GCE Mathematics (MEI) Unit 757: Further Applications of Advanced Mathematics Advanced GCE Mark Scheme for June 015 Oxford 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 will not enter into any discussion or correspondence in connection with this mark scheme. OCR 015
757 Mark Scheme June 015 Annotations and abbreviations Annotation in scoris Meaning and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, Method mark awarded 0, 1 A0, Accuracy mark awarded 0, 1 B0, Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations in Meaning mark scheme E1 Mark for explaining U1 Mark for correct units G1 Mark for a correct feature on a graph dep* Method mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working
757 Mark Scheme June 015 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. 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.
757 Mark Scheme June 015 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 e 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. 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 g 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 significant figures often being the norm. Small variations in the degree of accuracy to which an answer is given (e.g. or significant figures where 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. 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. 5
757 Mark Scheme June 015 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 mark in the question. Note that a miscopy of the candidate s own working is not a misread but an accuracy error. 6
757 Mark Scheme June 015 1 (i) 8 Any point is 5 15 5 8, 5 15, 5 6, 0 15, 9 5 AB 8, 5 15, 5 (,5, ) Distance AB = 6 0 15 9 5 15 866 598 17 0 0 1, B,10,18, C (0, 5, 7) Finding AB in terms of λ For B For C can also be given for verifying AB = 15 and showing B is on line (ii) AC [,10,11] 85 n 10 15 6 11 5 10 85x 6y 10z c Sub one value to give c 85x 6y 10z 100 6 Or any vector in plane other than BC Suitable vector product or other method for finding normal 7
757 Mark Scheme June 015 (iii) (iv) This plane contains the line BC and n 85 0 0 n' 6 15 165 5 10 5 199 5y z c Sub one value to give c 5y z 0, 15, 5 AD 1,, 1 BC 15 5 BC AD 85, 1,1 1 1 BC AD 85 1 1 7667 Appropriate vector product oe Allow uncancelled vector Must be seen! Finding vector product Finding magnitude, pythagoras must be seen N.b. Answer given AC =,10,11 Distance AC. BC AD 7667 170 10 11 51 7667 7667 Other vectors possible Accept 0.58... 7 8
757 Mark Scheme June 015 (v) A (, 5, ), B(, 10, 18), C(0,-5, -7), D(1, 1, ) AB =,5,1 AC = AD =, 10, 11 1,, 1 1 Volume = 6 AB ( CA DA 10 11 AC AD,9, 1 1 1 1 6 6,5,1,9, 68 5 8 51 17 6 Formula for volume Vector product 9
757 Mark Scheme June 015 (i) z x 1xy y 60 (ii)(a) (ii)(b) z 6x 1y 0 x y x z 1x 6y 0 y x y y y 0 or. x 0 or 8. z 60 or Stationary points at A(8,, ), and B(0, 0, 60) z x xy y 1 60 Substitute x 8 h, y k z 8 h 1(8 h)( k) ( k) 60 p h hk k 1k k h k k k 6 For all values of h and k, (h k) > 0 and k (k + 6) > 0 providing k > 6 So z > for all values of x and y close to A So is local minimum. 5 For finding both partials and setting = 0 Both Solving simultaneously to get x or y Also by verification For substitution and expansion For splitting k Or for small k 10
757 Mark Scheme June 015 (ii)(c) When x 0, z y 60 and so either side of (0, 0, 60) the value of z will be greater or less than 60. When y 0, z x 60 and so either side of (0, 0, 60) the value of z will always be greater than 60. So B is a saddle point. (iii) At (1, 1, 5), z 6x 1y 6 x z y 1x 6y 6 Equation of tangent plane is: z z z 5 ( x 1) ( y 1) x y z 5 6x 6y 1 6x 6y z 65 (iv) z z 6 and 6 x y x y y x y 1 and 1 y 0 y 1, Coordinates of R 5,,9 For obtaining functions For pt of inflexion - can be by sketch For minimum - can be by sketch Finding values of derivatives Eqn of plane ag Put partial derivatives = 6 Both correct Solve simultaneously Both values Or 6x 6y z = c 5 11
757 Mark Scheme June 015 (i) x a cos, x ' asin, x" a cos y bsin, y ' bcos, y" bsin derivatives (ii) r ( x ') ( y ') a sin b cos x ' y" x" y ' asin. bsin a cos. bcos a sin b ab At A, 0 r cos a sin 0 b cos 0 b b ab a The centre is on the x-axis r less than a b a b So centre of curvature is at a,0 i.e.,0 a a a ab Radius = b a b a Centre is at 0, b i.e. 0, b b (iii) a b a b b a The ellipse is a circle OR the centre of curvature for both points (and all points) is at (0, 0) 7 1 Apply formula (or for Set = 0 unsimplified ag 1
757 Mark Scheme June 015 (iv) s 0 dx dy d d d Applying formula 0 0 0 0 0 a b a sin cos d sin b cos d b sin b cos ( a b )sin d b ( a b )sin d b b ( a b ) 1 sin d ( a b ) b When a = b, = 0 1 s b d b 0 This is a quarter of the circumference of a circle Eliminate cos Or arc length of part of circle 7 1
757 Mark Scheme June 015 (v) Centre of curvature is at acos sin, bsin cos where giving a sin b ab cos ab a cos sin cos a b a b a sin b cos bcos xacos. sin b cos ax a a b cos cos cos 1 Parametric equations Deal with Or find equation of normal Or partial diffn of normal Eqn for x or y Similarly by sin b a ax by 1 a b a b oe i.e. ax by a b 7 1
757 Mark Scheme June 015 (i) The set is closed. b a b c d ac ad c i.e. 1 1 0 0 1 a c 0 ac Attempt to demonstrate closure Two general and different matrices Correct product therefore shows closure Identity is 1 0 0 1 or m(1,0) (ii) Each element has an inverse a b 1 b Inverse of 1 is a 0 a 0 a Which is in M e.g. product d c d a b ac cb a 1 1 0 0 1 c a 0 ac OR one numeric example. e.g. 1 1 7 but 0 10.5 0.5 0.50 1 0.5 7 Attempt to demonstrate inverse A general matrix and its inverse Demonstration by multiplying matrices each way round (one way might be quoted from (i)) correct (and different) products So No Dep on previous marks 15
757 Mark Scheme June 015 (iii) e.g. k b k c k... 1 1 1 0 0 0 k k k This is only in the set N if k Given k 0 k 1 only k k Multiplying matrices in N (Allow matrices the same) Sight of k (iv) 1 1 1 1 1 m(1,1) 0 1 0 1 0 1 (v) m(1,0) m(1,1) m(1,) m(1,) (vi) m(1,0) m(1,0) m(1,1) m(1,) m(1,) m(1,1) m(1,1) m(1,) m(1,) m(1,0) m(1,) m(1,) m(1,) m(1,0) m(1,1) m(1,) m(1,) m(1,0) m(1,1) m(1,) Group table for R OR Any argument that states that: a, b, c have order reason for this but only one element of P has order So No B B m(1,0) the identity 1 each error 1 each error Dependent on previous B 16
757 Mark Scheme June 015 5 (i) A B C B for two out of three columns correct 0 1 0 A P B 1 1 1 C 1 1 1 (ii)(a) P p 0.188 1 5 0 gives 0.188 6 0 (ii)(b) P = M 5 p... 5 P 0.85 0.86 Cube P Sight of matrix soi Using diagonal elements from P 5 Using probabilities from nd day Allow for P p 0.50.85 0.5 0.86 0.86 (iii) 0.1, 0.9, 0.9 1,, 7 7 7 Over a long period these are the probabilities that on any day at random the inspector is at these factories Ft Cao Obtaining equations Sight of x + y + z = 1 soi Or considering P n where n is large.(10 or more) A for probabilities, one error 17
757 Mark Scheme June 015 (iv) A B C 0.8 1 0 A Q B 0.1 1 1 C 0.1 1 1 0.55, 0.7, 0.7 5,, 11 11 11 (v) P( from A to A) 0.8. So 0.8 (vi) (vii) 0.8 1 0. New transition matrix: A B C 1 1 0 A R B 0 1 1 C 0 1 1 We need rd entry of nd row to be < 0.1.................. 9 10 R...... 0.116, R...... 0.0969.................. So 10 days later (which is day 5). A is the absorbing state. If it goes to A then it stays there. Solve or consider Q n for large n. 1 For using or 1 1 Method by trial with new matrix. For sight of one value oe 18
OCR (Oxford Cambridge and RSA Examinations) 1 Hills Road Cambridge C EU OCR Customer Contact Centre Education and Learning Telephone: 01 55998 Facsimile: 01 5567 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 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; 1 Hills Road, Cambridge, C EU Registered Company Number: 866 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: 01 5555 Facsimile: 01 5555 OCR 015