Download - Skema Add Math P2 Trial SPM ZON a 2010
3472/2 Matematik Tambahan Kertas 2 2 ½ jam Sept 2010
SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING
PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2010
MATEMATIK TAMBAHAN
Kertas 2
Dua jam tiga puluh minit
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
Skema Pemarkahan ini mengandungi 15 halaman bercetak
MARKING SCHEME
2
ADDITIONAL MATHEMATICS MARKING SCHEME
Zon A Kuching 2010 – PAPER 2
QUESTION
NO. SOLUTION MARKS
1
2 2
2
(2 ) 5 (2 ) 2 8 0
( 1)( 2) 0
x y
y y y y
y y
= +
+ − + + − =
+ + =
1, 2
@
1, 0
y y
x x
= − = −
= =
5
2
(a)
(b) (i)
(ii)
2
( ) ( )( 2)
( 2) 2
f x x p x
x p x p
= + −
= + − −
12 [ 2(3)]k= −
2k = −
2 2
2
2
1 12[ 6]
2 2
1 252( )
2 2
1 25 1 1max point , or ,12
2 2 2 2
y x x
x
= − + + − −
= − + +
∴ = − −
2
5
Solve the quadratic equation by using the factorization @ quadratic formula @ completing the square must be shown
Eliminate orx y
Note : OW−−−− 1 if the working of solving quadratic equation is not shown. 5
7
P1
K1
K1
N1
N1
K1
K1
N1
N1
K1
K1
N1
3
QUESTION NO.
SOLUTION MARKS
3
(a)
(b)
(c)
a = 40 and r = 0.8 |T6 = 40(0.8)5 = 13.1072
3
( )8.01
8.0140 5
5 −−=S
= 134.464
2
8.01
40
−=∞S
= 200
2
4
(a)
2 sincos2 2cos 1 or tan
cos
LHS to RHS or RHS to LHS
xx x x
x= − =
2
N1
K1
N1
K1
N1
P1
K1
7
N1
K1
4
QUESTION NO.
SOLUTION MARKS
(b)
(i)
(ii)
12 −= py
2
1=p
4
2
5 (a)
Refer to the graph Mode = 37
3
(b)
m = 29.5 +
10024
2 (10)30
−
=38.17
3
y ¼ π ½ π ¾ π π x O -1 Shape of sine curve Modulus Amplitude or period Translation
P1
P1
P1
P1
K1
6 N1
N1
8
K1
N1
P1
Lower boundary OR
10024
2 (10)30
−
5
QUESTION NO.
SOLUTION MARKS
6
(a)
(i) 3PR y x= +����
(ii) 4 3QS x y= − +����
2
(b)
3PT my mx= +����
STPSPT += = )34)(1(3 yxny −−+
= ynxn 3)44( +−
Compare coefficient of x and y
4
5n m= =
5
N1
N1
7
N1
K1
K1
N1
N1
6
19.5 29.5 39.5 49.5
5
20
30
0
10
15
marks
Number of students
Correct both axes (Uniform scale) K1 All points are plotted correctly N1
9.5 59.5 69.5
25
7
* y-intercept= 5 must be correct.
QUESTION NO.
SOLUTION MARKS
7
(a)
2500
4dA
rdr r
ππ= −
2500
4 0rr
ππ − =
r = 5 A = 678.58 m2
3
(b) (i)
(ii)
Finding area of trapezium
A1 = )1)(45(*2
1 + or A1 = 1
0
2
25
− x
x
Integrate dxxx )5( 2∫ −
A2 =
−
32
5 32 xx
Using limits ∫1
0 into A2
Area = A1 – A2 = 3
7@
3
12 @ 2.333 unit2
4
Integrate
dxxx 22 )5(∫ −π
+−
54
10
3
25 543 xxxπ
Use the limits ∫5
0into
+−
54
10
3
25 543 xxx
Volume generated = π6
625@ π
6
1104 @ 104.17π unit3
Note: OW − 1 once only for correct answer without showing the process of intergration.
3
N1
K1
K1
K1
K1
K1
N1
K1
K1
N1
10
8
Q8
2 3 4 5
0.2
×
0
×
x − 2
log10 y
×
2−x 3 4 5 7 8 10 log y
0.86 0.94 1.04 1.24 1.33 1.51
(a) Each set of values correct (log10 y must be at least 2 decimal places) N1, N1
Y = mX + c log 10 y = ( 2−x )log 10 b + log 10 a K1 where Y = log 10 y, X = (x − 2), m = log 10 b and c = log 10 a (c) log 10 b = gradient
log 10 b = 1.46 0.57
9.4 0
−−
= 0.09468 K1
b = 1.244 N1 log 10 a = Y-intercept log 10 a = 0.57 K1 a = 3⋅715 N1
Correct both axes (Uniform scale) K1 All points are plotted correctly N1 Line of best fit N1
1 6 7 8
×
×
10
N1
N1
0.4
0.6
0.8
1.0
1.2
1.4
1.6
×
9
QUESTION NO.
SOLUTION MARKS
9 (a) (i)
(ii)
3−r (r − 3)2 + 42 = r2 Using Pythagoras theorem on ∆BMO
r =6
25 (or 4.1667)
3
(b)
4.167 3 1.167
,
1 4tan
2 1.1671
1.287 rad.2
2.574 rad.
OM cm
let AOBθ
θ
θ
θ
= − == ∠
=
=
=
3
Area of sector OAB =21
(6
25)2 (2.574) = 22.34 cm2
Area OAB∆ =21
(8)( )36
25 − = 4.667 cm2
Area of shaded region =22.34 − 4.667 = 17.673 cm2
4
K1
P1
N1
K1
P1
N1
10
K1
K1
N1
K1
10
QUESTION NO.
SOLUTION MARKS
10 (a)
(b)
(c)
1
3
12
3
CD ABm m
y x
= =
= −
m = 6
3
3
1 3(3)
3 8
ADm
c
y x
= −
− = − +
= − +
Solving the equations y = −3x + 8 and 3 4 0x y− + = A(2, 2)
5
2 2
( 1) 41
3 8
11 3 20 0
y y
x x
x y x y
− − −× = −− −
+ − − + =
2
10
N1
K1
K1
K1
N1
N1
N1
K1
K1
N1
11
QUESTION
NO. SOLUTION MARKS
11
(a) (i)
(ii)
3
, 15
p p q= + =
52
( 0)5
P X = =
or IE
0.01024= ( 4) ( 4) ( 5)P X P X P X≥ = = + =
4 1 5
54
3 2 3
5 5 5C
= +
= 0.337
3
(b) (i)
(ii)
30 35 60 35
or 10 10
− −
( )0.5 2.5 1 ( 0.5) ( 2.5)P Z P Z P Z− ≤ ≤ = − ≥ − ≥
or 1 − 0.3085 − 0.00621 or R(−0.5) − R(2.5) = 0.6853 or 0.68525 Number of pupils = P( 60) 483X ≥ × = 3
7
N1
K1
K1
10
K1
N1
K1
K1
K1
N1
N1
12
QUESTION NO.
SOLUTION MARKS
12 (a)
(b)
(c)
(d)
v = 15 ms−1
1
a = 2t − 8 and a = 0 or dv
dt = 0
t = 4 vmin = −1 ms−1
3
(t − 3)(t − 5) > 0 0 ≤ t < 3, t > 5
2
324 15
3
ts t t= − +
| s3 − s2 | OR 3
2v dt∫
2
18 163
−
OR 3 3
2 23 24(3) 15(3) 4(2) 15(2)
3 3
− + − − +
11
3 m
4
10
K1
K1
K1
K1
N1
N1
P1
K1
N1
K1
13
QUESTION NO.
SOLUTION MARKS
13
(a) (i)
(ii)
(b)
(c) (i)
(ii)
Electricity : I = 2.1
1001.8
× = 116.7
Telephone and internet : I = 100
2
116.7(4) 133.3(2) 100(3) 120(6)
15
116.9
I+ + +=
=
3
110
210100
231
RM
RM
×
=
110 110(116.7 )(4) 133.3(2) 100(3) (120 )(6)
100 10015
124.8
× + + + ×
=
5
10
P2, 1
N1
K1
K1
K1
K1 K1
N1
N1
Any one of the price indices correct.
Any one of the price indices formula correct.
14
QUESTION
NO. SOLUTION MARKS
14 (a)
(b)
(c)
(d)
1
28 (14) (10) sin2
QRS= ∠
23 35'oQRS∴ ∠ =
2
2 2 2( ) 10 14 2(10)(14)cos23 35'
6.276
oQS
QS
= + −
=
2
6.276 7.5
sin sin30QPS=
∠ �
∠QPS = 24°44′ ∠PQS = 125°16′
3
Area of quadrilateral
128 (7.5)(6.276)sin125 15'
2
= 47.22
o
PQRS
= +
3
10
K1
N1
K1
N1
N1
K1
N1
K1 K1
N1
15
y
Answer for question 15
1 2 3 4 5 6 7 0 8
1
2
7
6
5
8
4
3
(3, 4) •
(a) I. 2y x≤
II. 7x y+ ≥ III. 3y x≥ −
(b) Refer to the graph, 1 or 2 graph(s) correct 3 graphs correct Correct area (c) i) xmin = 3 ii) max point (3, 4) k = 100x + 80y Maximum Profit = RM 100(3) + RM 80(4) = RM 620
10
N1
N1
N1
N1
N1
N1
K1
N1
K1
N1
R