Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

If $f(x)=\large\frac{1-\tan x}{4x-\pi}\qquad $$x\neq \large\frac{\pi}{4}$,$x \in [0,\large\frac{\pi}{2}]$ and f(x) is continuous in $[0,\large\frac{\pi}{2}]$ then $f(\large\frac{\pi}{4})$


Can you answer this question?

1 Answer

0 votes
Since $f(x)$ is continuous in $[0,\large\frac{\pi}{2}]$
$\therefore \lim\limits_{\large x\to \large\frac{\pi}{4}}\large\frac{1-\tan x}{4x-\pi}$$=f(\large\frac{\pi}{4})$
$\Rightarrow f(\large\frac{\pi}{4})=-\frac{1}{2}$
Hence (c) is the correct answer.
answered Dec 24, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App