Cambridge IGCSE Mathematics - Additional 0606 — 2014 May/June Paper 1 · Variant 3

This document consists of 15 printed pages and 1 blank page. DC (CW/CGW) 73329/3 © UCLES 2014 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 4 3 5 4 6 4 5 8 6 7 * ADDITIONAL MATHEMATICS 0606/13 Paper 1 May/June 2014 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 0606/13/M/J/14 © UCLES 2014 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 0606/13/M/J/14 © UCLES 2014 [Turn over 1 (i) Show that y x x 3 6 5 2 = - + can be written in the form ( ) y a x b c 2 = - + , where a, b and c are constants to be found. [3] (ii) Hence, or otherwise, find the coordinates of the stationary point of the curve y x x 3 6 5 2 = - + . [1] 2 Given that 2 4 8 1 x y x y 4 # # = - and 3 3 1 x y = + , find the value of x and of y. [4]

4 0606/13/M/J/14 © UCLES 2014 3 (i) Find, in terms of p, the remainder when x px p x 21 3 2 2 + + + is divided by x 3 + . [2] (ii) Hence find the set of values of p for which this remainder is negative. [3]

5 0606/13/M/J/14 © UCLES 2014 [Turn over 4 The diagram shows a thin square sheet of metal measuring 24 cm by 24 cm. A square of side x cm is cut off from each corner. The remainder is then folded to form an open box, x cm deep, whose square base is shown shaded in the diagram. 24 cm 24 cm x cm x cm (i) Show that the volume, V cm3, of the box is given by V x x x 4 96 576 3 2 = - + . [2] (ii) Given that x can vary, find the maximum volume of the box. [4]

6 0606/13/M/J/14 © UCLES 2014 5 (i) The first three terms in the expansion of ( )x 2 5 6 - , in ascending powers of x, are p qx rx2 + + . Find the value of each of the integers p, q and r. [3] (ii) In the expansion of ( ) ( ) x a bx 2 5 6 3 - + , the constant term is equal to 512 and the coefficient of x is zero. Find the value of each of the constants a and b. [4]

7 0606/13/M/J/14 © UCLES 2014 [Turn over 6 Find the equation of the normal to the curve ( ) y x x 12 2 3 1 = - at the point on the curve where x 2 = . [6]

8 0606/13/M/J/14 © UCLES 2014 7 (a) A 5-character password is to be chosen from the letters A, B, C, D, E and the digits 4, 5, 6, 7. Each letter or digit may be used only once. Find the number of different passwords that can be chosen if (i) there are no restrictions, [1] (ii) the password contains 2 letters followed by 3 digits. [2] (b) A school has 3 concert tickets to give out at random to a class of 18 boys and 15 girls. Find the number of ways in which this can be done if (i) there are no restrictions, [1] (ii) 2 of the tickets are given to boys and 1 ticket is given to a girl, [2]

9 0606/13/M/J/14 © UCLES 2014 [Turn over (iii) at least 1 boy gets a ticket. [2]

10 0606/13/M/J/14 © UCLES 2014 8 A particle moves in a straight line such that, t s after passing through a fixed point O, its velocity, v ms 1 - , is given by v 5 4e t2 = - - . (i) Find the velocity of the particle at O. [1] (ii) Find the value of t when the acceleration of the particle is 6ms 2 - . [3] (iii) Find the distance of the particle from O when . t 1 5 = . [5] (iv) Explain why the particle does not return to O. [1]

11 0606/13/M/J/14 © UCLES 2014 [Turn over 9 Solve (i) sin cos cos x x x 3 2 = for ° ° x 0 180 G G , [4] (ii) sin cos y y 10 8 2 + = for ° ° y 0 360 G G . [5]

12 0606/13/M/J/14 © UCLES 2014 10 The table shows experimental values of x and y. x 1.50 1.75 2.00 2.25 y 3.9 8.3 19.5 51.7 (i) Complete the following table. x2 1g y [1] (ii) By plotting a suitable straight line graph on the grid on page 13, show that x and y are related by the equation y Abx2 = , where A and b are constants. [2] (iii) Use your graph to find the value of A and of b. [4] (iv) Estimate the value of y when . x 1 25 = . [2]

13 0606/13/M/J/14 © UCLES 2014 [Turn over 2 1 –1 0 1 2 3 4 5 6

14 0606/13/M/J/14 © UCLES 2014 11 The diagram shows the graph of cos sin y x x 3 3 3 = + , which crosses the x-axis at A and has a maximum point at B. B A x y O cos sin y x x 3 3 3 = + (i) Find the x-coordinate of A. [3] (ii) Find x y d d and hence find the x-coordinate of B. [4]

15 0606/13/M/J/14 © UCLES 2014 (iii) Showing all your working, find the area of the shaded region bounded by the curve, the x-axis and the line through B parallel to the y-axis. [5]

16 0606/13/M/J/14 © UCLES 2014 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. University of 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

CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 2014 series 0606 ADDITIONAL MATHEMATICS 0606/13 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 2014 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 13 © Cambridge International Examinations 2014 1 (i) ( ) 2 1 3 2 + − = x y a = 3, b = 1, c = 2 B1, B1, B1 B1 for each, may be given in the form ( ) 2 1 3 2 + − = x y (ii) (1, 2) √B1 Follow through on their answers to (i) If using differentiation, follow through on their x only. 2 1 8 4 24 = × × −y x y x Considering powers of either 2, 4 or 8 0 7 = −y x 3 1 3 = + y x M1 M1 for considering powers of either 2, 4 or 8 and forming an equation using these powers Considering powers of 3 1 − = + y x B1 B1 for equation considering powers of 3 Solving both simultaneously gives 8 7 , 8 1 − = − = y x M1 A1 M1 for attempt to solve their equations A1 for both 3 (i) f(–3) = –27 + 9p – 3p2 + 21 = 9p – 3p2 – 6 M1 A1 M1 for substitution of 3 − = x A1 answer must be simplified (ii) 9p – 3p2 – 6 < 0 ( )( ) 0 2 1 > − − p p Critical values 1 and 2 2 ,1 > < p p M1 A1 A1 M1 for attempt to factorise A1 for critical values A1 for correct range 4 (i) ( )2 2 24 x x V − =     + − = 2 4 96 576 x x x = x x x 576 96 4 2 3 + − M1 A1 M1 for attempt at a product of 3 lengths, 2 of which must be the same A1 for expansion to reach given answer (ii) 2 d 12 192 576 d V x x x = − + M1 M1 for attempt to differentiate When d 0 d V x = , 2 12 192 576 0 x x − + = DM1 DM1 for equating d d V x to zero and attempt to solve leading to ( )( ) 0 12 4 = − − x x with 4 = x the only possible solution 1024 = V A1 A1 A1 for x = 4 A1 for 1024 V =

Page 3 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 13 © Cambridge International Examinations 2014 5 (i) 2 6000 960 64 x x + − B1, B1, B1 B1 for each correct term (ii) ( )( ) bx a a x x 2 3 2 3 6000 960 64 + + − , 2 , 512 64 3 = = a a B1 B1 B1 for first two terms of ( )3 bx a + B1 for equating constant term to 512 and obtaining 2 = a ( ) 0 64 3 960 2 3 = + − b a a M1 M1 for attempt to equate coefficient of x to zero, must have two terms involved leading to b = 10 A1 A1 for 10 = b 6 When 4 ,2 − = = y x ( ) ( )3 1 2 3 2 2 12 12 3 2 d d − + −       = − x x x x x y B1 M1, B1 A1 B1 for 4 − = y M1 for differentiation of a product B1 for ( ) 3 2 2 12 3 2 − − x x When 3 4 d d ,2 − = = x y x Normal: ( ) 2 4 3 4 − = + x y ( ) 22 3 4 − = x y M1 A1 M1 for attempt at normal equation A1 allow unsimplified 7 (a) (i) 15120 B1 (ii)         × × × × 2 3 4 4 5 480 M1 A1 M1 for attempt to multiply number of ways of getting 4 letters by the number of ways of getting 2 digits. (b) (i) 5456 B1 (ii) 15 2 18 × C 2295 M1 A1 M1 for attempt at an appropriate product, at least one term must be correct. (iii) 5456 – Number of ways only girls get tickets 5001 455 5456 = − M1 A1 M1 for a complete correct method their (i) – number of ways only girls get tickets Or 1B 2G 1890 2B 1G 2295 3B 816 Total 5001 M1 A1 M1 must be considering at least 2 of the cases shown

Page 4 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 13 © Cambridge International Examinations 2014 8 (i) 1 B1 (ii) a = 8e–2t M1 M1 for attempt to differentiate 8e –2t = 6, –2t = ln 4 3 DM1 DM1 for correct attempt to solve equation in the form 2 e t − = constant 144 .0 = t A1 A1 must be at least 3 sf (iii) ( )c t s t + + = − e 2 5 2 M1 M1 for attempt to integrate When t = 0, s = 0, so c = –2 DM1, A1 DM1 for attempt to find c, A1 c correct When 60 .5 ,5.1 = = s t M1, A1 M1 for substitution of t = 1.5 Alternative: 5.1 0 2 e 2 5     − + = t t s M1 DM1 A1 M1 M1 for attempt to integrate DM1 for attempt to use limits A1 all correct M1 for evaluation of square bracket notation Leading to s = 5.60 A1 (iv) Velocity is always +ve, so no change in direction B1 Allow any valid argument. 9 (i) ( ) 0 2 sin 3 cos = − x x o 90 ,0 cos = = x x , 3 2 sin = x x = 41.8°, 138.2° B1 M1 A1,√A1 B1 for 90° M1 for attempt to solve sinx = 3 2 Follow through on their first answer (ii) 8 cos sin 10 2 = + y y ( ) 8 cos cos 1 10 2 = + − y y 0 2 cos cos 10 2 = − − y y ( )( ) 0 2 cos 5 1 cos 2 = + − y y 5 2 cos , 2 1 cos − = = y y M1 M1 M1 M1 for use of correct identity M1 for attempt to reduce to a 3 term quadratic and attempt to solve quadratic M1 for attempt to solve using factors in terms of cos o o 300 , 60 = y and o o 4. 246 , 6. 113 = y A1, A1 A1 for any ‘pair’

Page 5 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 13 © Cambridge International Examinations 2014 10 (i) x2 2.25 3.06 4 5.06 lg y 0.59 0.92 1.29 1.71 B1 (ii) M1 A1, 0 M1 for plotting logy against x2 –1 each error, poor point plotting, poor line drawing (iii) Gradient: 5.2 ,4.0 lg = = b b (allow 2.45 to 2.55) Intercept : 5 0 ,3.0 lg . A A = − = (allow 0.4 to 0.6) M1 A1 M1 A1 M1 for correct use of gradient M1 for correct use intercept (iv) 2.1 (allow 2 to 2.2) M1, A1 1 2 3 4 5 6 −1 −0.5 0.5 1 1.5 2 x y

Page 6 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 13 © Cambridge International Examinations 2014 11 (i) at A 0 3 cos 3 sin 3 = + x x M1 M1 for equating to zero and attempt to solve using tan 6 5 3 , 3 1 3 tan π = − = x x 150° DM1 DM1 for dealing with 3x ( ) 873 .0 18 5π = x (allow o 50 ) A1 (ii) x x x y 3 sin 3 3 cos 3 3 d d − = When π x x x y 3 3 ,3 3 tan ,0 d d = = = or 3x = 60°,     = 349 .0 9 π x (allow o 20 ) B1, B1 M1 A1 B1 for x y d d M1 for attempt to solve 0 d d = x y (iii) Area = 18 5 9 3 sin 3 1 3 1 3 cos 3 3 π π         + + − x x x M1 A1, A1 M1 for attempt to integrate A1 for each term =       + − −       + − 3 sin 3 1 3 cos 3 3 6 5 sin 3 1 6 5 cos 3 3 π π π π DM1 DM1 for correct application of their limits 667 .0 or 3 2 = or better A1