Cambridge IGCSE Mathematics - Additional 0606 — 2013 May/June Paper 2 · Variant 3

This document consists of 16 printed pages. DC (JF/SW) 71796 © UCLES 2013 [Turn over * 8 4 8 5 2 9 6 2 9 8 * ADDITIONAL MATHEMATICS 0606/23 Paper 2 May/June 2013 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 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. UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education www.theallpapers.com

2 0606/23/M/J/13 © UCLES 2013 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 www.theallpapers.com

3 0606/23/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 1 Prove that cos sin 1 2 i i + c m + 2 4 cos sin tan 1 2 2 i i i - = + c m . [4] www.theallpapers.com

4 0606/23/M/J/13 © UCLES 2013 For Examiner’s Use 2 t s 24 20 5 16 O v ms–1 The velocity-time graph represents the motion of a particle moving in a straight line. (i) Find the acceleration during the first 5 seconds. [1] (ii) Find the length of time for which the particle is travelling with constant velocity. [1] (iii) Find the total distance travelled by the particle. [3] www.theallpapers.com

5 0606/23/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 3 Variables x and y are related by the equation sin y 10 4 2 = - x, where x 0 2 G G r . Given that x is increasing at a rate of 0.2 radians per second, find the corresponding rate of change of y when y = 8. [6] www.theallpapers.com

6 0606/23/M/J/13 © UCLES 2013 For Examiner’s Use 4 (i) Sketch the graph of y x 4 2 = - on the axes below, showing the coordinates of the points where the graph meets the axes. [3] x O y (ii) Solve the equation x x 4 2 - = . [3] www.theallpapers.com

7 0606/23/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 5 x cm R U Q P T S A piece of wire of length 96 cm is formed into the rectangular shape PQRSTU shown in the diagram. It is given that PQ = TU = SR = x cm. It may be assumed that PQ and TU coincide and that TS and QR have the same length. (i) Show that the area, A cm2, enclosed by the wire is given by A x x 2 96 3 2 = - . [2] (ii) Given that x can vary, find the stationary value of A and determine the nature of this stationary value. [4] www.theallpapers.com

8 0606/23/M/J/13 © UCLES 2013 For Examiner’s Use 6 Find the equation of the normal to the curve y x x 2 8 2 = - + at the point on the curve where x = 4. [6] www.theallpapers.com

9 0606/23/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 7 (i) Find the first four terms in the expansion of ( )x 2 6 + in ascending powers of x. [3] (ii) Hence find the coefficient of x3 in the expansion of x x x 1 3 1 2 6 + - + ^ ^ ^ h h h . [4] www.theallpapers.com

10 0606/23/M/J/13 © UCLES 2013 For Examiner’s Use 8 The line 2 8 y x = - cuts the curve 2 5 32 0 x y xy 2 2 + - + = at the points A and B. Find the length of the line AB. [7] www.theallpapers.com

11 0606/23/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 9 It is given that x ∈  and that  = {x : − 5 < x < 12}, S = {x : 5x + 24 > x2}, T = {x : 2x + 7 > 15}. Find the values of x such that (i) x S ! , [3] (ii) x S T j ! , [2] (iii) ( ) x S T k ! l. [3] www.theallpapers.com

12 0606/23/M/J/13 © UCLES 2013 For Examiner’s Use 10 A plane, whose speed in still air is 240 kmh–1, flies directly from A to B, where B is 500 km from A on a bearing of 032°. There is a constant wind of 50 kmh–1 blowing from the west. (i) Find the bearing on which the plane is steered. [4] www.theallpapers.com

13 0606/23/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (ii) Find, to the nearest minute, the time taken for the flight. [4] www.theallpapers.com

14 0606/23/M/J/13 © UCLES 2013 For Examiner’s Use 11 A one-one function f is defined by ( ) ( 1) 5 x x f 2 = - - for x k H . (i) State the least value that k can take. [1] For this least value of k (ii) write down the range of f, [1] (iii) find ( )x f 1 - , [2] www.theallpapers.com

15 0606/23/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (iv) sketch and label, on the axes below, the graph of ( ) y x f = and of ( ) y x f 1 = - , [2] x O y y = x (v) find the value of x for which ( ) ( ) x x f f 1 = - . [2] Question 12 is printed on the next page. www.theallpapers.com

16 0606/23/M/J/13 © UCLES 2013 For Examiner’s Use 12 The function ( )x x x ax b f 3 2 = + + + is divisible by x 3 - and leaves a remainder of 20 when divided by x 1 + . (i) Show that b 6 = and find the value of a. [4] (ii) Using your value of a and taking b as 6, find the non-integer roots of the equation ( ) 0 x f = in the form p q ! , where p and q are integers. [5] 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 0606/23 Paper 2, 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 23 © 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 23 © 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 23 © Cambridge International Examinations 2013 1 θ θ 2 2 cos sin 2 2 + B1 For all methods look for: – correct simplified expression – correct use of Pythagoras – use of tan = cos sin – use of cos 1 = sec Award first 3 then last B1 for final expression from fully correct method. Inconsistent no angle used then –1 (can recover). If start from RHS award similarly. θ θ sec 2 cos 2 2 = B1 θ θ θ 2 2 2 tan 2 cos sin = B1 θ θ 2 2 tan 2 2 sec 2 + = and completion B1 Or ( ) ( )2 2 tan sec tan sec θ θ θ θ − + + [B1, B1 θ θ 2 2 tan 2 sec 2 + B1 ( ) θ θ 2 2 tan 2 tan 1 2 + + and completion B1] Or θ θ 2 2 cos sin 2 2 + [B1 ( ) θ θ θ θ 2 2 2 2 cos sin 2 cos sin 2 + + B1 θ θ θ 2 2 2 tan 4 cos sin 4 = B1 2 cos cos 2 2 2 = θ θ and completion B1] 2 (i) 3.2 B1 (ii) 15 B1 (iii) uses area to find distance M1 If split 2 or 3 correct formulae and must be attempting total area two of 40, 240 and 32 A1 312 A1 or A2 for 312 from trapezium www.theallpapers.com

Page 5 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 23 © Cambridge International Examinations 2013 3 x x k x y cos sin d d = M1 k = –8 A1 Attempt to find x when y = 8 M1 Must get to x = numerical value ( ) 785 .0 4 x π = A1 45° = A0 (but can still gain next 2 marks) Uses t x x y t y d d d d d d × = M1 Must use numerical value for x and 0.2 for t x d d –0.8 (not rounded) A1 (condone poor notation if correct terms multiplied) 4 (i) Idea of modulus correct B1 Two straight lines above and touching x-axis 2 1 indicated on x-axis B1 Must be a sketch 2 indicated on y-axis B1 Must be a sketch (ii) 3 2 (0.667) B1 0.67 is B0 Solve 4x – 2 = –x or (4x – 2)2 = x2 M1 As far as x = numerical value 5 2 A1 SC: If drawn then B1, B2 for exact answers only 5 (i) (QR = PS =) 2 3 96 x − B1 Can be implied by next statement Area = x x ×       − 2 3 96 B1 AG (ii) 2 6 96 d d x x A − = or 48 – 3x o.e. B1 Solving 0 2 6 96 d d = − = x x A M1 As far as x = numerical value x = 16 A1 A = 384 and state maximum A1 www.theallpapers.com

Page 6 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 23 © Cambridge International Examinations 2013 6 Applies quotient rule correctly M1 or product rule ( ) ( ) ( )2 2 2 8 2 2 − + − − x x x x A1 2x (x – 2)–1 – (x2 + 8) (x – 2)–2 y = 12 B1 Uses m1m2 = –1 M1 (Gradient normal = 2 1 ) Uses equation of line for normal M1 If uses y = mx + c must find c for M1 ( ) 10 2 1 or 4 2 1 12 + = − = − x y x y A1 7 (i) 64 + 192x + 240x2 + 160x3 mark final answer B3, 2, 1, 0 3 terms correct earn B2; 2 terms correct earn B1 Can be earned in (ii); SC2 correct but unsimplified (ii) Multiply out (1 + 3x) (1 – x) M1 1 + 2x – 3x2 o.e. A1 (1) × (160) + (2) × (240) + (–3) ×(192) o.e. M1 3 terms 64 A1 Or Multiply out (1 – x) (64 + 192x + 240x2 + 160x3) [M1 May be other variations: for first M1 find x2 term or x3 term ...48x2 – 80x3...o.e. A1 Multiply by 1 + 3x M1 for second M1 must produce all relevant terms 64 A1] Or (1 + 3x) (64 + 192x + 240x2 + 160x3) [M1 ...816x2 + 880x3...o.e. A1 Multiply by 1 – x M1 64 A1] www.theallpapers.com

Page 7 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 23 © Cambridge International Examinations 2013 8 Eliminates y (or x) and full attempt at expansion M1 4x2 – 8x – 96 = 0 or y2 + 12y – 64 = 0 A1 Factorise 3 term relevant quadratic M1 Or use correct formula x = –4 and 6 or y = –16 and 4 A1 y = –16 and 4 or x = –4 and 6 A1√ Uses Pythagoras for relevant points M1 22.4 or 500 or 5 10 A1 cao 9 (i) Attempt to solve 3 term quadratic M1 –3 and 8 A1 –3 x 8 A1 Condone –3 x AND x 8 (ii) 4 x ( 12) B1 S ∪ T = –3 x 12 B1 (iii) S ∩ T = 4 x 8 or S′ = –5 x –3, 8 x 12 and T ′ = –5 x 4 B1 Penalise confusion over and (or and ) once only –5 x 4 B1√ their 4 8 x 12 B1√ their 8 (Ignore AND/OR etc.) www.theallpapers.com

Page 8 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 23 © Cambridge International Examinations 2013 10 (i) 240 58 sin 50 sin = α M1 A1 Use of sin rule/cosine rule/resolving with 50, 240 and 58/32/122/148. Must be correct for A1 2 10. α = A1 Bearing (0)21.8 or (0)22 A1√ √ for 32 – α (ii) V 2 = 2402 + 502 – 2 × 240 × 50 × cos (122 – α ) M1 Correct use of sin rule/cosine rule/resolving V = 263 awt A1 Can be in (i) V T 500 = M1 Only allow if V calculated from non right-angled triangle 114 or 1 hour 54 mins A1 Do not allow incorrect units Or 21.8 cos 240 32 cos 500 = T [M1 Alternative for part (ii) only Also can find distance for 240 (457) then 457/240 500 cos 32 B1 240 cos 21.8 B1 114 or 1 hour 54 mins A1] 11 (i) 1 B1 Not a range for k, but condone x = 1 and x 1 (ii) f –5 B1 Not x, but condone y (iii) Method of inverse M1 Do not reward poor algebra but allow slips 5 1 + + x A1 Must be f–1 = ...or y = ... (iv) f: Positive quadratic curve correct range and domain B1 Must cross x-axis f–1: Reflection of f in y = x B1√ √their f(x) sketch Condone slight inaccuracies unless clear contradiction. (v) Arrange f (x) = x or f–1 (x) = x to 3 term quadratic = 0 M1 4 only www A1 Allow x = 4 with no working. Condone (4, 4). Do not allow final A mark if –1 also given in answer www.theallpapers.com

Page 9 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 23 © Cambridge International Examinations 2013 12 (i) f(3) = (27 + 9 + 3a + b) = 0 or 3a + b = –36 M1 Equate f(3) to 0 f(–1) = (–1 + 1 – a + b) = 20 or –a + b = 20 M1 Equate f(–1) to 20 Solve equations M1 a = –14, b = 6 A1 If uses b = 6 then M0, A0 Need both values for A1 (ii) Find quadratic factor M1 If division, must be complete with first 2 terms correct If writes down, must be (x2 + kx – 2) x2 – 4x – 2 A1 Use quadratic formula or completing square on relevant 3 term quadratic M1 If completing square, must reach 2 2 2 2 2       ± =      + k k x 2 8 16 4 + ± − or better A1√ 6 2 ± − isw A1 cao www.theallpapers.com