GCE Mathematics (MEI) Mark Scheme for June Unit 4754A: Applications of Advanced Mathematics: Paper A. Advanced GCE PMT

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1 GCE Mathematics (MEI) Unit 75A: Applications of Advanced Mathematics: Paper A Advanced GCE Mark Scheme for June 06 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, 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 06

3 75A Mark Scheme June 06 Annotations and abbreviations Annotation in scoris and Meaning Benefit of doubt Follow through Ignore subsequent working,,, Method mark awarded 0, Accuracy mark awarded 0, Independent mark awarded 0, Special case Omission sign Highlighting Misread Other abbreviations in Meaning mark scheme E Mark for explaining U Mark for correct units G Mark for a correct feature on a graph M 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

4 75A Mark Scheme June 06 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 A 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.

5 75A Mark Scheme June 06 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

6 75A Mark Scheme June 06 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. 6

7 75A Mark Scheme June 06 cos sin R cos cos sin sin Rcos, Rsin M A Correct pairs. Condone sign errors for the M mark (so accept Rsin ) R 0 R 0 B Or. or better, not 0 unless 0 chosen tan.9 M A ft their pairs (condone sign errors but division must be the correct way round), A for.9 or better (accept.5), with no errors seen in method for angle Maximum value of cos sin is 0 B Or equivalent convincing numerical statement that no solutions exist e.g.. Maybe embedded in an attempt at a solution. Do not accept general 0 statements e.g. doesn t work must be clear why no solutions exist dependent on first B SC: If candidates state that cos,sin tan this could score M0A0BMAB (so max /6) [6] Note that those candidates who state R 0 and tan with no (wrong) working seen could go on to score full marks 7

8 75A Mark Scheme June 06 q x ( ) x q q x q p p! p M* One of q x p or q( q ) x! p q p qq ( ) A A Allow x s on both sides of equations (if correct) p q (soi), for example, scores M A p p( p) qq ( ) q p or Mdep* Eliminating p (or q ) from simultaneous equations (not involving x) involving p q both variables oe if MAA awarded followed by either p or q correct (www) this implies this M mark p A p www (or q ) q Aft q (or p ) for second value, ft their p or q (e.g. the negative of their p or q) provided first marks awarded and only a single computational error in the method so must be a correct method for solving their equation in p or q (ignore mention of p and/or q 0 ) x Valid for x A or x www, allow x but not say, x SC If M0 M0 awarded and no wrong working seen then B for p = and q = -, B for - < x < (oe) so max marks [7] Guidance for solving quadratics on this paper: use of correct quadratic equation formula (if formula is quoted correctly then only one sign slip is permitted, if the formula is quoted incorrectly M0, if not quoted at all substitution must be completely correct to earn the M) or factorising (giving their x term and one other term when factors multiplied out) or completing the square (must get to the square root stage involving and arithmetical errors may be condoned provided that perfect square term was correct) 8

9 75A Mark Scheme June 06 V x dy M M for 0 0 ( ) (d ) k x y with correct limits and k or, allow correct limits 0 seen or implied later, if formula not stated then must substitute for their correctly to imply this formula condone lack of for the M mark and dy throughout (condone incorrect use of dx too) V y dy A Correct (or with a /) limits may be seen or implied through later working y y 0 6 B A [] y or y (but only if k ), condone Exact mark final answer (so no isw if correct answer is subsequently halved) but if exact value seen and is then followed by then isw y.5.5 (oe) x 9

10 75A Mark Scheme June 06 sin cos cos M* Use of correct double angle formulae: sin sin cos and any one of cos cos sin or sin or cos cos sin cos 0 A Correct equation in solvable form e.g. sincos 0(oe) or 5sin 6sin 0 or 5cos cos 0 but not sin cos cos tan Mdep* Use of sin tan on their sin cos 0 cos or correct method for solving quadratic in eithersin or cos (See guidance in question for solving quadratics) 6.6 A www (6.6 or better) 90 B Not from incorrect working Ignore additional solutions outside the range. If any additional solutions given inside the range 0 80 and full marks would have been awarded then remove last mark (so /5) Both answers in radians: 0.6 (or better) and / scores B [5] Answers with no working scores B B(so max /5) 0

11 75A Mark Scheme June 06 5 (i) AC xsec B Accept any equivalent forme.g. ACcos x. If AC not seen then there must be a diagram as evidence of correct sides - xsec with no AC is B0 AD sec and AE sec x x B Accept x xsec (as AE = x) or any equivalent form. Otherwise there must be a corresponding diagram as evidence of correct sides. Accept cos x/ AC AC / AD AD / xfor the first two marks xsec x This line (oe) must be seen before the x s cancelled sec * B NB AG dependent on all previous marks [] 5 (ii) OR AD xcos B Same principles as above for each corresponding mark AC xcos and AB xcos B or x xcos (as AB = x) xcos x sec * B Must see xcos x(oe) before given answer ED xsin B oe e.g. ED x AD or ED ADtan with AD correctly expressed in terms of x and (or using 7.5 or better) - see (i) for alternatives for AD. Allow ED =.x (or better) but B0 if ED = missing CB xtan B oe e.g. CB AC x or CB ACsin with AC correctly expressed in terms of x and (or using 7.5 or better) - see (i) for alternatives for AC. Allow CB = 0.77x (or better) but B0 if CB = missing ED xsin cos CB x tan B www must come from exact working (so not using oe) - accept ED ED or sec (oe) (as from (i): sec ) CB sec CB / * B NB AG dependent on all previous marks in (ii) must be one step of intermediate working from cos to given answer []

12 75A Mark Scheme June 06 dy d y / dt / t M* M for their (d y/ d t ) / theird x/ dt in terms of t with at least one term correct dx d x / dt t 6 A A cao (oe) allow unsimplified even if subsequently cancelled incorrectly i.e. can isw with any non-zero gradient expressed as a function of t - or t any equivalent form (e.g. y = mx + c) but must have used the correct point, t t - if using y = mx + c must explicitly have c = before M can be awarded Mdep* y f ( t) x t y x t t t When x = 0, y t Aft Must be a function of t When y = 0, x = t Aft Must be a function of t A oe need not be simplified So area of triangle = t 8 (which is t independent of t) OR (for the first two marks) dy y x x dx x dy d x ( t) A [7] No ft on this mark an answer of 8 (www) with no additional comment is sufficient to award this mark M* Attempt to eliminate t and correctly differentiates their Cartesian equation A

13 75A Mark Scheme June 06 7 (i) B AG 5 Condone etc. if recovered = 50 5 B Correct answer implies both marks accept 7. or better [] 7 (ii) DP = i + j 5k or PD= i j + 5k B One correct direction vector in plane DPF (oe e.g. expressed as a column vector) DF= i + j (or PF = j + 5k) B Any other correct direction vector in plane DPF n DP = ( 5) = 0 B Scalar product with a direction vector in the plane (including evaluation and = 0) (OR M forms vector cross product with at least two correct terms in solution) n DF = 5 0 = 0 (or n P F 0 5 0) B Scalar product with a second direction vector in the plane (including evaluation and = 0) (following OR above, A all correct ie a multiple of 5i 0j k ) (NB finding only one direction vector and its scalar product is B B0 B B0) M 5x 0y z c oe (accept any non-cartesian form for M only) rn an 5x 0 y + z = 0 A MA0 for 5i 0j k 0 or 5x 0 y + z 0 SC: if states if 5i 0j k is normal then of the form5x 0y z c and substitutes one coordinate gets MA, then substitutes the other two coordinates A (not A, A, A). Then states so 5i 0j k is normal and states the correct equation of the plane this can get B provided that there is a clear argument ie MAAB. Without a clear argument this is B0 [6] SC: if finds two relevant direction vectors B B and then finds equation of the plane from vector form, r a b c gets B. Eliminating parameters B cao. If then states so 5i 0j k is normal can get B, and then a valid reason why 5i 0j k is normal scores the final B mark (each B mark is dependent on the previous one)

14 75A Mark Scheme June 06 7 (iii) x r = i + B Need r (or another single letter) = or y for first B z r j 5k i j5k - accept column.+ ( i + j + 5k) B NB answer is not unique e.g. vector form and condone row vectors (non-vector form scores B only) 5( ) 0() + (5) = 0 M Substituting their line in their plane equation from (ii) (condone a slip if intention clear) their line and plane must be of the correct form (e.g. the line must be of the form r a tb ) 0 = 00, = 0. A www cao NB is not unique as depends on choice of line seen in this part Q is (.,., ) A www - condone answer given as a position vector 7 (iv) AQ : QG = : A oe www [6] M* Selecting 5i 0j k and their direction vector from (iii) Angle between i j 5k and (5i 0j k) is where 5) (0) (5 cos Mdep* Correct formula (including cosine), with correct substitution, using their direction 50 6 vector from (iii) and the correct normal vector - condone either a single numerical slip or a single sign slip (but not one of each) if intention is clear. So it must be clear where the slip comes from e.g. if the magnitude of one vector is stated incorrectly with no working then this is M or.0 A www cao (accept 56 or better, or better,.6 radians or better, radians or better) angle between line and plane =.0 A www cao (accept or better, 0.59 radians or better) []

15 75A Mark Scheme June 06 8 (i) x x * B NB AG must be at least one intermediate step before given answer correct x x ( x)( x) ( x)( x) application of partial fractions is fine [] 8 (ii) x x ln 0 x x B 0 or e or ln( x) ln( x) x x x 0 B If only this line seen then award B0B 8 (iii) x t ln ln( x) ln( x) x dt dx x x [] B M SC: Allow B only for verifying that when x = 0, t = 0 Correct differentiation of their t OR for first two marks - If no subtraction law of logs seen e.g. dt ( x)() ( x)( ) d x x award B for correct first bracket ( x) x (reciprocal expression) and B for second correct bracket (quotient/chain rule)(oe) if additional constant(s) added (e.g. t = k ln( )) then award B only for a constant times a fully correct derivative dt dx ( x )( x ) d x ( x)( x) dt A dt dx and d x must be correctly attributed to the correct expression for this mark dt k A Explicitly stating (that the constant of proportionality is) possible to score A0Ain this part See next page for an alternative solution [] - therefore it is 5

16 75A Mark Scheme June 06 OR dx dx k( x)( x) k dt d t ( x)( x) B Allow omission of dx and/or dt ln( x) ln( x) kt( c) M Any non-zero constant, - further guidance in (v) for this and the next mark A www oe (condone absence of c) x ) ln( x ) kt ( c ) x 0, t 0 c 0 x ln kt k A www must include c and show that c = 0. Must explicitly state that k x 8 (iv) t x e x B t ( x)e x t t e xe x t t x( e ) e M Multiplying out, collecting x terms (condone sign slips and numerical errors (eg loss of a ) only but M0 if e t incorrectly replaced with e t ) and factorising their x terms correctly t t t t (e ) (e )e ( e ) A x t t t t * e ( e )e e ( x ) (as t ) B [] OR (for first three marks) t x e B x t ( x)e x t t e xe x t x( e ) e t ( e ) x t * e t M A www NB AG as AG must be an indication of how previous line leads to the required result (eg stating or showing multiplying by e t ) Multiplying out, collecting x terms (condone sign slips as above) and factorising their x terms correctly www NB AG 6

17 75A Mark Scheme June 06 8 (v) Separating variables - condone sign slips and issues with placement of k but M0 t dx k e dt xx M* for ( x)( x)d x... or equivalent algebraic error in separating variables unless recovered. If no subsequent work integral signs needed, but allow omission of dx and/or dt but must be correctly placed if present t ln( x) ln( x) e ( c) A Any non-zero constants,, - this line must be seen and cannot be implied by later working (as this is an AG) condone absence of c or if a constant present condone the use of k for their constant. Do not condone invisible brackets e.g. ln + x unless recovered before subtraction law of logs applied all of these points apply to the next A mark too t ln( x) ln( x) ke ( c) A www oe When t 0, x 0 c k Mdep* Substituting x0, t 0 into each term in an attempt to find their c (must get c = ) if they integrate and use k as their constant they must use x0, t 0 to find this single k term only x t ln k( e )* x A www NB AG must have obtained all previous marks in this part [5] OR (for first marks) final MA as above t dx k e dt xx M* Separating variables. If no subsequent work integral signs needed, but allow omission of dx or dt, but must be correctly placed if present x t t dx k e dt ln ke ( c) x x A Must see / x on lhs please note that one A mark cannot be awarded 8 (vi) as t, x.85 ln.85/0.5 = k M Sets e t to 0 and substitutes x =.85 condone substitution of a large value of t only if it leads to the correct value of k k = 0.8 A k 0.5ln 77 / or 0.8 or better [] 7

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