MCQ
If $f(x)=\left\{\begin{array}{cl}\frac{3 \sin \pi x}{5 x}, & x \neq 0 \\ 2 k, & x=0\end{array}\right.$ is continuous at $x=0$, then the value of k is equal to
  • $\frac{3 \pi}{10}$
  • B
    $\frac{3 \pi}{5}$
  • C
    $\frac{\pi}{10}$
  • D
    $\frac{3 \pi}{2}$

Answer

Correct option: A.
$\frac{3 \pi}{10}$
(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow 2 k =\lim _{x \rightarrow 0} \frac{3 \sin \pi x}{5 x}$
$=\lim _{x \rightarrow 0} \frac{3 \sin \pi x}{5(\pi x)} \times \pi=\frac{3 \pi}{5}$
$\therefore \quad k =\frac{3 \pi}{10}$

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