Question

The smallest and the largest values of
${\tan }^{-1}\left(\frac{1-x}{1+x}\right)$ , $0\le x\le 1$ are.

• A. $0,\pi$
• B. $0,\frac{\pi }{4}$
• C. $-\frac{\pi }{4},\frac{\pi }{4}$
• D. $\frac{\pi }{4},\frac{\pi }{2}$

Answer 1

Answer: (B)

$0,\frac{\pi }{4}$

$ta{n}^{-1}\left(\frac{1-x}{1+x}\right)$
$0\le x\le 1$
$1\le x+1\le 2$
$1\ge \frac{1}{1+x}\ge \frac{1}{2}$
$0\le \frac{1-x}{1+x}\le 1$
$\therefore xϵ[-1,1]byxϵ[0,1]$
So, $xϵ[0,1]$
$ta{n}^{-1}\left(\frac{1-x}{1+x}\right)ϵ[0,\frac{\pi }{4}]$