Question

Let $f\left(x\right)= \frac{1- \tan x }{4x -\pi} , x \ne \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) $ is

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

Answer 1

Answer: (C)

$f\left(x\right) = \frac{1- \tan x}{4x - \pi}$ is continuous in $\left[0, \frac{\pi}{2}\right]$
$ \therefore f\left(\frac{\pi}{4}\right) =^{\displaystyle\lim}_{x \to \frac{\pi}{4}} f\left(x\right) $
$=^{\lim}_{x \to \frac{\pi^{+}}{4}} f\left(x\right) $
$= \displaystyle\lim_{h \to0} f\left( \frac{\pi}{4} +h\right) $
$ = \displaystyle\lim_{ h\to0} \frac{1- \tan\left(\frac{\pi}{4} +h\right)}{4\left(\frac{\pi}{4} + h \right) - \pi}, h >0$
$ = \displaystyle\lim_{h \to0} \frac{1- \frac{1+ \tan h}{1- \tan h}}{4h} $
$ = \displaystyle\lim_{h \to0} \frac{-2}{1- \tan h} . \frac{\tan h}{4h} = \frac{-2 }{4} =- \frac{1}{2} $