June Core Mathematics C3 Mark Scheme

Transcription

1 June Cre Mathematics C Mark Questin. (a) Iterative frmula: n + +, 0. ( ) n + (.) An attempt t substitute 0. int the iterative frmula. Can be implied by. r.0 Bth.(0) and awrt.7 M A Bth awrt.6 and awrt.60 r.6 A cs [] (b) Let f f (.8) f (.9) Sign change (and f is cntinuus) therefre a rt α is α α.9 ( dp) such that (.8,.9) Chse suitable interval fr, e.g. [.8,.9] r tighter M any ne value awrt sf dm r truncated sf bth values crrect, sign change and cnclusin A [] 6 marks See appendi! At a minimum, bth values must be crrect t sf r truncated sf, candidate states change f sign, hence rt. 666/0 Cre Maths C 0 th June 009 L Cpe June 009 Advanced Level in GCE Mathematics Versin :

2 . (a) cs θ + sin θ ( cs θ) cs θ sin θ + cs θ cs θ cs θ Dividing cs θ + sin θ by cs θ t give underlined equatin. M + tan θ sec θ tan θ sec θ (as required) AG Cmplete prf. N errrs seen. A cs (see alternatives in appendi!) [] (b) tan θ secθ sec θ, ( eqn *) θ < 60 (sec θ ) + secθ + sec θ Substituting tan θ sec θ int eqn * t get a quadratic in secθ nly M sec θ + secθ + sec θ sec θ + secθ 0 Frming a three term ne sided quadratic epressin in sec θ. M ( θ )( θ ) sec + sec 0 secθ r secθ Attempt t factrise r slve a quadratic. (See rules fr factrising quadratics) M r csθ csθ cs θ ; r csθ csθ A; α 0 r α n slutins θ 0 0 A θ 0 0 r θ 60 θ when slving using cs θ... B θ { 0, 0 } Nte the final A mark has been changed t a B mark. [6] Nte: Please refer t the appendi if candidate ffers any etra slutins. 8 marks 666/0 Cre Maths C 0 th June 009 L Cpe June 009 Advanced Level in GCE Mathematics Versin :

3 . P 80e t (a) t 0 0 P 80e 80() B [] t 000 t (b) P e e t ln 80 Substitutes P 000 and rearranges equatin t make e t the subject. M t awrt.6 r years A [] dp t (c) 6e dt (d) 0 6e t ke t and k 80. 6e t M A [] 0 t ln { } dp Using 0 and dt an attempt t slve t t find the value f t r. M 0 ln 6 P 80e r ( ) P 80e Substitutes their value f t back int the equatin fr P. dm 80(0) P 0 0 r awrt 0 A 6 [] 8 marks Nte t r t awrt.6 t will scre A0. 666/0 Cre Maths C 0 th June 009 L Cpe June 009 Advanced Level in GCE Mathematics Versin :

4 . (i) (a) y cs Apply prduct rule: u v cs du dv sin d d Applies vu + uv crrectly fr their u, u, v, v AND gives an epressin dy f the frm α cs ± β sin cs sin d Any ne term crrect Bth terms crrect and n further simplificatin t terms in csα r sin β. M A A [] (b) y ln( + ) + u du ln( + ) smething + M + ln( + ) A + ln( ) d + Apply qutient rule: u ln( + ) v + du dv d + d ( + ) ln( + ) + d dy ( + ) vu uv Applying v Crrect differentiatin with crrect bracketing but allw recvery. M A [] y + d ( + ) d ln {Ignre subsequent wrking.} 666/0 Cre Maths C 0 th June 009 L Cpe June 009 Advanced Level in GCE Mathematics Versin :

5 . (ii) y +, > At P, y () + 9 At P, y 9 r B ± k(+ ) M* dy ( ) + () d ( + ) A aef dy d ( + ) At P, dy d Substituting int an equatin ()+ invlving d y ; d M Hence m(t) Either T: y ( ) ; r y + c and c c + ; Either T: y 9 ( ) ; T: y 9 y y m( ) r y y m( their stated ) with their TANGENT gradient and their y ; r uses y m+ c with their TANGENT gradient, their and their y. dm*; T: y+ 0 y+ 0 A r T: y + [6] T: y + T: y+ 0 marks Tangent must be stated in the frm a + by + c 0, where a, b and c are integers. 666/0 Cre Maths C 0 th June 009 L Cpe June 009 Advanced Level in GCE Mathematics Versin :

6 . (a) y Curve retains shape when > ln k B ( 0, k ) Curve reflects thrugh the -ais when < ln k B O ln k, 0 ( 0, k ) and ( ln k, 0) marked in the crrect psitins. B (b) ( k, 0) y ( 0, lnk ) Crrect shape f curve. The curve shuld be cntained in quadrants, and (Ignre asymptte) B [] O ( k, 0) and ( 0, lnk) B (c) Range f f: f > k r y> k r ( k, ) (d) y e k y+ k e Either f > k r y > k r ( k, ) r f > kr Range > k. Attempt t make (r swapped y) the subject ln y+ k ln y+ k Makes e the subject and takes ln f bth sides B M M [] [] Hence f ln( + k) ln( + k) A ca [] (e) f : Dmain: > k r ( k, ) Either > k r ( k, ) r Dmain > kr ft ne sided B inequality their part (c) RANGE answer (see appendi) [] 0 marks 666/0 Cre Maths C 6 0 th June 009 L Cpe June 009 Advanced Level in GCE Mathematics Versin :

7 A B cs A+ A csa cs Acs A sin Asin A 6. (a) Applies A B t cs( A+ B) t give the underlined equatin r cs A cs A sin A M cs A cs A sin A and cs A sin A + gives csa sin A sin A sin A (as required) Cmplete prf, with a link between LHS and RHS. N errrs seen. A AG [] (b) C C sin sin cs Eliminating y crrectly. M cs sin cs sin cs cs sin cs cs Using result in part (a) t substitute ± ± cs fr sin as r cs ksin as k ± ± t prduce an equatin in nly duble angles. M sin + cs Rearranges t give crrect result A AG [] (c) sin + cs Rcs( α ) sin + cs Rcscsα + Rsinsinα Equate sin : Rsinα Equate cs : Rcsα R + ; R B tanα α tanα ± r tanα ± r sinα ± r csα ± their R their R awrt 6.87 M A Hence, sin + cs cs( 6.87) [] 666/0 Cre Maths C 7 0 th June 009 L Cpe June 009 Advanced Level in GCE Mathematics Versin :

8 6. (d) sin + cs cs( 6.87) ( ) cs( their α ) cs 6.87 ± M their R awrt 66 A Hence,.69..., One f either awrt.6 r awrt.7 r awrt 6. r awrt 6. A Bth f awrt.6 r awrt 6. A [] marks If there are any EXTRA slutins inside the range 0 < 80 then withhld the final accuracy mark. Als ignre EXTRA slutins utside the range 0 < /0 Cre Maths C 8 0 th June 009 L Cpe June 009 Advanced Level in GCE Mathematics Versin :

9 7. 8 f + ( + ) ( )( + ) R,,. (a) f An attempt t cmbine t ne ( )( + ) ( ) + 8 fractin ( )( + ) Crrect result f cmbining all three fractins M A ( )( + ) + [( + )( ) ] Simplifies t give the crrect numeratr. Ignre missin f denminatr A ( + )( ) [( + )( ) ] An attempt t factrise the numeratr. See rules fr factrising quadratics. dm ( ) ( ) Crrect result A cs AG [] (b) g e e R, ln. Apply qutient rule: u e v e du dv e e d d g( ) e (e ) e (e ) (e ) vu uv Applying v Crrect differentiatin M A e e e + e (e ) e (e ) Crrect result A AG cs [] 666/0 Cre Maths C 9 0 th June 009 L Cpe June 009 Advanced Level in GCE Mathematics Versin :

10 e 7. (c) g( ) (e ) e (e ) e e e e + Puts their differentiated numeratr equal t their denminatr. M e e + 0 e e + A (e )(e ) 0 e r e Attempt t factrise r slve quadratic in e M ln r 0 bth 0, ln A [] marks 666/0 Cre Maths C 0 0 th June 009 L Cpe June 009 Advanced Level in GCE Mathematics Versin :

11 8. (a) sin sin cs sin cs B aef [] (b) csec 8cs 0, 0 < < π 8cs 0 sin Using csec M sin sin 8cs 8sincs ( ) sin cs sin sin k, where < k < M sin sin A { } { } Radians , Degrees.77..., 6... { } { } Radians ,... Degrees , Either arwt 7. r 8.76 r 0. r. r. r awrt 0.0π r awrt 0.6 π. Bth 0. and. A A ca [] 6 marks Slutins fr the final tw A marks must be given in nly. If there are any EXTRA slutins inside the range 0 < < π then withhld the final accuracy mark. Als ignre EXTRA slutins utside the range 0 < < π. 666/0 Cre Maths C 0 th June 009 L Cpe June 009 Advanced Level in GCE Mathematics Versin :