2012 Mathematics. Higher. Finalised Marking Instructions

Transcription

1 0 Mthemts Higher Finlised Mrking Instructions Scottish Quliftions Authority 0 The informtion in this publtion my be reproduced to support SQA quliftions only on non-commercil bsis. If it is to be used for ny other purposes written permission must be obtined from SQA s NQ Delivery: Em Opertions tem. Where the publtion includes mterils from sources other thn SQA (secondry copyright), this mteril should only be reproduced for the purposes of emintion or ssessment. If it needs to be reproduced for ny other purpose it is the centre s responsibility to obtin the necessry copyright clernce. SQA s NQ Delivery: Em Opertions tem my be ble to direct you to the secondry sources. These Mrking Instructions hve been prepred by Emintion Tems for use by SQA Appointed Mrkers when mrking Eternl Course Assessments. This publtion must not be reproduced for commercil or trde purposes.

2 Generl Comments These mrking instructions re for use with the 0 Higher Mthemts Emintion. For ech question the mrking instructions re in two sections, nmely Illustrtive Scheme nd Gener Scheme. The Illustrtive Scheme covers methods whh re commonly seen throughout the mrking. The Gener Scheme indtes the rtionle for whh ech mrk is wrded. In generl mrkers should use the Illustrtive Scheme nd only use the Gener Scheme where cndidte hs used method not covered in the Illustrtive Scheme. All mrkers should pply the following generl mrking principles throughout their mrking: Mrks must be ssigned in ccordnce with these mrking instructions. In principle, mrks re wrded for wht is correct, rther thn deducted for wht is wrong. Awrd one mrk for ech. There re no hlf mrks. The mrk wrded for ech prt of question should be entered in the outer right hnd mrgin, opposite the end of the working concerned. The mrks should correspond to those on the question pper nd these mrking instructions. Only the mrk, s whole number, should be written. / Mrks in this column - whole numbers only Do not record mrks on scripts in this mnner. Where cndidte hs not been wrded ny mrks for question, or prt of question, 0 should be written in the right hnd mrgin ginst their nswer. It should not be left blnk. Every pge of cndidte s script should be checked for working. Unless blnk, every pge whh is devoid of mrking symbol should hve tk plced in the bottom right hnd mrgin. Where the solution to prt of question is frgmented nd continues lter in the script, the mrks should be recorded t the end of the solution. This should be indted with down rrow ( ), in the mrgin, t the erlier stges. Working subsequent to n error must be followed through, with possible full mrks for the subsequent working, provided tht the level of diffulty involved is pproimtely similr. Where, subsequent to n error, the working for follow through mrk hs been esed, the follow through mrk cnnot be wrded. As indted on the front of the question pper, full credit should only be given where the solution contins pproprite working. Throughout this pper, unless speciflly mentioned in the mrking instructions, correct nswer with no working receives no credit. Pge

3 Mrking Symbols No comments or words should be written on scripts. Plese use the following nd the symbols indted on the welcome letter nd from comment on the previous pge. A tk should be used where piece of working is correct nd gins mrk. Mrkers must check through the whole of response, tking the work only where mrk is wrded. At the point where n error occurs, the error should be underlined nd cross used to indte where mrk hs not been wrded. If no mrk is lost the error should only be underlined, i.e. cross is only used where mrk is not wrded. A cross-tk should be used to indte correct working where mrk is wrded s result of follow through from n error. A double cross-tk should be used to indte correct working whh is irrelevnt or insuffient to score ny mrks. This should lso be used for working whh hs been esed. A tilde should be used to indte minor error whh is not being penlised, e.g. bd form. This should be used where cndidte is given the benefit of the doubt. A roof should be used to show tht something is missing, such s prt of solution or crucil step in the working. These will help mrkers to mintin consistency in their mrking nd will ssist the eminers in the lter stges of SQA procedures. The emples below illustrte the use of the mrking symbols. Emple Emple y dy d 0 y A(,,0), B(,,), C(,,0) AB b AC 0 Emple Emple (repeted error) sin cos k sin cos cos sin k cos, k sin 0 ( )( )( ) or or Since the reminder is 0, must be fctor. ( ) Pge

4 0 In generl, s consequence of n error perceived to be trivil, csul or insignifnt, e.g., cndidtes lose the opportunity of gining mrk. But note emple in comment nd comment. Where trnscription error (pper to script or within script) occurs, the cndidte should be penlised, e.g. This is trnscription error nd so the mrk is not wrded. Esed s no longer solution of qudrt eqution. 0 Eceptionlly this error is not treted s trnscription error s the cndidte dels with the intended qudrt eqution. The cndidte hs been given the benefit of the doubt. Cross mrking Where question results in two pirs of solutions, this technique should be pplied, but only if indted in the detiled mrking instructions for the question. Emple: Point of intersection of line with curve Illustrtive Scheme:, Cross mrked:, y y, y, y Mrkers should choose whhever method benefits the cndidte, but not combintion of both. In finl nswers, numerl vlues should be simplified s fr s possible. Emples: must be simplified to or must be simplified to 0 must be simplified to 0 must be simplified to must be simplified to Regulrly occurring responses (ROR) re shown in the mrking instructions to help mrk common nd/or non-routine solutions. RORs my lso be used s guide in mrking similr non-routine cndidte responses. Unless speciflly mentioned in the mrking instructions, the following should not be penlised: Working subsequent to correct nswer; Correct working in the wrong prt of question; Legitimte vritions in numerl nswers, e.g. ngles in degrees rounded to nerest degree; Omission of units; Bd form; Repeted error within question, but not between questions or ppers. 0 ( )( ) 0 or The squre root of perfect squres up to nd including 00 must be known. Pge

5 In ny Show tht... question, where the cndidte hs to rrive t formul, the lst mrk of tht prt is not vilble s follow through from previous error. All working should be crefully checked, even where fundmentl misunderstnding is pprent erly in the cndidte s response. Mrks my still be vilble lter in the question so reference must be mde continully to the mrking instructions. All working must be checked: the ppernce of the correct nswer does not necessrily indte tht the cndidte hs gined ll the vilble mrks. In the eceptionl circumstnce where you re in doubt whether mrk should or should not be wrded, consult your Tem Leder (TL). Scored out working whh hs not been replced should be mrked where still legible. However, if the scored out working hs been replced, only the work whh hs not been scored out should be mrked. 0 A vlid pproch, within Mthemtl problem solving, is to try different strtegies. Where this occurs, ll working should be mrked. The mrk wrded to the cndidte is from the highest scoring strtegy. This is distinctly different from the cndidte who gives two or more solutions to question/prt of question, delibertely leving ll solutions, hoping to gin some benefit. All such contrdtory responses should be mrked nd the lowest mrk given. It is of gret importnce tht the utmost cre should be eercised in totlling the mrks. The recommended procedure is s follows: Step Step Mnully clculte the totl from the cndidte s script. Check this totl using the grid issued with these mrking instructions. Step In EMC, enter the mrks nd obtin totl, whh should now be compred to the mnul totl. This procedure enbles mrkers to identify nd rectify ny errors in dt entry before submitting ech cndidte s mrks. The cndidte s script for Pper should be plced inside the script for Pper, nd the cndidte s totl score (i.e. Pper Section B + Pper ) written in the spce provided on the front cover of the script for Pper. In cses of diffulty, covered neither in detil nor in principle in these instructions, mrkers should contct their TL in the first instnce. A referrl to the Principl Assessor (PA) should only be mde in consulttion with the TL. Further detils of PA Referrls cn be found in The Generl Mrking Instructions. Pge

6 Pper Section A Question Answer C D B B A C A C A 0 B D B D A D C D B B 0 A Summry A B C D Pge

7 () (i) Show tht ( ) is fctor of (ii) Fctorise fully.. Pper Section B (iii) Solve 0. () Gener Scheme Illustrtive Scheme Method : Using synthet division ss know to use pd pd complete evlution stte conclusion find qudrt fctor fctorise completely stte solutions 0 'reminder is zero so ( ) is fctor' stted, or implied by ( )( )( ),, stted eplitly in ny order Method : Using substitution nd inspection know to use 0 0 ( ) is fctor ( )( ) stted, or implied by ( )( )( ) stted eplitly in ny order,,. is only vilble s consequence of the evidence for. Communtion t must be consistent with working t nd. i.e. cndidte s working must rrive legitimtely t zero before. is wrded. If the reminder is not 0 then n pproprite sttement would be '( ) is not fctor'.. Accept ny of the following for : ' f () 0 so ( ) is fctor ' ' since reminder is 0, it is fctor ' the 0 from tble linked to word ' fctor ' by e.g. ' so ', ' hence ',' ', ' ', ' '.. Do not ccept ny of the following for double underlining the zero or boing in the zero, without comment ' is fctor ', ' ( ) : the word ' fctor ' only, with no link.. To gin. ( )( )( ),,, must pper together in (). is fctor ', ' is root ', ' ( ) is root ' leding to, 0,, 0 nd, 0. ( )( ) only, leding to, does not gin only does not gin. s eqution solved is not cub.. Cndidtes who ttempt to solve the cub eqution subsequent to,, nd obtin different solutions, or no solutions, cnnot gin. Pge

8 (b) The digrm shows the curve with eqution y. The curve crosses the -is t P, Q nd R. Determine the shded re. Gener Scheme Illustrtive Scheme (b) 0 identify from working in () Q interpret pproprite limits ss know nd strt to integrte pd complete integrtion substitute limits pd stte re 0 0, integrte one term correctly (but see Note 0) or equivlent () () 0 or 0 but not deciml pproimtion. Evidence for nd my not pper until stge. 0. Where cndidte differentites one or more terms t. Cndidtes who substitute t. For cndidtes who mke n error in (), vlue is used for the upper limit.., without integrting t, then is only vilble where both limits re numerl vlues., 0,, do not gin nd, 0, re not vilble. nd. is only vilble if 0 is the lower limit nd positive integer. Cndidtes must show evidence tht they hve considered the lower limit 0 in their substitution t Regulrly occurring responses Response Response Cndidtes who use Q throughout Deling with negtives Cndidte A Cndidte B Q 0 ( ) d Q Q Q Q 0 However, if Q is replced by t this stge, nd working continues, ll mrks my still be vilble. Q 0 0 Q(, 0) ( ) ( ) ( ) ( ) ( ) 0 cnnot be negtive so but A d stge. Pge

9 () The epression cos sin cn be written in the form kcos( ) where k 0 nd 0. Clculte the vlues of k nd. Gener Scheme Illustrtive Scheme () ss pd pd use compound ngle formul compre coeffients process k process k k k cos cos k sin sin stted eplitly cos nd sin (do not ccept ) but must be consistent with stted eplitly. Tret kcos cos sin sin s bd form only if the equtions t the. cos cos sin sin or (cos cos sin sin ) is cceptble for. Accept kcos nd ksin for... is not vilble for kcos nd ksin is only vilble for single vlue of., however, stge both contin k. nd. is still vilble.. Cndidtes who work in degrees nd do not convert to rdin mesure in () do not gin. Cndidtes my use ny form of the wve eqution for vlue of is interpreted for the form kcos( ). Regulrly occurring responses Response : Missing informtion in working Cndidte A cos sin tn mrks out of, nd Cndidte B cos sin tn Response : Correct epnsion of kcos( ) nd possible errors for, however, 0 mrks out of nd. is only vilble if the Cndidte C Cndidte D Cndidte E kcos ksin tn so kcos ksin tn so kcos ksin Response : Lbelling incorrect using cos(a B) cos Acos B sin Asin B from formul list Cndidte F Cndidte G Cndidte H kcos Acos B ksin Asin B kcos ksin tn so kcos Acos B ksin Asin B kcos ksin tn so tn so kcos Acos B ksin Asin B k cos B k sin B tn B so B Not consistent with evidence t. Pge

10 (b) Find the points of intersection of the grph of y cos sin with the nd y es, in the intervl 0 Gener Scheme Illustrtive Scheme (b) ss interpret y-intercept strtegy for finding roots stte both roots or, e.g. cos 0 sin cos. Cndidtes should only be penlised once for leving their nswer in degrees in () nd (b).. If the epression used in (b) is not consistent with () then only 0. Correct roots without working cnnot gin but will gin. nd re vilble.. Cndidtes should only be penlised once for not simplifying in () nd (b). Regulrly occurring responses Response : Communtion for Cndidte I (, 0) without working. Cndidte J cos0 sin0 so (, 0). Response : Follow through from wrong vlue of Cndidte K From () then in (b), only Cndidte L From () 0 then in (b) 0, 0 only Note 0 Response : Root or grphl pproch Cndidte M Cndidte N Cndidte O () 0 nd nd (b) Response : Circulr rgument not leding nywhere Cndidte P 0 0 When 0, 0 moved 0 to left cuts -is t, 0 0 cos sin 0 cos sin 0 Response : Trnscription error in (b) Cndidte Q () correct (b) cos 0 so, y cos 0 cos is penlised s obtined in (). However nd re still vilble s follow through. See Note. Pge 0

11 () Find the eqution of, the perpendulr bisector of the line joining P(, ) to Q(, ). Gener Scheme Illustrtive Scheme () ss ss find midpoint of PQ find grdient of PQ interpret perpendulr grdient stte eqution of perp. bisector (, ) y ( ). is only vilble if midpoint nd perpendulr grdient re used.. Cndidtes who use y m c must obtin numerl vlue for c before is vilble. Regulrly occurring responses Response : Cndidtes who use wrong midpoint or no midpoint Cndidte A midpoint M(, ) m m MQ y ( ) Cndidte B m m PQ using R, y( ) (b) Find the eqution of whh is prllel to PQ nd psses through R(, ). P Gener Scheme Illustrtive Scheme (b) use prllel grdients stte eqution of line stted, or implied by y ( ) ( ). is only vilble to cndidtes who use R nd their grdient of PQ from (). Regulrly occurring responses Response : Not using prllel grdient for eqution Cndidte C Cndidte D Cndidte E y( ) Prllel so sme grdients m so m y( ) y( ) If m only do not wrd PQ Pge

12 (c) Find the point of intersection of nd. (c) ss pd pd Gener Scheme use vlid pproch solve for one vrible solve for other vrible Illustrtive Scheme y nd y e.g. e.g. e.g. y or or (y ) y. Neither y nd y nor y nd y re suffient to gin.., nd Equte zeros re not vilble to cndidtes who: Give nswers only, without working Use R for equtions in both () nd (b) Use the sme grdient for the lines in () nd (b). (d) Hence find the shortest distnce between PQ nd. Gener Scheme Illustrtive Scheme (d) 0 ss identify pproprite points 0 (, ) nd, pd clculte distnce 0 ccept or. 0 nd re only vilble for considering the distnce between the midpoint of PQ nd the cndidte s nswer from (c) or for considering the perpendulr distnce from P or Q to.. At lest one coordinte t 0 stge must be frction for. There should only be one clcultion of distnce to gin Regulrly occurring responses Response : Following through from correct (), (b) nd (c) Cndidte F (, ), (, ) d Response : Following through from correct (), (b) nd (c) Cndidte G (, ),, 0 PR, QR=, d so is shortest distnce. to be vilble.. 0 If reference ws mde to this being the perpendulr distnce then would be vilble. Pge

13 Pper Functions f nd g re defined on the set of rel numbers by f ( ) () Find epressions for: (i) f g( ) ; g( ) (ii) g f ( ). () Gener Scheme strt composite process correct substitution into epression complete second composite Illustrtive Scheme e.g. f( ) stted, or implied by ( ). Cndidtes must clerly identify whh of their nswers re f g( ) nd g f ( ) for this could be s little s using (i) nd (ii) s lbels. ; the minimum evidence. Cndidtes who interpret the composite functions s either f ( ) g( ) or f ( ) g( ), do not gin ny mrks. Regulrly occurring responses Response : The first two mrks re for either f g( ) or g f ( ) composite function. Cndidte A f g( ) ( ) Cndidte B f g( ) ( ) correct. The third mrk is for the other g f ( ) mrks out of g f ( ) mrks out of Response : Interpreting f g( ) s g f ( ) Cndidte C f g( ) g f ( ) ( ) nd ve vers. A mimum of mrks re vilble. Cndidte D f g( ) mrks out of Response : Identifying f g( ) nd g f ( ) mrk out of Cndidte E Cndidte F Cndidte G Cndidte H ( ) mrks out of ( ) mrk out of (i) ( ) ONLY (ii) or ( ) ONLY 0 mrks out of mrks out of Pge

14 (b) Show tht f g g f ( ) ( ) 0 hs no rel roots. (b) Gener Scheme Method : Discriminnt pd obtin qudrt epression ss know to nd use discriminnt interpret result Illustrtive Scheme Method : Discriminnt or stted, or implied by 0 or 0 so no rel roots Method : Qudrt Formul pd obtin qudrt epression ss know to nd use qudrt formul interpret result Method : Qudrt Formul stted, or implied by not possible so no rel roots. Cndidtes who use f ( ) g( ) cn gin no mrks in (b) s cub will be obtined.. Cndidtes who use f ( ) g( ) do not gin (esed) but nd. In method, ny other formul msquerding s discriminnt cnnot gin.., nd re only vilble if f( g( )) g( f ( )) b c, with b nd c both non-zero. re vilble s follow through mrks. nd. simplifies to qudrt epression of the form is only vilble for numerl vlue, clculted correctly from the cndidte s response t leding to no rel roots.. Do not ccept for : no roots in lieu of no rel roots mths error or m error.. Cndidtes who use the word derivtive insted of discriminnt should not be penlised. Regulrly occurring responses Response : Cndidtes who do not simplify the vlue of their discriminnt Cndidte I 0 0 so no rel roots Response : Acceptble communtion mrks Method Method Cndidte J Cndidte K Cndidte M, nd not vlid so no rel roots Cndidte L Discriminnt cn't find root of negtive so no rel roots no ve so no rel roots ( ) no rel roots if b 0 c 0 Pge

15 () Reltive to suitble set of coordinte es, digrm shows the line y 0 intersecting the circle y y 0 0 t the points P nd Q. Find the coordintes of P nd Q. P Digrm Q () ss ss Gener Scheme rerrnge liner eqution substitute into circle pd epress in stndrd form pd strt to solve stte roots pd determine corresponding y-coordintes Substituting for y Illustrtive Scheme y stted, or implied by... ( )... ( )... 0 e.g. ( )( ) nd y y nd = 0 must pper t the or stge to gin. Substituting for y stted, or implied by y y y y y y y y = 0 must pper t the e.g. ( )( ) or stge to gin nd nd.. At the qudrt must led to two rel distinct roots for. Cross mrking is vilble here for nd. nd. Cndidtes do not need to distinguish between points P nd Q. Regulrly occurring responses Response : Solving qudrt eqution to be vilble. Cndidte A Cndidte B Cndidte C 0 0 ( )( ) y y Cross mrking is not vilble here for ( ) ( ) y, nd, s there re no distinct roots. See Note ( ) ( ) y y Pge

16 (b) Digrm shows the circle from () nd second congruent circle, whh lso psses through P nd Q. Determine the eqution of this second circle. P Q Digrm (b) Gener Scheme centre of originl circle pd rdius of originl circle Illustrtive Scheme (, ) 0 Accept r 0 Method : Using midpoint 0 ss midpoint of chord ss evidence for finding new centre centre of new circle eqution of new circle Method : Using midpoint 0 (, ) e.g. stepping out or midpoint formul (, ) y ( ) ( ) 0 Method : Stepping out using P nd Q 0 ss evidence of C to P or C to Q ss evidence of Q to C or P to C centre of new circle eqution of new circle Method : Stepping out using P nd Q 0 e.g. stepping out or vector pproch e.g. stepping out or vector pproch (, ) y ( ) ( ) 0. The evidence for nd my pper in ().. Centre (, ) without working in method my still gin but not, 0 or. Any other centre without working in method does not gin, 0, or. but not 0,. The centre must hve been clerly indted before it is used t the. Do not ccept e.g. or 0 or stge. 0 or, or ny other deciml pproimtions for, in method my still gin, in method does not gin. The evidence for my not pper until the cndidte sttes the rdius or eqution of the second circle. Regulrly occurring responses Response : Emples of evidence for stepping out for, (, ) (, ) (, ) (, ), (, ) (, ) Response : Emples of evidence whh do not gin (, ) (, ) (, ) 0 in method or or. 0 in method for stepping out 0 in method (, ) (, ) (, ) (, ) (, ) (, ) For method (, ) P (, ) (, ) (, ) Q (, ) Pge

17 A function f is defined on the domin 0 by f ( ). Determine the mimum nd minimum vlues of f. Gener Scheme Illustrtive Scheme ss strt to differentite ss complete derivtive nd set to 0 pd strt to solve f( ) 0 pd solve f( ) 0 evlute f t relevnt sttionry point ss consider end-points stte m. nd min. vlues f nd f differentite or correctly e.g. ( )( ), f () (0) () m. nd min. = 0 must pper t or to gin.. The only vlid pproch is vi differentition. A numerl pproch cn only gin. Cndidtes who consider sttionry points only cnnot gin. Tret mimum (0, ) nd minimum (, ) s bd form.. Cross mrking is not pplble to or. or. Ignore ny nture tble whh my pper in cndidte s solution, however (, ) t tble is suffient for. Regulrly occurring responses Response : Algebr issues in working Cndidte A y' ()( ) Response : Derivtive not eplitly set to zero. Cndidte B 0 ()( ) Cndidte D Cndidte E Cndidte F Cndidte G '( ) f f'( ) 0, When, y f(0) nd f() m, min 0 Cndidte C ()( ) 0 0 f () f'( ) 0 '( ) f ( )( ) or so f (). Since is within the domin, f must lso be clculted to gin Ignore the vlue of if it is included. f'( ) 0 ( )( ) f here, f'( ) 0 only. Pge

18 continued Regulrly occurring responses Response : Solving qudrt eqution Cndidte H f ( ) 0 (), Response : Numerl pproch Cndidte J f (0) f () Cndidte I 0 ( ) () ( ) Due to method chosen re not vilble. This cndidte hs styed within the intervl 0., Ignore omission of negtive sign t squre here., nd Cndidte K f (0) f () f () f () Cndidte L f (0) f () f () f () f () This cndidte hs gone outwith the intervl 0. For, f () must come from clculus nd not from ny other pproch. Pge

19 The digrm below shows the grph of qurt y h( ), with sttionry points t 0 nd. On seprte digrms sketch the grphs of: () y h( ); O (b) y h( ). y y h( ) Gener Scheme Illustrtive Scheme () identify roots interpret point of inflection complete cub curve 0 nd only turning point t (, 0) cub, pssing through O with negtive grdient. All grphs must include both the nd y es (lbelled or unlbelled), however the origin need not be lbelled.. No mrks re vilble unless grph is ttempted.. No mrks re vilble to cndidte who mkes severl ttempts t grph on the sme digrm, unless it is cler whh is the finl grph.. A liner grph gins no mrks in both () nd (b). (b) reflection in -is 0 trnsltion nnottion of 'trnsformed' grph reflection of grph in () in -is grph moves prllel to y-is by units upwrds two 'trnsformed' points ppropritely nnotted (see Note ). Trnsformed here mens reflection followed by trnsltion.. nd pply to the entire curve.. In ech of the following circumstnces : Cndidtes who trnsform the originl grph Cndidtes who sketch prbol in () mrk the cndidte s ttempt s norml nd unless mrk of 0 hs been scored, deduct the lst mrk wrded. Indte this with (see Regulr occurring response G).. A reflection in ny line prllel to the y-is does not gin 0. A trnsltion other thn does not gin or. or. Grph for () Grph for (b) y y O O Pge reflection in is 0 trnsltion

20 Pge 0 continued Regulrly occurring responses In () In () nd (b) 0 O (0, ) (, ) (, ) A B C D E F G H J K L (, ) Tret cusp s turning point. y-vlues of should be roughly in line too fr prt here See Note See Note Curve does not only pss through 0 nd on -is. I See Note

21 A is the point (,, 0), B is (,, ) nd C is (, k, 0). () (i) Epress BA nd BC in component form. (ii) Show tht cosabc ˆ. ( k k) () Gener Scheme interpret vector pd process vector ss use sclr product pd find sclr product pd find BA 0 k Illustrtive Scheme ˆ BA. BC cos ABC see Note BA BC Tret 0 written s (, 0, ) s bd form. find epression for BC complete to result ( k ) ( ) or equivlent nd k k ( k k ) or BA BC k k nd ( k k ). If the evidence for. is dependent on gining Regulrly occurring responses does not pper eplitly, then is only wrded if working for is ttempted., Response : Clculting wrong ngle Cndidte A OA.OC cos AOC= OA OC nd. Cndidte B OA.OB cos AOB= OA OB OA.OC ( ) k 0 0 k OA.OB ( ) ( ) 0 OA OA OC k OB cos ABC= k k cos ABC= Pge

22 (b) If ngle ABC 0, find the possible vlues of k. Gener Scheme (b) Method : Squring first 0 link with () ss squre both sides pd rerrnge into 'non-frctionl' formt pd write in stndrd form pd solve for k Method : Deling with frctions first 0 link with () pd rerrnge into 'non-frctionl' formt ss squre both sides pd write in stndrd form pd solve for k Method : Squring first cos 0 ( k k) ( k k) 0 k k k k k or equivlent 0 or equivlent or Illustrtive Scheme Method : Deling with frctions first ( k k) cos 0 k k ( ) 0 k k ( ) k k k 0 or equivlent or = 0 must pper t this stge. = 0 must pper t this stge.. The evidence for. my pper in the working for 0 in both methods. is the only mrk vilble to cndidtes who replce cos 0 o by 0 in method nd 0 in method.. All mrks re vilble to cndidtes who use 0. for cos 0 o but 0. cn gin mimum of mrks. Regulrly occurring responses Response : Working with cos 0 o throughout the question Response : Using the wrong vlue for cos 0 o Cndidte C (Method ) Cndidte D (Method ) cos 0 ( k k) ( k k) (cos 0 ) (cos 0 ) ( k k) ( k k) (cos 0 ) ( k k ) 0 If cos 0 o is subsequently evluted then nd my still be vilble. ( k k ) ( k k ) k k k k 0 k ( ) 0, 0 Pge

23 For 0, sequences cn be generted using the recurrence reltion u (sin ) u cos, with u. n n 0 () Why do these sequences hve limit? () Gener Scheme condition on coeffient. For do not ccept: u n connect coeffient with given intervl sin lies between nd sin However, ccept ' sin greter thn nd less thn ' for. sin in intervl, 0 sin Illustrtive Scheme. Do not ccept for unless is clerly identified s sin, whh my not pper until (b).. 0 sin nd nothing else, does not gin but gins.. 0 sin nd nothing else, does not gin or. Regulrly occurring responses Response : Attempts t giving reson for limit Cndidte A Cndidte B This sequence hs limit becuse, sin within the domin. Since sin in this domin will lwys be greter thn 0 nd less thn. Cndidte C Cndidte D u n sin nd sin0 0 so the multiplier of is between 0 nd, so it hs limit. sin, for 0, 0 sin so limit eists Response : Minimum response for both mrks Cndidte E for 0, 0 sin so sin so limit Cndidte F if limit, sin for 0, 0 sin Pge

24 (b) The limit of one prtulr sequence generted by this recurrence reltion is sin. Find the vlue(s) of. (b) Gener Scheme ss pproprite limit method substitute for limit ss use pproprite double ngle formul pd epress in stndrd form pd strt to solve qudrt eqution pd reduce to equtions in sin only select vlid solution Illustrtive Scheme or or ( my be stted, or implied by in both methods) cos limit l sin l cos sin cos sin sin sin sin cos sin...sin... e.g. sin sin e.g. ( sin )(sin ) sin or sin 0 0 or outwith intervl = 0 must pper t to gin. or., nd re only vilble if qudrt eqution is obtined t. Cndidtes my epress the qudrt eqution t the. who do not solve trigonometr qudrt eqution t,. For. nd b L cos sin sin sin cos sin sin cos 0 sin 0 sin sin ( sin ) 0 sin nd sin 0, 0, stge. stge in the form s s 0. For cndidtes sin must pper eplitly to gin re not vilble to cndidtes who solve qudrt eqution in the form there must be one vlid solution, nd one solution outwith intervl whh is rejected. is not vilble to cndidtes who leve their nswer in degree mesure. 0. Cross mrking is vilble for Regulrly occurring responses nd. Response : Evidence for identiftion of ppering in (b) Cndidte G () (b) b cos L sin Response : Error in lgebr nd subsequent qudrt eqution solution Cndidte H Cndidte I cos sin sin ( sin ) sin sin sin 0 sin not possible (sin )(sin ) 0 sin nd sin. b c c, 0. See Note Pge

25 The digrm shows the curves with equtions y nd y. y y The grphs intersect t the point T. () Show tht the -coordinte of T cn be written in the form log p, for ll log q. T y O () Gener Scheme use lw of logs : log log log ss ss equte epressions for y tke logrithms of both sides n use lw of logs : log nlog pd gther like terms complete to required form In methods nd : If the first line of working is tht t the stge, then nd re wrded. If the first line of working is tht t the stge, then only wrded. nd re p q pq Illustrtive Scheme Method log ( ) log log log log log stted, or implied by (log log ) log log log stted eplitly Method Method log log log log log log log stted eplitly log log log log log log stted eplitly. In methods nd, if no bse is indted then In method, if no bse is indted then. In ll methods, if numerl bse is used then. In method, the omission of brckets t the. p nd q must be numerl vlues. Regulrly occurring responses Response : Omission of brckets round Cndidte A log log Response : Using different bses Cndidte C log log log ( ) log is not vilble, however is not vilble, however is not vilble., nd is still vilble. stge is treted s bd form, see Response. Cndidte B Response : Tking logs first Cndidte D y nd y re still vilble. log log (log log ) y y y y log log nd log log log log nd log ( ) log log ( ) log log log log log log log Pge

26 (b) Clculte the y-coordinte of T. Gener Scheme Illustrtive Scheme (b) substitute in for pd process y log log e.g. y stted, or implied by 0 y e.g. 0. Cndidtes must work to t lest two signifnt figures in (b) e.g. does not gin. vilble. is only vilble if the power used comes from log Regulrly occurring responses log Response : Using p nd q s integer vlues without working Cndidte E p q Response : Using integer vlues clculted in () Cndidte G p 0 q y y or p q Cndidte F p q y in (). 0 or, but is [END OF MARKING INSTRUCTIONS] Pge