The range of f(x)=4 sin ^{-1}left(frac{x^2}{x | vedclass

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(B) Let $u = \frac{x^2}{x^2+1}$.Since $x^2 \ge 0$,we have $x^2+1 \ge 1$,so $0 \le \frac{x^2}{x^2+1} < 1$.Thus,the range of $u$ is $[0, 1)$.Now,$f(x) =

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$[0, 2\pi)$

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(B) Let $u = \frac{x^2}{x^2+1}$.Since $x^2 \ge 0$,we have $x^2+1 \ge 1$,so $0 \le \frac{x^2}{x^2+1} Thus,the range of $u$ is $[0, 1)$.Now,$f(x) = 4 \sin^{-1}(u)$.Since $u \in [0, 1)$,$\sin^{-1}(u) \in [\sin^{-1}(0), \sin^{-1}(1)) = [0, \frac{\pi}{2})$.Multiplying by $4$,we get $f(x) \in [4 	imes 0, 4 	imes \frac{\pi}{2}) = [0, 2\pi)$.Therefore,the range of $f(x)$ is $[0, 2\pi)$.

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

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