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

α + 1

$\lim\limits_{x \rightarrow 0^{+}} \alpha+\frac{\sin [x]}{x}=\alpha$

f(0) = 2

$\lim\limits_{x \rightarrow 0^{-}} \beta+\frac{\sin [x-x]}{x^3}=\beta-1$

$\Rightarrow \beta-1=\alpha \Rightarrow \beta=\alpha+1$

Hence (1) is the correct answer.