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

* 3 9 3 6 1 3 8 4 3 1 * This document consists of 16 printed pages. DC (SC/SG) 145845/2 © UCLES 2018 [Turn over ADDITIONAL MATHEMATICS 0606/11 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 0606/11/M/J/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/M/J/18 © UCLES 2018 [Turn over 1 Solve the equations y x 4 - = , x y x y 8 4 16 0 2 2 + - - - = . [5]

4 0606/11/M/J/18 © UCLES 2018 2 Find the equation of the perpendicular bisector of the line joining the points (1, 3) and (4, -5). Give your answer in the form ax + by + c = 0, where a, b and c are integers. [5]

5 0606/11/M/J/18 © UCLES 2018 [Turn over 3 Diagrams A to D show four different graphs. In each case the whole graph is shown and the scales on the two axes are the same. y x A O y x B O y x C O y x D O Place ticks in the boxes in the table to indicate which descriptions, if any, apply to each graph. There may be more than one tick in any row or column of the table. [4] A B C D Not a function One-one function A function that is its own inverse A function with no inverse

6 0606/11/M/J/18 © UCLES 2018 4 (i) The curve sin y a b cx = + has an amplitude of 4 and a period of r 3 . Given that the curve passes through the point r , 12 2 J L KK N P OO, find the value of each of the constants a, b and c. [4] (ii) Using your values of a, b and c, sketch the graph of sin y a b cx = + for r x 0 G G radians. [3] y x 0 6 –6 r

7 0606/11/M/J/18 © UCLES 2018 [Turn over 5 The population, P, of a certain bacterium t days after the start of an experiment is modelled by P = 800ekt, where k is a constant. (i) State what the figure 800 represents in this experiment. [1] (ii) Given that the population is 20 000 two days after the start of the experiment, calculate the value of k. [3] (iii) Calculate the population three days after the start of the experiment. [2]

8 0606/11/M/J/18 © UCLES 2018 6 (a) Write log log log p q 2 2 3 3 + ^ ^h h as a single logarithm to base 3. [3] (b) Given that log log 5 4 5 3 0 a a 2 - + = ^ h , find the possible values of a. [3]

9 0606/11/M/J/18 © UCLES 2018 [Turn over 7 (i) Find the inverse of the matrix 4 5 2 3 - - J L KK N P OO. [2] (ii) Hence solve the simultaneous equations 8x - 4y - 5 = 0, -10x + 6y - 7 = 0. [4]

10 0606/11/M/J/18 © UCLES 2018 8 (a) Given that p = 2i - 5j and q = i - 3j, find the unit vector in the direction of 3p - 4q. [4] (b) B A 1.25 kmh–1 A river flows between parallel banks at a speed of 1.25 kmh-1. A boy standing at point A on one bank sends a toy boat across the river to his father standing directly opposite at point B. The toy boat, which can travel at v kmh-1 in still water, crosses the river with resultant speed 2.73 kmh-1 along the line AB. (i) Calculate the value of v. [2]

11 0606/11/M/J/18 © UCLES 2018 [Turn over The direction in which the boy points the boat makes an angle i with the line AB. (ii) Find the value of i. [2]

12 0606/11/M/J/18 © UCLES 2018 9 (i) Find the first 3 terms in the expansion of x x 2 16 1 8 - J L KK N P OO in descending powers of x. [3] (ii) Hence find the coefficient of x4 in the expansion of x x x 2 16 1 1 1 8 2 2 - + J L KK J L KK N P OO N P OO . [3]

13 0606/11/M/J/18 © UCLES 2018 [Turn over 10 Do not use a calculator in this question. (a) Simplify 6 5 5 6 5 + + . [3] (b) Show that 3 2 .0 5 7 #^ h can be written in the form a b, where a and b are integers and a > b. [2] (c) Solve the equation x x 2 4 + = , giving your answers in simplest surd form. [4]

14 0606/11/M/J/18 © UCLES 2018 11 y x A O y x x 16 27 2 = + The diagram shows part of the graph of y x x 16 27 2 = + , which has a minimum at A. (i) Find the coordinates of A. [4]

15 0606/11/M/J/18 © UCLES 2018 [Turn over The points P and Q lie on the curve y x x 16 27 2 = + and have x-coordinates 1 and 3 respectively. (ii) Find the area enclosed by the curve and the line PQ. You must show all your working. [6] Question 12 is printed on the next page.

16 0606/11/M/J/18 © UCLES 2018 16 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 A curve is such that x y x 2 5 d d 2 2 2 1 = - - ^ h . Given that the curve has a gradient of 6 at the point , 2 9 3 2 J L KK N P OO, find the equation of the curve. [8]

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 0606/11 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.

0606/11 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.

0606/11 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

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 4 of 10 Question Answer Marks Partial Marks 1 Substitution and simplification to obtain a 3 term quadratic in one variable M1 substitution of y = x + 4 or x = y – 4 and simplification to 3 terms. 2 2 8 0 − − = x x or 2 2 4 16 0 − − = x x or 2 10 16 0 − + = y y or 2 10 16 0 − + = y y A1 correct equation of the form 2 0 + + = ax bx c or 2 0 + + = ay by c Solution of quadratic equation M1 M1 dep 4, 8 = = x y 2, 2 = − = x y A2 A1 for each pair 2 Midpoint 5 , 1 2   −     B1 Gradient of line 8 3 = − B1 Gradient of perp 3 8 = M1 Equation of perp bisector: 3 5 1 8 2   + = −     y x M1 M1 dep Using their perpendicular gradient and their midpoint 6 16 31 0 − − = x y or 6 16 31 0 − + + = x y A1 3 A B C D 9 9 9 9 9 4 B1 for either each row correct or each column correct – mark to candidate’s advantage. 4(i) 4 = b B1 6 = c B1 π 2 4sin 2 = + a M1 Evaluation of a using their b and their c and the given point. 2 = − a A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 5 of 10 Question Answer Marks Partial Marks 4(ii) 3 B1 for 6 2 −- - y B1 for 3 complete cycles B1 for all correct 5(i) The number of bacteria at the start of the experiment B1 5(ii) 20 000 800e = kt so 2 20 000 e 800 = k or 2 ln 20 000 ln800 ln(e ) k = + M1 use of given equation and attempt to solve for e2k or use logs correctly 2 ln 25 = k M1 correct method to obtain 2k 1.61 A1 5(iii) 3ln5 800e = P M1 Substitution of t = 3 in formula using their k 100 000 = A1 answer in range 99800 to 100200 6(a) 3 3 3 3 log log 2 log log 2   × +     p q or 2 log 3 3 log 2 log + p q B1 3 3 log log + p q or 2 log 3 log (2 ) × p q B1 B1 dep 3 log pq B1 B1 dep 6(b) ( )( ) log 5 1 log 5 3 0 − − = a a M1 solution of quadratic equation log 5 1, 5 = = a a 3 log 5 3, 5 = = a a or 1.71 or 1 35 A2 A1 for 5 = a A1 for 3 5 = a or 1.71or 1 35 7(i) 3 2 1 5 4 2       2 B1 for 1 2 B1 for 3 2 5 4      

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 6 of 10 Question Answer Marks Partial Marks 7(ii) 5 4 2 2 − = x y 7 5 3 2 − + = x y B1 Relating solution of these equations to matrix in (i) B1 for adapted equation or 2.5 3.5       or 2      x y 3 2 2.5 1 5 4 3.5 2      =           x y or 3 2 5 1 2 5 4 7 2     =         x y M1 Correct method for pre-multiplication by their inverse matrix. 7.25 13.25     =         x y or 29 4 53 4             7.25, 13.25 x y = = A2 A1 for each. Condone in matrix form. 8(a) 3(2 5 ) 4( 3 ) − − − i j i j M1 For expansion and collection of terms 3 4 2 3 − = − p q i j A1 Magnitude of their 2 3 − i j 2 2 2 ( 3) + − M1 For method to find magnitude Unit vector = 2 3 13 − i j A1 8(b)(i) 2 2 2 2.73 1.25 = + v B1 Correct use of Pythagoras 3.00 = v B1 8(b)(ii) 1.25 tan 2.73 θ = oe M1 Use of a trig funtion to obtain a relevant angle Angle to AB= o 24.6 or 0.429 radians A1 9(i) 8 6 4 256 64 7 − + x x x 3 B1 for each term

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 7 of 10 Question Answer Marks Partial Marks 9(ii) 4 2 1 2 1 + + x x B1 ( ) 8 6 4 4 2 1 2 256 64 7 1   − + + +     x x x x x ( ) 4 256 1 64 2 7 1 × − × + × x M1 M1 for three correctly obtained products leading to terms in x4 using their 8 6 4 256 64 7 − + x x x and their 4 2 1 2 1 + + x x Coefficient of 4 x is 256 128 7 − + = 135 A1 10(a) 5 6 5 6 5 6 5 6 5 + − × + − M1 for rationalisation = 30 5 5 36 5 30 31 − + − M1 M1dep for expanding the numerator to obtain four terms. 31 5 31 = 5 = A1 A1 for 5 from correct working 10(b) ( ) 6 3 3 2 2 6 2 × × = × 8 6 B2 B1 for 6 from 3 2 × B1 for 8 from ( ) 6 2 or 3 2

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 8 of 10 Question Answer Marks Partial Marks 10(c) EITHER: 2 2 4 0 + − = x x B1 3 term quadratic equation equated to zero 2 18 2 − ± = x M1 use of the quadratic formula for use of 18 3 2 = M1 M1 dep 2, 2 2 − A1 For both from full working OR: 2 2 4 0 + − = x x B1 2 2 1 4 2 2   + = +       x 3 2 2 2 = ± − x M1 Correct use of completing the square method 4 2 , 2 2 x = − M1 M1dep for dealing with 2 in denominator 2, 2 2 = − x A1 11(i) 3 d 54 16 d = − y x x M1 for 3 d 16 d = ± y p x x Equating to zero and obtaining x3 M1 M1dep 3 , 36 2 = = x y A2 A1 for each

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 9 of 10 Question Answer Marks Partial Marks 11(ii) EITHER: When 1, 43 = = x y When 3, 51 = = x y B1 B1 for both ( ) 3 2 1 1 27 43 51 2 16 d 2   + × − +    ∫ x x x Area of trapezium = ( ) 1 43 51 2 2   + ×     oe B1 FT from their P and their Q Integration to find area under curve M1 for 2   +     q px x = 2 27 8   −     x x A1 Integration correct 2 2 27 27 8 3 8 1 3 1     = × − − × −         M1 M1dep for application of limits Required area = 94 – 82 =12 A1 OR: When 1, 43 = = x y When 3, 51 = = x y B1 B1 for both Equation of PQ: 4 39 = + y x B1 Equation of line FT from their P and their Q Integration of their 2 27 4 39 16 + − − x x x M1 for 2   + +     r px qx x = 2 27 39 6   − +     x x x A1 All correct 2 2 27 39 3 6 3 3 27 39 1 6 1 1   = × −× +       − × −× +     M1 M1dep for application of limits Required area = 72 – 60 = 12 A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 10 of 10 Question Answer Marks Partial Marks 12 ( ) ( ) 1 2 d 2 5 d = − + y x c x M1 for ( ) 1 2 2 5 − k x , for ( ) 1 2 2 5 − x A1 Substitution to obtain arbitrary constant M1 M1 dep Using d 6 d = y x when 9 2 = x ( ) 1 2 d 2 5 4 d  = − +     y x x A1 for correct d d y x Integration of their ( ) 1 2 2 5 − + k x c M1 M1 dep on first M1 for integration of ( ) 1 2 2 5 − k x to obtain ( ) 3 2 2 5 − m x ( ) ( ) 3 2 1 2 5 +4 3 = − + y x x d A1 for ( ) 3 2 1 2 5 +4 3 − x x FT their (non -zero) constant Finding constant M1 M1 dep for obtaining arbitrary constant for ( ) 3 2 2 5 − + + m x nx d using 9 2 , 2 3 x y = = ( ) 3 2 1 2 5 4 20 3 = − + − y x x A1 for correct equation