Question

The range of the function $f\left(x\right)=\ln \left({\sin }^{-1}\right(x^2+x\left)\right),$ is

• A. $[-\ln \left({\sin }^{-1}\frac{1}{4}\right),\ln \frac{\pi }{2}]$
• B. $[-\ln \frac{\pi }{2},\ln \frac{\pi }{2}]$
• C. $(0,\ln \frac{\pi }{2}]$
• D. $(-\infty ,\ln \frac{\pi }{2}]$

Answer 1

The correct option is D $(-\infty ,\ln \frac{\pi }{2}]$

$f\left(x\right)=\ln \left({\sin }^{-1}\right(x^2+x\left)\right)$
We know that, $x^2+x\in [\frac{-1}{4},\infty )$ for $x\in R$
But domain of ${\sin }^{-1}(x^2+x)$ is $[-1,1]$
$\therefore \frac{-1}{4}\le x^2+x\le 1$
$\Rightarrow {\sin }^{-1}\left(\frac{-1}{4}\right)\le {\sin }^{-1}(x^2+x)\le {\sin }^{-1}\left(1\right)$
We know that, input of the $\log$ function must be positive.
$\therefore 0<{\sin }^{-1}(x^2+x)\le \frac{\pi }{2}(\because {\sin }^{-1}\left(\frac{-1}{4}\right)<0)$
$\Rightarrow -\infty <\ln \left({\sin }^{-1}\right(x^2+x\left)\right)\le \ln \frac{\pi }{2}$

Hence, range of the $f\left(x\right)$ is $(-\infty ,\ln \frac{\pi }{2}].$