a
The given function $f$ is $f(x)\left\{ {\begin{array}{*{20}{l}}
{\frac{{k\cos x}}{{\pi - 2x}},}&{{\rm{ if }}\,x\, \ne \,\frac{\pi }{2}}\\
{3,}&{{\rm{ if }}\,x\, = \,\frac{\pi }{2}}
\end{array}} \right.$
The given function $f$ is continuous at $x=\frac{\pi}{2},$ it 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 $f\left(\frac{\pi}{2}\right)=3$
$\mathop {\lim }\limits_{x \to 2} \frac{\pi }{2}f(x) = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{k\cos x}}{{\pi - 2x}}$
Put $x=\frac{\pi}{2}+h$
Then, $x \rightarrow \frac{\pi}{2} \Rightarrow h \rightarrow 0$
$\therefore \mathop {\lim }\limits_{x \to \frac{\pi }{2}} f(x) = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{k\cos x}}{{\pi - 2x}} = \mathop {\lim }\limits_{h \to 0} \frac{{k\cos \left( {\frac{\pi }{2} + h} \right)}}{{\pi - 2\left( {\frac{\pi }{2} + h} \right)}}$
$ = k\mathop {\lim }\limits_{h \to 0} \frac{{ - \sin \,h}}{{ - 2h}} = \frac{k}{2}\mathop {\lim }\limits_{h \to 0} \frac{{\sin \,h}}{h} = \frac{k}{2} \cdot 1 = \frac{k}{2}$
$\therefore \mathop {\lim }\limits_{x \to \frac{\pi }{2}} f(x) = f\left( {\frac{\pi }{2}} \right)$
$\Rightarrow \frac{k}{2}=3$
$\Rightarrow k=6$
Therefore, the required value of $k$ is $6.$