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

JEE Main · 2004 · 24 Apr 2004 · Offline · Continuity and Differentiability

Let

f (x) = \frac{1 - t a n x}{4 x - π}, x \neq = \frac{π}{4}, x \in [0, \frac{π}{2}]

. If

f (x)

is continuous in

[0, \frac{π}{2}]

, then

f (\frac{π}{4})

Answer: (C)

Continuity at

x = \frac{π}{4}

requires

f (\frac{π}{4})

to equal the limit there. Put

x = \frac{π}{4} + h

; then

tan (\frac{π}{4} + h) = \frac{1 + t a n h}{1 - t a n h}

, so

1 - tan (\frac{π}{4} + h) = \frac{- 2 t a n h}{1 - t a n h}

, while

4 x - π

becomes

4 h

. The limit is

lim_{h \to 0} \frac{- 2 t a n h}{4 h ( 1 - t a n h )} = \frac{- 2}{4} = - \frac{1}{2}

, so

f (\frac{π}{4}) = - \frac{1}{2}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option