Cambridge A Level Mathematics 9709 — 2006 May/June Paper 3 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level MATHEMATICS 9709/03 Paper 3 Pure Mathematics 3 (P3) May/June 2006 1 hour 45 minutes Additional Materials: Answer Booklet/Paper Graph paper List of Formulae (MF9) READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. 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. 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. At the end of the examination, fasten all your work securely together. This document consists of 4 printed pages. © UCLES 2006 [Turn over

2 1 Given that x = 4
3−y, express y in terms of x. [3] 2 Solve the inequality 2x > |x −1|. [4] 3 The parametric equations of a curve are x = 2θ + sin 2θ, y = 1 −cos 2θ. Show that dy dx = tan θ. [5] 4 (i) Express 7 cos θ + 24 sin θ in the form R cos(θ −α), where R > 0 and 0◦< α < 90◦, giving the exact value of R and the value of α correct to 2 decimal places. [3] (ii) Hence solve the equation 7 cos θ + 24 sin θ = 15, giving all solutions in the interval 0◦≤θ ≤360◦. [4] 5 In a certain industrial process, a substance is being produced in a container. The mass of the substance in the container t minutes after the start of the process is x grams. At any time, the rate of formation of the substance is proportional to its mass. Also, throughout the process, the substance is removed from the container at a constant rate of 25 grams per minute. When t = 0, x = 1000 and dx dt = 75. (i) Show that x and t satisfy the differential equation dx dt = 0.1(x −250). [2] (ii) Solve this differential equation, obtaining an expression for x in terms of t. [6] 6 (i) By sketching a suitable pair of graphs, show that the equation 2 cot x = 1 + ex, where x is in radians, has only one root in the interval 0 < x < 1 2π. [2] (ii) Verify by calculation that this root lies between 0.5 and 1.0. [2] (iii) Show that this root also satisfies the equation x = tan−1 2 1 + ex . [1] (iv) Use the iterative formula xn+1 = tan−1 2 1 + exn , with initial value x1 = 0.7, to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] © UCLES 2006 9709/03/M/J/06

3 7 The complex number 2 + i is denoted by u. Its complex conjugate is denoted by u*. (i) Show, on a sketch of an Argand diagram with origin O, the points A, B and C representing the complex numbers u, u* and u + u* respectively. Describe in geometrical terms the relationship between the four points O, A, B and C. [4] (ii) Express u u* in the form x + iy, where x and y are real. [3] (iii) By considering the argument of u u*, or otherwise, prove that tan−1
4 3 = 2 tan−1
1 2. [2] 8 The diagram shows a sketch of the curve y = x 1 2 ln x and its minimum point M. The curve cuts the x-axis at the point (1, 0). (i) Find the exact value of the x-coordinate of M. [4] (ii) Use integration by parts to find the area of the shaded region enclosed by the curve, the x-axis and the line x = 4. Give your answer correct to 2 decimal places. [5] 9 (i) Express 10 (2 −x)(1 + x2) in partial fractions. [5] (ii) Hence, given that |x| < 1, obtain the expansion of 10 (2 −x)(1 + x2) in ascending powers of x, up to and including the term in x3, simplifying the coefficients. [5] © UCLES 2006 9709/03/M/J/06 [Turn over

4 10 The points A and B have position vectors, relative to the origin O, given by −−→ OA =  −1 3 5  and −−→ OB =  3 −1 −4  . The line l passes through A and is parallel to OB. The point N is the foot of the perpendicular from B to l. (i) State a vector equation for the line l. [1] (ii) Find the position vector of N and show that BN = 3. [6] (iii) Find the equation of the plane containing A, B and N, giving your answer in the form ax + by + c = d. [5] 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 University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. © UCLES 2006 9709/03/M/J/06

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced/Advanced Subsidiary Level MARK SCHEME for the May/June 2006 question paper 9709 MATHEMATICS 9709/03 Paper 3 Maximum raw mark 75 This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which Examiners were initially instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the Examination. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the Report on the Examination. The minimum marks in these components needed for various grades were previously published with these mark schemes, but are now instead included in the Report on the Examination for this session. • CIE will not enter into discussion or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2006 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

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.

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.

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