4037 ADDITIONAL MATHEMATICS - XtremePapers

CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge Ordinary Level

MARK SCHEME for the May/June 2015 series

4037 ADDITIONAL MATHEMATICS

4037/12

Paper 1, 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 2015 series for most

?

Cambridge IGCSE , Cambridge International A and AS Level components and some Cambridge O Level components.

? IGCSE is the registered trademark of Cambridge International Examinations.

Page 2

Mark Scheme Cambridge O Level ? May/June 2015

Abbreviations

awrt cao dep FT isw oe rot SC soi www

answers which round to correct answer only dependent follow through after error ignore subsequent working or equivalent rounded or truncated Special Case seen or implied without wrong working

Syllabus 4037

Paper 12

1

2

k

- 4(2k

+ 5)

(< 0)

2

k - 8k - 20

(< 0)

(k - 10) (k + 2) (< 0)

critical values of 10 and ?2 - 2 < k < 10

M1

M1 A1 A1

2

use of b - 4ac , (not as part of quadratic formula unless isolated at a later stage) with correct values for a, b and c Do not need to see < at this point attempt to obtain critical values correct critical values correct range

Alternative 1:

dy

= 2 ( 2k + 5) x + k

dx

When

dy = 0, x =

-k

2

8k + 20 - k

, y=

dx

2 ( 2k + 5)

4 ( 2k + 5)

When y = 0 , obtain critical values of 10 and ?2 - 2 < k < 10

Alternative 2:

2

2

y

=

(2k

+

5)

x

+

k

2(2k

+

5)

k -

+1

4(2k + 5)

2

k

Looking at 1 -

=0

4(2k + 5)

leads to

critical values of 10 and ?2 - 2 < k < 10

M1

attempt to differentiate, equate to zero and substitute x value back in to obtain a y value

M1

A1 A1

consider y = 0 in order to obtain critical values correct critical values correct range

M1

M1 A1 A1

attempt to complete the square and

2

k

consider '1 -

'

4(2k + 5)

attempt to solve above = to 0, to obtain critical values correct critical values correct range

? Cambridge International Examinations 2015

Page 3

Mark Scheme Cambridge O Level ? May/June 2015

Syllabus 4037

Paper 12

sin cos

tan + cot

+ cos sin

2

=

cosec

1

sin

2

2

sin + cos

sin cos =

1

sin

1 =

cos

= sec

Alternative:

2

tan + 1

tan + cot =

cosec

tan cosec

2

sec =

1 tan

sin

2

sec =

sec

= sec

1 3 -2

?1

3

A=

2 -5

4

x 1 3

=

y

2 -5

-2 8

4 9

M1

sin

cos

for tan =

, cot =

cos

sin

and

cosec =

1 ; allow when used

sin

M1

dealing correctly with fractions in the numerator; allow when seen

M1 A1

use of the appropriate identity; allow when seen

must be convinced it is from completely correct work ( beware missing brackets)

M1 M1

1

for either tan =

or

cot

1 cot =

tan

and

1

cosec =

; allow when used

sin

dealing correctly with fractions in numerator; allow when seen

M1 A1

use of the appropriate identity; allow when seen must be convinced it is from completely correct work

B1 B1

1 multiplied by a matrix

2

for matrix

M1

attempt to use the inverse matrix, must be pre-multiplication

x 1 6

=

y 2 -4

x = 3, y = -2

A1, A1

? Cambridge International Examinations 2015

Page 4

Mark Scheme Cambridge O Level ? May/June 2015

Syllabus 4037

Paper 12

4

(i)

Area =

1

2

? 12

? 1.7

+

1

2

? 12

sin (2

- 1.7 - 2.4)

2

2

(ii)

= awrt 181

2

BC

2

= 12

2

+ 12

- (2 ?12 ?12 cos 2.1832)

2 - 4.1

or BC = 2 ? 12 ? sin

2

BC = 21.296

Perimeter = (12 ?1.7 ) + 12 + 12 + 21.296

= 65.7

B1,B1 M1 A1

B1 for sector area, allow unsimplified B1 for correct angle BOC, allow unsimplified correct attempt at area of triangle, allow unsimplified using their angle BOC (Their angle BOC must not be 1.7 or 2.4)

M1 A1

correct attempt at BC, may be seen in (i), allow if used in (ii). Allow use of their angle BOC.

B1 M1

A1

for arc length, allow unsimplified for a correct `plan' (an arc + 2 radii and BC)

5

(a) (i) 20160

(ii)

6

3 ? P ?2 4

= 2160

(iii)

6

5? 2? P 4

= 3600

Alternative 1:

6

C ? 5!? 2 4

= 3600

Alternative 2:

( ) 7

6

P - P ?2

5

5

= 3600

Alternative 3:

( ( ) ( ) ( ) ) 6

6

5

6

4

6

3

6

2! P + P ? P + P ? P + P ? P + P

4

1

3

2

2

3

1

4

= 3600

B1

B1,B1

6

B1 for

P 4

(must

be

seen

in

a

product)

B1 for all correct, with no further

working

B1,B1 B1

6

B1 for

P 4

(must be seen in a

product)

B1 for 5 (must be in a product)

B1 for all correct, with no further

working

B2 B1

6

for

C ? 5! 4

6

for

C ? 5! ? 2 4

B2 B1

( ) 7

6

for

P- P

5

5

( ) 7

6

for

P - P ?2

5

5

B2 B1

4 terms correct or omission of 2! in each term all correct

? Cambridge International Examinations 2015

Page 5

Mark Scheme Cambridge O Level ? May/June 2015

Syllabus 4037

Paper 12

(b) (i)

14

10

C? C

4

4

or

14

8

C?C

8

4

(or numerical or factorial equivalent)

= 210210

(ii)

8

6

C?C

4

4

= 1050

6

(i)

10ln4 or 13.9 or better

B1,B1

14

14

B1 for either

C

4

or

C

8

as part

of a product

B1 for correct answer, with no

further working

B1,B1

8

6

B1 for either

C or

4

C

4

as part of a

product

B1 for correct answer with no

further working

B1

(ii)

dx 20t

= 2

-4

dt t + 4

When

dx

20t

= 0,

=4

2

dt

t +4

2

leading to t - 5t + 4 = 0 t = 1, t = 4

M1 B1 DM1

A1

attempt to differentiate and equate

to zero

20t or equivalent seen

2

t +4

dx

attempt to solve their

= 0 , must

dt

be a 2 or 3 term quadratic equation with real roots

for both

? Cambridge International Examinations 2015

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