JEE Main 2024 (Online) 1st February Evening ShiftQuestion 1
If the domain of the function $f(x)=\frac{\sqrt{x^{2}-25}}{(4-x^{2})}+\log_{10}\left(x^{2}+2x-15\right)$ is $(-\infty,\alpha)\cup[\beta,\infty)$, then $\alpha^{2}+\beta^{3}$ is equal to :
Functions
Solution
$x^2\ge25$ gives $x\le-5$ or $x\ge5$; $x^2+2x-15>0$ gives $x<-5$ or $x>3$. Domain $(-\infty,-5)\cup[5,\infty)$ (and $x\ne\pm2$ is automatic), so $25+125=150$.