ASRJC 2023 JC1 MYE Further Math (Solutions)

1 (a)       11f1 f 11 2rr rr r r        2 12 rr rr r     2 12rr r       11 11 f1 f12 2 nn rr rrrr r   =   f 0 f 1 1 2   f1 f2  f2 f3 + f 2 n  f1 n  f1 n f n              1 f0 f2 n  =   11 42 1 2nn  (b)(i)    11 11 lim12 12 n nrr rr r rr r       11lim 42 1 2n nn    1 4 (ii)     12 1 32 3 11 12 1 2 Ni N ri rr r ii i          3 1 1 12 N i ii i       11 42 1 2NN 

2 (a) (i) 2 1 aSa a r a r r     351 2 21 arSa r a r a r r    ๠erefore 1 2 211 aa r rr   1 2 (since exists, 1)1 12 31 r Srr rr r          1 3r  (ii)  2 22 11 93 22 221 8ar a r    22:, , , , , ,Haa ra r a a ra r  Sum to infinity of H 12 33 18 18 2711 a r   (b)  1000 1 1.4 0n  715.3 716nn  First negative term of series 1000 715 1.4 1   Sum of all positive terms of series 715 2 1000 714 1.42    357643 3 (a) Let F be the focus of the parabola. By the geometrical definition of parabola, AF AR and BF BS . Let 1AR k and 2BS k . ๠en 122( )nm k k  . ๠erefore, 12 2 nmkk  . Area of ABSR  12 1( ) 24 mn mmk k   (b)(i) Foci: (5, 2) and (5, 10). So centre: (5, 6). Conic is an ellipse with length of semi-minor axis, b = 3. 22 2 22 2 43ca b a   5a Cartesian equation in standard form: 22 22 (5 )(6 ) 135 xy  . x x x x (5, 6) (2, 6)

(ii) New equation: 22 2 2 22 2 (5 )( 56 ) (5 ) 1( 6 / 5 ) 135 3 xy x y     Centre: 65, 5   .  2 2231 8 82 2cc    Foci: 652 2 , 5   4 (a) (b)  32 2 sin 2cos sin coscos dcos sin d sin xxr        22 2 cos 2sin cos sin     dsin cos sin d yyr         3 22 2 2 ds i n d cos 2sin cos 1 cot 2 cot 1 (shown)(2 ) y x rr          Since cot is a decreasing function for 62   , 0, cot 0, 36r  . From the sketch of 2 1f( ) (2 )r rr  , 1f( ) f( 3) 53 r  . ๠erefore, d1 d 53 y x  . (c) cotr 

2 22 2 coscot sinr   22 2 2 2 2 22 22 sin cos () r xy r xy yx     42 22 24 2 2 0 4 2 yx yx x xxy     Since 2 0y  , choose 24 2 2 4 2 x xxy   . Also, since 0y , 24 2 4 2 x xxy   for 30 2x . 5 (a) A point on the plane is ( 1 ,0 ,0)B . ˆ 33 51 1 1 33 3 25 9 BP c c c                 .n .  2 5 9 or 2 5 9cc     7c or 2 (rejected, since 0)cc  (b) A line perpendicular to the mirror and passing through A is: 15 1 17 1 , 51 ss              r  Let F be the intersection between this line and the mirror. 15 1 17 1 1 51 15 17 5 1 73 1 2 s s s ss s s s                    

๠e intersection point has position vector 17 15 3 OF     and 2 2 2 AF     . ๠erefore, 17 2 19 15 2 13 32 1 OA                      . Coordinates of Ais (1 9 , 1 3 , 1 ) . (c) 15 6 6 AP     Let  be the acute angle between the line and the plane. 11 15 1 61 61 15sin sin 297 3 891 30.1668 30.2 (1 d.p.)                   6 (a)(i) 22 i( 2 i ) i ( 2 i )ww       44 i12 i1 46 i   22 i4 6 i 46 i46 i 64 i 46 31 i26 13 z       (ii) 23 i 0uui u z  3 iuui u u z    So 0u or 2 iui u z     2 iui u z   (considering conjugate) From (i), 2i 2iuu      .

So the two possible values of u are 0 and 2i . (b) 22 2 2sin i cos i cos i sin55 5 5p        2ii 25 i 10 ee e             i 10e n np  Equating arguments, (2 ),10 n kk     10(2 1)nk  When 1, 10kn  . When 2, 30kn  . When 3, 50kn  . ๠e 3 smallest positive integers n are 10, 30, 50. 7 (a)  1tan tan 88 xx    ; 120 20tan tan 13 13 xx     ๠erefore, 11 20tan tan81 3 x x   (b) Either   11 11 81 3 22 20 81 3 20 dtan tan81 3 d 11 xx xx x       

22 d8 1 3 d 64 569 40x xx x           22 22 8 569 40 13 64 64 569 40 xx x xx x          2 22 5 744 64 64 40 569 xx xx x     (c) Maximum light intensity at P occurs when d 0.dx   2744 64 0xx  10.05x (d) Using the graph of  against x , the minimum θ is 0.994 when 0x  . (e) When d2 010, d 44116x x   1dd d 2 0 0.5 0.000227 radsd d d 44116 x tx t    