JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Sets Relations and Functions

$f(x)=\ln x$ for $x>0$, $f(x)=e^{-x}$ for $x\le 0$; $g(x)=x$ for $x\ge 0$, $g(x)=e^x$ for $x<0$.

For $x\ge 1$: $f(x)=\ln x\ge 0\Rightarrow g(f(x))=\ln x$.
For $0<x<1$: $f(x)=\ln x<0\Rightarrow g(f(x))=e^{\ln x}=x$.
For $x\le 0$: $f(x)=e^{-x}\ge 1>0\Rightarrow g(f(x))=e^{-x}$.

Not one-one: At $x=2$ (branch $x\ge 1$), $g(f(2))=\ln 2$. At $x=\ln 2\approx 0.69$ (branch $0<x<1$), $g(f(\ln 2))=\ln 2$. Two different inputs give the same output, so $g\circ f$ is not one-one.

Not onto: The three branches give ranges $[0,\infty )$ (from $x\ge 1$), $(0,1)$ (from $0<x<1$), and $[1,\infty )$ (from $x\le 0$); the union is $[0,\infty )$, which is not all of $\mathbb{R}$ (no negative value is attained). So $g\circ f$ is not onto.

Hence $g\circ f$ is neither one-one nor onto.