Question

Let $f\left(x\right)=\frac{1-\tan x}{4x-\pi },x\ne \frac{\pi }{4},x\in [0,\frac{\pi }{4}]$. lf $f\left(x\right)$ 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

Answer: (A)

$-\frac{1}{2}$

Substitute $x=\frac{\pi }{4}+t$

$\begin{array}{c}lim\\ t\to o\end{array}\frac{1-tan(\frac{\pi }{4}+t)}{4t}$

$={lim}_{t\to 0}\frac{1-\frac{(1+tant)}{1-tant}}{4t}$

$={lim}_{t\to 0}\frac{-2tant}{(1-tant)4t}$

$=\frac{-1}{2}$