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SULIT 3472/2(PP)
Kertas ini mengandungi 8 halaman bercetak
3472/2(PP) © 2012 Hak Cipta BPSBPSK SULIT
BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH
DAN SEKOLAH KECEMERLNGAN
KEMENTERIAN PELAJARAN MALAYSIA
PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 2012
PEPERIKSAAN AKHIR TAHUN (TINGKATAN 4)
ADDITIONAL MATHEMATICS
Paper 2
MARKING SCHEME
SULIT 2 3472/2(PP)
3472/2(PP) SULIT
No Answer Marks
1 52 xy or 2
5 yx
P1
57)52(2 xxx or 572
5
2
5 2
y
yy K1
01253 2 xx or 023203 2 yy
0)3()43( xx or 0)1()233( yy K1
3,3
4 xx or 1,
3
23 yy N1
1,3
23 yy or 3,
3
4 xx N1 √
2 (a) 04222 pxpxx or
042)2(2 pxpx K1
0)42()1(4)2( 2 pp K1
01242 pp
0)2()6( pp
2,6 p N1
(b) 04)6*(22)6*(2 xxx and try to solve K1
4x N1
3 (a) (i) 30log 3 = )523(log 3
= 5log2log3log 333 K1
= qp 1 N1
(ii) 50log 9 = 9log
50log
3
3 K1
= 2
5log22log 33 or
2
5log5log2log 333 K1
= 2
2qp N1
SULIT 3 3472/2(PP)
3472/2(PP) SULIT
No Answer Marks
(b) 5log
33 2 x
5log 2 x K1
52x K1
32x N1
4 (a) Area = 0240
0630
2
1
=
)2)(0()4)(6()0)(3()0)(6()2)(3()4)(0(2
1 K1
= 9 unit 2 N1
(b) (i) 2
3m K1
c )6()2
3(*2 or )6()
2
3(*2 xy K1
Note : * accept any value of m except 3
2
112
3 xy or equivalent N1
(ii) See AB : BC = 2 : 3 (or implied) P1
32
)2(6)3(3
x or
32
)2(2)3(4
y
K1 √(with
diagram)
5
8,
5
3B N1
5 (a) 32
1
dt
dx K1
2
3 N1
SULIT 4 3472/2(PP)
3472/2(PP) SULIT
No Answer Marks
(b) dx
dp
dp
dy
dx
dyUse or Differentiate 1)12(3 2 xy
26 p
dx
dy or )12()2(3 x
dx
dy or
equivalent
K1
= 1224 x N1
(c) 24x ─12 = 0 K1
1
2x N1
(d) 36 0.02y K1
= ─ 0.72 N1
6 (a) 14.20)(15
680 2 mean K1
kgmean 019.5 N1
(b) (i)
Class 30 – 39 40 – 49 50 – 59 60 – 69
Cumulative
frequency 2 5 9 10
N1
(ii) 39.5 or 2 or 5 K1
10
5
2)10(2
1
5.39
Median K1
5.45Median N1
7 (a) 42)( 2 xxxf
422 xx
4)1(1 22 x or equivalent K1
SULIT 5 3472/2(PP)
3472/2(PP) SULIT
No Answer Marks
5)1( 2 x N1
p = 1 , q = 5 N1 N1
(b) (i) ( – 1 , 5 ) N1
(ii) x = – 1 N1
(c) Maximum shape
Maximum point * (- 1,
5)
( - 3, 1) and (3, –11)
N1
N1
N1
Range of 5)(11)( xfxf N1
8 (a) 612
ba
, 63
ba
K1 K1
Solve simultaneously K1
4
5a ,
5
18b N1 N1
(b)
mx
nxfg
9
)( K1
x
x
mxn
x
1
39 K1
Use comparison K1
3,3 nm N1 N1
( - 1, 5)
(- 3, 1)
( 3, - 11)
4
f(x)
x O
SULIT 6 3472/2(PP)
3472/2(PP) SULIT
No Answer Marks
9 (a) (i) c )2(116 or )2(116 xy K1
18 xy N1
(ii) 304)18(13 xx Solve simultaneous Equations K1
B(12, 6) N1
(iii) 2,1Midpoint P1
12
2
x or 2
2
16
y K1
)12,4( C N1
(b) PDPA2
1 or 2PA = PD K1
22 )16()2( yx or 22 )2()14( yx K1
2222 )2()14()16()2(2 yxyx
08401324433 22 yxyx N1
10 (a) 60BOC
)142.3(180
60 K1
= 1.047 rad N1
Note : accept answer without working
(b) 047.1142.3 AOC
)047.1*142.3(5 ACS K1
Length of chord AC = 60sin)5(2 K1
Perimeter of the shaded region =
60sin)5(2)047.1*142.3(5 K1
19.135 / 19.14 cm N1
SULIT 7 3472/2(PP)
3472/2(PP) SULIT
No Answer Marks
(c) Area of sector AOC )047.1*142.3()5(2
1 2 K1
Area of triangle AOC = 120sin)5(2
1 2 K1
Area of shaded region = )047.1*142.3()5(2
1 2 -
120sin)5(2
1 2
K1
= 15.36 cm2 N1
11 (a) 124 k K1
3k N1
(b) 15
1normalm P1
- c )1(15
13 or 1
15
13 xy
K1
15
44
15
1 xy or equivalent N1
(c) Use 3x(x + 4) = 0 to find x and substitute x into y K1
Turning points (0, –10) and (– 4, 22) (both) N1
Use 126
2
2
xdx
yd or other relevant method to determine
maximum or minimum point
K1
pointminimum)10,0( ais N1
pointmaximum)22,0( ais N1
12 (a) (i) 00 80sin
40
50sin
QT K1
cmQT 11.31 N1
(ii) Use PQ= QT=QR or SQ = 10.37 cm P1
SULIT 8 3472/2(PP)
3472/2(PP) SULIT
No Answer Marks
0222 100cos)11.31()37.10(211.3137.10 SR K1
SR = 34.46 cm N1
(b) (i)
P1
(ii) 01 20sin37.10)(37.10
2
1A K1
02 100sin11.31)(37.10
2
1A
K1
100sin)11.31()37.10(2
120sin)37.10()37.10(
2
121 AA
K1
= 177.24 cm 2
N1
13 (a) 1000
1 Q
QIUse K1
1400x , 1320y , 105z N2,1,0
(b) ,108,72,36,54,90:W P1
360
)108(130)72(105)36(110)54(115)90(120 * I K1
25.118 N1
(c) equivalentorP
Use100
25.118*
60025
2010
272302010 P
(d) 90130
100117
2010
2012 orequivalentorIUse
K1
m = 10 N1
S
V R