0606_s14_ms_23

CAMBRIDGE INTERNATIONAL EXAMINATIONS

International General Certificate of Secondary Education

MARK SCHEME for the May/June 2014 series

0606 ADDITIONAL MATHEMATICS

0606/23 Paper 2, maximum raw mark 80

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2014 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Mark Scheme Syllabus Paper

IGCSE May/June 2014 0606 23

Cambridge International Examinations 2014

1 (i)

(ii)

2

2

1500 r= (1.6)

25 only

their 25 + their 25 + their 25 1.6 or better

90

M1

A1

M1

A1

25 is A0

their 25 must be positive

2

xx

3log

13log = oe soi

u2 4u 12 = 0 oe

solve their 3 term quadratic in u

Solve log3 x = 6 or log3 x = 2 oe

729 and 9

1

B1

M1

M1

M1

A1

may be implied by u

x

13log = oe

condone sign errors

3 (i)

(ii)

031

413 and

1

3

5

or ( )135 and

04

41

13

Multiplication of compatible matrices

17

22 or ( )1722 as appropriate

( )11 with

17

22 or ( )1722 with

1

1

B1

M1

A1

B1

Must be correct shape from candidates

product




4 (a) (i)

B1

(ii)

or

B1

any Venn diagram showing three circles

which do not all overlap

(b) (i) 50 C B1

(ii)

64 S C B1ft

ft only on use of and instead of

and

(iii) n(S) = 90 B1

5 (i)

162168422

2

++=

+

Correct completion

B1

B1

(ii) Use a

acbb

2

42

Multiply top and bottom by 322

22

M1

M1

A1

+

+

=3222

422

Or 624

6 Eliminate x or y

Rearrange to quadratic in x or y

x2 27x + 72 = 0 or y2 + 9y 90 = 0

Factorise or solve 3 term quadratic

x = 3, x = 24 or y = 6, y = 15

y = 6, y = 15 or x = 3, x = 24

M1

M1

A1

M1

A1

B1




7 (a)

(b)

sin

1

cos

1

sin

cos

cos

sin

+

+

Clears the fractions in the numerator and

denominator using common denominator

cossin

cossin22

+

+ and completion

evidence of 13

13

5sin =x

13

12cos =x

B1

M1

A1

B1

B1

B1ft

ft on their 13

8 (i)

(ii)

(iii)

Attempt to find b2 4ac

Completely correct argument

m = 6(4) 8(2) + 3

y 10 = 11(x 2) or y = 11x 12

Integrate to 2x3 4x2 + 3x(+c)

10 = 2(2)3 4(2)2 + 3(2) + c

y = 2x3 4x2 + 3x + 4 soi

M1

A1

M1

A1

B2,1,0

M1

A1

may be in formula

or attempt to complete square

dep on c being a genuine constant of

integration




9 (i)

(ii)

(0, 7)

mAB = 2

perpendicular gradient 2

1=

72

1+= xy

mAB = 1

y = x + 13

Solve their y = x + 13 and 72

1+= xy

D(12,1)

Complete method for area

84

B1

B1

M1

A1

B1

B1

M1

A1

M1

A1

10 (i)

(ii)

21

21d

d

2

2

+=

+

x

x

x

x

Use of quotient rule

( )( )

( )2121

2212

2

2

2

+

+

+

x

x

x

xx

Multiply each term by ( )212 +x

( )

( )23

2

22

21

2212

+

+

x

xx

leading to k = 42

21

26

2+

x

x

k

Use limits in

21

2

2+

x

xC

55

8 or 0.145

B1

M1

A1

M1

A1

M1

M1

A1

Alt method using product rule

32

2

2121

1

d

d

+

=

+

x

x

xx

is B1

then M1 A1 as in quotient

k must be a constant




11 (i)

(ii)

(iii)

(iv)

a=OM

MB = 5b a

bON 3=

AP = (3b 2a)

APMAMP +=

a + (3b 2a)

Put MBMP =

Equate components

Solve simultaneous equations

7

5=

B1

B1

B1

B1

M1

A1

M1

M1

M1

A1

12 (i)

(ii)

(iii)

(iv)

3 < f < 7

f(12) = 5

(f(5) = ) 22 +

Clear indication of method

f1(x) = (x 2)2 + 3

gf (x) = ( ) 23

120

+x

Attempt to solve their gf (x) = 20

x = 19

B1,B1

B1

B1

M1

A1

B1

M1

A1

If B0 then SC1 for 3 I f I 7

f 2 (x) ( ) 2323 +

+x earns B1

condone y = (x 2)2 + 3

0606_s14_ms_23

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