If  $f(x)=\left\{\begin{array}{cc}\frac{k \cos x}{\pi-2 x}, & x \neq \frac{\pi}{2} \\ 3, & x=\frac{\pi}{2}\end{array}\right.$ is continuous at $x=\frac{\pi}{2}$, then k is :

6

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

f(x) is continuous at $x=\frac{\pi}{2}$

$f(\frac{\pi}{2})=3$

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

Using L' hospital rule (differentiating numeration and denominator separately)

$\Rightarrow \lim\limits_{x \rightarrow \frac{\pi}{2}} \frac{-k \sin x}{-2}$

$=\frac{k \sin \frac{\pi}{2}}{2}=\frac{k}{2}$

so  $\lim\limits_{x→\frac{\pi}{2}} f(x) = f(\frac{\pi}{2})$

⇒  $k = \frac{k}{2}=3$

k = 6