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If $2\sin^{-1}x=\sin^{-1}(2x\sqrt{1-x^2})$, then x belongs to: |
[–1, 1] $[-\frac{1}{\sqrt{2}},1]$ $[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$ None of these |
$[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$ |
$-\frac{π}{2}≤2\sin^{-1}x≤\frac{π}{2}⇒\frac{-π}{4}≤\sin^{-1}x≤\frac{π}{4}⇒x∈[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$ |
