Question

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

• A. $3$
• B. $\frac{8}{25}$
• C. $\frac{25}{8}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C $\frac{25}{8}$

$f\left(x\right)iscntinuousatx=o$

$\therefore Limf\left(x\right)existsandisequal$

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

$\therefore f\left(0\right)forx\ge 0=\alpha$

$hence,2+\beta =\alpha$

$Also,f\left(1/2\right)=2$

$\therefore 2=\alpha +1/4$

$\therefore \alpha =7/4$

$\therefore \beta =-1/4$

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