Question

If the function $f:R\to A$ defined as $f\left(x\right)={\left(tan\right)}^{-1}\left(\frac{2x^3}{1+x^6}\right)$ is a surjective function, then the set $A$ is equal to

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

Answer 1

Answer: (B)

Let, $x^3=t\in R$
Let, $y=\frac{2t}{1+t^2}$
$\Rightarrow y{t}^2-2t+y=0$
$\because t\in R$ , hence $D\ge 0$
$\Rightarrow {\left(-2\right)}^2-4y^2\ge 0\Rightarrow y\in \left[-1,1\right]$
${\left(tan\right)}^{-1}y\in \left[{\left(tan\right)}^{-1}\left(-1\right),{\left(tan\right)}^{-1}\left(1\right)\right]$
Hence, the range of $f\left(x\right)$ $\in \left[-\frac{\pi }{4},\frac{\pi }{4}\right]\equiv A$