Mark Scheme 4727 June 2006

Transcription

1 Mak Scheme 77 June 006

2 77 Mak Scheme June 006 (a) Identity = + 0 i Invese = + i i = + i i 0 0 (b) Identity = Invese = 0 0 i B Fo coect identity. Allow B Fo seen o implied + i = B Fo coect invese AEFcatesian B B Fo coect identity Fo coect invese (a) ( zz = )6e z z π i i π πi = e = e (b) ( w ) cis( π ) B Fo modulus = 6 B Fo agument = M A π Fo subtacting aguments Fo coect answe = M Fo use of de Moive 8 A Fo π seen o implied ( cos isin ) 8 8 = π+ π A Fo coect answe (allow and cis π ) 6

3 77 Mak Scheme June 006 EITHER c a = ± [,, ] B Fo vecto joining lines ( ca ) [8,, 6] M* Fo attempt at vecto poduct of c a and [8,, 6] n =± [,0,9] A Fo obtaining n. f.t. fom incoect c a n M Fo dividing n by magnitude of [8,, 6] d = [8,, 6] = 7 09 A Fo eithe magnitude coect (d = ) A Fo coect distance CAO OR c a = ± [,, ] B Fo vecto joining lines ( ca ).[8,, 6] M* Fo attempt at scala poduct of c a and [8,, 6] cos A Fo coect cosθ AEF. f.t. fom incoect c a M d = sin θ Fo using tigonomety fo pependicula distance A Fo coect epession fo d in tems of θ (d = ) A Fo coect distance CAO OR c a = ± [,, ] B Fo vecto joining lines ( ca ).[8,, 6] M* Fo attempt at scala poduct of c a and [8,, 6] = 09 = 09 Fo finding pojection of c aonto 09 A [8,, 6] f.t. fom incoect c a M d = 09 Fo using Pythagoas fo pependicula distance A Fo coect epession fo d (d = ) A Fo coect distance CAO OR CP =± [ + 8 t, + t,6 t] B Fo finding a vecto fom C(,, ) to a point on the line CP. [8,, 6] = 0 M* Fo using scala poduct fo pependiculaity t =± OR P = (9,, ) A Fo coect point. f.t. fom incoect CP d = M Fo finding magnitude of CP A Fo coect epession fo d (d = ) A 6 Fo coect distance CAO SR Obtain CP = [,, ] [8,, 6] =± [, 0, ] B Veify [, 0, ]. [8,, 6] = 0 M* 6 d = = M A A (maimum / 6) 7

4 77 Mak Scheme June 006 Integating facto e ln( + ) e + = ( ) d M Fo coect pocess fo finding integating facto = + A Fo coect IF, simplified (hee o late) d y( + ) = d ( + ) M Fo multiplying though by thei IF Fo integating RHS to obtain M k k y( + ) = ( + ) ( + c) A( + ) OR ln A( + ) A Fo coect integation (+c not equied hee) = + c c = M Fo substituting (0, ) into GS (including A + c) Fo coect c. f.t. fom thei GS ( ) ( ) y = A Fo coect solution. AEF in fom y = f( ) 8 (i) EITHER a = [,,], b =± [,, 0] B Fo stating vectos in the plane n = a b =± k [ 0,0, ] M Fo finding pependicula to plane A Fo coect n. f.t. fom incoect b Use (,, ) OR (0,, ) M Fo substituting a point into equation a + by + cz = d whee [ abc,, ] = thei n y+ z = 0 A Fo coect catesian equation AEF OR a = [,,], b =± [,, 0] B Fo stating vectos in the plane e.g. = [,,] +λ [,,0] +μ[,,] M Fo stating paametic equation of plane [, y, z ] = [+ λ+ μ,+ λ+ μ,+ μ] A Fo witing equations in, y, z f.t. fom incoect b M Fo eliminating λ and μ y+ z = 0 A Fo coect catesian equation AEF (ii) [ t, t, t9] B Fo stating a point A on l with paamete t AEF (iii) [, 7, ] M Fo finding diection of fom A and (,, ) ± [t+, t7, t ]. [,, ] = 0 M Fo using scala poduct fo pependiculaity with any vecto involving t t = A Fo coect value of t + z = = OR 9 A Fo a coect equation AEFcatesian y z = = 9 SR Fo p+ q+ = 0 and no futhe pogess awad B 0 8

5 77 Mak Scheme June 006 m + = 0 m= ± i B Fo coect solutions of auiliay equation (may be implied by coect CF) CF = Acos + Bsin B Fo coect CF i i (AEtig but not Ae + Be only) PI = psin ( + qcos ) B State a tial PI with at least psin 6 (i) ( ) psin ( qcos ) + psin ( + qcos ) = sin M Fo substituting PI into DE p =, q = 0 A Fo coect p and q (which may be implied) y = Acos + Bsin + sin B 6 Fo using GS = CF + PI, with abitay constants in CF and none in PI (ii) (0, 0) A = 0 B Fo coect equation in A and/o B f.t. fom thei GS dy Bcos cos B d = + = + M Fo diffeentiating thei GS and dy substituting values fo and d A= 0, B = A Fo coect A and B Allow A = i, B = i fom i i CF Ae + Be y = sin + sin A Fo stating coect solution CAO 7 (i) iθ iθ iθ iθ iθ C+ is = + e + e + e + e + e 6iθ e = iθ e iθ iθ iθ iθ iθ e e e e e =. = e iθ iθ iθ iθ iθ e e e e e isin (ii) C+ i S = θ. e iθ isin iθ 0 M Fo using de Moive, showing at least tems M Fo ecognising GP A Fo coect GP sum A M Fo obtaining coect epession AG Fo epessing numeato and denominato in tems of sines Fo k sin θ and k sin θ θ A Re C = sin θ cos θ cosec θ A Fo coect epession AG Im S = sin θsin θcosec θ B Fo coect epession C = S sin θ= 0, tan θ= Ignoe values outside 0 <θ<π,,, A (iii) M Fo eithe equation deduced AEF θ= π π A Fo both values coect and no etas θ= π π π Fo all values coect and no etas. Allow A fo any value OR all coect with etas 9

6 77 Mak Scheme June (i). a a. (ii) Possible subgoups ode, (iii) (a) {, ea} B B B B Fo stating the non-commutative poduct in the given table, o justifying anothe coect one Fo eithe ode stated Fo both odes stated, and no moe (Ignoe ) Fo coect subgoup (b) { e,,,, } B Fo coect subgoup (iv) ode of = B Fo coect ode ( a) = a. a = a. a = e M Fo attempt to find ( a) m = e OR ( a ) m = e ode of a = A Fo coect ode ( a ) = a a. = a a. = aa. = e ode of a = A Fo coect ode (v) If the bode elements a a a a ae a a a a not witten, it will be assumed that the poducts aise fom that ode a e a e a e a e B B B B B m Fo all 6 elements of the fom e o Fo all elements in leading diagonal = e Fo no epeated elements in any completed ow o column Fo any two ows o columns coect Fo all elements coect 0