Cambridge IGCSE Mathematics - Additional 0606 — 2018 May/June Paper 1 · Variant 2

* 1 2 3 9 0 4 9 8 8 9 * This document consists of 16 printed pages. DC (KN/SW) 145846/4 © UCLES 2018 [Turn over ADDITIONAL MATHEMATICS 0606/12 Paper 1 May/June 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 0606/12/M/J/18 © UCLES 2018 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x b b ac a = − − 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 © UCLES 2018 [Turn over 0606/12/M/J/18 1 It is given that tan y x 1 3 = + . (i) State the period of y. [1] (ii) On the axes below, sketch the graph of tan y x 1 3 = + for ° x 0 180 ° ° G G . [3] 0 30 60 90 120 150 180 y x

4 0606/12/M/J/18 © UCLES 2018 2 Find the values of k for which the line y kx 1 2 = - does not meet the curve ( ) y x k x 9 3 1 5 2 = - + + . [5]

5 0606/12/M/J/18 © UCLES 2018 [Turn over 3 The variables x and y are such that when e y is plotted against x 2, a straight line graph passing through the points (5, 3) and (3, 1) is obtained. Find y in terms of x. [5]

6 0606/12/M/J/18 © UCLES 2018 4 A particle P moves so that its displacement, x metres from a fixed point O, at time t seconds, is given by ( ) ln x t5 3 = + . (i) Find the value of t when the displacement of P is 3m. [2] (ii) Find the velocity of P when t = 0. [2] (iii) Explain why, after passing through O, the velocity of P is never negative. [1] (iv) Find the acceleration of P when t = 0. [2]

7 0606/12/M/J/18 © UCLES 2018 [Turn over 5 (i) The first three terms in the expansion of x 3 9 1 5 - J L KK N P OO can be written as a x b x c 2 + + . Find the value of each of the constants a, b and c. [3] (ii) Use your values of a, b and c to find the term independent of x in the expansion of ( ) x x 3 9 1 2 9 5 2 - + J L KK N P OO . [3]

8 0606/12/M/J/18 © UCLES 2018 6 Find the coordinates of the stationary point of the curve y x x 2 1 2 = - + . [6]

9 0606/12/M/J/18 © UCLES 2018 [Turn over 7 A population, B, of a particular bacterium, t hours after measurements began, is given by B 1000e t 4 = . (i) Find the value of B when t = 0. [1] (ii) Find the time taken for B to double in size. [3] (iii) Find the value of B when t = 8. [1]

10 0606/12/M/J/18 © UCLES 2018 8 (a) Solve cos sin 3 4 4 2i i + = for ° ° 0 180 G G i . [4] (b) Solve sin cos 2 3 2 z z = for 2 2 G G r r z - radians. [4]

11 0606/12/M/J/18 © UCLES 2018 [Turn over 9 (a) (i) Solve lgx 3 = . [1] (ii) Write lg lg a b 2 3 - + as a single logarithm. [3] (b) (i) Solve x x 5 6 0 - + = . [2] (ii) Hence, showing all your working, find the values of a such that log log a 5 6 4 0 a 4 - + = . [3]

12 0606/12/M/J/18 © UCLES 2018 10 Do not use a calculator in this question. All lengths in this question are in centimetres. 60° B A C 4 3 5 - 4 3 5 + The diagram shows the triangle ABC, where AB 4 3 5 = - , BC 4 3 5 = + and angle ° ABC 60 = . It is known that ° sin60 2 3 = , ° cos60 2 1 = , ° tan60 3 = . (i) Find the exact value of AC. [4]

13 0606/12/M/J/18 © UCLES 2018 [Turn over (ii) Hence show that ( ) cosecA q CB p 4 3 5 2 = + , where p and q are integers. [4]

14 0606/12/M/J/18 © UCLES 2018 11 O y x B A e y 8 3 x 4 = + The diagram shows the graph of the curve y 3 8 e x 4 = + . The curve meets the y-axis at the point A. The normal to the curve at A meets the x-axis at the point B. Find the area of the shaded region enclosed by the curve, the line AB and the line through B parallel to the y-axis. Give your answer in the form a e , where a is a constant. You must show all your working. [10]

15 0606/12/M/J/18 © UCLES 2018 [Turn over Question 12 is printed on the next page.

16 0606/12/M/J/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. 12 Do not use a calculator in this question. (a) Given that 6 8 3 9 p p q q 2 2 3 # # + - is equal to 2 3 7 4 # , find the value of each of the constants p and q. [3] (b) Using the substitution u x3 1 = , or otherwise, solve x x 4 3 0 1 3 2 3 + + = . [4]

IGCSE™ is a registered trademark. This document consists of 10 printed pages. © UCLES 2018 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/12 Paper 1 May/June 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 May/June 2018 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 3 of 10 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 May/June 2018 © UCLES 2018 Page 4 of 10 Question Answer Marks Partial Marks 1(i) π 3 or o 60 B1 1(ii) 3 B1 for 3 asymptotes at o o 30 , 90 = x and o 150 ; the curve must approach but not cross all 3 of the asymptotes and be in the 1st and 4th quadrants B1 for starting at ( ) 0,1 and finishing at ( ) 180,1 B1 for all correct 2 For an attempt to obtain an equation in x only M1 ( ) 2 9 1 4 0 − + + = x k x A1 correct 3 term equation ( ) ( ) 2 1 4 9 4 + − × × k M1 M1dep for correct use of 2 4 − b ac oe Critical values 11, 13 = = − k k A1 13 11 − < < k A1 For the correct range 3 2 e = + y ax b B1 may be implied, 0 ≠ b either 3 5 = + a b 1 3 = + a b or Gradient = 1, so 1 = a M1 correct attempt to find a or b by use of simultaneous equations or finding the gradient and equating it to a Coefficient of 2 x is 1 A1 Intercept is –2 A1 ( ) 2 ln 2 = − y x A1 For correct form 4(i) ( ) 3 ln 5 3 = + t 3e 5 3 = + t or better B1 3.42 = t B1 4(ii) d 5 d 5 3 = + x t t M1 for 1 5 3 + k t When 0 = t , d 5 d 3 = x t , 1.67 or better A1 all correct 45 90 135 180 −10 −5 5 10 x y

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 5 of 10 Question Answer Marks Partial Marks 4(iii) If 0 > t each term in 1 0 5 3 k t > + so never negative oe B1 dep on M1 in (ii) FT on their 1 5 3 + k t , provided 1 0 > k 4(iv) ( ) 2 2 2 2 d d 5 3 = + k x t t M1 ( ) 2 2 2 d 25 d 5 3 = − + x t t When 2 2 d 25 0, 9 d = = − x t t or 2.78 − A1 all correct 5(i) 10 243, 45, 3 = = − = a b c 3 B1 for each coefficient, must be simplified 5(ii) ( ) 2 2 45 10 243 4 36 81 3   − + + +     x x x x B1 For ( ) 2 4 36 81 + + x x for having 3 terms independent of x M1 Independent term is 972 1620 270 378 − + = − A1 6 attempt to differentiate quotient or equivalent product M1 ( ) ( ) 1 1 2 2 d 2 1 2 1 d − − = − x x x for a quotient ( ) ( ) 1 3 2 2 d 2 1 2 1 d − − − = − − x x x for a product B1 either ( ) ( ) ( ) 1 2 2 2 1 2 2 1 d d 2 1 −   −− + −     = − x x x y x x or ( ) ( ) ( ) 1 3 2 2 d 2 1 2 2 1 d − −   = − − + −     y x x x x A1 All other terms correct When d 0, d = y x 2 1 2 −= + x x M1 equate to zero and attempt to solve 3 = x A1 5 = y , 5 5 , 2.24 A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 6 of 10 Question Answer Marks Partial Marks 7(i) 1000 B1 7(ii) 4 2000 1000e = t B1 4ln2, ln16 = t M1 For 4ln k or 4 ln , 0 > k k 2.77 A1 7(iii) 2 1000e = B = 7389, 7390 B1 8(a) ( ) 2 3 1 sin 4sin 4 θ θ − + = M1 use of correct identity ( )( ) 3sin 1 sin 1 0 θ θ − − = 1 sin , sin 1 3 θ θ = = M1 For attempt to solve a 3 term quadratic equation in sinθ to obtain sinθ = o o 19.5 , 160.5 θ = A1 o 90 A1 8(b) tan2 3 φ = π 2π 2 , 3 3 φ = − M1 obtaining an equation in tan 2φ and correct attempt to solve for one solution to reach 2φ = k for one correct solution π , 6 φ = or 0.524 A1 for attempt at a second solution M1 π 3 φ = − , or 1.05 − A1 for a correct second solution and no other solutions within the range 9(a)(i) 1000 B1 9(a)(ii) for use of power rule M1 for addition or subtraction rule M1 dep on previous M1 2 1000 lg a b A1 Allow 3 2 10 lg a b 9(b)(i) 2 5 6 0 − + = x x M1 For attempt to obtain a quadratic equation and solve 3, 2 = = x x A1 for both

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 7 of 10 Question Answer Marks Partial Marks 9b(ii) ( ) 2 4 4 log 5log 6 0 − + = a a M1 For the connection with (i) and attempt to deal with at least one logarithm correctly, either t 3 4 heir or 2 4their 64 = a A1 16 = a A1 10(i) ( ) ( ) 2 2 2 4 3 5 4 3 5 = − + + AC M1 For attempt to use the cosine rule ( )( ) o 2 4 3 5 4 3 5 cos60 − − + A1 For all correct unsimplified 2 123 = AC M1 M1 dep for attempt to evaluate without use of calculator 123 = AC A1 ALTERNATIVE METHOD Taking D as the foot of the perpendicular from A: Find AD, BD, DC 2 2 2 = + AC AD DC M1 For a complete method to get AC2 2 2 2 12 5 3 15 4 3 2 2     − + = +             AC A1 For all correct unsimplified 2 123 = AC M1 M1dep for attempt to evaluate without use of calculator 123 = AC A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 8 of 10 Question Answer Marks Partial Marks 10(ii) o 4 3 5 sin sin60 − = AC ACB or sin = AD ACB AC M1 For attempt at the sine rule or trigonometry involving right-angled triangles For attempt at cosec M1 dep on first M mark ( ) 2 123 cosec 3 4 3 5 = − ACB or ( ) 2 41 4 3 5 − oe ( ) 2 123 4 3 5 cosec 3 4 3 5 4 3 5 ACB + = × + − M1 dep on previous M mark for a statement involving rationalisation using 3 + a b ( ) 2 41 4 3 5 23 = + A1 For rationalisation using 4 3 5 4 3 5 + + oe and simplification ALTERNATIVE METHOD ( )( ) 1 23 3 4 3 5 4 3 5 sin60 2 4 − + = M1 Area of ABC ( ) 1 23 3 123 4 3 5 sin 2 4 ACB + = M1 For attempt at a second area of ABC and equating to first area For attempt at cosec M1 dep on first 2 M marks ( ) 2 41 4 3 5 23 = + A1 Need to be convinced no calculator is being used in simplification

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 9 of 10 Question Answer Marks Partial Marks 11 When 1 0, 2 = = x y B1 For 1 2 = y 4 d 1 e d 2 = x y x B1 d 1 d 2 = y x , Gradient of normal = –2 B1 FT on their d d y x , must be numeric either: Normal 1 2 2 − = − y x or: Gradient of normal = −OA OB M1 For an attempt at a normal equation passing through their 1 0, 2       and a substitution of 0 = y When 1 0, 4 = = y x A1 EITHER: 1 4 4 0 1 3 e d 8 8 + ∫ x x M1 For attempt to integrate to obtain 4 1 3 e 8 + x k x , 1 1 8 ≠ k , 1 1 2 ≠ k 1 4 4 0 1 3 e 32 8   +     x x A1 For correct integration Use of limits M1 M1dep For area of triangle 1 16 = B1 FT on their 1 4 = x e 32 = A1 final answer in correct form OR: 1 4 4 0 1 3 1 e 2 d 8 8 2 + − + ∫ x x x M1 For attempt at subtraction and attempt to integrate to obtain 4 2 1 2 3 3 e 8 + + + x k x k x k x , 1 1 8 ≠ k 1 4 4 2 0 1 1 e 32 8   − +     x x x A2 –1 for each error for integration for use of limits M1 M1dep e 32 = A1 final answer in correct form

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 10 of 10 Question Answer Marks Partial Marks 12(a) 1 4 = p B1 4 6 4 + − + = p q q B1 FT on their p 3 4 = q B1 12(b) 1 1 3 3 3 1 0    + + =          x x M1 For attempt to factorise and solve, or solve using the quadratic formula oe, a quadratic in 1 3 x or u 1 3 1 = − x or 1 = − u 1 3 3 = − x or 3 = − u A1 For both 1 = − x A1 27 = − x A1