Mark Scheme (Pre-Standardisation) June 2010

Transcription

1 Mark (Pre-Standardisation) June 00 GCE GCE Core Mathematics C (666/0) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

2 General Marking Guidance All candidates must receive the same treatment. Eaminers must mark the first candidate in eactly 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. Eaminers 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. Eaminers should always award full marks if deserved, i.e. if the answer matches the mark scheme. Eaminers 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 eemplification may be limited. When eaminers 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.

3 June 00 Core Mathematics C 666 Mark Question ( 75 7 ) = 5 = A for 5 or seen A for ; allow or k =, = allow k =, =

4 c A = +, 5+ c A, A for some attempt to integrate: st A for correct, un-simplified n n + or term nd A for both and terms correct and simplified rd A for a fully correct integral including +c

5 6< 8 5< <.8 or or 5 5 A () Critical values are Choosing inside 7 = or 7 < < A () < <.8 ft () 6 for attempt to rearrange to k < m for both correct critical values ft their values and choose the inside region A for fully correct inequality 7 e.g. >, < scores A0 use of or is A0 7 BUT allow > and < to score A Allow.5 instead of 7 ft ft their answers to part and part. If their set is empty allow a suitable description in words or the symbol.

6 6 p = or ( + ) + or q = () y U shape with min in nd quad U shape ) only () b ac= 6 = 8 (-, ) is not required on the sketch A () 6 for some correct substitution into b ac

7 5 a = ( + ) = 7 a = "their 7" + = 0 ft () a = a 5 = + = * A cso () st for 7 only nd ft follow through their 7 in correct formula provided they have n,where n is an integer. for an attempt to find a Acso for a correct solution - no incorrect working seen

8 6 (-5, ) y Horizontal translation of ± (-5, ) marked on sketch or in tet (0,-5) and min visibly on y-ais A () (0, -5) (-, 6) y Correct shape and through (0,0) (-, 6) marked on graph or in tet (, -0) marked on graph or in tet () (, -0) ( a = ) 5 () 7 y for a horizontal translation of ± so accept i.e ma in st quad

9 7 + = + ( ) y =,, + + A A A A 6 st for attempting to divide(one term correct) st A for both terms correct, accept for nd n n for an attempt to differentiate for at least one term nd A for rd A for or allow for th A for allow for. Both terms needed.

10 8 m AB 0 = = 7 5 Equation of AB is: y 0 = ( ) 5 5y = 0 A () (d) AB = (7 ) + ( 0) = Using isos triangle with AB = AC then t = = = 8 Area of triangle = t (7 ) = 0 st for attempt at gradient of AB. Some correct substitution in correct formula. nd for an attempt at equation of AB. Follow through their gradient. A requires integer form. y A A () () A () 8 for an epression for AB or AB (d) for an epression for the area of the triangle - follow through their t.

11 9 a + 9d = 0.75 A () 0 0 S0 = a+ l a+ d So 005 = 5[ a ] * ( ) or ( 9 ) 67 = a so a = 6.5 9d = =.5 so d = 0.50 or 50p for attempt to use a + (n - )d with n =0 A as written A for an attempt to use an S n formula with n =0 for forming an equation with 005 and S n and simplifying to printed answer. st for an equation for a nd for an equation for d, can follow through their a Acso () A A () 8

12 0 y (i) shape Passing through origin and (,0) O 7 (ii) correct shape (-ve cubic) with minimum on -ais Minimum at (0,0) Passes through (7,0) (5) ( ) = ( 7 ) ( 0 8 ) 0 = [7 ( )] o.e. 0= 8+ * ( ) 8± 6 6 = + = 8± = = ± From sketch A is = So y = ( )( [ ] ) = + 8 for forming a suitable equation for a factor of taken out correctly. Acso no incorrect working seen st for some use of the correct formula or full attempt to solve leading to = st A for a fully correct epression - condone + instead of + for simplifying 8 =. Can be scored independently of this epression nd A for correct answer - can be + or + or - nd for selecting their answer in the interval (0,) rd for attempting y = using their rd A for correct answer only Acso () A A A (7) 5

13 5 ( y = ) ( + c) f() = 5 5= c f( ) 0 = m = = 7.5 or 5 y 5= Equation is: ( ) c = 9 y 5+ 50= 0 o.e. AA A (5) A A () st n n for an attempt to integrate + st A for at least correct terms in nd A for all terms in correct (condone missing +c at this point) nd for using the point (, 5) to form a linear equation for c. Must use = and y = 5 rd A for c = 9. The final epression is not required. st for an attempt to evaluate f() nd for using their value of m to form an equation of the line through (,5) st A for any correct epression for the equation of the line nd A for any correct equation with integer coefficients 9