Question
Find the value of k if f(x) is continuous at $\text{x}=\frac{\pi}{2},$ where
$\text{f}\text{(x)}=\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}, &\text{ x}\neq\frac{\pi}{2}\\3, &\text{ x}=\frac{\pi}{2}\end{cases}$

Answer

Since f(x) is continuous at $\text{x}=\frac{\pi}{2},$ L.H.Limit = R.H.Limit.
$\Rightarrow\lim\limits_{\text{x} \rightarrow \frac{\pi^-}{2}}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow \frac{\pi^+}{2}}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow \frac{\pi}{2}}\text{f}\text{(x)}=\text{f}\Big(\frac{\pi}{2}\Big)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow \frac{\pi^-}{2}}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}=3$
$\Rightarrow\text{k}\lim\limits_{\text{x} \rightarrow \frac{\pi}{2}}\frac{\sin\Big(\frac{\pi}{2}-\text{x}\Big)}{2\Big(\frac{\pi}{2}-\text{x}\Big)}=3$
$\Rightarrow\frac{\text{k}}{2}\lim\limits_{\text{x} \rightarrow \frac{\pi}{2}}\frac{\sin\Big(\frac{\pi}{2}-\text{x}\Big)}{2\Big(\frac{\pi}{2}-\text{x}\Big)}=3$
$\Rightarrow\frac{\text{k}}{2}=3$
$\Rightarrow\text{k}=6$

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