Let \[f(x)=\frac{1-\tan x}{4x-\pi },\ x\ne \frac{\pi }{4},\ \ x\in \left[ 0,\frac{\pi }{2} \right]\], If \[f(x)\]is continuous in \[\left[ 0,\frac{\pi }{2} \right]\], then \[f\left( \frac{\pi }{4} \right)\]is [AIEEE 2004]

Let \[f(x)=\frac{1-\tan x}{4x-\pi },\ x\ne \frac{\pi }{4},\ \ x\in \le

1.	Let \[f(x)=\frac{1-\tan x}{4x-\pi },\ x\ne \frac{\pi }{4},\ \ x\in \left[ 0,\frac{\pi }{2} \right]\], If \[f(x)\]is continuous in \[\left[ 0,\frac{\pi }{2} \right]\], then \[f\left( \frac{\pi }{4} \right)\]is [AIEEE 2004]
A.	-1
B.	\[\frac{1}{2}\]
C.	\[-\frac{1}{2}\]
D.	1
Answer» D. 1

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Let \[f(x)=\frac{1-\tan x}{4x-\pi },\ x\ne \frac{\pi }{4},\ \ x\in \left[ 0,\frac{\pi }{2} \right]\], If \[f(x)\]is continuous in \[\left[ 0,\frac{\pi }{2} \right]\], then \[f\left( \frac{\pi }{4} \right)\]is [AIEEE 2004]

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