CAMBRIDGE INTERNATIONAL EXAMINATIONS
Cambridge International General Certificate of Secondary Education
MARK SCHEME for the March 2016 series
0606 ADDITIONAL MATHEMATICS MATHEMATICS 0606/22
Paper 22, 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 March 2016 series for most Cambridge IGCSE and Cambridge International A and AS Level components.
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Page 3
Mark Scheme Cambridge IGCSE – March 2016
Question 4
(a)
Answer
Syllabus 0606 Marks
a = 10 b = 6 c = 4 or 10 cos 6 x + 4
Paper 22
Guidance
B2,1,0
for B1 allow correct FT of c from a e.g. their c = 14 – their a
B3,2,1,0
Correct shape; two cycles; both maximum at 1 and minimum at –5; starting at (0, –2) and ending at (180, –2)
(b) y
1
45°
O
90°
135°
180°
x
-2
-5
5
(i)
(ii)
6
2187
+ 5 103 kx + 5103 k
2
x
2
B3
1 for each term; ignore extra terms
2(5103k ) 5103k 2
M1
must not include x, x2
k = 2
A1
A0 if k = 0 also given as a solution
=
x
1+ 3 3
( x =)
(
x =
)
=
5−
3
oe soi
6+2 3
4 + 14 3
−
6+2 3 4 + 14 −
oe
3
6+2 3
×
M1
6−2 3 6−2 3
p = −27, q = 23 isw
M1
M1
A1 + A1
© Cambridge International Examinations 2016
allow ( x =)
27 + 23 3
−
6
Page 4
Mark Scheme Cambridge IGCSE – March 2016
Question 7
Answer
4 −2
(a)
6 0
−14 −23 (b) (i)
8
18 3 − 4 21 −6
3
2
6
1
Syllabus 0606 Marks
6
M1
3
Guidance
for attempt to multiply and subtract
A1
0 1 1 − oe 2 −4 −2
B1 + B1
1
1 mark for − 2
k
Valid method
−8 −6 13 7
and 1 mark
0
1
−4 −2
for (ii)
Paper 22
–1
M1
XD D = CD
A2,1,0
−1 each error If M0 then SC1 for
4
DC =
3
−14 −5
8
(i)
Eliminate x (or y)
M1
(
3 2 y − 2
)
2
+
(2 y − 2) y − y
2
= 12
2
3 x
13 y
(ii)
2
−
26 y
=
0
or
13 x
2
4
13 y( y – 2)
or x2 = 4
x = −2,
x = 2
y = 0
y = 2
their m AB
1 =
2
−
13
=
0
oe
x + 2 x + 2 + x − = 12 2 2
A1 M1
isw
or their m BC
2
2
= −
soi
use of (m AB ) × ( mBC ) = −1 and conclusion
A1 + A1FT
or for (−2, 0) or (2, 2) from correct working FT their x or y values to find their y or x values; or A1 for (−2, 0) and (2, 2)
M1
may be unsimplified or Pythagoras’ theorem correctly applied to their (0, −2), their (2, 2) and (0, 6)
A1
or use of h2 = a2 + b2 and conclusion
© Cambridge International Examinations 2016
Page 5
Mark Scheme Cambridge IGCSE – March 2016
Question 9
Answer =
tan θ
1
RS
x
=
1
=
or x
sin θ 1
−
=
1
2 tan θ
−
2 sin θ
cot θ
1
−
oe
cosecθ −
2
A = x +
(ii)
1
2
cot θ
2
cosecθ
or
RS
oe soi
=
cot θ
cosecθ
FT their RT and their RS , provided both are functions of trig ratios
M1 A
=
1
oe
cosecθ −
2
A1
A1
implies M1
(α + β )i – 20 j = 15i + (2α − 24) j
M1
implied by α + β = 15 or 2α − 24 = −20
α = 2
A1
β = 13
A1
π
=
(theirα
their β )
+
25
OC
3
cao
=
15i – 20 j
(b)
B1
=
equivalent must be exact
3
(ii)
RT
M1
θ
10 (a) (i)
2 3
or
oe
Paper 22
Guidance
B1
B1FT
correct completion to given answer
(iii)
Marks
1
RT
(i)
Syllabus 0606
2
+
( −20 )2 oe
M1
oe
A1FT
= OA +
λ AB
or
OC
FT their α + β provided nonzero
= OB +
(1 − λ ) BA
B1
[OC ] a + λ( b – a) or =
[OC ] b + (1 − λ)( a – b)
M1
=
[OC ] (1 − λ)a + λ b
A1
=
(c)
2 µ + 3
=
Solves
µ =
3
µ
9
µ
2
+ 3µ − 18 = 0
M1
or multiplies one of the vectors by a general scale factor and finds a pair of simultaneous equations to solve
M1
or solves their correct equation to find their scale factor and attempts to use it to find µ
A1
A0 if −6 not discarded
© Cambridge International Examinations 2016
Page 6
Mark Scheme Cambridge IGCSE – March 2016
Question 11
(i)
Answer d y d x
=
(
2
+
) ( x
1 (1) − ( x ) (2 x ) 2
+
)
1
oe
2
Syllabus 0606 Marks
Guidance
M1*
Attempts to differentiate using the quotient rule
A1 their (1
−
x
2
)
=
0
Paper 22
correct; allow unsimplified
M1 dep*
x = 1, x = −1
A1
from correct working only
y = 0.5 , y = −0.5 oe
A1
from correct working only or A1 for each of (1, 0.5), (−1, −0.5) oe from correct working; unsupported answers do not score
(ii)
d d x d
((
x
2
y
d x
2
=
2
)
+1
( x
2
2
) )
+1
=
2
( x
(
2
x
)(2
2
+1
x
) ( their
+1
) soi
−
2x
B1
) − ( their (1 − x
( x
2
)
2
+1
4
2
d y
Correct completion to given answer
When x = 1 their
d 2 y d x 2
=
(1
2
x =1
d x
2(1) 3 − 6(1) +
)
3
) ) 2 ( 2 x)
=
2
2x
( x
3
2
−
6x
)
3
M1
d 2 y d x 2
= x =−1
((
1)
−
+
)
3
4
+
2x
2
)
+1 =
4x
3
+
4x
Applies quotient rule and factors out
B1FT
Complete method including comparison to 0; FT their first or second derivative
B1FT
Complete method including comparison to 0; FT their first or second derivative
1
2
x
+1
oe < 0 therefore
2( −1) 3 − 6( −1)
d x
(
A1
maximum When x = −1 their
d
oe > 0
1
therefore minimum
© Cambridge International Examinations 2016
Page 7
Mark Scheme Cambridge IGCSE – March 2016
Question 12
Answer 9t
(i)
2
63t + 90 = 0
−
Syllabus 0606 Marks
Paper 22
Guidance
M1
( 9t − 18)( t − 5) showing that t = 2 is smaller value of t dv
(a )
(ii)
=
(iii)
attempted
A1
∫ ( 9
M1
2
( s =)
( s =)
−
63t
9t 3 3
−
9( 2)3 3
+
90 ) d t
63t 2 2 −
+
90t isw
63( 2)2 2
must see evidence of solving e.g. t = 5 and t = 2 or factors
M1
18(3.5) – 63 = 0 cao t
(iv) (a)
dt
A1
A2,1,0
−1 for each error or for +c left in 2
+
90(2)
M1
9t 3 63t 2 − + 90t or 2 3 0 FT their (iii)
78 [m] (b)
( s =)
A1
9(3)3 3
−
63(3)2 2
+
90(3) = 67.5
their 78 + 10.5 = 88.5 [m]
M1 A1FT
© Cambridge International Examinations 2016
FT their (iii)
