Modul Peningkatan Prestasi Matematik Tambahan …SULIT 1 3472/2 3472/2 (C) 2016 Hak Cipta SM Zon B...
Transcript of Modul Peningkatan Prestasi Matematik Tambahan …SULIT 1 3472/2 3472/2 (C) 2016 Hak Cipta SM Zon B...
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SULIT
Modul Peningkatan Prestasi Matematik Tambahan (Kertas 2)
SPM 2016 Zon B Kuching Sarawak
Question Solution and Answer Scheme Sub-
Mk
Full
Mk
1( )w f x
( )f w x
1 2w x
1 2x w
12
xw
1 1( ) 1
2f x x
Let
(Shown)
Then
K1
N1
2 (a)
4 2y x 4
2
yx
or
Substitute (1) into (2)
Note:
1. If the solutions of x and y are
matched wrongly, then SS-1
from full marks.
22 3 * 4 2 4 0x x x 2
4 42* 3* 4 0
2 2
y yy
2( 3) ( 3) 4(1)( 1)
2(1)x
2( 2) ( 2) 4(1)( 12)
2(1)y
3.303, 0.303x
2.606, 4.606y
or
or
5
P1
K1
K1
N1
N1
1
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Question Solution and Answer Scheme Sub-
Mk
Full
Mk
64a
14
r
23
1π (2) 1 256πR
1 2R
641
14
S
2563
or 1
853
or 85.33 6
K1
N1
K1
N1
K1
K1
3 (a)
(b)
1 1( ) 1
2gf x g x
1 14 1
2 2x
12 1
2x
12 2
2x
12 2
2x
1( ) ( )gh x f x
1 14 ( ) 1
2 2h x x
1 12 2
x
14 ( )
2h x x
1( )
8h x x
or 5
22
x 2.5 2xor
6
K1
N1
K1
N1
2 (b) (i)
(ii)
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Question Solution and Answer Scheme Sub-
Mk
Full
Mk
d0
d
y
x
23 6 dy x x x 3 23x x c
3 212 (1) 3(1) c
10c
3 23 10y x x
Equation of the curve
At turning point,
23 6 0x x
3 2 0x x
0, 2x x
When
3 2(0) 3(0) 10y
14
0,x 2,x
3 2(2) 3(2) 10y
10
When
2
2
d6 6
d
yx
x Second derivative,
When 0,x 2,x When
2
2
d6(0) 6
d
y
x
6 0
0, 10
2, 14
6 0
2
2
d6(2) 6
d
y
x
is the maximum point.
is the minimum point. 8
K1
N1
K1
N1
K1
K1
K1
N1
4 (a)
(b)
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Question Solution and Answer Scheme Sub-
Mk
Full
Mk
Shape of sine graph
Number of solutions = 3
sin sintan tan
cos cosA B
A BA B
sin( )
cos cos
A B
A B
sin cos cos sincos cos
A B A BA B
Amplitude = 2
1.5 cycle for 3π
02
x
22
πy x or equivalent
Correct gradient or correct y -intercept
y
xπ
π2
O 3π2
2sin2y x2
π4
3π4
5π4
2
22
πy x
1
8
K1
P1
N1
K1
P1
K1
P1
P1
5 (a)
(b) (i)
(ii)
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SULIT
Question Solution and Answer Scheme Sub-
Mk
Full
Mk
40 2cos
2 60 3
1 2cos 48.19
2 3
1.682 rad
2 48.19 96.38
3.14296.38
180
21 1
60 2 3.142 *1.682 60 60 sin*96.382 2
Area of major sector POR
Area of triangle POR
21
60 2 3.142 *1.6822
160 60 sin*96.38
2
Area of immersed cross-sectional region
210072.45 cm 7 N1
N1
K1
K1
K1
K1
K1 6 (a)
(b)
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Question Solution and Answer Scheme Sub-
Mk
Full
Mk
2(5) 1(2) 2(4) 1( 2),
1 2 1 2T
4,2 (shown)
Area of QOR 0 2 5 01
2 0 2 4 0
10 8 0 0 10 0
2
29 unit
1
4 ( 2)2
5 2m
212
m
12 4
2y x
14
2y x
2 2( 4) ( 2) 3x y
2 28 16 4 4 9x x y y
2 2 8 4 11 0x y x y
2
22 1m
10
N1
N1
K1
K1
K1
K1
N1
N1
K1
K1
7 (a)
(b)
(c)
(d)
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Question Solution and Answer Scheme Sub-
Mk
Full
Mk
2,4B
4,0A
*2
20
( 4) dA x x x
Area under the curve from 0x to 2x
*2 2
04 dx x x
*232
0
23x
x
3
2(*2)2(*2) 0
3
163
Area of the region required by the question
16 1* *2 *4
3 2
43
4 2
2π dy x
Volume generated
4 22
2π 4 dx x x
4 4 3 2
2π 8 16 dx x x x
5 3 5 34 4(4) 16(4) (2) 16(2)
π 2(4) 2(2)5 3 5 3
256π
15 or
117 π
15
45 34
2
16π 2
5 3x x
x
10
K1
K1
N1
N1
N1
K1
K1
K1
K1
N1
8 (a)
(b)
(c)
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Question Solution and Answer Scheme Sub-
Mk
Full
Mk
Range 41 50 1 10
2 2
The height of the bars are proportional to respective frequencies
45.5 5.5
40
Midpoints 5.5, 15.5, 25.5, 35.5, 45.5
3(5.5) 8(15.5) 10(25.5) 12(35.5) 7(45.5)
3 8 10 12 7x
Mean,
114028.5
40
Variance,
2 2 2 2 22 23(5.5) 8(15.5) 10(25.5) 12(35.5) 7(45.5)
28.53 8 10 12 7
23813028.5
40
141
New variance *141
Boundaries or midpoints or class intervals are correctly labeled
(bars are of the same width)
Correct way of finding mode
Mode 33.5 10
K1
N1
N1
K1
P1
P1
K1
N1
P1
P1
9 (a)
(b)
(ii)
(i)
(c)
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SULIT
0
f
x0.5 10.5 20.5 30.5 40.5 50.5 60.5
2
4
6
8
10
12
14
mode 33.5
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Question Solution and Answer Scheme Sub-
Mk
Full
Mk
10n 25
p 35
q or or
( 8) ( 8) ( 9) ( 10)P X P X P X P X
8 2 9 1 10 0
10 10 108 9 10
2 3 2 3 2 35 5 5 5 5 5
C C C
0.01229
0.010616832 0.001572864 0.000104857
50 62( 50)
8P X P Z
1.5P Z
0.9332
1 1.5P Z
1 1.5P Z
1 0.0668
Number of students 0.9332 400
373
( ) 0.1P X a
620.1
8a
P Z
621.281
8a
72.25a
1.281 8 62a
The minimum marks to get grade A, 10
N1
K1
K1
K1
P1
K1
N1
K1
K1
N1
10 (a)
(b)
(ii)
(i)
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Question Solution and Answer Scheme Sub-
Mk
Full
Mk
1x 2 3 4 5 6 7
10log y 1.0792 1.3802 1.6812 1.9823 2.2833 2.5843
1.075 1.375 1.675 1.975 2.275 2.575
Note:
1. In the table, the decimal number being rounded off must not be less precise
than the nearest 0.025 for Y.
2. The N mark is given for all correct X and Y.
3. If the value(s) of X and Y are not shown through a table, then the N mark is
given for all the correct points on the graph.
At least one *point correct, using correct axes
10log against 1y x in correct direction and orientation,
and with uniform scales
All 6 *points plotted accordingly
Line of best fit
Notes:
1. Graph can only be read with precision until half of the smallest grid
division.
2. Do not accept readings obtained by calculation or from the wrong graph.
3. SS-1 from full marks if not using the given scale.
(or equivalent) 10 10 10log log 1 logy a x b
*0.475
*0.47510a
2.985
10 10log log -intercepta y
Note: 2.985 3.162a
10log b m
0.3
0.310b
1.995
2.7 0.67.4 0.4
10
N1
N1
K1
K1
N1
K1
N1
P1
K1
N1
11 (a)
(b)
(ii)
(i)
(c)
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0
10log y
1x
0.475c
1 2 3 4 5 6 7
0.5
1.0
1.5
2.0
2.5
3.0
3.5
0.4, 0.6
7.4, 2.7
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Question Solution and Answer Scheme Sub-
Mk
Full
Mk
ddv
at
1 4t
0 1 4t
14
t
2
max1 1
6 24 4
v
16
8
At maximum velocity, 0a .
or 498
6.125or
0 1 4(0)a
1
At initial acceleration, 0t .
When stopping instantaneously, 0v .
22 6 0t t
2 3 2 0t t
3, 2
2t t
0t
2t 2
0 dv t
22 3
0
26
2 3t t
t
2 3(2) 2(2)
6(2) 02 3
263
Distance from point O
28
3or 8.667or 10
K1
K1
K1
N1
K1
N1
K1
N1
K1
K1
12 (a)
(b)
(c)
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Question Solution and Answer Scheme Sub-
Mk
Full
Mk
2 2 27 18 2(7)(18)cos75PR
17.54PR
*17.54 8.5sin110 sin QPR
1 8.5 sin110sin
17.54QPR
27.09
Note: 27.08 27.09QPR
or 27 5'
180 110 *27.09QRP
42.91
Area of PQR 1
(8.5)(17.54)sin 42.912
Area of PRS 1
(7)(18)sin752
Area of quadrilateral PQRS
1 1(8.5)(17.54)sin 42.91 (7)(18)sin75
2 2
50.75386719 60.85332706
111.61
Let the shortest distance from point Q to the line PR be h .
sin 42.918.5h
8.5sin42.91h
5.787
OR 1 1
(17.54) * (8.5)(17.54)sin 42.912 2
h
Note: 111.61 Area 111.63
Note: 5.787 5.788h 10
K1
K1
K1
N1
K1
N1
K1
N1
K1
N1
13 (a)
(b)
(c)
(i)
(ii)
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Question Solution and Answer Scheme Sub-
Mk
Full
Mk
I: 2y x
1II:
4y x
At least one straight line correct from the *inequalities
involving x and y
Draw all three lines correctly from the *inequalities
Shade the correct region
min 8y
35 25x y k
Objective function
Optimum point
max 35 43 25 11k
RM 1780
43,11
When 30, 7.5 20x y
Note: N0 if answer min 7.5y
Notes: 1. For (c)(i) and (c)(ii), accept the coordinates only if they are integers.
2. SS-1 if
in (a), symbol "<", ">" is used instead of "≤", "≥" or more than three
inequalities are given (except x ≥ 0 or y ≥ 0)
OR
in (b), not using the given scale or axes are reversed or not using graph
paper
2 3 120x y or III: 20 30 1200x y
10
N1
N1
N1
N1
K1
K1
N1
N1
N1
N1
14 (a)
(b)
(c) (i)
(ii)
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0
y
x10 20 30 40 50 60 70
5
10
15
20
25
30
35
40 2y x
14
y x
2 3 120x y
R
35 25 875x y
max35 25x y k
43,11
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Question Solution and Answer Scheme Sub-
Mk
Full
Mk
7.50100
6.00x
125x
150 1004.00
y
6.00y
7.70140 100
z
5.50z
48(125) 24(150) 16(125) 12(140)
48 24 16 12I
13280100
132.8
14Food132.8 100
60.00
14Food RM 79.68
16,10 152.72I
16,10
132.8 115100
I
or equivalent
10
N1
N1
N1
K2
K1
N1
N1
K1
N1
15 (a)
(b)
(c) (i)
(ii)