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Question:

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 $f\left(\frac{\pi}{4}\right)=$

Solution:

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 $f\left(\frac{\pi}{4}\right)=$ $-\frac{1}{2}$

 

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