JEE Main 2022 (Online) 26th June Evening ShiftQuestion 1
Let $f:\mathbf{R}\to\mathbf{R}$ be defined as $f(x)=x-1$ and $g:\mathbf{R}-\{1,-1\}\to\mathbf{R}$ be defined as $g(x)=\frac{x^{2}}{x^{2}-1}$. Then the function $f\circ g$ is :
Functions
Solution
$f\circ g(x)=\frac{x^2}{x^2-1}-1=\frac1{x^2-1}$, even so not one-one; $x^2-1\in[-1,0)\cup(0,\infty)$ so range excludes $(-1,0)$, hence not onto.