June 006 6663 Core Mathematics C Mark Scheme Question number (+c) Scheme Marks = 3 n n for some attempt to integrate Total 4 marks st 6 3 A for either or or better 3 for all terms in correct. Allow and. for + c, when first seen with a changed epression. Critical Values 7 49 7 ( a)( b) with ab=8 or or 7 7 8 7 7 ( 9)( + ) or or Solving Inequality > 9 or < Choosing outside For attempting to find critical values. Factors alone are OK for, = appearing somewhere for the formula and as written for completing the square st A Factors alone are OK. Formula or completing the square need = as written. nd For choosing outside region. Can f.t. their critical values. They must have two different critical values. > > 9 is A0 but ignore if it follows a correct version < < 9 is M0A0 whatever the diagram looks like. nd A Use of > in final answer gets A0 Total 4 marks
estion number Scheme Marks y U shape touching -ais B ( 3, 0) B 3 9 B (3) y Translated parallel to y-ais up (0, 9 + k) () 9 + k nd B They can score this even if other intersections with the - ais are given. nd B & 3 rd B The -3 and 9 can appear on the sketch as shown Follow their curve in (a) up only. If it is not obvious do not give it. e.g. if it cuts y-ais in (a) but doesn t in (b) then it is M0. Bf.t. Follow through their 9 Total 5 marks
estion number Scheme Marks (a) a = 4 a 3 = 3 a 5 = 7 Bf.t. () (b) a4 3a3 5( 6) and a5 3a4 5( 43) 3 + 4 + 7 + 6 + 43 = 73 Ac.a.o. (3) nd Bf.t. Follow through their a but it must be a value. 3 4 5 is B0. Give wherever it is first seen. st For two further attempts to use of a n+ = 3a n 5, wherever seen. Condone arithmetic slips nd For attempting to add 5 relevant terms (i.e. terms derived from an attempt to use the recurrence formula) or an epression. Follow through their values for a a 5 Use of formulae for arithmetic series is M0A0 but could get st if a 4 and a 5 are correctly attempted. Total 5 marks
Question number Scheme Marks 5. (a) 4 3 ( y 6 y ) 4 3 or (b) 4 8 6 4 8 6 (allow 4+4 for 8) 4 ( y y ) 6 o.e. (a) For some attempt to differentiate st A For one correct term as printed. nd A For both terms correct as printed. 3 4 3 n 3 4 3 A (3) n c scores AA0 (b) st For attempt to epand 4, must have 0,, terms and at least correct e.g. 8 8 or 6 st ( 4) A Correct epression for. As printed but allow 6 0 and 8. nd For some correct differentiation, any term. Can follow 8 6 through their simplification. N.B. giving rise to ( + 8)/ is M0A0 ALT Product or Quotient rule (If in doubt send to review) M For correct use of product or quotient rule. Apply usual rules on formulae. st ( 4) A For ( 4) or nd 4 A for A (4) Total 7 marks
Question number Scheme Marks 6. (a) 6 4 3 4 3 3 or 6 3 = 3 Ac.a.o () (b) 6 4 3 4 3 4 3 6(4 3) = = 8 3 3 or 8 ( ) 3 or a= 8 and b = - (a) For 4 terms, at least 3 correct e.g. 8 4 3 4 3 3 or 6 8 3 3 4 instead of 6 is OK or 6 + 3 ( 4 3)(4 3) scores M0A0 (b) For a correct attempt to rationalise the denominator can be implied 4 3 NB is OK 4 3 A () Total 4 marks
Question number Scheme Marks 7. a + (n )d = k k = 9 or ( u ) a 0d 9 Ac.a.o. n ( a l) [a ( n ) d] 77 or n 77 l = 9 or ( a 9) ( S ) (a 0 d) 77 or 77 A e.g. a + 0d = 9 a + 5d = 7 or a + 9 = 4 a = 5 and d = 0.4 or eact equivalent A A Total 7 marks st Use of u n to form a linear equation in a and d. a + nd =9 is M0A0 st A For a + 0d = 9. nd Use of S n to form an equation for a and d (LHS) or in a (RHS) nd A A correct equation based on. For st Ms they must write n or use n =. 3 rd Solving (LHS simultaneously) or (RHS a linear equation in a) Must lead to a = or d =. and depends on one previous M 3 rd A for a = 5 4 th A for d = 0.4 (o.e.) ALT Uses ( a l ) n 77 to get a = 5, gets second and third A i.e. 4/7 n Then uses [ a ( n ) d] 77 to get d, gets st A and 4 th A MR Consistent MR of for 9 leading to a = 3, d = 0.8 scores A0A0AftAft S n
Question number Marks Scheme 8. (a) b 4ac 4 p 4(3p 4) 4 p p 6 (=0), A or p p p p p 0) ( ) (3 4) 0 3 4( (p 4)(p + ) =0 p = (- or) 4 Ac.s.o. (4) (b) b a or ( p)( p) 0... (= -p) = - 4 Af.t. () 6 (a) st For use of b 4ac or a full attempt to complete the square leading to a 3TQ in p. May use b 4ac. One of b or c must be correct. st A For a correct 3TQ in p. Condone missing =0 but all 3 terms must be on one side. nd For attempt to solve their 3TQ leading to p = nd A For p = 4 (ignore p = -). scores 4/4. b 4ac leading to p 4(3p 4) and then "spotting" p 4 (b) For a full method leading to a repeated root = Af.t. For = -4 (- their p)
Trial and Improvement M factorize. Ac.s.o. For substituting values of p into the equation and attempting to (Really need to get to p = 4 or -) Achieve p = 4. Don t give without valid method being seen.
Question number Marks Scheme 9. (a) f() = [( 6)( ) 3] or 3 6 3 ( f() = ( 8 5) b= - 8 or c = 5 A = A (3) both and a (b) ( 8 5) ( 5)( 3) f() = ( 5)( 3) A () (c) B y their 3 or their 5 Shape Bf.t. 0 3 5
implication both their 3 and their 5 Bf.t. (3) and (0,0) by 0 3 5 8 (a) for a correct method to get the factor of. ( as printed is the minimum. st A for b = -8 or c = 5. -8 comes from -6- and must be coefficient of, and 5 from 6+3 and must have no s. nd A for a =, b = -8 and c = 5. Must have ( 8 5). (b) for attempt to factorise their 3TQ from part (a). A for all 3 terms correct. They must include the. For part (c) they must have at most non-zero roots of their f() =0 to ft their 3 and their 5. (c) st B for correct shape (i.e. from bottom left to top right and two turning points.) nd Bf.t. for crossing at their 3 or their 5 indicated on graph or in tet. 3 rd Bf.t. if graph passes through (0, 0) [needn t be marked] and both their 3 and their 5.
Question number Marks Scheme 3 3 0.(a) f() = ( c) is OK A 5 3 (3, 7 ) gives 9 3 c 3 or 3 are OK instead of 9 or 3 Af.t. c = A (5) 3 (b) f(-) = 4 (*) Bc.s.o. () (c) m = - 4 + 3 4, = -3.5,A Equation of tangent is: y 5 = -3.5( + ) 4y + 3 +6=0 A (4) o.e. 0 (a) st n n for some attempt to integrate st A for both terms as printed or better. Ignore (+c) here. nd for use of (3,7 ) or (-, 5) to form an equation for c. There must be some correct substitution. No +c is M0. Some changes in terms of function needed. nd Af.t. for a correct equation for c. Follow through their integration. They must tidy up fraction/fraction and signs (e.g. - - to +). (b) Bcso If (-, 5) is used to find c in (a) B0 here unless they verify f(3)=7.5. (c) st for attempting m = f( ) st 3 A for or 3.5 4
nd for attempting equation of tangent at (-, 5), f.t. their m, based on d y d. nd A o.e. must have a, b and c integers and = 0. Treat (a) and (b) together as a batch of 6 marks.
Question number.(a) Marks 8 m ( ) A Scheme y = ( ) or y 8 ( ) o.e. y = 5 accept eact equivalents e.g. 6 Ac.a.o. (4) (b) Gradient of l Equation of l : y 0 = -( 0) [y = - + 0] 5 0 = 7 and y = 6 depend on all 3 Ms A, A (5) (c) RS (0 7) (0 6) ( 3 6 ) RS 45 3 5 (*) Ac.s.o. () (d) PQ = 6, 6 5 or 80 or PS= 4 5 and SQ= 5,A Area = PQ RS 6 5 3 5 d
= 45 A c.a.o. (4) 5 (a) st y y for attempting, must be y over. No formula condone one sign slip, but if formula is quoted then there must be some correct substitution. st A for a fully correct epression, needn t be simplified. nd for attempting to find equation of l. (b) st for using the perpendicular gradient rule nd for attempting to find equation of l. Follow their gradient provided different. 3 rd for forming a suitable equation to find S. (c) for epression for RS or RS. Ft their S coordinates (d) st for epression for PQ or PQ. PQ 6 is but PQ 6 is M0 Allow one numerical slip. nd d for a full, correct attempt at area of triangle. Dependent on previous.