GCE Further Mathematics B (MEI) Y435/01: Extra pure Advanced GCE Mark Scheme for November 2020

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Oxford Cambridge and RSA Examinations GCE Further Mathematics B (MEI) Y435/01: Extra pure Advanced GCE Mark Scheme for November 2020Oxford Cambridge and RSA Examinations OCR (Oxford Cambridge an... d 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 2020Y435/01 Mark Scheme November 2020 2 Text Instructions 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.Y435/01 Mark Scheme November 2020 3 Subject-specific Marking Instructions for AS Level Mathematics B (MEI) 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.Y435/01 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. 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 suchY435/01 Mark Scheme November 2020 5 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 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.Y435/01 Mark Scheme November 2020 6 Question Answer Marks AOs Guidance 1 2 det( ) (1 ) 6 3 1 λ λ λ λ λ − − = = + − − − A I M1 1.1a For ch eqn in any form Can be implied by correct evals Allow one sign error λ2 + λ – 6 [= 0 ] So the eigenvalues are 2 and –3 A1 1.1 For both e-vals correct 0 2 2 2 2 : 3 1 3 2 1 1 x y x e y x y y x y        =        = =        − −   ⇒ = ⇒     M1 A1 1.1 1.1 Either equation correct in any form FT Or any non-zero multiple 0 2 2 3 3: 3 1 3 3 2 3 2 3 x y x e y x y y x y        − = − = =               − − −   − ⇒ = − ⇒     A1 1.1 Or any non-zero multiple If each e-vec is not paired with its e-val (either explicitly or in the working) or if they are wrongly assigned then SC1 if they are both correct [5] 2 1 8 (1 )! a t b = = + soi B1 3.1a Using the initial condition to obtain an equation in a and b 1 3 ( 1 )! ( 3)( )! n n t a a t + = ⇒ = n n b n n b + + + + + M1 3.1a Substituting solution formula into recurrence relation so we need n + 3 = n + 1 + b M1 1.1 Cancelling a and (n + b)! => b = 2 A1 1.1 and a = 8×3! = 48 � tn = ( 2)! n48 + � A1 [5] 3.2aY435/01 Mark Scheme November 2020 7 2 Alternative Method: 1 8 (1 )! a t b = = + �2 = � (2+�)(1+�)! = 2 Solving to give a = 48, b = 2 � �+1 = 48 (� + 2)!(� + 3) � �+1 = 48 (� + 3)! M1 A1A1 M1 A1 [5] 3.1a 3.1a 1.1 3.2a Attempting to find two terms eg t1 and t2 or recognising a general pattern Using ��+1 = �� �+3 to verify solution CompletionY435/01 Mark Scheme November 2020 8 3 (a) un = kpn ⇒ p2 – 4p + 5 [= 0] M1 1.1 Auxiliary equation One sign error => p = 2 ± i M1 1.1 BC. Solving their auxiliary equation r = = 5, tan 0.5 θ soi M1 1.1 Finding mod/arg of at least one of their roots Could be seen later un = A(2 + i)n + B(2 – i)n M1 1.1 General solution in any form (can be implied by correct real form) FT un = rn(αcos(nθ) + βsin(nθ)) soi A1 1.1 General solution in real form with r and θ either specified or in situ FT their A and B. r could be seen as eg 2.24 and θ as eg 0.46 or 26.6 here eg n = 0 [ ⇒ α = 0] M1 1.1 Substituting either initial condition into their GS n = 1 => β = 1 so 5 sin 2 n u n n = θ oe A1 1.1 allow eg u n n = 5 sin tan 2 n ( −1( 12)) One sign error Could be seen later FT their A and B. r could be seen as eg 2.24 and θ as eg 0.46 or 26.6 here Alternative Method un = kpn ⇒ p2 – 4p + 5 = 0 => p = 2 ± i r = = 5, tan 0.5 θ soi un = A(2 + i)n + B(2 – i)n 0 = A + B ⇒ A = -B A(2 + i)n + B(2 – i)n = 1 ⇒ A = -½i, B = ½i � � = − 1 2 �√5�(���� + �����)� + 1 2 �√5�(���� − �����)� n25 sin u n n = θ oe [7] M1 M1 M1 M1 M1 A1 A1 [7] 1.1 1.1 1.1 1.1 1.1 1.1 1.1 Auxiliary equation BC. Solving their auxiliary equation Finding mod/arg of at least one root General solution in any form (can be implied by correct real form) Using one initial condition to find an arbitrary constant Solution given in mod/arg form (Could also see ���)Y435/01 Mark Scheme November 2020 9 3 (b) If a = 0.1 then vn converges to 0 as as n→∞. B1 2.5 No need to mention oscillatory but must give the limit Diagrams only not sufficient for all three cases If a = 0.2 then vn [does not converge...] B1 2.2b Not “diverges” ...and is bounded and oscillatory. B1 2.2b Allow descriptions (eg “the sign changes regularly” or “it goes positive and negative” and “it is bounded or “always between –1 and 1”). Ignore “periodic” B1 bounded B1 oscillatory If a = 1 then vn diverges... B1 2.2b Allow eg “the terms get bigger (in size)”. ...and is oscillatory. B1 2.2b Allow descriptions (eg “the sign changes regularly” or “it goes positive and negative”). [5]Y435/01 Mark Scheme November 2020 10 4 (a) (i) 2n + 1 + 2m + 1 = 2(n + m + 1) so not closed B1 2.1 Must show some working [0 ∉ G so] no identity. B1 2.2a Could be seen with next B1 Since no identity the inverse property cannot be satisfied. B1 2.2a Cannot gain this B1 without previous B1 [3] 4 (a) (ii) ( 2)( 2) 2 ( ) 2 a b c d ac bd bc ad + + = + + + so closed B1 2.1 Must show some working Elements must be general and distinct a = 1, b = 0 ⇒ 1 [∈ G] so identity exists B1 2.2a 1 must be seen eg 1 1 1 2 2 2 2 = − ∉G + so inverse property not satisfied B1 2.2a Single numerical counter example is sufficient or 0−1 ∉ � For �2−�2�2 + �2�−√22�2 need to justify answer eg � �2−2�2 is not always in G [3] 4 (a) (iii) a b ab , ∈ ⇒ ∈ R R so closed 1 ∈ ℝ ��� � × 1 = 1 × � = � ∈ ℝ so identity exists B1 2.1 Need justification B1 2.2a 1 must be seen 0−1 ∉ ℝ so inverse property not satisfied B1 2.2a [3] 4 (b) (i) 1 0 0 1   −     − B1 3.1a 1 1 i 2 i 1   − −     − − B1 1.1 0 i i 0   −     − B1 1.1 [3] 4 (b) (ii) They are not isomorphic because M contains only one element of order 2 while N is known to contain at least 3. B1 2.4 Or other valid reason (eg M is cyclic while N is not since it requires more than 1 element to generate it) [1]Y435/01 Mark Scheme November 2020 11 5 (a) 1 1 3 3 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 a a b c b a b c c a b c  − − − −      − − = − + −           − −    − − +  M1 3.1a Finding vector A.f e.f = 0 ⇒ a + b + c = 0 M1 3.1a Using perpendicularity condition to find a relationship between a, b and c 13 2 2 2( ) 2 2( ) 2 b c b c b c b c b c b b c b c c  − − − − − −    ∴ = + + − =          + − +    A.f M1 1.1 Eliminating a, b or c consistently in all 3 components to derive A.f in two unknowns or eliminating b or c in x, a or c in y and a or b in z. soi abc   ∴ = =       A.f f so f is an e-vec of A A1 [4] 3.2a Completing substitution and correct conclusion Alternative method: 1 0 1 1 0 1 λ µ     = − +             − f M1 3.1a Expressing a general f in terms of two non-parallel vectors which are both perpendicular to e Showing that a specific perpendicular vector is an e-vec SC2 or M1SC1 13 1 1 3 3 1 2 2 1 0 2 1 2 1 1 2 2 1 0 1 1 2 2 1 1 2 2 0 2 1 2 1 2 1 2 1 2 2 1 0 2 2 1 1 λ µ λ µ       − −           ∴ = − − − + =                       − −   −  − −    − −           − − − + − −            − −    − − −   A.f M1 1.1 Opening brackets 1 1 3 3 3 0 3 3 0 3 λ µ         = − +             − M1 3.1a Multiplying vectors into matrixY435/01 Mark Scheme November 2020 12 1 0 1 1 0 1 λ µ         ∴ = − + =             − A.f f so f is an e-vec of A A1 3.2a Completing and correct conclusion Alternative Method 2: To find e-vals put 1 2 2 3 3 3 2 1 2 3 2 3 3 3 2 2 1 3 3 3 0 1 0 λ λ λ λ λ λ   − − −     − − − = ⇒ − − + =     − − − (λ = –1 gives e) so consider λ = 1: 1 2 2 3 3 3 2 1 2 3 3 3 2 2 1 3 3 3 1 2 2 3 3 3 1 2 1 2 3 3 3 3 2 2 1 3 3 3 If then we need 1 2 2 1 or 2 1 2 or 1 2 2 1 1 a a a b b b c c c a a a b b b c c c         − −         =     − − =           − −       − − −           − −           − − = − − − =         − − −   − −     f 0 ⟹ � + � + � = 0 But � = �� ������ � = �1 1 1�so � + � + � = 0⟹ e.f =0 ⟹ f must be perpendicular to e [4] M1 M1 M1 A1 3.1a 1.1 3.1a 3.2a [4] For attempt at ch eqn eg ���|� − ��| seen 5 (b) λf = 1 B1 2.2aY435/01 Mark Scheme November 2020 13 5 (c) Since the e-val of any vector f is 1 then f must be parallel to (or lie in) the mirror plane. B1 2.4 SC1 using the word line instead of plane Since the e-val of e is –1 then e must be perpendicular to the mirror plane. B1 2.4 Needs more than invariant line/line of invariant points [2] 5 (d) Since e is the normal to the mirror plane and O must be in the plane the equation is 1 0 1 1 0 1 0 0 1 0 1 x y z             = = ⇒ + + =             r. . B1 3.1a [1]Y435/01 Mark Scheme November 2020 14 6 (a) (i) f 3 2 16 34 x xy x ∂ = − ∂ f 3 2 16 34 y x y y ∂ = − ∂ B1 1.1 for both 16x3 – 34xy2 = 0 and 16y3 – 34x2y = 0 M1 1.1 both So x = 0 or 16x2 – 34y2 = 0 (or equivalent for f 0 y ∂ = ∂ ) M1 1.1 For 16�M0 2 −here 34�SC1 2 = 0for and subs x2 or y2 into the other equation But both x = 0 and 16�2 − 34�2 = 0 ⇒ y = 0 when substituted into the other equation so x = 0 and y = 0 [is the only solution]. M1 A1 1.1 1.1 For M0 here SC1 for x = y = 0 only [5] 6 (a) (ii) (s =) 0 B1 1.1 [1] 6 (a) (iii) 4x4 + 4y4 – 17x2y2 = (4x2 – y2)(x2 – 4y2) M1 1.1 (2x – y)(2x + y)(x – 2y)(x + 2y) or y = (+/-)2x and y = (+/-) ½x A1 1.1 A1 1.1 Sketch of the four complete (ie each line going across two quadrants) lines y = ±2x and y = ±½x. No scale necessary. [3] 6 (a) (iv) The z = 0 plane is divided into positive and negative ‘wedges’ so it is not the case that z > 0 at all points near P (the stationary point) so it is not a minimum and similarly it is not the case that z < 0 at all points near P so it is not a maximum. So P must be a saddle point. B1 2.4 or equivalent explanation eg moving eg along x-axis, through P, z is +ve, 0, +ve while along eg y = x, through P, z is –ve, 0, –ve or z is positive on the negative xaxis and negative on the positive branch of y = x etc SC1 No appeal to diagram but correctly finding two z coordinates, one positive and one negative and stating that there are no other SPs No FT for 0/3 in 6 (a) (iii)Y435/01 Mark Scheme November 2020 15 6 (b) (i) 3 2 3 3 2 3 16 34 18 16 34 18 1 1 a a a a a a a a  − × −        = − × = −          − −    n M1 1.1 or any non-zero multiple FT from (a)(i) 3 3 4 4 2 2 18 18 4 4 17 1 a a p a a a a a a     −     =     −     + − × −     . M1 1.1 their n 3 3 4 18 18 27 a 1 a a       =     r. oe A1 1.1 cao isw 6 (b) (ii) 27a4 d = n soi M1 3.1a FT their (i) n = + ≈ 2 18 1 18 2 ( a a 3 3 )2 for large a M1 3.1a FT their (i) Or equivalent argument using limits No limits discussion SC1 4 3 1 27 3 2 18 2 4 d a a a a ∴ → × = A1 3.2a AG [3] 6 (b) (iii) Below. If x = y = 0 in equation for Π then z = 27a4 > 0 so the z-intercept is positive so the origin is below the plane. E1 2.4 Or in the equation 3 3 4 18 18 27 a 1 a a       =     r. the z component of n is positive and so n is pointing upward but p > 0 so O is on the other side of the plane ie below (or equivalent argument with –ve signs). Needs more than z > 0 [1]OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA OCR Customer Contact Centre [Show More]

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