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(A) The given function $f$ is defined as:$f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & 	ext{if } x 
eq \frac{\pi}{2} \ 3, & 	ext{if } x = \fr

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$6$

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(A) The given function $f$ is defined as:$f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & 	ext{if } x 
eq \frac{\pi}{2} \ 3, & 	ext{if } x = \frac{\pi}{2} \end{cases}$Since the function $f$ is continuous at $x = \frac{\pi}{2}$,the limit of $f(x)$ as $x 	o \frac{\pi}{2}$ must be equal to $f\left(\frac{\pi}{2}ight)$.Given $f\left(\frac{\pi}{2}ight) = 3$.Now,calculate the limit:$\lim_{x 	o \frac{\pi}{2}} f(x) = \lim_{x 	o \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x}$Let $x = \frac{\pi}{2} + h$. As $x 	o \frac{\pi}{2}$,$h 	o 0$.$\lim_{h 	o 0} \frac{k \cos(\frac{\pi}{2} + h)}{\pi - 2(\frac{\pi}{2} + h)} = \lim_{h 	o 0} \frac{k(-\sin h)}{\pi - \pi - 2h} = \lim_{h 	o 0} \frac{-k \sin h}{-2h} = \frac{k}{2} \lim_{h 	o 0} \frac{\sin h}{h}$Since $\lim_{h 	o 0} \frac{\sin h}{h} = 1$,we have:$\lim_{x 	o \frac{\pi}{2}} f(x) = \frac{k}{2} \cdot 1 = \frac{k}{2}$Equating the limit to the function value:$\frac{k}{2} = 3 \implies k = 6$.Therefore,the required value of $k$ is $6$.

Find the values of $k$ so that the function $f$ is continuous at the indicated point.
$f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases}$ at $x = \frac{\pi}{2}$

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Find the values of $k$ so that the function $f$ is continuous at the indicated point.$f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases}$ at $x = \frac{\pi}{2}$