Mark Scheme (Results) January GCE Mathematics 6666 Core Mathematics 4

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1 Mark Scheme (Resuls) January 0 GCE Mahemaics 6666 Core Mahemaics 4

2 Edecel and BTEC Qualificaions Edecel and BTEC qualificaions come from Pearson, he world s leading learning company. We provide a wide range of qualificaions including academic, vocaional, occupaional and specific programmes for employers. For furher informaion, please visi our websie a Our websie subjec pages hold useful resources, suppor maerial and live feeds from our subjec advisors giving you access o a poral of informaion. If you have any subjec specific quesions abou his specificaion ha require he help of a subjec specialis, you may find our Ask The Eper service helpful. Pearson: helping people progress, everywhere Our aim is o help everyone progress in heir lives hrough educaion. We believe in every kind of learning, for all kinds of people, wherever hey are in he world. We ve been involved in educaion for over 50 years, and by working across 70 counries, in 00 languages, we have buil an inernaional repuaion for our commimen o high sandards and raising achievemen hrough innovaion in educaion. Find ou more abou how we can help you and your sudens a: January 0 Publicaions Code UA0468 All he maerial in his publicaion is copyrigh Pearson Educaion Ld 0

3 General Marking Guidance All candidaes mus receive he same reamen. Eaminers mus mark he firs candidae in eacly he same way as hey mark he las. Mark schemes should be applied posiively. Candidaes mus be rewarded for wha hey have shown hey can do raher han penalised for omissions. Eaminers should mark according o he mark scheme no according o heir percepion of where he grade boundaries may lie. There is no ceiling on achievemen. All marks on he mark scheme should be used appropriaely. All he marks on he mark scheme are designed o be awarded. Eaminers should always award full marks if deserved, i.e. if he answer maches he mark scheme. Eaminers should also be prepared o award zero marks if he candidae s response is no worhy of credi according o he mark scheme. Where some judgemen is required, mark schemes will provide he principles by which marks will be awarded and eemplificaion may be limied. When eaminers are in doub regarding he applicaion of he mark scheme o a candidae s response, he eam leader mus be consuled. Crossed ou work should be marked UNLESS he candidae has replaced i wih an alernaive response.

4 General Insrucions for Marking EDEXCEL GCE MATHEMATICS. The oal number of marks for he paper is 75.. The Edecel Mahemaics mark schemes use he following ypes of marks: M marks: mehod marks are awarded for knowing a mehod and aemping o apply i, unless oherwise indicaed. A marks: Accuracy marks can only be awarded if he relevan mehod (M) marks have been earned. B marks are uncondiional accuracy marks (independen of M marks) Marks should no be subdivided. In some insances, he mark disribuions (e.g. M, B and A) prined on he candidae s response may differ from he final mark scheme.. Abbreviaions These are some of he radiional marking abbreviaions ha will appear in he mark schemes. bod benefi of doub f follow hrough he symbol will be used for correc f cao correc answer only cso - correc soluion only. There mus be no errors in his par of he quesion o obain his mark isw ignore subsequen working awr answers which round o SC: special case oe or equivalen (and appropriae) dep dependen indep independen dp decimal places sf significan figures or AG: The answer is prined on he paper dm denoes a mehod mark which is dependen upon he award of he previous mehod mark. ddm denoes a mehod mark which is dependen upon he award of he previous mehod marks. dm* denoes a mehod mark which is dependen upon he award of he M* mark. 4. All A marks are correc answer only (cao.), unless shown, for eample, as A f o indicae ha previous wrong working is o be followed hrough. Afer a misread however, he subsequen A marks affeced are reaed as A f, bu incorrec answers should never be awarded A marks.

5 Use of a formula Where a mehod involves using a formula ha has been learn, he advice given in recen eaminers repors is ha he formula should be quoed firs. Normal marking procedure is as follows: Mehod mark for quoing a correc formula and aemping o use i, even if here are misakes in he subsiuion of values. Where he formula is no quoed, he mehod mark can be gained by implicaion from correc working wih values, bu may be los if here is any misake in he working. Eac answers Eaminers repors have emphasised ha where, for eample, an eac answer is asked for, or working wih surds is clearly required, marks will normally be los if he candidae resors o using rounded decimals. Answers wihou working The rubric says ha hese may no gain full credi. Individual mark schemes will give deails of wha happens in paricular cases. General policy is ha if i could be done in your head, deailed working would no be required. Misreads A misread mus be consisen for he whole quesion o be inerpreed as such. These are no common. In clear cases, please deduc he firs A (or B) marks which would have been los by following he scheme. (Noe ha marks is he maimum misread penaly, bu ha misreads which aler he naure or difficuly of he quesion canno be reaed so generously and i will usually be necessary here o follow he scheme as wrien). Someimes following he scheme as wrien is more generous o he candidae han applying he misread rule, so in his case use he scheme as wrien.

6 Quesion Number. ( ) January Core Mahemaics C4 Mark Scheme Scheme ( + ) = + = + 8 ( )( 4) ( )( 4)( 5) = + ( )( k) + ( k) + ( k) !! () or 8 see noes B Marks M A ( )( 4) ( )( 4)( 5) = + ( ) !! = ; See noes below! = ; A; A B: () or 8 ouside brackes or 8 as consan erm in he binomial epansion. M: Epands (... + k) o give any erms ou of 4 erms simplified or un-simplified, ( )( 4) Eg: + ( )( k) or ( ) ( k) ( k) ( )( 4) + or ( k)!! ( )( 4) ( )( 4)( 5) or ( k) + ( k) where k are ok for M.!! ( )( 4) ( )( 4)( 5) A: A correc simplified or un-simplified + ( )( k) + ( k) + ( k)!! epansion wih consisen ( k ) where k. ( )( 4) ( )( 4)( 5) Incorrec brackeing + ( ) is MA0 8!! unless recovered. A: For (simplified fracions) or also allow Allow Special Case A for eiher SC: 9 ;... 8 or SC: K eiher a simplified fracion or a decimal. (where K can be or omied), wih each erm in he [ ] A: Accep only 7 5 or or [5] 5

7 . cd ( )( 4) ( )( 4)( 5) Candidaes who wrie = + ( ) !! k = and no and achieve will ge BMAA0A Alernaive mehod: Candidaes can apply an alernaive form of he binomial epansion. 4 ( )( 4) 5 ( )( 4)( 5) 6 ( + ) = () + ( )() ( ) + () ( ) + () ( )!! B: 8 or () where M: Any wo of four (un-simplified) erms correc. A: All four (un-simplified) erms correc. A: A: + 6 Noe: The erms in C need o be evaluaed, so 4 5 C 6 0() + C() ( ) + C() ( ) + C() ( ) wihou furher working is B0M0A0.

8 Quesion Number Scheme. (a) du u = ln = d ln d, dv = v = = d (b) = ln. d = ln + d = ln + + { c} ± λ In he form ln ± µ. M ln ln ln ln 4 = () 4() () 4() simplified or un-simplified. A. simplified or un-simplified. A ± µ ± β Correc answer, wih/wihou.. Applies limis of and o heir par (a) answer and subracs he correc way round. dm + c A 8 = ln or ln or ( ln ), ec, or awr 0. or equivalen. A ± λ (a) M: Inegraion by pars is applied in he form ln ± µ. or equivalen. A: ln simplified or un-simplified. (b) A :. or equivalen. You can ignore he d. dm: Depends on he previous M. ± µ ± β.. A: ln + { + c} or ln { c} = 4 + or ln { + c } 4 ln or { + c} or equivalen. 4 You can ignore subsequen working afer a correc saed answer. M: Some evidence of applying limis of and o heir par (a) answer and subracs he correc way round. ln ( 8 4 ) + A: Two erm eac answer of eiher ln or ln or ( ln ) or or ln. Also allow awr 0.. Also noe he fracion erms mus be combined. Noe: Award he final A0 in par (b) for a candidae who achieves awr 0. in par (b), when heir answer o par (a) is incorrec. M [5] [] 7

9 . (b) cd Noe: Decimal answer is in par (b). Alernaive Soluion du 4 u = = d ln d, dv = ln v = ln d ln d = ( ln ) ( ln ) d 4 ln d = ( ln ) d ln d = ( ln ) + + c { } ln d = ( ln ) + c 4 = ln 4 + { c} { } λ k ln d = ( ln ) ± d where k Any one of ( ln ) or d ( ln ) d and k = ± µ ± β. ( ln ) 4 or equivalen wih/wihou + c. M A A dm A

10 Quesion Number Scheme Marks. Mehod : Using one ideniy B C A + + ( + )( ) ( + ) ( ) A = heir consan erm = B A( + )( ) + B( ) + C( + ) Forming a correc ideniy. B Eiher : 9 = A, : 0 = 5A + B + C Aemps o find he value of consan: 0 = A B + C eiher one of heir B or heir C M or from heir ideniy. = = 7B 4 = 7B B = Correc values for heir B and heir C, which are A = + 0 = C = C C = found using a correc ideniy. Mehod : Long Division ( + )( ) ( + )( ) heir consan erm = B 5 4 B C So, + ( + )( ) ( + ) ( ) 5 4 B( ) + C( + ) Forming a correc ideniy. B [4] Eiher : 5 = B + C, consan: 4 = B + C or = 0 4 = 7B 4 = 7B B = = 4 = C = C C = So, ( + )( ) ( + ) ( ) Aemps o find he value of eiher one of heir B or heir C from heir ideniy. Correc values for heir B and heir C, which are found using 5 4 B( ) + C( + ) M A [4] 4 s B: Their consan erm mus be equal o for his mark. nd B (M on epen): Forming a correc ideniy. This can be implied by laer working. M (A on epen): Aemps o find he value of eiher one of heir B or heir C from heir ideniy. This can be achieved by eiher subsiuing values ino heir ideniy or comparing coefficiens and solving he resuling equaions simulaneously. A: Correc values for heir B and heir C, which are found using a correc ideniy A B Noe : +, leading o A( ) + B( + ), leading o ( + )( ) ( + ) ( ) A = and B = will gain a maimum of B0B0MA0.

11 . cd Noe: You can imply he nd B from eiher 5 4 B( ) + C( + ) or ( + )( ) ( + )( ) A( + )( ) + B( ) + C( + ) ( + )( ) ( + )( ) Alernaive Mehod : Iniially dividing by ( + ) "( + )"( ) ( ) ( + )( ) ( ) ( + )( ) B: heir consan erm = 4 B C So, + ( + )( ) ( + ) ( ) 4 B( ) + C( + ) B: Forming a correc ideniy. B =, C = 6 M: Aemps o find eiher one of heir B or heir C from heir ideniy So, + + ( + )( ) ( ) ( + ) ( ) and ( + )( ) ( + ) ( ) Alernaive Mehod : Iniially dividing by ( - ) ( + )"( )" ( + ) ( + )( ) 5 7 A: Correc answer in parial fracions. + B: heir consan erm = ( + ) ( + )( ) 7 B C So, + ( + )( ) ( + ) ( ) 7 B( ) + C( + ) B: Forming a correc ideniy. B =, C = So, + + ( + )( ) ( + ) ( + ) ( ) and ( + )( ) ( + ) ( ) M: Aemps o find eiher one of heir B or heir C from heir ideniy. A: Correc answer in parial fracions.

12 Quesion Scheme Marks Number 4. (a).098 B cao [] (b) Area ; ( heir.098) +. B; M = =.845 =.84 ( dp).84 or awr.84 A [] du d (c) { u = + } = or ( u ) d du = B (a) ( u ) d =.( u ) du + u u {} u ( u ) ( u u + u ) = d = du u = {} u u + du u u u = {} + u lnu u Area( R ) = u + 6u lnu () () = () + 6() ln () + 6() ln 9 = + ln ln or + ln or ln, ec 4 ( u )... u ( ) u.( u ) u Epands o give a four erm cubic in u. Eg: ± Au ± Bu ± Cu ± D M A M An aemp o divide a leas hree erms in heir cubic by u. See noes. M ( u ) u u + u lnu u A Applies limis of and in u or 4 and in and subracs eiher way round. Correc eac answer or equivalen. B:.098 correc answer only. Look for his on he able or in he candidae s working. (b) B: Ouside brackes or M: For srucure of rapezium rule[... ] A: anyhing ha rounds o.84 Noe: Working mus be seen o demonsrae he use of he rapezium rule. Noe: acual area is Noe: Award BM A for (0.5 +.) + ( heir.098 ) =.845 Brackeing misake: Unless he final answer implies ha he calculaion has been done correcly Award BM0A0 for ( heir.098 ) +. (nb: answer of 6.86). Award BM0A0 for (0.5 +.) + ( heir.098 ) (nb: answer of ). M A [8]

13 4. (b) cd Alernaive mehod for par (b): Adding individual rapezia Area + + =.845 B: and a divisor of on all erms inside brackes. M: Firs and las ordinaes once and wo of he middle ordinaes wice inside brackes ignoring he. A: anyhing ha rounds o.84 (c) du d B: = or du = d or du = d or d = ( u )du or ( u ) d du = oe. s ( u ) M: becoming (Ignore inegral sign). + u s ( u ) ( u ) A (B on epen): d becoming.( u ) { d u} or. { du}. + u u ( u ) You can ignore he inegral sign and he d u. nd M: Epands o give a four erm cubic in u, ± Au ± Bu ± Cu ± D where A 0, B 0, C 0 and D 0 The cubic does no need o be simplified for his mark. rd M: An aemp o divide a leas hree erms in heir cubic by u. ( u u + u ) Ie. u u + u u nd ( u ) u u A: du + u lnu u 4 h M: Some evidence of limis of and in u and subracing eiher way round. rd 9 A: Eac answer of + ln ln or + ln or ln 4 or + ln ln 6 or + ln, ec. Noe: ha fracions mus be combined o give eiher or or 6 6 Alernaive mehod for nd M and rd M mark ( u ) ( u u + ).( u ) du =.( u ) du u u {} {} {} ( ) = {} u +.( u ) du = u... du u = {} u u + u + du u = {} u u + du u An aemp o epand ( u ), hen divide he resul by u and hen go on o muliply by ( u ). o give hree ou of four of D ± Au, ± Bu, ± C or ± u nd M rd M

14 4. (c) cd Final wo marks in par (c): u = + ( + ) Area( R ) = ln+ ( + 4 ) = ( + 4) + 6 ( + 4) ln ( + 4 ) ( + ) ( + ) + 6 ( + ) ln ( + ) 6 = ( ln ) + ln 9 = + ln ln or + ln or ln, ec 4 ( ) ( ) ( ) 4 M: Applies limis of 4 and in and subracs eiher way round. A: Correc eac answer or equivalen. Alernaive mehod for he final 5 marks in par (b) d" u" ( u ) " u" = u = u d d u, 4 u d v ( u ) = ( u ) v = d ( u ) ( u ) = d u 4u 4 u 4 4 ( u ) u 4u + 6u 4u + M: Applies inegraion by pars and = + d u 4u 4 u epands o give a five erm quaric. 4 ( u ) 4 = + u 4u du M: Dividing a leas 4 erms. 4u 4 u u 4 ( u ) u = + u + 6u 4lnu A: Correc Inegraion. 4u 4 u 4 ( u ) ( u ) u u u du = + + lnu u 4u 4u = + + ln + + ln = ( 7 ln) ln = + ln 6 ( u ) Area( R) = du = + ln A u 6 M

15 or equivalen. A Quesion Number Scheme Marks 5. Working paramerically: ln =, y = or y = e (a) { = 0 } 0 = = Applies = 0 o obain a value for. M When =, y = = Correc value for y. A [] (b) Applies y = 0 o obain a value for. { y = 0 } 0 = = 0 (Mus be seen in par (b)). M When = 0, = (0) = = A [] (c) d d dy ln d dy ln e ln d B dy ln d Aemps heir d y divided by heir d. d d M A A, = "", so m( T) = 8ln m( N ) = Applies = "" and m( N) = 8ln m( T) M M A oe y = ( 0) or y = + or equivalen. See noes. 8ln 8ln cso [5] (d) Complee subsiuion Area( R) = ( ). d for boh y and d M = = 4 and = = 0 B Eiher ln ( ) or ( ) M* = ± α (ln ) ln or ( ) ± α (ln)( ) ( ) ln A 0 6 Depends on he previous mehod mark. = 4 ln ln ln Subsiues heir changed limis in and 4 subracs eiher way round. dm* = 5 5 ln ln [6] 5

16 5. (a) M: Applies = 0 and obains a value of. A: For y = = or y = 4 = Alernaive Soluion : M: For subsiuing = ino eiher or y. A: = () = 0 and y = = Alernaive Soluion : M: Applies y = and obains a value of. A: For = () = 0 or = = 0. Alernaive Soluion : M: Applies y = or = 0 and obains a value of. A: Shows ha = for boh y = and = 0. (b) M: Applies y = 0 and obains a value of. Working mus be seen in par (b). A: For finding =. Noe: Award MA for =. (c) B: Boh d d and d y d correc. This mark can be implied by laer working. M: Their d y divided by heir d dy or heir. Noe: heir d y mus be a funcion of. d d d d d heir d M: Uses heir value of found in par (a) and applies m( N) =. m( T) M: y = (heir normal gradien) or y = (heir normal gradien) + or equivalen. A: y = ( 0) or y = + or y = ( 0) or ( 8ln) y 4ln = 8ln 8ln ln 56 y or =. You can apply isw here. ( 0) 8ln Working in decimals is ok for he hree mehod marks. B, A require eac values. d. heir d B: Changes limis from. = = 4 and = = 0. Noe = 4 and = 0 seen is B. (d) M: Complee subsiuion for boh y and d. So candidae should wrie down ( ) M*: Inegraes correcly o give... or inegraes ( ) ln o give eiher ( ) ± α (ln ) or ± α(ln )( ). A: Correc inegraion of ( ) wih respec o o give. ln dm*: Depends upon he previous mehod mark. Subsiues heir limis in and subracs eiher way round. 5 A: Eac answer of ln or 5 5 4ln 7.5 or or ln 4 ln ln or 5 log e or equivalen.

17 Quesio n Scheme Number 5. Alernaive: Convering o a Caresian equaion: = y = (a) { } y = 0 = Applies = 0 in heir Caresian equaion... y =... o arrive a a correc answer of. A (b) { } y = 0 0 = 0 = =... Applies y = 0 o obain a value for. (Mus be seen in par (b)). = = dy d (c) = ( ) ln M M A ± λ, λ M lnor equivalen A ( ) A A, = 0, so m( T) = 8ln m( N ) = Applies = 0 and m( N) = 8ln m( T) y = ( 0) or y = + or 8ln 8ln As in he original scheme. equivalen. (d) R = ( ) (d) Area( ) d ( ) = d = ln Form he inegral of heir Caresian equaion of C. For wih limis of = and =. Ie. ( ) Eiher or ( ) or ( ) ln ± α (ln ) α(ln )( ) ± ( ) ln M Marks [] [] M A oe or equivalen. A 6 Depends on he previous mehod = + ln mark. ln ln Subsiues limis of - and heir dm* B and subracs eiher way round. = 5 5 ln ln [6] 5 Alernaive mehod: In Caresian and applying u = M B M* A [5]

18 u ( ){ }, where u = M0: Unless a candidae wries ( ){ d } 0 u ( )( ){ du} Area( R) = d = 4 Then apply he working paramerically mark scheme.

19 Quesio n Scheme Number 5. (d) Alernaive mehod: For subsiuion u= Area( R) = ( ). d du du where u = ln uln d = d = = = 4 u = 6 and = = 0 u = So u area( R) = du u ln = d ln u ln u u lnu = ln ln ln 6 ln6 u u ln ln = 6 ln ln ln 5 ln6 5 = or ln ln ln Complee subsiuion for boh y and d Boh correc limis in or boh correc limis in u. If no awarded above, you can award M for his inegral M B Marks u Eiher ln ln or ( ) u u M* ± α (ln ) ln ln u or ( ) ± α (ln )( u) ln u ln u ( ) A ln ln Depends on he previous mehod mark. dm* Subsiues heir changed limis in u and subracs eiher way round. 5 ln6 5 or ln ln ln A or equivalen. [6]

20 Quesio n Number 6. (a) { y } (b) Scheme Marks = 0 cos = 0 cos = 0, seen or implied. M A leas one correc value of. (See noes). A π 5π =, π 5π Boh and A cso { } ( cos ) d V 5π π = π ( cos ) d = ( 4cos + 4cos )d + cos = 4cos + 4 d ( ) = 4cos + cos d For π ( cos ). Ignore limis and d See noes. cos = cos Aemps y o give any wo of ± A ± A, ± Bcos ± Bsin or sin = 4sin + ± λ cos ± µ sin. Correc inegraion. ( ) { } ( ) ( ) 0 π ( ) ( ) ( ) π sin 5 5 sin π π π π V π 4sin 4sin Applying limis = + + he correc way round. Ignore π. = π 5π + π + (( ) ( )) = π = 7.765π = ( ) = π 4π + or 4π + π Two erm eac answer. A B M M A ddm [] [6] 9

21 6. (a) M: cos = 0. This can be implied by eiher degrees. s π A: Any one of eiher nd π 5π A: Boh and. cos = or any one of he correc values for in radians or in 5π or or 60 or 00 or awr.05 or 5. or awr 5.4. (b) B: (M on epen) For π ( cos ). Ignore limis and d. s M: Any correc form of cos = cos used or wrien down in he same variable. + cos This can be implied by cos = or 4cos + cos or cos A = cos A. nd M: Aemps y o give any wo of ± A ± A, ± Bcos ± Bsin or ± λ cos ± µ sin. Do no worry abou he signs when inegraing cos or cos for his mark. Noe: ( cos ) = + 4cos y. is ok for an aemp a s sin sin A: Correc inegraion. Eg. 4sin + or 4sin + + oe. rd ddm: Depends on boh of he wo previous mehod marks. (Ignore π ). 5π π Some evidence of subsiuing heir = and heir = and subracing he correc way round. You will need o use your calculaor o check for correc subsiuion of heir limis ino heir inegrand if a candidae does no eplicily give some evidence. π = 7.765π = Noe: For correc inegral and limis decimals gives: (( ) ( )) nd A: Two erm eac answer of eiher ( ) π 4π + or 4π + π or equivalen. Noe: The π in he volume formula is only required for he B mark and he final A mark. Noe: Decimal answer of wihou correc eac answer is A0. Noe: Applying ( cos) d will usually be given no marks in his par.

22 Quesio n Number 7. (a) (b) (c) Scheme Marks i: 9 + λ = + µ () j: + 4λ = + µ () Any wo equaions. (Allow one slip). k: λ = + µ () M Eg: () () :6+ 6λ = or An aemp o eliminae () 4 () : = 9 7µ one of he parameers. dm Leading o λ = or µ = Eiher λ = or µ = A l : r = 4 = or l : r = + = See noes ddm A [5] Realisaion ha he do produc is required d = 4, d = 4 beween ± Ad and ± Bd. M + 4 cos θ =± Correc equaion. A () + (4) + ( ). () + () + () 4 cos θ = θ = = 69. ( dp) awr 69. A λ uuur uuur OA = 6, OP = + λ 4 = + 4λ λ 9 + λ 4 λ + 5 uuur AP = + 4λ 6 = 4λ λ λ λ + 5 uuur AP d = 0 4λ 4 = λ λ + 4λ = 0 λ leading o { λ 7 = 0 } λ = uuur 4 Posiion vecor OP = 4 4 or + = λ = M A dm A ddm A [] [6] 4

23 7. (a) M: Wries down any wo equaions. Allow one slip. dm: Aemps o eliminae eiher λ or µ o form an equaion in one parameer only. A: For eiher λ = or µ =. Noe: candidaes only need o find one of he parameers. ddm: For eiher subsiuing heir value of λ ino l or heir µ ino l. 6 nd A: For eiher or 6i + j+ k or ( 6 ). Noe: Each of he mehod marks in his par are dependen upon he previous mehod marks. M: Realisaion ha he do produc is required beween ± Ad and ± Bd. Allow one slip in (b) d = i + 4j k. d d A: Correc applicaion of he do produc formula d d = ± d d cosθ or cosθ =± d d The do produc mus be correcly applied and he square roos alhough hey can be un-simplified mus be correcly applied. c A: awr 69.. This can be also be achieved by = awr 69.. θ = is A Common response: cos θ = = is MA... ( ) + ( ) + (6). (4) + () + () Alernaive Mehod: Vecor Cross Produc Only apply his scheme if i is clear ha a candidae is applying a vecor cross produc mehod. i j k M: Realisaion ha he vecor cross d d = 4 = 4 = 6i 5j 7k produc is required beween ± Ad and ± Bd. Allow one slip in d = i + 4j k. (6) + (5) + ( 7) sin θ = A: Correc applied equaion. () + (4) + ( ). () + () + () (c) 0 sin θ = θ = = 69. ( dp) A: awr uuur uuur M: Aemps o find AP in erms of he parameer by subracing he componens of OP from l and uuur uuur uuur uuur OA. Ignore he direcion of subracion and ignore any confusion beween OP and PO or beween OA and uuur AO. The correc subracion of wo componens is enough o esablish ha subracion is inended. The coordinaes or posiion vecor of P mus be given in erms of a parameer. Taking P: (, y, z ) gains no marks alhough his can be recovered laer. uuur See Addiional Soluions. A: (M on epen) A correc epression for AP. Again accep he reverse direcion. uuur dm: Depends on he previous M. Taking he scalar produc of heir epression for AP wih d or a muliple of d and equaing o 0 and obaining an equaion for λ. The equaion mus derive from an epression of he form + yy + zz = 0. Differeniaion can be used. See Addiional Soluions. A: Solving o find λ =. ddm: Depends on boh previous Ms. Subsiues heir value of he parameer ino heir epression for uuur uuur OP. Subsiuing ino AP is a common error which loses he mark. Noe: Needs correc co-ordinaes if λ = found and hen P saed wihou mehod o gain ddm.

24 A: 9 i + 4 j k. Accep vecor noaion or coordinaes. Mus be eac.

25 7. (c) Addiional Soluion : uuur Taking OP = y, in iself, can gain no marks bu his may be convered o a parameer a a laer z sage in he soluion and, a ha sage, any relevan marks can be awarded. 4 4 uuur For eample, AP = y 6 = y 6 z z+ 4 leading o: y 6 4 = 4 + 4y 64 z 6 = 0 No marks gained a his sage. z λ uuur Using, OP = + λ 4 = + 4λ on + 4y z = 74 λ which gives: A his sage award MA and dm 9 + λ + 4(+ 4 λ) ( λ) = 74 (which is implied by an equaion) λ + 67 = 74 λ = A: Solving o find λ =. Posiion vecor uuur 4 OP = + 4 = 4 or ddm A Addiional Soluion : Using Differeniaion 9 + λ 4 λ + 5 uuur AP = + 4λ 6 = 4λ MA: As main scheme λ λ { } = ( + 5) + (4 ) + ( ) = AP λ λ λ λ λ d dλ ( AP ) = 4λ 4 = 0 M leading o λ = A: Solving o find λ =.... hen apply he main scheme.

26 Quesion Number 8. (a) Scheme or B Marks d θ ( θ) = d d d 5 = θ 5 5 d θ = θ d ln ( ) { } 5 c ln θ = 5 { + c} See noes. M A ln ( ) 5 c Correc compleion 5 5 e + c or e c θ = e o θ = Ae θ = Ae + * A [4] = 0, θ = (0) 6 = Ae + ; A = See noes. M; A Subsiues θ = 0 ino an equaion = e + of he form θ = Ae +, M or equivalen. See noes. Correc algebra o = ln k, e = = ln where k is a posiive value. See noes. M 7 ln = = ( 0.008) = 77 ( neares minue) awr 77 A [5] 9 (b) { } 8. (a) B: (M on epen) Separaes variables as shown. dθ and d should be in he correc posiions, hough his mark can be implied by laer working. Ignore he inegral signs. ± λln θ or ± λln θ and ± µ where λ and µ are consans. M: Boh ( ) ( ) A: For ln ( θ ) = or ( ) ln θ = or 5ln ( θ ) = or 5ln ( θ ) 5 5 = Noe: + c is no needed for his mark A: Correc compleion o θ = Ae +. Noe: + c is needed for his mark. Noe: ln ( ) A0. θ = 5 + c leading o 5 c θ = e + e or Noe: From ln ( θ ) = +, hen ( ) ln 5 c 5 + c 5 θ = e or θ = e e c θ = 5 + c Noe: From ln ( θ ) = +, hen ( ) ln θ 5 c 5 + c 5 c θ = e or θ = e e θ = Ae + is required for A. = 5 + c Noe: The jump from θ = Ae o θ = Ae + is fine. θ = e + A, would be final θ = Ae + is sufficien for A.

27 5 Noe: ln ( θ ) = + c θ = Ae, where candidae wries A = e c is also 5 accepable.

28 8. (b) M: (B on epen) Subsiues θ = 6, = 0, ino eiher heir equaion conaining an unknown consan or he prined equaion. Noe: You can imply his mehod mark A: (M on epen) A =. Noe: θ = e + wihou any working implies he firs wo marks, MA M: Subsiues θ = 0 ino an equaion of he form θ = Ae +, or equivalen. where A is a posiive or negaive numerical value and A can be equal o or -. M: Uses correc algebra o rearrange heir equaion ino he form = ln k, where k is a posiive numerical value. A: awr 77 or awr hour 7 minues. Alernaive Mehod for par (b) d d θ = θ 5 ln ( θ ) = 5 + c M: Subsiues = 0, θ = 6, ln ( 6 ) = (0) + c { = 0, θ = 6 } 5 ino ln ( θ ) = 5 + c c = ln ln ( θ ) = ln or ( ) A: c = ln 5 ln = 5 + ln ln ( 0 ) ln 5 form ln ( ) 5 ± µ M: Subsiues θ = 0 ino an equaion of he where λ, µ are numerical values. M: Uses correc algebra o rearrange heir ln ln 7 5 equaion ino he form ± = ln C ln D, where C, D are posiive numerical values. = = 77 ( neares minue) A: awr 77. Alernaive Mehod for par (b) d d = θ 5 ln 5 + c M: Subsiues = 0, θ = 6, ln 6 = (0) + c { = 0, θ = 6 } 5 ino ln ( ) c = ln 5 c A: c = ln ln ln 5 ln 5 + ln ln ( 0) ln 5 form ln ( ) 5 ± µ M: Subsiues θ = 0 ino an equaion of he where λ, µ are numerical values. ln ln 7 5 equaion ino he form ± = ln C ln D,

29 where C, D are posiive numerical values. = = A: awr ( neares minue) 8. (b) Alernaive Mehod for par (b) 0 dθ = 6 θ 0 d 5 0 = ln θ = MA: ln M: Subsiues limi of θ = 0 correcly. ln 7 ln = 5 M: Uses correc algebra o rearrange heir own equaion ino he form ± = ln C ln D, where C, D are posiive numerical values. = = A: awr ( neares minue) Alernaive Mehod 4 for par (b) { θ = 6 } { θ = 0 } = Ae = Ae = ln A or = ln A () 7 ln ln A = A and () = ln ln A A = () () = ln = = = 77 neares minue ( 0.008) ( ) M*: Wries down a pair of equaions in A and, for θ = 6 and θ = 0 wih eiher A unknown or A being a posiive or negaive value. A: Two equaions wih an unknown A. M: Uses correc algebra o solve boh of heir equaions leading o answers of he form = ln k, where k is a posiive numerical value. M: Finds difference beween he wo imes. (eiher way round). A: awr 77. Correc soluion only.