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If sin-1 (1 - x) - 2 sin-1 x = \(\frac{π}{2}\) then x is(A) \(-\frac{1}{2}\)(B) 1(C) 0(D) \(\frac{1}{2}\) |
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Answer» (C) sin-1 (1-x) -2 sin-1 x = \(\frac{π}{2}\) Let x = sin θ ∴ sin-1 (1 - sin θ) - 2 sin-1 (sin θ) = \(\frac{π}{2}\) ∴ sin-1 (1 - sin θ) - 2θ = \(\frac{π}{2}\) ∴ sin-1 (1 - sin θ) = \(\frac{π}{2}\) + 20 ∴ 1 - sin θ = sin (\(\frac{π}{2}\) + 20) ∴ 1 - sin θ = cos 2θ ...[∵ sin (\(\frac{π}{2}\) + θ) = cons θ] ∴ 1 - sin θ = 1 -2sin2 θ ∴ 2sin2θ - sin θ = 0 ∴ sin θ (2 sin θ - 1) = 0 ∴ sin θ = 0 or sin θ = \(\frac{1}{2}\) ∴ x = 0 or x = \(\frac{1}{2}\) For x = \(\frac{1}{2}\), sin-1 (1 - x) -2 sin-1 x = sin-1 (1 - \(\frac{1}{2}\)) - 2sin-1 (\(\frac{1}{2}\)) sin-1(\(\frac{1}{2}\)) - 2 sin-1(\(\frac{1}{2}\)) = - sin-1(\(\frac{1}{2}\)) = - \(\frac{π}{6} \) ≠ \(\frac{π}{2}\) ∴ x ≠ \(\frac{1}{2}\) ∴ x = 0 |
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