OCR A Level Further Mathematics A Y540 SAM
A Level Further Mathematics AY540 Pure Core 1Sample Question PaperDate – Morning/AfternoonTime allowed: 1 hour 30 minutesSpecimYou must have: Printed Answer Booklet Formulae A Level Further Mathematics A Scientific or graphical calculatorenOCR supplied materials: Printed Answer Booklet Formulae A Level Further Mathematics A*000000*INSTRUCTIONS Use black ink. HB pencil may be used for graphs and diagrams only.Complete the boxes provided on the Printed Answer Booklet with your name, centre numberand candidate number.Answer all the questions.Write your answer to each question in the space provided in the Printed AnswerBooklet.Additional paper may be used if necessary but you must clearly show your candidatenumber, centre number and question number(s).Do not write in the bar codes.You are permitted to use a scientific or graphical calculator in this paper.Final answers should be given to a degree of accuracy appropriate to the context.The acceleration due to gravity is denoted by g m s-2. Unless otherwise instructed, when anumerical value is needed, use g 9.8.INFORMATION The total number of marks for this paper is 75.The marks for each question are shown in brackets [ ].You are reminded of the need for clear presentation in your answers.The Printed Answer Booklet consists of 12 pages. The Question Paper consists of 8 pages. OCR 2017603/1325/0Y540B10030/2.2Turn over
2Answer all the questions.5 1 i.2 4i 21Show that[2]2In this question you must show detailed reasoning.The equation f ( x) 0 , where f ( x) x4 2 x3 2 x2 26 x 169 , has a root x 2 3i.[4](ii) Hence write down all the roots of the equation f( x) 0 .[1]In this question you must show detailed reasoning.en3(i) Express f ( x) as a product of two quadratic factors.SpecimThe diagram below shows the curve r 2cos 4 for k k where k is a constant to be determined.θ 0Calculate the exact area enclosed by the curve.[6]4Draw the region in an Argand diagram for which z 2 and z z 3i .[3]5(i) Show that d2sinh 1 2 x .dx4 x2 1[2] 1dx .(ii) Find 2 2 x x26[3]The equation x3 2 x2 x 3 0 has roots , and .The equation x3 px 2 qx r 0 has roots , and .Find the values of p, q and r. OCR 2017[5]Y540
378The lines l1 and l2 have equationsx 3 y 5 z 2x 4 y 2 z 7and. 3 11224(i) Find the shortest distance between l1 and l2 .[5](ii) Find a cartesian equation of the plane which contains l1 and is parallel to l2.[2](i) Find the solution to the following simultaneous equations.x y z 32 x 4 y 5z 97 x 11y 12 z 20[2]en(ii) Determine the values of p and k for which there are an infinity of solutions to the followingsimultaneous equations.x y z 32 x 4 y 5z 97 x 11y pz kSpecim9[6]Prove by induction that, for all positive integers n,5 4r n n.r5r 1 5n OCR 2017Y540[5]Turn over
410The Argand diagram below shows the origin O and pentagon ABCDE, where A, B, C, D and E are thepoints that represent the complex numbers a, b, c, d and e, and where a is a positive real number.You are given that these five complex numbers are the roots of the equation z 5 a5 0 .BCOenSpecimDAE(i) Justify each of the following statements.(a) A, B, C, D and E lie on a circle with centre O.[1](b) ABCDE is a regular pentagon.[2](c) b e2i 5 c(d) b* e[1](e) a b c d e 0[2](ii) The midpoints of sides AB, BC, CD, DE and EA represent the complex numbers p, q, r, s and t.Determine a polynomial equation, with real coefficients, that has roots p, q, r, s and t. OCR 2017[1]Y540[3]
511A company is required to weigh any goods before exporting them overseas. When a crate is placed on a setof weighing scales, the mass displayed takes time to settle down to its final value.The company wishes to model the mass, m kg, which is displayed t seconds after a crate X is placed on thescales.dm m with respect toFor the displayed mass it is assumed that the rate of change of the quantity 0.5dt time is proportional to 80 m .(i) Show thatd 2mdm 2 2km 160k , where k is a real constant.2dtdt[2]It is given that the complementary function for the differential equation in part (i) ise t ( A cos 2t B sin 2t ), where A and B are arbitrary constants.(ii) Show that k 52 and state the value of the constant .en[4]SpecimWhen X is initially placed on the scales the displayed mass is zero and the rate of increase of the displayedmass is 160 kg s-1.(iii) Find m in terms of t.[7](iv) Describe the long term behaviour of m.[1](v) With reference to your answer to part (iv), comment on a limitation of the model.[1](vi) (a) Find the value of m that corresponds to the stationary point on the curve m f (t ) with thesmallest positive value of t.[2](b) Interpret this value of m in the context of the model.[1]d 2mdm 2 5m 400 to model the mass displayed t seconds after a2dtdtcrate Y, of mass 100 kg, is placed on the scales.[1](vii) Adapt the differential equationEND OF QUESTION PAPER OCR 2017Y540Turn over
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Specimen8Copyright Information:OCR is committed to seeking permission to reproduce all third-party content that it uses in the assessment materials. OCR has attempted toidentify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of answer-related information tocandidates, all copyright acknowledgements are reproduced in the OCR Copyright Acknowledgements booklet. This is produced for eachseries of examinations and is freely available to download from our public website (www.ocr.org.uk) after the live examination series.If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correctits mistake at the earliest possible opportunity.For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local ExaminationsSyndicate (UCLES), which is itself a department of the University of Cambridge. OCR 2017Y540
day June 20XX – Morning/AfternoonA Level Further Mathematics AenY540 Pure Core 1ec75SpMAXIMUM MARKimSAMPLE MARK SCHEMEThis document consists of 16 pagesB10030/2.2Duration: 1 hour 30 minutes
Y540Mark SchemeText InstructionsMeaningOther abbreviations inmark schemeE1dep*caooerotsoiwwwAGawrtBCDRMeaningBenefit of doubtFollow throughIgnore subsequent workingMethod mark awarded 0, 1Accuracy mark awarded 0, 1Independent mark awarded 0, 1Special caseOmission signMisreadimAnnotation in scoris and BODFTISWM0, M1A0, A1B0, B1SC MRHighlightingen1. Annotations and abbreviationsSpecMark for explaining a result or establishing a given resultMark dependent on a previous mark, indicated by *Correct answer onlyOr equivalentRounded or truncatedSeen or impliedWithout wrong workingAnswer givenAnything which rounds toBy CalculatorThis question included the instruction: In this question you must show detailed reasoning.2June 20XX
Y5402.Mark SchemeJune 20XXSubject-specific Marking Instructions for A Level Further Mathematics AAnnotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts forresponses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks 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 assist in markingincorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answersthat are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must beinvestigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method.Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, escalate the question to yourTeam Leader who will decide on a course of action with the Principal Examiner.If you are in any doubt whatsoever you should contact your Team Leader.cThe following types of marks are available.enaAecimMA suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lostfor 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 methodor 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.In some cases the nature of the errors allowed for the award of an M mark may be specified.SpAccuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method markis 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 an unknown result.Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimesthis is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correctanswer as part of a wrong argument.3
Y540Mark SchemeJune 20XXWhen a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; andsimilarly where there are several B marks allocated. (The notation ‘dep*’ is used to indicate that a particular mark is dependent on an earlier, asterisked,mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on isworthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, theearlier 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 Bmarks are given for correct work only – differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained fromincorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equallyacceptable. In such cases, what is acceptable will be detailed in the mark scheme. If this is not the case please, escalate the question to your TeamLeader 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 ‘follow through’. In suchcases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find iteasier 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 and expressed either inSI or in the units of the question. (e.g. lengths will be assumed to be in metres unless in a particular question all the lengths are in km, when this would beassumed to be the unspecified unit.) We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usuallyonly penalised where the scheme explicitly says so. When a value is given in the paper only accept an answer correct to at least as many significantfigures as the given value. This rule should be applied to each case. When a value is not given in the paper accept any answer that agrees with the correctvalue to 2 s.f. Follow through should be used so that only one mark is lost for each distinct accuracy error, except for errors due to prematureapproximation which should be penalised only once in the examination. There is no penalty for using a wrong value for g. E marks will be lost except whenresults agree to the accuracy required in the question.gRules for replaced work: if a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examinersshould do as the candidate requests; if there are two or more attempts at a question which have not been crossed out, examiners should mark whatappears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those 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, mark according to thescheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some papers.This is achieved by withholding one A mark in the question. Marks designated as cao may be awarded as long as there are no other errors. E marks arelost unless, by chance, the given results are established by equivalent working. ‘Fresh starts’ will not affect an earlier decision about a misread. Note that amiscopy of the candidate’s own working is not a misread but an accuracy error.iIf a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers (provided, of course, that thereis nothing in the wording of the question specifying that analytical methods are required). Where an answer is wrong but there is some evidence ofmethod, 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.Specimend4
Y540Mark Scheme12(i)Answer552 4i 10 20i 1 2 i2 4i 2 4i 2 4i20MarksM1AO2.1A1[2]1.1DR 2 3i 2 3iB12.2a x2 4 x 13 0 is a quadratic factorM1 f ( x) x 2 4 x 13 x 2 6 x 13(ii)DR Roots are 2 3i, 3 2i i by 2 4iFor multiplying numerator anddenominator by 2 4iClear demonstrationor multiplying1.1Attempt to create first quadraticMust come from x x M1A1[4]1.11.1Attempt to derive second quadraticA11.1Depends on M2 awarded in part (i)12im2 GuidanceenQuestionJune 20XX[1]DRr 0 k 18 1 A 12 81 r 2 d 12 81 2cos 4 2 d 8 1 81 8 2 1 81 8 8 cos 2 4 d 1 sin8 8 1 cos8 d 8 1 8sin 81 8 2.2aec1B1M13.1aLimits not requiredA11.1Correct form to be integratedA1M11.13.1aCorrect indefinite integralUsing correct limitsMust be seenA11.1Must show f ( 18 ) f ( 18 )A0 for decimal answerSp3sin( ) 1 1 8 8 4 [6]5Must be seen
Y540Mark iGuidanceCircle centre 0, radius 2Line parallel to x axis Through 1.5iCorrect shading with continuous arc anddashed lineMust state which region is therequired locusReen2June 20XXimThe shaded regionis the required locus[3] dsinh 1 2 x dx1 x 2 ( 12 ) 2 5(ii)1x 2 ( 12 )2 ec(i)M1Sp522 24 x2 12 2 x x 1 x 1 22 1 dx sinh 1 x 1 c 2 1 x 1 1.1aUse formulae from formulae booklet2.1AG At least one intermediate step mustbe seen[2]B13.1aComplete square soiM1A11.11.1Use correct formoe6M1 sinh y 2 xdycosh y 2dxdy22 E1dx cosh y4 x2 1soE1[3]OR
Y540Mark SchemeQuestion6Answer 2 , 1, 3MarksB1AO1.1p 1B11.1q M1A12.11.1A11.1 6r 9June 20XXGuidancesoiManipulation of roots 1 2 5 n 2 1 10 3 4 5 4 3 1 b a 2 5 7 7 2 9 Shortest distance is given by 67(ii)x 2 y z 51.1a1.1Finding a mutual perpendicularB11.1Any vector between the linesM1Sp 1 5 7 10 b a n 9 5 D n150M1A1Accept any ng formula from the formula bookletUsing their mutual perpendicular frompart (i)oeOr any other valid method
Question(i)Answer x 1 1 1 y 2 4 5 z 7 11 12 1 3 9 20 x 5, y –9, z 7 1 1 1 M 2 4 5 7 11 p AOM11.2A1[2]M11.1BC or in vector form1.1calculating det MDet M 4 p 35 22 28 2 p 55M is singular, so 2p – 26 0 p 13The third equation must be a linearcombination of the first two k k 27M12.1A11.1ec(ii)MarksGuidanceOR elimination methodM1 reduce to two equations intwo unknownssetting their 2p – 26 equal to 0 andsolvingconclude that p 13E12.2aDeducing R3 3R1 2R2M1A1[6]3.1a1.127 from correct reasoningSp8June 20XXim8Mark SchemeenY5408ORM1 state that singular matrixrequires each row to be acombination of the other twoM1 state that R3 3R1 2R2or equivalentA1 conclude that p 13
Y540Mark SchemeQuestionAnswer9Formula is true for n 1 since1 1 and51 5June 20XXMarksB1AO2.5GuidanceMust include an arithmetic justificationM13.1aAdd extra termM12.1ManipulateA12.2a 5 4r k 5 4(k 1) k 555k 1r 1 5k 5 4(k 1)5k 1k 15k 1So if true for n k then true also for n k 1So true generally (i)(a)All roots satisfy z a z a5E12.4Needs conclusion[5]E12.2aClear justification that all points areequidistant from OB11.1Using the answer to part (b)E12.4ec105imk 1Thenen5 4 1 1 55Assume that the formula is true for n k[1](i)(b)All points are distance a from OSp10 2k hence 5 2 Each root of the form a cis spaced at angular intervals of5around thiscircle[2]9
Y540Mark SchemeQuestion10 (i)(c)Answer arg b e2 i5 arg b arg e MarksAOE12.2a2 i5 54 arg (c)June 20XXGuidanceMultiplying any complex number by2 i 2 ,e 5 rotates it through an angle of5so b c[1](ii)a, b, c, d and e are the roots of z a 00so sum of a, b, c, d and e is 0155All roots of equal magnitude a cos 5All roots spaced at angles of2 5r is negative real root z (a cos 5 )5 05E12.4[1]M12.1en(e) 2 2 * 8 cis e cis 5 5 5 b* cis im10(i)(d)E11.1[2]M13.1aec10(i)From any sensible right-angled triangleB13.1aMust justify equally spread argumentsA12.2aAllow correct eqn. from 1st M1 onlySp10[3]10Or by noting that complex rootsof a real polynomial occur inconjugate pairs (the pairing isobvious)Accept any completejustification
Y540Mark SchemeQuestion11 (i)Answerd dm m k 80 m 0.5dt dt June 20XXMarksB1AO3.3GuidanceTranslate the given context into a correctmathematical modelE12.1AGd 2 m dm k (80 m)dtdt 2d2mdm 2 2 2k (80 m)dtdt d 2mdm 2 2km 160k2dtdt[2] A.E. n2 2n 2k 0 2 4 8k 1 1 2k2CF in the form A cos 2t B sin 2t 1 2k 0 n so 2k 1 2 k 52 1imd mdm 2 2km 160k2dtdtM11.1For auxiliary equationA11.1Need not be simplifiedec(ii)E1Sp112en0.5A1ft[4]2.2aAG1.1Follow through their auxiliary equation11Or complete the square
Y540Mark SchemeAnswerMarksAOGuidancedm d 2m 2 0dtdtSubstituting 5 400 80M13.1aCorrect Particular Integral andsubstituting G.S. m e t A cos 2t B sin 2t 80A11.1General Solution including corr
A Level Further Mathematics A . Y540 Pure Core 1 . Sample Question Paper . Date – Morning/Afternoon. Time allowed: 1 hour 30 minutes . OCR supplied materials: Printed Answer Booklet Formulae A Level Further Mathematics A . You must have: Printed Answer Booklet Formulae A Level Further Mathematics A Scientific or graphical .
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