Let $f(x)=\left\{\begin{array}{cl}\alpha+\frac{\sin [x]}{x} & , ~x>0 \\ 2 & , ~x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right] & , ~x<0\end{array}\right.$, where [ ] denotes the greatest integer function. If f(x) is continuous at x = 0, then $\beta$ is equal to

$\alpha+1$

We know that

$x-\frac{x^3}{6}<\sin x<x$ for all $x>0$

$\Rightarrow -\frac{x^3}{6}<\sin x-x<0$ for all $x>0$

$\Rightarrow -\frac{1}{6}<\frac{\sin x-x}{x^3}<0$ for all $x>0$

Also, $\frac{\sin x-x}{x^3}$ is an even function. Therefore, $-\frac{1}{6}<\frac{\sin x-x}{x^3}<0$ for all x.

Thus, $\left[\frac{\sin x-x}{x^3}\right]=-1$ for all x.

It is given that f(x) is continuous at x = 0.

∴  $\lim\limits_{x \rightarrow 0^{+}} f(x)=f(0)=\lim\limits_{x \rightarrow 0^{-}} f(x)$

$\Rightarrow \lim\limits_{x \rightarrow 0^{+}} \alpha+\frac{\sin [x]}{x}=2=\lim\limits_{x \rightarrow 0^{-}} \beta+\left[\frac{\sin x-x}{x^3}\right] $

$\Rightarrow \alpha+0=2=\beta-1 \Rightarrow \alpha=2, \beta=3 \Rightarrow \beta=\alpha+1$