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Bahan Pecutan Akhir Add Math SPM
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2
No Solution and Mark Scheme Sub
Marks
Total
Marks
1
24 3 3 2 3y xy x x y
3 2x y
3
2
xy
2
2
4 0
4 3 2 3 2 0
y xy x
y y y y
2
2
4 0
3 34 0
2 2
y xy x
x xx x
2 1 3 0y y 2 9 0x x
1@ 3
2
2@9
y
x
2@9
1@ 3
2
x
y
1
1
1
1
1
5
2(a)
2 2
4 2 3 2
81 27
3 3
8 3 6
6
5
x x
x x
x x
x
1
1
1
7
3
(b)
2 2
2 2
2
4
2
32
3
log 16log 1
log 2
log 216
216
64
4
xx
x
x
x
x
x
1
1
1
1
3(a)
2
3
2220500 1.05
200000
T ar
a
a
2
7
(b)
(ii)
1
1
400000
200000 1.05 400000
1.05 2
1 log1.05 log 2
1 14.21
15.21
16
n
n
n
T
n
n
n
n
2008 415
2009 419.5
2010 423.34
2011 427.57
2012 431.85
2013 436.17
2014 440.53
2993.62
7
7
415 1.01
415 1.01 1
1.01 1
2993.62
a r or
S
5
4
4(a)
(b)
2
sin 2sin cos
1 cos 2cos 1
sin 1 2cos
cos 1 2cos
tan
x x x
x x
x x
x x
x
Proven / Terbukti
Bentuk tan x (di graf) P1
Period (di graf) P1
Straight Line (di graf) P1
1x
y
P1
No Of Solution = 2 N1
2
5
7
2
5
5(a)
(b)
2
3
3
3
y x x
y x x
2
2
3 3 0
1 0
1
1, 2
1, 2
dyx
dx
x
x
x y
x y
2
26
1, 6 min
1, 6 max
d yx
dx
dyx imum
dx
dyx imum
dx
Maximum point is 1,2
0.02x
22, 3 2 3 9dy
xdx
9 0.02
0.18
y
y
4
3
7
6(a)
Find the coordinates of B(6,0) or coordinates C( 0,4)
Find midpoint of BC 3,2
Gradient of BC = 2
3 .
Gradient of the line perpendicular to BC = 2
3
2m
The equation of perpendicular bisector to BC is :
3
2 32
3 5
2 2
2 3 5
y x
y x
or
y x
7
7 N1
N1
K1
N1
6
(b)
Area of triangle ABC
2
14 0 6 4 0 2 2 6 0 4 4
2
124 28
2
26
area
unit
No Solution and Mark Scheme Sub
Marks
Total
Marks
7(a)
(b)
(c)
x
1
1.5
2
2.5
3
3.5
log10 y
1.50
1.22
0.93
0.66
0.34
0.10
Graf = Rujuk Lampiran
Plot 10log y against x and correct axes and uniform scales.
6 points plotted correctly N1
Line of best fit N1
* If table not shown but all points are correctly plotted , award N1 mark.
10 10 10log 3log logy k x p
(Can be implied)
(i) 103log 0.56k
0.65k
K1
K1
P1
N1
K1
N1
N1
K1
7
(ii) 10 log 2.04p
109.65p
(iii) 10 1.38
23.99
loy y
y
8(a)
(b)
10tan
6
59.04
1.031
2 1.031 6
31.518
CD
CD
11.66 1.031
12.02
AB
AB
2 210 6
11.66
OB
OB
11.66 6
5.66
BD AC
Perimeter of shaded region
31.518 5.66 5.66 12.02
54.858
2
5
10
(c )
Area of sector
2111.66 1.031
2
70.09
Or
3
K1
N1
N1
8
Area of right angle triangle
16 10
2
30
Area of shaded region
70.09 30
40.09
9(a)
(i)
(ii)
(b)
Use the triangle law for or PS QR
9
9
OP PS OS
y PS x
PS x y
5 3
3 5
OQ QR OR
x QR y
QR y x
=m 9 ST y x
=n 5 3 RT x y
3
7
10
K1
N1
N1
N1
N1
9
Use the triangle law to find TQ :
4 9 5 3 5 3
4 9 5 3 5 3
4 5 5 9 3 3
1 5 9 1 5 3 3
11 5 9 45 3 3
7
242 6
7
1
7
ST TQ SQ RT TQ RQ
TQ x m y x TQ x y n x y
x m y x x y n x y
compare
m n m n
m n n n
m n n
m n
n
10
(a)
2dy
xdx
2(1) 2Tm
Equation of tangent
2 2 1
2
y x
y x
3
10
(b) Area of triangle Area under the curve
1
1 22
1
1
2
0
13
0
1
3
11 0
3
4
3
x dx
xx
4
N1
K1
K1
N1
K1
10
Area of shaded region = 24 11
3 3unit
(c)
2
1
22
1
2 2
2
1
2
(2) (1)2 1
2 2
1
2
y dy
yy
unit
3
11
(a)
(i)
(ii)
(b)
(i)
8, 0.8, 0.2n p q
8 0 8 8 1 7 8 2 6
0 1 2
2 0 1 2
0.8 0.2 0.8 0.2 0.8 0.2
0.7969
P X P X P X P X
C C C
250 0.8 0.2
6.325
npq
110, 4
100 120
100 110 120 110
4 4
2.5 2.5
0.9876
P X
P Z
P Z
5
5
10
11
No Solution and Mark Scheme Sub
Marks
Total
Marks
12
(a)
(b)
2
2
32
32
2 18 28
4 18
2 18 28
29 28
3
0, 0, 0
29 28
3
v t t
dva t
dt
s vdt t t dt
ts t t c
s t c
ts t t
2
, 0
18
at A t
a ms
2
1
2 18 28
, 0
28
initial
v t t
v t
v ms
2
2
10
(ii)
100
100 110
4
2.5
0.00621
P X
P Z
P Z
0.00621480
2.98
n AP A
n s
n A
n A
Bilangan helaian yang ditolak ialah 3
12
(c)
2
2
2 18 28 0
9 14 0
2 7 0
2 7
v t t
t t
t t
t
3
(d)
distance s
3 2
3 2
2
2
29 28
3
22 9 2 28 2
3
125
3
t
t
s t t t
s
s
dt
3 2
7
2
27 9 7 28 7
3
116
3
t
t
s
s
The distance is
2 72
1 12 25 16
3 3
67
t ts s s
s
s
3
72
13
13(a)
10
10
116 100250
290
Q
Q
2
10
(b)
10/ 08
10/ 08
115
130 100
149.5
I
I
2006 2008 2010
2006 100 115 10/ 08I
2008 - 100 130
2010 - - 100
2
(c)
I w Iw
110 13 1430
116 25 2900
130 40 5200
x 10 103x
120 12 1440
100 10970 + 103 x
10970 10120
100
103
xI
x
4
14
(d)
08
08
1599120 100
1332.5
Q
Q
2
14a
(i)
2 2 210.5 12.5 2 10.5 12.5 cos80
14.86
QS
QS
2 10
(ii)
0
0
sin sin 35
14.86 9.5
sin 0.8972
63.79
116.21
R
R
QRS
QRS
2
b(i)
1
RQ'
Q
P
15
(ii)
*
2
sin sin80
12.5 14.86
sin 0.8284
55.94
'
1(10.5)(10.5)sin 68.12
2
51.15
Find PQS
Q
Q
PQS
Area of PQQ
cm
0110.5 12.5 sin 80
2
64.63
'
64.63 51.15
13.48
Find area of PQS
Area of PQ S
5
15(a)
(b)
(c)
1.8 2 240x y
3 2 600x y
50y x
Rujuk Lampiran
(i) Jambu batu sebanyak 66 pokok
(ii) Maximum point = (100 ,150)
Maximum total fees = 120 ( 100 ) + 180 (150 )
= 39000
16
Soalan 7(a)
2.0
1.8
1.6
1 0.5 1.5 2 2.5 3
0.2
3.5
0.4
0.6
0.8
1.0
1.2
1.4
X
X
X
X
X
X
log10 y
20
17
Soalan 15(b)
40 20 60 80 100 120 x
20
140
60
80
100
120
140
y
160
160
180
R
(100, 150)
0
40
180