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Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function given by
$f(x)=\left\{\begin{array}{ll}
\frac{1-\cos 2 x}{x^2}, & x < 0 \\
\alpha, & x=0, \\
\frac{\beta \sqrt{1-\cos x}}{x}, & x>0
\end{array}\right.$
where $\alpha, \beta \in \mathbf{R}$. If $f$ is continuous at $x=0$, then $\alpha^2+\beta^2$ is equal to :
Options:
$f(x)=\left\{\begin{array}{ll}
\frac{1-\cos 2 x}{x^2}, & x < 0 \\
\alpha, & x=0, \\
\frac{\beta \sqrt{1-\cos x}}{x}, & x>0
\end{array}\right.$
where $\alpha, \beta \in \mathbf{R}$. If $f$ is continuous at $x=0$, then $\alpha^2+\beta^2$ is equal to :
Solution:
1958 Upvotes
Verified Answer
The correct answer is:
12
$\begin{aligned} & f\left(0^{-}\right)=\lim _{x \rightarrow 0^{-}} \frac{2 \sin ^2 x}{x^2}=2=\alpha \\ & f\left(0^{+}\right)=\lim _{x \rightarrow 0^{+}} \beta \times \sqrt{2} \frac{\sin \frac{x}{2}}{2 \frac{x}{2}}=\frac{\beta}{\sqrt{2}}=2 \\ & \Rightarrow \beta=2 \sqrt{2} \\ & \alpha^2+\beta^2=4+8=12\end{aligned}$
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