skema percubaan mt_k2 spm 2015 jpnk
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Matematik
Tambahan
Kertas 2
2 jam 30 minit
Ogos 2015
JABATAN PELAJARAN NEGERI KELANTAN
PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2015
ADDITIONAL MATHEMATICS
Paper 2
Skema Pemarkahan ini mengandungi 11 halaman bercetak
MARKING SCHEME
2
MAJLIS PENGETUA KELANTAN
PEPERIKSAAN PERCUBAAN SPM 2015
SKEMA PERMARKAHAN MATEMATIK TAMBAHAN KERTAS 2
NO SOLUTIONS MARKS TOTAL
1. y = 2x – 4 atau
4
2
yx
(2x – 4)2 = 2 (x
2 + 2) atau
2
2 42 2
2
yy
x2 – 8x + 6 = 0 atau y
2 – 8y – 20 = 0
2( 8) ( 8) 4(1)(6)
2(1)x
atau
2( 8) ( 8) 4(1)( 20)
2(1)y
x = 7.162 , 0.838 (both)
y = 10.325 , – 2.325 (both)
P1
K1
K1
N1
N1
5
2(a)
(b)
(0, 3)
2 2
2
2 2
32
2 4 4 2
2 34 8
k k kx x
k kx
2
3 58
4, 1
k
k p
N1
K1
K1
K1
N1, N1
6
3(a)
(b)
(c)
3x, 3x+6, 3x+12, …
15
765 2(3 ) (14)(6)2
x
x = 3
159 (14)(6)
93
T
2(9) ( 1)(6) 1320
2
nn
2 2 440 0n n atau setara
(n + 22)(n – 20) = 0
n = 20
P1
K1
N1
N1
K1
K1
N1
7
3
4(a)
(b)
23(1- cos )
sin cos
x
x x
23sin
sin cos
x
x x
3 tan x
shape of cosine graph
amplitud = 3
Kala = 2
straight line
2
xy
No. of solutions = 4
K1
N1
P1
P1
P1
K1
N1
N1
8
5(a)
(b)
(c)
Saiz Ikan Bil. Ikan
25 – 29 8
30 – 34 7
35 – 39 10
40 – 44 6
45 – 49 5
50 – 54 4
8(27) 7(32) 10(37) 6(42) 5(47) 4(52)
40
37.625
m
2 2 2 2 2 22
2
8(27) 7(32) 10(37) 6(42) 5(47) 4(52)37.625
40
5913537.625
40
7.9205
P1
P1
K1
N1
K1
K1
N1
7
3 2
-3
2 2 3
2
P1 untuk saiz kelas betul
P1 untuk semua kekerapan
betul
P1 jika 10 dilihat
4
6(a)
(b)
orBC BA AC
AD AB BD
~ ~
3 2BC x y
~ ~
9 1
4 2AD x y
AF k AD
~ ~
9 1
4 2k x k y
AF AE k AD
~~ ~
2 3( (2 ) 3 )
3 2y h y x
~ ~
2 23 ( )
3 3h x h y
Compare and solve the equation
9 2 2 13 or
4 3 3 2h k h k
2 2 3 1( )
3 3 4 2k k
2 1 , ( )
3 2k h both
K1
N1
N1
P1
K1
N1
K1
N1
8
5
7 (a)
(b)
x 1 2 4 6 8 9
Log10 y 0.84 0.99 1.29 1.57 1.87 2.00
Paksi betul dan skala seragam
Plot 5 titik betul
Garis lurus penyuaian terbaik
10 10 10log log logy V x T
(i)
10log 0.1411
1.384 (1.35 - 1.43)
T
T
(ii)
10log 0.73
5.370
V
V
(iii) 10log 1.7 , 6.9y x
N1
K1
N1
N1
P1
K1
N1
K1
N1
N1
10
8(a)
(b)
(i)
(ii)
0.8, 02p q
110
10
1019
9
10 )2.0()8.0()2.0()8.0()9( CCXP
= 0.3758
68 70 75 70
2.5 2.5P Z
=0.76539
70( ) 0.25
2.5
mP z
70( ) ( 0.674)
2.5
mP z p z
700.674
2.5
68.315
m
m
P1
K1K1
N1
K1
N1
K1
K1
K1
N1
10
6
9 (a)
(b)
(c)
2
1
22
k
h
x
dxxArea
1
0
23)42)(1(2
1
1
0
3
333
xx
=3
1
3
2
2
1(1) (2) (3 )
3
1
2
Volume y dy
K1
N1
N1
K1
K1
K1
N1
K1 K1
N1
10
10(a)
(b)
(c)
(i) x = 4
(ii) 3y +2 (4) = 11
y = 1
C(4, 1)
)2
5,10(
15
20
45
20
D
y
x
FD = 4
2 2
2 2
2 2
5( 10) ( ) 4
2
5( 10) ( ) 16
2
4 4 80 20 361 0
x y
x y
x y x y
P1
K1
N1
K1
K1
N1
K1
K1
K1
N1
10
7
12
(a)
(b)
(c)
(d)
11
11
275000100 110
= 250 000
Q
Q
1311
10 10
92 115,
100 100
115 100
100 100 92
125
Q Q
p
p
110(4) 125(6) 125(10)
20
122
I
11
11
805200122 100
660,000
Q
Q
K1
N1
P1
K1
N1
K1
N1
K1
K1
N1
11(a)
(b)
(c)
θ = 600
θ = 1.047
Perimeter = 14(1.047)+14(1.047) + 14
= 43.32
2 21 1(14) (1.047) 2 (14) sin 60
2 2
120.34
Area
P1
N1
K1K1K1
N1
K1K1K1
N1
8
13
(a) i)
ii)
(b) i)
ii)
9
sin 56 sin 68
RU
RU
2TU = 7
2 + 8.0473
2 – 2 (7)(8.0473) kos 124
0
0
13.2951 8.0473
sin124 sinT
'30.12 @30 7o oSTU
VM = 5 dan VL = 89 dan JL = 208
208 = 132 + 89 – 2(13)( 89 ) kos JVL
JVL = 78.24o @ 78
o14
’
Luas = '1
(13)( 89)sin 78 142
o
= 60.03
K1
N1
K1
K1
N1
P1
K1
N1
K1
N1
10
14
(a)
(b)
(c) (i)
(ii)
y 100
200 500 x
700 yx
2002 yx
Graf
Bilangan maksimum bekas plastik bulat = 450
k = 0.8x + 0.6y
= 0.8(500)+0.6(200)
= RM 520
N3,2,1,0
K1(satu
garis
betul)
N1(semua
garis
betul)
N1 rantau
betul
N1
N1 (500,
200)
K1
N1
9
15
(a)
(b)
(c)
26 2 4 0
2
3
8
3Y
t t
t
v
2
12 2
8 4
X
Y
a t
S t t
8t-4t2 = 0
t = 2
Ya = 26 ms-1
3 2 22 4 8 4
3
2
t t t t t
t saat
2 3 23 3 3 3 3
8( ) 4( ) atau 2( ) ( ) 4( )2 2 2 2 2
Y Xs s
= 3 = 3
K1
N1
N1
K1
K1
N1
N1
K1
N1
N1
10
10
y
0
800
1
R
x
11
X