Cambridge International AS & A Level

MATHEMATICS 9709/15
Paper 1 Pure Mathematics 1 May/June 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 for
Teachers.

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

Cambridge International is publishing the mark schemes for the May/June 2025 series for most
Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level
components.

This document consists of 24 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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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

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

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

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

Judgment made by the PE

Premature approximation

Special case

Indicates that work/page has been seen

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Annotation Meaning

Error in number of significant figures

Correct

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 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 B1 B1 OE
 y =  ( 2 x − 5)3 
12 8 2
+ x   +c  Terms may be unsimplified.
 3  2  2 
 
May see  y = 16 x3 −116 x 2 +300 x+c.
4 = 2×(2×2 – 5)3 + 4×22 + c [⇒ c = –10] M1 Sub (2, 4) correctly into their integrated expression
with a ‘+ c’.

y or f or f ( x ) = 2 ( 2 x − 5) + 4 x2 − 10 A1 May see y = 16 x3 − 116 x 2 + 300 x − 260.

3

' y = or 'f ( x ) = or 'f = ' can be implied if seen in

working.

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Question Answer Marks Guidance

2 x2 coeff = 10  33  a2 and 6  62  ( −1) allow correct terms with x2 B1 Expect 270a 2 and 216.
2

Terms may be seen separately.

Allow if seen in an expansion. Combinations must be
evaluated.

x coeff = 5  34  a and 4  63  ( −1) allow correct terms with x B1 Expect 405a and − 864.
Terms may be seen separately.
Allow if seen in an expansion. Combinations must be
evaluated.

270a 2 + 216 = 6 ( 405a − 864 ) M1 OE

For forming a correct quadratic equation using their 4
terms only.

45a2 − 405a + 900 = 0 ⇒ 45 ( a − 4 )( a − 5) = 0 DM1 For evidence of a correct method of solving.

If quadratic formula used a full substitution must be
seen.

a = 4 or a = 5 A1 Only dependent on the first M1.

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Question Answer Marks Guidance

3(a)  1
2

2
1  1  M1 OE
4  x −   −2 = 0 or ( 2 x − 1)  −2 = 0 or  x −   − = 0
2
Must deal with the coefficient of x2 and x correctly to
 2  2  2 
produce an ( ax + b ) term.
2

( )
1 1 A1 1
x=  OE, e.g. x = 1 2 .
2 2 2
SC B1 only for correct solutions from another method.

M1 Setting tan = their x for at least one value of their x.

( )
3(b) 1
tan = 1 2 May restart and solve the quadratic in tan .
2

[ =] 50.4, 168.3 AWRT and no other answers in the range 0    180 A1 A1 SC A1 only for AWRT 0.879 and 2.94 radians.
SC M0 B1 B1 for answers only.
SC M0 B1(only) for both answers only in radians.

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Question Answer Marks Guidance

4 Intersects x-axis when x4 = 1  x = 1 B1 WWW

May be seen without working.
May be seen as a limit in the integration.
Allow ±1 if x = 1 is seen as the lower limit in the
integration.

 2 1 
2
 4 1  *M1 1
V =  π  y dx =  π   x − 2  dx =  π   x − 2 + 4  dx with attempt at
With some attempt at squaring (accept x 4 
   as
2

 x   x  x4
integration evidence).

1 5 1 A1
x − 2x − 3
5 3x

 1 5 1  DM1 Use of limits, x = 2 and their x = 1, min evidence if

 π 
1  1
 2 − 2 2 − − − 2 − 
3   283 32
 5 3 2   5 3  correct answer: + . If answer incorrect,
120 15
substitution of limits must be clear.
DM0 for use of the limit x = 0.

π × 4.49… = 14.11 A1 AWRT, WWW

539
Allow π or 14.1, dependent on the first M1 and
120
A1.

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Question Answer Marks Guidance

5(a) π B1 OE
Angle CAD = or 60o, CD = 75 or 5 3 ˆ or CD. Allow CAD
ˆ = 1.05, CD = 8.7
3 For either CAD
or 8.66.

B1 B1 OE
 π

[Perimeter =] 10   + 75 + 5 or 
 3

 60
 360


 2π (10 )  + 75 + 5

 B1 for the two sides and B1 for arc length (allow 10.5).

= 24.1 cm B1

5(b) 1 π 1 60 1 M1 OE
Area =  102  −  5  75 or  π  102 −  5  75 Use of sector area formula minus triangle area
2 3 2 360 2
formulae, with their angle CAD and their side CD.

= 30.7 cm2 A1

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Question Answer Marks Guidance

6(a) Eleventh birthday: Father 10+10×5 [= 60] M1 For correct use of AP formula OE.

Mother 10×1.210 [= 61.9(2)] M1 For correct use of GP formula OE.

60 and 61.9 A1 Both answers. Accept 2 sf answers.

6(b) 18 M1A1 A1 may be implied by a correct final answer.

Father ( 2  10 + 17  5) = 945
2

Mother
(
10 1 − 1.218 ) = 1281.17
M1A1 A1 may be implied by a correct final answer.

(1 − 1.2 )
Total = 2226.17 A1 Accept 3 or more sf answers.
Ignore S17 ( = 1909.3) as an extra answer.

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Question Answer Marks Guidance

7(a) 7− p 2 2 M1 OE
= − or use of straight line equations with m = − , ( x, y ) = ( 3,7 )
3−6 3 3  2
Expect y − 7 =  −  ( x − 3) with (6, p) substituted or
with (6, p) substituted.  3
−2
y= x + 9 with (6, p) substituted.
3

p=5 A1

7(b) ({4}, {their 2p-7}) B1 WWW

B1FT Extra solutions lose both marks.

7(c) Mid-point AB is (4.5, 6) B1

  B1
 −1  3
Gradient of perp bisector  =  =
 −2 2
 3

3 M1 OE. Must be using their mid-point and their

Equation y − 6 = ( x − 4.5) perpendicular gradient.
2

Crosses axes at (0, − 0.75), (0.5, 0) DM1 Correct use of their perpendicular bisector equation to
find the x- and y-intercept.

Area = 0.1875 (accept 3 sf accuracy) A1 3

OE, e.g. . Last 2 marks can be gained by integrating
16
the line equation between zero and 0.5.

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Question Answer Marks Guidance

8(a) 9 =1+ a + b + 5 B1

dy B1
= 3x 2 + 2ax + b
dx

Gradient = 0 at (1, 9) so 0 = 3 + 2a + b M1 dy
Setting their to zero and substituting x = 1.
dx

Attempt to solve their linear equations simultaneously DM1 Can be implied by their answers.

a = −6, b = 9 A1 WWW

8(b) dy M1 dy
= 3x 2 − 12 x + 9 = 0 Setting their to zero.
dx dx

Solution [leading to x = 1 or x = 3] DM1 Solving their 3-term quadratic

(3, 5) or x = 3, y = 5 A1 WWW
Ignore (1, 9 ) if given as a second answer.
Only dependent on the first M1.

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Question Answer Marks Guidance

8(c) dy M1 dy
At x = 5, = 3  52 − 12  5 + 9 Substituting x = 5 into their . May be implied.
dx dx

dx dt M1 OE
6 = their 24  or their 24 = 6 
dt dx  dy dy dx 
Use of chain rule SOI  =   .
 dt dx dt 
dx dt
Linking correctly (or ), their 24 and 6.
dt dx

dx 1 A1 OE
=
dt 4

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Question Answer Marks Guidance

9(a) {Stretch}{factor 3} {in y-direction} B2,1,0 If 2 or more stretches or extra transformations, give B0
for stretches.

 π  B2,1,0 π

{Translation or shift}   Translation may be split to   before the stretch and
 2  0
0
  after the stretch.
 2
If both vectors correct but ‘translation’/’shift’ not
stated, then B1 only for the translations.

Alternative Method for Question 9(a)

 π  0 B2,1,0 If two or more stretches or any extra incorrect

  π
{Translation or shift}   2   or   and  2  transformation is given, then B0 for the stretches.
0   If order incorrect, maximum of 3/4.
 3  3
  

followed by {stretch} {factor 3} {in y direction} B2,1,0

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Question Answer Marks Guidance

9(b) 7 y
B1† Graph of f(x) with correct domain.
6

4 B1† For g(x) being a decreasing function in the domain π to

3 2π.
2

1
x B1 Domain π to 2π for g(x).
π/2 π 3π/2 2π
−1

−2
B1 Range for g approximately correct (should be from 5 to
−1).
−3

4 †Sketches must be curves and have zero gradient at the

ends of the given domains.

9(c) π 1 B1 May be implied by correct substitution later.

[g-1] f   = [g-1 ]  
3
  2

 x−2 −1  y − 2 
M1 Finding inverse of g.
[g −1 ( x ) or y =] cos −1   + π or  x =  cos  +π Allow one sign error.
 3   3 
1
Alt: 3cos ( x − π ) + 2 =
2

1  DM1 1
 −2 Substituting their x = into their expression for
 1 
[g −1   =] cos −1  2 +π
2
2  3  g −1 ( x ) .
 
1 2π
Alt: Use of arccos for cos ( x − π ) = − ⇒ x−π=
2 3

2π 5π A1
[g −1 ( x ) =] +π= or 5.24 AWRT
3 3

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Question Answer Marks Guidance

9(d) The domain of f does not include the whole of the range of g B1 OE, but must mention range of g and domain of f.

Alternative Method for Question 9(d)

Show clearly that a particular value or set of values in the domain of g gives a B1 Value of x for substitution into g ( x ) must be in the
value of g ( x ) which is outside the domain of f ranges: π x  4.322 or 5.447  x 2π.

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Question Answer Marks Guidance

10(a) Centre is (–2, 4) B1

8−4 M1 OE, e.g. using equation of the radius.

Gradient of perpendicular = [ = 1]
2 − ( −2 )

Equation of tangent is y – 8 = –1(x – 2) M1 Using a correct point and their tangent gradient from
dy
an attempt to find , or the negative reciprocal of
dx
their perpendicular gradient.

x + y − 10 = 0 or –x – y + 10 = 0 A1 CAO – this form is required.

Answer without working, allow maximum of 2/4.

Alternative Method for Question 10(a)

1 1 B1 OE; a correct expression for y.

(
y =    28 − x 2 − 4 x ) 2
(
+ 4 or y =    32 − ( x + 2 ) )
2 2
+4

−1 M1 OE
dy
dx
1
2
(
=    28 − x 2 − 4 x ) 2
( −2 x − 4 ) [= − 1 at x = 2] Differentiating y with no more than one sign error.

Equation of tangent is y – 8 = –1(x – 2) M1 Using a correct point and their tangent gradient from
dy
an attempt to find , or the negative reciprocal of
dx
their perpendicular gradient.

x + y − 10 = 0 or –x – y + 10 = 0 A1 CAO – this form is required.

Answer without working, allow maximum of 2/4.

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Question Answer Marks Guidance

Alternative Method 2 for Question 10(a)

dy dy B1 Differentiating implicitly.
2x + 2 y + 4−8 = 0
dx dx

dy ( 2 x + 4 ) M1 dy
 −1 at ( 2, 8)
= = Rearranging to make the subject,
dx ( 8 − 2 y ) dx
dy
May substitute (2, 8) and then rearrange to find .
dx

Equation of tangent is y – 8 = –1(x – 2) M1 Using a correct point and their tangent gradient from
dy
an attempt to find , or the negative reciprocal of
dx
their perpendicular gradient.

x + y − 10 = 0 or –x – y + 10 = 0 A1 CAO – this form is required.

Answer without working, allow maximum of 2/4.

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Question Answer Marks Guidance

10(b) ( k − 3 y )2 + y 2 + 4 ( k − 3 y ) − 8 y − 12 = 0 M1* k−x

Sub x = k − 3 y or y = into circle equation.
or ( k − 3 y + 2) + ( y − 4) = 32
2 2 3

y=
k−x ( k − x ) 2 − 8 k − x + x 2 + 4 x − 12 = 0
gives:
3 9 3
2
k−x 
or ( x + 2 ) + 
2
− 4  = 32.
 3 

9 y 2 − 6ky − 20 y + y 2 + k 2 + 4k − 12 = 0 DM1 OE
All squared brackets expanded to give a quadratic
equation in y (or x, which gives:
9x2 + 60x − 2kx + k 2 + x2 − 24k −108 = 0 or
x 2 + k 2 − 2kx 8k − 8 x
− + 16 + x 2 + 4 x + 4 = 32. )
9 3
2
(
b 2 − 4ac = ( 6k + 20 ) − 4  10  k 2 + 4k − 12 ) M1** OE
Factorising out y (or x) from their quadratic (this is the
quadratic where the previous DM1 or possibly DM0
was awarded) to identify ‘b’ and then finding the
discriminant. From equation in x, this is:
(
( 60 − 2k )2 − ( 4 )(10 ) k 2 − 24k − 108 . )
−4k 2 + 80k + 880  0 DM1 OE
Simplify the discriminant and setting it to less than
zero. From eliminating y this is:
−36k 2 + 720k + 7920  0
Only dependent on the previous M1.

k 2 − 20k − 220  0 A1 AG
WWW

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Question Answer Marks Guidance

10(b) Alternative Method for Question 10(b)

Centre of circle (–2, 4) and radius of circle is 32 B1

1( −2 ) + 3( 4 ) − k M1* Correct use of ‘distance of a line from a point formula’

Distance of x + 3 y − k from ( −2,4 ) = with their centre coordinates.
(12 + 32

1( −2 ) + 3 ( 4 ) − k DM1 Setting distance to be greater than their radius.

Setting  32
(12 + 32 )

Squaring both sides DM1 Removing the square roots.

Rearranging to k 2 − 20k − 220  0 A1 AG

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