Let$f(x) = \frac{1 - an x}{4x - \pi},\ x
eq \frac{\pi}{4},\ \ \ x \in \left\lbrack 0,\frac{\pi}{2} ight brack$. If $f(x)$ is continuous in $\left\lbrack 0,\frac{\pi}{2} ight brack$, then $f\left( \frac{\pi}{4} ight)$ is

Question

Let$f(x) = \frac{1 - 	an x}{4x - \pi},\ x 
eq \frac{\pi}{4},\ \ \ x \in \left\lbrack 0,\frac{\pi}{2} ightbrack$. If $f(x)$ is continuous in $\left\lbrack 0,\frac{\pi}{2} ightbrack$, then $f\left( \frac{\pi}{4} ight)$ is

Dr. Abubacker · Accepted Answer

(c)
Since, $f(x) = \frac{1 - 	an x}{4x - \pi}$
Now, $\lim_{x ightarrow \pi/4}{f(x) = \lim_{x ightarrow \pi/4}\left( \frac{1 - 	an x}{4x - \pi} ight)}$
$= \lim_{x ightarrow \pi/4}\ \left( \frac{–\sec^{2}x}{4} ight) = - \frac{1}{2}$
Since, $f(x)$ is continuous at
$x = \frac{\pi}{4}$
$	herefore\ \ \ \lim_{x ightarrow \pi/4}{f(x)} = f\left( \frac{\pi}{4} ight) = - \frac{1}{2}$

Q. Let$f(x) = \frac{1 - \tan x}{4x - \pi},\ x \neq \frac{\pi}{4},\ \ \ x \in \left\lbrack 0,\frac{\pi}{2} \right\rbrack$. If $f(x)$ is continuous in $\left\lbrack 0,\frac{\pi}{2} \right\rbrack$, then $f\left( \frac{\pi}{4} \right)$ is

Explanation

Reviewed by Dr. Abubacker

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