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7 06 H Mah 9758 Promos Markers Scheme Qn Soluions Remarks Le a, b and c represen he amoun of A, B and C o be mied Majori go he marks. respecivel. Then - Some wrong inerpreaions of he quesion (confusion a b c 00 beween % and volume). 0.a 0.7b 0.4c Carelessness wih he % a 0.008b 0.007c A ver small minori did no Solving he simulaneous equaions, know how o use he GC o a 7.5 lires, b 96 lires, c 76.5 lires. obain he answer. Onl a minori go he full marks. 7 Among hose who go i righ, mos did i wr (long mehod). Onl a ver small minori O inegraed wr. 7 Among hose who go i wrong, - A significan minori could no idenif he hperbola, leading o a wrong graph. Poins of inersecion are:.40589,.88 and - Those who skeched he , graphs correcl, and acuall found inersecion poins, did no inerpre he area o be.88 7 found correcl. Majori of Required area d hese area-idenificaion misakes resuled in hem.8 square unis ( s.f.) finding on he area above he - ais bounded b he curves. - Some did no use he GC unil he laer sages, and among hese include he sudens who ried o inegrae he roo of (+ ). - For he man sudens who spli he area up ino smaller porions, a grea man had problems wih he signs (or modulus) as well as he limis. - There were quie a number who paid a flippan aenion o accurac, rounding off inermediae values o heir convenience, leading o a small error in heir final answer. d A good majori obained a 4 leas -5 marks. Among hose who go i righ, d 4 here were quie a few cases of sudens aemping o make he subjec, and going wrong wih careless misakes. These d 4 were no penalised bu sudens should avoid he errors idenified. Some seps were also unclear: The inegraion sep in he separaing of he variables

8 4(i) ln c, where c is an arbirar consan. 4 / Addiional working if sudens make he subjec: ln c 4 ln 4 4 c e e c 4 c 4c 4 e Ae 4c or where A e. 4 c 4 e Ae when, 4 / 4 ln c 4 / 4 c ln 4 ln ln. 4 4 should be presened clearl. Also, he final answer, he paricular soluion, should be presened clearl as requesed b he quesion. Among hose who go i wrong hough: - Some had problems separaing he variables, and came up wih a varie of innovaive bu wrong seps forward. These are he ones who did no demonsrae an knowledge in erms of solving differenial equaions. - The inegraion of he reciprocal roo of (-4 ) also proved o be a problem for some. The mos common error among hose who could ge he logarihmic funcion was, b far, he wrong consan muliplier of 0.5. This usuall semmed from he failure o uilise he MF6 accurael. Majori of sudens obained full marks for (i). O A small number of sudens eiher did no ranslae he graph of f ( ) o f ( ), or ranslaed he graph of f ( ) in he wrong direcion. Some sudens carelessl wroe equaion of asmpoe as = 0 insead of = 0. There were some who wrongl regarded = 0 as an asmpoe despie he graph no ending owards i. O

9 Pariall correc answers: O O No ranslaion Translaion in wrong direcion 4(ii) f ( ) d f ( ) f ( ) Skech he graph of he derivaive of f ( ) This was badl done. Onl a handful of sudens undersood he graph as he derivaive graph of. f ( ) Of hose who skeched he derivaive graph, some failed o realise ha here is a horizonal asmpoe of =. O 5(i) Le he heigh of he clinder be h cm. Hence heigh of he cone is H h cm. Considering he pained surface area, we have πr πrh H π. H r H r h. r r Volume of conainer: R: Sudens misread H in various was. Mos commonl, he assumed he quani referred o he oal surface area of he enire objec, or he curved surface area of he cone insead of he clinder. Some also assumed wrongl ha he clinder and he cone had he same heighs.

10 5(ii) V πr h πr H h πr h πr H H r πr πr H r π H r r Hr (shown). π V H r r H r dv π H rh r dr dv 0 H Hr r 0. dr H 4H H r H H. 6 Since H 0, r 0, r H H H. Therefore, r H gives a saionar value of V and here is onl one possible answer. d V π H 6 r. dr For r H, d V π H 6 H dr 4Hπ 0 as H 0. Therefore, r H gives a maimum value of V b he second derivaive es. P: Sudens are o be reminded ha sufficien working needs o be shown in a show quesion, Mos sudens were able o use he epression for V o answer (ii) despie geing suck in (i), a good echnique o adop in an eam. K: Sudens mus disinguish wha are he variables and consans in an epression. A number of sudens differeniaed wih respec o H insead of r. K: The answer r H / can be immediael rejeced as he radius is a posiive quani, bu here were sudens who subsiued i ino he second derivaive anwa, wasing valuable ime. K: In he second derivaive es, some sudens subsiued H r o ge an epression in r raher han H. Again, sudens need o know which a variable is and which a consan is. P: Sudens should simplif heir epressions for he nd derivaive afer subsiuion. K: The nd derivaive es is recommended for his quesion. To use he s derivaive es, sudens would need o use concree values such as 0.99H and.0h and find he corresponding values of d V / dr.

11 6(a) A This was generall well-done, hough some used sine rule and resuled in a much longer soluion. Man did no wrie down he values of a and b. B C 6(b) Since cos and is small,. Hence a and b. A N Majori of sudens were able o show he resul for s par of 6(b). Those who couldn had assumed angle ANC o be a righ angle. BC 4 B C CN BN BC (shown). 5 B Cosine Rule, AN AC CN CN AC cos ACN cos ACN cos ACN Hence c and d 0.0. Of hose who were able o appl cosine rule, (i) some wrongl wroe he lengh of CN as insead of (ii) Some considered lengh of AN as 7 insead of 5. (iii) Some did no know how o handle (iv) Some evaluaed as 4, ciing being small as he reason. I is imporan o noe ha is no negligible so binomial series epansion workings need o be shown o jusif he resuls. Again, man did no wrie down he values of c and d.

12 7(i) a a 7r 7 r r Hence r k. 7 7r 7r 4 7r 7r 4 7r 7r 4 7r 7r 4 7r 7r 4 7r 7r 4 7r 7r 4 7 Some weak sudens hough his was an AP and wroe a few lines of redundan working like a = 4, d = 7, ec. A few sudens did no show proper proof from LHS o RHS. The simpl had k 7r 7r 4 7r, which did no deserve full marks. Man careless misakes were seen: 7 7r 7r 4 r r r r r r (ii) n a r r r a n 7r 7r 4 7 r 7 7 n n 7n 4 I is necessar o sae k as required k is 7 a consan o be deermined. As his was anoher proof quesion, missing a sep such as n 7r 7r 4 7 r would miss mark. Marks were deduced for: n r LHS Ans = 7 r 4 7 7r 7r 4 7 Cancellaion of erms is

13 7(iii) n a a a a a a n n n a a Now r4 r r n n a a a a a a r4 r r r r r r r r n n n n. 7 7 Hence 7n 5 n 7 7 7n n 4 n 5n 0 From G.C, 8.8 n 6.6 Therefore he minimum possible value of n is 9, and he maimum possible value of n is 6. required for Mehod of Differences. Some ignoran sudens simpl used (ii) as he LHS of he inequali, no realising ha he needed o eclude a oal of 4 erms, i.e. firs erms and he las erm! Ecluding hese 4 erms can be done quie quickl from MOD in (ii), i.e. LHS = 5 7n. 7 Some sudens oall ignored he insrucion o deduce a quadraic inequali! The mos common misake o reach he required quadraic inequali was squaring individual erms insead of squaring boh sides of 7n n 4. Oher variaions were n 5n 0 (wrong manipulaion wih negaive erms) & n 5n 0 (furher 7 7 simplificaion needed). A minori of sudens wrongl gave decimals for he values of n, which mus be posiive inegers.

14 8(i) is no well- The horizonal line cus he graph f ( ) a 6 O 5 and. 6 Hence f is no a one-o-one funcion, and so he inverse funcion defined. To ensure eisence of f f, resric domain of f o,. P: A significan number of sudens correcl idenified ha f is no one-o-one, leading o non-eisence of he inverse. The presenaion of his reason was no alwas precise; he mos common eample is sudens saing he line k for k cus he graph more han once. K: The oher common reasons given were: The graphs of f ( ) and reflecions in he line f ( ) were no The domain of f is no equal o he range of f. Marks were no awarded for hese reasons as he were no he mos fundamenal. 8(ii) Le ( ) e ln ( ) ( ) ln ln Since, rejec negaive square roo ln O K: Despie he skech, man sudens were unable o pick a suiable domain resricion for f such ha i is one-one. For eample, having picked 0 as a line cuing hrough hree poins, he hen go on o resric he domain o (, ), wihou ha he res of he funcion is sill no one-one. This migh impl ha sudens do no acuall undersand visuall he condiion for a funcion o be one-one, or wha he are acuall wriing down when jusifing if an inverse funcion eiss. K: Mos sudens were able o aemp o make he subjec of he formula. Some were challenged b he prospec of square rooing ln, and chose o eiher inven new algebraic manipulaion rules or skip he res of he quesion enirel. K: Sudens are sill forgeing o include he smbol when aking square roos, and he are sill unsure abou how o jusif choosing he correc square roo. g : ln,, e K: There were several incorrec answers for D g, which usuall arises from finding incorrecl. R g

15 8(iii) R D f f,, D R g g, e.57, 0.68 e Since R D, he composie funcion f g g f does no eis. P: Sudens are sill no ver good wih defining funcions properl. The would do well o use he forma of he funcions defined in he quesion as a guideline for epressing heir funcions in similar form. K: Onl a minori of sudens came up wih he wrong condiion. For hose who had he correc condiion, errors usuall arose from wriing down he wrong ses (usuall ou of confusion). P: Sudens are reminded o use proper se noaion when describing ses like D. f 8(iv) f g ( ) D f g g R f g f f ln sin ln D, e R wih resriced domain R, 0 g O K: Mos sudens could produce he composie funcion in he correc sequence, based on whaever epression he go for g previousl. A minori of sudens are sill working ou he composie funcion in he wrong order. K: Mos sudens compleed he las par of he quesion b working ou gf and hen comparing he coefficiens of gf and he RHS o solve for, an eas ask o do. gf ( ) e f ( ) g e f g e sin ln e 0 sin 0

16 9(a) 9(b) d d d d sin sin sin ln 4 7 ( ) ln 4 7 an c Some sudens failed o subsiue he limis or replace in erms of d, or did so incorrecl. Some also reversed he order of he limis. Afer performing he subsiuion of he limis, he large value ma no alwas be he upper limi he limis mus be subsiued accordingl and should be subsiued a he poin when has been replaced b d. The limis canno be replaced as and when deemed convenien. The ke was o simplif and inegrae he epression and man sudens go suck a d or did an incorrec inegraion. Some sudens ook he negaive sign ou of he sqr which is no valid, e.g.: Some sudens were no able o recognise he special angles and failed o correcl obain he final required simplified answer. This is a Show quesion and since he answer is given, sudens should show sufficien workings. There were quie some poor presenaions (seps skipped, sudden change in coefficiens jus o mach he answer, ec) and credi was no given if seps were no clear/ ambiguous, essenial seps/ workings were no shown or if he

17 F( ) F( ) F() F( ) F() F( ) F( ) ln an ln 4 an 0 ln an ln previous epression was no equivalen o he subsequen one. This par was quie badl done and mos sudens sill seem o have a problem handing inegrals wih modulus epressions. Some sudens simpl removed he modulus or simplified incorrecl o ge he following inequivalen inegrals: d, d, d, d, Spliing mus be done o solve his modulus inegraion. Some sudens seem o have a misconcepion ha he spli mus alwas be done a 0 : d d d Some sudens also had incorrec epressions, e.g. i should be or 4 7 and no 4 7, i.e. he 4 7 negaive sign should be for he enire epression and no jus for. For his paricular quesion, because he denominaor happens o alwas be posiive, he below is rue and he laer epression is acceped: However, do noe ha i is no rue in general especiall if he

18 0(i) cos e d e d cos d d d d d d d d d d (shown). denominaor (or he oher accompaning epression) ma be negaive in he inerval of inegraion. Generall i was quie well done b man sudens, despie differen mehods which he migh have applied o solve his par. Those sudens who could no prove he resul were mainl due o: Wrong differeniaion Failed o simplif before he ne differeniaion Too careless wih ve sign. Besides, he presenaion of some sudens work need o be improved. 0(ii) 0(iii) d d d d d d d d. * For 0, π π π π d d d e, e, e, e from (*). Hence π π π π π cos e e e e e e.!! π a, b. cos sin d cos e e sin d Generall i was well done. Sudens were able d o differeniae up o, se = 0 find he corresponding values of f n (0) and appl Maclaurin series correcl. Quie a high percenage who scored full marks. However, a small group of d sudens did no ge correcl. Parl, he did no sar heir differeniaion using he proof from par 0(i). As a resul, he differeniaion was oo edious for some and he failed o coninue. No man scored full marks for his par, neverheless, quie a high percenage were able o a

19 , where k is a consan e sin d e sin e cos d k e sin e cos e sin d c, where c is a consan Hence, e sin d cos sin e c. Since cos sin cos sin, cos cos e e c cos cos e e c. 0(iv) (a) cos cos e e c When 0 π e c π, 4 e c ', 6 where c ' is a consan. π cos 0 c ' 0 0 e e. π cos 4 e e. 6 Heighs form an AP such ha a a 9d 6.5 Solving, d 0.5 Hence T6 a 5d Zoe is 4.5 cm all. leas geing up o sin and obained d correc inegral, e sin d afer he subsiuion. Quie a number of sudens were able o appl inegraion b pars wice and leading o he proof. However, sudens need o be reminded ha he have o wrie down heir soluion in deail o avoid losing marks, i.e. o show sin(cos ) clearl. A common misake in he las sep. Sudens overlooked and subsiued cos e e sin d, missing ou he -ve sign. Onl a small percenage of sudens who aemped par (iv). Among hose who aemped, quie a high percenage of sudens who oall forgo o wrie abou he consan, c of he indefinie inegral. A smaller group sudens did no evaluae and wrie down he value of c in he final soluion. Generall well done. Of he minori who did no ge full marks for his par, i was because of one of he following: No realising sequence is an AP Thinking sequence is a GP Carelessl hinking

20 (b)(i) Le h n denoe he heigh of he nh suden in B. Observing ha AGED and BHFE are squares, and ha ABG and BCH are similar riangles, hn hn hn hn hn hn hn hn hn hn hn hn. hn h Therefore h is in a geomeric progression. n n 6.5 d 0 Oher obvious careless misakes like reading he given numbers wrongl. This par was generall no well done. Amongs hose who go hem correc, here were ver good and creaive proofs. Those who did no, should have eplored he use of similar riangles or gradien of ABC o find a wa o prove CF BE. BE AD (b)(ii) Given ha ar 0.5 and Solving for r: r.005 n ar.6 n n ln n 0.55 ln.005 n ar 0.9, This par was supposed o be echnicall eas so mos sudens go some marks ou of he 4 marks allocaed. Man sudens had marks deduced for accurac problems: rounding o.0049,.00499, or worse,.00. Sudens should have a sense of how he numbers should be rounded in he cone of he quesion, especiall in siuaions when he quesion did no sae he level of accurac needed. (c) average heigh for A average heigh for B B has a greaer average heigh. Some sudens had problems recalling GP formulas for Sn and Tn. Oher han he usual wrong recollecion of AP and GP sum formulas, his quesion was generall well done. Marks were sill given if he had errors carried forward from he earlier pars.

21 (i) π π C: r( sin ), r cos, d d d d 4 4 r cos, r sin an d A poin, r( sin p), r cos p, an p Equaion of normal: r cos p co p r( sin p) co co p co p r r cos p r cos p p r Hence all normals pass hrough he poin r, 0 since i saisfies he equaion for all values of p. Majori of he sudens were able o obain he correc epression for d and sub in o form he equaion. There were however some glaring misakes commied b a surprisingl large number of sudens. differeniae wih respec o r insead of. d d r r( cos p). r cos p dr, a miure of inegraion and differeniaion 4. did no use gradien of normal, sign of gradien of normal is wrong Mos sudens were unable o ge he equaion in he form required/wihou simplificaion. Because of he required form, man sudens ried o facorise co p when moving rcosp from LHS o RHS, causing a lo of algebraic errors Common misakes are. Signs wrong when facorising a b c a ( b c). co p a co p( a co p (ii) π π r( sin ), r cos, sin, cos, r r r r r r r r r π π π π 4 4 r r cos 0 Mos sudens were no able o ge he las mark This par is slighl beer han par (i) wih some unepeced soluions. Man sudens sared wih finding. In his case, i works because i is a circle bu in oher cases i will no.. Some sudens make use of he fac ha i is a circle cenred a ( r, 0) wih radius r o obain he Caresian equaion.. Some sudens use he riangle and cos and r Phagoras heorem o ge he answer (iii) Volume generaed h 0 d Mos of he sudens were able o do he firs par wih a small number using he parameric equaion, insead of he Caresian equaion shown. There are a number of sudens

22 h d 0 r h r rh h Le h r, 4 r 8r r 0 Volume of sphere r( r) ( r) Le r 0, V 0h h dv 60h h 0h h dh dh dh dv 0 d dv d 0h h OR dv dh dh 60h h d d d dv dh 0h h d d Given d V dh 0 0, d d 0h h h increases slowes when h 0 (i.e. wides) dh 0 cm s d 0(0) 00 0 who wen o differeniae insead of inegrae he epression. Quie number of sudens sub h r which onl gives hem and muliplied he resul r b wihou much eplanaion. Mark was awarded as benefi of doub was given. Some inegraed he epression again wih he new limis as 0 o r. There were a number who redefined a circle wih cenre origin and his were marked wrong, as he were no deducions. This par was generall no done well. There were man sudens who wen o find he volume of he 4 sphere V r and differeniaed w.r.. r, where he quesion was asking hem o h find d d. Ohers jus replaced 4 he r wih h o ge V h, which shows lack of appreciaion for he quesion. There were a number of sudens who claims ha for d d o be smalles, d h d mus be equals o 0 which is fallacious. d h I should have been 0, d which anwa is no required of his quesion. A surprisingl simple soluion is obained b - sudens who acuall can do mos of he oher pars of he quesion. dh d is smalles when h 0. dh dv When h 0,. (0) dh 0 Hence. d (0) 0 h