Mark Scheme (Results) Summer 015 Pearson Edecel GCE in Ce Mathematics C1 (6663/01)
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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 f what they have shown they can do rather than penalised f omissions. Eaminers should mark accding to the mark scheme not accding 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 wthy of credit accding 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 wk should be marked UNLESS the candidate has replaced it with an alternative response.
PEARSON EDEXCEL GCE MATHEMATICS General Instructions f Marking 1. The total number of marks f the paper is 75. The Edecel Mathematics mark schemes use the following types of marks: M marks: Method marks are awarded f 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. 3. 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 f crect ft cao crect answer only cso - crect solution only. There must be no errs in this part of the question to obtain this mark isw igne subsequent wking awrt answers which round to SC: special case oe equivalent (and appropriate) d dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper ag- answer given d The second mark is dependent on gaining the first mark 4. All A marks are crect answer only (cao.), unless shown, f eample, as A1 ft to indicate that previous wrong wking 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. F misreading which does not alter the character of a question materially simplify it, deduct two from any A B marks gained, in that part of the question affected. 6. If a candidate makes me 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 none are crossed out, mark all the attempts and sce the highest single attempt. 7. Igne wrong wking increct statements following a crect answer.
General Principles f Ce Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark f solving 3 term quadratic: 1. Factisation ( b c) ( p)( q), where pq c, leading to = ( a b c) ( m p)( n q), where pq c and mn a, leading to =. Fmula Attempt to use the crect fmula (with values f a, b and c). 3. Completing the square b Solving b c 0 : q c 0, q 0, leading to = Solving a b c 0 : b c a p 0, p 0, leading to = a a Method marks f differentiation and integration: 1. Differentiation Power of at least one term decreased by 1. ( n n1 ). Integration Power of at least one term increased by 1. ( n n1 )
Use of a fmula Where a method involves using a fmula that has been learnt, the advice given in recent eaminers repts is that the fmula should be quoted first. Nmal marking procedure is as follows: Method mark f quoting a crect fmula and attempting to use it, even if there are small errs in the substitution of values. Where the fmula is not quoted, the method mark can be gained by implication from crect wking with values, but may be lost if there is any mistake in the wking. Eact answers Eaminers repts have emphasised that where, f eample, an eact answer is asked f, wking with surds is clearly required, marks will nmally be lost if the candidate rests to using rounded decimals.
Question Number Scheme Marks 1.(a) (b) 0 Sight of 0. (4 5 is not sufficient) B1 5 3 5 3 5 3 Multiplies top and bottom by a crect epression. This statement is sufficient. NB 5 3 0 18 (Allow to multiply top and bottom by k( 5 3 ) ) Obtains a denominat of sight of... 5 3 5 3 with no errs seen in this epansion. May be implied by... k Note that M0A1 is not possible. The must come from a crect method. Note that if is sced there is no need to consider the numerat. e.g. Numerat = ( 5 3 ) 10 6 ( MR?) 5 3... sces A1 5 3 5 3 An attempt to multiply the numerat by 5 3 and obtain an epression of the fm p q 10 where p and q are integers. This may be implied by e.g. 10 3 4 by their final answer. (Allow attempt to multiply the numerat by k( 5 3 ) ) 10 6 3 10 5 3 Cso. F the answer as written 10 3 a statement that a = 3 and b = 10. Sce when first seen and igne any subsequent attempt to simplify. Allow 1 10 f 10 Alternative f (b) A1 A1 (1) (4) (5 marks) 1 5 3 10 3 10 6 1 10 3 10 3 10 3 : Divides multiplies top and bottom by k A1: k 10 3 : Multiplies top and bottom by 10 3 3 10 A1 A1. y 4 0, 4 y 0 0 8
Question Number y 4 4 ( 4) 0 0 y 4 y 4 ( y 4) y 10( y 4) 0 Scheme Attempts to rearrange the linear equation to y =... =... =... and attempts to fully substitute into the second equation. Marks 8 3616 0 y y 4 0 4 ( 1)( 4) 0... ( y 4)( y 3) 0 y... 0.5, 4 y 4, y 3 Sub into y 4 y 4 Sub into y 3, y 4 and 4, 0.5 y 4 4 ( 4) 0 0 : Collects terms together to produce quadratic epression = 0. The = 0 may be implied by later wk. A1: Crect three term quadratic equation in y. The = 0 may be implied by later wk. Attempt to factise and solve complete the square and solve uses a crect quadratic fmula f a 3 term quadratic. Crect answers f either both values of both values of y (possibly un-simplified) Substitutes at least one of their values of into a crect equation as far as y =... substitutes at least one of their values of y into a crect equation as far as y =... Fully crect solutions and simplified. Pairing not required. If there are any etra values of y, sce A0. Special Case: Uses y = - - 4 A1 A1 cso A1 (7 marks) 8 36 16 0 A1 4 ( 1)( 4) 0... 0.5, 4 A0 Sub into y 4 Sub into y 4is M0 y 3, y 4 and A0 4, 0.5 9
Question Number Scheme Marks 3. 3 5 y4 n n 1 : 3 1 e.g. Sight of 3 10 3 1 A1: 34 5 (oe) 3 (a) (Igne + c f this mark) 10 3 A1: 1 1 10 all on one line and no 3 + c Apply ISW here and award marks when first seen. (b) 4 5 c 4 1 5 c : n n 1. 4 1 1 e.g. Sight of 1 Do not award f integrating their answer to part (a) 4 1 A1: 4 5 4 1 A1: F fully crect and simplified answer with + c all on one line. Allow 5 c Allow 1 4 f 4 Apply ISW here and award marks when first seen. Igne spurious integral signs f all marks. 4 1 A1A1 (3) A1A1 (3) (6 marks) 10
Question Number Scheme Marks 4(i).(a) U3 4 cao B1 (b) n0 n1 U n 4 4 4... 4 0 4 (ii)(a) V3 3 k, V4 4k (b) n5 V k k 3k 4k 5k 165 n1 n F realising that all 0 terms are 4 and that the sum is required. Possible ways are 4+4+4..+4 0 4 1 1 0 4 19 0 0 4 4 (Use of a crect sum fmula with n = 0, a = 4 and d = 0 n = 0, a = 4 and l = 4) = 80 cao A1 Crect answer with no wking sces A1 1 5 k 4 k 165 1 5 k 5 k 165 15k 165 k.. May sce in (b) if clearly identified V and V as 3 4 Attempts V 5, adds their V1, V, V3, V4, V5 AND sets equal to 165 Use of a crect sum fmula with a = k, d = k and n = 5 a = k, l = 5k and n = 5 AND sets equal to 165 Attempts to solve their linear equation in k having set the sum of their first 5 terms equal to 165. Solving V 5 = 165 sces no marks. k 11 cao and cso A1 B1, B1 (1) () () (3) (8 marks) 11
Question Number 5(a) (b) b 4ac 0 e.g. 4 4( p1)( p5) 0 0 4 4( p 1)( p 5) 4 4( p 1)( p 5) 4( p1)( p5) 4 4 p 6p 5 p 6p1 0 p p p 6 1 0... Scheme : Attempts to use b 4ac with at least two of a, b c crect. May be in the quadratic fmula. Could also be, f eample, comparing equating b and 4ac. Must be considering the given quadratic equation. Inequality sign not needed f this.there must be no terms. A1: F a crect un-simplified inequality that is not the given answer Crect solution with no errs that includes an epansion of (p 1)(p 5) F an attempt to solve p 6p1 0 (not their quadratic) leading to solutions f p (do not allow attempts to factise must be using the quadratic fmula completing the square) p 3 any equivalent crect epressions e.g. p 3 8 6 3 p (May be implied by their inequalities) Discriminant must be a single number not e.g. 36-4 Allow the A1 to sce anywhere f solving the given quadratic : Chooses outside region not dependent on the previous method mark A1: p 3 8, p 3 8 equivalent e.g. p 3 8 p 3 8 6 3 6 3 p, p, 3 83 8, Allow,, a space between the answers but do not allow p 3 8 and p 3 8 (this sces A0) Apply ISW if necessary. Marks A1 A1* (3) A1 A1 A crect solution to the quadratic followed by p 3 8sces A1M0A0 3 8 p 3 8 sces A0 Allow candidates to use rather than p but must be in terms of p f the final A1 (4) (7 marks) 1
Question Number 6(a) 3 ( 4)( 3) 3 4 1 3 3 4 1 3 6 dy 3 6 d oe e.g. 3 3 1 Scheme 1 Attempt to multiply out the numerat to get a cubic with 4 terms and at least crect : Attempt to divide each term by. The powers of of at least two terms must follow from their epansion. Allow an attempt to multiply by -1 A1: Crect epression. May be un-simplified but powers of must be combined 3 e.g. not n n 1 dd: 0 Dependent on both previous method marks. 3 6 A1: oe and isw Accept 1 even 1 1 but not and not 0. If they lose the previous A1 because of an increct constant only then allow recovery here and in part (b) f a crect derivative. Marks A1 dda1 See appendi f alternatives using product/quotient rule (b) At 1, y 10 Crect value f y B1 : Substitutes = 1 into their dy 3 6 1 3.5 epression f dy/d A1 d 1 A1: 3.5 oe cso Uses their tangent gradient which must come from calculus with 1and their numerical y with a y '10' '3.5' 1 crect straight line method. If using y = m + c, this mark is awarded f crectly establishing a value f c. y 7 7 0 k y 7 7 0cso A1 (5) (5) (10 marks) 13
Question Number 7.(a) 4 y (b) Scheme 8y 9y 1 (8y 1)( y 1) 0 y... (8( ) 1)(( ) 1) 0... 1 ( y), 1 8 3 0 Allow y y y y squared "4 " not required Must be seen in part (a) F attempting to solve the given equation as a 3 term quadratic in y as a 3 term quadratic in leading to a value of y (Apply usual rules f solving the quadratic see general guidance) Allow ( any other letter) instead of y f this mark e.g. an attempt to solve 8 91 0 Both crect answers of 1 8 (oe) and 1 f y their letter but not unless ( y) is implied later : A crect attempt to find one numerical value of from their ( y) which must have come from a 3 term quadratic equation. If logs are used then they must be evaluated. A1: Both 3 and/ 0 May be implied by e.g. 3 1 0 and 1 and no 8 etra values. B1 A1 Marks A1 (1) (4) (5 marks) 14
Question Scheme Number 8(a) 3 Takes out a common fact of 9 4 (9 4 ) (4 9) crectly. 9 4 (3 )(3 ) 4 9 ( 3)( 3) 3 9 4 (3 )(3 ) 9 4 ( 3 )( 3 ) 4 9 ( 3)( 3) Cao but allow equivalents e.g. ( 3 )( 3 ) ( 3)( 3) Note: 4 3 9 (4 9) 3 3 so 9 4 3 3 3 Note: Crect wk leading to 91 1 (b) B1 A1 Marks would sce full marks would sce full marks 3 3 Allow ( ± 0) (- ± 0) instead of and - A cubic shape with one maimum and one minimum (3) (c) Any line curve drawn passing through (not touching) the igin Must be the crect shape and in all four quadrants and pass through (-1.5, 0) and (1.5, 0) (Allow (0, -1.5) and (0, 1.5) just -1.5 and 1.5 provided they are positioned crectly). Must be on the diagram (Allow f 1.5) B1: y = 14 y = 5 A (, 14), B (1, 5) B1: y = 14 and y = 5 These must be seen used in (c) Crect use of Pythagas including AB ( 1) (14 5) the square root. Must be a crect 90 epression f their A and B if a crect fmula is not quoted 9 4 B1 A1 B1 B1 (3) E.g. AB ( 1) (14 5) sces M0. However AB ( y y ) ( ) ( 1) (14 5) sces 1 1 AB 3 10 cao A1 (4) (10 marks) Special case: Use of 4 3 9 f the curve gives (-, -14) and (1, -5) in part (c). Allow this to sce a maimum of B0B0A1 as a special case in part (c) as the length AB comes from equivalent wk. 15
Question Number 9.(a) Scheme 3000 17000 ( k1) 1500 k... Use of 3000 with a crect fmula in an attempt to find k. A crect fmula could be implied by a crect answer. (k =) 11 Cso (Allow n = 11) A1 Accept crect answer only. 3000 = 17000 + 1500k k 10 is M0A0 (wrong fmula) 3000 17000 10 k 11 is A1 (crect fmula implied) 1500 Listing: All terms must be listed up to 3000 and 11 crectly identified. A solution that sces if fully crect and 0 otherwise. (b) : k S (17000 k1 1500) k k1 (17000 3000) S (17000 k 1500) k1 (17000 30500) A1: 11 11 S (17000 101500) (17000 3000) 10 S ( 17000 9 1500) 10 (17000 30500) (= 69 500 37 500) 3000 88 000 + 69 500 = 557 500 30 000 + 37 500 = 557 500 : Use of crect sum fmula with their integer n = k k 1 from part (a) where 3 k 0 and a = 17000 and d = 1500. See below f special case f using n = 0. A1: Any crect unsimplified numerical epression with n = 11 n = 10 3000 where is an integer and 3 < < 18 : Attempts to add their two values. It is dependent upon the two previous M s being sced and must be the sum of 0 terms i.e. k 0 A1: 557 500 Special Case: If they just find S 0 ( 65 000) in (b) sce the first otherwise apply the scheme. Marks A1 dda1 () (5) (7 marks) Listing: n 1 3 4 5 6 7 8 9 10 u n 17000 18500 0000 1500 3000 4500 6000 7500 9000 30500 n 11 1 13 14 15 16 17 18 19 0 u n 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 Look f a sum befe awarding marks. Award the M s as above then A f 557 500 If they sum the parts separately then apply the scheme. 16
Question Number 10(a) (b) Scheme n n 1 : A1: Two terms in crect, simplification is not required in 3 1 9 coefficients powers f ( ) c A1: All terms in crect. Simplification not required in coefficients powers and + c is not required : Sub = 4, y = 9 into f () to Sub 4, y9 into f( ) c... obtain a value f c. If no + c then M0. Use of = 9, y = 4 is M0. Accept equivalents but must be 3 1 9 f ( ) simplified e.g. f ( ) 4.5 Must be all on one line and simplified. Allow f 3 3 Marks A1A1 : Gradient of 1 Gradient of nmal is 1 dy 1 y 0 is ( m) 1 d Gradient of tangent = + A1: Gradient of tangent = + (May be A1 implied) The A1 may be implied by 1 1 3 9 3 9 0 4 4 4 6 9 0.. 1.5 1 Beware 3 9 4 3 9 4 Sets the given f their f = their changed m and not their m where m has come from y + = 0 4 equivalent crect algebraic processing (allow sign/arithmetic errs only) and attempt to solve to obtain a value f. If f they need to be solving a three term quadratic in crectly and square to obtain a value f f f this. Must be using the given mark. 3 (1.5) Accept equivalents e.g. 9 6 If any etra values are not rejected, sce A0. 1 4 1 1 etc. leads to the crect 3 9 answer and could sce A1M0(increct processing)a0 A1 A1 (5) (5) (10 marks) 17
Way Quotient Way 3 Product Way 4 Product Appendi 6(a) Attempt to multiply out the 3 ( 4)( 3) 3 4 1 numerat to get a cubic with 4 terms and at least crect 3 : Crect application of dy 3 6 4 3 4 1 quotient rule d A1: Crect derivative 3 4 6 4 3 6 4 4 4 oe e.g. 3 3 1 1 3 3 y 4 dy 1 3 d dy 3 1 3 4 d 3 6 3 6 3 oe e.g. 3 3 1 3 ( 4)( 3) 3 4 1 : Collects terms and divides by denominat. Dependent on both previous method marks. 3 6 A1: oe and isw Accept 1 even 1 1 but not and not 0. Divides one bracket by : Crect application of product rule A1: Crect derivative : Epands and collects terms. Dependent on both previous method marks. 3 6 A1: oe and isw Accept 1 even 1 1 but not and not 0. Attempt to multiply out the numerat to get a cubic with 4 terms and at least crect dy 1 3 1 3 4 1 1 3 6 4 d : Crect application of product rule A1: Crect derivative dy 3 6 3 3 6 3 d dd: Epands and collects terms Dependent on both previous method marks. 3 3 6 3 1 A1: oe e.g. and isw. Accept 1 even 1 1 but not and not 0. A1 dda1 A1 dda1 A1 dda1 18
1 3 3 y 4 3 6 1 Divides one bracket by : Epands A1: Crect epression A1 Way 5 dy 3 6 d oe e.g. 3 3 1 n n 1 dd: 0 Dependent on both previous method marks. 3 6 A1: oe and isw Accept 1 even 1 1 but not If they lose the previous A1 because of an increct constant only then allow recovery here f a crect derivative. dda1 19
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