Search any question & find its solution
Question:
Answered & Verified by Expert
The range of the function $\mathrm{f}(x)=\frac{x^2}{x^2+1}$ is
Options:
Solution:
1328 Upvotes
Verified Answer
The correct answer is:
$[0,1)$
$$
\begin{aligned}
& \text { Let } y=\frac{x^2}{x^2+1} \\
& \Rightarrow y x^2+y=x^2 \\
& \Rightarrow x^2(y-1)+y=0 \\
& \Rightarrow x^2=\frac{y}{1-y}
\end{aligned}
$$
For $x$ to be real,
$$
\begin{aligned}
& y(1-y) \geq 0 \text { and } 1-y \neq 0 \\
& \Rightarrow y(y-1) \leq 0 \text { and } y \neq 1 \\
& \Rightarrow 0 \leq y < 1
\end{aligned}
$$
\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}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.