Cambridge A Level Mathematics (US) 9280 — 2015 May/June Paper 2 · Variant 1

*8983155712* Cambridge International Examinations Cambridge International Advanced Subsidiary Level MATHEMATICS (US) 9280/21 Paper 2 Pure Mathematics 2 (P2) May/June 2015 1 hour 15 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 50. 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_21/FP © UCLES 2015 [Turn over

2 1 (i) Solve the equation 3x + 4 = 3x −11. [3] (ii) Hence, using logarithms, solve the equation 3 × 2y + 4 = 3 × 2y −11, giving the answer correct to 3 significant figures. [2] 2 x ln y O (2, 1.60) (5, 2.92) The variables x and y satisfy the equation y = Aepx−1, where A and p are constants. The graph of ln y against x is a straight line passing through the points 2, 1.60 and 5, 2.92, as shown in the diagram. Find the values of A and p correct to 2 significant figures. [5] 3 The equation of a curve is y = 6 sin x −2 cos 2x. Find the equation of the tangent to the curve at the point 1 60, 2. Give the answer in the form y = mx + c, where the values of m and c are correct to 3 significant figures. [5] 4 The polynomials fx and gx are defined by fx = x3 + ax2 + b and gx = x3 + bx2 −a, where a and b are constants. It is given that x + 2 is a factor of fx. It is also given that, when gx is divided by x + 1, the remainder is −18. (i) Find the values of a and b. [5] (ii) When a and b have these values, find the greatest possible value of gx −fx as x varies. [2] 5 (i) Given that Ó a 0 3e 1 2x + 1 dx = 10, show that the positive constant a satisfies the equation a = 2 ln @16 −a 6 A . 5 (ii) Use the iterative formula an+1 = 2 ln @16 −an 6 A with a1 = 2 to find the value of a correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] © UCLES 2015 9280/21/M/J/15

3 6 (i) Prove that 2 cosec 21 tan 1  sec2 1. [3] (ii) Hence (a) solve the equation 2 cosec 21 tan 1 = 5 for 0 < 1 < 0, [3] (b) find the exact value of Ó 1 60 0 2 cosec 4x tan2x dx. [4] 7 The equation of a curve is y3 + 4xy = 16. (i) Show that dy dx = − 4y 3y2 + 4x. [4] (ii) Show that the curve has no stationary points. [2] (iii) Find the coordinates of the point on the curve where the tangent is parallel to the y-axis. [4] © UCLES 2015 9280/21/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 9280/21/M/J/15

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Subsidiary and Advanced Level MARK SCHEME for the May/June 2015 series 9280 MATHEMATICS (US) 9280/21 Paper 2, maximum raw mark 50 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 AS/A Level – May/June 2015 9280 21 © 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 AS/A Level – May/June 2015 9280 21 © 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 AS/A Level – May/June 2015 9280 21 © Cambridge International Examinations 2015 1 (i) State or imply equation 2 2 (3 4) (3 11) x x + = − or 3 4 (3 11) x x + = − − B1 Attempt solution of ‘quadratic’ equation or linear equation M1 Obtain 7 6 x = or equivalent (and no other solutions) A1 [3] (ii) Use logarithms to solve equation of form 2y = their answer to (i) ( must be + ve) M1 Obtain 0.222 (and no other solutions) A1 [2] 2 State or imply that ln ln ( 1) y A p x = + − B1 Equate gradient to p or obtain two equations for ln A and p M1 Obtain 0.44 p = A1 Substitute values correctly, to find value of ln A DM1 Obtain 3.2 A = A1 [5] Alternative: Obtain an equation either 1.6 e e p A = or 2.92 4 e e p A = M1 Obtain both equations correctly A1 Solve to obtain 0.44 p = A1 Substitute value correctly to find A DM1 Obtain 3.2 A = A1 [5] 3 Differentiate to obtain form cos sin 2 p x q x + or equivalent M1 Obtain correct 6cos 4sin 2 x x + or equivalent A1 Substitute 1 6π to obtain derivative equal to 5 3 or 8.66 A1 Form equation of tangent (not normal) using numerical value of gradient obtained by differentiation M1 Obtain 8.66 2.53 y x = − cao A1 [5] 4 (i) Substitute 2 x = − in f( )x and equate to zero to obtain 8 4 0 a b −+ + = or equiv B1 Substitute 1 x = − in g( ) x and equate to –18 M1 Obtain 1 18 b a −+ − = − or equivalent A1 Solve a pair of linear equations for a or b DM1 Obtain 5 a = , 12 b = − A1 [5] (ii) Simplify g( ) f ( ) x x − to obtain form 2 kx c + where 0 k < M1 Obtain 2 17 7 x − + and state 7, following their value of c A1√ [2]

Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9280 21 © Cambridge International Examinations 2015 5 (i) Obtain integral of form 1 2e x k mx + M1 Obtain correct 1 2 6e x x + A1 Apply limits and obtain correct 1 2 6e 6 a a + − A1 Equate to 10 and introduce natural logarithm correctly DM1 Obtain given answer ( ) 16 6 2ln a a − = correctly A1 [5] (ii) Use the iterative formula correctly at least once M1 Obtain final answer 1.732 A1 Show sufficient iterations to justify accuracy to 3 d.p. or show sign change in interval (1.7315, 1.7325) A1 [3] 6 (i) State or imply 1 cosec2 sin 2 θ θ = B1 Express left-hand side in terms of sinθ and cosθ M1 Obtain given answer 2 sec θ correctly A1 [3] (ii) (a) State or imply 1 cos 5 θ = or tan 2 θ = at least B1 Obtain 1.11 or awrt 1.11, allow 0.353π B1 Obtain 2.03 or awrt 2.03 , allow 0.648π and no other values between 0 and π B1 [3] (b) State integrand as 2 sec 2x B1 Integrate to obtain expression of form tan k mx M1 Obtain correct 1 2 tan 2x A1 Obtain 1 2 3 or exact equivalent A1 [4] 7 (i) Obtain 2 d 3 d y y x as derivative of 3 y B1 Obtain d 4 4 d y y x x + as derivative of 4xy B1 Equate derivative of left-hand side to zero and solve for d d y x , must be from implicit differentiation M1 Confirm given answer 2 d 4 d 3 4 y y x y x = − + correctly A1 [4] (ii) State or imply 0 y = B1 Substitute in equation of curve and show contradiction B1 [2] (iii) State or imply 2 3 4 0 y x + = B1 Eliminate one variable from equation of curve using 2 3 4 0 y x + = M1 Obtain 2 y = − A1 Obtain 3 x = − A1 [4]