* * MATHEMATICS (MEI) 4756 Further Methods for Advanced Mathematics (FP2) ADVANCED GCE. Monday 11 January 2010 Morning. Duration: 1 hour 30 minutes

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1 ADVANCED GCE MATHEMATICS (MEI) 76 Furher Mehods for Advanced Mahemaics (FP) Candidaes answer on he Answer Bookle OCR Supplied Maerials: 8 page Answer Bookle MEI Eaminaion Formulae and Tables (MF) Oher Maerials Required: None Monda Januar 00 Morning Duraion: hour 30 minues * * * * INSTRUCTIONS TO CANDIDATES Wrie our name clearl in capial leers, our Cenre Number and Candidae Number in he spaces provided on he Answer Bookle. Use black ink. Pencil ma be used for graphs and diagrams onl. Read each quesion carefull and make sure ha ou know wha ou have o do before saring our answer. Answer all he quesions in Secion A and one quesion from Secion B. Do no wrie in he bar codes. You are permied o use a graphical calculaor in his paper. Final answers should be given o a degree of accurac appropriae o he cone. INFORMATION FOR CANDIDATES The number of marks is given in brackes [ ] a he end of each quesion or par quesion. You are advised ha an answer ma receive no marks unless ou show sufficien deail of he working o indicae ha a correc mehod is being used. The oal number of marks for his paper is 7. This documen consiss of pages. An blank pages are indicaed. OCR 00 [H/0/66] OCR is an eemp Chari 3R 9F Turn over

2 Secion A ( marks) Answer all he quesions (a) Given ha = arcan, find d, giving our answer in erms of. Hence show ha d 0 ( + ) d = π. [6] (b) A curve has caresian equaion + = +. (i) Show ha he polar equaion of he curve is r = sin θ. [] (ii) Deermine he greaes and leas posiive values of r and he values of θ beween 0 and π for which he occur. [6] (iii) Skech he curve. [] (a) Use de Moivre s heorem o find he consans a, b, c in he ideni cos θ a cos θ + b cos 3 θ + c cos θ. [6] (b) Le C = cos θ + cos(θ + π n ) + cos(θ + π n and S = sin θ + sin(θ + π n ) + sin(θ + π n where n is an ineger greaer han. (n )π ) cos(θ + ), n (n )π ) sin(θ + ), n B considering C + js, show ha C = 0 and S = 0. [7] (c) Wrie down he Maclaurin series for e as far as he erm in. Hence show ha, for close o zero, e. [] OCR Jan0

3 3 3 (i) Find he inverse of he mari where a. a 3 Show ha when a = he inverse is 0. [6] 3 (ii) Solve, in erms of b, he following ssem of equaions. [] + = + = b 3 + = (iii) Find he value of b for which he equaions + + = + = b 3 + = have soluions. Give a geomerical inerpreaion of he soluions in his case. [7] Secion B (8 marks) Answer one quesion Opion : Hperbolic funcions (i) Prove, using eponenial funcions, ha cosh = + sinh. Differeniae his resul o obain a formula for sinh. [] (ii) Solve he equaion cosh + 3 sinh = 3, epressing our answers in eac logarihmic form. [7] (iii) Given ha cosh =, show b using eponenial funcions ha = ± ln. Find he eac value of he inegral d. [7] 6 OCR Jan0 Turn over

4 Opion : Invesigaion of curves This quesion requires he use of a graphical calculaor. A line PQ is of lengh k (where k > ) and i passes hrough he poin (, 0). PQ is inclined a angle θ o he posiive -ais. The end Q moves along he -ais. See Fig.. The end P races ou a locus. P O Q Fig. (i) Show ha he locus of P ma be epressed paramericall as follows. [3] = k cos θ = k sin θ an θ You are now required o invesigae curves wih hese parameric equaions, where k ma ake an non-zero value and π < θ < π. (ii) Use our calculaor o skech he curve in each of he cases k =, k =, k = and k =. [] (iii) For wha value(s) of k does he curve have (A) an asmpoe (ou should sae wha he asmpoe is), (B) a cusp, (C) a loop? (iv) For he case k =, find he angle a which he curve crosses iself. [] (v) For he case k = 8, find in an eac form he coordinaes of he highes poin on he loop. [3] (vi) Verif ha he caresian equaion of he curve is [3] = ( ) (k ). [3] Coprigh Informaion OCR is commied o seeking permission o reproduce all hird-par conen ha i uses in is assessmen maerials. OCR has aemped o idenif and conac all coprigh holders whose work is used in his paper. To avoid he issue of disclosure of answer-relaed informaion o candidaes, all coprigh acknowledgemens are reproduced in he OCR Coprigh Acknowledgemens Bookle. This is produced for each series of eaminaions, is given o all schools ha receive assessmen maerial and is freel available o download from our public websie ( afer he live eaminaion series. If OCR has unwiingl failed o correcl acknowledge or clear an hird-par conen in his assessmen maerial, OCR will be happ o correc is misake a he earlies possible opporuni. For queries or furher informaion please conac he Coprigh Team, Firs Floor, 9 Hills Road, Cambridge CB GE. OCR is par of he Cambridge Assessmen Group; Cambridge Assessmen is he brand name of Universi of Cambridge Local Eaminaions Sndicae (UCLES), which is iself a deparmen of he Universi of Cambridge. OCR Jan0

5 76 Mark Scheme Januar (FP) Furher Mehods for Advanced Mahemaics (a) = arcan u =, = arcan u du d d du u d = d + u = = + + ( ) OR an = sec d d = sec = + an = + d + d = ( ) MA A M A A Using Chain Rule Correc derivaive in an form Correc derivaive in erms of Rearranging for or and differeniaing implicil M Inegral in form k arcan arcan 0 ( ) d = + 0 A k = = arcan arcan 0 π π = = A (ag) (b)(i) = r cos θ, = r sin θ, + = r M Using a leas one of hese + = + r = r cos θ sin θ + A LHS A RHS r = ½r sin θ + r = r sin θ + r ( sin θ) = r = sinθ A (ag) Clearl obained SR: = r sin θ, = r cos θ used MAA0A0 ma. (ii) Ma r is B Occurs when sin θ = M Aemping o solve π π Boh. Accep degrees. θ =, A A0 if eras in range Min r = 6 B 3 3 Occurs when sin θ = M Aemping o solve (mus be ) θ = 3 π 7π Boh. Accep degrees., A A0 if eras in range 6

6 76 Mark Scheme Januar 00 (iii) G Closed curve, roughl ellipical, wih 0. no poins or dens G Major ais along = 8 (a) cos θ + j sin θ = (cos θ + j sin θ) M Using de Moivre = cos θ + cos θ j sin θ + 0 cos 3 θ j sin θ + 0cos θ j 3 sin 3 θ + cos θ j sin θ + j sin θ M Using binomial heorem appropriael = cos θ 0cos 3 θ sin θ + cos θ sin Correc real par. Mus evaluae θ + j( ) A powers of j cos θ = cos θ 0cos 3 θ sin θ + cos θ sin θ M Equaing real pars = cos θ 0cos 3 θ( cos θ) + cos θ( cos θ) M Replacing sin θ b cos θ = 6cos θ 0cos 3 θ + cos θ A a = 6, b = 0, c = 6 (b) C + js M Forming series C + js as eponenials (c) π ( n ) π jθ+ j θ+ jθ = e + e n e n A Need no see whole series This is a G.P. M Aemping o sum finie or infinie G.P. jθ a = e, r = e j π Correc a, r used or saed, and n erms n A Mus see j Sum = e jθ e e e π n j n π j n jθ j Numeraor = ( e π ) and j e π = so sum = 0 E Convincing eplanaion ha sum = 0 C = 0 and S = 0 E C = S = 0. Dep. on previous E Boh E marks dep. on marks above 7 e + + B Ignore erms in higher powers M Subsiuing Maclaurin series e + A ( ) = = + = +... Suiable manipulaion and use of M + + binomial heorem OR + + = = Hence e OR ( e )( ) (...)( ) M A A (ag) = + + M Subsiuing Maclaurin series A Correc epression + erms in 3 M Mulipling ou e A Convincing eplanaion 8

7 76 Mark Scheme Januar 00 M Evaluaing deerminan 3 (i) a + a A a M = 3a a M Finding a leas four cofacors a 3 A Si signed cofacors correc M Transposing and dividing b de When a =, (ii) 0 = b z 3 0 = 3 M A M 6 M correc (in erms of a) and resul for a = saed SR: Afer 0 scored, SC for M when a =, obained correcl wih some working Aemping o mulipl ( b ) T b given mari (M0 if wrong order) M Mulipling ou 3 8 =, = b, z = b A A for one correc OR + = b = b o.e. M Eliminaing one unknown in was Or e.g. 3 + z = b, = 3 Or e.g. 3 z = b, z = 7 M Solve o obain one value. Dep. on M above 3 One unknown correc = A Afer M0, SC for value of M Finding he oher wo unknowns 8 = b, z = b A Boh correc (iii) e.g. 3 3 = b + M Eliminaing one unknown in was = AA Two correc equaions Or e.g z = b, + 0z = 3 Or e.g z = b, + 0z = 7 Consisen if b + = 3 M Aemping o find b b = Soluion is a line A B 7 8 6

8 76 Mark Scheme Januar 00 (i) e e sinh = e + = e sinh + = ( sinh e e = ) e + e + B e + e e + e = = cosh B Correc compleion sinh = sinh cosh B Boh correc derivaives sinh = sinh cosh B Correc compleion (ii) cosh + 3 sinh = 3 ( + sinh ) + 3 sinh = 3 M Using ideni sinh + 3 sinh = 0 A Correc quadraic ( sinh )(sinh + ) = 0 M Solving quadraic sinh = ¼, A Boh = arsinh(¼) = ln( + 7 ) A Mus evaluae = arsinh( ) = ln( + ) A OR e e e e = 0 ( e e )( e e ) M Use of arsinh = ln( + + ) o.e. Mus obain a leas one value of + = 0 MA Facorising quaric e + ± 7 = or ± MA Solving eiher quadraic = ln( + 7 ) or ln( + ) MAA Using ln (dependen on firs M) 7 (iii) cosh = e + e = e e + = 0 M Forming quadraic in e ( e )( e ) = 0 M Solving quadraic e =, = ±ln A d = arcosh 6 B A (ag) Convincing working = arcosh arcosh M Subsiuing limis = ln A A0 for ±ln OR d = ln ( + 6 ) 6 B = ln 8 ln M Subsiuing limis = ln A 7 8 7

9 76 Mark Scheme Januar 00 (i) Horz. projecion of QP = k cos θ B Ver. projecion of QP = k sin θ B Subrac OQ = an θ B Clearl obained 3 (ii) k = k = k = ½ k = G G Loop Cusp G G (iii)(a) for all k, ais is an asmpoe B Boh (B) k = B (C) k > B 3 (iv) Crosses iself a (, 0) k = cos θ = ½ θ = 60 M Obaining a value of θ curve crosses iself a 0 A Accep 0 (v) = 8 sin θ an θ d dθ = 8 cos θ sec θ 8 cos θ = 0 a highes poin cos θ cos 3 θ = 8 cos θ = ± M Complee mehod giving θ θ = 60 a op A = = 3 3 A Boh 3 (vi) RHS = ( ) k cosθ ( k k cos θ) k cos θ M Epressing one side in erms of θ = ( k cosθ ) k sin θ k cos θ = ( ) k cosθ an θ M Using rig ideniies = (( k cosθ anθ ) ) = ( sin θ an θ) k = LHS E 3 8 8

10 Repors on he Unis aken in Januar Furher Mehods for Advanced Mahemaics (FP) General Commens I is pleasing o repor an increase of abou a quarer in he number of candidaes for his paper, as compared wih Januar 009. There was no evidence ha hese era candidaes had been incorrecl enered: again, he number of candidaes scoring under 0 marks was well below 0% of he oal enr, and he mean mark was almos idenical o Januar 009, alhough he sandard deviaion was a lile greaer. There was a grea deal of ver good work, wih nearl a quarer of candidaes scoring 60 marks or more. Even compeen candidaes someimes succumbed o quie frighening errors in elemenar algebra or calculus. These included: confusion of differeniaion and inegraion (he derivaive of in Q(i) ofen appeared as ); elemenar differeniaion errors (e.g. he derivaive of e appeared as e ); poor use of he laws of logarihms (e.g. e + e =. in Q was followed b = ln(.) or + = ln(.)), ignorance of he laws of algebraic fracions (e.g. in Q(c) + was ver frequenl followed b + = + or similar) and oher wishful hinking (e.g. in Q(a) d = d d or + ( + ) d ). On he oher hand, some processes were handled wih remarkable efficienc: hese included he inverse mari in Q3(i) and he hperbolic equaion in Q(ii). Quesions 3 and were slighl beer done han Quesions and ; Quesion (Invesigaions of Curves) was aemped b onl one candidae. There was lile evidence of ime rouble, alhough as usual some candidaes used ver inefficien mehods o answer some pars of quesions: his was paricularl eviden in Q(b)(ii), Q(b) and in some pars of Q. Presenaion was generall good alhough once again here were candidaes who spli quesions up and scaered hem around he paper, and ohers who used up o hree eigh-page answer books. Candidaes who used supplemenar answer shees ofen agged hem in he middle of heir main answer book: i is much easier o mark he paper (and someimes o urn he pages) if hese are agged a he end. 0 Commens on Individual Quesions Calculus of inverse rigonomeric funcions, polar curves The mean mark for his quesion was jus under. (a) While here were correc and efficien soluions, ver man candidaes hough he derivaive of arcan was. This caused problems wih he inegral, wih man + candidaes blundering ahead in he was described above. The fac ha he answer was given did no deer some candidaes from producing π or ln. 8

11 Repors on he Unis aken in Januar 00 (b) The majori of candidaes knew how o conver from Caresian o polar co-ordinaes, which was an improvemen on previous series. Some of hose candidaes could no hen handle he algebra required o obain he prined answer, wih s becoming s and he like. The responses o he second par were ver varied. Man candidaes did use properies of he sine funcion, bu man of hese hough ha he minimum value of sin θ was 0, and ohers confused θ and θ. Worring saemens like sin θ =, someimes followed b or following sin θ =, were seen fairl frequenl. Man gave values of r raher han of r. Oher candidaes differeniaed, eiher implicil or (worse) he square roo of he given righ hand side: his someimes led o he correc answers, bu more frequenl o a mess which someimes covered wo pages. Ohers produced a able: if he correc incremen for θ was chosen (b chance?) his someimes led o an answer which could be credied. The skech was usuall well done, alhough some of he ellipses acquired dens or sharp poins, and oher candidaes were deermined o draw well-known polar curves such as cardioids. Comple numbers The mean mark for his quesion was. (a) (b) (c) This par was well answered b he majori of candidaes, and i was pleasan o see he efficien mehods man of hem used. I is quie accepable o replace cos θ b c and sin θ b s. One common saring poin was cos θ = (cos θ + j sin θ) : his was condoned, bu i is no correc. Some candidaes were deermined o use he z + mehod, which was no appropriae. Errors included: minor misakes wih z binomial coefficiens or signs; ignoring he sine erms compleel when equaing real pars; including he sine erms as par of he consans a, b and c, and omiing j alogeher from he epansion. Candidaes generall did a lo more work han was required in his par. Mos recognised wha was required a he beginning, wriing he series C + js in eponenial form and recognising a geomeric series, and man obained a correc epression for is sum, alhough finding a sum o infini was ver common and a subsanial number eperienced rouble in finding he common raio, which ofen los is j. Those who obained an sor of answer hen ofen spen several pages ring o realise he denominaor : his ofen included reconversion o rigonomerical form, he use of addiion formulae and appeals o he periodici of sine and cosine. All ha was required was o show ha he sum was 0, so i was sufficien o show ha he numeraor was 0, which followed sraigh from e πj = : onl a small minori of candidaes appreciaed his. Alhough he required Maclaurin series appears in he formula book (and was ofen correcl quoed) and he sem of he quesion was Wrie down, man candidaes spen ime deriving i, someimes incorrecl. Then mos candidaes could ge as far as subsiuing he series ino he given epression, obaining or + +. A minori were able o proceed appropriael, eiher using he binomial heorem or an elegan argumen involving mulipling b or equivalen. Unforunael he majori, having reached his sage, sared blundering abou (see above) or jus 9

12 Repors on he Unis aken in Januar 00 wroe down he given answer wih no linking saemens. Some candidaes jus subsiued a few numbers, or omied he par alogeher. Quie a few convered he epression o one involving e : his required grea care in implemenaion, bu was someimes successful. 3 Marices The mean mark for his quesion was jus under. (i) was done ver well, and (ii) was ver ofen done well, while (iii) proved much more of a challenge. (i) (ii) (iii) Finding he inverse of a 3 3 mari caused lile problem o he vas majori of candidaes: indeed, some were able o almos wrie down he answer, which was mos impressive. There were he usual minor sign errors: for some reason he op elemen of he middle column was mos ofen incorrec. A few candidaes muliplied he cofacors b he elemens of he original mari, while one or wo se a = a he sar, and derived he given inverse wihou involving a: his araced one mark if done correcl. Man candidaes used he inverse in (i) and obained he answer wih he minimum of fuss, alhough some forgo o divide b or made oher slips. Those who ried o solve he equaions from scrach were ofen successful as well, alhough as usual man eliminaed, hen, hen z, obaining hree equaions in wo unknowns from which he could no make progress. This par was much less well done. Some ried o use he inverse mari, seing a = and ignoring he inconvenien deerminan. Ohers ried o use he soluion o par (ii), ignoring he fac ha he ssem of equaions had changed: man of hese appeared o believe ha, because he second and hird equaions had no changed, neiher would he values of or z, so if were eliminaed, he could proceed o find b. Ohers sill were deermined o use heir knowledge of eigenvalues and eigenvecors somewhere, and proceeded o r o find and solve he characerisic equaion. Those who did r o eliminae one unknown in wo differen was frequenl made slips, ofen when subracing negaive quaniies or in working wih fracions, so he correc value of b was no seen ver ofen. The geomerical inerpreaion was raher badl done. Line or sheaf were seen (wih sheaf someimes becoming sheah or even shaf ) bu more frequenl candidaes assered ha he planes inerseced a a unique poin, or formed a riangular prism, or made self-conradicor saemens such as he planes all inersec along a line and wo of hem are parallel. Man candidaes omied his par. Hperbolic funcions This quesion was, b a small margin, he bes answered, wih a mean mark of. (i) The proof was done well, alhough weaker candidaes did no alwas prove from definiions involving eponenials and used oher (unproven) ideniies. The second par required candidaes o differeniae he ideni wih respec o and obain sinh = sinhcosh in he firs insance. This rarel occurred. Man wen back o he eponenial definiions and differeniaed or oherwise manipulaed hose: hose who did aemp o differeniae he ideni ofen los he on he lef, and he righhand side defeaed man compleel. Man candidaes omied his par alogeher. 0

13 Repors on he Unis aken in Januar 00 (ii) (iii) This equaion was ofen solved ver efficienl and was a ver good source of marks for candidaes. Mos picked up he clue from (i), alhough a small minori convered everhing o eponenials, obaining a quaric which could no be facorised, alhough several almos managed o. Once values of arsinh had been obained, man candidaes wen back o firs principles, obaining and solving quadraics in e o find ; i would have been enough o quoe and appl he logarihmic form of arsinh from he formula book. This par gave candidaes a value of cosh and asked hem o show using eponenial funcions ha was ±ln ; i was no accepable here o quoe from he formula book (wih or wihou remarks abou cosh being an even funcion) bu several did. The vas majori proceeded in a more appropriae wa, alhough he manner in which ln was obained was someimes raher opaque. A few candidaes verified he given resul, which was quie accepable as long as i involved eponenial funcions. The inegral was usuall done well alhough he final answer was given as ±ln raher frequenl. One small poin is ha, alhough he inegral has wo forms in he formula book, i is no rue ha arcosh = ln( + 6 ), which was commonl assered. Invesigaions of Curves Onl one candidae aemped his quesion his series.

Morning Time: 1 hour 30 minutes Additional materials (enclosed):

Morning Time: 1 hour 30 minutes Additional materials (enclosed): ADVANCED GCE 78/0 MATHEMATICS (MEI) Differenial Equaions THURSDAY JANUARY 008 Morning Time: hour 30 minues Addiional maerials (enclosed): None Addiional maerials (required): Answer Bookle (8 pages) Graph