Mark Scheme (Results) Summer GCE Core Mathematics C1 (6663) Paper 1

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

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3 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.

4 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. 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 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.

6 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.. Note: If a candidate scores MAAA and their answer is NOT ON THE SAME LINE then withhold the final accuracy mark.

7 Question Number. (a) ( ) 5 ( ) 5 ( ) (b) Scheme 5 5 = or or or 768 M = 8 A 5 5 = or or 5 5 or 5 5 = Notes (a) M: for a full correct interpretation of the fractional power. Note: 5 ( ) is M0. (b) See notes below See notes for other alternatives A: for 8 only. Note: Award MA for writing down 8. M: For use of OR use of Use of : Candidate needs to demonstrate the they have rooted all three elements in their bracket. C A B Use of -: Either Candidate has or becomes. C Bracket B A Allow M for... 5 or or or or or 5 or 5 or or or or or K C 5 or where K, Care any powers including A: for either or or 0. or Note: is not enough work by itself for the method mark. Note: A final answer of or or is A Note: Also allow or or 0. or 5 5 ± ± ± ± for A. Marks M A [] []

8 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 8 = seen or implied in candidate s working. nd B: for 8 = or = 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: = or 8 = are not sufficient for the B marks. Note: A candidate who writes down (by misread) 8 for 8 can potentially obtain MA0BBA0, where 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 ( ) 5 ( + ) ( ) ( + ) {( + )} = = M = + A A for Alternative solution = = 8 ( ) ( ) = +, with no incorrect working seen is awarded MABBA. ( ) Please record the marks in the relevant places on the mark grid. or = +

9 Question Number. (a) (b) (a) (b) Scheme y = dy = 5() 6 + d M Marks = A A A [] d y 8 = 0 d M A [] 6 M: for an attempt to differentiate st A: for nd A: for Notes n n to one of the first three terms of 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 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 d y 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

10 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} = " " = 7 M ai = i = + (6 c) + ( c) + ( 7 c) M "5 c" or "5 c" = M c or c A cso Notes Marks [] [] [] 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 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 a only. rd M: for their sum (of or 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: 5 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 or a, aand a or a, a, aand a 5 or a, a, a, aand a 5

11 Question Number Scheme Boy s Sequence: 0,5, 0, 5,... a = 0, d = 5 T = a + d = 0 + (5); = 80 or 0. + (0.05); = 0.80 M; A 6. (a) { } (b) { } [ ] 5 60 S 60 = (0) + 59(5) M A = 0(5) = 950 or 9.50 A Boy s Sister s Sequence: 0, 0, 0, 0,... 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 + d 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. [] [] [] [] 0 (b) 60 5 (0) 59(5) (0) (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 + d = 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 950 or 950p or 9.50 and apply ISW. Otherwise, 950 or 9.50 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 950 will then be awarded all three marks of MAA.

12 (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.

13 Question Number 7. (a) Scheme 6 P(, ) lies on C where f( ) = +, > 0 6 f() = () + ; = M; A T: y = ( ) dm T: y = 9 A = + + or equivalent. M A () ( ) (b) f( ) ( c) { } 6 f () = () + () + c = dm + + c = c = 7 So, { } { } 6 f( ) = A cso () ( ) NB: f ( ) = Notes (a) st M: for clear attempt at f(). st A: for obtaining from f(). nd y dm: for y = ( their f ())( ) or = ( their f ()) or full method of y = m + c, with =, y = and their f() 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 6 + on 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 = and y = in an integrated equation to form a linear equation in c equal to -. ie: applying f() =. 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). [] [] 8

14 Question Number 5 ( ), Scheme = q p p, q are integers. 8. (a) { } Marks 5 = + 5 = ( ) + 5 = ( ) + M = ( ) A A [] " b ac" = ( )( 5) = 6 0 M (b) { } { } (c) y = A [] O Correct shape M - 5 Maimum within the th quadrant A Curve cuts through -5 or (0, 5) marked on the y-ais B (a) Notes M: for an attempt to complete the square eg: ± ( ± ± ) ± k 5, k 0 or seen or implied in working. st A: for nd A: for q = p = or for α ( ) ±, α can be 0. Note: Allow MAA for a correct written down epression of ( ) Ignore 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. ± ± ± ± λ λ [] 8 ( ), 5 = ( ) 0. ± then imply the M mark. = = gets M0A0A0. Alternative to (a) { 5 = } 5 = ( ) 5 = ( ) + 5 = ( ) Alternative to (a) q ( + p) = q ( + p + p ) = p + q p Compare terms: p = p = Compare constant terms: M: Either ± p = or q q p q q = 5 = 5 = p ± = 5 ; st A: for p = ; nd A: for q =

15 Alternative to (a) Negating 5 gives + 5 So, 5 ( ) 5 { ( ) } + = + = + M for 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 = 5 ± ± ± ± k + ( ) 5 p + q p = q p + = + p q p + = + p ( ) q p + 5 = p + p scores Special Case MAA only once p achieved. (b) M: for correctly substituting any two of a =, b =, c = 5 into b ac if this is quoted. If b ac is not quoted then the substitution must be correct. Substitution into b < ac or b = ac or b > ac is M0. A: for only. If they write < 0 treat the < 0 as ISW and award A. If they write 0 then score A0. So substituting into b ac < 0 leading to < 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 ac. 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).

16 Question Number 9. (a) { } Scheme (b) { } (c) { L L } Marks + = A L :y + = y = ; A( p,)lieson L. p = 9 9 or or 9.5 B [] y + = y = m( L ) = or M A So ml ( ) = Bft L : y ( ) M L : 8 0 or : y 8 0 [5] = (8 ) + = or + 8 = M =.5, y = A, A cso [] (d) (".5" ) ("" ) CD = + M (".5" ) ("" ) 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) B = 5 A Notes 9. (a) B: 9.5 oe. (b) st M: for an attempt to rearrange y + = 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 ( ) = 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 = ± λ( ) where λ is a numerical value, λ 0. or full method of y = m + c, with =, y = and (their ± λ ) to find c. nd A: + y 8 = 0 or y + 8 = 0 or y + 8 = 0 or + 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

17 (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" ) + ("" ) or simplified. This mark can be implied by finding CD. ".5" + "" or correctly simplified. st Aft: for finding their CD - ft their point D. Eg: ( ) ( ) 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. (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 = 00 = 0 or 5 5 = 0 Note: This mark can be implied. A: for 5 only. Alternative to part (e): Area = ( 80 ) = ( ) = 5 M: ( AB)( CE ). B: Evidence of correct surd removal. A: for 5. 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) = = 5 M: For finding the area of one of the four triangles. B: Evidence of correct surd removal. A: for 5. 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. = 5 A: for 5 only.

18 Question Number 0. (a) {Coordinates of A are} ( ) (b)(i) Scheme.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 =.5 or or etc. or A =.5 or or (.5, 0 ). Can be written on graph. Allow ( 0,.5 ) marked on curve for B. Otherwise ( 0,.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. their in part ( a).5 9 Aft: for ; 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.

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