Question

Find the values of k so that the function f is continuous at the indicated point.
f(x)=\left\\{\begin{matrix} \frac{k\,cos\,x}{\pi-2x}, &if\,x\neq\frac{\pi}{2} \\\ 3,&if\,x=\frac{\pi}{2} \end{matrix}\right.at\, x=\frac{\pi}{2}

Answer 1

f(x)=\left\\{\begin{matrix} \frac{k\,cos\,x}{\pi-2x}, &if\,x\neq\frac{\pi}{2} \\\ 3,&if\,x=\frac{\pi}{2} \end{matrix}\right.

The given function f is continuous at x=$\frac{\pi}{2}$ if f is defined at x=$\frac{\pi}{2}$ and if the value of the f at x=$\frac{\pi}{2}$ equals the limit of f at x=$\frac{\pi}{2}$.
It is evident that f is defined at x=π/2 and f(π/2)=3
$\lim_{x\rightarrow \frac{\pi}{2}}$ f(x)=$\lim_{x\rightarrow \frac{\pi}{2}}$ $\frac{k\,cos\,x}{\pi-2x}$
put x=$\frac{\pi}{2}$+h
Then,x\rightarrow$$\frac{\pi}{2}$$\Rightarrowh$\rightarrow$0

∴limx\rightarrow$$\frac{\pi}{2} f(x)=\lim_{x\rightarrow \frac{\pi}{2}}$$\frac{k\,cos\,x}{\pi-2x}=\lim_{h\rightarrow 0}$$\frac{k\,cos(\frac{\pi}{2}+h)}{\pi-2(\frac{\pi}{2}+h)}
=k$\lim_{x\rightarrow 0^-}\frac{sin\,h}{2h}$=\frac{k}{2}$$\lim_{x\rightarrow 0^-}\frac{sin\,h}{2h}=$\frac{k}{2.1}$=$\frac{k}{2}$
∴$\lim_{x\rightarrow \frac{\pi}{2}}$ f(x)=f($\frac{\pi}{2}$)
\Rightarrow$$\frac{k}{2}=3

$\Rightarrow$k=6
Therefore, the required value of k is 6.