Cambridge IGCSE Mathematics - Additional 0606 — 2023 Oct/Nov Paper 1 · Variant 2

This document has 16 pages. [Turn over Cambridge IGCSE™ * 2 5 3 4 1 2 2 8 6 6 * DC (LK/JG) 317827/2 © UCLES 2023 ADDITIONAL MATHEMATICS 0606/12 Paper 1 October/November 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 [ ].

2 0606/12/O/N/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/O/N/23 © UCLES 2023 [Turn over 1 0 x y 3 1 - -15 2 5 1 The diagram shows the graph of the cubic polynomial ( ) f y x = . (a) Find an expression for ( ) f x in factorised form. Write each linear factor with its coefficients as integers. [3] (b) Write down the values of x such that ( ) f x 0 1 . [2]

4 0606/12/O/N/23 © UCLES 2023 2 The function g is defined by ( ) g sin x x 5 4 3 2 = - for all values of x. (a) Write down the amplitude of g. [1] (b) Write down the period of g in degrees. [1] (c) On the axes, sketch the graph of ( ) g y x = , for ° ° x 180 180 G G - . [3] – 8 – 5 – 6 – 7 – 2 – 1 – 3 – 4 2 3 1 5 4 8 7 6 0 x y – 180° – 120° – 60° 60° 120° 180°

5 0606/12/O/N/23 © UCLES 2023 [Turn over 3 When ( ) ln y 2 + is plotted against x2 a straight line graph is obtained. The line passes through the points ( . , . ) 2 25 9 37 and ( . , . ) 4 75 3 92 . Find y in terms of x. [5]

6 0606/12/O/N/23 © UCLES 2023 4 (a) It is given that the first four terms, in ascending powers of x, in the expansion of x 1 2 n - b l can be written in the form x px qx 1 8 2 3 - + + , where n, p and q are integers. Find the values of n, p and q. [5] (b) Find the term independent of x in the expansion of x x 2 2 6 + 3 e o , giving your answer as a rational number. [2]

7 0606/12/O/N/23 © UCLES 2023 [Turn over 5 Solve the equation r sec 3 2 6 4 2 i+ = b l for r r 2 2 1 1 i - , giving your answers in terms of r. [5]

8 0606/12/O/N/23 © UCLES 2023 6 The polynomial ( ) p x is such that ( ) p x ax bx cx 5 3 2 = + + - , where a, b and c are integers. It is given that ( ) p 0 12 = l . It is also given that ( ) p x has a factor of x 3 1 - and a remainder of 95 when divided by x 2 - . (a) Find the values of a, b and c. [7] (b) Show that the equation ( ) p x 0 = has only one real root. [3]

9 0606/12/O/N/23 © UCLES 2023 [Turn over 7 (a) A 6-digit number is to be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. Each digit can be used only once in any 6-digit number. A 6-digit number cannot start with 0. (i) Find how many 6-digit numbers can be formed. [1] (ii) Find how many of these 6-digit numbers are divisible by 5. [3] (b) A committee of 7 people is to be chosen from 6 doctors, 10 nurses and 8 dentists. (i) Find the number of committees that can be chosen. [1] (ii) Find the number of committees that can be chosen if all the doctors have to be on the committee. [1] (iii) Find the number of committees that can be chosen if there has to be at least one dentist on the committee. [2]

10 0606/12/O/N/23 © UCLES 2023 8 (a) It is given that : ( ) f x x 3 1 4 2 " + - for x a H , and that f 1 - exists. (i) Find the least possible value of a. [1] (ii) Using this value of a, write down the range of f. [1] (iii) Using this value of a, sketch the graphs of ( ) f y x = and ( ) f y x 1 = - on the axes, stating the intercepts with the coordinate axes. [4] y x O

11 0606/12/O/N/23 © UCLES 2023 [Turn over (b) It is given that ( ) ( ) g ln x x 2 5 2 = + for x 0 H , ( ) h x x 3 2 = - for x 0 H . Solve the equation ( ) hg x 4 = giving your answer in exact form. [3] 9 Solve the equation x x 12 5 11 0 3 2 3 2 - - = - for x 0 2 . Give your answer correct to one decimal place. [4]

12 0606/12/O/N/23 © UCLES 2023 10 In this question all lengths are in centimetres and all angles are in radians. C A D B O 12 27 The diagram shows a badge which consists of a minor sector, OAB, of the circle with centre O and radius 12, and a kite OBCD, where OB OD = and CD CB = . The arc AB has length 27. The line OB is perpendicular to the line CB, and COA is a straight line. (a) Find the perimeter of the badge. [4]

13 0606/12/O/N/23 © UCLES 2023 [Turn over (b) Find the area of the badge. [3]

14 0606/12/O/N/23 © UCLES 2023 11 A O X B Y Z In the triangle OAB, a OA = and b OB = . The mid-point of the line OB is X, and the mid-point of the line AB is Y. The lines OY and AX intersect at the point Z. It is given that AZ AX m = and OZ OY n = where m and n are rational numbers. (a) Find OZ in terms of a, b and m. [3] (b) Find OZ in terms of a, b and n. [2]

15 0606/12/O/N/23 © UCLES 2023 [Turn over (c) Find the values of m and n. [3] (d) Hence find OZ in terms of a and b only. [1] Question 12 is printed on the next page.

16 0606/12/O/N/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. 12 A curve has equation y x x 5 2 3 = - - . (a) Explain why the curve does not exist when x 5 2 1 . [1] (b) Show that d d x y can be written in the form ( ) ( ) x x Ax B 2 3 5 2 2 - - - + , where A and B are positive integers. [5]

This document consists of 9 printed pages. © UCLES 2023 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 October/November 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 October/November 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 October/November 2023 © UCLES 2023 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 3 of 9 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. 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 2023 © UCLES 2023 Page 4 of 9 Question Answer Marks Guidance 1(a) ( ) ( ) ( )( )( ) f or 3 3 1 1 2 5 = − + − − x y x x x 3 B1 for ( ) 1 5 1 3 2     + − −         k x x x and no other work that would gain marks. B2 for ( )( )( ) 3 1 1 2 5 + − − m x x x and no other work that would gain marks. 1(b) 1 1 3 −   x B1 Must be in terms of x 5 2  x B1 Must be in terms of x 2(a) 5 B1 2(b) o 480 B1 2(c) 3 To obtain any marks the graph must be a curve with one min in the third quadrant and one max in the first quadrant. B1 for the shape, starting in the 3rd quadrant and ending in the 1st quadrant. Must cross the x-axis only once, between 0o and 60o. Must extend for the complete domain starting with 6 5 − − y and ending with 1 2   y B1 for passing through( ) 0, 2 − B1 for passing though ( ) o 120 , 3 and ( ) o 120 , 7 − − soi 3 ( ) 2 ln 2 + = + y mx c soi B1 Either of: 9.37 2.25 = + m c 3.92 4.75 = + m c M1 For at least one correct equation involving m and c 109 2.18, oe 50 = − − m 571 14.3, 14.28, 14.275, 40 = c 2 Dep M1 for attempt to solve for at least one unknown. A1 for both. ( ) 2 14.3 2.18 e 2 − = − x y oe A1 FT on the first M1 for their m and c

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 5 of 9 Question Answer Marks Guidance 3 Alternative ( ) 2 ln 2 + = + y mx c soi (B1) Gradient = –2.18, 109 oe 50 − (B1) 9.37 2.25 = + m c 3.92 4.75 = + m c (M1) Use of a correct equation with their gradient and c 571 14.3, 14.28, 14.275, 40 = c (A1) ( ) 2 14.3 2.18 e 2 − = − x y oe (A1) FT on their m and c 4(a) 16 = n B1 ( ) ( ) 2 1 2! 2 −   + −     n n x , ( ) 2 2 C 2   −     n x oe ( )1 , 30 8 − = = n n p p 2 M1 for attempt at third term allow unsimplified in terms of n or their n, but not just as part of an expansion unless used to find p A1 for p. ( )( ) ( ) 3 1 2 3! 2 − −   + −     n n n x , ( ) 3 3 C 2   −     n x ( )( ) 1 2 , 70 48 − − = = − n n n q q 2 M1 for attempt at fourth term, allow unsimplified in terms of n or their n, but not just as part of an expansion unless used to find q A1 for q. 4(b) 2 4 6 4 2 2 3          x C x B1 For identifying the correct term and attempting to evaluate. 20 27 B1 5 ( ) π 3 cos 2 6 2    + =      oe or ( ) π 1 tan 2 6 3    + =      oe B1 π π , 0, 6 3 = − oe 4 M1 for a correct order of operations, may be implied by one correct solution. A1 for 1 correct solution. A1 for a 2nd correct solution A1 for a 3rd correct solution with no extra solutions in the range. All solutions must be from correct working.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 6 of 9 Question Answer Marks Guidance 6(a) 12 = c B1 1 p : 5 0 3 27 9 3   + + − =     a b c soi M1 Allow one arithmetic or sign error, may substitute in their c. ( ) p 2 : 8 4 2 5 95 + + − = a b c M1 Allow one arithmetic or sign error, may substitute in their c. 3 27 + = a b or 1 27 9 + = a b oe A1 Allow multiples but c needs to have been eliminated and terms with powers evaluated. 2 19 + = a b oe A1 Allow multiples but c needs to have been eliminated 6, 7 = = a b 2 M1, dep on at least one previous M1, for attempt to solve their equations in a and b only, to find a or b. A1 for both a and b. 6(b) ( )( ) 2 3 1 2 3 5 − + + x x x cao 2 M1 for attempt at 2 terms in their quadratic factor. A1 for both factors. For 2 2 3 5 0, + + = x x discriminant is less than zero, so no solutions. [Only solution is 1. 3 = x ] B1 Allow other valid arguments, but must be using a correct quadratic factor and an attempt to evaluate the discriminant 7(a)(i) 136 080 B1 7(a)(ii) (End in 0) 15 120 or 9 5P or 9 8 7 6 5  B1 (End in 5) 13 440 or 8 4 8 P  or 8 8 7 6 5  B1 Total: 28 560 B1 Alternative 1 (Does not start with 5: ) 26 880 or 8 4 16 P  (B1) (Starts with 5:) 1680 or 8 4 P (B1) Total: 28 560 (B1)

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 7 of 9 Question Answer Marks Guidance 7(a)(ii) Alternative 2 (Number not divisible by 5:) 107 520 or 8 4 8 8 P  (B1) 136 080 – 107 520 (B1) FT on their 136 080 Total: 28 560 (B1) 7(b)(i) 346 104 B1 7(b)(ii) 18 B1 Do not isw subsequent work 7(b)(iii) the number of committees with no dentists: 11 440 M1 Allow attempts at 7 options, but must have all of them: 8, 448, 6720, 39 200, 101 920, 122 304 and 64 064 334 664 A1 May come from 24 16 7 7 C C − 8(a)(i) 1 3 = − a or 1 3 − x B1 Allow 0.333 − or better Allow a correct recurring decimal 8(a)(ii) f 4 − B1 8(a)(iii) 4 B1 for ( ) f , = y x must have a correct shape (right hand side from the vertex of a quadratic curve), must be a 1:1 function, intersecting each of the x and y axes once, in quadrants 1, 3 and 4. B1, dependent on previous B for passing through( ) 0, 3 − and 1 , 0 3       . B1 dependent on first B1 for ( ) 1 f , − = y x being a correct reflection of their ( ) f , = y x intersecting each of the x and y axes and ( ) f = y x once. B1 dependent on previous B for passing through ( ) 3, 0 − and 1 0, 3       . 8(b) ( ) ( ) ( ) 2 3 ln 2 5 2 4 + − = x M1 For correct order 2e 5 2 − = x or exact equivalent 2 M1 dep for a correct attempt to deal with logarithms and obtain x = … Allow one arithmetic or sign slip.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 8 of 9 Question Answer Marks Guidance 9 2 2 2 3 3 12 11 5 0     − − =             x x B1 For recognition of a 3-term quadratic equation in terms of 2 3 x or a suitable substitution 2 2 3 3 3 1 4 5 0    + − =          x x 2 2 3 3 1 5 , 3 4 = − = x x 2 M1 for attempt to solve a 3-term quadratic equation in the form 2 12 11 5 0   = u u to obtain at least one solution in the form 2 3 ... = x or ‘u’ = … A1 for at least one correct solution. 1.4 = x only A1 10(a) 27 12 = 9 4 = oe B1   = AOB Either ( ) tan π 12  − = CB soi Or ( ) 12 π sin π sin 2   = −   −     CB M1 Allow with their .  Perimeter = 24 27 2(14.86..) + + M1 Allow with their CB. Perimeter = awrt 80.7 A1 From correct working only 10(b) ( ) 2 1 12 12 2      +      their their CB 340 oe or 341 oe 3 M1 for each area A1 for awrt 340 or 341 11(a) 2 = − AX b a B1 Allow unsimplified 2   = + −     OZ b a a 2 M1 for 2    = +  −     OZ their b a a Allow unsimplified Mark final answer 11(b) ( ) 1 2 = + OY a b oe B1 Allow unsimplified ( ) 2  = + OZ a b B1 FT on their OY , allow unsimplified Mark final answer

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 9 of 9 Question Answer Marks Guidance 11(c) 2   + −     b a a = ( ) 2  + a b M1 For equating their final answer for (a) and their final answer for (b) and attempt to equate like vectors at least once to obtain a scalar equation 2 3   = = 2 M1 dep for solving their simultaneous equations to obtain at least one unknown. Each equation must be in terms of and  A1 for both. 11(d) ( ) 1 3 = + OZ a b oe B1 Must be from correct work 12(a) Cannot have the square root of a negative number. oe B1 Must be a correct statement related to the question. Allow a numerical argument. 12(b) ( ) ( ) ( ) ( ) 1 1 2 2 2 5 3 5 2 5 2 2 3 −   −  − − −     − x x x x or ( ) ( ) 1 1 2 5 3 5 2 2 x x − − − − ( ) ( )( ) 1 2 2 3 5 2 x x − + − − − 3 B1 for ( ) 1 2 5 5 2 2 −  − x seen M1 for an attempt to differentiate a quotient or correct product A1 for all other terms correct. ( ) ( ) ( ) ( ) ( ) 1 2 2 5 2 5 3 2 5 2 2 3 − − − − − − x x x x oe M1 Dep for an attempt to simplify to the given form, allow a sign error and an arithmetic error (e.g. Omission of a factor of 2 in the linear term). ( ) ( ) 2 5 11 2 3 5 2 − + − − x x x cao A1