Let $$\alpha \in R$$ be such that the function $$f(x) = \begin{cases} \frac{\cos^{-1}(1-\{x\}^2)\sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, & x \neq 0 \\ \alpha, & x = 0 \end{cases}$$ is continuous at $$x = 0$$, where $$\{x\} = x - [x]$$, $$[x]$$ is the greatest integer less than or equal to $$x$$. Then:

We need to check continuity at $$x = 0$$ for $$f(x) = \frac{\cos^{-1}(1 - \{x\}^2)\sin^{-1}(1 - \{x\})}{\{x\} - \{x\}^3}$$ where $$\{x\} = x - [x]$$.

For the right-hand limit ($$x \to 0^+$$): $$\{x\} = x$$, so $$f(x) = \frac{\cos^{-1}(1 - x^2)\sin^{-1}(1 - x)}{x - x^3} = \frac{\cos^{-1}(1 - x^2)\sin^{-1}(1 - x)}{x(1 - x^2)}$$.

As $$x \to 0^+$$: Using the approximation $$\cos^{-1}(1 - t) \approx \sqrt{2t}$$ for small $$t$$, we get $$\cos^{-1}(1 - x^2) \approx x\sqrt{2}$$. Also, $$\sin^{-1}(1 - x) \to \sin^{-1}(1) = \frac{\pi}{2}$$ and $$1 - x^2 \to 1$$.

So the right-hand limit $$= \frac{x\sqrt{2} \cdot \frac{\pi}{2}}{x \cdot 1} = \frac{\pi}{\sqrt{2}}$$.

For the left-hand limit ($$x \to 0^-$$): Let $$x = -\varepsilon$$ where $$\varepsilon \to 0^+$$. Then $$[x] = -1$$, so $$\{x\} = x - (-1) = 1 - \varepsilon$$.

Now $$1 - \{x\}^2 = 1 - (1-\varepsilon)^2 = 2\varepsilon - \varepsilon^2 \to 0$$, so $$\cos^{-1}(1 - \{x\}^2) \to \cos^{-1}(0) = \frac{\pi}{2}$$.

Also $$1 - \{x\} = \varepsilon$$, so $$\sin^{-1}(1 - \{x\}) = \sin^{-1}(\varepsilon) \approx \varepsilon$$.

The denominator: $$\{x\} - \{x\}^3 = (1-\varepsilon)(1-(1-\varepsilon)^2) = (1-\varepsilon)\varepsilon(2-\varepsilon) \approx 2\varepsilon$$.

So the left-hand limit $$= \frac{\frac{\pi}{2} \cdot \varepsilon}{2\varepsilon} = \frac{\pi}{4}$$.

Since the right-hand limit $$\frac{\pi}{\sqrt{2}} \neq \frac{\pi}{4}$$ (the left-hand limit), the function cannot be made continuous at $$x = 0$$. Therefore, no such $$\alpha$$ exists.