$\because x \to \frac{\pi }{2}$
$ \Rightarrow x \ne \frac{\pi }{2}$
Put $x = \frac{\pi }{2} + h$ where $h \to 0$ , we get
$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} f\left( x \right) $$ = \mathop {\lim }\limits_{h \to 0} \frac{{k\cos \left( {\frac{\pi }{2} + h} \right)}}{{\pi - 2\left( {\frac{\pi }{2} + h} \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{{ - k\sinh }}{{\pi - \pi - 2h}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{k\sin \,h}}{{ 2h}} = \frac{k}{2} \times \mathop {\lim }\limits_{h \to 0} \frac{{\sin h}}{h}$
$ = \frac{k}{2}$…….(i)
And $f\left( {\frac{\pi }{2}} \right) = 3$ ……….(ii)
$\because f\left( x \right) = 3$ when $x = \frac{\pi }{2}$ [Given]
Since f(x) is continuous at $x = \frac{\pi }{2}$
$\therefore \mathop {\lim }\limits_{x \to \frac{\pi }{2}} f\left( x \right) = f\left( {\frac{\pi }{2}} \right)$
$\therefore$ From eq. (i) and (ii),
$\frac{k}{2} = 3$
$ \Rightarrow k = 6$
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