Question

Let $f\left(x\right)={\cos }^{-1}\left(\frac{x^2}{1+x^2}\right).$ The range of $f$ is

• A. $[0,\frac{\pi }{2}]$
• B. $[-\frac{\pi }{2},\frac{\pi }{2}]$
• C. $[-\frac{\pi }{2},0]$
• D. none of these

Answer 1

Answer: (D)

none of these

Let $g\left(x\right)=\frac{x^2}{1+x^2}$
$=1-\frac{1}{1+x^2}$
Now the minimum value of $g\left(x\right)=0$ for $x=0$
$li{m}_{x\to \infty }\left(g\right(x\left)\right)$
$=li{m}_{x\to \infty }(1-\frac{1}{1+x^2})$
$=1$
Hence the range of $co{s}^{-1}\left(\frac{x^2}{1+x^2}\right)$ will be
$(0,\frac{\pi }{2}]$
Hence, option 'D' is correct.