Mark Scheme (Results) Jauary 013 Iteratioal GCSE Further Pure Mathematics (4PM0/01)
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Geeral Markig Guidace All cadidates must receive the same treatmet. Examiers must mark the first cadidate i exactly the same way as they mark the last. Mark schemes should be applied positively. Cadidates must be rewarded for what they have show they ca do rather tha pealised for omissios. Examiers should mark accordig to the mark scheme ot accordig to their perceptio of where the grade boudaries may lie. There is o ceilig o achievemet. All marks o the mark scheme should be used appropriately. All the marks o the mark scheme are desiged to be awarded. Examiers should always award full marks if deserved, i.e. if the aswer matches the mark scheme. Examiers should also be prepared to award zero marks if the cadidate s respose is ot worthy of credit accordig to the mark scheme. Where some judgemet is required, mark schemes will provide the priciples by which marks will be awarded ad exemplificatio may be limited. Whe examiers are i doubt regardig the applicatio of the mark scheme to a cadidate s respose, the team leader must be cosulted. Crossed out work should be marked UNLESS the cadidate has replaced it with a alterative respose. Types of mark o M marks: method marks o A marks: accuracy marks. Ca oly be awarded if the relevat method mark(s) has (have) bee gaied. o B marks: ucoditioal accuracy marks (idepedet of M marks) Abbreviatios o cao correct aswer oly o ft follow through o isw igore subsequet workig o SC - special case o oe or equivalet (ad appropriate) o dep depedet o idep idepedet o eeoo each error or omissio No workig If o workig is show the correct aswers may score full marks If o workig is show the icorrect (eve though early correct) aswers score o marks.
With workig If there is a wrog aswer idicated always check the workig ad award ay marks appropriate from the mark scheme. If it is clear from the workig that the correct aswer has bee obtaied from icorrect workig, award 0 marks. Ay case of suspected misread which does ot sigificatly simplify the questio loses two A (or B) marks o that questio, but ca gai all the M marks. Mark all work o follow through but eter A0 (or B0) for the first two A or B marks gaied. If workig is crossed out ad still legible, the it should be give ay appropriate marks, as log as it has ot bee replaced by alterative work. If there are multiple attempts show, the all attempts should be marked ad the highest score o a sigle attempt should be awarded. Follow through marks Follow through marks which ivolve a sigle stage calculatio ca be awarded without workig sice you ca check the aswer yourself, but if ambiguous do ot award. Follow through marks which ivolve more tha oe stage of calculatio ca oly be awarded o sight of the relevat workig, eve if it appears obvious that there is oly oe way you could get the aswer give. Igorig subsequet work It is appropriate to igore subsequet work whe the additioal work does ot chage the aswer i a way that is iappropriate for the questio: eg. icorrect cacellig of a fractio that would otherwise be correct. It is ot appropriate to igore subsequet work whe the additioal work essetially shows that the cadidate did ot uderstad the demad of the questio. Liear equatios Full marks ca be gaied if the solutio aloe is give, or otherwise uambiguously idicated i workig (without cotradictio elsewhere). Where the correct solutio oly is show substituted, but ot idetified as the solutio, the accuracy mark is lost but ay method marks ca be awarded. Parts of questios Uless allowed by the mark scheme, the marks allocated to oe part of the questio CANNOT be awarded i aother
Questio Number Scheme Marks 1. 8 6 R (a) B1 B1 B1 (3) O 4 4 6 3 (b) B1 (1) [4]. b 4ac > 0 16 p 36 0 4 p 9 0 > > or 9 p > A1 4 ( p )( p ) 3 + 3 > 0 3 3 p< ad p > A1 [4] 3. (a) ( ) ( f x = 3 x + x) + 7 ( ) A x + Bx+ B + C ( x ) = 3 + 1 + 7 3 1 A AB AB C = 3 = 6 + = 7 ( x ) = 3 + 1 + 4 A= 3 B = 1 C = 4 A,1,0 (3) (b) (i) x = 1 (ii) 1 4 B1ft B1ft () [5]
Questio Number 4(a) ( ) r= 1 1 Scheme 3r 4 = 3 r 4 Marks = 3 ( 1+ ) 4 or = 3 ( + ( 1) 1) 4 A1 = ( 3 5) * A1 (3) 50 50 10 (b) ( 3r 4) = ( 3r 4) ( 3r 4) r= 11 r= 1 r= 1 (c) ( ) 50 10 3 50 5 3 10 5 = 3500 A1 () = ( ) ( ) 3 5 = 186 3 5 37= 0 ( )( ) 3 + 31 1 = 0 or formula dep = 1 A1 (3) [8] 5(a) v= 0= 5cost π cos t = 0 t = ( = 0.7853... ) s 4 A1 () (b) dv a = = 10si t dt mag a max = 10 m/s depa1 (3) (c) v= 5cost s = 5costdt 5 s = si t ( + c) A1 t = 0, s = c = dep π t = s =.5 + = 4.5 m A1 (4) [9] 4
Questio Number 6 (a) Scheme si C si 8 = 10 6 Marks si 8 si C = 10 C = 51.49 A1 6 DBC = 180 51.49 = 77.0 (or DBC = (90 51.49) = 77.0 ) A1 (4) (b) ABD = 3.49 B1 AD 10 6 = or = si 3.49 si18.51 si 8 or AD = 6 + 10 6 10cos 3.49 AD = 5.093... = 5.09 (cm) A1 (3) (c) 1 Area = 10 6si100.51, = 9.49... = 9.5 (cm ) A1,A1 (3) [10] 7 (a) radius = ( 3 ) ( 3 1) 5 (.36... ) + = = A1 () (b) B is ( 1, 1) B1, B1 () (c) DE = + 4 = 0 = 5 diameter (or fid the mid-poit of DE) A1 () (d) CP ( x ) ( y 1) = + A1 () (e) CP 5 ( x ) ( y 1) = = + x x y y 4 + = 0 * A1 () [10]
Questio Number Scheme Marks 8 (a) 5siθ = 1 1 siθ = 5 θ = 0.01,.94 A1A1 (3) π (b) ta θ + = 0.4 3 π θ + = ( 0.3805... ), 3.5..., 6.663... 3 θ = 1.4,.81 41 cos 7cosθ = (c) ( θ) depa1a1 (4) 9 4cos θ + 7 cosθ = 0 ( θ )( θ ) 4cos 1 cos + = 0 A1 cosθ = 0.5 dep θ = 1.3 A1 (4) [11] (a) S = ( + 3) ( ) = 1 S = a= 1 + 3 = 5 * A1 () (b) S = 4 ( + 3) = 14 A1 14 5 ( 5 d) d 4 = + + = A1 (3) (c) 1th term = a+ 11d = 5 + 11 4 = 49 A1 () (d) S + p 4 1 = + S p ( p 4)( p 8 3) 1 p( p 3) + + + + = + + 19 + 45= 4 + 6 p p p p p 13p 45 0 = A1 ( p )( p ) + 5 9 = 0 dep p = 9 A1 (4) [11]
Questio Number 10 (a) α + β = 5 Scheme 1 αβ = B1 Marks 5 1 1 ( ) α + β = α + β αβ = = 4 A1 (3) 4 4 1 1 α + β = α + β α β = 4 433 A1 = * 16 () (b) ( ) (c) 1 1 1 1 α + + β + = α + β + + α β α β β + α αβ 1 1 105 4 1 4 = α + β +, = + 4 =,A1 1 1 β α 1 + + = + + + α β α β α β α β α β 4 4 β + α 1 1 433 5 α β + + = + + = α β α β 16 1 1 oe A1 105 5 x sum x+ product ( = 0) x x+ 4 ( = 0) 4x 105x+ 450 = 0 A1 (7) [1]
Questio Number Scheme Marks 11 (a) r = B1 (1) x 1 x 3 x+ = x + px + qx+ 6 (b) ( )( )( ) 3 ( )( ) 3 x 4x+ 3 x+ = x x 5x+ 6 p =, q = 5 A1A1 (3) (c) = 5 + 6 3 y x x x dy 3x 4x 5 dx = dy x = = 1 8 5= 1 A1ft dx y-coord of B is 8 8 10+ 6= 4 B1ft Equatio taget: y+ 4= 1( x ) ( y x ) = A1 (5) (d) A is (,0) B1ft 0+ 4= 1( ) passes through A B1ft () { } 3 (e) = ( + ) ( ) x x x x x area 5 6 d 3 ( + ) x x 4x 8 dx 1 = + 8 4 3 4 3 x x x x A1ft 16 16 = 4 8 + 16 4 + 8 16 3 3 1 = 1 (accept awrt 1.3) 3 dep A1 (5) [16]
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