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

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

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

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

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

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Basic Mathematics

Let

A = {x : x^{2} - 10 \leq 6}

and

B = {x : ∣ x - 2∣ > 1}

.

Then

Answer: (D)

From

∣ x^{2} - 10∣ \leq 6

4 \leq x^{2} \leq 16

, so

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

. From

∣ x - 2∣ > 1

x < 1

x > 3

, so

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

. To form

B - A

, remove the points of

A

from

B

: from

(- \infty, 1)

delete

[- 4, - 2]

to leave

(- \infty, - 4) \cup (- 2, 1)

, and from

(3, \infty)

delete

(3, 4]

to leave

(4, \infty)

. Therefore

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

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Single correct option