HCI 2023 JC1 MYE 9649 (Solutions)

1 Let nP be the statement 22 n nnab a b  for n  When 1,n  LHS = 2 ab R H S 2 ab 1 is true.P Assume kP is true for for some k  , i.e. 22 k kkab a b  Want to prove that 1kP is true, i.e. 1 11 22 k kkab a b       11 11 11 11 11 11 11 22 --- *22 22 4 244 24 4 24 2 k kk k kk kk k k kk kk k k kk kk k k kkkk kkk ab a b ab ab a b ab aa b a b b ab ab a b a b ab ab a b a b aba bab abab                                 04 k ab  Hence kP is true 1kP is true. Since 1P is true and kP is true 1kP is true, by mathematical induction nP is true for all .n   2(a) (**)

2(b) Let the polar coordinates of N be11,r  and the polar coordinates of P be 2c o s,a   1 2 OPC OC PC PON OPC CON               2 1 cos 2 cos cos 2c o s ON OP a ra      22   1     2 11 2 cos cos 2 1 1 cosra a a      (Shown) 3(a) Characteristic equation:   2 56 0 32 0 3o r 2 mm mm mm       32 nn nuA B  1ua ,  32aA B 2ub ,  94bA B N P O C C O N P

23 3 2 B ab abB    32 32 22 33 aA B aba baA     2 3 baA , 3 2 abB       1 1 23 3232 23 3 2 n n n n n n ba a bu ub a a b        3(b)         1 1 2 2 1 23 3 2 23 3 2 n n n n n n ba a bu u ba a b               1 1 1 223 3 11 223 32 3 n n n n ba a bu u ba a b                If 2ba , as n, 1 2 03 n ,    1 lim 3 n n n u u     If 2ba , as n, 1 lim 2 n n n u u     Or   2 Let the limit be 56 0 32 0 3o r 2 L LL LL LL     

4(a) The polar equation form of a conic is 1c o s kr e   . Let F be the pole and ,AB have polar coordinates  ,AF  and  ,BF  respectively Since ,AB lies on the conic, 1c o s kAF e   and 1c o s kBF e   1c o s ke AF and 1c o s ke BF  1c o s ke BF Summing, 1 cos 1 cos 2kk eeAF BF      11 2 AF BF k 4(b) B y (i), F A B O

11 1 1 1 11 2 21 ----------------(1) PF F Q k PF PFFQ k   Similarly 22 2 2 2 11 2 21 -------------(2) PF F R k PF PFFR k   Summing (1) and (2) ,  12 12 12 12 12 12 2211 22 PF PF PF PFFQ F R k k PF PF PF PFFQ F R k              Since this is an ellipse, 12PFP F is a constant, let 12PFP F C , 12 12 2 2 constantPF PF C FQ F R k   (Shown) 5 (a)    :, 0 :, 0 :, Aa B a Px y  22 22() ()AP BP x a y x a y                 2 22 22 4 22 2 22 2 4 222 22 2 22 2 2 2 4 () () 22 2 22 2 AP BP a xa y xa y a xa x a y xa x a ya xy xy a x a x ya x a a x aa x a a                    222 22 24 2 24 222 2 2 2 24 20 x yx y a a a x a xy a yx       

    222 2 2 2 2 22 2 2 2 2s i n c o s0 2 cos sin 2 cos 2 [Shown] ra r r ra a        Alternative Solution, by cosine rule  22 2 22 2 22 2c o s 2c o s 2c o s AP a r ar BP a r ar ar a r               2 22 22 2 222 44 2 2 4 2 2 2 22 4 22 2 42 2 2 2 2 22 2 22 22 2c o s 2c o s 2c o s 24 c o s 24 c o s 0 4c o s2 =2 2cos 1 =2 cos 2 2c o s 2 AP BP ar a r ar a r ar a r aa a rr a r ar r ar ra r a r ar ar ra                        5(b) ( 2 ,0)a (2 ,)a  O  ,0Aa  ,Ba   ,Pr  π 2  /g3 0  /g3 3π 2  /g3 π  /g3

5(c) For cos 2 0, 0 4   . Area 24 0 2 4 0 42 0 2 22 142 c o s 2 d2 4c o s 2 d sin 24 2 140 2 2 units a a a a a                  6(a) Using Shell method, Volume of S rotating about x-axis 1 0 1 21 0 2 d 2e d ( S h o w n )y xy y yy        where  21Fe yyy  . 6(b) 1 11 21 21 21 00 0 1 21 0 1 1 11e d = e e d 22 11ee24 11 1ee e24 4 11ee44 yy y y y yy y             

6(c) When y = 0, x-intercept = 1e Using Disk method, Volume of T rotating about the x-axis  1 1 1 e 2 e 2 e e e 2 e d 1l n d2 ln 1 d4 yx x x x x              Volume of cylinder  2 1e e     Volume of S + Volume of T = Volume of cylinder 1 1e 221 0e 2e d l n 1 de 4 yyy x x        1 e1 2 21 e0 1 1 ln 1 d 4 e 2 e d 24e e e4 114e e22 12e e yx xy y           6(d) 1l n 2 d1 d2 xy y x x   Surface area when rotated about the x-axis 1 2 e e 2 2 1l n 121 d 22 10.287666 units 10.3 units x xx            

The area of the surface generated when the arc of a curve with equation 21e xy  between the y-intercept and ey is rotated through 2 radians about the y-axis is also 10.3 unit2. 21e xy  is the inverse function of 21 l nyx . Hence the area of the surface generated when the arc of the curve with equation 21e xy  between the y-intercept and ey is rotated through 2 radians about the y-axis is the same as the area of the surface generated when the arc of the curve with equation 21 l nyx between the x-intercept and ex is rotated through 2 radians about the x-axis. 7(a) The number of cross-sectional areas given must be odd 7(b) Using Simpson’s rule on the first 5 surface areas    4 1 63 4 190 4 127 100 100 4 76 20 103 16.293 10 m V      Using Trapezoidal Rule on the last 2 surface areas  4 2 63 3 20 14 102 0.51 10 m V     4 63 63 Total Volume 16.293 0.51 10 16.803 10 m 16.8 10 m (3 s.f.)     1 2 ra ah r    21 21 1 1 21 ar a h r ar r h r hra rr    

      22 21 22 2 21 2 1 2121 2 21 2 12 1 2 2 12 1 2 12 1 2 11ππ33 11ππ33 11π33 11 1ππ33 3 1 ππ3 1 3 Vr a h r a ar r rh hr rrrr A hrr rrh r h A h hr r h r A hA AAA              7(c)     1 4 63 4 190 190 127 127 127 100 127 1003 +100+ 100 76 76 76 76 20 20 10 16.136 10 m V             46 3 2 3 20 20 14 14 10 0.39831 10 m3V       6 63 63 16.136 0.39831 10 16.53431 10 m 16.5 10 m (3 s.f.) V     7(d)  6 6 Difference 16.803 16.534 10 0.269 10    7(e) 80% of water = 660.8 16.803 10 13.442 10  Number of days the reservoir can last 6 6 13.442 10 19.2 190.7 10   days Since the difference between answer in part (b) and (c) is less than half of 60.7 10 , there will be no change in the estimated number of days. 8 Method 1