HKALE Pure Maths 2001 Paper01

r cos r sin r cos sin dk

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Unformatted text preview: 1 where k = 0, 1, 2, = cos + i sin z +1 n n (2k + 1)π For k = 0, 1, 2, , n−1 , let θ k = . n z − 1 = z cos θ k + iz sin θ k + cos θ k + i sin θ k z= 1 + cos θ k + i sin θ k 1 − cos θ k − i sin θ k θk 2 (2k + 1)π = i cot 2n (2k + 1)π Putting α k = cot , the result follows. 2n = i cot , n−1 . 2001 Paper 1 Section B (c) Expanding the LHS, (*) becomes n n [ z n + C1n z n −1 + C 2 z n − 2 + + 1] + [ z n − C1n z n −1 + C 2 z n − 2 − n n z n + C 2 z n− 2 + C 4 z n −4 + + (−1) n ] = 0 =0 Using the relations between the roots and coefficients, we have and ∑ ∑ n −1 k =0 n −1 iα k = 0 j ,k =0 j≠k n (iα j )(iα k ) = C 2 ∑ Hence (iα k ) 2 = k =0 n −1 ∑ 2 n −1 iα k − 2 k =0 ∑ n −1 (iα j )(iα k ) j ,k =0 j≠k n(n − 1) = 0 − 2 2 ∑ (d) ∑ n −1 k =0 2 n −1 k =0 α k 2 = n(n − 1) ∑ iα − r (cos β + i sin β ) = ∑ [(r cos β ) + (α − r sin β ) ] = r ∑ (cos β + sin β ) + ∑ α dk = n −1 k =0 n −1 k =0 n −1 2 k =0 2 2 k 2 2 k 2 = nr + n(n − 1) which is independent of β . 2 n −1 k =0 k 2 − 2r sin β ∑ n −1 k =0 αk 2001 Paper 1 Section B 12. (a) Let x, y and z be real numbers such that xa + yb + zc = 0 . 8 3 x 0 7 The equation is equivalent to 2 − 7 13 y = 0 . 17 − 6 3 z 0 7 8 3 Since 2 − 7 13 = −147 + 1768 − 36 + 357 + 546 − 48 = 2440 ≠ 0 , 17 − 6 3 ∴ The system has the trivial solution only, i.e. x = y = z = 0. Hence a , b and c are linearly independent. (b) Let Then v = λ a+ µ b =α c+ β d . (λ a + µ b) ⋅ (c × d) = (α c + β d) ⋅ (c × d) = 0 λ a ⋅ (c × d ) + µ b ⋅ (c × d ) = 0 λ : µ = −b ⋅ c × d : a ⋅ c × d i k c × d = 17 −6 3 2r Since j − 4r − 5r = r (42i + 91j − 56k ) ∴ = 7 r (6i + 13j − 8k ) λ : µ = −7r (2 × 6 − 7 × 13 − 13 × 8) : 7r (7 × 6 + 8 × 13 − 3 × 8) = 183r : 122r = 3: 2 2001 Paper 1 Section B (c) (i) If A , B , C and D are coplanar, then (d − a) ⋅ (b − a) × (c − a) = 0 2 r − 7 − 4 r − 8 − 5r − 3 2−7 −7 −8 13 − 3 = 0 17...
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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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