Let $\mathrm{f}(x)=\frac{1- an x}{4 x-\pi}, x
eq \frac{\pi}{4}, x \in\left[0, \frac{1}{2} ight], \quad \mathrm{f}(x)$ is continuous in $\left[0, \frac{\pi}{2} ight]$, then $f\left(\frac{\pi}{4} ight)$ is

Question

Let $\mathrm{f}(x)=\frac{1-	an x}{4 x-\pi}, x 
eq \frac{\pi}{4}, x \in\left[0, \frac{1}{2}ight], \quad \mathrm{f}(x)$ is continuous in $\left[0, \frac{\pi}{2}ight]$, then $f\left(\frac{\pi}{4}ight)$ is

Aicharya · Accepted Answer

$\begin{aligned} & f(x)=\frac{1-	an x}{4 x-\pi} \ & 	ext { Let } x=\frac{\pi}{4}+\mathrm{t} \ & 	herefore \quad 	ext { when } x ightarrow \frac{\pi}{4}, \mathrm{t} ightarrow 0 \ & 	herefore f(x)=\frac{1-	an \left(\frac{\pi}{4}+tight)}{4\left(\frac{\pi}{4}+tight)-\pi} \end{aligned}$ $\begin{aligned} 	herefore \quad f(t) & =\frac{1-	an \left(\frac{\pi}{4}+tight)}{4 t} \ & =\frac{1-\left(\frac{1+	an t}{1-	an t}ight)}{4 t} \ & =\frac{-2 	an t}{(1-	an t) 4 t}=\frac{-1}{2} 	imes \frac{	an t}{t} 	imes \frac{1}{1-	an t} \ 	herefore \quad f\left(\frac{\pi}{4}ight) & =\lim _{t ightarrow 0} f(t)=\frac{-1}{2}\end{aligned}$

Let $\mathrm{f}(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{1}{2}\right], \quad…

Solution