Let $f(x)=\cos^{-1}\left(\frac{x^2}{1+x^2}\right)$. The range of f is
Answer & explanation
Correct answer: option 4
$\left(\frac{x^2}{1+x^2}\right)≤1$
This is true for all x ∈ R. So, the domain = R
Now, $\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}$
$∴ 0≤\frac{x^2}{1+x^2}<1;\cos^{-1}0≥\cos^{-1}\frac{x^2}{1+x^2}>\cos^{-1}1⇒\frac{π}{2}≥\cos^{-1}\frac{x^2}{1+x^2}>0$
∴ the range = $\left(0,\frac{π}{2}\right]$