Cambridge A Level Mathematics (US) 9280 — 2015 May/June Paper 3 · Variant 1

*3984395455* Cambridge International Examinations Cambridge International Advanced Level MATHEMATICS (US) 9280/31 Paper 3 Pure Mathematics 3 (P3) May/June 2015 1 hour 45 minutes Additional Materials: Answer Booklet/Paper Graph Paper List of Formulas (MF9) (US) READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Center number, candidate number, and name on 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 75. Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying larger numbers of marks later in the paper. This document consists of 3 printed pages and 1 blank page. JC15 06_9280_31/FP © UCLES 2015 [Turn over

2 1 Use logarithms to solve the equation 25x = 32x+1, giving the answer correct to 3 significant figures. [4] 2 Use the trapezium rule with three intervals to find an approximation to Ó 3 0 3x −10 dx. 4 3 Show that, for small values of x2, 1 −2x2−2 −1 + 6x22 3 ≈kx4, where the value of the constant k is to be determined. [6] 4 The equation of a curve is y = 3 cos 2x + 7 sin x + 2. Find the x-coordinates of the stationary points in the interval 0 ≤x ≤0. Give each answer correct to 3 significant figures. [7] 5 (a) Find Ó 4 + tan22x dx. [3] (b) Find the exact value of Ô 1 20 1 40 sinx + 1 60 sin x dx. [5] 6 The straight line l1 passes through the points 0, 1, 5 and 2, −2, 1. The straight line l2 has equation r = 7i + j + k + -i + 2j + 5k. (i) Show that the lines l1 and l2 are skew. [6] (ii) Find the acute angle between the direction of the line l2 and the direction of the x-axis. [3] 7 Given that y = 1 when x = 0, solve the differential equation dy dx = 4x3y2 + 10y + 3, obtaining an expression for y in terms of x. [9] 8 The complex number w is defined by w = 22 + 4i 2 −i2 . (i) Without using a calculator, show that w = 2 + 4i. [3] (ii) It is given that p is a real number such that 1 40 ≤argw + p ≤3 40. Find the set of possible values of p. [3] (iii) The complex conjugate of w is denoted by w*. The complex numbers w and w* are represented in an Argand diagram by the points S and T respectively. Find, in the form Ï −a = k, the equation of the circle passing through S, T and the origin. [3] © UCLES 2015 9280/31/M/J/15

3 9 x y O M The diagram shows the curve y = x2e2−x and its maximum point M. (i) Show that the x-coordinate of M is 2. [3] (ii) Find the exact value of Ó 2 0 x2e2−x dx. [6] 10 x y O P The diagram shows part of the curve with parametric equations x = 2 lnt + 2, y = t3 + 2t + 3. (i) Find the slope of the curve at the origin. [5] (ii) At the point P on the curve, the value of the parameter is p. It is given that the slope of the curve at P is 1 2. (a) Show that p = 1 3p2 + 2 −2. [1] (b) By first using an iterative formula based on the equation in part (a), determine the coordinates of the point P. Give the result of each iteration to 5 decimal places and each coordinate of P correct to 2 decimal places. [4] © UCLES 2015 9280/31/M/J/15

4 BLANK PAGE 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. © UCLES 2015 9280/31/M/J/15

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the May/June 2015 series 9280 MATHEMATICS (US) 9280/31 Paper 3 (Paper 3), maximum raw mark 75 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 International A Level – May/June 2015 9280 31 © Cambridge International Examinations 2015 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 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. The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. • Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. • For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.

Page 3 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9280 31 © Cambridge International Examinations 2015 The following abbreviations may be used in a mark scheme or used on the scripts: AEF Any Equivalent Form (of answer is equally acceptable) 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) CWO Correct Working Only – often written by a ‘fortuitous’ answer 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) SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance) 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. An MR –2 penalty may be applied in particular cases if agreed at the coordination meeting. PA –1 This is deducted from A or B marks in the case of premature approximation. The PA –1 penalty is usually discussed at the meeting.

Page 4 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9280 31 © Cambridge International Examinations 2015 1 Use law for the logarithm of a power at least once *M1 Obtain correct linear equation, e.g. In3 )1 2 ( 2 In 5 + = x x A1 Solve a linear equation for x M1 dep *M Obtain 0.866 x = A1 [4] 2 Attempt calculation of at least 3 ordinates M1 Obtain 9, 7, 1, 17 A1 Use trapezium rule with 1 = h M1 Obtain 1 2 ( ) 9 14 2 17 + + + or equivalent and hence 21 A1 [4] 3 Either Obtain correct (unsimplified) version of 2 x or 4 x term in 2 2) 2 1( − −x M1 Obtain 2 4 1 x + A1 Obtain … 4 12x + A1 Obtain correct (unsimplified) version of 2 x or 4 x term in 3 2 2) 6 1( x + M1 Obtain 4 2 4 4 1 x x − + A1 Combine expansions to obtain 16 = k with no error seen A1 Or Obtain correct (unsimplified) version of 2 x or 4 x term in 3 2 2) 6 1( x + M1 Obtain 2 4 1 x + A1 Obtain … 4 4x − A1 Obtain correct (unsimplified) version of 2 x or 4 x term in 2 2) 2 1( − −x M1 Obtain 4 2 12 4 1 x x + + A1 Combine expansions to obtain 16 = k with no error seen A1 [6] 4 Differentiate to obtain form x b x a cos 2 sin + M1 Obtain correct x x cos 7 2 sin 6 + − A1 Use identity x x x cos sin 2 2 sin = B1 Solve equation of form 0 cos cos sin = + x d x x c to find at least one value of x M1 Obtain 0.623 A1 Obtain 2.52 A1 Obtain 1.57 or 1 2 π from equation of form 0 cos cos sin = + x d x x c A1 Treat answers in degrees as MR – 1 situation [7]

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9280 31 © Cambridge International Examinations 2015 5 (a) Use identity 1 2 sec 2 tan 2 2 − = x x B1 Obtain integral of form x b ax 2 tan + M1 Obtain correct 3x + 1 2 tan 2x , condoning absence of + c A1 [3] (b) State sin cos x 1 2 cos sin x π + 1 6 π B1 Simplify integrand to cos 1 6 1 6 cos sin sin x x π π + or equivalent B1 Integrate to obtain at least term of form ) sin In( x a *M1 Apply limits and simplify to obtain two terms M1 dep *M Obtain 1 8 3 π −1 2 ln( 1 2 ) or equivalent A1 [5] 6 (i) Obtain           − − ± 4 3 2 as direction vector of 1l B1 State that two direction vectors are not parallel B1 Express general point of 1l or 2l in component form, e.g. ) 4 5 , 3 1 , 2 ( λ λ λ − − or ) 5 1 , 2 , 7 ( µ µ µ + + + l B1 Equate at least two pairs of components and solve for λ or for µ M1 Obtain correct answers for λ and µ A1 Verify that all three component equations are not satisfied (with no errors seen) A1 [6] (ii) Carry out correct process for evaluating scalar product of           5 2 1 and           0 0 1 M1 Use correct process for finding modulus and evaluating inverse cosine M1 Obtain ° 5. 79 or 1.39 radians A1 [3] 7 Separate variables and factorise to obtain x x y y y d 4 )3 )( 1 3 ( d = + + or equivalent B1 State or imply the form 3 1 3 + + + y B y A and use a relevant method to find A or B M1 Obtain A = 3 8 and B = −1 8 A1 Integrate to obtain form 1 2 ln(3 1) ln( 3) k y k y + + + M1 Obtain correct 1 8 ln(3 1) y + −1 8 2 ln( 3) 2 y x = = or equivalent A1 Substitute 0 = x and 1 = y in equation of form 2 1 2 3 ln(3 1) ln( 3) k y k y k x c + + + = + to find a value of c M1 Obtain 0 = c A1 Use correct process to obtain equation without natural logarithm present M1 Obtain 2 2 16 16 3 1 3 x x e e y − − = or equivalent A1 [9]

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9280 31 © Cambridge International Examinations 2015 8 (i) Either Expand 2)i 2 ( − to obtain i4 3 − or unsimplified equivalent B1 Multiply by 4i 3 i4 3 + + and simplify to y x i + form or equivalent M1 Confirm given answer i4 2 + A1 Or Expand 2 i) 2 ( − to obtain i4 3 − or unsimplified equivalent B1 Obtain two equations in x and y and solve for x or y M1 Confirm given answer i4 2 + A1 [3] (ii) Identify 4 4 + or i4 4 + − as point at either end or state 2 = p or state 6 − = p B1 Use appropriate method to find both critical values of p M1 State –6 Y p Y 2 A1 [3] (iii) Identify equation as of form a a z = − or equivalent M1 Form correct equation for a not involving modulus, e.g. 2 2 2 4 ) 2 ( a a = + − A1 State 5 5 = − z A1 [3] 9 (i) Use product rule to find first derivative M1 Obtain x x x x − −− 2 2 2 e e 2 A1 Confirm 2 = x at M A1 [3] (ii) Attempt integration by parts and reach ∫ − −± ± x x x x x d e 2 e 2 2 2 *M1 Obtain ∫ − −+ − x x x x x d e 2 e 2 2 2 A1 Attempt integration by parts and reach x x x x x − − − ± ± ± 2 2 2 2 e 2 e 2 e *M1 Obtain x x x x x − − − − − − 2 2 2 2 e 2 e 2 e A1 Use limits 0 and 2 having integrated twice M1 dep *M Obtain 10 e 2 2 − A1 [6] 10 (i) Obtain 2 2 d d + = t t x and 2 3 d d 2 + = t t y B1 Use t x t y x y d d d d d d ÷ = M1 Obtain d d y x = 1 2 2 (3 2)( 2) t t + + A1 Identify value of t at the origin as –1 B1 Substitute to obtain 5 2 as gradient at the origin A1 [5] (ii) (a) Equate derivative to 1 2 and confirm 2 2 3 1 2 − + = p p B1 [1] (b) Use the iterative formula correctly at least once M1 Obtain value 924 .1 − = p or better (–1.92367…) A1 Show sufficient iterations to justify accuracy or show a sign change in appropriate interval A1 Obtain coordinates (–5.15, –7.97) A1 [4]