Question

If $f\left(x\right)=x^2+\alpha$, for $x\ge 0$
$=2\sqrt{x^2+1}+\beta$, for $x<0$ and $f\left(1/2\right)=2$ is continuous at $x=0$, find $\alpha$ and $\beta$.

Answer 1

$\begin{array}{ccc}f\left(x\right)=x^2+a& & x\ge 0\\ =2\sqrt{x^2+1}+\beta & & x<0\end{array}$

Since $f\left(x\right)$ is continuous LHL = RHL at $x=0$

$\begin{array}{c}{0}^2+\alpha =2\sqrt{1}+\beta \\ or,\alpha =2+\beta \\ andf\left(1/2\right)=2=\frac{1}{4}+\alpha \\ \alpha =\frac{7}{4}\\ \beta =\frac{-1}{4}\end{array}$