Let $\mathrm{\alpha }\in \mathrm{R}$ be such that the function 

$\text{⨍ (x)=}\left\{\begin{array}{ll}\frac{{\cos }^{-1}\left(1-{\left\{\mathrm{x}\right\}}^2\right){\sin }^{-1}\left(\left\{\mathrm{x}\right\}\right)}{\left\{\mathrm{x}\right\}\text{-}{\left\{\mathrm{x}\right\}}^3}\text{,}& \mathrm{x}\ne 0\\ \mathrm{\alpha },& \mathrm{x}=0\end{array}\right.$

is Continuous at $\mathrm{x}=0,$ where $\left\{\mathrm{x}\right\}=\mathrm{x}-\left[\mathrm{x}\right],\left[\mathrm{x}\right]$
is the greatest integer less than or equal to $\mathrm{x}.$
Then:

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{{\cos }^{-1}\left(1-{\left\{\mathrm{x}\right\}}^2\right){\sin }^{-1}\left(1-{\left\{\mathrm{x}\right\}}^2\right)}{\left\{\mathrm{x}\right\}-{\left\{\mathrm{x}\right\}}^3} ; \mathrm{x}\ne 0\\ \mathrm{\alpha } ; \mathrm{x}=0\end{array}\right. \\ \mathrm{is} \mathrm{continuous} \mathrm{at} \mathrm{x}=0 \\ \mathrm{f}\left(\mathrm{x}\right)=\frac{{\cos }^{-1}\left(1-{(\mathrm{x}-\left[\mathrm{x}\right])}^{2 }\right){\sin }^{-1}\left(1-\left(\mathrm{x}-\left[\mathrm{x}\right]\right)\right)}{\left(\mathrm{x}-\left[\mathrm{x}\right]\right)-{\left(\mathrm{x}-\left[\mathrm{x}\right]\right)}^3} \end{gathered}$$

$$\begin{gathered} \underset{\mathrm{x}\to 0^{-}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 0^{-}}{\lim }\frac{{\cos }^{-1}\left(1-{(\mathrm{x}+1)}^2\right){\sin }^{-1}\left(1-\left(\mathrm{x}+1\right)\right)}{\left(\mathrm{x}+1\right)-{\left(\mathrm{x}+1\right)}^3}, \mathrm{x}+1=\mathrm{t} \\ =\underset{\mathrm{t}\to 1}{\lim }\frac{{\cos }^{-1}\left(1-{\mathrm{t}}^2\right){\sin }^{-1}\left(1-\mathrm{t}\right)}{\mathrm{t}\left(1+\mathrm{t}\right)\left(1-\mathrm{t}\right)}=\frac{{\cos }^{-1}0}{2}=\frac{\mathrm{\pi }}{4} \\ \underset{\mathrm{x}\to 0^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 0^{+}}{\lim }\frac{{\cos }^{-1}\left(1-{\mathrm{x}}^2\right){\mathrm{son}}^{-1}\left(1-\mathrm{x}\right)}{\mathrm{x}\left(1-{\mathrm{x}}^2\right)}=\underset{\mathrm{x}\to 0}{\lim }\frac{{\cos }^{-1}\left(1-{\mathrm{x}}^2\right)}{\mathrm{x}}\left(\frac{\mathrm{\pi }}{2}\right) \end{gathered}$$$$\begin{gathered} \mathrm{Sub} \mathrm{x}=\sqrt{2} \sin \mathrm{\theta }\Rightarrow \underset{\mathrm{\theta }\to 0}{\lim }\frac{{\cos }^{-1}\left(1-2{\sin }^2\mathrm{\theta }\right)}{\sqrt{2}\sin \mathrm{\theta }}.\frac{\mathrm{\pi }}{2} \\ \underset{\mathrm{x}\to 0^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{\theta }\to 0}{\lim }\frac{{\cos }^{-1}\cos 2\mathrm{\theta }}{\sqrt{2} \sin \mathrm{\theta }}.\frac{\mathrm{\pi }}{2} \\ =\frac{\mathrm{\pi }}{2\sqrt{2}}\underset{\mathrm{\theta }\to 0}{\lim }\frac{2\mathrm{\theta }}{\sin \mathrm{\theta }}=\frac{\mathrm{\pi }}{\sqrt{2}} \\ \mathrm{LHL}\ne \mathrm{CHL} \\ \therefore \mathrm{f} \mathrm{is} \mathrm{discontinuous} \mathrm{at} \mathrm{x}=0 \\ \mathrm{no} \mathrm{such} \mathrm{x} \mathrm{exists} \end{gathered}$$