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2 GCE Edecel GCE Core Mathematics C(666) Summer 005 Mark Scheme (Results) Edecel GCE Core Mathematics C (666)

3 June Core Mathematics C Mark Scheme Question Number. (a) Scheme Penalise ± B Marks () (b) 8 = 64 or ( a) or 8 or Allow ± 8 M = 4 or 0.5 A () () (b) M for understanding that - power means reciprocal 8 = 4 is M0A0 and - is MA0 4. (a) (b) dy = n n d 6 ( 6 4 ) d = c both M A () M A A () (5) (b) st A for one correct term in : 6 or + 4 (or better simplified versions) nd A for all terms as printed or better in one line. 666 Core Mathematics C June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

4 Question Number. (a) 8 9 ( 4) 45 Scheme ( ± 4) ( 4) 6 + ( 9) ( ± 4) 45 M A A Marks () ALT Compare coefficients 8 = a equation for a a = 4 AND a + b = 9 b = 45 M A A () (b) ( 4) = 45 (follow through their a and b from (a)) 4 = ± 45 c = 4 = 4 ± 5 d = M A A () (6) (a) M for ( ± 4) or an equation for a. (b) M for a full method leading to 4 =... or =... A for c and A for d 8 ± 6 5 Note Use of formula that ends with scores M A A0 (but must be 5) i.e. only penalise non-integers by one mark. 666 Core Mathematics C June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

5 Question Number 4. (a) Scheme Shape Points B B Marks () (b) M Marks for shape: graphs must have curved sides and round top. - and 4 ma A A () (5) (a) (b) st B for shape through (0, 0) and ( (k,0) where k >0) nd B for ma at (, 5) and 6 labelled or (6, 0) seen Condone (5,) if and 5 are correct on aes. Similarly (5,) in (b) M for shape NOT through (0, 0) but must cut -ais twice. st A for - and 4 labelled or (-, 0) and (4, 0) seen nd A for ma at (, 5). Must be clearly in st quadrant 5. = + y and sub ( + y ) + y = 9 5y + 4y 8( = 0) i.e. ( 5y + 4)( y ) = 0 ( y =) or 4 (o.e.) (both) 5 M A M A y = = + 4 = 5 ; 4 y = 5 = (o.e) 5 MA f.t. (6) st M Attempt to sub leading to equation in variable st A Correct TQ (condone = 0 missing) nd M Attempt to solve TQ leading to values for y. nd A Condone mislabelling = for y = but then M0A0 in part (c). rd M Attempt to find at least one value r d A f.t. f.t. only in = + y (sf if not eact) Both values N.B. False squaring (e.g. y = + 4y = ) can only score the last marks. 666 Core Mathematics C June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

6 Question Number 6. (a) Scheme 6 + > 5 8 > > or 0.5 or 4 8 M A Marks () (b) ( )( ) (> 0) Critical values =, (both) M A Choosing outside region > or < M A f.t. (4) ( c ) > or < < 4 Bf.t. Bf.t. () (8) (a) M Multiply out and collect terms (allow one slip and allow use of = here) (b) st M Attempting to factorise TQ =... nd M Choosing the outside region nd A f.t. f.t. their critical values N.B.(>, > is M0A0) For p < < q where p > q penalise the final A in (b). (c) f.t. their answers to (a) and (b) st B a correct f.t. leading to an infinite region nd B a correct f.t. leading to a finite region Penalise or once only at first offence. e.g. (a) (b) (c) Mark > 4 > 4 < < < < B0 B >, > > B B0 666 Core Mathematics C June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

7 Question Number 7. (a) ( ) = by Scheme Marks M A c.s.o. () (b) 9 (9 6 + ) d = 6 + ( + c) M A//0 use y = and = : = c So y = c = - M A c.s..o. Af.t. (6) (8) (a) M Attempt to multiply out ( ). Must have or 4 terms, allow one sign error A cso Fully correct solution to printed answer. Penalise wrong working. (b) n+ st n M Some correct integration: A At least correct unsimplified terms A All terms correct (unsimplified) Ignore + c nd M Use of y = and = to find c. No + c is M0. Ac.s.o. for -. (o.e.) Award this mark if c = " stated i.e. not as part of an epression for y Af.t. for simplified terms with y = and a numerical value for c. Follow through their value of c but it must be a number. 666 Core Mathematics C June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

8 Question Number 8. (a) Scheme y ( 4) = ( 9) y + = 0 (o.e.) (condone terms with integer coefficients e.g. y+=) Marks M A A () (b) Equation of l is: y = (o.e.) Solving l and l : 6 + = 0 p is point where =, = 6 p y p p or y p y p or p B M A Af.t. (4) (c ) ( l is y = 7 ) C is (0, -7) or OC = 7 Area of OCP = OC p, = 7 = 0. 5 or Bf.t. M Ac.a.o. () (0) (a) M for full method to find equation of l sta any unsimplified form (b) (c ) M Attempt to solve two linear equations leading to linear equation in one variable nd A f.t. only f.t. their p or y p in y = Bf.t. Either a correct OC or f.t. from their l M for correct attempt in letters or symbols for OCP A c.a.o. 7 scores M A0 666 Core Mathematics C June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

9 Question Number Scheme Marks 9 (a) ( = ) a + ( a + d) a + ( n ) d ( S = )[ a + ( n ) d] a S = [ a + ( n ) d] [ a + ( n ) d] } either S = n[ a + ( n ) d] n S = [ a + ( n ) d] S [ ] B M dm A c.s.o (4) (b) ( a = 49, d = ) u = ( ) = 09 M A () (c ) n S n = 5000 n 50n = 0 (*) S n = [ 49 + ( n )( ) ] ( = n( 50 n) ) M A A c.s.o () (d) ( n 00)( n 50) = 0 n = 50 or 00 M A//0 () (e) u < 0 n = 00 not sensible 00 B f.t. () () (a) B requires at least terms, must include first and last terms, an adjacent term dots and + signs. st M for reversing series. Must be arithmetic with a, d (or a, l) and n. nd dm for adding, must have S and be a genuine attempt. Either line is sufficient. Dependent on st M (NB Allow first marks for use of l for last term but as given for final mark ) (b) M for using a = 49 and d = ± in a + ( n ) d formula. (c) M for using their a, d in S n A any correct epression Acso for putting S n =5000 and simplifying to given epression. No wrong work (d) M Attempt to solve leading to n =... A//0 Give AA0 for correct value and AA for both correct (e) B f.t. Must mention 00 and state u 00 < 0 (or loan paid or equivalent) If giving f.t. then must have n Core Mathematics C June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

10 Question Number Scheme Marks 0 (a) =, y = = 0 ( =0 is OK) B () (b) (c) dy = (= 8 + 8) d dy When =, = m = 7 d Equation of tangent: y 0 = 7( ) y = 7 + dy = m gives = 7 d ( = 0) ( 5)( ) = 0 = () or 5 = 5 M A M M A c.a.o M M A (5) y = y = 5 or M A (5) () (b) st M some correct differentiation ( n n for one term) st A correct unsimplified (all terms) nd M substituting (= P ) in their d dy clear evidence rd M using their m to find tangent at p. st dy M forming a correct equation their = gradient of their tangent d (c) nd M dy for solving a quadratic based on their leading to = d rd M for using their value in y to obtain y coordinate MR For misreading (0, ) for (, 0) award B0 and then MA as in scheme. Then allow all M marks but no A ft. (Ma 7) 666 Core Mathematics C June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

11 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. 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. 666 Core Mathematics C June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

12 GCE Edecel GCE Core Mathematics C (666) January 006 Mark Scheme (Results) Edecel GCE Core Mathematics C (666)

13 January Core Mathematics C Mark Scheme Question Scheme Marks number. ( 4 + ) Factor of. (Allow ( 0) ) M = ( )( ) Factorise term quadratic M A Total marks () Alternative: ( )( ) or ( )( ) scores the second M (allow ± for each sign), then ( )( ) scores the first M, and A if correct. Alternative: Finding factor ( ) or ( ) by the factor theorem scores the second M, then completing, using factor, scores the first M, and A if correct. Factors split : e.g. ( 4 + ) ( )( ). Allow full marks. Factor not seen: e.g. Dividing by ( )( ). M0 M A0. If an equation is solved, i.s.w.

14 Question Scheme Marks number. (a) u = ( ) 4 B = =, u4 4 u = For u, ft ( u Bft, B ) (b) u = 0 4 Bft () () Total 4 marks (b) ft only if sequence is oscillating. Do not give marks if answers have clearly been obtained from wrong working, e.g. u = ( ) 0 = u = (4 ) = u4 = (5 ) = 4

15 Question Scheme Marks number. (a) y = 5 ( ) = (or equivalent verification) (*) B (b) Gradient of L is B () y ( ) = ( ) (ft from a changed gradient) M Aft y 5 = 0 (or equiv. with integer coefficients) A (4) Total 5 marks (a) y ( ) = ( ) y = 5 is fine for B. Just a table of values including =, y = is insufficient. (b) M: eqn of a line through (, ), with any numerical gradient (ecept 0 or ). y ( ) For the M Aft, the equation may be in any form, e.g. =. Alternatively, the M may be scored by using y = m + c with a numerical gradient and substituting (, ) to find the value of c, with Aft if the value of c follows through correctly from a changed gradient. Allow y = 5 or equiv., but must be integer coefficients. The = 0 can be implied if correct working precedes.

16 Question Scheme Marks number 4. (a) dy d = M: or M A () (b) 6 + C M: or or + C M A A () = + + C First A: + C Second A: 6 Total 5 marks In both parts, accept any correct version, simplified or not. Accept 4 for 4. + C in part (a) instead of part (b): Penalise only once, so if otherwise correct scores M A0, M A A.

17 Question Scheme Marks number 5. (a) 5 (or a = ) B () (b) ( + 5) ( + 5) ( 5) ( + 5) ( 5)( + 5) = 9 5 ( = 4) M (Used as or intended as denominator) B ( + 5)( p ± q 5) =... 4 terms ( p 0, q 0) or ( 6 + 5)( p ± q 5) =... 4 terms ( p 0, q 0) (Independent) M [Correct version: ( + 5)( + 5) = , or double this.] ( ) 4 = st A: b = 7, nd A: c = A A (5) Total 6 marks (b) nd M mark for attempting ( + 5)( p + q 5) is generous. Condone errors.

18 Question Scheme Marks number 6. (a) (See below) M Clearly through origin (or (0, 0) seen) A labelled (or (, 0) seen) A () (b) 6 Stretch parallel to y-ais M and 4 labelled (or (, 0) and (4, 0) seen) A 4 6 labelled (or (0, 6) seen) A () (c) Stretch parallel to -ais M and 8 labelled (or (, 0) and (8, 0) seen) A 8 labelled (or (0, ) seen) A () Total 9 marks (a) M: (b) M: with at least two of: (, 0) unchanged (4, 0) unchanged (0, ) changed (c) M: with at least two of: (, 0) changed (4, 0) changed (0, ) unchanged Beware: Candidates may sometimes re-label the parts of their solution.

19 Question Scheme Marks number S = (*) B () (b) Using a = 500, d = 00 with n = 7, 8 or 9 a + ( n ) d or listing M 7. (a) ( ) = 00 or = { } (7 00) = ( )900 A () + or n { a + l}, or listing and summing terms M S 8 = 8{ } or S 8 = 8{ }, or all terms in list correct A = ( ) 9600 A () (d) n { ( n ) 00} = 000 M: General S n, equated to 000 M A (c) Using n { a ( n ) d} n + 4n 0 = 0 (or equiv.) M: Simplify to term quadratic M A ( n + 0)( n 6) = 0 n =... M: Attempt to solve t.q. M n = 6, Age is 6 Acso,Acso (7) (b) Correct answer with no working: Allow both marks. (c) Some working must be seen to score marks: Minimum working: (+ 900) = scores M (A). (d) Allow or > throughout, apart from Age 6. A common misread here is 00. This gives n = 4 and age 4, and can score M A0 M A0 M A A with the usual misread rule. Alternative: (Listing sums) (500, 00, 00, 00, 4500, 6000, 7700, 9600,) 700, 4000, 6500, 900, 00, 500, 8500, 000. List at least up to 000 M All values correct A n = 6 (perhaps implied by age) Acso Age 6 Acso If there is a mistake in the list, e.g. 6 th sum = 00, possible marks are: M A0 A0 A0 Alternative: (Trial and improvement) Use of S n formula with n = 6 (and perhaps other values) M Accurately achieving 000 for n = 6 A Age 6 A Total marks

20 Question Scheme Marks number = 5 + M: One term correct. M A A: Both terms correct, and no etra terms. 5 5 f ( ) = + + ( + C) (+ C not required here) M Aft 5 6 = C Use of = and y =6 to form eqn. in C M C = Acso (simplified version required) A (ft C) (7) 5 [or: or equiv.] Total 7 marks For the integration: M requires evidence from just one term (e.g. ), but not just +C. Aft requires correct integration of at least terms, with at least one of these terms having a fractional power. For the final A, follow through on C only.

21 Question Scheme Marks number 9. (a) (P), (Q) (± scores B B) B, B (b) y = (May be seen earlier) Multiply out, giving 4 terms M y d d = 4 (*) M Acso (c) At = : d y = ( ) ( ) 4 = d Eqn. of tangent: y 6 = ( ( )), y = + 7 (*) M Acso () () () (d) 4 = (Equating to gradient of tangent ) M 5 = 0 ( 5)( + ) = 0 = M 5 = or equiv. A 5 5 y = 4, = = or equiv M, A (5) Total marks (b) Alternative: Attempt to differentiate by product rule scores the second M: y = d d {( 4) } + {( ) } Then multiplying out scores the first M, with A if correct (cso). dy (c) M requires full method: Evaluate and use in eqn. of line through (,6), d (n.b. the gradient need not be for this M). Alternative: Gradient of y = + 7 is, so solve 4 =, as in (d) M to get =. Acso (d) nd and rd M marks are dependent on starting with k is a constant. 4 = k, where

22 Question Scheme Marks number 0. (a) + + = ( + ), + (a =, b = ) B, B () 6 y (b) U -shaped parabola M 5 4 Verte in correct quadrant (ft from ( a, b) Aft (0, ) (or on y-ais) B () 4 (c) b 4ac = 4 = 8 B Negative, so curve does not cross -ais B () (d) b 4ac = k (May be within the quadratic formula) M k < 0 (Correct inequality epression in any form) < k < (or < k < ) M A (4) A Total marks (b) The B mark can be scored independently of the sketch. (, 0) shown on the y-ais scores the B, but if not shown on the ais, it is B0. (c). no real roots is insufficient for the nd B mark.. curve does not touch -ais is insufficient for the nd B mark. (d) nd M: correct solution method for their quadratic inequality, e.g. k < 0 gives k between the critical values α < k < β, whereas k > 0 gives k < α, k > β. k > and k < scores the final M A, but k > or k < scores M A0, k >, k < scores M A0. N.B. k < ± does not score the nd M mark. k < does not score the nd M mark. instead of <: Penalise only once, on first occurrence.

23 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. 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 (See the net sheet for a simple eample). 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.

24 MISREADS Question 8. 5 misread as = M A0 7 5 f ( ) = + + ( + C) M Aft = C M 7 7 C = 7 0 7, f ( ) = A0, A

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

26 June Core Mathematics C Mark Scheme Question Scheme Marks number (+c) M A = + + A +c B 4 M 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. 666 Core Mathematics C June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

27 Question Scheme Marks number. Critical Values ( ± a)( ± b) with ab=8 or ( 9)( + ) or 7± = or 7± = or ( ) ( ) ± M 7 = ± A Solving Inequality > 9 or < - Choosing outside M A 4 st M For attempting to find critical values. Factors alone are OK for M, = 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 M For choosing outside region. Can f.t. their critical values. They must have two different critical values. - > > 9 is MA0 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 666 Core Mathematics C June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

28 Question Scheme Marks number. (a) y U shape touching -ais B - (-, 0) B (0, 9) B 9 () (b) y Translated parallel to y-ais up M 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) M 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 Core Mathematics C 4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

29 Question Scheme Marks number 4. (a) a = 4 B a = a 5= 7 Bf.t. (b) a4 a a5 a4 = 5( = 6) and = 5( = 4) M () M = 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 M For two further attempts to use of an+ = an 5, wherever seen. Condone arithmetic slips nd M 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 M if a 4 and a5 are correctly attempted. 666 Core Mathematics C 5 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

30 Question Scheme Marks number 4 5. (a) ( y = + 6 y = ) 4 + or 4 + MAA () + 4 = M (b) ( ) ( + 4) = (allow 4+4 for 8) A ( + 4) ( y = y = ) 6 o.e. MA (4) 7 (a) M For some attempt to differentiate st A For one correct term as printed. nd A For both terms correct as printed c scores MAA0 (b) st M 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 M 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) 666 Core Mathematics C 6 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

31 Question Scheme Marks number 6. (a) ( ) or 6 M = Ac.a.o () (b) M = 6(4 ) = 8 or 8 + ( ) or a= 8 and b = - A () 4 (a) M For 4 terms, at least correct e.g ( ) or 6 8 ( ) 4 instead of 6 is OK ( 4 + )(4 + ) scores M0A0 ± or 6 + (b) M For a correct attempt to rationalise the denominator Can be implied NB is OK 666 Core Mathematics C 7 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

32 Question Scheme Marks number 7. a + ( n ) d = k k = 9 or M ( 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 M or a + 9 = 4 M a = 5 and d = 0.4 or eact equivalent st M 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 M 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 M 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 MA i.e. 4/7 n Then uses [ a+ ( n ) d] = 77 to get d, gets st MA and 4 th A MR Consistent MR of for 9 leading to a =, d = 0.8 scores MA0MA0MAftAft 666 Core Mathematics C 8 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

33 Question Scheme Marks number 8. (a) 4 = 4 4( + 4) = 4 6 (=0) M, A b ac p p p p or ( + p) p + (p+ 4) = 0 p p 4( = 0) (p 4)(p + ) =0 M p = (- or) 4 Ac.s.o. (4) b (b) = or ( + p)( + p) = 0 =... M a (= -p) = - 4 Af.t. () (a) st M 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 M 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) M 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. 666 Core Mathematics C 9 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

34 Question Scheme Marks number 9. (a) f() = [( 6)( ) + ] or = ( M f() = ( 8 5) + b= - 8 or c = 5 A both and a = A () (b) ( 8 5) ( 5)( ) + = M 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) M 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) M 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 Core Mathematics C 0 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

35 Question Scheme Marks number 0.(a) f() = + ( + c) is OK MA 5 (, 7 ) gives 9 = + c or are OK instead of 9 or MAf.t. c = A (5) (b) f(-) = 4 + (*) Bc.s.o. () (c) m = , = -.5 M,A Equation of tangent is: y 5 = -.5( + ) M 4y + +6=0 o.e. A (4) 0 (a) st M n n for some attempt to integrate + st A for both terms as printed or better. Ignore (+c) here. nd M 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 M for attempting m = f ( ± ) st A for or.5 4 nd M 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. 666 Core Mathematics C June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

36 Question Scheme Marks number.(a) 8 m = ( = ) + M A y = ( ) or y 8 = ( ) o.e. M y = accept eact equivalents e.g. (b) Gradient of l = Equation of l : y 0 = -( 0) [y = - + 0] M Ac.a.o. (4) 5 + = + 0 M = 7 and y = 6 depend on all Ms A, A (5) M (c) RS = (0 7) + (0 6) ( = + 6 ) M RS = 45 = 5 (*) Ac.s.o. () (d) PQ = + 6, = 6 5 or 80 or PS= 4 5 and SQ= 5 M,A Area = PQ RS = dm = 45 A c.a.o. (4) 5 (a) st M 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 M for attempting to find equation of l. (b) st M for using the perpendicular gradient rule nd M for attempting to find equation of l. Follow their gradient provided different. rd M for forming a suitable equation to find S. (c) M for epression for RS or RS. Ft their S coordinates (d) st M for epression for PQ or PQ. PQ = + 6 is M but PQ = + 6 is M0 Allow one numerical slip. nd dm for a full, correct attempt at area of triangle. Dependent on previous M. 666 Core Mathematics C June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

37 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. 666 Core Mathematics C June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

38 Mark Scheme (Results) January 007 GCE GCE Mathematics Core Mathematics C (666) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

39 January Core Mathematics C Mark Scheme Question Scheme Marks number. 4 k or k (k a non-zero constant) M, +..., ( 0) A, A, B (4) 4 Accept equivalent alternatives to, e.g. 0.5,,. M: 4 differentiated to give k, or differentiated to give k (but not for just 0 ). st A: (Do not allow just 4 ) nd A: or equivalent. (Do not allow just, but allow or ). B: differentiated to give zero (or disappearing ). Can be given provided that at least one of the other terms has been changed. Adding an etra term, e.g. + C, is B0.

40 Question Scheme Marks number. (a) 6 (a = 6) B () (b) Epanding 7, 4 ( ) to get or 4 separate terms M ( = 7, c = 4) (a) ± 6 also scores B. (b) M: The or 4 terms may be wrong. st A for 7, nd A for 4. b A, A () Correct answer 7 4 with no working scores all marks with or without working scores M A A0. Other wrong answers with no working score no marks. 4

41 Question Scheme Marks number 8 y. (a) Shape of f() B Moved up Asymptotes: y = M B = 0 (Allow y-ais ) B (4) ( y is B0, 0 is B0). (b) + = 0 No variations accepted. M 4 = (or 0. ) Decimal answer requires at least d.p. A () 6 (a) B: Shape requires both branches and no obvious overlap with the asymptotes (see below), but otherwise this mark is awarded generously. The curve may, e.g., bend away from the asymptote a little at the end. Sufficient curve must be seen to suggest the asymptotic behaviour, both horizontal and vertical. M: Evidence of an upward translation parallel to the y-ais. The shape of the graph can be wrong, but the complete graph (both branches if they have branches) must be translated upwards. This mark can be awarded generously by implication where the graph drawn is an upward translation of another standard curve (but not a straight line). The B marks for asymptote equations are independent of the graph. Ignore etra asymptote equations, if seen. (b) Correct answer with no working scores both marks. The answer may be seen on the sketch in part (a). Ignore any attempts to find an intersection with the y-ais. e.g. (a) This scores B0 (clear overlap with horiz. asymp.) M (Upward translation bod that both branches have been translated). B0 M B0 M B0 M0 No marks unless the original curve is seen, to show upward translation.

42 Question Scheme Marks number 4. ( ) = or ( y + ) = y + 4y + 4 M: or 4 terms M ( ) + = 0 or y + ( y + ) = 0 M: Substitute M 4 6 = 0 or y + 4y 6 = 0 Correct terms A ( )( + ) = 0, =... or ( y + )( y ) = 0, y =... M (The above factorisations may also appear as ( 6)( + ) or equivalent). = = or y = y = A y = y = or = = M A (7) 6 (Allow equivalent fractions such as: = for = ). 7 st M: Squaring a bracket, needs or 4 terms, one of which must be an or y term. nd M: Substituting to get an equation in one variable (awarded generously). st A: Accept equivalent forms, e.g. 4 = 6. rd M: Attempting to solve a -term quadratic, to get solutions. 4 th M: Attempting at least one y value (or value). If y solutions are given as values, or vice-versa, penalise at the end, so that it is possible to score M MA M A M0 A0. Strict pairing of values at the end is not required. Non-algebraic solutions: No working, and only one correct solution pair found (e.g. =, y = ): M0 M0 A0 M0 A0 M A0 No working, and both correct solution pairs found, but not demonstrated: M0 M0 A0 M A M A Both correct solution pairs found, and demonstrated, perhaps in a table of values: Full marks Squaring individual terms: e.g. y = + 4 M = 0 M A0 (Eqn. in one variable) = M0 A0 (Not solving -term quad.) y = + 4 = 7 y = 7 M A0 (Attempting one y value) 4

43 Question Scheme Marks number 5. Use of b 4ac, perhaps implicit (e.g. in quadratic formula) M ( ) ( 9 + 8( k + ) < 0) 4 ( k + ) < 0 A 8k < 7 (Manipulate to get pk < q, or pk > q, or pk = q ) M 7 k < Or equiv : k < or k <. 5 Acso (4) 8 8 st M: Could also be, for eample, comparing or equating b and 4 ac. Must be considering the given quadratic equation. There must not be terms in the epression, but there must be a k term. st A: Correct epression (need not be simplified) and correct inequality sign. Allow also 4 ( k + ) < 0. nd M: Condone sign or bracketing mistakes in manipulation. Not dependent on st M, but should not be given for irrelevant work. M0 M could be scored: e.g. where b + 4ac is used instead of b 4 ac. Special cases:. Where there are terms in the discriminant epression, but then division by gives an inequality/equation in k. (This could score M0 A0 M A).. Use of instead of < loses one A mark only, at first occurrence, so an 7 otherwise correct solution leading to k scores M A0 M A. 8 N.B. Use of b = instead of b = implies no A marks. 4 5

44 Question Scheme Marks number 6. (a) ( 4 )(4 + ) k 0. M (The k value need not be eplicitly stated see below) , or k = 4 Acso () + seen, or a numerical value of k seen, ( ) (b) 6 c or k c or 9 c M 9 ( ) d 6 C, = A, Aft () 5 (a) e.g. ( 4 + )(4 + ) alone scores M A0, (but not e.g scores M A0. ( 4 + ) alone). k = 4 or ,with no further evidence, scores full marks M A. Correct solution only (cso): any wrong working seen loses the A mark. 9 9 (b) A: C. Allow 4.5 or 4 as equivalent to. k Aft: (candidate s value of k, or general k). For this final mark, allow for eample 48 as equivalent to 6, but do not allow unsimplified double fractions such as ( ) not allow unsimplified products such as 4. 4, and do A single term is required, e.g is not enough. An otherwise correct solution with, say, C missing, followed by an incorrect solution including + C can be awarded full marks (isw, but allowing the C to appear at any stage). 6

45 Question Scheme Marks number 7. (a) c or 6 c or 8 c M f ( ) = 8 6 ( + C) Substitute = and y = into a changed function to form an equation in C. A A = C C = Acso (5) 8 (b) 6 M = 4 A Eqn. of tangent: y = 4( ) M y = 4 7 (Must be in this form) A (4) M (a) First A marks: + C is not required, and coefficients need not be simplified, but powers must be simplified. All terms correct: A A Two terms correct: A A0 Only one term correct: A0 A0 Allow the M A for finding C to be scored either in part (a) or in part (b). (b) st M: Substituting = into 6 (must be this function). nd M: Awarded generously for attempting the equation of a straight line through (, ) or (, ) with any value of m, however found. nd M: Alternative is to use (, ) or (, ) in y = m + c to find a value for c. 8 9 If calculation for the gradient value is seen in part (a), it must be used in part (b) to score the first M A in (b). Using (, ) instead of (, ): Loses the nd method mark in (a). Gains the nd method mark in (b). 7

46 Question Scheme Marks number 8. (a) 4 k or k or k M d y 9 = A A () d (b) For = 4, y = ( 4 4) + ( 4 4) ( 6) = = 8 (*) B () dy (c) = = d Gradient of normal = M: Evaluate their dy at = 4 M d Aft Equation of normal: y 8 = ( 4), y = + 0 (*) M, A (4) (d) = 0 : =... ( 0) y and use PQ = or ( ) + ( y y) M PQ = Follow through from (k, 0) Aft 4 and 8. May also be scored with ( ) ( ) = 8 0 A () (a) For the A marks coefficients need not be simplified, but powers must be simplified. For eample, is acceptable. All terms correct: A A Two terms correct: A A0 Only one term correct: A0 A0 (b) There must be some evidence of the 4 value. (c) In this part, beware working backwards from the given answer. d y Aft: Follow through is just from the candidate s value of. d nd M: Is not given if an m value appears from nowhere. nd M: Must be an attempt at a normal equation, not a tangent. nd M: Alternative is to use (4, 8) in y = m + c to find a value for c. (d) M: Using the normal equation to attempt coordinates of Q, (even if using = 0 instead of y = 0), and using Pythagoras to attempt PQ or PQ. Follow through from (k, 0), but not from (0, k) A common wrong answer is to use = 0 to give 0. This scores M A0 A0. For final answer, accept other simplifications of 640, e.g. 60 or

47 Question Scheme Marks number 9. (a) Recognising arithmetic series with first term 4 and common difference. B (If not scored here, this mark may be given if seen elsewhere in the solution). a + ( n ) d = 4 + ( n ) = n + M A () ( ) n 0 S n = a + ( n ) d = + =, M A, A () (b) { } { 8 (0 ) }, 75 k k + S k M (c) S k < 750 : { 8 + ( k ) } < 750 or > 750 : { 8 + k} > ( or k + 49 > 0) k + 5k 500 < 0 k M A (Allow equivalent -term versions such as k + 5k = 500). ( k 00)( k + 5) < 0 Requires use of correct inequality throughout.(*) Acso (4) (d) or equiv. seen or, k = (and no other values) M, A () (a) B: Usually identified by a = 4 and d =. M: Attempted use of term formula for arithmetic series, or answer in the form (n + constant), where the constant is a non-zero value. Answer for (a) does not require simplification, and a correct answer without working scores all marks. (b) M: Use of correct sum formula with n = 9, 0 or. A: Correct, perhaps unsimplified, numerical version. A: 75 Alternative: (Listing and summing terms). M: Summing 9, 0 or terms. (At least st, nd and last terms must be seen). A: Correct terms (perhaps implied by last term ). A: 75 Alternative: (Listing all sums) M: Listing 9, 0 or sums. (At least 4, 7,.., last ). A: Correct sums, correct finishing value 75. A: 75 Alternative: (Using last term). n M: Using S n = ( a + l) with T 9, T 0 or T as the last term. 0 A: Correct numerical version (4 + ). A: 75 Correct answer with no working scores mark:,0,0. (c) For the first marks, allow any inequality sign, or equals. st M: Use of correct sum formula to form inequality or equation in k, with the 750. nd M: (Dependent on st M). Form -term quadratic in k. st A: Correct terms. Allow credit for part (c) if valid work is seen in part (d). (d) Allow both marks for k = seen without working. Working for part (d) must be seen in part (d), not part (c). 9

48 Question Scheme Marks number 0 y 0. (a) (i) Shape or or B Ma. at (0, 0). B (, 0), (or shown on -ais). B () (ii) Shape B (It need not go below -ais) Through origin. B 0 (6, 0), (or 6 shown on -ais). B () (b) ( ) = (6 ) M 6 = 0 Epand to form -term cubic (or -term quadratic if divided by ), with all terms on one side. The = 0 M may be implied. ( )( + ) = 0 =... Factor (or divide by ), and solve quadratic. M = and = A = : y = 6 Attempt y value for a non-zero value by M substituting back into ( ) or ( 6 ). = : y = 9 Both y values are needed for A. A (, 6) and (,9) (0, 0) This can just be written down. Ignore any method shown. (But must be seen in part (b)). B (7) (a) (i) For the third shape shown above, where a section of the graph coincides with the -ais, the B for (, 0) can still be awarded if the is shown on the -ais. For the final B in (i), and similarly for (6, 0) in (ii): There must be a sketch. If, for eample (, 0) is written separately from the sketch, the sketch must not clearly contradict this. If (0, ) instead of (, 0) is shown on the sketch, allow the mark. Ignore etra intersections with the -ais. (ii) nd B is dependent on st B. Separate sketches can score all marks. (b) Note the dependence of the first three M marks. A common wrong solution is (-, 0), (, 0), (0, 0), which scores M0 A0 B as the last marks. A solution using no algebra (e.g. trial and error), can score up to marks: M0 M0 M0 A0 M A B. (The final A requires both y values). Also, if the cubic is found but not solved algebraically, up to 5 marks: M M M0 A0 M A B. (The final A requires both y values). 0

49 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. 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 (See the net sheet for a simple eample). 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.

50 MISREADS Question 7. misread as 4 8 (a) f ( ) = 6 4 M A A0 = C C = M A0 8 (b) m = 6 = 6 M A Eqn. of tangent: y = 6( ) M y = 6 A

51 Mark Scheme (Results) Summer 007 GCE GCE Mathematics Core Mathematics C (666) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

52 June Core Mathematics C Mark Scheme Question Scheme Marks number. 9 5 or or 5 or ( ) 5 5 M = 4 Acso () M for an attempt to multiply out. There must be at least correct terms. Allow one sign slip only, no arithmetic errors. e.g ( 5) is MA0 ( ) is MA0 as indeed is 9± BUT ( = 4) is M0A0 since there is more than a sign error is M0A0 since there is an arithmetic error. If all you see is that is M but please check it has not come from incorrect working. Epansion of ( + 5)( + 5) is M0A0 Acso for 4 only. Please check that no incorrect working is seen. Correct answer only scores both marks.

53 Question Scheme Marks number. (a) Attempt 8 or ( 8 4 ) M = 6 A () (b) 5 5, B, B () 4 (a) M for: (on its own) or ( ) 4 or ( ) or 5 or ( 4096) is M0 8 or 8 or or 8 4 or 4096 A for 6 only (b) st B for 5 on its own or something. nd B for So e.g. 5 4 is B But 5 is B0 An epression showing cancelling is not sufficient (see first epression of QC the mark is scored for the second epression) Can use ISW (incorrect subsequent working) e.g 4 5 scores BB0 but it may lead to 4 5 which we ignore as ISW. Correct answers only score full marks in both parts.

54 Question Scheme Marks number d 4 y. (a) = 6 d + or 6 + M A () (b) 6 + or 6 + M Aft () (c) C A: or 4 A: both, simplified and + C M A A () 7 (a) M for some attempt to differentiate: Condone missing d y d n n A for both terms correct, as written or better. No + C here. Of course is acceptable. (b) M for some attempt to differentiate again. Follow through their d y, at least one term correct d or correct follow through. Af.t. as written or better, follow through must have distinct terms and simplified e.g. 4 4 =. (c) M for some attempt to integrate: (+C alone is not sufficient) 4 n n + dy. Condone misreading or d d y for y. d st A for either or (or better) 4 is OK here too but not for nd A. 8 and 8 nd A for both or i.e. simplified terms and +C all on one line. 8 instead of is OK

55 Question Scheme Marks number 4. (a) Identify a = 5 and d = (May be implied) B ( ) u00 = a+ (00 ) d (= 5 + (00 ) ) M = 40(p) or ( ) 4.0 A () (b) ( S ) [ a (00 ) d] or ( a "their 40" ) = + + M = [ 5 + (00 ) ] or ( 5 + "their 40" ) A = or 408 A () (a) B can be implied if the correct answer is obtained. If 40 is not obtained then the values of a and d must be clearly identified as a = 5 and d =. This mark can be awarded at any point. M for attempt to use nth term formula with n = 00. Follow through their a and d. Must have use of n = 00 and one of a or d correct or correct follow through. Must be 99 not 00. A for 40 or 4.0 (i.e. condone missing sign here). Condone 40 here. N.B. a =, d = is B0 and a + 00d is M0 BUT + 00 is BM and A if it leads to 40. Answer only of 40 (or 4.0) scores /. 6 (b) M for use of correct sum formula with n = 00. Follow through their a and d and their 40. Must have some use of n = 00,and some of a, d or l correct or correct follow through. st A for any correct epression (i.e. must have a = 5 and d = ) but can f.t. their 40 still. nd A for or 408 (i.e. the sign is required before we accept 408 this time) p is fine for A but is A0. ALT Listing (a) They might score B if a =5 and d = are clearly identified. Then award MA together for 40. (b) 00 r= (r + ). Give M for 00 (0) + k (with k >), A for k = 00 and A for

56 Question Scheme Marks number 5. (a) 6 y Translation parallel to -ais M Top branch intersects +ve y-ais Lower branch has no intersections A No obvious overlap 0, or marked on y- ais B () (b) =, y = 0 B, B () S.C. [Allow ft on first B for = when translated the wrong way but must be compatible with their sketch.] 5 (a) M for a horizontal translation two branches with one branch cutting y ais only. If one of the branches cuts both aes (translation up and across) this is M0. A for a horizontal translation to left. Ignore any figures on aes for this mark. B for correct intersection on positive y-ais. More than intersection is B0. =0 and y =.5 in a table alone is insufficient unless intersection of their sketch is with +ve y-ais. A point marked on the graph overrides a point given elsewhere. (b) st B for =. NB is B0. Can accept = + if this is compatible with their sketch. Usually they will have MA0 in part (a) (and usually B0 too) nd B for y = 0. S.C. If = - and y =0 and some other asymptotes are also given award BB0 The asymptote equations should be clearly stated in part (b). Simply marking =- or y =0 on the sketch is insufficient unless they are clearly marked asymptote = - etc.

57 Question Scheme Marks number 6. (a) ( 4) = 8 M (b) = 0 4 ± 4 ( 4 8) = or ( ) = - + (any correct epression) (*) Acso () + ± 4 8= 0 M 48 = 6 = 4 or = 4 = B ( ± ) 4 y = M: Attempt at least one y value M = +, y = 6 + =, y = 6 A (5) 7 (a) M for correct attempt to form an equation in only. Condone sign errors/slips but attempt at A (b) this line must be seen. E.g. ± 4= 8 is OK for M. Acso for correctly simplifying to printed form. No incorrect working seen. The = 0 is required. These two marks can be scored in part (b). For multiple attempts pick best. st M for use of correct formula. If formula is not quoted then a fully correct substitution is required. Condone missing = or just + or instead of + for M. For completing the square must have as printed or better. If they have 4 8 then M can be given for ( ) =0 ± 4 8= 0. st A for - + any correct epression. (The + is required but = is not) B for simplifying the surd e.g. 48 = 4. Must reduce to b so 6 or 4 are OK. nd M for attempting to find at least one y value. Substitution into one of the given equations and an attempt to solve for y. nd A for correct y answers. Pairings need not be eplicit but they must say which is and which y. Mis-labelling and y loses final A only.

58 Question Scheme Marks number 7. (a) Attempt to use discriminant b 4ac M k 4( k + ) > 0 k 4k > 0 (*) Acso () (b) k 4k = 0 4± 4 4 ± ±, with = or = or ± M ( k a)( k b) ab ( k ) ( k ) k = and 6 (both) A k <, k > 6 or (, );( 6, ) M: choosing outside M Aft (4) 6 (a) M for use of b 4ac, one of b or c must be correct. Or full attempt using completing the square that leads to a TQ in k k k e.g. + = ( k+ ) 4 Acso Correct argument to printed result. Need to state (or imply) that b 4ac >0 and no incorrect working seen. Must have >0. If > 0 just appears with k 4 k+ > 0that is OK. If >0 appears on last line only with no eplanation give A0. b 4ac followed by k 4k > 0 only is insufficient so M0A0 ( ) e.g. k 4 k+ (missing brackets) can get MA0 but Using b 4a c > 0 is M0. k + 4( k+ ) is M0A0 (wrong formula) (b) st M for attempting to find critical regions. Factors, formula or completing the square. st A for k = 6 and only nd M for choosing the outside regions nd Af.t. as printed or f.t. their (non identical) critical values 6 < k < is MA0 but ignore if it follows a correct version < k < 6 is M0A0 whatever their diagram looks like Condone use of instead of k for critical values and final answers in (b). Treat this question as two mark parts. If part (a) is seen in (b) or vice versa marks can be awarded.

59 Question Scheme Marks number 8. (a) ( = )k + 5 [must be seen in part (a) or labelled a = ] B () a (b) ( a = )(k + 5) + 5 M = 9k + 0 (*) Acso () (c)(i) a = (9k + 0) + 5 ( = 7k 65) M 4 = r 4 + a r = k + (k + 5) + (9k + 0) + (7k + 65) (ii) = 40k + 90 A = 0(4k + 9) (or eplain why divisible by 0) Aft (4) 7 M (b) M for attempting to find a, follow through their a k. Acso for simplifying to printed result with no incorrect working seen. (c) st M for attempting to find a 4. Can allow a slip here e.g. (9k + 0) [i.e. forgot +5] nd M for attempting sum of 4 relevant terms, follow through their (a) and (b). Must have 4 terms starting with k. Use of arithmetic series formulae at this point is M0A0A0 st A for simplifying to 40k + 90 or better nd Aft for taking out a factor of 0 or dividing by 0 or an eplanation in words true k. Follow through their sum of 4 terms provided that both Ms are scored and their sum is divisible by 0. A comment is not required. e.g. 40 k + 90 = 4k +9 is OK for this final A. 0 S.C. 5 a r r= = 0k + 90 = 0(k + 9) can have MM0A0Aft.

60 Question Scheme Marks number (a) f ( ) = ( + C) M A = 5: C = 65 C = 0 M A (4) (b) ( 5 ) or ( + )( 4) or ( + )( 4) M = ( + )( 4) (*) Acso () (a) (c) 0 y Shape B Through origin B,0 and (4,0) B () st n n M for attempting to integrate, + st A for all terms correct, need not be simplified. Ignore + C here. nd M for some use of = 5 and f(5)=65 to form an equation in C based on their integration. There must be some visible attempt to use = 5 and f(5)=65. No +C is M0. nd A for C = 0. This mark cannot be scored unless a suitable equation is seen. 9 (b) M for attempting to take out a correct factor or to verify. Allow usual errors on signs. They must get to the equivalent of one of the given partially factorised epressions or, if verifying, ( + 8 ) i.e. with no errors in signs. Acso for proceeding to printed answer with no incorrect working seen. Comment not required. This mark is dependent upon a fully correct solution to part (a) so MAM0A0MA0 for (a) & (b). Will be common or MAMA0MA0. To score in (b) they must score 4 in (a). (c) st B for positive shaped curve (with a ma and a min) positioned anywhere. nd B for any curve that passes through the origin (B0 if it only touches at the origin) rd B for the two points clearly given as coords or values marked in appropriate places on ais. Ignore any etra crossing points (they should have lost first B). Condone (.5, 0) if clearly marked on ve -ais. Condone (0, 4) etc if marked on +ve ais. Curve can stop (i.e. not pass through) at (-.5, 0) and (4, 0). A point on the graph overrides coordinates given elsewhere.

61 Question Scheme Marks number 0. (a) = : y = =, = : y = 6 + = 4 (can be given st B for ( ) + ( 4 ( ) ) = 70 in (b) or (c)) nd B for - 4 PQ = (*) M Acso (4) (b) y = M dy d = 4 dy = : = 4 = M: Evaluate at one of the points M d M A dy = : = 4 = Parallel A: Both correct + conclusion A (5) d (c) Finding gradient of normal m = M y = ( ) M Aft y 4 = 0 o.e. Acso (4) (a) M for attempting PQ or PQ using their P and their Q. Usual rules about quoting formulae. We must see attempt at ( ) + yp yq for M. PQ =... etc could be MA0. Acso for proceeding to the correct answer with no incorrect working seen. (b) st M for multiplying by, the or 6 must be correct. nd M for some correct differentiation, at least one term must be correct as printed. st A for a fully correct derivative. These marks can be awarded anywhere when first seen. rd M for attempting to substitute = or = in their derivative. Substituting in y is M0. nd A for - from both substitutions and a brief comment. The must come from their derivative. (c) st M for use of the perpendicular gradient rule. Follow through their. nd M for full method to find the equation of the normal or tangent at P. If formula is quoted allow slips in substitution, otherwise a correct substitution is required. st Aft for a correct epression. Follow through their and their changed gradient. nd Acso for a correct equation with = 0 and integer coefficients. This mark is dependent upon the coming from their derivative in (b) hence cso. Tangent can get M0MA0A0, changed gradient can get M0MAA0orMMAA0. Condone confusion over terminology of tangent and normal, mark gradient and equation. MR 4 Allow for or (+6) but not omitting 4 or treating it as 4.

62 Question Scheme Marks number. (a) y = ( + 4) Gradient = 4 (b) + = + 4 =, 9 M A () M, A 4 0 y = + = = A () 9 (c) Where y =, l : A = l : B = M: Attempt one of these M A Area = ( B A )( y P ) M = = = o.e. A (4) 8 8 (a) M for an attempt to write + y 8 =0 in the form y = m + c y y or a full method that leads to m =, e.g find points, and attempt gradient using e.g. finding y = alone can score M (even if they go on to say m = 4) A dy for m = (can ignore the +c) or = d (b) M for forming a suitable equation in one variable and attempting to solve leading to =..or y= st A for any eact correct value for nd A for any eact correct value for y (These marks can be scored anywhere, they may treat (a) and (b) as a single part) (c) st M for attempting the coordinate of A or B. One correct value seen scores M. st A for A = and B = nd M for a full method for the area of the triangle follow through their A, B, y P. 4 9 e.g. determinant approach =... (...) 0 nd A for 49 or an eact equivalent. 8 All accuracy marks require answers as single fractions or mied numbers not necessarily in lowest terms. 9

63 Mark Scheme (Results) January 008 GCE GCE Mathematics (666/0) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

64 January Core Mathematics C Mark Scheme Question Scheme Marks number. k or k or 7 k (k a non-zero constant) M or (Either of these, simplified or unsimplified) A or equivalent unsimplified, such as A 6 + C (or any other constant, e.g. + K) B (4) M: Given for increasing by one the power of in one of the three terms. A marks: Ignore subsequent working after a correct unsimplified version of a term is seen. B: Allow the mark (independently) for an integration constant appearing at any stage (even if it appears, then disappears from the final answer). This B mark can be allowed even when no other marks are scored Core Mathematics C January 008 Advanced Subsidiary/Advanced Level in GCE Mathematics

65 Question Scheme Marks number. (a) B () (b) 9 seen, or ( ) answer to (a) seen, or ) ( seen. M 9 8 A () (b) M: Look for 9 first if seen, this is M., e.g. this would score M even if it does not subsequently become 8. (Similarly for other answers to (a)). If not seen, look for ( answer to (a)) In ) (, the is implied, so this scores the M mark. Negative answers: (a) Allow. Allow ±. Allow or. (b) Allow 9 9 ± 8. Allow 8 or 9 8. N.B. If part (a) is wrong, it is possible to restart in part (b) and to score full marks in part (b). 666 Core Mathematics C January 008 Advanced Subsidiary/Advanced Level in GCE Mathematics

66 Question Scheme Marks number ( 5 ). ( + ) 0 = ( ) ( ) 5... ( = 7 ) + ( ) =... 7 Allow (a = ) A st M: Multiplying top and bottom by ( ). (As shown above is sufficient). nd M: Attempt to multiply out numerator ( 5 ) ( ). Must have at least terms correct. M M 7 (b = 7) A (4) Final answer: Although denominator = may be implied, the 7 must obviously be the final answer (not an intermediate step), to score full marks. (Also M0 M A A is not an option). The A marks cannot be scored unless the st M mark has been scored, but this st M mark could be implied by correct epansions of both numerator and denominator. It is possible to score M M0 A A0 or M M0 A0 A (after correct terms in the numerator). Special case: If numerator is multiplied by ( + ) instead of ( ), the nd M can still be scored for at least of these terms correct: ( ). The maimum score in the special case is mark: M0 M A0 A0. Answer only: Scores no marks. Alternative method: 5 = ( a + b )( + ) ( a + b )( + ) = a + a + b + M: At least terms correct. 5 = a + b = a + b a = or b = M: Form and attempt to solve simultaneous equations. a =, b = 7 A, A Core Mathematics C 4 January 008 Advanced Subsidiary/Advanced Level in GCE Mathematics

67 Question Scheme Marks number 4 ( ) (a) m = or, = or = M, A ( 6) 4 4 Equation: y 4 = ( ( 6)) or y ( ) = ( 8) M + y = 0 (or equiv. with integer coefficients must have = 0 ) A (4) (e.g. 4 y = 0 and 4 7 4y = 0 are acceptable) (b) ( 6 8) + (4 ( )) M or ( 4) + 7 or 4 + ( 7) (M A may be implied by 45) A AB = or 7 ( + ) or Acso () (a) st M: Attempt to use y y m = (may be implicit in an equation of L). nd M: Attempting straight line equation in any form, e.g y y = m( ), y y = m, with any value of m (ecept 0 or ) and either ( 6, 4) or (8, ). N.B. It is also possible to use a different point which lies on the line, such as the midpoint of AB (, 0.5). Alternatively, the nd M may be scored by using y = m + c with a numerical gradient and substituting ( 6, 4) or (8, ) to find the value of c. Having coords the wrong way round, e.g. y ( 6) = ( 4), loses the nd M mark unless a correct general formula is seen, e.g. y y = m ). ( 7 (b) M: Attempting to use ( ) + ( y y). Missing bracket, e.g implies M if no earlier version is seen with no further work would be M A followed by recovery can score full marks. 666 Core Mathematics C 5 January 008 Advanced Subsidiary/Advanced Level in GCE Mathematics

68 Question Scheme Marks number 5. (a) (b) + p =, q = B, B () y = d y = 5 ( 0 or 5 ) ( 5 7 correctly differentiated) B d Attempt to differentiate either or p with a fractional p, giving k p ( k 0), (the fraction p could be in decimal form) q with a negative q, giving k q ( k 0). M (b): =, Aft, Aft (4) N.B. It is possible to start again in (b), so the p and q may be different from those seen in (a), but note that the M mark is for the attempt to p q differentiate or. However, marks for part (a) cannot be earned in part (b). st Aft: ft their nd Aft: ft their p, but p must be a fraction and coefficient must be simplified (the fraction p could be in decimal form). q, but q must be negative and coefficient must be simplified. 6 'Simplified' coefficient means b a where a and b are integers with no common factors. Only a single + or sign is allowed (e.g. must be replaced by +). Having +C loses the B mark. 666 Core Mathematics C 6 January 008 Advanced Subsidiary/Advanced Level in GCE Mathematics

69 Question Scheme Marks number 6. (a) (, 0) Shape: Ma in st quadrant and intersections on positive -ais B 4 and 4 labelled (in correct place) or clearly stated as coordinates B (, 0) labelled or clearly stated B () (b) Shape: Ma in nd quadrant and intersections on negative -ais (, 5) and 4 labelled (in correct place) or clearly stated as coordinates B B 4 (, 5) labelled or clearly stated B () (c) (a = ) May be implicit, i.e. f ( + ) B () Beware: The answer to part (c) may be seen on the first page. (a) and (b): st B: Shape is generous, providing the conditions are satisfied. nd and rd B marks are dependent upon a sketch having been drawn. nd B marks: Allow (0, ), etc. (coordinates the wrong way round) if the sketch is correct. Points must be labelled correctly and be in appropriate place (e.g. (, 5) in the first quadrant is B0). (b) Special case: If the graph is reflected in the -ais (instead of the y-ais), B B0 B0 can be scored. This requires shape and coordinates to be fully correct, i.e. Shape: Minimum in 4 th quadrant and intersections on positive -ais, 7 and 4 labelled (in correct place) or clearly stated as coordinates, (, 5) labelled or clearly stated. 666 Core Mathematics C 7 January 008 Advanced Subsidiary/Advanced Level in GCE Mathematics

70 Question Scheme Marks number 7. (a) ( p + ) or p + B () (b) (( a ))( p + ( a) ) [(a) must be a function of p]. [( + )( p + p + ) ] p M = + p + p (*) Acso () = (c) + p + p M p ( p + ) = 0 p =... M p = (ignore p = 0, if seen, even if chosen as the answer) A () (d) Noting that even terms are the same. M This M mark can be implied by listing at least 4 terms, e.g.,,,, = A () 8 (b) M: Valid attempt to use the given recurrence relation to find. Missing brackets, e.g. p + ( p + p + ) Condone for the M, then if all terms in the epansion are correct, with working fully shown, M A is still allowed. Beware working back from the answer, e.g. + p + p = ( + p)( + p) scores no marks unless the recurrence relation is justified. (c) nd M: Attempt to solve a quadratic equation in p (e.g. quadratic formula or completing the square). The equation must be based on =. The attempt must lead to a non-zero solution, so just stating the zero solution p = 0 is M0. A: The A mark is dependent on both M marks. (d) M: Can be implied by a correct answer for their p (answer is p + ), and can also be implied if the working is obscure ). Trivialising, e.g. p = 0, so every term =, is M0. If the additional answer 008 = (from p = 0) is seen, ignore this (isw). 666 Core Mathematics C 8 January 008 Advanced Subsidiary/Advanced Level in GCE Mathematics

71 Question Scheme Marks number 8. (a) + k + (8 k) ( = 0) 8 k need not be bracketed M b 4ac = k 4(8 k) M b 4ac < 0 k + 4k < 0 (*) Acso () (b) ( k + 8)( k 4) = 0 k =... M k = 8 k = 4 A Choosing 'inside' region (between the two k values) M 8 < k < 4 or 4 > k > 8 A (4) (a) st M: Using the k from the right hand side to form -term quadratic in ('= 0' can be implied), or k k attempting to complete the square k ( = 0) or equiv., 4 using the k from the right hand side. For either approach, condone sign errors. st M may be implied when candidate moves straight to the discriminant. nd M: Dependent on the st M. Forming epressions in k (with no s) by using b and 4 ac. (Usually seen as the discriminant b 4ac, but separate epressions are fine, and also allow the use of b + 4ac. (For 'completing the square' approach, the epression must be clearly separated from the equation in ). If b and 4 ac are used in the quadratic formula, they must be clearly separated from the formula to score this mark. For any approach, condone sign errors. If the wrong statement b 4ac < 0 is seen, maimum score is M M A0. (b) Condone the use of (instead of k) in part (b). st M: Attempt to solve a -term quadratic equation in k. It might be different from the given quadratic in part (a). Ignore the use of < in solving the equation. The st M A can be scored if 8 and 4 are achieved, even if stated as k < 8, k < 4. Allow the first M A to be scored in part (a). N.B. k > 8, k < 4 scores nd M A0 k > 8 or k < 4 scores nd M A0 k > 8 and k < 4 scores nd M A k = 7, 6, 5, 4,,,, 0,,, scores nd M0 A0 Use of (in the answer) loses the final mark Core Mathematics C 9 January 008 Advanced Subsidiary/Advanced Level in GCE Mathematics

72 Question Scheme Marks number 8 k 9. (a) 4 k or 6 k or (k a non-zero constant) M f ( ) =, 4, 8 (+ C) (+ C not required) A, A, A At = 4, y = : = ( 6) 4 4 ( 8 4 ) + C Must be in part (a) M C = A (6) (b) 8 9 f (4) = 6 (6 ) + = 6 ( m) Gradient of normal is 9 = M: Attempt f (4) with the given f. M = m Must be in part (b) M: Attempt perp. grad. rule. M Dependent on the use of their f ( ) y Eqn. of normal: y = ( 4) (or any equiv. form, e.g. = ) M A (4) Typical answers for A: y = + ( + 9y 7 = 0) ( y = 0. &.8& ) Final answer: gradient or is A0 (but all M marks are available) ( ) (a) The first A marks are awarded in the order shown, and the terms must be simplified. 'Simplified' coefficient means b a where a and b are integers with no common factors. Only a single + or sign is allowed (e.g. + must be replaced by ). nd M: Using = 4 and y = (not y = 0) to form an eqn in C. (No C is M0) (b) nd M: Dependent upon use of their f ( ). rd M: eqn. of a straight line through (4, ) with any gradient ecept 0 or. Alternative for rd M: Using (4, ) in y = m + c to find a value of c, but an equation (general or specific) must be seen. Having coords the wrong way round, e.g. y 4 = ( ), loses the rd M 9 mark unless a correct general formula is seen, e.g. y y = m ). ( N.B. The A mark is scored for any form of the correct equation be prepared to apply isw if necessary Core Mathematics C 0 January 008 Advanced Subsidiary/Advanced Level in GCE Mathematics

73 Question Scheme Marks number 0. (a) y Shape (drawn anywhere) B 0 5 Minimum at (, 0) B (perhaps labelled on -ais) (,0) (or shown on ve -ais) B 4 4 (0, ) (or shown on +ve y-ais) B (4) N.B. The ma. can be anywhere. (b) y = ( + )( + ) Marks can be awarded if M = (k = ) this is seen in part (a) Acso () (c) d y = + 5 d + 5 = or + 8 = 0 M M A ( 4)( + ) = 0 =... M 4 = (or eact equiv.), = A, A (6) (a) The individual marks are independent, but the nd, rd and 4 th B s are dependent upon a sketch having been attempted. B marks for coordinates: Allow (0, ), etc. (coordinates the wrong way round) if marked in the correct place on the sketch. (b) M: Attempt to multiply out ( ) and write as a product with ( + ), or attempt to multiply out ( + )( ) and write as a product with ( ), or attempt to epand ( + )( )( ) directly (at least 7 terms). The ( ) or ( + )( ) epansion must have (or 4) terms, so should not, for eample, be just +. A: It is not necessary to state eplicitly 'k = '. Condone missing brackets if the intention seems clear and a fully correct epansion is seen. (c) st M: Attempt to differentiate (correct power of in at least one term). nd M: Setting their derivative equal to. rd M: Attempt to solve a -term quadratic based on their derivative. d y The equation could come from = 0. d N.B. After an incorrect k value in (b), full marks are still possible in (c). 666 Core Mathematics C January 008 Advanced Subsidiary/Advanced Level in GCE Mathematics

74 Question Scheme Marks number. (a) u = a + 4d = (.5) M 5 = 6 A () (b) a + ( n ) d = 0.5( r ) = 0 M 0 r = A () (c) S = { (.5) } or S = { (.5) } or = { 0 + 0} 0 S M Aft = 5 A () (a) M: Substitution of a = 0 and d = ±. 5 into (a + 4d). Use of a + 5d (or any other variations on 4) scores M0. (b) M: Attempting to use the term formula, equated to 0, to form an equation in r (with no other unknowns). Allow this to be called n instead of r. Here, being one off (e.g. equivalent to a + nd), scores M. (c) M: Attempting to use the correct sum formula to obtain S 0, S, or, with their r from part (b), S r or S r. st A(ft): A correct numerical epression for S 0, S, or, with their r from part (b), S r or S r. but the ft is dependent on an integer value of r. Methods such as calculus to find a maimum only begin to score marks after establishing a value of r at which the maimum sum occurs. This value of r can be used for the M Aft, but must be a positive integer to score A marks, so evaluation with, say, n = 0.5 would score M A0 A0. Listing and other methods (a) M: Listing terms (found by a correct method), and picking the 5 th term. (There may be numerical slips). (b) M: Listing terms (found by a correct method), until the zero term is seen. (There may be numerical slips). Trial and error approaches (or where working is unclear or non-eistent) score M A for, M A0 for 0 or, and M0 A0 otherwise. (c) M: Listing sums, or listing and adding terms (found by a correct method), at least as far as the 0 th term. (There may be numerical slips). A (scored as A A) for 5 (clearly selected as the answer). Trial and error approaches essentially follow the main scheme, beginning to score marks when trying S 0, S, or, with their r from part (b), S r or S r. If no working (or no legitimate working) is seen, but the answer 5 is given, allow one mark (scored as M A0 A0). For reference: Sums: 0, 58.5, 85.5,, 5, 57.5, 78.5, 98, 6,.5, 47.5, 6, 7, 8.5, 9.5, 00, 06, 0.5,.5, 5, Core Mathematics C January 008 Advanced Subsidiary/Advanced Level in GCE Mathematics

75 666 Core Mathematics C January 008 Advanced Subsidiary/Advanced Level in GCE Mathematics

76 Mark Scheme (Results) Summer 008 GCE Mathematics (666/0) GCE s

77 June Core Mathematics C Mark Scheme Question Scheme Marks number c MAA () M for an attempt to integrate n n +. Can be given if +c is only correct term. st A for 5 or + c. Accept 5 for. Do not accept or as final answer nd A for as printed (no etra or omitted terms). Accept 5 or.6 & for but not.6 or.67 etc Give marks for the first time correct answers are seen e.g. 5 that later becomes.67, the.67 is treated as ISW NB MA0A is not possible s Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

78 Question Scheme Marks number. ( 9) or ( ± 0)( 9) or ( )( + ) or ( + )( ) B ( )( + ) MA () B M for first factor taken out correctly as indicated in line above. So for attempting to factorise a relevant quadratic. ( + 9) is B0 Ends correct so e.g. ( 9) = ( ± p)( ± q) where pq = 9 is OK. This mark can be scored for ( 9) ( )( ) = + seen anywhere. A for a fully correct epression with all factors. Watch out for ( )( + ) which scores A Treat any working to solve the equation 9 as ISW. s

79 y Question Scheme Marks number (a) y 0 (b) (7, ) BBB () y 7 (a) (.5, 0) Allow stopping at (0, 0) or (0, 7) instead of cutting BB () st B for moving the given curve up. Must be U shaped curve, minimum in first quadrant, not touching -ais but cutting positive y-ais. Ignore any values on aes. nd B for curve cutting y-ais at (0, 0). Point 0(or even (0, 0) marked on positive y-ais is OK) rd B for minimum indicated at (7, ). Must have both coordinates and in the right order. 5 0 this would score B0BB0 If the curve flattens out to a turning point like this penalise once at first offence ie st B in (a) or in (b) but not in both. The U shape mark can be awarded if the sides are fairly straight as long as the verte is rounded. (b) st B for U shaped curve, touching positive -ais and crossing y-ais at (0, 7)[condone (7, 0) if marked on positive y ais] or 7 marked on y-ais nd B for minimum at (.5, 0) or.5 or 7 marked on -ais. Do not condone (0,.5) here. Redrawing f() will score BB0 in part (b). Points on sketch override points given in tet/table. If coordinates are given elsewhere (tet or table) marks can be awarded if they are compatible with the sketch. s 4

80 Question Scheme Marks number 4. (a) [ f ( ) = ] + MA () (b) + = 5 and start to try and simplify M = k = k (ignore +) M = (ignore = ) A () 5 (a) M for attempting to differentiate n n. Just one term will do. A poor integration attempt that gives +... (or similar) scores M0A0 A for a fully correct epression. Must be not. If there is a + c they score A0. 0 (b) st M for forming a correct equation and trying to rearrange their f ( ) = 5 e.g. collect terms. 5 e.g. = 5 or + = 5 or even + = or + = 5 6= 5 (i.e algebra can be awful as long as they try to collect terms in their f ( ) = 5 equation) nd M this is dependent upon their f ( ) being of the form a+ b and attempting to solve a+ b = 5 For correct processing leading to = Can condone arithmetic slips but processes should be correct so e.g = 5 = = scores MM0A0 + = 5 = = 9 = scores MM0A0 + = 5 = = = scores MM0A0 s 5

81 Question Scheme Marks number 5. (a) [ = ] a B () (b) [ = ] a or aa ( ) M = aa ( ) both lines needed for A = a a (*) Acso () (c) a a = 7 a a 0= 0 or a a = 0 M ( a 5)( a+ ) = 0 dm a = 5 or A () 6 (a) B for a or better. Give for a in part (a) or if it appears in (b) they must state = a This must be seen in (a) or before the aa ( ) step. (b) M for clear show that. Usually for aa ( ) but can follow through their and even allow a A for correct processing leading to printed answer. Both lines needed and no incorrect working seen. (c) st M for attempt to form a correct equation and start to collect terms. It must be a quadratic but need not lead to a TQ=0 nd dm This mark is dependent upon the first M. for attempt to factorize their TQ=0 or to solve their TQ=0. The =0 can be implied. 9 4 ( ± p)( ± q) = 0, where pq= 0 or ( ± ) ± 0 = 0 or correct use of quadratic formula with + They must have a form that leads directly to values for a. Trial and Improvement that leads to only one answer gets M0 here. A for both correct answers. Allow = Give / for correct answers with no working or trial and improvement that gives both values for a. s 6

82 Question Scheme Marks Number 6. (a) BMA () 5 y -.5 (b) + 5= M + 5 [=0] or + 5= A ( )( + ) [=0] M = or A y = or ( ) + 5 or y = or ( ) + 5 Points are ( ) M, and (,6) (correct pairings) Aft 9 (a) B for curve of correct shape i.e branches of curve, in correct quadrants, of roughly the correct shape and no touching or intersections with aes. Condone up to inward bends but there must be some ends that are roughly asymptotic. M for a straight line cutting the positive y-ais and the negative -ais. Ignore any values. A for (0,5) and (-.5,0) or points correctly marked on aes. Do not give for values in tables. Condone miing up (, y) as (y, ) if one value is zero and other value correct. (b) st M for attempt to form a suitable equation and multiply by (at least one of or +5) should be multiplied. st A for correct TQ - condone missing = 0 nd M for an attempt to solve a relevant TQ leading to values for = nd A for both = - and 0.5. T&I for values may score st MA otherwise no marks unless both values correct. Answer only of = - and = scores 4/4, then apply the scheme for the final MAft rd M for an attempt to find at least one y value by substituting their in either or + 5 rd Aft follow through both their values, in either equation but the same for each, correct pairings required but can be = -, y = - etc s 7

83 Question Scheme Marks number 7. (a) 5, 7, 9, or 5+++= or 5+6= use a = 5, d =, n = 4 and t 4 = 5+ = B () (b) t = a+ ( n ) d with one of a = 5 or d = correct (can have a letter for the other) M n = 5 + (n ) or n + or + (n + ) A () n (c) S = [ 5+ ( n ) ] or use of ( 5 "their n " ) n n + + (may also be scored in (b)) MA = {n(5 + n - )} = n(n + 4) (*) Acso () (d) 4 = n + M [n] = 0 A () (e) S 0 = 0 4, = 480 (km) MA () (a) B Any other sum must have a convincing argument 0 (b) M for an attempt to use a + (n - )d with one of a or d correct (the other can be a letter) Allow any answer of the form n + p (p 5) to score M. A for a correct epression (needn t be simplified) [ Beware 5 + (n ) scores A0] Epression must be in n not. Correct answers with no working scores /. (c) M for an attempt to use S n formula with a = 5 or d = or a = 5 and their n + st A for a fully correct epression nd A for correctly simplifying to given answer. No incorrect working seen. Must see S n used. Do not give credit for part (b) if the equivalent work is given in part (d) (d) M for forming a suitable equation in n (ft their (b)) and attempting to solve leading to n = A for 0 Correct answer only scores /. Allow 0 following a restart but check working. eg 4 = n + 5 that leads to 40 = n and n =0 should score MA0. (e) M for using their answer for n in n(n + 4) or S n formula, their n must be a value. A for 480 (ignore units but accept m etc)[ no matter where their 0 comes from] NB attempting to solve eg part (d) means we will allow sign slips and slips in arithmetic but not in processes. So dividing when they should subtract etc would lead to M0. Listing in parts (d) and (e) can score (if correct) or 0 otherwise in each part. Poor labelling may occur (especially in (b) and (c) ). If you see work to get n(n + 4) mark as (c) s 8

84 Question Scheme Marks number 8. (a) [No real roots implies b 4ac< 0.] b 4ac= q 4 q ( ) M q 4 q < 0 i.e. q + 8q< 0 (*) Acso () So ( ) (b) q(q + 8) = 0 or ( q ± 4) ± 6= 0 M (q) = 0 or 8 ( cvs) A 8< q < 0 or q (-8, 0) or q < 0 and q > -8 Aft () 5 (a) M for attempting b 4ac with one of b or a correct. < 0 not needed for M This may be inside a square root. Acso for simplifying to printed result with no incorrect working or statements seen. Need an intermediate step e.g. q 8q< 0 or q 4 q < 0 or q 4( q)( ) < 0 or q 8 q( ) < 0 or q 8q < 0 i.e. must have or brackets on the 4ac term < 0 must be seen at least one line before the final answer. (b) M for factorizing or completing the square or attempting to solve q ± 8q= 0. A method that would lead to values for q. The = 0 may be implied by values appearing later. st A for q = 0 and q = 8 nd A for 8< q < 0. Can follow through their cvs but must choose inside region. q < 0, q > 8 is A0, q < 0 or q > 8 is A0, (-8, 0) on its own is A0 BUT q < 0 and q > 8 is A Do not accept a number line for final mark s 9

85 Question Scheme Marks number 9. (a) dy = k + d MA () (b) Gradient of line is 7 B 7 = : +, = M, M 4 When k ( ) ( ) k = k = A (4) 4 (c) y k ( ) ( ) = = = M, A () 8 4 5, 6 (a) M for attempting to differentiate A all correct. A + c scores A0 n n (or -5 going to 0 will do) 8 (b) B for m = 7. Rearranging the line into 7 y = + c does not score this mark until you are sure they are using 7 as the gradient of the line or state m = 7 st M for substituting dy = into their, some correct substitution seen d nd M for forming a suitable equation in k and attempting to solve leading to k = Equation must use their d y and their gradient of line. Assuming the gradient is 0 or 7 scores d M0 unless they have clearly stated that this is the gradient of the line. A for k = (c) M for attempting to substitute their k (however it was found or can still be a letter) and A for - 6 = y (some correct substitution) into s 0

86 Question Scheme Marks number 0. (a) ( 7 ) ( 0 ) QR = + M = or 45 (condone + ) A = 5 or a = ( ± 5 etc is A0) A () (b) Gradient of QR (orl ) = 0 or, = M, A Gradient of l is or 7 6 Equation for y = or = [or y = + ] M Aft (5) (c) P is (0, ) (allow = 0, y = but it must be clearly identifiable as P) B () l is: ( ) P P (d) PQ ( ) ( y ) y = + Determinant Method M e.g(0+0+7) - (++0) PQ = + = 5 A = - 5 (o.e.) Area of triangle is 5 QR PQ = 5 5, = or 7.5 Area = 5,= 7.5 dm, A (4) Rules for quoting formula: For an M mark, if a correct formula is quoted and some correct substitutions seen then M can be awarded, if no values are correct then M0. If no correct formula is seen then M can only be scored for a fully correct epression. (a) M for attempting QR or QR. May be implied by 6 + st A for as printed or better. Must have square root. Condone + M (b) st M for attempting gradient of QR st A for or, can be implied by gradient of l = nd M for an attempt to use the perpendicular rule on their gradient of QR. rd M for attempting equation of a line using Q with their changed gradient. nd Aft requires all Ms but can ft their gradient of QR. y = + with no working. Send to review. (d) st M for attempting PQ or PQ follow through their coordinates of P st A for PQ as one of the given forms. nd dm for correct attempt at area of the triangle. Follow through their value of a and their PQ. This M mark is dependent upon the first M mark nd A for 7.5 or some eact equivalent. Depends on both Ms. Some working must be seen. ALT Use QS where S is (, 0) st M for attempting area of OPQS and QSR and OPR. Need all. st A for OPQS = ( + ) =, QSR = 9, OPR = 7 nd dm for OPQS + QSR OPR = Follow through their values. nd A for 7.5 Determinant Method M for attempt -at least one value in each bracket correct. A if correct (+ 5) M for correct area formula A for 7.5 MR Misreading -ais for y-ais for P. Do NOT use MR rule as this oversimplifies the question. They can only get M marks in (d) if they use PQ and QR. s

87 Question Scheme Marks number. (a) ( ) 4 + = M ( ) = = (*) Acso () 9 y = c MAA c M c = 4 A [ y = ] Aft (6) 8 (b) ( ) (a) M for attempting to epand ( ) A + and having at least (out of the 4) correct terms. at least this should be seen and no incorrect working seen. 9 If they never write as 9 they score A0. (b) st M for some correct integration, one correct term as printed or better u Trying loses the first M mark but could pick up the second. v st A for two correct terms, un-simplified, as printed or better nd A for a fully correct epression. Terms need not be simplified and +c is not required. No + c loses the net marks nd dy M for using = and y = 0 in their epression for f() d rd A for c = 4 4 th Aft for an epression for y with simplified terms: Condone missing y = Follow through their numerical value of c only. 9 for 9 to form a linear equation for c is OK. s

88 Mark Scheme (Results) January 009 GCE GCE Mathematics (666/0) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

89 January Core Mathematics C Mark Scheme Question Number Scheme Marks (a) (b) 5 ( ± 5 is B0) ( their 5) or their 5 = 5 or 0.04 ( ± is A0) 5 B M A () () [] (b) M follow through their value of 5. Must have reciprocal and square. 5 is not sufficient to score this mark, unless follows this. 5 A negative introduced at any stage can score the M but not the A, e.g. 5 = = scores M A = = scores M A Correct answer with no working scores both marks. Alternative: 5 or = 565 ( 5 ) = A 5 M (reciprocal and the correct number squared) 666/0 GCE Mathematics January 009

90 Question Number Scheme Marks ( I = ) + + c = + + c M AAA [4] n M for an attempt to integrate 6 4 (i.e. a or a or a, where a is any non-zero constant). Also, this M mark can be scored for just the + c (seen at some stage), even if no other terms are correct. st 6 A for nd 4 A for rd A for + c (or + k, etc., any appropriate letter can be used as the constant) n+ Allow + c, but not + c. Note that the A marks can be awarded at separate stages, e.g scores nd A c 6 scores rd A scores st A (even though the c has now been lost). Remember that all the A marks are dependent on the M mark. If applicable, isw (ignore subsequent working) after a correct answer is seen. Ignore wrong notation if the intention is clear, e.g. Answer c d. 666/0 GCE Mathematics January 009

91 Question Number Scheme Marks , or 7 4 or an eact equivalent such as = 49 M A [] M for an epanded epression. At worst, there can be one wrong term and one wrong sign, or two wrong signs. e.g is M (one wrong term ) is M (two wrong signs + 7 and + 4 ) is M (one wrong term +, one wrong sign + 7 ) is M (one wrong term 7, one wrong sign + 4 ) is M0 (two wrong terms 7 and ) is M0 (two wrong terms 4 and 4 ) If only terms are given, they must be correct, i.e. (7 4) or an equivalent unsimplified version to score M. The terms can be seen separately for the M. Correct answer with no working scores both marks. 666/0 GCE Mathematics January 009 4

92 Question Number Scheme Marks 4 ( f( ) = ) 7( + c) = 7 ( + c) f(4) = = c c = st M for an attempt to integrate ( or seen). The term is insufficient for this mark and similarly the + c is insufficient. st A for or (An unsimplified or simplified correct form) nd A for all three terms correct and simplified (the simplification may be seen later). The + c is not required for this mark. Allow 7, but not 7. nd M for an attempt to use = 4 and y = in a changed function (even if differentiated) to form an equation in c. rd A for c = with no earlier incorrect work (a final epression for f() is not required). M AA M Acso (5) [5] 666/0 GCE Mathematics January 009 5

93 Question Number Scheme Marks 5 (a) y Shape maimum., touching the -ais at its M - Through (0,0) & marked on -ais, or (,0) seen. Allow ( 0, ) if marked on the -ais. Marked in the correct place, but, is A0. A (-,-) Min at (, ) A () (b) y Correct shape (top left - bottom right) B - Through and ma at (0, 0). Marked in the correct place, but, is B0. Min at (, ) B B () (-,-) [6] (a) (b) M as described above. Be generous, even when the curve seems to be composed of straight line segments, but there must be a discernible 'curve' at the ma. and min. st A for curve passing through and the origin. Ma at (,0) nd A for minimum at (, ). Can simply be indicated on sketch. st B for the correct shape. A negative cubic passing from top left to bottom right. Shape: Be generous, even when the curve seems to be composed of straight line segments, but there must be a discernible 'curve' at the ma. and min. nd B for curve passing through (,0) having a ma at (0, 0) and no other ma. rd B for minimum at (, ) and no other minimum. If in correct quadrant but labelled, e.g. (,), this is B0. In each part the (0, 0) does not need to be written to score the second mark having the curve pass through the origin is sufficient. The last mark (for the minimum) in each part is dependent on a sketch being attempted, and the sketch must show the minimum in approimately the correct place (not, for eample, (, ) marked in the wrong quadrant). The mark for the minimum is not given for the coordinates just marked on the aes unless these are clearly linked to the minimum by vertical and horizontal lines. 666/0 GCE Mathematics January 009 6

94 Question Number Scheme Marks 6 (a) (b) (a) (b) or p = (Not ) or or q = d y = 0 + d = 0 + st B for p =.5 or eact equivalent nd B for q = n n M for an attempt to differentiate (for any of the 4 terms) st A for 0 (the must 'disappear') nd Aft for or. Follow through their p but they must be differentiating p, where p is a fraction, and the coefficient must be simplified if necessary. rd 0 Aft for (not the unsimplified ), or follow through for correct q q differentiation of their (i.e. coefficient of is ). If ft is applied, the coefficient must be simplified if necessary. 'Simplified' coefficient means b a where a and b are integers with no common factors. Only a single + or sign is allowed (e.g. must be replaced by +). If there is a 'restart' in part (b) it can be marked independently of part (a), but marks for part (a) cannot be scored for work seen in (b). Multiplying by 4 : (assuming this is a restart) e.g. y = 5 + dy = d Etra term included: This invalidates the final mark. 4 scores M A0 A0 (p not a fraction) Aft. e.g. y = 5 + dy = scores M A A0 (p not a fraction) A0. d Numerator and denominator differentiated separately: For this, neither of the last two (ft) marks should be awarded. Quotient/product rule: Last two terms must be correct to score the last marks. (If the M mark has not already been earned, it can be given for the quotient/product rule attempt.) B B () M AAftAft (4) [6] 666/0 GCE Mathematics January 009 7

95 Question Number 7 (a) ( ) Scheme b 4ac> 0 6 4k 5 k > 0 or equiv., e.g. 6 > 4k(5 k) So k 5k+ 4> 0 (Allow any order of terms, e.g. 4 5k + k 0 ) (*) > MA Marks Acso () (b) (a) Critical Values ( k )( k ) 4 = 0 k =. k = or 4 k < or k > 4 Choosing outside region For this question, ignore (a) and (b) labels and award marks wherever correct work is seen. M A M A (4) [7] M for attempting to use the discriminant of the initial equation (> 0 not required, but substitution of a, b and c in the correct formula is required). If the formula b 4ac is seen, at least of a, b and c must be correct. If the formula b 4ac is not seen, all (a, b and c) must be correct. This mark can still be scored if substitution in b 4ac is within the quadratic formula. This mark can also be scored by comparing b and 4 ac (with substitution). However, use of b + 4ac is M0. st A for fully correct epression, possibly unsimplified, with > symbol. NB must appear before the last line, even if this is simply in a statement such as b 4ac > 0 or discriminant positive. Condone a bracketing slip, e.g. 6 4 k 5 k if subsequent work is correct and convincing. nd A for a fully correct derivation with no incorrect working seen. Condone a bracketing slip if otherwise correct and convincing. Using b 4ac > 0 : Only available mark is the first M (unless recovery is seen). (b) st M for attempt to solve an appropriate TQ st A for both k = and 4 (only the critical values are required, so accept, e.g. k > and k > 4). ** nd M for choosing the outside region. A diagram or table alone is not sufficient. Follow through their values of k. The set of values must be 'narrowed down' to score this M mark listing everything k <, < k < 4, k > 4 is M0. nd A for correct answer only, condone "k <, k > 4" and even "k < and k > 4", but " > k > 4 " is A0. ** Often the statement k > and k > 4 is followed by the correct final answer. Allow full marks. Seeing and 4 used as critical values gives the first M A by implication. In part (b), condone working with 's ecept for the final mark, where the set of values must be a set of values of k (i.e. marks out of 4). Use of (or ) in the final answer loses the final mark. 666/0 GCE Mathematics January 009 8

96 Question Number Scheme Marks 8 (a) (b) ( ) ( ) ( ) a = + =4 (, 4) or y = 4 is also acceptable y (i) Shape or anywhere B () B - Min at (,0) can be on -ais. Allow ( 0, ) if marked on the -ais. Marked in the correct place, but, is B0. (, 0) and (0, ) can be on aes B B (ii) Top branch in st quadrant with intersections Bottom branch in rd quadrant (ignore any intersections) B B (5) (c) ( intersections therefore) (roots) Bft () [7] (b) st B for shape or Can be anywhere, but there must be one ma. and one min. and no further ma. and min. turning points. Shape: Be generous, even when the curve seems to be composed of straight line segments, but there must be a discernible 'curve' at the ma. and min. nd B for minimum at (,0) (even if there is an additional minimum point shown) rd B for the sketch meeting aes at (, 0) and (0, ). They can simply mark on the aes. The marks for minimum and intersections are dependent upon having a sketch. Answers on the diagram for min. and intersections take precedence over answers seen elsewhere. 4 th B for the branch fully within st quadrant having intersections with (not just touching ) the other curve. The curve can touch the aes. A curve of (roughly) the correct shape is required, but be very generous, even when the arc appears to turn 'inwards' rather than approaching the aes, and when the curve looks like two straight lines with a small curve at the join. Allow, for eample, shapes like these: 5 th B for a branch fully in the rd quadrant (ignore any intersections with the other curve for this branch). The curve can touch the aes. A curve of (roughly) the correct shape is required, but be very generous, even when the arc appears to turn 'inwards' rather than approaching the aes. (c) Bft for a statement about the number of roots - compatible with their sketch. No sketch is B0. The answer incompatible with the sketch is B0 (ignore any algebra seen). If the sketch shows the correct intersections and, for eample, one other intersection, the answer here should be, not, to score the mark. 666/0 GCE Mathematics January 009 9

97 Question Number Scheme Marks 9 (a) (b) (c) a + 7 d = 5 or equiv. (for st B), a + 0 d =. 5 or equiv. (for nd B), Solving (Subtract) d = 7.5 so d =.5 a = so a = -7.5 (*) n = 5 + ( n ) { = n(5n 75) } = n n 5 n ( ) 5n= (*) B, B () M Acso () MAft M Acso (4) (d) (a) (b) (c) (d) n 5n = 0 or n 5n 00 = 0 ( n 55)( n+ 40) = 0 n = n = 55 (ignore - 40) M M A () [] Mark parts (a) and (b) as one part, ignoring labelling. Alternative: st.5 5 B: d =.5 or equiv. or d =. No method required, but a = 7. 5 must not be assumed. nd B: Either a + 7 d = 5 or a + 0 d =. 5 seen, or used with a value of d or for listing terms or similar methods, counting back 7 (or 0) terms. M: In main scheme: for a full method (allow numerical or sign slips) leading to solution for d or a without assuming a = 7. 5 In alternative scheme: for using a d value to find a value for a. 5 A: Finding correct values for both a and d (allowing equiv. fractions such as d = ), with no 6 incorrect working seen. In the main scheme, if the given a is used to find d from one of the equations, then allow MA if both values are checked in the nd equation. st M for attempt to form equation with correct S n formula and 750, with values of a and d. st Aft for a correct equation following through their d. nd M for epanding and simplifying to a term quadratic. nd A for correct working leading to printed result (no incorrect working seen). st M forming the correct TQ = 0. Can condone missing = 0 but all terms must be on one side. First M can be implied (perhaps seen in (c), but there must be an attempt at (d) for it to be scored). nd M for attempt to solve TQ, by factorisation, formula or completing the square (see general marking principles at end of scheme). If this mark is earned for the completing the square method or if the factors are written down directly, the st M is given by implication. A for n = 55 dependent on both Ms. Ignore 40 if seen. No working or trial and improvement methods in (d) score all marks for the answer 55, otherwise no marks. 666/0 GCE Mathematics January 009 0

98 Question Number 0 (a) (b) (c) (d) (a) (b) (c) (d) Scheme y 5 y 5= ( ) or equivalent, e.g. =, y = + = y = ( ) + 6 = 7 (therefore B lies on the line) (or equivalent verification methods) AB = ( ) + (7 5), = = 0, AB = 0 = 5 ( ) ( p ) C is (p, p + 6 ), so AC = p Therefore 5 = p 4 p+ 4 + p p =.5p 5p + 5 or 00 = 5p 0 p+ 0 (or better, RHS simplified to terms) Leading to: 0= p 4p 6 (*) M A The version in the scheme above can be written down directly (for marks), and M A0 can be allowed if there is just one slip (sign or number). If the 5 and are the wrong way round the M mark can still be given if a correct formula (e.g. y y = m( ) ) is seen, otherwise M0. If (, 5) is substituted into y = m + c to find c, the M mark is for attempting this and the st A mark is for c = 6. Correct answer without working or from a sketch scores full marks. A conclusion/comment is not required, ecept when the method used is to establish that the line through (,7) with gradient has the same eqn. as found in part (a), or to establish that the line through (,7) and (,5) has gradient. In these cases a comment 'same equation' or 'same gradient' or 'therefore on same line' is sufficient. M for attempting AB or AB. Allow one slip (sign or number) inside a bracket, i.e. do not allow ( ) (7 5). st A for 0 (condone bracketing slips such as = 4 ) nd A for 5 or k = (Ignore ± here). st M for ( p ) + (linear function of p). The linear function may be unsimplified but must be equivalent to ap + b, a 0, b 0. nd M (dependent on st M) for forming an equation in p (using 5 or 5) and attempting (perhaps not very well) to multiply out both brackets. st A for collecting like p terms and having a correct epression. nd A for correct work leading to printed answer. Alternative, using the result: p = ± 5 and use one or both of the two solutions to find the Solve the quadratic ( ) length of AC or C C : e.g. AC = ( + 5 ) + ( 5 5 5) scores st M, and st A if fully correct. Finding the length of AC or AC for both values of p, or finding C C with some evidence of halving (or intending to halve) scores the nd M. Getting AC = 5 for both values of p, or showing C C = 5 scores the nd A (cso). Marks MA, Acao () B () M, A, A () M M A Acso (4) [] 666/0 GCE Mathematics January 009

99 Question Number (a) (b) (c) (a) (b) (c) Scheme dy = 4+ 8 (4 or 8 for M sign can be wrong) d = m = 4 + = 8 The first 4 marks could be earned in part (b) y = 9 8 = Equation of tangent is: y+ = ( ) y = (*) Gradient of normal = Equation is: y + = ( A :), ( B :) 8 or better equivalent, e.g. y = 4 MA M Marks Area of triangle is: ( B ± A ) yp with values for all of B, A and yp M 45 8 = or.5 A (4) 4 [] st M for 4 or 8 (ignore the signs). st A for both terms correct (including signs). nd M for substituting = into their d y (must be different from their y) d B for y P =, but not if clearly found from the given equation of the tangent. rd M for attempt to find the equation of tangent at P, follow through their m and y P. Apply general principles for straight line equations (see end of scheme). NO DIFFERENTIATION ATTEMPTED: Just assuming m = at this stage is M0 nd Acso for correct work leading to printed answer (allow equivalents with, y, and terms such as + y = 0 ). Bft for correct use of the perpendicular gradient rule. Follow through their m, but if m there must be clear evidence that the m is thought to be the gradient of the tangent. M for an attempt to find normal at P using their changed gradient and their y P. Apply general principles for straight line equations (see end of scheme). A for any correct form as specified above (correct answer only). st B for and nd B for 8. B M Acso (6) Bft MA B, B () M for a full method for the area of triangle ABP. Follow through their A, B and their y P, but the mark is to be awarded generously, condoning sign errors.. The final answer must be positive for A, with negatives in the working condoned. y Determinant: Area = y = =... (Attempt to multiply out required for M) y 8 0 Alternative: AP = ( 0.5) + ( ), BP = ( 8) + ( ), Area = AP BP =... M Intersections with y-ais instead of -ais: Only the M mark is available B0 B0 M A0. 666/0 GCE Mathematics January 009

100 Mark Scheme (Results) Summer 009 GCE GCE Mathematics (666/0)

101 Question Number June Core Mathematics C Mark Scheme Scheme Marks Q (a) ( 7) = 6 B () (b) ( 8 5)( 5) = (a) B (b) M + M =, 6 5 A, A for 6 only for an attempt to epand their brackets with > terms correct. They may collect the 5 terms to get or ( 5) Allow or 5 instead of the -5 These 4 values may appear in a list or table but they should have minus signs included The net two marks should be awarded for the final answer but check that correct values follow from correct working. Do not use ISW rule st A for from 6 5 or 6 5 from nd A for both and 6 5. S.C - Double sign error in epansion For leading to + allow one mark () [4] 666/0 GCE Mathematics June 009

102 Question Number Scheme Marks Q 5 = or 048 =, a = or 5 or = 048 B, B = or ( ) B [] st 5 B for = or 048 = This should be eplicitly seen: = a followed by nd B for 5 6 Even writing ( ) = = is OK but simply writing or ( ) 048 seen. This mark may be implied rd B for answer as written. Need a = so is B0 5 a = is OK 6 = is NOT a = or 5 or 5.5 with no working scores full marks. If a = 5.5 seen then award / unless it is clear that the value follows from totally incorrect work. Part solutions: e.g. 5 scores the first B. Special case: If = is not eplicitly seen, but the final answer includes, e.g. a =, a = 4, the second B is given by implication. 666/0 GCE Mathematics June 009

103 Question Number Q (a) dy = 6 6 d (b) ( + C) Scheme Marks M A A () M A 4 + C A () [6] (a) M n for an attempt to differentiate st A for 6 nd 6 A for 6 or Condone + n (b) M for some attempt to integrate an term of the given y. 6 here. Inclusion of +c scores A0 here. n n + st A for both terms correct but unsimplified- as printed or better. Ignore +c here nd A for both terms correct and simplified and +c. Accept but NOT + Condone the +c appearing on the first (unsimplified) line but missing on the final (simplified) line Apply ISW if a correct answer is seen If part (b) is attempted first and this is clearly labelled then apply the scheme and allow the marks. Otherwise assume the first solution is for part (a). 666/0 GCE Mathematics June 009 4

104 Question Number Scheme Marks Q4 (a) 5 > 0, > [Condone 0 > = for MA] M, A (b) ( + )( 4) = 0, Critical values are (c) < 4 and 4 M, A < < 4 M Aft (4) < Bft () [7] (a) M for attempt to collect like terms on each side leading to Must have a or b correct so eg > 4 scores M0 a > b, or a < b, or a = b (b) st M for an attempt to factorize or solve to find critical values. Method must potentially give critical values st A for and 4 seen. They may write <, < 4 and still get this A nd M for choosing the inside region for their critical values nd Aft follow through their distinct critical values Allow > with or, < 4 to score MA0 but and or score MA (,4) is MAbut [,4]is MA0. Score M0A0 for a number line or graph only (c) Bft Allow if a correct answer is seen or follow through their answer to (a) and their answer to (b) but their answers to (a) and (b) must be regions. Do not follow through single values. If their follow through answer is the empty set accept or {} or equivalent in words If (a) or (b) are not given then score this mark for cao () NB You may see <4 (with anything or nothing in-between) < -.5 in (b) and empty set in (c) for Bft Do not award marks for part (b) if only seen in part (c) Use of instead of < (or instead of >) loses one accuracy mark only, at first occurrence. 666/0 GCE Mathematics June 009 5

105 Question Number Scheme Marks Q5 (a) a + 9 d = 400 a + 9 d = 600 M 800 d = 0 d = 60 ( accept + 60 for A) M A () (b) a 540 = 400 a = 940 M A () (c) Total = n { a + ( n ) d} = 40 ( ) (ft values of a and d) M Aft = Acao () [8] Note: If the sequence is considered backwards, an equivalent solution may be given using d = 60 with a = 600 and l = 940 for part (b). This can still score full marks. Ignore labelling of (a) and (b) (a) st M for an attempt to use 400 and 600 in a+ ( n ) d formula. Must use both values i.e. need a + pd = 400 and a + qd = 600 where p = 8 or 9 and q = 8 or 9 (any combination) nd M for an attempt to solve their linear equations in a and d as far as d = A for d = Condone correct equations leading to d = 60 or a + 8d = 400 and a + 8d = 600 leading to d = They should get penalised in (b) and (c). NB This is a one off ruling for A. Usually an A mark must follow from their work. ALT st M for (0d) = + ( ) ( nd ) M for ( d = ) ± 0 A for d = + 60 a + 9d = 600, a + 9d = 400 only scores M0 BUT if they solve to find d = + 60 then use ALT scheme above. (b) M for use of their d in a correct linear equation to find a leading to a = A their a must be compatible with their d so d = 60 must have a = 600 and d = -60, a = 940 So for eample they can have 400 = a + 9(60) leading to a = for M but it scores A0 Any approach using a list scores MA for a correct a but M0A0 otherwise (c) M for use of a correct S n formula with n = 40 and at least one of a, d or l correct or correct ft. st Aft for use of a correct S 40 formula and both a, d or a, l correct or correct follow through n (ft value of a) M Aft ALT Total = { a + l} = 40 ( ) nd A for only 666/0 GCE Mathematics June 009 6

106 Question Number Scheme Marks Q6 b 4ac attempted, in terms of p. M (p) 4 p = 0 o.e. A Attempt to solve for p e.g. p( 9p 4) = 0 Must potentially lead to p = k, k 0 M 4 p = (Ignore p = 0, if seen) Acso 9 [4] st M for an attempt to substitute into b 4 ac or b = 4acwith b or c correct Condone s in one term only. This can be inside a square root as part of the quadratic formula for eample. Use of inequalities can score the M marks only st A for any correct equation: ( p) 4 p= 0 or better nd M for an attempt to factorize or solve their quadratic epression in p. Method must be sufficient to lead to their p = 4 9. Accept factors or use of quadratic formula or ( p ± ) = k (o.e. eg) ( p ) = k equivalent work on their eqn. ALT 9 ± or 9 p 9p = 4p = 4 which would lead to 9p = 4 is OK for this nd M p Comparing coefficients = + α + α and A for a correct equation eg p = p M for ( + α ) M for forming solving leading to p = or better Use of quadratic/discriminant formula (or any formula) Rule for awarding M mark If the formula is quoted accept some correct substitution leading to a partially correct epression. If the formula is not quoted only award for a fully correct epression using their values. 666/0 GCE Mathematics June 009 7

107 Question Number Scheme Marks Q7 (a) ( a = )k 7 B () (b) ( a = )(k 7) 7 or 4k 4 7, = 4k (*) M, Acso () (c) ( a4 = ) (4k ) 7 ( = 8k 49) M 4 ar = k+ "(k 7)" + (4k ) + "(8k 49)" M r= k + ( k 7) + (4k ) + (8k 49) = 5k 77 = 4 k = 8 M A (4) [7] (b) M must see (their a ) - 7 or (k 7) 7 or 4k 4 7. Their a must be a function of k. Acso must see the (k 7) 7 or 4k 4 7 epression and the 4k - with no incorrect working (c) st M for an attempt to find a 4 using the given rule. Can be awarded for 8k - 49 seen. Use of formulae for the sum of an arithmetic series scores M0M0A0 for the net marks. nd M for attempting the sum of the st 4 terms. Must have + not just, or clear attempt to sum. Follow through their a and a 4 provided they are linear functions of k. Must lead to linear epression in k. Condone use of their linear a 4k here too. rd M for forming a linear equation in k using their sum and the 4 and attempt to solve for k as far as pk = q A for k = 8 only so 0 k = is A0 5 Answer Only (e.g. trial improvement) Accept k = 8 only if = 4 is seen as well Sum a + a+ a4 + a5 or a + a + a4 Allow: M if 8k - 49 is seen, M0 for the sum (since they are not adding the st 4 terms) then M if they use their sum along with the 4 to form a linear equation and attempt to solve but A0 666/0 GCE Mathematics June 009 8

108 Question Number Scheme Marks Q8 (a) (b) (c) 7 5 AB: m =, = 8 6 B Using m m = : m = 5 M y 7 = ( 6), 5 5y + = 0 (o.e. with integer coefficients) M, A (4) Using = 0 in the answer to (a), y = or M, Aft () Area of triangle = = (o.e) e.g. 8, 8.4, M A () [8] (a) B for an epression for the gradient of AB. Does not need the =.5 st M for use of the perpendicular gradient rule. Follow through their m nd M for the use of (6, 7) and their changed gradient to form an equation for l. y 7 Can be awarded for = o.e. 6 5 Alternative is to use (6, 7) in y = m + c to find a value for c. Score when c = is reached. A for a correct equation in the required form and must have = 0 and integer coefficients (b) M for using = 0 in their answer to part (a) e.g. 5y + = 0 Aft for y = provided that = 0 clearly seen or C (0, 4.6). Follow through 5 their equation in (a) If =0, y = 4.6 are clearly seen but C is given as (4.6,0) apply ISW and award the mark. This A mark requires a simplified fraction or an eact decimal Accept their 4.6 marked on diagram net to C for MAft (c) M for 8 y C so can follow through their y coordinate of C. A for 8.4 (o.e.) but their y coordinate of C must be positive Use of triangles or trapezium and triangle Award M when an epression for area of OCB only is seen Determinant approach Award M when an epression containing 8 y C is seen 666/0 GCE Mathematics June 009 9

109 Question Number Scheme Marks Q9 (a) (b) (c) (a) ( ) ( 4 ) = A, A () f ( ) =, + M A, Aft () f (9) = + = + = M A () [8] M for an attempt to epand ( 4 ) or better with at least terms correct- as printed Or 9 k + 6 ( k 0). See also the MR rule below st A for their coefficient of = 6. Condone writing ( ) nd A for B = - 4 or their constant term = - 4 ( ) ± instead of 9 ± 9 M (b) M for an attempt to differentiate an term 9 n n st A for and their constant B differentiated to zero. NB 9 is A0 nd Aft follow through their A but can be scored without a value for A, i.e. for A (c) M for some correct substitution of = 9 in their epression for f( ) including an attempt ± at ( 9) k (k odd) somewhere that leads to some appropriate multiples of or A accept 5 6 or any eact equivalent of.5 e.g , or even Misread (MR) Only allow MR of the form ( k ) Score as MA0A0, MAAft, MAft N.B. Leads to answer in (c) of k 6 666/0 GCE Mathematics June 009 0

110 Question Number Scheme Marks Q0 (a) ( 6 + 9) B = ( )( ) M A () (b) 6 y Shape B Through origin (not touching) Touching -ais only once Touching at (, 0), or on -ais [Must be on graph not in a table] B B Bft (4) (c) 6 y 5 4 Moved horizontally (either way) (, 0) and (5, 0), or and 5 on -ais M A () [9] (a) B for correctly taking out a factor of M for an attempt to factorize their TQ e.g. ( + p)( + q) where pq = 9. S.C. So ( )( ) + will score M but A0 A for a fully correct factorized epression - accept ( ) If they solve use ISW If the only correct linear factor is ( - ), perhaps from factor theorem, award B0MA0 Do not award marks for factorising in part (b) For the graphs Sharp points will lose the st B in (b) but otherwise be generous on shape Condone (0, ) in (b) and (0, ), (0,5) in (c) if the points are marked in the correct places. (b) nd B for a curve that starts or terminates at (0, 0) score B0 4 th Bft for a curve that touches (not crossing or terminating) at (a, 0) where their a y = ( ) (c) M for their graph moved horizontally (only) or a fully correct graph Condone a partial stretch if ignoring their values looks like a simple translation A for their graph translated to the right and crossing or touching the ais at and 5 only Allow a fully correct graph (as shown above) to score MA whatever they have in (b) 666/0 GCE Mathematics June 009

111 Question Number Scheme Marks Q (a) = : y = = 7 (*) B () (b) d y = 4 M A d dy = : 8 ( ) d = = Aft y 7 = ( ), y = + M, A (5) (c) m = (for with their m) m Bft 4 =, 9 4 = 0 or = 0 (o.e.) 9 M, A = ( 6 = 6 6 = 6 6 ) or ( ) = 6 = ± 8 6 M = ( + 6) (*) Acso (5) [] (a) B there must be a clear attempt to substitute = leading to 7 e.g. + 9 = 7 (b) st M for an attempt to differentiate with at least one of the given terms fully correct. st A for a fully correct epression nd dy Aft for sub. = in their ( y) d accept for a correct epression e.g. ( ) 4 nd dy M for use of their (provided it comes from their ( y) and =) to find d equation of tangent. Alternative is to use (, 7) in y = m + c to find a value for c. Award when c = is seen. No attempted use of d y in (b) scores 0/5 d (c) st dy M for forming an equation from their ( y) d and their (must be m changed from m) st A for a correct TQ all terms on LHS (condone missing =0) nd M for proceeding to = or = by formula or completing the square for a TQ. Not factorising. Condone + nd A for proceeding to given answer with no incorrect working seen. Can still have +. ALT Verify (for MAMA) st M for attempting to square need > correct values in , st A for y nd M Dependent on st M in this case for substituting in all terms of their d d nd Acso for cso with a full comment e.g. the co-ord of Q is /0 GCE Mathematics June 009

112 Mark Scheme (Results) January 00 GCE Core Mathematics C (666) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

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114 January 00 Core Mathematics C 666 Mark Scheme Question number Scheme Q 4 k or k or 0 (k a non-zero constant) M Marks d y = 4..., with '' differentiated to zero (or 'vanishing') d A dy =... + d st A requires or equivalent, e.g. 4, and differentiated to zero. or ( ) A [] Having +C loses the st A mark. Terms not added, but otherwise correct, e.g. 4, loses the nd A mark. GCE Core Mathematics C (666) January 00

115 Question number Q Scheme (a) ( 7 5)( 5) = Epand to get or 4 terms M Marks (b) = 6, 4 5 ( st A for 6, nd A for 4 5) (i.s.w. if necessary, e.g ) (This is sufficient for the M mark) Correct denominator without surds, i.e. 9 5 or or 4 5 (a) M: Allowed for an attempt giving or 4 terms, with at least correct (even if unsimplified). e.g scores M. Answer only: scores full marks One term correct scores the M mark by implication, e.g scores M A0 A (b) Answer only: 4 5 scores full marks One term correct scores the M mark by implication, e.g scores M A0 A0 6 5 scores M A0 A0 Ignore subsequent working, e.g. 4 5 so a = 4, b = Note that, as always, A marks are dependent upon the preceding M mark, so that, for eample, = is M0 A Alternative ( a + b 5)( + 5) = 7 + 5, then form simultaneous equations in a and b. M Correct equations: a + 5b = 7 and b + a = A a = 4 and b = A A, A M A A () () [6] GCE Core Mathematics C (666) January 00 4

116 Question number Q Scheme (a) Putting the equation in the form y = m ( + c) and attempting to etract the m or m (not the c), or finding points on the line and using the correct gradient formula. Gradient = (or equivalent) 5 (b) Gradient of perp. line = (Using with the m from part (a)) "( )" m 5 5 y = " "( ) y = 4 (Must be in this form... allow y = but not y = ) This A mark is dependent upon both M marks. M A M M A Marks () () [5] (a) Condone sign errors and ignore the c for the M mark, so... both marks can be scored even if c is wrong (e.g. c = ) or omitted. 5 Answer only: scores M A. Any other answer only scores M0 A0. 5 y = + with no further progress scores M0 A0 (m or m not etracted). 5 5 (b) nd M: For the equation, in any form, of a straight line through (, ) with any numerical gradient (ecept 0 or ). (Alternative is to use (, ) in y = m + c to find a value for c, in which 5 case y = + c leading to c = 4 is sufficient for the A). (See general principles for straight line equations at the end of the scheme). GCE Core Mathematics C (666) January 00 5

117 Question number Q4 Scheme = (Seen, or implied by correct integration) B Marks k 5 or k (k a non-zero constant) M y =... + (+C) ( y = and +C are not required for these marks) 5 ( ) = C C = or equivalent 5 5 An equation in C is required (see conditions below). (With their terms simplified or unsimplified).,. 5 y = (Or equivalent simplified) I.s.w. if necessary, e.g. y = = The final A mark requires an equation y =... with correct terms (see below). B mark: often appears from integration of, which is B0. A A M A A ft [7] st A: Any unsimplified or simplified correct form, e.g nd ( ) A: Any unsimplified or simplified correct form, e.g. nd M: Attempting to use = 4 and y = 5 in a changed function (even if differentiated) to form an equation in C. rd A: Obtaining C = with no earlier incorrect work. 5 4th A: Follow-through only the value of C (i.e. the other terms must be correct). Accept equivalent simplified terms such as , 5 5. GCE Core Mathematics C (666) January 00 6

118 Question number Q5 = y ( ) 6 ( = 0) Scheme M Marks = = 0 (or equiv., e.g. = 4 ) ( )( 4) = 0 = = (or eact equivalent) = 4 M Acso M A y = y = 0 (Solutions need not be "paired") M A st M: Obtaining an equation in only (or y only). Condone missing = 0 Condone sign slips, e.g. ( + ) 6 = 0, but not other algebraic mistakes (such as squaring individual terms... see bottom of page). nd M: Multiplying out their ( ), which must lead to a term quadratic, i.e. a + b + c, where a 0, b 0, c 0, and collecting terms. [7] rd M: Solving a -term quadratic (see general principles at end of scheme). nd A: Both values. 4 th M: Using an value, found algebraically, to attempt at least one y value (or using a y value, found algebraically, to attempt at least one value) allow b.o.d. for this mark in cases where the value is wrong but working is not shown. rd A: Both values. If y solutions are given as values, or vice-versa, penalise at the end, so that it is possible to score M MA M A M0 A0. Non-algebraic solutions: No working, and only one correct solution pair found (e.g. = 4, y = 0): M0 M0 A0 M0 A0 M A0 No working, and both correct solution pairs found, but not demonstrated: M0 M0 A0 M A M A Both correct solution pairs found, and demonstrated: Full marks Alternative: + = y y + y + y 6 = 0 M y + y + 4y + 4 y 6 = 0 9 y 9y 0 = 0 M A ( y + )( y 0) = 0 y = y = y = 0 M A = = 4 M A Squaring each term in the first equation, e.g. y = 0, and using this to obtain an equation in only could score at most marks: M0 M0 A0 M A0 M A0. GCE Core Mathematics C (666) January 00 7

119 Question number Q6 (a) Scheme 5 4 y = = 5 4 (or equiv., e.g ) Marks M A dy = + 4 d or dy 4 = + d (b) = : y = 5 Allow if seen in part (a). M A B (4) dy d = 4 + = 7 4 Follow-through from candidate's non-constant This must be simplified to a single value. d y. d Bft y + 5 y + 5 = 7( ) (or equiv., e.g. y = 7 9 ) Allow = 7 M A (4) [8] (a) st M: Mult. out to get + b + c, b 0, c 0 and dividing by (not ). Obtaining one correct term, e.g... is sufficient evidence of a division attempt. nd M: Dependent on the st M: Evidence of n k n for one term (i.e. not just the constant term) is sufficient). Note that mark is not given if, for eample, the numerator and denominator are differentiated separately. A mistake in the 'middle term', e.g , does not invalidate the nd A mark, so M A0 M A is possible. d y (b) Bft: For evaluation, using =, of their, even if unlabelled or called y. d M: For the equation, in any form, of a straight line through (, 5 ) with dy candidate s value as gradient. d Alternative is to use (, 5 ) in y = m + c to find a value for c, in which case y = 7 + c leading to c = 9 is sufficient for the A).. (See general principles for straight line equations at the end of the scheme). Final A: 'Unsimplified' forms are acceptable, but y ( 5) = 7( ) is A0 (unresolved 'minus minus'). GCE Core Mathematics C (666) January 00 8

120 Question number Q7 Scheme (a) a + 9 d = = 40 M A (b) { a + ( n ) d} = { }, = 4900 Marks 0 n M A, A 0 Kevin's total = "4900" (or "4900" = Kevin's total) 0 { A + 9 0} = "4900" A = 05 (c) Kevin: n { a + ( n ) d} = { A + 9 0} (a) M: Using a + 9d with at least one of a = 50 and d = 0. Being one off (e.g. equivalent to a + 0d), scores M0. Correct answer with no working scores both marks. (b) M: Attempting to use the correct sum formula to obtain S 0, with at least one of a = 50 and d = 0. If the wrong value of n or a or d is used, the M mark is only scored if the correct sum formula has been quoted. st A: Any fully correct numerical version. (c) B: A correct epression, in terms of A, for Kevin's total. M: Equating Kevin's total to twice Jill's total, or Jill's total to twice Kevin's. For this M mark, the epression for Kevin's total need not be correct, but must be a linear function of A (or a). st A: (Kevin's total, correct, possibly unsimplified) = (Jill's total), ft Jill's total from part (b). Listing and other methods (a) M: Listing terms (found by a correct method with at least one of a = 50 and d = 0), and picking the 0 th term. (There may be numerical slips). (b) M: Listing sums, or listing and adding terms (found by a correct method with at least one of a = 50 and d = 0), far enough to establish the required sum. (There may be numerical slips). Note: 0 th term is 40. A (scored as A A) for 4900 (clearly selected as the answer). If no working (or no legitimate working) is seen, but the answer 4900 is given, allow one mark (scored as M A0 A0). (c) By trial and improvement: Obtaining a value of A for which Kevin's total is twice Jill's total, or Jill's total is twice Kevin's (using Jill's total from (b)): M Obtaining a value of A for which Kevin's total is twice Jill's total (using Jill's total from (b)): Aft Fully correct solutions then score the B and final A. The answer 05 with no working (or no legitimate working) scores no marks. B M Aft A () () (4) [9] GCE Core Mathematics C (666) January 00 9

121 Question number Q8 Scheme (a) (b) (c) Marks (a) (, 7), y = (Marks are dependent upon a sketch being attempted) See conditions below. (b) (, 0), y = 4 (Marks are dependent upon a sketch being attempted) See conditions below. (c) Sketch: Horizontal translation (either way) (There must be evidence that y = 5 at the ma and that the asymptote is still y = ) (, 5), y = Parts (a) and (b): (i) If only one of the B marks is scored, there is no penalty for a wrong sketch. (ii) If both the maimum and the equation of the asymptote are correct, the sketch must be correct to score B B. If the sketch is wrong, award B B0. The (generous) conditions for a correct sketch are that the maimum must be in the nd quadrant and that the curve must not cross the positive -ais... ignore other errors such as curve appearing to cross its asymptote and curve appearing to have a minimum in the st quadrant. Special case: 5 (b) Stretch instead of 4: Correct shape, with,, y = : B B Coordinates of maimum: If the coordinates are the wrong way round (e.g. (7, ) in part (a)), or the coordinates are just shown as values on the and y aes, penalise only once in the whole question, at first occurrence. Asymptote marks: If the equation of the asymptote is not given, e.g. in part (a), is marked on the y-ais but y = is not seen, penalise only once in the whole question, at first occurrence. Ignore etra asymptotes stated (such as = 0). B, B B, B B B, B () () () [7] GCE Core Mathematics C (666) January 00 0

122 Question number Q9 Scheme (a) ( 4) Factor seen in a correct factorised form of the epression. = ( )( + ) M: Attempt to factorise quadratic (general principles). Accept ( 0) or ( + 0) instead of at any stage. Factorisation must be seen in part (a) to score marks. (b) 4 y Shape ( turning points required) Through (or touching) origin 4 4 Crossing -ais or stopping at -ais (not a turning point) at (, 0) and (, 0). 4 Allow and on -ais. Also allow (0, ) and (0, ) if marked on -ais. Ignore etra intersections with -ais. (c) Either y = (at = ) or y = 5 (at = ) Allow if seen elsewhere. "5 " Gradient = ( ) "5 " For gradient M mark, if correct formula not seen, allow one slip, e.g. y " 5" = m( ) or y " " = m( ( )), with any value for m. y 5= ( ) or the correct equation in any form, 5 y 5 e.g. y = ( ( )), = ( ) + + y = + 6 (d) = (" 5 " ) + ( ( ) ) ( = ) Attempt correct grad. formula with their y values. AB (With their non-zero y values)... Square root is required. = 60 = 6 0 = 4 (Ignore ± if seen) ( 6 0 need not be seen). ( ) 0 (a) 4 ( 4) ( )( + ) scores B M A ( )( + ) scores B0 M A0 (dividing by ). 4 ( 4) ( 4) scores B0 M A0. 4 ( 4) ( ) scores B M A0 Special cases: 4 ( )( + ) scores B0 M A0. 4 ( ) (with no intermediate step seen) scores B0 M A0 (b) The nd and rd B marks are not dependent upon the st B mark, but are dependent upon a sketch having been attempted. (c) st M: May be implicit in the equation of the line, e.g. "5" "" = "5" "" nd M: An equation of a line through (, "5") or (, "" ) in any form, with any gradient (ecept 0 or ). nd M: Alternative is to use one of the points in y = m + c to find a value for c, in which case y = + c leading to c = 6 is sufficient for both A marks. st A: Correct equation in any form. B M A B B B B M M A Marks () () A (5) M A () [] GCE Core Mathematics C (666) January 00

123 Question number Q0 (a) ( + k) or Scheme 4k + ± F ± G ± ± (where F and G are any functions of k, not involving ) ( ) k ( k) 4k + ( + k ) + Accept unsimplified equivalents such as 4k 4k k, and i.s.w. if necessary. (b) Accept part (b) solutions seen in part (a). "4k k " = 0 ( 4k + )( k ) = 0 k =, [Or, starting again, b 4ac = (4k) 4( + k ) and proceed to k = ] and (Ignore any inequalities for the first marks in (b)). 4 Using b 4ac < 0 for no real roots, i.e. "4k k " < 0, to establish inequalities involving their two critical values m and n (even if the inequalities are wrong, e.g. k < m, k < n ). < k < (See conditions below) Follow through their critical values. 4 The final Aft is still scored if the answer m < k < n follows k < m, k < n. Using instead of k in the final answer loses only the nd A mark, (condone use of in earlier working). (c) Shape (seen in (c)) y 0 Minimum in correct quadrant, not touching the -ais, not on the y-ais, and there must 0 be no other minimum or maimum. (0, 4) or 4 on y-ais Allow (4, 0) marked on y-ais. n.b. Minimum is at (,0), (but there is no mark for this). (b) st M: Forming and solving a -term quadratic in k (usual rules.. see general principles at end of scheme). The quadratic must come from " b 4ac", or from the "q" in part (a). Using wrong discriminant, e.g. " b + 4ac" will score no marks in part (b). nd M: As defined in main scheme above. nd Aft: m < k < n, where m < n, for their critical values m and n. Other possible forms of the answer (in each case m < n): (i) n > k > m (ii) k > m and k < n In this case the word and must be seen (implying intersection). (iii) k ( m, n) (iv) { k : k > m} { k : k < n} Not just a number line. Not just k > m, k < n (without the word and ). (c) Final B is dependent upon a sketch having been attempted in part (c). Marks M M A () M A M Aft (4) B B B () [0] GCE Core Mathematics C (666) January 00

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127 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fa Order Code US0704 January 00 For more information on Edecel qualifications, please visit Edecel Limited. Registered in England and Wales no Registered Office: One90 High Holborn, London, WCV 7BH

128 Mark Scheme (Results) Summer 00 GCE Core Mathematics C (666) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

129 Edecel is one of the leading eamining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edecel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on , our GCSE team on , or visit our website at If you have any subject specific questions about the content of this Mark Scheme that require the help of a subject specialist, you may find our Ask The Epert service helpful. Ask The Epert can be accessed online at the following link: Summer 00 Publications Code UA0696 All the material in this publication is copyright Edecel Ltd 00

130 SOME GENERAL PRINCIPLES FOR C MARKING (But the particular mark scheme always takes precedence) Method marks Usually we would overlook simple arithmetic errors or sign slips but the correct processes should be used. So dividing by a number instead of subtracting would be M0 but adding a number instead of subtracting would be treated as the correct process but a sign error. 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 = a, leading to =. Formula Attempt to use correct formula (with values for a, b and c).. Completing the square b Solving + b + c = 0 : ± ± q ± c = 0, q 0, leading to = 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 (but refer to the mark scheme first the application of this principle may vary). 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. Equation of a straight line Apply the following conditions to the M mark for the equation of a line through ( a, b) : If the a and b are the wrong way round the M mark can still be given if a correct formula is seen, (e.g. y y = m( ) ) otherwise M0. If (a, b) is substituted into y = m + c to find c, the M mark is for attempting this and scored when c =... is reached. 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. If in doubt, send the response to Review. GCE Core Mathematics C (666) Summer 00

131 GCE Core Mathematics C (666) Summer 00

132 June 00 Core Mathematics C 666 Mark Scheme Question Number Scheme Marks. ( 75 7 ) = 5 M = A Notes for 5 from 75 or from 7 seen anywhere M or k =, = A for ; allow or allow k =, = Some Common errors 75 7 = 48 leading to 4 is M0A0 5 9 = 6 is M0A0 GCE Core Mathematics C (666) Summer 00

133 Question Number. 4 Scheme c M A 4 Marks = 4 + 4, 5+ c A A Notes M for some attempt to integrate a term in : n n + 4 st A for correct, possibly un-simplified nd A for both or term. e.g. or 4 and 4 terms correct and simplified on the same line N.B. some candidates write 4 or 4 which are, of course, fine for A rd A for 5+ c. Accept 5 + c. The +c must appear on the same line as the 5 N.B. We do not need to see one line with a fully correct integral Ignore ISW (ignore incorrect subsequent working) if a correct answer is followed by an incorrect version. Condone poor use of notation e.g c will score full marks. GCE Core Mathematics C (666) Summer 00

134 Question Number Scheme Marks. (a) 6< 8 5< 4 (Accept 5 4< 0 (o.e.)) M 4 <.8 or or (condone < ) A () (b) Critical values are Choosing inside 7 = and B 7 < < M A () (c) < <.8 Bft () (a) Accept any eact equivalents to -,.8,.5 6 Notes M for attempt to rearrange to k < m (o.e.) Either k = 5 or m = 4 should be correct Allow 5 = 4 or even 5 > 4 (b) B for both correct critical values. (May be implied by a correct inequality) M ft their values and choose the inside region A for fully correct inequality (Must be in part (b): do not give marks if only seen in (c)) Condone seeing < in working provided < is in the final answer e.g. >, < or > "or" < or > "blank space" < score MA0 7 BUT allow > and < to score MA (the and must be seen) 7, will score MA Also ( ) 7 NB <, < is of course M0A0 and a number line even with open ends is M0A0 Allow.5 instead of 7 (c) Bft for < <.8(ignoring their previous answers) or ft their answers to part (a) and part (b) provided both answers were regions and not single values. Allow use of and between inequalities as in part (b) If their set is empty allow a suitable description in words or the symbol. Common error: If (a) is correct and in (b) they simply leave their answer as <, <.5 then in (c) < would get Bft as this is a correct follow through of these inequalities. Penalise use of only on the A in part (b). [i.e. condone in part (a)] GCE Core Mathematics C (666) Summer 00

135 Question Number 4. (a) ( + ) + or Scheme 6 p = or q = B Marks B () (b) y U shape with min in nd quad (Must be above -ais and not on y=ais) B U shape crossing y-ais at (0, ) only (Condone (,0) marked on y-ais) B () (c) b 4ac= 6 4 M = 8 A () 6 Notes (a) Ignore an = 0 so ( + ) + = 0 can score both marks (b) (c) The U shape can be interpreted fairly generously. Penalise an obvious V on st B only. The U needn t have equal arms as long as there is a clear min that holds water st B for U shape with minimum in nd quad. Curve need not cross the y-ais but minimum should NOT touch -ais and should be left of (not on) y-ais nd B for U shaped curve crossing at (0, ). Just marked on y-ais is fine. The point must be marked on the sketch (do not allow from a table of values) Condone stopping at (0, ) M for some correct substitution intob 4ac. This may be as part of the quadratic formula but must be in part (c) and must be only numbers (no terms present). Substitution into b < 4 ac or b = 4 ac or b > 4 ac is M0 A for 8 only. If they write 8 < 0 treat the < 0 as ISW and award A If they write 8 > 0 then score A0 A substitution in the quadratic formula leading to 8 inside the square root is A0. So substituting into b 4ac< 0 leading to 8 < 0 can score MA. Only award marks for use of the discriminant in part (c) GCE Core Mathematics C (666) Summer 00

136 Question Number 5. Scheme (a) a = ( 4+ ) = 7 B Marks a = "their 7" + = 0 Bft () (b) a 4 = 0 + ( = ) M a 5 = + = 4 * A cso () Notes 4 (a) st B for 7 only nd Bft follow through their 7 in correct formula provided they have n,where n is an integer. (b) M for an attempt to find 4 a. Should see ( ) "their" a +. Must see evidence for M. a = 4 provided this follows from their a working or answer is sufficient Acso for a correct solution (M eplicit) must include the = 4. Ending at 6 only is A0 and ending with + 4 is A0. Ignore any incorrect statements that are not used e.g. common difference = Listing: A full list: ( = 4 ), 7, 0,, 6 = 4 is fine for MA ALT Formula: Some may state (or use) an = n+ leading to a5 = 5 + = 4. This will get marks in (a) [if correct values are seen] and can score the M in (b) if an = n+ or a4 = are seen. ± If ± appear any where ignore in part (a) and withhold the final A mark only GCE Core Mathematics C (666) Summer 00

137 Question Number 6. Scheme Marks (-5, ) y Horizontal translation of ± M (a) ( 5, ) marked on sketch or in tet B (0, -5) (0, 5) and min intentionally on y-ais Condone ( 5, 0) if correctly placed on negative y-ais A () (-, 6) y Correct shape and intentionally through (0,0) between the ma and min B (b) (, 6) marked on graph or in tet B (, -0) (, 0) marked on graph or in tet B () (c) (a = ) 5 B () Notes Turning points (not on aes) should have both co-ordinates given in form(,y). Do not accept points marked on aes e.g. 5on -ais and on y-ais is not sufficient. For repeated offenders apply this penalty once only at first offence and condone elsewhere. In (a) and (b) no graphs means no marks. In (a) and (b) the ends of the graphs do not need to cross the aes provided ma and min are clear y (a) M for a horizontal translation of ± so accept i.e ma in st quad and coordinates of (, ) or (6, 5) seen. [Horizontal translation to the left should have a min on the y-ais] If curve passes through (0,0) then M0 (and A0) but they could score the B mark. A for minimum clearly on negative y-ais and at least 5 marked on y-ais. Allow this mark if the minimum is very close and the point (0, 5) clearly indicated (b) st B Ignore coordinates for this mark Coordinates or points on sketch override coordinates given in the tet. Condone (y, ) confusion for points on aes only. So ( 5,0) for (0, 5) is OK if the point is marked correctly but (,0) is B0 even if in 4 th quadrant. (c) This may be at the bottom of a page or in the question...make sure you scroll up and down! GCE Core Mathematics C (666) Summer 00

138 Question Number 7. Scheme + = + M A ( ) Marks y = 4,, + M A AA 4 + Notes st M for attempting to divide(one term correct) st A for both terms correct on the same line, accept for or for These first two marks may be implied by a correct differentiation at the end. nd n n M for an attempt to differentiate for at least one term of their epression + 6 Differentiating and getting is M0 nd A for 4 only rd A for 4 allow. Must be simplified to this, not e.g. 4 th A for allow. Both terms needed. Condone + ( ) If +c is included then they lose this final mark They do not need one line with all terms correct for full marks. Award marks when first seen in this question and apply ISW. Condone a mied line of some differentiation and some division e.g can score st MA and nd MA 6 Quotient /Product Rule ( ) ( ) 6 + or 6 or ( ) + ( + )( ) st P Q M for an attempt: or R + ( S) with one of P,Q or R,S correct. st A for a correct epression (o.e.) 4 th A same rules as above GCE Core Mathematics C (666) Summer 00

139 Question Number 8. (a) m AB = = 7 5 Equation of AB is: y Scheme Marks M ( ) y 4= 7 M 5 5 (o.e.) 4-5y - 8 = 0 (o.e.) A () = or ( ) AB = M = 4 A () (b) ( ) ( ) ( ) (c) Using isos triangle with AB = AC then t = y A = 4= 8 B () (d) Area of triangle = t (7 ) M = 0 A () (a) Notes Apply the usual rules for quoting formulae here. For a correctly quoted formula with some correct substitution award M If no formula is quoted then a fully correct epression is needed for the M mark st M for attempt at gradient of AB. Some correct substitution in correct formula. nd M for an attempt at equation of AB. Follow through their gradient, not e.g. m Using y = m + c scores this mark when c is found. y y Use of = scores st M for denominator, nd M for use of a correct point y y A requires integer form but allow 5y + 8 = 4 etc. Must have an = or A0 8 (b) M for an epression for AB or AB. Ignore what is left of the equals sign (c) B for t = 8. May be implied by correct coordinates (, 8) or the value appearing in (d) (d) M for an epression for the area of the triangle, follow through their t ( 0) but must have the (7 ) or 5 and the. 7 Area = t+ + + t 0 4 t 0 Must have the for M DET e.g. ( ) GCE Core Mathematics C (666) Summer 00

140 Question Number Scheme Marks 9. (a) a + 9d = or a= d or 9d = a M A () = l or ( a ) or (a + (0 ) d) or 5(a + 9d) M So 005 = 5[ a ] * A cso () (b) ( S ) ( a + ) 05 (c) 67 = a so a = ( ) 6.5 or 65p or 6 NOT M A 4 4 9d = = 4.5 so d = ( )0.50 or 0.5 or 50p or M A (4) 8 (a) Notes M for attempt to use a + (n )d with n =0 to form an equation. So a + (0 )d = any number is OK A as written. Must see 9d not just (0 )d. Ignore any floating signs e.g. a+ 9 d = is OK for MA These two marks must be scored in (a). Some may omit (a) but get correct equation in (c) [or (b)] but we do not give the marks retrospectively. Parts (b) and (c) may run together (b) M for an attempt to use an S n formula with n =0. Must see one of the printed forms. ( S 0 = is not required) Acso for forming an equation with 005 and S n and simplifying to printed answer. Condone signs e.g. 5[a ]=005 is OK for A (c) st M for an attempt to simplify the given linear equation for a. Correct processes. Must get to ka = or k = a + m i.e. one step (division or subtraction) from a = Commonly: 5a = (= 9.75) st 05 A For a = 6.5 or 65p or 6 4 NOT 4 or any other fraction nd M for correct attempt at a linear equation for d, follow through their a or equation in (a) Equation just has to be linear in d, they don t have to simplify to d = nd A depends upon nd M and use of correct a. Do not penalise a second time if there were minor arithmetic errors in finding a provided a = 6.5 (o.e.) is used. Do not accept other fractions other than If answer is in pence a p must be seen. Sim Equ Use this scheme: st MA for a and nd MA for d. Typically solving: 005=0a + 45d and = a + 9d. If they find d first then follow through use of their d when finding a. GCE Core Mathematics C (666) Summer 00

141 Question Number 0. (a) y O 4 7 Scheme (i) shape (anywhere on diagram) Passing through or stopping at (0, 0) and (4,0) only(needn t be shape) (ii) correct shape (-ve cubic) with a ma and min drawn anywhere Minimum or maimum at (0,0) Passes through or stops at (7,0) but NOT touching. (7, 0) should be to right of (4,0) or B0 Condone (0,4) or (0, 7) marked correctly on -ais. Don t penalise poor overlap near origin. Points must be marked on the sketch...not in the tet B B B Marks B B (5) (b) ( 4 ) ( 7 ) = (0 = ) [7 (4 )] M (0 = ) [7 (4 )] (o.e.) Bft ( ) (c) ( ) 0= * A cso () 8± 64 6 = + = or ( ) ( ) ± ( = 0) 4 = 8± 4 = or ( 4) =± B = 4± A From sketch A is = 4 M So y = ( 4 )( 4 [4 ] ) (dependent on st M) M = + 8 A (7) 5 Notes (b) M for forming a suitable equation B for a common factor of taken out legitimately. Can treat this as an M mark. Can ft their cubic = 0 found from an attempt at solving their equations e.g. 8 4 = (... Acso no incorrect working seen. The = 0 is required but condone missing from some lines of working. Cancelling the scores B0A0. M A (c) st M for some use of the correct formula or attempt to complete the square st A for a fully correct epression: condone + instead of + or for ( 4) = B for simplifying 48 = 4 or =.Can be scored independently of this epression nd A for correct solution of the form p + q : can be + or + or nd M for selecting their answer in the interval (0,4). If they have no value in (0,4) score M0 rd M for attempting y = using their in correct equation. An epression needed for MA0 rd A for correct answer. If answers are given A0. GCE Core Mathematics C (666) Summer 00

142 Question Number. (a) (b) 5 ( y = ) ( + c) f(4) = 5 5= c f( ) 0 = m = 4 = 7.5 or 5 y 5= 4 Equation is: ( ) Scheme c = 9 y 5+ 50= 0 o.e. Marks MAA M A (5) M MA A (4) (9marks) (a) (b) st n n M for an attempt to integrate + st A for at least correct terms in (unsimplified) nd A for all terms in correct (condone missing +c at this point). Needn t be simplified nd M for using the point (4, 5) to form a linear equation for c. Must use = 4 and y = 5 and have no term and the function must have changed. rd A for c = 9. The final epression is not required. st M for an attempt to evaluate f(4). Some correct use of = 4 in f( ) but condone slips. 5 They must therefore have at least 4 or and clearly be using f( ) with = 4. Award this mark wherever it is seen. nd M for using their value of m [or their ] (provided it clearly comes from using = 4 in m f () ) to form an equation of the line through (4,5)). Allow this mark for an attempt at a normal or tangent. Their m must be numerical. Use of y = m +c scores this mark when c is found. st A for any correct epression for the equation of the line nd A for any correct equation with integer coefficients. An = is required. e.g. y = 5 50 etc as long as the equation is correct and has integer coefficients. Normal Attempt at normal can score both M marks in (b) but A0A0 GCE Core Mathematics C (666) Summer 00

143 GCE Core Mathematics C (666) Summer 00

144

145 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fa Order Code UA0696 Summer 00 For more information on Edecel qualifications, please visit Edecel Limited. Registered in England and Wales no Registered Office: One90 High Holborn, London, WCV 7BH

146 Mark Scheme (Results) January 0 GCE GCE Core Mathematics C (666) Paper Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

147 Edecel is one of the leading eamining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edecel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on , our GCSE team on , or visit our website at If you have any subject specific questions about the content of this Mark Scheme that require the help of a subject specialist, you may find our Ask The Epert service helpful. Ask The Epert can be accessed online at the following link: January 0 Publications Code US06 All the material in this publication is copyright Edecel Ltd 0

148 General Instructions for Marking. The total number of marks for the paper is 75.. The Edecel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark

149 January 0 Core Mathematics C 666 Mark Scheme Question Number. (a) Scheme Marks 4 6 = or or better M = or 0.5 (ignore ± ) A () (b) = 4 or = or equivalent 4 or 6 M A cao () 4 Notes (a) M for a correct statement dealing with the 4 or the power This may be awarded if is seen or for reciprocal of their s.c ¼ is M A0, also is M A0 ± is not penalised so M A (b) M for correct use of the power 4 on both the and the terms A for cancelling the and simplifying to one of these two forms. Correct answers with no working get full marks 6 4 GCE Core Mathematics C (666) January 0

150 Question Number ( = ),, +,( +c) = Scheme + + c A Marks MA,A,A 5 Notes M for some attempt to integrate: a non zero constant st 6 A for or better 6 nd A for or better rd A for or better n n + i.e 6 a or a or 4 a or a, where a is 4 th A for each term correct and simplified and the +c occurring in the final answer GCE Core Mathematics C (666) January 0

151 Question Number. ( + ) ( ) 5 + Scheme M Marks... = denominator of Numerator = A M So 5 = + + ( ) = 5, and form simultaneous M equations in p and q -p + q = 5 and p - q = - A Solve simultaneous equations to give p = and q =. M A Notes st M for multiplying numerator and denominator by same correct epression st A for a correct denominator as a single number (NB depends on M mark) Alternative: ( p q ) nd M for an attempt to multiply the numerator by ( ± ) and get 4 terms with at least correct. nd A for the answer as written or p = and q =. Allow 0.5 and.5. (Apply isw if correct answer seen, then slip writing p =, q = ) Answer only (very unlikely) is full marks if correct no part marks A 4 GCE Core Mathematics C (666) January 0

152 Question Number Scheme Marks 4 (a) ( a ) = c 6 B () (b) a = (their a) c (= 8-4c) M a+ a + a = + "(6 c)" + "(8 4 c)" M 6 5c = 0 Aft So c = 5. A o.a.e (b) Notes st M for attempting a. Can follow through their answer to (a) but it must be an epression in c. nd M for an attempt to find the sum a+ a + a must see evidence of sum st Aft for their sum put equal to 0. Follow through their values but answer must be in the form p + qc = 0 A accept any correct equivalent answer (4) 5 GCE Core Mathematics C (666) January 0 4

153 Question Number Scheme Marks 5. (a) y y= y= Correct shape with a single crossing of each ais y = labelled or stated B B = = = labelled or stated B () (b) (b) Horizontal translation so crosses the -ais at (, 0) ± ± New equation is ( ) ( ) When = 0 y = y = B M M = A Notes B for point (,0) identified - this may be marked on the sketch as on ais. Accept =. st M for attempt at new equation and either numerator or denominator correct nd M for setting = 0 in their new equation and solving as far as y = A for or eact equivalent. Must see y = or (0, ) or point marked on y-ais. Alternative f( ) = = scores MMA0 unless =0 is seen or they write the point as (0, ) or give y = / Answers only: =, y = / is full marks as is (,0) (0, /) Just and / is B0 M M A0 Special case : Translates unit to left (a) B0, B, B0 (b) Mark (b) as before May score B0 M M A0 so /7 or may ignore sketch and start again scoring full marks for this part. (4) 7 GCE Core Mathematics C (666) January 0 5

154 Question Number 6. (a) Scheme Marks 0 M S0 = [ a+ 9 d] or S0 = a+ a+ d+ a+ d+ a+ d+ a+ 4d+ a+ 5da+ 6d+ a+ 7d+ a+ 8d+ a+ 9d 6 = 0a + 45d * Acso () u = a+ ( n ) d 7= a+ 5d B (b) ( ) n () 0 ( b) gives 0a+ 50d = 70 M (a) is 0a + 45d = 6 Subtract 5d = 8 so d =.6 o.e. A Solving for a a = 7-5d M so a = 9 A (a) M for use of S n with n = 0 Notes (4) 7 (b) st M for an attempt to eliminate a or d from their two linear equations nd M for using their value of a or d to find the other value. GCE Core Mathematics C (666) January 0 6

155 Question Number 7. ( f( ) ) ( c) Scheme Marks 8 = + + M A A M ( f( ) = 0 ) 0= 4 ( ) 4 + c c = 9 A f( ) = Notes st n n M for an attempt to integrate + st A for at least terms in correct - needn t be simplified, ignore +c nd A for all the terms in correct but they need not be simplified. No need for +c nd M for using = - and y =0 to form a linear equation in c. No +c gets M0A0 rd A for c = 9. Final form of f() is not required. (a) b 4ac= ( k ) 4( k) ( ) ( ) k 6k+ 9 4 k > 0 or k + 8k > 0 or better M k + k > 0 * Acso (b) ( k )( k )[ 0] + = M Critical values are k = or - A (choosing outside region) M k > or k < - A cao Notes (a) st M for attempt to find b 4acwith one of b or c correct nd M for a correct inequality symbol and an attempt to epand. Acso no incorrect working seen (b) st M for an attempt to factorize or solve leading to k = ( values) nd M for a method that leads them to choose the outside region. Can follow through their critical values. nd A Allow, instead of or > loses the final A < k < - scores MA0 unless a correct version is seen before or after this one. M () (4) 7 GCE Core Mathematics C (666) January 0 7

156 Question Number 9. (a) ( 8 k 0) Scheme = so k = 5 B (b) y = + k M y = +... and so m= o.e. A (c) Perpendicular gradient = Equation of line is: y ( ) Marks Bft 4= MAft y+ 4 = 0 o.e. A () () (4) c (d) y = 0, B(7,0) or = 7 = 7 or a MAft () (e) ( 7 ) ( 4 0) AB = + M AB = 5 or A Notes (b) M for an attempt to rearrange to y = A for clear statement that gradient is.5, can be m =.5 o.e. (c) Bft for using the perpendicular gradient rule correctly on their.5 M for an attempt at finding the equation of the line through A using their gradient. Allow a sign slip st Aft for a correct equation of the line follow through their changed gradient () nd A as printed or equivalent with integer coefficients allow y+ = 4 or y = 4 (d) M for use of y = 0 to find = in their equation c Aft for = 7 or a (e) M A for an attempt to find AB or AB for any correct surd form- need not be simplified GCE Core Mathematics C (666) January 0 8

157 Question Number 0. Scheme (a) y (i) correct shape ( -ve cubic) B Crossing at (-, 0) B Through the origin B Crossing at (,0) B - O (ii) branches in correct quadrants not crossing aes One intersection with cubic on each branch B B Marks (6) (b) solutions Since only intersections Notes (b) Bft for a value that is compatible with their sketch dbft This mark is dependent on the value being compatible with their sketch. For a comment relating the number of solutions to the number of intersections. Bft dbft () 8. (a) (b) (c) [ Only allow 0, or 4] dy 7 = 8 d MAAA 8 M = 4 y = = = -8 * Acso = M y = A = 4 7 = 7 Gradient of the normal = " " M Equation of normal: y 8= ( 4) 7 7y + 64 = 0 A MAft (4) () (6) GCE Core Mathematics C (666) January 0 9

158 Question Number Scheme Notes (a) st M n n for an attempt to differentiate st A for one correct term in nd A for terms in correct rd A for all correct terms. No 0 term and no +c. Marks (b) M A (c) st M A for substituting = 4 into y = and attempting note this is a printed answer Substitute = 4 into y (allow slips) Obtains.5 or equivalent nd M for correct use of the perpendicular gradient rule using their gradient. (May be slip doing the division) Their gradient must have come from y rd M for an attempt at equation of tangent or normal at P nd Aft for correct use of their changed gradient to find normal at P. Depends on st, nd and rd Ms rd A for any equivalent form with integer coefficients 4 GCE Core Mathematics C (666) January 0 0

159 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fa Order Code US06 January 0 For more information on Edecel qualifications, please visit Edecel Limited. Registered in England and Wales no Registered Office: One90 High Holborn, London, WCV 7BH

160 Mark Scheme (Results) June 0 GCE Core Mathematics C (666) Paper

161 Edecel is one of the leading eamining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edecel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on , our GCSE team on , or visit our website at If you have any subject specific questions about the content of this Eaminers Report that require the help of a subject specialist, you may find our Ask The Epert service helpful. Ask The Epert can be accessed online at the following link: June 0 Publications Code UA07654 All the material in this publication is copyright Edecel Ltd 0

162 EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. The Edecel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark GCE Core Mathematics C (666) June 0

163 . Question Number June 0 Core Mathematics C 666 Mark Scheme Scheme (a) 5 (or ± 5 ) B (b) 5 = or 5 =5 or better M 5 or (or ± ) A 5 5 Notes (a) Give B for 5 or ± 5 Anything else is B0 (including just 5) (b) M: Requires reciprocal OR 5 = 5 Accept,,,, for M Marks () () Correct answer with no working ( or notation errors in working) scores both marks i.e. M A MA0 for - without +5 5 GCE Core Mathematics C (666) June 0

164 Question Number Scheme Marks. (a) (b) ( = ) dy = 0 or 0 4 d M A A () = M A A + C B (4) 7 Notes n n (a) M: Attempt to differentiate (for any of the terms) 4 4 i.e. a or a, where a is any non-zero constant or the 7 differentiated to give 0 is sufficient evidence for M st A: One correct (non-zero) term, possibly unsimplified. nd A: Fully correct simplified answer. n (b) M: Attempt to integrate 6 (i.e. a or a or a, where a is any non-zero constant). st A: Two correct terms, possibly unsimplified. nd A: All three terms correct and simplified. 6 Allow correct equivalents to printed answer, e.g. + 7 or Allow or 7 B: + C appearing at any stage in part (b) (independent of previous work) n+ 6 GCE Core Mathematics C (666) June 0

165 Question Number Scheme Marks. Mid-point of PQ is (4, ) B 0 6 PQ: m =, = 9 ( ) 5 B 5 Gradient perpendicular to PQ ( ) m M 5 y = ( 4) M 5 y = 0 or y 5 + = 0 or multiples e.g. 0 6y = 0 A (5) 5 Notes B: correct midpoint. B: correct numerical epression for gradient need not be simplified st M: Negative reciprocal of their numerical value for m nd M: Equation of a line through their (4, ) with any gradient ecept 0 or. If the 4 and are the wrong way round the nd M mark can still be given if a correct formula (e.g. y y = m( ) ) is seen, otherwise M0. If (4, ) is substituted into y = m + c to find c, the nd M mark is for attempting this. A: Requires integer form with an = zero (see eamples above) GCE Core Mathematics C (666) June 0

166 Question Number 4. Either y = Scheme Marks Or = 4 4y y M + 4(4 4 + or ) = = 4( ) or 4y y (4 4y + y y = 4 ( ) ) = M = 0 y + 4y 5 = 0 Correct terms A ( )( 5) = 0, =. ( y 5)( y + ) = 0, y =... M = = 5 5 y = y = A 5 y = y = = = 5 M A Notes st M: Squaring to give or 4 terms (need a middle term) nd M: Substitute to give quadratic in one variable (may have just two terms) rd M: Attempt to solve a term quadratic. 4 th M: Attempt to find at least one y value (or value). (The second variable) This will be by substitution or by starting again. If y solutions are given as values, or vice-versa, penalise accuracy, so that it is possible to score M MA M A0 M A0. Non-algebraic solutions: No working, and only one correct solution pair found (e.g. = 5, y = ): M0 M0 A0 M A0 M A0 No working, and both correct solution pairs found, but not demonstrated: M0 M0 A0 M A M A Both correct solution pairs found, and demonstrated: Full marks are possible (send to review) (7) 7 GCE Core Mathematics C (666) June 0 4

167 Question Number 5. = k + (a) ( ) 5 Scheme a B Marks () (b) ( = ) 5(5k + ) (c) a + M = 5k + 8 (*) A cso (i) = 5(5k + 8) + ( = 5k 9) a M = r a = k + (5k + ) + (5k + 8) + (5k + 9) M r = 56k + 4 A cao (ii) = 6(6k + 9) (or eplain each term is divisible by 6) A ft Notes (a) 5k + must be seen in (a) to gain the mark (b) st M: Substitutes their a into 5 a + - note the answer is given so working must be seen. (c) st M: Substitutes their a into 5 a + or uses 5k +9 nd M: for their sum k+ a + a+ a4 - must see evidence of four terms with plus signs and must not be sum of AP st A: All correct so far nd Aft: Limited ft previous answer must be divisible by 6 ( eg 56k + 4). This is dependent on second M mark in (c) Allow 56 k + 4 = 6k + 9 without eplanation. No conclusion is 6 needed. () (4) 7 GCE Core Mathematics C (666) June 0 5

168 Question Number 6. (a) p =, q = or (b) Scheme Marks, B, B = 4 + ( ) M Aft = 4, y = 90: C = 90 C = 6 M A 4 " 6" y = + + their A (5) 7 Notes (a) Accept any equivalent answers, e.g. p = 0.5, q = 4/ (b) st M: Attempt to integrate n n+ (for either term) st A: ft their p and q, but terms need not be simplified (+C not required for this mark) nd M: Using = 4 and y = 90 to form an equation in C. nd A: cao rd A: answer as shown with simplified correct coefficients and powers but follow through their value for C If there is a 'restart' in part (b) it can be marked independently of part (a), but marks for part (a) cannot be scored for work seen in (b). Numerator and denominator integrated separately: First M mark cannot be awarded so only mark available is second M mark. So out of 5 marks. () GCE Core Mathematics C (666) June 0 6

169 Question Number Scheme Marks 7. (a) Discriminant: b 4ac = ( k + ) 4k or equivalent M A (b) ( k + ) 4k = k + k + 9 = ( k + ) + 8 M A (c) For real roots, b 4ac 0 or b 4ac > 0 or ( k + ) + 8> 0 M ( ) k + 0 for all k, so b 4ac > 0, so roots are real for all k (or A cso equiv.) Notes (a) M: attempt to find discriminant substitution is required If formula b 4ac is seen at least of a, b and c must be correct If formula b 4ac is not seen all of a, b and c must be correct Use of b + 4ac is M0 A: correct unsimplified (b) M: Attempt at completion of square (see earlier notes) A: both correct (no ft for this mark) (c) M: States condition as on scheme or attempts to eplain that their ( k + ) + 8 is greater than 0 A: The final mark (Acso) requires ( k + ) 0 and conclusion. We will allow ( k + ) > 0 ( or word positive) also allow b 4ac 0and conclusion. () () () 6 GCE Core Mathematics C (666) June 0 7

170 Question Number 8. (a) 5 y 4 Scheme Shape through (0, 0) B Marks (, 0) B 5 0 (.5, ) B () (b) y 5 Shape B (0, 0) and (6, 0) B (, ) B () (c) y Shape, not through (0, 0) M Minimum in 4 th quadrant A ( p, 0) and (6 p, 0) B ( p, ) B Notes (4) 0 (a) B: U shaped parabola through origin B: (,0) stated or labelled on ais B: (.5, -) or equivalent e.g. (/, -) (b) B: Cap shaped parabola in any position B: through origin (may not be labelled) and (6,0) stated or 6 labelled on - ais B: (,) shown (c) M: U shaped parabola not through origin A: Minimum in 4 th quadrant (depends on M mark having been given) B: Coordinates stated or shown on ais B: Coordinates stated Note: If values are taken for p, then it is possible to give MAB0B0 even if there are several attempts. (In this case all minima should be in fourth quadrant) GCE Core Mathematics C (666) June 0 9

171 Question Number Scheme Marks 9. (a) Series has 50 terms B S = (50)( + 00) = 550 or S = (50)( ) = 550 M A () (b) (i) 00 k B (ii) Sum: ( k + 00) or k + k k k k M A 5000 = 50 + k (*) A cso (4) (c) 50 th term = a + ( n ) d = (k+ ) + 49"(k+ )" Or k + 49(k) () M = 00 k +48 = 00 k + 48 A () 9 Notes (a) B for seeing attempt to use n = 50 or n = 50 stated M for attempt to use na ( + l) or n( a+ ( n ) d) with a = and values for other variables (Using n = 00 may earn B0 MA0) (b) M for use of a = k and d = k or l = 00 with their value for n, could be numerical or even letter n in correct formula for sum. A: Correct formula with n = 00/k A: NB Answer is printed so no slips should have appeared in working (c) M for use of formula a+ 49d with a = k + and with d obtained from difference of terms A: Requires this simplified answer GCE Core Mathematics C (666) June 0 0

172 Question Number Scheme Marks 0. (a) y Shape (cubic in this orientation) Touching -ais at Crossing at on -ais Intersection at 9 on y-ais B B B B (4) (b) y = ( + )( ) = or equiv. (possibly unsimplified) B Differentiates their polynomial correctly may be unsimplified M d y = d (*) A cso (c) dy At = 5: = = 0 d B At = 5: y = 6 B y (" 6") = "0"( ( 5)) or y = 0 + c with (-5, - 6 ) used to find c M y = A (d) Parallel: = "0" M ( )( + 5) = 0 = M = A Notes (a) Crossing at is B0. Touching at is B0 (b) M: This needs to be correct differentiation here A: Fully correct simplified answer. (c) M: If the 5 and -6 are the wrong way round or omitted the M mark can still be given if a correct formula is seen, (e.g. y y = m( ) ) otherwise M0. m should be numerical and not 0 or infinity and should not have involved negative reciprocal. (d) st M: Putting the derivative epression equal to their value for gradient nd M: Attempt to solve quadratic (see notes) This may be implied by correct answer. () (4) () 4 GCE Core Mathematics C (666) June 0

173 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fa Order Code UA07654 June 0 For more information on Edecel qualifications, please visit Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Esse CM0 JE

174 Mark Scheme (Results) January 0 GCE Core Mathematics C (666) Paper

175 Edecel is one of the leading eamining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edecel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on , our GCSE team on , or visit our website at If you have any subject specific questions about the content of this Mark Scheme that require the help of a subject specialist, you may find our Ask The Epert service helpful. Ask The Epert can be accessed online at the following link: January 0 Publications Code US0004 All the material in this publication is copyright Pearson Education Ltd 0

176 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.

177 EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. The Edecel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark 4. All A marks are correct answer only (cao.), unless shown, for eample, as A ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks.

178 General Principals for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). 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 = a, leading to =. Formula Attempt to use correct formula (with values for a, b and c), leading to =. Completing the square Solving + b + c = 0 b : ( ) ± ± q ± c, q 0, leading to = Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( n. Integration Power of at least one term increased by. ( n ) + 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. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but may be lost if there is any mistake in the working.

179 January 0 C 666 Mark Scheme Question Scheme Marks. (a) 4 + MAA () (b) C Notes MAA 6 marks () (a) M for st A for n n i.e. 6 4 or or seen (o.e.) (ignore any + c for this mark) nd A for simplified terms i.e. both 4 and or and no +c Apply ISW here and award marks when first seen is A0 (b) M st A for for n n+ applied to y only so 5 or seen. Do not award for integrating their answer to part (a) 5 5 or 6 (or better). Allow 5 / 5 here but not for nd A / 5 nd A for fully correct and simplified answer with +C. Allow ( ) 5 If + C appears earlier but not on a line where nd A could be scored then A0

180 Question Scheme Marks. (a) = 4 or 8 = B ( + 8 = ) 7 B () (b) + or seen M + ( ) + 8 a a a = (or better) + [ 9 ] [ 9 ] dm =, A, A (4) ALT ( b + c)( + ) = 7 leading to: b + c = 7, c + b = 0 M e.g. (7 b) + b = 0 (o.e.) dm 6 marks Notes (a) st B for either surd simplified nd B for 7 or accept a = 7. Answer only scores BB NB Common error is + 8 = 50 = 5 this scores B0B0 but can use their 5 in (b) to get MM (b) st M for an attempt to multiply by (o.e.) Allow poor use of brackets nd dm for using a to correctly obtain a numerator of the form p + q where p and q are non-zero integers. Allow arithmetic slips e.g. 8 or = Follow through their a = 7 or a new value found in (b). Ignore denominator. Allow use of letter a. Dependent on st M So is M0 until they reduce p + q st A for or accept b = from correct working nd A for or accept c = from correct working ALT Simultaneous Equations st M for ( b + c)( + ) = 7 and forming simultaneous equations. Ft their a = 7 nd dm for solving their simultaneous equations: reducing to a linear equation in one variable

181 Question Scheme Marks. (a) 5 > 0 M > 4 A () (b) 4 = 0 ( + )( 6) [ = 0] M = 6, A <, > 6 M, Aft (4) (a) M A Notes for reducing to the form p > q with one of p or q correct Using p = q is M0 unless > appears later on > 4 only 6 marks (b) st M for multiplying out and attempting to solve a TQ with at least + 4 or + See General Principles for definitions of attempt to solve st A for 6 and seen. Allow > 6, > etc to score this mark. Values may be on a sketch. nd M for choosing the outside region for their critical values. Do not award simply for a diagram or table they must have chosen their outside regions nd Aft follow through their distinct critical values. Allow, or or a blank between answers. Use of and is MA0 i.e. loses the final A > > 6 scores MA0 i.e. loses the final A but apply ISW if > 6, < has been seen Accept (, ) ( 6, ) (o.e) Use of instead of < (or instead of >) loses the final A mark in (b) unless A mark was lost in (a) for > 4 in which case allow it here.

182 Question Scheme Marks 4. (a) ( ) (b) ( ) (c) 4 = = a + 5 B () = a"( a + 5)" + 5 M a + 5a + 5 = a + 5a + 5 (*) Acso () a a 6 = a + 5a M (a + 9)( a 4) = 0 M a = 4 or 9 A () 6 marks Notes + (etc) + 5 6( = 0) or (a) B accept a + 5 or a 5 (b) M must see a( their ) + 5 Acso must have seen a( a[] + 5) + 5 (etc or better) Must have both brackets (...) and no incorrect working seen (c) st M for forming a suitable equation using and 4 and an attempt to collect like terms and reduce to TQ (o.e). Allow one error in sign. Accept for eample a + 5a + 46( = 0) If completing the square should get to ( ) a ± = nd M Attempting to solve their relevant TQ (see General Principles) A for both 4 and 9 seen. If a = 4 and 9 is followed by 9 < a < 4 apply ISW. No working or trial and improvement leading to both answers scores / but no marks for only one answer. Allow use of other letters instead of a

183 Question Scheme Marks 5. (a) ( 5 ) = (5 + 4) (o.e.) M 5 4( 0).5 + = 0 A b + = (o.e.) e.g. ( ) ( ) 4ac = M = 5 < 0, so no roots or no intersections or no solutions A (4) (b) 8 y 6 4 Curve: shape and passing through (0, 0) shape and passing through (5, 0) B B Line : +ve gradient and no intersections with C. If no C drawn score B0 B 4 Line passing through (0, ) and ( 0.8, 0) marked on aes B (4) 8 marks Notes (a) st M for forming a suitable equation in one variable st A for a correct TQ equation. Allow missing = 0 Accept + 4 = 5 etc nd M for an attempt to evaluate discriminant for their TQ. Allow for b > 4 ac or b < 4ac 5 ± 5 Allow if it is part of a solution using the formula e.g. ( = ) 4 Correct formula quoted and some correct substitution or a correct epression False factorising is M0 nd A for correct evaluation of discriminant for a correct TQ e.g. 5 (or better) and a comment indicating no roots or equivalent. For contradictory statements score A0 a ± a q + c ALT nd b M for attempt at completing the square ( ) nd 5 A for( ) 7 4 = and a suitable comment 6 (b) SC Coordinates must be seen on the diagram. Do not award if only in the body of the script. Passing through means not stopping at and not touching. Allow (0, ) and (y, 0) if marked on the correct places on the correct ais. st B for correct shape and passing through origin. Can be assumed if it passes through the intersection of aes nd B for correct shape and 5 marked on -ais for shape stopping at both (5, 0) and (0, 0) award B0B rd B for a line of positive gradient that (if etended) has no intersection with their C (possibly etended). Must have both graphs on same aes for this mark. If no C given score B0 4 th B for straight line passing through 0.8 on -ais and on y-ais 4 Accept eact fraction equivalents to 0.8 or (e.g. )

184 Question Scheme Marks 6. (a) ( m = ) (or eact equivalent) B () (b) B: (0, 4) [award when first seen may be in (c)] B Gradient: = m M y 4 = or equiv. e.g. y = + 4, + y 8 = 0 A () (c) A: ( 6,0) [award when first seen may be in (b)] B ALT (a) B 8 C: = 4 = [award when first seen may be in (b)] Bft Area: Using ( C A ) y B M 8 5 = = = 7 A cso 4 BC = 6 5 (from similar triangles) (or possibly using C) nd Bft Area: Using ( AB BC) = 5 M 5 = 5 5 = = 7 A 8 marks Notes for seen. Do not award for and must be in part (a) (4) (b) B for coordinates of B. Accept 4 marked on y-ais (clearly labelled) M for use of perpendicular gradient rule. Follow through their value for m A for a correct equation (any form, need not be simplified). Answer only / (c) st B for the coordinates of A (clearly labelled). Accept 6 marked on -ais nd Bft for the coordinates of C (clearly labelled) or AC = 6. Accept = 8 marked on -ais. Follow through from l if >0 M for an epression for the area of the triangle (all lengths > 0). Ft their 4, - 6 and 8 A cso for 5 or eact equivalent seen but must be a single fraction or 7 or 7 6 etc 7 on its own can only score full marks if A, B and C are all correct. ALT nd Bft If they use this approach award this mark for C (if seen) or BC Use of Det nd M must get as far as: A yb C yb

185 Question Scheme Marks 7. [ f ( ) = ] + 5[ + c] or + 5 ( + c) MA 0 = c M c = A f() = + 5 " " = 5 (o.e.) Aft (5) 5 marks Notes st n n M for attempt to integrate + st A all correct, possibly unsimplified. Ignore +c here. nd M for using = and f() = 0 to form a linear equation in c. Allow sign errors. They should be substituting into a changed epression nd A for c = rd Aft for 9 + c Follow through their numerical c ( 0 ) This mark is dependent on st M and st A only.

186 Question Scheme Marks 8. (a) [ y = + ] so d y = 4 d + MA () (b) 6 y Shape Touching -ais at origin Through and not touching or stopping at on ais. Ignore etra intersections. B B B () (c) (d) At = : d y = ( ) + 4( ) = 4 d d y At = 0: = 0 d 4 y M (Both values correct) A () Horizontal translation (touches -ais still) k and k marked on positive -ais k ( k ) (o.e) marked on negative y-ais M B B 6 8 () 0 marks Notes n n (a) M for attempt to multiply out and then some attempt to differentiate Do not award for ( + ) or ( + ) etc Prod Rule Award M for a correct attempt: products with a + and at least one product correct A for both terms correct. (If +c or etra term is included score A0) (b) st B for correct shape (anywhere). Must have clear turning points. nd B for graph touching at origin (not crossing or ending) rd B for graph passing through (not stopping or touching at) on ais and marked on ais SC B0B0B for y = or cubic with straight line between (,0) and (0, 0) (c) M for attempt at y (0) or y ( ). Follow through their 0 or and their y ( ) or for a correct statement of zero gradient for an identified point on their curve that touches - ais A for both correct answers (d) For the M in part (d) ignore any coordinates marked just mark the shape. M for a horizontal translation of their (b). Should still touch ais if it did in (b) Or for a graph of correct shape with min. and intersection in correct order on +ve -ais st B for k and k on the positive -ais. Curve must pass through k and touch at k nd B for a correct intercept on negative y-ais in terms of k. Allow (0, k k ) (o.e.) seen in script if curve passes through ve y-ais

187 Question Scheme Marks 9. (a) 0 0 S0 = [ P + 9 T ] or ( [ 8 ]) P + P + T M e.g. 5[ P + 8 T ] = ( ) (0P + 90T) or ( ) 0P + 90T (*) Acso () (b) 0 Scheme : S0 = [ ( P + 800) + 9 T ] = { 0 P T} MA 0P + 90T = 0P T M 90T = T T = 400 (only) A (4) (c) Scheme, Year 0 salary: [ ] a + ( n ) d = ( P + 800) + 9T Bft P = 9850 M P = ( ) 4450 A () 9 marks Notes (a) M for identifying a = P or d = T and attempt at S 0. Using n = 0 and one of a or d correct. Must see evidence for M mark, at least one line before the answer. Acso for simplifying to given answer. No incorrect working seen. Do not penalise missing end bracket in working eg 5(P + 8T MA for a full list seen (with + signs or written in columns) and no incorrect working seen. List Any missing terms is M0A0 (b) st M for attempting S 0 for scheme (allow missing ( ) brackets e.g. P T) Using n = 0 and at least one of a or d correct. st A for a correct epression for S0 using scheme (needn t be multiplied out ) List Allow MA if they reach 0P T with no incorrect working seen 0P T with no working is MA nd M for forming an equation using the two sums that would enable P to be eliminated. Follow through their epressions provided P would disappear. nd A for T = 400 Answer only (4/4) (c) B for using u 0 for scheme. Can be 9T or follow through their value of T M for forming an equation based on u 0 for scheme and using 9850 and their value of T A for 4450 seen Answer only (/) MR If they misread scheme as scheme in part (c) apply MR rule and award B0MA0 ma for an equation based on u 0 for scheme and using 9850 and their value of T

188 Question Scheme Marks 0. (a), 0 B () (b) dy = d MA dy At =, = = 4 d (= m) A Gradient of normal = = m 4 M Equation of normal: y 0 = 4 M + 8y = 0 (*) Acso (6) (c) = M [ = = 0 ] or [ 8y 7 y = 0 ] ( )( + 8) = 0 leading to = M = or 8 A y = 7 Aft (or eact equivalent) 8 (4) marks Notes (a) B accept = if evidence that y = 0 has been used. Can be written on graph. Use ISW (b) st M for k even if the '' is not differentiated to zero. If no evidence of d y d st A for (o.e.) only seen then 0/6 nd A for using = 0.5 to get m = 4 (correctly) (or m = /0.5) To score final Acso must see at least one intermediate equation for the line after m = 4 nd M for using the perpendicular gradient rule on their m coming from their d y d Their m must be a value not a letter. rd M for using a changed gradient (based on y ) and their A to find equation of line rd Acso for reaching printed answer with no incorrect working seen. Accept + 8y = or equivalent equations with + and + 8y (c) Trial and improvement requires sight of first equation. st M for attempt to form a suitable equation in one variable. Do not penalise poor use of brackets etc. nd M for simplifying their equation to a TQ and attempting to solve. May be by = 8 st A for = 8 (ignore a second value). If found y first allow ft for if < 0 nd Aft for y = 7 Follow through their value in line or curve provided answer is > 0 8 This second A is dependent on both M marks

189 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fa Order Code US0004 January 0 For more information on Edecel qualifications, please visit Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Esse CM0 JE

190 Mark Scheme (Results) Summer 0 GCE Core Mathematics C (666) Paper

191 Edecel and BTEC Qualifications Edecel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Epert service helpful. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 50 years, and by working across 70 countries, in 00 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: Summer 0 Publications Code UA0950 All the material in this publication is copyright Pearson Education Ltd 0

192 Summer Core Mathematics C Mark Scheme 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.

193 EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. The Edecel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark 4. All A marks are correct answer only (cao.), unless shown, for eample, as A ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks.

194 General Principles for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). 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 = a, leading to =. Formula Attempt to use correct formula (with values for a, b and c), leading to =. Completing the square Solving + b + c = 0 b : ( ) ± ± q± c, 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. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but may 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 not gain full 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.

195 Summer Core Mathematics C Mark Scheme Question Number Scheme 6 M A d = ( + c) Marks = ; c A; A Notes n n M: for some attempt to integrate a term in : + So seeing either 6 ± λ or ± µ or 5 5 is M. st A: for a correct un-simplified or or term. nd A: for both and terms correct and simplified on the same line. Ie. or rd A: for c. Also allow c. This needs to be written on the same line. Ignore the incorrect use of the integral sign in candidates responses.. 4 Note: If a candidate scores MAAA and their answer is NOT ON THE SAME LINE then withhold the final accuracy mark.

196 Question Number. (a) ( ) 5 ( ) 5 ( ) (b) Scheme 5 5 = or or or 768 M = 8 A = or or or 5 5 = 4 Notes (a) M: for a full correct interpretation of the fractional power. Note: 5 ( ) is M0. (b) A: for 8 only. Note: Award MA for writing down 8. M: For use of OR use of See notes below See notes for other alternatives Use of : Candidate needs to demonstrate the they have rooted all three elements in their bracket. Use of -: Either Candidate has Allow M for... Bracket or C A B B becomes. C A or or or or or 5 or 5 or or K or 5 C or 5 5 or 4 5 or 4 where K, Care any powers including A: for either or or 0.4 or Note: is not enough work by itself for the method mark. 4 Note: A final answer of or or is A Note: Also allow or or 0.4 or 5 5 ± ± ± ± for A. Marks M A [] [] 4

197 Question Number. Scheme ( + 8) ( ) ( ) { ( + 8 )} = = = 8 ( + ) 8 Writing this is sufficient for M. For 8. This mark can be implied. M A B B Marks = + A cso Notes M: for a correct method to rationalise the denominator. st A: ( 8)( + 8) 8 or ( )( ) + st B: for = or 48 = 4 seen or implied in candidate s working. nd B: for 8 = or = 4 seen or implied in candidate s working. + nd A: for +. Note: as a final answer is A0. Note: The first accuracy mark is dependent on the first method mark being awarded. Note: + 8 = + with no intermediate working implies the BB marks. Note: = 4 or 8 = 4 are not sufficient for the B marks. Note: A candidate who writes down (by misread) 8 for the nd B will be awarded for 8 = or 7 = 6 Note: The final accuracy mark is for a correct solution only. Alternative solution = B B 8 ( ) 8 can potentially obtain MA0BBA0, where 5 ( + ) ( ) ( + ) {( + )} = = M A for = + A Alternative solution = = 8 = + ( ) ( ) with no incorrect working seen is awarded MABBA., ( ) Please record the marks in the relevant places on the mark grid. or = +

198 Question Number 4. (a) (b) (a) (b) 4 Scheme y = dy 4 = 5() 6 + d M Marks = A A A [4] d y 8 = 0 d M A [] 6 M: for an attempt to differentiate st A: for nd A: for n Notes n to one of the first three terms of 4 So seeing either 5 ± λ or 6 ± µ or is M. 5 only. 8 or 8 only. rd A: for + ( + c included in part (a) loses this mark). Note: M: For differentiating d y again to give either d a correct follow through differentiation of their β term 4 y = is A0 unless simplified to. or for ± α ±. A: for any correct epression on the same line (accept un-simplified coefficients). 8 For powers: is A0, but writing powers as one term eg: (5 ) is ok for A. dy 4 d y 4 Note: Candidates leaving their answers as = 5 + and = 0 d d 9 awarded MAA0A in part (a) and MA in part (b). 8 - Be careful: 0 will be A0. 8 Note: For an etra term appearing in part (b) on the same line, ie 0 + is MA0 8 Note: If a candidate writes in part (a) c and in part (b) 0 + c then award (a) MAAA0 (b) MA are

199 Question Number n+ 5. (a) { } Scheme a =, a = a c, n, c is a constant n a = c or () c or 6 c B a = "6 c" c M = c (*) A cso (b) { } ( ) (c) a ( c ) c { c} 4 = " " = 4 7 M 4 ai = i = + (6 c) + ( c) + (4 7 c) M "45 c" or "45 c" = M c or c A cso Notes Marks [] [] [4] 7 (a) The answer to part (a) cannot be recovered from candidate s working in part (b) or part (c). Once the candidate has achieved the correct result you can ignore subsequent working in this part. (b) M: For a correct substitution of their a which must include term(s) in c into a cgiving a result for a in terms of only c. Candidates must use correct bracketing for this mark. A: for correct solution only. No incorrect working/statements seen. (Note: the answer is given!) (c) st M: For a correct substitution of a which must include term(s) in c into a cgiving a result for a 4 in terms of only c. Candidates must use correct bracketing (can be implied) for this mark. nd M: for an attempt to sum their a, a, a and a4 only. rd M: for their sum (of or 4 or 5 consecutive terms) = or or > to form a linear inequality or equation in c. A: for c or c from a correct solution only. Beware: c c is A0. Note: 45 c c c would be A0 cso. n n Note: Applying either Sn = ( a + ( n ) d) or Sn = ( a + l) is nd M0, rd M0. Note: If a candidate gives a numerical answer in part (a); they will then get M0A0 in part (b); but if they use the printed result of a = c they could potentially get M0MMA0 in part (c) Note: If a candidate only adds numerical values (not in terms of c) in part (c) then they could potentially get only M0M0MA0. Note: For the rd M candidates will usually sum a, a, a and a 4 or a, aand a 4 or a, a, a4and a 5 or a, a, a, a4and a 5

200 Question Number Scheme Boy s Sequence: 0,5, 0, 5,... a = 0, d = 5 T = a + 4d = 0 + 4(5); = 80 or (0.05); = 0.80 M; A 6. (a) { } (b) { } [ ] 5 60 S 60 = (0) + 59(5) M A = 0(5) = 9450 or A Boy s Sister s Sequence: 0, 0, 0, 40,... m m a = 0, d = 0 Sm = (0) + ( m )(0) or 0( m + ) or 5 m( m+ ) M A m 6 or 600 = ( (0) + ( m )(0) ) dm m 600 = (0)( m + ) or 600 = 0 mm ( + ) 60 = mm ( + ) 5 6 = mm ( + ) (*) A cso (c) { } ( ) (d) { m = }5 B Marks Notes (a) M: for using the formula a + 4d with either a or d correct. A: for 80 or 80p or 0.80 or 0.80p and apply ISW. Otherwise, 80 or 0.80 or 0.80p would be A0. Award M0 if candidate applies a + 59d. Listing the first 5 terms and highlighting that the 5 th term is 80 or listing 5 terms with the final 5 th term aligned with 80 will then be awarded all two marks of MA. Writing down 80 with no working is MA. [] [] [4] [] 0 (b) 60 5 (0) 59(5) (0) 4(5) with a = 0, d = 5 and n = 60 or a = 0, d = 5 and n = 5. M: for use of correct [ + ] or ( + ) n If a candidate uses ( a + l ) with n = 60 or 5, there must be a full method of finding or stating l as either a + 59d = 05 or a + 4d = 80, respectively. ( ) ( ) st A: for a correct epression for S. ie. 60 [ ] 60 [ ] 60 (0) + 59(5) or (0.) + 59(0.05) or [ ] or [ ]. This mark can be implied by later working. nd A: for 9450 or 9450p or and apply ISW. Otherwise, 9450 or without sign is A0. Note: the bracketing error of 60 (0) + 59(5) is A0 unless recovered from later working. Adding together the first 60 terms to obtain 9450 will then be awarded all three marks of MAA.

201 (c) st M: for correct use of S m formula with one of a or d correct. st m m A: for a correct epression for S m. Eg: ( (0) + ( m )(0) ) or 0( m + ) or 5 m( m+ ) nd M: for forming a suitable equation using 6 or 600 and their S m. Dependent on st M. nd Acso: for reaching the printed result with no incorrect working seen. Long multiplication is not necessary for the final accuracy mark. m Note: ( (0) + ( m )(0) ) = 60 and not either 600 or 6 is dm0. Beware: Some candidates will try and fudge the result given on the question paper. Notes for awarding nd A Going from mm+ ( ) = 60 straight to mm+ ( ) = 5 6 is nd A. Going from mm+ ( ) = some factor decomposition of 600 straight to mm+ ( ) = 5 6 is nd A. Going from 0 mm+ ( ) = 600 straight to mm+ ( ) = 5 6 is nd A Going from mm+ ( ) = straight to mm+ ( ) = 5 6 is nd A0. 5 Alternative: working in an different letter, say n or p. n n MA: for ( (0) + ( n )(0) ) (although miing letters eg. ( (0) + ( m )(0) ) is M0A0). n dm: for 6 or 600 = ( (0) + ( n )(0) ) n Leading to 600 = (0)( n+ ) 60 = n( n+ ) 5 6 = n( n+ ) The candidate then needs to write either 5 6 = mm ( + ) or m n or m = n to gain the final A. (d) B: for 5 only.

202 Question Number 7. (a) Scheme 6 P(4, ) lies on C where f( ) = +, > 0 6 f(4) = (4) + ; = 4 M; A T: y = ( 4) dm T: y = 9 A = + + or equivalent. M A () ( ) (b) f( ) ( c) (a) { } 6 f (4) = () + (4) + c = dm c = c = 7 So, { } { } 6 f( ) = A cso () ( ) NB: f ( ) = st M: for clear attempt at f(4). st A: for obtaining from f(4). Notes nd y dm: for y = ( their f (4))( 4) or = ( their f (4)) 4 or full method of y = m + c, with = 4, y = and their f(4) to find a value for c. Note: this method mark is dependent on the first method mark being awarded. nd A: for y = 9 or y = 9 + Note: This work needs to be contained in part (a) only. (b) st M: for a clear attempt to integrate f( ) with at least one correct application of + n n on 6 f( ) = +. Marks 6 0 So seeing either or or + λ µ + + ± ± is M. st A: for correct un-simplified coefficients and powers (or equivalent) with or without + c. nd dm: for use of = 4 and y = in an integrated equation to form a linear equation in c equal to -. ie: applying f(4) =. This mark is dependent on the first method mark being awarded. 6 A: For { f( ) = } stated on one line where coefficients can be un-simplified or () ( ) simplified, but must contain one term powers. Note this mark is for correct solution only. Note: For a candidate attempting to find f() in part (a) If it is clear that they understand that they are finding f( ) in part (a); ie. by writing f ( ) =... or y =... then you can give credit for this working in part (b). [4] [4] 8

203 Question Number 8. (a) { } Scheme Marks 4 5 = q ( p), p, q are integers. 4 5 = = ( ) = ( ) + M = ( ) A A [] " b 4 ac" = 4 4( )( 5) = 6 0 M = 4 A [] (b) { } { } (c) y O Correct shape M - 5 Maimum within the 4 th quadrant A Curve cuts through -5 or (0, 5) marked on the y-ais B (a) [] 8 Notes M: for an attempt to complete the square eg: ± ( ± ± ) ± k 5, k 0 or ± ( ± ± ) ± λ, λ 5 seen or implied in working. st A: for p = or for ± α ( ), α can be 0. nd A: for q = Note: Allow MAA for a correct written down epression of ( ) Ignore ( ) = 0. Note: If a candidate states either p = or q = or writes k ( ) Note: A candidate who writes down with no working p, q ( a value which is not ) Note: Writing ( ), followed by p =, q = is MAA0. ± then imply the M mark. = = gets M0A0A0. Alternative to (a) { 4 5 = } 4 5 = ( ) 4 5 = ( ) = ( ) Alternative to (a) q ( + p) = q ( + p + p ) = p + q p Compare terms: p = 4 p = Compare constant terms: M: Either ± p = 4 or q q p q q = 5 4 = 5 = p ± = 5 ; st A: for p = ; nd A: for q =

204 Alternative to (a) Negating 4 5 gives So, { = ( ) = ( ) + } M for ± ( ± ± ) ± k + 5 then stating p = is st A and/or q = is nd A. or writing ( ) is AA. Special Case for part (a): q ( + p) = q ( + p + p ) = p + q p = 4 5 p + q p = 4 5 q p + 5= 4 + p q p + 5 = (4 + p) q p + 5 = p scores Special Case MAA only once p achieved. 4 + p (b) M: for correctly substituting any two of a =, b = 4, c = 5 into b 4ac if this is quoted. If b 4ac is not quoted then the substitution must be correct. Substitution into b < 4ac or b = 4ac or b > 4ac is M0. A: for 4 only. If they write 4< 0 treat the < 0 as ISW and award A. If they write 4 0 then score A0. So substituting into b 4ac < 0 leading to 4 < 0 can score MA Note: Only award marks for use of the discriminant in part (b). Note: Award M0 if the candidate uses the quadratic formula UNLESS they later go on to identify that the discriminant is the result of b 4ac. Beware: A number of candidates are writing up their solution to part (b) at the bottom of the second page. So please look! (c) M: Correct shape in any quadrant. A: The maimum must be within the fourth quadrant to award this mark. B: Curve (and not line!) cuts through -5 or (0, 5) marked on the y-ais Allow ( 5, 0) rather than (0, 5) if marked in the correct place on the y-ais. If the curve cuts through the negative y-ais and this is not marked, then you can recover (0, 5) from the candidate s working in part (c). You are not allowed to recover this point, though, from a table of values. Note: Do not recover work for part (a) in part (c).

205 Question Number 9. (a) { } Scheme (b) { } (c) { L L } Marks + = A L :4y + = y = ; A( p,4)lieson L. 4 p = 9 9 or or 9.5 B [] 4y + = y = m( L ) = or 4 4 M A So ml ( ) = Bft L : y 4 ( ) M L : 8 0 or : y 8 0 [5] = 4(8 ) + = or + 8 = 4 M =.5, y = A, A cso [] CD = ".5" + "" 4 M (d) ( ) ( ) (".5" ) ("" 4) CD = + A ft =.5 + =.5 + =.5 5 or 5 (*) A cso (e) Area = triangle ABC + triangle ABE = Finding the area of any triangle. M = = (0) + (0) 4 B = 45 A Notes 9. (a) B: 9.5 oe. (b) st M: for an attempt to rearrange 4y + = into y = m + c. This mark can be implied by the correct gradient of L or L. st A: for gradient of L = or. Stating ml ( 4 ) = without working is MA. Bft: for applying ml ( ) =. Need not be simplified. their ml ( ) Note: Writing down ml ( ) = with no earlier incorrect working gains MAB nd M: for applying y 4 = ± λ( ) where λ is a numerical value, λ 0. or full method of y = m + c, with =, y = 4 and (their ± λ ) to find c. nd A: + y 8 = 0 or y + 8 = 0 or y + 8 = 0 or 4 + y 6 = 0 or + y 8 = 0 etc. Must be = 0. So do not allow + y = 8 etc. Note: Condone the error of incorrectly rearranging L to give y = ml ( ) =. [] [] 5

206 (c) M: for an attempt to solve. Must form a linear equation in one variable. st A: for =.5 (correct solution only). nd A: for y = (correct solution only). Note: If =.5, y = is found from no working, then send to review. Note: Use of trial and error to find one of or y and then substitution into one of L or L can achieve MAA. (d) M: for an attempt at CD - ft their point D. Eg: (".5" ) ("" 4) This mark can be implied by finding CD. + or simplified. st Aft: for finding their CD - ft their point D. Eg: (".5" ) ("" 4) nd A:cso for no incorrect working seen. Note: A candidate initially writing down Alternatives part (d): Final accuracy = + 9 = + = = { }. { }.5 + =.5 =.5 5 = can be awarded MA. + or correctly simplified. (e) M: for an attempt at finding the area of either triangle ABC or triangle ABE. B: Correct method for removing a square root. Eg: 80 5 = 400 = 0 or = 0 Note: This mark can be implied. A: for 45 only. Alternative to part (e): Area = ( 80 ) = ( ) = 45 M: ( AB)( CE ). B: Evidence of correct surd removal. A: for 45. Note: Multiplying the diagonals (usually to find 90) is M0, B if surds are removed correctly, A0. Alternative to part (e): Area = triangle DAC + triangle DCB + triangle DEA + triangle DBE ( ) ( ) = = (5) + (5) + (5) + (5) = = 45 M: For finding the area of one of the four triangles. B: Evidence of correct surd removal. A: for 45. Alternative to part (e): 9 CE = CD + DE = = 5, { BD = DA + AB = = 7 5} Area = triangle BCE triangle ACE = ( CE)( BD) ( CE)( BD) 9 9 = M: for an attempt at the area of triangle BCE or triangle ACE. 6(5) 7(5) 6(5) = = = 9(5) B: Evidence of correct surd removal = 45 A: for 45 only.

207 Question Number 0. (a) {Coordinates of A are} ( ) (b)(i) Scheme 4.5, 0 See notes below B y Marks [] 7 Horizontal translation M - and their ft.5 on postitive -ais A ft Maimum at 7 marked on the y-ais B (ii) - y O.5 [] (, 7) Correct shape, minimum at (0, 0) and a maimum within the first quadrant. M.5 on -ais A ft Maimum at (, 7) B.5 O [] (c) { k = } 7 B [] 8 Notes (a) B: For stating either = 4.5 or or 4 etc. or A = 4.5 or or ( 4.5, 0 ). Can be written on graph. Allow ( 0, 4.5 ) marked on curve for B. Otherwise ( 0, 4.5 ) without reference to any of the above is B0. (b)(i) M: for any horizontal (left-right) translation where minimum is still on -ais not at (0, 0). Ignore any values. Aft: for - (NOT ) and.5 (or their in part (a) ) evaluated and marked on the positive -ais. Allow (0, ) and/or (0, ft.5) rather than (, 0) and (ft.5, 0) if marked in the correct place on the -ais. Note: Candidate cannot gain this mark if their in part (a) is less than. B: Maimum at 7 marked on the y-ais. Note: the maimum must be on the y-ais for this mark. (ii) M: for correct shape with minimum still at (0, 0) and a maimum within the first quadrant. Ignore values. Aft: for their in part ( a) ; as intercept on -ais eg : or.5 or or 6 Note: a generalised A is A0. Allow (0, ft.5) rather than (ft.5, 0) if marked in the correct place on the -ais. B: Maimum at (, 7) or allow marked on the -ais and the corresponding 7 marked on the y-ais. Note: Be careful to look at the correct graph. The candidate may draw another graph to help them to answer part (c). Note: You can recover (b)(i) (, 0) and (ft.5, 0) or in (b)(ii) (ft.5, 0) as correct coordinates only in candidate s working if these are not marked on their sketch(es). (c) B: for ( k = ) 7 only. BEWARE: This could be written in the middle or at the bottom of a page.

208

209 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fa Order Code UA0950 Summer 0 For more information on Edecel qualifications, please visit our website Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Esse CM0 JE

210 Mark Scheme (Results) January 0 GCE Core Mathematics C (666/0)

211 Edecel and BTEC Qualifications Edecel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at or for our BTEC qualifications. Alternatively, you can get in touch with us using the details on our contact us page at If you have any subject specific questions about this specification that require the help of a subject specialist, you can speak directly to the subject team at Pearson. Their contact details can be found on this link: You can also use our online Ask the Epert service at You will need an Edecel username and password to access this service. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 50 years, and by working across 70 countries, in 00 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: January 0 Publications Code US0459 All the material in this publication is copyright Pearson Education Ltd 0

212 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.

213 General Principles for Core Mathematics 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 = a, leading to =. Formula Attempt to use correct formula (with values for a, b and c).. Completing the square Solving + b + c = 0 : b ( ) q c, 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. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but may 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 not gain full 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.

214 January Core Mathematics C Mark Scheme Question Number Scheme Marks. Accept ( 4 + ) or ( 4 ) (4 ) or ( + 4 ) or even 4 ( ) or equivalent 4 quadratic (or initial cubic) into two brackets B M ( )(+ ) or ( )(+ ) or ( )( ) A [] marks Notes B: Takes out a factor of or or even 4. This line may be implied by correct final answer, but if this stage is shown it must be correct. So B0 for ( + 4 ) M: Factorises the quadratic resulting from their first factorisation using usual rules (see note in General Principles). e.g. ( 4) ( ). Also allow attempts to factorise cubic such as ( )(+ ) etc N.B. Should not be completing the square here. A: Accept either ( )( + ) or ( )( + ) or ( )( ). (No fractions for this final answer) Specific situations Note: ( 4 ) followed by ( ) scores BMA0 as factors follow quadratic factorisation criteria And ( 4 ) followed by (- 4)(+4) BM0A0. Answers with no working: Correct answer gets all three marks BMA : ( )(+ ) gets B0MA0 if no working as (4 ) would earn B0 Poor bracketing: e.g. ( + 4 ) gets B0 unless subsequent work implies bracket round the in which case candidate may recover the mark by the following correct work.

215 Question Number. Scheme Marks ( ) + + (+ ) a+ b (8 ) or = = with a = 6 or b = 9 M 6 9 = + or ( ) + = as final answer with no errors or ( y )6 9 = + or ( + ) A [] M: Uses A: Either Note: Eamples: Notes 8=, and multiplies powers ( + ). Does not add powers. ( Just 6 9 Special case: : + ( + or ) 6 Alternative method using logs: = or ( y = )6 + 9 or ( + ) + scores MA0 + : ( ) = = gets M0A0 6 9 = without seeing as single power MA0 + y + 8= or marks 8 = is M0 ) ( )log8 8 = (+ ) log8 = ylog y= log M So ( y = )6 + 9 or ( + ) A []

216 Question Number. (i) ( 5 8)( + ) (ii) Alternative Scheme Marks = M = =, seen or implied at any point. B = + + or a = and b=. A [] Method Method Method 0 5 Either Or 80 + = M = B = =.... = = 5 = = 0 5 A ( 5 )( ) for (i) (i) (ii) + This earns the B mark. = Multiplies out correctly with. This may be seen or implied and may be simplified M e.g. = o.e. For earlier use of B = + + or a = and b=. A [] 6 marks Notes M: Multiplies out brackets correctly giving four correct terms or simplifying to correct epansion. (This may be implied by correct answer) can appear as table B: 8 =, seen or implied at any point A: Fully and correctly simplified to + or a = and b=. M: Rationalises denominator i.e. Multiplies Method or similar e.g. = = k 5 5 by or 5, seen or implied or uses 5 5 B: (Independent mark) States 80 = 4 5 Or either 400 = 0 or 80 5 = 0 at any point if they use Method. A: 0 5 or 0. c = N.B There are other methods e.g = (B) then add = then M Aas before Those who multiply initial epression by 5 to obtain = = 50 earn M0 B A0 []

217 Question Number Scheme Marks 4. u = 9, un + = un, n (a) u = u = (9) ( = 7) u (b) 4 = u = (7) = 4 r = u = u + u + u + u r 4 u = (9). Can be implied by u = 7 M Both u = 7 and u 4 = A [] ( u ) = 5 ( u ) = 5 B 4 r = u r = "5" "7" + "" = 64 Adds their first four terms obtained M legitimately (see notes below) 64 A [] Notes (a) M: Substitutes 9 into RHS of iteration formula A: Needs both 7 and (but allow if either or both seen in part (b) ) (b) B: for u =5 ( however obtained may appear in (a)) May be called a=5 5 marks M: Uses their u found from u = u stated eplicitly, or uses u = 4 or 5, and adds it to u, their u and their u 4 only. (See special cases below). There should be no fifth term included. Use of sum of AP is irrelevant and scores M0 A: 64

218 Question Number 5. (a) Gradient of l is Scheme or 0.5 or B Marks 7 Either y 6 = " "( 5) or y = " " + c and 6 = " " ( 5 ) + c c = (" ") M. y+ 7= 0 or + y 7 = 0 or k ( y + 7) = 0 with k an integer A [] (b) Puts = 0, or y = 0 in their equation and solves to find appropriate co-ordinate -coordinate of A is -7 and y-coordinate of B is 7.. M A cao [] (c) Area OAB ( ) 7 49 = 7 = (units) 4 Applies ± (base)(height) M 49 4 Acso [] (a) (b) (c) 7 marks Notes B: Must have ½ or 0.5 or o.e. stated and stops, or used in their line equation M: Full method to obtain an equation of the line through (5,6) with their m. So y 6 = m( 5) with their gradient or uses y = m+ c with (5, 6) and their gradient to find c. Allow any numerical gradient here including or but not zero. (Allow (6,5) as a slip if y y = m( ) is quoted first ) A: Accept any multiple of the correct equation, provided that the coefficients are integers and equation = 0 e.g. +y 7 = 0 or k ( y + 7) = 0 or even y - 7 = 0 M: Either one of the or y coordinates using their equation A: Needs both correct values. Accept any correct equivalent.. Need not be written as co-ordinates. Even just 7 and.5 with no indication which is which may be awarded the A. M: Any correct method for area of triangle AOB, with their values for co-ordinates of A and B (may include negatives) Method usually half base times height but determinants could be used. A: Any eact equivalent to 49/4, e.g..5. (negative final answer is A0 but replacing by positive is A) Do not need units. c.s.o. implies if A0 is scored in (b) then A0 is scored in (c) as well. However if candidate has correct line equation in (a) of wrong form may score A0 in (a) and A in (b) and (c) Note: Special cases: ( ) (recovery) = (units) 4 is M A0 but changing sign to area = gets MA N.B. Candidates making sign errors in (b) and obtaining +7 and - 7. may also get 49 as their answer 4 following previous errors. They should be awarded A0 as this answer is not ft and is for correct solution only Special Case: In (a) and (b): Produces parallel line instead of perpendicular line: So uses m = - This is not treated as a misread as it simplifies the question. The marks will usually be B0 M A0, M A0, M A0 i.e. maimum of /7

219 Question Number 6. (a) y Scheme Marks y = is translated up or down. M O y = 5 is in the correct position. A Intersection with -ais at ( { }), 0 only 5 Independent mark. B y = 4 + : attempt at straight line, with positive gradient with positive y intercept. B Intersection with -ais at Check graph in question for possible answers (, { 0 }) and y-ais at ( {} ) B 0,. [5] and space below graph for answers to part (b) (b) Asymptotes : = 0 (or y-ais) and y = 5. An asymptote stated correctly. Independent of (a) B (Lose second B mark for etra asymptotes) These two lines only. Not ft their graph. B [] (c) Method : 5 4 = + y Method : = y + M = 0 = y + y 8= 0 y = dm =, 4 y = 6, A When =, y = 6,When 4 = When y = 6, = When y =, =. MA 4 [5] marks Notes (a) M: Curve implies y ais as asymptote and does not change shape significantly. Changed curve needs horizontal asymptote (roughly) Asymptote(s) need not be shown but shape of curve should be implying asymptote(s) parallel to ais. Curve should not remain where it was in the given figure. Both sections move in the same direction. There should be no reflection A: Crosses positive ais. Hyperbola has moved down. Both sections move by almost same amount. See sheet on page 9 for guidance. B: Check diagram and tet of answer. Accept /5 or 0.4 shown on -ais or = /5, or ( /5, 0) stated clearly in tet or on graph. This is independent of the graph. Accept (0, /5) if clearly on ais. Ignore any intersection points with y ais. Do not credit work in table of values for this mark. B: Must be attempt at a straight line, with positive gradient & with positive y intercept (need not cross ais) B: Accept = -/, or 0.5 shown on -ais or (-/, 0) or (-0.5,0) in tet or on graph and similarly accept on y ais or y = or (0, ) in tet or on graph. Need not cross curve and allow on separate aes. (b) B: For either correct asymptote equation. Second B: For both correct (lose this if etras e.g. =± are given also). These asymptotes may follow correctly from equation after wrong graph in (a) Just y = -5 is B B0 This may be awarded if given on the graph. However for other B mark it must be clear that = 0 (or the y-ais) is an asymptote. NB 0, y 5 is BB0 (c) M: Either of these equations is enough for the method mark (May appear labelled as part (b)) dm: Attempt to solve a term quadratic by factorising, formula, completion of square or implied by correct answers. (see note ) This mark depends on previous mark. A: Need both correct answers (Accept equivalents e.g. 0.5) or both correct y values (Method ) M: At least one attempt to find second variable (usually y) using their first variable (usually ) related to line meeting curve. Should not be substituting or y values from part (a) or (b). This mark is independent of previous marks. Candidate may substitute in equation of line or equation of curve. A: Need both correct second variable answers Need not be written as co-ordinates (allow as in the scheme) Note: Special case: Answer only with no working in part (c) can have 5 marks if completely correct, with both points found. If coordinates of just one of the points is correct with no working this earns M0 M0 A0 M A0 (i.e. / 5)

220 Question Number Scheme Marks 7. Lewis; arithmetic series, a = 40, d = 0. (a) T 0 = 40 + (0 )(0); = 50 Or lists 0 terms to get to 50 M; A OR 0 + (0)(0) [] Method Method (b) Either: Uses n( a + ( n ) d) Or: Uses na ( + l) M (c) 0 40 (0 )(0) "50" ft 50 A 6600 A ( + ) ( ) Sian; arithmetic series, a = 00, l = 700, S n = 8500 Or: May use both n Either: Attempt to use 8500 = ( a + l ) 8500 = n( a + ( n ) d) and M l = a + ( n ) d and eliminate d n = ( + ) 8500 ( ) n = + A n = 7 A [] [] 8 marks Notes (a) M: Attempt to use formula for 0 th term of Arithmetic series with first term 40 and d = 0. Normal formula rules apply see General principles at the start of the mark scheme re Method Marks Or: uses 0 + 0n with n = 0 Or: Listing method : Lists 40, 60, 80, 00, 0, 40, 60, 80, 50. MA if correct M0A0 if wrong. (So marks or zero) A: For 50 (b) M: An attempt to apply n( a + ( n ) d) or na ( + l) with their values for a, n, d and l A: Uses a = 40, d = 0, n = 0 in their formula (two alternatives given above) but ft on their value of l from (a) if they use Method. A: 6600 cao Or: Listing method : Lists 40, 60, 80, 00, 0, 40, 60, 80, 50 and adds 6600 gets MAA- any other answer gets M A0A0 provided there are 0 numbers, the first is 40 and the last is 50. (c) n First M: Attempt to use S n = ( a + l) with their values for a, and l and S =8500 method A: Uses formula with correct values A: Finds eact value 7 Alternative M: If both formulae 8500 = n( a + ( n ) d) and l = a + ( n ) d are used, then d must be eliminated method before this mark is awarded by valid work. Should not be using d = 400. This would be M0. A: Correct equation in n only then A for 7 eactly Trial and error methods: Finds d = 5 and n = 7 and list from 00 to 700 with total checked /

221 Question Number 8. Scheme dy 5 ( = ) + "" " " d M Marks "" 5 = ( ) ( ) 4 ( y ) " " ( c) 5 = ( ) ( ) 4 ( y ) ( c) Raises power correctly on any one term. Any two follow through terms correct. This is not follow through must be correct M Aft A 5 Given that y = 7, at =, then 7 = + + c c= M 4 4 So, 4 5 ( y = ) + + c, c =8 or = + + A ( y ) 8 [6] 6 marks Notes M: Epresses as three term polynomial with powers, - and. Allow slips in coefficients. This may be implied by later integration having all three powers 4, - and -. n n M: An attempt to integrate at least one term so + (not a term in the numerator or denominator) Aft: Any two integrations are correct coefficients may be unsimplified (follow through errors in coefficients only here) so should have two of the powers 4, - and after integration depends on nd method mark only. There should be a maimum of three terms here. A: Correct three terms coefficients may be unsimplified- do not need constant for this mark Depends on both Method marks M: Need constant for this mark. Uses y = 7 and = in their changed epression in order to find c, and attempt to find c. This mark is available even after there is suggestion of differentiation. A: Need all four correct terms to be simplified and need c = 8 here.

222 Question Number Scheme Marks 9. (a) Method : Attempts b 4ac for a = ( k + ), b = 6 and their c. c k M b 4ac = 6 4 k + k 5 A b ac ( 4 = ) As b Method : k k < or (b) Attempts to solve k ( )( ) 4k + 8k + 96 or ( b 4 ac = ) 4k 8k 96 B ( with no prior algebraic errors) 4ac > 0, then 4k + 8k + 96 > 0 and so, k k 4 < 0 A * Considers b > 4ac for a = ( k + ), b = 6 and their c. c k M 6 > 4 k + k 5 A ( )( ) 4k + 8k + 96 > 0 or 9> ( k + )( k 5) and so, k B (with no prior algebraic errors) k 4 < 0 following correct work A * k 4 = 0 to give k = ( Critical values, k = 6, 4. ) M [4] k k 4 < 0 gives - 4 < k < 6 M A (a) (b) [] 7 marks Notes Method : M: Attempts b 4ac for a = ( k + ), b = 6 and their c. c k or uses quadratic formula and has this epression under square root. (ignore > 0, < 0 or = 0 for first marks) A: Correct epression for b 4ac - need not be simplified (may be under root sign) B: Uses algebra to manipulate result without error into one of these three term quadratics. Again may be under root sign in quadratic formula. If inequality is used early in proof may see 4k 8k 96< 0 and B would be given for 4k 8k 96 correctly stated. A: Applies b 4ac > 0 correctly ( or writes b 4ac > 0 ) to achieve the result given in the question. No errors should be seen. Any incorrect line of argument should be penalised here. There are several ways of reaching the answer; either multiplication of both sides of inequality by, or taking every term to other side of inequality. Need conclusion i.e. printed answer. Method : M: Allow b > 4ac, b < 4ac or b = 4ac for a = ( k + ), b = 6 and their c. c k A: Correct epressions on either side (ignore >, < or =). B: Uses algebra to manipulate result into one of the two three term quadratics or divides both sides by 4 again without error A: Produces result with no errors seen from initial consideration of b > 4ac. M: Uses factorisation, formula, completion of square method to find two values for k, or finds two correct answers with no obvious method M: Their Lower Limit < k < Their Upper Limit. Allow the M mark mark for. (Allow k < upper and k > lower) A: -4 < k < 6 Lose this mark for Allow (-4, 6) [not square brackets] or k > -4 and k < 6 (must be and not or) Can also use intersection symbol NOT k > -4, k < 6 (MA0) Special case : In part (a) uses c = k instead of k scores 0. Allow k + 5 for method marks Special Case: In part (b) Obtaining 6 < k < 4 This is a common wrong answer. Give M M A0 special case. Special Case: In part (b) Use of instead of k MMA0 Special Case: -4 < k < 6 and k < -4, k > 6 both given is M0A0 for last two marks. Do not treat as isw.

223 Question Scheme Number 0. (a) This may be done by completion of square or by epansion and comparing coefficients a = 4 b = All three of a = 4, b = and c = - Marks B B B [] (b) y U shaped quadratic graph. M The curve is correctly positioned with the minimum in the third quadrant.. It crosses ais twice on negative ais and y ais once on positive y ais. A Curve cuts y-ais at ({ 0,.only } ) B Curve cuts -ais at, { 0 } (, 0 ). B ( ) and { } [4] 7 marks Notes (a) B: States a = 4 or obtains 4( + b) + c, B: States b = or obtains a ( + ) + c, B: States a = 4, b = and c = - or 4( + ) (Needs all correct for final mark) Special cases: If answer is left as ( + ) treat as misread BB0B0 or as ( + ) then the mark is B0BB0 from scheme (b) M: Any position provided U shaped (be generous in interpretation of U shape but V shape is M0) A : The curve is correctly positioned with the minimum in the third quadrant. It crosses ais twice on negative ais and y ais once on positive y ais. B: Allow on y ais and allow either y = or (0, ) if given in tet Curve does not need to pass through this point and this mark may be given even if there is no curve at all or if it is drawn as a line. B: Allow / and / if given on ais need co-ordinates if given in tet or = -/, = -/. Accept decimal equivalents. Curve does not need to pass through these points and this mark may be given even if there is no curve. Ignore third point of intersection and allow touching instead of cutting. So even a cubic curve might get M0A0 B B. A V shape with two ruled lines for eample might get M0A0BB

224 Question Number Scheme. C: y = 8 + 5, 0 (a) So, y = dy d { } Marks = ( > 0 ) M A A (b) (When, 4 y = so ) y = 4 4 B dy 4 (gradient = = ) { = 6} d M = ( ) ( ) ( 4 ) Either : y " " " 6" ( ) = 4 or: " 6" y = + c and " " = " 6" + c c = "" ( ) 4 So y = 6 + A [4] (c) Tangent at Q is parallel to y + 8 = 0 ( y = + 6 ) Gradient =. so tangent gradient is B 4 Sets their gradient function = their So, " " = " " numerical gradient. M 4 4 = = 9 Ignore etra answer = -9 A Substitutes their found into equation of curve. dm When = 9, y = 9 ( ) = y =. A [5] marks Notes (a) n n M: Evidence of differentiation, so 0 at least once so or or not just 5 0 A: Any two of the three terms correct do not need to see zero the 5 disappearing is sufficient; need not be simplified. A: 4 Both terms correct, and simplified. Do not need to include domain > 0 (b) B: Obtaining y = / or fractional or decimal equivalent (no working need be seen) M: An attempt to substitute = into d y to establish gradient. This may be implied by 6 or m = -6 but 4 d. not y = - 6. Can earn this M mark if they go on to use m = 6 or use their numerical value of d y d. dm: This depends on previous method mark. Complete method for obtaining the equation of the tangent, using their tangent gradient and their value for y (obtained from = ¼, allow slip) i.e. y y = mt ( 4 ) with their tangent gradient and their y or uses y m c,theiry and their tangent gradient. = + with ( 4 ) A: y = -6 + or y= 6 or a = -6 and b = (c) B: For the value / not / not -/ M: Sets their gradient function dy/d = their numerical gradient A: Obtains = 9 dm: Substitutes their (from gradient equation) into original equation of curve C i.e. original epression y = A: (9, -) or = 9, y = -, or just y = - Special Cases: In (b) Finds normal could get B M M0 A0 i.e. ma of /4 In (c) Uses perpendicular instead of parallel then award B0 M A0 M A0 i.e ma /5 see over dm []

225

226 Mark Scheme (Results) Summer 0 GCE Core Mathematics (666/0R)

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228 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. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

229 EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. The Edecel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes: bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark 4. All A marks are correct answer only (cao.), unless shown, for eample, as A ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks. 5. For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected. 6. If a candidate makes more than one attempt at any question: If all but one attempt is crossed out, mark the attempt which is NOT crossed out. If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt. 7. Ignore wrong working or incorrect statements following a correct answer. 8. In some instances, the mark distributions (e.g. M, B and A) printed on the candidate s response may differ from the final mark scheme.

230 General Principles for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). 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 a, leading to + + = + + = = =. Formula Attempt to use correct formula (with values for a, b and c).. Completing the square b Solving + b+ c= 0 : ± ± q± c, q 0, leading to =... Method marks for differentiation and integration:. Differentiation n n Power of at least one term decreased by. ( ). 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. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but may 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 not gain full 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.

231 Question Number Scheme Notes Marks. dy y = + 4+ = d substitute = gradient = ( ) n n M: including 0 A: Correct differentiation (Do not allow 4 0 unless 0 = is implied by later work) M: Substitutes = into their d y (not y) d Substitutes = into a changed function. They may even have integrated. A: cao MA MA [4]

232 Question Number Scheme Notes Marks. 5 5 = = 5 M: Attempts to multiply numerator and denominator by. This may be implied by a correct answer. A: 5 7 = B 5 7 = A Way = = Note that 6 6 = Correct answer only scores full marks Terms combined correctly with a common denominator (Need not be simplified) M: Attempts to multiply numerator and denominator by. This may be implied by a correct answer. A: = A = = 5 9 = 6 is quite common and scores MA0BA0 (i.e. 5 is never seen) MA B MA [4] [4]

233 Question Number Scheme Notes Marks. 4 = d 4 4 or 4 M: n n + 4 If they write here. for either term. as 4 allow A: or 4 (one correct term which may be un-simplified) A: and 4 (both terms correct which may be un-simplified) Note that MA0A is not possible = + + c + + c Fully correct simplified answer with + c all appearing on the same line. M,A,A A [4]

234 Question Number Scheme Notes Marks 4.(a) 4+ y = 0 y = + Attempt to write in the form y = M gradient = Accept any un-simplified form and allow even with an incorrect value of c A (a) Attempt to differentiate Way dy Alternative: 4+ = 0 dy d Allow p± q = 0, p, q 0 d M gradient = Accept any un-simplified form A Answer only scores MA (b) Using m N = m T Attempt to use m N = M gradient from ( a) [] y 5 = ' '( ) or Correct straight line method using a changed gradient and the point M Uses y = m + c in an attempt to find c (, 5) y= + 4 Cao (Isw) A () [5]

235 Question Number 5.(a) Scheme Notes Marks y = 8 y = Cao (Can be implied i.e. by ) B y (Alternative: Takes logs base : log = log8 ylog = log y = ) () (b) 8= Replaces 8 by (May be implied) M 4 + = ( ) + or ( + ) Replaces 4 by correctly. M (b) Way = + = = M: Adds their powers of on the lhs and puts this equal to leading to a solution for. A: = or = 0. or awrt 0. + = Obtains 4 + in terms of 4 correctly M MA 4 = 8 Combines their and 4 correctly M M: Solves 8 = k leading to a 4 8 = 8 8 = = solution for. A: = or = 0. or awrt 0. MA (4) [5]

236 Question Number Scheme Notes Marks 6.(a) = k Accept un-simplified e.g. - k B (b) k k k = ( ) ( ) Attempt to substitute their into = ( ) kwith their in terms of k. = k + k * Answer given A* (c) k + k = Setting (d) ( k k = 0 ) k(k ) = 0 k =.. k = n = Or = + ' ' n= ( k ) k k M + = M Solving their quadratic to obtain a value for k. Dependent on the previous M. Cao and cso (ignore any reference to k = 0) Writing out at least terms with the third term equal to the first term. Allow in terms of k as well as numerical values. Evidence that the sequence is oscillating between and k. This may be implied by a correct sum. 50 or or 50 ( ) = 5 An attempt to combine the terms correctly. Can be in terms of k here e.g 00 50k Allow an equivalent fraction, e.g. 50/ or 00/4 Note that the use of ( ) na+ l is acceptable here but ( ( ) ) n a+ n d is not. () dm A M M A () () () Allow correct answer only [9]

237 Question Number 7.(a) (b) (c ) Scheme Notes Marks Uses a+ ( n ) d with a=500, d=00 U 0 = ( 0 ) 00 M and n = 9,0 or =( )00 A If the term formula is not quoted and the numerical epression is incorrect score M0. A correct answer with no working scores full marks. Mark parts (b) and (c) together M: Attempt to use n S = { a+ ( n ) d} n { ( n ) 00} = 6700 with, MA S = 6700, a= 500 and d = 00 n A: Correct equation If the sum formula is not quoted and the equation is incorrect score M0. M: An attempt to remove brackets and collect terms. Dependent on the n + 4n 67 = 0 previous M A: A correct three term equation in any form E.g. allow n + 4n = 67, n = 67 4n, 67 4n n = 0, 00n + 800n = 4400 etc. Replaces 67 with 4 8 with the equation as printed (including = 0) with n + 4n 4 8= 0* no errors. (= 0 may not appear on the last line but must be seen at some point) ( n 4)( n+ 8) = 0 n =.. or nn ( + 4) = 4 8 n=.. 4 Allow correct answer only in (c) Solves the given quadratic in an attempt to find n. They may use the quadratic formula. States that n = 4, or the number of years is 4 dma A M A () (5) () [9]

238 Question Number Scheme Notes Marks Ignore any references to the units in this question 8.(a) length is + 4 May be implied B + ( ± 4) > 9.and proceeds to > > 9. >..... (Accept invisible brackets) Attempts widths + lengths > 9. leading to >... E.g > 9. >.9 scores B0MA0 >.8 * Achieves >.8 with no errors A(*) Mark parts (b) and (c) together (b)(i) + ( 4) < Cao B b(ii) Multiply out lhs, produce TQ = 0 and + 4 < 0 attempt to solve leading to =... according ( + 7)( ) < 0 =... to general guidelines M M: Attempts the inside for their critical values (may be from a TQ here) Either -7 < < or 0 < < A: Accept either -7 < < or 0 < < or ( > -7 and < ) or ( > 0 and < ) but not e.g. ( > -7, < ) or ( > -7 or < ) (There is no specific need for them to realise > 0) MA Note that many candidates stop here (c ).8 < < ( 7)( + ) < 0, = 7, = < < 7or0< < 7.8 < < 7 Scores B0MMA0Bft ( 4) < 4 < 0 Follow through their answers to (a) and (b) Provided their >.8 Eamples 4< 4 < 0 ( )(+ ) < 0, =± < < or0 < <.8 < < M Bft () (4) () [8] Scores B0M0MA0B0

239 Question Number Scheme Notes Marks 9.(a) (b) f( ) = ( + )( ) = + + = + ( )( 4 4) 4 M: Either stating or writing down that ( ± ) or ( ± ) is a factor may be implied by their f() A: Both ( + ) and ( - ) are factors - may be implied by their f() B: y or f()= ( + )( - ) M: Multiplying out a quadratic to get terms and then multiplying by the linear term to form a cubic. A: + 4or a = -, b = 0, c = 4 MAB MA (5) Same shape and position (ignore any coordinates) with the maimum on the y-ais B (a) Way = 0, y = 4 c = 4 =, y = 0 + a b+ c = 0 =, y = a+ b+ c = 0 a b = 4a+ b = a =.. or b =... y intercept = 4 or their c coordinates at - and 4 or marked as coordinates. Allow (0, -) and (0, 4) if they are marked in the correct position. The curve must cross or at least stop at = - Uses (0, 4) to obtain c = 4 (can be just stated) Uses both (-, 0) and (, 0) in y = + a + b + c to form simultaneous equations. Allow the equations to contain c here. Solves simultaneously with a value for c to obtain a value for a or a value for b Bft B B M M () [8] Either a = - or b = 0 A Both a = - and b = 0 A

240 Question Number 9.(a) Way Scheme Notes Marks dy a b d n n M: at least once including c 0 M dy = 0 = 0 b = 0 d Correct value for b A = 0, y = 4 c = 4 Uses (0, 4) to obtain c = 4 (can be just stated) B () + a() + b = 0 or ( ) + a( ) + b( ) + 4= 0 Obtains an equation in a M a = Correct value for a A Special case: A common incorrect approach is to assume the cubic is of the form e.g. f( ) = ( ± )( ± ) + 4 This scores B only for c = 4 (5) [8]

241 Question Number Scheme Notes Marks 0.(a) (b) Alternative to part (b) f'( ) = = + = + 9 f( ) = + 9 ( + c) ( 9) ( 9) c = 0 c =... f( ) = M: Correct attempt to split into separate terms or fractions. May be implied by one correct term. Divides by or multiplies by. A: + 9 or equivalent M: Independent method mark for n n+ on separate terms A: Allow un-simplified answers. No requirement for + c here Substitutes = 9 and y = 0 into their integrated epression leading to a value for c. If no c at this stage M0A0 follows unless their method implies that they are correctly finding a constant of integration. There is no requirement to simplify their f() so accept any correct un-simplified form. MA MA + 9 Sets f'( ) = = 0 and multiplies + 9 f'( ) = = 0 + 9= 0 by. The terms in must be in the + 9 numerator. E.g. allow = 0 M They must be setting either the original f'( ) = 0or an equivalent correct epression = 0 Correct attempt to solve a relevant ( 9)( ) = 0 =... TQ in leading to solution for. dm Dependent on the previous M. = 8, = Note that the = solution could be just written down and is Bbut must come from a correct equation. A, B + 9 = + + = ( ) ( 8)( ) = 0 =.. = 8, = + 9 Sets = 0, squares and multiplies by. They must be setting either the original f'( ) = 0or an equivalent correct epression = 0 Correct attempt to solve a relevant TQ leading to solution for. Dependent on the previous M. Note that the = solution could be just written down and is Bbut must come from a correct equation. M A M dm A, B (6) (4) [0]

242 Question Number. (a) Alternative to part (a) (b) Scheme Notes Marks y = + + 4( + ) = 5 Substitute y = ± ± into + 4y = 5to obtain an equation in only. M Alternative: = ( + ) or = ( + ) = 0 Multiply out and collects terms producing term quadratic in any form. M Solves their quadratic, usual rules, as (5+ 9)( ) = 0 =.. far as =... Dependent on the first M i.e. a correct method for eliminating dm y (or see below) 9 =, = 5 Both correct A for both 9, M: Substitutes back into either given y = y = 5 equation to find a value for y M Coordinates are ( ) and (, ) Correct matching pairs. Coordinates need not be given eplicitly but it must A be clear which goes with which y (6) = y ( y ) + 4y ( y ) = Substitutes + 4y = 5 M 5y 6y 7 = 0 Multiply out, collect terms producing term quadratic in any form. M Solves their quadratic, usual rules, as (5y+ 9)( y ) = 0 y =.. far as y =... Dependent on the first M i.e. a correct method for dm eliminating 9 y =, y = 5 Both correct A for both 9, M: Substitutes back into either given = = 5 equation to find a value for M Coordinates are ( ) and (,) d = 9 9 ( ) + ( ) 5 5 d = ( ) + ( ) or Correct matching pairs as above. M: Use of d = ( ) + ( y y ) or d = ( ) + ( y y ) where neither ( ) nor ( y y ) are zero. A MAft Aft: Correct ft epression for d or d (may be un-simplified) 4 d = 5 Allow 4.8 Acao () [9]

243 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fa Order Code UA0566 Summer 0 For more information on Edecel qualifications, please visit our website Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Esse CM0 JE

244 Mark Scheme (Results) Summer 0 GCE Core Mathematics (666/0)

245 Edecel and BTEC Qualifications Edecel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Epert service helpful. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 50 years, and by working across 70 countries, in 00 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: Summer 0 Publications Code UA05658 All the material in this publication is copyright Pearson Education Ltd 0

246 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. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

247 EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. The Edecel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes: bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark 4. All A marks are correct answer only (cao.), unless shown, for eample, as A ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks. 5. For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected. 6. If a candidate makes more than one attempt at any question: If all but one attempt is crossed out, mark the attempt which is NOT crossed out. If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt. 7. Ignore wrong working or incorrect statements following a correct answer. 8. In some instances, the mark distributions (e.g. M, B and A) printed on the candidate s response may differ from the final mark scheme.

248 General Principles for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). 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 a, leading to + + = + + = = =. Formula Attempt to use correct formula (with values for a, b and c).. Completing the square b Solving + b+ c= 0 : ± ± q± c, q 0, leading to =... Method marks for differentiation and integration:. Differentiation n n Power of at least one term decreased by. ( ). 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. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but may 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 not gain full 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.

249 Question Number Way 7+ 5 ( 5+ ) 5 ( 5+ ) Scheme Multiplies top and bottom by a correct epression. This statement is sufficient. k + ) Obtains a denominator of 4 or sight of ( 5 )( 5+ ) = 4 (Allow to multiply top and bottom by ( 5 )... = 4 Note that M0A is not possible. The 4 must come from a correct method. An attempt to multiply the numerator by ( ± 5± ) and (7 + 5)( 5 + ) = get 4 terms with at least correct for their ( ± 5± ). (May be implied) Answer as written or a = + 5 and b =. (Allow 5 + ) Correct answer with no working scores full marks 7+ 5 ( 5 ) 5 ( 5 ) Multiplies top and bottom by a correct epression. This statement is sufficient. M Marks Acso M Acso (Allow to multiply top and bottom by k( 5 ) )... = 4 Obtains a denominator of -4 Acso An attempt to multiply the numerator by ( ± 5± ) and (7 + 5)( 5 ) = get 4 terms with at least M correct for their ( ± 5± ). (May be implied) + 5 Answer as written or a = and b = Acso Correct answer with no working scores full marks Alternative using Simultaneous Equations: (7 + 5) = a+ b = ( a b) 5 + 5b a M 5 Multiplies and collects rational and irrational parts a b =, 5b a = 7 A Correct equations a =, b = M for attempt to solve simultaneous equations A both answers correct M [4] [4] `

250 Question Number ( = ), 5 = c Scheme M: Some attempt to integrate: n n + on at least one term. (not for + c) (If they think is you can still award the method mark for A: and or better 5 A: - or better Each term correct and simplified and the + c all appearing together on the same line. Allow for. Ignore any spurious integral or signs and/or dy/d s. Do not apply isw. If they obtain the correct answer and then e.g. divide by they lose the last mark. Marks MA, A A [4]

251 Question Number (a) (b) 8 = or 8 = Scheme A correct attempt to deal with the or the = 8 or 8 = = Cao A correct answer with no working scores full marks Alternative 5 8 = 8 8 = 8 = M (Deals with the /) = A = One correct power either or. on its own is not sufficient for this mark. Marks M A M () 8 4 = or M: Divides coefficients of and subtracts their powers of. Dependent on the previous M dma A: Correct answer Note that unless the power of implies that they have subtracted their 8 powers you would need to see evidence of subtraction. E.g. = 4 would score dm0 unless you see some evidence that / was intended for the power of. Note that there is a misconception that = - this scores 0/ 4 4 () [5]

252 Question Number Scheme For this question, mark (a) and (b) together and ignore labelling. Marks 4(a) ( a ) = k(4+) (= 6k) (b) a = k(their a + ) (= 6k + k) Any correct (possibly un-simplified) epression An attempt at a. Can follow through their answer to (a) but a must be an epression in k. a+ a + a = 4 + (6 k) + (6k + k) An attempt to find their a+ a + a M 4 + (6 k) + (6k + k) = A correct equation in any form. A Solves 6k + 8k+ = 0 to obtain k = (6k + 8k+ = (k+ )( k+ )) Solves their TQ as far as k =... according to the general principles. (An independent mark for solving their three term quadratic) k = / Any equivalent fraction A k = Must be from a correct equation. (Do not accept un-simplified) B Note that it is quite common to think the sequence is an AP. Unless they find a, this is likely only to score the M for solving their quadratic. B M M () (6) [7]

253 Question Number 5 (a) 6 + > 8 (b) Scheme Attempts to epand the bracket and collect terms on one side and constant terms on the other. Condone sign errors and allow one error in epanding the bracket. Allow <,,,= instead of >. > Cao A Do not isw here, mark their final answer. ( + )( )[ = 0] = and M: Attempt to solve the quadratic to obtain two critical values A: = and (may be implied by their inequality). Allow all equivalent fractions for - and /. (Allow 0. for /) M: Chooses inside region (The letter does not need to be used here) Aft: Allow < and > -or M Marks MA () < <, or < >. Follow through their critical values. (must be in terms of here) Allow all equivalent fractions for - and /. Both ( < or > -) and ( <, > -) as a final answer MAft score A0. Note that use of or appearing in an otherwise correct answer in (a) or (b) should only be penalised once, the first time it occurs. (4) [6]

254 Question Number Scheme 6 (-, ), (, ) (a) m = y y M:Correct method for the gradient =, = A: Any correct fraction or ( ) 4 decimal y = ¾ (+) or y = ¾ ( ) Correct straight line method using or y = ¾ + c with attempt at either of the given points and a substitution to find c numerical gradient. 4y 5 = 0 Or equivalent with integer coefficients (= 0 is required) This A should only be awarded in (a) (a) Way (b) y y y + = = y y + M: Use of a correct formula for the straight line A: Correct equation Marks M,A M A MA ( y ) = 9( + ) Eliminates fractions M 4y 5 = 0 Or equivalent with integer coefficients (= 0 is required) A Solves their equation from part (a) and L simultaneously to eliminate one variable Must reach as far as an equation in only or in y only. (Allow slips in the algebra) = or y = 6 One of = or y = 6 A Values can be un-simplified Both = and y = 6 fractions. Fully correct answers with no working can score / in (b) M A (4) (4) () (b) Way (,) a+ b+ c = 0 Substitutes the coordinates to (,) a+ b+ c = 0 obtain two equations M 4 Obtains sufficient equations to a = b, b = c 4 5 establish values for a, b and c A 4 e.g. c = b =, a = 5 5 Obtains values for a, b and c M 4 y+ = 0 4y 5= 0 Correct equation A 5 5 (4) [7]

255 Question Number 7(a) 600 = 00 + (N )0 N =... (b) Scheme Use of 600 with a correct formula in an attempt to find N. A correct formula could be implied by a correct answer. N = cso A Accept correct answer only. 600 = N N = 0 is M0A0 (wrong formula) = 0 N = is MA (correct formula implied) 0 Listing: All terms must be listed up to 600 and correctly identified. A solution that scores if fully correct and 0 otherwise. Look for an AP first: M: Use of correct sum formula S = ( )or ( ) with their integer n = N or N from part (a) where < N < 5 or and a = 00 and d = S = ( )or ( ) A: Any correct un-simplified numerical epression with n = 0 (= 8400 or 7800) or n = (No follow through here) Then for the constant terms: M: 600 k where k is an integer and < k < (5 " N ") (= 8600) A: A correct un-simplified follow through epression with their k consistent with n so that n + k = 5 So total is 7000 Cao A Note that for the constant terms, they may correctly use an AP sum with d = 0. There are no marks in (b) for just finding S 5 If they obtain N = 0 in (a) (0/) and then in (b) proceed with, S = 0 ( ) = = allow them to recover and score full marks in (b) Similarly If they obtain N = in (a) (0/) and then in (b) proceed with, S = ( ) = = allow them to recover and score full marks in (b) M Marks MA MAft () (5) [7]

256 Question Scheme Number 8 Horizontal translation does not have to cross the y-ais on the right but must at least reach the -ais. (a) (b) ( 5) ( ) + + Allow Touching at (-5, 0). This could be stated anywhere or -5 could be marked on the -ais. Or (0, -5) marked in the correct place. Be fairly generous with touching if the intention is clear. The right hand tail of their cubic shape crossing at (-, 0). This could be stated anywhere or - could be marked on the -ais. Or (0, -) marked in the correct place. The curve must cross the -ais and not stop at -. ( ) ( ) B B B B (c) When = 0, y = 5 M: Substitutes = 0 into their epression in part (b) which is not f(). This may be implied by their answer. Note that the question asks them to use part (b) but allow independent methods. A: y = 5 (Coordinates not needed) If they epand incorrectly prior to substituting = 0, score M A0 NB f( + ) = Marks M A () () () [6]

257 Question Number 9 (a) ( ) = Alternative : Writes Scheme An attempt to epand the numerator obtaining an epression of the form 4 9 ± p ± q, p, q Must come from -6 Must come from 6 ( ) as ( ) and attempts to epand = M then AA as in the scheme. 4 Alternative : Sets ( ) = 9 + A + B, epands ( ) and compares coefficients = M then AA as in the scheme. (b) 8 + (c) = + M: on separate terms at least once. Do not award for A 0 (Integrating is M0) Aft: 8 + " B" with a numerical B and no etra terms. (A may have been incorrect or even zero) (f ( ) 9 6 ) n n 4 M A A Marks M Aft n n M: + on separate terms at least once. (Differentiating is M0) f( ) = ( + c) B MAft Aft: 9 + A + ( + c) with numerical A and B, AB, 0 9 ( ) Uses = - and y =0 in what they think 0 = 6( ) + + c so c is f() (They may have differentiated here) but it must be a changed function =... i.e. not the original f( ), to form a linear M equation in c and attempts to find c. No + c gets M0 and A0 unless their method implies that they are correctly finding a constant. c = - cso A Follow through their c in an otherwise (possibly un-simplified) correct (f ( ) = ) their epression. Aft c 9 9 Allow for 9 or even. Note that if they integrate in (b), no marks there but if they then go on to use their integration in (c), the marks for integration are available. (5) [0] () ()

258 Question Number 0(a) (b) (b) Way Equal roots (b) Way (c) Scheme k k 4 ( ) + 5 ( = 0) So k k = 0* (8 k) 4 k k = (oe) k+ k = ( + k) 8k = k Marks Makes y the subject from the first equation and substitutes into the second equation (= 0 not needed here) or eliminates y by a correct method. Correct completion to printed answer. There must be no incorrect statements. M: Use of b 4ac(Could be in the quadratic formula or an inequality, = 0 not needed yet). There must be some correct substitution but there must be no s. No formula quoted followed by e.g. 8k 4k = 0is M0. A: Correct epression. Do not condone missing brackets unless they are implied by later work but can be implied by (8 k) > 4k etc. Cso (Ignore any reference to k = 0) but there must be no contradictory earlier statements. A fully correct solution with no errors. M: Correct strategy for equal roots A: Correct equation M Acso M A A MA k = (oe) Cso (Ignore any reference to k = 0) A 6 Completes the Square M: ( ± 4 k) ± p± k, p k+ k = ( + 4 k) 6k + k MA 6k k = 0 A: Correct equation k = (oe) Cso (Ignore any reference to k = 0) A = so ( + ) = 0 = =, y = 4 Substitutes their value of k into the given quadratic and attempt to solve their or term quadratic as far as = (may be implied by substitution into the quadratic formula) or starts again and substitutes their value of k into the second equation and solves simultaneously to obtain a value for. First A one answer correct, second A both answers correct. M AA () () () Special Case: + + = 0 =, y =, allow MAA () [8]

259 Question Number (a) Scheme,0 4. Accept = 4 B (b) y = 4 B: One correct asymptote = 0 or y-ais B: Both correct asymptotes and no etra ones. Special case 0 and y 4 scores BB0 (c) dy = d At =, gradient of curve = Gradient of normal = -/m Normal at P is ( y ) = ( + ) (d) (-4, 0) and (0, ). So AB has length 60 or AB has length 60 Marks BB dy dy = k (Allow = k + 4) M d d Cao (may be un-simplified but must be a fraction with no powers) e.g. ( ) scores A0 unless evaluated A as e.g. or is implied by their 9 normal gradient. Correct perpendicular gradient rule applied to a numerical gradient that must have come from substituting dm = - into their derivative. Dependent on the previous M. M: Correct straight line method using (-, ) and a changed gradient. A wrong equation with no dma formula quoted is M0. Also dependent on the first M. A: Any correct equation Both correct (May be seen on a sketch) M: Correct use of Pythagoras for their A and B one of which lies on the -ais and the other on the y-ais, obtained from their equation in (c). A correct method for AB or AB. A: 60 or better e.g. 4 0 with no errors seen B M Acso () () (5) () []

260 January 04 (IAL) Mark Scheme (Results) January 04 Pearson Edecel International Advanced Level Core Mathematics (666A/0)

261 January 04 (IAL) Edecel and BTEC Qualifications Edecel and BTEC qualifications are awarded by Pearson, the UK s largest awarding body. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at or Alternatively, you can get in touch with us using the details on our contact us page at Pearson: helping people progress, everywhere Pearson aspires to be the world s leading learning company. Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 50 years, and by working across 70 countries, in 00 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: January 04 Publications Code IA07647 All the material in this publication is copyright Pearson Education Ltd 04

262 January 04 (IAL) 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. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

263 January 04 (IAL) EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. The Edecel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark 4. All A marks are correct answer only (cao.), unless shown, for eample, as A ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks. 5. For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected. 6. If a candidate makes more than one attempt at any question: If all but one attempt is crossed out, mark the attempt which is NOT crossed out. If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt. 7. Ignore wrong working or incorrect statements following a correct answer.

264 January 04 (IAL) General Principles for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). 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 a, leading to =. Formula Attempt to use the correct formula (with values for a, b and c).. Completing the square b Solving b c 0 : q c 0, q 0, leading to = Method marks for differentiation and integration:. Differentiation. Integration Power of at least one term decreased by. ( n n ) 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. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but may 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 not gain full 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.

265 January 04 (IAL) Question Number Scheme Marks. (a) 4 B (b) (5 7) ( 7) ( 7) ( 7) () M, A 7 A () (4 marks) Notes (a) B 4.Accept alternatives such as 4, 4, 4 M For multiplying numerator and denominator by 7and attempting to epand the brackets. There is no requirement to get the epanded numerator or denominator correctseeing the brackets removed is sufficient. (b) A All four terms correct (unsimplified) on the numerator OR the correct denominator of - A Correct answer 7. Accept 7, 7 and other fully correct simplified forms

266 Question Number. (a) Scheme 4 4 January 04 (IAL) Marks dy 4 ( 0) n n M d dy 4 or 4 oe d A,A () n n (b) M 5 d y 4 or 4 5 d A () (5 marks) (a) M A A Notes n n for any term. The sight of sufficient A OR Do not follow through on an incorrect inde of 4 C D for this mark. One of the first two terms correct and simplified. Either 4 or Accept equivalents such as 4 and.5 Ignore +c for this mark. Do not accept unsimplified terms like OR 0 is A completely correct solution with no +c. That is 4 Accept simplified equivalent epressions such as 4 or 4 There is no requirement to give the lhs ie d y d. However if the lhs is incorrect withhold the last A (b) M For either 4 4 or answers in (a). A n n for a fractional term. Follow through on incorrect 5 4 A completely correct solution Award for epressions such as d y There is no requirement to give the lhs ie... d. However if the lhs is incorrect withhold the last A or 4 or 4 5

267 January 04 (IAL) Question Number. Scheme y y Marks (y ) 4y 0( y ) 9 0 ( ) M 8y 6y M,A 8 yy ( ) 0 ( )( 5) 0 M Alt y(8y 6) 0 Alt ( )( 5) 0 y 0, y, 5 y 0 in y y in y 5 in y 0 M 5 in y =,y=0 and =5,y= =,y=0 and =5,y= A,A (7 marks) Notes M Rearrange y 0 into.., or y.., or y.. and attempt to fully substitute into nd equation. It does not need to be correct but a clear attempt must be made. Condone missing brackets (y ) 4y 0 y 9 0 M Collect like terms to produce a quadratic equation in (or y) =0 A Correct quadratic equation in (or y)=0. Either A y ( y) 0 or B ( 6 5) 0 M Attempt to solve, with usual rules. Check the first and last terms only for factorisation. See appendi for completing the square and use of formula. Condone a solution from cancelling in a case like A( y y) 0. They must proceed to find at least one solution.. or y.. M Substitute at least one value of their to find y or vice versa. This may be implied by their solution- you will need to check! A Both s or both y s correct or a correct matching pair. Accept as a coordinate. Do not accept correct answers that are obtained from incorrect equations. A Both pairs correct. Accept as coordinates (,0) (5,) Special Cases where candidates write down answers with little or no working as can be awarded above. One correct solution B. Two correct solutions B, B To score all 7 marks candidates must prove that there are only two solutions. This could be justified by a sketch.

268 Question Number Scheme January 04 (IAL) Marks 4. (a) Horizontal translation of ±4 M Minimum point on they- ais at (0,) A () (b) Correct shape with P ' adapted M y intercept (0,6) and P (4,4) A () (4 marks) Notes (a) M A horizontal translationof 4. The y coordinate of P remains unchanged at. Look for P' 0, or (8,). Condone U shaped curves A The shape remains unchanged and has a minimum at (0,). Condone U shaped curves (b) M The curve remains in quadrant and quadrant with the minimum in quadrant. The shape must be correct. Condone U shaped curves. P must have been adapted. The mark cannot be scored for drawing the original curve with P =(4,). A Correct shape, condoning U shapes with the y intercept at (0, 6) and P =(4,4) The coordinates of the points may appear in the tet or besides the diagram. This is acceptable but if they contradict the diagram, the diagram takes precedence.

269 January 04 (IAL) Question Number 5. (a) Scheme Marks 5 ar M r = A (b) r 6 r 6 ar 4 6 M r a6 ar (part a) dm () a A () (5 marks) Notes (a) M Substitutes n=5 into the epression 4n and attempt to find a numerical answer A for 5 ar. r Accept as evidence epressions such as 4 5.., 4 (5).., even 0 4 Accept for this mark solutions which add 4, 4, 4, 4 4, 4 5 and as a result appears in a sum. cao. Accept this answer with no incorrect working for both marks. If it is consequently summed it will be scored A0 (b) M Substitutes n =6 into the epression 4n Accept as evidence 4 6.., 4 (6 ) or indeed 56. You can accept the appearance of in a sum of terms. dm Attempts to find their answer to A cao 44 6 r This is dependent upon the previous M mark. Also accept a restart where they attempt a their answer to part a r 6 5 a a r r r r Alternative to 5(b) M Writes down an epression for an 4n 4( n ) 4 n ( n ) 4( n ) dm Subs n 6 into the epression for an 4(n )... A cao 44

270 January 04 (IAL) Question Number Scheme 6. (a) (i) or equivalents such as.5 B (ii) (0,.5) Accept y= B (b) Perpendicular gradient l = Bft Equation of line is: y 5 ( ) MA y 7 0 A Marks (c ) Point C: y oe M, A () (4) AB= ( 0) (5.5) BC= (8.5 ) (5 0) 5 Area rectangle = AB BC= oe 4 M (either) dma (5) ( marks) (a) Notes B cao gradient =.5. Accept equivalences such as (b) B cao intercept =(0,.5). Accept.5, y=.5 and equivalences such as 7 Bft For using the perpendicular gradient rule, m on their.5. m... Accept or this as part of their equation for l Eg. '.5' y '.5'... M For an attempt at finding the equation of l using (,5) and their adapted gradient. Condone for this mark a gradient of going to.eg. Allow for y 5 If the form y m c is used it must be a full method to find c with (,5) and an adapted gradient. A For an a correct unsimplified equation of the line through (,5) with the correct gradient. y 5 7 Allow and 5 c c A cso (y 7) 0 y 5 An eample of BftM0A0A0 would be following a gradient of in part (a) y 5 An eample of BftMA0A0 would be following a gradient of in part (a) y 5 An eample of B0ftMA0A0 would be following a gradient of in part (a)

271 January 04 (IAL) Question Number Scheme Marks Notes for Question 6 continued (c ) M An attempt to use their equation found in part b to find the coordinate of C They must either use the equation of l and set y 0... or use its gradient A C= (8.5, 0). Allow equivalents such as =8.5 at C M An attempt to use ( ) ( y y ) for AB or BC. There is no need to calculate these. Evidence of an attempt would be AB.5 AB.. dm Multiplying together their values of AB and BC to find area ABCD It is dependent upon both M s having been scored. A cao6.5 or equivalents such as 65 4.

272 Question Number Scheme January 04 (IAL) Marks 7. (a) = M = 6000 A* n 9 (b) Sn ( a l) ( ) n 9 OR S9 ( a ( n ) d) ( ) M = A (c) Use a ( n ) d to find A () () A (0 ) M A 7000 A n n Use Sn ( a l) or Sn ( a ( n ) d) to find S for Anna 0 S0 ( ) 5000 or 0 S0 ( ) 5000 MA Shelim earns in 0 years =( 06000) Bft Difference= 9000 A (6) (0 marks) Notes (a) M Uses S a ( n ) d with a=4000, d=500 and n=8, 9 or 0 in an attempt to find salary in year 9 Accept a sequence written out only if all terms up to year 9 are included-allow no errors. A* csa It is acceptable to write a sequence for both the marks FYI the terms are 4000,5500,7000,8500,0000,500,000,4500,6000 Alt (a) Alternative working backwards M Uses S a ( n ) d with a=4000, d=500 and S =6000 in attempt to find n. It must reach n=.. A n=9 (b) n M Uses Sn ( a l) with a=4000, l=6000 and n=8, 9 or 0. Do not allow ft s on incorrect l s. n Alternatively uses Sn ( a ( n ) d) with a=4000, d=500 and n=8, 9 or 0. Weaker candidates may list the individual salaries. This is acceptable as long as all terms are included. For eample A Cao ( )

273 January 04 (IAL) Question Number Scheme Marks Notes for Question 7 continued (c ) M Use l a ( n ) d to find A. It must be a full method with d=000, l=6000a=a and n=9, 0 or leading to a value for A A A=7000. Accept A=7000 written down for marks as long as no incorrect work seen in its calculation. M n Use Sn ( a l) to find S for Anna. Follow through on their A, but l=6000 and n=9, 0 or n Alternatively uses Sn ( a ( n ) d) with their numerical value of A, d=000 and n=9, 0 or Accept a series of terms with their value of A, rising in 000 s up to a maimum of A Anna earns 0 S0 ( ) OR 0 S0 ( ) in 0 years This is an intermediate answer. There is no requirement to state the value Bft Shelim earns (b)+6000 in 0 years. This may be scored at the start of part c. A CAO and CSO Difference = 9000

274 January 04 (IAL) Question Number 8. (a) (b) b ac k k Scheme 4 ( ) 4 ( ) MA b ac k k k k ( ) A* Alt (a) Marks k k 4 0 ( k ) 5 M k 5 oe A k 5, k 5 dma (7 marks) b 4ac k 4 ( k ) MA k k 4 0 A* () (4) () Notes (a) M For attempting to use b 4ac with the values of a, b and c from the given equation. Condone invisible brackets. k 4 k could be evidence A Fully correct (unsimplified) epression for b 4ac k 4 ( k ) The bracketing must be correct. You can accept with or without any inequality signs. Accept a, b k, c k b 4ac k 4 ( k ) A* Full proof, no errors, this is a given answer. It must be stated or implied that b 4ac 0 Do not accept recovery from poor or incorrect bracketing or incorrect inequalities. Do not accept the answer written down without seeing an intermediate line such as 4k 4 ( k ) 0 k k 4 0 Or 4k 8k 8 0 k k 4 0 The inequality must have been seen at least once before the final line for this mark to have been awarded. Eg accept D 4k 8k 8 4k 8k 8 0 k k 0 (b) M Attempt to solve the given term quadratic (=0) by formula or completing the square. Do NOT accept an attempt to factorise in this question. If the formula is given it must be correct. ( ) ( ) 4 4 It can be implied by seeing either or 4 4 If completing the square is used it can be implied by ( k ) 4 0 k... A Obtains critical values of 5. Accept 0 dm Outsides of their values chosen. It is dependent upon the previous M mark having been awarded. States k their largest value, k their smallest value Do not award simply for a diagram or a table- they must have chosen their 'outside regions' A Correct answer only. Accept k 5or k 5, k 5 k 5,, 5 5, but not k 5and k 5, 5 k 5

275 January 04 (IAL) Question Number Scheme Marks Notes for Question 8 continued Also accept eact alternatives as a simplified form is not eplicitly asked for in the question 0 0 Accept versions such as k or k

276 January 04 (IAL) Question Number 9. (a) f '( ) ( )( 4) (b) Scheme 8 B 8d 8 y c MA, y c c.. M f ( ) 8 cso A f ( ) ( 4 4)( ) f ( ) ( ) ( p) f ( ) 4 4 f ( ) 8 (c ) p B cso MA Marks (5) () Shape B (0,) Min at (,0) B Crosses - ais at (-,0) Bft (-,0) (,0) Crosses y- ais at (0,) B (4) ( marks) Notes (a) B Writes ( )( 4) as 8 M n n in any one term. For this M to be scored there must have been an attempt to epand the brackets and obtain a quadratic epression A Correct (unsimplified) epression for f ( ), no need for +c. Accept 8 M Substitutes = and y=6 into their f() containing a constant c and proceed to find its value. A Cso f ( ) 8. Allow y =.. Do not accept an answer produced from part (b) (b) B States p This may be obtained from subbing (,6) into f ( ) ( ) ( p) M Multiplies out a pair of brackets first, usually ( ) and then attempts to multiply by the third.the minimum criteria should be the first multiplication is a T quadratic with correct first and last terms and the second is a 4T cubic with correct first and last terms. Accept an epression involving p for M ( ) ( p) (.. 4)( p).... 4p A cso f ( ) 8, which must be the same as their answer for part (a)

277 January 04 (IAL) Notes for Question 9 continued Candidates who have eperienced Core could take their answer to (a) and factorise. The mark scheme can be applied with M for division by ( )... and further factorisation of the quotient Alternatively the candidate could divide by to obtain (..) ( 4 4) The A is scored for f ( ) ( ) ( ) The B is awarded for a statement of p and not just ( ) ( ) (c) B Shape + graph with one maimum and one minimum. Its position is not important for this mark. It must appear to tend to + infinity at the rhs and infinity at the lhs. The curve must etend beyond its maimum point and minimum points. Eg. These are NOT acceptable. B There is a turning point at (, 0). Accept marked as a maimum or minimum on the - ais. Bft Graph crosses the - ais at (-, 0). Accept marked at the point where the curve crosses the -ais. You may follow through on their values of ' p ' as long as p B Graph crosses the y-ais at (0, ). Accept marked on the y- ais.

278 January 04 (IAL) Question Number n n 0. (a ) Sub = Scheme Marks dy d MA dy 4 () d M y dm y 5cso A* dy (b ) At Q 4 d M ( )( ) 0 dm A Sub into y dm 67 y 7 A (5) (0 marks) (5) (a) (b ) Notes M n n for any term including 0. A dy There is no need to see any simplification d M Sub = into their f '( ) dm Uses their numerical gradient with (, ) to find an equation of a tangent to y = f(). dy y It is dependent upon both M's. Accept their. Both signs must be d correct If y m c is used then it must be a full attempt to find a numerical c A* Cso y 5. This is a given answer and all steps must be correct. Look for gradient = having been achieved by differentiation. M Sets their d y d and proceeds to a TQ=0. Condone errors on dy d dm Factorises their TQ (usual rules) leading to a solution =. It is dependent upon the previous M. Award also for use of formula/ completion of square as long as the previous M has been awarded. A d M Sub their into y. It is dependent only upon the first M in (b) having been scored 67 A Correct y coordinate y or equivalent 7

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280 June 04 (R) Mark Scheme (Results) Summer 04 Pearson Edecel GCE in Core Mathematics R (666_0R)

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