Let A = {x : |x² − 10| ≤ 6}... |JEE Main 2026

JEE MAIN 2026

21-01-2026 S2

Question

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

Select the correct option:

$\mathrm{A} \cup \mathrm{B}=(-\infty, 1] \cup(2, \infty)$

$\mathrm{A}-\mathrm{B}=[2,3)$

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

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

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Mathematics

Easy

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