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