Version.0: 0608 abc General Certificate of Education Mathematics 660 MPC Pure Core Mark Scheme 008 eamination - June 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. Further copies of this Mark Scheme are available to download from the AQA Website: www.aqa.org.uk Copyright 008 AQA and its licensors. All rights reserved. COPYRIGHT AQA retains the copyright on all its publications. However, registered centres for AQA are permitted to copy material from this booklet for their own internal use, with the following important eception: AQA cannot give permission to centres to photocopy any material that is acknowledged to a third party even for internal use within the centre. Set and published by the Assessment and Qualifications Alliance. The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales (company number 6447) and a registered charity (registered charity number 074). Registered address: AQA, Devas Street, Manchester 5 6EX Dr Michael Cresswell Director General
MPC - AQA GCE Mark Scheme 008 June 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 0) 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.
MPC - AQA GCE Mark Scheme 008 June series MPC (a) = B OE; accept k =.5 (b)(i) dy d = B AF At least one inde reduced by and no term of the form a. For For.5 0.5.Ft on ans (a) non-integer k (ii) When = 4, y = 8 B y (4) = ; Attempt to find d y d when = 4 = (4).5( 4) = 5 Ft on one earlier error provided noninteger powers in (a) and AF (b)(i) Tangent: y 8= 5( 4) m y y(4) = y (4)[ 4] OE y = 5 A 5 CSO; must be y = 5 Total 9 (a) Arc PQ = rθ rθ = 6π (cm) A Condone missing units throughout the paper (b) π α + α + = π 7 OE α = π 7 A (c) Chord PQ = 4 cosα OE eg Accept equivalent fractions eg 4π 4 and condone 0.86π or better inclusive or Perimeter = 7.45 + 6π = 6.07 = 6. (cm) A Condone > sf Total 6 (a) r = 6 0 = 0.8 B OE π 4 sin or 7.45-7.5 4 π 4 + 4 4 cos 7 (b) a 0 r = OE Using a correct formula with a = 0 or 0.8 r = c s 0.8 = 00 AF ft on c s value of r provided r < (c) 0 a( r ) { S = } 0 r 0 = 00( 0.8 ) OE Using a correct formula with n = 0 = 98.847{07..} A Condone > dp (d) nth term = 0 r n = 0(0.8) n Ft on c s r. Award even if 6 n seen = 0 0.8 0.8 n = 5 0.8 n A CSO; AG Total 7 4
MPC - AQA GCE Mark Scheme 008 June series MPC (cont) 4(a) {BC =} 7.6 + 8. 7.6 8.cos65 RHS of cosine rule used.. = 57.76 + 68.89 5.75... m Correct order of evaluation BC = 7... = 8.56.. (= 8.56 m) A AG; must see 7... or > sf value (b) Area triangle = 7.6 8. sin 65 Use of bcsin A OE = 8.58 = 8.6 (m ) A Condone > sf (c) Area of triangle = 0.5 BC AD Or valid method to find sinb or sinc AD = [Ans (b)] [0.5 Ans (a)] m Or AD = 7.6sinB; Or AD = 8.sinC AD = 6.67.. = 6.7 (m) A If not 6.7 accept 6.65 to 6.69 inclusive. Total 8 5(a)(i) log a = 0 B (ii) log a a = B (b) loga = log a(5 6) loga.5 One law of logs used correctly 5 6 loga = log a.5 A second law of logs used correctly loga = loga0 = 0 A Total 5 6(a) 8= 8p + q Either equation. PI eg by combined eqn. 4 = 8p + q A Both (condone embedded values for the A) m Valid method to solve two simultaneous equations in p and q to find either p or q q = 6 A AG (condone if left as a fraction) p = 0.5 B 5 OE (b) u 4 = 5 BF Ft on (6 + 4p) (c)(i) L= pl+ q; (L = 0.5 L + 6) OE (ii) q L = m Rearranging p L = 6.5 = 4.8 AF Ft on Total 9 6 p Dependent on previous two marks 5
MPC - AQA GCE Mark Scheme 008 June series MPC (cont) 7(a) 4 + = 4 4 4 + ( ) () + + Any valid method as far as term(s) in / and term(s) in / 4 48 64 + + + = [] 4 6 A p = Accept even within a series 48 A q = 48 Accept 4 even within a series (b)(i) 4 + d = ( p q 64 4 6 + + + ) d Integral of an epansion, at least terms PI by the net line = = q 64 p 5 5 (+ c) 64 6 5 5 (+ c) m AF, 4 At least two powers correctly obtained Ft on c s non-zero integer values for p and q (AF for two terms correct; can be unsimplified) Condone missing c but check that signs have been simplified at some stage before the award of both A marks. (ii) p q 64 (8) 5() q 64 p F() F(), where F() is cand s answer 5 or the correct answer to (b)(i). =.4 A CSO Total 9 6
MPC - AQA GCE Mark Scheme 008 June series MPC (cont) 8(a)(i) h = 0.5 Integral = h/ { } B PI {..}=f(0)+[ f +f()+ f ] +f() OE summing of areas of the four traps. {}= + 6 6 6 6 + + + 6 Condone numerical slip. Accept sf = +[.449..+ 6 + 4.6969..] + 6 A values if not eact. = 7 +.46.. = 8.9 Integral = 0.5 8.9.. = 0.8 (sf) A 4 CAO; must be 0.8 (ii) Relevant trapezia drawn on a copy of Accept single trapezium with its sloping given graph side above the curve {Approimation is an}overestimate A Dep. on 4 trapezia with each of their upper vertices lying on the curve (b)(i) Stretch (I) in -direction (II) Need (I) and one of (II), (III) M0 if more than one transformation (scale factor) (III) A (ii) 6 = 84 PI Take logs of both sides of a = b, PI by log0 6 = log0 84 correct value(s) later or = log6 84 log0 6 = log0 84 m lg84 = lg6 = 0.849. = 0.84 (to dp) A 4 Use of log 6 = log 6 OE or = log6 84 seen Must see that logs have been used before any of the last marks are awarded in (b)(ii). Condone > dp (c) f() = 6 B, B for either 6 + or for 6 + Total 4 9(a) = 48 B PI by = 4º = 80 48 Accept equivalents for = 60 + 48 and = 60+80 48 Accept equivalents for = 4º, 66º, 04º, 46º A 4 CAO; need all four, no etras in given interval (b) sinθ = tanθ cosθ Stated or used sinθ cosθ = 0 tan θ =.5 A θ = 56.º A Condone > dp θ = 56.º + 80º = 6.º AF 4 Ft on c s PV+80º dep only on the provided no etra solutions in the given interval. Total 8 TOTAL 75 7