Version.: 6 General Certificate of Education abc Mathematics 66 MFP Further Pure Mark Scheme 6 eamination - January series Mark schemes are prepared by the Principal Eaminer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation meeting attended by all eaminers and is the scheme which was used by them in this eamination. The standardisation meeting ensures that the mark scheme covers the candidates responses to questions and that every eaminer understands and applies it in the same correct way. As preparation for the standardisation meeting each eaminer analyses a number of candidates scripts: alternative answers not already covered by the mark scheme are discussed at the meeting and legislated for. If, after this meeting, eaminers encounter unusual answers which have not been discussed at the meeting they are required to refer these to the Principal Eaminer. It must be stressed that a mark scheme is a working document, in many cases further developed and epanded on the basis of candidates 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 eamination paper. Copyright 6 AQA and its licensors. All rights reserved.
MFP AQA GCE Mark Scheme, 6 January series Key To Mark Scheme And Abbreviations Used In Marking M m or dm A B E mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for eplanation or ft or F follow through from previous incorrect result MC mis-copy CAO correct answer only MR mis-read CSO correct solution only RA required accuracy AWFW anything which falls within FW further work AWRT anything which rounds to ISW ignore subsequent work ACF any correct form FIW from incorrect work AG answer given BOD given benefit of doubt SC special case WR work replaced by candidate OE or equivalent FB formulae book A, or (or ) accuracy marks NOS not on scheme EE deduct marks for each error G graph NMS no method shown c candidate PI possibly implied sf significant figure(s) SCA substantially correct approach 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. However, there are situations in some units where part marks would be appropriate, particularly when similar techniques are involved. Your Principal Eaminer will alert you to these and details will be provided on the mark scheme. 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.
AQA GCE Mark Scheme, 6 January series MFP MFP (a) ( m + ) = M Completing sq or formula m = ± i A (b)(i) CF is e (A cos + B sin ) {or e A cos( + B) but not Ae ( +i) + Be ( i) } M A If m is real give M On wrong a s and b s but roots must be comple. {P.Int.} try y = p + q M OE p + ( p + q) = 4 p =, q = A A On one slip GS y = e (Acos + Bsin) + B 6 Their CF + their PI with two arbitrary constants. (ii) =, y= A = B Provided an Mgained in (b)(i) y () = e (Acos + Bsin) + M Product rule used + e ( Asin + Bcos) + A y () = = A+B+ B = A Slips (a) y = e (cos + sin) + 4 Total M e d d A = e - e Reasonable attempt at parts = e 4 e {+c} A Condone absence of +c a d e = ae a e a ( ) M F(a) F() 4 4 = 4 ae a 4 e a A 5 (b) lim a k a a e = B (c) e = d lim = { ae a e a } a 4 4 = = 4 4 M If this line oe is missing then / A On candidate s /4 in part (a). B must have been earned Total 8
MFP AQA GCE Mark Scheme, 6 January series MFP (a) y = y () = B Accept general cubic. y ( ) d + d y = + M Substitution into LHS of DE A ( ) = + = 5 d (b) [( ) y] d dy = y + ( ) d MA Completion. If using general cubic all unknown constants must be found Differentiating ( ) y = c wrt dy leads to y + ( ) = d c y = is a soln. of dy = d + y A Be generous SC Differentiated but not implicitly give ma of / for complete solution (c) c y = is a soln with one arb. dy constant of = d + y c y = is a CF of the DE GS is CF + PI c y = + M A Total 8 Must be using hence ; CF and PI functions of only CSO Must have eplicitly considered the link between one arbitrary constant and the GS of a first order differential equation. 4
AQA GCE Mark Scheme, 6 January series MFP MFP 4(a) 4 ln( ) =... 4 B sin (b)(i) f () = e f() = B f () = cos e f () = sin sin f () = sin e + cos e f () = sin MA MA Product rule used Maclaurin f ()= f()+f ()+ f () so st three terms are + + A 6 CSO AG (ii) f () = cos (cos sin sin) e + sin +{cos( sin) cos } e MA f () = so the coefficient of in the series is zero A CSO AG SC for (b): Use of series epansions.ma of 4/9 (c) sin. B Ignore higher power terms in sin epansion 4 sin + o( ) e + ln( ) = M Series from (a) & (b) used sin A Numerator k (+ ) lim e sin + o( ) = Condone if this step is missing + o( ) + ln( ) = A 4 On candidate s coefficient in (a) sin provided lower powers cancel Total 4 5
MFP AQA GCE Mark Scheme, 6 January series MFP 5(a)(i) y(.) = y() +.[ln+/] = +. =. MA A (ii) y(.) = y() + (.)[f(., y(.)] MA. = +(.)[.ln.+(.)/.] A On answer to (a)(i). = +..48498.. =.9684 =. to dp A 4 CAO (b)(i) IF is e d M = e ln A = ln e = = Condone e d for M mark A AG (be convinced) (b)(i) Solutions using the printed answer must be convincing before any marks are awarded (ii) d y = ln d MA y = ln d = ln d M Integration by parts for k ln y = ln + c A Condone missing c. y() = = ln + c m Dependent on at least one of the two previous M marks y c = y = ln + A 6 OE eg = ln + (iii) y(.) =.54 =. to dp B Total 7 6
AQA GCE Mark Scheme, 6 January series MFP MFP 6(a) + y y + 6 = 6 M Use of y = r sin θ ( = r cosθ PI) M Use of + y = r r r sinθ + 6 = 6 m r = sinθ A 4 CSO AG (b) Area = (sin + 5) θ dθ. M Use of r dθ. π.. = (4sin θ + sinθ + 5 )dθ B B Correct epn. of (sinθ +5) Correct limits π = (( cos θ ) + sinθ + 5 )d M Attempt to writesin θ in terms of cos θ. θ = [ 7θ sin θ cosθ ] A Correct integration ft wrong coeffs = 7π. A 6 CSO (c) At intersection sinθ = sinθ + 5 M OE eg r = 6( r 5) 5 sin θ = A OE eg r = 6 π Points 6, 6 and 5π 6, 6 A OE OPMQ is a rhombus of side 6 Or two equilateral triangles of side 6 Etra notes: The SC for Q4 π M Any valid complete method to find the Area = 6 6 sin oe A area (or half area) of quadrilateral. = 8 A 6 Accept unsimplified surd Total 6 Total 75 e sin = +...... +......! +!!!! M for st terms ignoring any higher powers than those shown. 7
MFP AQA GCE Mark Scheme, 6 January series A for all 4 terms (could be treated separately ie last term often only comes into (b)(ii) = + + (...) + (...) 6 6 = + + A (be convinced..ignore any powers of above power ) Coefficient of : + = A (be convinced..ignore any powers of above power ) 6 6 Quite often the nd A mark is awarded before the st A 8