Question

Find the range of

$f\left(x\right)={\sin }^{-1}\left(\sqrt{x^2+x+1}\right)$

• A. $\left[\frac{\pi }{3},\frac{\pi }{2}\right]$
• B. $\left[\frac{\pi }{3},\pi \right]$
• C. $\left[\frac{\pi }{2},\pi \right]$
• D. $\left[\frac{\pi }{6},\frac{\pi }{2}\right]$

Answer 1

The correct option is A $\left[\frac{\pi }{3},\frac{\pi }{2}\right]$

Consider the equation
$x^2+x+1=0$
It does not have any real roots as $b^2-4ac<0$
Since $b^2-4ac<0$, $x^2+x+1>0$ for all $xϵR$..(i)

${\sin }^{-1}\left(\sqrt{x^2+x+1}\right)ϵ(0,\frac{\pi }{2})$...(i)
${\sin }^{-1}\theta ϵ[\frac{-\pi }{2},\frac{\pi }{2}]$
Hence $\theta ϵ[-1,1]$
Since $x^2+x+1>0$
Hence, $0<x^2+x+1<1$
$x^2+x+1<1$ implies
$x^2+x<0$
$x(x+1)<0$
$x>-1$ and $x<0$

$\therefore xϵ(-1,0)$...(ii)
Now $f\left(x\right)=x^2+x+1$
$f^{\prime }\left(x\right)=2x+1=0$
$f^{\prime }\left(x\right)=x=\frac{-1}{2}$
Hence $f\left(x\right)$ attains a local minimum at $x=\frac{-1}{2}$.
Now ${\sin }^{-1}(\sqrt{x^2+x+1}{)}_{x=\frac{-1}{2}}$
$={\sin }^{-1}\left(\sqrt{\frac{1}{4}-\frac{1}{2}+1}\right)$
$={\sin }^{-1}\left(\sqrt{\frac{3}{4}}\right)$
$={\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)$
$=\frac{\pi }{3}$...(iii)
Hence from i ii and iii
$yϵ[\frac{\pi }{3},\frac{\pi }{2}]$