If the function f(x) = begin{cases} frac{kcos | vedclass

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(B) For the function $f(x)$ to be continuous at $x = \frac{\pi}{2}$,the limit of $f(x)$ as $x 	o \frac{\pi}{2}$ must equal $f\left(\frac{\pi}{2}igh

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$6$

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(B) For the function $f(x)$ to be continuous at $x = \frac{\pi}{2}$,the limit of $f(x)$ as $x 	o \frac{\pi}{2}$ must equal $f\left(\frac{\pi}{2}ight)$.Given $f\left(\frac{\pi}{2}ight) = 3$.We evaluate the limit: $\lim_{x 	o \frac{\pi}{2}} \frac{k\cos x}{\pi - 2x}$.Let $x = \frac{\pi}{2} + h$. As $x 	o \frac{\pi}{2}$,$h 	o 0$.Substituting this into the limit: $\lim_{h 	o 0} \frac{k\cos(\frac{\pi}{2} + h)}{\pi - 2(\frac{\pi}{2} + h)} = \lim_{h 	o 0} \frac{k(-\sin h)}{\pi - \pi - 2h} = \lim_{h 	o 0} \frac{-k\sin h}{-2h} = \frac{k}{2} \lim_{h 	o 0} \frac{\sin h}{h}$.Since $\lim_{h 	o 0} \frac{\sin h}{h} = 1$,the limit is $\frac{k}{2}$.Equating the limit to the function value: $\frac{k}{2} = 3 \implies k = 6$.

If the function $f(x) = \begin{cases} \frac{k\cos x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ 3, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then $k = $

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