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Cambridge International A Level

MATHEMATICS 9709/32
Paper 3 Pure Mathematics 3 February/March 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 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 2024 series f or most
Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level
components.

This document consists of 18 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 appli cation 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 Commence division and reach partial quotient of the form x 2  3 x M1

( )( )
or x4 − 3x3 + 9 x 2 − 12 x + 27 = x 2 + 5 Ax 2 + Bx + C + Dx + E
or Ax4 + Bx3 + (5A + C)x2 + 5Bx + 5C
and reach A = 1and B = 3

Obtain quotient x 2 − 3x + 4 A1 A = 1, B = −3
[5A + C = 9 so C = 4; 5B + D = − 12 so D = 3;
5C + E = 27 so E = 7].
A pair of incorrect statements ‘remainder
x 2 − 3x + 4 ’ and ‘quotient 3x + 7 ’ score
M1A1A0.

Obtain remainder 3x + 7 A1

x2 − 3x + 4 3
x2 + 5 x4 – 3x3 + 9x2 – 12x + 27
x4 + 5x2
− 3x + 4x2
3

− 3x3 − 15x
+ 4x2 + 3x
+ 4x2 + 20
+ 3x +7

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Question Answer Marks Guidance

2(a) State unsimplified term in x, or its coefficient, in the expansion of ( 4 − x ) 2

1
B1 1 1  −x  −x
42    = .
2  4  4

State unsimplifed term in x 2 , or its coefficient, in the expansion of ( 4 − x ) 2 B1

1 2
 −21  − x  x
2
1
1
− x2
42  2
  = . Allow   .
2  4  64 4
1 M1 Allow unsimplified 2x.
Multiply by ( 2 x − 5 ) and obtain 2 terms in x 2 , allow even if errors in 4 2 , signs, etc.
1  −x 
1
4 2     − 5.
2  4 
1 −1
 2
x
2
 −x 
1
4 
2 2 2    . Allow   .
2  4  4
 −x  −x  −1   −1 
2
2x    ( −5 )  or 2    ( −5)    .
 4  64  4  64 

27 54 A1 Allow in a full expansion up to x2 , ignore extra

Obtain − or –0.421875 or − terms even if they contain errors.
64 128

2(b) x 4 B1 or −4  x  4 .

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Question Answer Marks Guidance

(( − 3 ) )
Obtain r = 4
( )
3(a) B1 2 2
so r = z = − 3 + 12 .
2
z = + 12

Correct method for the argument M1 − 3 5π

 = 2 tan −1   or 2  .
 1  6

 A1 Arg with no working B1 instead of M1 A1.

Obtain  = − A0 if decimals.
3
Allow separate mod and arg to gain full marks

Alternative solution for Question 3(a)

( )
2 B1
z2 = 2 – 2 3 i so r = 22 + – 2 3 =4

Correct method for the argument M1 –2 3

arg z2 = tan −1
2

 A1 Arg with no working B1 instead of M1 A1.

Obtain  = − A0 if decimals.
3
Allow separate mod and arg to gain full marks

3(b) Use of  + their  = 0 or  + their  = −π or  + their  = π M1 Seen or implied.

Using their  or new value calculated in (b).

their r M1 Seen or implied.

Use of R =
12

1 −i
2π π
Obtain e 3
and 1 i3
e A1
3 3

3
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Question Answer Marks Guidance

4 Obtain ln p − ln q = a B1 p
= ea .
q

Obtain ln p + 2ln q = b B1 pq2 = eb .

Completed method to obtain ln p7 q ( ) M1

E.g. ln q =
b−a
, ln p =
2a + b
3 3
and attempt 7ln p + ln q.
All exponentials must be removed to obtain M1.

13a + 8b A1
Obtain
3

Alternative solution for Question 4

 p
x B1 p
( ) Or ln p 7 q = x ln + y ln q 2 p .
y
State p q =   q 2 p
7
q
q

Equate indices to form simultaneous equations in x and y, can have errors M1 x + y = 7 and − x + 2y = 1.

Obtain 7 = x + y and 1 = 2 y − x A1 Leading to x = 13

3
, y = 83 .

13a + 8b A1
Evaluate x  a + y  b to obtain
3

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Question Answer Marks Guidance

5(a) Show a circle with centre 4 + 2i B1

Im(z)

Show a circle with radius 3 and centre not at the origin B1

Show the straight line Re ( z ) = 5 B1 2i

Shade the correct region B1 O 4 5

Allow even if radius 3 mark not gained or shown incorrectly Re(z)

4 If 4 and 6 seen on diagram and line is at mid point,

but 5 not marked, allow final two B1 marks.

5(b) Carry out a complete method for finding the greatest value of arg z M1 2+2 2
e.g. tan −1 .
5
Allow 2√2 as √(32 − 12 ).

Obtain answer 0.768 radians or 44.0 A1

2 SC B1 tan−1 (2/4) + sin−1 (3/√(42 + 22 )) = 26.565° +

42.130° = 68.695°
68.7° or [1.19896] 1.20 radians.

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Question Answer Marks Guidance

6(a) State or imply 4 y ddyx as the derivative of 2 y 2 B1 SC If dy introduced instead of d then allow B1
dx dx
for both, followed by correct method M1 Max 2.

State or imply 3 y + 3x ddyx as the derivative of 3xy B1 Allow extra dy = correct expression to collect all
dx
marks if correct.

Complete the differentiation, all 4 terms, isolate 2 dy

terms on LHS or bracket dy M1
dx dx
dy
terms and solve for dx

dy 2 x − 3 y − 1 A1 Answer Given – need to have seen 4y dy + 3x dy

Obtain = dx dx
dx 4 y + 3x
= 2x −3y −1 or (4y + 3x) ddyx − 2x +3y = −1.
Need to see = 2x or = 0 consistently throughout
otherwise M1 A0.
No recovery allowed.
When all terms are included then must be an
equation.

4 Allow all marks if using dx and dy.

6(b) Equate numerator to zero, obtaining 2x = 3y + 1 or 3y = 2x −1 and form equation in M1* e.g. 2 ( 2x − 1)2 + x ( 2x − 1) + x = x2
9
x only or y only from 2y2 +3xy + x = x2
or 2 y 2 + 32 (1 + 3 y ) y + 12 (1 + 3 y ) = 14 (1 + 3 y ) .
2

Allow errors.

Obtain 92 ( 2 x − 1) = − x2 or a 3 term quadratic in one unknown and try to solve.

2
(
DM1 e.g. 17 x 2 − 8 x + 2 = 0 b2 − 4ac = −72 )
If errors in quadratic formulation allow solution, applying usual rules for solution of
quadratic equation, and allow M1
(
or 17 y + 6 y + 1 = 0 b2 − 4ac = −32 .
2
)
x = 4/17 ± (3√2/17)i, y = − 3//17 ± (2√2/17)i .

Conclude that the equation has no [real] roots A1 Given Answer. CWO

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Question Answer Marks Guidance

7(a) Use correct product rule M1

Obtain correct derivative in any form A1 e.g. dy

= e2 x + 2xe2 x − 5
dx

1  5  A1 Given answer – need to see e2 x= 5/(1 + 2x)

Equate derivative to zero and obtain  = ln   or ln e2 x = ln (5/(1 + 2x)) in working.
2  1 + 2 
Must be in terms of α not x.
Allow α to be used before equating to 0.

7(b) Calculate the value of a relevant expression or values of a pair of expressions at M1 Need to attempt BOTH values and have one
x = 0.4 and x = 0.5 correct.

Complete the argument correctly with correct calculated values A1 e.g. 0.4  0.51 08 and 0.5  0.458 or 0.46 or
0.45
or – 0.11[08] < 0 and 0.042 > 0
If use original derivative −0.994 (0.4) and
0.437 (0.5).

7(c) 1  5  M1 Obtain one value and then substitute it into the

Use the iterative process  n +1 = ln   correctly at least twice anywhere in formula to obtain a second value.
2  1 + 2 n 
iteration process

Obtain final answer 0.47 A1

Show sufficient iterations to 4 d.p. to justify 0.47 to 2 d.p. or show there is a sign A1 0.4,0.5108,0.4528,0.4823,0.4670,0.4749
change in the interval ( 0.465, 0.475 ) 0.45,0.4838,0.4663,0.4753,0.4707,0.4730
0.5,0.4581,0.4795,0.4685,0.4742
Allow self correction.

3 SC B1 No working 0.47

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Question Answer Marks Guidance

8(a) Use the correct expansion of cos ( x + 14 π ) to obtain sin x + 2cos x B1  1 1 

3sin x + 2 2  cos x − sin x  .
 2 2 

State R = 5 B1 FT ISW FT their a sin x + b cos x provided this

expression obtained by correct method.

Use correct trig formulae to find  M1  = tan−1 (b/a) from their a sin x + b cos x
or sin−1 or cos−1 provided this expression obtained
by correct method.
NB If cos  = 1 and sin  = 2 then M0 A0.

Obtain  = 1.107 A1 3 d.p. CAO

Treat answer in degrees as a misread ( 63.435 ) .

8(b)  1.5  B1 FT Follow their R.

sin −1  
 R 

Use a correct method to obtain an un-simplified value of  with their  M1   1.5    −1  1.5  
2  sin −1   −   or 2  π − sin   −  .
  R     R  

Obtain one correct answer e.g. −0.74 in the interval A1

Obtain second correct answer e.g. 2.60 (2.5986) or 4π – 0.74 = 11.8 A1 If uses 1.11° withhold first accuracy mark gained,
or 2.60 − 4π = −9.97 in the interval but allow rest of accuracy marks. Allow 2.6(0).

Obtain two more correct answers e.g. −9.97 and 11.8 and no others in the interval A1 Ignore answers outside the interval. Treat answers
in degrees as a misread.
( −571.1, − 42.6,148.9, 677.2 ) .
5

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Question Answer Marks Guidance

9(a) Find the scalar product of a pair of adjacent sides M1 OA = (5, − 2, 1), OB = (8, 2, − 6),
OC = (3, 4, −7), CB = (5, −2, 1),
AB = (3, 4, −7).

Show that the sides are perpendicular A1 e.g. OA.OC = 15 − 8 − 7 = 0 .

Need to see working of numerator, ignore
denominator.

Compare a pair of opposite sides M1 OA and CB or OC and AB .

Show that they are parallel and equal in length and hence OABC is a rectangle A1 e.g. AB = AO + OB = 3i + 4 j − 7k = OC .
If show AB = 3i + 4 j − 7 k = OC , then M1 A1
since this implies parallel and of equal length.
If only show lengths equal M1.
If repeat for other pair of opposite sides then A1.

Alternative solution for Question 9(a)

Show the diagonals OB and AC are equal in length ( 104 ) AC = ( −2, 6, −8).

Show the diagonals bisect each other at ( 4,1, −3) OB 1

= OC + ( OA − OC ) = (4, 1, −3).
2 2

Show the quadrilateral is a parallelogram e.g. OB =OA + OC .

Show both pairs of opposite sides are equal in length and a pair of adjacent sides are
perpendicular

4 Without calculation of scalar product max is M1

A1.

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Question Answer Marks Guidance

9(b) AC B1 Seen or implied using diagonals.

AC =  ( −2i + 6j − 8k ) or =  ( −1i + 3j − 4k )
2

Scalar product of a pair of relevant vectors M1 e.g. AC.OB = −16 + 12 + 48 .

Using the correct process for the moduli, divide the scalar product by the product of M1  44 
the moduli and obtain the inverse cosine of the result.  cos −1  .
 104 
For any two vectors.

Obtain answer 65. ( 0 )  A1 Accept 1.13 radians.

Alternative solution for Question 9(b)

Scalar product of a pair of relevant vectors M1 e.g. OA.OB = 40 − 4 − 6 using one side and a
diagonal.
or OC.OB = 24 + 8 + 42.
Must use scalar product.

Using the correct process for the moduli, divide the scalar product by the product of M1  30   74 
the moduli and obtain the inverse cosine of the result. Any two vectors.  cos −1  −1
 or cos   .
 104   104 

Required angle = 180 − 2 × 57.5° or 180 − 2 × 32.5° = 115° and 180° − 115° or B1 OE SOI
2× 32.5° Complete method to find the acute angle.

Obtain answer 65.0 A1 Accept 1.13 radians.

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Question Answer Marks Guidance

10(a) A B C B1 Allow if seen prior to assigning a value for a.

State or imply the form + +
2a + x 2a − x 5a − 2 x

Use a correct method for finding a coefficient M1

Obtain one of A = 1, B = 9, C = −16 A1

Obtain a second value A1

Obtain the third value A1

5 Dx + E C
SC + B0 M1 and C = −16
4a ^ 2 − x ^ 2 5a − 2 x
A1 Max 2/5.
SC Allow M1 only for other incorrect partial
fraction.

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Question Answer Marks Guidance

10(b) Integrate and obtain one of the terms ln 2a + x − 9ln 2a − x + 8ln 5a − 2 x B1 FT Condone missing modulus signs.
Use their A, B and C.

Obtain a second correct term B1 FT

Obtain the third correct term B1 FT Max 3/5 if value is assigned for a (award M0 A0).

Substitute limits correctly in an integral of the form M1 Either (i) collect terms with same coeeficient and
p ln 2a + x + q ln 2a − x + + r ln 5a − 2 x and remove all a’s remove all a’s
e.g. pln 3a −pln a + qln a − qln 3a + rln 3a –
rln7a
hence pln 3 − qln 3 + rln 3 − rln7
or
(ii) collect same ln terms and remove all a’s
e.g. (p – q + r) ln 3a – ( p – q) ln a – rln7a
and − (p − q) ln a = (−p + q − r ) ln a + r ln a
hence p ln 3 – q ln3 + rln 3 – r ln 7.

Obtain 18ln3 − 8ln7 from correct working A1 A0 if the solution involves logarithms of negative
numbers.

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Question Answer Marks Guidance

11 Separate variables correctly B1 −3 y 1

 (1 + y ) e dy =  1 + cos 2 d .
Allow 1/e3 y and missing integral signs.

Integrate to obtain p (1 + y ) e−3 y + qe−3 y dy


M1 Allow unless clear evidence that formula used has
a + sign.

−1 1 A1 Allow unsimplified.
Obtain (1 + y ) e−3 y + e−3 y dy

3 3

−1 1 A1 Condone no constant of integration.

Obtain (1 + y ) e−3 y − e−3 y ( + A)
3 9

1 B1
Use correct double angle formula to obtain  2cos  d
2

Obtain ktan  + B  B1 Condone no constant of integration.

π M1* 1 1 1  17 
Use y = 0,  = to evaluate a constant of integration in an expression of the form =− − +C C = 
4 2 3 9  18 
 ye−3 y ,  e −3 y and  tan  only. Allow ye3 y and e3 y. Must have integrated LHS
twice.

Use y = 1 DM1 − (1 + 1)
− 1( 9e3 ) = tan  − .
1 17
3
3e 2 18
Must have integrated LHS.

17 14 −3 A1 Or exact equivalent . Exact ISW.

Obtain tan  = − e
9 9  17 14 
Allow θ = tan −1  − e−3  .
 9 9 
If x instead of  then withhold final A1.

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