FSMQ Additional Mathematics ADVANCED FSMQ 699 Mark Scheme for the Unit June 7 699/MS/R/7 Oxford Cambridge and RSA Examinations
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CONTENTS Additional Mathematics FSMQ (699) MARK SCHEME FOR THE UNIT Unit Content Page 699 Additional Mathematics * Grade Thresholds
Mark Scheme 699 June 7
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 + 7 7 = 8 DM A Either sub to find c or sub and subtract from definite integral Alternatively: ( ) ( ) ( ) s = 6 + t dt = 6t+ t = 8 + 7 6 + = 8 M int A DM sub and sub A x + y x 6y+ = x x+ y 6y = + + 6 + 9= + 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 =± 75.96 x = 8 75.96 = and x = 6 75.96 = 8 Alternatively Use of s + c = M cos x = 7 x =± 75.96 A x = 8 75.96 = M A and x = 6 75.96 = 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
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.
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 8+ 8) + + 6 = 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 = 5 + 8 = 85 ( ) ( ) AC = 8 + = 85 (ii) 5 + 8 8 + M =, =, (iii) 8 5 Grad BC = = 5 8 9 9 Grad AM = = = = + 5 5 5 5 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
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
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
699 Mark Scheme June 7 (i) 8 9 P(All males) = =.68 (ii) 5 8 9 P(5 females) = C 5 =.568.57 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 7 7 = 8 =.9.76 =.97 A A A 6 Powers coefficient Ans Alternatively: Add 7 terms M 6 8 7 7 8 +... A powers A Coeffs =.97 A Ans S.C. Read At least two as exactly two 6 7 8 = 8.65 =.88 B 7
699 Mark Scheme June 7 (i) Pythagoras: (ii) (iii) OM = 7 OM = 5 CM = CM = 6 Use cosine rule on triangle OCM 6 + 5 cosc = C = 6. 5 6 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
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 +. +. 8. x + 76. x 768. = 76. ± 76. + 8. 768. 76. + 756. x = = = 5. 6. 6. 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
FSMQ Advanced Additional Mathematics 699 June 7 Assessment Session Unit Threshold Marks Unit Maximum Mark A B C D E U 699 7 6 5 The cumulative percentage of candidates awarded each grade was as follows: A B C D E U Total Number of Candidates 699 8.8 8.6 8. 57.5 66.8 55
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