JEE Main 2026 · 4 Apr · Shift 1 · Mathematics Q2

MathematicsSingle correct+4 / -1Question 2 of 75

If the set of all solutions of $|x^2+x-9|=|x|+|x^2-9|$ is $[\alpha,\beta]\cup[\gamma,\infty)$ , then $(\alpha^2+\beta^2+\gamma^2)$ is equal to:

Answer and worked solution

Official answer: B

Triangle equality. $x^2+x-9=x+(x^2-9)$ , so $|a+b|=|a|+|b|$ holds exactly when $x(x^2-9)\ge0$ .

Solve. $x\in[-3,0]\cup[3,\infty)$ , so $\alpha^2+\beta^2+\gamma^2=9+0+9=18$ .

(B)

Topics:Inequalities and modulus

JEE Main 2026 · 4 Apr · Shift 1, Mathematics question 2