Mark Scheme (Results) January 2008

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1 Mark Scheme (Results) January 008 GCE GCE Mathematics (6664/01) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WC1V 7BH

2 January Core Mathematics C Mark Scheme Question Number 1. a)i) ii) Scheme f() = ; = 5 ; A1 Marks f( ) = ( ) = 0 ( B1 on Epen, but A1 in fact) A1 () is for attempt at either f() or f( ) in (i) or f( ) or f() in (ii). [( + )]( 4 + 4) (= 0 not required) [must be seen or used in ] A1 ( + ) ( ) (= 0) ( can imply previous marks) Solutions: = or (both) or (,, ) A1 (4) [7] Notes: (a) No working seen: Both answers correct scores full marks One correct ; then A1B0 or A0B1, whichever appropriate. Alternative (Long division) Divide by ( ) OR ( + ) to get + a + b, a may be zero [] + 1 and + 5 seen i.s.w. (or remainder = 5 ) [A1] and 0 seen (or no remainder ) [B1] First requires division by a found factor ; e.g ( + ), ( ) or what candidate thinks is a factor to get ( + a + b), a may be zero. First A1 for [ ( + )] ( 4 + 4) or ( )( 4) Second :attempt to factorise their found quadratic. (or use formula correctly) [Usual rule: + a + b = ( + c)( + d), where cd = b. ] N.B. Second A1 is for solutions, not factors Alternative (first two marks) ( + )( + b + c) = + (+ b) + (b + c) + c = 0 and then compare = with to find b and c. [] b = 4, c = 4 [A1] Method of grouping = ( ),4( ± ) ; = ( ) 4( ) A1 [= ( 4)( )] = ( + )( ) Solutions: =, = both A1. (a) Complete method, using terms of form ar k, to find r [e.g. Dividing ar 6 = 80 by ar = 10 to find r ; r 6 r = 8 is M0] r = A1 () Complete method for finding a [e.g. Substituting value for r into equation of form ar k = 10 or 80 and finding a value for a. ]

3 (8a = 10 ) a = 5 1 = 1 (equivalent single fraction or 1.5) A1 () 4 4 Notes: (c). (a) Substituting their values of a and r into correct formula for sum. n a( r 1) 5 S = = ( 0 1) (= ) (only this) A1 () [6] r 1 4 (a) : Condone errors in powers, e.g. ar 4 = 10 and/or ar 7 = 80, A1: For r =, allow even if ar 4 = 10 and ar 7 = 80 used (just these) (M mark can be implied from numerical work, if used correctly) : Allow for numerical approach: e.g. 10 r c r c r c In (a) and correct answer, with no working, allow both marks. (c) Attempt 0 terms of series and add is (correct last term 65560) If formula not quoted, errors in applying their a and/or r is M0 Allow full marks for correct answer with no working seen = A1 1 = ; (or 11.5) + 15 ( coeffs need to be these, i.e, simplified) A1; A1 (4) [Allow A1A0, if totally correct with unsimplified, single fraction coefficients) 1 ( ( or 11.5) (0.01) A1 4 = = cao A1 () [7] ) 10 = 1 + 5(0.01) + ( ) Notes: (a) For first A1: Consider underlined epression only. Requires correct structure for at least two of the three terms: (i) Must be attempt at binomial coefficients. (ii) Must have increasing powers of, (iii) May be listed, need not be added; this applies for all marks. First A1: Requires all three correct terms but need not be simplified, allow 1 10 etc, 10 C etc, and condone omission of brackets around powers of ½ Second A1: Consider as B1 for For : Substituting their (0.01) into their (a) result First A1 (f.t.): Substitution of (0.01) into their 4 termed epression in (a) Answer with no working scores no marks (calculator gives this answer)

4 4. (a) sin θ cos θ = 1 sin θ (1 sin θ ) = 1 (: Use of sin θ + cos θ = 1) sin θ + sin θ = 1 5 sin θ = cso AG A1 () sin θ =, so sinθ = (±) Attempt to solve both sinθ = +.. and sinθ = (may be implied by later work) θ = o awrt θ = 50.8 (dependent on first only) A1 θ (= 180º c o ) ; = 19. o awrt 19.º [f.t. dependent on first M and rd M] sin θ = 0.6 ; A1 θ = o and o awrt 0.8º, 09.º (both) A1 (7) [9] Notes: (a) N.B: AG; need to see at least one line of working after substituting cos θ. First : Using 5 sin θ = to find value for sinθ or θ Second : Considering the value for sin θ. (usually later) First A1: Given for awrt Not dependent on second M. Third : For ( c ), need not see written down Final : Dependent on second M (but may be implied by answers) For (180 + candidate s 50.8) or ( c ) or equiv. Final A1: Requires both values. (no follow through) [ Finds cos θ = k (k = /5) and so cos θ = (±)..., then mark equivalently]

5 5. Method 1 (Substituting a = b into second equation at some stage) Using a law of logs correctly (anywhere) e.g. log ab = Substitution of b for a (or a/ for b) e.g. log b = Using base correctly on correctly derived log p= q e.g. b = First correct value b = (allow ½ ) A1 Correct method to find other value ( dep. on at least first M mark) Second answer a = b = or 7 A1 Method (Working with two equations in log a and log b) Taking logs of first equation and separating log a = log + log b ( = 1 + log b ) Solving simultaneous equations to find log a or log b [ log a = 1½, log b=½ ] Using base correctly to find a or b Correct value for a or b a = or b = A1 Correct method for second answer, dep. on first M; correct second answer [Ignore negative values] ;A1[6] Notes: Answers must be eact; decimal answers lose both A marks There are several variations on Method 1, depending on the stage at which a = b is used, but they should all mark as in scheme. In this method, the first three method marks on Epen are for (i) First : correct use of log law, (ii) Second : substitution of a = b, (iii) Third : requires using base correctly on correctly derived log p= q

6 N 6. C B θ º 500m 700m A 15 BC = cos 15 A1 ( = ) BC = 5 awrt A1 () (a) sin B sin15 = 700 candidate's BC sin B = sin /5 c = and giving an obtuse B ( 14. ) dep θ = 180º candidate s angle B (Dep. on first M only, B can be acute) θ = = (0)45.8 (allow 46 or awrt 45.7, 45.8, 45.9) A1 (4) [7] [46 needs to be from correct working] Notes: (a) If use cos 15º =.., then A1 not scored until written as BC = correctly Splitting into triangles BAX and CAX, where X is foot of perp. from B to AC Finding value for BX and CX and using Pythagoras BC o o = (500sin15 ) + ( cos15 ) A1 BC = 5 awrt A1 Several alternative methods: (Showing the M marks, rd M dep. on first M)) candidate'sbc 700 (i) cos B = or 700 = BCc 500BCc 500candidate'sBC Finding angle B, then as above (ii) triangle approach, as defined in notes for (a) 700 valueforax tan CBX = valueforbx Finding value for CBX ( 59 ) o o θ = [ 180 (75 + candidate' s CBX )] (iii) Using sine rule (or cos rule) to find C first: Correct use of sine or cos rule for C, Finding value for C Either B =180 (15º + candidate s C) or θ = (15º + candidate s C) o (iv) 700 cos15 = BC cosθ M {first two Ms earned in this case} Solving for θ ; θ = 45.8 (allow 46 or5.7, 45.8, 45.9 ;A1

7 7 (a) Either solving 0 = (6 ) and showing = 6 (and = 0) B1 (1) or showing (6,0) (and = 0) satisfies y = 6 [allow for showing = 6] Solving = 6 ( = 4) to =.. = 4 ( and = 0) A1 Conclusion: when = 4, y = 8 and when = 0, y = 0, A1 () (c) (Area =) (4) (0) Correct integration ) ( 6 Limits not required (+ c) A1 Correct use of correct limits on their result above (see notes on limits) [ ] 4 [ 1 ] 0 with limits substituted [= 48 1 = 6 ] Area of triangle = 8 =16 (Can be awarded even if no M scored, i.e. B1) A1 Shaded area = ± (area under curve area of triangle ) applied correctly ( = 6 16) = 10 (awrt 10.7) A1 (6)[10] Notes In scheme first A1: need only give = 4 If verifying approach used: Verifying (4,8) satisfies both the line and the curve (attempt at both), Both shown successfully A1 For final A1, (0,0) needs to be mentioned ; accept clear from diagram (4) (c) Alternative Using Area = ± {(6 ); } approach (0) (i) If candidate integrates separately can be marked as main scheme (4) If combine to work with = ± (4 ), first M mark and third M mark (0) = (±) [ (+ c) ] A1, Correct use of correct limits on their result second, Totally correct, unsimplified ± epression (may be implied by correct ans.) A1 10⅔ A1 [Allow this if, having given - 10⅔, they correct it] for correct use of correct limits: Must substitute correct limits for their { 4 0 strategy into a changed epression and subtract, either way round, e.g ± [] []} If a long method is used, e,g, finding three areas, this mark only gained for correct strategy and all limits need to be correct for this strategy. Use of trapezium rule: M0A0MA0,possibleA1for triangle (if correct application of trap. rule from = 0 to = 4) A0

8 8 (a) ( 6) + (y 4) = ; B1; B1 () Complete method for MP : = ( 1 6) + ( 6 4) = 40 (= 6.5) A1 [These first two marks can be scored if seen as part of solution for (c)] Complete method for cosθ, sinθ or tanθ MT e.g. cos θ = = MP candidate' s 40 (= ) ( θ = o ) [If TP = 6 is used, then M0] θ = rad AG A1 (4) (c) 1 Complete method for area TMP ; e.g. = 40 sin θ = 1 ( = ) allow awrt 8.5 A1 Area (sector)mtq = (= ) Area TPQ = candidate s ( ) Notes =.507 awrt A1 (5) [Note:.51 is A0] [11] (a) Allow 9 for. First can be implied by 40 For second : May find TP = ( 40) = 1, then either TP 1 1 sin θ = = ( = ) or tanθ = ( ) or cos rule MP 40 NB. Answer is given, but allow final A1 if all previous work is correct. 1 (c) First : (alternative) 40 9

9 9 (a) (Total area ) = y + (Vol: ) 100 y = 100 (y = Deriving epression for area in terms of only B1, y = 100 ) B1 (Substitution, or clear use of, y or y into epression for area ) 00 (Area =) + AG A1 cso (4) d A 00 = + 4 A1 da Setting = 0 and finding a value for correct power of, for cand. [ = 75] = 4.17 awrt 4. (allow eact 75 ) A1 (4) (c) d A 600 = + 4 = positive therefore minimum A1 () (d) Substituting found value of into (a) (Or finding y for found and substituting both in y + ) 100 [y = = 5.68] Area = awrt 107 A1 () [1] Notes (a) First B1: Earned for correct unsimplified epression, isw. (c) For : Find d A1: Candidate s A and eplicitly consider its sign, state > 0 or positive d must be correct for their A d and conclusion so minimum, (allow QED, ). ( may be wrong, or even no value of found) d A, sign must be + ve d da Alternative: : Find value of on either side of = 75 and consider sign d A A1: Indicate sign change of negative to positive for, and conclude minimum. OR : Consider values of A on either side of = 75 and compare with 107 A1: Both values greater than = 107 and conclude minimum. Allow marks for (c) and (d) where seen; even if part labelling confused.

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