Cambridge IGCSE Mathematics - Additional 0606 — 2023 May/June Paper 1 · Variant 3

This document has 16 pages. [Turn over ADDITIONAL MATHEMATICS 0606/13 Paper 1 May/June 2023 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 [ ]. Cambridge IGCSE™ * 5 4 3 4 4 6 4 5 0 0 * DC (PQ/CGW) 312489/1 © UCLES 2023

2 0606/13/M/J/23 © UCLES 2023 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 =

3 0606/13/M/J/23 © UCLES 2023 [Turn over 1 (a) Write down the period, in radians, of tan 3 2 3 i - . [1] (b) On the axes, sketch the graph of tan y 3 2 3 i = - for r r G G i - , stating the coordinates of the points where the graph meets the axes. [3] y 0 r r - i

4 0606/13/M/J/23 © UCLES 2023 2 (a) Write x x 2 5 3 2 + + in the form ( ) x a b 2 2 + + , where a and b are rational numbers. [2] (b) Hence write down the coordinates of the stationary point on the curve y x x 2 5 3 2 = + + . [2] (c) Solve the inequality x x 2 5 3 8 15 2 1 + + . [3]

5 0606/13/M/J/23 © UCLES 2023 [Turn over 3 (a) Write lg lg a b 3 2 2 1 4 2 + - ` j, where a and b are both positive, as a single logarithm to base 10. Give your answer in its simplest form. [3] (b) Given that log log c 2 3 7 4 c 3 = + , find the possible values of the positive constant c, giving your answers in exact form. [5]

6 0606/13/M/J/23 © UCLES 2023 4 The straight line y x 3 11 = - and the curve xy x x 4 3 2 2 = - - intersect at the points A and B. The point C, with coordinates (a, 8 - ) where a is a constant, lies on the perpendicular bisector of the line AB. Find the value of a. [8]

7 0606/13/M/J/23 © UCLES 2023 [Turn over 5 (a) Find the first three terms in the expansion of x x 4 2 2 10 - e o in descending powers of x. Give each term in its simplest form. [3] (b) Hence find the coefficient of x16 in the expansion of x x x x 4 2 2 2 10 2 2 2 - + e e o o . [3]

8 0606/13/M/J/23 © UCLES 2023 6 In this question lengths are in centimetres and angles are in radians. A B O r i The diagram shows a circle with centre O and radius r. The points A and B lie on the circumference of the circle. The area of the minor sector OAB is 25 cm2. The angle AOB is i. (a) Find an expression for the perimeter, P, of the minor sector AOB, in terms of r. [3]

9 0606/13/M/J/23 © UCLES 2023 [Turn over (b) Given that r can vary, show that P has a minimum value and find this minimum value. [4]

10 0606/13/M/J/23 © UCLES 2023 7 The table shows values of the variables x and y which are related by an equation of the form y Axb = , where A and b are constants. x 1.5 2 2.5 3 4 y 13.8 27.5 46.9 72.6 145 (a) Use the data to draw a straight line graph of lny against lnx. [3] 1 0 2 3 0.5 1 2 ln x 1.5 4 5 ln y 6

11 0606/13/M/J/23 © UCLES 2023 [Turn over (b) Use your graph to estimate the values of A and b. [5] (c) Estimate the value of x when y 100 = . [2]

12 0606/13/M/J/23 © UCLES 2023 8 B Y X A O The diagram shows the triangle OAB with OA a = and OB b = . The point X lies on the line OA such that OX a 5 3 = . The point Y is the mid-point of the line AB. Find, in terms of a and b, (a) AB [1] (b) XY . [2] B Y Z X A O The lines OB and XY are extended to meet at the point Z. It is given that YZ XY m = and BZ b n = .

13 0606/13/M/J/23 © UCLES 2023 [Turn over (c) Find XZ in terms of m, a and b. [2] (d) Find XZ in terms of n, a and b. [2] (e) Hence find the values of m and n. [3]

14 0606/13/M/J/23 © UCLES 2023 9 In this question lengths are in centimetres and time is in seconds. A particle P moves in a straight line such that its displacement s, from a fixed point at a time t, is given by ( )( ) s t t 3 2 4 2 = + - for t 0 5 G G . (a) Find the values of t for which the velocity, v, of P is zero. [4] (b) On the axes below, sketch the displacement–time graph of P, stating the intercepts with the axes. [3] s t 0 5

15 0606/13/M/J/23 © UCLES 2023 [Turn over (c) On the axes below, sketch the velocity–time graph of P, stating the intercepts with the axes. [2] v 0 5 t (d) (i) Find an expression for the acceleration of P at time t. [1] (ii) Hence, on the axes below, sketch the acceleration–time graph of P, stating the intercepts with the axes. [2] acceleration 0 5 t Question 10 is printed on the next page.

16 0606/13/M/J/23 © UCLES 2023 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. 10 (a) Show that cos sin cos 1 2 4 4 2 i i i - + = . [3] (b) Solve the equation cos sin 3 3 1 2 1 4 4 z z - + = , for r r 3 3 1 1 z - , giving your answers in terms of r. [5]

This document consists of 9 printed pages. © UCLES 2023 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/13 Paper 1 May/June 2023 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 2023 series for most Cambridge IGCSE, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 2 of 9 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 descriptors 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 May/June 2023 © UCLES 2023 Page 3 of 9 Maths-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, 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 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 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.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 4 of 9 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 Question Answer Marks Guidance 1(a) 2π B1 1(b) 3 B1 for the correct shape, must be tending correctly towards the asymptotes B1 for   0, 3  B1 for π , 0 2       2(a) 2 5 1 2 4 8 x         oe 2 B1 for 2 5 2 4 x        B1 for 1 8  2(b) 5 1 , 4 8         oe 2 B1 FT for each on their (a).

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 5 of 9 Question Answer Marks Guidance 2(c) Use of their (a) or expansion to 3 term quadratic (= 0), to obtain two critical values. M1 9 1 , 4 4   A1 For both critical values 9 1 4 4 x    A1 3(a) 2 500 lg a b oe 3 B1 for 2 lga , lg2b  or lg1000 B2 for 2 lg 2 a b , 2 lg1000a or 1000 lg 2b 3(b) 3 1 log log 3 c c  B1     2 2 log 3 7log 3 4 0 c c        2log 3 1 log 3 4 0 c c    1 log 3 , log 3 4 2 c c   M1 Attempt to obtain a 3 term quadratic equation (= 0) and attempt to solve to obtain log 3 ... c  Allow one sign error 1 2 3 c  4 3 c  M1 Dep for attempt to solve at least one of their log equations 1 4 1 , 3 9 c c   or exact equivalents 2 A1 for each Alternative Method 3 1 log 3 log c c  B1     2 3 3 4 log 7log 2 0 c c        3 3 4log 1 log 2 0 c c    3 3 1 log , log 2 4 c c   M1 Attempt to obtain a 3 term quadratic equation = 0 and attempt to solve to obtain 3 log ... c  Allow one sign error 1 4 1 3 , c 9 c   or exact equivalents 3 M1 dep for attempt to solve at least one of their log equations. A2 for both or A1 for either

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 6 of 9 Question Answer Marks Guidance 4   2 5 8 4 0 x x    or   2 5 36 305 0 y y    M1 For attempt to eliminate one variable to obtain a 3 term quadratic equation (= 0). Allow one sign error. 2 , 2 5 x x   61, 5 5 y y   3 Dep M1 for attempt to solve their quadratic equation A1 for any correct pair A1 for a second correct pair. Mid-point 4 43 , 5 5        M1 For attempt to find mid-point using their coordinates Gradient of perpendicular = 1 3  B1 43 1 4 5 3 5 y x          M1 For attempt to find the equation of the perpendicular bisector using their perpendicular gradient and their midpoint. Allow alternative methods 1 a  A1 5(a) 20 16 12 40 720 x x x   3 B1 for each correct term 5(b) 4 4 4 4 x x         B1 Allow unsimplified   4 40 4 720 their their     soi M1 Must have 3 terms 564 A1 6(a) 2 1 25 2 r  2 50 r  B1 50 2 P r r   2 M1 for use of 2 P r r   with attempt to eliminate 

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 7 of 9 Question Answer Marks Guidance 6(b) 2 d 50 2 d P r r   M1 For attempt to differentiate their P, must be in terms of r. When d 0, 5 d P r r   A1 2 2 3 d 100 d P r r  When 2 2 d 5, d P r r  is positive so a minimum oe. B1 Allow alternative valid methods Minimum P = 20 A1 7(a) ln x 0.41 0.69 0.92 1.1 1.4 ln y 2.6 3.3 3.8 4.3 5 3 M1 for attempt to find ln values of all and plotting the graph. A1 for 4 correct points. A0 for fewer than 4 correct points. 7(b) ln ln ln y A b x   soi B1 Allow if seen in (a) Gradient = b = 2.4 Allow 2.3 to 2.5 2 Dep M1 for attempt to find the gradient of their graph, must have M1 in part(a) Must be using points on their graph. ln 1.6 A  5 A  Allow awrt 4.8 to awrt 5.5 2 Dep M1 for attempt to find the intercept on the vertical axis and equate to ln A. must have M1 in part (a). 7(c) 3.42 x  Allow 3.2 to 3.5 2 M1 for a correct attempt to find the estimate either using their graph or their equation. 8(a) b – a B1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 8 of 9 Question Answer Marks Guidance 8(b)     2 1 5 2 their   a b a soi oe M1 1 1 2 10  b a A1 Allow unsimplified 8(c)   1 1 1 2 10           b a oe 2 M1 for   1 1 1 2 10 their           b a A1 allow unsimplified 8(d)   3 1 5     a b oe 2 B1 for each vector, allow unsimplified. 8(e)     1 1 3 1 1 2 10 5               b a a b 5  2  3 M1 for equating   1 1 1 2 10 their           b a and their   3 1 5     a b and then equating like vectors at least once. A1 for each. 9(a)   9 4 v t t   oe 2 M1 for correct attempt to differentiate, allow one arithmetic error. 0 t  , 4 t  2 Dep M1 for equating their v to zero and attempt to solve. A1 for both. 9(b) 3 B1 for correct cubic curve for given domain B1 for   0, 96 and no other intercept on the y-axis B1 for touching at   4, 0 and no other intercept on the x-axis 9(c) 2 B1 for a correct quadratic curve for the given domain, starting from the origin. B1 for   4, 0 and no other x- intercept 9(d)(i) 18 36 t  B1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 9 of 9 Question Answer Marks Guidance 9(d)(ii) 2 B1 for a correctly positioned straight line graph for the given domain. Dep B1 for   0, 36  and   2, 0 10(a) 4 4 cos sin        2 2 2 2 cos sin cos sin       soi B1 2 2 cos sin 1     M1 Must show sufficient detail to show the given result. 2 2cos  A1 Alternative 1   2 4 2 cos 1 cos 1      (B1)   4 4 2 cos cos 2cos 1 1        (M1) Must show sufficient detail to show the given result. 2 2cos  (A1) Alternative 2   2 2 2 1 sin sin 1      (B1)   2 4 4 1 2sin sin sin 1        2 2 2sin   (M1) Must show sufficient detail to show the given result. 2 2cos  (A1) 10(b) 1 cos 3 2        soi B1 2π, π, π, 2π   4 M1 for obtaining one correct solution. A1 for obtaining 2 correct solutions. A1 for obtaining a third correct solution. A1 for a fourth correct solution and no extras within the range.