UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 0 question paper for the guidance of teachers 9709 MATHEMATICS 9709/ Paper, 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 nsidered the acceptability of alternative answers. Mark schemes must be read in njunction with the question papers and the report on the examination. Cambridge will not enter into discussions or rrespondence in nnection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 0 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.
Page Mark Scheme: Teachers version Syllabus Paper Mark Scheme Notes Marks are of the following three types: M A B 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 rrect answer. Accuracy mark, awarded for a rrect answer or intermediate step rrectly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). Mark for a rrect 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 rrectly following on from previously inrrect results. Otherwise, A or B marks are given for rrect work only. A and B marks are not given for fortuitously rrect answers or results obtained from inrrect working. Note: B or A means that the candidate can earn or 0. B//0 means that the candidate can earn anything from 0 to. 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 rrect 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 rrect to s.f., or which would be rrect to s.f. if rounded ( d.p. in the case of an angle). As stated above, an A or B mark is not given if a rrect numerical answer arises fortuitously from inrrect working. For Mechanics questions, allow A or B marks for rrect answers which arise from taking g equal to 9.8 or 9.8 instead of 0.
Page Mark Scheme: Teachers version Syllabus Paper The following abbreviations may be used in a mark scheme or used on the scripts: AEF AG BOD CAO CWO ISW MR PA SOS SR Any Equivalent Form (of answer is equally acceptable) Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) Correct Answer Only (emphasising that no follow through from a previous error is allowed) Correct Working Only often written by a fortuitous answer Ignore Subsequent Working Misread Premature Approximation (resulting in basically rrect work that is insufficiently accurate) See Other Solution (the candidate makes a better attempt at the same question) 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 PA A penalty of MR 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 beme follow through marks. MR is not applied when the candidate misreads his own figures this is regarded as an error in accuracy. An MR penalty may be applied in particular cases if agreed at the ordination meeting. This is deducted from A or B marks in the case of premature approximation. The PA penalty is usually discussed at the meeting.
Page 4 Mark Scheme: Teachers version Syllabus Paper x + dx x 4 x x = + + c 4 B [] Allow unsimplified, mark for each term, including c x ( ) 6 (i) Term in x² Term in x³ 6 6 ± x C = ± x C = 5x 4 540x 8 [] For either unsimplified term (omission or error with can still gain out of ) (ii) Term in x³ = k =. 70x 5kx 4 [] nsiders exactly terms in x³ (i) x + px + q = ( x + )( x 5) p =, q = 5. (any other method ok) [] Must be (x + ) and (x 5). (ii) x + px + q + r = 0 Use of b 4ac Uses a, b and c rrectly r = 6 or = (x + k)² k = p () k² = q + r () k = r = 6 () D [] Any use of b 4ac c must include both q and r. 4 4 y = x 4 (i) dy = 4(x 4) dx If x =, m = Eqn of tangent y = ( x ) B B [4] Correct without. For. Correct line eqn. (for normal M0A0) (ii) tanθ = ±( ) θ = ±08.4º (or ±7.6º) or scalar product, tanθ = y-step x-step or use of tan (A B) for each [] Correct link with (± his gradient) (accept acute or obtuse) or 7.6º or radians
Page 5 Mark Scheme: Teachers version Syllabus Paper 5 (i) sθ s θ tanθ ) sinθ ) sin θ = sinθ ) + sinθ = = + sinθ sinθ [] Use of t = s c Replaces s²θ with sin²θ to form f(sinθ). AG. Ensure all ok. Must show difference of squares. (ii) sθ = 4 + = 4 tanθ ) sinθ sinθ = ⅓ θ = 9.5º, 60.5º x + 6 (i) f(x) = x x + + ff(x) = x = ( x + ) x 7x = x 7 [] B [] Linking up to obtain sinθ = k.. 80º st answer providing there are no other solutions in the range 0º to 60º. Replacing x twice - must be rrect Correct algebra clearing (x ) AG all rrect. x + (ii) y = x xy y = x + x ( y ) = y + x + f (x) = x [] Attempt to make x the subject and mplete method or since ff(x) = x, x + f (x) = f(x) = x (, ) 7 (i) (, 5) to (0, 9) gradient = ½ Equation of L y = x. Gradient of perpendicular = Eqn of Perp y 5 = ( x ) Sim Eqns C(.6,.8) B B [5] on gradient of L Use of m m = Correct form of line eqn (ii) d² =.6² +.² d =.58 [] Correct method for AC (accept with 5 in answer)
Page 6 Mark Scheme: Teachers version Syllabus Paper 8 (i) BA. BC or AB. CB 6 BA =, BC = BA. BC = 8 = 7 sθ θ =.4º or.96 radians (ii) OD = OA + AD = OA + BC 8 = 8 9 (i) (a) f(x) = 4s²x. One limit is Other limit is (b) 4s²x = s²x = ½ sx = ± x = ¼ π or ¾ π B [6] B B [] [] [] Correct two vectors for angle ABC. Correct method for one of the sides. Correct use for any pair of vectors. Correct method for moduli. All linked rrectly. (67.6º usually gets 4/6) Correct method. (allow for d = a + b c or for d = a + c b or for d = b + c a) for his BC. irrespective of inequalities irrespective of inequalities Makes s x the subject. (radians). for π ( st answer) ( exact means that decimal answers only earn A0 ) (ii) (a) B B [] Joins (0, ) to (π, 7), providing increasing function Not a line, flattens at extremities-needs inflexion. (b) f has an inverse since it is : or increasing or no turning points. B [] independent of part (i)
Page 7 Mark Scheme: Teachers version Syllabus Paper ( a + 4a) 0 (a) a + 5 d = 4a or 6 6 ( a + 4a) (a + 5d) or 6 = 60 B Correct left-hand side. All rrect. Sim Eqns a = 4º or Arc length = 5θ Perimeter =.. π rads 5 [6] Either answer. Correct use of arc length with θ in rads. (b) (i) k + 6 k = k + k + 6 k 9k 6 = 0 k = (NB stating a, ar, ar² as f(k) gets ) [] Correct eqn for k. Co ndone inclusion of k =. y = 4 x x. (ii) r = ⅔, a = 7 S = 7 ⅓ = 8. (i) At A, 4 x x = 0 A(6, 0) dy = x dx = 0 when x = 4 (4, 4) (ii) Vol = π y dx = π ( 6x + x 8x ) dx 5 x x π [8x + 8 ] 5 Limits 0 to 6 6.5π. (or 7π) [] B B B [5] A,, D [6] Correct formula for S must have r.. independent of working. B for each part. Sets to 0 and solves his eqn. Use of rrect formula + attempt at integration One mark for each term unsimplified Correct use of his limits. (49 ok)