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MHT CET · Maths · Continuity and Differentiability

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

  1. A \(\frac{2 \pi}{5}\)
  2. B \(\frac{\pi}{5}\)
  3. C \(\frac{\pi}{10}\)
  4. D \(\frac{4 \pi}{5}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{2 \pi}{5}\)

Step-by-step Solution

Detailed explanation

(C)
Since \(f(x)\) is continuous at \(x=0\),
\(\begin{array}{l}
\lim _{x \rightarrow 0} f(x)=f(0) \Rightarrow \lim _{x \rightarrow 0} \frac{4 \sin \pi x}{5 x}=2 k \\
\therefore \lim _{x \rightarrow 0}\left(\frac{4 \sin \pi x}{\pi x}\right) \cdot \frac{\pi}{5}=2 k \Rightarrow(1) \cdot \frac{4 \pi}{5}=2 k \Rightarrow k=\frac{2 \pi}{5}
\end{array}\)