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

Answer: (C)

$\frac{25}{8}$

Answer 2

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): $\frac{25}{8}$