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Functions - JEE Mathematics question (JEE Main 2019 (Online) 11th January Morning Shift)

JEE Main 2019 (Online) 11th January Morning ShiftQuestion 1
Let $f:R\to R$ be defined by $f(x)=\frac{x}{1+x^{2}}$, $x\in R$. Then the range of $f$ is
Functions
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Solution

$yx^2-x+y=0$ has real $x$ iff $1-4y^2\ge0$, so range is $\left[-\frac12,\frac12\right]$.

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