spm trials 2010 - add maths p2 - putrajaya (scheme)
TRANSCRIPT
SULIT3472/2Additional MathematicsPaper 2Sept / Okt2010
PEJABAT PELAJARAN WILAYAH PERSEKUTUAN PUTRAJAYA KEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN PERCUBAAN SPM TINGKATAN 5
2010
ADDITIONAL MATHEMATICS
Paper 2
MARKING SCHEME
This marking scheme consists of 11 printed pages
SKEMA PERMARKAHAN ADDITIONAL MATHEMATICS KERTAS 2 PEPERIKSAAN PERCUBAAN SPM 2010
Number Solution and marking scheme Sub Marks
Full Marks
1
@
@
@
and and
and and
P1
K1
K1N1
N1
5
2 (a)
( b)
y=-5(x2 + )
Turning point = (
= 2 – 3k
k=
K1
N1
K1
N1
2
Number Solution and marking scheme Sub Marks
Full Marks
6
(c)
N1 shape
All correct
N1
3 (a)
(b)
x, x + y , x+2y, a = x , d = y given T = 200 x + 5y = 200 ……………………….(1)
and S = 2520
……………(2)
Solve (1) and (2) y = 20 , x = 100
T of Encik Ali = T of Encik Tan 100 + (n-1)20 = 200 + (n – 1)10 100 + 20n – 20 = 200 + 10n – 10 10n = 110 n = 11.
K1 ( either (1) or (2)
K1N1N1
K1K1
N1 7
4 (a)
K1
N1
3
(
3
-54
-9
-20
Number Solution and marking scheme Sub Marks
Full Marks
( b )
( c )
Sketch straight line
*to get N1 for no. of solution all marks for graph must 5 marks
P1
P1
P1
N1
K1
N1 8
5 (a)
(b )
The mean =
= 40.25 kg
(20) = 5th observation
First quartile’s class = 35 – 39
= 36.5
(20) = 15th observation
Third quartile’s class = 40 – 44
Third
= 43.875 The interquartile range = 43.875 – 36.5 = 7.375 kg
K1N1
P1 (34.5 or 39.5 )
K1
K1
K1N1
7
6 ( a) Gunakan
K1N1
7
(b) (i)
(ii)
Gunakan
K1N1
K1
K1
4
Shape of sin graph
No of solution = 2
Amplitude (max=4, min=0 )
1 periodic /cycle
Number Solution and marking scheme Sub Marks
Full Marks
N1
7 (a)
b (i)
( ii )
<4)
m = 6
equation of the curve,
K1
K1
K1
K1N1
K1
N1
K1
K1
N110
5
Number Solution and marking scheme Sub Marks
Full Marks
8 a)
b) Lihat graf
c) i)
ii)
iii)
x 15 20 25 30 35 40lg y -0.82 -0.42 -0.022 0.37 0.77 1.17
Using the correct axes and uniform scale6 points plotted correctly *Line of best fit
log10 y = x log10 c –k or –k log10 10y-intercept, c = -k or c = –k log10 10
- 2.00 = - k
k = 2.00
m = log10 c c = 1.2
x = 37.5
N1
N1N1N1
P1
K1N1
K1N1
N1 10
9 (a)
( b)
(c )
( d )
mPQ=
mAC=mPC=
or equivalent
m for perpendicular bisector of AC =
(x - 4)
5y = 3x - 2
KQ = 6
= 6
x2 + y2 - 16x - 8y + 44 = 0
K1
K1
K1
N1
K1
K1N1
K1K1N1
10
10. (a)
(b)Area of
N1N1
K1
K1
N1
6
Number Solution and marking scheme Sub Marks
Full Marks
10
( c ) K1
N1
K1 ( either one )
K1 N1
11 a) (i)
( ii )
P(X>3) = P(X=4) + P(X=5) + P(X=6)
=0.98415
P1K1
N1
K1N1
10
(b) i)
ii)
P(X>54) → P(Z>
= 0.2266
P(X<m) = 0.08
Z =
P(Z< ) = 0.08 From standard normal table,
P(Z<-1.406) = 0.08
Therefore, = -1.406
m = 28.13
K1
N1
K1
K1
N112 (a)
( b )
The initial velocity = 0 ms-1
K1N1
Limit K1
K1K1
7
0.08
z O
Number Solution and marking scheme Sub Marks
Full Marks
(c )
=
=
= 14.67 m Distance travelled during the third second = 14.67 m
When v < 0,
The object moves to the left when t > 4
K1
N1
K1
K1
N1
10
13(a) i)
(ii)
= 116.670 BD = 10.80 cm
cos BCD =
= -0.9221 BCD = 157º14’
K1
N1
K1
N1
10
8
Number Solution and marking scheme Sub Marks
Full Marks
( b )
i)
ii)
C
D’ B
BDC = 14º31’ DBC = 180º - 157º14’ - 14º31’ = 8º15’ BD’C = 180º - 14º31’ = 165º29’ BCD’= 180º - 165º29’ - 8º15’ = 6º16’
The area of ΔBCD’
= 1.528 cm2
N1
K1
K1
K1
K1
N1
1014. (a)
(b)Lihat graf
(c ) i)
ii)
Draw correctly at least one straight line
from the *inequalities which involves and
Draw correctly all the three *straight lines Note : Accept dotted lines
Region shaded correctly
N1N1N1
K1
K1N1
N1
K1N1
9
Number Solution and marking scheme Sub Marks
Full Marks
Use for point in the shaded region
Maximum point (33, 67)
RM 9 320
N1
15 (a)
(b )
(c ) (i)
( ii)
Use or b/20 x 100 = 125
,
b = 130
: 143
; RM28.60
K1
N1 N1
P1K1N1
K1 N1
K1 N1
10
10
11
100
90
80
70
60
50
40
30
20
10
R(33, 67)
75
50
NO 14
R
y = 30
x + y = 100
y = 2x
y
12
5 10 15 20 25 30
x
-2.0
-1.5
-1.0
-0.5
0
0.5
1.0
1.5
y10log
x
x
x
x
x
x
No.8(b)
35 40
-2.5
10 20 30 40 50 60 70 80 90 100
x
13