If f(x) = begin{cases} frac{k cos x}{pi - 2x} | vedclass

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(C) For a function $f(x)$ to be continuous at $x = a$,the limit $\lim_{x 	o a} f(x)$ must equal $f(a)$.Given $f(x)$ is continuous at $x = \frac{\pi}{

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$1$

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(C) For a function $f(x)$ to be continuous at $x = a$,the limit $\lim_{x 	o a} f(x)$ must equal $f(a)$.Given $f(x)$ is continuous at $x = \frac{\pi}{2}$,we have $\lim_{x 	o \frac{\pi}{2}} f(x) = f(\frac{\pi}{2}) = \frac{1}{2}$.Now,calculate 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 this to $f(\frac{\pi}{2}) = \frac{1}{2}$,we get $\frac{k}{2} = \frac{1}{2}$,which implies $k = 1$.

If $f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ \frac{1}{2}, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then the value of $k$ is . . . . . . .

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