General Marking Guidance •All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. •Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. •Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. •There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. •All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate’s response is not worthy of credit according to the mark scheme. •Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. •When examiners are in doubt regarding the application of the mark scheme to a candidate’s response, the team leader must be consulted. •Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response. www.mymathscloud.com
Mark scheme version dated 20/01/2009 (final) 1EDEXCEL January 2009 Core Mathematics C2 Mark SchemeQuestion number Scheme Marks 1. 243)23(5=−x, ......()xx810)2(354−=−×+ ...... B1, B1()2231080)2(3245xx+=−×+M1 A1 (4) 4Notes First term must be 243 for B1, writing just 53 is B0 (Mark their final answers except in second line of special cases below). Term must be simplified to –810x for B1 The x is required for this mark. The method mark (M1)is generous and is awarded for an attempt at Binomial to get the third term. There must be an 2x (or no x- i.e. not wrong power) and attempt at Binomial Coefficient and at dealing with powers of 3 and 2. The power of 3 should not be one, but the power of 2 may be one (regarded as bracketing slip). So allow 52⎛⎞⎜⎟⎝⎠ or 53⎛⎞⎜⎟⎝⎠ or 52C or 53C or even 52⎛⎞⎜⎟⎝⎠ or 53⎛⎞⎜⎟⎝⎠or use of ‘10’ (maybe from Pascal’s triangle) May see 53 22(3) ( 2 )Cx− or 5322(3) ( 2 )Cx−or 552223(3) ()Cx− or 3210(3) (2 )xwhich would each score the M1A1is c.a.o and needs 21080x (if 21080xis written with no working this is awarded both marks i.e. M1 A1.)Special cases: 2243 810 1080xx++ is B1B0M1A1 (condone no negative signs) Follows correct answer with 227 90 120xx−+ can isw here (sp case)– full marks for correct answer Misreads ascending and gives 54332 240 720xxx−+ − is marked as B1B0M1A0 special case and must be completely correct. (If any slips could get B0B0M1A0)Ignores 3 and expands 5(1 2 )x± is 0/4 243, -810x, 1080x2 is full marks but 243, -810, 1080 is B1,B0,M1,A0 NB Alternative method ()()25555522233353(1 ) 3 5 33..3xxx⎛⎞−=−××+ −+⎜⎟⎝⎠ is B0B0M1A0– answers must be simplified to 243 –810x 21080x+for full marks (awarded as before)Special case ()()2522233353(1 ) 3 5 33..3xxx⎛⎞−=−×× + − +⎜⎟⎝⎠ is B0, B0, M1, A0 Or 53(1 2 )x− is B0B0M0A0www.mymathscloud.com
