If $\mathrm{\alpha }$ and $\mathrm{\beta }$ are such that the function 

$\text{⨍ (x)=}\left\{\begin{array}{ll}{\mathrm{ax}}^2-\mathrm{\beta },& for \left|\mathrm{x}\right|<1\\ \frac{-1}{\left|\mathrm{x}\right|}\text{,}& for\left| \mathrm{x}\right|\text{≤1}\end{array}\right.$  is differentiable
everywhere then the ordered pair $\left(\mathrm{\alpha },\mathrm{\beta }\right)$

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{1}{\mathrm{x}} ; \mathrm{x}\le -1\\ {\mathrm{\alpha x}}^2-\mathrm{\beta } ;-1<\mathrm{x}<1\\ \frac{-1}{\mathrm{x}} ; \mathrm{x}>1\end{array}\right. \\ \mathrm{is} \mathrm{differentiable} \\ \Rightarrow \mathrm{f} \mathrm{is} \mathrm{continuous} \\ \mathrm{f}\left(-1^{-}\right)=\mathrm{f}\left(-1^{+}\right)\Rightarrow -1=\mathrm{\alpha }-\mathrm{\beta } \end{gathered}$$

$$\begin{gathered} {\mathrm{f}}^1\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{-1}{{\mathrm{x}}^2} ; \mathrm{x}<-1\\ 2\mathrm{\alpha x} ; -1<\mathrm{x}<1\\ \frac{1}{{\mathrm{x}}^2} ; \mathrm{x}>1\end{array}\right. \\ \therefore {\mathrm{f}}^1\left(1^{-}\right)={\mathrm{f}}^1\left(1^{+}\right)\Rightarrow 2\mathrm{\alpha }=1\Rightarrow \mathrm{\alpha }=\frac{1}{2} \\ \Rightarrow \mathrm{\beta }=\mathrm{\alpha }+1=\frac{3}{2}, \\ \Rightarrow \left(\mathrm{\alpha }, \mathrm{\beta }\right)=\left(\frac{1}{2}, \frac{3}{2}\right) \end{gathered}$$