Question

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

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

Answer 1

The correct option is B

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

Explanation for the correct answer:

Compute the required values:

Let $x=\tan A$ $\Rightarrow A={\tan }^{-1}x$

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

$\Rightarrow$ $f\left(x\right)={\tan }^{-1}\left(\frac{1-\tan A}{1+\tan A}\right)$

$\Rightarrow$ $f\left(x\right)={\tan }^{-1}\left(\tan \left(\frac{\mathrm{\pi }}{4}-A\right)\right)$ $\left(\because \tan \left(A-B\right)=\frac{\tan A-\tan B}{1+\tan A\tan B}\right)$, $\left(\because \tan \frac{\mathrm{\pi }}{4}=1\right)$

$\Rightarrow$ $f\left(x\right)=\frac{\mathrm{\pi }}{4}-A$

$\Rightarrow$ $f\left(x\right)=\frac{\mathrm{\pi }}{4}-{\tan }^{-1}x$

$0\le x\le 1$ …(given)

$f\left(0\right)=\frac{\mathrm{\pi }}{4}-{\tan }^{-1}\left(0\right)$

$=\frac{\mathrm{\pi }}{4}$ (maximum)

$f\left(1\right)=\frac{\mathrm{\pi }}{4}-{\tan }^{-1}\left(1\right)$

$=\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4}=0$ (minimum)

Hence, the maximum and minimum values of ${\tan }^{-1}\left(\frac{1-x}{1+x}\right)$ are $\frac{\mathrm{\pi }}{4}$ and $0$ respectively.

Hence, option $\left(B\right)$ is the correct answer.