Cambridge IGCSE Mathematics - Additional 0606 — 2025 May/June Paper 1 · Variant 1

This document has 16 pages. [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/11 Paper 1 Non-calculator May/June 2025 2 hours You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer all questions. ● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ● Write your name, centre number and candidate number in the boxes at the top of the page. ● Write your answer to each question in the space provided. ● Do not use an erasable pen or correction fluid. ● Do not write on any bar codes. ● Calculators must not be used in this paper. ● You must show all necessary working clearly. INFORMATION ● The total mark for this paper is 80. ● The number of marks for each question or part question is shown in brackets [ ]. * 5 6 3 8 6 5 7 0 6 7 * DC (CJ/CGW) 346450/1 © UCLES 2025 ,










, * 0000800000001 * ¬OŠ. 4mHuOªEŠ^z6€W ¬€2qN¨vU‚‡’<W¦”†‚ ¥Uu•5U¥U u UE5•U

2 0606/11/M/J/25 © UCLES 2025 List of formulas Equation of a circle with centre (a, b) and radius r. (x – a)2 + (y – b)2 = r 2 Curved surface area, A, of cone of radius r, sloping edge l. r A rl = Surface area, A, of sphere of radius r. r A r 4 2 = Volume, V, of pyramid or cone, base area A, height h. V Ah 3 1 = Volume, V, of sphere of radius r. r V r 3 4 3 = Quadratic equation For the equation ax 2 + bx + c = 0, x a b b ac 2 4 2 ! = - - Binomial theorem ( ) … … a b a n a b n a b n r a b b 1 2 n n n n n r r n 1 2 2 + = + + + + + + - - - J L KK J L KK J L KK N P OO N P OO N P OO , where n is a positive integer and ( )! ! ! n r n r r n = - J L KK N P OO Arithmetic series un = a + (n – 1)d Sn = 2 1 n(a + l) = 2 1 n{2a + (n – 1)d} Geometric series un = arn – 1 Sn = ( ) r a r 1 1 n − − (r ≠ 1) S∞ = r a 1 − (|r| < 1) Identities sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A Formulas for ∆ABC sin sin sin A a B b C c = = a2 = b2 + c2 – 2bc cos A ∆ = 2 1 ab sin C * 0000800000002 * ,










, ĬÍĊ®Ġ´íÈõÏĪÅĊßü¸þ× ĬĀ´òÏĤòúÞïñíĄÝÖìþĂ ĥąĥĕõµåµµµÕÅąĕåõµÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

3 0606/11/M/J/25 © UCLES 2025 [Turn over Calculators must not be used in this paper. 1 (a) Given that , PQ PR RQ and find 3 7 4 2 8 = - = - e e o o . [2] (b) The vectors a, b and c are such that , a i j 6 a = + b i j 4 b = + and ( ) c i j 2 5 20 a b = + + , where a and b are scalars. Given that c 3 2 a b = - , find the values of a and b. [3] * 0000800000003 * ,










, ĬÏĊ®Ġ´íÈõÏĪÅĊßú¸þ× ĬĀ³ñ×ĦîĊÛĉĀ¼ØÕĂìîĂ ĥąĕÕµÕÅÕåÅąÅąõŵåÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

4 0606/11/M/J/25 © UCLES 2025 2 Solve the inequality ( )( ) x x x 3 5 8 9 3 H - + - . [4] * 0000800000004 * ,










, ĬÍĊ®Ġ´íÈõÏĪÅĊÝü¸Ā× ĬĀ³ô×ĠĀÿàćć³öùä¼ĆĂ ĥµÅÕõÕÅõÅĥõÅÅõĥµµÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

5 0606/11/M/J/25 © UCLES 2025 [Turn over 3 Point A has coordinates ( , ) 3 1 - . A circle has equation x y 4 3 5 2 2 - + + = ` ` j j . (a) Show that A lies on the circumference of the circle. [1] (b) Given that AB is a diameter of the circle, find the coordinates of B. [2] (c) Find the equation of the tangent to the circle at A. [3] * 0000800000005 * ,










, ĬÏĊ®Ġ´íÈõÏĪÅĊÝú¸Ā× ĬĀ´óÏĪĄïÙñúöâāø¼öĂ ĥµµĕµµåĕÕĕåÅÅĕąõåÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

6 0606/11/M/J/25 © UCLES 2025 4 (a) Solve the equation x x 2 3 1 6 1 - = . [4] (b) Solve the simultaneous equations ( ) . lg x y x xy y 2 0 4 1 2 + = + + = [5] * 0000800000006 * ,










, ĬÑĊ®Ġ´íÈõÏĪÅĊàúµþ× ĬĀ³óÔĤĬõÑöċÿÂùĕĔĆĂ ĥĥĕĕµĕåµĕõąÅąĕåµõÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

7 0606/11/M/J/25 © UCLES 2025 [Turn over 5 (a) Solve the equation x 5 2 1 8 23 - + = . [3] (b) On the axes, sketch the graph of sin for y x x 5 2 0 360 ° ° G G = - . [2] y x 0 180° 360° 8 – 8 * 0000800000007 * ,










, ĬÓĊ®Ġ´íÈõÏĪÅĊàüµþ× ĬĀ´ôÜĦĨąèĄöÊĖāÁĔöĂ ĥĥĥÕõõÅÕąąÕÅąõÅõĥÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

8 0606/11/M/J/25 © UCLES 2025 6 A curve has equation y x x 1 1 2 2 4 = + - f p . (a) Show that x y d d can be written as x Ax x 1 1 2 5 2 3 + - ` ` j j , where A is a positive integer to be found. [5] * 0000800000008 * ,










, ĬÑĊ®Ġ´íÈõÏĪÅĊÞúµĀ× ĬĀ´ñÜĠĖĄÓþíÁ¸ÝģĄþĂ ĥÕµÕµõÅõĥååÅÅõĥõõÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

9 0606/11/M/J/25 © UCLES 2025 [Turn over (b) (i) Show that the curve has stationary points where , and x x x 1 0 1 =- = = . [1] (ii) Use the first derivative test to determine which two stationary points have the same nature and state whether they are maximum or minimum points. [2] * 0000800000009 * ,










, ĬÓĊ®Ġ´íÈõÏĪÅĊÞüµĀ× ĬĀ³òÔĪĚôæüĄĈĤÕ·ĄîĂ ĥÕÅĕõĕåĕõÕõÅÅĕąµĥÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

10 0606/11/M/J/25 © UCLES 2025 7 Solutions to this question by accurate drawing will not be accepted. Find the x-coordinates of the points where the curve ( ) y x x 2 9 5 42 2 = - + + ` j cuts the x-axis. [6] * 0000800000010 * ,










, ĬÑĊ®Ġ´íÈõÏĪÅĊßú·þ× ĬĀ±ñÑĞĎĖÍąöīÜßć¼öĂ ĥąÕĕµõĥõÅĕåąÅÕĥõÅÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

11 0606/11/M/J/25 © UCLES 2025 [Turn over 8 (a) Write down the set of values of x for which ( ) log x 12 4 5 - exists. [1] (b) Solve the equation ( ) log log x 12 4 125 6 1 x 5 - = + . [6] * 0000800000011 * ,









, ĬÓĊ®Ġ´íÈõÏĪÅĊßü·þ× ĬĀ²òÙĬĒĦìóċÞĀ×Ó¼ĆĂ ĥąåÕõĕąĕÕĥõąÅµąµÕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

12 0606/11/M/J/25 © UCLES 2025 9 y x A O y x 4 1 = + The point A with x-coordinate 2 lies on the curve y x 4 1 = + . The diagram shows part of this curve and the tangent to the curve at A. Find the area of the shaded region enclosed by the curve, the tangent and the x-axis. [10] * 0000800000012 * ,









, ĬÑĊ®Ġ´íÈõÏĪÅĊÝú·Ā× ĬĀ²óÙĢĤģÏíĄåÞûñìîĂ ĥµõÕµĕąµµÅąąąµåµÅÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

13 0606/11/M/J/25 © UCLES 2025 [Turn over Continuation of working space for Question 9. * 0000800000013 * ,









, ĬÓĊ®Ġ´íÈõÏĪÅĊÝü·Ā× ĬĀ±ôÑĨĠēêċíĤúăåìþĂ ĥµąĕõõĥÕåµÕąąÕÅõÕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

14 0606/11/M/J/25 © UCLES 2025 10 (a) Given that r 0 2 1 G i , show that cosec sin tan 1 1 1 2 2 i i i - + + can be written as seci. [4] (b) Given that secx a = , where r r x 2 3 2 1 G , find sinx in terms of a. [3] * 0000800000014 * ,









, ĬÍĊ®Ġ´íÈõÏĪÅĊÞû¶Ă× ĬĀ²òÔĪēīÜýú¿ÖÚáĜĆĂ ĥĕąÕµĕåõÅåÕÅÅĕĥµąÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

15 0606/11/M/J/25 © UCLES 2025 [Turn over 11 An arithmetic progression has common difference d. The 3rd term of this progression is 10. (a) Write down expressions for the 1st term and the 2nd term of this progression. Give your answers in terms of d only. [2] (b) When each of the first 3 terms is squared, the sum of these squares is 140 . There are two possible values for d. Using your answer to part (a), find the sum of the first 200 terms of the progression with the smaller value of d. [7] Question 12 is printed on the next page. * 0000800000015 * ,









, ĬÏĊ®Ġ´íÈõÏĪÅĊÞù¶Ă× ĬĀ±ñÜĠďěÝûćĊĂâõĜöĂ ĥĕõĕõõÅĕÕÕąÅÅõąõĕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

16 0606/11/M/J/25 © UCLES 2025 Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of Cambridge Assessment. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge. 12 In this question n 6 H . Use an algebraic method to show that C C n n 5 1 5 - - can be written as C n 1 4 - . [4] * 0000800000016 * ,









, ĬÍĊ®Ġ´íÈõÏĪÅĊàû¶Ą× ĬĀ±ôÜĦĝĎÚõĀāäþ×ČþĂ ĥååĕµõŵµõõÅąõåõąÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

This document consists of 18 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/11 Paper 1 Non-calculator May/June 2025 MARK SCHEME Maximum Mark: 80 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.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 2 of 18 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.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 3 of 18 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.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 4 of 18 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 Accuracy mark awarded two Accuracy mark awarded three Independent mark awarded zero Independent mark awarded one Independent mark awarded two Independent mark awarded three Benefit of the doubt Communication mark Incorrect Follow through Highlighter Highlight a key point in the working Ignore subsequent work Method mark awarded zero Method mark awarded one Method mark awarded two

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 5 of 18 Annotation Meaning Method mark awarded three Misread Omission Off-page comment Allows comments to be entered at the bottom of the RM marking window and then displayed when the associated question item is navigated to. On-page comment Allows comments to be entered in speech bubbles on the candidate response. Premature rounding/approximation Special case Indicates that work/page has been seen Transcription error Correct Correct answer from incorrect working

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 6 of 18 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 marks, awarded for a valid method applied to the problem. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. For accuracy marks to be given, the associated Method mark must be earned or implied. B Mark for a correct result or statement independent of Method marks. When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. The notation ‘dep’ is used to indicate that a particular M or B mark is dependent on an earlier mark in the scheme. Abbreviations awrt answers which round to cao correct answer only dep dependent FT follow through after error isw ignore subsequent working nfww not from wrong working oe or equivalent rot rounded or truncated SC Special Case soi seen or implied

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 7 of 18 Question Answer Marks Partial Marks 1(a) 2.5 5 −       oe B2 B1 for 2 3 1 8 7 4   − −     − +             oe, soi 1(b) 3 – 8 = 2 + 5 or 18 – 2 = 20 M1 Equates like vectors at least once  = 3  = –1 A2 A1 for either value correct 2 2 5 10 15*0 x x − − oe M1 where * is any inequality sign or = Factorises their 3-term quadratic expression or solves their 3-term quadratic equation M1 Critical values –1 and 3 A1 –1 ⩽ x ⩽ 3 A1 Do not accept separate inequalities unless connected with ‘and’ 3(a) 2 2 (3 4) ( 1 3) 5 − + −+ = or showing that distance between point A and centre of circle = radius e.g. 2 2 (3 4) ( 1 3) 5 − + −+ = B1 Accept if x-coordinate substituted to find y- coordinate 3(b) 4 1 3 2     +     − −     oe, soi or using centre and point A e.g. 3 1 4 and 3 2 2 x y + −+ = = − or solve simultaneously the line 2 5 y x = − + with the circle leading to 3 term quadratic   2 8 15 0 x x − + = with an attempt to solve M1 Award M1 for the 2 equations Award the M1 for the quadratic (5, –5) A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 8 of 18 Question Answer Marks Partial Marks 3(c) gradient of radius = 3 1 4 3 −+ = − –2 soi by gradient of tangent M1 gradient of tangent = ( ) 1 2 their − − M1 FT their –2 Allow if differentiation is used e.g.: y + 1 = 1 2 (x – 3) oe A1 FT 1 2 their − − ISW from a correct unsimplified answer Alternative use of differentiation e.g.: ( ) 2 2 2 2 3 5 ( 4) 8 11 3 d 2 8 d 2 8 11 y x y x x y x x x x + = − − = − + − − − + = − + − (M1) allow one error or use of implicit differentiation (not on syllabus) e.g.: d d 2 2 8 6 0 d d y y x y x x + −+ = leading to d 8 2 d 2 6 y x x y − = + Substitute (3, 1) − in their d d y x to get gradient = 1 2 (M1) Dep on first M1 y + 1 = 1 2 (x – 3) oe (A1) ISW from a correct unsimplified answer 4(a) Solves or factorises 1 1 3 6 2 0 x x − − = oe: ( )( ) 1 1 6 6 1 2 x x + − or when substituting 1 3 y x = factorises to ( )( ) 1 1 2 2 1 2 y y + − M1 A substitution may be used 6 6 1 1 6 3 6 2 x x     − =             scores 0 marks 1 1 log 2 3 6 1 1 then log 2 3 6 x x x x x     − = − =             scores 0 marks 1 1 6 6 1 , 2 x x   = − =   A1 x = 64 [x =1] A1 Rejects 1 6 1 x = −or x = 1 ignored at some stage, soi B1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 9 of 18 Question Answer Marks Partial Marks 4(b) 2 1 x y + = B1 Correctly eliminates one unknown M1 Dep on B1 allow unsimplified e.g. 2 1 1 2 or 4 1 x x y y x − = − = + Correct quadratic in solvable form:   2 4 0 y y − = oe or   2 2 3 1 0 x x − + = oe A1 Factorises or solves their quadratic to get two solutions M1 Dep on previous M1 y = 0, x = 1 and y = 1 4 , x = 1 2 A1 5(a) 2x – 1 = –3 and 2x – 1 = 3 M1 x = 2, x = –1 A2 A1 for either correct Alternative 2 4 4 8 0 x x − − = oe (B1) Solves or factorises their 3-term quadratic (M1) x = –1, 2 (A1) 5(b) Correct graph 0 −8 8 180° 360° 2 B1 for a correct sine curve one cycle and with correct amplitude at 3 and 7 − B1 for a correct sine curve one cycle and with correct midline y = –2

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 10 of 18 Question Answer Marks Partial Marks 6(a) Correct derivative 2 2 2 2 2 2 d 1 ( 1)(2 ) ( 1)(2 ) d 1 ( 1) x x x x x x x x   − + − − =   + +   or ( ) ( )( ) 1 2 2 2 2 2 1 2 1 1 x x x x x − − + − − + 2 M1 for an attempt at the quotient or product rule A1 fully correct Correct derivative 3 2 2 2 2 2 2 d 1 ( 1)(2 ) ( 1)(2 ) 4 d 1 ( 1) y x x x x x x x x    − + − − =    + +    or ( ) ( )( ) ( ) 3 2 1 2 2 2 2 2 1 4 2 1 2 1 1 1 x x x x x x x − −   − + − − +   +   2 Dep M1 for 3 2 2 d 1 f( ) d 1 y x k x x x   − =    +   where f(x) is their attempt to differentiate A1 fully correct Correct completion to 2 3 2 5 16 ( 1) ( 1) x x x − + A1 Alternative Correct derivatives 2 4 ( 1) u x = − 2 3 d 4( 1) (2 ) u x x = −  oe and 2 4 ( 1) v x = + 2 3 d 4( 1) (2 ) v x x = +  oe (2) M1 for either correct OR for 2 3 d 4( 1) g( ) u x x = −  where g(x) is an attempt to differentiate 2 1 x − and 2 3 d 4( 1) h( ) v x x = +  where h(x) is an attempt to differentiate 2 1 x + Correct derivative ( ) 2 4 2 3 2 4 2 3 2 2 4 d d ( 1) 4( 1) (2 ) ( 1) 4( 1) (2 ) ( 1) y x x x x x x x x = +  − − −  + + or ( )( ) ( ) ( )( ) ( ) 3 4 2 2 4 5 2 2 4 2 1 1 4 2 1 1 x x x x x x − − − + − − + (2) M1 for an attempt at the product or quotient rule Correct completion to 2 3 2 5 16 ( 1) ( 1) x x x − + (A1)

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 11 of 18 Question Answer Marks Partial Marks 6(b)(i) 2 3 ( 1) 16 0 their x x − = therefore x = 0, 2 1 0 x −= → x = 1 oe B1 FT their 16 Allow for 2 16 ( 1 ) 0 their x x − = 6(b)(ii) Any two of x –1 < x < 0 0 0 < x < 1 d d y x positive 0 negative x 0 < x < 1 1 x > 1 d d y x negative 0 positive x x < –1 –1 –1 < x < 0 d d y x negative 0 positive M1 or equivalent. Must show change in the sign of first derivative around the stationary points M0 only if second derivative is used x = 1 and x = –1 are minimum points A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 12 of 18 Question Answer Marks Partial Marks 7   3 2 2 9 10 3 0 x x x − + − = B1 Finds a correct linear factor x – 1 or x – 3 or 2x – 1 M1 Finds the correct corresponding quadratic factor For (x – 1) gives 2 2 7 3 x x − + or For (x – 3) gives 2 2 3 1 x x − + or For (2x – 1) gives 2 4 3 x x − + 2 M1 for a corresponding quadratic factor with 2 terms correct Factorises their 3-term quadratic factor or solves their 3-term quadratic equation M1 x = 1 , 2 1, 3 A1 Alternative 3 2 2 9 10 3 0 x x x − + − = (B1) Finds a correct linear factor x – 1 or x – 3 or 2x – 1 (M1) Finds a second correct linear factor x – 1 or x – 3 or 2x – 1 (M1) Finds a third correct linear factor x – 1 or x – 3 or 2x – 1 (M1) x – 1, x – 3 and 2x – 1 (A1) x = 1 , 2 1, 3 (A1) 8(a) x > 4 12 oe B1 allow 0.3 • for 4 12

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 13 of 18 Question Answer Marks Partial Marks 8(b) 5 5 5 125 log 125 3 log 125 log log 1 log log 125 x x x x x = = = or 5 log (12 4) log (12 4) log 5 x x x x − − = B1 for a correct and relevant change of base seen 5 5 log (12 4) 2log 1 x x − = + or log (12 4) 2 log 5 x x x − = + B1 Dep on correct change of base 5 2 12 4 log 1 x x − = oe or 12 4 log 2 5 x x − = oe or 2 5 5 log (12 4) log (5 ) x x − = B1 Dep on correct change of base 2 5 12 4 0 x x − + = B1 Dep on correct change of base (5 2)( 2) 0 x x − − = oe M1 dep on at least one previous B1 awarded x = 0.4, 2 A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 14 of 18 Question Answer Marks Partial Marks 9 [When x = 2] y = 3 B1 1 2 d 1 (4 1) 4 d 2 y x x − = +  M1 for 1 2 d 1 (4 1) d 2 y x k x −   = +      , k ≠ 0 [gradient of tangent =] 1 2 1 (4(2) 1) 4 2 − +  soi M1 FT their d d y x must be in the form 1 2 (4 1) k x − + , k ≠ 0 ( ) 2 3 2 3 − = − y their their x M1 Dep on first M1 [x-intercept for tangent =] 5 2 − soi A1 Must be from correct straight line equation 1 5 2 3 2 2    +      soi B1 For area of triangle – allow unsimplified [x-intercept for curve =] 1 4 − soi B1 Area below curve: ( ) 2 3 2 0.25 4 1 3 4 2 x −     +        oe M1 M1 for integration in the form of ( ) 3 2 4 1 k x + , k ≠ 0 or 4 correct use of upper and lower limits: ( ) ( ) 3 3 2 2 4(2) 1 4( 0.25) 1 6 6   + − +   −       oe M1 Dep on previous M1 for area under the curve Allow if using wrong limits but must see correct substitution Shaded area = 27 9 9 4 2 4 − = oe A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 15 of 18 Question Answer Marks Partial Marks 10(a) 2 2 sin 1 cot sec    + B1 for correct use of the trigonometry identity 2 2 cosec 1 cot   −= sin tan cos    + oe B1 sin sin cos cos      + B1 2 2 sin cos 1 sec cos cos      + = = B1 Dep on all 3 previous marks Last mark is not available if persistent missing of θ Alternative 2 2 sin 1 cot sec    + (B1) for correct use of the trigonometry identity 2 2 cosec 1 cot   −= sec sin cot cot sec      + (B1) sin cos cos sin cos 1 sin cos        +  (B1) 2 2 sin cos cos sin 1 sin      + leading to 1 cos (B1) Dep on all 3 previous marks Last mark is not available if persistent missing θ

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 16 of 18 Question Answer Marks Partial Marks 10(b) 1 cos x  = B1 could be implied by 2 2 1 1 sin x  = − 2 2 1 sin 1 x    + =     oe OR [opposite 2=] 2 1 − soi and use of opposite sin hypotenuse x = at some stage leading to  2 1 sin 1 x  =  − oe M1 2 1 sin 1 x  = − − oe A1 11(a) 10 – 2d and 10 – d B2 B1 for either ignore labels 11(b) Squares each term and forms a sum e.g.   2 2 100 40 4 100 20 100 140 d d d d − + + − + + = or   2 2 2 2 100 40 4 100 20 40 20 100 40 4 d d d d a a a − + + − + = + + + = oe M1 FT from their two terms in (a), must be both 2 terms expression in terms of d only or Correct expressions in terms of a only, must see substitution and expansion Correct equation in solvable form 2 2 5 60 160 0 5 20 60 0 d d a a − + = + − = oe A1 Factorises their 3-term quadratic expression or solves their 3-term quadratic equation M1 8 d = d = 4 A1 Finds the first term when d = 4: a = 2 M1 FT 10 – 2(their d)     200 200 4 199 4 2 S = +  oe M1 FT their a and their d 80 000 A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 17 of 18 Question Answer Marks Partial Marks 12 ! ( 1)! ( 5)!5! ( 6)!5! ( 1)! [ ( 5)] ( 5)!5! n n n n n n n n − − − − −  − − − M1 For substituting 1 5 5 and n n C C − ( 1)! 1 ( 6)!5! 5 n n n n −   −   − −   oe M1 dep on previous M1 awarded For Factorising ( 1)! ( 6)!5! n n − − or ( 1)! ( 5)!5! n n − − Do not allow if using 1 4 n C − to simplify the left-hand side ( 1)! 5 ( 6)!5! 5 n n n −     − −   oe M1 dep on at least one previous M1 awarded Simplifying fraction Completes argument: 1 4 ( 1)! 5 ( 1)! ( 6)!5! 5 ( 1 4)!4! n n n C n n n − − −  = =   − − −−   A1 Alternative 1 ! ( 1)! ( 5)!5! ( 6)!5! n n n n − − − − (M1) For substituting 1 5 5 and n n C C − 1 5 ( 6)! ( 5)! n n n − = − − or ! ( 1)! n n n =  − (B1) dep on previous M1 awarded ( 1)! [ ( 5)] ( 5)!5! n n n n −  − − − (M1) dep on previous M1 awarded factorising ( 1)! n − in the fraction 1 4 ( 1)! 5 ( 1)! ( 5)!5! ( 5)!4! n n n C n n − −  − = = − − (A1) Alternative 2 ! ( 1)! ( 5)!5! ( 6)!5! n n n n − − − − (M1) For substituting 1 5 5 and n n C C − could be implied by expansions of the factorial ( 1)( 2)( 3)( 4) 5! ( 1)( 2)( 3)( 4)( 5) 5! n n n n n n n n n n − − − − − − − − − − (B1) dep on previous M1 awarded ( 1)( 2)( 3)( 4)[ ( 5)] 5! n n n n n n − − − − − − (M1) dep on previous M1 awarded factorising ( 1)( 2)( 3)( 4) n n n n − − − − in the fraction

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 18 of 18 Question Answer Marks Partial Marks 12 Completes argument: 1 4 ( 1)( 2)( 3)( 4) ( 1)! 4! 4!( 5)! n n n n n n C n − − − − − − = = − (A1)