Cambridge IGCSE Mathematics - Additional 0606 — 2021 Oct/Nov Paper 1 · Variant 3

This document has 16 pages. Any blank pages are indicated. Cambridge IGCSE™ DC (PQ/SG) 212423/2 © UCLES 2021 [Turn over * 2 7 0 9 1 9 0 5 2 6 * ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 2021 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 0606/13/O/N/21 © UCLES 2021 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/21 © UCLES 2021 [Turn over 1 On the axes below, sketch the graph of ( )( )( ) y x x x 4 1 2 1 3 4 =- + - + stating the intercepts with the coordinate axes. [3] 0 y x 2 A particle moves in a straight line such that its velocity, v ms 1 - , at time t seconds after passing through a fixed point O, is given by v 25 e t3 = - . Find the speed of the particle when t 1 = . [2]

4 0606/13/O/N/21 © UCLES 2021 3 Solve the equation cot x 2 3 3 1 2 r - = b l , where x is in radians and x 0 1 G r. [5]

5 0606/13/O/N/21 © UCLES 2021 [Turn over 4 (a) Find the first three terms, in ascending powers of x2, in the expansion of x 2 1 3 2 2 8 - e o . Write your coefficients as rational numbers. [3] (b) Find the coefficient of x2 in the expansion of x x x 2 1 3 2 2 1 2 8 2 - + e b o l . [3]

6 0606/13/O/N/21 © UCLES 2021 5 A geometric progression is such that its sum to 4 terms is 17 times its sum to 2 terms. It is given that the common ratio of this geometric progression is positive and not equal to 1. (a) Find the common ratio of this geometric progression. [3] (b) Given that the 6th term of the geometric progression is 64, find the first term. [2] (c) Explain why this geometric progression does not have a sum to infinity. [1]

7 0606/13/O/N/21 © UCLES 2021 [Turn over 6 (a) A 5-digit number is made using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. No digit may be used more than once in any 5-digit number. Find how many such 5-digit numbers are odd and greater than 70 000. [3] (b) The number of combinations of n objects taken 3 at a time is 2 times the number of combinations of n objects taken 2 at a time. Find the value of n. [4]

8 0606/13/O/N/21 © UCLES 2021 7 P B A O Q The diagram shows a circle, centre O, radius 10 cm. The points A, B and P lie on the circumference of the circle. The chord AB is of length 14 cm. The point Q lies on AB and the line POQ is perpendicular to AB. (a) Show that angle POA is 2.366 radians, correct to 3 decimal places. [2] (b) Find the area of the shaded region. [3]

9 0606/13/O/N/21 © UCLES 2021 [Turn over (c) Find the perimeter of the shaded region. [5]

10 0606/13/O/N/21 © UCLES 2021 8 The curves y x x 1 2 = + - and y x x 2 6 2 2 = + - intersect at the points A and B. (a) Show that the mid-point of the line AB is (2, 9). [5] The line l is the perpendicular bisector of AB. (b) Show that the point C (12, 7) lies on the line l. [3]

11 0606/13/O/N/21 © UCLES 2021 [Turn over (c) The point D also lies on l, such that the distance of D from AB is two times the distance of C from AB. Find the coordinates of the two possible positions of D. [4]

12 0606/13/O/N/21 © UCLES 2021 9 When e y 2 is plotted against x2, a straight line graph passing through the points (4, 7.96) and (2, 3.76) is obtained. (a) Find y in terms of x. [5] (b) Find y when x 1 = . [2] (c) Using your equation from part (a), find the positive values of x for which the straight line exists. [3]

13 0606/13/O/N/21 © UCLES 2021 [Turn over 10 A curve with equation ( ) y x f = is such that d d x y x 2 3 5 2 2 2 1 = + + - ` j for x 0 2 . The curve has gradient 10 at the point ,3 2 19 b l. (a) Show that, when x 11 = , x y 52 d d = . [5] (b) Find ( )x f . [4]

14 0606/13/O/N/21 © UCLES 2021 11 A curve has equation y x x 1 5 2 3 1 = + - ` j for x 1 2- . (a) Show that x y x x Ax Bx C 3 1 5 d d 2 2 2 3 2 = + - + + ` ` j j where A, B and C are integers. [6]

15 0606/13/O/N/21 © UCLES 2021 (b) Find the x-coordinate of the stationary point on the curve. [2] (c) Explain how you could determine the nature of this stationary point. [2] [You are not required to find the nature of this stationary point.]

16 0606/13/O/N/21 © UCLES 2021 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 the Cambridge Assessment Group. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which itself is a department of the University of Cambridge. BLANK PAGE

This document consists of 11 printed pages. © UCLES 2021 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 2021 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 2021 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 2021 © UCLES 2021 Page 2 of 11 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 2021 © UCLES 2021 Page 3 of 11 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 2021 © UCLES 2021 Page 4 of 11 Question Answer Marks Guidance 1 3 B1 for a cubic shape with a maximum in the first quadrant, a minimum in the third quadrant, extending into the second and 4th quadrants. The extensions must not curve incorrectly and not lead to a complete stationary point. B1 for x-intercepts 1 4, ,3 2 − − either on diagram or stated but must be with a cubic graph. B1 for y-intercept 3 either on diagram or stated but must be with a cubic graph. 2 4.91 =− v soi B1 Speed = 4.91 B1 3 π tan 2 3 3   − = ±     x soi or π 3 sin 2 3 2   − = ±     x soi B1 B0 if negative root is rejected Allow truncated decimals May be implied by subsequent work From use of 2 2 π π cosec 2 1 cot 2 3 3     − − = −         x x π π π 2π 4π 2 , , , 3 3 3 3 3 x − = − 2π 5π 2 0, , π, 3 3 x = π π 5π 0, , , 3 2 6 x = or 0, 1.05, 1.57, 2.62 or greater accuracy 4 M1 for correct order of operations to obtain one solution in the range using π tan 2 3   − =     x k or π sin 2 , 1 3   − = <     x m m Dep M1 for correct order of operations to obtain a second solution in the range using ( ) 1 π 2 tan π 3 −   − = ±     x k or ( ) 1 π 2 π sin , 1 3 −   − = − <     x m m oe ( ) 1 π 2 sin , 1 3 −   − = − <     x m m oe A1 for any pair of correct solutions A1 for remaining pair of solutions, with no extra solutions within the range 4(a) 2 4 1 7 256 24 36 − + x x 3 B1 for 1 256 B1 for 2 24 −x B1 for 4 7 36 x

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 5 of 11 Question Answer Marks Guidance 4(b) 2 2 1 4 4 + + x x soi B1 Coefficient of 2 x 1 4 256 1 4 24 7 36 their their their their their   ×       + × −       +    M1 Allow one sign error, but must have 3 terms in 2 x only, with an attempt at addition. 25 576 A1 5(a) ( ) ( ) 4 2 1 1 17 1 1 − − = − − a r a r r r M1 Allow equivalents Allow if ‘a’ terms missing (assume to have been cancelled) ( )( ) ( ) 2 2 2 1 1 17 1 − + = − r r r or better 4 2 17 16 0 − + = r r oe 3 2 16 16 0 + − − = r r r oe M1 Dep M1 for a correct simplified equation in r only r = 4 only, from correct working A1 5(b) 5 64 = ar M1 For use of 5 a r with their positive r 0.0625 = a or 1 16 A1 Must be exact A0 if 4 = r not from correct working in (a) 5(c) Because 1 > r oe B1 FT on their 1 > r Must have a value for r

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 6 of 11 Question Answer Marks Guidance 6(a) Either Starts with 8: 1680 B1 1680 must not be part of a product. May be implied by final answer Starts with 7 or 9: 2688 B1 May be implied by final answer Total: 4368 B1 Or Alternative 1 Starts with 7, 8 or 9 and ends in 1, 3 or 5: 3024 (B1) Allow for 1008 three times May be implied by final answer Starts with 8 or 9 and ends in 7: 672 Starts with 7 or 8 and ends in 9: 672 (B1) For both May be implied by final answer Total: 4368 (B1) Or Alternative 2 13 ways of obtaining odd 5-digit numbers which start with 7, 8 or 9 (B1) Needs to be part of a product. May be implied by final answer 8 3P ways of arranging the remaining 3 digits: 336 (B1) Needs to be part of a product. May be implied by final answer Total = 13 336 4368 × = (B1) Or Alternative 3 Last digit is 7 or 9: 1344 B1 May be implied by final answer Last digit is 1, 3 or 5: 3024 B1 May be implied by final answer Total: 4368 B1 Or Alternative 4 ( ) ( ) ( ) 10 9 8 8 5 4 3 3 P P 7 P 5 P 4 − × − × − × ( ) 8 3P 5 − × B2 Must be complete Total: 4368 B1 6(b) ( ) ( ) ! 2 ! 3 !3! 2 !2! = − − n n n n soi B1 ( )2 6 − = n soi B2 Dep B1 on first B for ( )2 − n soi Dep B1 on first B for 6 soi 8 = n B1 Dep on previous B marks

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 7 of 11 Question Answer Marks Guidance 7(a) 7 sin 10 = AOQ π = − POA AOQ or 2 2 2 14 10 10 200 cos = + − AOB oe 2π 2 − = AOB POA M1 Allow alternatives, but must be a complete method to find POA 2.366195157 POA = = 2.366 to 3 dp A1 Must see an angle correct to more than 3dp used in order to justify 3 dp 7(b) Area of sector = ( ) 2 110 2.366 2 ( ) 118.3 B1 Allow unsimplified. Also allow use of 2.37 Area of triangle = 2 110 sin2.366 2 ( ) 35 B1 Allow unsimplified. Also allow use of 2.37 Total area = awrt 153 B1 Allow greater accuracy 7(c) Major arc PB = 10 2.366 × B1 Allow unsimplified. Also allow use of 2.37 2 sin 2 10 = AP POA or 2 2 2 10 10 200 cos = + − AP POA M1 For a valid attempt to find AP – may be seen in (a) or (b) but AP must be stated in this part. 18.5 = AP A1 Allow awrt 18.5 Perimeter: major arc PB + 20 + their AP B1 For plan, may be implied, but must have an attempt to calculate AP Total perimeter = 62.2 A1 Allow awrt 62.2 8(a) 2 2 2 2 2 6 2 + − = + − x x x x 2 4 0 − = x x ( )4 0 − = x x 0, 4 = = x x M1 For obtaining an equation in one variable M1 Dep for a correct attempt to obtain at least one solution ( ) 0, 1 − A1 nfww ( ) 4, 19 A1 nfww Mid-point ( ) 2, 9 with sufficient detail B1 AG

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 8 of 11 Question Answer Marks Guidance 8(b) Either Gradient of perpendicular 1 5 = − M1 ( ) 1 9 2 5 − = − − y x M1 Dep on previous M mark for perpendicular bisector using their mid- point and their perpendicular gradient ( ) 1 7 9 12 2 5 − = − − oe A1 For checking by substitution, must see evidence. Or Alternative 1 Gradient of perpendicular 1 5 = − (M1) ( ) 1 7 12 5 − = − − y x (M1) Dep on previous M mark for perpendicular bisector using ( ) 12, 7 and their perpendicular gradient ( ) 1 9 7 2 12 5 − = − − oe (A1) For checking by substitution, must see evidence Or Alternative 2 Gradient of perpendicular 1 5 = − (M1) Gradient of line joining their ( ) 2, 9 to ( ) 12, 7 = 1 5 = − (M1) ( ) 2, 9 is a common point and gradients of perpendicular bisector and l are the same so C lies on l. (A1) 8(c) ( ) 22,5 2 B1 for 22 B1 for 5 ( ) 18,13 − 2 B1 for 18 − B1 for 13

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 9 of 11 Question Answer Marks Guidance 9(a) 2 2 e = + y mx c B1 May be implied by later work Either 7.96 4 = + m c 3.76 2 = + m c M1 2.1 = m oe A1 0.44 =− c oe A1 ( ) 2 1 ln 2.1 0.44 2 = − y x oe A1 Do not isw Or gradient = 2.1 oe (B1) Use of either 7.96 4 = + m c or 3.76 2 = + m c (M1) For use with their m 0.44 =− c oe (A1) ( ) 2 1 ln 2.1 0.44 2 = − y x oe (A1) Must be bracketed correctly 9(b) ( ) 2 1 ln 2.1 0.44 2 = − y their x their oe M1 Must use the form ( ) 2 ln = ± y k px q 1 ≠ p and 0 ≠ q or 2 2 e = + y mx c 0.253 A1 9(c) 2 2.1 0.44 0 their x their − > or = 0 or ≥ 0 soi B1 Correct attempt to obtain the critical value using 2 2.1 0.44 0 their x − = M1 Must be from the form ( ) 2 ln = − y k px q , 1 ≠ p and 0 > q 0.458 > x or 22 105 x > oe A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 10 of 11 Question Answer Marks Guidance 10(a) ( ) ( ) 1 2 d 2 3 5 d y x x c x = + + + B1 For ( ) 1 2 2 3 x + , allow unsimplified M1 For ( ) 1 2 2 3 5 k x x + + 10 3 15 = + +c M1 Dep for use of 10 and 3 = x in their d d y x to obtain c ( ) 1 2 d 2 3 5 8 d y x x x = + + − soi A1 When d 11, 5 55 8 d = = + − y x x oe = 52 A1 AG – need to see sufficient detail 10(b) ( ) ( ) ( ) 2 3 2 1 5 f 2 3 8 3 2 x x x x d = + + − + B1 For ( ) 3 2 1 2 3 3 + x , must be ( ) 1 2 2 3 d +  x x M1 For ( ) 2 3 2 5 2 3 2 x k x + + 19 27 45 24 2 3 2 d = + − + d = 2 M1 For use of 19 2 = y and x = 3 in their y ( ) ( ) ( ) 2 3 2 1 5 f 2 3 8 2 3 2 x x x x = + + − + A1 Allow 8 − if obtained from using d 52 d = y x in (a) rather than d 10 d = y x 11(a) ( ) ( ) ( ) ( ) 2 1 2 2 3 3 2 d d 1 1 2 5 5 3 1 − =   + × × − − −       + y x x x x x x or ( ) ( ) ( ) ( ) ( ) 2 1 2 3 1 2 2 3 1 1 2 5 3 5 1 − − −   + × × −       + − − + x x x x x 3 B1 for ( ) 2 2 3 1 2 5 3 x x − × × − M1 for an attempt at a quotient or a correct product A1 for all other terms correct ( ) ( ) 2 2 2 2 3 2 15 3 1 5 − + + + − x x x x 3 Dep on first 3 marks A1 for 2 −x in a quadratic numerator A1 for 2x in a quadratic numerator A1 for 15 in a quadratic numerator

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 11 of 11 Question Answer Marks Guidance 11(b) 2 2 15 0 x x − + + = M1 For attempt to solve their 2 2 15 0 − + + = x x to obtain x = .. Must be a quadratic equation. x = 5 only A1 11(c) Either Find the gradient either side of the stationary point B1 If gradient changes from +ve to -ve: max If gradient changes from -ve to +ve: min B1 Dep on previous B1 Or Alternative 1 Take the second derivative and substitute in the value of x obtained in (b) (B1) Allow alternative valid methods Allow a general method as this is not dependent on (a) or (b) If second derivative is + ve, then a min If second derivative is – ve, then a max (B1) Dep on previous B1 Or Alternative 2 Consider a y-value to one side of the stationary point (B1) Allow alternative valid methods Allow a general method as this is not dependent on (a) or (b) If y value of stationary point is greater, then a max. If y value of stationary point is less, then a min. (B1) Dep on previous B1