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    UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2011 question paper for the guidance of teachers 9709 MATHEMATICS   9709/32 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 notindicate 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 must be read in conjunction with the question papers and the report on theexamination. ã Cambridge will not enter into discussions or correspondence in connection with these mark schemes.Cambridge is publishing the mark schemes for the October/November 2011 question papers for mostIGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Levelsyllabuses.   w  w  w  . X   t  r  e  m  e  P  a   p  e  r  s  . c  o  m    Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9709 32 © University of Cambridge International Examinations 2011 Mark Scheme Notes Marks are of the following three types:M Method mark, awarded for a valid method applied to the problem. Method marks arenot lost for numerical errors, algebraic slips or errors in units. However, it is notusually 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 inhand, e.g. by substituting the relevant quantities into the formula. Correct applicationof a formula without the formula being quoted obviously earns the M mark and in somecases 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.ã When a part of a question has two or more “method” steps, the M marks are generallyindependent unless the scheme specifically says otherwise; and similarly when there areseveral B marks allocated. The notation DM or DB (or dep*) is used to indicate that aparticular 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 andfull credit is given.ã The symbol √ implies that the A or B mark indicated is allowed for work correctly followingon from previously incorrect results. Otherwise, A or B marks are given for correct workonly. A and B marks are not given for fortuitously “correct” answers or results obtained fromincorrect working.ã 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. Unlessotherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong workingfollowing a correct form of answer is ignored.ã Wrong or missing units in an answer should not lead to the loss of a mark unless thescheme specifically indicates otherwise.ã 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 statedabove, an A or B mark is not given if a correct numerical answer arises fortuitously fromincorrect working. For Mechanics questions, allow A or B   marks for correct answers whicharise from taking g  equal to 9.8 or 9.81 instead of 10.  Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9709 32 © University of Cambridge International Examinations 2011 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 thatthe detailed working leading to the result is valid)BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutelyclear)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 WorkingMR MisreadPA Premature Approximation (resulting in basically correct work that is insufficientlyaccurate)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 acase where some standard marking practice is to be varied in the light of aparticular 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 questionremain 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 isregarded 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. ThePA –1 penalty is usually discussed at the meeting.  Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9709 32 © University of Cambridge International Examinations 2011 1 Rearrange as e 2  x – e  x – 6 = 0, or u 2 –  u – 6 = 0, or equivalent B1Solve a 3-term quadratic for e  x or for u M1Obtain simplified solution e  x = 3 or u = 3 A1Obtain final answer  x = 1.10 and no other A1 [4] 2    EITHER : Use chain rule M1obtain t t t  x cossin6dd = , or equivalent A1obtain t t t  y sincos6dd 2 −= , or equivalent A1Use t  xt  y x y dddddd ÷= M1Obtain final answer t  x y cosdd −= A1 OR : Express  y in terms of   x and use chain rule M1Obtain 21 )32(dd xk  x y −= , or equivalent A1Obtain 21 )32(dd x x y −−= , or equivalent A1Express derivative in terms of  t  M1Obtain final answer t  x y cosdd −= A1 [5] 3 (i)    EITHER : Attempt division by  x 2 –   x + 1 reaching a partial quotient of   x 2 + kx M1Obtain quotient  x 2 + 4  x + 3 A1Equate remainder of form lx to zero and solve for a , or equivalent M1Obtain answer a = 1 A1 OR : Substitute a complex zero of   x 2 –   x + 1 in p(  x ) and equate to zero M1Obtain a correct equation in a in any unsimplified form A1Expand terms, use i 2 = –1 and solve for a M1Obtain answer a = 1 A1 [4][SR: The first M1 is earned if inspection reaches an unknown factor  x 2 +  Bx + C  and anequation in  B and/or C  , or an unknown factor  Ax 2 +  Bx + 3 and an equation in  A and/or  B .The second M1 is only earned if use of the equation a =  B –  C  is seen or implied.] (ii) State answer, e.g.  x = –3 B1State answer, e.g.  x = –1 and no others B1 [2] 4 Separate variables and attempt integration of at least one side M1Obtain term ln(  x + 1) A1Obtain term k  ln sin 2 θ  , where k  = ±1, ±2, or ± 21 M1Obtain correct term 21 ln sin 2 θ  A1Evaluate a constant, or use limits θ  = 121 π  ,  x = 0 in a solution containing terms a ln(  x + 1) and b ln sin 2 θ  M1Obtain solution in any form, e.g. ln(  x + 1) = 212121 ln2sinln − θ  (f.t. on k  = ±1, ±2, or ± 21 ) A1√Rearrange and obtain 1)2sin2( −= θ   x , or simple equivalent A1 [7]
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