PhysicsAndMathsTutor.com. Mark Scheme (Results) Summer Pearson Edexcel GCE in Core Mathematics C2 (6664/01)

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1 Mak Scheme (Results) Summe 05 Peason Edexcel GCE in Coe Mathematics C (6664/0)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications ae awaded by Peason, the UK s lagest awading body. We povide a wide ange of qualifications including academic, vocational, occupational and specific pogammes fo employes. Fo futhe infomation visit ou qualifications websites at o Altenatively, you can get in touch with us using the details on ou contact us page at Peason: helping people pogess, eveywhee Peason aspies to be the wold s leading leaning company. Ou aim is to help eveyone pogess in thei lives though education. We believe in evey kind of leaning, fo all kinds of people, wheeve they ae in the wold. We ve been involved in education fo ove 50 yeas, and by woking acoss 0 counties, in 00 languages, we have built an intenational eputation fo ou commitment to high standads and aising achievement though innovation in education. Find out moe about how we can help you and you students at: Summe 05 Publications Code UA0496 All the mateial in this publication is copyight Peason Education Ltd 05

3 Geneal Making Guidance All candidates must eceive the same teatment. Examines must mak the fist candidate in exactly the same way as they mak the last. Mak schemes should be applied positively. Candidates must be ewaded fo what they have shown they can do athe than penalised fo omissions. Examines should mak accoding to the mak scheme not accoding to thei peception of whee the gade boundaies may lie. Thee is no ceiling on achievement. All maks on the mak scheme should be used appopiately. All the maks on the mak scheme ae designed to be awaded. Examines should always awad full maks if deseved, i.e. if the answe matches the mak scheme. Examines should also be pepaed to awad zeo maks if the candidate s esponse is not wothy of cedit accoding to the mak scheme. Whee some judgement is equied, mak schemes will povide the pinciples by which maks will be awaded and exemplification may be limited. Cossed out wok should be maked UNLESS the candidate has eplaced it with an altenative esponse.

4 PEARSON EDEXCEL GCE MATHEMATICS Geneal Instuctions fo Making. The total numbe of maks fo the pape is 5. The Edexcel Mathematics mak schemes use the following types of maks: M maks: Method maks ae awaded fo knowing a method and attempting to apply it, unless othewise indicated. A maks: Accuacy maks can only be awaded if the elevant method (M) maks have been eaned. B maks ae unconditional accuacy maks (independent of M maks) Maks should not be subdivided.. Abbeviations These ae some of the taditional making abbeviations that will appea in the mak schemes. bod benefit of doubt ft follow though the symbol will be used fo coect ft cao coect answe only cso - coect solution only. Thee must be no eos in this pat of the question to obtain this mak isw ignoe subsequent woking awt answes which ound to SC: special case oe o equivalent (and appopiate) d o dep dependent indep independent dp decimal places sf significant figues The answe is pinted on the pape o ag- answe given o d The second mak is dependent on gaining the fist mak 4. All A maks ae coect answe only (cao.), unless shown, fo example, as A ft to indicate that pevious wong woking is to be followed though. Afte a misead howeve, the subsequent A maks affected ae teated as A ft, but manifestly absud answes should neve be awaded A maks.

5 5. Fo miseading which does not alte the chaacte of a question o mateially simplify it, deduct two fom any A o B maks gained, in that pat of the question affected. 6. If a candidate makes moe than one attempt at any question: If all but one attempt is cossed out, mak the attempt which is NOT cossed out. If eithe all attempts ae cossed out o none ae cossed out, mak all the attempts and scoe the highest single attempt.. Ignoe wong woking o incoect statements following a coect answe.

6 Geneal Pinciples fo Coe Mathematics Making (But note that specific mak schemes may sometimes oveide these geneal pinciples). Method mak fo solving tem quadatic:. Factoisation ( x bx c) ( x p)( x q), whee pq c, leading to x = ( ax bx c) ( mx p)( nx q), whee pq c and mn a, leading to x =. Fomula Attempt to use the coect fomula (with values fo a, b and c).. Completing the squae b Solving x bx c 0 : x q c 0, q 0, leading to x = Method maks fo diffeentiation and integation:. Diffeentiation Powe of at least one tem deceased by. ( x n x n ). Integation Powe of at least one tem inceased by. ( x n x n )

7 Use of a fomula Whee a method involves using a fomula that has been leant, the advice given in ecent examines epots is that the fomula should be quoted fist. Nomal making pocedue is as follows: Method mak fo quoting a coect fomula and attempting to use it, even if thee ae small eos in the substitution of values. Whee the fomula is not quoted, the method mak can be gained by implication fom coect woking with values, but may be lost if thee is any mistake in the woking. Exact answes Examines epots have emphasised that whee, fo example, an exact answe is asked fo, o woking with suds is clealy equied, maks will nomally be lost if the candidate esots to using ounded decimals.

8 May Coe Mathematics C Mak Scheme Question Numbe. Way 0 Scheme x x x x 0x Way 0 x x k 09 x (...) 04 80x 0x Eithe 80x o Fo eithe the x tem o the x tem including a coect binomial coefficient M with a coect powe of x Fist tem of 04 B 0x (Allow +-80x hee) Both 80x and 0x (Do not allow +-80x hee) A A M Maks [4] BA A [4] Notes M: Fo eithe the x tem o the x tem having coect stuctue i.e. a coect binomial coefficient in any fom with the coect powe of x. Condone sign eos and condone missing backets and allow altenative foms fo binomial 0 0 coefficients e.g. 0 C o o even o 0. The powes of o of ¼ may be wong o missing. B: Awad this fo 04 when fist seen as a distinct constant tem (not 04x 0 ) and not + 04 A: Fo one coect tem in x with coefficient simplified. Eithe -80x o 0x ( allow +-80x hee) x Allow 0x to come fom with no negative sign. So use of + sign thoughout could give M B A A0 4 A: Fo both coect simplified tems i.e. -80x and 0x (Do not allow +-80x hee) Allow tems to be listed fo full maks e.g. 04, 80 x, 0x N.B. If they follow a coect answe by a facto such as 5 640x 60x then isw Tems may be listed. Ignoe any exta tems. Notes fo Way M: Coect stuctue fo at least one of the undelined tems. i.e. a coect binomial coefficient in any fom with the coect powe of x. Condone sign eos and condone missing backets and allow altenative foms fo binomial coefficients 0 0 e.g. 0 C o o even o 0. k may even be 0 o k may not be seen. Just conside the backet fo this mak. B: Needs 04(. To become 04 A, A: as befoe 8

9 Question Numbe Scheme Maks Way Way (a) ( x m ) ( y ) k, k > 0 x y m 4x y c 0 M (b) Way Attempts to use Obtains (4 ) ( 5 ) ( x ) ( y ) 0 4 ( 5) c 0 M A x y 4x y 5 0 N.B. Special case: ( x ) ( y ) 0 is not a cicle equation but eans M0MA0 Gadient of adius fom cente to (4, -5) = - (must be coect) B Tangent gadient = M thei numeical gadient of adius Equation of tangent is ( y5) ' '( x 4) M So equation is x y 4 = 0 (o y x +4 = 0 o othe intege multiples of this answe) A (4) b)way Quotes xx yy ( x x) ( y y) 5 0 and substitutes (4, -5)_ B 4x 5y ( x 4) ( y 5) 5 0 so x 4y 8 = 0 (o altenatives as in Way ) M,MA (4) b)way Use diffeentiation to find expession fo gadient of cicle dy d y ( x ) Eithe ( x ) ( y ) 0 o states y 0 ( x ) so dx dx 0 ( x ) B Substitute x = 4, y = -5 afte valid diffeentiation to give gadient = Then as Way above ( y5) ' '( x 4) so x y 4 = 0 M A (4) [] Notes (a) M: Uses cente to wite down equation of cicle in one of these foms. Thee may be sign slips as shown. M: Attempts distance between two points to establish (independent of fist M)- allow one sign slip only using distance fomula with -5 o -, usually ( 5 ) in nd backet. Must not identify this distance as diamete. This mak may altenatively (e.g. way )be given fo substituting (4, -5) into a coect cicle equation with one unknown Can be awaded fo 0 o fo 0 stated o implied but not fo 0 o = 0 o 5 A: Eithe of the answes pinted o coect equivalent e.g. ( x ) ( y ) ( 5) is A but unless thee is ecovey Also ( x ) ( y ( )) ( 5) may be awaded MMAas a coect equivalent. N.B. ( x ) ( y ) 40 commonly aises fom one sign eo evaluating and eans MMA0 (b) Way : B: Must be coect answe - if evaluated (othewise may be implied by the following wok) M: Uses negative ecipocal of thei gadient y y m( x x ) with (4,-5) and thei changed gadient o uses y = mx + c and (4, -5) with thei M () 5 (no backet) is A0 M: Uses changed gadient (not gadient of adius) to find c A: answes in scheme o multiples of these answes (must have = 0 ). NB Allow x y 4 = 0 N.B. ( y 5) ' '( x 4) following gadient of is ½ afte eos leads to x y 4 = 0 but is woth B0M0M0A0 Way : Altenative method (b) is ae. Way : Some may use implicit diffeentiation to diffeentiate- othes may attempt to make y the subject and use chain ule B: the diffeentiation must be accuate and the algeba accuate too. Need to take (-) oot not (+)oot in the altenative M: Substitutes into thei gadient function but must follow valid accuate diffeentiation M: Must use thei tangent gadient and y 5 m( x 4) but allow ove simplified attempts at diffeentiation fo this mak. A: As in Way 9

10 Question Numbe Scheme Maks. f ( x) 6x x Ax B Way (a) Attempting f () = 45 o f ( ) 45 M f( ) 6 A B 45 o A B 45 B A 48 * (allow 48 = B A) A * cso () Way (b) Attempting f ( ) 0 M Way (a) A 6 B 0 o A B o A = B A o.e. Solve to obtain B = 48 and A = 96 0 Long Division (6x x Ax B) ( x ) 6x px q and sets emainde = 45 0 M A (4) Quotient is 6x x ( A ) and emainde is B A =45 so B A = 48 * A* Way (b) (6x x Ax B) (x ) x px q and sets emainde = 0 M A A Quotient is x and emainde is B = 0 A Then Solve to obtain B = 48 and A = 96 as in scheme above (Way ) M A (c) Obtain 48, 6, 6 96, A A B x x x x, x B, x o x 6 as facto o as quotient afte division by (x + ). Division by (x+4) o (x-4) see below Bft x 48, x 6, 48 x, 6 x o 6x 96 M Factoises = (x + )(x + 4)(x 4) (if this answe follows fom a wong A o B then awad A0) Acso isw if they go on to solve to give x = 4, -4 and -/ () [9] Notes (a) Way : M: o substituted into f(x) and expession put equal to ±45 A*: Answe is given. Must have substituted and put expession equal to +45. Coect equation with powes of evaluated and conclusion with no eos seen. Way : M: Long division as fa as a emainde which is set equal to ±45 A*: See coect quotient and coect emainde and pinted answe obtained with no eos (b) Way : M: Must see f ( ) and = 0 unless subsequent wok implies this. A: Give cedit fo a coect equation even unsimplified when fist seen, then isw. A coect equation implies MA. M: Attempts to solve the given equation fom pat (a) and thei simplified o unsimplified linea equation in A and B fom pat (b) as fa as A =... o B =...(must eliminate one of the constants but algeba need not be coect fo this mak). May just wite down the coect answes. A: Both A and B coect Way : M: Long division as fa as a emainde which is set equal to 0 A: See coect quotient and coect emainde put equal to 0 MA: As in Way Thee may be a mixtue of Way fo (a) and Way fo (b) o vice vesa. (c) B: May be witten staight down o fom long division, inspection, compaing coefficients o paiing tems x6 x could get MA0 M: Valid attempt to factoise a listed quadatic (see geneal notes) so Acso: (Cannot be awaded if A o B is wong) Needs the answe in the scheme o - (x+)(4+x)(4 x) o equivalent but facto must be shown and thee must be all the tems togethe with backets. 6 x x o 6x x fo B Way : A minoity might divide by (x- 4) o (x + 4) obtaining They then need to factoise 6 x x o 6x x fo M Then Acso as befoe Special cases: If they wite down f(x) = (x+)(x+4)(x - 4) with no woking, this is B M A But if they give f(x) = (x+)(x+4)(x - 4) with no woking (fom calculato?) give BM0A0 And f(x) = (x + )(x + )(x 4) o f(x) = (6x + )(x + 4)(x 4) o f(x) = (x + )(x + 4)(x ) is BMA0 M

11 Question Numbe 4.(a) (b) (c) Scheme In tiangle OCD complete method used to find angle COD so: Eithe µ cos COD o uses COD acsin 8 oe so COD 88 ( COD (894) ) = (sf) * accept awt Maks Uses s = 8 fo any in adians o 8 fo any in degees 60 M " COD" awt. o awt.4 and Peimete = +( 6 ) M accept awt 40.9 (cm) A () Eithe Way : (Use of Aea of two sectos + aea of tiangle) Aea of tiangle = 88 sin (o accept awt 5.)o 8 sin.8 o h afte h calculated fom coect Pythagoas o tig. M Aea of secto = (o accept awt 5.8 ) M 8 ".99" Total Aea = Aea of two sectos + aea of tiangle =awt 96. o 96.8 o 96.9 ( cm ) O Way : (Use of aea of semicicle aea of segment) Aea of semi-cicle = 8 8 ( o 00.5) M Aea of segment = 8 ("0.906" sin"0.906") M So aea equied = awt 96. o 96.8 o 96.9 ( cm ) A () [8] Notes (a) M: Eithe use coectly quoted cosine ule may quote as cos... O split isosceles tiangle into two ight angled tiangles and use acsin o longe methods using Pythagoas and acos (i.e. ). Thee ae many ways of showing this esult. accos COD.5 8 Must conclude that A*: (NB this is a given answe) If any eos o ove-appoximation is seen this is A0. It needs coect wok leading to stated answe of o awt fo A. The cosine of COD is equal to 9/8 o awt 0.6. Use of 0.6 (sf) does not lead to pinted answe. They may give 5.9 in degees then convet to adians. This is fine. The minimal solution cos (with no eos seen) can have MA but eos eaanging esult in MA0 (b) M: Uses fomula fo ac length with = 8 and any angle i.e. s = 8 if woking in ads o s = 8in degees 60 (If the fomula is quoted with the 8 may be implied by the value of thei ) M: Uses angles on staight line (o othe geomety) to find angle BOC o AOD and uses Peimete = + ac lengths BC and AD (may make a slip in calculation o miscopying) A: coect wok leading to awt 40.9 not 40.8 (do not need to see cm) This answe implies MMA (c) Way : M: Mak is given fo coect statement of aea of tiangle 88 sin (must use coect angle) o fo coect answe (awt 5.) Accept altenative coect methods using Pythagoas and ½ base height M: Mak is given fo fomula fo aea of secto (BOC + AOD) not COD. May use A with = 8 and thei angle BOC o AOD o 8 ".99" 8 if woking in degees 60 A: Coect wok leading to awt 96., 96.8 o 96.9 (This answe implies MMA) NB. Solution may combine the two sectos fo pat (b) and (c) and so might use BOC athe than BOC Way : M: Mak is given fo coect statement of aea of semicicle 8 8 o fo coect answe 00.5 M: Mak is given fo fomula fo aea of segment 8 ("0.906" sin"0.906") with = 8 o.8 A: As in Way M A * A () ()

12 Question Numbe 5.(i) Mak (a) and (b) togethe a( ) (a) a a 4 o 4 ( ) (Way ) (b) (Way ) Pat (b) fist Then pat (a) again Eliminate a to give ( )( ) o 8 (and so Substitute thei a = 8 and) only Eliminate to give o 8 ( 0 < < ) to give a = 9 Scheme a ( ) a 4 ; 6 ( ) 4.. (not a cubic) 6 Maks B; B am aa (4) bm ba () 4 a a bm a 6 gives a = 8 o 06 and ejects 06 to give a = 8 ba Substitute a = 8 to give = 8 9 aa (ii) 6 n 4( ) 90 6 to obtain So ( ) ( ) 6 n 4 94 o equivalent e.g. (Fo tial and impovement appoach see notes below) ( ) ( ) n o ( ) ( ) 6 n 4 4 log"( 94)" 4 So n > o log 6 6"( 94)" o equivalent but must be log of positive quantity M log( ) A (i.e. n >.9 ) so n = 8 (4) Notes (i) (a) B: Wites a coect equation connecting a and and 4 (allow equivalent equations may be implied) B: Wites a coect equation connecting a and and 6 (allow equivalent equation may be implied) 4 Way : am: Eliminates a coectly fo these two equations to give ( )( ) o ( )( ) o equivalent 8 6 not a cubic should have factoized ( - ) to give a coect quadatic aa: Coect value fo. Accept 0.8 ecuing o 8/9 (not 0.889) Must only have positive value. bm: Substitutes thei (0 < < ) into a coect fomula to give value fo a. Can be implied by a = 8 ba: must be 8 (not answes which ound to 8) Way : Finds a fist - B, B: As befoe then awad the (b) M and A maks befoe the (a) M and A maks bm: Eliminates coectly to give 4 a a o a 4a o equivalent a 6 ba: Coect value fo a so a = 8 only. (Only awad afte 06 has been ejected) am: Substitutes thei 8 to give = 8 aa: only 9 (ii) M: Allow n o n and any symbols fom >, <, o = etc A : Must be powe n ( not n ) with any symbol M: Uses logs coectly on () n o () n not on (6) n to get as fa as n Allow any symbol 6 6 A: n = 8 cso (any eos with inequalities ealie e.g. failue to evese the inequality when dividing by the negative log( ) o any contadictoy statements must be penalised hee) Those with equals thoughout may gain this mak if they 6 follow.9 by n=8. Just n = 8 without mention of.9 is only allowed following coect inequality wok. Special case: Tial and impovement: Gives n = 8 as S = awt 90. (MA)and when n = S = (awt) 89 so n = 8 (MA) n = 8 with no woking is MA0M0A0 and insufficient accuacy is MA0MA0 Uses nth tem instead of sum of n tems ove simplified do not teat as misead awad 0/4 am M A

13 Question Numbe Scheme Maks May mak (a) and (b) togethe 6. (a) Expands to give 0x 0x B 0 " 5 Integates to give " "0" x M Aft x (+ c ) 5 " " 5 Acao Simplifies to 4x 0x (+ c ) (4) (b) Use limits 0 and 4 eithe way ound on thei integated function (may only see 4 substituted) M Use limits 4 and 9 eithe way ound on thei integated function dm Obtains eithe o 94 needs at least one of the pevious M maks fo this to be awaded A (So aea = 4 9 ydx ydx ) i.e. + 94, = ddm,a (5) [9] Notes (a) B: Expands the backet coectly M: Coect integation pocess on at least one tem afte attempt at multiplication. (Follow coect expansion o k one slip esulting in 0x 0xwhee k may be o 5 o esulting in 0x Bx, whee B may be o 5) 5 x x 5 x x So x o x o x and/o x. 5 A: Coect unsimplified follow though fo both tems of thei integation. Does not need (+ c) A: Must be simplified and coect allow answe in scheme o 4x 0x. Does not need (+ c ) (b) M: (does not depend on fist method mak) Attempt to substitute 4 into thei integal (howeve obtained but must not be diffeentiated) o seeing thei evaluated numbe (usually ) is enough do not need to see minus zeo. dm: (depends on fist method mak in (a)) Attempt to subtact eithe way ound using the limits 4 and A9 B9 with A4 B4 is enough o seeing 6 ( -) {but not 6 } A: At least one of the values ( and 94) coect (needs just one of the two pevious M maks in (b)) o may see o o may be implied by coect final answe if not evaluated until last line of woking ddm: Adds and 94 (may see o may be implied by coect final answe if not evaluated until last line of woking). This depends on eveything being coect to this point. Acao: Final answe of 6 not ( - 6) 5 5 Common eos: obtains M M A0 (neithe no 94 seen and final answe incoect) then M0 A0 so /5 Uses coect limits to obtain = +/-6 is M M A ( seen) M0 A0 so /5 Special case: In pat (b) Uses limits 9 and 0 = = 6 M0 M A0 M0A0 scoes /5 This also applies if 4 neve seen.

14 Question Numbe. (i) 8 x 4 Scheme Maks (x ) log8 log 4 o o 8 and so ( x)log8 log o (x ) log8 4 ( x) log8 M log 4 log x o x log 8 4 x o x log 8 o.e. log8 log8 dm =0.64 A () log y log log y (ii) Notes (i) log y log log y. M y y log o log log dm y y y y log log o log log 6 (allow awt 6 if eplaced by 6 late) B y y Obtains 6y y 0 o.e. i.e. 6y y fo example A Solves quadatic to give y = ddm A y and (need both- one should not be ejected) (6) [9] M: Takes logs and uses law of powes coectly. (Any log base may be used) Allow lack of backets. dm: Make x subject of thei fomula coectly (may evaluate the log befoe subtacting and calculate e.g. (.58 -)/ ) A: Allow answes which ound to 0.64 (ii) M: Applies powe law of logaithms eplacing log y by log y dm: Applies quotient o poduct law of logaithms coectly to the thee log tems including tem in y. (dependent on fist M mak) o applies quotient ule to two tems and collects constants (allow tiple factions) log on RHS is not sufficient need log 6 o.58 log y log log y log becoming log y log 6y e.g. B: States o uses log o y y o fo at any point in the answe so may be given fo log y log log log, fo example (Sometimes this y mak will be awaded befoe the second M mak, and it is possible to scoe MM0Bin some cases) O may be given fo log o 6 A: This o equivalent quadatic equation (does not need to be in this fom but should be equation) ddm: (dependent on the two pevious M maks) Solves thei quadatic equation following easonable log wok using factoising, completion of squae, fomula o implied by both answes coect. A: Any equivalent coect fom need both answes- allow awt 0. fo the answe / *NB: If =0 is missing fom the equation but candidate continues coectly and obtains coect answes then allow the penultimate A to be implied (Allow use of x o othe vaable instead of y thoughout) 4

15 Question Numbe Scheme Maks Way : Divides by cos to give O Way : Squaes both sides, uses tan so cos sin, obtains 8. (i) M ( ) cos o sin so ( ) Adds o to pevious value of angle ( to give 4 o M 4 So,, (all thee, no exta in ange) A () (ii)(a) 4( cos x) cos x 4 k Applies sin x cos x M Attempts to solve 4cos x cos x k 0, to give cos x dm (b) 6k k cos x o cos x o othe coect equivalent A () cos x and (see the note below if eos ae made) M 8 4 Obtains two solutions fom 0, 9, (0 o.4 o.86 in adians) dm A x = 0 and 9 and (allow awt 9 and ) must be in degees Notes (i) M: Obtains. Allow x o even. Need not see woking hee. May be implied by in 9 final answe ( allow ( ).05 o 0.49 as decimals o ( ) 60 o 0 as degees fo this mak) Do not allow tan no tan M: Adding o to a pevious value howeve obtained. It is not dependent on the pevious mak. 4 (May be implied by final answe of o ). This mak may also be given fo answes as 9 9 decimals [4.9 o.], o degees ( 40 o 40). A: Need all thee coect answes in tems of and no extas in ange. Thee coect answes implies MMA NB : 0,80, 40 eans MMA0 and 0.49,.40 and.44 eans MMA0 (ii) (a) M: Applies sin x cos x(allow even if backets ae missing e.g. 4 cos x ). This must be awaded in (ii) (a) fo an expession with k not afte k = is substituted. dm: Uses fomula o completion of squae to obtain cosx = expession in k (Factoisation attempt is M0) A: cao - awad fo thei final simplified expession (b) M: Eithe attempts to substitute k = into thei answe to obtain two values fo cosx O estats with k = to find two values fo cosx (They cannot ean maks in ii(a) fo this) In both cases they need to have applied sin x cos x (backets may be missing) and coect method fo solving thei quadatic (usual ules see notes) The values fo cosx may be > o < - dm: Obtains two coect values fo x A: Obtains all thee coect values in degees (allow awt 9 and ) including 0. Ignoe excess answes outside ange (including 60 degees) Lose this mak fo excess answes in the ange o adian answes. () [9] 5

16 Question Numbe 9. (a) Scheme Maks Eithe: (Cost of polishing top and bottom (two cicles) is ) o (Cost of polishing cuved suface aea is) h o both - just need to see at least one of these poducts B 5 5 Uses volume to give ( h ) o ( h ) (simplified) (if V is misead see below) Bft 5 ( C) C 6 * dc 00 (b) d o 00 Substitutes expession fo h into aea o cost expession of fom A Bh M A* (4) (then isw) M A ft. 00 k 0 so value whee k,, 4 dm (c) Use cube oot to obtain = 00 Then C = awt 48 o 484 d 0 so minimum d C 600 thei.9 - allow =, and thus C =. ddm Acao (5) Notes (a) B: States o states h Bft: Obtains a coect expession fo h in tems of (ft only follows misead of V) M: Substitutes thei expession fo h into aea o cost expession of fom A Bh A*: Had coect expession fo C and achieves given answe in pat (a) including C = o Cost= and no eos seen such as C = aea expession without multiples of ( ) and ( ) at any point. Cost and aea must be pefectly distinguished at all stages fo this A mak. V N.B. Candidates using Cuved Suface Aea = - please send to eview (b) M: Attempts to diffeentiate as evidenced by at least one tem diffeentiated coectly Aft: Coect deivative allow 00 Bft () [0] then isw if the powe is misintepeted (ft only fo misead) dm: Sets thei d C k to 0, and obtains d value whee k =, o 4 (needs coect collection of powes of fom thei oiginal deivative expession allow eos dividing by π) ddm: Uses cube oot to find o see = awt as evidence of cube oot and substitutes into coect expession fo C to obtain value fo C A: Accept awt 48 o 484 d C d C (c) Bft: Finds coect expession fo and deduces value of > 0 so minimum ( may have been wong) d d OR checks gadient to left and ight of.9 and shows gadient goes fom negative to zeo to positive so minimum OR checks value of C to left and ight of.9 and shows that C > 48 so deduces minimum ( i.e. uses shape of gaph) Only ft on misead of V fo each ft mak (see below) N..B. Some candidates have misead the volume as 5 instead of 5π. PTO fo making instuction. 6

17 Following this misead candidates cannot legitimately obtain the pinted answe in pat (a). Eithe they obtain 00 C 6 o they fudge thei woking to appea to give the pinted answe. The policy fo a misead is to subtact maks fom A o B maks. In this case the A mak is to be subtacted fom pat (a) and the final A mak is to be subtacted fom pat (b) The maximum mak fo pat (a) following this misead is maks. The awad is B B M A0 as a maximum. (a) B: as befoe 5 B: Uses volume to give ( h ) 5 M: ( C) 6 4 A0: Pinted answe is not obtained without eo Most Candidates may then adopt the pinted answe and gain up to full maks fo the est of the question so 9 of the 0 maks maximum in all. 00 Any candidate who poceeds with thei answe C 6 may be awaded up to 4 maks in pat (b). These ae MAdMddMA0 and then the candidate may also be awaded the B mak in pat (c). So 8 of the 0 maks maximum in all. dc 00 (b) M A: o 00 (then isw) d 00 k 00 k dm: 0 so value whee k =, o 4 o 0 so value ddm: Use cube oot to obtain = C 6 A0: Cannot obtain C = 48 o 484 thei allow =, and thus C = must use d C 600 (c) B: 0 so minimum OR checks gadient to left and ight of.966 and shows gadient d goes fom negative to zeo to positive so minimum OR checks value of C to left and ight of.966 and shows that C > 5.4 so deduces minimum ( i.e. uses shape of gaph) Thee is an example in Pactice of this misead.

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