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Q.
The function $f: R \rightarrow\left[-\frac{1}{2}, \frac{1}{2}\right]$ defined as $ f\left(x\right) = \frac{x}{1+x^{2}}, $ is :
JEE MainJEE Main 2017Relations and Functions - Part 2
Solution:
$f\left(x\right) = \frac{x}{1+x^{2}}$
$f'\left(x\right) = \frac{\left(1+x\right)^{2}\cdot1-x\cdot2x}{\left(1+x^{2}\right)^{2}} = \frac{1-x^{2}}{\left(1+x^{2}\right)^{2}}$
f'(x) changes sign in different intervals.
$\therefore $ Not injective.
$y = \frac{x}{1+x^{2}}$
$yx^{2} - x+y = 0$
For y $\ne$ 0
$D = 1-4y^{2} \ge 0 \Rightarrow y \in\left[-\frac{1}{2}, \frac{1}{2}\right]-\left\{0\right\}$
For, y = 0 $\Rightarrow x = 0$
$\therefore $ Part of range
$\therefore $ Range : $\left[-\frac{1}{2}, \frac{1}{2}\right]$
$\therefore $ Surjective but not injective.