A) 0
B) \[\frac{1}{2}\]
C) 2
D) \[\frac{1}{4}\]
Correct Answer: D
Solution :
Since \[f(x)=\frac{\sqrt{2+\cos x}-1}{{{(\pi -x)}^{2}}}\]is Continuous at \[x=\pi \] \[\therefore \]L.H.L = R.H.L \[=f(\pi )\] Let \[(\pi -x)=\theta ,\theta \to 0\]when \[x\to \pi \] \[\therefore \]\[\underset{\theta \to 0}{\mathop{\lim }}\,\frac{\sqrt{2-\cos \theta }-1}{{{\theta }^{2}}}\] \[=\underset{\theta \to 0}{\mathop{\lim }}\,\frac{(2-\cos \theta )-1}{{{\theta }^{2}}}\times \frac{1}{\sqrt{2-\cos \theta }+1}\] \[=\underset{\theta \to 0}{\mathop{\lim }}\,\frac{1-\cos \theta }{{{\theta }^{2}}}.\frac{1}{2}\] \[(\because \cos 0=1)\] \[=\frac{1}{2}\underset{\theta \to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}\theta /2}{{{\theta }^{2}}}=\frac{2}{2}\underset{\theta \to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}\theta /2}{\frac{{{\theta }^{2}}}{4}.4}\] \[=\frac{1}{4}\] \[\left( \because \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1 \right)\]
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