Cambridge Assessment International Education

Cambridge International Advanced Subsidiary Level

MATHEMATICS 9709/21
Paper 2 October/November 2017
MARK SCHEME
Maximum Mark: 50

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 7 printed pages.

© UCLES 2017 [Turn over

9709/21 Cambridge International AS Level – Mark Scheme October/November
PUBLISHED 2017

Mark Scheme Notes

Marks are of the following three types:

M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for
numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a
candidate just to indicate an intention of using some method or just to quote a formula; the
formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant
quantities into the formula. Correct application of a formula without the formula being quoted
obviously earns the M mark and in some cases an M mark can be implied from a correct answer.

A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy
marks cannot be given unless the associated method mark is 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 generally
independent unless the scheme specifically says otherwise; and similarly when there are several
B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B
mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more
steps are run together by the candidate, the earlier marks are implied and full credit is given.

• The symbol FT implies that the A or B mark indicated is allowed for work correctly following on
from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and
B marks are not given for fortuitously “correct” answers or results obtained from incorrect
working.

• Note: B2 or A2 means that the candidate can earn 2 or 0.

B2/1/0 means that the candidate can earn anything from 0 to 2.

The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a
candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise
indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct
form of answer is ignored.

• Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme
specifically indicates otherwise.

• For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or
which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A
or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For
Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to
9.8 or 9.81 instead of 10.

© UCLES 2017 Page 2 of 7

9709/21 Cambridge International AS Level – Mark Scheme October/November
PUBLISHED 2017

The following abbreviations may be used in a mark scheme or used on the scripts:

AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent

AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed
working leading to the result is valid)

CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)

CWO Correct Working Only – often written by a ‘fortuitous’ answer

ISW Ignore Subsequent Working

SOI Seen or implied

SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where
some standard marking practice is to be varied in the light of a particular circumstance)

Penalties

MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or part

question are genuinely misread and the object and difficulty of the question remain unaltered.
In this case all A and B marks then become ‘follow through’ marks. MR is not applied when
the candidate misreads his own figures – this is regarded as an error in accuracy. An MR –2
penalty may be applied in particular cases if agreed at the coordination meeting.

PA –1 This is deducted from A or B marks in the case of premature approximation. The PA –1

penalty is usually discussed at the meeting.

© UCLES 2017 Page 3 of 7

9709/21 Cambridge International AS Level – Mark Scheme October/November
PUBLISHED 2017

Question Answer Marks Guidance

1 Use subtraction or addition property of *M1

logarithms

3x + 1 A1
Obtain = e or equivalent with no
x+2
presence of logarithm

Use correct process to solve equation DM1

2e − 1 A1
Obtain or exact equivalent
3−e

Question Answer Marks Guidance

2 Use cos 2θ = 2 cos 2 θ − 1 B1

Obtain 10cos3 θ = 4 or equivalent B1

Use correct process to find at least one value of M1

θ from equation of form k1 cos3 θ = k2

Obtain 42.5 A1

Obtain 317.5 and no others between 0 and 360 A1

© UCLES 2017 Page 4 of 7

9709/21 Cambridge International AS Level – Mark Scheme October/November
PUBLISHED 2017

Question Answer Marks Guidance

3 Take logarithms of both sides and apply power M1 Condone incorrect inequality
law signs until final answer. The
first 6 marks are for obtaining
the correct critical values.

ln 80 A1
Obtain 2 x < or equivalent using log10
ln1.3

Obtain x = 8.35... A1

State or imply non-modulus inequality B1

(3x − 1) 2 > (3 x − 10) 2 or corresponding equation
or linear equation 3 x − 1 = −(3 x − 10)

Attempt solution of inequality or equation M1

(obtaining 3 terms when squaring each bracket
or solving linear equation with signs of 3x
different)

Obtain x = 116 or x = 1.83... A1

Conclude 1.83 < x < 8.35 A1

Question Answer Marks Guidance

4(a) Obtain integrand of form a sec 2 θ + b M1

Obtain correct 5sec 2 θ − 1 A1

Integrate to obtain form a tan θ + bθ M1

Obtain 5 tan θ − θ + c A1

4(b) Obtain integral of form k ln(3 x + 1) *M1

Apply limits and obtain 2

ln(3a + 1) = ln16 A1
3

Obtain equation with no presence of ln DM1

Obtain 21 A1

© UCLES 2017 Page 5 of 7

9709/21 Cambridge International AS Level – Mark Scheme October/November
PUBLISHED 2017

Question Answer Marks Guidance

5(i) Substitute x = −2 and equate to zero *M1

Substitute x = 1
and equate to 40 *M1
2

Obtain −8a + 4b − 64 = 0 and 18 a + 14 b = 23

or A1
2
equivalents

Solve a pair of simultaneous equations for a or DM1 Needs at least one of the two
for b previous M marks

Obtain a = 12 and b = 40 A1

5(ii) Attempt division by ( x + 2) or inspection at M1

least as far as kx 2 + mx

Obtain 12 x 2 + 16 x + 5 A1

Conclude ( x + 2)(2 x + 1)(6 x + 5) A1

Question Answer Marks Guidance

6(i) Obtain dx
= 4e 2t + 4et B1
dt

Use product rule to find dy M1

dt

dy 5e 2t + 10te2t A1
Obtain = or equivalent
dx 4e2t + 4et

ae2t + bte2t M1
Equate first derivative of the form
ce2t + det
to zero and solve to find t

Obtain t = − 12 from completely correct work A1

Obtain (3.16, − 0.92) A1

© UCLES 2017 Page 6 of 7

9709/21 Cambridge International AS Level – Mark Scheme October/November
PUBLISHED 2017

Question Answer Marks Guidance

6(ii) Identify t = 0 B1

Substitute t = 0 in expression for first derivative M1

and find negative reciprocal

Obtain − 85 or equivalent A1

Question Answer Marks Guidance

7(i) Differentiate to obtain form k1 x + k2 + k3 sin 12 x *M1

Obtain correct 2 x + 3 − 52 sin 12 x and deduce or A1

imply gradient at P is 3

Equate first derivative to their −3 and rearrange DM1

Obtain x = 54 sin 12 x − 3 A1

7(ii) Consider sign of their 2 x + 6 − 52 sin 12 x at −4.5 M1

and −4.0 or equivalent

Complete argument correctly for correct A1

expression with appropriate calculations

7(iii) Use iteration formula correctly at least once M1

Obtain final answer −4.11 A1

Show sufficient iterations to justify accuracy to A1

3 sf or show sign change in interval
(−4.115, − 4.105)

© UCLES 2017 Page 7 of 7