A-LEVEL Mathematics. MFP2 Further Pure 2 Mark scheme June Version: 1.0 Final

Transcription

1 A-LEVEL Mathematics MFP Further Pure Mark scheme 660 June 0 Version:.0 Final

2 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

3 MARK SCHEME A LEVEL MATHEMATICS MFP JUNE 0 Key to mark scheme abbreviations M mark is for method dm mark is dependent on one or more M marks and is for method A mark is dependent on M or dm marks and is for accuracy B mark is independent of M or dm marks and is for method and accuracy E mark is for eplanation FT 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, the principal eaminer may suggest that we 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. Otherwise we require evidence of a correct method for any marks to be awarded. of

4 MARK SCHEME A LEVEL MATHEMATICS MFP JUNE 0 Q Solution Mark Total Comment (a) A B r r M A B 4 4 A and attempt to find A or B 4 4 r r OE (b) A B A B M clear attempt to use method of differences with their A and B n OE dm n k ( ) n ( ) 4(n ) OE A must have n Total 5 condone +,, or ( ) n instead of ( ) n ; may have r for n (b) For dm correct two remaining terms may be on separate lines with other terms crossed out Eample earns M dm (n ) Eample (n ) earns M dm0 Alternative for final A : n even ; n odd 4(n ) 4(n ) 4 of

5 MARK SCHEME A LEVEL MATHEMATICS MFP JUNE 0 Q Solution Mark Total Comment (a) 6 i OE M PI by correct A (b) (i) or i 6 i q or 6i OE i M or i A q 6i OE A q, q or correct unsimplified PI by correct q (ii) 9(6 i) p 6i 0 M correctly substituting their values for and q into equation p 8 i OE A (c) ( ) ( ) M correct identity = 6 6i 9 6 i 4i OE Acso Total 9 (b)(i) Withhold final A if 6 i and = q leads to correct answer and write FIW Do not treat for i as a misread, simply an error (b)(ii) " their " Alternative " their " i ; p " their "( ) M ; p 8 i A (c) Withhold Acso if 6 i is seen even if correct answer is given and write FIW 5 of

6 MARK SCHEME A LEVEL MATHEMATICS MFP JUNE 0 Q Solution Mark Total Comment (a) e e e e cosh and sinh B + (e e ) e e e e 0 ae be c 0 M obtaining term quadratic in e e e 0 OE A correct (e )(e ) 0 dm attempt at factors or correct use of formula or PI by both correct values and e ln A 5 and no other value given (b) ln ( cosh ) d 0 B correct epression for volume all on one line including limits, and d cosh cosh OE B or cosh e e 4 M their cosh term correctly integrated sinh sinh 4 A integral all correct ln sinh ln ln sinh ln 4 (Volume =) (6 48ln ) Allow (48ln 6) A 5 Total 0 (a) If using formula they must have a simplified correct discriminant for their quadratic so if correct they 6 must have e for dm ; if factorising, their factors must multiply out to give their e term and their constant term Alternatives & below involve squaring which if done correctly introduce spurious solutions that must be discarded by testing whether they satisfy the original equation and so c s are likely to lose final A Alternative : cosh sinh cosh cosh 4sinh 4(cosh ) B leading to quadratic in cosh M ; cosh cosh 5 0 (cosh 5)(cosh ) 0 A dm their factors must multiply out to give their cosh term and their constant term ; PI by correct values 5 and ; A for 5 5 cosh cosh (or ln ) eplaining why ln must be rejected and giving the single solution cosh (or ln ). Alternative : sinh cosh 4sinh 4sinh cosh sinh B leading to quadratic 4 in sinh M ; sinh 4sinh 0 A; sinh 0, sinh dm (both) ; final A includes eplaining why 0 must be rejected and giving the single solution 4 5 sinh (or ln ). (b) Allow missing brackets if epanded correctly, namely cosh cosh d for first B 0 Second B and M are available if they have different integrand (sometimes from using CSA formula) ln 6 of

7 MARK SCHEME A LEVEL MATHEMATICS MFP JUNE 0 Q 4 Solution Mark Total Comment (a) 9k k 6 B (b) When n LHS ; RHS Therefore (formula is) true when n = B must have this eplicit statement Assume result is true for n=k (*) Add (k + )th term to both sides k r(r )(r ) k( k )(9k k ) r 6 ( k )(k )(k ) M A adding correct (k+)th term to RHS both sides correct A0 if only RHS considered = ( ) (9 ) 6( )( ) 6 k k k k k k correct quartic is 4 k 44 k 5 k 5 k 6 ( ) k k k k cubic need not have all like terms A collected ( )( )(9 6) 6 k k k k must see this line to earn final A ( )( ) 9( ) ( ) 6 k k k k A from part (a) Hence formula is true for n=k+ (**) and since true for n=, formula is true for n =,,, [by induction] (***) E 6 must have (*), (**) and (***) and have earned previous 5 marks Total (b) For B, accept n= RHS=LHS= but must mention here or later that the result is true when n= Do not allow them to simply say true for all integers n at the end to earn this B mark. This is B0. Alternative to (***) is therefore true for all positive integers n or so true for all integers n etc However, true for all n is incorrect and immediately gets E0 Condone LHS= 5... ( k )( k )( k ) OE for first A but must have 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. of

8 MARK SCHEME A LEVEL MATHEMATICS MFP JUNE 0 Q 5 Solution Mark Total Comment (a)(i) tan arg (ii) i M A M i A or finding angle to Im(z) ais = 6 PI by correct answer PI by correct answer (b)(i) M arc of circle in second quadrant A circle centre at i ( marked on Im(z)-ais) and touching real ais at O Im(z) M A line from O to at least edge of circle inclined at roughly to negative real ais as drawn O Re(z) A 5 must have earned previous 4 marks correct shading of region bounded by line, imaginary ais and circular arc (ii) marked correctly B clear intention to be at intersection point B0 if only line or circle drawn (iii) i Ma value = 4i or 4i M or etc ACF 4 = A correct epression- distance from to 4i that could be evaluated to give correct ans d 4 4 cos 4 6 Total (a)(i) NMS M A; tan or sight of earns M (b)(i) Allow freehand circle and clear intention to touch real ais at origin First A : condone i or (0,) or clear dashes to indicate centre on Im(z) ais or radius indicated as and circle touching real ais at O Second A : award if angle made with negative Re(z) ais is greater than 4 Condone circle/line as dotted lines (iii) NMS ma = scores full marks 8 of

9 MARK SCHEME A LEVEL MATHEMATICS MFP JUNE 0 Q 6 Solution Mark Total Comment tan (d ) k M Integration by parts at least this far (denominator may be ) tan (d ) A or tan (d ) OE A A A OE BF or etc tan + tan A correct unsimplified I tan + tan A correct unsimplified sub of limits = 8 8 = A 6 4 Total 6 Do NOT allow misread of for ; it eases the question considerably Alternative : u ; I u tan u du = u tan u u + tan u A ; then A A as above. u u u tan u k du M; k A; u u u B; d Alternative : tan u ; sec u du ; I u tan usec u du I u tan u k tan u du M k A (correct) replacing tan u by sec u in integral dm ; I u tan u u tan u A ; then A A as above. 9 of

10 MARK SCHEME A LEVEL MATHEMATICS MFP JUNE 0 Q Solution Mark Total Comment (a) ( cosh )cosh sinh sinh M quotient rule correct ( cosh ) Numerator = cosh A correctly simplified cosh sinh dm f ' ( ) sinh A 4 AG no errors seen and f ' ( ) (b)(i) dy d M or or A s d A condone (d ) for M Allow this mark but withhold final A mark if d y or d y not seen d d AG (be convinced) - must have s =, limits and d and must have in numerator (ii) sinh d cosh d OE M or tan d sec d cosh cosh sec sec d (must have d ) sinh A d tan Attempt to split into two terms using cosh sinh M PI by correct split below sinh [d ] sinh A correct split and must have integral sign sinh ln cosh cosh dm integrating sinh correctly sinh s ln ln A correct unsimplified ln A AG partly so be convinced Total (a) Alternative: f ( ) ln(sinh ) ln( cosh ) and one term differentiated correctly M cosh sinh f ' ( ) sinh cosh A = ( cosh )cosh sinh dm (common denominator) sinh ( cosh ) f ' ( ) A (AG no errors seen and f ' ( ) sinh ) (b)(ii) In alternative on RHS; B for using sec tan used in numerator; dm for splitting integrand sectan and dm for integrating correctly sin NB (csc sec tan )d ln(csc cot ) sec 0 of

11 MARK SCHEME A LEVEL MATHEMATICS MFP JUNE 0 Q 8 Solution Mark Total Comment (a) cos isin cos isin B or sin Im part of cos isin PI by later work c c (i s) c (i s) 5 c (i s) c (i s) c (i s) c(i s) (i s) ( s ) (i s) 5( s ) (i s) ( s )(i s) (i s) 5 M A dm condone up to errors in imaginary part of epansion for M ignore real terms correct imaginary terms correct use of c s in at least two 6 imaginary terms i.e. c ( s ) etc 4 6 sin s( s s s ) 5 s ( s s ) s ( s ) s 4 5 A sin sin sin 64sin sin A 6 RHS correct unsimplified epansion and equated to sin AG be convinced terms must be in this order (b)(i) sin 0 n n M condone no mention of but sin 0 so ( ) sin seen or used sin is a root of cubic equation E must earn M and have/use and statement sin other roots are sin & sin OE B 4 accept sin & 5 sin etc but not sin & 4 sin etc (ii) Considering M must relate A = 56 / 64 8 / 64 A do not accept 56 sin etc to,, if using this approach Total (b)(i) Condone reverse argument namely sin 0 for M sin is a root of 0 earns M sin (ii) Alternative : put z / y M new equation z 56z z 64 0 A; sum of these roots = 56 NMS 8 scores no marks = 8 A of