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                        =  − + = − + + += − −  = − − = − − = − − − =+     (b)(ii) Perimeter circle arc cardioid arc 22 3 22 3 3 22 3 4(1 cos ) 4sin d3 4 4 1 2cos cos sin d 4 4 d (shown), 4 , 4, , 3r k p                       = + + +  = + + + + = + = = = =    3 2 3 3 3 4 4 2 1 cos d 4 4 2 1 2cos 1 d 2 4 8 cos d 2 4 16 sin 2 14 16 1 2 48                 = + + = + + − =+ =+  = + −  =+    

5 (a) 2 4y px= d d 224 dd y y pyp x x y=  = Equation of L: 222 ( ) 2 py pt x pt pt− = − ( ) 212 1 y pt x pt t y x ptt − = − =+ H has coordinates ( )0, pt . F1 has coordinates (p, 0). Gradient of line 1 0 0 ptF H t p −= =−− Since ( )( )1 1Gradient of Gradient of ( ) 1F H L t t = − =− , the lines are perpendicular. (b) Since 1RH F H= , R has coordinates ( ),2p pt− . The directrix equation is xp=− . Therefore, R lies on the directrix. (c) The points ( , )ab− and ( , )ab− lie on the directrix, since they are points of reflection of F in the y- and x-axis respectively. Gradient of directrix b b b a a a −−= =−+ . Equation of directrix: ( ) 0by b x a bx aya+ =− −  + = (shown) (d) Let the feet of perpendicular from D and E to the directrix be M and N respectively. Since distance from point on parabola to focus = distance from point on parabola to directrix, FD MD= and FE NE= 22 22 ( 1) aab ba + − = + 22 22 2( 2) bab ba − + = + 2 22 22( 1) aab ba+ − = + --- (1) 2 22 22 4( 2) bab ba− + = + --- (2)

22 2 2 2 2 22 444 (1) (2) : 4 4( 2 1) 4 4 aba b b a a b ba + + + − + + − + + = + 225 4 5 8 4 0a a b b− + − + = Completing the squares: 22 22 4 2 4 8 4 165 5 4 0 5 5 5 5 5 5a a b b        − + − + − + − + =                 22 24 055ab   − + − =       The only possible solution is 2 5a= and 4 5b= . 6 (a) Using the same people will eliminate any differences in grip strengths between different people and so will only compare the grip strengths of the dominant and non-dominant hands of the same person. (b) A 95% confidence interval for population mean grip strength 3.922.79 2.20099 12 =   ( ) 2.79 2.49065 0.29935, 5.28065 = = (c) The population of differences in grip strengths must be normally distributed. (d) Let d be the population mean difference in grip strengths (dominant – non-dominant) 0 1 H : 2 (physiotherapist's claim) H : 2 d d   =  Since the value 2 lies inside the confidence interval (0.29935, 5.28065), we do not reject 0H . Hence there is insufficient evidence at 5% significance level to refute the physiotherapist’s claim. (e) If samples of the same size are drawn repeatedly many times and a 95% confidence interval is computed for each sample in a similar manner, then about 95% of these intervals would contain the population mean difference in grip strength.. 7 The p.d.f. of V is ( ) 1 60 80f 20 0 otherwise vv  =  (a) ( ) ( )1P 60 20V v v = − (using area of rectangle)

( ) 0 60 60F 60 80 20 1 80 v vvv v   −=     (b) 240T V= For 60 80v , 34 t , ( ) ( ) 240 60240 240 240 12G P P P 1 P 1 4 20 tt T t t V V V t t t −     =  =  =  = −  = − = −           ( ) 0, 3 12G 4 , 3 4 1, 4 t tt t t   = −     (c) Let t hrs be the journey time from house to capital city. Then ( ) 3 4 12P 0.80 4 0.80 3 3 hrs 45 minsT t t t   −    =  the latest time to leave the house is 6.15 am. 8 (a) Flaws occur independently and at a constant mean rate. (b) ( )0no. of flaws in 200 m length of material ~P 3.2N = ( ) ( )P 5 1 P 4 0.219NN = −  = (c) ( ) ( )FPx X x= = P(at least 1 flaws in x metres) ( ) ( ) ( ) 00 0.016 0.8P 0 , where ~ P P 0.016 50 1 P 0 1 e for 0.x Y Y x x Y x− =  =  = − = = −   ( ) 0.0160.016e 0f 00 x xx x − =  X follows an exponential distribution with mean 1 62.50.016== (d) ( ) ( ) 0.8P 50 1 F 50 e 0.449X − = − = =

9 (a) The underlying population for the estimated time is normally distributed. (b) The hypotheses to be tested are: 0 1 H : Estimates are normally distributed. H : Estimates are not normally distributed. Level of significance 5% Using the unbiased estimates in (a), we test the fit of the data to the normal distribution ( ) 2N 61.1, 7.564 . Estimates x (seconds) Observed frequency iO Expected frequency iE ( ) 2 ii i OE E − 51.5x 4 6.1314 0.7409 51.5 55.5x 10 7.6412 0.7281 55.5 59.5x 9 11.2016 0.4327 59.5 63.5x 17 12.4951 1.6242 63.5 67.5x 15 10.6060 1.8204 67.5x 5 11.9247 4.02