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]