Question

Find the value of $K$ if $f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$,
$f\left(x\right)=\{\begin{array}{c}\frac{K.\cos x}{\pi -2x},ifx\ne \frac{\pi }{2}\\ 3ifx=\frac{\pi }{2}\end{array}$

Answer 1

Given that function is continous at $x=\frac{\pi }{2}$

$f$ is continuous at $x=\frac{\pi }{2}$

if L.H.L$=$R.H.L$=f\left(\frac{\pi }{2}\right)$

L.H.L

$=\underset{x\to {\frac{\pi }{2}}^{-}}{lim}f\left(x\right)$

$=\underset{x\to {\frac{\pi }{2}}^{-}}{lim}\frac{k\cos x}{\pi -2x}$

$=\underset{x\to {\frac{\pi }{2}}^{-}}{lim}\frac{k\sin (\frac{\pi }{2}-x)}{2(\frac{\pi }{2}-x)}$

$=\frac{k}{2}\underset{x\to {\frac{\pi }{2}}^{-}}{lim}\frac{\sin (\frac{\pi }{2}-x)}{(\frac{\pi }{2}-x)}$

Let $y=\frac{\pi }{2}-x$ as $x\to \frac{\pi }{2}$

$y\to \frac{\pi }{2}-\frac{\pi }{2}$

$y\to 0$

So our equation becomes

$=\frac{k}{2}\underset{y\to 0}{lim}\frac{\sin y}{y}$

$=\frac{k}{2}\times 1=\frac{k}{2}$

R.H.L$=\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)$

$=\underset{x\to {\frac{\pi }{2}}^{+}}{lim}\frac{k\cos x}{\pi -2x}$

$=\underset{x\to {\frac{\pi }{2}}^{+}}{lim}\frac{k\sin (\frac{\pi }{2}-x)}{2(\frac{\pi }{2}-x)}$

$=\frac{k}{2}\underset{x\to {\frac{\pi }{2}}^{+}}{lim}\frac{\sin (\frac{\pi }{2}-x)}{(\frac{\pi }{2}-x)}$

Let $y=\frac{\pi }{2}-x$ as $x\to \frac{\pi }{2}$

$y\to \frac{\pi }{2}-\frac{\pi }{2}$

$y\to 0$

So our equation becomes

$=\frac{k}{2}\underset{y\to 0}{lim}\frac{\sin y}{y}$

$=\frac{k}{2}\times 1=\frac{k}{2}$

Hence LHL$=$RHL$=\frac{k}{2}$

Now,LHL$=$RHL$=f\left(\frac{\pi }{2}\right)=3$

$\Rightarrow \frac{k}{2}=3$

$\therefore k=6$