A) 0
B) \[\pi \]
C) \[\pi \]
D) \[^{\text{235}}\text{U}\]
Correct Answer: D
Solution :
It is given that f (x) is continuous at \[x=\frac{\pi }{2}.\] \[\therefore \]\[k=\underset{x\to \pi /2}{\mathop{\lim }}\,f(x)\] \[\Rightarrow \]\[k=\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{\sin (\cos x)-\cos x}{{{(\pi -2x)}^{3}}}\] \[k=\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{\sin (\cos x)-\cos x}{{{\cos }^{3}}x}\times \frac{{{\sin }^{3}}\left( \frac{\pi }{2}-x \right)}{8\left( \frac{\pi }{2}-x \right)}\] \[\therefore \]\[k=-\frac{1}{6}\times \frac{1}{8}=-\frac{1}{48}\left[ \because \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x-x}{{{x}^{3}}}=-\frac{1}{6} \right]\]
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