Version.: 6 General Certificate of Education abc Mathematics 66 MPC Pure Core Mark Scheme 6 examination - January series Mark schemes are prepared by the Principal Examiner 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 examiners and is the scheme which was used by them in this examination. The standardisation meeting ensures that the mark scheme covers the candidates responses to questions and that every examiner understands and applies it in the same correct way. As preparation for the standardisation meeting each examiner 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, examiners encounter unusual answers which have not been discussed at the meeting they are required to refer these to the Principal Examiner. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded 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 examination paper.
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 explanation 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 ) accuracy marks NOS not on scheme x EE deduct x 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 Examiner 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.
AQA GCE Mark Scheme, 6 January series MPC MPC (a) ( 5) + 5 5 = A (b) 8 = ; 8 = Either correct Multiplying out or difference of two squares attempted Full marks for correct answer /no working Answer = 5 A Full marks for correct answer /no working Total (a)(i) 5 + k = 7 k = 8 k = B AG (condone verification or y = ) x x ( + ) or ( + ) y y Midpoint coordinates (, ) A One coordinate correct implies (b) Attempt at y/ x or y = x+ 7 Gradient AB = A (Not x over y)(may use M instead of A/B).75 etc any correct equivalent (c)(i) mm = used or stated Hence gradient AC = A Follow through their gradient of AB from part (b) y = ( x ) or y = x etc B Follow through their gradient of AC from part (c) (i) must be normal & (,) used (iii) y = x = x = Total (a)(i) (x ) B p = + 5 B q = 5 Putting y = in their AC equation and attempting to find x A CSO. C has coordinates, Minimum point (, 5) or x=, y = 5 B B for each coordinate correct or ft Alt method, A sketch, differentiation (b)(i) x= x x+ 9 x x = B Or + 9+ = x x x AG (be convinced) (must have = ) ( x )( x+ ) = Attempt at factors or quadratic formula or one value spotted x =, A Both values correct & simplified Substitute one value of x to find y May substitute into equation for L or C Points are (, 6) and (, ) A y-coordinates correct linked to x values
MPC AQA GCE Mark Scheme, 6 January series MPC (cont) (a) ( m+ ) = m + 8m+ 6 b ac= ( m+ ) (m+ ) = m + 8m+ 6 6m = m 8m+ = B A Condone m + m b ac(attempted and involving m s and no x s) or b ac= stated AG (be convinced all working correct- = appearing more than right at the end) (b) ( m )( m 6) = Attempt at factors or quadratic formula m =, m = 6 A SC B for or 6 only without working Total 5 5(a) ( x ) + ( y+ ) B B for one term correct = 6 B Condone 6 ( + 6 + 9 = 6) RHS (b)(i) Centre (, ) B Ft their a and b from part (a) Radius = 6 B Ft their r from part (a) (c)(i) ( ) CO = + Accept + or with numbers but must add CO = 5 A Full marks for answer only Considering CO and radius CO < r O is inside the circle A Ft outside circle when their CO > r or on the circle when their CO = r SC B if no explanation given 6(a)(i) p( ) = 8+ + 8 Finding p() M long division =, x is a factor A Shown = AND conclusion/ statement about x being a factor Attempt at quadratic factor or factor theorem again for nd factor x + x A or (x+) or (x ) proved to be a factor p( x) = ( x )( x+ )( x ) A (b) y B Graph through (,8) 8 marked 8 O x B Ft their factors roots marked on x- axis Cubic curve through their points A Correct including x- intercepts correct Condone max on y-axis etc or slightly wrong concavity at ends of graph
AQA GCE Mark Scheme, 6 January series MPC MPC (cont) 7(a)(i) dv 5 t 8t 6t One term correct unsimplified A Further term correct unsimplified A All correct unsimplified ( no + c etc) d V One term FT correct unsimplified = t t + 6 A CSO. All correct simplified (b) Substitute t = into their d V (= 6 6 + ) = A CSO. Rate of change of volume is m s (c)(i) dv t = = 8+ 6 Or putting their d V = Stationary value A CSO. Shown to = AND statement (If solving equation must obtain t = ) d V Sub t = into their second derivative or t = = 8 equivalent full test. Maximum value A Ft if their test implies minimum Total 8(a) y D = + = or y C = 8 = Attempt at either y coordinate Area ABCD = = A (b)(i) x x ( + C ) A A Increase one power by One term correct unsimplified All correct unsimplified (condone no +C) Sub limits and into their (b) (i) ans May use both, and, instead A [ 8 ] = 5 Shaded area = their (rectangle integral) = 5 = 6 A Alt method: difference of two integrals CSO. Attempted M, A (c)(i) dy 6x x dx One term correct A All correct ( no +C etc) When x =, y = when x =, dy = as their grad of tgt dx B May be implied by correct tgt equation Ft their derivative when x = Tangent is y = ( x ) A Any correct form y = x etc (iii) Decreasing when dy 6 x x dx = < ( x x ) < x x> A Watch no fudging here!! May work backwards in proof. AG (be convinced no step incorrect) (d) Two critical points and x >, x < ONLY A Total 8 TOTAL 75 Marked on diagram or in solution or A for < x < or > x > SC B for x > (or x < ) 5