Cambridge A Level Mathematics 9709 — 2014 Oct/Nov Paper 3 · Variant 3

*8698806885* Cambridge International Examinations Cambridge International Advanced Level MATHEMATICS 9709/33 Paper 3 Pure Mathematics 3 (P3) October/November 2014 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. 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 75. 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_9709_33/2R © UCLES 2014 [Turn over

2 1 Solve the inequality 3x −1 < 2x + 5. [4] 2 A curve is defined for 0 < 1 < 1 20 by the parametric equations x = tan 1, y = 2 cos21 sin 1. Show that dy dx = 6 cos51 −4 cos31. [5] 3 The polynomial 4x3 + ax2 + bx −2, where a and b are constants, is denoted by px. It is given that x + 1 and x + 2 are factors of px. (i) Find the values of a and b. [4] (ii) When a and b have these values, find the remainder when px is divided by x2 + 1. [3] 4 (i) Show that cos1 −60Å + cos1 + 60Å  cos 1. [3] (ii) Given that cos2x −60Å + cos2x + 60Å cosx −60Å + cosx + 60Å = 3, find the exact value of cosx. [4] 5 The complex numbers w and Ï are defined by w = 5 + 3i and Ï = 4 + i. (i) Express iw Ï in the form x + iy, showing all your working and giving the exact values of x and y. [3] (ii) Find wÏ and hence, by considering arguments, show that tan−13 5 + tan−11 4 = 1 40. 4 6 It is given that I = Ó 0.3 0 1 + 3x2−2 dx. (i) Use the trapezium rule with 3 intervals to find an approximation to I, giving the answer correct to 3 decimal places. [3] (ii) For small values of x, 1 + 3x2−2 ≈1 + ax2 + bx4. Find the values of the constants a and b. Hence, by evaluating Ó 0.3 0 1 + ax2 + bx4 dx, find a second approximation to I, giving the answer correct to 3 decimal places. [5] © UCLES 2014 9709/33/O/N/14

3 7 The equations of two straight lines are r = i + 4j −2k + ,i + 3k and r = ai + 2j −2k + -i + 2j + 3ak, where a is a constant. (i) Show that the lines intersect for all values of a. [4] (ii) Given that the point of intersection is at a distance of 9 units from the origin, find the possible values of a. [4] 8 The variables x and y are related by the differential equation dy dx = 1 5xy 1 2 sin 1 3x . (i) Find the general solution, giving y in terms of x. [6] (ii) Given that y = 100 when x = 0, find the value of y when x = 25. [3] 9 (i) Sketch the curve y = lnx + 1 and hence, by sketching a second curve, show that the equation x3 + lnx + 1 = 40 has exactly one real root. State the equation of the second curve. [3] (ii) Verify by calculation that the root lies between 3 and 4. [2] (iii) Use the iterative formula xn+1 = 340 −lnxn + 1, with a suitable starting value, to find the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] (iv) Deduce the root of the equation ey −13 + y = 40, giving the answer correct to 2 decimal places. [2] 10 By first using the substitution u = ex, show that Ô ln4 0 e2x e2x + 3ex + 2 dx = ln 8 5 . 10 © UCLES 2014 9709/33/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 9709/33/O/N/14

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the October/November 2014 series 9709 MATHEMATICS 9709/33 Paper 3, maximum raw mark 75 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 A Level – October/November 2014 9709 33 © 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 A Level – October/November 2014 9709 33 © 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 A Level – October/November 2014 9709 33 © Cambridge International Examinations 2014 1 Either State or imply non-modular inequality ( ) ( ) 2 2 3 1 2 5 x x − < + or corresponding quadratic equation or pair of linear equations ( ) 5 2 1 3 + ± = − x x B1 Solve a three-term quadratic or two linear equations 0 24 26 5 2 < − − x x M1 Obtain 4 5 − and 6 A1 State 4 6 5 x − < < A1 Or Obtain value 6 from graph, inspection or solving linear equation B1 Obtain value 4 5 − similarly B2 State 4 6 5 x − < < B1 [4] 2 Use correct product rule or correct chain rule to differentiate y M1 Use d d d d d d y x y x θ θ = M*1 Obtain θ θ θ θ 2 3 2 sec cos 2 sin cos 4 + − or equivalent A1 Express d d y x in terms of θ cos DM*1 Confirm given answer θ θ 3 5 cos 4 cos 6 − legitimately A1 [5] 3 (i) Either Equate p(–1) or p(–2) to zero or divide by ( )1 + x or ( ) 2 + x and M*1 equate constant remainder to zero. Obtain two equations 6 = −b a and 34 2 4 = −b a or equivalents A1 Solve pair of equations for a or b DM*1 Obtain 11 = a and 5 = b A1 Or State or imply third factor is 1 4 − x B1 Carry out complete expansion of ( )( )( )1 4 2 1 − + + x x x or M1 ( )( )( ) D Cx x x + + + 2 1 Obtain 11 = a A1 Obtain 5 = b A1 [4] (ii) Use division or equivalent and obtaining linear remainder M1 Obtain quotient a x + 4 , following their value of a A1 Indicate remainder 13 − x A1 [3]

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2014 9709 33 © Cambridge International Examinations 2014 4 (i) Either Use ( ) B A ± cos correctly at least once M1 State correct complete expansion A1 Confirm given answer θ cos with explicit use of 2 1 60 cos = o A1 SR: “correct” answer from sign errors in both expansions is B1 only Or Use correct B A cos cos + formula M1 State correct result e.g.      −       2 120 cos 2 2 cos 2 θ A1 Confirm given answer θ cos with explicit use of ( ) 2 1 60 cos = ± o A1 [3] (ii) State or imply 3 cos 2 cos = x x B1 Obtain equation 0 1 cos 3 cos 2 2 = − − x x B1 Solve a three-term quadratic equation for x cos M1 Obtain ( ) 17 3 4 1 − or exact equivalent and, finally, no other A1 [4] 5 (i) State or imply i 3 5i w = −+ B1 Carry out multiplication by i 4 i 4 − − M1 Obtain final answer 7 23i 17 17 − + or equivalent A1 [3] (ii) Multiply w by z to obtain i 17 17 + B1 State 1 3 arg tan 5 w − = or 1 1 arg tan 4 z − = B1 State z w wz arg arg arg + = M1 Confirm given result 1 1 3 1 1 tan tan 5 4 4π − − + = legitimately A1 [4] 6 (i) State or imply correct ordinates 1, 0.94259…, 0.79719…, 0.62000… B1 Use correct formula or equivalent with 1.0 = h and four y values M1 Obtain 0.255 with no errors seen A1 [3] (ii) Obtain or imply 6 − = a B1 Obtain 4 x term including correct attempt at coefficient M1 Obtain or imply 27 = b A1 Either Integrate to obtain 5 3 5 27 2 x x x + − , following their values of a and b B1 Obtain 0.259 B1 Or Use correct trapezium rule with at least 3 ordinates M1 Obtain 0.259 (from 4) A1 [5]

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2014 9709 33 © Cambridge International Examinations 2014 7 (i) State at least two of the equations µ λ + = + a 1 , µ 2 2 4 + = , µ λ a 3 2 3 2 + − = + − B1 Solve for λ or for µ M1 Obtain a = λ (or 1 − + = µ λ a ) and 1 = µ A1 Confirm values satisfy third equation A1 [4] (ii) State or imply point of intersection is( ) 2 3 ,4 ,1 − + a a B1 Use correct method for the modulus of the position vector and equate to 9, following their point of intersection M*1 Solve a three-term quadratic equation in a ( ) 0 6 2 = − −a a DM*1 Obtain –2 and 3 A1 [4] 8 (i) Sensibly separate variables and attempt integration of at least one side M1 Obtain ... 2 2 1 = y or equivalent A1 Correct integration by parts of x x 3 1 sin as far as ∫ ± x x b x ax d 3 1 cos 3 1 cos M1 Obtain ∫ + − x x x x d 3 1 cos 3 3 1 cos 3 or equivalent A1 Obtain x x x 3 1 sin 9 3 1 cos 3 + − or equivalent A1 Obtain 2 3 1 sin 10 9 3 1 cos 10 3       + + − = c x x x y or equivalent A1 [6] (ii) Use 0 = x and 100 = y to find constant M*1 Substitute 25 and calculate value of y DM*1 Obtain 203 A1 [3] 9 (i) Sketch increasing curve with correct curvature passing through origin, for 0 ≥ x B1 Recognisable sketch of 3 40 x y − = , with equation stated, for 0 > x B1 Indicate in some way the one intersection, dependent on both curves being roughly correct and both existing for some 0 < x B1 [3] (ii) Consider signs of ( ) 40 1 ln 3 − + + x x at 3 and 4 or equivalent or compare values of relevant expressions for 3 = x and 4 = x M1 Complete argument correctly with correct calculations (–11.6 and 25.6) A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 3.377 A1 Show sufficient iterations to justify accuracy to 3 d.p. or show sign change in interval (3.3765, 3.3775) A1 [3] (iv) Attempt value of ( )1 ln + x M1 Obtain 1.48 A1 [2]

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2014 9709 33 © Cambridge International Examinations 2014 10 State or imply d e d x u x = B1 Substitute throughout for x and dx M1 Obtain ∫ + + u u u u d 2 3 2 or equivalent (ignoring limits so far) A1 State or imply partial fractions of form 1 2 + + + u B u A , following their integrand B1 Carry out a correct process to find at least one constant for their integrand M1 Obtain correct 1 1 2 2 + − + u u A1 Integrate to obtain ( ) ( )1 ln 2 ln + + + u b u a M1 Obtain ( ) ( ) 1 ln 2 ln 2 + − + u u or equivalent, follow their A and B A1 Apply appropriate limits and use at least one logarithm property correctly M1 Obtain given answer 8 ln 5 legitimately A1 [10] SR for integrand ( )( ) 2 1 2 + + u u u u State or imply partial fractions of form 2 1 + + + + u C u B u A (B1) Carry out a correct process to find at least one constant (M1) Obtain correct 1 1 2 2 + − + u u (A1) …complete as above.