JEE Main 2025 (Online) 3rd April Morning ShiftQuestion 1
If the domain of the function $f(x)=\log_e\left(\frac{2x-3}{5+4x}\right)+\sin^{-1}\left(\frac{4+3x}{2-x}\right)$ is $[\alpha,\beta)$, then $\alpha^{2}+4\beta$ is equal to
Functions
Solution
$\frac{2x-3}{4x+5}>0\Rightarrow x<-\frac54$ or $x>\frac32$; $|3x+4|\le|2-x|\Rightarrow(4x+2)(2x+6)\le0\Rightarrow-3\le x\le-\frac12$. Domain $\left[-3,-\frac54\right)$, so $9-5=4$.