Let $\mathrm{f}(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right]$.…
Let $\mathrm{f}(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right]$. $f(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$, then $\mathrm{f}\left(\frac{\pi}{4}\right)$ is
$-\frac{1}{2}$
$\frac{1}{2}$
1
-1
Solution
Since $\mathrm{f}(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$.
$\therefore \quad$ it is continuous at $x=\frac{\pi}{4}$.
$\therefore \quad \mathrm{f}\left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} \mathrm{f}(x)=\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{4 x-\pi}$ Applying L'Hospital rule on R.H.S., we get
$\begin{aligned}
& \mathrm{f}\left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sec ^2 x}{4} \\
& \Rightarrow \mathrm{f}\left(\frac{\pi}{4}\right)=\frac{-2}{4}=\frac{-1}{2}
\end{aligned}$