JEE Main 2026 Apr 2 Shift 1, Mathematics Q21: Introduction Of Functions

The square root needs ${\log }_{0.6}\left|\frac{2x-5}{x^2-4}\right|\ge 0$. Since the base $0.6<1$, the inequality flips when the log is removed:

$0<\left|\frac{2x-5}{x^2-4}\right|\le 1$, with $x\ne \pm 2$ and $x\ne 5/2$.

Upper bound $|(2x-5)/(x^2-4)|\le 1$ means both $-1\le (2x-5)/(x^2-4)$ and $(2x-5)/(x^2-4)\le 1$.

From $(2x-5)/(x^2-4)+1\ge 0$: $\frac{x^2+2x-9}{x^2-4}\ge 0$, i.e. $\frac{(x+1{)}^2-10}{(x-2)(x+2)}\ge 0$. Sign analysis on $\mathbb{R}\setminus \{\pm 2\}$ gives $x\in (-\infty ,-1-\sqrt{10}]\cup (-2,2)\cup [-1+\sqrt{10},\infty )$.

From $(2x-5)/(x^2-4)-1\le 0$: $\frac{x^2-2x+1}{x^2-4}\ge 0\iff \frac{(x-1{)}^2}{(x-2)(x+2)}\ge 0$, i.e. $x\in (-\infty ,-2)\cup (2,\infty )\cup \{1\}$.

Intersecting both conditions and excluding $x=\pm 2$ and $x=5/2$ gives

$x\in (-\infty ,-1-\sqrt{10}]\cup \{1\}\cup [-1+\sqrt{10},5/2)\cup (5/2,\infty )$.

Matching with the given form: $a=-1-\sqrt{10}$, $b=1$, $c=-1+\sqrt{10}$, $d=5/2$, $e=5/2$. Therefore

$a+b+c+d+e=(-1-\sqrt{10})+1+(-1+\sqrt{10})+5/2+5/2=-1+5=4$.