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\(\rightarrow\)\(\frac{\pi}{2}\)\(\Rightarrow\)h\(\rightarrow\)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.