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CAMBRIDGE INTERNATIONAL EXAMINATIONS
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Cambridge International General Certificate of Secondary Education
MARK SCHEME for the May/June 2015 series
0606 ADDITIONAL MATHEMATICS
0606/21
Paper 2 (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 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 IGCSE – May/June 2015
Syllabus
0606
Paper
21
Abbreviations
awrt
cao
dep
FT
isw
oe
rot
SC
soi
www
1
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
log3 x
(a)
log3 27
log 3 x
isw
3
log a 15 − log a 3 = log a 5 soi
(b)
log a 5 or log a a
log a y = log a 125a ⇒ y = 125a
A1
NOT log3 x ÷ 3
M1
A1
[f ( x) =]2 x − 4 and [f ( x ) =] − 2 x + 4
(a)
Can use other interim bases if all correct
but M1 when in base 3 only
M1
3
2
M1
B1,B1
y
(b)
4
x
B1
B1
B1
O
3
Condone y =….
A=
(a)
1  51 − 8

2
4  31
8
A = 
4
(b) (i)
(ii)
−2
19   20
− 
65  15
− 5 

− 10 25 
0
6

10 
correct shape;
y intercept marked or seen nearby;
intent to tend to y = 3 (i.e. not tending to
or cutting x-axis)
M1
A1
Integer values
The (total) value of the stock in each of the 3
shops
B1
Must have “each” oe
The total value of the stock in all 3 shops
B1
Must have “total” oe
3
© Cambridge International Examinations 2015
Page 3
4
(i)
M1
PT =19.3
A1
awrt 19.3
M1
or
M1
8 × their PT – their sector
A1
awrt 79.1
 3π 
8   oe (18.8)
 4 

 3π  
6π + 16 tan  8   = 57.5



(iii)
(a)
Permutation because the order matters oe
(b) (i)
6
(ii)
(iii)
Syllabus
0606
PT
 3π 
= tan   oe
8
 8 
1 2 3π
×8 ×
oe (75.4)
2
4
 3π 
8 tan   × 8 – their sector oe (=154.5-‘75.4’)
 8 
79.1
(ii)
5
Mark Scheme
Cambridge IGCSE – May/June 2015
C4 + 5C4 + 7C4
55
2
C1 × 6C 1 × 5C1 × 7C1
420
6
C3 × 2C1 or 2C2 × 5C1 × 6C1
summation
70
Paper
21
PT
8
=
3π
π
sin
sin
8
8
1 2 3π
×8 ×
2
8
M1
A1
Accept 57.4 to 57.5
B1
M1
A1
3 correct terms added
M1
A1
4 correct terms multiplied
M1
M1
A1
for either correct product
adding two correct products
If 0 scored, then SC1for 1,1,1,0 and
0,0,2,1 seen
6
(i)
(ii)
2t 2 −14t +12 = 0
(t −1)(t − 6) oe
(t = ) 1
∫ ( 2t
2
( s =)
(iii)
−14t +12 ) dt
3
Can use formula, etc.
A1
If t = 1 with no working, then M1A1
M1
2
2t
14t
−
+12t
3
2
dv
( 4t − 14 )
dt
[4(3) – 14 =] −2 cao
( a =)
M1
A2,1,0
−1 for each error or for +c left in or
limits introduced
M1
A1
© Cambridge International Examinations 2015
Page 4
7
Mark Scheme
Cambridge IGCSE – May/June 2015
AB = 15b – 5a = 5(3b – a) or
BC = 24b – 3a – 15b = 3(3b – a) or
(a)
Syllabus
0606
Paper
21
B1
Any correct simplified vector
B1
Any second simplified vector
AC = 24b – 3a – 5a = 8(3b – a)
Comment: e.g. the vectors are scalar multiples of
each other AND they have a common point (A, B
or C as appropriate)
(b) (i)
2i + 11j soi
2
⇒
(ii)
(iii)
8
(a) (i)
(ii)
2 + 11
(ii)
B1
125 or 5 5 or 11.2 (3 s.f .) or better)
B1fT
ft their 2i +11j (not OP or OQ )
1
B1fT
ft their answers from (i)
5 5
(2i + 11j) isw
i − 4 j + 3i + 7 j
2i +11j
or i − 4 j +
or
2
2
2i +11j
3i + 7 j −
2
M1
2i + 1.5 j
A1
ke4 x + 3 (+c) oe
1
k = oe
4
M1
1 4(3) +3
e
− e4(2.5) +3 ) or better
(
4
 x
k sin   (+ c)
3
k=3
+c
−2
+ 2 + x 2 ) dx =
DM1
ft their integral attempt
1 15 13
(e − e )
4
A1
Accept
M1
any constant, non-zero k
A1
π 1
3sin  ×  − 3sin ( 0 )
 6 3
∫(x
any constant, non-zero k
A1
0.520 944… = 0.521 to 3 sf or better
(c)
Dep on both B marks being awarded.
2
706 650.99… = 707 000 to 3 sf or better
(b) (i)
B1dep
x −1
x3
+ 2x +
−1
3
DM1
A1
B1
M1
A1
B1
Dep on their integral attempt in sin;
condone omission of lower limit
π 
Accept 3 sin 
 18 
Expands – accept unsimplified
integration of their 3 term expansion
Fully correct
+c
© Cambridge International Examinations 2015
Page 5
9
Mark Scheme
Cambridge IGCSE – May/June 2015
(4 x −1) ( x + 5) [≤ 0]
1
critical values
and −5 soi
4
1
−5 ≤ x ≤
4
(a)
(b) (i)
(ii)
M1
Syllabus
0606
Paper
21
Solves quadratic
A1
A1
( x + 4)2 − 25 or a = 4 and b = −25
1

Accept:  −5,  ; –5 ≤ x AND x ≤ 0.25
4

B1, B1
(Greatest value =) 25
x = −4
B1ft
B1ft
Must be clear
y
9
(iii)
10
x
1
B1
Correct shape with maximum in second
quadrant and crossing positive and
negative axes correctly
B1
All 3 intercepts correctly shown on
graph
(i)
ln y = ln( Ab x ) ⇒ ln y = ln A + ln b x
⇒ ln y = ln A + x ln b
M1
A1
(ii)
ln A = 11.4 ⇒ A= etheir 11.4
M1
A = 90 000 cao
ln b = −1
b = 0.4 cao
A1
M1
A1
x = 2.5 ⇒ lny = 9
y = e9 or 8000 to 1 sf
M1
A1
7 − x, x, 6 − x oe
B1
their attempt at 7 − x + x + 6 − x + 16 = 25 oe
M1
x=4
A1
Condone x = 4 for all 3 marks
23 − y, y, 9 − y oe
B1
or n( A ∪ C ) = 48 −16 = 32
48 = 30 + 25 + 15 – 7 – 6 – (their 4 + y) + their 4
oe soi
M1
or 32 = 30 + 15 – (their 4 + y)
or 48 = (23 − y) + 3 + 16 + y + 4
+ 2 + (9 − y)
y=9
A1
Condone y = 9 for all 3 marks
n(C) = 15 and y + n( B ∩ C ) = 9 + 6 = 15
[and so A' ∩ B ' ∩ C = ∅].
B1
or equivalent deduction
(iii)
11
O
-9
(i)
(ii)
(iii)
condone misread of scale for M1 (11.2
only)
Allow awrt −1
Allow awrt 8100
© Cambridge International Examinations 2015