GCE Mathematics A - Revision Maths

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GCEMathematics AUnit H240/03: Pure Mathematics and MechanicsAdvanced GCEMark Scheme for June 2018Oxford Cambridge and RSA Examinations

OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range ofqualifications to meet the needs of candidates of all ages and abilities. OCR qualificationsinclude AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals ,Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications inareas such as IT, business, languages, teaching/training, administration and secretarial skills.It is also responsible for developing new specifications to meet national requirements and theneeds of students and teachers. OCR is a not-for-profit organisation; any surplus made isinvested back into the establishment to help towards the development of qualifications andsupport, which keep pace with the changing needs of today’s society.This mark scheme is published as an aid to teachers and students, to indicate the requirementsof the examination. It shows the basis on which marks were awarded by examiners. It does notindicate the details of the discussions which took place at an examiners’ meeting before markingcommenced.All examiners are instructed that alternative correct answers and unexpected approaches incandidates’ scripts must be given marks that fairly reflect the relevant knowledge and skillsdemonstrated.Mark schemes should be read in conjunction with the published question papers and the reporton the examination. OCR 2018

H240/03Mark SchemeAnnotations and abbreviationsAnnotation in scoris and BODFTISWM0, M1A0, A1B0, B1SC MRHighlightingMeaningBenefit of doubtFollow throughIgnore subsequent workingMethod mark awarded 0, 1Accuracy mark awarded 0, 1Independent mark awarded 0, 1Special caseOmission signMisreadOther abbreviations in mark MeaningschemeE1Mark for explaining a result or establishing a given resultdep*Mark dependent on a previous mark, indicated by *caoCorrect answer onlyoeOr equivalentrotRounded or truncatedsoiSeen or impliedwwwWithout wrong workingAGAnswer givenawrtAnything which rounds toBCBy CalculatorDRThis question included the instruction: In this question you must show detailed reasoning.3June 2018

H240/03Mark SchemeJune 2018Subject-specific Marking Instructions for A Level Mathematics AaAnnotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisationscripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how themarks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded.bAn element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assistin marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on theanswer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always belooked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by acorrect result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method whichdoes not correspond to the mark scheme, escalate the question to your Team Leader who will decide on a course of action with the PrincipalExaminer.If you are in any doubt whatsoever you should contact your Team Leader.cThe following types of marks are available.MA suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are notusually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate anintention 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. bysubstituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may bespecified.AAccuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless theassociated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded.BMark for a correct result or statement independent of Method marks.EMark for explaining a result or establishing a given result. This usually requires more working or explanation than the establishment of anunknown result.Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer isignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where acandidate passes through the correct answer as part of a wrong argument.4

H240/03Mark SchemeJune 2018dWhen a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically saysotherwise; and similarly where there are several B marks allocated. (The notation ‘dep*’ is used to indicate that a particular mark isdependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrongin a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two ormore steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.eThe abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results.Otherwise, A and B marks are given for correct work only – differences in notation are of course permitted. A (accuracy) marks are not givenfor answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may bevarious alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. If this is not the caseplease, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner.Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘followthrough’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within theimage zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.fUnless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct andexpressed either in SI or in the units of the question. (e.g. lengths will be assumed to be in metres unless in a particular question all thelengths are in km, when this would be assumed to be the unspecified unit.) We are usually quite flexible about the accuracy to which the finalanswer is expressed; over-specification is usually only penalised where the scheme explicitly says so. When a value is given in the paperonly accept an answer correct to at least as many significant figures as the given value. This rule should be applied to each case. When avalue is not given in the paper accept any answer that agrees with the correct value to 2 s.f. Follow through should be used so that only onemark is lost for each distinct accuracy error, except for errors due to premature approximation which should be penalised only once in theexamination. There is no penalty for using a wrong value for g. E marks will be lost except when results agree to the accuracy required in thequestion.gRules for replaced work: if a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, thenexaminers should do as the candidate requests; if there are two or more attempts at a question which have not been crossed out, examinersshould mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather thanthose in the assessor handbook.hFor a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, markaccording to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, thoughthis may differ for some units. This is achieved by withholding one A mark in the question. Marks designated as cao may be awarded as longas there are no other errors. E marks are lost unless, by chance, the given results are established by equivalent working. ‘Fresh starts’ willnot affect an earlier decision about a misread. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.5

H240/03Mark SchemeJune 2018iIf a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers (provided, ofcourse, that there is nothing in the wording of the question specifying that analytical methods are required). Where an answer is wrong butthere is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt,consult your Team Leader.jIf in any case the scheme operates with considerable unfairness consult your Team Leader.6

H240/031Question(i)Mark SchemeAnswer x 4 16 y 1 1 7 0 x 4 2 y 1 2 24222(ii)Radius AO1.1EGuidanceCorrect method to find centre of circlee.g. x 4 and2 y 1 2seen (orimplied)C ( 4,1)1MarksM1June 2018A1[2]B124Attempt process for finding both values[1]M11.1E1.1Eoe e.g. 2 61.1aEe.g. squaring both sides to obtain 3terms on both sides 4x3x 2 10 x 8( 0)Obtain 4 and 23A11.1EA11.1EBC2 4x 1 x2 6x 9 Or consider twolinear equations 2 x 1 x 3 1 correct solutionfor A1SC one correctsolution from onelinear equation B1[3]3DRx 3 14.5M13.1bEx 11.5A11.1EAccept any inequality or equals andany letter for the widthCorrect inequality (seen or implied)x x 3 180M13.1bEAccept any inequality or equalsx2 3x 180( 0) x 12 x 15 ( 0)M11.1E 15 x 1211.5 x 12A1B1[6]1.11.1CCCorrect expansion and attempt to solvethree term quadraticCorrect inequalities (seen or implied)7M1A1 correctanswer with noworkingSC B1: x 60B1: x 29 / 6

H240/034Mark SchemeQuestion(i)(a) Answer fg( x) f x 2 2 x 2 2MarksB1 3June 2018AO1.1EGuidance1.1ENo simplification required1.1EBinomial expansion of their x 2 2[1]4(i)(b) gf ( x) g x3 x32 2 x6 2 B1[1]4(ii)DR x2 33 2 x2 3 x22 (2) 3 x 2 (2)2 23M1 3Allow one slip- correct powers and coefficientsfg( x) x 6 x 12 x 8A11.1Cfg( x) gf ( x) 24 6 x4 12 x2 18 0A12.1Cx4 2 x2 3 0 x2 1 x2 3 0M11.1CCorrect method for solving theirquadratic in x 2x 2 3 0 has no real solutionsA1A1[6]B1M12.42.2aAAx 2 3 0 is acceptable for this mark1.12.1EEA11.1CM13.1aECorrect method for rationalising thedenominator of their surd together withcorrect simplificationA1[5]2.2aAAG – at least one step of intermediateworking (from application oftrapezium rule to given result)642 x2 1 0 x 15(i)h 2h 1 1 1 2 2 2 4 2 2 32I 4 2 2 2 212 2 22 22 2 2 2 311I 2 2 2448Use of correct formula with correct(exact) y-values with their hIf M0 next twomarks become BmarksCondone one errorin valuesMust be convincingas AG

H240/035Question(ii)Mark SchemeAnswerJune 2018MarksM1*AO3.1aEA11.1CDep*M12.1CRe-writes integral in the formb a 1 u duA1ft1.1ACorrectly integrates their a 2 2 2ln 2 2 0 2ln 2 0 M11.1C 2 2 2ln 2 A12.2aAUses correct limits correctly(dependent on both previous M marks)oe e.g. 4 – 4 ln 4 4 ln 2M11.1aCSetting the given result approx. equalto their (ii)A1[2]2.2aA5k 16x u dx 2u du2dx4 0 2 202udu2 ux2 2 u 222 2 du 2 1 du0 2 u02 u 2 u 2ln 2 u 02 GuidanceAn attempt at integration by sub allow any genuine attempt (as aminimum must differentiate their sub.and remove all x’s)Correct integral in terms of ubdu1 u[6]5(iii)11 2 2 2 2ln 2 4ln 2 165 429Limits not requiredfor first four marksOr use t 2 u toobtain integral ofbthe form a dtt4 2 t dt 2t 4ln t

H240/036Question(i)Mark SchemeMarksAOsin 2 cos 4 sin 4 cos 2 3cos 2 cos 4 3sin 2 sin 4M11.1ECorrect use of compound angleformulae at least once4sin 2 2cos 2 A1A11.12.2aEENot from incorrect workingAG – at least one step of intermediateworking seensin 2 Answer 3cos 2 4 4 ALT: tan 2 4 3(ii)Guidance sin 2 2 1 tan 2 12cos 2 6June 2018[3]B1tan 2 1 3 tan 2 1 3 1 tan 2 1 tan 2 tan 2 12M12 tan 11 tan 2 2tan 2 4tan 1 0M1*3.1aEDouble angle formula for tan 2 Dep*M11.1Etan 2 5A11.1CRearranges correctly to form 3-termquadratic in tanBC - One correct exact value 2 5 0 so tan 2 5 gives acute angleA12.3A tan 2 5A12.2aAtan 2 12 Correct use of compound angleformula for tan and removal of fractionA1[5]10Explicit rejection and reason forrejectionThis value onlyAllow one sign slipin formula

H240/03Mark SchemeQuestion7AnswerJune 2018MarksAOM12.5EA11.1EM1A11.11.1ECM1 for k 2 x 1 11 c , (1,1) c .24y4 2 x 1 M12.1C11 2y 2 x 1 2A12.2aAUse of (1, 1) to find c – dependent onthe previous two M marks andsubstituted into correct formOe1 2 2 x 1 1 y 2 x 1 2M13.1aAy 2 x 1 222 2 x 1 1M11.1Ay 4 x2 4 x 18x2 8x 1A12.2aAdy 2 x 1 3 4 y 2 0dx1 dydx 2 4 y 2 x 1 3dy1 y2 ydx 2 x 1 3 2 x 1 2(2)( 2)2[9]11GuidanceAttempt to separate variables 2Correct method for combining bothterms on rhs (dependent on previous Mmark) before taking the reciprocalTaking the reciprocal (dependent onprevious M marks) and making y thesubjecta 4, b 8Or re-write interms of yRemove tripledecker fractions

H240/0388Question(i)(ii)Mark SchemeAnswerMarksB1AO1.2E 15 7 0 5 5a 8 2 9.8 1.6 1.6 a or 11.8 g 2 M13.3EA1[3]3.4C1 1.6 2s 10 2 11.8 80 s 590 M13.4EUse of s ut 12 at 2 with t A1ft1.1Cmoments about D – correct number ofterms – oe (leading to an equation inW)Follow through their x onlyA1[3]1.1AAccept 23 or better 0 g 9.8 82 Position vector is 545 9(i)9(ii)25 N2(100) 75x x 0.5 (25)x 1.8759June 2018(iii) x 0.5 2 (100) W (4 0.5 x)W 23.1N[3]B1[1]M112GuidanceUse of F ma with correct m and twoterms of F correcteg moments about A – correct numberof termsFollow through their 25 only23.076923

H240/039910Mark SchemeQuestion(iv) (a)(iv)(b)AnswerModelling the stone as a particle assumes that theweight of the stone block acts exactly at B thereforethe block’s dimensions (or the distribution of themass of the block) have not been taken intoconsiderationModelling the plank as a rigid rod assumes that theplank remains in a straight line and does not bend4sin 2 6sin (i)8sin cos 6sin 3cos 41.4o410P 4cos 2 6cos (ii)P 510(iii)(a)3sin and P 3cos 3sin 2 P 3cos 2Magnitude is 3.39 NJune .2aEEECResolving horizontallyUse of double angle formulae[4]M13.3EA1[2]B11.1EResolving vertically – allow sin/coserrorsAccept 5.002 or better1.1EResolving horizontally and verticallyM11.1EA1[3]2.2aCPythagoras on two forces – both mustinclude 41.43.4 or better13GuidanceAccept ‘uniform’AG41.409622 Alt – M1 for cosinerule with their P, 3and 41.4, A1 for11.4966 or 11.53.3911649

H240/0310Question(iii) (b)Mark SchemeP 3cos tan 3sin Answer54.2o below the horizontal11(i)11(ii)June EE3.1aEA1ft1.1EB12.1AB1ftdep*M11.13.4EC252 142 2 1.3 s2M13.3ATotal distance s1 s2 265 mA12.2aAa k 0.06tk 0.06(20) 1.3k 1.3 1.2 0.1s 0.05t 2 0.01t 3 c t 0, s 0 c 0t 20, v 14s1 0.05 20 0.01 20 23[7]14GuidanceWhere is the angle below thehorizontal54.2 or better – must indicate ‘belowhorizontal’ or equivalent to the‘downward vertical’ (35.8) – directionmay be shown on diagram withminimum of arrow on resultant orarrows on both componentsAlt – M1 forsin sin 41.4 3'3.39'54.18696 Use of t 20 and a 1.3 in their aAttempt to integrate – all powersincreased by 1 (but not just multiplyingby t)1s kt 2 0.01t 32From a correct expression for sIf c 0 stated thenmust give a reason12 20kFinding distance travelled after 20 s(for reference s1 100 )Use of v2 u 2 2as with v 25 anda 1.3 and their uAll previous marks must have beenawarded

H240/0312Mark 3CN, II parallel to the plane – four termsAllow a2mg T 2m 14 gB13.3CN, II for BAllow a2mg F mg sin 30 34 mgA11.1CCorrect method for eliminating T2mg mg cos30 mg sin 30 34 mgA12.1ACorrect use of F R andR mg cos30A1[5]2.2aAR mg cos30T 14 mgT F mg sin30 0F mg cos30 mg 3 11mg mg 0 .4 2 2 12(i)(b)36F mg sin 30 mg 0 12T F mg sin 30 m 14 gF 0 12(ii)June 2018 32 15GuidanceResolving perpendicular to the planeResolving vertically for BResolving parallel to the plane – threeterms – allow signs and sin/cosconfusionUse of F RDeriving equation in (and m and g)and attempt to solve for –dependent on previous M marks andsecond B markResolving parallel to the plane with mg

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GCE Mathematics A Unit H240/03: Pure Mathematics and Mechanics Advanced GCE Mark Scheme for June 2018. OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of

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