JEE Main 2025 (Online) 7th April Evening ShiftQuestion 1
If the range of the function $f(x)=\frac{5-x}{x^{2}-3x+2}$, $x\neq1,2$, is $(-\infty,\alpha]\cup[\beta,\infty)$, then $\alpha^{2}+\beta^{2}$ is equal to :
Functions
Solution
$y(x^2-3x+2)=5-x$ gives $yx^2+(1-3y)x+2y-5=0$; real $x$ needs $y^2+14y+1\ge0$, so $y\le-7-4\sqrt3$ or $y\ge-7+4\sqrt3$. $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=196-2=194$.