PMT. GCE Edexcel GCE Mathematics Core Mathematics C1 (6663) June Mark Scheme (Results) Mathematics. Edexcel GCE

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1 GCE Edecel GCE Mathematics Core Mathematics C (666) June 006 Mark Scheme (Results) Edecel GCE Mathematics

2 June 006 Mark Scheme (+c) A = + + A +c B 4 for some attempt to integrate n n + st A for either 6 or or better nd A for all terms in correct. Allow and. B for + c, when first seen with a changed epression. June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

3 . Critical Values ( ± a)( ± b) with ab=8 or ( 9)( + ) or 7± = or 7± = or ( ) ( ) ± 7 = ± A Solving Inequality > 9 or < - Choosing outside A 4 st 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 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

4 . (a) y U shape touching -ais B - (-, 0) B (0, 9) B 9 () (b) y Translated parallel to y-ais up 9+k (0, 9+k) Bf.t. () 5 (a) nd B They can score this even if other intersections with the -ais are given. nd B & rd B The - and 9 can appear on the sketch as shown (b) 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 4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

5 4. (a) a = 4 B a = a 5= 7 Bf.t. (b) a4 a a5 a4 = 5( = 6) and = 5( = 4) () = 7 Ac.a.o. () 5 (a) nd Bf.t. Follow through their a but it must be a value. 4 5 is B0 Give wherever it is first seen. (b) st For two further attempts to use of an+ = an 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 a5 Use of formulae for arithmetic series is M0A0 but could get st if a 4 and a5 are correctly attempted. 5 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

6 4 5. (a) ( y = + 6 y = ) 4 + or 4 + AA () + 4 = (b) ( ) ( + 4) = (allow 4+4 for 8) A ( + 4) ( y = y = ) 6 o.e. A (4) 7 (a) For some attempt to differentiate st A For one correct term as printed. nd A For both terms correct as printed c scores AA0 (b) st For attempt to epand ( + 4), must have e.g or n n 0,, terms and at least correct st A ( + 4) Correct epression for. As printed but allow 6 and 0 8. nd For some correct differentiation, any term. Can follow 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 A For ( 4) + ( + 4) or nd A for ( + 4) 6 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

7 6. (a) ( ) or 6 = Ac.a.o () (b) = 6(4 ) = 8 or 8 + ( ) or a= 8 and b = - A () 4 (a) For 4 terms, at least correct e.g ( ) or 6 8 ( ) 4 instead of 6 is OK ( 4 + )(4 + ) scores M0A0 ± or 6 + (b) For a correct attempt to rationalise the denominator Can be implied NB is OK 7 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

8 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 eg.. a+ 0d= 9 a+ 5d = 7 or a + 9 = 4 a = 5 and d = 0.4 or eact equivalent st Use of u n to form a linear equation in a and d. a + nd =9 is M0A0 A A 7 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 S n. For st Ms they must write n or use n =. rd Solving (LHS simultaneously) or (RHS a linear equation in a) Must lead to a = or d =. and depends on one previous M 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 =, d = 0.8 scores A0A0AftAft 8 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

9 8. (a) 4 = 4 4( + 4) = 4 6 (=0), A b ac p p p p or ( + p) p + (p+ 4) = 0 p p 4( = 0) (p 4)(p + ) =0 p = (- or) 4 Ac.s.o. (4) b (b) = or ( + p)( + p) = 0 =... a (= -p) = - 4 Af.t. () (a) st For use of b 4ac or a full attempt to complete the square leading to a TQ in p. May use b = 4ac. One of b or c must be correct. st A For a correct TQ in p. Condone missing =0 but all terms must be on one side. nd For attempt to solve their TQ leading to p = nd A For p = 4 (ignore p = -). b = 4ac leading to p = 4(p + 4) and then "spotting" p = 4 scores 4/4. 6 (b) For a full method leading to a repeated root = Af.t. For = -4 (- their p) Trial and Improvement M Ac.s.o. For substituting values of p into the equation and attempting to factorize. (Really need to get to p = 4 or -) Achieve p = 4. Don t give without valid method being seen. 9 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

10 9. (a) f() = [( 6)( ) + ] or = ( f() = ( 8 5) + b= - 8 or c = 5 A both and a = A () (b) ( 8 5) ( 5)( ) + = f() = ( 5)( ) A () (c) y Shape B their or their 5 Bf.t. 0 5 both their and their 5 Bf.t. () and (0,0) by implication 8 (a) for a correct method to get the factor of. ( as printed is the minimum. st A for b = -8 or c = comes from -6- and must be coefficient of, and 5 from 6+ and must have no s. nd A for a =, b = -8 and c = 5. Must have ( 8 + 5). (b) for attempt to factorise their TQ from part (a). A for all terms correct. They must include the. For part (c) they must have at most non-zero roots of their f() =0 to ft their 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 or their 5 indicated on graph or in tet. rd Bf.t. if graph passes through (0, 0) [needn t be marked] and both their and their 5. 0 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

11 0.(a) f() = + ( + c) is OK A 5 (, 7 ) gives 9 = + c or are OK instead of 9 or Af.t. c = A (5) (b) f(-) = 4 + (*) Bc.s.o. () (c) m = , = -.5,A Equation of tangent is: y 5 = -.5( + ) 4y + +6=0 o.e. A (4) 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 (,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()=7.5. (c) st for attempting m = f ( ± ) st A for or.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. June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

12 .(a) 8 m = ( = ) + A y = ( ) or y 8 = ( ) o.e. y = accept eact equivalents e.g. (b) Gradient of l = Equation of l : y 0 = -( 0) [y = - + 0] Ac.a.o. (4) 5 + = + 0 = 7 and y = 6 depend on all Ms A, A (5) (c) RS = (0 7) + (0 6) ( = + 6 ) RS = 45 = 5 (*) Ac.s.o. () (d) PQ = + 6, = 6 5 or 80 or PS= 4 5 and SQ= 5,A Area = PQ RS = 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. 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. June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

13 GENERAL PRINCIPLES FOR C MARKING Method mark for solving term quadratic:. Factorisation ( + b + c) = ( + p)( + q), where pq = c, leading to = ( a + b + c) = ( m + p)( n + q), where pq = c and mn =. Formula Attempt to use correct formula (with values for a, b and c). a, leading to =. Completing the square Solving + b + c = 0 : ( ± p) ± q ± c, p 0, q 0, leading to = Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( n n ). Integration Power of at least one term increased by. ( n n+ ) Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent eaminers reports is that the formula should be quoted first. Normal marking procedure is as follows: Method mark for quoting a correct formula and attempting to use it, even if there are mistakes in the substitution of values. There must be some correct substitution. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but will be lost if there is any mistake in the working. Eact answers Eaminers reports have emphasised that where, for eample, an eact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals. Answers without working The rubric says that these may gain no credit. Individual mark schemes will give details of what happens in particular cases. General policy is that if it could be done in your head, detailed working would not be required. Most candidates do show working, but there are occasional awkward cases and if the mark scheme does not cover this, please contact your team leader for advice. Misreads A misread must be consistent for the whole question to be interpreted as such. These are not common. In clear cases, please deduct the first A (or B) marks which would have been lost by following the scheme. (Note that marks is the maimum misread penalty, but that misreads which alter the nature or difficulty of the question cannot be treated so generously and it will usually be necessary here to follow the scheme as written). Sometimes following the scheme as written is more generous to the candidate than applying the misread rule, so in this case use the scheme as written. June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics