JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Sets Relations and Functions

For the square root to be defined the logarithm must be non-negative, so ${\log }_{0.6}\left|\frac{2x-5}{x^2-4}\right|\ge 0$. Since the base $0.6<1$, this forces $0<\left|\frac{2x-5}{x^2-4}\right|\le 1$, which already requires $x\ne \pm 2$ and $x\ne \frac{5}{2}$. The inequality $\left|\frac{2x-5}{x^2-4}\right|\le 1$ is equivalent to $(2x-5{)}^2\le (x^2-4{)}^2$, i.e. $(x^2-4{)}^2-(2x-5{)}^2\ge 0$. Factorising, $[(x^2-4)-(2x-5)][(x^2-4)+(2x-5)]=(x-1{)}^2(x^2+2x-9)\ge 0$. As $(x-1{)}^2\ge 0$, this holds when $x^2+2x-9\ge 0$, that is $x\le -1-\sqrt{10}$ or $x\ge -1+\sqrt{10}$, together with the isolated point $x=1$ at which the square vanishes. Discarding $x=\frac{5}{2}$ (which lies in $x\ge -1+\sqrt{10}$) breaks that ray, giving the domain $(-\infty ,-1-\sqrt{10}]\cup \{1\}\cup [-1+\sqrt{10},\frac{5}{2})\cup (\frac{5}{2},\infty )$. Hence $a=-1-\sqrt{10},b=1,c=-1+\sqrt{10},d=e=\frac{5}{2}$, and $a+b+c+d+e=(-1-\sqrt{10})+1+(-1+\sqrt{10})+\frac{5}{2}+\frac{5}{2}=4$.