Question

Let $f:A\to B$ be a function defined by $y=f\left(x\right)$ where f is a bijective function, means f is injective (one-one) as well as surjective (onto), then there exist a unique mapping $g:B\to A$ such that $f\left(x\right)=y$ if and only if $g\left(y\right)=x\forall xϵA,yϵB$ Then function g is said to be inverse of f and vice versa so we write $g=f^{-1}:B\to A[\left\{f\right(x),x\}:\{x,f(x\left)\right\}ϵf^{-1}]$when branch of an inverse function is not given (define) then we consider its principal value branch.

For $0\le x\le 1$ the range of ${\tan }^{-1}\left(\frac{1+x}{1-x}\right)$ is?

• A. $(-\frac{\pi }{2},\frac{\pi }{2})$
• B. $(-\frac{\pi }{4},\frac{\pi }{4})$
• C. $(\frac{\pi }{4},\frac{\pi }{2})$
• D. None of these

Answer 1

The correct option is C $(\frac{\pi }{4},\frac{\pi }{2})$

Given ${\tan }^{-1}\left(\frac{1+x}{1-x}\right)$ for $0\le x\le 1$
Substitute $x=\tan \theta$
So,${\tan }^{-1}\left(\frac{1+x}{1-x}\right)={\tan }^{-1}\left(\frac{\tan \frac{\pi }{4}+\tan \theta }{1-tan\frac{\pi }{4}\tan \theta }\right)$
$={\tan }^{-1}\left(\tan (\frac{\pi }{4}+\theta )\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{1+x}{1-x}\right)=\frac{\pi }{4}+\theta$
Since, $0\le x\le 1$
$0\le \tan \theta \le 1$
$\Rightarrow 0\le \theta \le \frac{\pi }{4}$
$\Rightarrow \frac{\pi }{4}\le \frac{\pi }{4}+\theta \le \frac{\pi }{2}$
$\Rightarrow \frac{\pi }{4}\le {\tan }^{-1}\left(\frac{1+x}{1-x}\right)\le \frac{\pi }{2}$