Question

Let $f\left(x\right)=\frac{1-\tan x}{4x-\pi }$, $x\ne \pi /4$, $x\in [0,\frac{\pi }{2}]$. If $f\left(x\right)$ is continuous in $[0,\frac{\pi }{2}]$ then $f\left(\frac{\pi }{4}\right)$ is?

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $1$
• D. $-1$

Answer 1

The correct option is A $-\frac{1}{2}$

${lim}_{x\to \frac{\pi }{4}}\frac{1-\tan x}{4x-\pi }$

Put $4x-\pi =t$

$\therefore {lim}_{x\to \frac{\pi }{4}}\frac{(1-\tan x)(1+\tan x)}{(1+\tan x)[-4(\frac{\pi }{4}-x)]}$

$={lim}_{x\to \frac{\pi }{4}}\frac{(1-\tan x)}{(1+\tan x)}\frac{(1+\tan x)}{[-4(\frac{\pi }{4}-x)]}$

$={lim}_{x\to \frac{\pi }{4}}\frac{(\tan \frac{\pi }{4}-\tan x)}{(1+\tan \frac{\pi }{4}\tan x)}\times \frac{(1+\tan x)}{[-4(\frac{\pi }{4}-x)]}$ where $\tan \frac{\pi }{4}=1$

$=1\times \frac{1+\tan \frac{\pi }{4}}{-4}=\frac{1+1}{-4}=\frac{2}{-4}=-\frac{1}{2}$