Let k be a non-zero real number . If `f(x)={{:((e^x-1)^2/(sin(xpi/k)log(1+x/4)),"," x ne 0),(12,"," x=0):}` is a continuous function, then the value of k is

Options

A. 1 pi
B. 2 pi
C. 3 pi
D. 4 pi

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 0 \). This means that the limit of \( f(x) \) as \( x \) approaches 0 must equal \( f(0) \), which is given as 12.

### Step-by-step Solution:

1. **Identify the limit condition**: We need to find:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{(e^x - 1)^2}{\sin\left(\frac{\pi x}{k}\right) \log\left(1 + \frac{x}{4}\right)} = 12 \]

2. **Evaluate the numerator**: As \( x \to 0 \):
\[ e^x - 1 \approx x \quad \text{(using Taylor series expansion)} \]
Therefore,
\[ (e^x - 1)^2 \approx x^2 \]

3. **Evaluate the denominator**: We need to analyze each part of the denominator separately:
- For \( \sin\left(\frac{\pi x}{k}\right) \):
\[ \sin\left(\frac{\pi x}{k}\right) \approx \frac{\pi x}{k} \quad \text{(as \( x \to 0 \))} \]
- For \( \log\left(1 + \frac{x}{4}\right) \):
\[ \log\left(1 + \frac{x}{4}\right) \approx \frac{x}{4} \quad \text{(as \( x \to 0 \))} \]

4. **Combine the evaluations**: Plugging these approximations into the limit gives:
\[ \lim_{x \to 0} \frac{x^2}{\left(\frac{\pi x}{k}\right) \left(\frac{x}{4}\right)} = \lim_{x \to 0} \frac{x^2}{\frac{\pi x^2}{4k}} = \lim_{x \to 0} \frac{4k}{\pi} \]

5. **Set the limit equal to 12**: We set the limit equal to 12:
\[ \frac{4k}{\pi} = 12 \]

6. **Solve for \( k \)**: Rearranging gives:
\[ 4k = 12\pi \implies k = \frac{12\pi}{4} = 3\pi \]

### Conclusion:
The value of \( k \) is \( 3\pi \).