Cambridge IGCSE Mathematics - Additional 0606 — 2018 Oct/Nov Paper 1 · Variant 1

* 0 0 8 8 9 4 2 7 8 0 * This document consists of 15 printed pages and 1 blank page. DC (RW/SG) 153502/3 © UCLES 2018 [Turn over ADDITIONAL MATHEMATICS 0606/11 Paper 1 October/November 2018 2 hours Candidates answer on the Question Paper. Additional Materials: Electronic calculator READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 80. Cambridge International Examinations Cambridge International General Certificate of Secondary Education

2 0606/11/O/N/18 © UCLES 2018 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)n = an + ( n 1)an–1 b + ( n 2)an–2 b2 + … + ( n r)an–r br + … + bn, where n is a positive integer and ( n r) = n! (n – r)!r! . 2. TRIGONOMETRY Identities sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A Formulae for ∆ABC a sin A = b sin B = c sin C a2 = b2 + c2 – 2bc cos A ∆ = 1 2 bc sin A

3 0606/11/O/N/18 © UCLES 2018 [Turn over 1 (a) On the axes below, sketch the graph of cos y x 3 2 1 = - , for ° ° ° x 0 360 G G . 5 4 3 2 1 0 –1 –2 –3 –4 –5 –6 x y 90 180 270 360 [3] (b) Given that sin y x 4 6 = , write down (i) the amplitude of y, [1] (ii) the period of y. [1]

4 0606/11/O/N/18 © UCLES 2018 2 ( )x x x x a 2 5 4 p 3 2 = + + + ( )x x ax b 4 3 q 2 = + + Given that p(x) has a remainder of 2 when divided by x 2 1 + and that q(x) is divisible by x 2 + , (i) find the value of each of the constants a and b. [3] Given that ( ) ( ) ( ) x x x r p q = - and using your values of a and b, (ii) find the exact remainder when r(x) is divided by x 3 2 - . [3]

5 0606/11/O/N/18 © UCLES 2018 [Turn over 3 The coefficient of x2 in the expansion of ( )( ) x kx 2 3 6 - + is equal to 972. Find the possible values of the constant k. [6]

6 0606/11/O/N/18 © UCLES 2018 4 (i) Write x x 9 8 2 - + in the form ( ) x p q 2 - - , where p and q are constants. [2] (ii) Hence write down the coordinates of the minimum point on the curve y x x 9 8 2 = - + . [1] (iii) On the axes below, sketch the graph of y x x 9 8 2 = - + , showing the coordinates of the points where the curve meets the coordinate axes. 20 16 8 12 4 0 –8 –4 –12 –16 x y 10 8 6 4 2 –2 [3] (iv) Write down the value of k for which x x k 9 8 2 - + = has exactly 3 solutions. [1]

7 0606/11/O/N/18 © UCLES 2018 [Turn over 5 r cm O θ A B rad The diagram shows a circle with centre O and radius r cm. The minor arc AB is such that angle AOB is i radians. The area of the minor sector AOB is 48 cm2. (i) Show that r 96 2 i = . [2] (ii) Given that the minor arc AB has length 12 cm, find the value of r and of i. [3] (iii) Using your values of r and i, find the area of the shaded region. [2]

8 0606/11/O/N/18 © UCLES 2018 6 A curve has equation ( ) ln y x x 5 2 2 3 2 = + + . (i) Show that ln x y 4 5 3 d d =- when x 0 = . [4] (ii) Hence find the equation of the tangent to the curve at the point where x 0 = . [2]

9 0606/11/O/N/18 © UCLES 2018 [Turn over 7 (a) Express lg lg x y 2 3 + - as a single logarithm to base 10. [3] (b) (i) Solve x x 6 7 3 0 + - = . [2] (ii) Hence, given that log log a 6 3 7 3 0 a 3 + - = , find the possible values of a. [4]

10 0606/11/O/N/18 © UCLES 2018 8 (i) Find ( ) x x 5 4 d d 2 2 3 + . [2] (ii) Hence find ( ) x x x 5 4 d 2 2 1 + y . [2] Given that ( ) x x x 5 4 15 19 d a 2 0 2 1 + = y , (iii) find the value of the positive constant a. [4]

11 0606/11/O/N/18 © UCLES 2018 [Turn over 9 Variables s and t are such that s t4 3e t = + - . (i) Find the value of s when t 0 = . [1] (ii) Find the exact value of t when t s 2 d d = . [4] (iii) Find the approximate increase in s when t increases from ln5 to ln h 5+ , where h is small. [3]

12 0606/11/O/N/18 © UCLES 2018 10 Particle A is at the point with position vector 2 5 - J L KK N P OO at time t = 0 and moves with a speed of 10 ms-1 in the same direction as 3 4 J L KK N P OO. (i) Given that A is at the point with position vector a 38 J L KK N P OO when t = 6 s, find the value of the constant a. [3] Particle B is at the point with position vector 16 37 J L KK N P OO at time t = 0 and moves with velocity 4 2 J L KK N P OO ms-1. (ii) Write down, in terms of t, the position vector of B at time t s. [1]

13 0606/11/O/N/18 © UCLES 2018 [Turn over (iii) Verify that particles A and B collide. [4] (iv) Write down the position vector of the point of collision. [1]

14 0606/11/O/N/18 © UCLES 2018 11 (a) ( ) cos x x 3 2 f = - for r x 0 2 G G . (i) Write down the range of f. [2] (ii) Find the exact value of ( . ) 2 5 f 1 - . [3]

15 0606/11/O/N/18 © UCLES 2018 (b) ( )x x 3 g 2 = - for x R ! . Find the exact solutions of ( )x 6 g2 =- . [4]

16 0606/11/O/N/18 © UCLES 2018 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 International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. BLANK PAGE

This document consists of 9 printed pages. © UCLES 2018 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/11 Paper 1 October/November 2018 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 2018 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 2018 © UCLES 2018 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 2018 © UCLES 2018 Page 3 of 9 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 2018 © UCLES 2018 Page 4 of 9 Question Answer Marks Guidance 1(a) B3 B1 for 2 cycles, one max and 2 min points in the correct places and up to a max at each end B1 for going between 2 and –4 B1 for starting at ( ) 0,2 and finishing at ( ) 360,2 1(b)(i) 4 B1 1(b)(ii) o 60 or π 3 B1 2(i) 1 1 5 p 2 2 2 4 4 a     − = − + − + =         ( ) ( ) q 2 16 6 0 a b − = − + = M1 For either 1 p 2 2   − =     or ( ) q 2 0 − = 3 a = A1 2 b = A1 2(ii) ( ) 3 2 r 2 5 1 x x x x = + − + M1 For ( ) r x using their ( ) p x and ( ) q x 2 16 4 10 r 1 3 27 9 3  = + − +     M1 For 2 r 3       35 27 = − A1 Must be exact 90 180 270 360 −5 −4 −3 −2 −1 1 2 3 4 5 x y

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 5 of 9 Question Answer Marks Guidance 3 ( ) 6 3 kx + = 2 2 729 1458 1215 kx k x + + B2 B1 for 1458kx or 2 2 1215k x Terms in 2x for( )( ) 6 2 3 x kx − + = 2 1458 2430 k k − + 2 2430 1458 972 k k − = M1 For attempt at further expansion to obtain 2 terms in 2x and equating to 972 2 5 3 2 0 k k − − = ( )( ) 5 2 1 0 k k + − = M1 Dep for solution of resulting 3 term quadratic 2 5 k = − A1 1 k = A1 4(i) 2 9 49 2 4 x   − −     B2 B1 for 9 2 or 49 4 4(ii) 9 49 , 2 4   −     B1 FT their p and q 4(iii) B3 B1 for shape B1 for cusps at ( ) 1, 0 and ( ) 8, 0 B1 for all correct, passing through ( ) 0, 8 with maximum in correct position 4(iv) 49 4 B1 FT their q 5(i) 2 1 48 2 r θ = B1 2 96 r θ = B1 Dep Must have previous B1 5(ii) 12 rθ = B1 For statement of arc length 2 96 12 r r × = M1 For attempt to use part (i) to find either r or θ 8 r = and 1.5 θ = A1 For both 5 10 2 4 6 8 10 12 14 16 x y

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 6 of 9 Question Answer Marks Guidance 5(iii) Area = 2 1 48 sin 2 r θ   −    M1 For a complete method including a correct attempt for the area of the triangle using their r and θ =16.1 A1 6(i) For 2 4 2 3 x x + B1 M1 For attempt to differentiate a quotient or appropriate product d d y x = ( ) ( ) ( ) 2 2 2 4 5 2 5ln 2 3 2 3 5 2 x x x x x + − + + + A1 All other terms correct When 0 x = d 5ln3 d 4 y x − = A1 For given answer 6(ii) 1 ln3 2 y = or 0.549 B1 May be implied by tangent equation, allow 0.55 Equation of tangent 5 1 ln3 ln3 4 2 y x   = − +     or 1.37 0.549 y x = − + B1 7(a) lg100 2 = B1 3 3lg lg x x = B1 3 100 lg x y B1 7(b)(i) 2 6 7 3 0 x x + −= ( )( ) 2 3 3 1 0 x x + − = M1 For obtaining in suitable quadratic form and attempt to solve 3 2 x = − 1 3 x = A1 For both

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 7 of 9 Question Answer Marks Guidance 7(b)(ii) log 3 a x = 1 log 3 3 a = M1 For realising connection with part (i) and attempt to solve 1 log 3 3 a = or 3 log 3 2 a − = 27 a = A1 3 log 3 2 a − = M1 For realising connection with part (i) and attempt to solve 3 log 3 2 a − = or 1 log 3 3 a = 2 3 1 3 a   =     or 0.481or 1 3 1 9       oe A1 8(i) M1 For attempt to use chain rule to obtain ( ) 1 2 2 5 4 kx x + where k is a constant ( )( ) 1 2 2 3 10 5 4 2 x x + A1 Allow unsimplified 8(ii) M1 For attempt to use part (i) if in correct form of ( ) 3 2 2 5 4 m x + ( ) ( ) 3 2 2 1 5 4 15 x c + + A1 FT on their ( ) 3 2 2 1 5 4 x k + 8(iii) M1 For use of limits if their (ii) Must be in the form ( ) 3 2 2 5 4 m x + ( ) 3 3 2 2 2 1 19 5 4 4 15 15 a     + − =           A1 ( ) 3 2 2 5 4 27 a + = M1 Dep For complete and correct method to deal with the power of 3 2 leading to 1 a = A1 9(i) 3 B1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 8 of 9 Question Answer Marks Guidance 9(ii) M1 For attempt to differentiate to obtain e t a b − + d 4 3e d t s t − = − A1 All correct 2 4 3e t − = − M1 Dep for correct attempt to solve equation involving exponential where e 0 t −> leading to 3 ln 2 t = or 2 ln 3 − A1 Must be an exact form 9(iii) When d 17 ln5, d 5 s t t = = M1 For attempt to find value of d d s t when ln 2 t = M1 Dep for attempt to use method of small changes 17 5 h s ∂= A1 10(i) Velocity of A 6 8    B1 For velocity, may be implied by later work When 2 6 6, 6 5 8 A t    = = +    −    r M1 For a complete and correct method 38 43   =     A1 For 43 10(ii) 16 4 37 2 B t    = +       r B1 10(iii) 16 4 2 6 37 2 5 8 t t       + = +       −       M1 For equating position vectors at a time t 16 4 2 6 t t + = + or 37 2 5 8 t t + = −+ M1 Dep for equating like vectors at least once 7 t = A1 Allow from one correct equation Both equations lead to 7 t = B1 For showing that 7 t = satisfies both equations thus verifying collision, or equivalent 10(iv) 44 51       B1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 9 of 9 Question Answer Marks Guidance 11(a)(i) B1 For critical values 2 f 4 - - B1 Dep For correct inequality and notation 11(a)(ii) 3cos x y = cos2 0.5 y = M1 For attempt to find 1 f − and equate to 0.5 π 2 3 y = M1 Dep For correct attempt to solve, dealing with the double angle π 6 y = A1 11(b) ( ) ( ) 2 2 g g 3 x x = − ( ) 2 2 3 3 x = − − M1 For correct attempt at 2 g , allow unsimplified 2 4 6 6 6 x x −+ − = − 2 4 6 0 x x − = M1 Dep for equating to –6 and attempt to solve to obtain a non-zero root 0 x = B1 6 x = ± A1