Cambridge IGCSE Mathematics - Additional 0606 — 2020 Oct/Nov Paper 1 · Variant 3

DC (PQ/FC) 187995/4 © UCLES 2020 [Turn over This document has 16 pages. Blank pages are indicated. * 2 9 5 0 4 6 8 6 3 2 * ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 2020 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 [ ]. Cambridge IGCSE™

2 0606/13/O/N/20 © UCLES 2020 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x a b b ac 2 4 2 ! = - - Binomial Theorem ( ) … … a b a n a b n a b n r a b b 1 2 n n n n n r r n 1 2 2 + = + + + + + + - - - J L KK J L KK J L KK N P OO N P OO N P OO where n is a positive integer and ( )! ! ! n r n r r n = - J L KK N P OO Arithmetic series u a n d S n a l n a n d 1 2 1 2 1 2 1 n n − − = + = + = + ^ ^ ^ h h h # - Geometric series u ar S r a r r S r a r 1 1 1 1 1 n n n n 1 1 ! − − − = = = 3 − ^ ^ ^ h h h 2. TRIGONOMETRY Identities sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A Formulae for ∆ABC sin sin sin A a B b C c = = a2 = b2 + c2 – 2bc cos A ∆ = 2 1 bc sin A

3 0606/13/O/N/20 © UCLES 2020 [Turn over 1 (a) On the axes below, sketch the graph of ( )( )( ) y x x x 2 1 3 = - + - , stating the intercepts on the coordinate axes. O y x [3] (b) Hence write down the values of x such that ( )( )( ) x x x 2 1 3 0 2 - + - . [2]

4 0606/13/O/N/20 © UCLES 2020 2 (a) Given that y x 1 e x 2 2 3 = + - , find x y d d . [3] (b) Hence, given that y is increasing at the rate of 2 units per second, find the exact rate of change of x when x 2 = . [3]

5 0606/13/O/N/20 © UCLES 2020 [Turn over 3 (a) ( ) ( ) ln x x 4 2 1 f = - (i) Write down the largest possible domain for the function f. [1] (ii) Find ( )x f 1 - and its domain. [3] (b) ( )x x 5 g = + for x R ! ( )x x 2 3 h = - for x 2 3 H Solve ( )x 7 gh = . [3]

6 0606/13/O/N/20 © UCLES 2020 4 (a) x m t s 0 50 65 85 125 50 100 150 The diagram shows the x–t graph for a runner, where displacement, x, is measured in metres and time, t, is measured in seconds. (i) On the axes below, draw the v–t graph for the runner. [3] v ms–1 t s 0 1 2 3 4 - 4 - 3 - 2 - 1 50 65 85 125 (ii) Find the total distance covered by the runner in 125 s. [1]

7 0606/13/O/N/20 © UCLES 2020 [Turn over (b) The displacement, x m, of a particle from a fixed point at time t s is given by cos x t 6 3 3 r = + b l. Find the acceleration of the particle when t 3 2r = . [3] 5 Given that the coefficient of x2 in the expansion of ( )x x 1 1 2 n + - b l is 4 25, find the value of the positive integer n. [5]

8 0606/13/O/N/20 © UCLES 2020 6 It is known that y A 10bx2 # = , where A and b are constants. When lgy is plotted against x2, a straight line passing through the points (3.63, 5.25) and (4.83, 6.88) is obtained. (a) Find the value of A and of b. [4] Using your values of A and b, find (b) the value of y when x 2 = , [2] (c) the positive value of x when y 4 = . [2]

9 0606/13/O/N/20 © UCLES 2020 [Turn over 7 The polynomial ( )x ax bx x 19 4 p 3 2 = + - + , where a and b are constants, has a factor x 4 + and is such that ( ) ( ) 2 1 5 0 p p = . (a) Show that ( ) ( )( ) x x Ax Bx C 4 p 2 = + + + , where A, B and C are integers to be found. [6] (b) Hence factorise ( )x p . [1] (c) Find the remainder when ( )x pl is divided by x. [1]

10 0606/13/O/N/20 © UCLES 2020 8 In this question all lengths are in centimetres. C O B A r 0.5 r 1.45 r The diagram shows the figure ABC. The arc AB is part of a circle, centre O, radius r, and is of length 1.45r. The point O lies on the straight line CB such that . CO r 0 5 = . (a) Find, in radians, the angle AOB. [1] (b) Find the area of ABC, giving your answer in the form kr2, where k is a constant. [3]

11 0606/13/O/N/20 © UCLES 2020 [Turn over (c) Given that the perimeter of ABC is 12 cm, find the value of r. [4]

12 0606/13/O/N/20 © UCLES 2020 9 C O B b a X A Y The diagram shows the triangle OAC. The point B is the midpoint of OC. The point Y lies on AC such that OY intersects AB at the point X where : : AX XB 3 1 = . It is given that OA a = and b OB = . (a) Find OX in terms of a and b, giving your answer in its simplest form. [3] (b) Find AC in terms of a and b. [1]

13 0606/13/O/N/20 © UCLES 2020 [Turn over (c) Given that OY OX h = , find AY in terms of a, b and h. [1] (d) Given that AY AC m = , find the value of h and of m. [4]

14 0606/13/O/N/20 © UCLES 2020 10 (a) Show that x x 1 1 3 10 2 + + + can be written as x x x 3 13 10 5 12 2 + + + . [1] (b) y P O Q x = 2 x y x x x 3 13 10 5 12 2 = + + + The diagram shows part of the curve y x x x 3 13 10 5 12 2 = + + + , the line x 2 = and a straight line of gradient 1. The curve intersects the y-axis at the point P. The line of gradient 1 passes through P and intersects the x-axis at the point Q. Find the area of the shaded region, giving your answer in the form ln a b 3 2 3 + ` j, where a and b are constants. [9]

15 0606/13/O/N/20 © UCLES 2020 [Turn over Additional working space for question 10 Question 11 is printed on the next page.

16 0606/13/O/N/20 © UCLES 2020 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. 11 (a) Given that cos tan x x 2 3 = , show that sin sin x x 2 3 2 0 2 + - = . [3] (b) Hence solve cos tan 2 2 2 4 4 3 r r a a+ + = b b l l for 0 1 1 r a radians, giving your answers in terms of r. [4]

This document consists of 10 printed pages. © UCLES 2020 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 2020 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 2020 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 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 2020 © UCLES 2020 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 2020 © UCLES 2020 Page 4 of 10 Question Answer Marks Guidance 1(a) 3 B1 for a well-drawn cubic graph in correct orientation Both arms extending beyond x-axis Maximum above x-axis B1 for x-intercepts B1 for y-intercept 1(b) 1 x < − B1 Dep on a cubic curve in the correct orientation and –1 correct on x-axis 2 3 x < < or 3 2 x > > B1 Dep on a cubic curve in the correct orientation and 2 and 3 correct on x-axis 2(a) ( ) ( ) 2 2 3 2 3 2 2 1 2e 2 e d d 1 x x x x y x x − − + − = + oe or ( ) ( ) 2 3 2 3 2 2 2 d 2e 2 e d 1 1 x x y x x x x − − = − + + oe 3 B1 for ( ) 2 3 d e d x x − = 2 3 2e x−seen in a quotient rule or product rule expression M1 for correct method for differentiating a quotient or equivalent product A1 FT from their 2 3 2e x− 2(b) When 2 x = , d 6e d 25 y x = M1 For evaluation of their d d y x when 2 x = 6e d 2 25 d x t × = oe M1 For correct substitution of their evaluated d d y x and d 2 dt y = in a correct rates of change equation d 25 d 3e x t = , 50 6e A1 3(a)(i) 1 2 x > B1 Must be using x

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 5 of 10 Question Answer Marks Guidance 3(a)(ii) ( ) 4ln 2 1 x y = − 4 e 2 1 x y = − 4 1 1 e 2 x y   = +       M1 For full method for inverse using correct order of operations ( ) 1 4 1 f 1 e 2 x x −   = +       or ( ) ( ) 4 1 1 f 1 e 2 x x − = + A1 Must be using correct notation x∈ B1 3(b) 2 3 5 7 x − + = M1 For correct order 2 2 3 2 x + = M1 Dep on previous M mark, for obtaining x by simplifying and solving using correct order of operations, including squaring 7 2 x = or 3.5 A1 4(a)(i) 3 B1 For v = 2 for 0 ⩽ t ⩽ 50 B1 For v = 2.5 for 65 ⩽ t ⩽ 85 B1 For v = 3.75 for 85 ⩽ t ⩽ 125 and v = 0 for 50 ⩽ t ⩽ 65 4(a)(ii) 300 B1 4(b) d π 18sin 3 d 3 x t t   = − +     M1 π 18sin 3 3 t   ± +     2 2 d π 54cos 3 3 d x t t   = − +     M1 π 54cos 3 3 t   ± +     27 − nfww A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 6 of 10 Question Answer Marks Guidance 5 ( ) ( ) 2 1 1 1 ... 2 2 4 n n x x x n     −   + + − +               2 B1 For 1 2 n x   −      B1 For 2 2 4 n x         2 2 2 1 1 25 2 1 4 2 4 n n x x x   − =     M1 Correctly using two terms in n to obtain an 2 x term and equating to 2 25 4 x Dep on one B1 ( ) 1 25 8 2 4 n n n − − = oe A1 10 n = only A1 6(a) 2 lg lg y A bx = + B1 Stated or may be implied by later work If using 2 lg lg y A bx = + as a starting point 5.25 lg 3.63 A b = + and 6.88 lg 4.83 A b = + or 5.25 lg 1.358(3.63) A = + or ( ) 6.88 lg 1.358 4.83 A = + OR If finding the equation of the straight line and then finding lg A and b by inspection 2 lg 6.88 1.358( 4.83) y x − = − or 2 lg 5.25 1.358( 3.63) y x − = − or 2 lg 1.358 0.31..(or 0.32..) y x = + M1 For correctly finding required equation(s) 1.36, b = 163 43 or 1 120 120 B1 Must be b = and from correct working A in range 2.05 to 2.09 A1 6(b) ( ) lg 0.3132 4 1.36 y = + × 4 1.36 2.09 10 y × = × M1 For ( ) ( ) lg lg 4 y their A their b = + or ( )( ) 4( ) 10 their b y their A = Allow 553 000 to 576 000 A1 6 (c) ( ) 2 1.36 4 2.09 10 x = or 2 lg 4 0.3132 1.36x = + M1 ( )( ) 2 4 10their bx their A = or ( ) ( ) 2 lg4 lg their A their b x = + awrt 0.46 A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 7 of 10 Question Answer Marks Guidance 7(a) 4 5 0 a b − + + = oe B1 Allow multiples of equation 25 0 a b + − = oe B1 Allow multiples of equation 6, 19 a b = = 2 M1 for solving their 2 equations and obtaining two solutions A1 for both 6, 19 a b = = ( )( ) 2 4 6 5 1 x x x + − + 6, 5, 1 A B C = = − = 2 M1 for attempt to obtain quadratic factor by inspection or by algebraic long division A1 ( ) 2 6 5 1 x x − + or 6, 5, 1 A B C = = − = Alternative 25 0 a b + − = oe (B1) Allow multiples of equation Comparing coefficients 1 and C A a = = (B1) 4A B b + = (B1) Leading to 5 25 A B + = (M1) For use of their 25 0 a b + − = to obtain an equation in A and B 4 1 19 B + = − (B1) ( )( ) 2 4 6 5 1 x x x + − + 6, 5, 1 A B C = = − = (A1) 7(b) ( )( )( ) 4 3 1 2 1 x x x + − − B1 Must follow from a correct solution to (a) 7(c) –19 B1 8(a) 1.45 AOB ∠ = (radians) B1 8(b) Sector area ( )( ) 2 1 1.45 2 r = B1 For correct sector area. Allow unsimplified Area of triangle ( ) 1 0.5 sin π 1.45 2 r r their = × × × − B1 For correct area of triangle Allow unsimplified Total area = 2 0.973r B1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 8 of 10 Question Answer Marks Guidance 8(c) (AC2) = r2 + 0.25r2 – (2 × r × 0.5r cos(π – 1.45)) M1 For correct substitution in cosine rule using ( ) π 1.45 their − 1.17 AC r = A1 Perimeter = 2.95 1.17 r r + B1 FT on their AC 2.91 r = A1 9(a) AB = − b a  or BA = − a b  B1 3 4 OX AB = + a   or 1 4 OX BA = + b   ( ) 3 4 OX = + − a b a  or ( ) 1 4 OX = + − b a b  M1 For correct use of ratio, using their AB  or BA  3 4 4 OX = + a b  A1 9(b) 2 AC = − b a  B1 9(c) 3 4 4 AY h  = −+ +     a a b  B1 FT on their OX  9(d) ( ) 3 2 4 4 h m   −+ + = −     a a b b a M1 For equating appropriate vectors and attempt to equate like vectors 1 4 h m −+ = − A1 FT from their AY  and AC  3 2 4 h m = A1 FT from their AY  and AC  8 3 , 5 5 h m = = A1 For both 10(a) ( ) ( )( ) 3 10 2 1 1 3 10 x x x x + + + + + = ( )( ) 3 10 2 2 1 3 10 x x x x + + + + + = 2 5 12 3 13 10 x x x + + + B1 For expansion and simplification to obtain given answer

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 9 of 10 Question Answer Marks Guidance 10(b) 6 0, 5 P      and 6 , 0 5 Q  −     oe B1 Area of triangle = 18 25 or 0.72 B1 Area under curve = 2 0 1 2 d 1 3 10 x x x + + +  M1 For use of part (a) and attempt to integrate to obtain at least one ln term. ( ) ( ) 2 0 2 ln 1 ln 3 10 3 x x   = + + +     2 B1 For ( ) ln 1 x + B1 For ( ) 2 3 ln 3 10 x + 2 2 ln3 ln16 ln10 3 3 = + − M1 For correct use of limits. Dep on previous M mark. 2 2 2 ln3 3 ln16 ln10 3 3 3 = + − M1 For use of 2 ln3 ln3 3 3 = 2 2 16 ln3 3 ln 3 3 10   = +     2 48 3 ln 3 10   =       M1 For use of multiplication and division rule Total area = 18 25 2 24 3 ln 3 5   +       oe A1 For correct answer in the required form dep on the three preceding M marks Must not be obtained using a calculator 11(a) sin 2cos 3cos x x x = 2 2cos 3sin x x  = M1 For use of sin tan cos x x x = and multiplying by cosx ( ) 2 2 1 sin 3sin x x − = M1 For use of correct identity 2 2sin 3sin 2 0 x x + − = A1 For correct rearrangement to obtain the given answer Alternative 2 2sin 3sin 2 x x + − ( ) 2 2 1 cos 3sin 2 x x = − + − (M1) For use of correct identity 2cos cos 3sin x x x = − + 3tan cos 3sin x x x = − + (M1) For use of 2cos 3tan x x = 3sin 3sin 0 x x − + = (A1) For use of tan cos sin x x x = and answer 0

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 10 of 10 Question Answer Marks Guidance 11(b) π 1 sin 2 4 2 α   + =     only B1 For solution of quadratic from (a) to obtain π 1 sin 2 4 2 α   + =     only π π 5π 13π 2 , , 4 6 6 6 α + = 7π 23π 2 , 12 12 α = M1 For correct order of operations in attempt to solve π 1 sin 2 4 2 α   + =     , may be implied by one correct solution 7π 24 α = A1 23π 24 α = A1