- drveenasforte.com · (b) Draw correctly at least one (1) straight line of (a) with correct axes...
Transcript of - drveenasforte.com · (b) Draw correctly at least one (1) straight line of (a) with correct axes...
Question Solution and Marking Scheme Sub Marks 9
(a) (i) Median = = 52.7 (ii) Mean = [6(14.5) + 8(24.5) + 11(34.5) + 17(44.5) + 25(54.5) + 14(64.5) + 12(74.5) + 7(84.5)] = 51.7
(iii) Standard deviation =
301905100 51.72
= 18.61 *For 301905 or [6(14.5)2 + 8(24.5)2 + 11(34.5)2 + 17(44.5)2 + 25(54.5)2 + 14(64.5)2 + 12(74.5)2 + 7(84.5)2] and follow through for the mean (b) Axes with uniform scale and correct labelling for the boundaries or midpoints. All eight(8) bars correct. Correct method for determining the mode. 53.0 < mode < 54.0
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1002 42
25 1049.5
1100
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Question 9
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Question Solution and Marking Scheme Sub Marks
10 (a) p = 0.65 and q = 0.35 or seen P(X 3) = 1 - P(X 3) or 1 - [P(X = 0) + P(X = 1) + P(X = 2)] or P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) 1 - 10C0(0.65)0(0.35)10 - 10C1(0.65)1(0.35)9 - 10C2(0.65)2(0.35)8 or 10C3(0.65)3(0.35)7 + 10C4(0.65)4(0.35)6 + 10C5(0.65)5(0.35)5 + 10C6(0.65)6(0.35)4 + 10C7(0.65)7(0.35)3 + 10 (0.35)
C8(0.65)8
+
0.9
2 0 10C9(0.65)9(0.35)1 + 10C10(0.65)10(0.35)
952
) ( = 1.1
1 - [P(Z
0.7675
(b i) z (ii) x = 44.7 kg (iii)
> 1.1) + P(Z > 1.3)]
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x 5510 1.03
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4210 55
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Question Solution and Marking Scheme Sub Marks
11 (a) (i) 7.1375 or 7.138 cm2 (ii) 43.565 or 43.57 cm (iii) * 46.4125 or 46.41 cm (b) (i) or 1.2 rad (ii) (5)2(1.2*) - BR2(1.2*) = 12.6 QR = 2*(1.2*) 2.4 cm Note: 1.2* follow through on and 2* follow through on BR
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22 900sin
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K152 54 5
12 5
2 12 5
2 12 5
2 7.1375
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12
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Question Solution and Marking Scheme Sub Marks
Full Marks
12
(a) (i) RM2.00 (ii) 137.5 (iii) RM2.40 (b) 131.17 (c) * RM3279.25 Note: 131.17* follow through on the composite index. (d) or 100.27
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125100 12580 137.550 15060 12030100 80 50 60 30
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P2006
2500 100 131.17 K1
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0.600.55 100 2.45
2.40K1100
10280 102100 10260 0.600.55 10050 2.45
2.40 10030
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Question Solution and Marking Scheme Sub Marks
Full Marks
13
(a) AE2 = 202 + AE2 - 2(AE)(20)cos 30o 11.55 cm (b) (20)(11.55*)sin 30o 2(CE)(10)sin 60o = (20)(11.55*)sin 30o 6.668 cm Note: 11.55* follow through on AE (c) CD2 = 6.668*2 + 102 - 2(6.668*)(10)cos 60o 8.819 cm Note: 6.668* follow through on CE (d) (i) 40.90o (ii)
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12
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sinCDE6.668* 60o
8.819*sin
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EF
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Question Solution and Marking Scheme Sub Marks
Full Marks
14
(a) 6t - 8 -8 ms-2 (b) tA = 1 s or tB = s, 2 s 1 s t 2 s (c) for the correct limits 8 m (d) SA = 10t - 5t2 or SB = t3 - 4t2 + 4t t3 - 4t2 + 4t = 10t - 5t 2 s
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3P1
10t 10t2
2 K1
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Question Solution and Marking Scheme Sub Marks
Full Marks
15
(a) I : 2x + 3y 90 II : 3x + 2y 180 III : 2x - y 40
(b) Draw correctly at least one (1) straight line of (a) with correct axes and uniform scale. All three (3) lines of (a) correctly drawn. Region R correctly shaded and labelled. (c) (i) (ii) 3x + 12y = any constant or its line representation in the graph 3(37*) + 12(34*) RM519 Note: 37* and 34* follow through from any clear indication or reading of a maximum point using any line of 3x + 12y in the region of R in the graph.
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Question 15 (b)
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(SKEMA PERMARKAHAN MATEMATIK TAMBAHAN) PERCUBAAN 2007 KERTAS 2
NO.
SOALAN PENYELESAIAN PER
MARKAHAN 1 Menyatakan atau 3 2m n 7 723 6m mn
7 2
3
nm
atau
7 3
2
mn
atau setara
Menghapuskan m atau n
*2 7 3
3 62
mm m
7
atau * 2 *
7 2 7 23 6 7
3 3
n nn
Selesaikan persamaan kuadratik Menggunakan rumus atau Penyempurnaan kuasa dua m = 0.257 , 2.591 atau n = 3.114 , 7.387 ATAU n = 3.114 , 7.387 atau m = 0.257 , 2.591
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2
NO. SOALAN
PENYELESAIAN PER MARKAHAN
2 (a)
(b)
2( ) 2 ( 1) 1 3f x x atau setara
2( ) 2( 1) 1f x x
Nilai maksimum f(x) ialah -1
2 4 21x x 0 Memfaktorkan ungkapan kuadratik dan mencari julat dengan kaedah yang sesuai 7 3x
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3(a)
(b)(i)
(ii)
Menggunakan rumus sudut berganda
sin 2 2sin cosx x x dan 2cos 2 1 2sinx x
memansuhkan sebutan dalam 2
2sin cos
2sin
x x
x
menyatakan cos
cotsin
xx
x
Mendapatkan tan2
p
Menggantikan ke dalam 2
2 tan2tan
1 tan
2
2tan
1
p
p
2
2sin
1
p
p
atau
2
2
1cos
1
p
p
menggantikan p = 2 ke dalam
** 2
2 2
2 12
1 1
p p
p p
24
25
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3
NO. SOALAN
PENYELESAIAN PER MARKAHAN
4(a)
(b)
Mencari
dy
dx dengan betul dan
menggantikan 1x
kaMengguna n * 1m m1 2
2y x
34dy
xdx
3
4( 0.1)
4y
0.00625
1
1
1
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5 (a)
(b)
Kamirkan
2(3 2 2)x x d x
3 2 2x x x c
-3) ke dalam
)
Gantikan (1,
* 3( 2 2y x x x c untuk mencari nilai c
3 2 2 3y x x x
Kamirkan
24
dyy
1
161
y
Gunakan had ke dalam *hasil kamiran
4
2
4
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4
6(a)
(b)(i)
(ii)
, , , ....2 16 128
xy xy xy
atau mana-mana tiga sebutan berturutan.
apatkan sekurang-kurangnya dua nisbah bagi
D
16 128
2 16
,xy xy
xy xy atau setara
1
8r
kan
Guna 1 146nar
1** 1 125600 6
4 4
n
atau Senaraikan setiap sebutan sehingga 1
46
25600, 3200, 400, 50, 146
5n
Gunakan 1
aS
r
*
*
25600, 1
11
8
r
1
292577
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5
7 (a)
(b)
(c)
y 3.3 2.4
1.8 1.5 1.3 1.1
xy 1.65 3.60 4.50 5.25 5.85 6.05
Plo e y b n se )
Garis lurus penyuaian terbaik
t xy m lawan (paksi etul da skala ragam
6 *titik ditanda betul . xy by a
8.3a
kecerunan b 2.0b
1 b1
xy a a
ecerunan = 12
1
1
1
1
k 0.
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6
8(a)(i)
(ii)
(b)(i)
(ii)
1 7 3 11
242 8 22 k
0
enggunakan
6 12 48k
6 1k or k
M 1 2 1m m
2
3
2m
*5y
3
4 2
x
311
2y x atau setara
Menggunakan
10AP 2 2( 1) ( 2) 10x y
2 2 2 4 5x y x y 0
Gantikan titik (4,1) ke dalam *persamaan lokus.
okus melalui titik (4,1)
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