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JEE Main 20264 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
4 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Basics of Mathematics

JEE Main 2026 · 4 April · Morning Shift · Mathematics · Basics of Mathematics

If the set of all solutions of |x^2 + x - 9| = |x| + |x^2 - 9| is...

If the set of all solutions of ∣x2+x−9∣=∣x∣+∣x2−9∣|x^2 + x - 9| = |x| + |x^2 - 9| is [α,β]∪[γ,∞)[\alpha, \beta] \cup [\gamma, \infty), then (α2+β2+γ2)(\alpha^2 + \beta^2 + \gamma^2) is equal to:

Options

  1. A99
  2. B1818
  3. C3636
  4. D7272

Correct answer

B. 1818

Step-by-step solution

The given equation is ∣x2+x−9∣=∣x∣+∣x2−9∣|x^2 + x - 9| = |x| + |x^2 - 9|.

Let A=xA = x and B=x2−9B = x^2 - 9. Then A+B=x2+x−9A + B = x^2 + x - 9.

The equation is of the form ∣A+B∣=∣A∣+∣B∣|A + B| = |A| + |B|, which holds true if and only if A⋅B≥0A \cdot B \ge 0.

Substituting the values of AA and BB:

x(x2−9)≥0x(x^2 - 9) \ge 0

x(x−3)(x+3)≥0x(x - 3)(x + 3) \ge 0

The critical points are −3,0,3-3, 0, 3. Using the wavy curve method, the solution set is:

x∈[−3,0]∪[3,∞)x \in [-3, 0] \cup [3, \infty)

Comparing this with the given solution set [α,β]∪[γ,∞)[\alpha, \beta] \cup [\gamma, \infty), we get:

α=−3\alpha = -3, β=0\beta = 0, γ=3\gamma = 3

We need to find the value of α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2:

α2+β2+γ2=(−3)2+02+32=9+0+9=18\alpha^2 + \beta^2 + \gamma^2 = (-3)^2 + 0^2 + 3^2 = 9 + 0 + 9 = 18

Answer: 1818
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