Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)-2 \tan ^{-1} x$ and $g(x)=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)+4 \tan ^{-1} x$ then range of $( f ( x )- g ( x ))$ for $x \in(-\infty,-1]$, is

Inverse Trigonometric Functions

Solution:

$ \forall x \in(-\infty,-1]$
$f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)-2 \tan ^{-1} x $
$=-\pi-2 \tan ^{-1} x -2 \tan ^{-1} x$
$=-\pi-4 \tan ^{-1} x $
$g(x)=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)+4 \tan ^{-1} x $
$=\frac{\pi}{2}-\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)+4 \tan ^{-1} x $
$=\frac{\pi}{2}-\left(-2 \tan ^{-1} x\right)+4 \tan ^{-1} x $
$=\frac{\pi}{2}+6 \tan ^{-1} x$
Now, $f(x)-g(x)=\frac{-3 \pi}{2}-10 \tan ^{-1} x$
$\forall x \in(-\infty,-1], \tan ^{-1} x \in\left(\frac{-\pi}{2}, \frac{-\pi}{4}\right]$
So, $\left.( f ( x )- g ( x )) \in\left[\pi, \frac{7 \pi}{2}\right)\right]$