Mark Scheme (Results) Summer 2015 - Edexcel

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Mark Scheme (Results)Summer 2015Pearson Edexcel GCEin Core Mathematics C2 (6664/01)

Edexcel and BTEC QualificationsEdexcel and BTEC qualifications are awarded by Pearson, the UK’s largest awardingbody. We provide a wide range of qualifications including academic, vocational,occupational and specific programmes for employers. For further information visit ourqualifications websites at www.edexcel.com or www.btec.co.uk. Alternatively, you canget in touch with us using the details on our contact us page atwww.edexcel.com/contactus.Pearson: helping people progress, everywherePearson aspires to be the world’s leading learning company. Our aim is to helpeveryone progress in their lives through education. We believe in every kind oflearning, for all kinds of people, wherever they are in the world. We’ve been involvedin education for over 150 years, and by working across 70 countries, in 100languages, we have built an international reputation for our commitment to highstandards and raising achievement through innovation in education. Find out moreabout how we can help you and your students at: www.pearson.com/ukSummer 2015Publications Code UA041196All the material in this publication is copyright Pearson Education Ltd 2015

General Marking Guidance All candidates must receive the same treatment. Examinersmust mark the first candidate in exactly the same way as they markthe last. Mark schemes should be applied positively. Candidates must berewarded for what they have shown they can do rather than penalisedfor omissions. Examiners should mark according to the mark scheme notaccording to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the markscheme should be used appropriately. All the marks on the mark scheme are designed to be awarded.Examiners should always award full marks if deserved, i.e. if theanswer matches the mark scheme. Examiners should also be preparedto award zero marks if the candidate’s response is not worthy of creditaccording to the mark scheme. Where some judgement is required, mark schemes will providethe principles by which marks will be awarded and exemplification maybe limited. Crossed out work should be marked UNLESS the candidate hasreplaced it with an alternative response.

PEARSON EDEXCEL GCE MATHEMATICSGeneral Instructions for Marking1. The total number of marks for the paper is 752. The Edexcel Mathematics mark schemes use the following types of marks: M marks: Method marks are awarded for ‘knowing a method and attempting toapply it’, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) markshave been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided.3. AbbreviationsThese are some of the traditional marking abbreviations that will appear in the markschemes. bod – benefit of doubt ft – follow through the symbol cao – correct answer only cso - correct solution only. There must be no errors in this part of the question toobtain this mark isw – ignore subsequent working awrt – answers which round to SC: special case oe – or equivalent (and appropriate) d or dep – dependent indep – independent dp decimal places sf significant figures The answer is printed on the paper or ag- answer given will be used for correct ftor d The second mark is dependent on gaining the first mark4. All A marks are ‘correct answer only’ (cao.), unless shown, for example, as A1 ft toindicate that previous wrong working is to be followed through. After a misreadhowever, the subsequent A marks affected are treated as A ft, but manifestly absurdanswers should never be awarded A marks.

5. For misreading which does not alter the character of a question or materially simplifyit, 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 crossedout. If either all attempts are crossed out or none are crossed out, mark all theattempts and score the highest single attempt.7. Ignore wrong working or incorrect statements following a correct answer.

General Principles for Core Mathematics Marking(But note that specific mark schemes may sometimes override these general principles).Method mark for solving 3 term quadratic:1. Factorisation( x 2 bx c) ( x p)( x q), where pq c , leading to x (ax 2 bx c) (mx p)(nx q), where pq c and mn a , leading to x 2. FormulaAttempt to use the correct formula (with values for a, b and c).3. Completing the squareSolving x bx c 0 :22b x q c 0, q 0 , leading to x 2 Method marks for differentiation and integration:1. DifferentiationPower of at least one term decreased by 1. ( x xnn 1)2. IntegrationPower of at least one term increased by 1. ( x xnn 1)

Use of a formulaWhere a method involves using a formula that has been learnt, the advice given in recentexaminers’ 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 smallerrors in the substitution of values.Where the formula is not quoted, the method mark can be gained by implication from correctworking with values, but may be lost if there is any mistake in the working.Exact answersExaminers’ reports have emphasised that where, for example, an exact answer is asked for,or working with surds is clearly required, marks will normally be lost if the candidate resortsto using rounded decimals.7

May 20156664 Core Mathematics C2Mark SchemeQuestionNumberSchemeMarks10x 2 4 1. 10 1 10 1 2210 29 x 28 x . 1 4 2 4 Way 1For either the x term or the x 2 termincluding a correct binomial coefficientwith a correct power of xM1First term of 1024 B1A1Either 1280x or 720x (Allow -1280x here)2 1024 1280 x 720 x2Way 2Both 1280x and 720x 2 (Do not allow -1280xhere)A1[4] x 10 9 x 2 x k 2 2 1 10 4 82 8 1024(1 .)M1 1024 1280 x 720 x2B1A1 A1[4]10NotesM1: For either the x term or the x term having correct structure i.e. a correct binomial coefficient in any form with the2correct power of x. Condone sign errors and condone missing brackets and allow alternative forms for binomialcoefficients e.g.10 10 10 or even or 10. The powers of 2 or of ¼ may be wrong or missing. 1 1 C1 or B1: Award this for 1024 when first seen as a distinct constant term (not 1024x0) and not 1 1024A1: For one correct term in x with coefficient simplified. Either -1280x or 720x 2 ( allow -1280x here)2Allow 720x2 x to come from with no negative sign. So use of sign throughout could give M1 B1 A1 A0 4 A1: For both correct simplified terms i.e. -1280x and 720x 2 (Do not allow -1280x here)Allow terms to be listed for full marks e.g. 1024, 1280 x, 720 x2N.B. If they follow a correct answer by a factor such as 512 640 x 360 x2 then iswTerms may be listed. Ignore any extra terms.Notes for Way 2M1: Correct structure for at least one of the underlined terms. i.e. a correct binomial coefficient in any form with the correctpower of x. Condone sign errors and condone missing brackets and allow alternative forms for binomial coefficientse.g.10 10 10 or even or 10. k may even be 0 or 2k may not be seen. Just consider the bracket for 1 1 C1 or this mark.B1: Needs 1024(1 . To become 1024A1, A1: as before8

QuestionNumberSchemeWay 1( x m2) ( y 1) k , k 02 (a)2Way 2x y m4 x 2 y c 0M14 ( 5) 4 4 2 5 c 0M1A122Attempts to use r (4 2) ( 5 1)2Marks22222x2 y 2 4 x 2 y 15 0Obtains ( x 2)2 ( y 1)2 20(3)N.B. Special case: ( x 2) ( y 1) 20 is not a circle equation but earns M0M1A02(b)Way 12Gradient of radius from centre to (4, -5) -2(must be correct)B11their numerical gradient of radiusEquation of tangent is ( y 5) ' 12 '( x 4)Tangent gradient M1So equation is x – 2y –14 0 (or 2y – x 14 0 or other integer multiples of this answer)b)Way 2Quotes xx yy 2( x x ) ( y y ) 15 0 and substitutes (4, -5)4 x 5 y 2( x 4) ( y 5) 15 0 so 2x – 4y – 28 0 (or alternatives as in Way 1)b)Way 3Use differentiation to find expression for gradient of circleEither 2( x 2) 2( y 1)dydx2 0 or states y 1 20 ( x 2) sody( x 2) dx20 ( x 2)2Substitute x 4, y -5 after valid differentiation to give gradient M1A1(4)B1M1,M1A1(4)B1M1M1 A1Then as Way 1 above ( y 5) ' '( x 4) so x – 2y –14 012(4)[7]Notes(a) M1: Uses centre to write down equation of circle in one of these forms. There may be sign slips as shown.2M1: Attempts distance between two points to establish r (independent of first M1)- allow one sign slip only using distanceformula with -5 or -1, usually (– 5 – 1) in 2nd bracket. Must not identify this distance as diameter.This mark may alternatively (e.g. way 2)be given for substituting (4, -5) into a correct circle equation with one unknownCan be awarded for r 20 or for r 2 20 stated or implied but not for r 2 20 or r 20 or r 52A1: Either of the answers printed or correct equivalent e.g. ( x 2)2 ( y 1) 2 (2 5) 2 is A1 but 2 5 (no bracket) is A0unless there is recoveryAlso ( x 2)2 ( y ( 1))2 (2 5) 2 may be awarded M1M1A1as a correct equivalent.N.B. ( x 2) ( y 1) 40 commonly arises from one sign error evaluating r and earns M1M1A0(b) Way 1:B1: Must be correct answer -2 if evaluated (otherwise may be implied by the following work)M1: Uses negative reciprocal of their gradientM1: Uses y y1 m( x x1 ) with (4,-5) and their changed gradient or uses y mx c and (4, -5) with theirchanged gradient (not gradient of radius) to find cA1: answers in scheme or multiples of these answers (must have “ 0” ). NB Allow 1x – 2y –14 0N.B. ( y 5) ' 12 '( x 4) following gradient of is ½ after errors leads to x – 2y –14 0 but is worth B0M0M0A022Way 2: Alternative method (b) is rare.Way 3: Some may use implicit differentiation to differentiate- others may attempt to make y the subject and use chain ruleB1: the differentiation must be accurate and the algebra accurate too. Need to take (-) root not ( )root in the alternativeM1: Substitutes into their gradient function but must follow valid accurate differentiationM1: Must use “their” tangent gradient and y 5 m( x 4) but allow over simplified attempts at differentiation for this mark.A1: As in Way 19

QuestionNumberSchemeMarks3.Way 1 (a)f ( x) 6 x3 3x2 Ax BAttempting f (1) 45 or f ( 1) 45f ( 1) 6 3 A B 45 or 3 A B 45 B A 48 * (allow 48 B – A)Way 1 (b)Attempting f ( 12 ) 0M1A1 * cso(2)M16 12 3 12 A 12 B 0 or 12 A B 0 or A 2B3Way 2 (a)Way 2 (b)2A1 o.e.Solve to obtain B 48 and A 96M1 A1Long Division(6 x3 3x2 Ax B) ( x 1) 6 x2 px q and sets remainder 45M1Quotient is 6 x 2 3x ( A 3) and remainder is B – A – 3 45 so B – A 48 *A1*(6 x 3x Ax B) (2 x 1) 3x px q and sets remainder 0M1AAand remainder is B 022A132(4)22Quotient is 3 x Then Solve to obtain B 48 and A 96 as in scheme above (Way 1)(c)Obtain 3x2 48 , x2 16 , 6x2 96 , 3x2 A A M1 A1B 2 22 , 3x B , x or x as2 6 3 B1ftfactor or as quotient after division by (2x 1) . Division by (x 4) or (x-4) see below 22222Factorises 3x 48 , x 16 , 48 3x , 16 x or 6 x 96 M1 3 (2x 1)(x 4)(x 4) (if this answer follows from a wrong A or B then award A0)A1csoisw if they go on to solve to give x 4, -4 and -1/2(3) [9]Notes(a) Way 1: M1: 1 or –1 substituted into f(x) and expression put equal to 45A1*: Answer is given. Must have substituted 1 and put expression equal to 45.Correct equation with powers of 1 evaluated and conclusion with no errors seen.Way 2: M1: Long division as far as a remainder which is set equal to 45A1*: See correct quotient and correct remainder and printed answer obtained with no errors(b) Way 1: M1: Must see f ( 12 ) and “ 0” unless subsequent work implies this.A1: Give credit for a correct equation even unsimplified when first seen, then isw.A correct equation implies M1A1.M1: Attempts to solve the given equation from part (a) and their simplified or unsimplified linearequation in A and B from part (b) as far as A . or B .(must eliminate one of the constants butalgebra need not be correct for this mark). May just write down the correct answers.A1: Both A and B correctWay 2: M1: Long division as far as a remainder which is set equal to 0A1: See correct quotient and correct remainder put equal to 0M1A1: As in Way 1There may be a mixture of Way 1 for (a) and Way 2 for (b) or vice versa.(c) B1: May be written straight down or from long division, inspection, comparing coefficients or pairing termsM1: Valid attempt to factorise a listed quadratic (see general notes) so 3x 16 x 3 could get M1A0A1cso: (Cannot be awarded if A or B is wrong) Needs the answer in the scheme or -3 (2x 1)(4 x)(4 x) orequivalent but factor 3 must be shown and there must be all the terms together with brackets. 22Way 2: A minority might divide by (x- 4) or (x 4) obtaining 6 x 27 x 12 or 6 x 21x 12 22They then need to factorise 6 x 27 x 12 or 6 x 21x 12 for B1for M1Then A1cso as beforeSpecial cases:If they write down f(x) 3 (2x 1)(x 4)(x - 4) with no working, this is B1 M1 A1But if they give f(x) (2x 1)(x 4)(x - 4) with no working (from calculator?) give B1M0A0And f(x) (2x 1)(3x 12)(x 4) or f(x) (6x 3)(x 4)(x 4) or f(x) (2x 1)(x 4)(3x 12) is B1M1A010

QuestionNumber4.(a)SchemeIn triangle OCD complete method used to find angle COD so:µ Either cos COD82 82 72or uses COD 2 arcsin 3.58 oe2 8 8( COD 0.9056(331894) )(b) 0.906 (3sf) *Uses s 8 for any in radians or (c)Marks " COD "2 360so COD M1accept awrt 0.906A1 * 2 8 for any in degreesM1 awrt 1.12 or 2 awrt 2.24 and Perimeter 23 ( 16 )accept awrt 40.9 (cm)Either Way 1: (Use of Area of two sectors area of triangle)Area of triangle 12 8 8 sin 0.906 (or 25.1781155 accept awrt 25.2)or12M1A1 8 7 sin1.118 or 12 7 h after h calculated from correct Pythagoras or trig.2Area of sector 12 8 "1.117979732" (or 35.77535142(2)(3)M1accept awrt 35.8 )M1Total Area Area of two sectors area of triangle awrt 96.7 or 96.8 or 96.9 ( cm 2 )A1(3)Or Way 2: (Use of area of semicircle – area of segment)Area of semi-circle 12 8 8 ( or 100.5)Area of segment 12M182 ("0.906" sin"0.906") ( or 3.807)M1A1So area required awrt 96.7 or 96.8 or 96.9 ( cm 2 )(3)[8]Notes(a) M1: Either use correctly quoted cosine rule – may quote as 7 8 8 2 8 8cos .Or split isosceles triangle into two right angled triangles and use arcsin or longer methods using Pythagorasand arcos (i.e. 2 arccos 3.58 ). There are many ways of showing this result.222Must conclude that COD A1*: (NB this is a given answer) If any errors or over-approximation is seen this is A0. It needs correct work leading tostated answer of 0.906 or awrt 0.906 for A1. The cosine of COD is equal to 79/128 or awrt 0.617. Use of 0.62 (2sf)does not lead to printed answer. They may give 51.9 in degrees then convert to radians. This is fine.The minimal solution 7 8 8 2 8 8cos . 0.906 (with no errors seen) can have M1A1 but errorsrearranging result in M1A0222(b) M1: Uses formula for arc length with r 8 and any angle i.e. s 8 if working in rads or s 2 8 in degrees360(If the formula is quoted with r the 8 may be implied by the value of their r )M1: Uses angles on straight line (or other geometry) to find angle BOC or AOD and usesPerimeter 23 arc lengths BC and AD (may make a slip – in calculation or miscopying)A1: correct work leading to awrt 40.9 not 40.8 (do not need to see cm) This answer implies M1M1A1(c) Way 1: M1: Mark is given for correct statement of area of triangle 12 8 8 sin 0.906 (must use correctangle) or for correct answer (awrt 25.2) Accept alternative correct methods using Pythagoras and ½ base heightM1: Mark is given for formula for area of sector(BOC AOD) not COD . May use A 81228 "1.117979732 " with r 8 and their angle BOC or AOD or2if working in degrees360A1: Correct work leading to awrt 96.7, 96.8 or 96.9 (This answer implies M1M1A1)NB. Solution may combine the two sectors for part (b) and (c) and so might use 2 BOC rather than BOCWay 2: M1: Mark is given for correct statement of area of semicircleM1: Mark is given for formula for area of segment1212 8 8 or for correct answer 100.58 (" 0.906" sin" 0.906") with r 8 or 3.81 A1: As in Way 1211

QuestionNumber5.(i)(a)(Way 1)SchemeMark (a) and (b) togethera ar 34 ora (1 r 2 )(1 r ) 34 orEliminate a to give (1 r )(1 r ) (and so(b)Marks1781a(r 2 1)(r 1)ora 34 ;1 r2 1 r34162 162aM1.(not a cubic)aA18r 6481 and) r 9 only2Substitute theirr 89(4)bM1bA1( 0 r 1) to give a a 18(Way 2)Part (b)firstThen part(a) again(2)Eliminate r to givegives34 aa 1 abM1162bA1a 18 or 306 and rejects 306 to give a 18Substitute a 18 to give r aM1r 89(ii)aA142(1 76 )M1n1 76to obtain SoSo n B1; B1 290(For trial and improvement approach see notes below)6 n42( 76 )n ( 294) or equivalent e.g. ( 76 )n ( 2944 ) or ( 7 ) ( 147 )4log"( 294)"4)" or equivalent but must be log of positive quantityor log 6 "( 29467log( 7 )A1M1A1(i.e. n 27.9 ) so n 28(4)Notes(a) B1: Writes a correct equation connecting a and r and 34 (allow equivalent equations – may be implied)B1: Writes a correct equation connecting a and r and 162 (allow equivalent equation – may be implied)1734Way 1: aM1: Eliminates a correctly for these two equations to give (1 r )(1 r ) or (1 r )(1 r ) or equivalent –81162not a cubic – should have factorized (1 - r) to give a correct quadraticaA1: Correct value for r. Accept 0.8 recurring or 8/9 (not 0.889) Must only have positive value.(i)bM1: Substitutes their r (0 r 1) into a correct formula to give value for a. Can be implied by a 18bA1: must be 18 (not answers which round to 18)Way 2: Finds a first - B1, B1: As before then award the (b) M and A marks before the (a) M and A marks34 aa 1 bM1: Eliminates r correctly to giveor a 2 324a 5508 0 or equivalenta162bA1: Correct value for a so a 18 only. (Only award after 306 has been rejected)aM1: Substitutes their 18 to give r aA1: r 89 only(ii) M1: Allow n or n – 1 and any symbols from “ ”, “ ”, or “ ” etc A1 : Must be power n ( not n – 1) with any symbolM1: Uses logs correctly on( 76 ) n or ( 76 ) n not on (36) n to get as far as n Allow any symbolA1: n 28 cso (any errors with inequalities earlier e.g. failure to reverse the inequality when dividing by the negativelog( 76 ) or any contradictory statements must be penalised here) Those with equals throughout may gain this mark if theyfollow 27.9 by n 28. Just n 28 without mention of 27.9 is only allowed following correct inequality work.Special case: Trial and improvement: Gives n 28 as S awrt 290.1 (M1A1)and when n 27 S (awrt) 289 so n 28 (M1A1)–n 28 with no working is M1A0M0A0 and insufficient accuracy is M1A0M1A0Uses nth term instead of sum of n terms – over simplified – do not treat as misread – award 0/412

QuestionNumberSchemeMarksMay mark (a) and (b) together3Expands to give 10 x 2 20 x6. (a)B1M1 A1ft10 " 52 " "20" x( c )x " 52 "22Integrates to give5Simplifies to 4 x 2 10 x 2 ( c )Use limits 0 and 4 either way round on their integrated function (may only see 4 substituted)(b)4needs at least one of the previous M marks for this to be awarded9 ydx ydx(So area 0)(4)M1dM1

Pearson Edexcel GCE in Core Mathematics C2 (6664/01) Edexcel and BTEC Qualifications Edexcel 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 www.edexcel.com or www.btec.co.uk .

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