C4: QUESTIONS FROM PAST PAPERS - PARAMETRICS

Transcription

1 C: QUESTIONS FROM PAST PAPERS - PARAMETRICS.. The iagram above shows a skech of he curve C wih parameric equaions = 5, = (9 ) The curve C cus he -ais a he poins A an B. (a) Fin he -coorinae a he poin A an he -coorinae a he poin B. () The region R, as shown shae in he iagram above, is enclose b he loop of he curve. (b) Use inegraion o fin he area of R. () (Toal 9 marks) The iagram above shows a skech of he curve wih parameric equaions Ci of Lonon Acaem

2 = cos, = sin, 0 (a) Fin he graien of he curve a he poin where. () (b) Fin a caresian equaion of he curve in he form = f(), k k, (c) saing he value of he consan k. Wrie own he range of f (). () () (Toal 0 marks). (a) Using he ieni cosθ = sin θ, fin sin. () The iagram above shows par of he curve C wih parameric equaions = anθ, = sinθ, 0 θ < The finie shae region S shown in he iagram is boune b C, he line = an he -ais. This shae region is roae hrough π raians abou he -ais o form a soli of revoluion. (b) Show ha he volume of he soli of revoluion forme is given b he inegral k sin 0 where k is a consan. (5) (c) Hence fin he eac value for his volume, giving our answer in he form pπ + qπ, where p an q are consans. () (Toal 0 marks) Ci of Lonon Acaem

3 . The curve C shown above has parameric equaions 8, where is a parameer. Given ha he poin A has parameer =, (a) fin he coorinaes of A. () The line l is he angen o C a A. (b) Show ha an equaion for l is 5 9 = 0. (5) The line l also inersecs he curve a he poin B. (c) Fin he coorinaes of B. () (Toal marks) 5. l P C R O The iagram above shows he curve C wih parameric equaions 8cos, sin, 0. Ci of Lonon Acaem

4 The poin P lies on C an has coorinaes (, ). (a) Fin he value of a he poin P. () The line l is a normal o C a P. (b) Show ha an equaion for l is = +. () The finie region R is enclose b he curve C, he -ais an he line =, as shown shae in he iagram above. (c) Show ha he area of R is given b he inegral sin cos. () () Use his inegral o fin he area of R, giving our answer in he form a + b, where a an b are consans o be eermine. () (Toal marks). C R O ln ln The curve C has parameric equaions ln( ),, ( ) The finie region R beween he curve C an he -ais, boune b he lines wih equaions = n an = n, is shown shae in he iagram above. (a) Show ha he area of R is given b he inegral (b) 0. ( )( ) Hence fin an eac value for his area. () () (c) Fin a caresian equaion of he curve C, in he form = f(). () Ci of Lonon Acaem

5 () Sae he omain of values for for his curve. () (Toal 5 marks) 7. A curve has parameric equaions an, sin, 0. (a) Fin an epression for in erms of. You nee no simplif our answer. () (b) Fin an equaion of he angen o he curve a he poin where. Give our answer in he form = a + b, where a an b are consans o be eermine. (5) (c) Fin a caresian equaion of he curve in he form = f(). () (Toal marks) O 0.5 The curve shown in he figure above has parameric equaions sin π π π, sin ( ),. (a) Fin an equaion of he angen o he curve a he poin where. () (b) Show ha a caresian equaion of he curve is ( ), () (Toal 9 marks) Ci of Lonon Acaem 5

6 9. a A O B a The curve shown in he figure above has parameric equaions a cos, asin, 0. The curve mees he aes a poins A an B as shown. The sraigh line shown is par of he angen o he curve a he poin A. Fin, in erms of a, (a) an equaion of he angen a A, () (b) an eac value for he area of he finie region beween he curve, he angen a A an he -ais, shown shae in he figure above. (9) (Toal 5 marks) 0. R O The curve shown in he figure above has parameric equaions = sin, = cos, 0 (a) 5 Show ha he curve crosses he -ais where an. () The finie region R is enclose b he curve an he -ais, as shown shae in he figure above. Ci of Lonon Acaem

7 (b) Show ha he area of R is given b he inegral (c) Use his inegral o fin he eac value of he shae area. 5 cos. () (7) (Toal marks). C O The curve C has parameric equaions =, =, <. (a) Fin an equaion for he angen o C a he poin where =. (b) Show ha C saisfies he caresian equaion =. (7) () The finie region beween he curve C an he -ais, boune b he lines wih equaions = an =, is shown shae in he figure above. (c) Calculae he eac value of he area of his region, giving our answer in he form a + b ln c, where a, b an c are consans. () (Toal marks). A curve has parameric equaions = co, = sin, 0 <. (a) Fin an epression for in erms of he parameer. Ci of Lonon Acaem 7

8 () (b) Fin an equaion of he angen o he curve a he poin where =. () (c) Fin a caresian equaion of he curve in he form = f(). Sae he omain on which he curve is efine. () (Toal marks). C P O R a The iagram above shows a skech of he curve C wih parameric equaions = sin, = sec, 0 <. The poin P(a, ) lies on C. (a) Fin he eac value of a. () The region R is enclose b C, he aes an he line = a as shown in he iagram above. (b) Show ha he area of R is given b (an + ). 0 (c) Fin he eac value of he area of R. () () (Toal marks) Ci of Lonon Acaem 8

9 . (meres) C R A B (meres) The iagram above shows a cross-secion R of a am. The line AC is he verical face of he am, AB is he horizonal base an he curve BC is he profile. Taking an o be he horizonal an verical aes, hen A, B an C have coorinaes (0, 0), (, 0) an (0, 0) respecivel. The area of he cross-secion is o be calculae. Iniiall he profile BC is approimae b a sraigh line. (a) Fin an esimae for he area of he cross-secion R using his approimaion. () The profile BC is acuall escribe b he parameric equaions. =, = 0 sin,. (b) Fin he eac area of he cross-secion R. (7) (c) Calculae he percenage error in he esimae of he area of he cross-secion R ha ou foun in par (a). () (Toal 0 marks) 5. The curve C is escribe b he parameric equaions = cos, = cos, 0. (a) Fin a caresian equaion of he curve C. (b) Draw a skech of he curve C. () () (Toal marks). The caresian equaion of he circle C is = 0. (a) Fin he coorinaes of he cenre of C an he raius of C. Ci of Lonon Acaem 9

10 () (b) Skech C. (c) Fin parameric equaions for C. () () () Fin, in caresian form, an equaion for each angen o C which passes hrough he origin O. (5) (Toal marks) 7. The curve shown in he iagram above has parameric equaions = cos, = sin, 0 <. (a) Fin an epression for in erms of he parameer. (b) Fin he values of he parameer a he poins where = 0. () () (c) () Hence give he eac values of he coorinaes of he poins on he curve where he angens are parallel o he -ais. Show ha a caresian equaion for he par of he curve where 0 < is = ( ). () () (e) Wrie own a caresian equaion for he par of he curve where <. () (Toal marks) Ci of Lonon Acaem 0

11 MARK SCHEME. (a) = 0,, An one correc value B A = 0, = 5 (0) = A =, = 5 () = A, 5 Meho for fining one value of M A A, = ; a B, = Boh A (b) 0 Seen or implie B M A A = =8 (unis ) M A [9]. (a) = sin, cos B, B cos = M sin sin A =, m accep equivalens, awr 0.87 A Alernaives o (a) where he parameer is eliminae ( 8 9) (8 9) ( 9) B Ci of Lonon Acaem

12 A, cos B 9 (7) M A = B A, sin B 9 M A (b) Use of cos = sin M cos,sin M Leaing o ( 8 9)( ( )) cao A k = B (c) 0 f() eiher 0 f() or f() B Full correc. Accep 0, [0, ] B [0]. (a) sin ( cos ) sin ( C) M A (b) an sec (sin cos) cos (sin ) sec M A M Ci of Lonon Acaem

13 sin k = π A = 0 anθ = 0 θ = 0, an B5 V sin 0 (c) sin V sin (0 0) 0 M Use of correc limis *M p, q 8 A [0]. (a) A A, = + 8 = 7 & = ( ) = A(7,) A(7, ) B (b) = 8, =, 8, 8 Their ivie b heir M Correc A A A, m(t) = ( ) ( ) Subsiues for o give an of he four unerline oe: T: (heir ) = m r ( (heir 7)) Fining an equaion of a angen wih heir poin an heir angen graien or = 9 ( 7) cc or fins c an uses M = (heir graien) + c. Hence T: gives T: 5 9 = 0 AG A cso5 (c) ( 8) 5 9 = 0 Subsiuion of boh = 8 an Ci of Lonon Acaem

14 = ino T M 5 9 = 0 ( + ){( 7 9) = 0} ( + ){( + )( 9) = 0} A realisaion ha ( + )is a facor. M 9 { = (a A) = a B} 9 A = or awr Caniae uses heir value of o fin eiher he or coorinae M or awr 0. One of eiher or correc. Boh an correc. A A Hence B 8, 8 awr [] 5. (a) A P(, ) eiher = 8cos or sin onl soluion is where 0 = 8cos or sin M or awr.05 (raians) onl sae in he range 0 A (b) = 8 cos, = sin 8sin, 8cos 8 cos A P, 8sin N: N: AG or 8 awr 0.58 ( 8) Hence m(n) = or ( ) () c c Ci of Lonon Acaem

15 so N: Aemp o iffereniae boh an wr o give ±p sin an ±q cos respecivel M Correc an A Divies in correc wa roun an aemps o subsiue heir value of (in egrees or raians) ino heir epression. M* You ma nee o check caniae s subsiuions for M* Noe he ne wo meho marks are epenen on M* Uses m(n) = heir m( T) M* Uses = (heir m N )( ) or fins c using = an = an uses = (heir m N ) + c. M* A cso AG (c) A A A A 0 sin.( 8sin ) sin.sin.sin cos.sin cos ( sin cos).sin aemp a A = M correc epression (ignore limis an ) Seeing sin = sin cos anwhere in his par. A M Correc proof. Appreciaion of how he negaive sign affecs he limis. A AG Noe ha he answer is given in he quesion. Ci of Lonon Acaem 5

16 Ci of Lonon Acaem () {Using subsiuion u = sin = cos} {change limis: when =, u = & when =, u = } (Noe ha a =, b = 8) k sin or ku wih u = sin M Correc inegraion ignoring limis. A Subsiues limis of eiher ( = an = ) or (u = an u = ) an subracs he correc wa roun. M A aef isw Aef in he form a +, wih awr. an anhing ha cancels o a = an b = 8. []. (a) Area (R) = Changing limis, when: = ln ln = ln( + ) = + = 0 = ln ln = ln( + ) = + = Hence, Area (R) = Mus sae B u or sin A A u A A 8 b, ), ln( ; 0 ln ln ) )( ( 0

17 Area =. Ignore limis. M;. Ignore limis. A AG changes limis so ha ln 0 an ln B (b) A B ( )( ) ( ) ( ) = A( + ) + B( + ) Le =, = A() A = Le =, = B( ) B = ( )( ) 0 ( ) ( = 0 ln( ) ln( ) 0 = (ln ln ) (ln ln ) ) Takes ou brackes: = ln ln + ln = ln ln = ln A B wih A an B foun ( ) ( ) Fins boh A an B correcl. Can be implie. Wriing own ( )( ) ( ) ( ) Wriing own ( )( ) ( ) ( ) Eiher ± a ln( + ) or ± b ln( + ) Boh ln erms correcl f. means firs MA0. means firs MA. M A M Af Subsiues boh limis of an 0 an subracs he correc wa roun. M ln ln + ln or (mus eal wih ln ) ln ln or ln ln or ln A aef isw (c) = ln( + ), = e = + = e = e e Aemp o make = he subjec giving = e M A Ci of Lonon Acaem 7

18 Eliminaes b subsiuing in M giving = e A Alier Wa + = or ( + ) = + = = = ln or ln ln e e e Aemp o make = he subjec M Giving eiher = or = A Eliminaes b subsiuing in M giving = A e Alier Wa e = + + = e = e Aemp o make + = he subjec giving + = e Eliminaes b subsiuing in M A M giving = A e Ci of Lonon Acaem 8

19 Alier Wa + = ln e e e or ln or ln Aemp o make + = he subjec M Eiher + = or A Eliminaes b subsiuing in M giving = A e () Domain: > 0 > 0 or jus > 0 B [5] 7. (a) = an, = sin (an) sec, cos (*) cos cos an sec sin B M Correc cos heir an Af (*) cos heir Ci of Lonon Acaem 9

20 (b) When =,, (nee values) cos, m( T) When = an sec.().().()() 8 B, B The poin, or (, awr 0.7) These coorinaes can be implie. ( = sin is no sufficien for B) B aef an of he five unerline epressions or awr 0.8 T: ( ) (Noe: The an coorinaes mus be he righ wa roun.) T: or (*) 8 8 or () c c (*) A caniae who incorrecl iffereniaes an o give = sec or = sec is hen able o fluke he correc answer in par (b). Such caniaes can poeniall ge: (a) B0MAf (b) BBBMA0 cso. Noe: cso means correc soluion onl. Noe: par (a) no full correc implies caniae can achieve a maimum of ou of 5 marks in par (b). Hence T: or Mf aef Fining an equaion of a angen wih heir poin an heir angen graien or fins c using = (heir graien) + c. A aef cso Correc simplifie EXACT equaion of angen (c) Wa Ci of Lonon Acaem 0

21 = an sin = sin cos = sin = = sin ( ) = = = + = ( + ) = M Uses cos = sin M Eliminaes o wrie an equaion involving an. M Rearranging an facorising wih an aemp o make he subjec. A Alier Wa + co = cosec = sin Hence, + Hence, = ( or ) M Uses + co = cosec M implie Uses cosec = sin M Eliminaes o wrie an equaion involving an. A ( or ) Alier Wa = an + an = sec = sin Ci of Lonon Acaem

22 cos sin Hence, + = Hence, = ( or ) M Uses + an = sec M Uses sec = cos M Eliminaes o wrie an equaion involving an. A ( or ) Alier Wa = sin = cos sec ( an ) Hence, ( ) or M Uses sin = cos M Uses cos = sec M hen uses sec = + an A ( or ) is an accepable response for he final accurac A mark. Alier Wa 5 = an = sin Ci of Lonon Acaem

23 = an an = ( + ) Hence, = sin = M Draws a righ-angle riangle an places boh an on he riangle M Uses Phagoras o euce he hpoenuse M Eliminaes o wrie an equaion involving an A is an accepable response for he final accurac A mark. There are so man was ha a caniae can procee wih par (c). If a caniae prouces a correc soluion hen please awar all four marks. if he use a meho commensurae wih he five was as eaile on he mark scheme hen awar he marks appropriael. If ou are unsure of how o appl he scheme please escalae our response up o our eam leaer. [] 8. (a) = sin = sin( + ) M Aemp o iffereniae boh an wr o give wo erms in cos cos, cos Correc an A When, cos( ) cos( ) awr 0.58 Divies in correc wa an subsiues for o give an of he four unerline oe: Ignore he ouble negaive if caniae has iffereniae sin cos A Ci of Lonon Acaem

24 when,, The poin, or (, 0.87 ) awr B T: ( ) Fining an equaion of a angen wih heir poin an heir angen graien or fins c an uses = (heir graien) + c. M Correc EXACT equaion of angen oe. A oe or c c or T: (b) = sin ( + ) = sin cos + cos sin M Use of compoun angle formula for sine. Nb: sin + cos = cos sin = sin gives cos = Use of rig ieni o fin cos in erms of or cos in erms of. M sin cos gives ( ) AG A cso. Alier Wa Subsiues for sin, cos, cos an sin o give in erms of (a) = sin = sin ( + ) = sin + cos + cos sin M (Do no give his for par (b)) Aemp o iffereniae an wr o give an in he form ± a cos ± b sin in erms of cos [9] = cos ; cos cos sin sin A Correc an Ci of Lonon Acaem

25 Ci of Lonon Acaem 5 When =, A Divies in correc wa an subsiues for o give an of he four unerline oe When = B The poin or T: Fining an equaion of a angen wih heir poin an heir angen graien or fins c an uses = (heir graien) + c. M Correc EXACT equaion of angen oe. A oe or or T: Alier Wa (a) Aemp o iffereniae wo erms using he chain rule for he secon erm. M Correc A A Correc subsiuion of = ino a correc When = =, = B The poin or T: cos sin sin cos cos,,,,awr 0.87 c c ) ( ) ( ) ( (0.5)) ( (0.5),, ) 0.87, ( awr ( )

26 or Fining an equaion of a angen wih heir poin an heir angen graien or fins c an uses M = (heir graien) + c Correc EXACT equaion of angen A oe oe. or T: Alier Wa c c (b) = sin gives = sin ( sin ) M Subsiues = sin ino he equaion give in. Nb: sin + cos cos sin Cos = ( sin ) M Use of rig ieni o euce ha cos = ( sin ) gives = sin cos Hence = sin cos cos sin sin ( ) A cso Using he compoun angle formula o prove = sin ( ) 9. (a) a sin, a cos herefore cos M A sin When = 0, = B Graien is M Line equaion is ( a) = ( 0) M A (b) Area beneah curve is a sin ( a sin ) M = a (cos cos) M a [ sin sin ] M A a Uses limis 0 an o give A Area of riangle beneah angen is a a a M A Ci of Lonon Acaem

27 a a a Thus require area is A 9 N.B. The inegraion of he prouc of wo sines is worh marks (lines an of o par (b)) If he use pars sin sin = cos sin + cos cos M 8I = cos sin cos sin = cos sin + cos sin + 9sin sin M A [5] 5 0. (a) Solves = 0 cos = o obain = or (nee boh for A) M A Or subsiues boh values of an shows ha = 0 (b) = cos M A 5 Area = = ( cos ) ( cos ) 5 = ( cos) AG B (c) Area = cos + cos erms M cos + (cos + ) (use of correc ouble angle formula) M = cos + cos M A = [ sin + sin ] M A Subsiues he wo correc limis = 5 an an subracs. M = π + AA 7 []. (a) an B, B ( ) ( ) ( ) an a = ½, graien is 9 M Acao ( ) M requires heir / / heir / an subsiuion of. Ci of Lonon Acaem 7

28 A he poin of conac = an = B Equaion is = -9 ( ) M A 7 (b) Eiher obain in erms of an i,e, = or = (or boh) M Then subsiue ino oher epression = f() or = g() an rearrange M (or pu an rearrange) To obain = (*) a ( ) Or Subsiue ino M = A ( ) = (*) M (c) Area = B u u = u M u u puing ino a form o inegrae = u ln u M A = ln M = ln or an correc equivalen. A [] Or Area = B = M puing ino a form o inegrae Ci of Lonon Acaem 8

29 = ln( ) M A ln ln = M A Or Area = B ( ) A B C = M ( ) ( ) ( ) puing ino a form o inegrae = ln( ) ln( ) ( ) M af = Using limis 0 an ½ an subracing (eiher wa roun) M = ln or an correc equivalen. A Or Area = hen use pars B = ln( ) ( ) M = ln( ) ( ) ln( ) MA = ln ln DM = ln A. (a) = cosec, = sin cos boh M A sin cos (= sin cos ) cosec M A (b) A =, =, = B boh an Subsiues = ino an aemp a o obain graien M Equaion of angen is = ( ) M A Accep + = or an correc equivalen (c) Uses + co = cosec, or equivalen, o eliminae M Ci of Lonon Acaem 9

30 + A correcl eliminaes 8 = cao A The omain ir 0 B Alernaive for (c): sin = ; cos = sin sin + cos = = M A 8 Leaing o = A []. (a) = sec cos =, = M, A a = sin = B a 0 (b) A = = M Change of variable = sec [sin + cos] M Aemp = ( an, ) (*) A, Acso 0 Final A requires limi sae (c) A = [ ln sec + ] M, A 0 Some inegraion (M) boh correc (A) ignore lim. 9 = ( ln + ) (0) Use of M = ln + A []. (a) Area of riangle = 0 ( =.) B Accep 0 or 50 Ci of Lonon Acaem 0

31 (b) Eiher Area shae = 0sin. M A = [ 80 cos 80cos] M A = [ 80 cos 0sin ] A f = 0( ) M A 7 or 0cos.( ) M A = [(0 sin ( ) 80 cos + 80cos] M A = [ 80 cos 0sin ] A f = 0( ) M A 7 (c) 0( ) esimae Percenage error = 00 =.% 0( ) M A (Accep answers in he range.% o.%) [0] 5. (a) Aemp o use correcl sae ouble angle M formula cos = cos, or complee meho using oher ouble angle formula for cos wih cos + sin = o eliminae an obain = = or an correc equivalen.(even =cos(cos ) A (b) shape B posiion incluing resrice omain < < B [] Ci of Lonon Acaem

32 . (a) = ( ) + ( ) 9 + ( ) ( ) = 9 M A Cenre (, ), raius A A (b) (, ) O B, B (c) = + cos = + sin (0 < ) M A A (, ) () Line hrough origin = m -coorinae of poins where his line cus C saisfies ( + m ) 8 m + = 0 M A As line is angen his equaion has repeae roos (8 + m) = ( + m ) + 9m + m = + m m = 7m M A m = 0, m = 7 Equaions of angens are = 0, = A5 7 Ci of Lonon Acaem

33 [] 7. (a) cos = sin, = cos = sin M A A (b) cos = =,,, M 5 7 So =,,, A A (c),), ),), ) M A () = sin cos M = cos cos = M A (e) = B [] ========================================================================= EXAMINERS REPORTS. Par (a) was well one. The majori of caniaes correcl foun he -coorinaes of A an B, b puing = 0, solving for an hen subsiuing in = 5. Full marks were common. Par (b) prove ifficul. A subsanial minori of caniaes faile o subsiue for he when subsiuing ino or use raher han. A surprising feaure of he soluions seen was he number of caniaes who, having obaine he correc 9, were unable o remove he brackes correcl o obain. Weaknesses in elemenar algebra flawe man oherwise correc soluions. Anoher source of error was using he -coorinaes for he limis when he variable in he inegral was. A he en of he quesion, man faile o realise ha 90 0 gives onl half of he require area. 0 Some caniaes mae eiher he whole of he quesion, or jus par (b), more ifficul b eliminaing parameers an using he caresian equaion. This is a possible meho bu he inices involve are ver complicae an here were ver few successful soluions using his meho.. Nearl all caniaes knew he meho for solving par (a), alhough here were man errors in iffereniaing rig funcions. In paricular (cos) was ofen incorrec. I was clear from boh his quesion an quesion ha, for man, he calculus of rig funcions was an area of weakness. Nearl all caniaes were able o obain an eac answer in sur form. In par (b), he majori of caniaes were able o eliminae bu, in manipulaing rigonomeric ieniies, man errors, paricularl wih signs, were seen. The answer was given in a varie of forms an all eac equivalen answers o ha prine in he mark scheme were accepe. The value of k was ofen omie an i is possible ha some simpl overlooke his. Domain an range remains an unpopular opic an man i no aemp par (c). In his case, inspecion of he prine figure gives he lower limi an was inene o give caniaes a lea o ienifing he upper limi.. The responses o his quesion were ver variable an man los marks hrough errors in manipulaion or noaion, possibl hrough menal ireness. For eamples, man mae errors in manipulaion an coul no procee correcl from he prine cos θ = sin θ o sin an he answer was ofen seen, cos sin Ci of Lonon Acaem

34 insea of. In par (b), man never foun or realise ha he appropriae form for he volume was However he majori i fin a correc inegral in erms of θ alhough some were unable o use he ieni sin θ = sinθ cosθ o simplif heir inegral. The incorrec value k = 8π was ver common, resuling from a failure o square he facor in sin θ = sinθ cosθ. Caniaes were epece o emonsrae he correc change of limis. Minimall a reference o he resul an or an equivalen, was require. Those who ha complee soluions usuall gaine he wo meho marks in par (c) bu earlier errors ofen le o incorrec answers.. Par (a) was answere correcl b almos all caniaes. In par (b), man caniaes correcl applie he meho of fining a angen b using parameric iffereniaion o give he answer in he correc form. Few caniaes rie o eliminae o fin a Caresian equaion for C, bu hese caniaes were usuall no able o fin he correc graien a A. In par (c), full correc soluions were much less frequenl seen. A significan number of caniaes were able o obain an equaion in one variable o score he firs meho mark, bu were hen unsure abou how o procee. Successful caniaes mosl forme an equaion in, use he fac ha + was a facor an applie he facor heorem in orer for hem o fin a he poin B. The hen subsiue his ino he parameric equaions o fin he coorinaes of B. Those caniaes who iniiall forme an equaion in onl wen no furher. A common misconcepion in par (c), was for caniaes o believe ha he graien a he poin B woul be he same as he graien a he poin A an a significan minori of caniaes aempe o solve o fin a he poin B. 5. In par (a), man caniaes were able o give =. Some caniaes gave heir answer onl in egrees insea of raians. Oher caniaes subsiue he -value of P ino = sin an foun wo values for, namel =. The majori of hese caniaes i no go on o rejec =. In par (b), man caniaes were able o appl he correc formula for fining in erms of, alhough some caniaes erroneousl believe ha iffereniaion of sin gave eiher 8 cos, 8cos or cos. Some caniaes who ha iffereniae incorrecl, subsiue heir value of ino an rie o fuge heir answer for as, afer realising from he given answer ha he graien of he normal was. The majori of caniaes unersoo he relaionship beween he graien of he angen an is normal an man were able o prouce a full correc soluion o his par. A few caniaes, however, i no realise ha parameric iffereniaion was require in par (b) an some of hese caniaes rie o conver he parameric equaions ino a Caresian equaion. Alhough some caniaes hen wen on o aemp o iffereniae heir Caresian equaion, his meho was rarel eecue correcl. Few convincing proofs were seen in par (c) wih a significan number of caniaes who were no aware wih he proceure of reversing he limis an he sign of he inegral. Therefore, some caniaes convenienl iffereniae 8 cos o give posiive 8sin, even hough he ha previousl iffereniae 8 cos correcl in par (a). Afer compleing his par, some caniaes ha a crisis of confience wih heir iffereniaion rules an hen wen on o amen heir correc soluion o par (a) o prouce an incorrec soluion. Oher caniaes iffereniae 8 cos correcl bu wroe heir limis he wrong wa roun giving A = sin.( 8 sin ) an afer saing sin = cos sin (as man caniaes were able o o in his par) wroe A = sin. sin cos, 8 5,. These caniaes hen wroe own he given answer b arguing ha all areas shoul be Ci of Lonon Acaem

35 posiive. Par () was unsrucure in he sense ha he quesion i no ell he caniaes how o inegrae he given epression. Onl a minori of caniaes spoe ha a subsiuion was require, alhough some caniaes were able o inegrae sin cos b inspecion. Man caniaes replace sin wih ( cos ) bu hen muliplie his ou wih cos o give cos cos cos. Ver few caniaes correcl applie he sum-prouc formula on his epression, bu mos caniaes usuall gave up a his poin or wen on o prouce some incorrec inegraion. Oher caniaes replace sin wih cos, bu i no make much progress wih his. A significan number of caniaes use inegraion b pars wih a surprising number of hem persevering wih his echnique more han once before eciing he coul make no progress. I is possible, however, o use a loop meho bu his was ver rarel seen. I was clear o eaminers ha a significan number of sronger caniaes spen much ime ring o unsuccessfull answer his par wih a few caniaes proucing a leas wo pages of working.. The majori of caniaes were able o show he inegral require in par (a). Some caniaes, however, i no show evience of convering he given limis, whils for oher caniaes his was he onl hing he were able o o. In par (b), i was isappoining o see ha some srong caniaes were unable o gain an marks b failing o recognise he nee o use parial fracions. Those caniaes who spli he inegral up as parial fracions usuall gaine all si marks, while hose caniaes who faile o use parial fracions usuall gave answers such as ln( + + ) or ln( + ) ln( + ) afer inegraion. A few caniaes onl subsiue he limi of, assuming ha he resul of subsiuing a limi of 0 woul be zero Few caniaes gave a ecimal answer insea of he eac value require b he quesion. Par (c) was well answere b caniaes of all abiliies wih caniaes using a varie of mehos as ienifie in he mark scheme. Occasionall some caniaes were able o eliminae bu hen faile o make he subjec. The omain was no so well unersoo in par (), wih a significan number of caniaes failing o correcl ienif i. 7. In par (a), a significan number of caniaes sruggle wih appling he chain rule in orer o iffereniae an wih respec o. Some caniaes replace an wih an proceee o iffereniae his epression using boh he chain rule an he quoien rule. Ver few caniaes incorrecl iffereniae sin o give cos. A majori of caniaes were hen able o fin b iviing heir b heir. In par (b), he majori of caniaes were able o wrie own he poin, an fin he equaion of he angen using his poin an heir angen graien. Some caniaes foun he incorrec value of he angen graien using = even hough he ha correcl foun correcl wrien own he equaion of he angen as equaion ino he form. = a + b. cos in par (a). There were a significan number of caniaes, who having ( ) were unable o correcl rearrange his In par (c), here were man was ha a caniae coul ackle his quesion an here were man goo soluions seen. Errors usuall arose when caniaes wroe own an use incorrec rigonomeric ieniies. I was isappoining o see a number of caniaes who use rigonomeric ieniies correcl an reache = ( ) bu were hen unable o rearrange his or, worse sill, hough ha his was he answer o he quesion. 8. Par (a) was surprisingl well one b caniaes wih par (b) proviing more of a challenge even for some caniaes who ha prouce a perfec soluion in par (a). In par (a), man caniaes were able o appl he correc formula for fining sin in erms of, alhough some caniaes erroneousl believe ha iffereniaion of a sine funcion prouce a negaive cosine funcion. Oher 8, Ci of Lonon Acaem 5

36 misakes inclue a few caniaes who eiher cancelle ou cos in heir graien epression o give subsiue = ino heir an epressions before proceeing o iffereniae each wih respec o. Oher caniaes mae life more ifficul for hemselves b epaning he epression using he compoun angle formula, giving hem more work, bu for he same crei. Man caniaes were able o subsiue = ino heir graien epression o give, bu i was no uncommon o see some caniaes who simplifie incorrecl o give The majori of caniaes wroe own he poin, an unersoo how o fin he equaion of he angen using his poin an heir angen graien. Whils some caniaes omie par (b) alogeher, mos realise he neee o use he compoun angle formula, hough i was common o see ha some caniaes who believe ha sin ( + )coul be rewrien as sin + sin. Man caniaes o no appreciae ha a proof requires evience, as was require in esablishing ha cos =, an so los he final wo marks. There were, however, a significan number of caniaes who successfull obaine he require Caresian equaion. 9. This quesion prove a significan es for man caniaes wih full correc soluions being rare. Man caniaes were able o fin an, alhough confusing iffereniaion wih inegraion ofen le o inaccuracies. Some caniaes aempe o fin he equaion of he angen bu man were unsuccessful because he faile o use in orer o fin he graien as. Those caniaes who aempe par (b) rarel progresse beon saing an epression for he area uner he curve. Some aemps were mae a inegraion b pars, alhough ver few caniaes wen furher han he firs line. I was obvious ha mos caniaes were no familiar wih inegraing epressions of he kin hose who were ofen spen ime eriving resuls raher han using he relevan formula in he formulae book. or. Even Those caniaes who were successful in par (a) frequenl wen on o fin he area of a riangle an so were able o gain a leas wo marks in par (b). 0. The majori of caniaes gaine he marks in par (a) an a goo proporion manage o prouce he given resul in 5 par (b). Some caniaes suggese ha he area of R was, which mae he quesion raher rivial; alhough ha happene o be rue here as, working was neee o prouce ha saemen. The inegraion in par (c), alhough well one b goo caniaes, prove a challenge for man; weaker caniaes inegraing (-cos) as or somehing similar. I ma have been ha some caniaes were presse for ime a his poin bu even hose who knew ha a cosine ouble-angle formula was neee ofen mae a sign error, forgo o mulipl heir epression for cos b, or even forgo o inegrae ha epression. I has o be menione again ha he limis were someimes use as hough = 80, 5 so ha cos sin sin became [ 900.] [ 80.].. (a) A number of caniaes los marks in his quesion. Some confuse iffereniaion wih inegraion an obaine a logarihm, ohers mae sign slips iffereniaing, an a number who obaine he correc graien faile o coninue o fin he equaion of he angen using equaions of a sraigh line. sin asin b Ci of Lonon Acaem

37 (b) (c) There was a lack of unersaning of proof wih a number of caniaes merel subsiuing in values. Beer caniaes were able o begin correcl bu some i no realise ha if he answer is given i is necessar o show more working. Ver few go his correc. There was a enenc o use pars an o be unable o eal wih he inegral of ln( ). The mos successful mehos involve iviing ou, or subsiuing for (-). Those who rie a parameric approach rarel recognise he nee for parial fracions.. This prove a esing quesion an few coul fin boh an correcl. A common error was o inegrae, ln sin giving. Mos knew, however, how o obain from an an were able o pick up marks here an in par (b). In par (b), he meho for fining he equaion of he angen was well unersoo. Par (c) prove ver emaning an onl a minori of caniaes were able o use one of he rigonomeric forms of Phagoras o eliminae an manipulae he resuling equaion o obain an answer in he require form. Few even aempe he omain an he full correc answer, 0, was ver rarel seen. Par (a) was ofen answere well bu some caniaes who worke in egrees gave he final answer as 90. Par (b) prove more challenging for man; some i no know how o change he variable an ohers faile o realize ha require he chain rule. Mos caniaes mae some progress in par (c) alhough a surprising number hough ha an was sec require) raher han reaching for heir calculaors.. The eaminers were encourage o see mos caniaes ring o give an eac answer (as. This was foun o be he mos ifficul quesion on he paper. Some ecellen caniaes i no appear o have learne how o fin he area using parameric coorinaes an coul no even wrie own he firs inegral. A few caniaes use he formula from he formula book. The inegraion b pars was ackle successfull, b hose who go o ha sage an here were few errors seen. The percenage a he en of he quesion was usuall answere well b he few who complee he quesion. 5. Par (a) was generall answere quie well an mos of he caniaes use he correc ouble angle formula, as on he mark scheme. Quie a few use sin + cos = o obain /9 + ( )/ =. A sizeable minori also eliminae o obain =cos(cos - (/)). Some of hese caniaes arrive a a sraigh line equaion, e.g. =. If he ouble angle formula was mis-quoe i ene o be b saing cos = cos or cos = sin. Quie a few caniaes, however foun an an hen soppe. Presumabl he iffereniae before he rea he quesion! These caniaes seeme o have no concep of ring o eliminae. In Par (b) ver few correc graphs were seen, an a large number of answers were on graph paper. Man caniaes achieve a B for he shape, bu ver few realise he imporance of he resrice omain.. Some caniaes skeche heir erive Caresian equaion, an ohers worke successfull from he original parameric equaions.. No Repor available for his quesion. 7. No Repor available for his quesion. Ci of Lonon Acaem 7