Upload
ainun-syakirah
View
31
Download
6
Embed Size (px)
DESCRIPTION
ADDMATH
Citation preview
1
3472/1
Matematik Tambahan Kertas 12 jam Ogos 2012
BAHAGIAN PENGURUSANSEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN
KEMENTERIAN PELAJARAN MALAYSIA
PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 2012 PERCUBAAN SIJIL PELAJARAN MALAYSIA
ADDITIONAL MATHEMATICS
Paper 1
Skema Pemarkahan ini mengandungi 6 halaman bercetak
MARKING SCHEME
Muat turun (percum
a) skema lain di : w
ww
.myschoolchildren.com
Muat turun (percum
a) skema lain di
2
PERATURAN PEMARKAHAN- KERTAS 1
No. Solution and Mark Scheme Sub Marks
Total Marks
1(a) 25 1 2
(b) 4 1
2(a) 2 1 3
(b) 4 2
B1:2x
OR3y
2 [use k 1(2) y].x 3 y 2
3(a) x 3 2 3
B1:5
5
f (x) x 3
(b) 1 1
4 m 1
B2: (6)2 4(2 m)(3) 0
B1: 2 mx 2 6x 3 0
3 3
5 2 p 1
B2 : ( p 2)( p 1) 0 or -2 -1
B1: p 2 3 p 2 0
3 3
6(a) x 1 1 3
(b) 1 1
(c) (1, 4) 1
7 x 2
B2: 2(2x 3) 1 5 6x or 2x 3 5 6x
2
2(2 x3) 1
B1: 3 2 35 36 x
3 3
Muat turun (percum
a) skema lain di : w
ww
.myschoolchildren.com
Muat turun (percum
a) skema lain di
MMuuaatt ttuurruunn ((ppeerrccuummaa)) sskkeemmaa llaaiinn ddii ::
wwwwww..mmyysscchhoooollcchhiillddrreenn..ccoomm MMuuaatt ttuurruunn ((ppeerrccuummaa)) sskkeemmaa llaaiinn ddii wwwwww..mmyysscchhoooollcchhiillddrreenn..ccoomm
www.myschoolchildren.com
6
25
or
B1 log5
4B
oreq
uiva
lent
1
log5
9
(
1
(*
B1
B2
2
3
4 3 3 3
4 3 1 2 3
1
6253
n or6
25n3
m
n 3
m n 3
B1:
log5
m(f
or c
hang
e ba
se)
20) =
– k3(
20
20
- 7 +
7)) =
1
] OR
h –
(-
1)
]
d
h
11 [b
oth
k =
11
d =
20
- 7 +
3(
1 2
3
1)
9(h
1 or
h =
2 a
nd k
=
B2:
h =
2 o
r
B1:
- 7 +
3 O
R
r23
3 10
S
75h
55 10
[2(3
h
B1:
a
3h
8 10 (b)
11
4
12
(a)
(b)
p 3
q
B1:y pq
1 p or
x x
5
pq 3
2
1
3
13 y 3
x 5
or 4y 6x 5 or equivalent2 4
B3 : y 1 3 x
3
2 2
B2 : m 3
or 3
,12 2 2
B1 : m2
OR S(3,0) and T(0, 2)1
3
4 4
14 x2 y2 4x 4 y 92 0
B2 : (x 2)2 ( y 2)2 (6 (2))2 (4 2)2
B1 : PS = PQ OR (6 (2))2 (4 2)2
3 3
15(a) 5i 12 j
1 513 12
1 3
(b)
5 i
12 j or
13 135i 12 j
13or 2
B1 : | OR |=13
Muat turun (percum
a) skema lain di : w
ww
.myschoolchildren.com
Muat turun (percum
a) skema lain di
5
16 k 3
4
B2: 4k 3 0
B1: (4k 3)i (4k 2) j
3 3
17(a)
(b)
0.9506 rad / 0.9505 rad / 0.9507 rad
15.36 or 15.355
B1 : arc OS = 5 (*0.9506) or PS = 3.602
1
2
3
18 x =15°,75°,195°,255° 3 3
B2 : 2x 300 ,1500 , 3900 , 5100
or sin 2x = 1
2
B1 : 2(2sin x cos x) 1
19(a) 80 32x 1 3
(b)x 2.5 or x
5 2
2B1 : 80 32x 0
20 y 4x 5
or equivalent2
B2 : y 3 4x 1 or equivalent
2
B1 :dy
4 ordy
(1) 3dx dx
3 3
MM
uuaatt ttuurruunn ((ppeerrccuumm
aa)) sskkeemm
aa llaaiinn ddii :: ww
ww
ww
..mm
yysscchhoooollcchhiillddrreenn..ccoomm
6
21(a) 10
3
B2 : hx3
5
71 2 2
3 3 f (x) 7B1: hdx 2
dx 21 1
1 4(b)
3
22(a) 3
1.648 or 1.6475
B2:163
( 51
)2 or equivalent20 20
51 2(0)2 3(1)2 2(2)2 8(3)2 5(4)2
B1 : x or 2 20 20
1 4
(b) 3
23(a) 252 1 3
(b) 66 2B1 : 4C 6C or 60 OR 4C 6C or 6
3 2 4 1
24(a) 110
B1 : 3
1
2
5 4 3
1920B1 : 1
3
1
1 5 4 3
2 4
(b)2
25(a) 0.1741 1 4
(b) 49.69 3
B2 : 0.938 k 45
5B1 : z = 0.938
MM
uuaatt ttuurruunn ((ppeerrccuumm
aa)) sskkeemm
aa llaaiinn ddii :: ww
ww
ww
..mm
yysscchhoooollcchhiillddrreenn..ccoomm
1
3472/2
Matematik Tambahan Kertas 2Ogos 20122 ½ jam
BAHAGIAN PENGURUSANSEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN
KEMENTERIAN PELAJARAN MALAYSIA
PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 2012 PERCUBAAN SIJIL PELAJARAN MALAYSIA
ADDITIONAL MATHEMATICS
Paper 2
Skema Pemarkahan ini mengandungi 10 halaman bercetak
MARKING SCHEME
MM
uuaatt ttuurruunn ((ppeerrccuumm
aa)) sskkeemm
aa llaaiinn ddii :: ww
ww
ww
..mm
yysscchhoooollcchhiillddrreenn..ccoomm
Muat turun (percum
a) skema lain di : w
ww
.myschoolchildren.com
Muat turun (percum
a) skema lain di
2
No Solution and Mark Scheme Sub Marks
Total Marks
1x
8 3yOR y
8 2xP12 3
2
y2 3y 8 3y 6 0 OR 8 2x
3x 8 2x 6 0 K1 2 3 3
Replace a, b & c into formula K1
(24) (24)2 4(7)(12) (20) (20)2 4(7)(59)y OR x
2(7) 2(7)
y 0.443, 3.871 OR x 4.664, 1.807 N1x 4.664, 1.807 OR y 0.443, 3.871 N1
5 5
2 (a)
y x2 2x 3k
y (x 1)2 1 3k K11 3k 4 K1
k 1 N1
(b)
(1,4)3
-1 3
-Maximum shape P1-*Maximum point K1-Another 1 point y-intercept / x-intercept K1
3
3
6
ww
w.m
yschoolchildren.com
kkkeemm
aaa lllaaiinnn dddiii ::: aaaiinnn dddiii
oM
MM
uuaattt tttuurruu)
uuunnn cccuuumm
aa))
3
3(a) 16,8, 4,...... OR r 1
2
1 n
16 1 a(1 rn ) 2 30.5 Use Sn
1 1 r1
2
n 4.416
n 5
64 ,16 , 4 ,......OR r 1
4
S64
1 a1 Use S
4 1 r
= 85 1 or 85.33
3
P1
K1
K1
N1
P1
K1
N1
4 7
(b) 3
4(a)(i) Change 3y x 6 0to y
1 x 2 or m
3 BC
OR mAB 3
y 5 3 x (6) OR any correct method
y 3x 23(ii) Use simultaneous equation to find point B*y 3x 23 and 3y x 6 0 or y
1 x 2
3
B = 15
, 1
2 2
* 15
, 1
2(x) 3(6) , 2( y) 3(5)
2 2 5 5
D = 39
, 25
4 4
13
K1
K1
N1
K1
N1
K1
N1
5 7
(b)2
Muat turun (percum
a) skema lain di : w
ww
.myschoolchildren.com
Muat turun (percum
a) skema lain di
4
5(a)y y
5x 2
3
y 3sin 2x
O 3 2 x
2 2
–3
Amplitude = 3 [ Maximum = 3 and Minimum = − 3 ] P1 Sine shape correct P1Two full cycle in 0 x 2 P1Negative sine shape correct(reflect) P1
3sin 2x 5x
2 or y 5x
2 N1
Draw the straight line y 5x
2 K1
Number of solutions is 3 N1
4 7
(b)3
6(a) L = 79.5 OR F = 24 OR fm = 4 P1 3
(36) 24 79.5 4 10 K1
4
87 N1
3 8
(b) (i) 5
X (44.5 4) (54.55) (64.5 6) (74.59) (84.5 4)
36
OR 2602K1
36= 72.28 N1
MM
uuaatt ttuurruunn ((ppeerrccuumm
aa)) sskkeemm
aa llaaiinn ddii :: ww
ww
ww
..mm
yysscchhoooollcchhiillddrreenn..ccoomm
5
(ii)(44.5)2 4 (54.5)2 5 (64.5)2 6 (74.5)2 9 (84.5)2 4 (94.5)2
8 K1
197689
(*72.28)2
36 K1 16.34 N1
7 Rujuk Lampiran
8(a)4 2 K1
y x3 dx
x2 c
(4) 2
c , c 2 K1
(1)2
y 2
2 N1
x2
2 2 x2 2 dx
52
2x1 K1 1
2x 5 2(2)1 2(5)1 K1
1 2(2) 1
2(5)
33
or 6.6N1
52 2(i) Volume ( 2)2 dxx2 K15
2 4x3 8x1
4x K1 3 1 5
2 4 ( 2 ) 3 8 ( 2 ) 1
2 4 5 3 8 5 1 3
1
4 2 3
1
4 5 K1 5 5
14.56 N1
3 10
(b)
3
(c )4
MM
uuaatt ttuurruunn ((ppeerrccuumm
aa)) sskkeemm
aa llaaiinn ddii :: ww
ww
ww
..mm
yysscchhoooollcchhiillddrreenn..ccoomm
6
9(a) AC 7 x 5 y
OD 7x 3
(7 x 5 y)4
7 x
15 y4 4
3 7 x 5 y h 5 y k 7
x 15
y 4
4 4
21 7k
4 4k 3
15
5h 15
k4 4
h 3
50 1 5 4 t
2t 5
N1
K1
N1
K1
K1
N1
K1
N1
K1
N1
3 10
(b)5
(c)
2
10(a) (i) P X 6 10C 0.36 0.74K1 5 10
= 0.03676 N1
(ii) 10C 0.39 0.71OR 10C 0.310 0.70 K1
9 10
P X 9 10C 0.39 0.71 10C 0.310
K19 10
= 0.0001437 N1
5
MM
uuaatt ttuurruunn ((ppeerrccuumm
aa)) sskkeemm
aa llaaiinn ddii :: ww
ww
ww
..mm
yysscchhoooollcchhiillddrreenn..ccoomm
7
(b) (i) P40 X 48 P 40 45
Z 48 45
3.5 3.5
= 0.7278
(ii) P X m 0.7m 45
0.524 3.5
m 43.166
K1
N1
K1
K1
N1
11(a) OR RQ PR 7cm tan 1
rad 0.7855 rad
4
7(1.571) OR 7(2.3565)
72 72 2(7)(7)(cos135o )
Perimeter
7 7 7(1.571) 7(2.3565) ( 72 72 2(7)(7)(cos135o )
54.4268
1 72
4
1 72 2.3565
1 72 sin135o
2 2
Area 1 72
1 72 2.3565
1 72 sin135o
4 2 2
78.8996
K1
N1
K1 K1
K1
N1
K1
K1
K1
N1
2 10
(b)4
(c)
4w
ww
.myschoolchildren.com
Muat turun (percum
a) skema lain di : w
ww
.myschoolchildren.com
Muat turun (percum
a) skema lain di
8
No Solution and Mark Scheme Sub Marks
Total Marks
12(a)a = 10 - 5t = 0 Use a = 0 K1t = 2 sv 10t
5 t 2 c Integrate a to K12 find v
30 10(0) 5
(0)2 c2
c = 30v 10t
5 t 2 30
2 Integrate and K1v 10(2)
5 (2)2 30
substitute t = 2
2 N1= 40 ms- 1
v 10t 5
t 2 30 0 Use v > 0 K12
t 2t 6 0 K1
0 t 6 N1
s 5t 2 5
t 3 30t c Integrate v dt K1
6s = 0, t = 0 , c = 0
s 5t 2 5
t 3 30t 6
s 5(6)2 5
(6)3 30(6) or s 5(8)2 5
(8)3 30(8) K16 6
= 180 = 133.33
Total distance = 180 + 180 133.33= 226.67 m N1
OR
4 10
(b) 3
(c) 3
P
MMMuuaattt tttuurruunnn (((ppeerrccuummaa))) ssskkkeemmaaa lllaaiinnn dddiii :::
wwwwwww..mmyysscchhoooollcchhiillddrrr
nn..ccoommm MMMuuaattt tttuurruuunnn (((ppeerrcccuuummaa))) ssskkeemmaaa lllaaiinnn dddiii wwwwwww..mmyysscchh
llcchhiillddrrreenn..ccoommm
55
1
0t t
2
30
dt
1
0t t
2
30
dtIn
tegr
ate
v2
20
6
K
2K
8 8
I
9
102 2 2 4
6
8
K1
6 8
1
80
46.
67
+
K1
06
2
26.6
7N1
55
100
125
K1
P10
= R
M 4
4N1
110
h
125
4
140
h
3
88 5
h
4
h
3
5* h
= 1
N1
P11
10
0
115
P07
= R
M 2
3N1
See
125P
1I S
=
125
110
K1
110
*1
1
25
4
140
*1
3
1
10
5 K
11
4
4
5=
122
.86N
1
13(a
)
(b)
(c)
(d)
Muat turun (percum
a) skema lain di : w
ww
.myschoolchildren.com
Muat turun (percum
a) skema lain di
14 Rujuk Lampiran
15(a) Using sine rule to find BAC .sin BAC sin 30o
27 14
BAC 74.64o
BAC (obtuse) 180o 74.64o
105.36o
DCB 105.36o 30o or DC 6 cm
Use cosine rule to find BD.
BD2 62 272 2 627cos135.36
BD 31.55
Use formula correctly to find area of triangle ABC or ACD.ABC 180o 30o 105.36o
44.64o
AC 2 272 142 2 2714cos 44.64
AC 19.67
Area ABC 1
(14)(27) sin 44.64o or2
Area ACD 1
(6)(19.67) sin105.36o
2
Use Area ABCD = sum of two areas
Area ABCD = 189.7 cm2 .
K1
N1
N1
P1
K1
N1
K1
K1
K1
N1
3 10
(b) 3
(c) 4
END OF MARKING SCHEME
10
m x
Pentaksiran Diagnostik Akademik SBP 2012 Paper 2 ADM
No.7(a) 0.1 0.2 0.25 0.4 0.5 0.8 N1
y 62 54 50 38 29 4
y n 1 2m P1
Plot against K1x
(at least one point)
6 points plotted correctly K170
Line of best fit N1
x60
x50 x
ii.
iii.
N1
m 80
n K1
n 2800 N1
1 0.37
x K1
x 2.703 N1
40x
30 x
20
10
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
x
0.8
MM
uuaatt ttuurruunn ((ppeerrccuumm
aa)) sskkeemm
aa llaaiinn ddii :: ww
ww
ww
..mm
yysscchhoooollcchhiillddrreenn..ccoomm
8
MMuuaatt ttuurruunn ((ppeerrccuummaa)) sskkeemmaa llaaiinn ddii ::
wwwwww..mmyysscchhoooollcchhiillddrreenn..ccoomm MMuuaatt ttuurruunn ((ppeerrccuummaa)) sskkeemmaa llaaiinn ddii wwwwww..mmyysscchhoooollcchhiillddrreenn..ccoomm
N 1
N 1 N 1 N 1
(I.
N
x
y
II
.N
Ans
w
y 2
x
I
(R
efer
to th
e
1 gr
aph
co
rrec
tK
N
Cor
rect
are
a
(c)
max
poi
nt (
i)k
= 10
0x +
80
y M
ax
= 10
0(50
)+ 8
0(12
0)K
= R
M 1
4,
N10
80
y
ii
(50,
120)
x0
23
41
56
78
1 1 1 1 8
www.myschoolchildren.comAns
w
1 6 4 2