiles = 400 – 190
= 210
OR
10
102 3 9 4
2
210
S
Difference = 210 – 190
= 20
P1 any 2 correct
K1 use S10
N1
N1
N1
5
4
(a)
(i)
6 5
2 4
1
2
2 4
QT QP PT
QR i j
(ii)
1 6 2
5 2 4
3
SR SP PQ QR
i j
K1
N1
K1
N1
5
(b)
3 6
1 2
1
2
SR k PQ
k
k
SR is parallel to PQ
K1
N1
N1
7
5
(a)
(b)
1 11
1 1 11
2 3 18
2 ,3 ,18
2 3 3 18
1 2 1
2
x y q
y qx
y y qx
k
k k k
k k k k
x y q
xyq
x y
3
11
2
11 )
2( 2
2
log 5, 3 5
9 9
3
3
45
x
xk
x
x
x
K1
K1
N1
K1
K1
N1
6
6
N0. SOLUTION MARKS
6
(a)
(b)
(c)
3 cos 2y x
1 2
2
xf x
xy
Number of solutions = 2
P1 graph cosine curve
P1 amplitude 3
P1 2 cycle 0 to 2
P1 shifted graph
1y f x
P1 1y f x
K1 line 2x
y
N1 equation 2x
y
N1
8
7
N0. SOLUTION MARKS
7
(a)
(i)
(ii)
(b)
(i)
(ii)
9 1 1010 10 0
9 10
0 4 0 6
8
9 10
0 4 0 6 0 4 0 6
0 001573 0 0001049
0 001678
p q
P X
P X P X
C C
0.4 0.6 12
600
n
n
72 55
5
3 4
Z
0.185
550 185
5
P X k
kP z
550 896
5
59 48
k
k
P1
K1 use n r n r
rC p q
N1
K1
N1
K1
N1
K1 use score-z
Z =
X
K1
N1
10
8
N0. SOLUTION MARKS
8
(a)
(b)
(c)
7/ 0 7 / 40 10 / 40 11
10SWT
250 4364 / 0 4363
180rad rad
0 7 0 4363 1 1363
65 10
1 136 3 . .
rad
rad d p
8 1 136 9 088ArcQV
8 8 9 088
25 088
Perimeter
2
2
18 1 136 36 352
2
110 0 7 35
2
QPV
SWT
A
A
71 352 63 88
7 472
Area
K1 use s r
K1
N1
K1
K1
N1
K1
K1
use formula 21
2A r
to find ,QPV SWTA A
K1
N1
10
9
N0. SOLUTION MARKS
9
(a)
(b)
(i)
1
1
QR
RS
m
m
0,6
,0
R
S h
61
6
h
h
(ii)
18 2 4, ,
4 4
4,1
x y
(iii)
4 2 4 41
1 4 2 12
116 4 4 2 16 8
2
116 26
2
21
A
2 2
2 2
41
2 6
4 4 12
4 4 12 0
y y
x x
y y x x
x y x y
K1 for using 1 2 1m m
K1
N1
K1
N1
K1 use formula of Area
N1
K1 for using 1 2 1m m to
form equation
K1
N1
10
10
N0. SOLUTION MARKS
10
(a)
22 1
9
2 3
y x
y
x or r
Volume of cylinder
23 9
81
Volume
9
1
92
1
1
2
2 2
81 19 1
2 2 2
322
16
ydy
yy
Volume of metal
81 16
65
K1 Substitute
9
2 3
y to find
x or r
K1
K1 integrate and use the
limit correctly
K1
N1
1
2 x
y
O
9
3
9 cm
1 cm1 cm
metallogam
11
(b)
(i)
2 3 6 0
,4
2 6
3
x y
h
h
h
OR
2
2
4
2
, 4
2 1
3
yx
h
h
h
(ii)
Area of trapezium
1
2 4 32
9
Area
5 2
3
5
3
4 2
14
2
1 14
3 1
24
3
8
3
x dx
x
N1
K1
K1 integrate and use the
limit correctly
12
Area of the shaded region
89
3
35 2/ 11 / 11 67
3 3
K1
N1
10
13
N0. SOLUTION MARKS
11
(a)
(b)
(c)
(i)
(ii)
(iii)
1
x
1 25 1 00 0 80 0 50 0 40 0 20
1
y 2 78
2 22
1 69 0 96 0 70 0 20
1 1k h
y x
0 3h
2 48k
11 25 1.30
0 8 0.77
y
y
N1 6 correct values
N1 6 correct values
K1 plot / correct axes &
uniform scale
N1 6 points plotted
correctly
N1 line of best-fit
P1
N1 for y-intercept
N1 finding gradient
K1
N1
10
14
N0. SOLUTION MARKS
12
(a)
(b)
(c)
(d)
18ms
2 2 8 0
( 2)( 4) 0
0 4
t t
t t
t
32
32
32
83
(4)4, (4) 8(4)
3
(5)5, (5) 8(5)
3
tS t t
t S
t S
Distance=
2 2 126 (26 23 )
3 3 3
30
N1
K1
K1
N1
K1
N1
K1
K1
K1
N1
10
t
v
8
0
15
N0. SOLUTION MARKS
13
(a)
(b)
(i)
2 2 28 3 71 2 8 3 71 cos112
10
5
PR
PR
SR
(ii)
5
sin120 sin30
8 66
PT
PT
8 3 71 5 5 8.66
30 37
Perimeter
(i)
(ii)
1
8 66 10 sin302
21 65
Area
OR
1
8 66 52
21 65
Area
K1 use cosine rule
N1
K1 use sine rule
N1
K1
N1
N1
K1 use 1
sin2
A ab c
K1 P’S’ = 10
N1
10
P' S'
T'
16
N0. SOLUTION MARKS
14
(a)
(b)
(c)
80
4
2 60
x y
y x
x y
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
35 50 3550
7 10 710
x y
or
x y
Draw a straight line in the graph.
X=30
Maximum .number of cake A = 30
N1
N1
N1
N1
N1
N1
K1
K1
K1
N1
10
R
60 80 0
30 7x + 10 y = 710
x + 2y = 60
y = 4x
x
y
80
x + y = 80
17
N0. SOLUTION MARKS
15
(a)
(b)
(c)
(d)
(i)
(ii)
135x
1 50 100115
1 30
Q
120140
100
116 67
I
I
120 2 115 3 135 1 110116
2 3 1
4
100 11620
23 2
50021 55
23 2
max 21
y
y
y
p
p
N1
K1
N1
K1
N1
K1
N1
K1
K1
N1
10
END OF MARKING SCHEME