JEE Main 2026 — Mathematics Algebra

For the function $f(x) = \sqrt{\log_{0.6}\left(\left|\dfrac{2x-5}{x^2-4}\right|\right)}$f(x)=log0.6​(​x2−42x−5​​)​ to be defined, two conditions must be satisfied:

1. The argument of the logarithm must be strictly positive:

$$\left|\dfrac{2x-5}{x^2-4}\right| > 0 \implies 2x - 5 \neq 0 \implies x \neq \dfrac{5}{2}$$​x2−42x−5​​>0⟹2x−5=0⟹x=25​

Also, the denominator cannot be zero:

$$x^2 - 4 \neq 0 \implies x \neq \pm 2$$x2−4=0⟹x=±2

1. The expression inside the square root must be non-negative:

$$\log_{0.6}\left(\left|\dfrac{2x-5}{x^2-4}\right|\right) \ge 0$$log0.6​(​x2−42x−5​​)≥0

Since the base of the logarithm is $0.6 < 1$0.6<1, the inequality sign reverses when removing the logarithm:

$$\left|\dfrac{2x-5}{x^2-4}\right| \le (0.6)^0$$​x2−42x−5​​≤(0.6)0

$$\left|\dfrac{2x-5}{x^2-4}\right| \le 1$$​x2−42x−5​​≤1

Since both sides are non-negative, we can square both sides:

$$(2x-5)^2 \le (x^2-4)^2$$(2x−5)2≤(x2−4)2

$$(x^2-4)^2 - (2x-5)^2 \ge 0$$(x2−4)2−(2x−5)2≥0

Using the identity $A^2 - B^2 = (A-B)(A+B)$A2−B2=(A−B)(A+B):

$$(x^2 - 4 - 2x + 5)(x^2 - 4 + 2x - 5) \ge 0$$(x2−4−2x+5)(x2−4+2x−5)≥0

$$(x^2 - 2x + 1)(x^2 + 2x - 9) \ge 0$$(x2−2x+1)(x2+2x−9)≥0

$$(x - 1)^2 (x^2 + 2x - 9) \ge 0$$(x−1)2(x2+2x−9)≥0

Since $(x - 1)^2 \ge 0$(x−1)2≥0 for all real $x$x, the inequality holds if:

$$x - 1 = 0 \implies x = 1$$x−1=0⟹x=1

or

$$x^2 + 2x - 9 \ge 0$$x2+2x−9≥0

Finding the roots of $x^2 + 2x - 9 = 0$x2+2x−9=0 using the quadratic formula:

$$x = \dfrac{-2 \pm \sqrt{4 - 4(1)(-9)}}{2} = \dfrac{-2 \pm \sqrt{40}}{2} = -1 \pm \sqrt{10}$$x=2−2±4−4(1)(−9)​​=2−2±40​​=−1±10​

Thus, $x^2 + 2x - 9 \ge 0$x2+2x−9≥0 gives:

$$x \in (-\infty, -1 - \sqrt{10}] \cup [-1 + \sqrt{10}, \infty)$$x∈(−∞,−1−10​]∪[−1+10​,∞)

Combining this with $x = 1$x=1, the solution to the inequality is:

$$x \in (-\infty, -1 - \sqrt{10}] \cup \{1\} \cup [-1 + \sqrt{10}, \infty)$$x∈(−∞,−1−10​]∪{1}∪[−1+10​,∞)

Now, we must exclude the restricted values $x = \pm 2$x=±2 and $x = \dfrac{5}{2}$x=25​.

Note that $-1 - \sqrt{10} \approx -4.16$−1−10​≈−4.16 and $-1 + \sqrt{10} \approx 2.16$−1+10​≈2.16.

The values $-2$−2 and $2$2 do not fall in the above intervals, so they are already excluded.

However, $\dfrac{5}{2} = 2.5$25​=2.5, which lies in the interval $[-1 + \sqrt{10}, \infty)$[−1+10​,∞). We must exclude it by splitting the interval:

$$[-1 + \sqrt{10}, \dfrac{5}{2}) \cup (\dfrac{5}{2}, \infty)$$[−1+10​,25​)∪(25​,∞)

The final domain of the function is:

$$(-\infty, -1 - \sqrt{10}] \cup \{1\} \cup [-1 + \sqrt{10}, \dfrac{5}{2}) \cup (\dfrac{5}{2}, \infty)$$(−∞,−1−10​]∪{1}∪[−1+10​,25​)∪(25​,∞)

Comparing this with the given domain $(-\infty, a] \cup \{b\} \cup [c, d) \cup (e, \infty)$(−∞,a]∪{b}∪[c,d)∪(e,∞), we get:

$$a = -1 - \sqrt{10}$$a=−1−10​

$$b = 1$$b=1

$$c = -1 + \sqrt{10}$$c=−1+10​

$$d = \dfrac{5}{2}$$d=25​

$$e = \dfrac{5}{2}$$e=25​

We need to find the value of $a + b + c + d + e$a+b+c+d+e:

$$a + b + c + d + e = (-1 - \sqrt{10}) + 1 + (-1 + \sqrt{10}) + \dfrac{5}{2} + \dfrac{5}{2}$$a+b+c+d+e=(−1−10​)+1+(−1+10​)+25​+25​

$$= -1 - \sqrt{10} + 1 - 1 + \sqrt{10} + 5$$=−1−10​+1−1+10​+5

$$= 4$$=4

Answer: $4$4