Cambridge IGCSE Mathematics - Additional 0606 — 2022 Oct/Nov Paper 1 · Variant 3

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge IGCSE™ DC (CE/SG) 303100/3 © UCLES 2022 ADDITIONAL MATHEMATICS 0606/13 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 [ ]. * 3 3 6 2 7 1 6 2 3 3 *

2 0606/13/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 0606/13/O/N/22 © UCLES 2022 [Turn over 1 On the axes, sketch the graph of sin y x 4 3 2 = - for x 3 3 G G r r - . [3] – 6 – 4 – 2 2 4 6 0 x y 3 r - 6 r - 6 r 3 r

4 0606/13/O/N/22 © UCLES 2022 2 (a) Show that x x 2 15 2 + - can be written in the form x a b 2 2 + + ` j , where a and b are exact constants to be found. [2] (b) Hence write down the coordinates of the stationary point on the curve y x x 2 15 2 = + - . [2] (c) On the axes, sketch the graph of y x x 2 15 2 = + - , stating the coordinates of the points where the graph meets the coordinate axes. [3] O x y (d) Write down the value of the constant k for which the equation x x k 2 15 2 + - = has 3 distinct solutions. [1]

5 0606/13/O/N/22 © UCLES 2022 [Turn over 3 (a) Solve the following simultaneous equations. y x 3 2 2 0 - + = xy 2 1 = [3] (b) Solve the equation log log x 3 10 3 x 3 + = , giving your answers as powers of 3. [4]

6 0606/13/O/N/22 © UCLES 2022 4 The polynomial ( )x p is such that ( )x ax x bx c 13 p 3 2 = + + + , where a, b and c are integers. It is given that ( ) 0 9 p =- l . (a) Show that b 9 =- . [1] It is also given that x 3 2 + is a factor of ( )x p and that when ( )x p is divided by x 1 + the remainder is 6. (b) Find the values of a and c. [4] (c) Find the quadratic ( )x q such that ( ) ( ) ( ) x x x 3 2 p q # = + . [1] (d) Hence find ( )x p as a product of linear factors with integer coefficients. [1]

7 0606/13/O/N/22 © UCLES 2022 [Turn over 5 A geometric progression is such that the fifteenth term is equal to 8 1 of the twelfth term. The sum to infinity is 5. (a) Find the first term and the common ratio. [4] (b) Find the least number of terms needed for the sum of the geometric progression to be greater than 4.999. [3]

8 0606/13/O/N/22 © UCLES 2022 6 A function ( )x f is such that ( )x 4 f e x 3 = - , for x R d . (a) Find the range of f. [1] (b) Find an expression for ( )x f 1 - . [2] (c) On the axes, sketch the graphs of ( ) y x f = and ( ) y x f 1 = - stating the exact values of the intercepts with the coordinate axes. [4] O x y

9 0606/13/O/N/22 © UCLES 2022 [Turn over 7 Find the exact value of ( ) cos sin x x x 3 4 2 1 d 0 r 2 + + y . [5]

10 0606/13/O/N/22 © UCLES 2022 8 In this question all lengths are in metres. 7 24 O B C A 12.25 The diagram shows a circle, centre O, radius 7. The points A and B lie on the circumference of the circle. The line BC is a tangent to the circle at the point B such that the length of BC is 24. The length of the minor arc AB is 12.25. (a) Find the obtuse angle AOB, giving your answer in radians. [1] (b) Find the perimeter of the shaded region. [4]

11 0606/13/O/N/22 © UCLES 2022 [Turn over (c) Find the area of the shaded region. [2] 9 A 6-character password is to be formed from the following characters. Letters A B C D Numbers 1 2 3 4 Symbols * # $ £ No character may be used more than once in any password. (a) (i) Find the number of different 6-character passwords that can be formed. [1] (ii) How many of these 6-character passwords end with a symbol? [1] (b) Find the number of different 6-character passwords that include all the symbols, but do not start or end with a symbol. [2]

12 0606/13/O/N/22 © UCLES 2022 10 Solve the equation ( . ) ( . ) cos sin x x 2 3 1 2 2 3 1 2 + = + , where x is in radians, for . . x 1 5 1 5 G G - . [5]

13 0606/13/O/N/22 © UCLES 2022 [Turn over 11 It is given that ln x x x x 3 2 3 2 1 2 1 5 1 d a 1 + - + - = e o y , where a 1 2 . Find the exact value of a. [6]

14 0606/13/O/N/22 © UCLES 2022 12 It is given that y x x 3 2 1 2 3 2 = - - ` j , for x 1 2 . (a) Write x y d d in the form x x x Ax B 3 2 1 2 2 2 3 1 - - + + - ` ` ` j j j, where A and B are integers. [5] (b) Find the approximate increase in y as x increases from 2 to p 2+ , where p is small. [2]

15 0606/13/O/N/22 © UCLES 2022 13 The points P and Q have coordinates ( , ) 5 12 - and ( , ) 15 6 - respectively. The point R lies on the line l, the perpendicular bisector of the line PQ. The x-coordinate of R is 7. (a) Find the y-coordinate of R. [4] (b) The point S lies on l such that its distance from PQ is 3 times the distance of R from PQ. Find the coordinates of the two possible positions of S. [3]

16 0606/13/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 0606/13 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.

0606/13 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.

0606/13 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

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 4 of 10 Question Answer Marks Guidance 1 3 B1 for a curve starting at , 2 3    − −     and finishing at , 2 3    −     B1 for a curve, must have implied symmetry about π 6 and π 6 − , one complete cycle only. B1 for a curve passing through (0, –2) and distinct maximum at π , 2 6       and distinct minimum at only π , 6 6   − −     2(a) 2 1 121 2 4 8 x   + −     2 B1 for 1 4 a = B1 for 121 8 b = − 2(b) 1 121 , 4 8   − −     2 FTB1 for each, follow through on their a and b from (a) or SC1 if differentiation is used ie d 4 1 0 d y x x = + = then 1 121 , 4 8   − −     2(c) 3 B1 for a correct shape. Must have the parabola part of the curve in the first and second quadrant with cusps and correct curvature and a max in the 2nd quadrant. Ignore labelling of their maximum point if incorrect coordinates B1 for a curve 5 ,0 2       and (–3, 0) B1 for a curve (0, 15) 2(d) 121 8 k = B1 FT Follow through on their b −

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 5 of 10 Question Answer Marks Guidance 3(a)     2 2 3 2 1 0 oe or 4 4 3 0 oe y y x x + − = − − = M1 M1 for obtaining a 3 term quadratic equation in y or x and an attempt to solve 3 1 , oe 2 2 x x = = − A1 1, 1 oe 3 y y = = − A1 Allow A1 for the one correct pair e.g. 3 1 1 , or , 1 2 3 2     − −         3(b)     3 3 10 log 3 oe log 1 or 3 10log 3 oe log 3 x x x x + = + = B1 For change of base ( ) 2 3 3 log 3log 10 0 x x + − = or ( ) 2 10 log 3 3log 3 1 0 x x − −= 3 log 5 x = − 3 log 2 x = or 1 log 3 5 x = − 1 log 3 2 x = M1 Dep on previous B mark, for attempt to obtain a 3-term quadratic equation and attempt to solve to obtain 2 solutions of the form or 3–5 32 isw 2 A1 for each 4(a) ( ) 2 p 3 26 x ax x b  = + + ( ) p 0 b  = B1 Must see at least ( ) 2 p 3 26 x ax x b  = + + to award the mark 4(b) 2 p :8 27 318 3 a c   − − =     oe M1 For use of 2 3 x = − , at least once and attempt at simplification leading to an equation in a and c only Allow one sign error. ( ) p 1 : 16 a c − − = oe M1 For use of x = –1 and attempt at simplification leading to an equation in a and c only 6, 10 a c = = − 2 M1 dep on both previous M marks and attempt to solve simultaneously to obtain both a and c A1 for both 4(c) 2 2 3 5 x x + − B1 Allow if seen embedded i.e.: ( )( ) 2 3 2 2 3 5 x x x + + − or as a quotient in long division 3 log x p = log 3 x q =

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 6 of 10 Question Answer Marks Guidance 4(d) ( )( )( ) 3 2 1 2 5 x x x + − + B1 5(a) 3 1 soi 8 r = M1 Allow unsimplified 14 11 1 8 ar ar = or 11 12 14 15 5 5 8 5 5 r r r r − = − oe 1 2 r = A1 5 1 a r = − M1 For use of sum to infinity with their r, must be –1 < r < 1 5 2 a = A1 5(b) ( ) ( ) ( ) ( ) 1 1 n their r their their r −  − a M1 For use of the sum to n terms ( ) 0.0002 n their r = (12.29) M1 M1 dep For simplification and attempt to obtain the critical value using either an equation or an inequality leading to n = or n > 13 A1 Accept n ⩾ 13 6(a) f > –4 B1 Allow y > –4 or –4 < f <  ( ) or f 4, −  6(b) ( ) 1 1 f ln( 4) 3 x x −   = +   2 M1 for a correct method to find the inverse, allow one sign error Must be in the form of 3 ln( 4) or 3 ln( 4) x y y x =  =  A1 allow y =

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 7 of 10 Question Answer Marks Guidance 6(c) 4 B1 for f(x) with correct shape in quadrant 1, 3 and 4 and appropriate asymptotic behaviour B1 for –3 on the y-axis and 1 ln 4 3 on the x-axis for f(x) must have the correct shape B1 for f–1(x)with correct shape in quadrant 1, 2 and 3 and appropriate asymptotic behaviour B1 for –3 on the x-axis and 1 ln 4 3 on the y-axis for f–1(x) must have correct shape and intersect at least once 7 1sin3 2cos2 3 x x x − + 2 M1 for sin3 cos2 a x b x x + + , 3 and 8 a b   A1 all correct ( ) 1 3π π sin 2cosπ 2 3 2 2   − + −−     M1 Dep on previous M mark for correct substitution (seen or implied) of both limits in x 11 3 A1 π 2 B1 From correct substitution (seen or implied) of both limits in x 8(a) 1.75 B1 8(b) 7 24 24 cos , tan , sin 25 7 25 BOC BOC BOC = = = BOC = 1.287 soi B1 Arc length = r  their 1.287 B1 Follow through on their BOC Perimeter = 12.25 + their 9.009 + 14 M1 For a complete method 35.3 A1 8(c) 2 2 1 1 7 1.75 7 2 2 their BOC       +           oe or 2 2 1 π 7 7 (2π 1.75 1.287) 2 their  −   − − M1 For a complete method 74.4 A1 9(a)(i) 665 280 B1 9(a)(ii) 221 760 B1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 8 of 10 Question Answer Marks Guidance 9(b) 8  4  3  2  1  7 M1 For either 8 7  or 4! or 24 as part of a product 1344 A1 10 ( ) 1 3 1.2 tan 2 x   + =     or ( ) 2 2 cos 3 1.2 3 x   + =     or ( ) 2 1 sin 3 1.2 3 x   + =     M1 For an attempt to obtain an equation in ( ) 3 1.2 sin x + , ( ) 3 1.2 cos x + or ( ) 3 1.2 tan x + x = –1.24, –0.195, 0.852 or better 4 M1 dep for a correct attempt to obtain one correct solution A1 for one correct solution in the range M1 dep for an attempt to obtain another solution within the range A1 for 2 more correct solutions within the range and no extra solutions within the range 11 ( ) ( ) ln 3 2 ln 2 1 ln x x x + − + − 2 B1 for 1 correct term B1 for the other two terms correct ( ) ( ) 3 2 5 ln ln 2 1 3 a a a + − + 3(3 2) 1 ln ln 5 (2 1) 5 (3 2) 1 or ln ln (2 1) 3 a a a a a a +   =   +   + = + 2 M1 for application of limits correctly, dep on at least one B mark M1 for application of log laws to obtain a single logarithm, dep on at least one B mark a2 – 4a – 3 = 0 2 7 a = + 2 M1 for equating to 1 ln 5 and attempt to solve resulting 3-term quadratic equation, dep on at least one B mark A1 for 2 7 + must reject 2 7 −

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 9 of 10 Question Answer Marks Guidance 12(a) ( ) ( ) ( ) 1 2 2 2 3 3 2 d d 2 1 6 3 2 3 2 3 ( 1) y x x x x x x −   =     −   − − − − ( ) ( ) ( ) ( ) 1 2 1 2 2 2 3 3 d or d 2 1 6 3 2 1 3 2 3 y x x x x x x − − −   =     −   − − − − 3 B1 for ( ) 1 2 3 2 6 3 2 3 x x −  − M1 for differentiation of a quotient or product A1 for all terms other than ( ) 1 2 3 2 6 3 2 3 x x −  − correct ( ) ( ) ( ) 1 2 3 2 2 3 2 4 2 1 x x x x − − − + − 2 M1 dep for attempt to factorise, must be in the form ( ) ( ) ( ) ( ) 1 2 3 2 2 3 2 1 3 2 1 x ax x x x − −   − − −   − A1 all correct 12(b) When 3 d 2 2, oe d 10 y x x = = − M1 Dep on the differentiation M mark from part (a) For attempt to find the value of their d d y x when x = 2 3 2 or 0.928 10 p p − − A1 13(a) Midpoint (10, 9) B1 Gradient of l = 5 3 − B1 Equation of l: ( ) 5 9 10 3 y x + = − − oe M1 Must be using their perpendicular gradient and their mid-point y = –4 A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 10 of 10 Question Answer Marks Guidance 13(b) Attempt to use their R and displacement vectors or Pythagoras to find S M1 May be implied by one correct coordinate If Pythagoras is used: M1 for an attempt to reach to a 3- term quadratic with one variable using their equation and their midpoint from (a) e.g. 2 34 680 646 0 x x − + = (1, 6) A1 (19, –24) A1