Cambridge A Level Mathematics (US) 9280 — 2014 Oct/Nov Paper 2 · Variant 1

*6815098295* Cambridge International Examinations Cambridge International Advanced Subsidiary Level MATHEMATICS (US) 9280/21 Paper 2 Pure Mathematics 2 (P2) October/November 2014 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. JC14 11_9280_21/FP © UCLES 2014 [Turn over

2 1 Use the trapezium rule with four intervals to find an approximation to Ó 5 1 2x −8 dx. 3 2 x ln y O (0.75, 1.70) (1.53, 2.18) The variables x and y satisfy the equation y = abx, where a and b are constants. The graph of ln y against x is a straight line passing through the points 0.75, 1.70 and 1.53, 2.18, as shown in the diagram. Find the values of a and b correct to 2 decimal places. [5] 3 (a) Find Ó 4 cos21 21 d1. [3] (b) Find the exact value of Ô 6 −1 1 2x + 3 dx. [4] 4 For each of the following curves, find the exact slope of the tangent to the curve at the point indicated: (i) y = 3 cos 2x −5 sin x at 1 60, −1, [3] (ii) x3 + 6xy + y3 = 21 at 1, 2. [5] © UCLES 2014 9280/21/O/N/14

3 5 (i) Given that x + 2 and x + 3 are factors of 5x3 + ax2 + b, find the values of the constants a and b. [4] (ii) When a and b have these values, factorize 5x3 + ax2 + b completely, and hence solve the equation 53y+1 + a × 52y + b = 0, giving any answers correct to 3 significant figures. [5] 6 x y O M The diagram shows part of the curve y = x2 1 + e3x and its maximum point M. The x-coordinate of M is denoted by m. (i) Find dy dx and hence show that m satisfies the equation x = 2 31 + e−3x. [4] (ii) Show by calculation that m lies between 0.7 and 0.8. [2] (iii) Use an iterative formula based on the equation in part (i) to find m correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] 7 The angle ! lies between 0Å and 90Å and is such that 2 tan2! + sec2! = 5 −4 tan !. (i) Show that 3 tan2! + 4 tan ! −4 = 0 and hence find the exact value of tan !. [4] (ii) It is given that the angle " is such that cot! + " = 6. Without using a calculator, find the exact value of cot ". [5] © UCLES 2014 9280/21/O/N/14

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. 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 2014 9280/21/O/N/14

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Subsidiary Level MARK SCHEME for the October/November 2014 series 9280 MATHEMATICS 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 October/November 2014 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 Level – October/November 2014 9280 21 © Cambridge International Examinations 2014 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 Level – October/November 2014 9280 21 © Cambridge International Examinations 2014 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 Level – October/November 2014 9280 21 © Cambridge International Examinations 2014 1 State or imply correct y–values 6, 4, 0, 8, 24 B1 Use correct formula, or equivalent, with h = 1 and five y–values M1 Obtain 27 A1 [3] 2 State or imply lny = lna + xlnb B1 Equate lnb to numerical gradient of line M1 Obtain b = 1.85 A1 Substitute to find value of a M1 Obtain a = 3.45 A1 [5] 3 (a) Express integrand in the form 2 cos + θ p M1 State correct 2 cos 2 + θ A1 Integrate to obtain θ θ 2 sin 2 + (+ c) A1 [3] (b) Integrate to obtain form ( ) 3 2 ln + x k M1 Obtain correct ( ) 3 2 ln 2 1 + x A1 Apply limits correctly DM1 Obtain 15 ln 2 1 A1 [4] 4 (i) Differentiate to obtain form x k x k cos 2 sin 2 1 + M1 Obtain correct x x cos 5 2 sin 6 − − A1 Substitute π 6 1 to obtain 3 2 11 − or exact equivalent A1 [3] (ii) Obtain x y x y d d 6 6 + as derivative of 6xy B1 Obtain x y y d d 3 2 as derivative of y3 B1 Obtain 0 d d 3 d d 6 6 3 2 2 = + + + x y y x y x y x or equivalent B1 Substitute 1 and 2 to find value of gradient dependent on at least one B1 M1 Obtain gradient 18 15 − or 6 5 − A1 [5] 5 (i) State 0 4 40 = + + − b a or equivalent B1 State 0 9 135 = + + − b a or equivalent B1 Solve a pair of linear simultaneous equations M1 Obtain 19 = a and 36 − = b A1 [4] (ii) Identify 6 5 − x as a factor B1 State ( )( )( ) 6 5 3 2 − + + x x x B1 State or imply 5 6 5 = y , following a positive value from factorisation B1√ Apply logarithms and use power law M1 Obtain 0.113 only A1 [5]

Page 5 Mark Scheme Syllabus Paper Cambridge International AS Level – October/November 2014 9280 21 © Cambridge International Examinations 2014 6 (i) Use quotient rule or equivalent M1 Obtain ( ) ( ) 2 3 3 2 3 1 3 1 2 x x x e e x e x + − + or equivalent A1 Equate first derivative to zero and attempt rearrangement to x = … DM1 Obtain ( ) x e x 3 1 3 2 − + = with sufficient detail and no errors seen (AG) A1 [4] (ii) Consider sign of ( ) x e x 3 1 3 2 − + − at 0.7 and 0.8 or equivalent M1 Obtain correct values (–0.05 and 0.07 or equivalents) and conclude appropriately A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 0.739 A1 Show sufficient iterations to 5 decimal places to justify result or show a sign change in the interval (0.7385, 0.7395) A1 [3] 7 (i) Use α α 2 2 tan 1 sec + = B1 Confirm 0 4 tan 4 tan 3 2 = − + α α B1 Solve quadratic equation for α tan M1 Obtain, finally, 3 2 tan = α only A1 [4] (ii) State or imply ( ) 6 1 tan = + β α B1 State 6 1 tan 1 tan 3 2 3 2 = − + β β , following their value of α tan B1√ Solve equation of form dt c bt a + + for t M1 Obtain 20 9 tan − = β A1 Conclude with 9 20 cot − = β or exact equivalent A1 [5]