Cambridge IGCSE Mathematics - Additional 0606 — 2015 May/June Paper 1 · Variant 2

ADDITIONAL MATHEMATICS 0606/12 Paper 1 May/June 2015 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. This document consists of 16 printed pages. DC (CW/JG) 91463/2 © UCLES 2015 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 9 9 3 7 8 4 8 4 1 4 *

2 0606/12/M/J/15 © UCLES 2015 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 0606/12/M/J/15 © UCLES 2015 [Turn over 1 Given that the graph of ( ) y k x kx 2 5 1 2 = + + + does not meet the x-axis, find the possible values of k. [4] 2 Show that cosec tan cot sec i i i i + = . [4]

4 0606/12/M/J/15 © UCLES 2015 3 Find the inverse of the matrix 4 5 2 3 J L KK N P OO and hence solve the simultaneous equations x y 4 2 8 0 + - = , x y 5 3 9 0 + - = . [5]

5 0606/12/M/J/15 © UCLES 2015 [Turn over 4 1.7 rad 2.4rad A C B 12 cm O The diagram shows a circle, centre O, radius 12 cm. The points A, B and C lie on the circumference of this circle such that angle AOB is 1.7 radians and angle AOC is 2.4 radians. (i) Find the area of the shaded region. [4] (ii) Find the perimeter of the shaded region. [5]

6 0606/12/M/J/15 © UCLES 2015 5 (a) A security code is to be chosen using 6 of the following: • the letters A, B and C • the numbers 2, 3 and 5 • the symbols * and $. None of the above may be used more than once. Find the number of different security codes that may be chosen if (i) there are no restrictions, [1] (ii) the security code starts with a letter and finishes with a symbol, [2] (iii) the two symbols are next to each other in the security code. [3]

7 0606/12/M/J/15 © UCLES 2015 [Turn over (b) Two teams, each of 4 students, are to be selected from a class of 8 boys and 6 girls. Find the number of different ways the two teams may be selected if (i) there are no restrictions, [2] (ii) one team is to contain boys only and the other team is to contain girls only. [2]

8 0606/12/M/J/15 © UCLES 2015 6 A particle moves in a straight line such that its displacement, x m, from a fixed point O after t s, is given by ln x t t 10 4 4 2 = + - ^ h . (i) Find the initial displacement of the particle from O. [1] (ii) Find the values of t when the particle is instantaneously at rest. [4]

9 0606/12/M/J/15 © UCLES 2015 [Turn over (iii) Find the value of t when the acceleration of the particle is zero. [5]

10 0606/12/M/J/15 © UCLES 2015 7 A X C D 7a 4a B b In the diagram AB = 4a, BC = b and DC = 7a. The lines AC and DB intersect at the point X. Find, in terms of a and b, (i) DA, [1] (ii) DB. [1] Given that AX AC m = , find, in terms of a, b and m, (iii) AX , [1] (iv) DX . [2]

11 0606/12/M/J/15 © UCLES 2015 [Turn over Given that DX DB n = , (v) find the value of m and of n. [4]

12 0606/12/M/J/15 © UCLES 2015 8 (i) Find x 10e e d x x 2 2 + - ^ h y . [2] (ii) Hence find ( ) x 10e e d x x k k 2 2 + - -y in terms of the constant k. [2] (iii) Given that ( ) x 10 60 e e d x x k k 2 2 + =- - -y , show that 11 11 120 0 e e k k 2 2 - + = - . [2]

13 0606/12/M/J/15 © UCLES 2015 [Turn over (iv) Using a substitution of y e k 2 = or otherwise, find the value of k in the form ln a b, where a and b are constants. [3]

14 0606/12/M/J/15 © UCLES 2015 9 A curve has equation cos y x x 4 3 2 = + . The normal to the curve at the point where x 4 r = meets the x- and y-axes at the points A and B respectively. Find the exact area of the triangle AOB, where O is the origin. [8]

15 0606/12/M/J/15 © UCLES 2015 [Turn over 10 (a) Solve cos sec x x 2 3 3 = for ° ° x 0 120 G G . [3] (b) Solve cosec cot y y 3 5 5 0 2 + - = for ° ° y 0 360 G G . [5] Question 10(c) is printed on the next page.

16 0606/12/M/J/15 © UCLES 2015 (c) Solve sin z 2 3 1 r + = J L KK N P OO for z 0 2 G G r radians. [4] 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.

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International General Certificate of Secondary Education MARK SCHEME for the May/June 2015 series 0606 ADDITIONAL MATHEMATICS 0606/12 Paper 1, maximum raw mark 80 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 will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2015 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 12 © Cambridge International Examinations 2015 Abbreviations awrt answers which round to cao correct answer only dep dependent FT follow through after error isw ignore subsequent working oe or equivalent rot rounded or truncated SC Special Case soi seen or implied www without wrong working 1 ( ) ( ) 0 5 2 4 2 < + − k k ( ) 0 20 8 2 < − −k k ( )( ) ( ) 0 2 10 < + − k k critical values of 10 and –2 10 2 < < − k Alternative 1: ( ) d 2 2 5 d y k x k x = + + When ( ) ( ) 2 d 8 20 0, , d 2 2 5 4 2 5 y k k k x y x k k − + − = = = + + When 0 y = , obtain critical values of 10 and –2 10 2 < < − k Alternative 2: ( ) ( ) ( ) 1 5 2 4 5 2 2 5 2 2 2 +         + −       + + + = k k k k x k y Looking at ( ) 0 5 2 4 1 2 = + − k k leads to critical values of 10 and –2 10 2 < < − k M1 M1 A1 A1 M1 M1 A1 A1 M1 M1 A1 A1 use of ac b 4 2 − , (not as part of quadratic formula unless isolated at a later stage) with correct values for a, b and c Do not need to see < at this point attempt to obtain critical values correct critical values correct range attempt to differentiate, equate to zero and substitute x value back in to obtain a y value consider 0 = y in order to obtain critical values correct critical values correct range attempt to complete the square and consider ( )' 5 2 4 1' 2 + − k k attempt to solve above = to 0, to obtain critical values correct critical values correct range

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 12 © Cambridge International Examinations 2015 2 sin cos tan cot cos sin 1 cosec sin θ θ θ θ θ θ θ θ + + = θ θ θ θ θ sin 1 cos sin cos sin 2 2 + = θ cos 1 = θ sec = Alternative: 2 tan 1 tan cot tan cosec cosec θ θ θ θ θ θ + + = θ θ θ sin 1 tan sec2 = θ θ sec sec2 = θ sec = M1 M1 M1 A1 M1 M1 M1 A1 for sin cos tan , cot cos sin θ θ θ θ θ θ = = and 1 cosec sin θ θ = ; allow when used dealing correctly with fractions in the numerator; allow when seen use of the appropriate identity; allow when seen must be convinced it is from completely correct work ( beware missing brackets) for either 1 tan cot θ θ = or θ θ tan 1 cot = and θ θ sin 1 cosec = ; allow when used dealing correctly with fractions in numerator; allow when seen use of the appropriate identity; allow when seen must be convinced it is from completely correct work 3 A–1       − − = 4 5 2 3 2 1             − − =       9 8 4 5 2 3 2 1 y x 6 1 4 2 x y     =     −     2 ,3 − = = y x B1 B1 M1 A1, A1 2 1 multiplied by a matrix for matrix attempt to use the inverse matrix, must be pre-multiplication

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 12 © Cambridge International Examinations 2015 4 (i) Area = ( )      − − × +       × × 4.2 7.1 2 sin 12 2 1 7.1 12 2 1 2 2 π = awrt 181 B1,B1 M1 A1 B1 for sector area, allow unsimplified B1 for correct angle BOC, allow unsimplified correct attempt at area of triangle, allow unsimplified using their angle BOC (Their angle BOC must not be 1.7 or 2.4) (ii) ( ) 1832 .2 cos 12 12 2 12 12 2 2 2 × × − + = BC or       − × × = 2 1.4 2 sin 12 2 π BC 296 . 21 = BC Perimeter = ( ) 296 . 21 12 12 7.1 12 + + + × = 65.7 M1 A1 B1 M1 A1 correct attempt at BC, may be seen in (i), allow if used in (ii). Allow use of their angle BOC. for arc length, allow unsimplified for a correct ‘plan’ (an arc + 2 radii and BC) 5 (a) (i) 20160 B1 (ii) 6 4 3 2 P × × = 2160 B1,B1 B1 for 4 6P (must be seen in a product) B1 for all correct, with no further working (iii) 4 6 2 5 P × × = 3600 Alternative 1: 6 4 5! 2 C × × = 3600 Alternative 2: ( ) 2 5 6 5 7 × −P P = 3600 Alternative 3: ( ) ( ) ( ) ( ) 4 6 1 3 3 6 2 4 2 6 3 5 1 6 4 6 !2 P P P P P P P P + × + × + × + = 3600 B1,B1 B1 B2 B1 B2 B1 B2 B1 B1 for 4 6P (must be seen in a product) B1 for 5 (must be in a product) B1 for all correct, with no further working for !5 4 6 × C for 2 !5 4 6 × × C for ( ) 5 6 5 7 P P − for ( ) 2 5 6 5 7 × −P P 4 terms correct or omission of 2! in each term all correct

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 12 © Cambridge International Examinations 2015 (b) (i) 4 10 4 14 C C × or 4 8 8 14 C C × (or numerical or factorial equivalent) = 210210 B1,B1 B1 for either 4 14C or 8 14C as part of a product B1 for correct answer, with no further working (ii) 4 6 4 8 C C × = 1050 B1,B1 B1 for either 4 6 4 8 or C C as part of a product B1 for correct answer with no further working 6 (i) 10ln4 or 13.9 or better B1 (ii) 2 d 20 4 d 4 x t t t   = −   +   When 2 d 20 0, 4 d 4 x t t t = = + leading to 0 4 5 2 = + −t t 1, 4 t t = = M1 B1 DM1 A1 attempt to differentiate and equate to zero 4 20 2 + t t or equivalent seen attempt to solve their d 0 d x t = , must be a 2 or 3 term quadratic equation with real roots for both

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 12 © Cambridge International Examinations 2015 (iii) If ( ) 4 4 20 2 − + = t t v ( ) ( ) ( ) ( ) 2 2 2 20 4 20 2 4 t t t a t + − = + ( ) 2 4 20 t − or 2 20 80 t − or 2 4 t − or equivalent expression involving 2t − When acceleration is 0, 2 t = only Alternative 1 for first 3 marks: If( ) 4 16 4 20 2 2 + − − = t t t v ( )( ) ( ) ( ) ( ) ( ) 2 2 2 2 4 20 8 20 4 16 2 4 t t t t t a t + − − − − = + Alternative 2 for M1 mark: If ( ) ( ) 1 2 20 4 4 v t t − = + − ( ) ( ) ( ) ( ) 2 1 2 2 20 2 4 20 4 a t t t t − − = − + + + Alternative 3 for the first 3 marks If ( ) ( )( ) 1 2 2 4 16 4 20 − + − − = t t t v ( )( ) ( ) ( )( ) 1 2 2 2 2 4 8 20 4 2 16 4 20 − − + − +     + − − − = t t t t t t a Numerator = ( ) ( ) ( ) 2 2 2 20 4 16 20 8 4 t t t t t − − − + − + M1 A1 A1 A1 B1 M1 A1 A1 M1 M1 A1 A1 attempt to differentiate their d d x t ( ) 4 20 2 + t ( )t t 2 20 ( ) 2 4 20 t − or 2 20 80 t − or 2 4 t − 2 = t , dependent on obtaining first and second A marks attempt to differentiate their d d x t for ( )( )t t 8 20 4 2 − + for ( )( ) 2t 16 4 20 2 − −t t attempt to differentiate their d d x t attempt to differentiate their d d x t for ( ) 15 4 20 2 2 − −t t t for ( )( ) 4 8 20 2 + − t t 7 (i) b a − = 3 DA B1 mark final answer, allow unsimplified (ii) b a − = 7 DB B1 mark final answer, allow unsimplified (iii) ( ) b a + = 4 λ AX B1 mark final answer, allow unsimplified (iv) ( ) b a b a + + − = 4 3 λ DX M1 A1 their (i) + their (iii) or equivalent valid method or b a − 3 + their (iii) Allow unsimplified

Page 7 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 12 © Cambridge International Examinations 2015 (v) ( ) ( ) b a b a b a − = + + − 7 4 3 µ λ Equate like vectors: µ λ 7 4 3 = + µ λ − = + −1 leads to 11 7 , 11 4 = = µ λ M1 DM1 A1,A1 equating their (iv) and × µ their (ii) for an attempt to equate like vectors and attempt to solve 2 linear equations for λ and µ A1 for each 8 (i) ( ) 2 2 1 5e e 2 x k c − − + B1, B1 B1 for each term, allow unsimplified (ii) 2 2 2 2 1 1 5e e 5e e 2 2 k k k k − −     − − −         M1 A1 use of limits provided integration has taken place. Signs must be correct if brackets are not included. allow any correct form (iii) 2 2 2 2 1 1 5e e 5e e 60 2 2 k k k k − −    − − = −       or 2 2 11 11 e e 60 2 2 k k − − = − or equivalent leading to 0 120 11 11 2 2 = + − −k k e e B1 DB1 correct expression from (ii) either simplified or unsimplified equated to 60 − , must be first line seen. must be convinced as AG (iv) 0 11 120 11 2 = − + y y ( )( ) 0 11 1 11 = + − y y leading to 11 1 n 1 2 1 = k , 1 ln , 1n 11 11 − , 11 n 1 2 1 − M1 DM1 A1 attempt to obtain a quadratic equation in y or 2e k and solve to get y or 2e k ( only need positive solution) attempt to deal with e to get k =. any of given answers only.

Page 8 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 12 © Cambridge International Examinations 2015 9 d 4 6sin2 d y x x = − When π , 4 π = = y x d 2 d y x = − so gradient of normal = 2 1 Normal equation       − = − 4 π 2 1 π x y When x = 0, 8 π 7 = y When 4 π 7 ,0 − = = x y Area = 2 1 7π 7π 49π 2 4 8 64 × × = M1,A1 B1 DM1 DM1 A1 A1 B1ft M1 for attempt to differentiate A1 for all correct for y for substitution of 4 π = x into their d d y x and use of '1 ' 2 1 − = m m , dependent on first M1 correct attempt to obtain the equation of the normal, dependent on previous DM mark must be terms of π must be terms of π Follow through on their x and y intercepts; must be exact values 10 (a) 2 1 3 cos2 = x , ( ) 2 1 3 cos ± = x 3 45 ,135 , 225 , 315 x = ° ° ° ° 15 , 45 , 75 ,105 x = ° ° ° ° M1 A1,A1 complete correct method, dealing with sec and 3, correctly A1 for each correct pair (b) ( ) 0 5 cot 5 1 cot 3 2 = − + + y y Leading to 0 2 cot 5 cot 3 2 = − + y y or 0 3 tan 5 tan 2 2 = − − y y ( )( ) 0 2 cot 1 cot 3 = + − y y or ( )( ) 0 1 tan 2 3 tan = + − y y 2 1 tan ,3 tan = = y y 71.6 , 251.6 153.4 , 333.4 y = ° ° ° ° M1 M1 M1 A1,A1 use of a correct identity to get an equation in terms of one trig ratio only for y y tan 1 cot = to obtain either a quadratic equation in tan y or solutions in terms of tan ;y allow where appropriate for solution of a quadratic equation in terms of either y tan or y cot A1 for each correct ‘pair’ (c) 2 1 3 π sin =       + z π π 5π 13π , , 3 6 6 6 z + = π 11π , 2 6 z = (allow 76 .5 , 57 .1 ) M1 A1 M1 A1 completely correct method of solution one correct solution in range correct attempt to obtain a second solution within the range second correct solution (and no other)