52760792-jawapan-add-maths-t4-kertas-2-2010
TRANSCRIPT
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3472/2
Form4
Additional
Mathematics
Paper
2
2010
PEPERIKSAANAKHIRTAHUN2010
TINGKATAN4
ADDITIONAL
MATHEMATICS
Paper2
MARKINGSCHEME
This
marking
scheme
consists
of
11
printed
pages.
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3
4
a)p3,h9
dy
0
dx
9p3h0
@
ph=12
2 2dy ( 3x1)( 2x)(
x
)( 3)b)
2 2dx (3x1)
23x2x
(3x1)2
2,
2
B1
B1
1
1
6
5
(
a
)
y
+2x
-
4=
0
y
=
-2x
+4
Gradientofatangent=-2
1
1
2
3
2
(
b
)
y
=
x
+
3x
-11x
+
9
dy
2
dx=3x+6x-11
2-2=3x+6x
-
11
2
3x
+
6x
-
9=
0
2
x+2x
-
3=0
(
x-1)(x+3)
=0
x=1orx=-3
When
x=
1
,
y=2
When
x=-3,
y=42
Substitut(1,2)intoy=-2x+4
2=2
Substitut
(-3,42)
into
y
=-2x+4
42
10
1
1
1
4
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ThusthecoordinateofpointPis(1,2)
1
(c)y-42=-2(x+3)
y=-2x+36
1
1 2
6 (a)log3mn=log3m+log3n
=x+y
1
1 2
x
x
2x6
(
b
)
2
3
=
6
x 2x6
(
2
X
3
)
=
6
x
2x
66
=
6
x=6
1
1
2
(c
)
2
log9
x=log34
2
log9x=log34
log3x2
=log34log39
log3x2
2=log34
log3
3
2
2
log3x
=
log34
2 2
x=4
x
=4
1
1
1
1
4
Answer four quest ions f romth is sect ion
7
(a)i. PG=GA2 2 2 2
(x2)(y3)= (22)(30)2
2
x
+y-4x
-6x
12
=0
2
2
ii. B(5,t),5+t-4(5)6t-12=0
2t-6t7=0
(t+1)(t7)=0t=-1or7
1
1
1
1
1,1
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5
0
3 3(b) GradientGA= =
22 4
4
The
equation
of
tangent
;
y
0
=
(x+2)
3
4 8y= x-
3 3
8 8x=0,y=- T(0, - )
3 3
Areaoftriangle0AH=x2x83
8 2= unit3
1
1
1
1
10
8 (a) GradientTU=75
60
=13
GradientUV=-3
7
q
3
6
p
3p+q-25=0.(1)
(b) AreatriangleTUV=0 6 p 0
5 7 q 5
=(30
7p)(6q
5p)
15
+
p
3q
shown
(c) AreaofTUVW=40units2
AreatriangleTUV=20units2
15+p3q=20
p=5+3q.(2)
sub(2)into(1)
3
[
5
+3q
]
+
q
25
=
0
q=1
, p=
8
KoordinatV=(8,1)
1
1
1
1
1
1
1
1
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6
(d) GradientTW=UV
=-3
Equation
TW
;
y
=
-3x
+
5
y
+
3x
=
5
y 3x 1
5 5
1
1
9
a) rad4
2 2
b)OA22=2.828cm
c)OQ=3OB=3(2)=6
BQ
=
6
2
=
4
AP=OPOA=
6
2.828
=
3.172
cm
LengthofarcPQ=6 =4.713cm
4
Perimeteroftheshadedregion
=AP+AB+BQ+ArcPQ=
3.172
+
2
+
4
+
4.713
=13.89cm
d)Areaoftheshadedregion=
Area
sector
OPQ
Area
of
OAB
1 2 1
= (6) (2)(2) 2
4 2=12.14cm
2
1
1
1
1
1
1
1
1,1
1
10
10 a)5x+5x+6x+y+y=120
y=608x..(1)
1
b)A
=
6x(
y)
(6x)(4x)
22
=6xy12x(2) Substitude(1)into(2)
1
1
1
1
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A6x(608x)12x2
36x(10x)
dA
c) 36072x
dxFor
A
is
a
maximum
dA
0
dx360
72x
=
0
x=5
whenx=5theny
=
60
8(5)
=
20
Maximum
area
=
36(5)(10
5)
=
900cm2
1
1
1
1
1
1
11 (a)(i) =48.25
p=17
(ii)
m=
=47.23
(b)Graph
Drawthehistogramwiththeuniformscalex-axis&y-axis
2
1
2
1
2
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Drawfindmode
Mode
=
45.5
1
1
Answer
two
quest ions
f rom
th is
sect ion
12 a)
x
=
140,
y
=
RM
11.25
,
z
=
RM8.00
b)131.50
120 (30 )
140 ( 20 )
125 (10 )
130 ( 25)
150 (15)
100
1,
1,
1
3
B2
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a)RM110.46
P05 P07X
X100
157.870
P05
120
I05/07
131.50
X
157.80
100
4
B3
B2
13 a)
(i) 220
x
x100110
x=RM200
(ii)y=187 . 5150
x100
y
=
125
(iii) z
400x100130
z
=
RM
520
(b) IWI
W
( 110
x100 )
( 125x
80 )
( 105
x
60 )
( 130
x
40 )
280
116.07
c) U
75000 x100116.07
U =RM87053.57
(d) 140
100x116.07
162.5
1
1
1
1
1
1
1
1,1
1
10
14
a)
2 2 2AC
AB
BC
2(
AB(BC)
cos
ABC
2 2 0
12
8
2(12)(8)
cos
65
126.86
AC11.26cm
1
1
10
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ADC5811
ACD180 75 5811
=
4649
(12)(8)sin
65
(11.26)(12.8)sin
46
49
10
b)i)
ii)
sin
ADCsin
750
11.2612.8
0
0
0
1
1
1
1
c)AreaofquadrilateralABCD
=
Area
ofABC+
Area
ofACD
1 1022
1,1,1
=
96.05
cm
2
1
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15Sin ADC=
3
2
1
ADC
=
600
(a)
AC2=
(6.6)2+(5.4)
2-2(6.6)(5.4)cos60
0
=
43.56+29.16-
71.28(0.5)
=37.08
AC=
6.089cm
1
1
1
(b)
sinsin500
=
6.0899.3
=
30.100
1
1
(c)AB
sin99.9=
9.3
sin501
10
AB =11.96cm
XtX11.96=
X6.089X9.3Xsin99.9
t
=4.664cm
OR
EC=ACsin50
=
4.664cm
1
1
1
1
1