Let f(x) = frac{1-tan x}{4x-pi},where x neq f | vedclass

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(A) Since $f(x)$ is continuous in $[0, \frac{\pi}{2}]$,it must be continuous at $x = \frac{\pi}{4}$.Therefore,$f(\frac{\pi}{4}) = \lim_{x 	o \frac{\p

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$-\frac{1}{2}$

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(A) Since $f(x)$ is continuous in $[0, \frac{\pi}{2}]$,it must be continuous at $x = \frac{\pi}{4}$.Therefore,$f(\frac{\pi}{4}) = \lim_{x 	o \frac{\pi}{4}} f(x) = \lim_{x 	o \frac{\pi}{4}} \frac{1-	an x}{4x-\pi}$.This is a $\frac{0}{0}$ indeterminate form,so we apply $L$'Hospital's rule by differentiating the numerator and denominator with respect to $x$:$f(\frac{\pi}{4}) = \lim_{x 	o \frac{\pi}{4}} \frac{\frac{d}{dx}(1-	an x)}{\frac{d}{dx}(4x-\pi)} = \lim_{x 	o \frac{\pi}{4}} \frac{-\sec^2 x}{4}$.Substituting $x = \frac{\pi}{4}$:$f(\frac{\pi}{4}) = \frac{-\sec^2(\frac{\pi}{4})}{4} = \frac{-(\sqrt{2})^2}{4} = \frac{-2}{4} = -\frac{1}{2}$.

Let $f(x) = \frac{1-\tan x}{4x-\pi}$,where $x \neq \frac{\pi}{4}$ and $x \in [0, \frac{\pi}{2}]$. If $f(x)$ is continuous in $[0, \frac{\pi}{2}]$,then $f(\frac{\pi}{4})$ is:

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