$\text{ If }f\left(x\right)=\left\{\begin{array}{l}{x}^2+\alpha \text{ for }x\ge 0\\ 2\sqrt{x^2+1}+\beta \text{ for }x<0\end{array}\text{ is continuous at }x=0\text{ and }f\left(\frac{1}{2}\right)=2\text{ then }{\alpha }^2+{\beta }^2=\right.$

$\text{ R.H.L }=\alpha ,\text{ L. }H.\mathrm{L}=2+\beta$

$f\left(0\right)=\alpha \therefore \alpha =2+\beta$

$\begin{array}{l}f\left(\frac{1}{2}\right)=2\\ \Rightarrow \frac{1}{4}+\alpha =2\Rightarrow \alpha =2-\frac{1}{4}=\frac{7}{4}\end{array}$

$\Rightarrow \beta =\alpha -2=\frac{7}{4}-2=\frac{-1}{4}$

$\therefore {\alpha }^2+{\beta }^2=\frac{50}{16}=\frac{25}{8}=3.125$