Cambridge IGCSE Mathematics - Additional 0606 — 2012 May/June Paper 1 · Variant 2

This document consists of 20 printed pages. DC (SLM) 57493 © UCLES 2012 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education * 4 5 2 0 6 0 3 8 8 4 * ADDITIONAL MATHEMATICS 0606/12 Paper 1 May/June 2012 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. For Examiner’s Use 1 2 3 4 5 6 7 8 9 10 11 12 Total www.theallpapers.com

2 0606/12/M/J/12 © UCLES 2012 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 0606/12/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use 1 (i) Find  7x – 5 dx. [3] (ii) Hence evaluate 3 2 7x – 5 dx. [2] www.theallpapers.com

4 0606/12/M/J/12 © UCLES 2012 For Examiner’s Use 2 Using the substitution u = 2x, find the values of x such that 22x+2 = 5 (2x) – 1 . [5] www.theallpapers.com

5 0606/12/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use 3 Show that cot A + sin A 1 + cos A = cosec A . [4] www.theallpapers.com

6 0606/12/M/J/12 © UCLES 2012 For Examiner’s Use 4 Solve the simultaneous equations 5x + 3y = 2 and 2 x – 3 y = 1 . [5] www.theallpapers.com

7 0606/12/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use 5 Differentiate the following with respect to x. (i) (2 – x2)1n(3x + 1) [3] (ii) 4 – tan 2x 5x [3] www.theallpapers.com

8 0606/12/M/J/12 © UCLES 2012 For Examiner’s Use 6 You must not use a calculator in this question. (i) Express 8 3 + 1 in the form a( 3 – 1), where a is an integer. [2] An equilateral triangle has sides of length 8 3 + 1 . (ii) Show that the height of the triangle is 6 – 2 3 . [2] www.theallpapers.com

9 0606/12/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use (iii) Hence, or otherwise, find the area of the triangle in the form p 3 – q, where p and q are integers. [2] www.theallpapers.com

10 0606/12/M/J/12 © UCLES 2012 For Examiner’s Use 7 (i) Sketch the graph of y = |x2 – x – 6|, showing the coordinates of the points where the curve meets the coordinate axes. [3] (ii) Solve |x2 – x – 6| = 6. [3] www.theallpapers.com

11 0606/12/M/J/12 © UCLES 2012 For Examiner’s Use [Turn over 8 O A B X 2π 3 rad 10 cm The figure shows a circle, centre O, with radius 10 cm. The lines XA and XB are tangents to the circle at A and B respectively, and angle AOB is 2π 3 radians. (i) Find the perimeter of the shaded region. [3] (ii) Find the area of the shaded region. [4] www.theallpapers.com

12 0606/12/M/J/12 © UCLES 2012 For Examiner’s Use 9 Variables N and x are such that N = 200 + 50e x 100 . (i) Find the value of N when x = 0. [1] (ii) Find the value of x when N = 600. [3] www.theallpapers.com

13 0606/12/M/J/12 © UCLES 2012 For Examiner’s Use [Turn over (iii) Find the value of N when dN dx = 45. [4] www.theallpapers.com

14 0606/12/M/J/12 © UCLES 2012 For Examiner’s Use 10 (a) It is given that f(x) = 1 2 + x for x  –2, x. (i) Find f ″(x). [2] (ii) Find f –1 (x). [2] (iii) Solve f 2(x) = –1. [3] www.theallpapers.com

15 0606/12/M/J/12 © UCLES 2012 For Examiner’s Use (b) The functions g, h and k are defined, for x, by g(x) = 1 x + 5, x  –5, h(x) = x2 –1, k(x) = 2x + 1. Express the following in terms of g, h and/or k. (i) 1 (x2–1) + 5 [1] (ii) 2 x + 5 + 1 [1] [Turn over www.theallpapers.com

16 0606/12/M/J/12 © UCLES 2012 For Examiner’s Use 11 The point P lies on the line joining A(–1, –5) and B(11, 13) such that AP = 1 3AB. (i) Find the equation of the line perpendicular to AB and passing through P. [5] The line perpendicular to AB passing through P and the line parallel to the x-axis passing through B intersect at the point Q. (ii) Find the coordinates of the point Q. [2] www.theallpapers.com

17 0606/12/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use (iii) Find the area of the triangle PBQ. [2] www.theallpapers.com

18 0606/12/M/J/12 © UCLES 2012 For Examiner’s Use Answer only one of the following two alternatives. 12 EITHER At 12 00 hours, a ship has position vector (54i + 16j) km relative to a lighthouse, where i is a unit vector due East and j is a unit vector due North. The ship is travelling with a speed of 20 km h–1 in the direction 3i +4j. (i) Show that the position vector of the ship at 15 00 hours is (90i + 64j) km. [2] (ii) Find the position vector of the ship t hours after 12 00 hours. [2] A speedboat leaves the lighthouse at 14 00 hours and travels in a straight line to intercept the ship. Given that the speedboat intercepts the ship at 16 00 hours, find (iii) the speed of the speedboat, [3] (iv) the velocity of the speedboat relative to the ship, [1] (v) the angle the direction of the speedboat makes with North. [2] OR R A O B P Q The position vectors of points A and B relative to an origin O are a and b respectively. The point P is such that OP = 5 4 OB . The point Q is such that AQ = 1 3 AB . The point R lies on OA such that RQP is a straight line where OR =λOA and QR = μPR . (i) Express OQ and PQ in terms of a and b. [2] (ii) Express QR in terms of λ, a and b. [2] (iii) Express QR in terms of μ , a and b. [3] (iv) Hence find the value of λ and of μ. [3] www.theallpapers.com

19 0606/12/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use Start your answer to Question 12 here. Indicate which question you are answering. EITHER OR … … … … … … … … … … … … … … … … … … … … … … … … … www.theallpapers.com

20 0606/12/M/J/12 © UCLES 2012 For Examiner’s Use Continue your answer here if necessary. … … … … … … … … … … … … … … … … … … … … … … … … … 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

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 2012 question paper for the guidance of teachers 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 must be read in conjunction with the question papers and the report on the examination. • Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2012 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.theallpapers.com

Page 2 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2012 0606 12 © University of Cambridge International Examinations 2012 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: Teachers’ version Syllabus Paper IGCSE – May/June 2012 0606 12 © University of Cambridge International Examinations 2012 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: Teachers’ version Syllabus Paper IGCSE – May/June 2012 0606 12 © University of Cambridge International Examinations 2012 1 (i) 2 3 ) 5 7 ( 21 2 − x (+ c) B1 B1, B1 B1 for multiplication by 3 2 , or division by 2 3 B1 for 2 3 ) 5 7 ( − x , B1 for 1 7 (ii)       − 2 3 2 3 9 16 21 2 ( = ) 27 64 ( 21 2 − ) = 21 74 or awrt 3.52 or 21 11 3 M1 A1 [5] M1 for correct use of limits, must have attempted integration, must be using their 2 1 2 ) 5 7 ( + − n x from (i) 2 4u2 – 5u + 1 = 0 (4u – 1) (u – 1) = 0 or (4.2x – 1)(2x – 1) = 0 2x = 4 1 , 2x = 1 leading to x = –2, 0 Alternate scheme for one correct factor: 2x = 4 1 , leading to x = –2 2x = 1, leading to x = 0 B1, M1 DM1 A1 A1 [A1] [A1] [5] B1 for 22x + 2 = 4u2 or 4 × 22x or 22 × 22x or 22u2 M1 for attempt to obtain a 3 term quadratic equation in terms of either or, equated to zero. DM1 for solution of quadratic equation A1 for both A1 for both 3 A A A A cos 1 sin sin cos + + = ) cos 1( sin sin cos cos 2 2 A A A A A + + + = ) cos 1( sin ) cos 1( A A A + + = A sin 1 = cosecA Alternate solution: ) cos 1 )( cos 1( ) cos 1( sin cot A A A A A − + − + = ) sin ) cos 1( sin cot 2 A A A A − + = A A A sin cos 1 cot − + = A A A sin 1 cot cot + − leading to cosecA B1 M1 M1 A1 [M1] [M1] [M1] [A1] [4] B1 for A A A sin cos cot = M1 for obtaining as a single fraction M1 for use of cos2 A + sin2 A = 1 A1 for correct simplification – answer given. M1 for multiplying by (1 – cos A) M1 for use of cos2 A + sin2 A = 1 anywhere M1 for cancelling sin A A1 for subtraction and simplification www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2012 0606 12 © University of Cambridge International Examinations 2012 4 Using 3 5 2 x y − = or, using 5 3 2 y x − = 5x2 – 21x + 4 = 0 or 3y2 + 17y – 6 = 0 (5x – 1)(x – 4) = 0 or (3y – 1)(y + 6) = 0 x = 5 1 , y = 3 1 x = 4, y = –6 Alternate substitutions: y y x + = 3 2 or x x y − = 2 3 M1 M1 DM1 A1, A1 [5] M1 for substitution to get an equation in terms of one variable M1 for attempt to form a 3 term quadratic equation = 0 DM1 for solution of quadratic equation A1 for each ‘pair’ 5 (i) )1 3 ln( 2 )1 3 ( 3 ) 2 ( 2 + − + − x x x x (ii) 2 2 25 ) 2 tan 4 ( 5 ) 2 sec 2 ( 5 x x x x − − − or 2 2 ) 5 ( ) 2 tan 4 ( 5 ) 2 sec 2 ( 5 x x x x − − − B1 M1 A1 B1 M1 A1 [6] B1 for differentiating ln(3x + 1) correctly M1 for correct attempt at product A1 for all else correct B1 for differentiating tan(4 – 2x) correctly M1 for correct attempt at quotient or product A1 for all else correct 6 (i) )1 3 ( 4 )1 3 )( 1 3 ( ) 1 3 ( 8 − = − + − or ), 1 3 ( 1 3 8 − = + a 8 = a )1 3 )( 1 3 ( + − a = 4 M1 A1 M1 for rationalisation or attempt to form equation (ii) )1 3 ( 4 2 3 60 sin − = = h )1 3 ( 2 3 60 tan − = = h Or 2 2 2 )) 1 3 ( 2 ( )) 1 3 ( 4 ( − + = − h 3 2 6 − = h ANSWER GIVEN M1 A1 M1 for use of sine or tangent and their value of a from (i) or 1 3 8 + or Pythagoras, A1 for rearranging and simplifying correctly to obtain given answer. (iii) Area = ) 3 2 6 )( 1 3 ( 4 2 1 − − or ° − − 60 sin )1 3 ( 4 )1 3 ( 4 2 1 = 24 3 16 − M1 A1 [6] M1 for valid method for area using their a from (i) or 1 3 8 + A1 working must be seen www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2012 0606 12 © University of Cambridge International Examinations 2012 7 (i) −3 −2 −1 1 2 3 4 −1 1 2 3 4 5 6 7 x y B1 B1 B1 B1 for shape B1 for x = –2, 3 B1 for y = 6 (ii) x2 – x – 6 = 6, leading to x = –3, 4 (www) x2 – x – 6 = –6, leading to x = 0, 1 (www) B1 B1 B1 [6] B1 for one correct answer B1 for a second correct answer B1 for a third and fourth correct answer 8 (i) arc AB = 3 π 20 or 20.94, 20.9 10 3 π tan AX = , AX = 10 3 , 17.3 (or XB) Perimeter = awrt 55.6 or 3 π 20 3 20 + B1 B1 B1 B1 for arc length correct B1 for AX/XB B1 for final answer (ii) Area of sector AOB 3 π 2 10 2 1 2 = or 104.7 or 105 Area of OAXB = 3 100 or 173.2 Shaded area = awrt 68.5 or 3 π 100 3 100 − B1 M1 M1 A1 [7] B1 for sector area correct M1 for valid attempt at area OAXB, using their BX from part (i) (10 × their BX ) M1 for area OAXB – sector area used (independent) Must be considering a quadrilateral, not a triangle. 9 (i) 250 B1 B1 for 250 (ii) 100 e 8 x = 100 x = ln ‘their 8’ or x = 100 ln their 8 x = 208 or awrt 208 B1 M1 A1 B1 for 100 e 8 x = M1 for dealing with e correctly, using ln A1 for awrt 208 (iii) 100 e 2 1 d d x x N = 100 e 2 1 45 x = 90 e100 = x , so N = 4700 (awrt 4700) B1, B1 M1 A1 [8] B1 for 100 e x , B1 for 100 e 2 1 x or 100 e 100 50 x M1 for equating their d d N x to 45 and attempt to solve A1 for 4700 www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2012 0606 12 © University of Cambridge International Examinations 2012 10 (a) (i) f ' (x) = –(2 + x)–2 f '' (x) = 2(2 + x)–3 (ii) x y + = 2 1 , 2 1 − = y x f–1 (x) = x 1 – 2 or x x 2 1− (iii) f 2 (x) =           + + x 2 1 2 1 = x x 2 5 2 + + Equating to – 1 leads to 3 7 − = x or –2.33 (b) (i) gh (x) or gh (ii) kg (x) or kg B1 B1 M1 A1 M1 DM1 A1 B1 B1 [9] First B1 may be implied by a correct answer for f '' (x) If done by quotient rule, allow unsimplified M1 for a valid attempt at the inverse A1 must be in correct form, allow y = … M1 for correct attempt at f 2 (x) DM1 for attempt at solution of f 2 (x) = –1 A1 for 3 7 − = x or equivalent B1 for either form B1 for either form 11 (i) P (3, 1) Grad 12 18 = AB ⊥ grad 3 2 − PQ: y – 1 = – ) 3 ( 3 2 − x (2x + 3y = 9) (ii) Q (–15, 13) (iii) Area = 2 2 12 18 2 1 + 2 2 12 8 + or Area = 1 3 13 15 13 11 1 3 2 1 − or Area = 12 26 2 1 × × = 156 B1, B1 B1 B1 B1 M1 A1 M1 A1 [9] B1 for each coordinate B1 for gradient of AB B1 for perpendicular gradient B1 on their perp gradient and their point P Must be y = … M1 for use of y = 13 and their PQ equation. A1 for both coordinates (can be implied) M1 for a valid attempt at area PB PQ× × 2 1 Matrix method using their coordinates correctly × ×QB 2 1 vertical perp height www.theallpapers.com

Page 8 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2012 0606 12 © University of Cambridge International Examinations 2012 12 EITHER (i) velocity = 12i + 16j position = ( 54i + 16j) + ( 36i + 48j) = 90i + 64j ANSWER GIVEN (ii) (54i + 16j) + (12ti + 16tj) (iii) At 16 00, ship has ‘travelled’ (102i + 80j) boat needs to do this in 2 hours so velocity of boat (51i + 40j) speed 2 2 40 51 + = 64.8 (iv) (51i + 40j) – (12i + 16j) = 39i +24j (v) 40 51 tan = α angle = 51.9 M1 A1 M1, A1 B1 M1 A1 B1 M1 A1 [10] M1 for (3 × their velocity ( must in numeric vector form)) + (54i + 16j) M1 for position vector + (their numeric velocity vector × time) B1 for (102i + 80j) M1 for attempt at velocity of boat and speed B1, allow unsimplified but must be correct M1 for use of tan and their velocity vector www.theallpapers.com

Page 9 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2012 0606 12 © University of Cambridge International Examinations 2012 12 OR (i) OQ a + 3 1 (b – a) = 3 2 a + 3 1 b 4 5 − = PQ b + a + 3 1 (b – a) = 3 2 a – 12 11 b (ii) λ = QR a – (a + 3 1 (b – a)) = λ a – 3 2 a – 3 1 b (iii) ) ( QR PQ QR + = µ PQ QR µ µ = − ) 1(       − − = b a 12 11 3 2 1 µ µ QR (iv) Equating b’s 3 1 1 12 11 − = − − µ µ 15 4 = µ 11 10 = λ B1 B1 M1 A1 M1 M1 A1 M1 A1 A1 [10] Allow unsimplified Follow through on theirOQ , allow unsimplified M1 for λa – their OQ A1 – allow unsimplified M1 for attempt to obtain QR in terms of PQ M1 for attempt to simplifiy M1 for equating like vectors and attempt to solve A1 for each www.theallpapers.com