Cambridge Assessment International Education

Cambridge International General Certificate of Secondary Education

ADDITIONAL MATHEMATICS 0606/23

Paper 2 October/November 2017
MARK SCHEME
Maximum Mark: 80

Published

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 International will not enter into discussions about these mark schemes.

Cambridge International is publishing the mark schemes for the October/November 2017 series for most
Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level
components.

® IGCSE is a registered trademark.

This document consists of 6 printed pages.

© UCLES 2017 [Turn over

0606/23 Cambridge IGCSE – Mark Scheme October/November
PUBLISHED 2017

MARK SCHEME NOTES

The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes
may include marks awarded for specific reasons outside the scope of these notes.

Types of mark

M Method marks, awarded for a valid method applied to the problem.

A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. For accuracy marks
to be given, the associated Method mark must be earned or implied.

B Mark for a correct result or statement independent of Method marks.

When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the
scheme specifically says otherwise; and similarly where there are several B marks allocated. The notation ‘dep’
is used to indicate that a particular M or B mark is dependent on an earlier mark in the scheme.

Abbreviations

awrt answers which round to

cao correct answer only
dep dependent
FT follow through after error
isw ignore subsequent working
nfww not from wrong working
oe or equivalent
rot rounded or truncated
SC Special Case
soi seen or implied

© UCLES 2017 Page 2 of 6

0606/23 Cambridge IGCSE – Mark Scheme October/November
PUBLISHED 2017

Question Answer Marks Guidance

1(a) B2 B1 for each

1(b) n ( P ') = 18 B1

n ((Q ∪ R ) ∩ P ) = 11 B1

n ( Q′ ∪ P ) = 29 B1

2 3x − 1 = 5 + x x=3 B1

3x − 1 = −5 − x oe M1 M1 not earned if incorrect

equation(s) present

x = –1 A1

3 p ( ) ( 3 − 1) = q + 3
3 +1 +
3
M1 on LHS take common denominator
or rationalise each term or multiply
( 3 − 1)( 3 + 1) throughout

p ( ) (
3 +1 + )
3 − 1 = 2q + 6 3 oe A1 correct eqn with no surds in
denominators of LHS

equate surd/non surd parts M1 equate and solve for p or q (≠0)

p = 5 and q = 2 A1

4 log3 3 = 1 or log3 9 = 2 B1 implied by one correct equation

x + 1 = 3y B1

x− y =9 B1

solve correct equations for x or y M1

x = 14 and y = 5 A1
JJJG
5(i) OX = λ (1.5b + 3a ) B1

JJJG JJJG
5(ii) AB = b − a or BA = a –b B1
JJJG
OX = a + µ ( b − a ) B1

JJJG JJJG
5(iii) 1.5λ = µ or 3λ = 1 − µ M1 OX = OX and equate for a or b

µ=
1
λ=
2 A2 A1 for each
3 9

© UCLES 2017 Page 3 of 6

0606/23 Cambridge IGCSE – Mark Scheme October/November
PUBLISHED 2017

Question Answer Marks Guidance

5(iv) AX 1 B1 1
= Accept 1 : 2 but not :1
XB 2 2

5(v) OX 2 B1 2
= Accept 2 : 7 but not :1
XD 7 7

6(i) f 2 = f(f) used M1 numerical or algebraic

algebraic ([(x + 2)2 + 1] + 2)2 + 1

17 A1

6(ii) y−2 M1 change x and y

x=
2 y −1

2 xy − x = y − 2 → y ( 2 x − 1) = x − 2 M1 M1dep
multiply, collect y terms, factorise

x−2 A1 correct completion

y=
2x −1  g ( x ) 
=

6(iii) ( x + 2 )2 + 1 − 2 B1
gf ( x ) =   oe
2 ( x + 2 ) + 1 − 1
 2
 

( x + 2 )2 − 1 = 8 M1
their gf =
8
and simplify to
19
2 ( x + 2 ) + 1 19
2

quadratic equation
3 ( x + 2 ) = 27 oe
2
3x2 + 12x – 15 = 0

solve quadratic M1 M1dep

Must be of equivalent form

x =1 x = −5 A1

7(i) 1 M1 set v = 0 and solve for cos2t

v = 0 → cos2t =
3

→ t = 0.615 or 0.616 A1

7(ii) 3 M1A1 M1 for sin2t and ± t

s = sin2t − t ( +c )
2

π π A1
t= → s = 1.5 − ( = 0.715 )
4 4

7(iii) a = −6sin2t M1A1 M1 for −sin2t

t = 0.615 → a = –5.66 or –5.65 or −2 8 A1 condone substitution of degrees

© UCLES 2017 Page 4 of 6

0606/23 Cambridge IGCSE – Mark Scheme October/November
PUBLISHED 2017

Question Answer Marks Guidance

8(i) 1 M1
cosα = oe
3

α = 70.5° A1

8(ii) speed = 32 − 12 M1 Pythagoras/trig ratio/cosine rule

8 or 2 2 or 2.83 m s–1 A1

8(iii) 50 M1
time =
their 8

25 2 A1
or 17.7s
2

8(iv) their 8 ( iii ) seen B1

BC = 10 2 or 14.1 m or 14.2 m B1

9(i) d 1 B1 seen
(ln x) = and
dx x
d 3 d −3
x = 3x 2 or x = −3x −4
dx dx

Substitution of their derivatives into quotient rule M1

1 A1 correct completion
x3 × − 3 x 2 lnx
d  lnx  x
 = oe
dx  x 3  x6

9(ii) dy 1 M1 dy
= 0 → 1 − 3lnx = 0 lnx = equate given to zero and solve
dx 3 dx
for lnx or x

1 A1 seen
x = e3

1 A1 seen
y=
3e

9(iii) lnx 1 − 3lnx M1 use given statement in (i)

x 3
= ∫
x4
dx oe

1 −1 B1 seen anywhere
∫x 4
dx =
3 x3

lnx 1 lnx A2 A1 for each term

∫x 4
dx = − 3
− 3
9 x 3x
(+C) oe

© UCLES 2017 Page 5 of 6

0606/23 Cambridge IGCSE – Mark Scheme October/November
PUBLISHED 2017

Question Answer Marks Guidance

10(a) sin 2 x + (1 + cosx ) B1 correct addition of fractions

2

LHS =
sinx (1 + cosx )

1 + 2cosx + 1 B1 expansion and use of identity

=
sinx (1 + cosx )

2 (1 + cosx ) B1 factorisation and completion

= = 2cosecx
sinx (1 + cosx )

10(b)(i) cosec 2 y − 1 + cosecy − 5 = 0 M1 use of identity for cot2y to obtain

cosec2y + cosecy – 6 = 0 quadratic in cosecy

( cosecy − 2 )( cosecy + 3) = 0 M1 solve 3 term quadratic for cosecy

1 1 M1 obtain values for siny

siny = , siny = −
2 3

y = 30°, 150°, 199.5°, 340.5° A2 A1 for 2 values

10(b)(ii) π 5π 7π M2 5π
2z + = or (2.6…, 3.6…) M1 equate to
4 6 6 6
7π
M1 equate to
6

7π 11π A2 A1 for 1 value

z= or (0.916, 1.44)
24 24

11(i) Other root = 4 B1

f ( x) = ( x − 3)( x − 3)( x − 4 ) M1 multiply out (x – 3)(x – 3)(x ± p)

= x3 − 10 x 2 + 33x − 36

a = –10 b = 33 A2 A1 for each

Can be implied by correct cubic

11(ii) x = 6, x = 6, x = 1 B4 B1 for each of first two sets

x = 2, x = 2, x = 9 B2 for third set

x = 1, x = 1, x = 36

© UCLES 2017 Page 6 of 6