Q27 / 75 · JEE Main 2026 (21 January Shift 2) · Mathematics

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

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

Solution

$A = \{x : |x^2 - 10| \le 6\}$: $-6 \le x^2 - 10 \le 6 \Rightarrow 4 \le x^2 \le 16$, so $A = [-4, -2] \cup [2, 4]$.
$B = \{x : |x - 2| > 1\} = (-\infty, 1) \cup (3, \infty)$.
$A^c = (-\infty, -4) \cup (-2, 2) \cup (4, \infty)$.
$B - A = B \cap A^c = [(-\infty, 1) \cap A^c] \cup [(3, \infty) \cap A^c]$
$= [(-\infty, -4) \cup (-2, 1)] \cup [(4, \infty)]$
$= (-\infty, -4) \cup (-2, 1) \cup (4, \infty)$.

Hence, the correct answer is option: 'B'.

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