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

We need the range of $$f(x) = 4\sin^{-1}\left(\frac{x^2}{x^2+1}\right)$$.

Let $$t = \frac{x^2}{x^2+1} = 1 - \frac{1}{x^2+1}$$.

Since $$x^2 \geq 0$$: when $$x = 0$$, $$t = 0$$. As $$|x| \to \infty$$, $$t \to 1$$.

So $$t \in [0, 1)$$.

$$\sin^{-1}(t)$$ for $$t \in [0, 1)$$ gives values in $$[0, \pi/2)$$.

$$f(x) = 4\sin^{-1}(t)$$ ranges over $$[0, 2\pi)$$.

Note: $$t$$ never equals 1 (since $$x^2/(x^2+1) < 1$$ for all finite $$x$$), so $$\sin^{-1}(t)$$ never equals $$\pi/2$$, hence $$f(x)$$ never reaches $$2\pi$$.

The range is $$[0, 2\pi)$$.

This matches option 3.