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

\(\mathrm{f}(x)\) is continuous at \(x=0\).
\(\begin{aligned}
\therefore \quad \mathrm{f}(0) & =\lim _{x \rightarrow 0} \mathrm{f}(x) \\
\Rightarrow 12 & =\lim _{x \rightarrow 0} \frac{\left(\mathrm{e}^x-1\right)^2}{\sin \left(\frac{x}{\mathrm{k}}\right) \cdot \log \left(1+\frac{x}{4}\right)} \\
\Rightarrow 12 & =\lim _{x \rightarrow 0} \frac{\frac{\left(\mathrm{e}^x-1\right)^2}{x^2}}{\frac{\sin \left(\frac{x}{\mathrm{k}}\right)}{x} \cdot \log \frac{\left(1+\frac{x}{4}\right)}{x}}
\end{aligned}\)
\(\begin{aligned} & \Rightarrow 12=\frac{\lim _{x \rightarrow 0}\left(\frac{\mathrm{e}^x-1}{x}\right)^2}{\lim _{x \rightarrow 0} \frac{\sin \frac{x}{\mathrm{k}}}{\frac{x}{\mathrm{k}} \mathrm{k}} \cdot \log \lim _{x \rightarrow 0}\left(1+\frac{x}{4}\right)^{\frac{1}{x}}} \\ & \Rightarrow 12=\frac{\mathrm{k} \lim _{x \rightarrow 0}\left(\frac{\mathrm{e}^x-1}{x}\right)^2}{\lim _{x \rightarrow 0} \frac{\sin \frac{x}{\mathrm{k}}}{\frac{x}{\mathrm{k}}} \cdot \log _{x \rightarrow 0} \lim _{x \rightarrow 0}\left[\left(1+\frac{x}{4}\right)^{\frac{4}{x}}\right]^{\frac{1}{4}}}\end{aligned}\)
\(\begin{aligned} & \Rightarrow 12=\frac{\mathrm{k}}{1 \times \log ^{\frac{1}{4}}} \quad \cdots\left[\begin{array}{l}\lim _{x \rightarrow 0} \frac{\mathrm{e}^x-1}{x}=1 \\ \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ \lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}=\mathrm{e}\end{array}\right] \\ & \Rightarrow 12=\frac{\mathrm{k}}{\frac{1}{4}} \\ & \Rightarrow 12=4 \mathrm{k} \\ & \Rightarrow \mathrm{k}=3\end{aligned}\)