f is continuous at x=frac{pi}{2} where,f(x)=b | vedclass

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(C) Since $f(x)$ is continuous at $x=\frac{\pi}{2}$,we have $\lim_{x 	o \frac{\pi}{2}} f(x) = f(\frac{\pi}{2})$.Given $f(\frac{\pi}{2}) = 2024$.Now,$

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$2024$

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(C) Since $f(x)$ is continuous at $x=\frac{\pi}{2}$,we have $\lim_{x 	o \frac{\pi}{2}} f(x) = f(\frac{\pi}{2})$.Given $f(\frac{\pi}{2}) = 2024$.Now,$\lim_{x 	o \frac{\pi}{2}} \frac{2k \cos x}{\pi-2x} = 2024$.Let $x = \frac{\pi}{2} + h$. As $x 	o \frac{\pi}{2}$,$h 	o 0$.$\lim_{h 	o 0} \frac{2k \cos(\frac{\pi}{2} + h)}{\pi - 2(\frac{\pi}{2} + h)} = \lim_{h 	o 0} \frac{2k(-\sin h)}{\pi - \pi - 2h} = \lim_{h 	o 0} \frac{-2k \sin h}{-2h} = k \lim_{h 	o 0} \frac{\sin h}{h} = k(1) = k$.Equating the limit to the function value,we get $k = 2024$.

$f$ is continuous at $x=\frac{\pi}{2}$ where,
$f(x)=\begin{cases}\frac{2 k \cos x}{\pi-2 x}, & x \neq \frac{\pi}{2} \\ 2024, & x=\frac{\pi}{2}\end{cases}$ then,the value of $k$ is . . . . . .

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$f$ is continuous at $x=\frac{\pi}{2}$ where,$f(x)=\begin{cases}\frac{2 k \cos x}{\pi-2 x}, & x \neq \frac{\pi}{2} \\ 2024, & x=\frac{\pi}{2}\end{cases}$ then,the value of $k$ is . . . . . .