Question

If $f(x)=\dfrac{4\sin \pi x}{5x}$ for $x\neq 0$ and $f(x)=2k$ for $x=0$, and $f(x)$ is continuous at $x=0$, then the value of $k$ is

• A. $\dfrac{2\pi}{5}$
• B. $\dfrac{\pi}{5}$
• C. $\dfrac{\pi}{10}$
• D. $\dfrac{4\pi}{5}$

Hint:

For continuity problems at a point, always equate the limit value with the function value.

Answer 1

The Correct Option is A

Step 1: Using the condition of continuity at $x=0$.
For continuity at $x=0$, \[ \lim_{x\to 0} f(x)=f(0) \]
Step 2: Evaluating the limit.
\[ \lim_{x\to 0}\frac{4\sin \pi x}{5x} =\frac{4\pi}{5}\lim_{x\to 0}\frac{\sin \pi x}{\pi x} =\frac{4\pi}{5} \]
Step 3: Equating with $f(0)$.
\[ 2k=\frac{4\pi}{5} \Rightarrow k=\frac{2\pi}{5} \]
Step 4: Conclusion.
The value of $k$ is $\dfrac{2\pi}{5}$.