HCI 2023 JC1 MYE 9649 (Solutions)
Uploaded by fwyr · 14 September 2024
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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 be11,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
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