Question

The values of $x$ for which $2{\tan }^{-1}x+{\sin }^{-1}\frac{2x}{1+x^2}$ is a constant are

• A. $x\in [-1,1]$
• B. $x\in [1,\infty )$
• C. $x\in (-\infty ,1]$
• D. none of these

Answer 1

The correct option is B $x\in [1,\infty )$

Let $\theta ={\tan }^{-1}x,\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$
Let $f\left(x\right)=2{\tan }^{-1}x+{\sin }^{-1}\frac{2x}{1+x^2}=2{\tan }^{-1}\tan \theta +{\sin }^{-1}\sin 2\theta$

As $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})\Rightarrow 2\theta \in (-\pi ,\pi )$

There will be three cases
Case $1:$
$-\frac{\pi }{2}<2\theta <\frac{\pi }{2}$
$2{\tan }^{-1}x+{\sin }^{-1}\frac{2x}{1+x^2}=2\theta +2\theta =4\theta =4{\tan }^{-1}x$
So its dependent on $x\in (-1,1)(\because -\frac{\pi }{2}<2\theta <\frac{\pi }{2})$

Case $2:$
$\frac{\pi }{2}\le 2\theta <\pi$
$\Rightarrow 0<\pi -2\theta \le \frac{\pi }{2}$
$2{\tan }^{-1}x+{\sin }^{-1}\frac{2x}{1+x^2}=2\theta +{\sin }^{-1}\sin (\pi -2\theta )=2\theta +\pi -2\theta =\pi$
It is a constant $\forall x\in [1,\infty )$

Case $3:$
$-\pi <2\theta \le -\frac{\pi }{2}$
Similar to case $2$
$f\left(x\right)=-\pi ,\forall x\in (-\infty ,-1]$

$f\left(x\right)=\{\begin{array}{c}-\pi ,x\le -1\\ 4{\tan }^{-1}x,-1<x<1\\ \pi ,x\ge 1\end{array}$