Question

if $f(x)=\{\begin{array}{ll}\frac{3\sin \pi x}{5x}& x\ne 0\\ 2K& x=0\end{array}$ is continuous at $x=0$, then the value of $K$ is

• A. $\frac{3\pi }{10}$
• B. $\frac{3\pi }{5}$
• C. $\frac{\pi }{10}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

Given, $f(x)=\{\begin{array}{ll}\frac{3\sin \pi x}{5x}& x\ne 0\\ 2k& x=0\end{array}$
Now, $\underset{x\to 0}{\lim }f(x)=\underset{x\to 0}{\lim }\left(\frac{3\sin \pi x}{5x}\right)$
$=\frac{3}{5}\underset{x\to 0}{\lim }\left(\sin \frac{\pi x}{\pi x}\right)\times \pi$
$=\frac{3}{5}\times 1\times \pi$
$=\frac{3}{5}\pi$
Also, $f\left(0\right)=2k$
Since, $f\left(x\right)$ is continuous at $x=0.$
$\therefore f\left(O\right)=\underset{x\to 0}{\lim }f(x)$
$\Rightarrow 2k=\frac{3}{5}\pi$
$\Rightarrow k=\frac{3\pi }{10}$