HKALE Pure Maths 2001 Paper01

1 2 1 2 a b 2 ii clearly 1 2 1 1 1 k k k k

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Unformatted text preview: sfies ( x − p ) 2 + y 2 + z 2 = 1 , then (t − p ) 2 + (t + 1) 2 + t 2 = 1 3t 2 − 2( p − 1)t + p 2 = 0 For t to have real solution, 4( p − 1) 2 − 12 p 2 ≥ 0 2 p 2 + 2 p −1 ≤ 0 −1 − 3 −1+ 3 ≤ p≤ 2 2 2001 Paper 1 Section B 10. (a) (i) Clearly, a1 , b1 > 0 and a1 2 − 2b1 2 = (−1)1 . Assume a k , b k > 0 and a k 2 − 2bk 2 = (−1) k where k ∈ Z+ . Then a k +1 = a k + 2bk > 0 , bk +1 = a k + b k > 0 and a k +1 2 − 2bk +1 2 = (a k + 2bk )2 − 2(a k + b k )2 ( = − a k 2 − 2bk 2 k +1 ) = (−1) By the principle of mathematical induction, the result follows. ( ) = a +b 2 . 1 Assume ( + 2 ) = a + b 2 where k is a positive integer. (1 + 2 ) = (1 + 2 )(a + b 2 ) (ii) Clearly, 1 + 2 1 1 1 k k k k +1 k k = (a k + 2bk ) + (a k + b k ) 2 = a k +1 + bk +1 2 By the principle of mathematical induction, the result follows. (b) Let n = 1, 2, 3, (i) u n +1 = an +2 bn u +2 =n = an un +1 +1 bn a n +1 a + 2bn =n b n +1 a n + bn (ii) From (a)(i), ∴ . 2 2 a n − 2bn = (−1) n . an b n 2 (−1) n = 2+ 2 bn un = 2 + (−1) n bn 2 Hence u 2 n −1 < 2 and u 2 n > 2 . [since u n = an >0] bn [since bn > 0 ] 2001 Paper 1 Section B (iii) Using (b)(i), Hence u n+2 un + 2 +2 un +1 u n +1 + 2 3u n + 4 = = = un + 2 u n +1 + 1 2u n + 3 +1 un +1 u n+ 2 − u n = 3u n + 4 − un 2u n + 3 2 = = ⇒ − 2u n + 4 2u n + 3 − 2 un − 2 u n + 2 2un + 3 ( u 2 n +1 − u 2 n −1 = and u 2 n + 2 − u 2 n = )( ( ) )( − 2 u2 n −1 − 2 u2 n −1 + 2 2u2 n −1 + 3 ( )( ) >0 [since u 2 n −1 < 2 ] − 2 u 2n − 2 u 2n + 2 2u 2 n + 3 ) <0 [since u 2 n > 2 ] i.e. the sequence the sequence { u1 , u 3 , u 5 , } is strictly increasing and { u 2 , u 4 , u 6 , } is strictly decreasing. { u1 , u 3 , u 5 , } is strictly increasing and bounded above by { u 2 , u 4 , u 6 , } is strictly decreasing and bounded below by (iv) Since and therefore both the sequences converge. Let lim u 2 n −1 = a and lim u 2 n = b . n →∞ n →∞ From (b)(iii), ∴ a= 3a + 4 3b + 4 and b = 2a + 3 2b + 3 a 2 = b 2 = 2 ⇒ a = b = 2 [since u n > 0] 2 2, 2001 Paper 1 Section B 11. (a) θ θ θ 1 + 2 cos 2 − 1 + i 2 sin cos 2 2 2 θ θ θ 1 − 1 − 2 sin 2 − i 2 sin cos 2 2 2 θ θ θ 2 cos cos + i sin 2 2 2 = θ θ θ − 2i sin cos + i sin 2 2 2 1 + cos θ + i sin θ = 1 − cos θ − i sin θ = i cot θ 2 n z −1 (b) Clearly, z ≠ −1 . (*) can be rewritten as = −1 . z +1 Using De Moivre’s theorem, (2k + 1)π (2k + 1)π z...
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This note was uploaded on 06/07/2012 for the course MATHS 2002 taught by Professor Lo during the Fall '02 term at Wisc Oshkosh.

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