spm percubaan 2007 sbp add maths paper 2 marking scheme
TRANSCRIPT
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Matematik
Tambahan
Kertas 2
2 jam
Ogos 2007
SEKTOR SEKOLAH BERASRAMA PENUH
BAHAGIAN SEKOLAH
KEMENTERIAN PENDIDIKAN MALAYSIA
PEPERIKSAAN PERCUBAAN
SIJIL PELAJARAN MALAYSIA 2007
MATEMATIK TAMBAHAN
Kertas 2
Dua jam tiga puluh minit
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
Skema Pemarkahan ini mengandungi 16 halaman bercetak
MARKING SCHEME
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ADDITIONAL MATHEMATICS MARKING SCHEME
TRIAL SBP 2007 PAPER 2
QUESTION
NO. SOLUTION MARKS
1
2 *
2
66 4
4
( 6 4 ) 2
6 6
24 4
y y x or x
x x x
or
y y
y
1, 4
2, 10
2, 10
1, 4
x
or
y
OR
y
or
x
5
2(a)
1
3 2 2
0
2
5 4 2
4 3 0
( 1)( 3) 0
1, 3
x kx k x
k k
k k
k
3
2
Solve the quadratic equation
by using the factorization @
quadratic formula @
completing the square
Eliminate x or y
Integration(all terms correctly
integrated)
Solve quadraticequation
Note :OW-1 if the working of solving
quadratic equation is not shown.
5
P1
K1
K1
N1
N1
K1
K1
N1
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QUESTION
NO.SOLUTION MARKS
(b)
+= dxxdxdy )512(xx 56
2 +=
2
5
2
52
2
522
2
52
)56(
23
3
2
+=
++=
++=
+=
xxy
c
cxx
dxxxy
5
3
Integrate2
2
d y
dx
Integratedy
dx
Find the value
ofc
8
K1
K1
K1
N1
N1
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QUESTION
NO.SOLUTION MARKS
3(a)
(b)
(c)
3
2@
2
3=
= SRPS mm
*
24 ( 2)
3
3 2 16 0
y x
y x
= +
=
3
*R (4, 8)
*Q (6, 5)
Area =0 6 4 2 01
1 5 8 4 12
= 26 unit2
3
2
4
Solve simultaneous equation between SR andPR
P1
Use straight line formula
Or equivalent
* equation ofSR
*R coordinate
Use the formula ofpolygon
* R and Q coordinate
8
K1
N1
K1
N1
N1
K1
N1
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QUESTION
NO.SOLUTION MARKS
4
(a)
(b)
(c)
20 115cos n= @ 2sin 25 1 m
2 2
cos 40 cos(25 15 )
cos 25 cos15 sin 25 sin15
1 1m n n m
3
2
sin 50 sin 2(25 )
2sin 25 cos25
2 1m m
2
21 1cos 2(12 ) 2 cos 12 12 2
1 1cos12
2 2
m
2
5
(a) x = 5 1
QUESTION
NO.SOLUTION MARKS
5
Use theadditional
foemulae
Use the double-
angle formulae
Use the double-angle formulae
7
K1
K1
K1
N1
N1
N1
N1
P1
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(b)
(c)
4 12 6 4 53
13
13
x
x
x
2
Use any acceptable method
60 x
OR
Listing method
1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 4, .,5
60 x
Note : Accept x = 0, 1, 2, 3, 4, 5, 6
2
6 (a)
(b)
r = 0.8
0
8
8
47.270
8.01
)8.01(65
=
=S
3
4
6
Use the mean formula
Use formula of Sn for GP
Use the formula of Tn with
a = -11 and d = 6
5
K1
N1
K1
K1
K1
P1
N1
N1
N1
K1
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7
[ 11 ( 1)6] [ 11 ( 1 1)6] [ 11 ( 2 1)6] 93
33 18 93
18 126
7
11 ( 7 1)6
25, 31, 37
n n n
n
n
n
T
+ + + + + + + =
+ =
=
=
= +
OR
Listing method
-11, -5, 1, 7, 13, 19, 25, 31, 37,
25, 31, 37
OR
(m) + ( m + 6 ) + ( m + 6 + 6 ) = 93
m = 25
25, 31, 37
Note : Accept answer without working, but OW -1
7
1 2 93n n nT T T
N1N1
Use d = 6
1 2 93n n nT T T
7
K1
K1
N1
N2,1,0
N1
K1
N1
N1
N1
K1
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QUESTION
NO.SOLUTION MARKS
7
(a)
(b)(i)
(ii)
(iii)
x 42.50 38.50 32.01 25.50 13.00 7.98
log y 0.15 0.20 0.30 0.40 0.58 0.65
log 10 y = (log 10 p) x - log 10 q
Refer to the graph.
m = * log 10 p
p = 0.9667 0.05
3.23
10
QUESTION SOLUTION MARKS
8
All values of log10 y correct
(accept correct to 2 decimal places)
Plot log 10 y against x
6 points mark correctly
Line of best fit
c = - log 10 q
q = 0.17 0.05
10
N1
P1
K1
N1
K1
N1
N1
N1
K1
N1
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NO.
8(a)
(b)
(c)
(d)
BOC = 1.942 rad
0 '1801.942 x = 111 15
1
0 '= cos 55 385
= 2.822 cm
x
x
2
0 '= sin 55 385
= 4.127 cm
y
y
The length of BC = 2 x 4.127
= 8.254 cm
The length of the arc BC
=s r
5 x 1.942
= 9.71
Therefore the perimeter of the shaded region8.254 + 9.71
= 17.964 cm
4
Area =21
2r
21 (5) (1.942)
2
= 24.28
Area of triangle =
1
x 8.254 x 2.8222 = 11.65
Therefore the area of the shaded region24.28 -11.65
= 12.63
3
QUESTION SOLUTION MARKS
9
Use trigo to find BC
or1
2BC
*
BOC
KK
Find area of triangle or
sector
* BOC 10
K1
N1
N1
K1
K1
K1
N1
N1
K1
K1
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NO.
9
(a)
(b)(i)
(ii)
(c)
(d)
+= APOAOP
~~ca
2
1+=
1
OBOQ =
)ba(~~
+=
CPCQ =
)OPCO(OQCO +=+
COOPCOOQ -+=
~~~~
c)ca(c +2
1++-=
~~
c)(a2
-1+=
3
From (b), = and2
-1=
2
-1=
3
2==
3
2
3=
18
OBCofArea
227= unitOBCofArea
2
54= unitramlogparalleofArea
3
QUESTION
NO.SOLUTION MARKS
10
Compare the coefficient of
~
a and~
c
Solve
simultaneous
equation
10
Use CQ = CO OQ@ CP = CO OP
N1
N1
K1
N1
K1
K1
K1
N1
N1
N1
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10
(a)
(b)
(c)
2 20 33
2
2
2
xx
x
k
2
22
10
32
0
( 3)
33
L x dx
xx
21 14
72 3
L
Area = L 1 + L 2
= 25
4
1
7
3
72
3
32
( 3)V
yy
y dx
221
(2) 33
V
Volume = V1 + V2
= 12
4
QUESTION
NO.SOLUTION MARKS
11
Solve the simultaneous
equation
Integrate and use the
limit 0 and 2
Find the area of
triangle
L 1 +L2
Integrate and use thelimit 3 and 7
Find the volume of
cone
V1 + V2
10
N1
N1
N1
K1
K1
K1
K1
K1
K1
K1
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11
(a) (i)
(ii)
(b) (i)
(ii)
np = 6
npq = 2.4
q = 0.4*p = 0.6* n = 10
( 2) ( 0) ( 1) ( 2)P x P X P X P X
= 10C0(0.6)0(0.4)10 + 10C1(0.6)
1(0.4)9
+ 10C2(0.6)2(0.4)8
= 0.01230
5
P ( 280 < X < 350) = )Z(P25
325-350315) = )Z(P25
325-320>
= P(Z > -0.2)
= 1-P( Z < - 0.2)
= 0.5793
The percentage of battery that has a life-span
of more than 315 hours is 57.93 %
5
QUESTION
NO.SOLUTION MARKS
12
Both p and q
* n and p
( 0) @ ( 1) @ ( 2) P X P X P X
( 0) ( 1) ( 2 P X P X P X
KK
10
K1
N1
N1
K1
K1
K1
K1
K1
N1
N1
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12
(a)
(b)
(c)
a = 2t 7 = 0
t = 3.5 s
22
7
2
tv t c
10 = (0)2 7(0) + c
c = 10
v = t2 7t+ 10
v = (3.5)2 7(3.5) + 10
= - 2 .25 ms- 1
3
v = t2 7t+ 10 < 0
(t 2)(t 5) < 0
2 < t < 5
3
3 27
103 2
t t s t c
s = 0, t = 0 , c = 0
3 2
7 103 2t ts t
3 22 7(2)10(2)
3 2s or
3 25 7(5)10(5)
3 2s
=26
3=
25
6
Total distance =26
3+
26 25
3 6
= 796
m
OR
4
QUESTION
NO.SOLUTION MARKS
13
Integrate and
substitute t = 3.5
N1
K1
K1
K1
K1
K1
K1
K1
N1
N1
Use a = 0
Use v < 0
Integrate v dt
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2 52 2
0 2( 7 10) ( 7 10)
26 93 2
79
6
S t t dt t t dt
13
(a)
(b)
(c)
(d)
06
50x 100 = 125
P
P06 = RM 40
2
125 4 120 80 5 150 3
12012 2
p p
p
*p = 3
2
07 x 100 = 12015
P
P07 = RM 18
2
I C =80
x 115 = 92100
125 4 120 3 92 5 150 6
18I
= 123.33
4
14
Answer for question 14
(a) I.170x y
II. 80x y
III. 2 20y x
(b) Refer to the graph,
1 graph correct
3 graphs correct
Correct area
(c) max point ( 50,120 )
i) k = 100x + 80y
Max fees = 100(50) + 80(120)
= RM 14, 600
ii) 60 150y 10 20 30 40 50 60 700 8020401401201001601808060
Use the limit
2 5
0 2and
Integrate v
2 5
0 2
See 115
*p
10
10
K1
K1
K1
K1
K1
K1
K1
K1
N1
N1
N1
N1
P1
N1
N1
N1
N1
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15
(50,120)
1
N1
N1
N1
K1
N1
K1
N1
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15
(a) (i.)
(a) (ii)
(b) (i)
(b) (ii)
82 = 52 + (AC)2 - 2 (5)(AC) cos 60
(AC)2 5 (AC) 39 = 0
AC =
2( 5) ( 5) 4(1) 39
2 1
AC = 9.227 , ( - 4. 227 )
EC = 9.227 8.5 = 0.727 cm
OR
8 5
sin 60 sin c
c = 32.77
8sin 87.32 sin 60
AC
AC = 9.227 cm
EC = 9.227 - 8.5 = 0.727 cm
1
(8.5)(15.6) sin 542
54.54
180 54.54 125.46AED
5
7tan
10
= 34.99 @ 35
AE = AF = 149
2 2 28 ( 149) ( 149) 2 149 149 cos
cos = 0.7852
= 38.26 @ 38 15 @ 38 16
5
16
Use cos rule inABC
Use sin rule in ABC
Use cosrule in
AFE
10
K1
K1
N1
P1
N1
N1
N1
P1
N1
K1
K1
K1
K1
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17
0 5 10 15 20 25 30 35 40
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
x
x
x
x
x
x
yLog10
x
(a)
x 42.50 38.50 32.01 25.50 13.00 7.98
Log y 0.15 0.20 0.30 0.40 0.58 0.65
Answer for question 7