Cambridge IGCSE Mathematics - Additional 0606 — 2022 Oct/Nov Paper 1 · Variant 2

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge IGCSE™ DC (CE/SG) 303194/2 © UCLES 2022 ADDITIONAL MATHEMATICS 0606/12 Paper 1 October/November 2022 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 [ ]. * 9 1 2 9 4 4 4 5 5 1 *

2 0606/12/O/N/22 © UCLES 2022 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/O/N/22 © UCLES 2022 [Turn over 1 – 2 – 3 – 1 1 0 – 4 – 5 – 6 – 7 r - r x y 2 r - 2 r The diagram shows the graph of sin y a bx c = + , where a, b and c are integers. Find the values of a, b and c. [3]

4 0606/12/O/N/22 © UCLES 2022 2 (a) On the axes, draw the graph of y x x 3 13 10 2 = + - , stating the coordinates of the points where the graph meets the axes. [4] O x y (b) Find the set of values of the constant k such that the equation x x k 3 13 10 2 = + - has exactly 2 distinct roots. [4]

5 0606/12/O/N/22 © UCLES 2022 [Turn over 3 Write ( ) p q r p q r 2 9 3 1 5 2 3 # - - ` j in the form kp q r a b c, where k, a, b and c are constants. [4] 4 Solve the equation sin cos x x 3 2 4 3 2 4 r r + = + b b l l, for x 0 G G r. [5]

6 0606/12/O/N/22 © UCLES 2022 5 (a) Find the vector with magnitude 200 in the direction of 24 - 7 e o. [2] (b) a b O C B A The diagram shows triangle AOB such that OA a = , and OB b = . The point C lies on the line AB such that AC AB 1 3 | | = . Find the vector OC in terms of a and b, giving your answer in its simplest form. [3]

7 0606/12/O/N/22 © UCLES 2022 [Turn over (c) Given the vector equation p q p 2 1 2 4 5 + = - + 1 p q + e e e o o o, find the values of p and q. [3] 6 A group of 15 people includes 3 brothers. A team of 6 people is to be chosen from this group. The three brothers must not be separated. Find the number of possible teams that can be chosen. [3]

8 0606/12/O/N/22 © UCLES 2022 7 i rad O B A C 10 cm The diagram shows a circle, centre O, radius 10 cm. The points A and B lie on the circumference of the circle. The tangent at A and the tangent at B meet at the point C. The angle AOB is i radians. The length of the minor arc AB is 28 cm. (a) Find the value of i. [1] (b) Find the perimeter of the shaded region. [3]

9 0606/12/O/N/22 © UCLES 2022 [Turn over (c) Find the area of the shaded region. [3]

10 0606/12/O/N/22 © UCLES 2022 8 A function ( )x f is such that ( ) ( ) ln ln x x 2 3 4 f = + + , for x a 2 , where a is a constant. (a) Write down the least possible value of a. [1] (b) Using your value of a, write down the range of f. [1] (c) Using your value of a, find ( )x f 1 - , stating its range. [4]

11 0606/12/O/N/22 © UCLES 2022 [Turn over (d) On the axes below, sketch the graphs of ( ) y x f = and ( ) y x f 1 = - , stating the exact intercepts of each graph with the coordinate axes. Label each of your graphs. [4] O x y

12 0606/12/O/N/22 © UCLES 2022 9 (a) Show that ( ) ( ) ( ) . x x x x x x x 2 1 1 2 1 1 4 1 4 2 1 4 1 24 14 4 2 2 2 + - + + - = + - + + [2] (b) Hence find ( ) ( ) , x x x x x 2 1 4 1 24 14 4 d 2 2 1 2 1 + - + + y giving your answer in the form lnp q 2 1 + , where p and q are rational numbers. [7]

13 0606/12/O/N/22 © UCLES 2022 [Turn over 10 The first three terms of an arithmetic progression are lgx, lgx5, lgx9, where x 0 2 . (a) Show that the sum to n terms of this arithmetic progression can be written as ( )lg n pn x 1 - , where p is an integer. [4] (b) Hence find the value of n for which the sum to n terms is equal to lgx 4950 . [2] (c) Given that this sum to n terms is also equal to -14850, find the exact value of x. [2]

14 0606/12/O/N/22 © UCLES 2022 11 A particle P moves in a straight line such that, t seconds after passing through a fixed point O, its displacement, s metres, is given by s t t 2 1 1 1 2 3 = + + - ` j . (a) Show that the velocity of P at time t can be written in the form ( ) ( ) t t bt a 1 2 1 2 2 1 + + + ` j , where a and b are integers to be found. [5] (b) Show that P is never at instantaneous rest after passing through O. [1]

15 0606/12/O/N/22 © UCLES 2022 12 The first three terms, in descending powers of x, of the expansion of ax x b 5 2 1 5 2 + - b b l l , can be written as x x cx 32 160 5 4 3 - + , where a, b and c are constants. Find the exact values of a, b and c. [9]

16 0606/12/O/N/22 © UCLES 2022 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

This document consists of 10 printed pages. © UCLES 2022 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 October/November 2022 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 2022 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 October/November 2022 © UCLES 2022 Page 2 of 10 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 October/November 2022 © UCLES 2022 Page 3 of 10 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 October/November 2022 © UCLES 2022 Page 4 of 10 Question Answer Marks Guidance 1 2 = a B1 3 = b B1 4 = − c B1 2(a) 4 B1 for a correct basic shape, allow ‘construction curve’ Dep B1 for (0, 10) must have correct basic shape, must be convinced that this is the vertical intercept B1 for (–5, 0) and 2 ,0 3       or (0.667, 0) or better Dep B1 on all previous B marks for all correct with cusps and the correct shape for 5 − x and 2 3  x 2(b) Stationary point when 13 6 = − x soi M1 For differentiation or completing the square or use of symmetry ( ) 289 12 − or ( )24.1 − or better A1 For y-value of stationary point, allow +ve or –ve value. 289 12  k or 24.1  k or better A1 0 = k B1 Alternative ( ) 2 3 13 10 + − + x x k Using discriminant, ( ) 169 12 10 + + k (M1) Allow a sign error in ( ) 2 3 13 10 + − + x x k , but must have a term in k not 2 k Critical value ( ) 289 12 − or ( )24.1 − or better (A1) 289 12  k or 24.1  k or better (A1) One solution only from correct work 0 = k (B1)

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 5 of 10 Question Answer Marks Guidance 3 16 3 2 5 2 3 8 − − p q r 4 B1 for 3 8 = k or 0.375 B1 for 2 = − a B1 for 3 2 = b oe B1 for 16 5 = − c , 1 3.2. 35 − − 4 π 1 tan 2 4 3   + =     x or 2 π 1 sin 2 4 4   + =     x or 2 π 3 cos 2 4 4   + =     x B1 Must be from correct working Allow if π 2 4 = + x oe π π 7π 13π 2 , , 4 6 6 6 + = x π 24 = − x M1 Dep on previous B1 For attempt at the correct order of operations, may be implied by a correct solution or π 24 = − x . 11π 24 = x or 23π 24 oe 0.458π or 0.958π 1.44 or 3.01 2 Dep M1 for an attempt to find a solution within the given range. Must be working with 7π 13π or 6 6 A1 for either 11π 24 = x or 23π 24 oe 0.458π or 0.958π 1.44 or 3.01 A1 For a second solution within the given range with no extra solutions within the range. 5(a) 25 B1 soi 56 192     −   or 7 8 24     −   B1 5(b) ( ) 1 3 = − AC b a or ( ) 2 3 = − CB b a oe B1 ( ) 1 3 = + − OC a b a or ( ) 2 3 − − b b a oe M1 For using + OA their AC or + OB theirBC oe 2 1 3 3 + a b A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 6 of 10 Question Answer Marks Guidance 5(c) 2 2 5 5 + = − + p q p or 4 5 5 + = + p q p q M1 For equating like vectors to obtain at least one equation 5, 20 = − = p q 2 Dep M mark for attempt to solve their equations to obtain both p and q A1 for both 6 1144 3 B1 With the brothers: 220 or 12 3 C B1 Without the brothers: 924 or 12 6 C 7(a) 2.8 oe B1 7(b) ( ) 10tan1.4 = = BC AC or 10sin1.4 sin0.1708 M1 Perimeter = ( ) ( ) 10 2.8 2 or + their their AC BC M1 144 A1 7(c) Area of triangle AOC or BOC = ( ) 1 or 10 2  their AC BC or 1 10sin1.4 2  their OC soi M1 Allow premature approximation for OC Area of minor sector AOB = 140 B1 FT on 50  their 2.8 Shaded area = 439 to 440 A1 Must have 579 ⩽ kite area ⩽ 580 8(a) 1.5 − B1 8(b) f  B1 Allow  y , , ( ) f −  x oe, ( ) f  x

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 7 of 10 Question Answer Marks Guidance 8(c) ( ) ln 8 12 + x or ( ) ( ) ln 4 2 3 + x B1 May be implied ( ) 1 e 12 f 8 − − = x x oe 2 M1 for attempt to find the inverse, allow one sign error A1 allow y = … Range: ( ) 1 f 1.5 − − their B1 Must be correct notation, follow through on their (a) ( ) ( ) 1 f 1.5 −  − x their , ( ) 1.5  − y their Alternative ( ) ln4 1 e 3 f 2 − − − = x x oe (3) B1 for ln 4 e − x or ln 4 e − y M1 for attempt to find the inverse, allow one sign error A1 allow y = … Range: ( ) 1 f 1.5 − − their (B1) Must be correct notation, follow through on their (a) ( ) ( ) 1 f 1.5 −  − x their , ( ) 1.5  − y their 8(d) 4 B1 for correct shape of ( ) f x in quadrants 1, 2 and 3, with asymptotic behaviour B1 for ln12 and 11 8 − or 1.375 − in correct position, must have a correct shape. B1 for correct shape of ( ) 1 f − x in quadrants 1, 3 and 4, with asymptotic behaviour B1 for ln12 and 11 8 − or 1.375 − in correct position, must have a correct shape and intersect at least once with ( ) f = y x 9(a) ( )( ) ( ) ( ) ( ) ( ) 2 2 4 1 2 1 4 1 4 2 1 2 1 4 1 − + − − + + + − x x x x x x M1 For attempt to obtain a single fraction An extra term of ( ) 2 1 + x throughout must be dealt with correctly before awarding M1 ( ) ( ) 2 2 24 14 4 2 1 4 1 + + + − x x x x A1 Must see sufficient detail of expansion and collecting terms cso as AG

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 8 of 10 Question Answer Marks Guidance 9(b) ( ) 1 ln 2 1 2 + x B1 ( ) 1 2 2 1 + x B1 Allow ( ) 1 2 1 1 2 − − + − x oe ( ) ln 4 1 − x B1 1 1 1 1 ln3 ln3 ln 2 2 6 2 4     + + − +         M1 For correct application of limits, must have at least one log term. Must be using individual fractions from (a) Fractions and log terms must be bracketed correctly and manipulated correctly 1 27 1 ln 2 2 12 − 3 M1 for application of log laws using 1 1 ln3 ln3 ln2 2 2 + − to obtain the correct form A1 for 1 27 ln 2 2 B1 for 1 12 − 10(a) Common difference = 4lg x B1 Sum to n terms = ( )( ) ( ) 2lg 1 4lg 2 + − n x n x M1 For use of the sum formula with their common difference ( ) 2 1 lg − n n x 2 Dep M1 for a correct attempt to rearrange to the required form A1 cao 10(b) ( ) 2 1 4950 − = n n M1 For ( ) ( ) 1 4950 − = n their p n together with an attempt to solve to obtain n 50 A1 cao 10(c) ( ) 50 99 lg 14850 = − x M1 For use of their n and p in a complete method to find x or use of part (b) 3 10− or equivalent A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 9 of 10 Question Answer Marks Guidance 11(a) ( ) ( ) ( ) ( ) 1 3 2 2 2 3 1 2 2 1 2 1 2 1   +   + − +     + t t t t 3 B1 for ( ) 1 2 3 2 2 1 2  + t M1 for a correct attempt at a quotient or a product A1 for all terms apart from ( ) 1 2 3 2 2 1 2  + t correct ( ) ( ) ( ) 1 2 2 2 1 2 1 + + + t t t 2 M1 dep on previous M mark for attempt to obtain in the required form Alternative ( ) ( ) 3 2 2 1 1 1 + −− = + t t s t ( ) ( ) ( ) ( ) 1 3 2 2 1 3 2 1 1 2 1 1 2 1 +   + − − + − − +                   t t t t t (3) B1 for ( ) 1 2 3 2 2 1 2  + t M1 for a correct attempt at a quotient or a product A1 for all terms apart from ( ) 1 2 3 2 2 1 2  + t correct ( ) ( ) ( ) 1 2 2 2 1 2 1 + + + t t t (2) M1 dep on previous M mark for attempt to obtain in the required form 11(b) ( ) ( ) 1 2 2 1 2 0 + + = t t oe has no real positive solutions so velocity is never zero B1 FT on their positive linear factor Reference needs to be made to both factors.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 10 of 10 Question Answer Marks Guidance 12 5 5 4 4 3 3 8 2 5 + + a x a x a x 3 B1 for each correct term, allow when first seen 2 2 2 1− + b b x x B1 2 = a B1 32 64 160 − = − b M1 For using their expansions and their value for a to obtain two terms involving 4 x 3 = b A1 64 192 288 5 − + = c M1 For using their expansions and their value for a to obtain three terms involving 3x 544 5 = c oe A1 Alternative ( ) 5 32 = a (B1) 2 = a (B1) 5 4 2 2 160 − + = − ba a soi 32 64 160 − = − b (2) B1 for 4 2a soi M1 For using their expansions and their value for a to obtain two terms involving 4 x 3 = b (A1) 3 4 5 2 8 4 5 − + = a a b a b c 64 192 288 5 − + = c (3) B2 for both 3 8 5 a and 5 2 a b B1 for either 3 8 5 a or 5 2 a b if only one correct M1 for using their expansions and their value for a to obtain three terms involving 3x 544 5 = c oe (A1)