The range of the function $\mathrm{f}(x)=\frac{x^2}{x^2+1}$ is
The range of the function $\mathrm{f}(x)=\frac{x^2}{x^2+1}$ is
- $(0,1)$
- $[0,1)$
- $(0,1]$
- $[0,1]$
Solution
$\begin{aligned}
& \text { Let } y=\frac{x^2}{x^2+1} \\
& \Rightarrow y x^2+y=x^2 \\
& \Rightarrow x^2(y-1)+y=0 \\
& \Rightarrow x^2=\frac{y}{1-y}
\end{aligned}$
For $x$ to be real,
$\begin{aligned}
& y(1-y) \geq 0 \text { and } 1-y \neq 0 \\
& \Rightarrow y(y-1) \leq 0 \text { and } y \neq 1 \\
& \Rightarrow 0 \leq y < 1
\end{aligned}$
Asked in: MHT CET 2023 (14 May Shift 1)
Practice more Functions questions on Aicharya