Cambridge International AS & A Level
FURTHER MATHEMATICS
9231/11
Paper 1 Further Pure Mathematics
May/June 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 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 16 printed pages.
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Cambridge International AS & A Level – Mark Scheme
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Generic Marking Principles
May/June 2024
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.
Mathematics Specific Marking Principles
© Cambridge University Press & Assessment 2024
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Cambridge International AS & A Level – Mark Scheme
PUBLISHED
May/June 2024
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.
© Cambridge University Press & Assessment 2024
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Cambridge International AS & A Level – Mark Scheme
PUBLISHED
Mark Scheme Notes
May/June 2024
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.
© Cambridge University Press & Assessment 2024
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Cambridge International AS & A Level – Mark Scheme
PUBLISHED
May/June 2024
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
© Cambridge University Press & Assessment 2024
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Cambridge International AS & A Level – Mark Scheme
PUBLISHED
Question
1(a)
Answer
Marks
May/June 2024
Guidance
B1
 12 p
1
1(b)
 2   2    2           52   12 
M1 Factorises.
 54
A1
2
1(c)
B1 FT OE
FT on their value of      (in terms of p).
4 z 3  2 pz 2  5 z  25  0
1
1(d)
1
 14  p
3
M1 Uses  2   2   2       2  2(     )
p  121
A1 CAO
2
© Cambridge University Press & Assessment 2024
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PUBLISHED
Question
2
Answer
May/June 2024
Marks
Guidance
64  38  2  1332 is divisible by 74.
B1 Checks base case.
Assume that 64 k  38k  2 is divisible by 74 for some positive integer k.
B1 States inductive hypothesis.
M1 A1 Separates 64 k  38k  2 or considers difference.
Then 64 k  4  38k 1  2  (1295  1)  64 k  (37  1)  38k  2
Is divisible by 74 because 1295  64 k  37  38k is divisible by 74.
A1
Hence, by induction, 64 k  38k  2 is divisible by 74, is true for every positive integer
A1
n.
6
Question
3(a)
Answer
N

r 1
1
N
12
r3  7
r 1
N

r2  4
r 1
B1 Expands.
N

Guidance
r
r 1
 N  12  76 N ( N  1)(2 N  1)  2 N  N  1
M1 Substitutes formulae from MF19.
 9 N  N  2 N  1  14  2 N  3N  1  24 N  24
A1
3
N2
4
1
N
12
N

r (r  1)  3r  4   3
Marks
2
2
 9 N  46 N  75N  38  N ( N  1)( N  2)(9 N  19)
3
2
AG
1
12
3
© Cambridge University Press & Assessment 2024
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Cambridge International AS & A Level – Mark Scheme
PUBLISHED
Question
3(b)
Answer
Marks
3r  4 4
1
 
r (r  1) r r  1
N
3r  4  1 
 
r 1 r ( r  1)  4 

r 1
May/June 2024
Guidance
M1 A1 Finds partial fractions.
1 
 1  4 1   1  4 1 
 1  4
  2      3        N 1  

 4  1 2   4  2 3 
 4  N N  1 
M1 Writes at least three terms, including last.
A1
1
1
 N 1
4 4 ( N  1)
4
3(c)
B1 CAO
1
4
1
Question
4(a)
Answer
Marks
[One way] stretch, rotation
B1
Stretch followed by rotation
B1 Correct order.
Stretch parallel to the x-axis, scale factor 14
B1
Rotation, 13 π [anticlockwise] about the origin.
B1
4
© Cambridge University Press & Assessment 2024
Page 8 of 16
Guidance
9231/11
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
Question
4(b)
Answer
1
 12
 1
 12 3 

  2
1
1 3

 1 3
2
2

 2
1
M 1   14
0
0   12

1    12 3
1
2
1
2
May/June 2024
Marks
Guidance
B1 Both inverses correct.
1
3  14 0 
1 1 0 
 ,
  

1 
14  0 14 
 0 1
2 
B1 Correct order.
3

1 
2 
2
4(c)
B1
 7
 12 3 

M 
1
7 3

2


 7

7 3

B1
 12 3   x   7 x  12 3 y 

   
1
  y   7 3x  1 y 
2

2 


7 3x  12 mx  m 7 x  12 3mx

x
X
Transforms   to  
 y
Y 
M1 Uses y  mx and Y  mX .
7 3  12 m  7 m  12 3m 2  12 3m 2  132 m  7 3  0
A1
y  2 3 x and y  73 3 x
A1
5
4(d)
28  14  ABC
M1 Uses DEF  det M ABC .
2 cm 2
A1 Allow with units missing.
2
© Cambridge University Press & Assessment 2024
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Cambridge International AS & A Level – Mark Scheme
PUBLISHED
Question
5(a)
Answer

AB  2 j  5k
Marks


AC  5i  5 j BC  5i  7 j  5k
May/June 2024
Guidance
B1 Finds direction vectors of two lines in the plane.
 25   5 

  
0 2 5   25  ~  5 
5 5 0  10   2 
M1 A1 Finds normal to the plane ABC.
5(2)  5(2)  2(4)  8  5 x  5 y  2 z  8
M1 A1 Substitutes point. AEF for final answer.
i
j
k
5
5(b)
8   2i  j  3k  5i  5 j  2k 
52  5 2  2 2

M1 A1 Correct formula for distance from point to plane.
7
 0.953
54
2
5(c)
 2   3   5 
      
CD   1    3    4 
 3   4   1 
     
B1
i j k  18 


0 2 5   25 
5 4 1  10 
M1 A1 Find common perpendicular.
 5   18 
35
 

5  25  
 1.08

1049   
1049

 0   10 
M1 A1 Uses formula for shortest distance.
1
5
© Cambridge University Press & Assessment 2024
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Cambridge International AS & A Level – Mark Scheme
PUBLISHED
Question
6(a)
Answer
Marks
x  2
y
May/June 2024
Guidance
B1
 x  2  ( x  a  2)  2a  5  x  a  2  5  2a
x2
M1
x2
y  xa2
A1
3
6(b)


dy  x  2  2 x  a   x  ax  1

dx
 x  2 2
x 2  4 x  2a  1  0
2
M1 Differentiates.
A1
 dy
2a  5 
 or

1
2 
 dx
x

2




dy
.
dx
Not from wrong working.
16  4  2a  1  20  8a  0 (or y '  0 )  No stationary points
M1 A1 Consideration of discriminant or sign of y ' with
correct conclusion.
4
© Cambridge University Press & Assessment 2024
Forms quadratic equation or simplifies
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Cambridge International AS & A Level – Mark Scheme
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Question
Answer
Marks
y
6(c)
May/June 2024
Guidance
B1 Axes and asymptotes labelled.
y  xa2
B1 Branches correct. (Asymptotes may cross above, on
or below the x-axis.)
x
x  2
 0, 12 
B1 May be seen on their diagram.
3
© Cambridge University Press & Assessment 2024
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Cambridge International AS & A Level – Mark Scheme
PUBLISHED
Question
Answer
6(d)(i)
y
Marks
May/June 2024
Guidance
B1 FT from their attempt in part 6(c).
B1 Everything correct (approach to vertical asymptote
and cusps correct).
x
2
y
6(d)(ii)
B1
ya
x
1
© Cambridge University Press & Assessment 2024
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Cambridge International AS & A Level – Mark Scheme
PUBLISHED
Question
6(d)(iii)
Answer
Marks
May/June 2024
Guidance
M1 Or direct use of x  3
x 2  ax  1
x 2  ax  1
 a or
 a
x2
x2
x2  1  2a  0 or x 2  2ax  1  2a  0
 a  a 2  2 a  1  x   2 a  1,  a  a 2  2 a  1  x  2 a  1
a5
A1
2
Question
Answer
Marks
7(a)
Guidance
B1 Correct shape.
B1 Section 12 π    π correct.
 0
 π tan π , 0  (1.99, 0)
B1 May be seen on their diagram.
1
3
© Cambridge University Press & Assessment 2024
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Cambridge International AS & A Level – Mark Scheme
PUBLISHED
Question
7(b)
Answer
A  12
1 1
2 2
π
0
1
2 π
Marks
0
π
π
 u 2 tan 1  u  0  14 0
1
or A  12


A1
π
π
π
A1

A1
2
A  12  12  u  tan 1  u    14 u  tan 1  u  
0

0

 
A  14 π 2 tan 1 π  14 π  tan 1 π  14 π 2  1 tan 1 π  14 π  2.65
7
© Cambridge University Press & Assessment 2024
1
0
M1 Rearranges integral into form which can be
integrated.

u  tan 1 u 

0
.

  u  tan  u  du
M1 A1 Integrates by parts.
u2
du
1  u2
π 1  u2 1
π
u2
1
d
u

du  1 
du
2
2
0
0 1 u
0 1 u
1  u2
π
Guidance
M1 Uses correct formula.
  π    tan  π    d or A     u  tan  u  du
1
May/June 2024
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Cambridge International AS & A Level – Mark Scheme
PUBLISHED
Question
7(c)
Answer
y
dy

d
Marks
 π    tan  π    cos 
 π    1   π   2 
1
 tan 1  π   
2  π    tan 1  π   
  0 , π  2  π    tan 1  π    cot  
2  π  1.2  tan 1  π  1.2  cot1.2 
π 
1 π  
π  1.2
1   π  1.2 
and 2  π  1.3 tan 1  π  1.3 cot1.3 
2
2
M1 A1 Finds derivative.
sin 
 tan 1  π     0
 tan 1  π  1.2   0.151
π  1.3
1   π  1.3
2
A1
 tan 1  π  1.3  0.395
Page 16 of 16
Puts
dy
 0 and forms equation. AG.
d
B1 Shows sign change.
5
© Cambridge University Press & Assessment 2024
Guidance
B1 Uses y  r sin 
 π    tan 1  π    sin 
1
May/June 2024