JEE Main 2025 — 3 April, Morning Shift — Question 27

Given function is

$f(x)=\log _{e}\left(\frac{2 x-3}{5+4 x}\right)+\sin ^{-1}\left(\frac{4+3 x}{2-x}\right)$

For domain, the conditions are

$\frac{2 x-3}{5+4 x}>0$ and $\left|\frac{4+3 x}{2-x}\right| \leq 1$

Now, $\frac{2 x-3}{5+4 x}>0 \Rightarrow x \in\left(-\infty,-\frac{5}{4}\right) \cup\left[\frac{3}{2}, \infty\right)$

and $-1 \leq \frac{4+3 x}{2-x} \leq 1$

$\Rightarrow\left(-1 \leq \frac{4+3 x}{2-x}\right) \cap\left(\frac{4+3 x}{2-x} \leq 1\right)$

$\Rightarrow\left(\frac{6+2 x}{2-x} \geq 0\right) \cap\left(\frac{2+4 x}{2-x} \leq 0\right)$

$\Rightarrow \frac{6+2 x}{2-x} \cdot \frac{2+4 x}{2-x} \leq 0$

$\Rightarrow \mathrm{x} \in\left[-3,-\frac{1}{2}\right]$

Hence, we get the domain of f as $\mathrm{x} \in\left[-3,-\frac{5}{4}\right)$

This means that $\alpha=-3, \beta=-\frac{5}{4}$

Thus, $\alpha^{2}+4 \beta=9-5=4$