Question

If the function $f:R \rightarrow A$ defined as $f\left(x\right)=\left(tan\right)^{- 1} \left(\frac{2 x^{3}}{1 + x^{6}}\right)$ 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, $y=\frac{2 t}{1 + t^{2}}$
$\Rightarrow yt^{2}-2t+y=0$
$\because t\in R$ , hence $D\geq 0$
$\Rightarrow \left(- 2\right)^{2}-4y^{2}\geq 0\Rightarrow y\in \left[- 1,1\right]$
$\left(tan\right)^{- 1} y\in \left[\left(tan\right)^{- 1} ⁡ \left(- 1\right) , \left(tan\right)^{- 1} ⁡ \left(1\right)\right]$
Hence, the range of $f\left(x\right)$ $\in \left[- \frac{\pi }{4} , \frac{\pi }{4}\right]\equiv A$