Trial Exam Mark Scheme Paper 13
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7/31/2019 Trial Exam Mark Scheme Paper 13
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SRI KL INTERNATIONAL SCHOOL
International General Certificate of Secondary Education
MARK SCHEME for IGCSE Trial Examination August 2012 question paper
0606 ADDITIONAL MATHEMATICS0606/13 Paper 13, maximum raw mark 80
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Page 4 Mark Scheme : Teachers Version Syllabus Paper
IGCSE Trial Examination August 2012 0606 13
1 (i) Initial temperature, sub x = 0,T = 60(0.95) 0
= 60o
C A1 [1]
(ii) x = 10, T = 60(0.95) 10 M1= 35.92= 35.9 oC A1 [2]
(iii) T = 27, 27 = 60(0.95) x
0.95 x = 0.45lg 0.95 x = lg 0. 45
x =lg0.45lg0.95
M1
= 15.56= 15.6 min A1 [2]
Total : [5]
2 (i) Let f( x) = 3 23 11ax x x b ,f( 2 ) = 0
3 2( 2) 3( 2) 11( 2)a b = 0 M18 10 0a b
f( 1) 12 3 2( 1) 3( 1) 11( 1) 12a b M1
4a b
Solving both equations, a = 2 and b = 6 M1, A1, A1 [5]
(ii) 3 2
f 2 3 11 6 x x x x = 2( 2)(2 7 3) x x x , using long division or polynomial identities M1, A1= ( 2)(2 3)( 3) x x x A1 [3]
Total : [8]
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3 (i) P and Q are subsets of U A1P and Q intersect A1 [2]
(ii)
Numbers Nature of numbers Region227
Positive rational number P Q
25 10 Positive rational number P Q sin(60 ) Positive irrational number P only
0 Neutral number Q only3 8 Negative rational number Q only
e Negative irrational number ( ) 'P Q All 6 correct 4 marks
5 correct 3 marks3 & 4 correct 2 marks1 & 2 correct 1 mark
A0,1,2,3,4 [4]Total : [6]
4 (i) ( 2)( 3) y x x M1
= 2 2 3 6 x x x
= 2 6 x x By comparison, p = 1 , q = 6 A0,1,2 [3]
(ii) To find minimum point, complete the square for the function.2 6 y x x
2 2
1 1 62 2
y x
=2
1 252 4
x
M1
Min point =1 25
,2 4
A1 [2]
(iii) Using answer from part (ii),
y =2
1 25
2 4
x
a = 1, b =12
, c =254
A0,1,2 [2]
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(iv)
Correct shape A1Correct values A1 [2]
(v) k = 254
A1 [1]
Total : [10]
5 (i) Maximum value occurs when cos x is 1, 10 = a b(1) M1a b = 10
Minimum value occurs when cos x = 1 , 2 = a b( 1 ) M1a + b = 2
Solving both equations, a = 4, b = 6 A1 [3](ii)
Correct shape A1Correct values A1 [2]
Total : [5]
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6 (i) M =900 450 600
500 1000 700
A0,1,2 [2]
(ii) P =
50
80
40
A1 [1]
(iii) Total profit =1
1
900 450 600
500 1000 700
50
80
40
M1
=1
1
105000 133000 A1
= 238000 A1 [3]Total : [6]
7 (i) 2 12 1
y ym
x x From the graph, Y-intercept = 1
=5 18 0
=12
M1
Y = mX + clg y = m lg x + c
lg y =12
lg x + 1
3 =12
lg x + 1 M1
lg x = 2 2
lg x = 4 x = 10 4 = 10000 A1 [3]
(ii) From part (i), lg y =12 lg x + 1
= lg12 x + lg 10 M1
= lg (1012 x )
y = 1012 x
y = ax n
a = 10, n =12
A0,1,2 [3]
Total : [6]
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8 (i)Arrangement Principal 5 other teachers Total number of ways
Number of ways 1 105C
1051 C = 252
M1 + A1 = [2]
(ii)Arrangement 2 female
teachers chosenfrom 4
4 maleteachers/principal
chosen from 7
Total number of ways
Number of ways 42C
74C
4 72 4C C = 210
M1 + A1 = [2] (iii)
Arrangement Number of ways Total number of ways2 females + 4 males 4 7
2 4C C = 210210 + 140 + 21 = 371 ways3 females + 3 males 4 7
3 3C C 140
4 females + 2 males 4 74 2C C 21
M1 + A1 = [2]Total : [6]
9 (i) 3 2d d
(sin 2 ) 3sin 2 (sin2 )d d
x x x x x
M1
= 23sin 2 (2cos2 ) x x
= 26sin 2 cos 2 x x A1 [2]
(ii)
2 24 4
0 0
1sin 2 cos 2 d 6sin 2 cos 2 d
6 x x x x x x M1
=
3 40
1 sin 26
x
= 3 31
sin 2( ) sin 2(0)6 4
=16
A1 [2]
(iii) 2 2sin 2 cos2 cos2 (1 cos 2 ) x x x x M1
= 3cos2 cos 2 x x A1 [2]
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(iv)
2 34 4
0 0sin 2 cos 2 d cos 2 cos 2 d x x x x x x
34
0
1cos2 cos 2 d
6 x x x M1
34 4
0 0
1cos 2 d cos 2 d
6 x x x x
34 4
0 01cos 2 d cos 2 d6
x x x x M1
=
4
0
sin 2 12 6
x
=
sin 2( ) sin 2(0) 14
2 2 6
=1 12 6 A1
=13
A1 [4]
Total : [10]
10 (i) Initial distance, t = 0,s = 6 t 2 2t 3 + 30
= 6(0) 2 2(0) 3 + 30= 30 m
Initial distance from O = 30 m A1 [1]
(ii)dd
sv
t ,
= 12 t 6t 2 A1When v = 0 m 1s , 0 = 12 t 6t 2
6 (2 )t t = 0 t = 0 s, t = 2 s A1
dd
va
t
= 12 12 t A1t = 2 s , a = 12 12(2)
= 12 m 2s A1 [4]
(iii)total distance
average speedtotal time taken
=48 m4 s
= 12 m 1s A1
Integrate v with respect to t from t = 0 to t = 2 s gives distance travelled = 8 m.from t = 2 to t = 4 s gives distance travelled = 40 m A1
Total distance = 48 m A1 [3]Total : [8]
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11 EITHER(i) A1 = Area of minor segment
= 21
( sin )2
r M1
=21
(10) ( sin )2 A1
= 50( sin ) A1 [3]
(ii) A2 = Area of major segment= Area of circle - Area of minor segment
= 2 50( sin )r M1= 100 50 50sin A1 [2]
(iii)
3
, A1 =
50( sin )
3 3
= 3
50( )3 2
= 9.058607
A2 =
100 50( ) 50sin3 3
= 3
100 50( ) 50( )3 2
= 305.100
Percentage of 12
A A
= 9.058607 100%305.100
= 2.969% = 2.97 % A0,1,2 [2]
(iv) Chord AB = 2 sin2
r
=
32(10)sin2
M1
= 10 cm A1 [2]
(v) Perimeter AOB = 10 + 10 + 10= 30 cm A1 [1]
Total : [10]
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11 OR
(i) Volume of cone = 21
3
r h
= 21 1
( )3 5
h h
= 31
75
h A1
Using similar triangle concept,4
20r
h
=15
r h A1
Given 1d
0.1 cm sdh
t ,
d d dd d dV V h
t h t
= 21
0.0125
h
=21
(6) 0.0125 A1
= 0.0452389= 0.0452 cm 3 s
1 A1 [4]
(ii) (a) To find the maximum value of V ,d
0dV
x
31 (147 )2
V x x
3147 12 2V x x
2d 147 3d 2 2V
x x
M1
2147 3 02 2
x
7 x cm A1
Maximum V occurs at x = 7 cm = 31
(147(7) (7) )2
V
= 343 cm 3 A1 [3](b) 8.9 9 0.1r cm,
d dV V
x x,
d
dV
V x x
2147 3 0.12 2
V x
M1, A1
Sub x = 9 cm, 2147 3
(9) 0.12 2
V
= 4.8 cm 3 A1 [3]Total : [10]