JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Sets Relations and Functions

For $x\ge 0$: $g(x)=e^x\ge 1$, so $f(g(x))=|e^x-1|=e^x-1\ge 0$, an increasing function.
For $x\le 0$: $g(x)=x+1\le 1$, so $f(g(x))=|x+1-1|=|x|=-x\ge 0$.
Both branches take values in $[0,\infty )$ (never negative), so $f\circ g:\mathbb{R}\to \mathbb{R}$ misses every negative value — not onto.
Also, e.g. $x=1$ gives $f(g(1))=e-1$, and a suitable negative $x=-(e-1)$ gives the same value $e-1$ from the other branch — two different inputs, same output, so not one-one.
Hence $f(g(x))$ is neither one-one nor onto.