$\underset{x\to 0^{+}}{\lim }f\left(x\right)=f\left(0\right)={Lim}_{x\to 0^{-}}\left(x\right)$

$=\underset{x\to 0^{+}}{\lim }\frac{{\cos }^{-1}\left(1-x^2\right)·{\sin }^{-1}\left(1-x\right)}{x\left(1-x\right)\left(1+x\right)}$

$=\underset{x\to 0^{+}}{\lim }\frac{{\cos }^{-1}\left(1-x^2\right)}{x·1·1}·\frac{\pi }{2}$

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

$=\frac{\pi }{2}\underset{\theta \to 0^{+}}{\lim }\frac{\theta }{\sqrt{1-\cos \theta }}$

$=\frac{\pi }{2}\underset{\theta \to 0^{+}}{\lim }\frac{\theta }{\sqrt{2}\sin \frac{\theta }{2}}=\frac{\pi }{\sqrt{2}}$

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

$\underset{x\to 0^{-}}{\lim }\frac{\frac{\pi }{2}\left(-{\sin }^{-1}x\right)}{\left(1+x\right)\left(2+x\right)\left(-x\right)}$

$\underset{x\to 0^{-}}{\lim }\frac{\frac{\pi }{2}}{1·2}·\frac{{\sin }^{-1}x}{x}=\frac{\pi }{4}$

$\Rightarrow \mathrm{RHL}\ne \mathrm{LHL}$

Function can't be continuous

$\Rightarrow$ No value of $\alpha$ exist