A-LEVEL Mathematics MPC UNIT: Pure Core Mark scheme 660 June 07 Version:.0 Final
MARK SCHEME A LEVEL MATHEMATICS MPC JUNE 07 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 eamination. 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 epanded 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 eamination paper. Further copies of this mark scheme are available from aqa.org.uk
MARK SCHEME A LEVEL MATHEMATICS MPC JUNE 07 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 eplanation 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 EE deduct 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.
MARK SCHEME A LEVEL MATHEMATICS MPC JUNE 07 Q Solution Mark Total Comment (a) dy d sin cos Asin B Asin sec tan Bcos sec cos cos AB, 0 A, B (b) k ln( ) ( c ) Where k is a constant ln( ) c Must have +c as part of final answer Total (a) Do not allow ( ) for cos cos ( sin sin ) Candidate using the quotient rule correctly, then SC for cos (b) Condone poor use of brackets for but only allow if recovered. Condone 6/ for / or better
MARK SCHEME A LEVEL MATHEMATICS MPC JUNE 07 Q Solution Mark Total Comment All 5 correct values (and no etra used) (a) y PI by 5 correct y values 0.6.80.6 e.87 0.8.0.8 e 6.606 At least correct y values in eact form or e 7.8906 decimals, rounded or truncated to dp or..6. e 6.508 better (in table or formula)... (PI by correct answer) e.8877 (b) 0. (.87.. 6.606......) 5.9 dy d Equating e ( ) OE d Correct sub into formula with h = 0. OE and 5 correct y values either listed, with + signs, or totalled. (PI by correct answer) CAO, must see this value eactly and no error seen Do not condone poor use of brackets for this mark, unless written correctly later dy their 0 d FT their d y, PI by further working d d y d, y e From 0 oe, y e And no etra answers, coming from eponential terms e ( 6 ) ( ) e OE Do not condone poor use of brackets for, y "() e ( 6) ;, y"( ) e (6) this mark, unless written correctly later d y Sub both correct values into their d 6e 0 so maimum at 6e 0 so minimum at 7 Total Including inequalities (symbol or wording), and both conclusions. Must have scored nd Final A mark can be earned even if A0A0 earlier (b) May have used quotient rule to find dy e e e e for d ( e ) The values of the nd derivative may be evaluated (to.[] and 0.8[]), for the final If a candidate has values for, then they lose the nd A mark, but subsequent marks are available 5
MARK SCHEME A LEVEL MATHEMATICS MPC JUNE 07 Q Solution Mark Total Comment du sin OE d k u ( u )du m u u du d Condone omission of du Condone omission of brackets and du 5 u u ( c ) 5 OE Must have seen du on an earlier line where all terms are in u only 5 cos cos ( c) OE Condone omission of + c 0 6 5 Withhold final for poor notation eg Can score A0 at end, if there is an omission of du Total 5 5 cos but may have cos 5 for 5 cos 6
MARK SCHEME A LEVEL MATHEMATICS MPC JUNE 07 Q Solution Mark Total Comment (a) 0 f ( ) ln f () 0.(78...) f ().(7...) (or reverse) Both values rounded or truncated to at least sf Change of sign(or different signs) Must have both statement and interval in words or symbols or comparing sides: at, ln(/ ).(79..) ; at, ln(6 / 7) = 0.8( ) () Conclusion as before () (b)(i) 0.87.77 Ignore further values (ii) Vertical line from to the curve, seen or implied, and then horizontal to y All correct with nd vertical and horizontal lines (only required above the y = line), and, labelled on the -ais Total 6 (a) Condone less than or equal to ; allow, root for α but not it 0 Candidates could change f() into eponentials eg f ( ) e leading to f()= 0.5 and f()=5 7
MARK SCHEME A LEVEL MATHEMATICS MPC JUNE 07 Q5 Solution Mark Total Comment Either order for : (a) ln(y ) Interchange and y e y Correctly converting to e form. [f ( )] (e ) ACF [g( ) ] (b) (e ) 9 e 0 e 0 ln Correctly isolating term in e from their f ( ) and their g() Must see an intermediate line AG All correct and no errors seen Total 6 (a) Condone poor use of brackets if recovered (b) Do not condone poor use of brackets even if recovered for If a candidate has equation in terms of e then they must isolate correctly to score 8
MARK SCHEME A LEVEL MATHEMATICS MPC JUNE 07 Q6 Solution Mark Total Comment 0.5 ( ) d u dv (d ) 0.5 ( ) u and attempted dv (d ) correct, with d u d and dv du (d ) ( ) 0.5 v All correct 0.5 0.5 ( ) ( ) (d ) d Correct substitution of their terms into the 0.5.5 ( ) ( ) OE parts formula 5 [ ] (5 7) ( ) d F(5) F(), correct from A( ) B( ) =6 Check that an answer of 6 follows correct working 6 Total 6 0.5.5 9
MARK SCHEME A LEVEL MATHEMATICS MPC JUNE 07 Q7 Solution Mark Total Comment V-shaped mod graphs, one with verte on positive -ais and other with verte on negative -ais 5k Critical values 5 k ( k) OE 9k [ ] 8 PI And no other values 9k 5k, 8 May have OR between two inequalities but not AND 5 Total 5 For first condone line(s) etended, bending to show intersection of lines For, condone other symbols for = To find the cv s, a candidate might have squared and factorised, the is earned for (8 9 k)( 5 k), the accuracy marks are as above. Mark last line as final answer 0
MARK SCHEME A LEVEL MATHEMATICS MPC JUNE 07 Q8 Solution Mark Total Comment (a) tan p sec p [ sec p] sec p sec sec p sec p 0 (sec p )(sec p )[ 0] sec p, p p.[...],.8[...], 5.05[...],.59[...] Correct use of trig identity PI Factorisation or correct use of formula PI Both correct and no errors seen Sight of any of these values correct to dp of these values correct to dp 0.88,.7,.9,.79 7 correct (must be to dp) All correct (must be to dp) and no etras in interval (ignore answers outside interval) (b) Stretch (I) (Parallel to) -ais (or line y = 0) (II) SF (III) (followed by) k Translation through 0 π 6 0 I and (II or III) I + II + III OR k Translation through 0 π 0 (followed by) Stretch ( I ) Parallel to -ais (or line y = 0) (II) SF ( III) () () () () Total As above (a) May use cos and sin leading to cos p /, / for first,, Condone using as p For first mark, this can be implied by one correct final value For second mark, this can be implied by three correct final values, but do not accept values in terms of π
MARK SCHEME A LEVEL MATHEMATICS MPC JUNE 07 Q9 Solution Mark Total Comment (a) Modulus graph, sections Correct on outside sections Correct on inside sections, ma, one on the y-ais (appro.), and correct cusps Ignore any dotted sections (b) Graph with eactly ma and min All correct, symmetrical about y-ais (c)(i) a yb Each value may be stated or given as coordinates (ii) 0.5a y 7b Each value may be stated or given as coordinates Total 9 (b) The min must be at the same depth appro for (c) Condone coordinates written in columns (c)(i) Do not allow 0 a for a, nor b + b for b
MARK SCHEME A LEVEL MATHEMATICS MPC JUNE 07 Q0 Solution Mark Total Comment (a)(i) ln ln, y e (ii) y ln6 e 6 dy e d y6 ( ln ) OE With no eponentials 6 [ y 0] ln ln or [ ln 0.5] y 6 (ln 0.5 ln ) y 0.5 6 0 Must see this line oe (b) [Cone ] π 6 (ln (ln 0.5)) FT their y=6 8 π ln [For curve, Vol ] π (e ) d [ e d ] e 0 AG All correct and no errors seen. Must be using a correct equation from (i), ln (condone e unsimplified) Correct including π, limits, d ln 0 [Vol ] π (e e ) d Correct substitution of correct limits, including π (PI by net ) π (56 ) Correct unsimplified eact value, no eponentials 55 [ π] 55 8 Required ' vol' π F Vol = vol under curve vol of cone, must have scored 5 = π 8 Total (c) Condone poor use of brackets for d, if recovered Condone omission of π for d if recovered