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Mathematics · JEE | RELATIONS AND FUNCTIONS
हिन्दीLet \( f : \mathbb{R} \to \mathbb{R} \) be defined by \( f(x) = \frac{x}{x^2 + 1} \). Then the range of \( f \) is:
View Answer & Solution
Correct answer · Option B
\( [-1/2, 1/2] \)
Explanation
Setting \( y = \frac{x}{x^2+1} \implies yx^2 - x + y = 0 \). For real \( x \), discriminant \( \Delta = 1 - 4y^2 \ge 0 \implies y^2 \le 1/4 \implies y \in [-1/2, 1/2] \).
