# Find the value of k if f(x) is continuous

Question:

Find the value of $k$ if $f(x)$ is continuous at $x=\pi / 2$, where

$f(x)=\left\{\begin{array}{rr}\frac{k \cos x}{\pi-2 x}, & x \neq \pi / 2 \\ 3 & , x=\pi / 2\end{array}\right.$

Solution:

Given:

$f(x)=\left\{\begin{array}{l}\frac{k \cos x}{\pi-2 x}, x \neq \frac{\pi}{2} \\ 3, x=\frac{\pi}{2}\end{array}\right.$

If $f(x)$ is continuous at $x=\frac{\pi}{2}$, then

$\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right)$

$\Rightarrow \lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}=3$       ....(1)

Putting $\frac{\pi}{2}-x=h$, we get

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}$

From (1), we have

$\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}=3$

$\Rightarrow \lim _{h \rightarrow 0} \frac{k \sin h}{2 h}=3$

$\Rightarrow \lim _{h \rightarrow 0} \frac{k \sin h}{h}=6$

$\Rightarrow k \lim _{h \rightarrow 0} \frac{\sin h}{h}=6$

$\Rightarrow k \times 1=6$

$\Rightarrow k=6$

Hence, for $k=6, f(x)$ is continuous at $x=\frac{\pi}{2}$.