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If $\begin{aligned} f(x) &=\frac{4 \sin \pi x}{5 x} \text { for } x \neq 0 \\ &=2 \mathrm{k} \quad \text { for } x=0 \end{aligned}$
is continuous at $x=0$, then the value of $k$ is
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is continuous at $x=0$, then the value of $k$ is
Solution:
2733 Upvotes
Verified Answer
The correct answer is:
$\frac{2 \pi}{5}$
(C)
Since $f(x)$ is continuous at $x=0$,
$\begin{array}{l}
\lim _{x \rightarrow 0} f(x)=f(0) \Rightarrow \lim _{x \rightarrow 0} \frac{4 \sin \pi x}{5 x}=2 k \\
\therefore \lim _{x \rightarrow 0}\left(\frac{4 \sin \pi x}{\pi x}\right) \cdot \frac{\pi}{5}=2 k \Rightarrow(1) \cdot \frac{4 \pi}{5}=2 k \Rightarrow k=\frac{2 \pi}{5}
\end{array}$
Since $f(x)$ is continuous at $x=0$,
$\begin{array}{l}
\lim _{x \rightarrow 0} f(x)=f(0) \Rightarrow \lim _{x \rightarrow 0} \frac{4 \sin \pi x}{5 x}=2 k \\
\therefore \lim _{x \rightarrow 0}\left(\frac{4 \sin \pi x}{\pi x}\right) \cdot \frac{\pi}{5}=2 k \Rightarrow(1) \cdot \frac{4 \pi}{5}=2 k \Rightarrow k=\frac{2 \pi}{5}
\end{array}$
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