Let $$k$$ be a non-zero real number. If $$f(x) = \begin{cases} \frac{(e^x - 1)^2}{\sin\left(\frac{x}{k}\right)\log\left(1 + \frac{x}{4}\right)}, & x \neq 0 \\ 12, & x = 0 \end{cases}$$ is a continuous function at $$x = 0$$, then the value of $$k$$ is

To ensure the function $$f(x)$$ is continuous at $$x = 0$$, the limit as $$x$$ approaches 0 must equal the function value at $$x = 0$$, which is 12. So, we need to evaluate:

$$\lim_{x \to 0} \frac{(e^x - 1)^2}{\sin\left(\frac{x}{k}\right) \log\left(1 + \frac{x}{4}\right)} = 12.$$

As $$x$$ approaches 0, both the numerator $$(e^x - 1)^2$$ and the denominator $$\sin\left(\frac{x}{k}\right) \log\left(1 + \frac{x}{4}\right)$$ approach 0, resulting in a $$\frac{0}{0}$$ indeterminate form. We can resolve this using standard limits:

Recall that $$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$$, $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$, and $$\lim_{x \to 0} \frac{\log(1 + x)}{x} = 1$$.

Rewrite the expression by multiplying numerator and denominator by $$x^2$$:

$$\lim_{x \to 0} \frac{(e^x - 1)^2}{\sin\left(\frac{x}{k}\right) \log\left(1 + \frac{x}{4}\right)} = \lim_{x \to 0} \frac{(e^x - 1)^2 \cdot x^2}{x^2 \cdot \sin\left(\frac{x}{k}\right) \log\left(1 + \frac{x}{4}\right)}.$$

Rearrange the denominator:

$$x^2 \cdot \sin\left(\frac{x}{k}\right) \log\left(1 + \frac{x}{4}\right) = \left[ \frac{\sin\left(\frac{x}{k}\right)}{\frac{x}{k}} \cdot \frac{x}{k} \right] \cdot \left[ \frac{\log\left(1 + \frac{x}{4}\right)}{\frac{x}{4}} \cdot \frac{x}{4} \right] \cdot x^2.$$

Simplify the fractions:

$$\frac{x}{k} \cdot \frac{x}{4} = \frac{x^2}{4k}.$$

So the denominator becomes:

$$\frac{x^2}{4k} \cdot \frac{\sin\left(\frac{x}{k}\right)}{\frac{x}{k}} \cdot \frac{\log\left(1 + \frac{x}{4}\right)}{\frac{x}{4}}.$$

Now the limit is:

$$\lim_{x \to 0} \frac{(e^x - 1)^2 \cdot x^2}{\frac{x^2}{4k} \cdot \frac{\sin\left(\frac{x}{k}\right)}{\frac{x}{k}} \cdot \frac{\log\left(1 + \frac{x}{4}\right)}{\frac{x}{4}}} = \lim_{x \to 0} \frac{(e^x - 1)^2 \cdot 4k}{\frac{\sin\left(\frac{x}{k}\right)}{\frac{x}{k}} \cdot \frac{\log\left(1 + \frac{x}{4}\right)}{\frac{x}{4}}}.$$

Factor out constants and rewrite:

$$4k \cdot \lim_{x \to 0} \frac{(e^x - 1)^2}{x^2} \cdot \frac{1}{\frac{\sin\left(\frac{x}{k}\right)}{\frac{x}{k}} \cdot \frac{\log\left(1 + \frac{x}{4}\right)}{\frac{x}{4}}}.$$

Note that $$\frac{(e^x - 1)^2}{x^2} = \left( \frac{e^x - 1}{x} \right)^2$$. So:

$$4k \cdot \lim_{x \to 0} \left( \frac{e^x - 1}{x} \right)^2 \cdot \frac{1}{\frac{\sin\left(\frac{x}{k}\right)}{\frac{x}{k}} \cdot \frac{\log\left(1 + \frac{x}{4}\right)}{\frac{x}{4}}}.$$

As $$x \to 0$$, each standard limit applies:

$$\lim_{x \to 0} \frac{e^x - 1}{x} = 1 \quad \Rightarrow \quad \lim_{x \to 0} \left( \frac{e^x - 1}{x} \right)^2 = 1^2 = 1.$$

Let $$t = \frac{x}{k}$$. As $$x \to 0$$, $$t \to 0$$:

$$\lim_{x \to 0} \frac{\sin\left(\frac{x}{k}\right)}{\frac{x}{k}} = \lim_{t \to 0} \frac{\sin t}{t} = 1.$$

Let $$u = \frac{x}{4}$$. As $$x \to 0$$, $$u \to 0$$:

$$\lim_{x \to 0} \frac{\log\left(1 + \frac{x}{4}\right)}{\frac{x}{4}} = \lim_{u \to 0} \frac{\log(1 + u)}{u} = 1.$$

Substituting these limits:

$$4k \cdot \frac{1}{1 \cdot 1} = 4k.$$

Set this equal to 12 for continuity:

$$4k = 12.$$

Solving for $$k$$:

$$k = \frac{12}{4} = 3.$$

Hence, the value of $$k$$ is 3.

Therefore, the correct answer is Option C.