If the function $\left\{\begin{matrix}\frac{3}{x^2}\sin2x^2;&if\,\,x<0\\0;&if\,\,x=2\\\frac{x^2-3x+k}{1-3x^2};&if\,\,x≥0,x≠2\end{matrix}\right\}$ is continuous at x = 0, then the value of k is

6

$\underset{x→0-}{\lim}f(x)=\underset{x→0}{\lim}\frac{3}{x^2}\sin(2x^2)$

$=\underset{x→0}{\lim}\frac{6}{2x^2}\sin(2x^2)=6$

$\underset{x→0+}{\lim}f(x)=\underset{x→0}{\lim}\frac{x^2-3x+k}{1-3x^2}=k$

Since f (x) is continuous at x = 0

$∴\underset{x→0-}{\lim}f(x)=\underset{x→0+}{\lim}f(x)=f(0)$

$∴k = 6$