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Range of the function $f(x)=\frac{x^2}{x^2+1}$ is
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$[0,1)$
Let $y=\frac{x^2}{x^2+1}$
$\Rightarrow(y-1) x^2+0 x+y=1, y \neq 1$ for real values of $x$,
we have $D \geq 0 \Rightarrow-4 y(y-1) \geq 0 \Rightarrow y(y-1) \leq 0 \Rightarrow y \in[0,1)$
$0 \leq \frac{x^2}{x^2+1} \lt 1$
$\Rightarrow(y-1) x^2+0 x+y=1, y \neq 1$ for real values of $x$,
we have $D \geq 0 \Rightarrow-4 y(y-1) \geq 0 \Rightarrow y(y-1) \leq 0 \Rightarrow y \in[0,1)$
$0 \leq \frac{x^2}{x^2+1} \lt 1$
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