6. SULIT 3472/2 2014 BAHAN KECEMERLANGAN (BK 9) [Lihat sebelah
No. Penyelesaian & Markah Jumlah Markah
9
(a) m = 2 K1
y – 0 = 2( x (6) ) K1
y = 2x + 12 N1
(b) 0
3
2 6
x
atau 12
3
2 0
y
K1
(3, 18) N1
(c) Area of Δ = 9 36
2
1
K1
= 22.5 unit2 N1
(d) 2 2 (x 6) (y 0) or 2 2 (x 0) (y 3) P1
2 2 2 2 (x 6) (y 0) 2 (x 0) (y 3) K1
3x2 + 3y2 – 12x +24y = 0 N1
10
10 (a) tan OAC =
4
8
K1
OAC = 2657
AOC = 12686 or 12687 P1
= 2214 rad or 2215 rad N1
(b) Length of arc ABC = 8(4070) or 8(4069) K1
4.070 rad or 4.069 rad P1
= 3256 cm or 3255 cm N1
(c) Area of sector OABC = 1 2
(8)
2
(2214) or 1 2
(8)
2
(2215) K1
Area of OAC = 1 2
(8) sin126 86
2
K1
Area of the shaded region = 1 2
(8) (2 214)
2
1 2
(8) sin126 86
2
= 4525 cm2 or 4528 cm2 N1
10
K1 for subtraction
7. SULIT 3472/2 2014 BAHAN KECEMERLANGAN (BK 9) [Lihat sebelah
No. Penyelesaian & Markah Jumlah Markah
11
(a) (i) 8 7 1 8 8 0
7 8 C 0.35 0.65 C 0.35 0.65 K1
0.00357 N1
(ii) P(X 2) ; p = 0.65 q = 0.35 K1
8 0 8 8 1 7 8 2 6
0 1 2 C p q C p q C p q K1
0.02532 N1
(b) (i)
70 55
16
P Z K1
2625 N1
(ii)
55
0.915
16
m
P1 ( - 0.915) K1
m = 40.36 N1
10
12 (a) use 2011
2009
100
Q
Q
K1
x = 120 N1
y = RM 0.85 N1
z = RM 2.20 N1
(b) (i)
(120)(110) (160)(45) (150)(120) (80)(85)
360
I
K1
= 125.56 N1
(ii) Corresponding expenditure 2011 =
2985
100
125.56
K1
= RM 2377.35 N1
(c)
125.56
150 100
I
K1
I 17/11 = 188.34 N1
10
8. SULIT 3472/2 2014 BAHAN KECEMERLANGAN (BK 9) [Lihat sebelah
No. Penyelesaian & Markah Jumlah Markah
13
(a) (i) 122 = 112 + 62 – 2(8)(6)Cos ADC K1
ADC = 84.35o N1
(ii) CDB = 95.65o atau CAD = 29.84o P1
8
95.65
6
Sin CBD Sin
atau
8
29.84
12
Sin CBD Sin
K1
48.28o N1
(iii) Luas ΔACD = (11)(6) 84.35
2
1
Sin = 32.84
atau (11)(12) 29.84
2
1
Sin = 32.84 K1
atau Luas ΔCDB = (8)(6) 35.79
2
1
Sin = 14.04
Luas ΔABC = Luas ΔACD + Luas ΔCDB K1
46.88 cm2 N1
(b)
N1
CB'D = 131.44o N1
10
D
B
C
B’
9. SULIT 3472/2 2014 BAHAN KECEMERLANGAN (BK 9) [Lihat sebelah
14 (a) x y 200 OR equivalent N1
y 3x OR equivalent N1
y x 10 OR equivalent N1
(b) Draw correctly at least one straight line K1
Draw correctly all the three straight lines K1
Region R shaded correctly N1
(c) (i) 50 y 120 N1
(ii) Maximum fee = 30x 35y
Maximum point ( 50,150 ) N1
= 30(50) + 35(150) K1
= RM 6750 N1
10
15 (a) v = (8 2t)dt = 2 8t t 20
(i) 8 – 2t = 0 ( Using a = 0 ) K1
vmax = 8(4) – (4)2 + 20 K1
= 36 ms-1 N1
(ii) 8t – t2 + 20 = 0 (Using v = 0 ) K1
z = 10 s N1
(b)
(c) s =
10
2
0
(8t t 20)dt K1 ( integrate )
=
10
2 3
0
8
20
2 3
t t
t
K1 ( using limit )
=
2
266
3
m N1
10
0 t
v
10
20
36
4
P1 ( max shape )
N1 ( max point and other two points )
10. SULIT 3472/2 2014 BAHAN KECEMERLANGAN (BK 9) [Lihat sebelah