Cambridge IGCSE Mathematics - Additional 0606 — 2018 Oct/Nov Paper 1 · Variant 2

* 7 2 0 6 8 4 0 8 6 8 * This document consists of 16 printed pages. DC (ST/JG) 153501/2 © UCLES 2018 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/12 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.

2 0606/12/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 0606/12/O/N/18 © UCLES 2018 [Turn over 1 Solve ( °) sin x 1 2 50 0 + + = for ° ° x 180 180 G G - . [4]

4 0606/12/O/N/18 © UCLES 2018 2 Find the equation of the curve which has a gradient of 4 at the point ( , ) 0 3 - and is such that x y 5 d d e x 2 2 2 = + . [5]

5 0606/12/O/N/18 © UCLES 2018 [Turn over 3 (i) On the axes below, sketch the graph of y x 6 3 = - , showing the coordinates of the points where the graph meets the coordinate axes. [2] 10 12 8 4 6 2 0 –4 –2 –6 –8 –10 –12 x y 5 6 4 3 2 1 –1 –2 (ii) Solve x 6 3 2 - = . [3] (iii) Hence find the values of x for which x 6 3 2 2 - . [1]

6 0606/12/O/N/18 © UCLES 2018 4 ( ) ln y x x 1 2 3 = + (i) Find the value of x y d d when . x 0 3 = . You must show all your working. [4] (ii) Hence find the approximate increase in y when x increases from 0.3 to . h 0 3+ , where h is small. [1]

7 0606/12/O/N/18 © UCLES 2018 [Turn over 5 The 7th term in the expansion of ( ) a bx 12 + in ascending powers of x is x 924 6. It is given that a and b are positive constants. (i) Show that b a 1 = . [2] The 6th term in the expansion of ( ) a bx 12 + in ascending powers of x is x 198 5. (ii) Find the value of a and of b. [4]

8 0606/12/O/N/18 © UCLES 2018 6 (i) Find ( ) x x 5 125 d d 2 3 2 - . [2] (ii) Using your answer to part (i), find ( ) x x x 5 125 d 2 3 1 - - y . [2] (iii) Hence find ( ) x x x 5 125 d 2 6 10 3 1 - - y . [2]

9 0606/12/O/N/18 © UCLES 2018 [Turn over 7 (a) The vector v has a magnitude of 39 units and is in the same direction as 12 5 - e o. Write v in the form a b e o, where a and b are constants. [2] (b) Vectors p and q are such that r s r p 6 = + + e o and r s q 5 1 2 1 = + - e o, where r and s are constants. Given that p q 2 3 0 0 + = e o, find the value of r and of s. [4]

10 0606/12/O/N/18 © UCLES 2018 8 a a A 4 3 4 = + e o (i) Find the values of the constant a for which A 1 - does not exist. [3] (ii) Given that a 4 = , find A 1 - . [2] (iii) Hence find the matrix B such that AB 2 4 3 5 = - e o. [3]

11 0606/12/O/N/18 © UCLES 2018 [Turn over 9 The polynomial ( )x ax bx cx 9 p 3 2 = + + - is divisible by x 3 + . It is given that ( ) 0 36 p = l and ( ) 0 86 p = ll . (i) Find the value of each of the constants a, b and c. [6] (ii) Using your values of a, b and c, find the remainder when ( )x p is divided by x 2 1 - . [2]

12 0606/12/O/N/18 © UCLES 2018 10 y x 0 y = a + 4 cos bx 6 6 M P r The diagram shows part of the curve cos y a bx 4 = + , where a and b are positive constants. The curve meets the y-axis at the point ( , ) 0 6 and the x-axis at the point , 6 0 r b l. The curve meets the x-axis again at the point P and has a minimum at the point M. (i) Find the value of a and of b. [3]

13 0606/12/O/N/18 © UCLES 2018 [Turn over Using your values of a and b find, (ii) the exact coordinates of P, [2] (iii) the exact coordinates of M. [2]

14 0606/12/O/N/18 © UCLES 2018 11 r cm O rad i P Q The diagram shows the sector OPQ of a circle, centre O, radius r cm, where angle POQ i = radians. The perimeter of the sector is 10 cm. (i) Show that area, A cm2, of the sector is given by ( ) A 50 2 2 i i = + . [5]

15 0606/12/O/N/18 © UCLES 2018 [Turn over It is given that i can vary and A has a maximum value. (ii) Find the maximum value of A. [5] Question 12 is printed on the next page.

16 0606/12/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. 12 The line y x 2 5 = + intersects the curve y xy 5 + = at the points A and B. Find the coordinates of the point where the perpendicular bisector of the line AB intersects the line y x = . [9]

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 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 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 October/November 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 October/November 2018 © UCLES 2018 Page 4 of 10 Question Answer Marks Guidance 1 ( ) o 1 sin 50 2 x + = − ( ) o o o 50 45 , 225 x + = − M1 For order of operations – subtraction of 1, division by 2 ± and attempt at 1 sin− M1 Dep For obtaining a solution by subtracting o 50 o o 95 , 175 x = − A2 A1 for one correct solution A1 for a second correct solution and no others within the range 2 ( ) 2 d 1 5 e d 2 x y x c x = + + M1 For attempt to integrate to get d d y x in the form 2 5 e x x p + . Condone omission of + c When d 0, 4 d y x x = = so 7 2 c = M1 Dep For attempt to get value of c ( ) 2 2 5 1 7 e 2 4 2 x x y x d = + + + M1 Dep on first M1 only For attempt to get y in the form including 2 2 5 e 2 x x p + . Condone omission of + d. When 0, 3 x y = = − so 13 4 d = − M1 Dep on previous DepM1 For attempt to obtain d, allow if c not found 2 2 5 1 7 13 e 2 4 2 4 x x y x = + + − A1 Must have an equation 3(i) B2 B1 for correct shape with vertex at ( ) 2,0 Dep B1 for passing through or starting at ( ) 0,6 −2 −1 1 2 3 4 5 6 −10 −5 5 10 x y

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 5 of 10 Question Answer Marks Guidance 3(ii) Either 6 3 2 x − = 4 3 x = B1 For 4 3 x = 6 3 2 x − = − M1 For considering – 2 8 3 x = A1 Or 2 9 36 32 0 x x − + = M1 For squaring each side and attempt to solve a 3 term quadratic = 0 4 3 x = A1 8 3 x = A1 3(iii) 4 8 , 3 3 x x < > B1 FT on their solutions in part (ii), must both be positive and written as 2 separate statements 4(i) B1 For 2 2 1 x + M1 For attempt to differentiate a product ( ) 3 2 d 2 3 ln 2 1 d 2 1 y x x x x x = + + + A1 For all other terms correct When d 0.3, 0.161 d y x x = = A1 For awrt 0.161 4(ii) 0.161h B1 FT on their numerical answer to part (i) 5(i) 7th term: 6 6 6 6 924 924 a b x x = 6 6 924 924 a b = ( ) 6 6 6 924 924 a bx x = B1 For any correct statement ( ) 6 1 ab = or 1 ab = so 1 b a = B1 Dep on first B1 Must be convinced, nfww

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 6 of 10 Question Answer Marks Guidance 5(ii) 6th term: 7 5 5 5 792 198 a b x x = 7 5 792 198 a b = ( ) 5 7 5 792 198 a bx x = B1 For any correct statement use of 1 ab = to obtain 2 ... a = or 2 ... b = M1 For attempt to solve their equations simultaneously to obtain an equation in 2 a or 5 b 1 2 a = A1 2 b = A1 6(i) M1 For ( ) 1 3 5 125 kx x − − ( ) 1 2 3 2 10 5 125 3 x x − × − ( ) 1 2 3 20 5 125 3 x x −   −       A1 Allow unsimplified 6(ii) M1 For ( ) ( ) 2 2 3 5 125 m x c − + ( ) ( ) 2 2 3 3 5 125 20 x c − + A1 FT on their k from part (i) 6(iii) ( ) ( ) 2 2 3 3 3 375 55 20   −     M1 Dep on previous M1 For use of limits in their answer to part (ii), must be in the form ( ) ( ) 2 2 3 5 125 m x c − + , 5.63 = A1 Allow greater accuracy

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 7 of 10 Question Answer Marks Guidance 7(a) 12 13 5 −  =     B1 For magnitude, may be implied by a correct v 36 15 −   =     v or 12 3 5 −       B1 Must be a vector 7(a) Alternative If 12 39 5 t −  =     , 3 = t B1 For value of t, may be implied by a correct v 36 15 −   =     v or 12 3 5 −       B1 7(b) M1 For equating like vectors at least once 17 2 3 0 r s + + = 2 6 9 0 r s + + = M1 Dep For solution of resulting equations to obtain 2 solutions 0 r = A1 3 2 s = − oe A1 8(i) ( ) 4 12 0 a a + − = M1 For correct use of det = 0 2 4 12 0 a a + − = M1 Dep For solution of resulting quadratic equation leading to 6, 2 a a = − = A1 For both 8(ii) 1 8 3 1 4 4 20 − −   =   −   A oe B2 B1 for 1 20 B1 for 8 3 4 4 −     −   8(iii) 1 2 3 4 5 −  =   −   B A M1 For pre-multiplication by their A-1 M1 Dep For multiplication of 2 matrices – need to see at least 2 correct elements – may be unsimplified 4 39 1 8 32 20   =   −   A1 For final matrix oe

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 8 of 10 Question Answer Marks Guidance 9(i) ( ) p 3 0 − = leading to 27 9 3 9 0 a b c − + − − = M1 For substitution of 3 = − x and equating to zero ( ) 2 p 3 2 x ax bx c ′ = + + ( ) p 0 36 ′ = M1 For differentiation in the form 2 + + rx sx t and substitution of 0 x = 36 c = A1 nfww ( ) p 6 2 x ax b ′′ = + ( ) p 0 2b ′′ = M1 For further differentiation in the form + vx w of their ( ) p x ′ and substitution of 0 x = 43 b = A1 nfww 10 a = A1 nfww 9(ii) 1 p 2       M1 For use of 1 2 x = in their ( ) p x from part (i) 21 A1 10(i) 2 a = B1 1 cos 2 bx = − M1 For a correct attempt to solve π cos 6 4 a b = ± provided 0 4 a < ≤ to get ... = b leading to 4 b = A1 10(ii) 1 cos4 2 x = − M1 Dep For attempt to solve their cos 4 a bx = ± provided 0 4 a < ≤ or use of symmetry to get ... = x π 3 x = so π , 0 3       A1 10(iii) At M, 2 y = − B1 π 4 x = B1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 9 of 10 Question Answer Marks Guidance 11(i) 2 10 r rθ + = M1 For use of arc length and attempt to get perimeter, must have 2 terms involving r M1 Dep For attempt to get r in terms of θ 10 2 r θ = + A1 2 1 10 2 2 A θ θ   =   +   M1 For attempt to obtain the area of the sector in terms of θ only, using their r ( ) 2 50 2 A θ θ = + A1 For manipulation to get the required answer nfww AG 11(ii) M1 For attempt to differentiate a quotient or an equivalent product ( ) ( ) ( ) 2 4 50 2 100 2 d d 2 A θ θ θ θ θ + − + = + or ( ) ( ) 2 3 d 50 2 100 2 d θ θ θ θ − − = + − + A A1 All correct, allow unsimplified When d 0 d A θ = M1 For equating their d dθ A to 0 and attempt to solve – need to see at least one line of working 2 θ = A1 Condone inclusion of 2 − 25 4 A = A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 10 of 10 Question Answer Marks Guidance 11(ii) Alternative Starting again using 10 2 2 θ − = r so 2 5 = − A r r M1 A complete method to obtain d d A r d 5 2 d = − A r r A1 When d 0 d A r = M1 For equating to zero and attempt to solve 2.5 r = A1 25 4 A = A1 12 2 2 7 0 x x + = or 2 3 10 0 y y − − = M1 For attempt to obtain a simplified quadratic equation in one variable equated to 0 M1 Dep For solution of quadratic ( ) 0, 5 A1 7 , 2 2   − −     A1 Midpoint 7 3 , 4 2   −     B1 Gradient of AB = 2 ∴⊥ gradient = 1 2 − M1 For attempt to obtain gradient of line perpendicular to AB using their coordinates ⊥ bisector: 3 1 7 2 2 4 y x   − = − +     M1 For a correct attempt to obtain equation of perpendicular bisector using their midpoint and their perpendicular gradient Consideration of when y x = M1 Dep on previous M1 For attempt to find intersection with the line y x = 5 12 x y = = A1 For both