Question

In the following, determine the value of constant involved in the definition so that the given function is continuou: 

\[f\left( x \right) = \begin{cases}\frac{k \cos x}{\pi - 2x} , & x < \frac{\pi}{2} \\ 3 , & x = \frac{\pi}{2} \\ \frac{3 \tan 2x}{2x - \pi}, & x > \frac{\pi}{2}\end{cases}\]

Answer 1

 Given: 

\[f\left( x \right) = \begin{cases}\frac{k \cos x}{\pi - 2x}, x < \frac{\pi}{2} \\ 3 , x = \frac{\pi}{2} \\ \frac{3 \tan 2x}{2x - \pi}, x > \frac{\pi}{2}\end{cases}\] 

If  \[f\left( x \right)\]  is continuous at x = \[\frac{\pi}{2}\] , then 

\[\lim_{x \to \frac{\pi}{2}^-} f\left( x \right) = f\left( \frac{\pi}{2} \right)\]

\[\Rightarrow \lim_{h \to 0} f\left( \frac{\pi}{2} - h \right) = f\left( \frac{\pi}{2} \right)\]
\[ \Rightarrow \lim_{h \to 0} f\left( \frac{\pi}{2} - h \right) = 3\]
\[ \Rightarrow \lim_{h \to 0} \left[ \frac{k \cos \left( \frac{\pi}{2} - h \right)}{\pi - 2\left( \frac{\pi}{2} - h \right)} \right] = 3\]
\[ \Rightarrow \lim_{h \to 0} \left[ \frac{k \sin h}{\pi - \pi + 2h} \right] = 3\]
\[ \Rightarrow \lim_{h \to 0} \left[ \frac{k \sin h}{2h} \right] = 3\]
\[ \Rightarrow \frac{k}{2} \lim_{h \to 0} \left[ \frac{\sin h}{h} \right] =3\]
\[ \Rightarrow \frac{k}{2} = 3\]
\[ \Rightarrow k = 2(3) = 6\]