PMT GCE. Mathematics (MEI) Advanced GCE Unit 4753: Methods for Advanced Mathematics. Mark Scheme for June Oxford Cambridge and RSA Examinations

Transcription

1 GCE Mathematics (MEI) Advanced GCE Unit 4753: Methods for Advanced Mathematics Mark Scheme for June 0 Oxford Cambridge and RSA Examinations

2 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, OCR Nationals, 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 will not enter into any discussion or correspondence in connection with this mark scheme. OCR 0 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG5 0DL Telephone: Facsimile: publications@ocr.org.uk

3 4753 Mark Scheme June 0 Annotations and abbreviations Annotation in scoris Meaning and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, Method mark awarded 0, A0, Accuracy mark awarded 0, B0, B Independent mark awarded 0, SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations in mark scheme E U G Meaning Mark for explaining Mark for correct units Mark for a correct feature on a graph dep* Method mark dependent on a previous mark, indicated by * cao oe rot soi www Correct answer only Or equivalent Rounded or truncated Seen or implied Without wrong working

4 4753 Mark Scheme June 0 Subject-specific Marking Instructions for GCE Mathematics (MEI) Pure strand a. Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses 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. 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 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, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader. 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, eg 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 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 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks.

5 4753 Mark Scheme June 0 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, eg 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, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. 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. Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. Candidates are expected to give numerical answers to an appropriate degree of accuracy, with 3 significant figures often being the norm. Small variations in the degree of accuracy to which an answer is given (e.g. or 4 significant figures where 3 is expected) should not normally be penalised, while answers which are grossly over- or under-specified should normally result in the loss of a mark. The situation regarding any particular cases where the accuracy of the answer may be a marking issue should be detailed in the mark scheme rationale. If in doubt, contact your Team Leader. g. Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests. If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook. 3

6 4753 Mark Scheme June 0 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; mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Note that a miscopy of the candidate s own working is not a misread but an accuracy error. 4

7 4753 Mark Scheme June 0 Question Answer Marks Guidance [k (3x ) / ] A k = /3 / d (3 ) 3 x x 3 x = = /3* OR u = 3x du/dx = 3 4 d. d 3 x x u 3 u 4 / 3 u = = /3* x + > 4 x + > 4, x > 3/ or x + < 4, 3 y x < ½ e 5 e x y d y x e e d x x d y e y dx e d y e At (0, ln) dx e 8 0 ln dep dep [5] B [3] B B dep cao [4] substituting limits dep st NB AG u /3 (du) / 3 u o.e. substituting correct limits dep st NB AG x > 3/ mark final ans; if from x > 3 B0 o.e., e.g. (x + ) > 4 (or x + = 4) if x + < 4, M0 x < ½ mark final ans allow 3 for correct unsupported answers y d y e... d x = e x substituting x = 0, y = ln into their dy/dx dep st B allow one slip or w = 3x w /3 w (dw) 3 w upper lower, to 4 for u or to for w or substituting back (correctly) for x and using to by squaring: 4x + 4x 5 > (or =) 0 x > 3/ x < ½ Penalise and once only 3/ < x < ½ SC (mark final ans) or y = ln (5 e x ) o.e (e.g.½ ln(5 e x ) )B dy/dx = e x /[(5 e x )] o.e. B (but must be correct) or substituting x = 0 into their correct dy/dx 5

8 4753 Mark Scheme June 0 Question Answer Marks Guidance 4 (i) 9a = 0 a = /9 a = /3 or 9x = 0 or 0.33 or better (/9) is A0 ( 9a ) = 3a is M0 not a = ± /3 nor x = /3 [] 4 (ii) Range 0 y B [] 4 (iii) 5 (i) When t = 0, P = 7 = 5, so 5 (million) In the long term e kt 0 So long-term population is 7 (million) 5 (ii) P = 7 e kt When t =, P = = 7 e k e k = (7 5.5)/ = 0.75 k = ln((7 5.5)/) k = 0.88 (3 s.f.) [3] B [3] [3] or 0 f(x) or 0 f, not 0 x 0 y is B0 curve goes from x = 3a to x = 3a (or to ) vertex at origin curve, centre (0, ), from (, ) to (, ) (y-coords of can be inferred from vertex at O and correct scaling) allow substituting a large number for t (for both marks) re-arranging and anti-logging allow slip (e.g. arith of 7 5.5, or k for k) or ln k = ln.5 o.e. 0.3 or better allow ln(4/3) or ln(3/4) if final ans allow also [0,], or 0 to inclusive, but not 0 to or (0,) must have evidence of using s.f. 3 allow also if s.f.3 is stated and stretch is reasonably to scale allow from ( 3a, ) to (3a, ) provided a = /3 or x = [±] /3 in (i) A0 for badly inconsistent scale(s) allow 7 unsupported but penalise negative lns, e.g. ln(.5) = ln( ) k rounding from a correct value of k = , penalise truncation, and incorrect work with negatives 6

9 4753 Mark Scheme June 0 Question Answer Marks Guidance 6 (i) y = arc sin ½ = /6 y = arcsin ½ = /3 must be in terms of can isw approximate answers.047 implies [] 6 (ii) y = arcsin x x y x = arcsin y x/ = arcsin y y = sin (x/) [so g(x) = sin (x/)] dy/dx = ½ cos(½ x) At Q, x = /3 dy/dx = ½ cos /6 = ½ 3/ = 3/4 gradient at P = 4/3 7 (i) s(x) = f(x) + g(x) = f(x) + g(x) = (f(x) + g(x)) = s(x) ( so s is odd ) 7 (ii) p(x) = f(x)g(x) =( f(x)) (g(x)) = f(x)g(x) = p(x) so p is even cao B ft [6] [] [] or y/ = arcsin x but must interchange x and y at some stage substituting their /3 into their derivative must be exact, with their cos(/6) evaluated o.e. e.g. 4 3/3 but must be exact ft their 3/4 unless must have s( x) = must have p( x) = Allow SC for showing that p( x) = p(x) using two specific odd functions, but in this case they must still show that p is even or f (x) = / ( x ) f ( ½ ) = / ¾ = 4/ 3 cao e.g. f(x) = x, g(x) = x 3, p(x) = x 4 p( x) = ( x) 4 = x 4 = p(x), so p even condone f and g being the same function 7

10 4753 Mark Scheme June 0 Question Answer Marks Guidance 8 (i) d y sin x xcos x d x d/dx(sin x) = cos x soi cao, mark final answer dy/dx = 0 when sin x + x cos x = 0 equating their derivative to zero, sin x xcos x 0 provided it has two terms cos x tan x + x = 0 * [4] must show evidence of division by cos x 8 (ii) At P, x sin x = 0 sin x = 0, x = (0), x = / x = / At P, dy/dx = sin + (/) cos = Eqn of tangent: y 0 = (x /) y = x + / x +y = * When x = 0, y = /, so Q is (0, /) 8 (iii) Area = triangle OPQ area under curve Triangle OPQ = ½ / / [= 3 /8] Parts: u = x, dv/dx = sin x du/dx =, v = ½ cosx π/ π/ π/ xsin xd x xcos x cos xd x π/ xcos x sin x 4 0 π cosπ sin π( 0cos 0 sin 0) π[ 0] So shaded area = 3 /8 /4 = ( )/8* B ft [7] Bcao ft cao [7] ft their / and their derivative substituting 0, their / and their into y y = m(x x ) NB AG can isw inexact answers from / can be inferred from dy/dx = x cos x e.g. dy/dx = tan x + x is A0 Finding x = / using the given line equation is M0 or their into y = mx+c, and then evaluating c: y = ( )x + c, 0 = ( ) (/) + c c = / y = x + / x + y = * soi (or area under PQ area under curve area under line may be expressed in integral form allow art 3.9 or using integral: condone v = k cosx soi ft their v = ½ cosx, ignore limits [ ½ x cos x + ¼ sin x] o.e., must be correct at this stage, ignore limits (so dep previous ) NB AG must be from fully correct work π/ π/ π π π π πx x π x πx ( )d [ ] v can be inferred from their uv 8

11 4753 Mark Scheme June 0 Question Answer Marks Guidance 9 (i) (A) (0, 6) and (, 4) (B) (, 5) and (0, 4) BB BB Condone P and Q incorrectly labelled (or unlabelled) 9 (ii) ( x). x( x 3). f'( x) ( x ) 9 (iii) 9 (iv) f(x) = 0 x (x + ) (x + 3) = 0 x + x 3 = 0 (x )(x + 3) = 0 x = or x = 3 When x = 3, y = /( ) = 6 so other TP is ( 3, 6) ( x ) 3 f(x ) = x = x x3 x x x4 4 x * x x b 4 ( x )dx x x 4lnx a x a ( b b4ln b) ( a a4ln a ) Area is f( x )d x 0 So taking a = and b = area = ( 4 + 4ln ) ( ½ + 4ln ) = 4 ln ½ b [4] dep BBcao [6] [3] B cao [5] Quotient or product rule consistent with their derivatives, condone missing brackets correct expression their derivative = 0 obtaining correct quadratic equation (soi) dep st but withhold if denominator also set to zero must be from correct work (but see note re quadratic) substituting x for both x s in f NB AG 4ln x x x F(b) F(a) condone missing brackets oe (mark final answer) must be simplified with ln = 0 PR: (x +3)( )(x+) + x(x+) If formula stated correctly, allow one substitution error. condone missing brackets if subsequent working implies they are intended Some candidates get x + x + 3, then realise this should be x + x 3, and correct back, but not for every occurrence. Treat this sympathetically. Must be supported, but 3 could be verified by substitution into correct derivative allow slip for F must show evidence of integration of at least one term or f(x) = x + + 4/(x+) A = f( x)d x x x 4ln( x) 0 0 = ½ + 4ln = 4ln ½ 9