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