Cambridge IGCSE Mathematics - Additional 0606 — 2013 May/June Paper 1 · Variant 3

This document consists of 16 printed pages. DC (TL/SW) 71814 © UCLES 2013 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education 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 a pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, 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. * 6 6 0 7 5 1 7 7 6 4 * ADDITIONAL MATHEMATICS 0606/13 Paper 1 May/June 2013 2 hours Candidates answer on the Question Paper. Additional Materials: Electronic calculator www.theallpapers.com

2 0606/13/M/J/13 © UCLES 2013 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 ΔABC a sin A = b sin B = c sin C a2 = b2 + c2 – 2bc cos A Δ = 1 2 bc sin A www.theallpapers.com

3 0606/13/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 1 On the axes below sketch, for 0 2r x G G , the graph of (i) cos y x 1 = - , [2] (ii) sin y x 2 = . [2] –2 –1 0 y x 1 2 › 2 › 3› 2 2› (iii) State the number of solutions of the equation cos sin x x 2 1 - = , for r 0 2 x G G . [1] www.theallpapers.com

4 0606/13/M/J/13 © UCLES 2013 For Examiner’s Use 2 Variables x and y are such that y Abx = , where A and b are constants. The diagram shows the graph of ln y against x, passing through the points (2, 4) and (8, 10). (2, 4) (8, 10) x O ln y Find the value of A and of b. [5] www.theallpapers.com

5 0606/13/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 3 A committee of 6 members is to be selected from 5 men and 9 women. Find the number of different committees that could be selected if (i) there are no restrictions, [1] (ii) there are exactly 3 men and 3 women on the committee, [2] (iii) there is at least 1 man on the committee. [3] www.theallpapers.com

6 0606/13/M/J/13 © UCLES 2013 For Examiner’s Use 4 (i) Given that log x 2 1 4 = , find the value of x. [1] (ii) Solve log log y y 2 5 12 2 1 4 4 - - = ^ h . [4] www.theallpapers.com

7 0606/13/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 5 (i) Find x x 1 6 d 2 - c m y . [2] (ii) Hence find the value of the positive constant k for which 2 x x 1 6 d k k 2 3 - = c m y . [4] www.theallpapers.com

8 0606/13/M/J/13 © UCLES 2013 For Examiner’s Use 6 (i) Given that A 2 3 1 5 = - c m, find A 1 - . [2] (ii) Using your answer from part (i), or otherwise, find the values of a, b, c and d such that a c b d 1 7 17 5 A - = c c m m. [5] www.theallpapers.com

9 0606/13/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 7 Calculators must not be used in this question. B 5 – 2 A C 5 + 1 The diagram shows a triangle ABC in which angle 90c A = . Sides AB and AC are 5 2 - and 5 1 + respectively. Find (i) tan B in the form a b 5 + , where a and b are integers, [3] (ii) sec B 2 in the form c d 5 + , where c and d are integers. [4] www.theallpapers.com

10 0606/13/M/J/13 © UCLES 2013 For Examiner’s Use 8 ș rad M A D B F E C 6 cm 16 cm 16 cm The diagram shows a square ABCD of side 16 cm. M is the mid-point of AB. The points E and F are on AD and BC respectively such that AE BF = = 6 cm. EF is an arc of the circle centre M, such that angle EMF is i radians. (i) Show that .1 855 i = radians, correct to 3 decimal places. [2] (ii) Calculate the perimeter of the shaded region. [4] www.theallpapers.com

11 0606/13/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (iii) Calculate the area of the shaded region. [3] www.theallpapers.com

12 0606/13/M/J/13 © UCLES 2013 For Examiner’s Use 9 B A C O P Q a b c The figure shows points A, B and C with position vectors a, b and c respectively, relative to an origin O. The point P lies on AB such that 3:4 : AP AB = . The point Q lies on OC such that 2:3 : OQ QC = . (i) Express AP in terms of a and b and hence show that 3 OP 4 1 a b = + ^ h. [3] (ii) Find PQ in terms of a, b and c. [3] www.theallpapers.com

13 0606/13/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (iii) Given that PQ BC 5 6 = , find c in terms of a and b. [2] www.theallpapers.com

14 0606/13/M/J/13 © UCLES 2013 For Examiner’s Use 10 The point A, whose x-coordinate is 2, lies on the curve with equation y x x x 4 1 3 2 = - + + . (i) Find the equation of the tangent to the curve at A. [4] This tangent meets the curve again at the point B. (ii) Find the coordinates of B. [4] www.theallpapers.com

15 0606/13/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (iii) Find the equation of the perpendicular bisector of the line AB. [4] Question 11 is printed on the next page. www.theallpapers.com

16 0606/13/M/J/13 © UCLES 2013 For Examiner’s Use 11 (a) Solve r sin x 2 3 1 + = - ` j for r 0 2 x G G radians. [4] (b) Solve tan cot y y 2 - = for y 0 180 c c G G . [6] 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. www.theallpapers.com

CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 2013 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 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components. www.theallpapers.com

Page 2 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 13 © Cambridge International Examinations 2013 Mark Scheme Notes Marks are of the following three types: M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). B Accuracy 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 generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. • The symbol √ implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously “correct” answers or results obtained from incorrect working. • Note: B2 or A2 means that the candidate can earn 2 or 0. B2, 1, 0 means that the candidate can earn anything from 0 to 2. www.theallpapers.com

Page 3 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 13 © Cambridge International Examinations 2013 The following abbreviations may be used in a mark scheme or used on the scripts: AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) CAO Correct Answer Only (emphasising that no “follow through” from a previous error is allowed) ISW Ignore Subsequent Working MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) Penalties MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become “follow through √ ” marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. OW –1,2 This is deducted from A or B marks when essential working is omitted. PA –1 This is deducted from A or B marks in the case of premature approximation. S –1 Occasionally used for persistent slackness – usually discussed at a meeting. EX –1 Applied to A or B marks when extra solutions are offered to a particular equation. Again, this is usually discussed at the meeting. www.theallpapers.com

Page 4 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 13 © Cambridge International Examinations 2013 1 (i) B1 correct shape for y = cos x – 1 (ii) B1 all correct B1 correct shape for y = sin 2x B1 all correct (iii) 3 B1 2 Either gradient = 1 B1 intercept = 2 B1 ln b = gradient or ln A = intercept M1 M1, need to equate either gradient to ln b or intercept to ln A b = e or 2.72 A1 A = e2, A = 7.39 A1 Or e4 = Ab2 and e10 = Ab8 [B1 B1 B1 for each equation leading to b6 = e6 or e4 = e2 A or e10 = e8A M1 M1 for attempt to solve for either A or b b = e or 2.72 A1 A = e2, A = 7.39 A1] Or 10 = 8 ln b + ln A [B1 4 = 2 ln b + ln A B1 leading to ln b = 1 or 6 = 3 ln A M1 M1 for attempt to solve for either A or b b = e or 2.72 A1 A = e2, A = 7.39 A1] 2 4 6 −3 −2 −1 1 2 x y www.theallpapers.com

Page 5 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 13 © Cambridge International Examinations 2013 3 (i) 14C6 = 3003 B1 (ii) 5C3 × 9C3 = 840 M1 A1 M1 for product of 2 combinations (iii) Either 3003 – 9C6 = 2919 M1 B1 A1 M1 for 3003 – number of committees containing no men B1 for 9C6 Or 1M + 5W: 5 × 9C5 = 630 2M + 4W: 5C2 × 9C4 = 1260 3M + 3W: 840 (part (ii)) 4M + 2W: 5C4 ×9C2 = 180 5M + 1W: 1 × 9C1 = 9 Total: 2919 [B2 1 0 B1] –1 each error B1 for correct final answer 4 (i) 2 B1 (ii) log4 y2 – log4 (5y – 12) (=log4 2) B1 B1 for power log4       −12 5 2 y y = (=log4 2) M1 correct division y2 – 10y + 24 = 0 M1 attempt at solution of a 3 term quadratic y = 4, 6 A1 A1 for both 5 (i) x + ( ) c x + 6 B1 B1 B1 for each term (ii)       + k k 3 6 3 –      + k k 6 (=2) M1 correct use of limits 2k2 – 2k – 4 = 0 M1 attempt to obtain a 3 term quadratic from 2 brackets equated to 2 leading to k = 2 DM1 A1 DM1 or solution of quadratic dependent on 2nd M1 www.theallpapers.com

Page 6 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 13 © Cambridge International Examinations 2013 6 (i) A–1 =       − 2 3 1 5 13 1 B1 B1 B1 for matrix, B1 for multiplying by a correct determinant (ii) Either       −1 c b a =             − d 17 5 7 2 3 1 5 13 1 M1 evidence of multiplication of both sides by A-1 =       + − + d d 2 15 13 25 52 13 1 leading to a = 4, c = 1 DM1 DM1 for attempt to equate like elements and b = 2, d = 1 A3,2,1,0 –1 each error Or       − 5 3 1 2       −1 c b a =       d 17 5 7 [M1 M1 for evidence of matrix multiplication 2a – c = 7, 3a + 5c = 17, a = 4, c = 1 2b + 1 = 5, 3b – 5 = d, b = 2, d = 1 DM1 A3,2,1,0] DM1 for attempt to equate like elements –1 each error 7 (i) tan B = 2 5 1 5 − + B1 = 2 5 2 5 2 5 1 5 + + × − + M1 attempt at rationalisation (Allow if inverse is used) = 7 + 3 5 A1 (ii) (7 + 3 5 )2 + 1 = sec2 B M1 M1 M1 for attempt to use the correct identity M1 for simplification to give 3 or 4 terms sec2 B = 95 + 42 5 √A1 √A1 cao A1 for 95, A1 for 42 5 Or sec2 B = B 2 cos 1 = ( ) ( ) ( ) 2 2 2 2 5 2 5 1 5 − − + + [M1 M1 for attempt to use to find BC2 sec2 B = 5 4 9 5 4 9 5 4 9 5 2 15 + + × − − M1 M1 for use of sec B = B cos 1 sec2 B = 95 + 42 5 A1 A1] A1 for 95, A1 for 52 5 www.theallpapers.com

Page 7 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 13 © Cambridge International Examinations 2013 8 (i) Either tan 6 8 2 = θ M1 M1 for use of trig to obtain half angle ... . θ 927 0 2 = Can use sin 10 8 2 = θ or cos 10 6 2 = θ 855 .1 = θ A1 A1 Allow if done in degrees and converted Or Area of triangle MEF = 48 [M1 M1 for a complete method to find the obtuse angle 48 sin 10 2 1 2 = × × θ 287 1 287 1 . , . θ − π = 855 1. θ = A1] Or 162 = 102 + 102 – (2 × 10 × cos θ) [M1 M1 for use of the cosine rule, need to see working as answer given θ =1.855 A1] (ii) radius = 10 B1 B1 for the radius, allow anywhere P = (10 × 1.855) + 10 + 10 + 16 M1 M1 M1 for use of arc length M1 for method, must be arc +3 sides = 54.6 or 54.5 or 54.55 A1 (iii) A =256 – 2 ( ) 855 1 10 2 1 6 8 2 1 2 . −       × × M1 M1 M1 for area of sector M1 for a correct plan to obtain the required area = 115.25 or 115.3 or 115 A1 awrt 115 www.theallpapers.com

Page 8 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 13 © Cambridge International Examinations 2013 9 (i) ( ) a b − = 4 3 AP B1 ( ) a b a − + = 4 3 OP , or M1 M1 for attempt at vector addition ( ) a b a − − = 4 1 OP , ( ) b a 3 4 1 + = A1 Answer given (ii) c 5 2 = OQ , or c 5 3 = QC or c 5 3 − = CQ B1 B1 for OQ , QC or CQ PQ = OQ – OP M1 M1 for correct vector addition/subtraction = 4 3 4 5 2 b a c − − A1 (iii) ( ) b c b a c − = − − 6 4 15 4 5 2 M1 M1 for use of their vectors and attempt to get k c 16 5 9 a b c − = A1 10 (i) When x = 2, y = –5 B1 B1 for y = –5 1 8 3 d d 2 + − = x x x y M1 M1 for attempt to differentiate when x = 2, 3 d d − = x y DM1 DM1 for attempt at tangent equation – must be tangent with use of x = 2 Tangent: y + 5 = –3 (x – 2) (y = 1 –3x) A1 allow unsimplified (ii) 1 – 3x = x3 – 4x2 + x + 1 M1 M1 for equating tangent and curve equations x (x – 2)2 = 0 DM1 DM1 for attempt to solve resulting cubic equation Meets at (0, 1) A1 A1 A1 for each coordinate www.theallpapers.com

Page 9 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 13 © Cambridge International Examinations 2013 (iii) Grad of perp = 3 1 √B1 √B1 on their gradient in (i) only Midpoint (1, –2) M1 M1 for attempt to find the midpoint Perp bisector y + 2 = 3 1 (x – 1) M1 A1 M1 for attempt at line equation – must be perp bisector A1 allow unsimplified 11 (a) sin 2 1 3 − =       π + x B1 6 11 6 7 3 π π = π + , x B1 B1 for 6 7π and 6 11π 2 3 6 5 π π = , x B1 B1 B1 for first correct solution B1 for a second correct solution with all solutions in radians and with no excess solutions within the range (b) tan y – 2 = y tan 1 B1 B1 for a correct equation tan2 y – 2 tan y – 1 = 0 M1 A1 M1 for attempt to obtain a 3 term quadratic equation A1 for a correct equation equated to zero tan y = 1 ± 2 DM1 DM1 for solution of quadratic y = 67.5°, 157.5° A1 A1 A1 for first correct solution A1 for a second correct solution with all solutions in degrees and with no excess solutions within the range. www.theallpapers.com