9709 MATHEMATICS. 9709/13 Paper 1, maximum raw mark 75

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1 CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 0 series 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 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 May/June 0 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

2 Page Mark Scheme: Teachers version Syllabus Paper GCE AS/A LEVEL May/June 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 correct answer. 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). 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: 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 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 s.f., or which would be correct 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 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 9.8 instead of 0. University of Cambridge International Examinations 0

3 Page Mark Scheme: Teachers version Syllabus Paper GCE AS/A LEVEL May/June 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 correct 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 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- penalty may be applied in particular cases if agreed at the coordination meeting. This is deducted from A or B marks in the case of premature approximation. The PA - penalty is usually discussed at the meeting. University of Cambridge International Examinations 0

4 Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL May/June dy = x + 5 ( x + 5) ( +c) Everything without. Uses (, 5) c = 4 [4] Uses point in an integral. (i) ½.²π = ½9²θ - ½²θ θ = 4 π (ii) P = π + 9 ¼π =.4 cm. or + π cos²θ = tan²θ [] [] needs ½r²θ once. all correct. Answer given is for use of s=rθ once. (i) sin θ cos²θ = cos θ Uses c²+ s² = c 4 = c² [] Use of t² = s² c² or alternative. Correct eqn. (ii) (c² )(c² + ) = 0 c = ± Method of solving for -term quadratic. θ = ¼π or ¾π. [] (in terms of π). for π st ans. Cannot gain if other answers given in the range. 4 (i) 5 ( + ax) = + 80ax + 80a²x² [] for each term. (ii) ( + x) 40 = 80a² + 60a a = or a =. D [] Realises need to consider terms. Solution of -term quadratic. 5 (i) + 0 π π D D y =sinx has cycles, starts and finishes on the x-axis, max comes first. From + to. Smooth curves. y = cosx has one cycle, starts and finishes on x-axis, with a minimum pt. From 0 to, smooth curve, flattens. [4] (ii) (a) sinx = ½ 4 solutions [] for their curve. (b) sinx+ cosx+=0 solutions. [] for intersections of their curves. Cambridge International Examinations 0

5 Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL May/June u = x y y + x = 9 9 u = x (9 x) or y y Expressing u in terms of variable du du D Knowing to differentiate. =8x 9x² or =7 y + y² dy = 0 when x = or y = u = D Setting differential to 0. d u =8 8x ve D [7] Any valid method 7 A (, 4), B (4, 6) and C (7, ). (i) m of AB = ⅔ m of perpendicular = For use of m m = eqn of AB y 4 = ( x ) Allow for unsimplified eqn eqn of CX y = ( x 7) Allow for unsimplified eqn Sim Eqns X (, 8) [6] For solution of sim eqns. (ii) AX : XB = 4 8 : 8 6 = : Or (9²+6²) : (²+²) = : [] Vector steps or Pythagoras. 8 OA = and 4 5 OB = 0. (i) OC = AB = b a = 6 Knowing how to find OC Uses OC and OB Using OC.OB or CO.BO OC.OB = = 7 9 cos BOC Use of x x +... for modulus Angle BOC = 54.⁰ (or rad) [6] everything linked. (nb uses BO.OC loses ) (nb uses other vectors max ) Cambridge International Examinations 0

6 Page 6 Mark Scheme Syllabus Paper GCE AS/A LEVEL May/June (ii) Modulus of OC = 7 Vector = 5 7 OC Knows to scale by factor of 5 Mod ± (a) S n = n + 8n [] For their OC. S = 0 = a S = 4 = a + (a + d) d = 4 [] correct use of S n formula. (b) GP a = 64 ar = 48 r = ¾ rd term is ar² = 6 ar² numerical for their r AP a = 64, a + 8d = 48 d = 6 = 64 + (n )( ) correct use of a+(n )d n = 5. [5] 0 f : x a x + k, g : x a x 6x + 8, (i) (x + ) + = 5 x = 4 or {f() = 5, f(4) = } [] ff(x) needs to be correctly formed (ii) x 6x + 8 = x + k x 8x + 8 k = 0 Uses b² 4ac < 0 k < 8 [] Eliminates y to form eqn in x. Uses the discriminant even if =0.>0 (iii) x 6x + 8 = (x )² y = (x )² Makes x the subject ± (x + ) + Needs specifically to lose the. [4] For and Makes x the subject, in terms of x and without or ±. Cambridge International Examinations 0

7 Page 7 Mark Scheme Syllabus Paper GCE AS/A LEVEL May/June y = 8 x x (i) dy = 4x = when x = 4. Eqn of BC y = ( x 4) C (, 4½) 0 [4] needs both Subs x = 4 into dy/ Must be using differential + correct form of line at B(4,0). (ii) 8 area under curve = ( x) x = 8x ½x² Limits to 4 8½ Area under tangent = ½ 4½ = 6¾ Shaded area = ¾ [5] (both unsimplified) Using correct limits. Or could use calculus) Cambridge International Examinations 0

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