MARK SCHEME for the October/November 2012 series 9709 MATHEMATICS. 9709/12 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 October/November 01 series 9709 MATHEMATICS 9709/1 Paper 1, maimum raw mar 75 This mar scheme is published as an aid to teachers and candidates, to indicate the requirements of the eamination. It shows the basis on which Eaminers were instructed to award mars. It does not indicate the details of the discussions that too place at an Eaminers meeting before maring began, which would have nsidered the acceptability of alternative answers. Mar schemes should be read in njunction with the question paper and the Principal Eaminer Report for Teachers. Cambridge will not enter into discussions about these mar schemes. Cambridge is publishing the mar schemes for the October/November 01 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level mponents and some Ordinary Level mponents.

2 Page Mar Scheme Syllabus Paper GCE AS/A LEVEL October/November Mar Scheme Notes Mars are of the following three types: M A B Method mar, awarded for a valid method applied to the problem. Method mars 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 mar and in some cases an M mar can be implied from a rrect answer. Accuracy mar, awarded for a rrect answer or intermediate step rrectly obtained. Accuracy mars cannot be given unless the associated method mar is earned (or implied). Mar for a rrect result or statement independent of method mars. When a part of a question has two or more method steps, the M mars are generally independent unless the scheme specifically says otherwise; and similarly when there are several B mars allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mar is dependent on an earlier M or B (asterised) mar in the scheme. When two or more steps are run together by the candidate, the earlier mars are implied and full credit is given. The symbol implies that the A or B mar indicated is allowed for wor rrectly following on from previously inrrect results. Otherwise, A or B mars are given for rrect wor only. A and B mars are not given for fortuitously rrect answers or results obtained from inrrect woring. Note: B or A means that the candidate can earn or 0. B/1/0 means that the candidate can earn anything from 0 to. The mars indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mar, allow the candidate the benefit of the doubt. Unless otherwise indicated, mars once gained cannot subsequently be lost, e.g. wrong woring following a rrect form of answer is ignored. Wrong or missing units in an answer should not lead to the loss of a mar unless the scheme specifically indicates otherwise. For a numerical answer, allow the A or B mar if a value is obtained which is rrect to 3 s.f., or which would be rrect to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mar is not given if a rrect numerical answer arises fortuitously from inrrect woring. For Mechanics questions, allow A or B mars for rrect answers which arise from taing g equal to 9.8 or 9.81 instead of 10. Cambridge International Eaminations 01

3 Page 3 Mar Scheme Syllabus Paper GCE AS/A LEVEL October/November The following abbreviations may be used in a mar 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 etra checing is needed to ensure that the detailed woring 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 Woring Only often written by a fortuitous answer Ignore Subsequent Woring Misread Premature Approimation (resulting in basically rrect wor that is insufficiently accurate) See Other Solution (the candidate maes a better attempt at the same question) Special Ruling (detailing the mar to be given for a specific wrong solution, or a case where some standard maring practice is to be varied in the light of a particular circumstance) Penalties MR 1 PA 1 A penalty of MR 1 is deducted from A or B mars 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 mars then beme follow through mars. 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 mars in the case of premature approimation. The PA 1 penalty is usually discussed at the meeting. Cambridge International Eaminations 01

4 Page 4 Mar Scheme Syllabus Paper GCE AS/A LEVEL October/November a Term in 5 is 7 C 3 (²) 4 ( a/)³ This term isolated Equated to 80 a =. Allow on own or in an epansion. Correct term in 5 selected. Equated to (i) f() = + 1, for 3 Mae the subject or interchanges,y ( 1) Attempt at as subject and removes +1 Squares both sides and deals with "+3" and " ". (ii) domain of f 1 is 1. [1]. ndone >1 3 (i) A = 400 0(60 ) (40 ) 30 A = ² (uld be trapezium triangle) (ii) da = 30 or ( 15)² d = 0 when = 15 or Min at = 15 A = 975. [] Needs attempts at all areas answer given - either method oay Sets differential to 0 + solution.. 4 y = + (i) 4 4y = = + 4 Uses b² 4ac = 1 4 = 0 1 (calculus = 4 1 =, y = = 1 ) (ii) y = 1, 4y = ² = 0 P(, 1) Eliminates or y mpletely. Uses b² 4ac for a quadratic = 0 nb a,b,c must not be f() Elimination of or y Soln of eqn.. Cambridge International Eaminations 01

5 Page 5 Mar Scheme Syllabus Paper GCE AS/A LEVEL October/November A (1, 3), B (5, 11), X (4, 4) (i) Gradient of AB = Gradient of BC = ½ Eqn of BC is y 11 = 1 ( 5) (ii) gradient of AC (or AX) is ⅓ eqn of AC is y = 1 ( 1) = 1 3 ( 4 or y ) Sim equations C (13,7) For use of m 1 m = 1 unsimplified is fine Correct form of line equation + sim eqns answer only -0/3- assumed AB = BC. Uses graph or table and gets eactly (13,7) allow the 3 mars for (ii). 6 s = 3tan (i) Replaces tan by sin s c² =3s s² +3s = 0 Uses t = s c Uses s² + c² = 1. Correct eqn. (ii) Soln of quadratic y = 15 º y can also be y = 75 º. D Method for quadratic = 0 and Wors with y first before for 90º 1 st answer. (loses mar if etra soln in range) 7 1 OA = 0 OB = 1 (i) 0. = 10 4 = 5 4 s θ θ = 4.1º Use of 1 + y 1 y + z 1 z Product of moduli All nnected rrectly. (ii) 1 AB = allow each cpt ± ( 1) + + ( ) = 0 = 1 or ⅔ Correct for either AB or BA. Sum of 3 squares (doesn't need =1) Correct quadratic Cambridge International Eaminations 01

6 Page 6 Mar Scheme Syllabus Paper GCE AS/A LEVEL October/November (a) (i) ar = 4, ar³ = 13½ Eliminates a (or r) r = ¾ a = 3 (ii) sum to infinity = 3 ¼ = 18 [] Both needed Method of Solution. Correct formula used. on value of r (b) a = 3, d = n (6 + ( n 1) ) ( = 360) n + 4n 70=0 n = 18 Correct value for d Correct S n used. no need for 360 here. Correct quadratic 9 9 y = 3 A (3, 1) B (0, 3) (i) d y 9 = d ( + 3) m = 9 y = ( 3) 1 9 Correct without the. For, independent of first part. Correct form of tan - numerical dy/d For his m following use of dy/d. (normal ma /4, no calculus 0/4) (ii) Meets the y-ais when = 0, y = 1⅔ This is nearer to B than to O. [1] Sets to 0 in his tangent. The 1⅔ and part (i) must be rrect (iii) Integral of = ( + 3) Uses limits 0 to 3 = 9π 6 Correct without the. For, Use of limits with integral of y² only no π ma ¾. Use of area - 0/4, Cambridge International Eaminations 01

7 Page 7 Mar Scheme Syllabus Paper GCE AS/A LEVEL October/November dy 4 = + and P (4, 8) d 4 (i) y = + (c) Uses (4, 8) c = 1 d y (ii) d = = 0 when =. (ignore +c at this stage) Uses the point after integration for c Co Sets to 0 + solution or verifies and states a nclusion (stationary or min) gradient of d/d( 1 ) = 3 4 +ve Min. Allow for = into dy/d. Any valid method - 3rd differential +ve nd diff goes 0+, or 1st goes >3,3,>3 11 (i) OQ = + OC = 0 sin 0.6 = OC + sin 0.6 OC = sin 0.6 = 0 = 7.18 Used somewhere needs 0. Use of trig in 90º triangle Soln of linear equation. (answer given, ensure there is a rrect method) (ii) Area = ½. 0² 1. π 7.18² = 76.3 (iii) Angle PCR = π 1. Arc PR = 7.18 (π 1.) = (14.01) OP = OR = tan 0.6 Perimeter of 35.1 cm [] Use of ½r²θ - needs r=0 and θ = 1. Use of s=rθ with r = any θ -even π/3 Correct use of trig or Pythagoras Cambridge International Eaminations 01

9709 MATHEMATICS 9709/11 Paper 11, maximum raw mark 75

9709 MATHEMATICS 9709/11 Paper 11, maximum raw mark 75 UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2009 question paper for the guidance of teachers 9709 MATHEMATICS