w
w
ap
eP
m
e
tr
.X
w
CAMBRIDGE INTERNATIONAL EXAMINATIONS
om
.c
s
er
Cambridge International General Certificate of Secondary Education
MARK SCHEME for the May/June 2015 series
0606 ADDITIONAL MATHEMATICS
0606/12
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 IGCSE – May/June 2015
Syllabus
0606
Paper
12
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
k 2 − 4(2k + 5)
k 2 − 8k − 20
(< 0)
(< 0)
M1
(k − 10 )(k + 2 ) (< 0 )
critical values of 10 and –2
− 2 < k < 10
Alternative 1:
dy
= 2 ( 2k + 5 ) x + k
dx
When
dy
−k
8k + 20 − k 2
= 0, x =
,y=
dx
2 ( 2k + 5 )
4 ( 2k + 5 )
When y = 0 , obtain critical values of 10 and –2
− 2 < k < 10
Alternative 2:
2



k
k2
 +1
 −
y = (2k + 5)   x +

2(2k + 5) 
4(2k + 5) 


k2
= 0 leads to
4(2k + 5)
critical values of 10 and –2
− 2 < k < 10
Looking at 1 −
M1
A1
A1
use of b 2 − 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
M1
attempt to differentiate, equate to
zero and substitute x value back in
to obtain a y value
M1
consider y = 0 in order to obtain
critical values
correct critical values
correct range
A1
A1
M1
M1
A1
A1
© Cambridge International Examinations 2015
attempt to complete the square and
k2
'
consider '1 −
4(2k + 5)
attempt to solve above = to 0, to
obtain critical values
correct critical values
correct range
Page 3
2
Mark Scheme
Cambridge IGCSE – May/June 2015
sinθ cosθ
tan θ + cot θ cosθ + sinθ
=
1
cosecθ
sinθ
Syllabus
0606
Paper
12
sin θ
cos θ
, cot θ =
and
cosθ
sin θ
1
cosecθ =
; allow when used
sin θ
M1
for tan θ =
M1
dealing correctly with fractions in
the numerator; allow when seen
1
cosθ
M1
use of the appropriate identity;
allow when seen
= sec θ
A1
must be convinced it is from
completely correct work ( beware
missing brackets)
M1
for either tan θ =
sin 2 θ + cos 2 θ
= sin θ cosθ
1
sin θ
=
Alternative:
tan 2 θ + 1
tan θ + cot θ
= tan θ
cosecθ
cosecθ
1
or
cot θ
1
and
tan θ
1
cosecθ =
; allow when used
sin θ
dealing correctly with fractions in
numerator; allow when seen
cot θ =
sec 2 θ
1
tan θ
sin θ
2
sec θ
=
sec θ
=
= sec θ
3
A–1 =
1  3 −2 


2  −5 4 
 x  1  3 − 2  8 
  = 
 
 y  2  −5 4  9 
 x
 y =
 
M1
M1
A1
B1
B1
M1
1 6 
2  −4 
x = 3, y = −2
A1, A1
© Cambridge International Examinations 2015
use of the appropriate identity;
allow when seen
must be convinced it is from
completely correct work
1
multiplied by a matrix
2
for matrix
attempt to use the inverse matrix,
must be pre-multiplication
Page 4
4
(i)
Mark Scheme
Cambridge IGCSE – May/June 2015
Area =
1
 1

2
2
 × 12 × 1.7  +  × 12 sin (2π − 1.7 − 2.4 )
2
 2

Syllabus
0606
B1,B1
M1
(ii)
= awrt 181
A1
BC 2 = 122 + 122 − (2 × 12 × 12 cos 2.1832 )
M1
 2π − 4.1 
or BC = 2 × 12 × sin 

2


BC = 21.296
A1
Perimeter = (12 × 1.7 ) + 12 + 12 + 21.296
B1
M1
= 65.7
5
(a) (i)
Paper
12
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)
correct attempt at BC, may be seen
in (i), allow if used in (ii). Allow
use of their angle BOC.
for arc length, allow unsimplified
for a correct ‘plan’
(an arc + 2 radii and BC)
A1
20160
B1
(ii)
3 × 6 P4 × 2
= 2160
B1,B1
B1 for 6 P4 (must be seen in a
product)
B1 for all correct, with no further
working
(iii)
5 × 2 × 6P4
= 3600
B1,B1
B1
B1 for 6 P4 (must be seen in a
product)
B1 for 5 (must be in a product)
B1 for all correct, with no further
working
Alternative 1:
6
C4 × 5! × 2
= 3600
B2
for 6C4 × 5!
B1
for 6C 4 × 5! × 2
Alternative 2:
( P − P )× 2
B2
= 3600
B1
7
5
6
5
( P− P)
for ( P − P ) × 2
for
7
7
5
5
6
6
5
5
Alternative 3:
(
(
) (
) (
)
2! 6 P4 + 6 P1 ×5 P3 + 6 P2 × 4 P2 + 6 P3 ×3 P1 + 6P4
= 3600
)
B2
B1
© Cambridge International Examinations 2015
4 terms correct or omission of 2! in
each term
all correct
Page 5
(b) (i)
(ii)
6
Mark Scheme
Cambridge IGCSE – May/June 2015
14
8
Syllabus
0606
Paper
12
C4 ×10 C4 or 14C8 ×8 C 4
(or numerical or factorial equivalent)
= 210210
B1,B1
B1 for either 14C4 or 14C8 as part
of a product
B1 for correct answer, with no
further working
C 4 ×6 C 4
= 1050
B1,B1
B1 for either 8C 4 or 6C 4 as part of a
product
B1 for correct answer with no
further working
(i)
10ln4 or 13.9 or better
B1
(ii)
 dx

 dt
M1
 20t
= 2
−4
t +4
B1
When
dx
20t
= 0, 2
=4
dt
t +4
leading to t 2 − 5t + 4 = 0
t = 1, t = 4
DM1
A1
© Cambridge International Examinations 2015
attempt to differentiate and equate
to zero
20t
or equivalent seen
2
t +4
dx
= 0 , must
dt
be a 2 or 3 term quadratic equation
with real roots
attempt to solve their
for both
Page 6
(iii)
Mark Scheme
Cambridge IGCSE – May/June 2015
Syllabus
0606
Paper
12
20t
−4
t +4
If (v = )
2
20 ( t 2 + 4 ) − 20t ( 2t )
dx
dt
M1
attempt to differentiate their
A1
A1
20 t 2 + 4
20t (2t )
A1
20 4 − t 2 or 80 − 20t 2 or 4 − t 2
B1
t = 2 , dependent on obtaining first
and second A marks
M1
attempt to differentiate their
A1
A1
( )
for (20t − 4t
M1
attempt to differentiate their
dx
dt
M1
attempt to differentiate their
Numerator = −2t ( 20t − 4t − 16 ) + ( 20 − 8t ) ( t + 4 )
A1
for 2t 20t − 4t 2 − 15
dx
dt
A1
for (20 − 8t ) t + 4
(i)
DA = 3a − b
B1
mark final answer, allow
unsimplified
(ii)
DB = 7a − b
B1
mark final answer, allow
unsimplified
(iii)
AX = λ (4a + b )
B1
mark final answer, allow
unsimplified
(iv)
DX = 3a − b + λ (4a + b )
M1
their (i) + their (iii) or equivalent
valid method or 3a − b + their (iii)
Allow unsimplified
( a =)
(t
(
2
+ 4)
2
)
20 4 − t 2 or 80 − 20t 2 or 4 − t 2 or equivalent
(
)
(
)
expression involving − t 2
When acceleration is 0, t = 2 only
Alternative 1 for first 3 marks:
20t − 4t 2 − 16
If (v = )
t2 + 4
( a =)
(t
2
+ 4 ) ( 20 − 8t ) − ( 20t − 4t 2 − 16 ) ( 2t )
(t
+ 4)
2
2
dx
dt
for t 2 + 4 (20 − 8t )
2
)
− 16 (2t )
Alternative 2 for M1 mark:
(
If ( v = ) 20t t + 4
( a = ) 20t
2
)
−1
−4
( −2t (t + 4) ) + 20 (t + 4)
−2
2
2
−1
Alternative 3 for the first 3 marks
(
)(
If (v = ) 20t − 4t 2 − 16 t 2 + 4
(a = ) (20t − 4t
2
)
(
−1
− 16  − 2t t + 4

2
7
)
2
)
−2
(
 + (20 − 8t ) t + 4

2
2
)
−1
A1
© Cambridge International Examinations 2015
(
(
2
)
)
Page 7
(v)
Mark Scheme
Cambridge IGCSE – May/June 2015
3a − b + λ (4a + b ) = µ (7a − b )
Equate like vectors: 3 + 4λ = 7 µ
−1 + λ = − µ
leads to λ =
8
4
7
, µ=
11
11
Syllabus
0606
M1
DM1
A1,A1
(i)
1
5e 2 x − e −2 k
2
( +c )
B1, B1
(ii)
 2 k 1 −2 k 
 −2 k 1 2 k 
 5e − e  −  5e − e 
2
2




M1
A1
(iii)
 2 k 1 −2 k   −2 k 1 2 k 
 5e − e   5e − e  = −60
2
2

 

or
11 2 k 11 −2 k
e − e = −60
2
2
or equivalent
leading to 11e 2 k − 11e −2 k + 120 = 0
(iv)
11 y 2 + 120 y − 11 = 0
(11 y − 1)( y + 11) = 0
leading to
1
1 1
,
k = 1n , ln
2 11
11
1
− 1n 11 , − 1n11
2
B1
Paper
12
equating their (iv) and µ × their (ii)
for an attempt to equate like vectors
and attempt to solve 2 linear
equations for λ and µ
A1 for each
B1 for each term, allow
unsimplified
use of limits provided integration
has taken place. Signs must be
correct if brackets are not included.
allow any correct form
correct expression from (ii) either
simplified or unsimplified equated
to − 60 , must be first line seen.
DB1
must be convinced as AG
M1
attempt to obtain a quadratic
equation in y or e 2 k and solve to
get y or e 2 k ( only need positive
solution)
attempt to deal with e to get k =.
any of given answers only.
DM1
A1
© Cambridge International Examinations 2015
Page 8
Mark Scheme
Cambridge IGCSE – May/June 2015
dy
= 4 − 6sin 2 x
dx
π
When x = , y = π
4
dy
1
= −2 so gradient of normal =
2
dx
9
1
π
Normal equation y − π =  x − 
2
4
7π
8
7π
When y = 0, x = −
4
When x = 0, y =
Area =
10 (a)
(b)
1 7π 7π
49π 2
× ×
=
2 4 8
64
1
1
cos 3x = (± )
,
2
2
3 x = 45°, 135°, 225°, 315°
x = 15°, 45°, 75°, 105°
cos 2 3x =
(
)
3 cot 2 y + 1 + 5 cot y − 5 = 0
Leading to
3 cot 2 y + 5 cot y − 2 = 0 or
2 tan 2 y − 5 tan y − 3 = 0
(3 cot y − 1)(cot y + 2 ) = 0 or
(tan y − 3)(2 tan y + 1) = 0
tan y = 3,
tan y =
y = 71.6°, 251.6°
(c)
1
2
Syllabus
0606
M1,A1
B1
DM1
π 1

sin  z +  =
3 2

π π 5π 13π
z+ = , ,
3 6 6 6
π 11π
z= ,
2 6
(allow 1.57, 5.76 )
M1 for attempt to differentiate
A1 for all correct
for y
for substitution of x =
π
into their
4
dy
and use of ' m1m2 = −1' ,
dx
dependent on first M1
DM1
correct attempt to obtain the
equation of the normal, dependent
on previous DM mark
A1
must be terms of π
A1
must be terms of π
B1ft
Follow through on their x and y
intercepts; must be exact values
M1
complete correct method, dealing
with sec and 3, correctly
A1 for each correct pair
A1,A1
M1
M1
M1
153.4°, 333.4°
Paper
12
A1,A1
use of a correct identity to get an
equation in terms of one trig ratio
only
1
to obtain either a
tan y
quadratic equation in tan y or
solutions in terms of tan y; allow
where appropriate
for cot y =
for solution of a quadratic equation
in terms of either tan y or cot y
A1 for each correct ‘pair’
M1
completely correct method of
solution
A1
one correct solution in range
M1
correct attempt to obtain a second
solution within the range
second correct solution (and no
other)
A1
© Cambridge International Examinations 2015