If `f(x)={((3 sin pi x)/(5x)"," , xne 0),(2k",", x =0):}` is continuous at x = 0, then the value of k is equal to

Options

A. `(3pi)/(10)`
B. `(3pi)/(5)`
C. `(pi)/(10)`
D. `(3pi)/(2)`

Explanation

To determine the value of \( k \) such that the function

\[ f(x) = \begin{cases} \frac{3 \sin(\pi x)}{5x} & \text{if } x \neq 0 \\ 2k & \text{if } x = 0 \end{cases} \]

is continuous at \( x = 0 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 equals \( f(0) \).

### Step-by-Step Solution:

**Step 1: Find the limit of \( f(x) \) as \( x \) approaches 0.**

We need to calculate:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{3 \sin(\pi x)}{5x} \]

**Hint:** Use the limit property \( \lim_{x \to 0} \frac{\sin(ax)}{ax} = 1 \).

**Step 2: Rewrite the limit using the known limit property.**

We can rewrite the limit as follows:

\[ \lim_{x \to 0} \frac{3 \sin(\pi x)}{5x} = \lim_{x \to 0} \frac{3}{5} \cdot \frac{\sin(\pi x)}{\pi x} \cdot \pi \]

**Hint:** Factor out constants from the limit.

**Step 3: Apply the limit property.**

Now we can apply the limit property:

\[ \lim_{x \to 0} \frac{\sin(\pi x)}{\pi x} = 1 \]

Thus, we have:

\[ \lim_{x \to 0} f(x) = \frac{3}{5} \cdot \pi \cdot 1 = \frac{3\pi}{5} \]

**Hint:** Remember that the limit of a product is the product of the limits.

**Step 4: Set the limit equal to \( f(0) \).**

For the function to be continuous at \( x = 0 \), we need:

\[ \lim_{x \to 0} f(x) = f(0) \]

This gives us:

\[ \frac{3\pi}{5} = 2k \]

**Hint:** Set the limit equal to the function value at that point.

**Step 5: Solve for \( k \).**

Now, we solve for \( k \):

\[ k = \frac{3\pi}{10} \]

### Final Answer:

The value of \( k \) is

\[ \boxed{\frac{3\pi}{10}} \]