A) ?1
B) 1
C) 0
D) 2
Correct Answer: C
Solution :
\[f(x)\] is continuous at \[x=\frac{\pi }{2}\], then \[\underset{x\to \pi /2}{\mathop{\lim }}\,f(x)=f(0)\] or \[\lambda =\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{1-\sin x}{\pi -2x}\], \[\left( \frac{0}{0}\text{ form} \right)\] Applying L-Hospital?s rule, \[\lambda =\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{-\cos x}{-2}\] Þ \[\lambda =\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{\cos x}{2}=0.\]You need to login to perform this action.
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