Cambridge IGCSE Mathematics - Additional 0606 — 2023 Feb/March Paper 1 · Variant 2

This document has 16 pages. [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 February/March 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 [ ]. * 7 0 2 9 1 0 6 8 3 8 * DC (PQ/CT) 312456/2 © UCLES 2023

2 0606/12/F/M/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 0606/12/F/M/23 © UCLES 2023 [Turn over 1 Find the exact values of k such that the straight line y k x 1 = - - is a tangent to the curve y kx x k 2 2 = + + . [4]

4 0606/12/F/M/23 © UCLES 2023 2 A curve has equation ( )( ) y x x 5 2 2 = - + . (a) Find the x-coordinates of the stationary points on the curve. [4] (b) On the axes below, sketch the graph of ( )( ) y x x 5 2 2 = - + , stating the coordinates of the points where the curve meets the axes. [3] y x O

5 0606/12/F/M/23 © UCLES 2023 [Turn over (c) Find the values of k for which the equation ( )( ) k x x 5 2 2 = - + has one distinct root only. [3]

6 0606/12/F/M/23 © UCLES 2023 3 Find the coefficient of x8 in the expansion of x x x 1 2 1 2 10 - - ` bj l . [5]

7 0606/12/F/M/23 © UCLES 2023 [Turn over 4 (a) Write lg lg x y 3 2 3 2 - - as a single logarithm to base 10. [3] (b) Solve the equation log log x 3 2 5 x 3 + = . [5]

8 0606/12/F/M/23 © UCLES 2023 5 The table shows values of the variables x and y, which are related by an equation of the form y Abx2 = , where A and b are constants. x 1 1.5 2 2.5 y 2.0 11.3 128 2896 (a) Use the data to draw a straight line graph of lny against x2. [2] ln y x2 0 7 8 6 5 4 3 2 1 1 2 3 4 5 6 7 – 1

9 0606/12/F/M/23 © UCLES 2023 [Turn over (b) Use your graph to estimate the values of A and b. Give your answers correct to 1 significant figure. [5] (c) Estimate the value of y when . x 1 75 = . [2] (d) Estimate the positive value of x when y 20 = . [2]

10 0606/12/F/M/23 © UCLES 2023 6 Given that ( ) ( ) x x 5 2 f 5 2 = + - m , ( ) 6 3 17 f = l and ( ) 6 3 26 f = , find an expression for ( )x f . [8]

11 0606/12/F/M/23 © UCLES 2023 [Turn over 7 (a) A 5-character password is to be formed from the following 13 characters. Letters A B C D E Numbers 9 8 7 6 5 Symbols * # ! No character may be used more than once in any password. (i) Find the number of possible passwords that can be formed. [1] (ii) Find the number of possible passwords that contain at least one symbol. [2] (b) Given that ( ) n 16 10 C C n n 12 1 11 # # = - + , find the value of n. [3]

12 0606/12/F/M/23 © UCLES 2023 8 y x O A C B y x 2 1 3 = - - y x 6 9 2 = - The diagram shows part of the curve y x 2 1 3 = - - and the straight line y x 6 9 2 = - . The curve intersects the x-axis at point A and the line at point B. The line intersects the x-axis at point C. Find the area of the shaded region ABC, giving your answer in the form ln p q + , where p and q are rational numbers. [11]

13 0606/12/F/M/23 © UCLES 2023 [Turn over Additional working space for Question 8.

14 0606/12/F/M/23 © UCLES 2023 9 In this question, all lengths are in metres. (a) A particle P has position vector t t 5 5 2 12 - + J L KK N P OO at a time t seconds, t 0 H . (i) Write down the initial position vector of P. [1] (ii) Find the speed of P. [2] (iii) Determine whether P passes through the point with position vector 48 - 158 J L KK N P OO. [2]

15 0606/12/F/M/23 © UCLES 2023 [Turn over (b) O A B C a b c The diagram shows the triangle OAC. The point B lies on AC such that : : AB AC 1 4 = . Given that OA a = , OB b = and OC c = , find c in terms of a and b. [3] Question 10 is printed on the next page.

16 0606/12/F/M/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) It is given that cos x 2 i + = for x 1 3 1 1 and cosec y 2 i = for y 2 2 . Find y in terms of x. [4] (b) Solve the equation cos sin 3 2 3 2 z z = for r r 4 4 1 1 z - . [5]

This document consists of 11 printed pages. © UCLES 2023 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 February/March 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 February/March 2023 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 2 of 11 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 3 of 11 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. 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

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 4 of 11 Question Answer Marks Guidance 1   2 2 3 1 0 + + − = kx x k 2 M1 for attempt to equate equations of the line and curve, re-arrange and equate to zero. Allow one sign error. ( ) 4 4 3 1 =  − k their k oe M1 Dep for attempt to use the discriminant of their quadratic equation and solve to obtain k. 1 13 6  = k isw A1 Alternative method 2 2 3 1 0 + + −= kx x k 2 M1 for attempt to equate equations of the line and curve, re-arrange and equate to zero. Allow one sign error. Grad of straight line = 1 − Gradient function of curve = 2 1 + kx Substitution to obtain 2 3 1 0 − −= k k oe with attempt to solve to obtain k M1 Dep 1 13 6  = k isw A1 2(a) ( )( ) ( )( ) 2 d 2 2 5 1 2 d = + − + − + y x x x x or 3 2 16 20 = − + + + y x x x 2 d 3 2 16 d = − + + y x x x 2 M1 for attempt at differentiation of a product, or expansion and then differentiation. A1 for all correct ( )( ) 2 8 3 0 + − = x x M1 Dep for attempt to solve their quadratic d 0 d = y x 8 2, 3 = − x A1 For both. 2(b) 3 B1 for correctly shaped curve, with maximum point in the first quadrant B1 for ( ) 5, 0 and a stationary point at ( ) 2, 0 − , must have a cubic graph. B1 for ( ) 0, 20 , must have a cubic graph

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 5 of 11 Question Answer Marks Guidance 2(c) When 8 1372 , 3 27 = = x y or awrt 50.8 M1 For attempt to find the value of y using their 8 3 . 1372 27  k or awrt 50.8 A1 0  k B1 3 ( ) 2 8 10 2 1 C 2   −     x x or ( ) 1 9 10 1 1 C 2   −     x x M1 For attempting to find terms which will give terms of 8x or 6 x , allow coefficients. Allow as part of an expansion 6 45 256      x oe A1  8 5120   −   x A1 ( ) ( ) 11520 5120 their their − + − M1 Dep –16 640 A1 Condone inclusion of 8x 4(a) 3 4 lg 1000 x y oe 3 B1 for 3 lg1000 = M1 for correct use of power rule at least once and division rule at least once A1 cao 4(b) 3 1 log 3 log = x x soi B1 For change of base. ( ) ( ) 2 3 3 2 log 5 log 2 0 − + = x x M1 For attempt to obtain a 3-term quadratic equation, equated to zero. Allow one sign error. May be using a substitution. 3 2log 1 = x 3 log 2 = x M1 Dep for attempt to solve their quadratic equation and a correct attempt to obtain at least one value of x. 3 = x oe A1 Allow 1.73(2…) 9 = x A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 6 of 11 Question Answer Marks Guidance 4(b) Alternative method 1 3 1 log log 3 = x x soi B1 For change of base. ( ) ( ) 2 2 log 3 5 log 3 2 0 − + = x x M1 For attempt to obtain a 3-term quadratic equation, equated to zero. Allow one sign error. 2log 3 1 = x log 3 2 = x M1 Dep for attempt to solve their quadratic equation and a correct attempt to obtain at least one value of x. 3 = x oe A1 Allow 1.73(2…) 9 = x A1 Alternative method 2 3 lg log lg3 = x x and lg3 log 3 lg = x x oe B1 For a consistent change of base. ( ) ( ) ( ) 2 2 2 lg 5 lg 2 lg3 0 − + = x x oe M1 For attempt to obtain a 3-term quadratic equation, equated to zero. Allow one sign error. 2lg lg3 = x lg 2lg3 = x oe M1 Dep for attempt to solve their quadratic equation and a correct attempt to obtain at least one value of x. 3 = x oe A1 Allow 1.73(2…) 9 = x oe A1 5(a) 2 B1 for 3 or 4 correctly plotted points

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 7 of 11 Question Answer Marks Guidance 5(b) 2 ln = + y mx c soi B1 ln , ln   m A c b Gradient = lnb M1 For attempt to find the numerical gradient of their straight line graph and equate to lnb . May be implied by later work 4 = b A1 Intercept on vertical axis = ln A M1 For use of their intercept on the vertical axis of their straight line graph oe. 0.5 = A A1 Alternative method 2 ln = + y mx c soi B1 ln , ln   m A c b Forming 2 equations correctly using points on their graph M1 Solving the equations to obtain either A or b M1 Dep 4 = b A1 0.5 = A A1 Special case 0.5 = A not using transformed data B1 4 = b not using transformed data B1 5(c) 35 nfww Allow answers between 33 and 37 2 M1 for attempt at a complete method using their straight line graph or equation 5(d) 1.63 nfww Allow answers between 1.5 and 1.7 2 M1 for attempt at a complete method using their straight line graph or equation

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 8 of 11 Question Answer Marks Guidance 6 ( ) 3 5 5 2 + k x M1 ( ) ( ) ( ) 3 5 1 f 5 2 3  = + + x x c A1 Condone omission of c ( ) 3 5 17 1 32 3 3 = + c oe M1 Dep for use of ( ) f 6  and attempt to evaluate c 3 = c A1 ( ) 8 5 5 2 + k x M1 ( ) 8 5 1 5 2 24 + + x cx A1 FT on their c ( ) ( ) ( ) 8 5 26 1 32 3 6 3 24 = + +  d oe M1 Dep for use of ( ) f 6 and attempt to evaluate d. ( ) ( ) 8 5 1 f 5 2 3 20 24  = + + −   x x x A1 7(a)(i) 154 440 B1 7(a)(ii) 124 200 2 B1 for 10 5P Alternative method 124 200 2 B1 for 1 symbol: 75 600 2 symbols: 43 200 3 symbols: 5400 7(b) ( ) ( ) 16 11 12 1 − = + n n oe B2 B1 for correct numbers or correct factors must be using combinations 47 = n B1 Dep on both previous B marks Must be the only solution

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 9 of 11 Question Answer Marks Guidance 8 ( ) 2.5, 0 A soi B1 ( ) 4.5, 0 C soi B1 2 2 21 0 + − = x x M1 For a correct attempt to find the intersection of the straight line and the curve. Must have attempt to solve the resulting quadratic equation to obtain x = . 7 3, 2   = −     x A1 1 3, 2       B soi A1 ( ) 3 2 d 2 3ln 1 1   − = − −   −    x x x x B1 ( ) ( ) 3 5 2 e.g. 2 3ln 1 3 6 3ln2 5 3ln 2  − − =     − − −     x x M1 Dep for application of appropriate limits e.g. 5 2 = x their and 3 = x their 5 2 = x their and 9 2 = x their 3 = x their and 9 2 = x their Integral must be in the form ( ) ln 1 + − ax b x 3 1 3ln 4 + oe A1 Area of an appropriate triangle B1 FT on 1 1 3 2 2 2   their their oe 1 2 2 2 3   their their oe Must be appropriate for their method Area = 11 27 ln 8 64 + 2 B1 for each correct term 9(a)(i) 2 5    B1 9(a)(ii) Velocity vector = 12 5     −   soi by correct speed B1 Speed = 13 B1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 10 of 11 Question Answer Marks Guidance 9(a)(iii) 2 12 158 + = t and 5 5 48 − = − t 13, 10.6 = = t t soi M1 Either for finding two values of t or for finding one value of t and substitute to obtain a position vector. Times are different so P does not pass through the given point or time calculated gives an inconsistent position vector A1 For a valid conclusion 9(b) = − AB b a and = − AC c a B1 ( ) 4 = − b - a c a oe M1 For substitution into a valid equation from their ratio. FT on their AB and their AC 4 3 = − c b a A1 Alternative method = − AB b a and 4 4 = − AC b a oe B1 ( ) 4 4 = = + − OC c a b a M1 FT on their AB and their AC 4 3 = − c b a A1 10(a) cos 2 = − x and 2 sin= y soi B1 ( ) 2 2 4 2 1 − + = x y M1 For a correct attempt to use 2 2 cos sin 1   + = or other relevant identity ( ) 2 2 4 1 2 = − − y x oe M1 Dep for attempt to rearrange to obtain 2 y ( ) 2 2 1 2 = − − y x or 2 2 4 3 − − x x oe A1 Must be positive Alternative method ( ) 1 cos 2  − = − x and 1 2 sin  −  =     y B1 ( ) 1 1 2 cos 2 sin − −  − =     x y M1 ( ) ( ) 1 2 sin cos 2 − = − y x 2 Dep M1 for correct attempt to rearrange to obtain y = ...

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 11 of 11 Question Answer Marks Guidance 10(b) tan 3 2 = or 3 sin 2 2 = or 1 cos 2 2 = B1 π 2 3 = or awrt 1.05 M1 Dep for a correct attempt to solve their equation, must be using 2 . 2π 3 = or awrt 2.09 M1 Dep for correct order of operations, may be implied by one correct solution. 10π 4π 2π 8π , , , 3 3 3 3 = − − or 10.5, 4.19, 2.09, 8.38 − − A2 A1 for a correct pair of solutions. A1 for a second correct pair of solutions and no extra solutions within the range. Allow greater accuracy if decimals used.