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

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/11 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 [ ]. DC (LK/SG) 303197/2 © UCLES 2022 * 3 6 9 8 3 1 2 9 8 5 *

2 0606/11/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 0606/11/O/N/22 © UCLES 2022 [Turn over 1 (a) On the axes, sketch the graphs of x y 2 1 + = and y x 5 3 = - for x 2 8 G G - . State the coordinates of the points where these graphs meet the coordinate axes. [3] – 2 8 0 x y (b) Solve the equation x x 2 1 5 3 + = - . [3]

4 0606/11/O/N/22 © UCLES 2022 2 (a) On the axes, sketch the graph of sin y x 5 2 1 = + for x 2 2 G G r r - . [3] – 2r – 2 2 4 6 0 – 4 – 6 – r r 2r x y (b) Write down the amplitude of sin x 5 2 1 + . [1] (c) Write down the period of sin x 5 2 1 + . [1]

5 0606/11/O/N/22 © UCLES 2022 [Turn over 3 When y3 is plotted against lnx, a straight line graph is obtained, passing through the points (1, 5) and (6, 15). Find y in terms of x. [4]

6 0606/11/O/N/22 © UCLES 2022 4 DO NOT USE A CALCULATOR IN THIS QUESTION. Solve the equation x x 5 1 2 5 1 0 2 - - - + = ` ` j j , giving your answers in the form a b 5 + , where a and b are constants. [6]

7 0606/11/O/N/22 © UCLES 2022 [Turn over 5 An arithmetic progression is such that the fourth term is 25 and the ninth term is 50. (a) Find the first term and the common difference. [3] (b) Find the least number of terms for which the sum of the progression is greater than 25 000. [3]

8 0606/11/O/N/22 © UCLES 2022 6 The first three terms, in ascending powers of x, in the expansion of ( ) x x 1 9 2 1 3 18 3 - + e o are written in the form ax bx 1 2 + + , where a and b are constants. Find the exact values of a and b. [7]

9 0606/11/O/N/22 © UCLES 2022 [Turn over 7 A O D C B x i rad r The diagram shows a circle with centre O and radius r. OAB and OCD are sectors of a circle with centre O and radius x, where x r 0 1 G . Angle AOB = angle COD i = radians, where r 0 1 1 i . (a) Find, in terms of r, x and i, the perimeter of the shaded region. [3] (b) Find, in terms of r, x and i, the area of the shaded region. [1] It is given that x can vary and that r and i are constant. (c) Write down the least possible area of the shaded region in terms of r and i. [2]

10 0606/11/O/N/22 © UCLES 2022 8 Find d x x x 1 2 2 1 a 0 + - + e o y , where a is a positive constant. Give your answer, as a single logarithm, in terms of a. [5]

11 0606/11/O/N/22 © UCLES 2022 [Turn over 9 Solve the equation log log y p 2 10 9 0 p y + - = , where p is a positive constant, giving y in terms of p. [5] 10 Given that ( ) C C n 65 2 1 n n 5 1 6 # # = - + , find the value of n. [3]

12 0606/11/O/N/22 © UCLES 2022 11 a b c O B C A The diagram shows a triangle OAC. The point B lies on AC such that : : AB AC 2 5 = . It is given that ,a b OA OB = = and c OC = . (a) Show that b a c 5 3 2 - = . [4]

13 0606/11/O/N/22 © UCLES 2022 [Turn over a b c O B C A X Y The diagram now includes points X and Y, such that OX OA 4 3 = and OY mOB = , where m is a constant. It is also given that : : XY XC 1 m = , where m is a constant. (b) Using part (a), find XC in terms of a and b. [2] (c) Hence find the values of m and m. [4]

14 0606/11/O/N/22 © UCLES 2022 12 (a) Show that sin sec cosec cosec 1 1 2 1 1 2 i i i i + = - + . [3] (b) Hence solve the equation cosec cosec sin 2 1 1 2 1 1 4 2 z z z - + + = , for ° ° 90 90 G G z - . [6]

15 0606/11/O/N/22 © UCLES 2022 13 Given that ( ) ( ) x x 6 3 4 = + - f 2 1 ll , ( ) f 4 18 = l and ( ) f 4 9 512 = , find ( ) f x . [8]

16 0606/11/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 9 printed pages. © UCLES 2022 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/11 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.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 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.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 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. 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

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 4 of 9 Question Answer Marks Guidance 1(a) 3 B1 for 2 V-shaped graphs with vertices in the 1st and 2nd quadrants, intersecting twice in the first quadrant. Dep B1 for ( ) 0,1 and ( ) 0,5 B1 for 1 ,0 2   −     and 5,0 3       1(b) 4 5 = x B1 2 1 5 3 + = −+ x x oe M1 For considering the negative for one of the functions 6 = x A1 Alternative 2 5 34 24 0 − + = x x (2) M1 for squaring each function and attempt to form a 3-term quadratic equation = 0. Allow one error. A1 for a correct equation 4 5 = x , 6 = x (A1) For both 2(a) 3 B1 for a complete cycle starting and finishing at ( ) 2π, 1 − and ( ) 2π, 1 B1 for intercept at 1 = y B1 for a maximum when 6 = y and a minimum when 4 = − y 2(b) 5 B1 2(c) 4π or o 720 B1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 5 of 9 Question Answer Marks Guidance 3 3 ln = + y m x c B1 May be implied by subsequent work 5 = + m c 15 6 = + m c 2, 3 = = m c 2 B1 for 2 = m B1 for 3 = c 3 2ln 3 = + y x B1 Alternative 3 ln = + y m x c (B1) May be implied by subsequent work Gradient = 2 (B1) For finding the gradient and equating to m 5 = + m c 15 6 = + m c 3 = c (B1) For at least one correct equation and finding c 3 2ln 3 = + y x (B1) 4 ( )( ) ( ) 2 4 4 5 1 5 1 2 5 1  + − + = − x M1 For a correct use of the quadratic formula with sufficient detail ( ) 2 2 5 2 5 1  = − x or ( ) 1 5 5 1  = − x 2 Dep M1 for attempt to simplify to obtain 2 real roots A1 for either ( ) ( ) ( ) ( ) 5 1 5 1 5 1 5 1 + + =  − + x M1 For attempt at rationalisation 3 5 2 2 = + x A1 1 = − x B1 5(a) 3 25 + = a d 8 50 + = a d M1 For at least one correct equation and attempt to solve to find at least one unknown 10 = a A1 5 = d A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 6 of 9 Question Answer Marks Guidance 5(b) ( ) ( ) ( ) 20 1 5 25 000 2 + − = n n M1 For attempting the sum to n terms using their a and d 2 5 15 50 000 0 + − = n n 98.5... = n A1 99 = n A1 6 2 68 1 4 9 − + x x 2 B1 for 1 4 −x B1 for 2 68 9 x or 2 7.56x 2 1 9 27 + + x x B1 Term in x: 4 9 5 − + = x x x or coefficients of x: 4 9 −+ M1 For ( ) ( ) ( ) ( ) 4 9 − + their x their x 5 = a A1 Term in 2 2 2 2 68 : 27 36 9 + − x x x x or coefficients of 2 68 : 27 36 9 + − x M1 For ( ) ( ) ( ) 68 27 9 their x their x  + +     ( ) ( ) ( ) ( ) ( ) 4 9 their x their x −  13 9 = − b A1 Must be exact 7(a) 2π 4 2  + + r x x 3 B1 for 2πr B1 for 4 + x B1 for 2  x 7(b) 2 2 π  − r x B1 7(c) Least value when = x r B1 Least value = ( ) 2 π  − r oe B1 8 ( ) ( ) 2ln 1 ln 2 + − + x x 2 B1 for ( ) 2ln 1 + x B1 for ( ) ln 2 − + x ( ) ( ) ( ) 2ln 1 ln 2 ln2 + − + + a a M1 For attempt to apply limits correctly, dependent on having 2 log terms. ( ) ( ) 2 2 1 ln 2 + + a a 2 M1 for use of either power rule or the division rule.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 7 of 9 Question Answer Marks Guidance 9 10 2log 9 0 log + − = p p y y or 2 10log 9 0 log + − = y y p p B1 For a change of base ( ) 2 2 log 9log 10 0 − + = p p y y or ( ) 2 10 log 9log 2 0 − + = y y p p M1 For attempt to obtain a 3-term quadratic equation = 0, in either log p y or log y p 5 log , log 2 2 = = p p y y or 2 1 log , log 5 2 = = y y p p M1 Dep M mark for attempt to solve the quadratic to obtain 2 solutions 5 2 = y p A1 2 = y p A1 10 ( ) ( )( ) ( ) 2 1 1 ! 65 ! 5 !5! 5 !6! − + = − − n n n n n 2 1 65 3 − = n 2 B1 for simplifying numerical factorials to 3 B1 for simplifying algebraic factorials to either ( )( ) 1 1 − + n n or 2 1 − n 14 = n B1 11(a) = − AC c a B1 ( ) 2 5 = − AB c a or ( ) 3 5 = − BC c a B1 ( ) 2 5 − = c a − b a or ( ) 3 5 − = c a − c b M1 For equating two different forms of AB or 2 different forms of BC 5 3 2 − = b a c A1 Simplification to obtain the given answer 11(b) 3 4 = − XC a c B1 5 9 2 4 = − XC b a B1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 8 of 9 Question Answer Marks Guidance 11(c) 3 5 9 4 2 4   − = −     m b a b a B1 1 5 , 3 6 = = m 3 M1 for equating like vectors at least once A1 for 1 3 = A1 for 5 6 = m 12(a) 2 cosec 1 cosec 1 cosec 1    + + − − B1 Allow denominator unsimplified 2 2cosec cot   B1 2 2 2 sin sin cos     2 2sin sec   B1 Sufficient detail must be seen 12(b) 2 2sin 2 sec 2 4sin 2    = Leading to sin2 0 = o o 90 , 0 =  2 M1 for attempt to solve sin2 0 = obtaining at least one correct solution A1 for all solutions 2 2sin 2 sec 2 4sin 2    = ( ) 1 cos2 2 =  M1 For dealing with 2 sec 2to obtain cos2= k , where 0 1   k o o 67.5 , 22.5 =   3 M1 for solution to obtain at least one correct solution A1 for a correct pair of solutions A1 for a second correct pair of solutions with no extra solutions within the range

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 9 of 9 Question Answer Marks Guidance 13 ( ) ( ) ( ) 1 2 f 4 3 4  = + + x x c 2 M1 for ( ) 1 2 3 4 + a x A1 for ( ) 1 2 4 3 4 + x ( ) 18 4 4 = + c M1 Dep M mark for attempting correctly to find the value of the arbitrary constant 2 = c A1 ( ) ( ) ( ) 3 2 8 f 3 4 2 9 = + + + x x x d M1 For ( ) 3 2 3 4 + b x ( ) ( ) ( ) 3 2 8 f 3 4 2 9 = + + + x x x d A1 Allow unsimplified ( ) 64 64 8 9 9 = + + d M1 Dep M mark for attempt to find a second arbitrary constant ( ) ( ) 3 2 8 f 3 4 2 8 9 = + + − x x x A1