If f(x) = begin{cases} x^2 + alpha, & x ge 0 | vedclass

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(C) For $f(x)$ to be continuous at $x = 0$,the left-hand limit must equal the right-hand limit and the function value at $x = 0$.$\lim_{x 	o 0^+} (x^

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$\frac{25}{8}$

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(C) For $f(x)$ to be continuous at $x = 0$,the left-hand limit must equal the right-hand limit and the function value at $x = 0$.$\lim_{x 	o 0^+} (x^2 + \alpha) = \lim_{x 	o 0^-} (2\sqrt{x^2 + 1} + \beta)$$0^2 + \alpha = 2\sqrt{0^2 + 1} + \beta$$\alpha = 2 + \beta \implies \alpha - \beta = 2 . . . (1)$Given $f(\frac{1}{2}) = 2$. Since $\frac{1}{2} \ge 0$,we use the first part of the function:$f(\frac{1}{2}) = (\frac{1}{2})^2 + \alpha = 2$$\frac{1}{4} + \alpha = 2$$\alpha = 2 - \frac{1}{4} = \frac{7}{4}$Substitute $\alpha = \frac{7}{4}$ into equation $(1)$:$\frac{7}{4} - \beta = 2$$\beta = \frac{7}{4} - 2 = -\frac{1}{4}$Now,calculate $\alpha^2 + \beta^2$:$\alpha^2 + \beta^2 = (\frac{7}{4})^2 + (-\frac{1}{4})^2$$\alpha^2 + \beta^2 = \frac{49}{16} + \frac{1}{16} = \frac{50}{16} = \frac{25}{8}$

If $f(x) = \begin{cases} x^2 + \alpha, & x \ge 0 \\ 2\sqrt{x^2 + 1} + \beta, & x < 0 \end{cases}$ is continuous at $x = 0$ and $f(\frac{1}{2}) = 2$,then $\alpha^2 + \beta^2$ is

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