Let α ∈ R be such that the function f(x) = \ cases ^ - 1 (1 - \ x\...

$\lim_{x \to {0}^{+}}f\left(x\right) = f\left(0\right) = \lim_{x \to {0}^{-}}\left(x\right)$

$= \lim_{x \to {0}^{+}}\dfrac{\cos^{- 1}\left(1 - {x}^{2}\right) \cdot \sin^{- 1}\left(1 - x\right)}{x\left(1 - x\right)\left(1 + x\right)}$

$= \lim_{x \to {0}^{+}}\dfrac{\cos^{- 1}\left(1 - {x}^{2}\right)}{x \cdot 1 \cdot 1} \cdot \dfrac{\pi}{2}$

Let $1 - {x}^{2} = cos\theta$

$= \dfrac{\pi}{2}\lim_{\theta \to {0}^{+}}\dfrac{\theta}{\sqrt{1 - cos\theta}}$

$= \dfrac{\pi}{2}\lim_{\theta \to {0}^{+}}\dfrac{\theta}{\sqrt{2}sin\dfrac{\theta}{2}} = \dfrac{\pi}{\sqrt{2}}$

Now, $\lim_{x \to {0}^{-}}\dfrac{\cos^{- 1}\left(1 - {\left(1 + x\right)}^{2}\right)\sin^{- 1}\left(- x\right)}{\left(1 + x\right) - {\left(1 + x\right)}^{3}}$

$\lim_{x \to {0}^{-}}\dfrac{\dfrac{\pi}{2}\left(- \sin^{- 1}x\right)}{\left(1 + x\right)\left(2 + x\right)\left(- x\right)}$

$\lim_{x \to {0}^{-}}\dfrac{\dfrac{\pi}{2}}{1 \cdot 2} \cdot \dfrac{\sin^{- 1}x}{x} = \dfrac{\pi}{4}$

$⇒ RHL \ne LHL$

Function can't be continuous

$⇒$ No value of $\alpha$ exist