Q.2. If the set of all solutions of is , then is equal to :

Sol. We use the property $|a+b| = |a| + |b|$ if and only if $ab \ge 0$. Here, $|(x^2-9) + x| = |x^2-9| + |x|$. So, we must have $x(x^2-9) \ge 0$. This implies $x(x-3)(x+3) \ge 0$. By analyzing the sign of the product, we get $x \in [-3, 0] \cup [3, \infty)$. Comparing this with $[\alpha, \beta] \cup [\gamma,\infty)$, we have $\alpha = -3$, $\beta = 0$, $\gamma = 3$. Therefore, $\alpha^2 + \beta^2 + \gamma^2 = (-3)^2 + 0^2 + 3^2 = 9 + 0 + 9 = 18$.

What is the correct answer for JEE Main Mathematics Q2?

The correct answer is B. Sol. We use the property $|a+b| = |a| + |b|$ if and only if $ab \ge 0$. Here, $|(x^2-9) + x| = |x^2-9| + |x|$. So, we must have $x(x^2-9) \ge 0$. This implies $x(x-3)(x+3) \ge 0$. By analyzing the sign of the product, we get $x \in [-3, 0] \cup [3, \infty)$. Comparing this with $[\alpha, \beta] \cup [\gamma,\infty)$, we have $\alpha = -3$, $\beta = 0$, $\gamma = 3$. Therefore, $\alpha^2 + \beta^2

Q.2. If the set of all solutions of is , then is equal to :

The correct answer is B. Sol. We use the property $|a+b| = |a| + |b|$ if and only if $ab \ge 0$. Here, $|(x^2-9) + x| = |x^2-9| + |x|$. So, we must have $x(x^2-9) \ge 0$. This implies $x(x-3)(x+3) \ge 0$. By analyzing the sign of the product, we get $x \in [-3, 0] \cup [3, \infty)$. Comparing this with $[\alpha, \beta] \cup [\gamma,\infty)$, we have $\alpha = -3$, $\beta = 0$, $\gamma = 3$. Therefore, $\alpha^2 + \beta^2

What are the options for this JEE Main Mathematics question?

A. 9 · B. 18 · C. 36 · D. 72

JEE Main 2026… PYQ · With Answer · Free

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Q.2. 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 :

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