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,
$x^2+x\in [\frac{-1}{4},\infty )\forall x\in R$
But for sine inverse to exist
$(x^2+x)\in [-1,1]$
Therefore,
$\frac{-1}{4}\le x^2+x\le 1\Rightarrow {\sin }^{-1}\left(\frac{-1}{4}\right)\le {\sin }^{-1}(x^2+x)\le {\sin }^{-1}\left(1\right)\Rightarrow -{\sin }^{-1}\left(\frac{1}{4}\right)\le {\sin }^{-1}(x^2+x)\le \frac{\pi }{2}$
As the input of $\log$ is positive, so
$0<{\sin }^{-1}(x^2+x)\le \frac{\pi }{2}\Rightarrow -\infty <\ln \left({\sin }^{-1}\right(x^2+x\left)\right)\le \ln \frac{\pi }{2}$

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