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

*5474892099* Cambridge International Examinations Cambridge International Advanced Subsidiary Level MATHEMATICS (US) 9280/21 Paper 2 Pure Mathematics 2 (P2) May/June 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 06_9280_21/FP © UCLES 2014 [Turn over

2 1 Solve the inequality 3x −2 ≥x + 4. [4] 2 Find the slope of each of the following curves at the point for which x = 0. (i) y = 3 sin x + tan 2x [3] (ii) y = 6 1 + e2x [3] 3 (i) Find the quotient when 6x4 −x3 −26x2 + 4x + 15 is divided by x2 −4, and confirm that the remainder is 7. [3] (ii) Hence solve the equation 6x4 −x3 −26x2 + 4x + 8 = 0. [3] 4 (i) By sketching a suitable pair of graphs, show that the equation 3 ln x = 15 −x3 has exactly one real root. [3] (ii) Show by calculation that the root lies between 2.0 and 2.5. [2] (iii) Use the iterative formula xn+1 = 315 −3 ln xn to find the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] 5 (i) Prove that tan 1 + cot 1  2 sin 21. [3] (ii) Hence (a) find the exact value of tan 1 80 + cot 1 80, [2] (b) evaluate Ô 1 20 0 6 tan 1 + cot 1 d1. [3] 6 (a) Show that Ô 16 6 6 2x −7 dx = ln 125. [5] (b) Use the trapezium rule with four intervals to find an approximation to Ó 17 1 log10 x dx, giving your answer correct to 3 significant figures. [3] © UCLES 2014 9280/21/M/J/14

3 7 The equation of a curve is 2x2 + 3xy + y2 = 3. (i) Find the equation of the tangent to the curve at the point 2, −1, giving your answer in the form ax + by + c = 0, where a, b and c are integers. [6] (ii) Show that the curve has no stationary points. [4] © UCLES 2014 9280/21/M/J/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/M/J/14

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level MARK SCHEME for the May/June 2014 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 2014 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE AS LEVEL – May/June 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 GCE AS LEVEL – May/June 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 GCE AS LEVEL – May/June 2014 9280 21 © Cambridge International Examinations 2014 1 Either State or imply non-modular inequality ( )2 2 3 − x >( )2 4 + x or corresponding equation or pair of linear equations B1 Attempt solution of 3-term quadratic equation or of 2 linear equations M1 Obtain critical values 2 1 − and 3 A1 State answer x < 2 1 − , x > 3 A1 [4] Or Obtain critical value x = 3 from graphical method, inspection, equation B1 Obtain critical value 2 1 − = x similarly B2 State answer x < 2 1 − , x > 3 B1 [4] 2 (i) Differentiate to obtain form x k x k 2 sec cos 2 2 1 + M1 Obtain correct second term x 2 sec 2 2 A1 Obtain x x 2 sec 2 cos 3 2 + and hence answer 5 A1 [3] (ii) Differentiate to obtain form ( ) 2 2 2 e 1 e − + x x k M1 Obtain correct ( ) 2 2 2 e 1 e 12 − + − x x or equivalent (may be implied) A1 Obtain –3 A1 [3] 3 (i) Divide at least as far as x term in quotient, use synthetic division correctly or make use of an identity M1 Obtain at least x x − 2 6 A1 Obtain quotient 2 6 2 − −x x and confirm remainder is 7 (AG) A1 [3] (ii) State equation in form ( )( ) 0 2 6 4 2 2 = − + − kx x x , any constant k (may be implied) M1 Obtain two of the roots 3 2 2 1 , 2, ,2 − − A1 Obtain remaining two roots and no others A1 [3] 4 (i) Sketch, showing the correct shape of each, x y ln 3 = and 3 15 x y − = B1 Indicate the correct intercepts (1,0) and (0,15) B1 Indicate one real root from two correct sketches B1 [3] (ii) Consider sign of 15 ln 3 3 − + x x for 2.0 and 2.5 or equivalent M1 Justify conclusion with correct calculations (–4.9 and 3.4 or equivalents) A1 [2] (iii) Use the iteration process correctly at least once M1 Obtain final answer 2.319 A1 Show sufficient iterations to 5 decimal places to justify answer or show a sign change in the interval (2.3185, 2.3195) A1 [3] 5 (i) Express left-hand side as a single fraction M1 Use θ θ θ cos sin 2 2 sin = at some point B1 Complete proof with no errors seen (AG) A1 [3]

Page 5 Mark Scheme Syllabus Paper GCE AS LEVEL – May/June 2014 9280 21 © Cambridge International Examinations 2014 (ii) (a) State π 4 1 sin 2 or equivalent B1 Obtain 2 2 or exact equivalent (dependent on first B1) B1 [2] (b) State or imply θ 2 sin k for any k B1 Integrate to obtain θ 2 cos 2 3 − B1 Substitute both limits correctly to obtain 3 B1 [3] 6 (a) Integrate to obtain form ( ) 7 2 ln − x k M1 Obtain correct ( ) 7 2 ln 3 − x A1 Substitute limits correctly (dependent on first M1) DM1 Use law for logarithm of a quotient or power (dependent on first M1) DM1 Confirm 125 ln following correct work and sufficient detail (AG) A1 [5] (b) Evaluate y at (1), 5, 9, 13, 17 M1 Use correct formula, or equivalent, with h = 4 and five y-values M1 Obtain 13.5 A1 [3] 7 (i) Obtain x y x y d d 3 3 + as derivative of 3xy B1 Obtain x y y d d 2 as derivative of y2 B1 State 0 2 3 3 4 d d d d = + + + x y x y y x y x B1 Substitute 2 and –1 to find gradient of curve (dependent on at least one B1) M1 Form equation of tangent through (2, 1) − with numerical gradient (dependent on previous M1) DM1 Obtain 0 6 4 5 = − + y x or equivalent of required form A1 [6] (ii) Use 0 d d = x y to find relation between x and y (dependent on at least one B1 from part(i)) M1 Obtain 4 3 0 x y + = or equivalent A1 Substitute for x or y in equation of curve M1 Obtain 3 2 8 1 = − y or 3 2 9 2 = − x or equivalent and conclude appropriately A1 [4]