Cambridge International A Level

MATHEMATICS 9709/32
Paper 3 Pure Mathematics 3 February/March 2025
MARK SCHEME
Maximum Mark: 75

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 f or
Teachers.

Cambridge International will not enter into discussions about these mark schemes.

Cambridge International is publishing the mark schemes f or the February/March 2025 series f or most
Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level
components.

This document consists of 23 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 point s.

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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Annotations guidance for centres

Examiners use a system of annotations as a shorthand for communicating their marking decisions to one another. Examiners are trained during the standardisation
process on how and when to use annotations. The purpose of annotations is to inform the standardisation and monitoring processes and guide the supervising
examiners when they are checking the work of examiners within their team. The meaning of annotations and how they are used is specific to each component and
is understood by all examiners who mark the component.

We publish annotations in our mark schemes to help centres understand the annotations they may see on copies of scripts. Note that there may not be a direct
correlation between the number of annotations on a script and the mark awarded. Similarly, the use of an annotation may not be an indication of the quality of the
response.

The annotations listed below were available to examiners marking this component in this series.

Annotations

Annotation Meaning

More information required

Accuracy mark awarded zero

Accuracy mark awarded one

Independent accuracy mark awarded zero

Independent accuracy mark awarded one

Independent accuracy mark awarded two

Benefit of the doubt

Blank Page

Incorrect point

Dep Used to indicate DM0 or DM1

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Annotation Meaning

DM1 Dependent on the previous M1 mark(s)

Follow through

Indicate working that is right or wrong

Highlighter Highlight a key point in the working

Ignore subsequent work

Judgement

Judgement

Method mark awarded zero

Method mark awarded one

Method mark awarded two

Misread

Omission or Other solution

Off-page Allows comments to be entered at the bottom of the RM marking window and then displayed when the associated question item is
comment navigated to.

On-page Allows comments to be entered in speech bubbles on the candidate response.

comment

Judgment made by the PE

Premature approximation

Special case

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Annotation Meaning

Indicates that work/page has been seen

Error in number of significant figures

Correct point

Transcription error

Correct answer from incorrect working

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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 specificall y 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 sign ificant 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 State that 1 – e–2 x = e–3 B1 OE, with ln removed.

Use correct method to solve an equation of the form e±2 x = a, where a > 0, and a M1 E.g. [e–2 x = 1 − e–3 ]
reasonable attempt at the B1, for ± 2x ln e or ± x ln e −2x = ln(1 − e–3 ) … OE.
Can be numerical ln (1 − 0.049787) = ln 0.9502.
Evidence of method must be seen.

Obtain answer 0.0255 A1 CAO

Must be 4 decimal places.
No working seen scores 0.

Alternative Method for Question 1

State that 1 – e–2 x = e–3 B1 OE, without ln.

Rearrange to obtain an expression for ex and solve an equation of the form e±x = a, M1  1 e3 
where a > 0, and a reasonable attempt at the B1, for x E.g. e x = , e x
= ,
 1 − e −3 e3 − 1 
1
x = ln
1 − e −3
Can be numerical.
Evidence of method must be seen.

Obtain answer 0.0255 A1 CAO

Must be 4 decimal places.
No working seen scores 0.

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Question Answer Marks Guidance

2 dy B1
1+ 2
State or imply dx as the derivative of ln (x + 2y)
x + 2y

dy B1 dy
State derivative of xy2 is x2y + y2 x2 y + y2
dx dx

dy B1 OE
1+ 2 May be implied by correct final answer.
dy dx = 0
Obtain y 2 + 2 xy +
dx x + 2 y

1 B1 OE
Obtain y = e when x = 0 Allow 12  2.718 or 1.36 or better e.g. 1.359…
2
May be implied by correct final answer.

( )
dy 1 B1 OE
Obtain = − e3 + 4 Accept AWRT –3.01. ISW.
dx 8

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Question Answer Marks Guidance

3(a) Obtain Im (z) ⩽ –1 B1 Condone strict inequalities throughout (a).

Obtain answer of the form z − a b M1 Accept equation or any inequality sign.

a = ± 2 ± i and b = 3,
e.g. z + 2 − i = 3 or z + 2 − i  3.

Obtain answer z + 2 − i  3 A1 Accept z − ( −2 + i ) 3 as final answer.

Do not ISW.

3(b) Identify the coordinates of correct point M1*

( −2 − )
5, −1 , if correct.
From solving (x ± 2)² + (y ± 1)² = 3² (or = 3)
with y = –1, or attempt to get 2 + 5 using a right-
angled triangle.

Carry out a correct method for finding the greatest value of | z | DM1

A1 AWRT 4.35, e.g. 4.3525…

Obtain answer 4.35 or 10 + 4 5

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Question Answer Marks Guidance

4 tan x − 3 B1 OE
State Allow decimals throughout.
1 + 3 tan x

2 B1 SOI
2 cot x replaced by
tan x

Reduce the equation to tan² x – 3 3 tan x – 2 = 0, or three-term equivalent B1 May be implied by further work.

Solve a three-term quadratic in tan x, for x M1 FT their 3-term quadratic. Allow tan–1 (..).

Obtain answer, e.g. 79.8° A1 AWRT 79.8.

Obtain the second answer, e.g. 160.2° and no other in the interval A1 Allow 160, or AWRT 160.2.
Treat answers in radians as a misread.
Ignore answers outside the given interval.

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Question Answer Marks Guidance

5 Square x + iy and equate real and imaginary parts to – 4 and 6 5 respectively M1

Obtain equations x2 – y2 = – 4 and 2xy = 6 5 A1 Or x2 + y2 = 14.

( )
2
( −4 )
2
or x2 + y 2 = + 6 5

Eliminate one variable and find a horizontal equation in the other M1 Allow slips in e.g. signs, powers etc.

Obtain x4 + 4x2 – 45 = 0 or y4 – 4y2 – 45 = 0 or three-term equivalents, A1 May be implied by further work.

or 2x2 = 10 or 2y2 = 18

Obtain answers  ( 5 + 3i ) A1 Accept e.g. x = 5, y = 3 and x = − 5 , y = –3

or  ( )
5,3 , but must be clearly paired. Can be
implied by (e.g.) column working.

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Question Answer Marks Guidance

6 Use correct double angle formula to express sin 2 2θ in terms of cos 4θ B1 1

sin2 2θ = (1 – cos 4θ)
2

Separate variables correctly and reasonable attempt at integration of at least one side M1 Position of (x + 5) or ( 15 x + 1) and sin2 2θ
sufficient for correct separation.

1  B1 OE
Obtain term 5ln  x + 1 May see 5ln ( x + 5 ) .
5 

1 1 B1 FT 1 1  1
Obtain term ( − sin 4 ) Allow    sin 4  from = ( 1 ± cos 4θ).
2 4 2 4  2

Use x = 5 when θ = 0 to evaluate a constant or as limits in a solution containing M1 OE

1 
terms of the form ln  x + 1 ,  and sin 4
5 

Obtain correct answer in any form A1 1 1

E.g. 5ln ( x + 5 ) = ( − sin 4 ) + 5 ln 10
2 4

1  A1 FT 1  
Obtain final answer x = 10exp   − sin 4  − 5 or equivalent
1 1
x = 10exp     sin 4  − 5
10  4  2 4 
with ln removed Must remove ln from
1 1
ln ( x + 5 ) = (  sin 4 ) +  ln 10.
2 4

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Question Answer Marks Guidance

7(a) Use correct product rule M1  d 

3x2 cos (2x) + x3  (cos 2x)  OE
 dx 

Obtain derivative 3x2 cos (2x) – 2x3 sin (2x) A1 OE

1  3  A1 AG
Equate correct derivative to zero and obtain p = tan −1   after full and correct Must reach tan 2p = … or tan 2x = … before given
2  2p  answer.
working

7(b) Calculate the values of a relevant expression or pair of expressions at x = 0.5 and M1 1  3 
x = 0.7 Must have at least 1 of 2 values correct or 3 of 4 values correct f(p) = p − tan −1  
2  2p 
f(0.5) = –0.124… < 0, f(0.7) = 0.132.. > 0
1  3 
or comparing p and tan −1  .
2  2p 
1 −1  3 
0.5 tan   = 0.624 0.5 < 0.624…
2  2p 
1 −1  3 
0.7 tan   = 0.567 0.7 > 0.567
2  2p 

Complete the argument correctly with correct calculated values A1

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Question Answer Marks Guidance

7(c) Use the iterative formula correctly at least twice M1

Obtain final answer 0.596 A1

Show sufficient iterations to at least 5 decimal places to justify 0.596 to 3 decimal A1 E.g. 0.5, 0.62452, 0.58814, 0.59856, 0.59556,
places or show there is a sign change in the interval (0.5955, 0.5965) 0.59642

0.6, 0.59514, 0.59654, 0.59614, 0.59626

0.7, 0.56708, 0.60467, 0.59380, 0.59693, 0 59603,

0.59629

Can recover from wrong values.

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Question Answer Marks Guidance

8(a) Express general point of a line in component form, B1

e.g. (–1 + 2λ, 3 + 3λ, – 4 – λ) or (2 – μ, –3 – 2μ, –1 + μ)

Equate at least two pairs of components and solve for λ or for μ M1

Obtain correct answer for λ or for μ A1 Possible answers are 6, 12, 0 for λ and
–9, –21, –3 for μ.

Verify that one component equation is not satisfied A1 E.g. show 21 ≠ 15 for (11, 21, –10) and
Can show by correctly obtaining 2 values of λ or 2 values of μ (11, 15, –10),
or show –16 ≠ –22 for (23, 39, –16) and
(23, 39, –22),
or show –1 ≠ 5 for (–1, 3, – 4) and (5, 3, – 4).

Show that the lines are not parallel B1  2  −1 

E.g.  3   k  −2  at least 2 components
 
 −1   1
   
required.

Just a statement that direction vectors are not

scalar multiple of each other insufficient, if
direction vectors have not been clearly identified.
Also, told answer is skew.

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Question Answer Marks Guidance

8(b)  2  −1  M1 E.g. (2 × –1) + (3 × –2) + (–1 × 1) or –2 – 6 – 1

Carry out correct process for evaluating the scalar product of  3  and  −2 
  or –9.
 −1   1
   

Using the correct process for the moduli, divide the scalar product by the product of M1 Allow for any pair of vectors here but must be
the moduli and evaluate the inverse cosine of the result consistent between scalar product and magnitudes.

Obtain answer AWRT 169.1° or 2.95c A1 Allow 169°.

Question Answer Marks Guidance

9(a) Substitute x = 3 or –3 into p (x) and equate to 0 or into p' (x) and equate to 72 M1*

Obtain 162 + 9a + 3b + 9 = 0 A1 OE

Obtain 162 + 6a + b = 72 A1 OE

Solve simultaneous equations to obtain either a or b after using p (±3) = 0 and DM1
pꞌ (±3) = 72

Obtain a = –11 and b = –24 A1

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Question Answer Marks Guidance

9(b) Equate (x – 3)(6x2 + Ax + B) to 6x3 – 11x2 – 24x + 9 and obtain equations to solve for M1 Using their a and their b.
A and B A – 18 = a = − 11, B – 3A = − b = − 24,
− 3B = 9.
A = 7 and B = –3.

or divide 6x3 – 11x2 – 24x + 9 by x – 3 and reach 6x2 ± 7x Or reach 6x2 ± (their a + 18) x.

(x – 3)(6x2 + 7x – 3) A1 SOI

Obtain (x – 3)(2x + 3)(3x – 1) A1

Special Case: If only ( x – 3) ( x + 32 )( x – 13 )

or (x – 3)(2x + 3)(3x – 1) seen, SC B1 only
(but can gain two marks in (c)).

9(c) 3 1 B1 FT Must be final answer not in working.

Obtain one correct region x  − or  x  3 FT is on the last two brackets (not ( x – 3)).
2 3

3 1 B1 FT 3 1
Obtain both regions x  − ,  x  3 Allow x  − and  x  3.
2 3 2 3
3 1
SC B1 for x − , x 3.
2 3
FT is on the last two brackets (not ( x – 3)).
If incorrect factor or factors in (b) but correct
regions here, allow SC B1 only.

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Question Answer Marks Guidance

10(a) A Bx + C B1
State or imply the form +
1 + x (
( ) 4 + x2 )
Use a correct method for finding a constant Even with incorrect PF denominators M1 A( 4 + x 2 ) + (Bx + C)(1 + x) + = −7 x 2 +2x − 6

Obtain one of A = –3, B = – 4 and C = 6 A1

Obtain a second value A1

Obtain a third value A1

A C
Special Case 1: + + Find A,
( )
(1 + x ) 4 + x 2
M1 A1. Max 2/5.
A Bx
Special Case 2: + + Find A,
( )
(1 + x ) 4 + x 2
M1 A1. Max 2/5.

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Question Answer Marks Guidance

10(b) Obtain term –3ln (1 + x) B1 FT OE

FT A ln (1 + x)

Obtain term –2ln (4 + x2 ) B1 FT OE

B
FT ln (4 + x2 )
2

Obtain integral of the form c tan −1 dx with d ≠ 1 following separation into two M1 d = 1
2
or 2 only.
expressions

x A1 FT C −1 x
Obtain 3tan −1 FT tan
2 2 2

Substitute correct limits correctly in an expression (obtained correctly) of the form M1 a ln (3) + b ln (8) − b ln (4) + c ( 1 π ) , where
a ln (1 + x), b ln (4 + x2 ), and c tan −1 ( 12 x ) , where a, b, c ≠ 0.
4
a, b, c ≠ 0.
Do not allow slips, and must get to c ( 14 π ).

3 A1 Must be in the form aπ – ln b.

Obtain answer π − ln108
4

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Question Answer Marks Guidance

11 1 1 *M1 OE
Commence integration by parts and reach Ax 2 sin x   Bx sin xdx BOD on ± otherwise scores 0/6.
3 3

1 1 A1 OE
Obtain 3x 2 sin x −  6 x sin x dx Allow 3×2 for 6.
3 3

1 1 1 *DM1 OE
Complete integration by parts and reach Ax 2 sin x + Bx cos x + C sin x
3 3 3

1 1 1 A1 Allow 6 × 3 for 18, 9 × 6 for 54 OE.

Obtain 3x2 sin x + 18 x cos x − 54sin x oe
3 3 3

Substitute limits correctly in an expression of the form DM1 Dependent on both previous M1 marks.
1 1 1
Ax 2 sin x + Bx cos x + C sin x , where ABC ≠ 0 π 3 π 1
Need to use sin = , and cos = to obtain
3 3 3 3 2 3 2
3 Bπ C 3
Aπ 2 + + .
2 2 2

3 3 2 A1 27 3 3 18 54
Obtain answer π + 9π − 27 3 or exact equivalent ISW Allow for , for 9, for 27,
2 2 2 2 2
2187 for 27 3 etc.

Alternative Method for first 4 marks:

1 1 *M1
Commence integration by parts and reach Ax 2 sin x + Bx cos x
3 3

1 1 A1 OE
Obtain 3x 2 sin x + 18x cos x Allow 6 × 3 for 18.
3 3

1 1 1 *DM1
Complete integration by parts and reach Ax 2 sin x + Bx cos x + C sin x
3 3 3

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Question Answer Marks Guidance

11 1 1 1 A1 OE
Obtain 3x2 sin x + 18 x cos x − 54sin x Allow 6 × 3 for 18, 9 × 6 for 54, OE.
3 3 3

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