Question

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

Answer 1

Given definition of $f$ is

$f\left(x\right)=\{\begin{array}{c}\frac{k\cos x}{\pi -2x},ifx\ne \frac{\pi }{2}\\ 3,ifx=\frac{\pi }{2}\end{array}$ .... $\left(1\right)$

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

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

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

$=\underset{h\to 0}{lim}f(\frac{\pi }{2}-h)$

$=\underset{h\to 0}{lim}\frac{k\cos (\frac{\pi }{2}-h)}{\pi -2(\frac{\pi }{2}-h)}$

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

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

$=\frac{k}{2}$ $\left(\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1\right)$

Also, by $\left(1\right)$

$f\left(\frac{\pi }{2}\right)=3$

Substituting these values in $\left(2\right),$ we get

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

$\Rightarrow k=6$