Question:
Let \(f : \mathbb{R} \to \mathbb{R}\) be defined by \(f(x) = \frac{x}{1+x^2},\ x \in \mathbb{R}\). Then the range of f is :
Let \(f : \mathbb{R} \to \mathbb{R}\) be defined by \(f(x) = \frac{x}{1+x^2},\ x \in \mathbb{R}\). Then the range of f is :
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Use \(1 + x^2 \ge 2|x|\) to bound the function, or treat \(y = f(x)\) as a quadratic in x and demand a real root.
Updated On: Nov 5, 2026
- \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
- \(\mathbb{R} - \left[-\frac{1}{2}, \frac{1}{2}\right]\)
- \((-1, 1) - \{0\}\)
- \(\mathbb{R} - [-1, 1]\)
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The Correct Option is A
Solution and Explanation
Since \(1 + x^2 \ge 2|x|\) (AM-GM), \(\left|\frac{x}{1+x^2}\right| \le \frac12\), with equality at \(x = \pm1\).
f is continuous and odd, \(f(1)=\frac12,\ f(-1)=-\frac12\), so by the intermediate value theorem every value in \(\left[-\frac12, \frac12\right]\) is attained.
Range \(= \left[-\frac12, \frac12\right]\).
Answer: (1) \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
f is continuous and odd, \(f(1)=\frac12,\ f(-1)=-\frac12\), so by the intermediate value theorem every value in \(\left[-\frac12, \frac12\right]\) is attained.
Range \(= \left[-\frac12, \frac12\right]\).
Answer: (1) \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
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