ASRJC_9649_2024_Prelim P2 Solutions (updated)

1 12 1 3 2 3 4 5 1 1 0 3 1 1 0 3 1 1 0 3 4 8 1 0 4 4 13 0 4 4 13 5 9 8 0 4 8 4 15 0 0 8 2 RR R R R R k k k k k k k k k k k −+ − + − + − − −            ⎯⎯⎯⎯ → − + ⎯⎯⎯→ − +           − + −      For all real values of k, there are exactly 3 pivots in the REF. Therefore, the only possible value for dim( )kS = 4 – 3 = 1. For 4 0x = in the solution, we require 8k = . A basis for kS when 8k = is 2 2 1 0  − . 2 (a) det 0 15 ( ) ( 3 ) 0a a b a=  − − + − =M 215 3 5 3 a ab ab a += =+ (b) 1 2 2 0 3 1 1 1 0 5 2 2 ab a           =         −−     2 2 2 3 2 1 2 10 2 ab a + − = −= − =− 4, 2ab = = (c) Let the characteristic equation of M be f ( ) 0 = . f ( ) det( ) ( 1)( 3)( 5) 3 0    = − = − − − − =IM f (2) (1)( 1)( 3) 3 0= − − − = . Since 2 = satisfies the characteristic equation, 2 is an eigenvalue. 32 9 23 18 0  − + − = 2( 2)( 7 9) 0  − − + = (by comparing coefficients) 2 7 49 4(9) 7 137 9 0 22   − − + =  = = (d) Solve 1 3 0 0 0 1 1 0 03 0 3 x y z  −        − − =       −−     3 3 2 1 1 2 3 3 1 22 3 3 3 1 3 0 1 3 0 1 0 3 0 1 1 0 1 1 0 1 1 0 3 3 0 0 03 0 3 R R R R R R R R R RR →− →− →+ →−      −−     − − ⎯⎯⎯⎯⎯ → − − ⎯⎯⎯⎯⎯ →            −−−−     

3 1, 1 x y z z z −  = −    . An eigenvector corresponding to 2 is 3 1 1    − . Alternatively, use cross-product: 1 03 3 1 1 0 1 1         −  − =        −−    3 (a) 3 i( 1) 2 i( (1 3i) 2 ( (1 3i) 2z z z+ −   − +   − +  (1 3i) ( 1 5i)zz− +  − − + (b) 13tan tan 3 1APF APF − =  = 3tan 34CPF CPF  =  = . 22 22sin 333 CPE = = + . 11 2sin arg( 2) tan 343 z −−−  +  0.295 arg( 2) 1.25 (3 s.f.)z +  Im Re C(1, 3) x x P(–2, 0) D(0, 4) A(–1, 3) B(1, 5) E F 2 x (–1, 5)

(c) 1tan 4DGC= 22 2 1 3sin 1714 CGH + = = + 1113tan sin24 17 0.510991 0.511 (3 s.f.) k  −−= − − = 4 (a) dd cos sindd xr r=− , dd sin cosdd yr r=+ . 2 2 2 22 22 22 2 2 2 2 d d d d1 d 1 dd d d d dd ddd dd cos sin sin cos ddd d d d dcos 2 sin cos sin sin 2 sin cd d d d y y x xxx x x xy rr rr r r r r r r r                       + = +                 =+           = − + +          = − + + +          ( ) 22 2 2 2 2 2 2 os cos d d sin cos dd d dd r rr rr        +  = + +  =+     (b)(i) Find intersections: 13 2(1 cos ) cos 23   = +  =  = Exact area Im Re C(1, 3) x A(–1, 3) B(1, 5) H F G(0, –1) D k 2 1 Q

  22 3 2 3 3 3 3 12 3 4(1 cos ) d2 9 4(1 2cos cos ) d 1 cos 25 8cos 4 d 2 3 8cos 2cos 2 d 3 8sin sin 2 33 4 3 2 932 2               