JEE Main 2026 — 2 April, Morning Shift — Question 37

For domain $\log _{0.6}\left|\frac{2 \mathrm{x}-5}{\mathrm{x}^{2}-4}\right| \geq 0$ $\left|\frac{2 x-5}{x^{2}-4}\right| \leq 1 \& x \neq \frac{5}{2} \ldots$. $-1 \leq \frac{2 \mathrm{x}-5}{\mathrm{x}^{2}-4} \leq 1$ $\frac{2 \mathrm{x}-5}{\mathrm{x}^{2}-4}+1 \geq 0$ $\frac{\mathrm{x}^{2}+2 \mathrm{x}-9}{\mathrm{x}^{2}-4} \geq 0$ $\frac{(x+1)^{2}-10}{(x-2)(x+2)} \geq 0$

\mathrm{x} \in(-\infty,-1-\sqrt{10}] \cup(-2,2) \cup[-1+\sqrt{10}, \infty) \end{gathered}$$ $\frac{2 x-5}{x^{2}-4}-1 \leq 0$ $\frac{2 x-5-x^{2}+4}{x^{2}-4} \leq 0$ $\frac{\mathrm{x}^{2}-2 \mathrm{x}+1}{\mathrm{x}^{2}-4} \geq 0$ $\frac{(\mathrm{x}-1)^{2}}{(\mathrm{x}-2)(\mathrm{x}+2)} \geq 0$ $$\begin{gathered} \mathrm{x} \in(-\infty,-2) \cup(2, \infty) \cup\{1\} \end{gathered}$$ $(1) \cap(2) \cap(3)$ $\mathrm{x} \in(-\infty,-1-\sqrt{10}] \cup\{1\} \cup\left[-1+\sqrt{10}, \frac{5}{2}\right) \cup\left(\frac{5}{2}, \infty\right)$ $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}=-2+1+5=4$