PEPERIKSAAN PERCUBAAN SPM TINGKATAN 5 2015 SP… · 1" " SULIT 3472/1 Additional Mathematics Paper...
Transcript of PEPERIKSAAN PERCUBAAN SPM TINGKATAN 5 2015 SP… · 1" " SULIT 3472/1 Additional Mathematics Paper...
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SULIT 3472/1 Additional Mathematics Paper 1 August 2015
BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH
DAN SEKOLAH KECEMERLANGAN KEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN PERCUBAAN SPM
TINGKATAN 5
2015
ADDITIONAL MATHEMATICS
Paper 1
MARKING SCHEME
This marking scheme consists of 6 printed pages
2
MARKING SCHEME
Question Answers Sub Mark
Mark
1 (a) p =10
(b) h : x → 1x,x ≠ 0 or equivalent
Fungsi kerana hubungan satu dengan satu
1 1 1
3
2
(a) q = 3
B1: 2 = 45−q
(b) h x( ) = 3x + 4x ,x ≠ 0
B1: y =4x −3
or y x −3( )= 4
2 2
4
3 −1.637 or 2.137
B2: −(−1)± (−1)2 − 4(2)(−7)2(2)
B1: 2x 2 − x −7 = 0
3
3
4 p = 8
B1: 3 53!
"#$
%&
2
−53p +5= 0
2
2
5 (a) 3=p (b) 2=q (c) 3=x
1 1 1
3
6 21 <<− x B2: (x − 2)(x +1) < 0 or B1: 022 <−− xx
3
3
3
7 n 3 − n −2 or n 3 − 1
n 2
B2: (23)m − (22 )−m B1: 23 or 22
3 3
8 y = 8x −1
B2: y +1x
= 23 or y +18
= x
B1: 3log2 2 ory +13
or y +18
3 3
9 n = 6 B2: log(1.05)n >log1.3 B1: 300000(1.05)n >390000
3 3
10 (a) 12.6, 25.2, 37.8, 50.4 (b) 12.6
1 1
2
11 (a) 9 (b) 9207
B1: 9(210 −1)2−1
1 2
12 (a)
y 2
x= −2x +10
(b) p = 5 and q = 4 B2: p = 5 or q = 4 B1: 0 = −2 p +10 or q = −2(3)+10
1 3
4
4
13 −150 = −2(100)+50 Tiang bendera perlu dipindahkan sebab terletak diatas lorong (mesti tunjukkan coordinat diuji) B3: y = −2x +50 Equation of perpendicular bisector Canteen and Block B B2: y −150 = −2* x − (−50)( )
B1: Gradient of canteen and blok B, mCB =225−75
100− (−200)=12
4 4
14 (a) 3
(b) 103
B1: − q2= −53
or m = −53
1 2
3
15 (a) − p
p 2 +1
(b) 1− p1+ p
B1: tan45°− tanθ1+ tan45° tanθ
1 2
3
16 82.22
B2: 126( )2 π2
!
"#
$
%&−12×6×6
B1: 126( )2 π2
!
"#
$
%& or
12×6×6
3 3
17 (a) −3
2
"
#$$
%
&''
(b) 3i - 7j
1 1
2
5
18 (a) unit vector PQ
! "!!!=
113
2s −3r( )
B1: PQ! "!!
= 2s −3r or 22 + −3( )2
(b) a = 5
b = −8
2 1 1
4
19 −4323125 or −0.1382
B2: f ''(x ) =432
(3x −5)5 or equivalent
B1: −36(3x −5)4
or equivalent
3
3
20 (2.5, 8.75) B3: x = 2.5 B2: −2x +6 =1
B1: dydx
= −2x +6 or gradient, m =1
4 4
21 V = 2t 3 − 3t
2
2+ t −1
B2: 11= 2 2( )3−
3 2( )2
2+ 2+c or c = −1
B1: dAdt
= 6t 2 −3t +1 or A = 6t 2 −3t +1dt∫
3
3
22 k = 4
B2: 42 = 39.5+
24+ k2
−11
12
"
#
$$$$
%
&
''''×10
B1: 39.5 or 11 or 12