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CAMBRIDGE INTERNATIONAL EXAMINATIONS
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er
Cambridge Ordinary Level
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MARK SCHEME for the October/November 2014 series
4037 ADDITIONAL MATHEMATICS
4037/22
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
Mark Scheme
Cambridge O Level – October/November 2014
Syllabus
4037
Paper
22
(a)
B1
B1
(b)
No.in H only = 50 − x ; No in F only = 60 − x
Sum: 50 − x + 60 − x + x + 30 − 2 x = 98
A1
x = 14
2
3
B1
M1
9 x 2 + 2 x − 1 < ( x + 1)
2
M1
8 x 2 < 2 oe isw
1
1
− <x<
2
2
A1
log 2 ( x + 3) = log 2 y + 2 → x + 3 = 4 y
B1
Expand and collect terms
A1
log 2 ( x + y ) = 3 → x + y = 8
x + 3 = 4(8 − x )
B1
M1
5 x = 29 → x = 5.8, oe
y = 2.2 oe
Both written or on diagram
Add at least 3 terms each with x
involved and equate to 98 soi
Eliminate y or x from two linear three
term equations
A1
A1
© Cambridge International Examinations 2014
Page 3
4
(i)
Mark Scheme
Cambridge O Level – October/November 2014
f ( 37 ) = 3 or gf ( x ) =
gf ( 37 ) =
(ii)
y=
( x + 3)
(iii)
(
)
x −1 − 3 − 3
3−2 1
=
6−3 3
+1= f
−1
( y + 3)
( x)
2
= x −1
oe isw
B1
x−2
2x − 3
2 xy − 3 y = x − 2 → 2 xy − x = 3 y − 2
M1
Rearrange and square in any order
A1
Interchange x and y and complete
M1
Multiply and collect like terms
A1
Interchange and complete
Mark final answer
y=
3x − 2
= g −1 ( x ) oe
2x −1
5
(i)
B = 900
B1
(ii)
B = 500 + 400e2 = 3455 or 3456 or 3460
B1
(iii)
 dB 
=  80e 0.2t

 dt 
dB
t = 10 →
= 80e 2 = 591 ( /day )
dt
(iv)
10000 = 500 + 400e0.2t
0.2t = ln 23.75
t = 15.8 ( days )
Paper
22
B1
x −1 − 3 →
2
2
x −1 − 3 − 2
Syllabus
4037
→ e0.2t = ( 23.75)
3455.6 scores B0
B1
B1
awrt
M1
DM1
e0.2t = k
take logs: 0.2t = ln k
A1
awrt
© Cambridge International Examinations 2014
Page 4
6
(i)
Mark Scheme
Cambridge O Level – October/November 2014
(x + 2)2 + x 2 = 10
M1
A1
A1
or elimination of x leads to y2 – 2y – 3 = 0,
then as above
m 2 x 2 + 10mx + 25 + x 2 = 10
(m
2
)
+ 1 x 2 + 10mx + 15 = 0
b − 4ac = ( 0 ) → 100m 2 − 60 ( m 2 + 1) = 0
B1
attempt to use discriminant on three
term quadratic. Allow unsimplified
A1
cao ± is required
B1
allow unsimplified
Attempt to solve with x 2 + y 2 = 10
M1
Eliminate x or y
y = 2, x = ± 6
3
oe
m=±
6
A1
both
(i)
v = 2cost + 1
B1
mark final answer
(ii)
2cost + 1 = 0
M1
A1
equate their v to zero (must be a
differential) and attempt to solve to find
an angle
awrt
B1
awrt
B1ft
ft their v (2nd differential)
DB1ft
ft using their angle t in correct a awrt
2
2
dy ( 2 + x ) × 2 x − x × 2 x
4x
=
=
2
2
dx
( 2 + x2 )
( 2 + x2 )
M1
A1
apply quotient or product rule
unsimplified
k =4
A1
k=4 does not need to be specifically
identified
1
× original function
their k
m=±
3
oe isw
2
Alternative solution:
dy
x
dy
−x
=−
or
=
2
dx
y
dx
10 − x
Result:
y 2 = x 2 + 5 y after inserted in y = mx + 5
t=
(iii)
8
3 term quadratic with attempt to solve
both x or a pair
both y or second pair
M1
A1
2
7
Paper
22
B1
x 2 + 2 x − 3 = 0 → ( x + 3)( x − 1) = 0
Points (1, 3), (–3, –1) isw
(ii)
Syllabus
4037
(i)
(ii)
A1
2π
or 2.09
3
2π
 2π  2π
→ x = 2sin 
= 3.83m
+
3
 3  3
a = −2 sin t
2π
1.73 −2
t=
a=− 3=−
ms
3
4
t=
x2
1
d
x
=
×
+ ( c ) isw
∫ 2 + x2 2
4 2 + x2
(
)
x
B1
B1
© Cambridge International Examinations 2014
Page 5
(a + 3 5 )
9
10
Mark Scheme
Cambridge O Level – October/November 2014
2
(i)
= a 2 + 3 5a + 3 5a + 45 oe
(i)
anywhere
B1
B1
( a + 3)( a − 2 ) = 0
M1
a = –3, 2
b = –18, 12
A1
A1
Attempt to solve three term quadratic
with integer coefficients obtained by
equating coeffs
Both as correct or one correct pair
Both bs correct
B1
anywhere
B1
anywhere
B1ft
correct addition of their terms
B1
use of identity and cancel
3cot x − cot x = tan x → 2cot x = tan x
M1
tan 2 x = 2 oe
x = 54.7, 125.3, 234.7, 305.3
A1
A1
A1
equate and collect like terms, allow sign
errors
secxcosecx =
1
cos xsin x
cos x
sin x
1 − cos 2 x
LHS =
oe
cos x sin x
sin 2 x
=
= tan x
cos x sin x
11
AG
1 2
× x × 0.8 ( = 0.4 x 2 cm 2 )
2
SR = 5sin 0.8 ( = 3.59 ) or
2 values
only 2 more values. awrt
B1
anywhere
B1
SR may be seen in stated
Area of triangle =
1
5cos 0.8 × 5sin 0.8 = 6.247 cm 2
2
0.08 x 2 = 6.247
M1
A1
insert correct terms into correct area
formulae
x = 8.837 cm
A1
Area of sector =
OR = 5cos0.8 ( = 3.48)
(ii)
AG
1
absin C
2
SQ = 8.84 − 5 ( = 3.84cm )
PR = 8.84 − 5cos0.8 ( = 5.35 or 5.36cm )
PQ = 8.84 × 0.8 ( = 7.07cm )
Perimeter = 19.84 to 19.86 cm or rounded to
19.8 or 19.9
(iii)
Paper
22
Equate: a 2 + a + 45 = 51
and 6a − b = 0
cot x =
(ii)
B1
Syllabus
4037
Area PQSR = 4 × 6.247
= 25cm
2
B1
two lengths from SQ, PR, PQ awrt
B1
B1
third length awrt
sum
M1
A1
24.95 to 25
© Cambridge International Examinations 2014
Page 6
12
(i)
Mark Scheme
Cambridge O Level – October/November 2014
f ( 2 ) = 3 ( 23 ) − 14 ( 22 ) + 32 = 0
Syllabus
4037
B1
Or complete long division
f ( x ) = ( x − 2 ) ( 3 x 2 − 8 x − 16 )
f ( x ) = ( x − 2 )( x − 4 )( 3x + 4 )
M1
A1
M1
A1
(iii)
x = 2, 4
B1
(iv)
∫ 3x − 14 + x
(ii)
32
2
dx = 1.5 x 2 − 14 x −
4
32 

Area = 1.5 x 2 − 14 x − 
x 2

= (–) 2
32
(+ c)
x
3x2 and 16
8x and correct signs
Factorise three term quadratic
B1
B1
first 2 terms
third term correct unsimplified
M1
Limits of 2 and 4 and subtract
A1
© Cambridge International Examinations 2014
Paper
22