Find the values of k so that the function f is continuous at the indicated point.

The given function $f$ is $f(x)= \begin{cases}\frac{k \cos x}{\pi-2 x}, & \text { if } x \neq \frac{\pi}{2} \\ 3, & \text { if } x=\frac{\pi}{2}\end{cases}$

The given function $f$ is continuous at $x=\frac{\pi}{2}$, if $f$ is defined at $x=\frac{\pi}{2}$ and if the value of the $f$ at $x=\frac{\pi}{2}$ equals the limit of $f$ at $x=\frac{\pi}{2}$.

It is evident that $f$ is defined at $x=\frac{\pi}{2}$ and $f\left(\frac{\pi}{2}\right)=3$

$\lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}$

Put $x=\frac{\pi}{2}+h$

Then, $x \rightarrow \frac{\pi}{2} \Rightarrow h \rightarrow 0$

$\therefore \lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}+h\right)}{\pi-2\left(\frac{\pi}{2}+h\right)}$

$=k \lim _{h \rightarrow 0} \frac{-\sin h}{-2 h}=\frac{k}{2} \lim _{h \rightarrow 0} \frac{\sin h}{h}=\frac{k}{2} .1=\frac{k}{2}$

$\therefore \lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right)$

$\Rightarrow \frac{k}{2}=3$

$\Rightarrow k=6$

Therefore, the required value of k is 6.