ASRJC 2023 JC1 Promos 9649 (Solutions)

1 1 (a) Length of semi-major axis, 13a . Length of semi-minor axis, 5b . Distance from centre to focus, . 22 22 13 5 12ca b    . Centre of ellipse: (12, 0) Since the principal axis is along the x-axis, one of the foci has to be (0, 0), the origin. (b) 12 13 ce a . Then the polar equation of the ellipse is 121c o s13 kr    for some k. When 0  , 13 1313 12(1) krk  . When  , 13 13 13 12( 1) 25 krk  . So 13 2513 2 2625 13kk a k   . Hence, the polar equation of the ellipse is 25 13 12cosr   . 2 42ln( 1) e 1 yyx x    Volume of bowl ln 2 22 0 1(1 ) ln 2 d5 xy     ln 2 0 1 ln 2 e 1 d5 y y   Let 2 de12 e d y yuuu y   . When 0, 0.yu When ln 2, 1.yu Then ln 2 1 00 2e1 d d e y y uyu u 21 20 1 2 0 11 0 2 d 1 121 d 1 2t a n 21 0 2 42 u uu uu uu                3 1Volume of bowl ln 2 252 9ln 2 units25           

2 3 (a) 22d (1 cos ) cos sin sin cos cos sin d (1 cos )sin sin cos sin 2sin cos y x                 When d 1d y x  , cos cos 2 sin sin 2   (double-angle formulas). (b) Using factor formula, 332cos cos 2sin cos22 22 33cos sin cos 022 2        33cos sin or cos 022 2    (no solutions for 0 22  ) 33 5tan 1 or 22 4 4     5 or 66   4 (a)  th 1 width of height ofrectangle rectangle 22 f n n r r rU nn    Total area of the rectanglesn Since f is a strictly decreasing function, the total area will be an underestimate of the actual area under the curve from 0 to 2x . So nUI  . Also, the total area of the n rectangles approaches the actual area, which is denoted by the definite integral 2 0 f( ) dxx . Therefore, for the limiting case, UI  . (b) nV is an over-estimate of I. Therefore, nVI  . (c) Let 1f( ) 2( 1)x x  . y O x …

3 21 1 1 1 246 221 21 21 2 1 1 111 1 246 2111 1 111 1 246 2 nU nn nnn n nn nnn n nnn n n                                  Therefore, 2 0 111 1lim 246 3 1 d2( 1) 1 ln 32 n Unnn n I xx             5 (a) (b) ii iizk zk zk  22 2 2 min i (2 ) 2 2 2 44 42 2 4 8 2 2 zk k kk kk          C(−2, −2) A S O T

4 (c) 1142tan tan 0.588 rad63OAT   1121tan tan42OAC   11 1 22 22 2sin sin sin 1042 CSCAS AC      Therefore, 11 21sin tan 0.221 210 OAS CAS OAC      So 0.588 arg( 6) 0.221 (3 s.f.)z  . (d) Multiplying w with ie rotates the point W representing w about the origin anticlockwise by an angle of . If the p