GCE Mathematics B (MEI) H640/01: Pure Mathematics and Mechanics Advanced GCE Mark Scheme for Autumn 2021

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GCE Mathematics B (MEI) H640/01: Pure Mathematics and Mechanics Advanced GCE Mark Scheme for Autumn 2021 Oxford Cambridge and RSA Examinations GCE Mathematics B (MEI) H640/01: Pure Mathemati ... cs and Mechanics Advanced GCE Mark Scheme for Autumn 2021Oxford 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 2021H640/01 Mark Scheme November 2021 1. Annotations and abbreviations Annotation in scoris 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 E Explanation mark 1 SC Special case ^ Omission sign MR Misread BP Blank page Highlighting Other abbreviations in mark scheme Meaning E1 Mark for explaining a result or establishing a given result dep* Mark dependent on a previous mark, indicated by *. The * may be omitted if only 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 indicates that the instruction In this question you must show detailed reasoning appears in the question.H640/01 Mark Scheme November 2021 2. Subject-specific Marking Instructions for AS/A Level Further Mathematics B (MEI) 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.H640/01 Mark Scheme November 2021 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. E A given result is to be established or a result has to be explained. 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. 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.H640/01 Mark Scheme November 2021 f Unless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct and expressed 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 the lengths are in km, when this would be assumed to be the unspecified unit.) 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. • When a value is not given in the paper accept any answer that agrees with the correct value to 2 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 A the rubric specifies 3 s.f. as standard, so this statement reads “3 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. E marks are lost unless, by chance, the given results are established by equivalent working. 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” and “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.H640/01 Mark Scheme November 2021 Question Answer Marks AOs Guidance 1 For example ( 3) 9 2 4 − = > = 2 2 and ( 3) 2 − < So Beth is not correct M1 E1 [2] 2.1 2.2a Stating any pair of numbers where 2 2 p q > and p q < Fully convincing argument – do not allow for only disproving the converse Also accept general statement about a negative number [for p] 2 Mass of craft is 5200 [530.612...] 9.8 = kg Weight on Mars is 3.7 mass 3.7 530.6122 × = × =1963 N (1960 to 3s.f.) M1 M1 A1 [3] 3.1b 1.1b 1.1b Attempt to find the mass Use of ‘mg’ with Mars gravity Cao 3 (a) y x > + 5 y x x ≤ − − 8 2 2 B1 B1 [2] 2.5 2.5 Allow interchange of > and ≥ or < and ≤ for one inequality as long as the direction is correct Both inequalities fully correct. Allow x y x x + < ≤ − − 5 8 2 2 oe 3 (b) DR Boundary values when 8 2 5 − − = + x x x 2 x x 2 + − = 3 3 0 giving x = 3 21 2 − ± From the graph, the line lies below the curve for 3 21 3 21 2 2 x − − − + < < or 3 21 3 21 : : 2 2 x x x x     − −     − +   > ∩ <           M1 A1 B1 [3] 2.1 2.1 2.1 A correct three term quadratic equation (or inequality) must be seen Correct roots of the equation soi Must be exact Shows correct inequality FT their roots Any method including BC is acceptable for solving the quadratic equation clearly seen in the form 2 ax bx c + + = 0 . Allow B1 for − < < 3.79 0.79 x wwwH640/01 Mark Scheme November 2021 Question Answer Marks AOs Guidance 4 (a) (i) Sequence is periodic [with period 4] B1 [1] 1.2 Do not allow repeating, recurring etc. The sequence can also be described as oscillating 4 (a) (ii) Total of 200 terms is 50 (2 3 0 3) 400 × + + + = B1 [1] 1.1b cao 4 (b) (i) Sequence divergent for Either b >1 or b ≤ −1 B1 B1 [2] 1.1b 1.1b Allow for one correct inequality Must have “or” or the union of sets Condone b < −1 or b >1 Note for b =1 , the sequence is convergent, but the corresponding series is divergent 4 (b) (ii) Infinite sum of geometric series with a r = = 1 1 3 3 , 13 13 1 1 1 2 a S r = = = − − M1 A1 [2] 1.1b 1.1b Using the sum of geometric series with 13 r = www 5 (a) F = 50 Take moments about A (or D) Fx = × + × 20 30 20 10 OR Take moments about B (or C) F x (30 20 10 30 30 − + × = × ) 800 16 50 x = = B1 M1 A1 [3] 1.1b 1.1a 1.1b cao Forming an equation for moments about any point –allow their value of F used. Allow one incorrect distance or a missing term Allow 0.16 m if unit given Do not allow for moment = force 5 (b) Forces have a resultant moment which have a clockwise turning effect on the lamina E1 E1 [2] 2.4 2.4 Allow ‘resultant moment’ or ‘not in equilibrium’ soi Clockwise must be indicated Do not allow for ‘a resultant force’ on its ownH640/01 Mark Scheme November 2021 Question Answer Marks AOs Guidance 6 (a) Asymptotes x = 0 (or y-axis) x = π and x = 2π B1 [1] 1.2 Must have all three 6 (b) When , 2 3 3 3 x y π = = When 3 x π = , d 2 cosec cot d 3 3 3 yx π π = − = − Equation of the tangent is 2 3 2 3 3 3 y x   π − = − −     When y = 0 , 2 3 2 3 3 3 x   π − = − −     giving 3 3 x π = + (AG) B1 M1 A1 M1 M1 A1 [6] 1.1b 3.1a 1.1b 2.1 2.1 2.1 soi; any exact form eg 2 3 Uses the derivative when 3 x π = May be embedded in the tangent equation Uses both their coordinates and their gradient to find the equation of the tangent Substituting y = 0 into their tangent Working must be correct and exact throughout If 2 3 y x c = − + used, there must be an attempt to find c using both their coordinates and their gradientH640/01 Mark Scheme November 2021 Question Answer Marks AOs Guidance 7 (a) DR Midpoint of AB is (3, 1) Centre C of the circle is (3, 1) and radius (7 3 2 1 5 − + − − = )2 2 ( ) So circle is (x y − + − = 3 1 25 )2 2 ( ) B1 M1 M1 A1 [4] 3.1a 3.1a 1.1b 1.1b soi Attempt to find length of AB, AC or BC Uses their midpoint and radius (do not allow for diameter used) Need not be simplified 7 (b) DR Crosses y x = + 2 5 where (x x − + + − = 3 2 5 1 25 )2 2 ( ) 2 5 10 0 x x + = giving x = − 2, 0 So points are (–2, 1) and (0, 5) M1 A1 A1 [3] 1.1b 1.1b 1.1b Substituting y x = + 2 5 and attempting to collect terms oe Both values correct Correct y coordinates FT their xcoordinates Allow for a quadratic solved BC providing it is seen in form 2 ax bx + = 0 or 2 ay by c + + = 0 7 (c) DR AQ = 2 and BQ = 7 7 7 2 2 2 + = Triangle ABQ has a right angle at Q (angle in a semicircle) So area of triangle is 12 × AQ×BQ Area = 7 M1 M1 A1 [3] 3.1a 2.1 1.1b Attempt to find two lengths to be used in their area calculation (excluding AB) Correct method for finding the area FT their Q Note QAB=81.9° and QBA=8.1°H640/01 Mark Scheme November 2021 Question Answer Marks AOs Guidance 8 (a) Karim has a valid argument that there is a root between 5 and 6 because there is a change of sign on his table E1 [1] 2.3 Argues from change of sign that this argument is valid Allow argument is not valid as he does not state that the function is continuous 8 (b) There are two roots between 2 and 3 (and/or between 4 and 5) so there is no change of sign in the table E1 [1] 2.3 Allow for a comment that implies changes of sign are missed 8 (c) (i) f 4cos 4 e ′( x x ) = − −x So N-R formula is 1 sin 4 e 0.75 4cos 4 e n n x n n n x n x x x x − + − + + = − − M1 A1 [2] 1.1b 1.1b Attempt to differentiate cao, but condone missing subscripts in the fraction 8 (c) (ii) The root is 2.908 to 4 s.f. x0 3 x1 2.920853… x2 2.908274… x3 2.907846… M1 A1 A1 [3] 1.1b 1.1b 2.2a Produces at least two iterations Three iterations with correct values either rounded or truncated to at least 3 decimal places Correct to at least 3 s.f. FT their values if their sequence seems to converge (root is 2.907845109 to 10 sf) 8 (d) (i) B1 B1 [2] 1.1b 1.1b Attempt to draw a tangent at x = 5 as far as the x-axis Drawing the second tangent approximately at the point where x = 3.97 as far as the x-axis. 8 (d) (ii) The start value is close to a stationary point, [so the gradient is very small] and the tangent meets the x-axis far away from the required root The sequence converges to a root, but not the required root B1 B1 [2] 2.4 2.4 Conveys the idea that the stationary point or the value of the gradient causes the problem Conveys the idea that the wrong root is found 8 (d) (iii) Use x0 any value [between 5.28 and 5.85] which is nearer to the required root. E1 [1] 2.4 Allow for a ‘starting value between 5 and 6’ oeH640/01 Mark Scheme November 2021 Question Answer Marks AOs Guidance 9 (a) (i) Total mass 390 g (0.390 kg) 12 4 2.5 0.390 − × = a M1 A1 [2] 3.3 3.3 Newton’s second law; condone missing or incorrect resistance or incorrect total mass Fully correct equation need not be simplified Allow for T – R = ma if the correct substitution of values seen in part 9aii 9 (a) (ii) 5.13 m s–2 B1 [1] 1.1b cao 9 (b) B1 B1 B1 [3] 1.1b 1.1b 1.1b Correct vertical forces 12 N and 2.5 N correctly drawn and labelled with no extra forces Tension in the ribbon correctly drawn and labelled Allow ‘weight’ or 120g N oe for 0.12g N Allow ‘resistance’ for 2.5 N and ‘tension’ for 12 N providing it is clear that the tension in the string is distinct from the tension in the ribbon 9 (c) N2L for the head only 12 2.5 0.12 − − = T a OR N2L for the three body sections together T a − × = 3 2.5 0.270 T = 8.88 N M1 A1 [2] 1.1b 1.1b N2L with correct mass and their acceleration. Allow missing or incorrect resistance cao Allow for mass in grams if same error seen in part (a) Allow A1 for correct answer from consistent use of grams Tension R [in the ribbon] 2.5 N 12 N 0.12g NH640/01 Mark Scheme November 2021 Question Answer Marks AOs Guidance 10 (a) Time when v = 0 given by 0 29.4 9.8 , = − t so t = 3 s E1 [1] 2.1 Using suvat equation(s) leading to correct value for t with v = 0 Allow for verifying that t = 3 gives v = 0 if identified as the maximum point oe 10 (b) B1 B1 [2] 1.1b 1.1b straight line with negative gradient through either (3, 0) or (0, 29.4) Both (3, 0) and (0, 29.4) clearly seen Must include negative values of v for t > 3 10 (c) When t = 5, v = − × 29.4 9.8 5 v = −19.6 Speed is 19.6 m s–1 M1 A1 A1 [3] 1.1a 1.1b 1.1b Using suvat equation(s) leading to a value for v with t = 5. Allow sign errors May be implied by 19.6 seen FT their negative velocity If motion from the highest point considered u = 0, t = 2, g = +9.8 then v = 19.6 is fully correct. Allow M1A1A0 if 29.4 9.8 5 19.6 − × = seen 10 (d) Max height unchanged so uy = 29.4 m s–1 B1 3.1b Allow if calculated from y = 44.1 m Time to max height unchanged, so 3 s B1 [2] 3.3 10 (e) ux × = 3 48 2 2 2 2 16 29.4 33.5 u u u = + = + = x y 29.4 tan 16 y x u u α = = giving α = ° 61.4 M1 M1 A1 [3] 1.1a 1.1b 1.1b Using (their) t = 3 to find ux Combining their components to find either one of u and α Both values correctH640/01 Mark Scheme November 2021 Question Answer Marks AOs Guidance 11 (a) d d V k t x = When 0, 5 and 21, d Vd t x t = = = so d 105 Vd t x = d 2 4 Vd x x = π , so the chain rule gives d d d d 4 2 d d d d V V x x x t x t t = × = π Hence 2 3 d 1 105 105 d 4 4 x t x π π x x = × = AG M1 A1 B1 M1 A1 [5] 3.3 3.3 2.1 2.1 3.3 Expresses inverse proportionality with a constant Evaluating k, oe (may be done later) may be embedded in chain rule Use of the chain rule Convincing argument 11 (b) ∫ ∫ 4 d 105 d π x x t 3 = 4 π x t c = + 105 When t x = = 0, 5 so c = 625π When t = 120 x 4 105 120 625 8.25 π = × + = cm M1 A1 M1 A1 A1 [5] 3.1a 1.1b 3.3 3.3 3.4 Separating the variables Condone missing +c here Using initial conditions Correct value for c cao 11 (c) As t gets very large, the volume gets very large so the balloon will get beyond the maximum it can be without bursting and so burst. E1 [1] 3.5b Conveys the idea that t V → ∞ ⇒ → ∞ or x → ∞ Indicates a practical problem with very large volumeH640/01 Mark Scheme November 2021 Question Answer Marks AOs Guidance 12 Normal reaction mg cos15° Max friction µN mg = 0.4 cos15° Resolve down the slope mg F ma sin15° − = mg mg ma sin15 0.4 cos15 ° − ° = giving a = −1.25 m s–2 Using v u as 2 2 = + 2 0 1.2 2 1.25 2 2 = + × − ( ) s giving s = 0.576 m B1 M1 B1 M1 A1 M1 A1 [7] 3.1b 1.1b 1.1b 3.1b 1.1b 3.1b 1.1b Correct normal reaction Attempt to evaluate friction FT their normal reaction. Correct component of weight seen (2.536m) All terms present; allow sign errors, sin/cos interchange for weight and their F Correct equation (a need not be explicitly evaluated here) Use of suvat equation(s) leading to a value for s using v = 0 FT their negative a Only allow 0.4mg if it is clear that mg is their normal reaction and not just weightH640/01 Mark Scheme November 2021 Question Answer Marks AOs Guidance 13 (a) ( ) ( 2 ) 3 d 3 7 d 7 t t t t t = = + = + + r v i j ∫ ∫ i j c When t = = − + = 0, 2 r i j c So r i j = − + + (t t 3 1 7 2 ) ( ) When t = 2, r i j i j = − + × + = + (2 1 7 2 2 7 16 3 ) ( ) distance = 7 16 305 2 2 + = distance = 17.5 m M1* M1 dep* M1 dep* A1 M1 A1 [6] 3.4 1.1b 3.1b 1.1b 3.1b 3.2a Attempt to integrate both components; condone missing +c Using initial conditions Using t = 2 to find position vector or values for x and y Accept vector form or two clear components. Using Pythagoras FT their components 13 (b) using x t y t = − = + 3 1, 7 2 Substitute 2 7 y t − = into equation for x 3 2 1 7 y x   − = −     AG M1 M1 dep A1 [3] 3.1a 1.1b 1.1b Extracting equations for x and y from their displacement vector Attempt to eliminate t cao Equivalent form y x = + + 7 1 2 ( )13 for M1M1A0 13 (c) d 6 d t t = = v a i When t = = 2, 12 a i The force must be in that direction, so F i a = = 48 m So m = 4 kg M1* M1* M1* A1 dep* [4] 3.1b 3.4 3.1b 1.1b Must be vector acceleration Evaluating when t = 2 Newton’s second law in vector form, or in x-direction only cao a =12 is sufficient here If their a has two nonzero components, allow for dividing 48 by the magnitude of their aOCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]

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