If ${sin}^{- 1} x + {sin}^{- 1} (1 - x) = {cos}^{- 1} x$ , then $x$ equals-

1, - 1

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1, 0

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0, \frac{1}{2}

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None of these

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Solution

The correct option is C $0, \frac{1}{2}$
Given, ${sin}^{- 1} x + {sin}^{- 1} (1 - x) = {cos}^{- 1} x \Rightarrow {sin}^{- 1} (x \sqrt{1 - {(1 - x)}^{2}} + (1 - x) \sqrt{1 - x^{2}}) = {cos}^{- 1} x \Rightarrow x \sqrt{{- x}^{2} + 2 x} + (1 - x) \sqrt{1 - x^{2}} = sin ({cos}^{- 1} x) \Rightarrow x \sqrt{2 x - x^{2}} + (1 - x) \sqrt{1 - x^{2}} = sin ({sin}^{- 1} \sqrt{1 - x^{2}}) = \sqrt{{1 - x}^{2}} \Rightarrow x \sqrt{2 x - x^{2}} = x \sqrt{1 - x^{2}}$
$\Rightarrow 2 x - x^{2} = 1 - x^{2}$
$\Rightarrow x = \frac{1}{2}, 0$

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