Cambridge International AS Level

MATHEMATICS 9709/22
Paper 2 Pure Mathematics 2 May/June 2024
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 May/June 2024 series for most
Cambridge IGCSE, Cambridge International A and AS Level and Cambridge Pre-U components, and some
Cambridge O Level components.

This document consists of 12 printed pages.

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Generic Marking Principles

These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the
mark scheme or generic level descriptions for a question. Each question paper and mark scheme will also comply with these marking principles.

GENERIC MARKING PRINCIPLE 1:

Marks must be awarded in line with:
 the specific content of the mark scheme or the generic level descriptors for the question
 the specific skills defined in the mark scheme or in the generic level descriptors for the question
 the standard of response required by a candidate as exemplified by the standardisation scripts.

GENERIC MARKING PRINCIPLE 2:

Marks awarded are always whole marks (not half marks, or other fractions).

GENERIC MARKING PRINCIPLE 3:

Marks must be awarded positively:
 marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the
syllabus and mark scheme, referring to your Team Leader as appropriate
 marks are awarded when candidates clearly demonstrate what they know and can do
 marks are not deducted for errors
 marks are not deducted for omissions
 answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as
indicated by the mark scheme. The meaning, however, should be unambiguous.

GENERIC MARKING PRINCIPLE 4:

Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors.

GENERIC MARKING PRINCIPLE 5:

Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited
according to the quality of the candidate responses seen).

GENERIC MARKING PRINCIPLE 6:

Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in
mind.

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Mathematics Specific Marking Principles

1 Unless a particular method has been specified in the question, full marks may be awarded for any correct method. However, if a calculation is required
then no marks will be awarded for a scale drawing.

2 Unless specified in the question, non-integer answers may be given as fractions, decimals or in standard form. Ignore superfluous zeros, provided that the
degree of accuracy is not affected.

3 Allow alternative conventions for notation if used consistently throughout the paper, e.g. commas being used as decimal points.

4 Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).

5 Where a candidate has misread a number or sign in the question and used that value consistently throughout, provided that number does not alter the
difficulty or the method required, award all marks earned and deduct just 1 A or B mark for the misread.

6 Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.

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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 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.

DM or DB 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 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.

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 or B marks are given for correct work only (not for results obtained from incorrect working) unless follow through is allowed (see abbreviation FT above).
 For a numerical answer, allow the A or B mark if the answer is correct to 3 significant figures or would be correct to 3 significant figures if rounded (1
decimal place for angles in degrees).
 The total number of marks available for each question is shown at the bottom of the Marks column.
 Wrong or missing units in an answer should not result in loss of marks unless the guidance indicates otherwise.
 Square brackets [ ] around text or numbers show extra information not needed for the mark to be awarded.

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Abbreviations

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

ISW Ignore Subsequent Working

SOI Seen Or Implied

SC Special Case (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)

WWW Without Wrong Working

AWRT Answer Which Rounds To

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Question Answer Marks Guidance

1 Solve 5x  7  2 x  3 to obtain  103 B1 Or inequality.

Attempt solution of linear equation where 5x and 2x have M1 Or inequality.

different signs

Obtain  74 A1

State x   103 , x   74 A1 A0 if ‘… and …’ used.

Alternative Method for Question 1

State or imply non-modulus equation (5 x  7) 2  (2 x  3) 2 (B1) Or inequality.

Attempt solution of three-term quadratic equation (M1) Or inequality.

Obtain  103 and  74 (A1)

State x   103 , x   74 (A1) A0 if ‘… and …’ used.

Question Answer Marks Guidance

2  2 x  1 ln 6 M1

ln5  3x  2 *M1

Attempt solution of linear equation DM1

Obtain 9.256 A1 Or greater accuracy.

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Question Answer Marks Guidance

3(a) Differentiate to obtain form k1e  x  k2 e 2 x M1 Where k1k 2  0 , k1  8 and k 2  1 .

Obtain 8e  x  2e 2 x A1

Substitute x  0 to obtain –10 A1

3(b) Attempt to find x-coordinate of B M1 8e  x  e 2 x  0 .

Obtain e3 x  8 and hence x  ln 2 A1 AG so necessary detail needed.

A0 if decimals used.

Integrate to obtain 8e  x  12 e 2 x B1

Use limits 0 and ln 2 correctly to find area M1 For integral of form k e  x  k e 2 x where k k  0 .
3 4 3 4
k1  8 and k 2  1 .

Obtain 5
2
A1 OE

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Question Answer Marks Guidance

4 dx dy *M1
Obtain forms  k1 cos t sin t or  k2 cos 2t
dt dt

Obtain correct 8cos t sin t or 4sin 2t and 2 3 cos 2t A1

dy *DM1 Need to see attempt at substitution.

Attempt value of when t  16 π
dx

dy 1 A1
Obtain 
dx 2

State or imply gradient of normal is 2 **M1FT Following their value of the first derivative.

Attempt equation of normal **DM1  3

Not tangent and with attempt to find coordinates  3,  .
 2

Obtain 4 x  2 y  9  0 A1 Or equivalent of requested form.

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Question Answer Marks Guidance

5(a) Carry out division at least as far as 3x 2  k1 x M1 Or equivalent (inspection, …).

Obtain quotient 3 x 2  4 x  1 A1

Confirm remainder is 6 A1 Answer given – necessary detail needed.

SC B1 for use of the factor theorem to show remainder is 6 if no

other marks are awarded.

Alternative Method for Question 5(a)

Synthetic division (M1)

–2/3 9 18 5 4

–6 8 –2

9 12 –3 6

Obtain quotient 3 x 2  4 x  1 (A1)

Confirm remainder is 6 (A1)

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Question Answer Marks Guidance

5(b) 6 B1FT Following their quotient.

Identify integrand as 3x 2  4 x  1 
3x  2

Integrate to obtain at least x 3 and k2 ln(3 x  2) terms *M1

Obtain x 3  2 x 2  x  2ln(3 x  2) A1

Apply limits and appropriate logarithm properties DM1

Obtain 14  ln16 A1

Question Answer Marks Guidance

6(a) Attempt use of quotient rule M1 Or equivalent method.

( x  3) 2
2 x 1
 ln(2 x  1) A1 OE
Obtain 2
( x  3)

6(b) Equate first derivative to zero and arrange as far as 2 x  1  ... M1

x3 A1 Answer given – necessary detail needed.

Confirm x   0.5
ln(2 x  1)

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Question Answer Marks Guidance

6(c) x3 M1
Consider sign of x   0.5 or equivalent for 2.5 and 3.0
ln(2 x  1)

Obtain 0.07  0.0696... and 0.4  0.4166... and justify A1 Answer given – necessary detail needed.
conclusion

Alternative Method for Question 6(c)

x3 (M1)
Consider the values of f  x    0.5 and obtain
ln  2 x  1
f  2.5   2.57 (2.5696…) and f  3  2.58 (2.58339…)

Conclude f  2.5   3 and f  3  2.5 so root lies in given interval (A1) Answer given – necessary detail needed.

6(d) Use iterative process correctly at least once M1

Obtain final answer 2.569 A1 Answer required to exactly 4sf.

Show sufficient iterations to 6 sf to justify answer or show sign A1

change in interval [2.5685, 2.5695]

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Question Answer Marks Guidance

7(a) Express left-hand side in terms of sin  and cos using M1

1
cosec 2 
sin 2

1 A1 Answer given – necessary detail needed.

Obtain and confirm sec
cos

7(b) Attempt to obtain quadratic equation in sec or cos only *M1

Obtain sec 2   1  72 sec  8 involving one trigonometric ratio A1 Or equivalent, may be unsimplified, but reduce to
2sec 2   7 sec   18  0
18cos 2   7 cos   2  0 .

Attempt to solve 3-term quadratic equation for sec , using a DM1 Or equivalent using cos .
correct method, to find at least one value of 

Obtain any two of the four correct solutions 0.952,  1.76 A1 Or greater accuracy.

Obtain remaining two correct solutions A1 Or greater accuracy; and no others between  π and π .

7(c) Identify integrand as 2sec 2 12 x B1

Integrate k sec 2 12 x to obtain 2k tan 12 x M1

Obtain correct 4 tan 12 x A1 Condone omission of ...  c .

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