Cambridge IGCSE Mathematics - Additional 0606 — 2024 May/June Paper 1 · Variant 2

This document has 16 pages. [Turn over * 7 0 5 8 4 4 3 8 6 1 * Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 May/June 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 [ ]. DC (DE/CGW) 332388/3 © UCLES 2024 ,

 

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2 0606/12/M/J/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 = * 0019655405202 * ,

 

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3 0606/12/M/J/24 © UCLES 2024 [Turn over 1 y x – 360° – 180° 180° 0 360° – 8 – 6 – 4 – 2 2 The diagram shows the graph of sin y a bx c = + for x 360 360 c c G G - , where a, b and c are constants. Find the values of a, b and c. [3] 2 Given that log log r s 2 8 3 9 + = , find the value of rs. [3] * 0019655405203 * ,

 

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4 0606/12/M/J/24 © UCLES 2024 3 Given that tan y x 2 = , find the exact value of d d x y when r x 3 = . [4] * 0019655405204 * ,

 

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5 0606/12/M/J/24 © UCLES 2024 [Turn over 4 A team of 8 people is to be formed from 6 teachers, 5 doctors and 4 police officers. (a) Find the number of teams that can be formed. [1] (b) Find the number of teams that can be formed without any teachers. [1] (c) Find the number of teams that can be formed with the same number of doctors as teachers. [4] * 0019655405205 * ,

 

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6 0606/12/M/J/24 © UCLES 2024 5 DO NOT USE A CALCULATOR IN THIS QUESTION. In this question, all lengths are in centimetres. B C A D 8 7 7 - 7 2 + 9 7 9 - The diagram shows the trapezium ABCD. The lengths of AB, BC and CD are 8 7 7 - , 7 2 + and 9 7 9 - respectively. The line BC is perpendicular to the lines AB and CD. (a) Find the perimeter of the trapezium, giving your answer in its simplest form. [3] (b) Find the area of the trapezium, giving your answer in the form p q 7 + , where p and q are rational numbers. [3] * 0019655405206 * ,

 

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7 0606/12/M/J/24 © UCLES 2024 [Turn over (c) Find cotDBC, giving your answer in the form r s 7 + , where r and s are simplified rational numbers. [3] * 0019655405207 * ,

 

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8 0606/12/M/J/24 © UCLES 2024 6 In this question, all lengths are in metres and all angles are in radians. A D B C O 5 i The diagram shows a circle with centre O and radius 5. The points A, B, C and D lie on the circumference of the circle. Angle DOC i = . Angle . AOD COB 0 5 angle = = . The length of the minor arc DC is 3.75. (a) Show that .0 75 i = . [1] (b) Find the perimeter of the shaded region. [5] * 0019655405208 * ,

 

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9 0606/12/M/J/24 © UCLES 2024 [Turn over (c) Find the area of the shaded region. [3] * 0019655405309 * ,

 

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10 0606/12/M/J/24 © UCLES 2024 7 (a) The line y x 3 2 = - intersects the curve x xy y 2 2 2 2 - + = at the points A and B. The point C with coordinates ,k 8 7 b l lies on the perpendicular bisector of the line AB. Find the exact value of k. [9] * 0019655405310 * ,

 

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11 0606/12/M/J/24 © UCLES 2024 [Turn over (b) The point D lies on the perpendicular bisector of AB such that D is a reflection of C in the line AB. Find the coordinates of D. [2] * 0019655405311 * ,

 
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12 0606/12/M/J/24 © UCLES 2024 8 A curve has equation ( ) y x x 4 3 5 2 3 1 = + - . (a) Show that d d x y can be written in the form ( ) ( ) x x Ax Bx C 3 5 4 2 3 2 2 2 - + + + , where A, B and C are integers. [5] (b) Hence find the x-coordinates of the stationary points on the curve. Give your answers in their simplest exact form. [3] * 0019655405312 * ,

 
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13 0606/12/M/J/24 © UCLES 2024 [Turn over 9 In this question, all distances are in metres and time, t, is in seconds. A particle P moves with a speed of 14.5 parallel to the vector 20 21 - e o. (a) Find the velocity vector of P. [2] Initially, P has position vector 3 5 e o. (b) Write down the position vector of P at time t. [2] A second particle Q has position vector . t 1 3 5 7 5 - + - e e o o at time t. (c) Find, in terms of t, the distance between P and Q at time t. Simplify your answer. [4] (d) Hence show that P and Q never collide. [2] * 0019655405313 * ,

 
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14 0606/12/M/J/24 © UCLES 2024 10 (a) The first 3 terms of an arithmetic progression are sin x 3 2 , sin x 2 5 , sin x 2 7 . (i) Show that the sum to n terms of this arithmetic progression can be written in the form ( )sin n n a x 2 + , where a is a constant. [3] (ii) Given that r x 3 2 = , find the exact sum of the first 20 terms. [2] * 0019655405314 * ,

 
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15 0606/12/M/J/24 © UCLES 2024 [Turn over (b) The first 3 terms of a geometric progression are ln y 2 , ln y 4 2, ln y 16 4. (i) Find the nth term of this geometric progression. [2] (ii) Find the sum to n terms of this geometric progression, giving your answer in its simplest form. [2] (c) The first 3 terms of a different geometric progression are w 2 4 1 - b l, w 2 4 1 2 - b l , w 2 4 1 3 - b l . Find the values of w for which this geometric progression has a sum to infinity. [3] Question 11 is printed on the next page. * 0019655405315 * ,

 
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16 0606/12/M/J/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. 11 (a) Given that ln y x x 2 = , find d d x y. [2] (b) Hence find ln x x x d y . [3] * 0019655405316 * ,

 
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This document consists of 12 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 May/June 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 May/June 2024 series for most Cambridge IGCSE, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 2 of 12 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 3 of 12 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 4 of 12 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 3  a B1 2 3  b oe B1 4  c B1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 5 of 12 Question Answer Marks Guidance 2 6561  rs isw 3 B2 for 3 log 8  rs or   2 3 log 16  rs B1 for 3 9 3 2log 2log log 9  s s soi Alternative 1 6561  rs isw (3) B2 for 9 log 4  rs or   2 9 log 8  rs B1 for 9 3 9 log log log 3  r r soi Alternative 2 6561  rs isw (3) B2 for log 8log3  rs or log 4log9  rs for any base B1 for log 2log 8 log3 log9   r s soi for any base 3 2 d 1 sec d 2 2  y x x or 2 1 2cos 2       x 2 M1 for 2 sec 2 x k or 2 cos 2       k x , 1 2  k 2 π 4 sec 6 3  or 2 3 cos 6 4        soi B1 d 2 d 3  y x A1 4(a) 6435 B1 4(b) 9 B1 4(c) 6 5 2 2 C C  (150) B1 6 5 4 3 3 2 C C C   (1200) B1 6 5 4 4 C C  (75) B1 Total: 1425 B1 Dep on all previous B marks

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 6 of 12 Question Answer Marks Guidance 5(a)     2 2 2 7 2 7 2     AD soi, oe B1 2 11 4 7 11 4 7     AD or 11 4 7 11 4 7     AD B1 This is the minimum detail that is acceptable for this mark Perimeter = 18 7 22 14   isw B1 Allow if previous B mark not awarded because of lack of detail. 5(b)    1 8 7 7 9 7 9 7 2 2     oe B1   1 119 18 7 32 2   B1 This is the minimum detail that is acceptable 87 9 7 2  oe isw B1 Allow if previous B mark not awarded because of lack of detail. Must be 2 separate terms. Alternative 1 Area of triangle = 3 2 (B1) Area of rectangle =    8 7 7 7 2   56 9 7 14    (B1) This is the minimum detail that is acceptable 87 9 7 2  oe isw (B1) Allow if previous B mark not awarded because of lack of detail. Alternative 2 Area of triangle = 3 2 (B1) Area of outer rectangle =    9 7 9 7 2   63 9 7 18    (B1) This is the minimum detail that is acceptable 87 9 7 2  oe isw Allow if previous B mark not awarded because of lack of detail.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 7 of 12 Question Answer Marks Guidance 5(c)   7 2 cot 9 7 1     or   1 9 3 7  oe B1 May have factors of 9 multiplied out         7 2 7 1 9 7 1 7 1      or      3 7 9 3 7 3 7    M1 For rationalisation of their cot 7 1 18 6  A1 Alternative   7 2 cot 9 7 1     (B1)      7 2 7 9 7 1     r s 1 9 9   r s 2 63 9   r s (M1) Obtain 2 simultaneous equations from their cot. Allow one error 7 1 18 6  (A1) 6(a) 5 3.75  so 0.75  AG B1 Allow 3.75 5  as a starting point

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 8 of 12 Question Answer Marks Guidance 6(b)   2 2 2 5 5 2 5 5cos1.75     AB   0.5 sin 0.875 5  AB 5 sin1.75 sin0.696  AB or   2 2 2 5 5 2 5 5cos0.75     DC   0.5 sin 0.375 5  DC 5 sin0.75 sin1.196  DC M1 For attempt to find the length of at least one chord. May be implied by a correct length.  7.67 5  AB soi A1  3.66 3  DC soi A1 Arc length AD or BC :5 0.5  soi oe B1 Allow unsimplified 16.3 A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 9 of 12 Question Answer Marks Guidance Choose one scheme to mark to the advantage of the candidate 6(c)   2 1 5 1.75 sin1.75 2           2 1 5 0.75 sin0.75 2          M2 M2 for a fully correct method with no extra terms M1 for attempt at the area of at least one segment – may be implied by different methods 8.7 A1 Alternative 1 Triangle ODC + sectors OCB and AOD only 2 2 1 1 5 sin0.75 2 5 0.5 2 2                  (M1) Triangle OAB 2 1 5 sin1.75 2   (M1) 8.7 (A1) Alternative 2 Area of trapezium formed by Triangle ODC – triangle OPQ = 2 2 1 1 5 sin0.75 sin0.75 2 2                  OP (M1) The points P and Q are where the lines OD and OC meet AB. Candidates may have different labels Area of APD and BQC = 2 1 1 2 5 0.5 5 sin0.5 2 2                         OP (M1) 8.7 (A1) Alternative 3 Area of trapezium ABCD =   1 1.45 2   AB CD (M1) Allow 8.21 area of 2 remaining shaded segments 2 2 1 1 2 5 0.5 5 sin0.5 2 2                         (M1) Allow M1 for one segment 8.7 (A1)

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 10 of 12 Question Answer Marks Guidance 7(a)     2 2 2 3 2 3 2 2      x x x x or 2 2 2 2 2 2 3 3                  y y y y M1 For a quadratic equation in one variable 2 4 5 1 0   x x oe or 2 4 5 0    y y oe A1 Allow multiples but must be a 3 term equation = 0 1 1 4   x x 5 1 4   y y 2 A1 for each correct pair Midpoint: 5 1 , 8 8        M1 Allow for use of their coordinates Gradient of AB = 3 M1 M1 for gradient of AB = 3 Allow for use of their coordinates Perpendicular gradient: 1 3  M1 Dep on previous M mark Perpendicular equation: 1 1 5 8 3 8          y x M1 Allow for their perp gradient and their midpoint When 7 19 , 8 8   y k or 2.375  oe A1 7(b) 29 9 , 8 8        or   3.625, 1.125  oe 2 B1 for each coordinate 8(a)         2 1 2 2 3 3 2 1 6 3 5 4 3 5 3 4       x x x x x or            2 1 2 3 1 2 2 3 1 6 3 5 4 3 3 5 4          x x x x x 3 B1 for    2 2 3 1 6 3 5 3   x x soi M1 for attempt at differentiation of a quotient or valid product A1 for all other terms correct      2 2 4 3 5    x x x M1 M1 dep on previous M mark for simplification of their numerator Allow one sign slip but must be dealing with two quadratic terms.    2 2 2 2 3 8 5 3 5 4      x x x x A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 11 of 12 Question Answer Marks Guidance 8(b) 2 8 5 0     x x oe, together with attempt to solve M1 Allow for use of their quadratic numerator from part (a) 4 21  2 A2 for both A1 for 8 84 2  or 8 84 2  oe 9(a) Velocity vector = 10 10.5        or 20 1 21 2        B2 B1 for 20 29 21        soi or one correct element 9(b) 3 10 5 10.5            t oe 2 M1 FT on their velocity vector from part (a) if not correct. 9(c) 5 4 3 2           t PQ t oe 2 Dep M1 must have M1 in part (b) for 1 5 3 7.5                       t their answer to part (b) or their answer to part b 1 5 3 7.5                       t     2 2 5 4 3 2     PQ t t M1 Dep on previous M mark for using their  PQ 2 34 28 20    PQ t t A1 9(d) For collision 2 34 28 20 0    t t soi M1 Need to be using their quadratic 2 PQ from part (c) or forming a quadratic from their  PQ from part (c) Discriminant = 1936  So no real roots so no collision oe A1 Discriminant must be correct, but allow unsimplified 10(a)(i) 2sin 2  d x soi B1     6sin 2 2 1 sin 2 2    n n S x n x or     3sin 2 2 1 sin 2 2    n n S x n x M1 For correct use of the sum formula with their d   2 sin2  n n x A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 12 of 12 Question Answer Marks Guidance 10(a)(ii) 220 3  isw 2 Allow ‘starting again’ M1 for use of their answer to (a), but must be a correct method, 4π sin 3       needs to be evaluated to give an exact answer. 10(b)(i) 1 2 ln2  n y or   1 ln 2 2  n y or   1 2 ln 2  n y B2 B1 for 2  r 10(b)(ii)     1 ln 2 1   n their r y their r M1 For correct use of sum formula with their non-logarithmic r   2 1 ln 2  n y or   1 1 2 ln 2   n y or   2 1 ln 2  n y or ln2 2 ln2   n y y oe isw when appropriate A1 A0 if denominator is still present. 10(c) 3 5 8 8    w 0.375 0.625    w 5 8  w and 3 8  w 3 B2 for 1 2 1 4   w or 1 1 2 1 4    w or 5 8 and 3 8  seen B1 for 1 2 4   r w . May be seen in the sum to infinity formula 11(a) 2 ln  x x x 2 M1 for attempt at product rule Allow unsimplified for A1 11(b) 2 ln mx x M1 Mark final answer Dep on M1 in part (a) 2 2 1 ln d ln 2 4     x x x x x x c oe 2 A1 for 2 1 ln 2 x x oe A1 for 2 4   x c oe