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If the function $\; f(x) = \{ \begin{array} \large\frac{\sqrt{2+cos x}-1}{(\pi-x)^{2}} ,& x \neq \pi \\ k ,& x=\pi \end{array} \;$ is continuous at $\;x=\pi \;$ , then k equals :

$(a)\;0\qquad(b)\;\large\frac{1}{2}\qquad(c)\; \normalsize 2\qquad(d)\;\large\frac{1}{4}$

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