MM2 Marking Sbh
Transcript of MM2 Marking Sbh
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MARKING SCHEME PAPER 2 EXCEL 2
Question Solution and Mark Scheme Marks
1 (a)
(b)
P1
P2 3
2
( )
23 5 2 0
3 1 ( 2) 0 or equivalent.
2
1or 0.3333
3
n n
n n
n
n
=
+ =
=
=
Note:
1. Accept without = 02. Do not accept solutions solved not using factorization.
K1
K1
N1
N1
4
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3 13 6 or 7 or equivalent
2x y x y+ = =
55 20 or 5 or equivalent
6y x= =
or
1414 2 or or equivalent
2
55 20 or 5 or equivalent
6
xx y y
y x
= + =
= =
(K1)
(K1)
or
3 2 141
1 1 61(3) ( 2)(1)
6
4
x
y
x
y
=
=
=
(K2)
Note:
3 211.
1 11(3) ( 2)(1)
62. as final answer, award N1
4
x
y
=
K1
K1
N1
N1
4
4 Identify orTRS SRT
0'13.53853
6
8tan
or
equivalentor
o=
=
P1
K1
N1
3
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5(a)
2 1
6 ( 3) 3
t=
t = 5
(b)
2 0
3 ( 4)RSm
=
y = 2 x 7
(c) y-intercept is 7 or c = - 7.
K1
N1
K1
N1
N1
5
6 (a)
135 22 45 222 14 or 2 21
360 7 360 7
135 22 45 222 14 + 2 21 7 21 14
360 7 360 7
or equivalent
1 18391 or 91.5 or
2 2
+ + +
(b)
2
2
45 22 121 or 14 14
360 7 2
45 22 121 14 14
360 7 2
1 30175 or 75.25 or4 4
Note:
1. Accept for K mark.2. Correct answer from incomplete working, award KK2.
K1
K1
N1
K1
K1
N1
6
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7 (a) 3 is an odd number or 14 ( 2) 7 = .
(b) sin 135o is positive.
(c) A + B = 90oA and B are complementary angles
P2
K1
P1
P1 5
8 (a) n = 3
1 (4) ( - 3)(2) = m(3) 2
m = 4(b)
4 3 61
2 1 21(4) ( 3)(2)
3
1
3
1
p
q
p
q
p
q
=
=
=
=
Note:
1.
*p 6
q 2
inverse
matrix
=
or
4 31seen, award K1.
2 11(4) ( 3)(2)
2. Do not accept
* *1 3 1 0
or .2 4 0 1
inverse inverse
matrix matrix
= =
3.3
as final answer, award N11
p
q
=
4. Do not accept any solutions solve not using matrices.
P1
K1
N1
K2
N1
N1
7
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9 (a) (4,8), (4,9)
2 1or
24 12
(b) (4, G), (4, J), (4, K)
(B, 8), (B, 9), (F, 8), (F, 9)Note: Any two groupings correct, award K1.
7
24
K1
N1
K2
N1
5
10 (a) 12s
(b)
2 2
10 3
4
7or 1.75
4ms ms
(c)
1 1(3 10)4 (16 4)10 (10 22)( 16) 306
2 2
26
t
t s
+ + + + =
=
P1
K1
N1
K2
N1
6
112
3
2 3
223
7
1 4 223
2 3 7
22 1 4 22 63 3 282
7 2 3 7 7
12
t
t
cm
=
Note:
1. Accept for K mark.2. Correct answer from incomplete working, award KK2.
K1
K1
K1
N1
4
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Question Solution and Mark Scheme Marks
12(a) (i) (7,5)
Note :
Award P1 if coordinate (7,5) marked on diagram orcoordinate (4,7) indicated in answer space.
[P2]
(ii) (1,5)
Note :Award P1 if coordinate (1,5) marked on diagram or
coordinate (5,5) indicated in answer space.
[P2] 4
(b) (i)
(ii)
U : Reflection in the line GH ory = 9-x
Note :
1. Award P1, if the word reflection seen.
V : Enlargement at centre (9,4) with scale factor of 2.
Note :
1. Award P2, if Enlargement at centre (9,4) orEnlargement scale factor of 2 seen.
2. Award P1, if the word Enlargement seen.
[P2]
[P3] 5
(c) Area of image = scale factor2 x Area of object
336 = 22 x Area of objectArea of object = 336 4
= 84 unit2.
[K2]
[N1] 3
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Question Solution and Mark Scheme Marks
13(a) 2
5.5
[K1]
[K1] 2
(b) Graph
1. Axes drawn in correct direction with uniformscales in
-2.5x 3 and -8y 30.5
2. All points correctly plotted or curve passesthrough these points for -2.5x 3
3.
Smooth and continuous curve without anystraight line and passes through all correct
points
Note :
1. 6 or 7 points correctly plotted, award K1
[P1]
[K2]
[N1] 4
(c) (i) 9 y 10 [P1]
(ii) -2.2x -2.1 [P1] 2
(d)y = 4x2 5x 7
- 0 = 4x2 + 2x 2
y = 7x 5
Straight line y = 7x 5 correctly drawn
0.4x 0.6
-1.1x -0.9
Note :1. Allow P mark or N mark if values ofy andxshown on graph.
2. Values of y and x obtained by computations,award P0 or N0.
[K1]
[K1]
[N1]
[N1] 4
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2Graph for Question 13
Graph for Question 13
1 2 3-1-2-3 0
5
10
15
20
25
30
-5
-10
X
Y
y = 4x2
-5x-7
y = -7x-5
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Question Solution and Mark Scheme Marks
14(a) Mass
(kg)
Frequency Mid Point
1-5 5 3 I
6-10 4 8 II
11-15 5 13 III
16-20 6 18 IV
21-25 9 23 V
26-30 7 28 VI
31-35 3 33 VII
36-40 1 38 VIII
Class interval : II - VIIIFrequency : II VIII
Midpoint : II - VIII
Note :
Allow 1 mistake in Frequency for P1
[P1]
[P2]
[P1]
4
(b) (3x5)+(8x4)+(13x5)+(18x6)+(23x9)+(28x7)+(33x3)+(38x1)40
760
40
= 19
Note :1. Allow 2 mistakes for K1
Award KK2, if incomplete working shown.
[K2]
[N1]
3
(c) Graph
1. Axes drawn in correct direction with uniform scalesfor
0.5x 40.5 and 0y 9
Accept midpoint/class interval for the horizontalaxis.
2. All bar correctly drawnNote :
6 or 7 bar correctly drawn, award K1
[P2]
[K2]
4
(d) The modal class is 21-25 [P1] 1
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Graph for Question 14
Graph for Question 14
5.5 105 15.5 20.5 25.5 30.5 35.5 40.5
1
2
3
4
5
6
7
Mass (kg)
Frequency
0.5
8
9
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Question Solution and Mark Scheme Marks
15(a)
J /F I H/Q/G P/ N
6cm
E/A D C/M/ B L/ K
2cm 2cm 3 cm
Correct shape with rectanglesEDIJ,DCHIand CLPH.
All solid lines.
ED=DC, EL>JE>CL>ED
Measurement correct to 0.2 cm (one way) and all anglesat the vertices of rectangles = 90o 1o.
[K1]
[K1] dep
K1
[N1] dep
K1K13
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Question Solution and Mark Scheme Marks
15(b)(i)E C
D
6cm
5cm
R7cm
M
2cm
L
4 cm
A 4cm B 3 cm K
Correct shape with pentagonABCDEandBKLRM.All solid lines.
AE>CM>AB>CD>BK >RL
BK=LK
Measurement correct to 0.2 cm (one way) and angle A,
angle B and angle K= 90o 1o.
[K1]
[K1] depK1
[N2] dep
K1K1 4
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Question Solution and Mark Scheme Marks
15(b)(ii)
C H
1cmD I
2cm
R S
2 cm
L P
2cm
K N6cm
Correct shape with rectangleKLPN,RLPSand CRSH
All solid lines.
Note :
Ignore dotted lineDI
D and I joined with dotted line to form rectangle CDIH
CK>KN, IS>PN>PS >IH
Measurement correct to 0.2 cm (one way) and all angle at
the vertices of rectangles = 90o
1o.
[K1]
[K1]
dep K1
[K1]
dep
K1K1
[N2]
dep
K1K1
K1
5
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Question Solution and Mark Scheme Marks
16(a) 175oE
Note :
a. 175o
or E, award P1
[P2]
2
(b) (180o-54o-54o)x60
=72 x 60=4320 n.m.
[K2
]
[N1
]
3
(c) (5+25) x 60 x Cos 54o
= 1058 n.m
Note:
Award K1, ifa. (5+25) orb. Cos 54o used correctly.
[K2
]
[N1
]
3
(d) 550 = Distance
6
Distance = 550 x 6= 3300 n.m
330060
=55o
55o-54o
=1
o
N
[K2
]
[K1
]
[K1
]
4
Ju
mla
h
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