Find the value

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

Let $h=x-\frac{\pi}{2}$

$\rightarrow \mathrm{x}=\mathrm{h}+\frac{\pi}{2}$

$\mathrm{x} \rightarrow \frac{\pi}{2}$

or, $\mathrm{h}+\frac{\pi}{2} \rightarrow \frac{\pi}{2}$

or, $h \rightarrow 0$

Putting this in the original sum,

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

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

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

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

[ Applying formula $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$ ]

$=-k \times 1$

$=-\mathrm{k}$

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

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

$\therefore-\mathrm{k}=3$

$\rightarrow \mathrm{k}=-3$