If the set of all solutions of x 2 + x - g = x + x 2 - g is [α ,β ]cup [γ ,∞) , then (α 2 +β 2 +γ 2) is equal to:

Mathematics · Basic Maths

JEE Main 2026 — 4 April, Morning Shift — Question 25

If the set of all solutions of $\left|\mathbf{x}^2 +\mathbf{x} - \mathbf{g}\right| = \left|\mathbf{x}\right| + \left|\mathbf{x}^2 -\mathbf{g}\right|$ is $[\alpha ,\beta ]\cup [\gamma ,\infty)$ , then $(\alpha^2 +\beta^2 +\gamma^2)$ is equal to:

Option A:
9
Option B:
18
Correct
Option C:
36
Option D:
72

Answer: B

Step-by-step solution

Given: $|x^2 + x - 9| = |x| + |x^2 - 9|$ . Recall: $|a| + |b| = |a + b|$ if and only if $a \cdot b \ge 0$ . Here, $a = x$ , $b = x^2 - 9$ . Condition: $x(x^2 - 9) \ge 0$ . Factor: $x(x-3)(x+3) \ge 0$ . Solve inequality using sign chart: intervals $(-\infty, -3]$ , $[-3,0]$ , $[0,3]$ , $[3,\infty)$ . Signs: $x(x-3)(x+3) \ge 0$ for $x \in [-3,0] \cup [3,\infty)$ . Thus, $\alpha = -3$ , $\beta = 0$ , $\gamma = 3$ . Compute: $\alpha^2 + \beta^2 + \gamma^2 = 9 + 0 + 9 = 18$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 4 April, Morning Shift
Subject: Mathematics
Chapter: Basic Maths
Topic: Modulus Function