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CAMBRIDGE INTERNATIONAL EXAMINATIONS
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Cambridge International General Certificate of Secondary Education
MARK SCHEME for the October/November 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 October/November 2014 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
1
(i)
AG
(i)
12
C 4 = 495
B1
(ii)
7
C2 × 5C 2 = 21×10
M1
A1
not K and B = 6C 2 × 4C1 =15 × 4 = 60
K and not B = 6C1 × 4C2 = 6 × 6 = 36
60 + 36
96
B1
B1
M1
A1
OR
K and B = 6C1× 4C1 = 6×4 = 24
not K and not B = 6C 2 × 4C2 =15×6 = 90
210 – 90 – 24
96
B1
M1
A1
(i)
C is (1, 6)
D is (1, 6) + (12, 9)
= (13, 15)
B1
M1
A1ft
(ii)
gradient of CD =
(iii)
area =
1
1
× AB × CD = × 10 × 15
2
2
=75
or array method
Paper
23
factorise or solve quadratic factor = 0
A1
= 210
15 − 6  3 
= 
13 − 1  4 
10 − 2  8 −4 
=
gradient of AB =
=

−2 − 4  −6 3 
3 −4
×
= − 1 lines are perpendicular
4 3
Syllabus
0606
M1
A1
M1
A1
M1
(iii)
3
f(2)=0 → 3(2)3 +8(2) 2 − 33(2) + p = 0
correct working to p = 10
method for quadratic factor
f( x ) = ( x − 2)(3x 2 + 14 x − 5)
f( x) = ( x − 2)(3x − 1)( x + 5)
1
f(x) = 0 → x = 2, − 5,
3
(ii)
2
Mark Scheme
Cambridge IGCSE – October/November 2014
B1
B1ft
B1
B1
correct completion
M1
good attempt at two relevant lengths
for 12 base × height method
A1
© Cambridge International Examinations 2014
www
Page 3
4
5
(i)
2000 = 1000e a + b
(ii)
3297 = 1000e 2 a − b →
= ln 3.297
oe
→
a + b = ln 2
B1
2a + b
M1
A1
(iii)
Solve for one value
a = 0.5 and b = 0.193 or 0.19
M1
A1
(iv)
n = 10 P = 1000e5.193
= $180 000.
M1
A1
uuur
OX = µ ( a + b )
B1
(i)
(ii)
(iii)
6
Mark Scheme
Cambridge IGCSE – October/November 2014
(i)
(ii)
(iii)
uuur
uuur
RP = b − 3a or RX = λ ( b − 3a )
uuur
OX = 3a + λ ( b − 3a )
uuur uuur
OX = OX and equate both coefficients
µ = 3 − 3λ
µ =λ
µ = λ = 0.75
RX
= 3 or 3:1
XP
m=4
39 − 19
ln y − 39
equation of line is
=
x
9−4
3 −9
x
ln y = 4(3 ) + 3
oe
Syllabus
0606
Paper
23
substitution of 2, 3297 and
rearrange
B1
B1
M1
A1
A1ft
λ
1− λ
B1
M1
forms equation of line
A1ft
ft only on their gradient
x = 0.5 → ln y = 4 3 + 3 = 9.928
y = 20 500
M1
correct expression for lny
Substitutes y and rearrange for 3x
Solve 3x = 1.150
x = 0.127
M1
M1
A1
A1
© Cambridge International Examinations 2014
Page 4
7
(i)
Mark Scheme
Cambridge IGCSE – October/November 2014
2
+1 →
y
2
f −1 ( x) =
x −1
x=
y=
2
x −1
M1
Paper
23
any valid method
A1
2
(ii)
2 
gf ( x ) =  + 1 + 2
x 
(iii)
fg ( x ) =
2
+1
x +2
(iv)
ff ( x ) =
2
2x
+1 =
+1
2
x
+
2
+1
x
2
B2/1/0
–1 each error
B2/1/0
–1 each error
M1
correct starting expression
3x + 2
x+2
3x + 2
= x → x2 − x − 2 = 0
x+2
(x – 2)(x + 1) = 0
x = 2 only
A1
correct algebra to given answer
M1
form and solve 3 term quadratic
(i)
v = C +Ksin 2t
v = 5 + 6sin2t
a = 12cos2t
M1
A1
A1ft
(ii)
a = 0 → cos 2t = 0 and solve
π
t=
or 0.785 or 0.79
4
π
v = 5 + 6sin = 11
2
=
8
Syllabus
0606
(iii)
C≠0
v = 2 → sin 2t = −
t=
7π
12
1
2
and solve
or 1.83 − 1.84
a = 12cos
7π
= −6 3 or
6
A1
M1
set a = 0 and solve for t
A1
A1ft
ft only on K
M1
set v = 2 and solve for t
A1
− 10.4
A1
© Cambridge International Examinations 2014
Page 5
9
(i)
Mark Scheme
Cambridge IGCSE – October/November 2014
dy
1
=4–
dx
( x − 2) 2
dy
1
2
= 0 → ( x − 2) =
dx
4
2
(4x – 16x + 15 = 0)
x = 2.5 or 1.5
y = 12 or 4
d2 y
−3
= 2 ( x − 2)
2
dx
d2 y
> 0 → minimum
dx 2
d2 y
x = 1.5 →
< 0 → maximum
dx 2
x = 2.5 →
(ii)
dy
=3
dx
Use m1m2 = –1 for gradient normal from gradient
tangent
1
y − 13
Eqn of normal :
=–
3
x−3
x=3 →
Intersection of norm and curve
x
1
14 –
= 4x +
3
x−2
13x2 – 68x + 87 = 0
29
x=
or 2.23
13
10
(i)
LHS =
1 + cos x + 1 − cos x
(1 − cos x )(1 + cos x )
2
1 − cos 2 x
2
= 2 = RHS
sin x
=
(ii)
2cosec 2 x = 8
1
4
1
sin x = ±
2
o
x = 30 , 150o , 210o , 330o
sin 2 x =
Syllabus
0606
Paper
23
B1
M1
solve 3 term quadratic from
dy
=0
dx
A1
A1
M1
x values or 1 pair
y values or 1 pair
d2 y
use 2 with solution from
dx
dy
=0
dx
A1
both identified
www
B1
M1
must use numerical values
A1ft
M1
equation and attempt to simplify
DM1
attempt to solve 3 term quadratic
A1
B1
correct fraction
B1
correct evaluation
B1
use of 1 − cos 2 x = sin 2 x and
completion of fully correct
proof
M1
identity used
A1
A1
A1
© Cambridge International Examinations 2014