Question

If $f(x) = \begin{cases} x^2 + \alpha & \text{for } x \ge 0 \\[2ex] 2\sqrt{x^2 + 1} + \beta & \text{for} x < 0 \end{cases}$ 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

we have
$f(x) =
\begin{cases}
x^2 + \alpha & \text{for } x \ge 0 \\[2ex]
2\sqrt{x^2 + 1} + \beta & \text{for} x < 0
\end{cases}$

is continuous at \(x = 0\)
\(\therefore \displaystyle\lim_{x \to 0^-} f(x) = \displaystyle\lim_{x \to 0^+} f(x) = f(0)\)
\(\Rightarrow \displaystyle\lim_{x \to 0^-} x^2 + \alpha = \displaystyle\lim_{x\to0^+} 2\sqrt{x^2 + 1} + \beta = \alpha\)
\([ \because f(0) = \alpha]\)
\(\Rightarrow \alpha = 2 + \beta\)
\(\Rightarrow \alpha - \beta = 2\,\dots (i)\) 
Given \(f(\frac{1}{2}) = 2\)
\(f(\frac{1}{2}) = (\frac{1}{2})^2 + \alpha \,\,\left[\because \frac{1}{2} > 0\right]\)
\(\Rightarrow 2 = \frac{1}{4} + \alpha\)
\(\Rightarrow \alpha = \frac{7}{4}\) 
On putting the value of a in E (i), we get 
\(\beta = \frac{7}{4} - 2 = -\frac{1}{4}\) 
Hence \(\alpha^2 + \beta^2 = (\frac{7}{4})^2 + (\frac{-1}{4})^2\)
\(= \frac{49+1}{16} = \frac{50}{16} = \frac{25}{8}\)

Therefore, the correct option is (C): \(\frac{25}{8}\)