Let $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{\frac{\sin \mathrm{x}}{\mathrm{x}}-\cos 2\mathrm{x}}{{\mathrm{x}}^2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\ne 0\\ \mathrm{k},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}=0\end{array}\right.$ If f(x) is continuous at x = 0, then the value of 6k is _______ .

f(x) is continuous at x - 0.
$\begin{array}{l}\therefore \mathrm{k}=\underset{\mathrm{x}\to 0}{lim} \frac{\sin \mathrm{x}-\mathrm{xcos}2\mathrm{x}}{{\mathrm{x}}^3}\\ =\underset{\mathrm{x}\to 0}{lim} \left(\frac{\sin \mathrm{x}-\mathrm{x}}{{\mathrm{x}}^3}+\frac{1-\cos 2\mathrm{x}}{{\mathrm{x}}^2}\right)\\ =\underset{\mathrm{x}\to 0}{lim} \left(\frac{\cos \mathrm{x}-1}{3{\mathrm{x}}^2}+\frac{2\sin 2\mathrm{x}}{2\mathrm{x}}\right) (\mathrm{Using} \mathrm{L}\text{'}\mathrm{Hospital} \mathrm{rule}) \\ =\underset{\mathrm{x}\to 0}{lim} \left(\frac{-\sin \mathrm{x}}{6\mathrm{x}}+2\right)\\ =\frac{-1}{6}+2=\frac{11}{6}\end{array}$