Mark Scheme (Results) Summer 2008

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Gordon Welch
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1 (Results) Summer 8 GCE GCE Mathematics (75/) Edecel Limited. Registered in England and Wales No. 975 Registered Office: One9 High Holborn, London WCV 7BH

2 Further Pure Mathematics FP d. ( ln( tanh ) ) sech = tanh = = = cosech sinh cosh sinh (*) () M Any valid differentiation attempt including ln( e e ) ln( e + e ) cosh sinh A c.a.o. (o.e e.g. ) sinh cosh M Proceeding to a hyperbolic epression in A c.s.o.

3 Further Pure FP e + e e e. 8 = e + e e + e = e e + = ( e )(e ) = (or equiv.) e =, e = ft = ln ( or ln ), = ln A () M A M Aft A Correctly substituting eponentials for all hyperbolics To a three term quadratic in c.a.o. (o.e.) Solving their equation to e = f.t. their equation. c.a.o. e

4 Further Pure FP = arcosh + 9 A = + () + 9 ln = ln( ) + 7 ln( ) = ln + 7 = ln + 9 (*) A (7) 7 Correctly changing to an integrable form. M Complete attempt to integrate at least one bit. A One term correct A All correct DM Substituting limits in all.must have got first M A Correctly (no follow through) A c.s.o.

5 Further Pure FP. (a) =, At = = = +, A y arsinh ( ) = ( ) M ( ) y = + ln + (*) A (5) a (b) = 9a = ( + a ) + a a 9a + = ( a )(a a ) = A ± + + a = a =.9 (5) at least (a)m Attempt to differentiate need ( ) A correct A c.a.o. M Substituting into straight line equation (linear). Must use = A c.s.o. (b)m Their derivative = their gradient (condone throughout) M= A mark cao, any form A quartic cao M Solving their quartic to a = A c.a.o. (a.w.r.t..9 to dp) 5

6 Further Pure FP π n n n 5. (a) I = e sin = [ e sin ] e nsin cos n n n n n [ e sin ne sin cos ] + n e ( sin + ( n )cos sin cos ) n n π [ sin ne sin cos ] = e I n = n e n sin + n( n ) sin n ( sin ) M I ) n = nin + n( n ) In n( n In n = I n n( n ) I (*) (8) n + (b) I = I, = I M, A π π π I = [ e ]..., = ( e ) e = = I M, A () 85 (a)m Complete attempt to use parts once in the right direction need A cao M Attempt to use parts again with sensible choice of parts, not reversing. Need to be differentiating a product. A cao both = at some point. (doesn t need to be correct, must must =) k DM I n = epressions in e sin Depends on nd M DMEpresssion in In and In to I n =. Depends on rd M A c.s.o. (b)m I in terms of I A I correctly in terms of I [ o.e.] M e A c.a.o for I. sin n

7 Further Pure FP. cosh (a) cosh arctan(sinh ) = sinh arctan(sinh ) sinh + sinh A = sinh arctan(sinh ) ln( + sinh ) ( + C) (5) Or:... tanh = sinh arctan(sinh ) ln(cosh ) ( + C) Alternative: Let t = sinh, t = cosh, arctant = t arctant + t A =... ln ( + t ) M = sinh arctan(sinh ) ln( + sinh ) ( + C) (or equiv.) A = (b) [ sinh arctan(sinh ) ln(cosh )]...,. (*) M, A () (a) Alternative: Let tant = sinh, sec t = cosh, t sec t = t tant tant A =... ln(sect) M = sinh arctan(sinh ) ln + sinh ( + C) (or equiv.) A (a)m Complete attempt to use parts A One term correct. A All correct. M All integration completed. Need a ln term. A c.a.o. ( in ) o.e, any correct form, simplified or not (b)m Use of limits and and /. A c.s.o. 7 7

8 Further Pure FP 7. y d (a) = 9 = sect tant, = sec t 9 sect = = = y 8tant sint sint y tant = ( sect) M sint + y = 5tant (*) A () (b) Using ( 5 b = a e ) : ae = a + b = 5 or e = P: sect = 5 cost = M 5 Coordinates of P: ( sec,tant) = 5, (c) R: 9 t (5) 5tant 5 = = M sint SR SP = 5 = = 8 8 () Area of PRS: ( ) (a)m Differentitating A c.a.o. M in terms of t. A c.a.o. M Substituting gradient of normal into straight line equation. A c.s.o. b = a ( e ) (b)m Use of A c.a.o. for ae or for e M Using coordinate of focus= coordinate of P, to get single term f(t)= constant. (Allow recovery in (c) ) M Substituting into P coordinates to a for and for y. A c.a.o. (c)m Attempt to find coordinate of R. M Substituting into correct template i.e. ½ their R - their H their P y A c.a.o. s.f. or better. 8

9 Further Pure FP 8. (a) & = + cost y& = sint (b) (c) y& sin t = = & + cost sin t cos t = cos t t = tan (*) () s = & + y& = + cost ft t t t = cos = sin (Limits or establish C = for A) (*) () t t tan ψ = tan ψ = s = sinψ (d) Surface area = t () π y & + y& = 8 π ( cost) + cost ft t t = 7π sin cos M =... sin t t s L π L π L But sin = =, so surface area = = (*) (7) (a) both M Attempt at y / A cso on paper need to see half angles (b)m Attempt at arc length, integral formula A cao follow through on their and y one variable only M Integrating A cso on paper (c) cao (d) M Attempt at Surface area, integral formula.condone lack of π. A cao follow through on their and y condone lack of π. one variable only DMGetting to integrable form condone lack of π. Depends on previous M mark. DMintegrating condone lack of π. Depends on previous M mark. A cao DMEliminating t to give epression in L only Depends on previous M mark. A cso on paper. 9

FP3 mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers (back to June 2002)

FP3 mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers (back to June 2002) FP mark schemes from old P, P5, P6 and FP, FP, FP papers (back to June ) Please note that the following pages contain mark schemes for questions from past papers. Where a question reference is marked with