Aspire Faculty ID #17922 · Topic: JEE Main 2026 (21 January Evening Shift) · Just now
JEE Main 2026 (21 January Evening Shift)

Let $A = {x : |x^2 - 10| \le 6}$ and $B = {x : |x - 2| > 1}$. Then

Solution

$|x^2 - 10| \le 6$

$-6 \le x^2 - 10 \le 6$

$4 \le x^2 \le 16$

$\Rightarrow A = [-4, -2] \cup [2, 4]$

$|x - 2| > 1$

$\Rightarrow B = (-\infty, 1) \cup (3, \infty)$

$A \cup B = (-\infty, 1) \cup (2, \infty)$

$A \cap B = [-4, -2] \cup [3, 4]$

$A - B = [2, 3]$

$B - A = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)$

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