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JABATAN PELAJARAN NEGERI TERENGGANU
PENTAKSIRAN SUMATIF 3 SPM 2013MATEMATIK TAMBAHANKertas 2
Peraturan Pemarkahan
Ogos 2013
Peraturan pemarkahan ini mengandungi 15 halaman bercetak.
3472/2
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SULIT 2 3472/2(PP)
INSTRUCTIONS FOR EXAMINERS
1. MARKING GUIDE
1.1 Mark all the answers.
1.2 Do not mark working / answer that has been cancelled.1.3 Give the mark P/ K/ N in line with steps ofcalculation given by thestudents.1.4 Give the mark PO / KO / NO for the incorrect working / answer.1.5 Ifmore than one final answer is given, mark all the solution and choose the answer with the
highest mark.
1.6 Accept other corect methods which are not given in the marking scheme.
2. NOTATION
P - The mark is given ifthe working / answer in accordance with the Knowledge assessedas stated in the marking scheme.
K - The mark is given ifthe working / answer in accordance with the Skills assessed asstated in the marking scheme.
N - The mark is given ifthe working / answer in accordance with the Values assessed asstated in the marking scheme.
PA - Subtract 1mark (only once) from the Nmark when students make an early rounding ofnumbers.
KP - Subtract 1mark (only once) from the P mark orNmark when students do not write theimportant steps of the calculations.
3. Accept answers correct to 4significant figures unless stated otherwise in the marking scheme.
4. Accept other correct methods which are not given in the marking scheme.
5. Accept answers in Bahasa Melayu.
6. Calculating total marks.
EScore for Paper 1+Y Score for Paper 2! ^=* x 100%
180
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SULIT3472/2(PP)
PENTAKSIRAN SUMATIF 3 SPM 2013PERATURAN PEMARKAHAN UNTUK MATEMATIKTAMBAHAN KERTAS 2
No.
SECTION A I40MARKSJ
MARK SCHEME
j-x
x = 3 - 2y or y - PI
(3 - lyf + y2 +2(3 - 2y)y =5 or x2 +2x
/ -6y +4 =0 or x2 +6x- 11 =0
j-x
v 2
j-x
-(-6)±x/(-6)2-4(l)(4) x=~ m:-6±J62-4(1)(-11)y =
2(1) 2(1)
;; = 5-24, 0-76 Nl (both)
x = -7-47, 1-47 Nl(both)
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= 5
Kl
SMARK
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No. MARK SCHEME EMARK
2 (a) j»nd(b)
.V;
4
3
*•»
s ^X 2
"/ •—- \ o x/ X y-3--1 \ 7t
I 1 T *"
7
0 / n x
-4
- shape of cosine graph PI
- amplitude (max = 4) PI
- negative graph PI
- period/cycle in 0 <x < n PI (ignore range greater than /z)
y = 3 Kl (equation of straight line)
Kl (any straight line with negative gradient or ^-intercept of 3^
No. of solutions = 2 Nl (without any mistake done)
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No. MARK SCHEME EMARK
3(a) 4 + («-l)(2) Kl
2« + 2 Nl
(b)7>30
6+(p -1)3= 30 PI (d =3) Kl (useformula)
n= 9 Nl7
(c)y[2(6) +(20-l)3] Kl
690 cm Nl
4 (a) (i) ^ =3.r-10x Kl<ix
Use 5v = — x §x and substitute Sx = - 0-01dx
= (3jc2-10:c)(-0-01) Kl
= 0-03 Nl
(ii) Approximate value -10 + 0-03
-9-97 Nl
(b) ^ =13 , —=0-4 (both) PIdx dt
Use chain rule, — = — x —dt dx dt
= 0-4 x 13 Kl
= 5-2 Nl
7
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No. MARK SCHEME SMARK
5 (a) x = — or 26-8 Klv ' 25
19670 Nl
6
«=̂ ™-(2^? K,= 8-28 Nl
(b) (8-28 x 2)2 or (8-28)2 x (2)2 Kl
= 274-2 Nl
6
(a) (i) BD = BA + AD = -$y + -AC
= -8v + -(8y + 4x)2 ~
= -Ay + 2x
Use triangle law of vectorKl
-> -> ->
(ii) CE = G4 + ;4£ = -8>> - 4x + 2y
=-4x-6yJ
Nl
(b) (i) £F = hBD = h(-4y +2x) = 2hx - 4hy Nl
(ii) CF = kCE =k(-4x-6y)= -4kx-6ky Nl8
(c) BF = BC + CF = 4x-4kx-6ky
= (4- 4*)x - 6£y K1 for fmding BFand compare
2/7 = 4- 4£
- 4/j = -6A
Solve Kl
7
A=6 Nl both7
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No. 7
|..:_j..q4^H^^
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No. MARK SCHEME
2xl(a) v = 8x + c Kl for integration
9 -'(Give Kl for — or -8x)
2
2(1)39 = vJ -8(l) +c Kl for substitution to find c
2 v 'v = .y2-8.v + 16
y - (*-4)2 Nl
(b) -(7+9)(1) + \{x-4fdx Kl for area of trapezium
=8 +(*-4)3
-T»
= 17 Nl
(c) Volume = iv \{x-4)* dx
= TV(x-4f
-4
- 1
Kl for integration of curve
Kl for Al+A2
Kl
•i
for limit '
Kl for integration (Ignore tc)
= 48-* Nl (Accept 152-7)
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IMARK
10
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No. MARK SCHEME
(a) m = —v } 2
PI
y-X- = -(x-2) OR,2 2
2y = 3x-5 or equivalent
2 2
Nl
Kl
(6) Tr>- to solve simultaneous equations :
2x+ 3?-12 = 0
2>; = 3.x:-5
13^ = 26 Kl (until 1 unknown left)
y —2, x = 3
(3,2) Nl
(c) Q(0, 4),R\ -, o] Both coordinates seen anywhere PI
Area = —2
s
0 33
0
4 2 0 4
= ±|0(2) +3(0)+f(4)-3(4)--(2)-0| Kl2 j J
i 1*3
= 4-unit2 Nl (Accept4-333 or —)3 J
{d) 3x+2& =o or 3y +2(2) =4 or equivalent Kl
-2,yl
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EMARK
10
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No.
10
10
MARK SCHEME
_, (MON\ 12 __t(a) tan1)—-J=y Kl
Z.MON = 1-176 x 2 = 2-352 radian
ZQPR = 0-395 x2= 0-79 radian
(b) Perimeter = 2-352(5) + 0-79(8) + 2(12-8) Kl Kl
= 26-08 cm Nl
(c) Area = Area of2 triangles -[Area ofMON+ area of PQR ]
=2x1x12x5 -[ix(5)2(2-352) +ix(8)2(0-79)] for2 2 2
triangle & sector K1K1
= 60- [29-4+ 25-28] Kl
= 5-32 cm Nl
Nl
Nl
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EMARK
10
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SULIT 11
No.
11
MARK SCHEME
(a) (i) 2280 = «(0-75) Kl
n = 3040 Nl
(ii) P(X>2)
= \-P(X=0)-P(X=\)
=1- 8C0 (0 -75)° (0 -25)8 - 8C, (0 •75)' (0 -25)*
= 0-9996 Nl
(b) (i) P(58<X<67)
'58-64 _. 67-64= P
12<Z<-
12Kl for using Z =
Kl
X-no
= P(- 0-5 <Z< 0-25)
= 1-P(Z<- 0-5)-P(Z> 0-25) or 1-0-3085-0-4013 Kl
= 0-2902
= 29-02% Nl
(ii) P(X>m) = 0-978
P\Z>m-64
12= 0-978 Kl
m - 64
12= -2-014 Kl
w = 39-83 Nl
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SULIT 12 3472/2(PP)
SECTION C |20 MARKS|
No. MARK SCHEME SMARK
12(a) v= \2t-\0dt =r-\0t + c Kl
24 = (0)2-I0(0) + c; c = 24 PI
v = /2-10/ + 24
a = 0, 2/- 10 =0
t =5s Nl
v = (5)2-10(5) + 24= -1 Nl
(b) v = /2-10/ + 24 < 0 PI v /
(r-6)(/-4)<0 Kl
104 < t< 6 Nl "<m/" "
4 6
(c) Dis= \t2 -\Qt +24dt + J?-10/+ 24*// Kl for integrate and0 4
limit
= \'--^ +24,3 2
4
+
03 2
6
Kl
4
= 38-m Nl3
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No.
13
13
MARK SCHEME
(a) l32xQ* =142 Kl100
£, = 107-6 Nl
(b) (i)/7=ii^xl00 Klv J 12-10
= 120-7 Nl
(ii)—^—xlOO =116 Kl [give once-(i) or (ii)|12-10
i>« = RM14-04 Nl
(132x40) +(U6x25) +(A:x35) = m K1KC) 100
k = 109-1 Nl
(^0 4012/120x110
100
= 132 Nl
^xlOO = 132 Kl38
Pmi = RM50-16 Nl
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10
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y
400
360
320
280
240
200
160
120
MO
40
No. 14
(a)
(b)
(c)
14
20x + 25y < 10 000 or equivalent N1
x<2y or equivalent Nl
y —x > 40 or equivalent N1
IvI
Nl
Nl
(i)
(ii)
draw correctlyat least one straight linedraw correctly three straight linesshaded region
y maximum- 240 Nl
Use3Qx + 40> i.e.
30(200) + 40(240) Kl Kl
= RM 15 600 Nl
x
400
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SULIT 15
No.
15
MARK SCHEME
(tf)(rMC2 = 8-72+ 12-22-2(8-7)(12-2) cos 125 Kl
= 18-61 cm Nl
AC = 18-61 cm Nl
,,„sinC _12-2
sinl25Kl
18-61
ZBCA = 32-48° Nl
OR
12-22 = 8-72~ 18-612 —2(8-7)(l 8-61) cos C
ZBCA = 32-48° Nl
C
(b) (i)
BB' 1-7(i')~
sin 70 sin 55
BB' = 9-98 cm Nl
Area = -(8-7)(9-98 + 12-2)sin 55 Kl
= 79-03 cm2 Nl
Kl
OR
BB'
sin 70- 87 K!
sin55
BB' = 9-98 cm Nl
Area = - (18-61)(9-98 + 12-2)sin 22-52 Kl
= 79-05 cm2 Nl
OR
Area = - (8-7)(18-61) sin (70 + 32-48) Kl
= 79-04 cm2 Nl
PI for 102-48c
PI for 70°
Kl
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2 MARK
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