GCE Mathematics A H240/01: Pure Mathematics Advanced GCE Mark Scheme for November 2020

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GCE Mathematics A H240/01: Pure Mathematics Advanced GCE Mark Scheme for November 2020 Oxford Cambridge and RSA Examinations GCE Mathematics A H240/01: Pure Mathematics Advanced GCE Mark S ... cheme for November 2020Oxford Cambridge and RSA Examinations OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, 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 requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. © OCR 2020H240/01 Mark Scheme November 2020 2 Text Instructions 1. Annotations and abbreviations Annotation in RM assessor Meaning and  BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread BP Blank Page Seen Highlighting Other abbreviations in mark scheme Meaning dep* Mark dependent on a previous mark, indicated by *. The * may be omitted if only one previous M mark cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working AG Answer given awrt Anything which rounds to BC By Calculator DR This question included the instruction: In this question you must show detailed reasoning.H240/01 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.H240/01 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.H240/01 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.H240/01 Mark Scheme November 2020 6 Question Answer Marks AO Guidance 1 (a) 2 1 1 ( − + − 12θ θ 2) ( )2 B1 2.1 Correct statement First term could possibly already be expanded 2 2 2 1 2 − + − + θ θ θ M1 2.1 Attempt to expand and simplify given expression = 3 2 − θ A.G. A1 2.1 Obtain given answer Max of M1A1 if neither (1−θ )2 nor 1 – 2tanθ + tan2θ seen [3] (b) 3 2 28 − = θ θ M1 1.1a Use 28sin 28 θ θ ≈ and attempt to solve θ = 0.1 A1 1.1 Obtain 0.1 oe BOD if 0.1o ISW once 0.1 seen [2]H240/01 Mark Scheme November 2020 7 Question Answer Marks AO Guidance 2 (a) 6a3 B1 1.1 Obtain 6 B1 only for ± 6a3 B1 1.1 Obtain a3 [2] (b) 1 (64 ) 4 b b 3 3 = or (4 ) 4 12 1 2 2 b b − = B1 1.1a Correct simplification of either term Allow (2 ) b2 1 − for the second term 1 2b− or 2 b B1 1.1 Correct final answer [2] (c) 9 3 3 6 c c = 27 3 2 6 c c = B1 1.1 Either 9c or 272c correct as a power of 3 (or 729) Ignore coefficient Index must be simplified M1 1.1a Attempt to write the other one of 9c and 272c with the same base Ignore coefficient Allow unsimplified index B2 for 272c = 93c 7 3 4 3 3 3 × − × = × 6 6 6 c c c A1 3.1a Combine to obtain correct single term Allow equiv eg 3 729 × c or 2 3 27 × c or 3 9 × 3c = 36 1 c+ A1 1.1 Obtain correct final answer Must be single power of 3 [4] OR B1 92c(7 × 9c – 4 × 32c) M1 92c(7 × 32c – 4 × 32c) 92c × 3 × 32c A1 3 × 272c A1 36 1 c+H240/01 Mark Scheme November 2020 8 Question Answer Marks AO Guidance 3 (a) π 16000π r h 2 = B1 3.1b Correct equation for volume seen or used h likely to be used, but could be other variable 2 A r rh = + 2π 2π B1 3.1b Correct expression for surface area seen Two terms may be seen at separate stages of the proof If alternative formula used eg 2 1 2π 2 r Vr + − then this must be clearly derived Allow BOD for 2 2 2π 2π 16000 r r r + × − as long as h seen explicitly in terms of r first 2 2 = + × 2π 2π 16000 r r r− M1 1.1a Eliminate h from expression for surface area Allow if just attempt at curved surface area 2 1 = + 2π 32000π r r− A.G. A1 1.1 Obtain given answer If 2πr 2 is first seen in the final answer then it must be justified eg ‘plus two ends’, otherwise max B1B0M1A0 [4]H240/01 Mark Scheme November 2020 9 Question Answer Marks AO Guidance (b) d 2 dAr = 4π 320 r r − 00π − M1 1.1a Attempt differentiation Both powers decrease by 1 4π 32000π 0 r r − = −2 r3 = 8000 r = 20 M1 3.1b Equate derivative to 0 and attempt to solve for r (or h) 12 2 -32000π π 16000 h h − + − A1 1.1 Obtain correct r, units not needed h 32 = 64000 Surface area = 2400π cm2 / 7540 cm2 A1 1.1 Obtain correct A, units not needed Allow exact or decimal (3sf or better) 2 2 d 3 d A 4π 64000π r r − = + when r = 20, d2 2 d A 12π r = (or 37.7) M1 2.1 Attempt method to justify minimum, including substitution or consideration of sign Could also test first derivative, or A, on both sides of r = 20 2 2 dd A 0 r > , hence minimum A1 2.2a Correct conclusion, with justification, from correct working If second derivative is evaluated, it must be correct (condone truncated decimal of 37.6) [6] 4 Assume that there is a greatest multiple of 5 ie N = 5k B1* 2.1 Assumption for contradiction Some indication that they are starting with the greatest multiple of 5 N + 5 = 5k + 5 = 5(k + 1) M1 2.1 Add on 5, or a multiple of 5 Or any equiv operation that would result in a larger multiple of 5 M0 if just numerical example This is a multiple of 5, and N + 5 > N which contradicts the assumption Hence there is no greatest multiple of 5 A1d* 2.4 Statement denying assumption Need justification about why it is a multiple of 5, why it is greater, as well as ‘contradiction’ or clear equiv such as ‘initial assumption is incorrect’ [3]H240/01 Mark Scheme November 2020 10 Question Answer Marks AO Guidance 5 (a) 1 BQ = 2  (a – b) B1 1.1a Correct  BQ or QB  Or any correct vector involving Q, but must be clear which vector it is 14 PQ =  b + 1 2 (a – b) = 12 a – 14 b B1 1.1 Correct  PQ Must be simplified to two terms SC Allow B1 if correct unsimplified PQ is seen but individual vectors not explicit [2] (b)  PR has the same direction as  PQ , so vector must be a multiple of PQ  B1 2.4 Explain parallel (or collinear) vectors have direction vectors that are multiples of each other Allow ‘gradient’ for ‘direction’, or ‘they are on the same straight line’, but must state or use ‘multiple’ So PR  = λ( 12 a – 14 b) = 14 λ (2a – b) = k (2a – b) A.G. B1 2.1 Show given answer convincingly Clear detail of scaling factor [2] (c)  AR = – a + 3 4 b + k (2a – b) B1 1.1 Correct expression for  AR (or OR  ), in terms of k Could use A to Q to R (condone if k still used) AR  multiple of a only, 34 b – k b = 0 M1 3.1a Use coefficient of b = 0 Must be used in  AR or OR  Obtain k = 3 4 A1 1.1 Obtain correct value for k May get different value for their k ratio OA : AR = 2:1 A1 1.1 Correct ratio (allow 1: 12 ) oe Answer only is 0, as question says ‘determine’ [4]H240/01 Mark Scheme November 2020 11 Question Answer Marks AO Guidance 6 (a) log10S = log10(abt) log10S = log10a + log10bt log10S = tlog10b + log10a M1 2.1 Attempt to show reduction to linear form Introduce logs on both sides, and correctly split to the sum of two terms A1 2.1 Obtain correct equation Condone no base; any bases seen must be 10 A0 for log10bt unless previously seen as tlog10b which is of the form Y = mX + c A1 2.4 Link to equation of straight line Base of 10 must now be explicit throughout Could instead refer to a linear relationship [3] (b) log10a = 0.583 ⇒ a = 100.583 = 3.8 B1 1.1 Obtain a = 3.8, or better, from either eqn Must clearly be value for a log10b = 0.146 ⇒ b = 100.146 = 1.4 B1 1.1 Obtain b = 1.4, or better, from either eqn Must clearly be value for b [2] (c) 3.8 × 1.4t = 200 1.4t = 52.63 M1 3.1a Link their model to 200 and attempt to solve for t Or use linear equation and attempt to solve for t Must use correct solution method Allow M1 if using S = 200,000,000 Allow if their a and b transposed t = 11.8 A1 1.1 Obtain t = 11.8, or better, www (allow t = 12) Condone 11.7 as truncated value for t so year is 2027 A1FT 3.2a FT their value for t Answer in context, so not just ‘12 years later’ FT on 2015 + integer number of years, from rounding up their t [3]H240/01 Mark Scheme November 2020 12 Question Answer Marks AO Guidance (d) Unlikely that sales will continue at same rate Finite market B1 3.5b Any sensible reason – eg pattern not necessarily continuing or the market being limited by no. of customers Allow ‘extrapolation unreliable’ Reason needed not just eg ‘other external factors’ [1] 7 (a) Anna = 30 + 9 × 15 = 165 mins M1 3.3 Attempt u10 for AP, using correct eqn Method must be seen Ben = 30 × 1.19 = 71 mins M1 3.3 Attempt u10 for GP, using correct eqn Method must be seen 165 – 71 = 94 minutes A.G. A1 3.4 Obtain given answer of 94 minutes www AG so both terms need to be explicitly evaluated for the A1 Show subtraction, or give more accurate value before 94 [3] (b) Anna: uX = 30 + 15(X – 1) Ben: uX = 30 × 1.1X – 1 B1 3.4 Both uX terms correct Condone unknown other than X 30 × 1.1X – 1 > 30 + 15(X – 1) 30 × 1.1X – 1 > 15X + 15 1.1X – 1 > 0.5X + 0.5 X – 1 > log1.1(0.5X + 0.5) X > log1.1(0.5X + 0.5) + 1 A.G. M1 2.1 Link correct expressions and attempt to rearrange Condone an incorrect linking sign eg = not > Must be using a correct process, so allow sign errors only A1 2.1 Show given answer convincingly Must now be >, with justification if = used in proof, and with X [3] (c) 18.9 B1 1.1a Correct first iterate Allow 19 or 18.8… 25.1, 28.0, 29.0, 29.4, 29.6, 29.6,… M1 1.1 Use correct iterative process to find at least two further values Allow integer values Could be truncated not rounded X = 30 A1 3.4 Obtain X = 30 Must be an integerH240/01 Mark Scheme November 2020 13 Question Answer Marks AO Guidance [3] (d) (i) Eventually there will not be enough hours in the day for revision B1 3.5b Comment on long-term behaviour Allow other sensible reason [1] (ii) Increasing by 10% will involve decimals of minutes so will no longer be accurate B1 3.5b Comment on not being able to measure time that accurately Allow long-term behaviour if not already given in (i) Allow other sensible reason B0 if referring to reasons that may prevent revision from happening eg illness If a correct reason is given then ISW an incorrect numerical valueH240/01 Mark Scheme November 2020 14 Question Answer Marks AO Guidance [1] 8 (a) ddx(e 2e 2 2 x x ) = B1 1.1 Seen anywhere in solution 6xe2x + (2 + 3x2)(2e2x) e2x(6x2 + 6x + 4) M1 1.1a Attempt product rule Could expand first A1 1.1 Obtain any fully correct expression [3] (b) e2x > 0 for all x B1 2.1 B0 if clearly considering f(x) or f ( ) ′′ x and not f ( ) ′ x 6x2 + 6x + 4 = 6(x + + 12 2 )2 5 minimum value is 5 2 so > 0 for all x M1 2.1 Attempt to show that their 3 term quadratic factor is > 0 for all x Complete the square or consider discriminant Could be multiple or fraction of their quadraticH240/01 Mark Scheme November 2020 15 Question Answer Marks AO Guidance A1 2.4 Full justification that quadratic factor is always positive Show minimum point > 0, or show that quadratic is always positive Gradient e2x(6x2 + 6x + 4) > 0 for all x so it is increasing for all x A1 2.4 Justify increasing function as f ( ) 0 ′ x > for all x [4] OR B1 e2x ≠ 0 M1 Show that quadratic ≠ 0 (detail required) M1 Show gradient is positive at one point, as part of attempt to show f ( ) ′ x ≠ 0 A1 Conclude that gradient must hence be positive for all x, so increasing function 9 (a) (1.5, 0) B1 1.1 Allow x = 1.5 Unless contradicted with non-zero y-coord (0, 3) B1 1.1 Allow y = 3 Unless contradicted with non-zero x-coord [2] (b) (i) a < 2 B1 3.1a Allow for answer of form k < a < 2H240/01 Mark Scheme November 2020 16 Question Answer Marks AO Guidance 0 = 1.5a + 2 M1 3.1a Attempt to find value of a at their x intersection eg Use equation of line to find a Use gradient of line to find a Use a point of intersection of the two lines = their 1.5 Equate two points of intersection and solve for a Square both sides and link discriminant to 0 43 a = − A1 1.1 Obtain 4 − 3 (condone any inequality sign, an equals sign or no sign) Question is ‘determine’ so method required for this value of a 43 − < < a 2 A1 1.1 Correct final inequality Formal set notation not required [4] (ii) 2x – 3 = ax + 2 5 2 x a = − B1 1.1 Correct point of intersection – allow any exact equiv OR M1 – square both sides and attempt to solve – as far as substituting into quadratic formula A1 A1 for each root 3 – 2x = ax + 2 (2 + a)x = 1 M1 1.1a Attempt to solve linear equation with 2x and ax of different signs Method may be seen in (i), only credit if answers seen in (ii) 1 2 x a = + A1 1.1 Correct point of intersection – allow any exact equiv Max of 2 out of 3 if additional roots as well. [3] 10 (a) 0.25{sin 0 sin 0.25 sin 0.5 + + ( 1 1 2 2 ) ( ) + sin 0.75 } ( 12 ) M1 1.2 Attempt four rectangles of width 0.25, with height on left-hand side No need to see sin0 Allow M1 if evaluated in degrees (0.00452) Lower bound = 0.253 A1 1.1 Obtain 0.253, or better soi as lower boundH240/01 Mark Scheme November 2020 17 Question Answer Marks AO Guidance 0.25{sin 0.25 sin 0.5 sin 0.75 ( 1 1 1 2 2 2 ) + + ( ) ( ) 12 +sin } M1 1.1a Attempt rectangles of width 0.25, with height on right-hand side Or subtract sin0 from part (ii) and add sin0.5 Allow M1 if evaluated in degrees (0.00670) M0 if sin0 explicitly included Upper bound= 0.373 or 0.374 A1 1.1 Obtain 0.373 or 0.374 (from rounding upper bound up), or better soi as upper bound [4] (b) (i) t x 2 = −1 2tdt = dx M1 2.1 Attempt to link dt and dx A1 2.1 Obtain correct equation linking dt and dx Allow for d 1 d t x x = − 12 ( )− 12 oe ∫ ∫ sin 1 d sin 2 d ( 1 1 2 2 x x t t t − = ) ( ) = 1 ∫2 sin( )d t t t 2 A.G. A1 2.5 Attempt integrand in terms of t to obtain given answer Award A1 once all elements are correct, even if not in same order as given answer BOD if no brackets [3] (b) (ii) − + 4 cos 4cos d t t t t ( 1 1 2 2 ) ∫ ( ) M1* 3.1a Attempt integration by parts Correct parts As far as first stage − + 4 cos 8sin t t t ( 1 1 2 2 ) ( ) A1 2.1 Correct integral 1 1 ( 4cos 8sin ) ( 0 0) − + − − + 2 2 M1d* 2.1 Attempt use of limits Using either t or x, but must be consistent Condone no clear use of the lower limit for M1 1 1 8sin 4cos 2 2 − AG A1 [4] 2.4 Obtain given answer Must see some indication that lower limit considered 11 DRH240/01 Mark Scheme November 2020 18 Question Answer Marks AO Guidance (a) (i) x2 + (mx + 2)2 – 10x – 14(mx + 2) + 64 = 0 M1 1.1a Substitute eqn of tangent into eqn of circle Could work backwards, eliminating m to obtain equation of circle x2 + m2x2 + 4mx + 4 – 10x – 14mx – 28 + 64 = 0(m 2 + 1)x2 – 10(m + 1)x + 40 = 0 A.G. A1 1.1 Expand and tidy to given answer, including ‘= 0’ in final answer AG so unsimplified expansion needs to be seen [2] (ii) 100(m + 1)2 – 160(m2 + 1) = 0 M1* 3.1a Use b2 – 4ac = 0 M1 only awarded when ‘= 0’ soi 60m2 – 200m + 60 = 0 A1 1.1 Obtain correct equation Any correct 3 term equation (3m – 1)(m – 3) = 0 m = 3, m = 13 M1d* 1.1a Attempt to solve quadratic DR so method for solving the quadratic must be shown y = 3x + 2 A1 1.1 Obtain correct equation SC B1for correct equation if roots not justified A0 if second equation also given [4] OR (for first 2 marks) M1 – Attempt two equations in m and x (eg use lengths and gradients) and eliminate one variable A1 – correct quadratic in m or x (b) radius = 10 , PC = 5 2 , PA= PB = 2 10 , AB = 4 2 M1 3.1a Attempt (at least 2) useful lengths NB points of intersection are (2, 8) and (6, 4) tan( 1 1 2 2 APB) = A1 1.1 Obtain a correct related trig ratio cos APB = 53 , from cosine rule tan APB 14 1 1 = − M1 3.1a Attempt tan APB DR so need to see use of identity or relevant triangle to find tan APBH240/01 Mark Scheme November 2020 19 Question Answer Marks AO Guidance tan 4 APB = 3 A1 1.1 Obtain 34 From explicit, exact, working [4] OR M2 – attempt 1 m n mn ± ± ± with their values for m and n A1FT – correct 1 m n mn − + for their values of m and n A1 – obtain tan 4 APB = 3 12 3 2 1 20 35 d d 2 3 11 6 x y x y x x x − = ∫ ∫ − − + M1 1.1 Separate variables Correct process to deal with algebraic fractions, with BOD on integral notation f(x) = 2 3 11 6 x x x 3 2 − − + = (x – 3)(2x2 + 3x – 2) M1 3.1a Attempt to factorise cubic Possibly BC, so correct factorised cubic implies M1A1 If incorrect factorised cubic then method must be seen for M1 Allow M1A0 for (x – 3)(x + 2)(x – 0.5) = (x – 3)(x + 2)(2x – 1) A1 1.1 Correct factorised cubic 3 2 20 35 + 2 3 11 6 2 3 2 1 x A B C x x x x x x − = + − − + + − − M1 1.1a Attempt partial fractions, using their 3 linear factors Must be correct structure, attempting at least one numerator 3 1 4 + x x x 2 3 2 1 = − + + − − A1 1.1 Obtain any one correct fraction www Possibly implied by eg A = –3 A1 1.1 Obtain fully correct partial fractions Could be implied by A = –3 etc, if subsequent slip when writing out partial fractions 1 d ln y y y ∫ = B1 1.1 Correct integration of 1y Condone no modulus signH240/01 Mark Scheme November 2020 20 Question Answer Marks AO Guidance − + + − + − + 3ln 2 ln 3 2ln 2 1 ln x x x A A1FT 1.1 Obtain correct integral following their 3 linear partial fractions Condone no constant of integration Condone brackets and not modulus FT from point that partial fractions were credited, and not on subsequent errors 2 3 ( 3)(2 1) ( 2) A x x y x − − = + A1 1.1 Obtain correct equation Any correct form not involving ln May be e c not A, but A0 if fraction +c Could have (x + 2)-3 in a product [9]OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]

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