Cambridge International A Level

MATHEMATICS 9709/32
Paper 3 Pure Mathematics 3 October/November 2024
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 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 20 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 Obtain a correct unsimplified version of the x or x 2 term of the expansion of M1 1 1 1

1 1
1 E.g. −  x or − 2 2  x 2 or
( 9 − 3 x ) 2 or 1 − x 
2
1
1 2 3 2 9
−1 −1 −3
 3  1 2 1

9 ( −3x ) or 2 2 9 2 ( −3 x ) .
1 2

2 2
n
Not for symbolic coefficients in the form C r .

State correct first term 3 B1

1 1 A1 A1 A1 for each term correct.

Obtain the next two terms − x − x 2 Do not ISW.
2 24

1 1
SC M1A1 for 1 − x − x 2 seen on its own or
6 72
as a factor.

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Question Answer Marks Guidance

2(a) Sketch a relevant graph, e.g. y = cot 2x B1 Alt: use tan 2x and cos x . And only one root in
π
range. (also cross at 2 )

Sketch a second relevant graph on the same axes, e.g. y = sec x and justify the given B1 Need to mark intersection with a dot, a cross, or
statement say roots at points of intersection, OE.

2(b) 1 B1 Should see tan 2x = cos x before the given

State x = tan −1 ( cos x )
2 conclusion.
1
and rearrange to the given equation cot 2x = sec x Or rearrange cot 2x = sec x to x = tan −1 ( cos x )
2
and state iterative formula
1
Note: If using the alternative approach in (a), can stop at tan 2x = cos x xn+1 = tan −1 ( cos xn ) .
2

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Question Answer Marks Guidance

3 Square x + iy and equate real and imaginary parts to 6 and −8 respectively *M1 Condone +8 in place of -8 and/or i 2 = 1.

Obtain equations x2 – y2 = 6 and 2xy = –8 A1 OE

Eliminate one variable and find an equation in the other (from 2 equations each in 2 DM1 Condone a slip but not seriously incorrect
unknowns) algebra, e.g. use of x = −4 y is M0.

Obtain x4 – 6x2 – 16 = 0 or y4 + 6y2 – 16 = 0 A1 Accept 3-term equivalents e.g. x 4 = 6 x 2 + 16.

Condone missing ‘= 0’ if implied by subsequent
work.

( )
Obtain answers  2 2 − 2i or exact equivalents A1 Allow if values of x and y stated separately but
the pairing is clear.
Ignore additional correct solutions for x and y
not real, but A0 if any additional incorrect
answers.

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Question Answer Marks Guidance

4 Use laws of indices correctly and solve for 5 x *M1 5

E.g. obtain 5 x = OE.
12
Allow for y = … if they have previously stated
y = 5 x.
Could be implied if they have a correct
simplified equation in 5x, e.g. 12  5x = 5.

Use a correct method for solving an equation of the form 5 x = a, where a > 0 DM1 Allow x ln5 = ln  10 
 .
 24 

Obtain answer – 0.544 A1 CWO. If no working shown, 0/3.

Note: 3 dp required.

5(a) 4  M1 SOI
and/or − 4  1  4 1
7 7 Allow  −  − π  or  − π
7  7  7 7

Note: Many multiply top and bottom by the

conjugate, which is fine, but to score the M1
they need to state or imply the argument of a
complex number.

5 A1 Do not accept degrees.

Obtain arg u = π
7

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Question Answer Marks Guidance

5(b) Reflection (in the) real axis B1 Correct non-contradictory statement.

Condone x–axis or horizontal axis.
Need ‘reflection’.
Not ‘mirror’, ‘flip’.

5 B1FT FT their exact (a). Accept 2π − their exact (a).

arg u* = − π
7 Accept an ‘exact’ expression in place of an exact
value.
Need to see a value or an expression. Do not
accept arg u* = − arg u without a value seen.

6 Form a pair of equations in a and b *M1 Condone sign slips but must be using the given
coordinates correctly.
ln a + 8.27 = 3.4ln b
e.g. 
ln a + 2.24 = 0.5ln b
ae2.24 = b0.5
or 

 ae
8.27
= b3.4

Carry out a correct method for finding lna or lnb or a or b DM1 Condone sign slip.

Obtain value a = 0.3 A1 (0.30109…)

Obtain value b = 8 A1 (7.99895…)

Allow A0A1 if both values ‘correct’ but not
rounded to 1 sf.
Allow 4/4 for 0.3 y = 8 x with correct working
shown.

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Question Answer Marks Guidance

6 Alternative Method for Question 6:

Carry out a correct method for finding lnb or b *M1 Condone sign slips but must be using the given
(Need to link the gradient to lnb at some point) coordinates correctly.
8.27 − 2.24
ln b = ( = 2.079....)
3.4 − 0.5

Obtain value b = 8 A1

Correct method to find  ln a or a DM1 Condone sign slip ( ln a = −1.200...) .

Obtain value a = 0.3 A1 Allow A0A1 if both values ‘correct’ but not
rounded to 1 sf.
Allow 4/4 for 0.3 y = 8 x with correct working
shown.

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Question Answer Marks Guidance

7(a) Use correct double angle formula to obtain an equation in tan x M1 2  2 tan x
e.g. tan 3 x + − tan x ( = 0 ) .
1 − tan 2 x
Allow if the correct formula is quoted but then
they lose the 2 from the numerator when they
use the formula.

Obtain a correct equation in tan x in any form without fractions A1 E.g. tan3 x − tan5 x + 4tan x − tan x + tan3 x ( = 0).
Condone if ‘= 0’ is missing here.

Reduce to the given answer of tan 4 x − 2tan 2 x − 3 = 0 correctly A1 Obtain given answer from correct working but
condone if never mention tan x  0.
Condone the right terms in a different order
‘Show that’ so each line must be correct.

7(b) A complete correct method to solve the equation to obtain a value for  M1
( tan 2 =  3 )
Allow if they make a slip in copying the
equation but do have a complete method to
obtain a value of  .
M0 if they get a value for 2 but never halve it.

1 1 2 5 A1
Obtain two of (  =) , ,  and 
6 3 3 6

1 1 2 5 A1 Exact, ignore any answers outside interval

Obtain the other two of ( =) , ,  and  and no others in the interval
6 3 3 6 2 1 4 2
Accept  for  and  for 
6 3 6 3

Do not need to see  = (from tan 2 = 0).
2

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Question Answer Marks Guidance

8(a) Use correct product rule or chain rule to find derivative of x with respect to t M1 Obtain k tan 2t sec2 2t.

dx A1
Obtain = 4 tan 2t sec2 2t oe
dt

dy B1
= −2sin 2t
dt

dy dy dt dy 1 B1 dy
Use =  to obtain given answer = − cos3 2t Condone if missing.
dx dt dx dx 2 dx

8(b) 2 B1 Accept y = 0.707…

Obtain x = 1 and y =
2

− 2 B1 5
State or imply gradient of tangent is or gradient of normal is 4 2 Any equivalent form, e.g. 2 2 .
8
Accept – 0.177 or 5.66.

Use correct method to find equation of normal using their values M1 Need a fully substituted equation for the normal
(in any form) or to get at least as far as finding
value for m and expression for c.

7 2 A1 E.g., y = 5.66x – 4.95.

Obtain equation of normal is y = 4 2 x − or equivalent 3 term equation Must be y = …
2

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Question Answer Marks Guidance

9(a) Use a correct method to find OD (

M1 E.g. OC + 3 OA − OB = )
(–3i – 2j + 2k) + 3((2i + j – 3k) – (4j + k))
( AB = −2i + 3j + 4k)
Accept column vectors throughout.

Obtain position vector of D is 3i – 11j – 10k A1 Accept coordinates.

9(b) Carry out correct method for finding a vector equation for AC or BD *M1 E.g. 2i + j – 3k + λ (5i + 3j – 5k)
or 4j + k + µ (3i – 15j – 11k).
Condone missing r = …

Both diagonal equations correct. A1ft Seen or implied.

Follow their D. Condone missing r = …

Equate at least two pairs of corresponding components and solve for λ or for µ DM1 Dependent on using relevant lines and two
different parameters.

1 1 A1 The values will depend on the directions of their

Obtain λ = – or µ = lines
4 4

3 1 7 A1 OE
Obtain position vector of P is i+ j– k Accept coordinates.
4 4 4
Do not ISW.

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Question Answer Marks Guidance

9(b) Alternative Method for Question 9(b):

State or imply AC = 5i − 3j + 5k B1 FT Or BD = 3i − 15 j − 11k

Follow their D if used.

Identify similar triangles with ratio 1 : 3 M1

1 M1 Must be correct fraction.

Use similar triangles to obtain OP , e.g. OP = OA + AC
4

3 1 7 A2 OE
Obtain position vector of P is i+ j– k Allow A1A0 if any two values are correct.
4 4 4

9(c) Find direction vector BA = 2i – 3j – 4k and BC = –3i – 6j + k or equivalent B1FT Or AB and CB.
FT if using an incorrect AB from earlier work.

Carry out correct process for evaluating the scalar product of two relevant vectors M1 Allow if one is going in the negative direction,
e.g. AB and BC .

Using the correct process for the moduli, divide their scalar product by the product M1 Independent of the first M1.
of their moduli and evaluate the inverse cosine of the result to obtain an angle For their two vectors
8
 = cos −1 = ...
29 46

Obtain answer 77.3° (or 1.35 radians) A1 77.347…

Correctly rounded to more than 3 sf or
AWRT 77.3.

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Question Answer Marks Guidance

10(a) dV B1 Need a complete correct statement seen or

Obtain = 40π − 0.8πr or equivalent implied.
dt

dV dV dr B1 Need a complete correct statement seen or

Obtain = 4πr 2 or equivalent e.g. = 4r 2 implied.
dr dt dt

Use the chain rule to obtain given answer (including the derivative) B1 dr 50 − r dr 40 − 0.8r
Allow if = 2
follows =
dt 5r dt 4r 2
without further explanation (π already cancelled)
and no incorrect statements seen.

10(b) Commence division and reach quotient of the form M1 Allow M1 if divide by r − 50 to obtain
–5r ± 250 5r  250 .
or 5r2 = (50 – r)(Ar + B) + C and reach A = –5 and B = ± 250

Obtain quotient –5r – 250 A1 Do not need to state which is quotient and which
is remainder. However, if clearly muddled, then
M1A1A0 for both expressions correct.

Obtain remainder 12 500 A1 Note: 12 500 following division by r – 50 is

correct and scores this A1 ISW.

SC B1 only for correct use of remainder

theorem to obtain correct remainder.

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Question Answer Marks Guidance

10(c) Prepare to integrate e.g. separate variables correctly B1FT

 5r dr = 1dt
2

 50 − r 
dt 5r 2  12500 
Or express in the form =  = − ( 5r + 250 ) +  Condone missing dr, dt or missing integral
dr 50 − r  50 − r  signs, but not both.
Follow their division in (b) if substitute before
separating.

Obtain term t DB1

A 2 M1 C
Obtain terms r + Br − Cln ( 50 − r ) From their Ar + B + in (b) where
2 50 − r
ABC ≠ 0.
Allow a single slip in the coefficients.

5 A1FT FT their (b), provided of the correct form.

Obtain terms − r 2 − 250r − 12500ln(50 − r )
2

Use t = 0, r = 0 to evaluate a constant or as limits in a solution containing terms of M1

the form r2, r, ln(50 – r) and t

5 A1 OE
Obtain final answer t = − r 2 − 250r − 12500ln(50 − r ) + 12500ln 50 Must be t = …..
2
Allow with 12500ln50 = 48900 or better.

10(d) Obtain t = 70.5 B1 May be more accurate (70.4605…).

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Question Answer Marks Guidance

11(a) Use correct quotient rule NB the question asks for f  ( x ) so need complete form M1 Or correct product rule.

4e 2 x (e 2 x − 3e x + 2) − 2e 2 x (2e 2 x − 3e x ) A1
Obtain correct derivative in any form, e.g.
(e )
2
2x
− 3e x + 2

Equate their derivative to zero *M1 Can be implied by numerator equated to zero for
(
quotient rule. 8 = 6e
x
)
Solve for x to obtain x = lna DM1 a positive.

4 A1 No errors seen.
Obtain x = ln and y = –16
3 8
Accept equivalent exact forms, e.g. x = ln .
6

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Question Answer Marks Guidance

11(a) Alternative Method for Question 11(a)

Complete method to express f ( x ) in partial fractions M1

As far as p +
q
+ x
r
with values for p,
e − 2 e −1
x

q and r  2 + 8
− x
2 
.
 e − 2 e −1 
x

(
Allow in u u = e x . )
se x te x *M1 Note: the question requires f’(x) so if they have
Differentiate to obtain f  ( x ) = + substituted for ex, they will also need chain rule.
(e ) (e )
2 2
x
−2 x
−1

−8e x 2e x A1 From correct work.

Obtain f  ( x ) = +
(e ) (e )
2 2
x
−2 x
−1

Equate derivative to zero and solve for x to obtain x = lna DM1 Must follow correctly to give a positive value of
a.

4 A1 No errors seen.
Obtain x = ln and y = –16
3 8
Accept x = ln , or equivalent.
6

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Question Answer Marks Guidance

11(b) du B1
State or imply = ex
dx

2u B1 Correct expression in u.
Obtain u 2
− 3u + 2
du or equivalent
Condone missing du or missing integral but not
both.
  2 +  8 − 2 du
Or1    Allow FT if using their partial fractions from
  u  u ( u − 2 ) u ( u − 1) (a).

A B B1 FT Complete reduction to partial fractions.

+
State or imply partial fractions of the form
u −1 u − 2 Correct form for their integrand.
C D E 2u − 3 3 2u − 3 F G
Or1 + + Or2 2 + 2 = 2 + +
u u − 2 u −1 u − 3u + 2 u − 3u + 2 u − 3u + 2 u − 2 u −1

Use a correct method for finding a constant M1 Available if they have incorrect form.

−2 4 A1
Obtain correct +
u −1 u − 2
2u − 3 3 3
Or2 2 + −
u − 3u + 2 u − 2 u − 1

Integrate to obtain a ln (u – 1) + b ln (u – 2) or equivalent *M1 M0 if they have additional terms that do not
cancel out.

Obtain correct –2 ln (u – 1) + 4 ln (u – 2) or equivalent A1FT FT values of their partial fraction coefficients.

Correctly use limits u = 5 and 3 in an expression of the form a ln (u – 1) + b ln (u – 2) DM1

or x = ln 5 and ln 3 in an expression of the form a ln (ex – 1) + b ln (ex – 2)

81 A1 Accept ln 20.25.
Obtain ln
4

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