If $\alpha$ and $\beta$are such that the function f(x) defined by $f\left(x\right)=\left\{\begin{array}{l}\alpha x^2-\beta ,for\left|x\right|<1\\ \frac{-1}{\left|x\right|},for\left|x\right|\ge 1\end{array}\right.$ is differentiable everywhere then the ordered pair $(\alpha , \beta )$ =

$\mathrm{f}\text{'}(1-)=\frac{d}{dx}\left(\alpha x^2-\beta \right)\text{ at }1=2\alpha x\text{ at }1=2\alpha .f\text{'}(1+)=\frac{d}{dx}\left(-\frac{1}{x}\right)\text{ at }1=\left(\frac{1}{x^2}\right)\text{ at }1=1$

$f\text{ is differentiable at }1\Rightarrow f^{\mathrm{\prime }}\left(1^{-}\right)=f^{\mathrm{\prime }}\left(1^{+}\right)=2\alpha =1\Rightarrow \alpha =1/2$

$f\text{ is differentiable at }1\Rightarrow f\text{ is continous at }1$

$\therefore \underset{x\to 1^{-}}{lim} f\left(x\right)=\underset{x\to 1^{+}}{lim} f\left(x\right)\Rightarrow \underset{x\to 1^{-}}{lim} \alpha x^2-\beta =\underset{x\to 1^{+}}{lim} -\frac{1}{\left|x\right|}\Rightarrow \alpha -\beta =-1\Rightarrow \frac{1}{2}-\beta =-1\Rightarrow \beta =\frac{3}{2}$

$\therefore (\alpha ,\beta )=(1/2,3/2)$