Cambridge International AS & A Level

MATHEMATICS 9709/12
Paper 1 Pure Mathematics 1 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 22 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

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Annotation Meaning

Dep Used to indicate DM0 or DM1

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

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Annotation Meaning

Premature approximation

Special case

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 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
[kx + 13 = 5 + 3x − 2x2 ] 2x2 + ( k − 3) x + 8 = 0 B1 OE
Eliminate y to obtain a three-term quadratic.

Use of b2 − 4ac  0 or b2 − 4ac = 0 with their coefficients of their new M1 OE

quadratic equation. Condone  errors only. Use of ‘  0 ’ scores M0, unless recovered.

–5 and 11 A1 Identification of correct critical values, may only be seen in

their final answer.

−5  k  11 A1 CWO
Do not allow ‘or’.
A0 if ⩽ sign or signs used.

Question Answer Marks Guidance

2(a) 5 B1 OE
4x +
x2

 dy  B1 FT Correct use of x = 1 in their two-term differentiated

 dx = 9
  expression, defined as an expression with one correct
power.

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Question Answer Marks Guidance

2(b)  5  M1 B
Their  4 x + 2  = 0 and valid method as far as ' x = ...' Equate their derivative of the form Ax  to zero, where
 x  x2
A, B ≠ 0, and solve. If no working is seen, this can be
implied by a correct answer for x.

x =−1.08 A1 AWRT

y = 9.96 A1 AWRT

Question Answer Marks Guidance

3(a)  3 
4 B1 May be seen in a full expansion.
( 2x ) 4
and  
This can be implied by 16x 4 and +
81
unless they are
 x x4
3
clearly using + throughout.
x

3  −3  2  −3 
2
 −3 
3 B1 Correct combination of numerical coefficients for the
4 ( 2x )   + 6 ( 2x )   + 4 ( 2x )   middle three terms.
 x  x  x Can be implied by a correct full expansion.

16 x 4  + k1 x 2 + k2  x0  + k3 x −2  +81x −4  M1 OE
      Powers now simplified correctly with their k1 , k2 , k3  0.

16 x 4 − 96 x 2 + 216 − 216 x −2 + 81x − 4 A1 OE

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Question Answer Marks Guidance

3(b) Use of (their 216) + 5×(their –96) only, to arrive at the coefficient of x 2 M1 Other terms may be seen.

–264 A1 Accept −264 x 2 as the final answer.

Question Answer Marks Guidance

4(a) 6  0.8 B1 45.8

Accept  12π.
360

AB 2 =102 + 102 − 2  10  10cos 0.8 or AB = 2 (10sin 0.4 ) or M1 Allow angles correctly converted to degrees for this mark.
ˆ = OBA
ˆ = 67.1.
0.8 rad = 45.8, OAB
10sin 0.8
AB =  π − 0.8 
 π − 0.8    = 1.17
sin    2 
 2  This mark can be implied by AWRT 7.8.

20.6 A1 AWRT

4(b) 1 2 B1
[Area of sector =]  6  0.8
2

1 M1 OE
[Area of triangle =]  102  sin 0.8 or 10sin0.4 10cos0.4
2 Allow use of their value of  or
1
2
 in degrees.
or other complete method.

21.5 A1 AWRT

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Question Answer Marks Guidance

5 Attempt to solve either: M1 Attempt to determine positions of first and last terms
5 + ( n − 1) 6 = 150 or 5 + ( n − 1) 6 = 400 involved

151 401 A1 OE
and Can be implied by 25 or 26 and 66 or 67.
6 6

Correct use of Sn formula with their 66 and their 25 M1

S66 or 67 − S25 or 26 M1

S66 – S25 A1

11275 A1

Alternative Method for Question 5:

Attempt to find new a and l for reduced series M1

155 and 395 A1

their 395 = their 155 + ( n − 1) 6 M1 Attempt to find n, which results in 40, 41 or 42.

n = 41 A1 CWO

their 41 M1 OE
Stheir 41 = ( their 155 + their 395)
2

11275 A1

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Question Answer Marks Guidance

6(a) ( r + 5, r + 8) B1 OE
Allow x = r + 5, y = r + 8.
If values are stated without reference to x and y, take the
first value to be their x.

6(b) B1 FT OE
their ( r + 5) + their ( r + 8) = 152
2 2
Following their answers to (a), which must both contain r.

 r 2 + 13r − 68 = 0  ( r + 17 )( r − 4 )  = 0 M1 Or other valid method of solution for their three-term

  quadratic.

 r = 4 A1 CWO
r = −4  r = 4 scores A0.

Special Case: After B1M0, r = 4 scores SCB1, but after

B1M0, r = −4  r = 4 scores B0.

6(c) their ( r + 8 )
from (a) with their r from (b), or
( their r from ( b ) ) + 8 M1 r  0 only.
their ( r + 5 ) ( their r from ( b ) ) + 5

−
3 A1 FT
OE, i.e. −
their ( r + 5)
or −
( their r from ( b ) ) + 8 .
4 their ( r + 8) ( their r from ( b ) ) + 5
2

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Question Answer Marks Guidance

7(a) M1
Use of tan 2  = sin 2 
2

cos 

Relevant use of cos2  = 1 − sin 2  at least once M1

8sin 2  − 5sin 4  A1 AG
1 − sin 2  All necessary detail needed.

7(b) Attempt to solve their 5sin 4  − 17sin 2  + 9 = 0 using a “correct” method *M1 Allow  errors in arriving at their quadratic in sin 2  . This
can be implied by either sin 2  = 2.744 or 0.6559.

A1 Condone inclusion of sin = 1.65 for this mark.

17 − 109
sin  =    0.81 0  or
10 17  109
Allow .
10
This mark can be implied by correct values.

Their 54.1, and 180 − their 54.1 or 180 + their 54.1 DM1 A correct method for obtaining a second angle within the
range 0  their 54.1  90.

54.1, 125.9, 234.1 A1 AWRT

A0 for additional values between 0 and 270.

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Question Answer Marks Guidance

8(a) a B1
ar = −120 and = 160
1− r

120 1 a M1 Elimination of either a or r.

−  = 160 or = 160
r 1− r 120 Condone  errors for this mark.
1+
a

4r 2 − 4r − 3= 0 or a2 −160a −19200= 0 A1 OE

Rearrange to arrive at a three-term quadratic.

r = − 12 only A1

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Question Answer Marks Guidance

8(b)  a = 240 B1 FT −120  ( their r ) , where −1  r  1, r  0.

 1  M1 With (their 240) and ( their r ) as long as −1  r  1, r  0.

240 1 − (− )9 
 2  Condone reversed subtraction.
160 −
 1
1−  − 
 2

Alternative Method 1 for first two marks of Question 8(b)

 a = 240 B1 FT −120  ( their r ) , where −1  r  1, r  0.

1 M1 Correctly using the 10th term as ‘a’ and the sum to infinity.
240  (− )9
2 With (their 240) and ( their r ) as long as −1  r  1, r  0.
 1 
1−  − 
 2 

Alternative Method 2 for first two marks of Question 8(b)

B1 FT
15 −120  ( their r ) , where −1  r  1, r  0.
8
[10th term =] − , 0.46875 OE
32

15 M1 With ( their a and r ) , where −1  r  1, r  0.

−
32
 1
1−  − 
 2

5 A1 5
− or − 0.3125 A0 for −0.313 without sight of − or −0.3125.
16 16
3
Condone use of r = to provide a second solution.
2

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Question Answer Marks Guidance

9(a) 6 5 M1 Substitute x = 1
and evaluate second derivative.
4
− 3
2
1 1
   
2 2

96 − 40  = 56   0  Minimum A1 CWO
Evidence and conclusion.
2
SC B1 for d y2  0 without sight of 96 − 40 or 56.
dx

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Question Answer Marks Guidance

9(b)  6 −3   5 −2  B1 B1 OE
 x  − x   +c1 
 −3   −2  B1 for each correct { }.

6 1 5 1
−3 −2 M1 Substitute x = 12 into two terms of an integrated expression
0=   −   + c1 (at least one correct power), now with c1 , and equate to 0 to
−3  2  −2  2 
find c1 .

c1 = 6 A1

k1 x −2 + k2 x −1 + k3 x  +c2  M1
Integration of their
dy
to produce at least two terms with
dx
correct powers, k1 , k2  0.

1 5 1 M1 OE
9= − + 6   + c2
1 Substitute ( 12 , 9 ) into integrated expression (at least two
2   
2
1 2
  2 correct powers) to find c2 .
2

5 −1 A1 OE
y = x −2 − x + 6x + 7 Condone their final answer being c2 = 7 if a completely
2
correct simplified expression for the equation containing
c 2 has been stated previously.

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Question Answer Marks Guidance

10(a) 2 4 3 
 2 x2  B1 B1 B1 for each correct { }.
  ( 3x + 4 ) 2 − − 6x
3 3  2 

8 3
 8 3  M1 Correct use of 7 and 0 in an expression with at least two
 A =  ( 21 + 4 )   ( 4)2 
2 − 7 − 67 −
2
terms with two correct powers.
9  9 

13 A1

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Question Answer Marks Guidance

10(b)  1
 B1
 2 ( 3 x + 4 ) −
2 3 − 2

 

 dy  *M1 dy
y − 0 =  their with x = 7  ( x − 7 ) Using x = 7, their value for and then any form of the
 dx  dx
equation of a straight line using ( 7, 0 ) .

Either:

 28  DM1 Use x = 0 in their equation of PQ.

Q is   0, 
 5 

[Area of OPQ =] 1  7   their 28  DM1

2 5  

1  28   33 A1 FT Only FT, following B0M1DM1DM1, from their (a) if

 2  7   their 5  − ( their 13 from ( a ) ) = 5
    (their 13 from ( a ))  98
5
and the area is > 0.

Or:

  4  DM1
  their  − ( x − 7)   dx
  5 

 4  x2  DM1
Evaluating their  −  – 7 x   with limits 7 and 0
 5 2 
  

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Question Answer Marks Guidance

10(b) A1 FT Only FT, following B0M1DM1DM1, from their (a) if

( their area of OPQ ) − ( their 13 from ( a ) ) =
33
5 (their 13 from ( a ))  98
5
and the area is > 0.

Question Answer Marks Guidance

11(a) 1 B1
State or imply g −1 ( x ) = ( x − k ) or equivalent
2

1 B1 OE
Obtain ( 3k + 1 − k ) = c and hence 2k + 1 = 2c
2

Or:

g ( c ) = 3k + 1 B1

[ 2c + k = 3k + 1  ] 2c = 2k + 1 B1 OE

Then:

( )
 gf ( x ) = 2 4 x 2 − c + k M1 Allow  errors only.

8x 2 − 2c + k  8 x 2 − ( 2k + 1) + k   gf ( x ) = 8 x 2 − k − 1 A1 AG
  All necessary detail needed.

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Question Answer Marks Guidance

11(b) B1 OE
8 ( x − 2) − k − 1 + 3
2
Translation coming before the stretch

( (
k their 8 ( x − 2 ) − k − 1 + 3
2
)) B1 FT OE
Stretch

(( (
h ( x ) = − k their 8 ( x − 2 ) − k − 1 + 3
2
))) B1 FT OE
Reflection

11(c) k 2 − 2k = 15 or k 2 − 2k 15 B1 OE

k = 5 only B1

11 B1 OE
c= only
2

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