Solution:

The provided function is

$f(x)=\left\{\begin{array}{ccc}\frac{k \cos x}{\pi-2 x}, & \text { if } x \neq \frac{\pi}{2} \\ 3, & \text { if } x=\frac{\pi}{2}\end{array}\right.$

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

Therefore, $x \rightarrow \frac{\pi}{2}$

Now this implies that, $x \neq \frac{\pi}{2}$

Putting the value of $x=\frac{\pi}{2}+h$ where $h \rightarrow 0$

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

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

$=\frac{k}{2} \times \lim _{h \rightarrow 0} \frac{\sin h}{h}$

$=\frac{k}{2} \ldots \ldots \ldots$..(1)

And $f\left(\frac{\pi}{2}\right)=3$…….(2)

Provided, $f(x)=3$ when $x=\frac{\pi}{2}$

Since we know, $f(x)$ is continuous at $x=\pi / 2$.

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

From eq. (1) and eq. (2), we now have

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

$k=6$

As a result, the value of $\mathrm{k}$ is 6