skema permarkahan matematik tambahan kertas 2 … · guna kaedah persamaan serentak x = - x + 12...
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SKEMA PERMARKAHAN MATEMATIK TAMBAHAN KERTAS 2
SET 5
BAHAGIAN A
NO SOLUTIONS MARKS TOTAL
1.
xyoryx 44
24242424 2222 xxxxoryyyy
(y - 1)(y - 7) = 0 or (x - 3)(x + 3)=0
y = 1 , 7 or x = 3, -3
A (1, 3) , B( 7, - 3) or A( 7, -3), B( 1 , 3)
P1
K1
K1
N1
N1
5
2.
2121
11
033 2
,and,
,x
xdx
dya
3
1
3
1
03
1
3
1
039
39336
6)(
2
22
2
2
x
xx
xii
xxxx
xdx
ydib
K1
N1
N1
K1
N1
K1
N1
7
Or
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3 (a)
,
,
,
[ (
) ]
P1
N1
K1
N1
4
3(b)
(
)
or (
)
K1
NI
2
3 (c)
K1
N1
2
4.
. (a)
y – 0 =
(x – 0) or y – 7 =
(x – 10)
y = x and y = -
x + 12
Guna kaedah persamaan serentak
x = -
x + 12
Mereka akan bertemu pada titik (8, 8). (b) Jarak antara titik (0, 0) dengan titik (8, 8)
= √
= √ unit
Masa yang diambil = √
= 22.6 minit Mereka akan bertemu pada jam 11.23 pagi
K1
N1
K1
N1
K1
N1
6
8 marks
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5.
(i) Min Aiman = 4
93768158
or
Min Badrul = 4
90708365
Min Aiman dan Badrul = 77
S. piawai = 2
2222
774
93708158
or
S. piawai = 2
2222
774
90708365
9.975 and 12.59 (ii) Pencapaian Badrul lebih konsisten kerana sisishan
piawainya lebih kecil.
K1
N1
KI
NI
N1
NI
6
6. 2 2 2
2
2cos tan sec
2cos 1
cos 2
LHS
x x
x
x
or another method
Shape of cosine
Amplitude (maximum and minimum)
Shifted
Straight line (gradient or y-intercept) for
23cos 1 3x
x
or 3x
y
Number of solutions = 4
K1
N1
P1
P1
P1
P1
K1
N1
8
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BAHAGIAN B
NO SOLUTIONS MARKS TOTAL
7 (a)
Tan θ =
or θ = 45º
or 0.7855
P1
N1
2
7 (b)
8 x 0.7855 or 6.284
or 4π + 6.284 + 8
26.85 cm
K1
K1
K1
N1
4
7 (c)
or
or
Luas BCD =
-
-
OR
25.136 - 8 - 12.568
Luas tembereng AD =
-
OR
12.568 - 8
OR
4.558
9.136
K1
K1
K1
N1
4
8.
3
54
153
4
,
ya
K1
N1
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083
4
23
curve under the Area
3
51
2
1 triangleof Areab
3
4
0
3
xx
Area under the curve – area of triangle
2unit2
57
3
4
2
2
4
2
2
42
442
2
16
22
2
unit
yy
dxyVc
K1
K1
K1
K1
N1
K1
K1
N1
10
9
(a) (i)
(ii)
(b)
(i)
(ii)
c)
Gunakan hukum segitiga utk mencari
= -5p +3q
=
= -5mp + 3mq
=
=
atau
= (
)
Bandingkan pekali p dan q
atau
Selesaikan
K1
N1
N1
N1
N1
K1
K1
K1
N1, N1
10
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10
(a) (i)
(ii)
(b)
(i)
(ii)
(kedua-duanya)
53
3
8 )5
3()
5
2( C
0.2787
1 - 80
0
8 )5
3()
5
2( C + 71
1
8 )5
3()
5
2( C + 62
2
8 )5
3()
5
2( C
Atau
53
3
8 )5
3()
5
2( C + 44
4
8 )5
3()
5
2( C +…+
80
8
8 )5
3()
5
2( C
0.6846
P(114<X<150)
= P(-1<Z<1.25)
= 1 – 0.1587 – 0.1056
= 0.7357
P(X>150)
= P(Z>1.25)
= 0.1056
n(S) =
P1
K1
N1
K1
N1
K1
N1
K1
K1
N1
10
11
(a)
2.25 4.0 6.25 9.0 12.25 16.0
0.20 0.27 0.38 0.50 0.64 0.83
(b) Refer the graf
Paksi dan skalar seragam dan 1 titik betul
Semua titik diplotkan betul
Garis lurus penyesuaian terbaik
(c) y = p
=
= +
(i) = 0.1
p = 1.259
N1
N1
P1
P1
P1
P1
K1
N1
10
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(ii) =
= 0.04493
y = 1.109
K1
N1
BAHAGIAN C
NO SOLUTIONS MARKS TOTAL
12
(a)
(b)
(c)
(d)
v = 8
a = 2 – 2t = 0
2t = 2
t = 1
v = 8 + 2(1) – (1)2
= 9
v = 8 + 2t – t2 = 0
( 4- t) (t + 2) = 0
t = 4
dt)28(s 2tt
c3
ttt8s
32
t = 0 , s = 0 c = 0
3
ttt8s
32
t = 4 , s =
3
80
3
4)4()4(8
32
t = 6 , s = 123
6)6()6(8
32
Total distance =
12
3
80
3
80
= 3
124
N1
K1
K1
N1
K1
N1
K1
K1
K1
N1
10
or
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13.
(a) 60056
30002530
yx
yx or equivalent
20x or equivalent
10 xy or equivalent
(b) draw correctly at least one straight line
draw correctly three straight lines
shaded region
(c) (i) y maximum = 96
(ii) Use 30x + 25y i.e.
= 30(20) + 25(30)
= RM1350
N1
N1
N1
N1
K1
K1 N1
10
14.
.
(a) (i)
=
Sin
(i) = + – 2 x 9.8 x 5.2 x kos PSR kos PSR = 0.2768
PSR = - =
(ii) - - =
Luas segitiga PQR, =
x 12.3 x 9.5 x sin
= 56.48
Luas segitiga PSQ, =
x 9.8 x 5.2 x sin
= Luas segiempat PQRS = = 56.48 + = 80.96
(a) (i)
(i) - =
K1
N1
K1
N1
K1
N1
K1
N1
N1
N1
10
P’ Q’ Q
R’
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15.
(a) 12010040.2
x
RM
x = 2.00
8010050.2
00.2y
11010000.3
z
z =3.30
(b) Indeks gubahan =
36.127360
)30(110)50(80)90(125)40(120)50(150
127.5
(c)
10.140
100
1105.127
10.1401006
14 P
40.814 RMP
N1
N1
N1
K1K1
N1
K1
N1
K1
N1
10
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No. 11
Log y
0
0.5
𝒙𝟐 2 4 6 8 10 12 14 16
0.1
0.2
0.3
0.4
0.6
0.7
0.8
0.9
x
x
x
x
x
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y
0
100
6x + 5y = 600
y=x+30
x
(a) or equivalent N1
or equivalent N1
or equivalent N1
(b) N1 draw correctly at least one straight line
N1 draw correctly three straight lines
N1 shaded region
(c) (i) y maximum = 96 N1
(ii) Use 30x + 25y i.e.
Titik (20,30) K1
= 30(20) + 25(30) K1
= RM1350 N1
R
x = 20
120
(20, 30)
20
40
60
80
20 40 60 80 100 120
30x + 25y = k
NO 13
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