7/27/2019 Add Maths Trial Exam P2 Marking Scheme 2013 set B.pdf
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Nama Pelajar : Tingkatan 5 : .
3472/2
Additional
Mathematics
Sept 2013
PROGRAM PENINGKATAN PRESTASI AKADEMIK SPM 2013
ADDITIONAL MATHEMATICS
Paper 2
(SET B)
.
MARKING SCHEME
SULIT 3472/2
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MARKING SCHEME
ADDITIONAL MATHEMATICS PAPER 2
N0. SOLUTION MARKS
1 3x y
or 3y x
2 2(3 ) (3 ) 6y y y 2(3 ) ( 1) 6x x x 22 7 6 0y y 22 5 3 0x x
(2 9)( 1) 0y y (2 3)( 1) 0x x
3
2x and 1x (both)
3
2y and 2y (both
P1K1 Eliminate x/y
K1 Solve quadratic equation
N1
N1
5
2
(a)
(b)
2
2
[ ( )] [(2 1) 4]
(2 1) 4
3, 12 11
1
f g x f x
m x m n
m n
n
( ) 5
1( )5
3, 5
f x p x y
x p y
p q
K1(find composite function)
N1
N1
K1(find inverse function)
N1N1
7
3
(a)
(b)
y =x
draw the straight line y =x
Number of solutions = 4
P1 cos shape correct.
P1 Amplitude = 6 [ Maximum = 3
and Minimum = -3 ]
P1 2 full cycle in 0 x 2 orP1 reflection the graph
N1 For equation
K1 Sketch the straight line
N1
6
2
3
-3
7/27/2019 Add Maths Trial Exam P2 Marking Scheme 2013 set B.pdf
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4
4
(a)
(i)
(ii)
(iii)
(b)
2( )
3
4 2
OQ OC CR
x y
1( )
2
3 2
BQ BO OC
x y
5
BR BQ QR
x y
.
BR hOC
cannot find h
not parallel
K1
N1
K1
N1
K1
N1
K1 find h
N1
8
5
(a)
(b)
i)6
10
60
x
x
ii)
22 2
2
7 610
850
x
x
new mean 3(6) 5 23
new standard deviation 3(7) 21
P1
K1
N1
K1 N1
K1 N1
7
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5
6
(a)
(b)
2 2 21 1 1, , ,...2 8 32
p p p
2 2
2 2
1 18 321 1
2 8
p p
p p
1
4r
(i)
11 253200( )
4 128
8
n
n
(ii)
3200
114
24266
3
S
K1
K1
N1
K1K1
N1
K1
N1
8
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7
(a)
(b)
(c)
(i)
(ii)
(iii)
x 1 2 3 4 5 6
10log y 0.26 0.41 0.53 0.66 0.80 0.94
10 10 10log log logy x h k h
10log h = *gradient
h = 1.37
10logk h = *y-intercept
= 0.88
y = 3.98
N1 6 correct
values of log y
K1 Plot10
log y vs
x.
Correct axes &
uniform scale
N1 6 points plotted
correctly
N1 Line of best-fit
P1
K1
N1
K1
N1
N1
10
10log y
0.12
0x
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N0. SOLUTION MARKS
8
(a)
(b)
(c)
21(6) 26.28
2
= 1.46 rad
6(1.46)AD
S = 8.76 cm
Perimeter = 8.76 + 5 + 5 + 8
= 26.76 cm
Area of triangle = 17.89 cm2
Area of rectangle = 40
Area of the shaded region = 17.89 + 4026.28= 31.61 cm
2
K1
N1
K1 Use s r
N1
K1
N1
K1
K1
K1
N1
10
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N0. SOLUTION MARKS
9(a)
(b)
c)
2
216
xx
dx
dy
At turning point, (p, 4) ,0
dx
dy
02
162
p
p
16p3
= 2
8
13 p ,2
1p
y = 16x 2x-2 dx
cx
x
2
2
16 2
= cx
x 2
8 2
At point A (2
1, 4 ) ,
4 = 8( )2
+ 4 + c
c = -2
The equation of the curve is , 22
82
xxy
32
24
16xdx
yd
At point A(2
1, 4) , 048)8(416
2
2
dx
yd
Thus,
4,
2
1A is a minimum point
P1
K1
N1
K1 use
dxyy )( 12
K1 integrate
correctly
K1
N1
K1
K1
N1
10
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N0. SOLUTION MARKS
10
(a)
(b)
(c)
(d)
1
68
yx
Thus,A( 0,6) and B( 8, 0)
LetD (x,y)
CD : DE= 1:3
x =
84
32
4
10321
y =
74
28
4
10321
ThusD 7,8
4
3
8
6
ABm
Thus ,
3
4
CEm
y - 6 = 03
4x
y = 63
4
x
Area ofAOB=
6706
0880
2
1
28562
1
N1N1
K1
K1
N1
K1
N1
K1 (use area formula)
K1N1
10
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N0. SOLUTION MARKS
11
(a)
(i)
(ii)
(b)
(i)
(ii)
X = Students have a laptop
p =10
7= 0.7 , n = 5
q = 3.010
3
10
71
P(X= 5) = 5055 7.03.07.05 c
= 0.1681
P( X< 3) = P( X =0) + P( X = 1) + P( X= 2)
= 32
2
41
1
50
0 3.07.053.07.053.07.05 ccc
= 0.1631
X =weights of pencil box
X 20,250N
P(X>265) =
20
250265ZP
=P(Z > 0.75)
= 0.2266
P(X < w ) = 0.3
P(Z > 20
250w= 0.3
From table,
20
250w= -0.524
W= 239.52 g
K1 Use P (X=r ) =
rnr
r
n qpC
N1
K1
K1 Use P (X=r ) =
rnr
r
n qpC
N1
K1 Use Z =
X
N1
K1
K1
N1
10
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N0. SOLUTION MARKS
12 (a)
(b)
(c)
(d)
vinitial= 5 m s-1
v= t26t +5
dt
dva
=2t -6
a = 0
t = 3
v = (3)2-6(3) + 5
= -4 ms-1
v < 0
t26t+5 < 0(t-1) (t-5) < 0
The range of values of t
1 < t < 5
5
1
23
1
0
23
5
1
2
1
0
2
1
0
5
1
533
533
)56()56(
ttt
ttt
dtttdttt
vdtvdt
)1(5)1(33
1)5(5)5(3
3
50)1(5)1(3
3
12
3
2
3
2
3
=13
N1
K1
K1
N1
K1
N1
K1 for
1
0
vdt
K1 (for
Integration;
either one)
K1 (for use and
summation)
N1
10
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N0. SOLUTION MARKS
13 (a)
i)
ii)
iii)
(b)
i)
(ii)
oSU 65cos)7)(8(2)7()8( 222
SU = 8.10 cm
Area of triangle SVU= O65sin)8)(7(2
1
= 25.38 cm2
SUTo
sin
6
75sin
1.8
Sin SUT= 0.7155
SUT=45.680
TSU=180O-75
O-45.68
O=59.32
O
STS=61.36O
oo SinSin
SS
32.59
6
36.61
'
= 6.123cm
K1
N1
K1
N1
K1
N1
P1
P1
(for 59.32o
OR
61.36o
K1
N1
10
T S
S
U
6 cm
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N0. SOLUTION MARKS
14
(a)
(b)
(c)
i)
ii)
x + y 5
10x +20y 150 or x + 2y 15
x 2y or y 2
x
At least one straight line is drawn correctly from inequalities involvingx and y.
All the three straight lines are drawn correctly. Region is correctly shaded.
3 kg
Minimum point (5,0)
Minimum cost = 10(5) + 20(0)
= RM 50
N1
N1
N1
K1
N1
N1
N1
N1
K1
N1
10
R
7.5
5
5 15
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N0. SOLUTION MARKS
15 (a)
i)
ii)
(b) (i)
(ii)
1005.162
20.511 xp
P11 = RM 3.20
100140
15012
13 xI
= 107.14
30+10+20+40 ( can be seen )
14540201030
)40135()20()10150()305.162(
XxXXX
X = 136.25
The usage of 136.25
56100
25.13613 xp
= RM 76.3
K1
N1
K1
N1
P1
K1
N1
P1
K1
N1
10
END OF MARKING SCHEME