A) \[\frac{3\pi }{10}\]
B) \[\frac{3\pi }{5}\]
C) \[\frac{\pi }{10}\]
D) \[\frac{3\pi }{2}\]
E) \[\frac{2\pi }{3}\]
Correct Answer: A
Solution :
Given, \[f(x)=\left\{ \begin{matrix} \frac{3\sin \pi x}{5x}, & x\ne 0 \\ 2k, & x=0 \\ \end{matrix} \right.\] \[\therefore \] \[LHL=\underset{x\to 0}{\mathop{\lim }}\,\frac{3\sin \pi x}{5x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{3}{5}.\frac{\sin \pi x}{\pi x}.\pi \] \[=\frac{3\pi }{5}\] Also, \[f(0)=2k\] Since, \[f(x)\]is continuous at\[x=0,\]then \[f(0)=LHL\] \[\Rightarrow \] \[2k=\frac{3\pi }{5}\Rightarrow k=\frac{3\pi }{10}\]
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