Question:

The range of $f(x)=4\sin^{-1}\!\left(\dfrac{x^2}{x^2+1}\right)$ is

[0,2\pi]

[0,\pi]

[0,2\pi)

[0,\pi)

Step-by-Step Solution

Key Concept: $\frac{x^2}{x^2+1}=1-\frac{1}{x^2+1}\in[0,1)$ for $x\in\mathbb{R}$ (equals $0$ when $x=0$, approaches but never reaches $1$).

Range $=[0,2\pi)$.

Correct Answer: 3