Y540/01 Mark scheme November 2020 3 2. Subject-specific Marking Instructions for A Level Mathematics A a Annotations must be used during your marking. For a response awarded zero (or full) marks a single appropriate annotation (cross, tick, M0 or ^) is sufficient, but not required. For responses that are not awarded either 0 or full marks, you must make it clear how you have arrived at the mark you have awarded and all responses must have enough annotation for a reviewer to decide if the mark awarded is correct without having to mark it independently. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. Award NR (No Response) - if there is nothing written at all in the answer space and no attempt elsewhere in the script - OR if there is a comment which does not in any way relate to the question (e.g. ‘can’t do’, ‘don’t know’) - OR if there is a mark (e.g. a dash, a question mark, a picture) which isn’t an attempt at the question. Note: Award 0 marks only for an attempt that earns no credit (including copying out the question). If a candidate uses the answer space for one question to answer another, for example using the space for 8(b) to answer 8(a), then give benefit of doubt unless it is ambiguous for which part it is intended. b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated 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 your Team 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. Y540/01 Mark scheme November 2020 4 c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually 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. In some cases the nature of the errors allowed for the award of an M mark may be specified. A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words “Determine” or “Show that”, or some other indication that the method must be given explicitly. A 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). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. d When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly 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 is worthless 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, the earlier marks are implied and full credit must be given. e The 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 given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. 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 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 ‘follow through’. 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 the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question. f We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only 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 significant figures as the given value. Y540/01 Mark scheme November 2020 5 • When a value is not given in the paper accept any answer that agrees with the correct value to 3 s.f. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range. NB for Specification B (MEI) the rubric is not specific about the level of accuracy required, so this statement reads “2 s.f”. Follow through should be used so that only one mark in any question is lost for each distinct accuracy error. Candidates using a value of 9.80, 9.81 or 10 for g should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric. g Rules for replaced work and multiple attempts: • If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others. • If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete. • if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately. h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors. If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error. i If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold “In this question you must show detailed reasoning”, or the command words “Show” or “Determine”. Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt, consult your Team Leader. j If in any case the scheme operates with considerable unfairness consult your Team Leader. Y540/01 Mark scheme November 2020 6 Question Answer Marks AO Guidance 1 3 0 3 2 0 1 Mean value = f ( )d 3 1 6 d 12 3 x x x x x = + = ∫ ∫ M1 A1 1.1 1.1 Use the correct formula BC [2] Question Answer Marks AO Guidance 2 ( ) ( ) ( )( ) 2 3 2 1 1 3 2 1 1 1 2 2 2 ( 1) ( 2 ) 2 1 1 1 ( 1) 2. ( 1)(2 1) ( 1) 4 6 2 1 ( 1) 3 ( 1) 4(2 1) 6 12 1 ( 1) 3 11 10 12 1 ( 1) 2 3 5 12 n n r r n n n r r r r r r r r r r r n n n n n n n n n n n n n n n n n n n n = = = = = = + = + + = + + = + + + + + + = + + + + + = + + + = + + + ∑ ∑ ∑ ∑ ∑ M1 A1 A1 1.1a 1.1 1.1 Correct split of terms and use of formulae Correct forms for each summation Can be earned even if 2 is dropped ( ) 4 3 2 1 3 14 21 10 12 n n n n + + + earns 2 marks Fully factorised form for this mark [3] Y540/01 Mark scheme November 2020 7 Question Answer Marks AO Guidance 3 (a) A4 = I B1 1.1 Accept 3×3 matrix [1] (b) Rotation Clockwise 900 about x-axis B1 B1 2.2a 2.2a Or 2700 anticlockwise Accept radians [2] (c) 1 0 0 0 1 0 0 0 1 −           B1 1.1 All correct [1] (d) (−2, 3, 4) B1 1.1 1 0 0 2 2 0 1 0 3 3 0 0 1 4 4 − −         =             Allow vector as answer [1] Y540/01 Mark scheme November 2020 8 Question Answer Marks AO Guidance 4 (a) i 2 25i 25e π = oe B1 3.3 Conversion soi i 2 5 i i 4 4 5 i i 4 4 25i 25e 25i 5e and 5e So the square roots are 5e and 5e π π π π π = ⇒ = M1 A1 3.1a 3.2a Attempt to find at least one root by rooting the modulus and halving the argument soi Both Alternate Method ( ) 1 5 4 4 25i i 5 2 2 5 1 1 25i 2 1 i 5 cos isin 2 4 4 25i 5e and 5e a b a b π π π π = + ⇒ = = ±   ⇒ = ± + = ± +     ⇒ = B1 M1 A1 Conversion and attempt to square a and b. Ignore ± Both [3] (b) B1 1.1 All three but no extras. Scales etc are not required but if no scale then the lines representing the roots should be at 45° to axis. Accept points. No extras Line representing 25i must be at least two times as long [1] Y540/01 Mark scheme November 2020 9 Question Answer Marks AO Guidance 5 3 2 6 2 2 2 6 1 3 1 3 z z z z z z   + = + + +     M1 A1 1.1 1.1 Use of binomial 3 2 2 3 2 2 2 6 2 6 1 1 2cos2 8cos 2 1 1 and 2cos2 and 2cos6 z z z z z z z z θ θ θ θ   + = ⇒ + =     + = + = M1 3.1a use of De Moivre for any one term 3 2 6 2 2 6 2 3 3 1 1 1 and 3 8cos 2 2cos6 6cos2 4cos 2 cos6 3cos2 z z z z z z θ θ θ θ θ θ       + = + + +             ⇒ = + ⇒ = + AG A1 A1 2.1 1.1 [5] Y540/01 Mark scheme November 2020 10 Question Answer Marks AO Guidance 6 2 1 2 1 1 10 , 2 4 3 113 1 2 1 2 2 0 1 3 2 1 2 0 . 10 2 3 8 0.753 to 3sf 113 d             = × − = −             − −       = −             − = − =             − −       −       −       −    ⇒ = = = n n b a n M1 A1 A1 M1 A1 3.1a 1.1 1.1 1.1 1.1 Cross product Modulus Using correct formula for d Accept exact or correct to 3sf (0.75257669…) FT an exact answer [5] Y540/01 Mark scheme November 2020 11 Question Answer Marks AO Guidance 7 ( ) ( ) ( ) ( ) 3 3 3 3 3 3 3 3 3 2 3 2 2 Basis case: 1 2 3 36 4 9 Consider the sum f ( ) + 1 + 2 Assume that f ( ) 9 for some Then f ( 1) f ( ) 3 f ( ) 3 3 3 9 27 f ( ) 9 27 27 = 9 9 3 3 9 ' for some ' So if t r r r r r k k r r r r r r r r r r r r k r r k k + + = = × = + + = ∈Ζ + = + + − = + + × + × + − = + + + + + + = ∈Ζ rue for then true also for 1 But it is true for 1 so is true for all integers r r r r + = B1 M1 M1 A1 A1 2.1 2.1 2.1 2.2a 2.5 Basis case – either 4 9 × or “36 is divisible by 9” given explicitly. For getting started (Sight of f(r) is not necessary) For finding f(r + 1) Each term as a multiple of 9 Conclusion dependant on all other marks being earned [5] Y540/01 Mark scheme November 2020 12 Question Answer Marks AO Guidance 8 (a) 2 2 e e 2sinh 1 2 1 2 u u u −   − + ≡ +     M1 2.1 Use of exponential form for sinh u 2 2 2 2 e 2 e e e 1 cosh 2 2 2 u u u u u − − − + + = + ≡ ≡ AG A1 2.1 [2] (b) 2 cosh 2 2sinh 1 2sinh 2 4sinh cosh sinh 2 2sinh cosh u u u u u u u u ≡ + ⇒ ≡ ⇒ ≡ AG B1 2.1 [1] (c) 2 2 2 sinh d 2sinh cosh d sinh d 2sinh cosh d 1 sinh 1 x u x u u u x u x u u u x u = ⇒ = ⇒ = + + ∫ ∫ M1 3.1a Attempt to find d d x u 2 2 sinh d u u = ∫ A1 1.1 ( ) cosh 2 1 d u u = − ∫ M1 1.1a Use double angle formulae and attempt to integrate. 1 1 sinh 2 sinh cosh 2 (1 ) sinh u u c u u u c x x x c − = − + = − + = + − + So f ( ) (1 ) , 1, 1 x x x c a b = + + = − = A1 A1 1.1 1.1 Ignore c. c must be included here as part of f(x) – allow a and b not being stated explicitly but f(x) must be Y540/01 Mark scheme November 2020 13 Alternative method 2 2 2 1 2 e e 2 d e 2 e d 2 u u u u u u − −   − = = − +     ∫ ∫ M1 1.1a Use exponentials and attempt to integrate. 2 2 1 1 1 4 4 2 1 e e sinh 2 (1 ) sinh u u u c u u c x x x c − − = − − + = − + = + − + So f ( ) (1 ) , 1, 1 x x x c a b = + + = − = A1 A1 1.1 1.1 Ignore c. c must be included here as part of f(x) – allow a and b not being stated explicitly but f(x) must be [5] (d) ( ) ( ) 2 1 1 Area = (1 ) sinh 6 ln( 2 3) 2 ln(1 2) 1 2 6 2 ln 2 3 x x x −   + −   = − + − − +   + = − +     +   1 2 So 6 2, 1, 2 3 p q r + = − = = + M1 A1 1.1 1.1 Correct limits substituted and subtracted into their answer to (c) soi p, q, r must be stated [2] Y540/01 Mark scheme November 2020 14 Question Answer Marks AO Guidance 9 (a) 1 i 2 β = − oe B1 2.2a [1] (b) ( )( ) 3 , 1 i 2 1 i 2 3 2 1 2 αβγ αβ γ = = + − = ⇒ = M1 A1 2.1 1.1 Use of 3 2 αβγ = to find the 3rd root. Alternatively, find 2 2 3 x x − + and divide [2] (c) ( ) ( ) ( ) ( )( ) ( )( ) 2 3 2 2 3 2 1 i 2 1 i 2 2 1 0 2 3 2 1 0 2 4 6 2 3 0 2 5 8 3 0 i.e. 5, 8 x x x x x x x x x x x x x x p q − + − − − = ⇒ − + − = ⇒ − + − + − = ⇒ − + − = = − = M1 A1 3.1a 1.1 Multiply out (can be seen in (b)) Alternate Method 5 5 2 2 a p α β γ + + = − = ⇒ = − ( ) ( )( ) 2 1 1 i 2 1 i 2 1 i 2 1 i 2 2 1 3 4 8 q q αβ βγ γα + + = = + + − + + − = + = ⇒ = M1 A1 Use of symmetry forms for roots [2] Y540/01 Mark scheme November 2020 15 (d) ( ) 1 2 1 2 1 i 2 3 i 3 cos isin 3 3 1 2 where cos , sin tan 2 3 3 α θ θ θ θ θ   = + = + = +       = = ⇒ = M1 2.1 Either α or β seen in mod/arg form ( ) 1 2 1 2 1 i 2 3 i 3 cos isin 3 3 β θ θ   = − = − = −       oe A1 1.1 For both of them – accept exponentials ( ) ( ) 2 2 2 2 3 cos isin , 3 cos isin 2 3 cos 1 2 3 cos 2 n n n n n n n n n n n n n n n n n n α θ θ β θ θ α β θ α β γ θ ⇒ = + = − ⇒ + = × × ⇒ + + = × × + AG M1 A1 2.1 2.1 Derivation of or n n α β [4] Y540/01 Mark scheme November 2020 16 Question Answer Marks AO Guidance 10 (a) F ma = (F1 is in the direction of motion and F2 is resisting motion) F t v λ µ = − M1 3.3 Use of Newton II with constants of proportionality. F = ma must be seen 1 d 2 d v t v t λ µ ⇒ = − AG A1 2.1 [2] (b) 2 1 d 2 d d 2 2 d I.F. e t v t v t v v t t µ λ µ µ λ = − ⇒ + = M1* A1 1.1a 1.1 Attempt to find IF ( ) 2 2 2 2 2 d e e 2 2 e d d e 2 e d t t t t t v v t t v t t µ µ µ µ µ µ λ λ ⇒ + = ⇒ = M1dep A1 3.1a 1.1 Getting DE in correct form 2 2 2 2 2 2 e 2 e d 1 1 e 2 e e 2 4 t t t t t v t t v t c µ µ µ µ µ λ λ µ µ ⇒ =   ⇒ = − +     ∫ 2 2 2 2 2 Given that 0, 0 4 e 2 2 t t v c v t µ λ µ λ λ λ µ µ µ − = = ⇒ = ⇒ = − + oe M1dep A1 A1 3.1a 1.1 3.4 Attempt integration by parts Y540/01 Mark scheme November 2020 17 Alternative method d 2 2 d v v t t µ λ + = M1 E 2 : 2 0 2 t AE m m CF Ae µ µ µ − + = ⇒ = − ⇒ = M1 A1 E E : PI v at b = + M1 C 2 2 2 d 2 2 2 d , 2 2 : 2 t v a a at b t t a a b GS v t Ae µ µ µ λ λ λ µ µ µ λ λ µ µ − ⇒ = ⇒ + + = ∴ = = − = − ∴ = − + M1 A1 A1 C C A [7] (c) 2 2 2 2 e 2 2 2, 1 2 1 e t t v t v t µ λ λ λ µ µ µ λ µ − − = − + = = ⇒ = −+ When t is large, 2 e t − is very small so 2 1 v t ≈ − M1 A1 3.4 3.3 Consider the behaviour of the exponential function in their equation from (b) soi or 2 v t ≈ [2] (d) 1 d 2 2 d v v t = − oe B1 3.5c [1] (e) d As approaches 2, 0 d i.e. approaches a constant value. v v t v → B1 3.4 [1] Y540/01 Mark scheme November 2020 18 Question Answer Marks AO Guidance 11 (a) ( ) ( ) ( ) 3 3 3 3 3 3 cos , sin cos sin 2 cos . sin cos sin 2cos sin 2cos sin cos sin x r y r r r r r r r θ θ θ θ θ θ θ θ θ θ θ θ θ θ = = ⇒ + = ⇒ + = ⇒ = + oe M1 A1 3.1a 1.1 Substitution May see “or r = 0” but not required. [2] (b) 3 3 3 3 1 1 2cos sin 1 2 2 f ( ) 1 1 2 cos sin 2 2 2sin cos sin cos π θ π θ π θ π θ π θ θ θ θ θ     − −         − =     − + −         = + M1 A1 1.1a 1.1 Correct substitution into their f(θ) [2] (c) So the line of symmetry is 4 π θ = B1 2.2a Allow y = x. Must have θ = [1] (d) 1 4 f ( ) 2 r π = = B1 1.1 BC [1] (e) 0 when 0. 0 also when 2 In range 0 , 0 and is continuous 2 So there is a loop r r r θ π θ π θ = = = = < > < B1 B1 3.1a 2.4 For both, ignore extras. Conclusion - both statements for r need to be mentioned [2] Y540/01 Mark scheme November 2020 19 Question Answer Marks AO Guidance 12 4 2 2 2 2 3 1 3 2 0 1 3 1 0 4 1 1 1 1 (1 )(1 ) (1 )(1 ) ( )(1 ) 4 1:4 4 1 1:4 4 1 0: 4 2 : 0 0 1 1 2 d 1 1 1 1 ln 2tan 1 3 1 ln 2t 3 1 A B Cx D x x x x A x x B x x Cx D x x A A x B B x A B D D x A B C C I x x x x x x x − + = + + − − + + ⇒ + + + − + + + − = = = ⇒ = = − = ⇒ = = + + = ⇒ = − − = ⇒ =   = + +   − + +    +    = +     −       + = +     −   ∫ ( ) ( ) 1 1 3 1 an 0 ln 3 3 3 1 ln 2 3 i.e. 2, 3, 3 3 a b c π π −   + − = +     −   = + + = = = M1 A1 A1 M1 M1 A1 3.1a 1.1 1.1 1.1 1.1 1.1 Proper split to produce integrable integrand For one of the terms /constants For all terms/constants Or 1 4 of these See below for other possibilities Correctly integration of their integrand without simplification – ignore limits Substitution – ignore – 0 Values must be stated [6] Alternatives: 2 1 D x + is M0 A0 A0 but consider integration for 3 marks. 4 2 2 2 1 1 1 1 1 2 1 1 1 1 1 1 1 2 2 1 1 1 x x x x x x   = +   − − +       = + +     − + +     is M1 A1 A1 Y540/01 Mark scheme November 2020 20 Question Answer Marks AO Guidance 12 4 2 4 1 1 1 1 A B D x x x x = + + − − + + M0 But consider last 3 marks for correct integration 4 2 2 4 1 1 1 A D x x x = + − − + with A = D = 2 by inspection Followed by 2 2 1 1 1 1 1 x x x + + + − + in integration section M1 A1 A1 Look to see the second split further on in question 4 2 2 1 1 3 3 4 2 2 0 0 1 1 1 1 1 3 0 4 1 1 1 4 2 2 d d 1 1 1 1 1 2tanh 2tan 2tanh 2tan 3 3 A D x x x x x x x x x x − − − − = + − − +   ⇒ = +   − − +       = + = +       ∫ ∫ ( ) 1 1 3 ln 2 0 1 6 1 3 3 1 1 2 3 3 4 2 3 ln ln ln 3 2 3 2 3 3 1 ln 2 3 3 π π π π π   +   = + −     −           + + + + = + = + = +             −       = + + M1 A1 A1 Look for the integration. If tanh-1 is used then give full marks here. OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA OCR Customer Contact Centre Education and Learning Telephone: 01223 553998 Facsimile: 01223 552627 Email: general.qualifications@ocr.org.uk www.ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored