HKALE Pure Maths 2001 Paper01

# Such that x1 1 x 2 and the equality holds iff x1 x 2

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Unformatted text preview: + s = −1 ps − pq = −2 Putting r = − p , ⇒ q + s = p 2 − 1 2 q − s = p ⇒ 2 2 2 q = p − 1 + p 2 s = p 2 − 1 − 2 p 2001 Paper 1 Section B Since qs = −1 , therefore 2 2 2 2 p − 1 + p − 1 − = −4 p p (p 3 )( ) − p + 2 p 3 − p − 2 = −4 p 2 6 4 2 p − 2p +5p −4 = 0 (p ) − 1 ( p 4 − p 2 + 4) = 0 p = ±1 Hence or 2 p = 1, q = 1 , r = − 1 , s = − 1 p = −1, q = −1 , r = 1 , s = 1 (iii) From (b)(ii), g(x) = ( x 2 − x − 1)( x 2 + x + 1) . − 1 ± 3i 1± 5 or . 2 2 Since f(x) = g(x + 2) , ∴ g(x) = 0 if x = ∴ f(x) = 0 if x = −3± 5 − 5 ± 3i or . 2 2 2001 Paper 1 Section B 14. (a) If a ≤ 1 ≤ b , then a + b − ab − 1 = (a − 1)(1 − b) .........(1) ≥0 ∴ a + b ≥ ab + 1 From (1), the equality holds iff a − 1 = 0 or 1 − b = 0 iff a = 1 or b = 1 . (b) If x1 , x 2 are positive real numbers such that x1 x 2 = 1 , without loss of generality, we may assume that x1 ≤ 1 ≤ x 2 . From (a), x1 + x 2 ≥ x1 x 2 + 1 = 2 and ∴ iff x1 = 1 or x 2 = 1 iff the equality holds x1 = x 2 = 1 (since x1 x 2 = 1 ) The statement holds for n = 2. Assume the statement holds for n = k where k ∈ N and k ≥ 2 . Let x1...
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