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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}$$\qquad(d)\;-1 Can you answer this question? ## 1 Answer 0 votes Since f(x) is continuous in [0,\large\frac{\pi}{2}] \therefore \lim\limits_{\large x\to \large\frac{\pi}{4}}\large\frac{1-\tan x}{4x-\pi}$$=f(\large\frac{\pi}{4})$
$\Rightarrow f(\large\frac{\pi}{4})=-\frac{1}{2}$
Hence (c) is the correct answer.