JEE Main 2026 — Functions Question with Solution

For the function $f(x)=\sqrt{{\log }_{0.6}\left(\left|\frac{2x-5}{x^2-4}\right|\right)}$ to be defined, two conditions must be satisfied:

1. The argument of the logarithm must be strictly positive:
$\left|\frac{2x-5}{x^2-4}\right|>0⟹2x-5\ne 0⟹x\ne \frac{5}{2}$
Also, the denominator cannot be zero:
$x^2-4\ne 0⟹x\ne \pm 2$

2. The expression inside the square root must be non-negative:
${\log }_{0.6}\left(\left|\frac{2x-5}{x^2-4}\right|\right)\ge 0$

Since the base of the logarithm is $0.6<1$, the inequality sign reverses when removing the logarithm:
$\left|\frac{2x-5}{x^2-4}\right|\le (0.6{)}^0$
$\left|\frac{2x-5}{x^2-4}\right|\le 1$

Since both sides are non-negative, we can square both sides:
$(2x-5{)}^2\le (x^2-4{)}^2$
$(x^2-4{)}^2-(2x-5{)}^2\ge 0$

Using the identity $A^2-B^2=(A-B)(A+B)$:
$(x^2-4-2x+5)(x^2-4+2x-5)\ge 0$
$(x^2-2x+1)(x^2+2x-9)\ge 0$
$(x-1{)}^2(x^2+2x-9)\ge 0$

Since $(x-1{)}^2\ge 0$ for all real $x$, the inequality holds if:
$x-1=0⟹x=1$
or
$x^2+2x-9\ge 0$

Finding the roots of $x^2+2x-9=0$ using the quadratic formula:
$x=\frac{-2\pm \sqrt{4-4(1)(-9)}}{2}=\frac{-2\pm \sqrt{40}}{2}=-1\pm \sqrt{10}$

Thus, $x^2+2x-9\ge 0$ gives:
$x\in (-\infty ,-1-\sqrt{10}]\cup [-1+\sqrt{10},\infty )$

Combining this with $x=1$, the solution to the inequality is:
$x\in (-\infty ,-1-\sqrt{10}]\cup \{1\}\cup [-1+\sqrt{10},\infty )$

Now, we must exclude the restricted values $x=\pm 2$ and $x=\frac{5}{2}$.
Note that $-1-\sqrt{10}\approx -4.16$ and $-1+\sqrt{10}\approx 2.16$.
The values $-2$ and $2$ do not fall in the above intervals, so they are already excluded.
However, $\frac{5}{2}=2.5$, which lies in the interval $[-1+\sqrt{10},\infty )$. We must exclude it by splitting the interval:
$[-1+\sqrt{10},\frac{5}{2})\cup (\frac{5}{2},\infty )$

The final domain of the function is:
$(-\infty ,-1-\sqrt{10}]\cup \{1\}\cup [-1+\sqrt{10},\frac{5}{2})\cup (\frac{5}{2},\infty )$

Comparing this with the given domain $(-\infty ,a]\cup \{b\}\cup [c,d)\cup (e,\infty )$, we get:
$a=-1-\sqrt{10}$
$b=1$
$c=-1+\sqrt{10}$
$d=\frac{5}{2}$
$e=\frac{5}{2}$

We need to find the value of $a+b+c+d+e$:
$a+b+c+d+e=(-1-\sqrt{10})+1+(-1+\sqrt{10})+\frac{5}{2}+\frac{5}{2}$
$=-1-\sqrt{10}+1-1+\sqrt{10}+5$
$=4$

Answer: $4$