A-LEVEL Mathematics Further Pure MFP Mark scheme 660 June 04 Version/Stage: Final
Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students scripts: alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students reactions to a particular paper. Assumptions about future mark schemes on the basis of one year s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Further copies of this Mark Scheme are available from aqa.org.uk Copyright 04 AQA and its licensors. All rights reserved. AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre.
MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 4 Key to mark scheme abbreviations M mark is for method m or dm mark is dependent on one or more M marks and is for method A mark is dependent on M or m marks and is for accuracy B mark is independent of M or m marks and is for method and accuracy E mark is for explanation or ft or F follow through from previous incorrect result CAO correct answer only CSO correct solution only AWFW anything which falls within AWRT anything which rounds to ACF any correct form AG answer given SC special case OE or equivalent A, or (or 0) accuracy marks x EE deduct x marks for each error NMS no method shown PI possibly implied SCA substantially correct approach c candidate sf significant figure(s) dp decimal place(s) No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded. of 4
MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 4 (a) r = 9 B θ = B condone.57 here only 9i = 9e i (b) r = B follow through ( their r ) 4 ; accept ( their θ ) / 4 M generous 5 7 θ =,,, 8 8 8 8 A two angles correct in correct interval A exactly four angles correct mod 4 9 etc i5 i i i7 8 8 8 8 e, e, e, e A 5 four correct roots in correct interval and in given form; accept for Total 7 (a) Accept correct values of r and θ for full marks without candidates actually writing Do not accept angles outside the required interval. Example θ = or θ = scores B0 9e i. (b) Condone r =.7 for B only. Do not follow through a negative value of r for B. 7 5 Example θ =,,, scores M A A 8 8 8 8 i ik + Example 8 e scores B M then k =, 0,, scores A A with final A only earned when four roots are written in given form 4 of
MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 4 (a) Straight line M Half line from on Im axis A not vertical or horizontal Making approx. 0 to positive Im axis & 60 to negative Re axis A (b)(i) Circle with centre on their L M Circle correct and touching Im z = A lowest point of circle at approx C d (b)(ii) d = tan M any correct expression for distance 6 b or = for M a a = A condone.7 or better b = 5 B centre is + 5 i Total 8 (a) The two A marks are independent. (b) (i) If candidate draws a horizontal line at Im z = then award A if there is a clear intention for their circle to touch this line. Allow freehand circle where centre is intended to be on their L for M but withhold A if L is in wrong quadrant or drawing of circle is very poor. Award A0 if candidate has not scored full marks in (a). 5 of
MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 4 (a) k + 7k + 4 B (b) When n = LHS = = RHS = 6 4 = Therefore true for n = B Assume formula is true for n=k (*) Add (k+)th term (to both sides) M (k+)th term must be correct k+ r r( r + ) ( ) r= = 6 k + 5k + 8 k A A0 if only considering RHS ( )( ) +( k + )( k + ) ( ) k = 6 ( ) k ( k + 0k + 6 k k ) k = 6 ( ) ( k + 7k + 4) = 6 (( k + ) + 5( k + ) + 8)( ) k A A from part (a) Hence formula is true when n = k+ (**) but true for n = so true for n =,, by induction (***) E 6 must have (*), (**) and (***) and must have earned previous 5 marks Total 7 (b) For B, accept n= RHS=LHS= but must mention here or later that the result is true when n= Alternative to (***) is therefore true for all positive integers n etc However, true for all n is incorrect and scores E0 May define P(k) as the proposition that the formula is true when n = k and earn full marks. However, if P(k) is not defined then allow B for showing P() is true but withhold E mark. 6 of
MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 4 4 (a) (i) α + β + γ = B αβ + βγ + γα = B (ii) α + β + γ ( ) = α + β + γ ( αβ + βγ + γα) M correct formula = 4 6 = Acso AG be convinced; must see 4 6 A0 if α + β + γ or αβ + βγ + γα not correct α + β β + γ = α + M or may use + 4 α + αβ = + 9 m ft their αβ + βγ + γα = 7 A (b) (i) ( )( ) (ii) αβγ = 4 B PI when earning m later ( α + β )( β + γ )( γ + α) = α αβ αβγ M or ( α)( β )( γ ) = 8 4 α αβ αβγ = 6 4 m Sub their α, αβ & αβγ = 0 A 4 (c) Sum of new roots = α = 4 B or NMS coefficient of z written as +4 ± 4 + " 7" " 0" (=0) M correct sub of their results from part (b) z z their z their New equation + 4 + 7 + 0 = 0 A z z z Total 4 Alternative y = z B ( y) + ( y) + ( y) 4 = 0 M y + 4 y + 7 y + 0 = 0 A NB candidate may do this first and then obtain results for part (b) (a)(ii) Accept ( α ) = α + αβ etc for M (b)(ii) If B not earned, award m for using αβγ = ± 4. (c) For M the signs of coefficients must be correct FT their results from (b) but condone missing = 0 However, for A the equation must be correct ( any variable) including = 0 7 of
MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 4 5(a) ( ) θ θ θ θ θ θ e e = e e + e e OE 4sinh θ + sinhθ = 4 e e + e e + 8 e e = ( e θ e θ ) = sinh θ θ θ θ θ θ θ ( ) ( ) B M correct expansion; terms need not be combined correct expression for sinhθ and attempt θ θ to expand ( e e ) A AG identity proved (b) 6sinh θ + sinhθ = 0 4sinh θ = 0 M attempt to use previous result sinh θ = 4 A 9 ( θ = ) ln + + 4 6 m correct ln form of sinh - for their 4 θ = ln A 4 (c) x = sinhθ = 4 M A correctly substituting their expression for θ into sinh θ removing any ln terms Total 9 θ θ (a) For M, must attempt to expand ( e e ) with at least terms and attempt to add expressions for two terms on LHS. For A, must see both sides of identity connected with at least trailing equal signs. (b) Withhold final A if answer is given as x = ln. θ 6 Alternative: e e θ θ θ θ θ = 0 e e = 0 so (e )(e + ) = 0 scores M for e kθ = p (quite generous) A for then m for correct ft from e kθ e θ = (and perhaps = p kθ = ln p and final A for e θ = 0.5 ) θ = ln and no other solutions 8 of
MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 4 6(a)(i) n z = cos nθ + isin nθ M n z = cos( nθ ) + isin( nθ ) = cos nθ isin nθ z n = isin nθ n z E cos nθ isin nθ or = cos nθ + i sin nθ cos nθ isin nθ shown not just stated A AG (ii) n z + = n z cos nθ B (b)(i) 4 z z + = z + 4 z z z B or z 4 4 + z (ii) ( ) ( ) isinθ cosθ = cos 4θ M using previous results 6sin θ cos θ = cos 4θ 8sin θ cos θ = cos 4θ Acso (c) x = sinθ dx = cosθ dθ M x 4 x dx 6sin θ cos θ dθ = A dx x = sinθ = k cosθ dθ = ( cos4θ ) ( dθ ) m correct or FT their (b)(ii) result = θ sin 4θ B FT integrand of form k( cos4 θ ) sin sin = x = θ = ; x = θ = ; 6 = + 4 Acso 5 Total n (a)(i) May score M E0 A if z = cos nθ i sin nθ merely quoted and not proved. Condone poor use of brackets for M but not for A. (b)(ii) For M, must use i sinθ and their cosθ on LHS but condone poor use of brackets etc when squaring. (c) For Acso, must simplify sin correctly in terms of. Allow first A for missing dθ or incorrect limits used/seen, but withhold final Acso. 9 of
MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 4 7 (a) d + x x + + x = = dx x ( x) ( x) B d y = d x + u M A ( x) ACF x where u = + x correct unsimplified = = ( ) ( ) x + + x + x A 4 AG be convinced d (b) either (tan x) = d x + x or + x + x tan = tan x + C x Putting x = 0 gives C = tan = 4 dx = tan x ( + c) + x x 4 tan tan x = B M A AG Total 7 (a) x Alternative tan y = + x d y sec y M = B d x ( x) + x d y + A with final A for proving given result x dx d (b) Must see (tan x) = within attempt at part (b) to award B d x + x 0 of
MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 4 8(a) d y y = ( x ) = ( x ) dx B d y + = + dx x ( s ) ( 9) d y = + d = ( ) dx ( x) ( ) M 9 x dx x A ft their d y d x 9 s = + dx x (be convinced) AG (must have limits & dx on final line) (b)(i) cosh = ln ( + 8) M ( + ) = + = + 8 need to see this line OE cosh = ln ( + ) = ln ( + ) A AG (be convinced) (ii) x = cosh θ dx = cosh θ sinh θ dθ M dx k coshθ sinhθ dθ = OE coshθ (s = ) cosh θ sinh θ d θ A including dθ on this or later line sinhθ cosh θ = + cosh θ OE B double angle formula or ( e θ e θ + + ) (s = ) θ + sinh θ θ θ e + θ e + cosh sinh(cosh ) ( ) cosh sinh cosh A or ( 4 4 ) m correct use of correct limits (s = ln ( + ) ln ( + ) + 6 must see this line OE = 5 + ln ( + ) A 6 partial AG (be convinced) Total TOTAL 75 (b)(i) SC for cosh ( ln ( + )) = ( + ) + ( + ) = + + = cosh = ln + ( ) ( ) ( ) (ii) Another possible correct form for m is ln( + ) ln ( + ) + sinh 4ln ( + ) sinh ln + ( ) ( ( )) of