Question:
The function \(f:R\to\left[-\frac{1}{2},\frac{1}{2}\right]\) defined as \(f(x)=\frac{x}{1+x^2}\), is
The function \(f:R\to\left[-\frac{1}{2},\frac{1}{2}\right]\) defined as \(f(x)=\frac{x}{1+x^2}\), is
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Check one-one using two different inputs with the same output, and find the range by treating \(y=\frac{x}{1+x^2}\) as a quadratic in \(x\).
Updated On: Nov 11, 2026
- invertible
- injective but not surjective.
- surjective but not injective
- neither injective nor surjective.
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The Correct Option is C
Solution and Explanation
Injectivity: \(f(2)=\frac{2}{5}\) and \(f\left(\frac{1}{2}\right)=\frac{1/2}{1+1/4}=\frac{2}{5}\), so two different inputs give the same output and \(f\) is not injective.
Range: let \(y=\frac{x}{1+x^2}\), so \(yx^2-x+y=0\). For real \(x\) the discriminant must satisfy \(1-4y^2\ge 0\), i.e. \(-\frac{1}{2}\le y\le \frac{1}{2}\).
Hence the range equals the codomain \(\left[-\frac{1}{2},\frac{1}{2}\right]\), so \(f\) is surjective.
Therefore \(f\) is surjective but not injective.
Answer: (3) surjective but not injective
Range: let \(y=\frac{x}{1+x^2}\), so \(yx^2-x+y=0\). For real \(x\) the discriminant must satisfy \(1-4y^2\ge 0\), i.e. \(-\frac{1}{2}\le y\le \frac{1}{2}\).
Hence the range equals the codomain \(\left[-\frac{1}{2},\frac{1}{2}\right]\), so \(f\) is surjective.
Therefore \(f\) is surjective but not injective.
Answer: (3) surjective but not injective
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