GCE. Mathematics. Mark Scheme for June Advanced GCE Unit 4723: Core Mathematics 3. Oxford Cambridge and RSA Examinations

Transcription

1 GCE Mathematics Advanced GCE Unit 7: Core 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 pupils 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 7 Mark Scheme June 0 (i) integral of form ke x M any non-zero constant k different from 6; using substitution u x to obtain ke u earns M (but answer to be in terms of x ) correct e x 6 A or equiv such as (ii) integral of form ln( ) M any non-zero constant k ; allow if brackets absent; ln (after sub n) earns M 0 correct 5l n(x ) A or equiv such as ln( ) brackets rather than modulus signs but brackets or modulus signs must be present (so that 5lnx earns A0) Include + c at least once B 5 anywhere in the whole of question ; this mark available even if no marks awarded for integration 5 Apply one of the transformations correctly to their equation B correct ln x ln B or equiv Show at least one logarithm property M correctly applied to their equation of resulting curve (even if errors have been made earlier) y ln( x ) A or equiv of required form; ln x earns A; correct answer only earns /; condone absence of y (a) State sin cos sin B or unsimplified equiv such as 7(sin cos ) sin Attempt to find value of cos M by valid process; may be implied A exact answer required; ignore subsequent work to find angle (b) Attempt use of identity for cos M of form cos ; initial use of cos sin needs attempt to express e x sin in terms of cos to earn M 6cos 9cos 0 A or unsimplified equiv or equiv involving sec Attempt solution of -term quadratic eqn M for cos or (after adjustment) for sec Use sec at some stage cos M or equiv A 5 or equiv; and (finally) no other answer 8

4 7 Mark Scheme June 0 (i) Draw sketch of y ( x ) *B touching positive x-axis and extending at least as far as the y-axis; no need for or 6 to be marked; ignore wrong intercepts Draw straight line with positive gradient *B at least in first quadrant and reaching positive y-axis; assess the two graphs independently of each other Indicate two roots B AG; dep *B *B and two correct graphs which meet on the y-axis; indicated in words or by marks on sketch [SC: Draw sketch of y ( x ) x 6 and indicate the two roots : B (i.e. max mark)] (ii) State 0 or x 0 B not merely for coordinates (0, 6) (iii) correct first iterate B to at least dp; any starting value (> 6) Show correct iteration process M producing at least iterates in all; may be implied by plausible converging values at least correct iterates A allowing recovery after error; iterates given to only d.p. acceptable; values may be rounded or truncated.8 A answer required to exactly dp; A0 here if number of iterates is not enough to justify.8; attempt consisting of answer only earns 0/ [ ; ; ; ; ; ] 8 5 Attempt use of product rule *M to produce ln( ) kx kx x x form xln(x ) A x x A or equiv Attempt second use of product rule *M Attempt use of quotient (or product) rule *M allow numerator the wrong way round 8x 8 x(x ) 6x ln(x ) x (x ) A or equiv Substitute into attempt at second deriv M dep *M *M *M 96 ln5 5 A 8 or exact equiv consisting of two terms 8

5 7 Mark Scheme June 0 6 Method : (Differentiation; assume value 0 ; eqn of tangent; through origin) Differentiate to obtain k(x 5) M any constant k ( x 5) A or equiv Attempt to find equation of tangent at P and attempt to show tangent passing through origin M assuming value 0 ; or equiv y x and confirm that 5 tangent passes through O A AG; necessary detail needed Method : (Differentiation; equate y change change x to deriv; solve for x) Differentiate to obtain k(x 5) M any constant k Equate y change change ( x 5) A or equiv x to deriv and attempt solution M x 5 x obtain 0 only (x 5) and solve to Method : (Differentiation; find x from y f( x) x and y x 5 ) Differentiate to obtain k(x 5) M any constant k A ( x 5) A or equiv State ( 5) x, y x 5, eliminate y and attempt solution M condone this attempt at eqn of tangent 0 only A Method : (No differentiation; general line through origin to meet curve at one point only) Eliminate y from equations y kx and y x 5 and attempt formation of quadratic eqn M k x x 5 0 A or equiv Equate discriminant to zero to find k M 0 k or equiv and confirm x A 5 Method 5: (No differentiation; use coords of P to find eqn of OP; confirm meets curve once) 0 5 Use coordinates (, 5) to obtain y x 0 or equiv as equation of OP B Eliminate y from this eqn and eqn of curve and attempt quadratic eqn M should be 9x 60x 00 0or equiv Attempt solution or attempt discriminant M Confirm 0 only or discriminant = 0 A

6 7 Mark Scheme June 0 Either: Integrate to obtain correct 9 k(x 5) *M any constant k ( x 5) A Apply limits 5 and 0 M dep *M; the right way round Make sound attempt at triangle area and calculate (triangle area) minus (their area under curve) M or equiv and hence A 9 or exact equiv involving single term Or: Arrange to x = and integrate to obtain ky ky form *M 5 y y A 9 Apply limits 0 and 5 M dep *M; the right way round Make sound attempt at triangle area and calculate (their area from integration) minus (triangle area) M and hence 5 5 A (9) or exact equiv involving single term (i) Either: Attempt solution of at least one linear eq n of form ax b M A and (finally) no other answer Or: Attempt solution of -term quadratic eq n obtained by squaring attempt at g( x ) on LHS and squaring or on RHS M A () and (finally) no other answer (ii) Either: (x 5) 5 for h B Attempt to find inverse function M of function of form ax b ( x 0) 9 A or equiv in terms of x Or: State or imply g is ( x 5) g g Attempt composition of with M 5 9 A () or more simplified equiv in terms of x (iii) State x 0 B give B for answer x 0 8 B

7 7 Mark Scheme June 0 0.0t k 8 (i) Differentiate to obtain form e M any constant k different from t 0.0t 5.6e or 5.6e A or (unsimplified) equiv.9 or.9 or.87 or.87 A but not greater accuracy; allow if final statement seems contradictory; answer only earns 0/ differentiation is needed (ii) Either: State or imply M 75e kt B or equiv Attempt to find formula for M M 0.07t ( ln 8 ) t M 75e A 0 or equiv such as 75e 5 Equate masses and attempt rearrangement M as far as equation with e appearing once 0.06t 6 e A 5 or equiv of required form which might involve 5. or greater accuracy on RHS; final two marks might be earned in part iii 0.t Or: State or imply M 75 r B for positive value r t B Attempt to find M in terms of e M Equate masses and attempt rearrangement M 0.06t 6 e A 5 or equiv of required form which might involve 5. or greater accuracy on RHS; final two marks might be earned in part iii (iii) Attempt solution involving logarithm of any equation of form e mt c M whether the conclusion of part ii or not 7. A or greater accuracy 7. ; correct answer only earns both marks 0 5

8 7 Mark Scheme June 0 9 (i) Use at least one identity correctly B angle-sum or angle-difference identity Attempt use of relevant identities in single rational expression M not earned if identities used in expression where step equiv to A B C A B C D E F D E F or similar has been carried out; condone (for MA) if signs of identities apparently switched (so that, for example, denominator appears as cos cos sin sin cos cos cos sin sin ) sin cos sin cos cos cos A or equiv but with the other two terms from each of num r and den r absent Attempt factorisation of num r and den r M sin and hence tan cos A 5 AG; necessary detail needed (ii) State or imply form k tan50 M obtained without any wrong method seen State or imply sin50 tan50 A or equiv such as 9cos50 A or exact equiv (such as ); correct 9 answer only earns / (iii) State or imply tan 6 k B State k 6 B Attempt second value of M using 6 tan k (multiple of 80) 6 tan k 0 A and no other value 6

9 OCR (Oxford Cambridge and RSA Examinations) Hills Road Cambridge CB EU OCR Customer Contact Centre 9 Qualifications (General) Telephone: Facsimile: general.qualifications@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; Hills Road, Cambridge, CB EU Registered Company Number: 866 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 0