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If $\sin^{- 1} (2 x \sqrt{1 - x^{2}}) - 2 \sin^{- 1} x = 0,$ then $x$ belongs to the interval

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[-1, 1]

[-1/2, 1/2]

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[-1/2, 1]

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detailed solution

Correct option is C

Let sin−1⁡x=θ. Then, x=sin⁡θ and 1−x2=cos⁡θNow, sin−1⁡2x1−x2=sin−1⁡(sin⁡2θ)=2θ, if −π2≤20≤π2=2sin−1⁡x, if −π4≤θ≤π4 i.e. if −12≤x≤12∴ sin−1⁡2x1−x2−2sin−1⁡x=0, if -12≤x≤12

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If sin−1⁡2x1−x2−2sin−1⁡x=0, then x belongs to the interval

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If $\sin^{- 1} (2 x \sqrt{1 - x^{2}}) - 2 \sin^{- 1} x = 0,$ then $x$ belongs to the interval