If the set of all solutions of $|{x}^{2}+x-9|=|x|+|{x}^{2}-9|$is $[\alpha ,\beta ]\cup [\gamma ,\infty ),$ then $\left({\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}\right)$ is equal to: [JEE Main 2026, 4 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: b) $18$. We know if $ |\mathrm{a}|+|\mathrm{b}|=|\mathrm{a}+\mathrm{b}|$ Then $ \mathrm{a} \cdot \mathrm{~b} \geq 0 $ $ \Rightarrow \mathrm{x}\left(\mathrm{x}^2-9\right) \geq 0$ $ \mathrm{x} \in[-3,0] \cup[3, \infty)$ $ \alpha^2+\beta^2+\gamma^2=18$

If the set of all solutions of \(|{x}^{2}+x-9|=|x|+|{x}^{2}-9|\)is \([\alpha ,\beta ]\cup [\gamma ,\infty ),\) then \(\l…

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