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JEE Main 2026 · 2 Apr · Shift 1, Mathematics question 21

If the domain of the function f(x)=log⁡(0.6)(∣2x−5x2−4∣)f(x)=\sqrt{\log_{(0.6)}\left(\left|\dfrac{2x-5}{x^2-4}\right|\right)} is (−∞,a]∪{b}∪[c,d)∪(e,∞)(-\infty,a]\cup\{b\}\cup[c,d)\cup(e,\infty), then the value of a+b+c+d+ea+b+c+d+e is ________.

Answer and worked solution

Official answer: 4

Condition. log⁡0.6t≥0\log_{0.6}t\ge0 means 0<t≤10<t\le1, with t=∣2x−5x2−4∣t=\left|\frac{2x-5}{x^2-4}\right|: so x≠±2,52x\ne\pm2,\frac52 and ∣2x−5∣≤∣x2−4∣|2x-5|\le|x^2-4|.

Solve the inequality.
(x2−4)2−(2x−5)2=(x2−2x+1)(x2+2x−9)=(x−1)2(x2+2x−9)≥0,(x^2-4)^2-(2x-5)^2=(x^2-2x+1)(x^2+2x-9)=(x-1)^2(x^2+2x-9)\ge0,
so x=1x=1 or x≤−1−10x\le-1-\sqrt{10} or x≥−1+10x\ge-1+\sqrt{10}.

Domain. (−∞,−1−10]∪{1}∪[−1+10,52)∪(52,∞)(-\infty,-1-\sqrt{10}]\cup\{1\}\cup[-1+\sqrt{10},\tfrac52)\cup(\tfrac52,\infty).

Sum. a+b+c+d+e=(−1−10)+1+(−1+10)+52+52=4a+b+c+d+e=(-1-\sqrt{10})+1+(-1+\sqrt{10})+\frac52+\frac52=4.

Answer: 4

Topics:Domain and rangeLogarithmsInequalities and modulus
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