Cambridge A Level Mathematics 9709 — 2016 Oct/Nov Paper 2 · Variant 2

*3536243891* Cambridge International Examinations Cambridge International Advanced Subsidiary Level MATHEMATICS 9709/22 Paper 2 Pure Mathematics 2 (P2) October/November 2016 1 hour 15 minutes Additional Materials: List of Formulae (MF9) READ THESE INSTRUCTIONS FIRST An answer booklet is provided inside this question paper. You should follow the instructions on the front cover of the answer booklet. If you need additional answer paper ask the invigilator for a continuation booklet. 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. This document consists of 3 printed pages, 1 blank page and 1 insert. JC16 11_9709_22/RP © UCLES 2016 [Turn over

2 1 Solve the equation 0.4x −0.8 = 2. [3] 2 (i) Given that 1 + 4y 3 + 2y = 5, find the value of 2y. [3] (ii) Use logarithms to find the value of y correct to 3 significant figures. [2] 3 The definite integral I is defined by I = Ó 2 0 4e 1 2x + 3 dx. (i) Show that I = 8e −2. [3] (ii) Sketch the curve y = 4e 1 2x + 3 for 0 ≤x ≤2. [2] (iii) State whether an estimate of I obtained by using the trapezium rule will be more than or less than 8e −2. Justify your answer. [1] 4 The polynomial px is defined by px = 4x3 + ax2 + ax + 4, where a is a constant. (i) Use the factor theorem to show that x + 1 is a factor of px for all values of a. [2] (ii) Given that the remainder is −42 when px is divided by x −2, find the value of a. [2] (iii) When a has the value found in part (ii), factorise px2 completely. [4] 5 x y O M The diagram shows the curve y = 4 ln x x2 + 1 and its stationary point M. The x-coordinate of M is m. (i) Find an expression for dy dx and hence show that m = e0.51+m−2. [5] (ii) Use an iterative formula based on the equation in part (i) to find the value of m correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3] © UCLES 2016 9709/22/O/N/16

3 6 (i) Show that cos 21 1 + cos 21  1 −1 2 sec21. [2] (ii) Solve the equation cos 2! 1 + cos 2! = 13 + 5 tan ! for 0 < ! < 0. [4] (iii) Find the exact value of Ô 1 20 −1 20 cosx 1 + cosx dx. [4] 7 x y O The diagram shows the curve with parametric equations x = 4 sin 1, y = 1 + 3 cos1 + 1 60 for 0 ≤1 < 20. (i) Show that dy dx can be expressed in the form k1 + ï3 tan 1 where the exact value of k is to be determined. [5] (ii) Find the equation of the normal to the curve at the point where the curve crosses the positive y-axis. Give your answer in the form y = mx + c, where the constants m and c are exact. [5] © UCLES 2016 9709/22/O/N/16

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. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. 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 2016 9709/22/O/N/16

® IGCSE is the registered trademark of Cambridge International Examinations. This document consists of 7 printed pages. © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International Advanced Subsidiary Level MATHEMATICS 9709/22 Paper 2 October/November 2016 MARK SCHEME Maximum Mark: 50 Published 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 2016 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 2016 9709 22 © UCLES 2016 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 2016 9709 22 © UCLES 2016 The following abbreviations may be used in a mark scheme or used on the scripts: AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) 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 SOI Seen or implied 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 2016 9709 22 © UCLES 2016 1 State non-modulus equation 2 (0.4 0.8) 4 x − = or equivalent or corresponding pair of linear equations Solve 3-term quadratic equation or pair of linear equations Obtain 3 − and 7 B1 M1 A1 [3] SR One solution only – B1 Must see some evidence of attempt to solve the quadratic for M1 for at least one value of x For a pair of linear equations, there must be a sign difference If extra solutions are given then A0 2 (i) Use 2 4 2 y y = Attempt solution of quadratic equation in 2y Obtain finally 2 7 y = only B1 M1 A1 [3] (ii) Apply logarithms to solve equation of form 2y k = where 0 k > Obtain 2.81 M1 A1 [2] Must be using their positive answer for (i) 3 (i) Obtain integral of form 1 2 1 2 e x k k x + Obtain correct 1 2 8e 3 x x + oe Use limits correctly to confirm 8e 2 − M1 A1 A1 [3] Allow 1 4 k = (ii) Draw increasing curve in first quadrant Draw more or less accurate sketch with correct curvature, gradient at 0 x = must be >0 M1 A1 [2] If incorrect y intercept used then M1 A0 Allow if no intercept stated (iii) State more and refer to top(s) of trapezium(s) above curve B1 [1] Can be shown using a diagram. Reference to a trapezium must be made

Page 5 Mark Scheme Syllabus Paper Cambridge International AS Level – October/November 2016 9709 22 © UCLES 2016 4 (i) Substitute 1 x = − and simplify Obtain 4 4 0 a a −+ − + = and conclude appropriately M1 A1 [2] Allow attempt at long division , must get down to a remainder Allow M1 if at least 2 numerical values of a are used May equate to ( )( ) 2 1 x Ax Bx C R + + + + - allow M1 if they get as far as finding R Must have a conclusion - allow ‘hence shown’, or made a statement of intent at the start of the question (ii) Substitute 2 x = and equate to 42 − and attempt to solve Obtain 13 a = − M1 A1 [2] May equate to ( )( ) 2 2 x Ax Bx C − + + , must have a complete method to get as far as a = ... to obtain M1 (iii) Divide p( ) x with their a at least as far as 2 4x kx + Obtain 2 4 17 4 x x − + Obtain ( 1)(4 1)( 4) x x x + − − or equivalent if 2x already involved Obtain 2 ( 1)(2 1)(2 1)( 2)( 2) x x x x x + − + − + M1 A1 A1 A1 [4] If ( 1)(4 1)( 4) x x x + − − seen with no evidence of long division then allow the marks 5 (i) Use quotient rule (or product rule) to find first derivative Obtain 2 4 2 2 ( 1) 8 ln ( 1) x x x x x + − + or equivalent State 2 4 ( 1) 8 ln 0 x x x x + − = or equivalent Carry out correct process to produce equation without ln, without any incorrect working Confirm 2 0.5(1 ) e m m − + = or 2 0.5(1 ) e x x − + = M1 A1 A1 M1 A1 [5] Quotient: Must have a difference in the numerator and 2 2 ( 1) x + in the denominator Product: Must see an application of the chain rule. Condone missing brackets if correct use is implied by correct work later

Page 6 Mark Scheme Syllabus Paper Cambridge International AS Level – October/November 2016 9709 22 © UCLES 2016 (ii) Use iterative formula correctly at least once Obtain final answer 1.895 Show sufficient iterations to 6 sf to justify answer or show sign change in interval (1.8945,1.8955) M1 A1 A1 [3] Should not be attempting to use 0 0 x = , but if used and ‘recovered’ then SC M1 A1- usually see 1 1.6487 m = 6 (i) Use 2 cos2 2cos 1 θ θ = − appropriately twice Simplify to confirm 2 1 2 1 sec θ − B1 B1 [2] Alternative method 2 2 2 2 1 2sin 1 sec tan 2 2cos θ θ θ θ − = − or 2 2 1 tan 2cos θ θ − B1 then as for 2nd B1 (ii) Use 2 2 sec 1 tan α α = + Obtain equation 2 tan 10tan 25 0 α α + + = or equivalent Attempt solution of 3-term quadratic equation for tanα and use correct process for finding value of α from negative value of tanα Obtain 1.77 B1 B1 M1 A1 [4] If quadratic is incorrect, need to see evidence of attempt to solve as required to obtain M1 Allow better or in terms of π 1013π 1800       (iii) State or imply integrand 2 1 1 2 2 1 sec x − Obtain integral of form 1 1 2 2 tan k x k x − Obtain correct 1 2 tan x x − Apply limits correctly to obtain 2 π − B1 M1 A1 A1 [4]

Page 7 Mark Scheme Syllabus Paper Cambridge International AS Level – October/November 2016 9709 22 © UCLES 2016 7 (i) Use correct addition formula for either 1 6 cos( ) θ π + or, after diffn, 1 6 sin( ) θ π + Differentiate to obtain d d y θ of form 1 2 sin cos k k θ θ + or 1 6 sin( ) k θ π + Divide attempt at d d y θ by attempt at d d x θ Obtain 3 3 3 2 2 sin cos 4cos θ θ θ − − or equivalent Simplify to obtain 3 8 (1 3 tan ) θ − + B1 M1 M1 A1 A1 [5] Condone ‘missing brackets’ (ii) Identify 0 θ = Substitute 0 into formula for d d y x and take negative reciprocal Obtain gradient of normal 8 3 Form equation of normal through point 3 3 2 (0,1 ) + Obtain 3 3 8 3 2 1 y x = + + or equivalent B1 M1 A1 M1 A1 [5] soi be implied by 3 3 1 2 y = + or 3.6 Must be from correct (i)