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 point upon rotation does not lie on the locus of z, we require w to be greater than max( 0 ) 4z . Hence 4w  . 6 (a)(i) 17 11 11P 4.940103168 10 (ii) 12 6P 5! 79833600 (the last row is filled by the 5 particular people in 5! ways; the 12 seats in the front 3 rows can be filled by the remaining 6 people in 12 6P ways) (iii) 37 21CC 2 1 (the 2 married couples ca n be chosen from the 3 in 3 2C ways; the 1 other person can be chosen from the remaining 7 people in 7 1C w a y s ) (b)(i) Treat the married coupl e and family of 4 as one unit each; together with the 2 women, there are 4 units to arrange in a circle. For each circular arrangement, the married couple and family of 4 can each be further arranged in 2! and 4! ways respectively. 41 !2 !4 ! 2 8 8 (ii) First, find the number of circular arrangements with the married couple seated together. Next, find the number of circular arrangements with the married couple seated together and the 2 women seated directly opposite each other. Subtract the second from the first. 7 1 ! 2! 1 4 2! 4! 1440 192 1248     7 (a) ln( 1) 1 sinyx  1d d cos 1 cos1d d yy x yxyx x         2 2 dd d cos 1 sin cos 1 sindd d yy y x yx x y xxx x        When 0,x 1s i n 01e e e 1yy   

5 d ed y x   2 2 d ee . 0 ed y x     Maclaurin’s series for y is  2211e1 e e e1e e 22x xx x    (b) Given x is small such that 3x and higher powers of x can be neglected, ln( 1) 1 sin 1yx x      2 11e e . e e 1 2 xx xyx            21e1e e 2yx x  (verified) (c) 111 1 sin 2 2 3 11 1 11 1 7ed e e e d e e e e 22 6 3 x xx x x x x x             (d) From the sketch, it is clear that 11 1s i n 2 11 1ed e e e d 2 x x xx x    therefore, the answer in (i ii) is an over-estimation. 8 (a) Possible sketch: A graph with a turning (or sharp) point as x-intercept (b) 3f ( ) 0.28383 0x   and f (3) 8.88751 0 . The root is in 3(, 3 )x . 33 4 3 f (3) 3f ( ) 1.47254 1.47 (2 d.p.)f( 3 ) f( ) xxx x    4f ( ) 0.16086 0x   . Root is in 4(, 3 )x . 44 5 4 f (3) 3f ( ) 1.499696 1.50 (2 d.p.)f( 3 ) f( ) xxx x    f (1.495) 0.1012 0  , f (1.505) 0.07407 0  . There is no change of sign. Therefore, the required accuracy correct to 2 decimal places have not been achieved. (c) The initial value needs to be larger than the x-coordinate of the turning point of f. 1 2f( ) ( 2 l n 1 ) 0 0 o r exx x x x        . Turning point occurs at 1 e x .

6 Set of values of 1u is 11 1 1:, e uu u   . (d) 1 f( ) f( ) n nn n uuu u    2 ln( ) 1 (2ln( ) 1) nn n nn uuu uu   (e) Using the recurrence relation, 23 4 5 62, 1.62859 1.63, 1.53734 1.54, 1.53161 1.53, 1.53158 1.53uu u u u        The required root is 1.53 (2 d.p.). 9 (a) Asymptote: byx a , Perpendicular line passing through F: ()ayx c b  . 22 22 22 2 22 22 () (since ) baxx cab ba a c xab b ac abx ba b ac a ca bab ab            (b) 22 22 22 2 22 2 22 22 2 2 2 2 2 2 () 1 (2 ) () 2 ( ) 0 xm x k ab b x a m xm k x ka b am b x am k x a b k        If the line is a tangent to H, discriminant = 0. 22 2 2 2 2 2 2 22 2 22 2 2 2 22 2 22 2 4 22 2 2 2 2 2 242 2 22 2 2 (2 ) 4( ) ( ) 0 () ( ) 0 () 0 0 (shown) am k am b a b k amk am b b k amk amb b amk bk a m bbb k ka m b              (b) A line passing through (, 0 )c and perpendicular to tangent is 1yx c m  . The foot of perpendicular is the intersection between this line and the tangent.

7  221 ()yx c m y x c m y x cm        --- (1) 222 2 2()y mx k y mx k a m b     --- (2) Adding (1) and (2), 22 2 2 22 2 2 2 2 22 2 2 2 2 2 2 2 22 222 22 ( 1)( ) [since ] (1 ) my m x y x y m x y mx am c b mx y a m a c a b am xya           The tangents cannot become the asymptotes, so we need to exclude this limiting case. The two points to be excluded both have x-coordinates 2 22 a ab . 10 (a)  1 2 3 1 2 3 4 2 3 4 5 3 4 5 6 Let T be Tim’s score.   18P 3 1 P 2 1 0.889 3sf 99TT     (b) (i)  1, 2, 3X     4 11P 1 P 1111 38 1X           44 4 123 4444 P 2 P 1112 P 1122 P 1222 P 2222 1111 4643333 4641 81 81 81 81 15 81 XC C C                        65P3 1 P2 81XX   x 1 2 3 P Xx 1 81 15 81 65 81 (ii)     3 1 1 30 195 226E P shown 81 81 81 81x Xx X x      

8    3 22 1 EP 1 60 585 646 7.98 3sf81 81 81 81 x Xx X x           22Var E E 0.191 3sfXX X 