Find the values of k so that the function f is continuous at the indicated point in$f\left(x\right)=\{\begin{array}{cc}\frac{k\cos x}{\pi -2x}& \text{if}x\ne \frac{\pi }{2}\\ 3& \text{if}x=\frac{\pi }{2}\end{array}$ at $x=\frac{\pi }{2}$

Solution: A function $f\left(x\right)$ is continuous at $x=c$, if
\displaystyle\Lim_{{{x}\to{c}}}{f{{\left({x}\right)}}}={f{{\left({c}\right)}}}
Here, $f\left(\frac{\pi }{2}\right)=3$
\displaystyle\therefore\Lim_{{{x}\to\frac{\pi}{{2}}}}\frac{{{k}{\cos{{x}}}}}{{\pi-{2}{x}}}={3}
As, left side of the equation is $\frac{0}{0}$ form. so, we will use $L^{\prime }$ hospital rule.
\displaystyle\Rightarrow\Lim_{{{x}\to\frac{\pi}{{2}}}}\frac{{-{k}{\sin{{x}}}}}{{-{2}}}={3}
$\Rightarrow \frac{k}{2}\sin \left(\frac{\pi }{2}\right)=3$
$\Rightarrow \frac{k}{2}=3$
$\Rightarrow k=6$
So, for $k=6$, $f\left(x\right)$ will be continuous at $x=\frac{\pi }{2}$.