JEE Main 2017 (Offline)Question 1
The function $f:\mathbb R\to\left[-\frac12,\frac12\right]$ defined as $f(x)=\frac{x}{1+x^2}$, is :
Sets, Relations and Function
Solution
$f'(x)=\frac{1-x^2}{(1+x^2)^2}$ gives max $f(1)=\frac12$, min $f(-1)=-\frac12$, and $f$ is continuous, so the range is $[-\frac12,\frac12]$ (surjective). $f(2)=f(\frac12)=\frac25$, so $f$ is not injective.