Question

$f\left(x\right)=\{\begin{array}{c}\frac{K\cos x}{\pi -2x};x\ne \frac{\pi }{2}\\ 5;x=\frac{\pi }{2}\end{array}$

Find the value of K so that the function is continuous at the point $x=\frac{\pi }{2}$.

Answer 1

$f\left(x\right)=\{\begin{array}{c}\frac{K\cos x}{\pi -2x};x\ne \frac{\pi }{2}\\ 5;x=\frac{\pi }{2}\end{array}$

If f(x) is continuous at $\frac{\pi }{2}$, it follows that

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

We have given,

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

Also, $\underset{x\to \frac{\pi }{2}}{lim}f\left(x\right)=5$

$\Rightarrow \underset{x\to \frac{\pi }{2}}{lim}f\left(x\right)\frac{K\cos x}{\pi -2x}=5$

Using L’hospitals rule

$\Rightarrow \underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{lim}\frac{f^{{}^{\prime }}\left(x\right)}{g^{{}^{\prime }}\left(x\right)}$

So LHS :

$\Rightarrow \underset{x\to \frac{\pi }{2}}{lim}\frac{K(-\sin x)}{0-2}=\frac{K}{2}\underset{x\to \frac{\pi }{2}}{lim}\sin x=\frac{K}{2}$

Now, $\frac{K}{2}=5$

$\Rightarrow k=10$