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Domain of a Log-Sqrt Function: a+b+c+d+e = 4 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
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
Solution
Answer: 4 (± 0.01)
Step 1: For the square root, need log⁡0.6∣2x−5x2−4∣≥0\log_{0.6}\left|\dfrac{2x-5}{x^2-4}\right|\ge0. Since base 0.6<10.6<1, this means
∣2x−5x2−4∣≤1,with x≠±2 and x≠52.\left|\frac{2x-5}{x^2-4}\right|\le1,\qquad\text{with}\ x\ne\pm2\ \text{and}\ x\ne\tfrac52.
 Step 2: Solve 2x−5x2−4≥−1\dfrac{2x-5}{x^2-4}\ge-1:
2x−5x2−4+1≥0 ⇒ x2+2x−9x2−4≥0 ⇒ (x+1)2−10(x−2)(x+2)≥0.\frac{2x-5}{x^2-4}+1\ge0\ \Rightarrow\ \frac{x^2+2x-9}{x^2-4}\ge0\ \Rightarrow\ \frac{(x+1)^2-10}{(x-2)(x+2)}\ge0.
 Sign analysis (critical points −1−10, −2, 2, −1+10-1-\sqrt{10},\ -2,\ 2,\ -1+\sqrt{10}) gives
x∈(−∞, −1−10 ]∪(−2,2)∪[ −1+10, ∞).x\in(-\infty,\,-1-\sqrt{10}\,]\cup(-2,2)\cup[\,-1+\sqrt{10},\,\infty).
 Step 3: Solve 2x−5x2−4≤1\dfrac{2x-5}{x^2-4}\le1:
2x−5x2−4−1≤0 ⇒ −x2+2x−1x2−4≤0 ⇒ (x−1)2(x−2)(x+2)≥0.\frac{2x-5}{x^2-4}-1\le0\ \Rightarrow\ \frac{-x^2+2x-1}{x^2-4}\le0\ \Rightarrow\ \frac{(x-1)^2}{(x-2)(x+2)}\ge0.
 This gives
x∈(−∞,−2)∪(2,∞)∪{1}.x\in(-\infty,-2)\cup(2,\infty)\cup\{1\}.
 Step 4: Intersect the two solution sets (and exclude x=52x=\tfrac52):
x∈(−∞, −1−10 ]∪{1}∪[ −1+10, 52)∪(52,∞).x\in\left(-\infty,\,-1-\sqrt{10}\,\right]\cup\{1\}\cup\left[\,-1+\sqrt{10},\,\tfrac52\right)\cup\left(\tfrac52,\infty\right).
 Step 5: Match to (−∞,a]∪{b}∪[c,d)∪(e,∞)(-\infty,a]\cup\{b\}\cup[c,d)\cup(e,\infty):
a=−1−10, b=1, c=−1+10, d=52, e=52.a=-1-\sqrt{10},\ b=1,\ c=-1+\sqrt{10},\ d=\tfrac52,\ e=\tfrac52.
 Sum:
a+b+c+d+e=(−1−10)+1+(−1+10)+52+52=−1+5=4.a+b+c+d+e=(-1-\sqrt{10})+1+(-1+\sqrt{10})+\tfrac52+\tfrac52=-1+5=4.
 Correct answer: 4
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