A) 1
B) \[-1\]
C) \[1/2\]
D) 0
Correct Answer: D
Solution :
\[f(x)=\left\{ \begin{matrix} \frac{1-\sin x}{\pi -2x}, & x\ne \frac{\pi }{2} \\ \lambda , & x=\frac{\pi }{2} \\ \end{matrix} \right.\] \[\because \]\[f(x)\]is continuous at\[x=\frac{\pi }{2},\] \[\therefore \] \[\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,f(x)=f\left( \frac{\pi }{2} \right)\] \[\Rightarrow \] \[\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{1-\sin x}{\pi -2x}=\lambda \] \[\Rightarrow \] \[\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{-\cos x}{-2}=\lambda \] [using L-Hospital rule] \[\frac{\cos \frac{\pi }{2}}{2}=\lambda \] \[\Rightarrow \] \[\lambda =0\]You need to login to perform this action.
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