Question

if $f(x)=\begin{cases}\frac{3 \sin \pi x}{5 x} & x \neq 0 \\ 2 K & x =0\end{cases}$ 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, $f(x) = \begin{cases} \frac{3 \sin \pi x}{5 x} & x \neq 0\\ 2k & x =0 \end{cases}$
Now, $\displaystyle \lim_{x \to\,0} f (x) = \displaystyle \lim_{x \to\, 0} \left(\frac{3\,\sin\,\pi x}{5\,x}\right)$
$= \frac{3}{5}\displaystyle \lim_{x \to\,0} \left(\sin \frac{\pi x}{\pi x}\right)\times\pi$
$=\frac{3}{5}\times1\times\pi$
$=\frac{3}{5}\,\pi$
Also, $f \left(0\right)=2\,k$
Since, $f\left(x\right)$ is continuous at $x =0.$
$\therefore f\left(O\right) = \displaystyle \lim_{x \to\,0} f (x)$
$\Rightarrow 2\,k=\frac{3}{5}\,\pi$
$\Rightarrow k=\frac{3\pi}{10}$