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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
0606/11
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 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
Mark Scheme
Cambridge IGCSE – October/November 2014
1
dy
16
= 2x − 2
dx
x
When
M1
A1
dy
= 0,
dx
DM1
x = 2, y = 12
2
(a)
4
Paper
11
for attempt to differentiate
all correct
dy
for equating
to zero and an
dx
attempt to solve for x.
A1
A1 for both, but no extra solutions
B1
for correct shape
B1
for max value of 2, starting at (0, 2)
and finishing at ( 180o , 2)
B1
for min value of –4
B1
must be positive
y
3
2
Syllabus
0606
2
1
x
1
2
3
−1
−2
−3
-4
−4
(b) (i)
(ii)
3
(i)
4
60o or
π
or 1.05 rad
3
1
y = 4(x + 3) 2 (+ c )
 1
10 = 4 9 2  + c
 
 
c = −2
1
y = 4(x + 3) 2 − 2
(ii)
B1
1
1
M1, A1
M1 for (x + 3) 2 , A1 for 4(x + 3) 2
M1
for a correct attempt to find c, but
must be from an attempt to
integrate
A1
Allow A1 for c = –2
1
6 = 4(x + 3) 2 − 2
x =1
A1 ft
ft for substitution into their
equation to obtain x; must have the
first M1
© Cambridge International Examinations 2014
Page 3
4
Mark Scheme
Cambridge IGCSE – October/November 2014
(i)
5y2 − 7 y + 2 = 0
B1, B1
(ii)
(5 y − 2)( y − 1) = 0
M1
(i)
x = −0.569
A1
must be evaluated to 3sf or better
y = 1, x = 0
B1
1
dy
= 3x 2 −
x
dx
M1
for attempt to differentiate
B1
for y = 1
dy
=2
dx
Tangent: y − 1 = 2(x − 1)
6
(i)
B1 for 5, B1 for –7
M1
2
ln 0.4
,x=
5
ln 5
When x = 1, y = 1 and
(ii)
Paper
11
for solution of quadratic equation
from (i)
for use of logarithms to solve
equation of the type 5 x = k
y=
5
Syllabus
0606
DM1
( y = 2 x − 1)
A1
Mid-point (5, 9)
B1
9 = 2(5) − 1
B1
Alternative Method:
Tangent equation y = 2 x − 1
Equation of line joining (–2, 16) and (12, 2)
y = − x + 14
Solve simultaneously x = 5, y = 9
B1
Mid-point (5, 9)
B1
(2 + px )6 = 64 + 192 px + 240 p 2 x 2 …
B1
for attempt to find equation of
tangent
allow equation unsimplified
for midpoint from given
coordinates
for checking the mid-point lies on
tangent
for a complete method to find the
coordinates of the point of
intersection
for midpoint from given
coordinates
for 240p2 or 240p2x2 or
2
6
C 2 × 2 4 × ( px ) or 6 C 2 × 2 4 × p 2
or 6 C 2 × 2 4 × p 2 x 2
240 p 2 = 60
1
p=
2
(ii)
M1
A1
(3 − x )(64 + 192 px + 240 p 2 x 2 ...)
Coefficient of x2 is 180 − 192 p
= 84
B1 ft
M1
A1
for equating their term in x2 to 60
and attempt to solve
ft for 192p, 96 or 192 × their p
for 180 – 192p
© Cambridge International Examinations 2014
Page 4
7
(i)
(ii)
Mark Scheme
Cambridge IGCSE – October/November 2014
A −1 =
1  b −2b 


5ab  a 3a 
X = BA −1
 1
 −a b   5a
=
 
 2a 2b   1

 5b
0
= 4

5
8
(i)
(ii)
2 
5a 

3 

5b 
−
1

2

5
uuur  12 
1
AB =   , at P, x = −2 + (12)
4
 16 
so at P , x = 1
1
y = 3 + (16 ) , y = 7
4
Gradient of AB =
Perp line:
(iii)
B1, B1
16
3
, so perp gradient = −
12
4
3
(x − 1)
4
(3x + 4 y = 31)
y−7 = −
 31 
Q  0, 
 4
Area AQB = 12.5
Syllabus
0606
B1 for
1
, B1 for
5ab
Paper
11
 b − 2b 


 a 3a 
M1
for post-multiplication by inverse
matrix
DM1
for correct attempt at matrix
multiplication, needs at least one
term correct for their BA −1 (allow
unsimplified)
A1
A1
for each correct pair of elements,
must be simplified
B1
for convincing argument for x = 1
B1
for y = 7
M1
for finding gradient of
perpendicular
M1
for equation of perpendicular
through their P
A1
Allow unsimplified
B1 ft
M1
A1
ft on their perpendicular line, may
be implied
for any valid method of finding the
area of the correct triangle, allow
use of their Q; must be in the form
( 0, q ) .
© Cambridge International Examinations 2014
Page 5
9
(i)
Mark Scheme
Cambridge IGCSE – October/November 2014
log y = log a + x log b
x
2
2.5
3
3.5
4
lg y
1.27
1.47
1.67
1.87
2.07
lny
2
2.93
2.5
3.39
3
3.84
logy
3
3.5
4.31
B1
for the statement, may be seen or
implied in later work,
M1
for attempt to draw graph of x
against log y
A2,1,0
–1 each error in points plotted
DM1
for attempt to find gradient and
equate it to log b, dependent on M1
in (i)
2
1
x
(ii)
2
3
4
x
Gradient = log b
lg b = 0.4 or ln b = 0.92
b = 2.5 (allow 2.4 to 2.6)
A1
Intercept = log a
lg a = 0.47 or ln a = 1.10
DM1
a = 3 (allow 2.8 to 3.2)
A1
Alternative method:
Simultaneous equations may be used provided
points that are on the plotted straight line are
used.
a = 3 (allow 2.8 to 3.2)
b = 2.5 (allow 2.4 to 2.6)
Paper
11
4
4.76
y
1
Syllabus
0606
DM1
DM1
A1
A1
for attempt to equate y-intercept to
log a or use their equation with
their gradient and a point on the
line, dependent on M1 in (i)
for a pair of equations using points
on the line, dependent on M1 in (i)
for solution of these equations,
dependent on M1 in (i)
A1 for each
© Cambridge International Examinations 2014
Page 6
10 (a) (i)
(ii)
(iii)
(b) (i)
(ii)
Mark Scheme
Cambridge IGCSE – October/November 2014
C5 × 12C5
B1, B1
B1
4 places are accounted for
Gender no longer ‘important’
M1
for realising that 4 places are
accounted or that gender is no
longer important
A1
for 8008
M1
for at least 5 of the 7 cases, allow
unsimplified
16
C6 = 8008
Alternative Method
6
C6 × 10C0 + 6C5 × 10C1
(
) (
) ... ( 6C0 × 10C6 )
1 + 60 + 675 + 2400 + 3150 + 1512 + 210 = 8008
cos 3 x
=0
sin 3 x
1 

cos 3x 2 −
=0
sin 3x 

2 cos 3 x −
1
sin 3 x = , 3 x = 30o , 150o
2
x = 10o , 50o
1
π

cos y +  = −
2
2

π 2π 4π
,
y+ =
2
3 3
so y =
for use of cot 3 x =
cos3 x
, may be
sin 3 x
implied
x = 30o , 90o
and
A1
M1
Leading to cos 3 x = 0, 3 x = 90o , 270o
(b)
B1 for each, allow unevaluated
with no extra terms
Final answer must be evaluated and
from multiplication
56 × 792 = 44352
Need
11 (a)
Paper
11
B1
B1
B1
360
60
36
8
Syllabus
0606
DM1
A1
DM1
for attempt to solve sin 3 x =
A1
correctly to obtain x
A1 for both, condone excess
solutions
M1
DM1
π 5π
(0.524, 2.62)
,
6 6
for attempt to solve cos 3x = 0
correctly from correct factorisation
to obtain x
A1 for both, no excess solutions in
the range
1
2
π

for dealing with sec  y + 
2

correctly
for correct order of operations,
must not mix degrees and radians
A1, A1
© Cambridge International Examinations 2014
Page 7
Mark Scheme
Cambridge IGCSE – October/November 2014
12 (i)
uuur
AQ = λ b − a
B1
(ii)
uuur
BP = µ a − b
B1
(iii)
(iv)
(v)
uuur
1
2
OR = a + ( λ b − a ) or λ b − ( λ b − a )
3
3
2
1
= a + λb
3
3
uuur
7
1
OR = b + ( µ a − b ) or µ a − ( µ a − b )
8
8
1
7
= b + µa
8
8
2
1
1
7
a + λb = b + µa
3
3
8
8
2 7
16
Allow 0.762
= µ, µ =
3 8
21
3
1
1
Allow 0.375
λ = ,λ =
8
3
8
Syllabus
0606
Paper
11
1
their (i)
3
M1
for a +
A1
Allow unsimplified
M1
for b +
A1
Allow unsimplified
M1
for equating (iii) and (iv) and then
equating like vectors
A1
A1
© Cambridge International Examinations 2014
7
their (ii)
8