Cambridge A Level Mathematics 9709 — 2004 Oct/Nov Paper 2 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level MATHEMATICS 9709/02 Paper 2 Pure Mathematics 2 (P2) October/November 2004 1 hour 15 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. 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. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. This document consists of 3 printed pages and 1 blank page. © UCLES 2004 [Turn over

2 1 Solve the inequality |x + 1| > |x|. [3] 2 Solve the equation x3.9 = 11x3.2, where x ≠0. [3] 3 Find the values of x satisfying the equation 3 sin 2x = cos x, for 0◦≤x ≤90◦. [4] 4 The cubic polynomial 2x3 −5x2 + ax + b is denoted by f(x). It is given that (x −2) is a factor of f(x), and that when f(x) is divided by (x + 1) the remainder is −6. Find the values of a and b. [5] 5 The curve with equation y = x2 ln x, where x > 0, has one stationary point. (i) Find the x-coordinate of this point, giving your answer in terms of e. [4] (ii) Determine whether this point is a maximum or a minimum point. [2] 6 (i) By sketching a suitable pair of graphs, show that there is only one value of x in the interval 0 < x < 1 2π that is a root of the equation cot x = x. [2] (ii) Verify by calculation that this root lies between 0.8 and 0.9 radians. [2] (iii) Show that this value of x is also a root of the equation x = tan−1
1 x. [1] (iv) Use the iterative formula xn+1 = tan−1
1 xn  to determine this root correct to 2 decimal places, showing the result of each iteration. [3] 9709/02/O/N/04

3 7 The diagram shows the curve y = 2ex + 3e−2x. The curve cuts the y-axis at A. (i) Write down the coordinates of A. [1] (ii) Find the equation of the tangent to the curve at A, and state the coordinates of the point where this tangent meets the x-axis. [6] (iii) Calculate the area of the region bounded by the curve and by the lines x = 0, y = 0 and x = 1, giving your answer correct to 2 significant figures. [4] 8 (i) Express cos θ + sin θ in the form R cos(θ −α), where R > 0 and 0 < α < 1 2π, giving the exact values of R and α. [3] (ii) Hence show that 1 (cos θ + sin θ)2 = 1 2 sec2(θ −1 4π). [1] (iii) By differentiating sin x cos x, show that if y = tan x then dy dx = sec2x. [3] (iv) Using the results of parts (ii) and (iii), show that  1 2π 0 1 (cos θ + sin θ)2 dθ = 1. [3] 9709/02/O/N/04

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level MARK SCHEME for the November 2004 question paper 9709 MATHEMATICS 9709/02 Paper 2, maximum raw mark 50 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. • CIE will not enter into discussion or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the November 2004 question papers for most IGCSE and GCE Advanced Level syllabuses.

Grade thresholds taken for Syllabus 9709 (Mathematics) in the November 2004 examination. minimum mark required for grade: maximum mark available A B E Component 2 50 33 29 15 The thresholds (minimum marks) for Grades C and D are normally set by dividing the mark range between the B and the E thresholds into three. For example, if the difference between the B and the E threshold is 24 marks, the C threshold is set 8 marks below the B threshold and the D threshold is set another 8 marks down. If dividing the interval by three results in a fraction of a mark, then the threshold is normally rounded down.

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.

November 2004 GCE AS LEVEL MARK SCHEME MAXIMUM MARK: 50 SYLLABUS/COMPONENT: 9709/02 MATHEMATICS Paper 2 (Pure Mathematics 2)

Page 1 Mark Scheme Syllabus Paper AS LEVEL – NOVEMBER 2004 9709 2 © University of Cambridge International Examinations 2005 1 EITHER: State or imply non-modular inequality (x + 1)2 > x2 or corresponding quadratic equation or linear equation x + 1 = −x B1 Obtain critical value 2 1 − B1 State answer x > 2 1 − B1 OR: Obtain critical value 2 1 − by solving a linear inequality or by graphical method or inspection B2 State answer x > 2 1 − B1 3 [For 2x + 1 > 0, x > + 2 1 , or similar reasonable method] M1 2 Use logarithms to obtain an equation in ln x M1 Obtain ln x = ( ) 2 . 3 9 . 3 11 ln − , or equivalent A1 Obtain answer x = 31 (accept 30.7, 30.74) A1 3 3 At any stage, state answer x = 90° (c.w.o) B1 Write the equation in the form 6sin x cos x = cos x B1 Remove factor of cos x and solve an equation in sin x for x M1 Obtain answer x = 9.59° and no others in the range (9.6° OK: rubric) A1 4 (Ignore answers outside the given range.) 4 State or obtain 16 – 20 + 2a + b = 0 B1 Substitute x = –1 and equate to –6 M1 Obtain a 3-term equation in any correct form A1 Solve a relevant pair of equations, obtaining a or b M1 Obtain a = 1 and b = 2 A1 5 5 (i) Use the product rule to obtain the first derivative (must involve 2 terms) M1 Obtain derivative 2x lnx + x2 x 1 or equivalent A1 Equate derivative to zero and solve for x M1 Obtain answer x = e−0.5 or e 1 or equivalent (e.g. 0.61) A1 4 (ii) Determine nature of stationary point using correct second derivative (3 + 2lnx) or correct first derivative or equation of the curve (3 y-values, central one y(exp (– 0.5)) M1 Show point is a minimum completely correctly A1 2 6 (i) Make recognisable sketch of an appropriate trig curve, e.g. y = cot x, for 0 < x < 2 1 π B1 Sketch the appropriate second curve e.g. y = x correctly and justify the given statement B1 2

Page 2 Mark Scheme Syllabus Paper AS LEVEL – NOVEMBER 2004 9709 2 © University of Cambridge International Examinations 2005 (ii) Consider sign of cot x – x at x = 0.8 and x = 0.9, or equivalent M1 Complete the argument correctly with appropriate calculations A1 2 (iii) Show, using cot x ≡ x tan 1 , that cot x = x is equivalent to x = arctan       x 1 (or vice versa) B1 1 (iv) Use the iterative formula correctly at least once M1 Obtain final answer 0.86 A1 Show sufficient iterations to justify its accuracy to 2 decimal places, or show that there is a sign change in (0.855, 0.865) B1 3 7 (i) State coordinates (0, 5) B1 1 (ii) State first derivative of the form k ex + m e−2x, where km ≠ 0 M1 Obtain correct first derivative 2 ex – 6 e−2x A1 Substitute x = 0, obtaining gradient of –4 A1√ Form equation of line through A with this gradient (NOT the normal) M1 Obtain equation in any correct form e.g. y – 5 = –4x A1 Obtain coordinates (1.25, 0) or equivalent A1 6 (iii) Integrate and obtain 2 ex – 2 3 e−2x, or equivalent B1 + B1 Use limits x = 0 and x = 1 correctly M1 Obtain answer 4.7 A1 4 8 (i) State answer R = 2 B1 Use trigonometric formulae to find α M1 Obtain answer α = 4 1 π (NOT 45°, unless 45° = c 4 π somewhere, later) A1 3 (ii) Use cos θ + sin θ = 2 cos(θ – 4 1 π) to justify the given answer B1 1 (iii) Differentiate using the quotient or product rule M1 Obtain derivative in any correct form A1 Obtain the given answer correctly A1 3 (iv) Convert integrand to give ∫2 1 sec2(θ – 4 π )dθ B1 Integrate, to obtain function 2 1 tan (θ – 4 π ) M1 Substitute (correct) limits correctly, to obtain given result A1 3