If the function $f(x)=\left\{\begin{matrix} \frac{sin\,5x}{3x}, & x≠0\\\frac{k}{3}, & x=0\end{matrix}\right.$ is continuous at $x=0, $ then $k^2-2k +10 $ is equal to :

25

The correct answer is Option (2) → 25

$f(0)=\frac{k}{3}$

$\underset{x→0}{\lim}f(x)=\underset{x→0}{\lim}\frac{\sin 5x}{3x}=\underset{x→0}{\lim}\frac{5\sin 5x}{3(5x)}$

$=\frac{5}{3}=\frac{k}{3}$ (as f is continuous)

so $k = 5⇒k^2-2k+10$

$=25-10+10=25$