JEE Main 2024 — Functions Question with Solution

$f(x)=\frac{(x+5)(x-3)}{x^2-4x+9}$
$\begin{array}{rl}& \text{ Let }g(x)=x^2-4x+9\\ & D<0\\ & g(x)>0\text{ for }x\in R\end{array}$
$\therefore\left[\begin{array}{l} \mathrm{f}(-5)=0 \\ \mathrm{f}(3)=0 \end{array}\right.$<br/>So,$\mathrm{f}(\mathrm{x})$ is many-one. again, $\begin{aligned} & y x^2-4 x y+9 y=x^2+2 x-15 \\ & x^2(y-1)-2 x(2 y+1)+(9 y+15)=0 \\ & \text { for } \forall x \in R \Rightarrow D \geq 0 \\ & D=4(2 y+1)^2-4(y-1)(9 y+15) \geq 0 \\ & 5 y^2+2 y+16 \leq 0 \\ & (5 y-8)(y+2) \leq 0 \end{aligned}$