FSMQ. Additional Mathematics. Mark Scheme for the Unit. June /MS/R/07 ADVANCED FSMQ Oxford Cambridge and RSA Examinations

Transcription

1 FSMQ Additional Mathematics ADVANCED FSMQ 699 Mark Scheme for the Unit June 7 699/MS/R/7 Oxford Cambridge and RSA Examinations

2 OCR (Oxford, Cambridge and RSA Examinations) is a unitary awarding body, established by the University of Cambridge Local Examinations Syndicate and the RSA Examinations Board in January 998. OCR provides a full range of GCSE, A level, GNVQ, Key Skills and other qualifications for schools and colleges in the United Kingdom, including those previously provided by MEG and OCEAC. It is also responsible for developing new syllabuses to meet national requirements and the needs of students and teachers. 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 7 Any enquiries about publications should be addressed to: OCR Publications PO Box 55 Annesley NOTTINGHAM NG5 DL Telephone: Facsimile: publications@ocr.org.uk

3 CONTENTS Additional Mathematics FSMQ (699) MARK SCHEME FOR THE UNIT Unit Content Page 699 Additional Mathematics * Grade Thresholds

4

5 Mark Scheme 699 June 7

6 699 Mark Scheme June 7 Q. Answer Mark Notes Section A ( x + ) > x M Expand and collect x + 6> x A Only terms x > A x > 6 6 v= + t s= t+ t + c M A Ignore c Take s = when t = c= 7 When t =, s = = 8 DM A Either sub to find c or sub and subtract from definite integral Alternatively: ( ) ( ) ( ) s = 6 + t dt = 6t+ t = = 8 M int A DM sub and sub A x + y x 6y+ = x x+ y 6y = = + 9 x x y y ( x ) ( y ) + = ( ) Centre (, ), radius.6... SC: Penultimate line M A S.C. Centre B Find a point on the circle and then use Pythagoras M to find radius A M B A Complete the square Centre Radius Accept correct answers with no working sin x= cosx tan x= x =± x = = and x = = 8 Alternatively Use of s + c = M cos x = 7 x =± A x = = M A and x = = 8 A S.C. Graphical method ± tolerance B B S.C. Answers with no working B for both. B B M A A 5 For either value from calculator For method to find a correct answer from that given on calculator - extra values Ignore values outside 6

7 699 Mark Scheme June 7 5 (i) Using v = u + as = + a. 6a = 8 a = M A A Got to be used! Ignore ve sign. (ii) Using v= u+ at = t M t = = 5 F From their a u+ v Or: s = t + = t This could be used in (i) to find t then a 6 t = = 5 6 dy = x dx B M Diff correctly Substitute in their gradient function dy At (, 6) = 9 y 6 = 9( x ) dx y = 9x DM A Set up equation with their gradient 7 dy = x x 5 dx = x x = ( x+ 5)( x ) = 5 x =, d y = 6x : dx d y When x =, > dx minimum x = when 5 M A M A M F A 7 = and attempt to solve Differentiate again and substitute Providing all other marks earned N.B. Any valid method is acceptable, but not that x = is the right hand value or that the y value is lower then for the other value of x.

8 699 Mark Scheme June 7 8 (i) (ii) x x = x x+ 6 x 8x+ 6= x x+ = ( x )( x ) = x =, Area = x x dx x x+ 6 dx ( ) ( ) x x = x x + 6x = ( 8 9) ( ) = 9 9+ = M M A M A DM A Equate and attempt to collect terms Solve a quadratic Ans only seen - B Integrate All terms; condone one slip Substitute and subtract (even if limits wrong) Alternatively: Area = 8x x 6 dx x = x 6x = ( 6 8 8) 6 = = ( ) M integrate A DM sub and sub A 9 (i) ( ) ( ) AB = = 85 ( ) ( ) AC = 8 + = 85 (ii) M =, =, (iii) 8 5 Grad BC = = Grad AM = = = = m.m =. = 5 Allow a geometric argument with reference to M being midpoint and the triangle isosceles. M A B E B For sight of Pythagoras used at least once Both gradients; AM ft from their M Both and demonstration (iv) y = + 5 5y = x+ 8 ( x ) M A Must use (, ) or their M and their g

9 699 Mark Scheme June 7 (i) B E One line Shading B E nd line Shading B 5 Other two lines and shading N.B. - no scales (ii) Maximum value on y-axis (, ) giving B B Allow B for 5

10 699 Mark Scheme June 7 Section B (a)(i) x x x= ( x ) x x = ( )( x ) x x + = x =,, S.C. just answers B M A A A Accept any valid method (ii) B Must have points on axes (b)(i) Remainder theorem or long division G( ) = M A For sub (ii) g() = B (iii) By continued trial or by division and quadratic factorisation g() =, g( ) = x =,, S.C. just answers B Alternatively: By division by (x ) and quadratic factorisation M (x )(x x 6) = A (x )(x + )(x ) = A x =,,. A M A A A For sub x = Final answer 6

11 699 Mark Scheme June 7 (i) 8 9 P(All males) = =.68 (ii) P(5 females) = C 5 = M A M M A A powers coefficient 56 (could be implied) (iii) 7 P(full-time) = Or P(PT) = P(at least two part-time) = P(all FT) P(7FT,PT) M A M probability correct terms = 8 =.9.76 =.97 A A A 6 Powers coefficient Ans Alternatively: Add 7 terms M A powers A Coeffs =.97 A Ans S.C. Read At least two as exactly two = 8.65 =.88 B 7

12 699 Mark Scheme June 7 (i) Pythagoras: (ii) (iii) OM = 7 OM = 5 CM = CM = 6 Use cosine rule on triangle OCM cosc = C = Sight of attempt to find base area Area = 6 = 9 Sight of attempt to find height h = sin6. 5 =. 8 Volume = 9. 8 = 8 cm M A A M M A A M A M A A 5 Correct use of Pythagoras for at least one Correct angle Correct use of cosine formula Ans Can be implied Can be implied 8

13 699 Mark Scheme June 7 (i) Apply Pythagoras to both triangles: x = y + (x +.95) = (y +.5) + (ii) Subtract:.95x +.95 =.5y +.5.y =.9x (.5.95 ).y =.9x. Alternatively: Multiply out one of the brackets Substitute for y Correct result B M A B B M A A (iii) Substitute for y: 9. x. x = +.. x = 9. x. 9. x x x 768. = 76. ± x = = = Substitute : 9. x. y = = 75.. M M A DM A DM F 7 Get y as subject Sub expression for y Correct quadratic Solve Ignore other root Withhold last mark if more than one answer given The quadratic in y is y + y 6 = Integer coefficients for x equation gives x + 9x = 9

14 FSMQ Advanced Additional Mathematics 699 June 7 Assessment Session Unit Threshold Marks Unit Maximum Mark A B C D E U The cumulative percentage of candidates awarded each grade was as follows: A B C D E U Total Number of Candidates

15

16 OCR (Oxford Cambridge and RSA Examinations) Hills Road Cambridge CB EU OCR Customer Contact Centre (General Qualifications) Telephone: Facsimile: 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: 5555 Facsimile: 5555 OCR 7