9231 Further Mathematics June 2004 FOREWORD... 1 FURTHER MATHEMATICS... 2

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1 9 Furher Mahemaics June CONTENTS FOREWORD... FURTHER MATHEMATICS... GCE Avance Level... Paper 9/ Paper... Paper 9/ Paper... 7 FOREWORD This bookle conains repors wrien b Eaminers on he work of caniaes in cerain papers. Is conens are primaril for he informaion of he subjec eachers concerne.

2 9 Furher Mahemaics June FURTHER MATHEMATICS GCE Avance Level Paper 9/ Paper General commens The overall sanar was high boh wih regar o presenaion an o he quali of he essenial mahemaics. Mos caniaes aempe all he quesions, an here were few misreas. There were some elemenar errors o be seen, bu hese, in mos cases, i no seriousl epress scrip oals. Caniaes generall showe high levels of knowlege an eperise wih such sllabus opics as summaion of series, eerminaion of asmpoes of curves efine b a raional funcion, reucion formulae, mean values, secon erivaive eerminaion an evaluaion of area of a surface of revoluion in he cone of parameric represenaion, an also soluion of sanar secon orer ifferenial equaions. In conras, inucion, comple numbers an linear algebra were clearl no well unersoo. There was a wie range of isribuion of marks. A he lower en, here were a few scrip oals which inicae serious lack of preparaion for his eaminaion whereas, a he oher ereme, a ver high sanar was aaine b a significan minori of caniaes. Commens on specific quesions Quesion Generall, his inroucor quesion was answere accurael. Some of he mehos aope were sub-opimal wih respec o use of eaminaion ime. Mos caniaes began b wriing S N 8 N (N + ) 6 N(N + )(N + ), an hen showe abou 6 he righ amoun of working o esablish he require resul. There were a few aemps o use inucion, bu wheher correc, or no, hese coul no be aware an crei, for he quesion specificall emans use of sanar resuls in he Lis of Formulae. In he secon par of he quesion, no all caniaes recognise ha he require sum is S N S N bu insea worke from incorrec forms such as S N S N+, or S N S N. N Moreover he correc ( n 6n ) N( N + )( 8N ) N( N + )( N ) n N+ ransforme o he isplae resul. Answer: ( N + N N ) N. 8 was no alwas accurael

3 9 Furher Mahemaics June Quesion The majori of caniaes obaine correc resuls for par. Responses o par, however, were generall incomplee an/or erroneous. Almos all responses sae, correcl, ha he verical asmpoe is. Beon ha, here generall appeare some aemp o obain he inermeiae resul a + a + ( a)/( ) (*) from which he oblique asmpoe can be obaine immeiael. Acuall a subsanial number of such ivision processes were erroneous bu, all he same, le o he require equaion. Also, a small minori of ivisions were incomplee an hese, more ofen han no, le o a resul such as a. In passing, i mus be remarke ha he sraeg base on he obaining of (*) is sub-opimal, for he oblique asmpoe is clearl of he form a + c so ha a a( ) + c( ) for large. Thus, equaing coefficiens of leas a once o c a. However, a small minori of caniaes argue successfull in his wa. a In conras, a few emploe he classical meho b seing m + c o obain (m + a) + (c m ) c. Tangenc a infini implies m + a, c m an hence m a, c a. Mos caniaes iffereniae he given funcion of, se he resul equal o zero an hen a ± evenuall obaine a a a. Few iffereniae he form (*) in par so as o obain he more useful resul ± for he a -coorinaes of he urning poins of C. To prove ha boh of hese values,,, sa, are posiive, an argumen such as he following was epece. a > an are real. Also, a > a < > an >. However onl abou half of all caniaes succeee in proving ha,, are real an onl a minori coul go on o show ha he are posiive. Answer: Asmpoes are an a + a. Quesion Mos caniaes mae some progress here, bu here were few ousaning responses. Almos all caniaes correcl ransforme he given equaion ino he polar form. There were few insances of omission of essenial eail. Mos responses showe, a leas, a pariall correc skech for Y θ Y π. However, in man cases here appeare eiher no oher loop or, a he oher ereme, loops for π Y θ Y π an/or π Y θ Y π where, in fac, C oes no eis. Thus hose responses which showe correc π π loops in Y θ Y an π Y θ Y, an no oher loops, were ver much in a minori. (iii) Mos responses showe an inelligen use of r sinθ o obain he maimum of r. There were few incorrec answers.

4 9 Furher Mahemaics June Quesion Mos caniaes foun his quesion beon hem an so here were ver few complee an correc responses. ln A his level i was o be epece ha he successive iffereniaion of wih respec o woul be a rouine ask for caniaes. Neverheless, a significan number of elemenar errors appeare an his inicae a eficienc in basic mahemaical echnique in he caniaure. Beon ha, some caniaes i no even comprehen ha hree iffereniaions were require in orer o esablish he values of a, a, a, an his lack of percepion ineviabl le o an incorrec conjecure for he form of a n, as require in he remainer of his quesion. A minori of caniaes wroe own a correc inucive hpohesis. Among hose who i work from H k : a k ( ) k k!, few wen on o prove convincingl ha H k a k+ ( ) k+ (k + )! an hence o complee he inucive argumen. Again, i was evien ha lack of echnique was he main cause of failure. Answers: a, a, a 6; a n () n n! Quesion This quesion enjoe onl moerael success aemps. unnecessaril prorace argumens. Man caniaes became involve in In mos cases, he correc eigenvalues appeare from he working. Neverheless, a significan minori of caniaes i no unersan ha as he mari A is riangular, hen he iagonal elemens are he eigenvalues, In consequence, he became involve in he labour of eriving an solving he characerisic equaion of A an his eercise mus have wase a lo of eaminaion ime. Mos caniaes emploe soun mehoolog for he obaining of he eigenvecors, bu a profusion of elemenar errors le o man failures in his respec. In paricular, i was no uncommon for o be obaine as he eigenvecor corresponing o he eigenvalue In he res of he quesion, mos caniaes prouce resuls which were consisen wih he resuls previousl obaine. Ver few wase ime b aemping o eermine P. Answers: The eigenvalues of A are,, : corresponing eigenvecors are:,,. D iag (,, ), P. Quesion 6 This quesion was well answere b he majori. The mos popular sraeg for esablishing he isplae reucion formula was b he meho inicae in he quesion. In his respec, working was generall accurae an complee. In conras, a minori of caniaes inegrae (ln ) n b pars o obain I n in erms of I n an hen upgrae n o n +. This meho was also applie correcl, in mos cases. For he final par, working was generall relevan, accurae an complee.

5 9 Furher Mahemaics June Quesion 7 This quesion is a sraighforwar eercise in comple numbers requiring onl a basic knowlege of he relae sllabus maerial. Neverheless he majori of caniaes appeare no o unersan he essence an no o have he necessar echnical eperise for answering quesions of his pe. Mos caniaes i, a leas, ge as far as obaining a correc resul for z. In conras, caniaes generall prouce hree possible values for arg z, bu i was common for a leas one of hem no o be wihin he sipulae range of θ, or o be compleel wrong. The ke o he remaining par of his quesion is o observe ha if z, sa, is an roo of he comple equaion k k k k z + i e. z π / 6 / 6 + i, hen [ ] ( ) [ ] i k πk i z e. The isplae resul is hen immeiae. However, a minori of caniaes coul prouce an equivalen argumen. π i 7πi Answer: z ep, ep 8 8 Quesion 8 9πi, ep 8. There were man responses o his quesion which evelope along he righ lines, bu few of hese coul be escribe as ousaning. As is so ofen he case wih quesions of his pe, almos all caniaes prouce a correc resul for in erms of, bu hen aempe o go on o obain he secon erivaive on he basis of erroneous formulae such as prevalen han he laer. In cases where an ( + ) caniaes epresse heir resul in he form ( ) complee argumen o prove ha < a all poins of C., he former being much more was correcl obaine in erms of, mos an supplemene his wih a vali an + + as he inegral represenaion of he require surface area. For he evaluaion, mos caniaes perceive ha use of he ieni [( ) + 6 ] / + is essenial an wen on o work accurael owars he final, numerical resul for S. Mos responses sare correcl wih S π ( ) ( ) 6 A small minori evaluae he corresponing arc lengh insea of S, an his migh sugges a lack of unersaning of he erminolog of he relae sllabus maerial. Answer: 88.

6 9 Furher Mahemaics June 6 Quesion 9 This quesion generae a lo of goo work, bu here were relaivel few compleel correc responses. In general, caniaes i bes wih he las par of he quesion. In he inroucor par, almos all responses showe a correc argumen o esablish he preliminar resul. In conras, less han half of all caniaes coul prove he associae secon orer resul. Man unersoo ha i coul, in principle, be erive from, bu lacke he necessar insigh o coninue furher. In fac, a caniae neee o provie he following: ( ) +, or equivalen. The mile secion requires no more han subsiuion of he resuls previousl obaine plus accurae simplificaion an in his cone here were few failures. In he final par, mos responses showe a soun sraeg. The correc complemenar funcion usuall appeare, bu here were a surprisingl large number of errors o be seen in he working for he paricular inegral. Answer: General soluion: Ae / + Be / + +. Quesion This quesion highlighe much erroneous hinking wih regar o he funamenals of linear spaces an ssems of linear equaions. Man responses lacke clari an frequenl he impression was given ha he caniae i no unersan wha was require. In fac, a, simple wa o procee is o obain an echelon form, such as θ. I is hen a simple maer o eal wih he cases 6 θ an 6 θ. In his respec wha was require were argumens such as ( ) r ) im( 6 A K θ an ( ) r ) im( 6 A K θ. However, few caniaes supplie such eail. For he mos par, caniaes working was accurae. (iii) A sanar proceure here is o epress,, z, in erms of eacl wo parameers, e.g., 7φ + 7ψ, 6φ ψ, z φ, ψ. This no onl provies a check of he resul for e obaine in par, bu also enables a resul for e o be wrien own. (iv) In general erms, i was epece ha he caniae woul verif clearl ha b A, an hen se ou working such as, A Ae o + k Ae + k Ae b + k + k b, for all k, k. Some responses appeare o procee along hese lines hough, i mus be sai, man such argumens were incomplee in some imporan wa. Answers: e 6 7 ; (iii) e 7.

7 9 Furher Mahemaics June Quesion EITHER Of hose who aempe his quesion, abou half worke hrough he firs hree secions wihou error. However, onl a small minori mae an progress in par (iv). (iii) (iv) Almos all responses showe accurae an relevan working. A few, however, gave he complemenar angle as he answer. The relevance of he vecor prouc (i k) (i + j + k) was perceive b mos an usuall, hough no alwas, his was evaluae accurael. Some caniaes mae no progress beon he en of par. The faile o realise ha he irecion of he line of Π Π is parallel o (i k) (i j + k) an his failure woul sugges lack of geomerical insigh. In his respec, he woul probabl have mae much beer progress ha he rawn an effecive iagram in he firs place. This woul also have mae clear o hem ha he hree vecors i k, i + j + k, i + j + k are parallel o he plane Π an are herefore linearl epenen. As i was, harl an caniaes argue in his simple wa, bu insea emploe eene algebraic mehos. A simple sraeg here begins wih he specificaion Q (u, v, u). Thus since l m, hen (u ) + (v ) + (u ) u + v 8 OQ u + 9 u. The minimum of OQ an hence of OQ can hen easil be foun. Answers:.6 ; + z ; (iii) r (i + j + k); (iv) 6.7. Quesion OR The quali of responses o his quesion was beer han ha for Quesion EITHER. However, elemenar errors owngrae a subsanial amoun of he work submie. (iii) There were ver few failures a his sage. Some caniaes evaluae S from he -equaion, bu hough he ha evaluae S. Ye ohers aempe o use S S S, an were generall successful. From his sage onwars man responses ran ino confusion an were frequenl incomplee, or simpl wrong. A popular bu uninspire meho was o generalise he previous mehoolog b wriing S n+ S n+ S n, an hen choose suiable values of n. This enjoe some success, bu even so, a more epeiious sraeg is o se an so obain in a ver similar wa o he erivaion of he -equaion from he given -equaion. As he -equaion has roos α, β, γ hen he sums of he squares, cubes an fourh powers of his equaion are S 8, S, S 6, respecivel, an so following he earlier mehoolog hese ma be obaine wihou ifficul. Answers: S ; (iii) S 8, S 9, S 6 9. Paper 9/ Paper General commens As is usuall he case, caniaes seeme on he whole o be more a ease wih he Saisics quesions. This was paricularl rue in he single quesion offering alernaive choices, namel Quesion, where viruall all caniaes elece o aemp he Saisics opion. Despie his general preference, mos caniaes ha sufficien ime o aemp all he quesions on he paper, inicaing ha he ime pressure was no unul grea. As inicae in more eail below, here was consierable variaion beween quesions in he level of success, wih almos no caniaes offering compleel correc answers o Quesions or 6, for eample. B conras, ohers such as Quesions an ofen prouce goo work. 7

8 9 Furher Mahemaics June Commens on specific quesions Quesion The grea majori of caniaes appreciae ha he change in momenum, foun from he prouc of he mass an he change in veloci, shoul be ivie b he ime of conac in orer o give he force causing he change, hough some overlooke he fac ha he wo given spees are in opposie irecions an herefore shoul be ae raher han subrace. Onl ver rarel, however, i a caniae realise ha he weigh of he hammer conribues o he force on he nail. Answer: N. Quesion This is clearl a quesion on angular moion, bu a large number of caniaes use a varie of sraagems, almos alwas invali, o ransform he quesion ino one on linear moion. Insea he prouc of he flwheel s momen of ineria an angular eceleraion shoul be equae o he couple acing on i, since his consan eceleraion acing for he require ime mus equal he given iniial angular spee. Some of hose caniaes who appreciae his general principle were unforunael raher cavalier wih he isincion beween acceleraion an eceleraion. The loss in energ equals he iniial roaional energ, foun from one half he prouc of he momen of ineria an he square of he iniial angular veloci, in appropriae unis. Answers: s, 6 kj. Quesion In each of he wo collisions he resuling velociies are reail foun b formulaing an hen solving he conservaion of momenum an he resiuion equaions. Where errors occurre, he were more ofen ue o faul arihmeic han confusion over signs, since he moion of all spheres is in he same irecion. This fac ogeher wih he final spees means ha i is immeiael clear ha here can be no furher impacs afer he secon collision. Answers: A res; ms ; ms. Quesion The easies wa of fining he verical reacions a A an B is o formulae one equaion, momens for he ssem abou B for eample, which iels one reacion, an hen a secon equaion for he oher, such as momens abou A or verical resoluion of forces for he ssem. Man caniaes neelessl complicae maers, hough, b inroucing unknown forces such as he fricion a A an B, or he forces beween he laers a C. The laer ma be reail foun b, for eample, resolving vericall for one of he laers, an aking momens abou he opposie en from C for eiher laer. The magniue of he resulan an is irecion ma hen be foun in he usual wa, hough i is necessar o specif he laer eplicil raher han simpl calculaing an unefine angle. Answers: W W, ; W, above horizonal. Quesion The firs par requires he formulaion of a conservaion of energ equaion relaing o he poin B, involving cos ACB. I is also necessar o resolve forces in he raial irecion, noing ha he cliner eers no force on he paricle as he laer leaves he surface a B. Combinaion of hese wo equaions iels he given spee an he value of cos ACB. While he Eaminers hough i preferable o euce cos ACB an he spee in his wa, he aware full crei o hose caniaes who emonsrae insea ha he given values saisf he wo equaions. There are a varie of was of fining he horizonal isance which he paricle moves beween leaving he surface an sriking he able. Fining he ime aken b consiering he verical componen of he moion, or using he sanar rajecor equaion, boh resul in a quaraic equaion for he ime an horizonal isance respecivel, while fining he verical spee a impac avois he nee o solve a quaraic. The horizonal moion is of course uner no acceleraion. The isance CD requires he aiion of he horizonal isance of B from AC. Answer:.a. 8

9 9 Furher Mahemaics June Quesion 6 Alhough his quesion was almos alwas aempe, compleel correc answers were rare. Some caniaes foun he cumulaive isribuion funcion of X, namel F(), an hen simpl replace b in an invali aemp o fin he cumulaive isribuion funcion G() of, giving an incorrec probabili ensi funcion g(). Ohers use he more vali approach of subsiuing for Y in P(Y < ), X which shoul iel P(X > ) an hence G() F, bu he mishanle he inequaliies an insea obaine P(X < ). Answer:. Quesion 7 Mos caniaes reail foun he value of E(N), bu a significan number miscalculae he probabili in he firs par of he quesion, aking i o be P(N Y E(N)) P(N Y ), for eample. The secon par seeme more challenging, bu simpl requires he summaion of an infinie geomeric series wih firs erm an raio equal o he prouc of an. 9 Answers:,. 7 Quesion 8 The require probabili is esseniall foun b calculaing F(6), where F() is he cumulaive isribuion funcion e /, bu man caniaes use insea a broal similar epression involving e. Correc answers were ver rarel seen, wih man caniaes aking P n o be F(n) or somehing similar, raher han he correc F(n) F(n ). The laer leas o a geomeric progression wih raio equal o e /. Answer: e. Quesion 9 The principal poenial sumbling block is obaining an unbiase esimae of he populaion variance, which 8. shoul be foun from ( m ), where m is he sample mean. The confience inerval ma hen 7 8 be foun in he usual wa, using he abular -value.6. Answer: [8., 9.9]. Quesion Calculaion of he epece values of he number of baches requires he correc probabili of a ulip being ellow, namel., hough oher values were seen quie ofen. Anoher frequen error was a failure o combine he firs wo cells as well as he las four ones. The calculae χ value. is hen compare wih he abular value 7.8 in orer o conclue ha he binomial isribuion fis he aa. 9

10 9 Furher Mahemaics June Quesion The wo prouc momen correlaion coefficiens (pmcc) were usuall eermine b a vali meho, hough arihmeical errors were someimes commie. Since he value.8 for sample A is greaer han he corresponing value.79 for B, he regression line mus be foun for he former, an c aken o be in orer o preic e. Alhough mos caniaes sae ha he pmcc of C is he same as ha of B, onl a minori gave a sufficien jusificaion, which hinges on he fac ha he square of he pmcc equals he prouc of he graiens of he wo regression lines, an hese graiens are of course equal for B an C. The pmcc for he combine sample is foun in he same wa as hose for A an B, an is value.768 is compare wih he abular value.77 o euce ha he populaion pmcc is no ifferen from. Answer: 67.. Quesion EITHER This was a ver unpopular alernaive, an no paricularl well answere b he few caniaes who chose i. Showing ha he moion is simple harmonic involves as usual relaing he acceleraion of he sone abou he cenre of moion o he ne force on i, arising from is weigh an he force eere b he compresse spring. The isance of he cenre of moion below T is reail foun b equaing hese wo opposing forces, an hence he isance.6 of his poin above he floor. Having eermine he square of he spee v of he sone when i makes conac wih he spring a T o be gh, his ma be insere ino he sanar formula v ω (a ) wih equal o in orer o fin he ampliue a, since he value of ω has alrea been foun. Conservaion of energ provies an alernaive meho. The sone will no hi he floor provie a is less han.6, ieling he given inequali for h. Answers:.6 m, ( + 8h). Quesion OR Mos of he man caniaes who aempe his quesion use he correc approach o eermining he confience inerval, hough no all chose he correc abular -value of.7, an some esimae he populaion variance inaccurael b using a roune value of he sample mean wih oo few significan figures. A frequen error when esimaing he common variance from he sanar poole formula was o confuse biase an unbiase esimaes when insering he previousl calculae esimaes of he wo populaion variances. The final es presene few ifficulies, wih comparison of he calculae value.8 of wih he abular value.8 leaing o he conclusion ha acciens o no occur more frequenl in he own han in he counr. 67 Answers: [6.,.],. 6

The Fundamental Theorems of Calculus

The Fundamental Theorems of Calculus FunamenalTheorems.nb 1 The Funamenal Theorems of Calculus You have now been inrouce o he wo main branches of calculus: ifferenial calculus (which we inrouce wih he angen line problem) an inegral calculus