Cambridge International AS Level

MATHEMATICS 9709/23
Paper 2 Pure Mathematics 2 October/November 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 October/November 2024 series for most
Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level
components.

This document consists of 12 printed pages.

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9709/23 Cambridge International AS Level – Mark Scheme October/November 2024
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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(a) 3 B1 AG – necessary detail needed.

State or imply 2 y ln a = 3x + k and conclude that gradient is
2ln a

1(b) 3 M1
Equate to gradient of line
2ln a

3 2.85 29 A1 Allow greater accuracy.

Obtain = or equivalent and hence obtain a = 4.6 or a = e 19
2ln a 2.9

Substitute appropriate values to find value of k M1

Obtain k = 1.7 A1

Alternative Method for Question 1(b)

Obtain 0.95 ( 2ln a ) = 3 ( 0.4 ) + k M1 OE

or a1.9 = e1.2+ k

Obtain 3.80 ( 2ln a ) = 3 ( 3.3) + k M1 OE

or a 7.6 = e9.9+ k
29
Obtain a = 4.6 or a = e 19 A1 Allow greater accuracy.

Obtain k = 1.7 A1

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Question Answer Marks Guidance

2 Attempt solution of equation or inequality, where signs of x and 4x are different M1

Obtain 4
5 … A1 OE

… and finally no other value A1

Conclude x  54 A1  4
Allow  −,  .
 5

Alternative Method for Question 2

State or imply non-modulus equation ( x − 7) 2 = (4 x + 3) 2 or inequality B1

Attempt solution of three-term quadratic equation or inequality M1

Obtain finally 4
only A1
5

Conclude x  54 A1  4
Allow  −, 
 5

Question Answer Marks Guidance

3(a) Differentiate to obtain form k tan 12 x sec2 12 x M1 OE. May use identities before differentiation.

Obtain correct tan 12 x sec2 12 x A1 OE. Allow unsimplified.

Substitute 23 π to obtain 4 3 A1

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Question Answer Marks Guidance

3(b) Express integrand as sec2 12 x − 1 + sin x B1

Integrate to obtain k1 tan 12 x − x + k2 cos x M1 Where k1k2  0.

Obtain correct 2 tan 12 x − x − cos x A1

Apply limits correctly to obtain 3 − 12 π or exact equivalent A1

Question Answer Marks Guidance

4(a) Substitute x = −2, equate to zero and attempt solution M1

Obtain a = 4 A1

4(b) Divide by x + 2 at least as far as k1 x 2 + k2 x M1

Obtain 4 x 2 − 12 x + 9 A1

Obtain ( x + 2)(2 x − 3) 2 or equivalent with integer coefficients only A1

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Question Answer Marks Guidance

4(c) Equate sin 2  to appropriate value from factorised form and attempt solution M1
Using their
2
.
3

Obtain 54.7 A1 Or greater accuracy.

Obtain –54.7 A1 Or greater accuracy.

No others in −90    90.

Question Answer Marks Guidance

5(a) Obtain integral of form k ln(2 x + 1) *M1

Obtain correct 5ln(2 x + 1) A1

Apply limits correctly and equate to 7 DM1

Apply appropriate logarithm property to reach at least a3 = ... DM1

Confirm a = 3 0.5e1.4 (2a + 1) − 0.5 A1 AG – necessary detail needed.

5(b) Use iterative process correctly at least once M1

Obtain final answer 2.18 A1 Answer required to exactly 3 sf.

Show sufficient iterations to 5 sf to justify answer or show a sign change in the A1

interval [2.175, 2.185]

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Question Answer Marks Guidance

6(a) Differentiate x using quotient rule or correct equivalent *M1

(e2t + 1) 2e2t − (e2t − 2) 2e2t A1

Obtain or equivalent
(e2t + 1)2

dy DM1
Attempt expression for in terms of t
dx

Obtain 1 t
2
e (e2t + 1)2 or (unsimplified) equivalent A1 No fractions within fractions.
Attempt to simplify (e 2t + 1) 2e 2t − (e 2t − 2) 2e 2t
must be seen.

6(b) Identify t = 12 ln 2 at point where curve crosses y-axis B1

dy M1
Substitute non-zero value of t in their expression for and attempt simplification
dx

Obtain 9
2 or exact equivalent A1
2

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Question Answer Marks Guidance

7(a) State (cos cos30 − sin  sin30)(cos cos60 − sin  sin 60) B1

Expand and use correct exact values M1

Obtain 1
4
3(cos2  + sin 2  ) − sin cos or similarly simplified equivalent A1

Conclude 1
3 − 12 sin 2 A1 AG – necessary detail needed.
4

7(b) Use identity to obtain value for sin 4 *M1

Obtain sin 4 = 12 3 − 52 or 0.466… A1

Show correct process to obtain one value of  DM1

Obtain 6.9 and 38.1 A1 Or greater accuracy; and no others between

0 and 90.

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Question Answer Marks Guidance

7(c) Substitute  = −10 to obtain cos20cos50 = 14 3 − 12 sin(−20) B1

3 1 B1
Substitute  = 10 to obtain cos 40cos 70 = − sin 20
4 2

Add and confirm 1

3 with clear indication that sin(−20) = − sin 20 B1 AG – necessary detail needed.
2

Alternative solution for Question 7(c)

Rewrite as sin70cos50 + sin50cos70 or cos20sin 40 + cos40sin 20 or B1

sin70sin 40 + cos70cos40

Obtain sin120 or sin 60 or cos30 B1

Confirm 1
3 B1 AG – necessary detail needed.
2

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