ASRJC FM 2023 Prelim P2 (Solutions)

1 1 2 *' kz z   2 *arg ' arg kz z          2*arg arg 0 arg arg( ) kz z z      Therefore the line PP’ passes through the origin O O,P,P’ are collinear and O,Q,Q’ are collinear. Let .OPQ  Area of triangle OP’Q’ is 4 times the Area of triangle OPQ. Then       22 ** 22 3 * * 4 3 4 4 2 11 ' ' sin 4 sin22 4 2 3 2 .3 since and 2 .323 2 .3 23 OP OQ OP OQ kk zw kk z z w wzw k k k           2 (a) 1 2 3 4 5 61, 1, 2, 3, 5, 8F F F F F F      1 2 3 4 5 61, 3, 4, 7, 11, 18L L L L L L      5 4 6 5 4 6 11 and 3 8 11 (verified) L F F L F F         (b) To prove: 11m n m n m nF F F F F   ,  2 fixed ,m m n  . For n = 1, LHS = 1mF  . RHS = 1 1 2 1 1m m m m mF F F F F F F      = LHS, implying the statement holds for n = 1. For n = 2, LHS = 2mF  .

2 RHS =  1 2 3 1 1 2m m m m m m mF F F F F F F F F        1 2 LHS, mm m FF F     implying the statement holds for n = 2. Assume statement holds for 1 and for some , 2n k n k k k     . That is, we have 1 1 1m k m k m kF F F F F    and 11m k m k m kF F F F F   . For n = k + 1, 11m k m k m kF F F         1 1 1 1 1 1 1 1 1 2 by (IH)m k m k m k m k m k k m k k m k m k F F F F F F F F F F F F F F F F F F                    implying that statement holds for n = k + 1. So 11m n m n m nF F F F F   for a fixed m  2 and for all n  . (c) Let ,mn   2 1 1 11 n n n n n n n n nn F F F F F F F F FL      3 (a) 11 1 1 1 2 2 2 2 2 1 1 1 1 11 cos sin cos sin 2 sin 2 2 2 2 2 i i i iie e e e e i i ie                                                2 21 4 sin . 2 iiee   (b)     2 3 2 2 2 2 2 2 21 .....1 2 3 1 4 sin 2 14 sin 2 i i i ni ni n i n in n n n nC iS e e e e e e e                                  

3 Equating real parts,   2 14 cos sin 2 n nCn  Equating imaginary parts,   2 14 sin sin 2 n nSn  2 6 631 , 0, 1, 2,3. kk ii w w e e k             3 66 2 3 6111 1 4 sin 64 sin 22 i i iw e e e           When   6 0, 1 0 w    When     6 6 1, 1 64 1 132 w       When     6 623 , 1 64 1 2732 w       When      66 , 1 64 1 1 64w     4 1 11 12 rr r r r r rr uu x x u u uu MM                        12 1 12 46r r r r rr rr u u u u uu uu M                     ie 4 1 6 1 0 0 0 1 0 M   4 1 6 1 1 1 1 0 0 1 1 1 1 0 1 0 1 1 1 4 1 6 4 8 4 1 0 0 2 4 2 2 0 1 0 1 2 1 4 1 6 9 27 9 1 0 0 3 9 3 3 0 1 0 1 3 1                                                                                                        

4 1 4 9 1 , 2 , 3 1 1 1                            are eigenvectors of M (verified) Take 1 4 9 1 0 0 1 2 3 , 0 2 0 1 1 1 0 0 3 P D                    1 4 9 1 4 9 1 0 0 1 2 3 1 2 3 0 2 0 1 1 1 1 1 1 0 0 3 M                         MP=PD 1M=PDP 1 1 3 3 22 1 2 2 2 1 2 3 1 1 ..... r r r r r r r r r r u u u u u u u u u u u u u u u r -2 -1M M M PD P                                                   Using GC, 1 5 1 12 12 2 12 133 1 1 1 4 4 2 -1P          2 1 2 2 1 1 5 1 12 12 21 4 9 1 0 0 1 121 2 3 0 2 0 1 1 331 1 1 0 0 3 7 1 1 1 4 4 2 r r r r r r u u u                                    

5           2 2 2 1 1 1 1 11 1 11 1 4 9 1 0 0 4 1 2 3 0 2 0 6 1 1 1 0 0 3 3 1 4 9 4 1 1 2 3 3.2 1 1 1 3 4 1 3.2 3 4 1 3.2 3 4 1 3.2 3 r r r r r r r rr r rr r rr                                                      4 1 3.2 3 for 1.      r rr rur 5(i)  f 0.5 0.21372 0   f 1 0.84147 0 Hence, there is a change in sign in the interval  0.5,1 .   1f ' cosxx x For 0.5 1 x , cos 0x and 1 0x   f ' 0x Hence, f is a strictly increasing function. Therefore there is exactly one root in the interval  0.5,1 . (ii)     1 11f 1 f22 1f 1 f 2 x      1 0.21372 0.841472 0.6010.84147 0.21372    correct to 3 s.f.   22 11f '' sin sinx x x xx      For 0.5 1 x , sin 0x and since 2 1 0x  for all real x,  f '' 0x   curve is concave downwards and from (i), f is a strictly increasing function, the estimate is an overestimation. (iii) Since 0.601 is an overestimate of the root, the root lies outside of the interval (0.601,1) and hence it is unwise to do so.

6 Alternative Method  f 0.601 0.056307 0  f 1 0.84147 0 Hence, there is no sign change in the interval  0.601,1 . Therefore the root does not lie in this interval and hence it is unwise to do so. (iv) Let   1 1F sin lnx x        2 22 1 11F' 1 1 ln11 ln xx xx xx          2 10.5, F' 2.77 1 0.5 1 ln 0.5 xx      Therefore the iterative formula 1 1 1sin lnnx x     may not work. (v) Using sin 1 e nx nx    , sin 0.5 1 e 0.61914x  2 0.55971x  3 0.58805x  4 0.57422x  5 0.58090x  6 0.57766x  0.58x Consider     f sin ln sin ln 1 1x x x x x      1 2 1x x x      f 0 2 1 0 0.5x x x      Hence, 0.5x is an appropriate initial estimate. Alternative method sine e 1xxx x x     0.5x Hence, 0.5x is an appropriate initial estimate.

7 Section B Statistics [50 marks ] 6(i)     0 F sin d 2 t t x x     0 1 cos2 t x  1 1 cos2 t      0, 0 1F 1 cos , 0 12 1, 1 x x x x x         (ii)     1P P 1 cos 2Y y X y       P cos 1 2Xy     11P cos 1 2Xy       1111 cos cos 1 22 y            11 1 cos cos 1 22 y    1 1 1 22 y    1 22 yy    dfF dyy y   d 1d yy 0 1 0 1xy      0,1YU , shown Alternative Method   1 0 sin d 12 xx          1 d d 21 cos 1 sin2 2 d d sin xxy x x y y x        0, 0xy and 1, 1xy     1 0 2sin d 12 sin xy x     

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