If $2\sin^{-1}x=\sin^{-1}(2x\sqrt{1-x^2})$, then x belongs to:
Answer & explanation
Correct answer: option 3
$-\frac{π}{2}≤2\sin^{-1}x≤\frac{π}{2}⇒\frac{-π}{4}≤\sin^{-1}x≤\frac{π}{4}⇒x∈[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$
If $2\sin^{-1}x=\sin^{-1}(2x\sqrt{1-x^2})$, then x belongs to:
Correct answer: option 3
$-\frac{π}{2}≤2\sin^{-1}x≤\frac{π}{2}⇒\frac{-π}{4}≤\sin^{-1}x≤\frac{π}{4}⇒x∈[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$