JEE Main 2026 — Apr 4, Shift 1 — Mathematics solution

JEE Main2026Apr 4, Shift 1Mathematics

If the set of all solutions of $∣ x^{2} + x - 9∣ = ∣ x ∣ + ∣ x^{2} - 9∣$ is $[α, β] \cup [γ, \infty)$ , then $(α^{2} + β^{2} + γ^{2})$ is equal to :

Solution

The identity $∣ a + b ∣ = ∣ a ∣ + ∣ b ∣$ holds if and only if $ab \geq 0$ . Setting $a = x$ and $b = x^{2} - 9$ , the condition becomes

$x (x^{2} - 9) \geq 0 \Rightarrow x (x - 3) (x + 3) \geq 0.$

A sign chart yields $x \in [- 3, 0] \cup [3, \infty)$ , so $α = - 3$ , $β = 0$ , $γ = 3$ .

$∴ α^{2} + β^{2} + γ^{2} = 9 + 0 + 9 = 18.$

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.