$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\cfrac{{k\cos x}}{{\pi - 2x}},}&{if}&{x \ne \cfrac{\pi }{2}}\\{3,}&{if}&{x = \cfrac{\pi }{2}}\end{array}} \right.$ at $x = \cfrac{\pi }{2}$ .

$\mathop {\lim }\limits_{x \to \cfrac{{{\pi ^ - }}}{2}} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to \cfrac{\pi }{2} - h\atop\scriptstyle h \to 0} \cfrac{{k\cos \left( {\cfrac{\pi }{2} - h} \right)}}{{\pi - 2\left( {\cfrac{\pi }{2} - h} \right)}}$

$= \mathop {\lim }\limits_{\scriptstyle x \to \cfrac{\pi }{2} - h\atop\scriptstyle h \to 0} \cfrac{{k\sinh }}{{\pi - \pi + 2h}}$

$= \mathop {\lim }\limits_{\scriptstyle x \to \cfrac{\pi }{2} - h\atop\scriptstyle h \to 0} \cfrac{{k\sinh }}{{2h}} = \cfrac{k}{2}\mathop {\lim }\limits_{\scriptstyle x \to \cfrac{\pi }{2} - h\atop\scriptstyle h \to 0} \cfrac{{\sinh }}{h} = \cfrac{k}{2}$

$\mathop {\lim }\limits_{x \to \cfrac{{{\pi ^ + }}}{2}} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to \cfrac{\pi }{2} + h\atop\scriptstyle h \to 0} \cfrac{{k\cos \left( {\cfrac{\pi }{2} + h} \right)}}{{\pi - 2\left( {\cfrac{\pi }{2} + h} \right)}}$

$= \mathop {\lim }\limits_{\scriptstyle x \to \cfrac{\pi }{2} + h\atop\scriptstyle h \to 0} \cfrac{{ - k\sinh }}{{ - 2h}}$

$= \cfrac{k}{2}\mathop {\lim }\limits_{\scriptstyle x \to \cfrac{\pi }{2} + h\atop\scriptstyle h \to 0} \cfrac{{\sinh }}{h} = \cfrac{k}{2}$

Also, $f\left( {\cfrac{\pi }{2}} \right) = 3.$ For continuity at $x = \cfrac{\pi }{2}$ , we have

$\mathop {\lim }\limits_{x \to \cfrac{{{\pi ^ - }}}{2}} f(x) = \mathop {\lim }\limits_{x \to \cfrac{{{\pi ^ + }}}{2}} f(x) = f\left( {\cfrac{\pi }{2}} \right) \Rightarrow \cfrac{k}{2} = 3 \Rightarrow k = 6$