Cambridge IGCSE Mathematics - Additional 0606 — 2024 Oct/Nov Paper 1 · Variant 3

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 2024 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. ● You should use a calculator where appropriate. ● You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator. ● Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question. INFORMATION ● The total mark for this paper is 80. ● The number of marks for each question or part question is shown in brackets [ ]. * 2 4 3 4 2 3 0 2 0 6 * DC (PQ/FC) 336598/2 © UCLES 2024 ,










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2 0606/13/O/N/24 © UCLES 2024 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax bx c 0 2 + + = , x a b b ac 2 4 2 ! = - - Binomial Theorem ( ) a b a a b a b a b n n n r b 1 2 n n n n n r r n 1 2 2 f f + = + + + + + + - - - e e e o o o where n is a positive integer and ( )! ! ! n r n r r n = - e o Arithmetic series ( ) u a n d 1 n = + - ( ) { ( ) } S n a l n a n d 2 1 2 1 2 1 n = + = + - Geometric series u ar n n 1 = - ( ) ( ) S r a r r 1 1 1 n n ! = - - ( ) S r a r 1 1 1 = - 3 2. TRIGONOMETRY Identities sin cos A A 1 2 2 + = sec tan A A 1 2 2 = + ec cos cot A A 1 2 2 = + Formulae for ∆ABC sin sin sin A a B b C c = = cos a b c bc A 2 2 2 2 = + - sin bc A 2 1 T = * 0000800000002 * ,










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3 0606/13/O/N/24 © UCLES 2024 [Turn over 1 (a) Find the coordinates of the stationary point on the curve ( )( ) y x x 3 4 = + - . [3] (b) On the axes, sketch the graph of ( )( ) y x x 3 4 = + - , stating the intercepts with the axes. [2] O y x (c) Given that k 0 2 , write down the values of k for which the equation ( )( ) x x k 3 4 + - = has exactly 2 distinct real roots. [1] * 0000800000003 * ,










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4 0606/13/O/N/24 © UCLES 2024 2 On the axes, sketch the graph of sin y 4 5 2 i = + , for ° ° 360 360 G G i - . State the intercept with the y-axis. [4] 0 y -360° -180° -10 180° 360° 10 i * 0000800000004 * ,










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5 0606/13/O/N/24 © UCLES 2024 [Turn over 3 Find the values of k for which the equation x k kx 4 4 2 2 - = - has no real roots. [4] * 0000800000005 * ,










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6 0606/13/O/N/24 © UCLES 2024 4 (a) Write log log a b 3 4 2 2 + - as a single base 2 logarithm. [3] (b) Solve the equation lg log x 4 10 x = . [4] * 0000800000006 * ,










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7 0606/13/O/N/24 © UCLES 2024 [Turn over 5 The polynomial p is such that ( )x ax bx x c 19 p 3 2 = + - + , where a, b and c are integers. It is given that x 2 + is a factor of ( )x p . When ( )x p is divided by x 1 + the remainder is 20. (a) Show that a b 7 3 39 - = . [3] It is also given that when ( )x pl is divided by x 1 - the remainder is 1. (b) Find the values of a, b and c. [3] * 0000800000007 * ,










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8 0606/13/O/N/24 © UCLES 2024 6 The table shows the variables x and y which are related by the equation y Abx2 = , where A and b are constants. x 1 1.5 2 2.5 3 y 14 33.3 112 532.8 3584 (a) Use the data to draw a straight line graph of lny against x2. [2] 0 1 1 2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 x2 ln y * 0000800000008 * ,










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9 0606/13/O/N/24 © UCLES 2024 [Turn over (b) Use your graph to estimate the values of A and b. Give your answers correct to 1 significant figure. [5] (c) Use your graph to estimate the value of x when y 200 = . Give your answer correct to 2 significant figures. [2] * 0000800000009 * ,










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10 0606/13/O/N/24 © UCLES 2024 7 (a) Given that ln y x x 3 = , find x y d d . [2] (b) Hence find ln x x x 3 d 2 1 2y , giving your answer in the form lna b + , where a is an integer and b is a rational number. [4] * 0000800000010 * ,










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11 0606/13/O/N/24 © UCLES 2024 [Turn over 8 The straight line y x 2 1 = + intersects the curve y xy x 3 15 2 + + = at the points A and B. The point C with coordinates , k 10 21 e o lies on the perpendicular bisector of AB. (a) Find the exact value of k. [8] (b) The point D lies on the perpendicular bisector of AB such that its perpendicular distance from AB is twice that of the point C from AB. Find the possible coordinates of D. [4] * 0000800000011 * ,









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12 0606/13/O/N/24 © UCLES 2024 9 A O X D B C 2a c 4a The diagram shows the trapezium OABC, where OA 4a = , OC c = , and CB 2a = . The point D lies on AB such that : : AD DB 2 1 = . The point X is the point of intersection of the lines OD and AC. It is given that AX AC m = and X O OD n = . Find in terms of a and c (a) AB [1] (b) OD. [2] (c) Find OX in terms of a, c and n. [1] (d) Find X A in terms of a, c and m. [2] * 0000800000012 * ,









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13 0606/13/O/N/24 © UCLES 2024 [Turn over (e) Hence find the values of m and n. [4] * 0000800000013 * ,









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14 0606/13/O/N/24 © UCLES 2024 10 (a) Solve the equation tan tan 7 5 2 0 2i i + - = , for 180 180 ° ° G G i - . [4] (b) Solve the equation ( . ) sin 3 3 1 5 2 0 z- - = , for 0 3 1 1 z , where z is in radians. [5] * 0000800000014 * ,









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15 0606/13/O/N/24 © UCLES 2024 11 (a) The first 3 terms of an arithmetic progression are log 3 x , log 81 x , log 2187 x . Find the sum to n terms, giving your answer in the form log k 3 x , where k is in terms of n. [3] (b) The first 3 terms of a geometric progression are 1, tan 3 2i, tan 9 4i, for r 0 2 1 1 i . Find the values of i for which this geometric progression has a sum to infinity. [4] * 0000800000015 * ,









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16 0606/13/O/N/24 © UCLES 2024 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. BLANK PAGE * 0000800000016 * ,









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This document consists of 8 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 2024 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 October/November 2024 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 2 of 8 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.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 3 of 8 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. 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

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 4 of 8 Question Answer Marks Guidance 1(a) 2 12 = − − y x x d 2 1 d = − y x x or ( ) ( ) 3 4 + + − x x or 2 1 49 2 4   = − −     y x or using symmetry 4 3 2 − = x M1 For expanding the brackets and differentiate with at least one correct term or for using the product rule or for completing the square or for using symmetry 1 2 = x A1 49 4 = − y oe A1 1(b) 2 B1 for the correct shape. Must have the parabola part of the curve with maximum in the first quadrant and cusps on the x- axis. Ignore labelling of their maximum point if incorrect coordinates B1 for correct intercepts. Must be correct shape 1(c) 49 4  k oe B1 FT on 49 4 − their excluding k >12 2 4 B1 for correct shape must be a curve with one min in 3rd quadrant and one max in first quadrant and correct endpoints ( 360,4) − and (360,4) Ignore labelling of their maximum point if incorrect coordinates. depB1 for intercept of 4 on y-axis. Must have the correct shape depB1 for max in correct position of ( ) o 180 , 9 . Must have the correct shape depB1 for min in correct position of ( ) o 180 , 1 − − . Must have the correct shape 3   2 4 4 2 0 − − + = x kx k B1 soi 2 2 + − k k Critical values 2, 1 − 2 M1 for use of discriminant on their three- term quadratic equation to obtain two critical values 2 1 −  k A1 Strict inequality

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 5 of 8 Question Answer Marks Guidance 4(a) 2 3 log 8 = B1 4 2 8 log a b 2 M1 for correct use of two operations from multiplication, division or power rule for logs to the base of 2. A1 for log to the base of 2 only 4(b) 4 lg lg = x x or 1 4log 10 log 10 = x x B1 Change of base ( ) 2 lg 4 = x or ( ) 2 1 log 10 4 = x B1 Dep on correct change of base Must work with ( ) 2 lg x or ( ) 2 log 10 x not 2 log x or ( ) log 100 x 100 = x B1 Dep on correct change of base 1 100 = x or 0.01 B1 Dep on correct change of base 5(a) ( ) p 2 : 8 4 38 0 − − + + + = a b c M1 For substitution of 2 − in p(x) and equating to zero. Allow one sign error in evaluating ( ) p 1 : 19 20 − −+ + + = a b c M1 For substitution of 1 − in p(x) and equating to 20. Allow one sign error in evaluating 7 3 39 − = a b A1 AG – must be from correct work 5(b) ( ) p 1 :3 2 19 1  + − = a b M1 For substitution of 1 in ( ) p x  and equating to 1 Allow one sign error in evaluating. Can be unsimplified 6, 1, 6 = = = a b c 2 M1 dep for solution of their equation with that from (a) to find at least one unknown. 6(a) x2 1 2.25 4 6.25 9 ln y 2.64 3.51 4.72 6.28 8.18 2 M1 for plotting points with one error

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 6 of 8 Question Answer Marks Guidance 6(b) 2 ln ln ln = + y x b A B1 May be seen in part (a) Gradient = lnb ( ) ln 0.7 = b 2 = b 2 M1 for attempt to find the gradient and equate to lnb Gradient must be from linear graph of lny vs x2 Intercept = ln A ( ) ln 1.95 = A 7 = A 2 M1 for attempt to use intercept 6(c) When 200, ln 5.3 = = y y 2 4.85 = x 2.2 = x (allow 2.1 or 2.3) 2 M1 for using their linear graph with ln 5.3 = y to obtain a value for 2 x A0 for 2.2 =  x if –2.2 is not rejected 7(a) 2 2 3 ln + x x x 2 M1 for attempt to differentiate a product Allow unsimplified for 2 marks 7(b) 2 3 2 3 ln d ln d = −   x x x x x x x B1 2 3 3 1 ln 3   −     x x x 8 1 8ln2 3 3 − + M1 Dep on B1 M1 for correct use of limits 7 ln256 3 − 2 A1 for one correct term 8(a) 2 5 3 14 0 + − = x x or 2 5 4 57 0 − − = y y M1 soi 7 19 , 5 5 = = x y 2, 3 = − = − x y 3 M1 for attempt to solve their quadratic to obtain either x = ... or y = … A1 for one correct pair both x or both y or one correct (x, y) point Midpoint 3 2 , 10 5   −     B1 Must be correct midpoint Gradient of perpendicular 1 2 − B1 Must be correct Perp bisector: 2 1 3 5 2 10   − = − +     y x M1 Must be using their midpoint and a gradient = 1 2 − 4 5 = − k A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 7 of 8 Question Answer Marks Guidance 8(b) 9 , 2 2   −     2 B1 for one correct FT on ( ) 2 2 5  − their k their 51 14 , 10 5   −     2 B1 for one correct FT on ( ) 2 3 2 5      −          their their k 9(a) 2 − c a B1 9(b) ( ) ( ) 2 4 2 3 + − their a c a oe M1 Alternative route: OC + CB + BD = c + 2a 1 3 − their AB 8 2 3 3 + a c A1 Allow unsimplified 9(c) 8 2 3 3     +         their a c B1 Must be in terms of a and c in a valid vector form. Allow unsimplified 9(d) 4 = − AC c a B1 ( ) ( ) 4  − their c a B1 Must be in terms of a and c in a valid vector form 9(e) 4 = a their (c) – their (d) oe M1 Must be in terms of a and c in a valid vector form 1 3 , 2 4   = = 3 M1 dep on first M1 for equating like vectors once M1 dep on first M1 for attempt to solve 2 simultaneous equations in and .  leading to or   = = A1 for both 10(a) 2 tan , tan 1 7   = = − 2 M1 for attempt to factorise or use formula to obtain tan ... = o o o o 15.9 , 164.1 , 45 , 135 − − 2 A1 for two correct solutions A1 for a further 2 correct solutions and no extras in the range

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 8 of 8 Question Answer Marks Guidance 10(b) ( ) 2 sin 3 1.5 3 − = 3 1.5 0.7297 − = M1 Correct order of the operation Do not accept in degrees    3 1.5 2.41 12 , 7.01 3 − = A1 soi by correct answers with no extras within the range 0.743, 1.30, 2.84 3 M1 dep on first M1 for correct order of operations or one correct solution A1 for one solution A1 for a further 2 correct solutions and no extras in the range Do not accept in degrees 11(a) 3 3log 3 or log 3 = x x d nfww B1 Must be exact Allow log 27 = x d nfww ( ) ( ) 2log 3 3 1 log 3 2 + − x x n n nfww M1 For use of sum formula with their d must be in the form of log 3 x ( ) 3 1 log 3 2 − x n n or 2 3 log 3 2 2   −     x n n A1 Must be in the form of log 3 x k 11(b) 2 3tan  = r B1 soi 2 3tan 1  or   2 1 3tan 1  −  or 2 1 1 tan 3 3    −       or   2 1 0 tan 3    B1 1 tan 3  or 1 0 tan 3    B1 π 0 6    B1