MATHEMATICS (MEI) 4753/01 Methods for Advanced Mathematics (C3)

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1 ADVANCED GCE MATHEMATICS (MEI) 4753/0 Mthods for Advancd Mathmatics (C3) QUESTION PAPER Candidats answr on th printd answr book. OCR supplid matrials: Printd answr book 4753/0 MEI Eamination Formula and Tabls (MF) Othr matrials rquird: Scintific or graphical calculator Monday 0 Jun 0 Morning Duration: hour 30 minuts INSTRUCTIONS TO CANDIDATES Ths instructions ar th sam on th printd answr book and th qustion papr. Th qustion papr will b found in th cntr of th printd answr book. Writ your nam, cntr numbr and candidat numbr in th spacs providd on th printd answr book. Plas writ clarly and in capital lttrs. Writ your answr to ach qustion in th spac providd in th printd answr book. Additional papr may b usd if ncssary but you must clarly show your candidat numbr, cntr numbr and qustion numbr(s). Us black ink. Pncil may b usd for graphs and diagrams only. Rad ach qustion carfully. Mak sur you know what you hav to do bfor starting your answr. Answr all th qustions. Do not writ in th bar cods. You ar prmittd to us a scintific or graphical calculator in this papr. Final answrs should b givn to a dgr of accuracy appropriat to th contt. INFORMATION FOR CANDIDATES This information is th sam on th printd answr book and th qustion papr. Th numbr of marks is givn in brackts [ ] at th nd of ach qustion or part qustion on th qustion papr. You ar advisd that an answr may rciv no marks unlss you show sufficint dtail of th working to indicat that a corrct mthod is bing usd. Th total numbr of marks for this papr is 7. Th printd answr book consists of 6 pags. Th qustion papr consists of 4 pags. Any blank pags ar indicatd. INSTRUCTION TO EXAMS OFFICER / INVIGILATOR Do not snd this qustion papr for marking; it should b rtaind in th cntr or dstroyd. OCR 0 [M/0/65] OCR is an mpt Charity R B3 Turn ovr

2 Sction A (36 marks) Solv th quation. Givn that f() ln and g(), find th composit function gf(), prssing your answr as simply as possibl. 3 (i) Diffrntiat ln, simplifying your answr. (ii) Using intgration by parts, show that ln d ( + ln ) + c. 4 Th hight h mtrs of a tr aftr t yars is modlld by th quation whr a, b and k ar positiv constants. h a b kt, (i) Givn that th long-trm hight of th tr is 0.5 mtrs, and th initial hight is 0.5 mtrs, find th valus of a and b. (ii) Givn also that th tr grows to a hight of 6 mtrs in 8 yars, find th valu of k, giving your answr corrct to dcimal placs. 5 Givn that y + 4, show that dy d (5 + ) A curv is dfind by th quation sin + cos y 3. (i) Vrify that th point P ( 6 π, π) lis on th curv. [] 6 (ii) Find dy d in trms of and y. Hnc find th gradint of th curv at th point P. 7 (i) Multiply out (3 n + )(3 n ). [] (ii) Hnc prov that if n is a positiv intgr thn 3 n is divisibl by 8. OCR /0 Jun

3 3 Sction B (36 marks) 8 y Fig. 8 shows th curv y f(), whr f() Fig (i) Show algbraically that f() is an vn function, and stat how this proprty rlats to th curv y f(). (ii) Find f (). (iii) Show that f() ( + ). [] (iv) Hnc, using th substitution u +, or othrwis, find th act ara nclosd by th curv y f(), th -ais, and th lins 0 and. (v) Show that thr is only on point of intrsction of th curvs y f() and y 4, and find its coordinats. [Qustion 9 is printd ovrlaf.] OCR /0 Jun Turn ovr

4 4 9 Fig. 9 shows th curv y f(). Th ndpoints of th curv ar P ( π, ) and Q (π, 3), and f() a + sin b, whr a and b ar constants. y 3 y f( ) Q (, 3) P (, ) Fig. 9 (i) Using Fig. 9, show that a and b. (ii) Find th gradint of th curv y f() at th point (0, ). Show that thr is no point on th curv at which th gradint is gratr than this. (iii) Find f (), and stat its domain and rang. Writ down th gradint of y f () at th point (, 0). [6] (iv) Find th ara nclosd by th curv y f(), th -ais, th y-ais and th lin π. Copyright Information OCR is committd to sking prmission to rproduc all third-party contnt that it uss in its assssmnt matrials. OCR has attmptd to idntify and contact all copyright holdrs whos work is usd in this papr. To avoid th issu of disclosur of answr-rlatd information to candidats, all copyright acknowldgmnts ar rproducd in th OCR Copyright Acknowldgmnts Booklt. This is producd for ach sris of aminations and is frly availabl to download from our public wbsit ( aftr th liv amination sris. If OCR has unwittingly faild to corrctly acknowldg or clar any third-party contnt in this assssmnt matrial, OCR will b happy to corrct its mistak at th arlist possibl opportunity. For quris or furthr information plas contact th Copyright Tam, First Floor, 9 Hills Road, Cambridg CB GE. OCR is part of th Cambridg Assssmnt Group; Cambridg Assssmnt is th brand nam of Univrsity of Cambridg Local Eaminations Syndicat (UCLES), which is itslf a dpartmnt of th Univrsity of Cambridg. OCR /0 Jun

5 4753 Mark Schm Jun 0, or ( ), /3 gf() ln 3(i) or ln. ln. d y 4 d ln 4 ln 3 d y 3 ln + ( ) d ln+ 3 3 (ii) ln d lt u ln, du/d / dv/d /, v ln. d + ln d + ln c + (ln + ) + c * MA MA M M A M B A A M B A A M A A A www www, or must b act for A (.g. not 0.33, but allow 0.3 ) condon doing both qualitis in on lin.g., tc Forming gf() (soi) quotint rul with u ln and v d/d (ln ) / soi corrct prssion (o..) o.. cao, mark final answr, but must hav dividd top and bottom by product rul with u and v ln d/d (ln ) / soi corrct prssion o.. cao, mark final answr, must simplify th.(/) trm. Intgration by parts with u ln, du/d /, dv/d /, v must b corrct, condon + c condon missing c NB AG must hav c shown in final answr allow unsupportd answrs or from graph or squaring M (3 )( ) 0 M factorising, formula or comp. squar, /3 A A allow M for sign rrors in factorisation if mor than two solutions offrd, but isw inqualitis Doing fg: ln( ) SC Allow (but not ) unsupportd Consistnt with thir drivativs. udv ± vdu in th quotint rul is M0 Condon ln. ln for this A (providd ln. is shown).g. ln, 3 3 or vic-vrsa Must b corrct 3 3 ln at this stag. Nd to s / 5

6 4753 Mark Schm Jun 0 4(i) h a b kt a 0.5 (thir)a b b 0 (ii) h kt Whn t 8, h k 6 0 8k 4.5 8k ln 0.45 k ln 0.45/( 8) y ( + 4) / dy. ( 4 ) ( + 4 ) d / / (5+ ) * + 4 / ( 4 ) ( 4 ) B M Acao M M A M B A M A a nd not b substitutd ft thir a and b (vn if mad up) taking lns corrctly on a corrct rarrangmnt - ft a, b if not asd cao (www) but allow 0. product rul with u, v ( + 4) ½ ( ) / soi corrct prssion factorising or combining fractions NB AG allow M for a b 8k 6 allow a and b unsubstitutd allow thir 0.45 (or 4.5) to b ngativ consistnt with thir drivativs; condon wrong ind in v usd for M only (nd not factor out th ) must hav vidnc of o or (5 + )( + 4) ½ or (5 + )/( + 4) ½ 6(i) sin(π/3) + cos(π/6) 3/ + 3/ 3 B [] (ii) d y cos sin y 0 d M A d y cos sin y d dy cos d sin y Acao Whn π/6, y π/6 Mdp dy cos π /3 A d sin π /6 7 (i) (3 n + )(3 n ) (3 n ) or 3 n B [] (ii) 3 n is odd 3 n + and 3 n both vn As conscutiv vn nos, on must b divisibl by 4, so product is divisibl by 8. M M A must b act, must show working Implicit diffrntiation corrct prssion substituting dp st M www mark final answr 3 n is odd 3 n + and 3 n both vn compltion Not just sin(π/3) + cos(π/6) 3, if substituting for y and solving for (or vv) must valuat sin π/3.g. not arcos( 3 sin π/3) allow on rror, but must hav (±) sin y dy/d. Ignor dy/d unlss pursud. cos d sin y dy 0 is MA (could diffrntiat wrt y, gt d/dy, tc.) cos is A0 sin y or 30 or 9 n ; pnalis 3 n if it looks lik 3 to th powr n. Induction: If tru for n, 3 n 8k, so 3 n + 8k, M 3 (n+) 9 (8k + ) 7k + 8 8(9k + ) so div by 8. A Whn n, 3 8 div by 8, tru A(or similar with 9 n ) 6

7 4753 Mark Schm Jun 0 8(i) f( ) ( ) + + f(), [ f is vn *] Symmtrical about Oy (ii) f( ) ( + + ) ( ) or ( + + ).0 ( ) ( + + ) ( ) ( + + ) (iii) (iv) f( ) + + * ( + ) A d 0 ( + ) lt u +, du d whn 0, u ; whn, u + + A d u u + u (v) Curvs intrsct whn f( ) 4 ( + ) 4 or 3 so as > 0, only on solution 0 whn 0, y ¼ M A B B M A M A [] B M A M Acao M M A B B substituting for in f() condon rflction in y-ais d/d ( ) and d/d( ) soi chain or quotint rul condon missing brackt on top if corrct thraftr o.. mark final answr top and bottom by (corrctly) condon for M but not A NB AG corrct intgral and limits (d u) u u substituting corrct limits (dp st M and intgration) o.. mark final answr. Must b act Don t allow. soi or quivalnt quadratic must b corrct gtting and discounting othr sol n 0 www (for this valu) y ¼ www (for th valu) Can imply that ( ) from f( ) + + Must mntion ais. If diffrntiating ( + ).g. ( ) ( + + ) or M, A ( + ) condon no ( ),for both M and A withhold A (unlss rsult in (iii) provd hr) condon no d, must us f(). Limits may b implid by ( + ) subsqunt work. If unsupportd, allow st B only or by inspction k M + A + uppr lowr; and + (or 3.7..)for u, or 0 and for if substitutd back (corrctly).g.. Can isw 0.3, which may b usd as vidnc of M. ( + ) Can isw numrical ans (.g. 0.3) but not algbraic rrors or ( ) With or ( ), condon, 0 [or + ] not possibl www unlss vrifid Do not allow unsupportd. A sktch is not sufficint 7

8 4753 Mark Schm Jun 0 9(i) Whn 0, f() a * Whn π, f(π) + sin bπ 3 sin bπ bπ ½ π, so b ½ * or a + sin ( πb) ( a sin πb) 3 a + sin (πb) sin πb, sin πb, πb π/, b ½ 3 a + or a a (o for b) B M A NB AG a is th y-intrcpt not nough but allow vrification (+sin 0 ) or whn π, f( π) + sin ( bπ) sin( bπ) condon using dgrs bπ ½ π, b ½ NB AG M for both points substitutd A solving for b or a A substituting to gt a (or b) or quiv transformation argumnts :.g. curv is shiftd up so a..g. priod of sin curv is 4π, or strtchd by sf. in -dirction (not squzd or squashd by ½ ) b ½ If vrifid allow MA0 If y + sin ½ vrifid at two points, SC A squnc of sktchs starting from y sin showing clarly th translation and th strtch (in ithr ordr) can arn full marks (ii) f () ½ cos ½ f (0) ½ Maimum valu of cos ½ is ma valu of gradint is ½ M A A M A ±k cos ½ cao www or f () ¼ sin ½ f () 0 0, so ma val of f () is ½ (iii) y + sin ½ y + sin ½ y sin ½ y arcsin( ) ½ y y f () arcsin( ) Domain 3 Rang π y π Gradint at (, 0) is (iv) π A ( + sin )d 0 π cos 0 π ( ) π + ( 8.83 ) M A A B B Bft [6] M M A Acao Attmpt to invrt formula or arcsin(y ) ½ must b y or f () or [, 3] or [ π, π] or π f () π ft thir answr in (ii) (cpt ±) /thir ½ corrct intgral and limits kcos whr k is positiv k answrs rounding to 8.3 viz solv for in trms of y or vic-vrsa on stp nough condon us of a and b in invrs function,.g. [arcsin( a)]/b or sin (y ) condon no brackt for st A only or sin ( ), condon f (), must hav brackt in final ans but not y 3 but not π π. Pnalis < s (or to 3, π to π ) onc only or by diffrntiating arcsin( ) or implicitly soi from subsqunt work, condon no d but not 80 Unsupportd corrct answrs scor st M only. 8

Note If the candidate believes that e x = 0 solves to x = 0 or gives an extra solution of x = 0, then withhold the final accuracy mark.

Note If the candidate believes that e x = 0 solves to x = 0 or gives an extra solution of x = 0, then withhold the final accuracy mark. . (a) Eithr y = or ( 0, ) (b) Whn =, y = ( 0 + ) = 0 = 0 ( + ) = 0 ( )( ) = 0 Eithr = (for possibly abov) or = A 3. Not If th candidat blivs that = 0 solvs to = 0 or givs an tra solution of = 0, thn withhold