Question

For ${\tan }^{-1}\left(\frac{1-x}{1+x}\right)$, $0\le x\le 1$.
What is the sum of the smallest and the largest values of function.

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{3\pi }{4}$
• D. $\frac{3\pi }{2}$

Answer 1

The correct option is A $\frac{\pi }{4}$

Let $f\left(x\right)={\tan }^{-1}\left(\frac{1-x}{1+x}\right)$, $0\le x\le 1$

Now $\frac{1-x}{1+x}=\frac{2}{1+x}-1$

Given $0\le x\le 1$

$\Rightarrow$ $\frac{2}{1+x}-1\in [0,1]$

$\Rightarrow$ ${\tan }^{-1}\left(\frac{1-x}{1+x}\right)\in [{\tan }^{-1}0,{\tan }^{-1}1]$ or $[0,\frac{\pi }{4}]$