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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 $f(\large\frac{\pi}{4})$

$(a)\;1\qquad(b)\;\large\frac{1}{2}$$\qquad(c)\;-\large\frac{1}{2}$$\qquad(d)\;-1$

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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
 

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