If the function $f(x)= \begin{cases}\frac{\sqrt{2+\cos x}-1}{(\pi-x)^{2}}, & x \neq \pi \\ k \quad, & x=\pi\end{cases}$ is continuous at $x=\pi$, then $\mathrm{k}$ equals:-
$\frac{1}{4}$
$\frac{1}{2}$
2
0
Correct Option: 1
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