Version 1.0. klm. General Certificate of Education June Mathematics. Pure Core 3. Mark Scheme

Transcription

1 Version.0 klm General Certificate of Education June 00 Mathematics MPC Pure Core Mark Scheme

2 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: Copyright 00 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 M5 6EX

3 MPC - AQA GCE Mark Scheme 00 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.

4 MPC - AQA GCE Mark Scheme 00 June series MPC (a) f ( ) = 0 + (or reverse) f() = 6 f( ) = 7 M Attempt to evaluate f() and f() Change of sign < α < A All working must be correct plus statement OR LHS () = RHS () = 9 LHS ( ) = 9 RHS ( ) = At LHS < RHS, < α < At LHS > RHS (M) (A) Must be these values (b)(i) = 0 = 0 This line must be seen = 0 B AG (ii) ( = ) =.9 M Sight of AWRT.9 or AWRT. =. A Both values correct Total 5 4

5 MPC - AQA GCE Mark Scheme 00 June series MPC Condone marked at A, A = etc (a)(i) ( y = ) B but not cos0, sec 0 (ii) y M Modulus graph y > 0 O A + sections roughly as shown, condone sections touching, variable minimum heights A Correct graph with correct behaviour at 4 asymptotes but need not show broken lines; and roughly same minima (b) cos = or cos seen M π or sight of ± 60 or ±, ±.05 (AWRT) = 60, 00 A Condone etra values outside 0 < < 60, but no etras in interval (c) sec( 0 ) =, sec( 0 ) = cos ( 0 ) = or cos ( 0 ) = 0 = 60, 00 or 0 = 0, 40 (ignore values outside 0 < < 60 ) M A Either of these, PI by further working Both correct values from one equation or correct values and no wrong values from both equations, but must have 0 = PI by = 70,0,50,0 = 5,65,5,55 B correct (and not more than etra value in 0 < < 80 ) B 4 All 4 correct (and no etras in interval) Total 0 5

6 MPC - AQA GCE Mark Scheme 00 June series MPC (cont) (a)(i) y= ln (5 ) M k 5 dy 5 5 = A No ISW, eg = d 5 5 (ii) y= sin dy = cos d M k cos A (b)(i) f ( ) ln0.5 or f( ) M ln A (ii) ( gf ( ) ) sin ln ( 5 ) or gf ( ) sin ln 5 = ( ) ( ) = B Condone sin ln 5 ( ) or sin ( ln ( 5 ) ) but not sin (ln 5 ) or sin ln 5 (iii) gf ( ) = 0 ( ) ln( 5 ) 0 sin ln 5 = 0 = M Correct first step from their (b)(ii) = m Their ( ) k ( ( ) ) 5 = A 5 f = from ln f = 0 Withhold if clear error seen other than omission of brackets (iv) = sin y sin ( g ( ) ) = y ( or sin y = ) M Correct equation involving = sin A Total sin 6

7 MPC - AQA GCE Mark Scheme 00 June series MPC (cont) 4(a) y = B values correct PI = 0.5 = = = = 0. 9 = B M = 0.5[ ] At least 5 y values that would be correct to sf or better, or eact values. May be seen within working. Clear attempt to use their y values within Simpson s rule = A 4 Answer must be with no etra sf (Note with no evidence of Simpson s rule scores 0/4) (b) 0 d + ln = ( + ) M kln ( ) = ln ( + ) ln A m + condone missing brackets Correct. A may be recovered for missing brackets if implied later F() ( F(0)) = ln A 4 ln must not be left in final answer Alternative u = l + du= d du = u (M) = [ ln u] (A) = ln ln (m) du du correct and integral of form k d Correct substitution of correct u limits or conversion back to and F() ( F(0)) u ln = (A) ln must not be left in final answer Total 8 7

8 MPC - AQA GCE Mark Scheme 00 June series MPC (cont) 5(a) 0cosec = 6 cot 0( + cot ) = 6 cot 0cot cot = B AG Must see evidence of correct identity and no errors. (b) Attempt at factors, giving when epanded. ± ± 0cot 6 M Use of formula: condone one error ( 5cot )( cot ) ( 0) + = A Correct factors cot =, 5 5 tan =, A,A 4 st A must be earned Condone AWRT 0.67 ISW if values attempted Alternative 0cot + cot 6 = 0 cos cos = 0 sin sin 0cos + cos sin 6sin = 0 5cos sin cos+ sin = 0 (M) ( )( ) ( ) ( 5cos = sin cos= sin ) 5 = tan = tan (A) (A), (A) Alternative 0 + tan 6 tan = 0 ( 5 tan)( + tan) ( = 0) (M) (A) 5 tan =, (A), (A) Total 5 Attempt at factors, gives ± 0cos ± 6sin when eplained As above st A must be earned Condone AWRT 0.67 ISW if values attempted Attempt at factors gives ± 0 ± 6tan st A must be earned Condone AWRT 0.67 ISW if values attempted 8

9 MPC - AQA GCE Mark Scheme 00 June series MPC (cont) 6(a) ln y = (when) y= 0 = or (, 0) Both coordinates must be stated, not B simply shown on diagram (b) dy = d ln ln = or ln ± ± ln M Quotient/product rule A OE must simplify At B, ln = 0 m Putting their d y = 0 or numerator = 0 d = e A CSO condone = e or e y = e A 5 CSO must simplify ln e (c) Gradient at = e lne = M (e ) Gradient of normal Substituting = (condone slip) but must have scored M in (b) = or e 6 6 e A PI e 6 = A CSO simplified to this Total 9 dy e into their d 9

10 MPC - AQA GCE Mark Scheme 00 June series MPC (cont) 7(a)(i) cos 4 d u= dv d = cos4 M cos4, ( ) d attempted d du sin 4 = v = d 4 A All correct sin 4 sin 4 Correct substitution of their terms into = d 4 4 parts formula sin 4 cos4 = + ( + c) A 4 OE with fractions unsimplified 4 6 m (ii) dv sin 4 d u = = sin 4 d du cos 4 = v= d 4 d attempted d M sin 4, ( ) cos 4 cos 4 = d 4 A 4 cos 4 = + cos 4 d 4 cos 4 = + [ ] 4 sin 4 cos4 Clear attempt to replace integral using m their answer from part (a)(i) cos4 sin4 cos4 = + + ( + c) A 4 OE with fractions unsimplified 4 8 ( 0.) (b) V ( π) ( 64) sin 4 ( d) = M ( 0) cos4 sin 4 cos4 Must see evidence of their (a)(ii) result or = ( π 64) starting again obtaining terms of the form ± A cos 4± Bsin 4± Ccos4 = m AND F(0.) F(0) attempted π[.0959 ] 0.99 = AWRT A Accept AWRT π Total 0

11 MPC - AQA GCE Mark Scheme 00 June series MPC (cont) 8(a) y = e e Stretch (I) scale factor (II) M I + (II or III) in -direction (III) A I + II + III Translation E Allow translate 0 B 4 OE unit down etc (b) = 0 y = 6 or (0, 6) B Both coordinates must be stated, not simply 6 marked on diagram (c)(i) e = 4e + 4 e e = 4+ e e e = 4+ e or ( ) (e ) e 4 0 M Multiplying both sides by = A AG With no errors seen (ii) ( e 4)( e + ) M ( e ± 4)( e ± ) = ln or ln 4 Reject e A = OE A eg e > 0, e e, impossible etc 4e d (I) (d) ( + ) 4e = + ln 0 ln 4e 4 = + ln + 0 = + ln++ = + ln ( e ) d (II) e = ln 0 ln e = ln 0 = ln = ln A = + ln ln M m A B I or II attempted and e correctly or e F[ their ln from (c)(ii)] F[0] Both I and II correctly integrated Attempt to find difference of their I their II integrated = ln or ln 8 or ln 4 OE A 5 CSO must be eact

12 MPC - AQA GCE Mark Scheme 00 June series MPC (cont) 8(d) Alternative A = 4e + d e d ( ) ( ) (B) Condone functions reversed ( ln ) ( 4e e ) d = + ( 0) ln 4e e = + 0 ln ln = e e + ln (M) (A) (m) e or e correctly integrated Correct substitution of their ln from (c)(ii) into their integrated epression = ln or ln 8 or ln 4 OE (A) CSO must be eact Total 5 TOTAL 75