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JEE Main 202621 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Sets and Relations

JEE Main 2026 · 21 January · Evening Shift · Mathematics · Sets and Relations

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

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

Options

  1. AA∪B=(−∞,1]∪(2,∞)\mathrm{A} \cup \mathrm{B}=(-\infty, 1] \cup(2, \infty)
  2. BB−A=(−∞,−4)∪(−2,1)∪(4,∞)\mathrm{B}-\mathrm{A}=(-\infty,-4) \cup(-2,1) \cup(4, \infty)
  3. CA−B=[2,3)\mathrm{A}-\mathrm{B}=[2,3)
  4. DA∩ B=[−4,−2]∪[3,4]\mathrm{A} \cap \mathrm{~B}=[-4,-2] \cup[3,4]

Correct answer

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

Step-by-step solution

A={x:∣x2−10∣≤6}A = \{x : |x^2 - 10| \le 6\}: −6≤x2−10≤6⇒4≤x2≤16-6 \le x^2 - 10 \le 6 \Rightarrow 4 \le x^2 \le 16, so A=[−4,−2]∪[2,4]A = [-4, -2] \cup [2, 4].
B={x:∣x−2∣>1}=(−∞,1)∪(3,∞)B = \{x : |x - 2| > 1\} = (-\infty, 1) \cup (3, \infty).
Ac=(−∞,−4)∪(−2,2)∪(4,∞)A^c = (-\infty, -4) \cup (-2, 2) \cup (4, \infty).
B−A=B∩Ac=[(−∞,1)∩Ac]∪[(3,∞)∩Ac]B - A = B \cap A^c = [(-\infty, 1) \cap A^c] \cup [(3, \infty) \cap A^c]
=[(−∞,−4)∪(−2,1)]∪[(4,∞)]= [(-\infty, -4) \cup (-2, 1)] \cup [(4, \infty)]
=(−∞,−4)∪(−2,1)∪(4,∞)= (-\infty, -4) \cup (-2, 1) \cup (4, \infty).
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