MHT CET · Maths · Functions
The range of the function \(\mathrm{f}(x)=\frac{x^2}{x^2+1}\) is
- A \((0,1)\)
- B \([0,1)\)
- C \((0,1]\)
- D \([0,1]\)
Answer & Solution
Correct Answer
(B) \([0,1)\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \text {Let } y=\frac{x^2}{x^2+1} \\
& \Rightarrow y x^2+y=x^2 \\
& \Rightarrow x^2(y-1)+y=0 \\
& \Rightarrow x^2=\frac{y}{1-y}
\end{aligned}
\)
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,
\(
\begin{aligned}
& y(1-y) \geq 0 \text { and } 1-y \neq 0 \\
& \Rightarrow y(y-1) \leq 0 \text { and } y \neq 1 \\
& \Rightarrow 0 \leq y < 1
\end{aligned}
\)
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