AS Further Mathematics

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1 AS Further Mathematics Paper Mark scheme Specime Versio.

2 Mark schemes are prepared by the Lead Assessmet Writer ad cosidered, together with the relevat questios, by a pael of subject teachers. This mark scheme icludes ay amedmets made at the stadardisatio evets which all associates participate i ad is the scheme which was used by them i this examiatio. The stadardisatio process esures that the mark scheme covers the studets resposes to questios ad that every associate uderstads ad applies it i the same correct way. As preparatio for stadardisatio each associate aalyses a umber of studets scripts. Alterative aswers ot already covered by the mark scheme are discussed ad legislated for. If, after the stadardisatio process, associates ecouter uusual aswers which have ot bee raised they are required to refer these to the Lead Assessmet Writer. It must be stressed that a mark scheme is a workig documet, i may cases further developed ad expaded o the basis of studets reactios to a particular paper. Assumptios about future mark schemes o the basis of oe year s documet should be avoided; whilst the guidig priciples of assessmet remai costat, details will chage, depedig o the cotet of a particular examiatio paper. Further copies of this mark scheme are available from aqa.org.uk Copyright 07 AQA ad its licesors. All rights reserved. AQA retais the copyright o all its publicatios. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their ow iteral use, with the followig importat exceptio: AQA caot give permissio to schools/colleges to photocopy ay material that is ackowledged to a third party eve for iteral use withi the cetre.

3 Mark scheme istructios to examiers Geeral The mark scheme for each questio shows: the marks available for each part of the questio the total marks available for the questio markig istructios that idicate whe marks should be awarded or withheld icludig the priciple o which each mark is awarded. Iformatio is icluded to help the examier make his or her judgemet ad to delieate what is creditworthy from that ot worthy of credit a typical solutio. This respose is oe we expect to see frequetly. However credit must be give o the basis of the markig istructios. If a studet uses a method which is ot explicitly covered by the markig istructios the same priciples of markig should be applied. Credit should be give to ay valid methods. Examiers should seek advice from their seior examier if i ay doubt. Key to mark types M dm R A B E F mark is for method mark is depedet o oe or more M marks ad is for method mark is for reasoig mark is depedet o M or m marks ad is for accuracy mark is idepedet of M or m marks ad is for method ad accuracy mark is for explaatio follow through from previous icorrect result Key to mark scheme abbreviatios CAO CSO ft their AWFW AWRT ACF AG SC OE NMS PI SCA sf dp correct aswer oly correct solutio oly follow through from previous icorrect result Idicates that credit ca be give from previous icorrect result aythig which falls withi aythig which rouds to ay correct form aswer give special case or equivalet o method show possibly implied substatially correct approach sigificat figure(s) decimal place(s)

4 Examiers should cosistetly apply the followig geeral markig priciples No Method Show Where the questio specifically requires a particular method to be used, we must usually see evidece of use of this method for ay marks to be awarded. Where the aswer ca be reasoably obtaied without showig workig ad it is very ulikely that the correct aswer ca be obtaied by usig a icorrect method, we must award full marks. However, the obvious pealty to cadidates showig o workig is that icorrect aswers, however close, ear o marks. Where a questio asks the cadidate to state or write dow a result, o method eed be show for full marks. Where the permitted calculator has fuctios which reasoably allow the solutio of the questio directly, the correct aswer without workig ears full marks, uless it is give to less tha the degree of accuracy accepted i the mark scheme, whe it gais o marks. Otherwise we require evidece of a correct method for ay marks to be awarded. Diagrams Diagrams that have workig o them should be treated like ormal resposes. If a diagram has bee writte o but the correct respose is withi the aswer space, the work withi the aswer space should be marked. Workig o diagrams that cotradicts work withi the aswer space is ot to be cosidered as choice but as workig, ad is ot, therefore, pealised. Work erased or crossed out Erased or crossed out work that is still legible ad has ot bee replaced should be marked. Erased or crossed out work that has bee replaced ca be igored. Choice Whe a choice of aswers ad/or methods is give ad the studet has ot clearly idicated which aswer they wat to be marked, oly the last complete attempt should be awarded marks.

5 Q Markig Istructios AO Mark Typical Solutio Circles correct aswer AO.b B y = 0 Total Circles correct aswer AO.b B 3 Total 3 Circles correct aswer AO.b B y x 3 Total 4(a) Fids k AO.b B k 0 k 4(b) States correct trasformatio AO. B Reflectio i the y-axis 4(c)(i) Fids product BA Allow oe slip Obtais iverse FT their BA provided awarded AO.b AF Fids A ad B AO.b B BA = k k BA k A k k A B Obtais correct ad shows that kk k thus completig verificatio Must clearly show A B method for this mark disallow if aswer simply stated AO. R 0 B 0 A B k k That is the same as (BA) k BA k A B 5 of 8

6 Q Markig Istructios AO Mark Typical Solutio 4(c)(ii) Uses equatio for idetity from defiitio AO3.a We require to demostrate that: (NM) {M N } = I Comeces argumet by maipulatig the matrix products withi the equatio with clear pairig AO. R (NM) M N = N(M M )N = N I N = NN Clearly demostrates that M M = I used AO.4 B = I Completes the argumet usig rigorous reasoig with defiitio of matrix iverse ad associativity metioed AO. R Usig defiitio of matrix iverse ad associativity of matrix multiplicatio Must see all workig with correct pairig of each matrix with iverse Hece true for all o-sigular matrices N ad M Total 0 6 of 8

7 Q Markig Istructios AO Mark Typical Solutio 5 Obtais y AO.b B y =9+6 x + x Itegrates their y withi the itegral to fid the volume of revolutio with at least two terms correct (codoe missig π) 4 Volume = ( 9+ 6 x + x)dx 3 x = 9x4x 4 Obtais all terms correctly AO.b AF FT their y, provided awarded Substitutig limits Substitutes correct limits ito their volume expressio FT provided previous awarded Completes a fully correct argumet to obtai correct expressio with y dx foud at some earlier stage i the workig AG AO. A Hece volume = 5 AG Total 5 7 of 8

8 Q Markig Istructios AO Mark Typical Solutio 6(a) Uses defiitios of sih x ad cosh x to obtai expressio for tah x AO. B x e e sih x x Multiplies by e x Obtais e x AO.b A e cosh x e tah x e x x e e e x x x x Completes a fully correct argumet by demostratig result by takig logs AO. R Multiplyig umerator ad deomiator by e x This mark is available oly if all previous marks have bee awarded e t e x x te t e x x [or multiplies by e x i te x te x e x e x ] t e x e x t t t t x l t hece t x l t 8 of 8

9 Q Markig Istructios AO Mark Typical Solutio 6(b)(i) Expresses cosh 3x ad cosh x i expoetial form See aywhere i solutio Expads LHS FT their LHS provided first awarded Allow oe slip AO. B To be prove: e x e x 3x e e 3x x x 3e e 4 4 LHS x e e x 3 3 Simplifies ad collects terms FT their expressio Allow oe slip Completes fully correct proof to reach the required result This mark is available oly if all previous marks have bee awarded AO. R (e 3 x x x x x x 3e. e 3e. e e 3 ) 8 3x 3x x x (e e ) e e cosh 3x cosh xrhs 4 4 From the defiitio of cosh x 6(b)(ii) Substitutes for cosh 3x i equatio from part (b)(i) Allow oe slip Obtais equatio i cosh x ad solves it Allow oe slip A03.a Elimiates 0 ad with reaso AO.4 E States correct solutio i exact log form AO.b A Total 3 3 cosh x 3cosh x cosh x cosh x 4cosh x 0 cosh x cosh x 4 0 Solutios are cosh x 0,, solutios 0 ad are ot possible sice rage of cosh x cosh x x l 3 9 of 8

10 Q Markig Istructios AO Mark Typical Solutio 7(a) Models light beams as straight lies ad forms vector equatios for straight lies usig a suitable origi Forms correct vector equatio for a lie. Allow oe slip Forms correct vector equatio for secod lie. Allow oe slip Forms equatios for two compoets usig their model FT their lies Solves their equatios correctly FT their lies Checks with third compoet ad cocludes that the beams of light itersect This mark is available oly if all previous marks have bee awarded AO3.3 AO.b A Modellig beams of light as straight lies takig the origi as poit A: r A AO.b A r B AO AO.b AF 3 ad 5 AO. R Itersect 0 of 8

11 Q Markig Istructios AO Mark Typical Solutio 7(b) Evaluates scalar product for their directio vectors. (PI) Sets up equatio to fid agle. (PI) FT oly if previous awarded AO3.a cos cos Obtais correct agle. AO.b A 7(c) States appropriate refiemet. AO3.5c E Take accout of the width of the beams. Total 0 of 8

12 Q Markig Istructios AO Mark Typical Solutio 8(a)(i) States max value for r AO.b B Maximum value of r = 5 States mi value for r AO.b B Miimum value of r = (ii) Draws simple closed curve eclosig pole Draws correct shape with dimple (ot cusp) whe θ = π AO.b A (b) Equates 3 + cosθ = 8cos θ 3 + cosθ = 8cos θ Solves their quadratic equatio FT their equatio oly if has bee awarded 8cos cos 3 = 0 4cos 3 cos = 0 Obtais values for for each value of cos FT their equatio oly if both marks have bee awarded Substitutes their cos ito a polar equatio to fid a value of r FT their cos oly if both marks have bee awarded AO.b AF cos = 3, cos or 5.56 or Obtais both values of r correct for their cos values FT their cos oly if both marks have bee awarded AO.b AF cos 3 r 9 4 cos r Deduces that four values for exist ad expresses poits i required form AO.a R Itersectio poits 9, 0.73, 9 π 4π, 5.56,,,, 3 3 Total 0 of 8

13 Q Markig Istructios AO Mark Typical Solutio 9(a) Draws circle with cetre + 0i Igore other features 5 Im (z) Draws circle passig through (0, 0), (4, 0), close to (, ) ad, with Imagiary axis tagetial AO.b A 5 O 5 Re (z) 5 (b) Uses mod/arg forms AO3.a z = cos isi 3 3 Substitutes exact values for cos ad si Allow oe slip 3 i Obtais result i exact form AO.b A z 3 3 i Total 5 3 of 8

14 Q Markig Istructios AO Mark Typical Solutio 0(a) Splits up the sum ito separate sums ar + br + ( c ) PI Substitutes for the two sums ad r r r r AO3.a r = r r r r r ( r 3r ) + r 3 r + r S = r Allow oe slip States or uses r PI AO. B 3 6 = Now 63 rr r Factorises out ( + ) Allow oe slip = Simplifies 9 ( ){ 6} to fid secod liear factor from their quadratic FT their quadratic provided all marks have bee awarded Allow oe slip 9 = 6( ) 9 ( ){ 6} = = 56 Completes a rigorous argumet to show the required result AO. R = )( 3 To obtai this mark factorisig must be clearly see ad all previous marks obtaied 4 of 8

15 Q Markig Istructios AO Mark Typical Solutio 0(b) Chooses a multiple of 4 for ad obtais a correct umerical value/expressio AO.4 E Whe = 4, r r r 63 = (5)(6)(7) Clear argumet with cocludig statemet AO.3 E = 0 which is ot a multiple of so Alex s statemet is false. Total 8 5 of 8

16 Q Markig Istructios AO Mark Typical Solutio Writes ad i the form p qi (see aywhere i the solutio) AO.5 B Real coefficiets β pqi ad γ pqi Uses sum of the roots = b/a together with a cojugate pair to determie the real part (p) of ad AO3.a α βγ 8 pqi pqi8 p 8 p 3 Uses (their p) ad the area of the triagle o a Argad diagram to determie the imagiary parts of ad AO3.a ( p) q8 q 8 Im q p Re q Uses a correct method to fid the value of c or d usig their values of p qi β 3 8i ad γ 3 8 i Obtais correct values for c ad d. CAO AO.b A d αβγ 46 c αβ 85 Total 5 6 of 8

17 Q Markig Istructios AO Mark Typical Solutio (a)(i) Elimiates y 5x x k x 4x4 Obtais a quadratic equatio i the form Ax Bx C 0, PI by later work AO3.a k x 4x4 5x x ( k5) x 4( k3) x4( k3) 0 = (A) Obtais b 4 ac i terms of k for their quadratic FT their quadratic provided first awarded AO.b AF y k itersects C so roots of (A) are real b 4 ac Obtais iequality, icludig 0, where k is the oly ukow for their discrimiat FT their discrimiat provided both marks have bee awarded [ 4( k3)] 4( k5)( 4( k3)) k k k k k k Completes a rigorous argumet to show that k 3k 0 This mark is available oly if all previous marks have bee awarded AO. R k k 3 0 k k 3 0 (a)(ii) Obtais critical values AO.b B Critical values are 3 ad Deduces that k 3 for AO.a R maximum k 3 (or k ) satisfy iequality For max pt, k 3 Substitutes for ito their quadratic from (a)(i) FT their quadratic oly if first awarded i (a)(i) Sub k = 3 i (A) gives 8 x =0 Max pt of C is ( 0, 3 ) States coordiates of max pt NMS 0/4 Must be usig (a)(i) AO.b A CAO 7 of 8

18 Q Markig Istructios AO Mark Typical Solutio (b) Uses discrimiat to determie solutio AO.4 E 4(5)() < 0 Deduces o vertical asymptotes with clear reasoig with referece to deomiator AO.a R k 0 Deomiator, 5x x of is ever 0 so C has o vertical asymptotes. f x (c) Obtais y = AO3.a B y = Total TOTAL 80 8 of 8

Version 1.0: abc. General Certificate of Education. Mathematics MFP2 Further Pure 2. Mark Scheme examination - January series

Version 1.0: abc. General Certificate of Education. Mathematics MFP2 Further Pure 2. Mark Scheme examination - January series Versio.0: 008 abc Geeral Certificate of Educatio Mathematics 660 MFP Further Pure Mark Scheme 008 examiatio - Jauary series Mark schemes are prepared by the Pricipal Examier ad cosidered, together with