Let $f(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right]$. If $f(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$, then $f\left(\frac{\pi}{4}\right)=$
The given function is $f(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right]$.
It is given that, the function $f(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$. So, the function is continuous at $x=\frac{\pi}{4}$
$\therefore f\left(\frac{\pi}{4}\right)$
$=\lim _{x \rightarrow \frac{\pi}{4}} f(x)$
$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{4 x-\pi}$
Put $x=\frac{\pi}{4}+h$
When $x \rightarrow \frac{\pi}{4}, h \rightarrow 0$
So,
$f\left(\frac{\pi}{4}\right)$
$=\lim _{h \rightarrow 0} \frac{1-\tan \left(\frac{\pi}{4}+h\right)}{4\left(\frac{\pi}{4}+h\right)-\pi}$
$=\lim _{h \rightarrow 0} \frac{1-\frac{\tan \frac{\pi}{4}+\tan h}{1-\tan \frac{\pi}{4} \tan h}}{4 h}$
$=\lim _{h \rightarrow 0} \frac{1-\tan h-1-\tan h}{4 h(1-\tan h)}$ $\left(\tan \frac{\pi}{4}=1\right)$
$=\lim _{h \rightarrow 0} \frac{-2 \tan h}{4 h(1-\tan h)}$
$=-\frac{1}{2} \times \lim _{h \rightarrow 0} \frac{\tan h}{h} \times \frac{1}{\lim _{h \rightarrow 0}(1-\tan h)}$
$=-\frac{1}{2} \times 1 \times \frac{1}{(1-0)}$ $\left(\lim _{x \rightarrow 0} \frac{\tan x}{x}=1\right)$
$=-\frac{1}{2}$
Thus, the value of $f\left(\frac{\pi}{4}\right)$ is $-\frac{1}{2}$.
Let $f(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right]$. If $f(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$, then $f\left(\frac{\pi}{4}\right)=$ $-\frac{1}{2}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.