w
w
ap
eP
m
e
tr
.X
w
CAMBRIDGE INTERNATIONAL EXAMINATIONS
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.
om
.c
MARK SCHEME for the May/June 2014 series
s
er
International General Certificate of Secondary Education
Page 2
1
(i)
(ii)
Mark Scheme
IGCSE – May/June 2014
500 =
1 2
r (1.6)
2
A1
± 25 is A0
their 25 + their 25 + their 25 × 1.6 or better
M1
their 25 must be positive
90
A1
3
(i)
1
oe soi
log 3 x
B1
may be implied by log x 3 =
u2 – 4u – 12 = 0 oe
M1
condone sign errors
solve their 3 term quadratic in u
M1
Solve log3 x = 6 or log3 x = –2 oe
M1
1
9
A1
 5
 
 3 1 4
 and  3 

1 3 0
1
 
B1
729 and
Paper
23
M1
25 only
log x 3 =
2
Syllabus
0606
1
oe
u
3 1


or (5 3 1) and  1 4 
 4 0


(ii)
Multiplication of compatible matrices
M1
 22 
  or (22 17 ) as appropriate
 17 
A1
 22 
1
 or (22 17 ) with  
 17 
1
(1 1) with 
Must be correct shape from candidates
product
B1
© Cambridge International Examinations 2014
Page 3
4
Mark Scheme
IGCSE – May/June 2014
or
(b) (i) 50 ∉ C
B1ft
(iii) n(S′) = 90
(ii)
6
B1
any Venn diagram showing three circles
which do not all overlap
B1
(ii) 64 ∈ S ∩ C
(i)
Paper
23
B1
(a) (i)
(ii)
5
Syllabus
0606
ft only on use of ⊄ and ⊂ instead of ∉
and ∈
B1
2
 2 2 + 4  = 8 + 16 2 + 16


B1
Correct completion
B1
− b ± b 2 − 4ac
Use
2a
M1
Multiply top and bottom by 2 2 − 3
M1
2− 2
A1
Eliminate x or y
M1
Rearrange to quadratic in x or y
M1
x2 – 27x + 72 = 0 or y2 + 9y – 90 = 0
A1
Factorise or solve 3 term quadratic
M1
x = 3, x = 24 or y = 6, y = –15
A1
y = 6, y = –15 or x = 3, x = 24
B1


=


 2 2 + 4  

 
2 2 2 + 3  


Or 4 2 − 6
© Cambridge International Examinations 2014
Page 4
7
(a)
Mark Scheme
IGCSE – May/June 2014
sin θ cosθ
+
cosθ sin θ
1
1
+
cosθ sin θ
A1
evidence of 13
B1
5
13
cos x = −
8
(i)
(ii)
(iii)
M1
sin 2 θ + cos 2 θ
and completion
sin θ + cosθ
sin x =
Paper
23
B1
Clears the fractions in the numerator and
denominator using common denominator
(b)
Syllabus
0606
B1
12
13
B1ft
ft on their 13
Attempt to find b2 – 4ac
M1
may be in formula
or attempt to complete square
Completely correct argument
A1
m = 6(4) – 8(2) + 3
M1
y – 10 = 11(x – 2) or y = 11x – 12
A1
Integrate to 2x3 – 4x2 + 3x(+c)
B2,1,0
10 = 2(2)3 – 4(2)2 + 3(2) + c
M1
y = 2x3 – 4x2 + 3x + 4 soi
A1
dep on c being a genuine constant of
integration
© Cambridge International Examinations 2014
Page 5
9
(i)
Mark Scheme
IGCSE – May/June 2014
(0, 7)
B1
mAB = 2
B1
1
2
perpendicular gradient = −
(ii)
10
(i)
1
y=− x+7
2
A1
mAB = –1
B1
y = –x + 13
B1
1
Solve their y = –x + 13 and y = − x + 7
2
M1
D(12,1)
A1
Complete method for area
M1
84
A1
x
d  2
 x + 21  =

dx 
(x
2
2
B1
2
x + 21
)
+ 21 − 2 x ×
(x
2
)
+ 21
x
2
Multiply each term by
(
)
2 x 2 + 21 − 2 x 2
(x
2
3
2
)
+ 21
)
+ 21
(x
2
8
or 0.145
55
)
+ 21
M1
A1
M1
2x
x 2 + 21
then M1 A1 as in quotient
A1
leading to k = 42
6
2x
×
k
x 2 + 21
Use limits in C ×
Alt method using product rule
d
1
−x
is B1
=
3
dx  x 2 + 21   2

  x + 21 

M1
(x
Paper
23
M1
Use of quotient rule
(ii)
Syllabus
0606
k must be a constant
M1
A1
© Cambridge International Examinations 2014
Page 6
11
Mark Scheme
IGCSE – May/June 2014
OM = a
B1
MB = 5b – a
B1
ON = 3b
B1
AP = λ (3b – 2a)
B1
(iii)
MP = MA + AP
a + λ (3b – 2a)
M1
A1
(iv)
Put MP = µ MB
M1
Equate components
M1
Solve simultaneous equations
M1
(i)
(ii)
5
7
A1
(i)
3<f<7
B1,B1
(ii)
f(12) = 5
B1
(f(5) = ) 2 + 2
B1
(iii)
Clear indication of method
f–1(x) = (x – 2)2 + 3
M1
A1
(iv)
gf (x) =
λ=
12
Syllabus
0606
120
(x − 3) + 2
If B0 then SC1 for 3 I f I 7
f 2 (x)
M1
x = 19
A1


(x − 3) + 2 − 3  + 2
condone y = (x – 2)2 + 3
B1
Attempt to solve their gf (x) = 20
Paper
23
© Cambridge International Examinations 2014

earns B1