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Let A= x: |x^2-10 | ≤ 6 and B=x:|x-2|>1. Then

JEE Main 2026 — Mathematics Algebra

2026mcqmedium

Let A={x:∣x2−10∣≤6}\mathrm{A}=\left\{x:\left|x^{2}-10\right| \leq 6\right\}A={x:​x2−10​≤6} and B={x:∣x−2∣>1}\mathrm{B}=\{x:|x-2|>1\}B={x:∣x−2∣>1}. Then

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Options

  1. A

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

  2. B

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

  3. C

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

  4. D

    A∩ B=[−4,−2]∪[3,4]\mathrm{A} \cap \mathrm{~B}=[-4,-2] \cup[3,4]A∩ B=[−4,−2]∪[3,4]

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