Question

Find the value of k if f(x) is continuous at x = π/2, where \[f\left( x \right) = \begin{cases}\frac{k \cos x}{\pi - 2x}, & x \neq \pi/2 \\ 3 , & x = \pi/2\end{cases}\]

Answer 1

Given: \[f\left( x \right) = \binom{\frac{k\ cosx}{\pi - 2x}, x \neq \frac{\pi}{2}}{3, x = \frac{\pi}{2}}\]

If f(x) is continuous at x =

\[\frac{\pi}{2}\]then

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

\[\lim_{x \to \frac{\pi}{2}} \frac{k\ cosx}{\pi - 2x} = 3\] 

Putting

\[\frac{\pi}{2} - x = h\]

\[\lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x} =\]

\[\lim_{h \to 0} \frac{k \cos \left( \frac{\pi}{2} - h \right)}{\pi - 2\left( \frac{\pi}{2} - h \right)}\]

From (1), we have

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

\[\Rightarrow \lim_{h \to 0} \frac{k \sin h}{2h} = 3\]

\[\Rightarrow \lim_{h \to 0} \frac{k \sin h}{h} = 6\]

\[\Rightarrow k \lim_{h \to 0} \frac{\sin h}{h} = 6\]

\[\Rightarrow k \times 1 = 6\]

\[\Rightarrow k = 6\]

Hence, for

\[k = 6\] , f(x) is continuous at x =

\[\frac{\pi}{2}\]