JEE Main 2024 — 6 April, Shift 1 — Question 5

$f(x)=\frac{(x+5)(x-3)}{x^{2}-4 x+9}$

Let $g(x)=x^{2}-4 x+9$

$\mathrm{D}<0$

$\mathrm{g}(\mathrm{x})>0$ for $\mathrm{x} \in \mathrm{R}$

$\therefore\left[\begin{array}{l}\mathrm{f}(-5)=0 , \mathrm{f}(3)=0\end{array}]\right.$

So, $\mathrm{f}(\mathrm{x})$ is many-one.

again,

$y x^{2}-4 x y+9 y=x^{2}+2 x-15$

$x^{2}(y-1)-2 x(2 y+1)+(9 y+15)=0$ for

$\forall \mathrm{x} \in \mathrm{R} \Rightarrow \mathrm{D} \geq 0$

$\mathrm{D}=4(2 \mathrm{y}+1)^{2}-4(\mathrm{y}-1)(9 \mathrm{y}+15) \geq 0$

$5 y^{2}+2 y+16 \leq 0$

$(5 y-8)(y+2) \leq 0$

$\mathrm{y} \in\left[-2, \frac{8}{5}\right]$ : range

Note : If function is defined from $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$

then only correct answer is option (3)