Let A = \ x : x 2 -10 leq 6 \ and B =\ x : x -2 1\ . Then

B - A =(-∞,-4) cup(-2,1) cup(4, ∞)

Mathematics · Sets and Relations

JEE Main 2026 — 21 January, Evening Shift — Question 13

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

Then

Option A:
$\mathrm{A} \cup \mathrm{B}=(-\infty, 1] \cup(2, \infty)$
Option B:
$\mathrm{A}-\mathrm{B}=[2,3)$
Option C:
$\mathrm{A} \cap \mathrm{B}=[-4,-2] \cup[3,4]$
Option D:
$\mathrm{B}-\mathrm{A}=(-\infty,-4) \cup(-2,1) \cup(4, \infty)$
Correct

$\left|x^{2}-10\right| \leq 6$

$-6 \leq \mathrm{x}^{2}-10 \leq 6$

$4 \leq \mathrm{x}^{2} \leq 16$

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

$|x-2|>1$

$\mathrm{B}=(-\infty, 1) \cup(3, \infty)$

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

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

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

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

Answer key and solution verified before publishing.

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