Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function given by

$$f(x)= \begin{cases}\frac{1-\cos 2 x}{x^2}, & x < 0 \\ \alpha, & x=0, \\ \frac{\beta \sqrt{1-\cos x}}{x}, & x>0\end{cases}$$

where $\alpha, \beta \in \mathbf{R}$. If $f$ is continuous at $x=0$, then $\alpha^2+\beta^2$ is equal to :

<p>$$f(x)=\left\{\begin{array}{cl} \frac{1-\cos 2 x}{x^2}, & x< \\ \alpha, & x=0 \\ \frac{\beta \sqrt{1-\cos x}}{x}, & x>0 \end{array}\right.$$</p> <p>$f(x)$ is continuous at $x=0$</p> <p>$$\Rightarrow f(0)=\lim _\limits{x \rightarrow 0^{-}} f(x)=\lim _\limits{x \rightarrow 0^{+}} f(x)$$</p> <p>$$\begin{aligned} &\begin{aligned} & \lim _{x \rightarrow 0^{-}} f(x)=\alpha \\ & \lim _{x \rightarrow 0^{-}}\left(\frac{1-\cos 2 x}{x^2}\right)=\alpha \\ & \Rightarrow \lim _{x \rightarrow 0^{-}} \frac{2 \sin ^2 h}{x^2}=\alpha \\ & \Rightarrow \lim _{h \rightarrow 0} \frac{2 \sin ^2}{h^2} h=\alpha \\ & \Rightarrow \alpha=2 \\ & \end{aligned}\\ &\begin{aligned} & \text { Also, } \lim _{x \rightarrow 0^{+}} f(x)=f(0) \\ & \Rightarrow \quad \lim _{x \rightarrow 0^{+}} \frac{\beta \sqrt{1-\cos x}}{x}=2 \end{aligned} \end{aligned}$$</p> <p>$$\Rightarrow \lim _\limits{h \rightarrow 0} \frac{\beta \sqrt{\frac{1-\cos h}{h^2}} h^2}{h}=2$$</p> <p>$$\begin{aligned} & \Rightarrow \quad \frac{\beta}{\sqrt{2}}=2 \\ & \Rightarrow \quad \beta=2 \sqrt{2} \\ & \Rightarrow \quad \alpha^2+\beta^2=4+8 \\ & \quad=12 \end{aligned}$$</p> <p>$\therefore$ Option (1) is correct</p>