JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Sets Relations and Functions

$f(x)=\frac{x^2+2x-15}{x^2-4x+9}$. The denominator has discriminant $<0$, so it is positive for all real $x$ and $f$ is defined on $\mathbb{R}$.
$f(-5)=0$ and $f(3)=0$, so $f$ is many-one.
Solving $y=f(x)$ for real $x$ requires $D=4(2y+1{)}^2-4(y-1)(9y+15)\ge 0\Rightarrow (5y-8)(y+2)\le 0\Rightarrow y\in \left[-2,\frac{8}{5}\right]$, so the range is a proper subset of $\mathbb{R}$ (not onto $\mathbb{R}$).
With codomain $\mathbb{R}$ the description is 'neither one-one nor onto'; Allen marks the question Bonus.