JEE Mains · Maths · STD 11 - 1. set theory
Let \( A=\{x :|x^{2}-10|\le6\} \) and \( B=\{x :|x-2|>1\}. \) Then
- A \( A\cup B=(-\infty,1]\cup(2,\infty) \)
- B \( A-B=[2,3) \)
- C \( A\cap B=[-4,-2]\cup[3,4] \)
- D \( B-A=(-\infty,-4)\cup(-2,1)\cup(4,\infty) \)
Answer & Solution
Correct Answer
(D) \( B-A=(-\infty,-4)\cup(-2,1)\cup(4,\infty) \)
Step-by-step Solution
Detailed explanation
\( |x^{2}-10|\le6 \) \( -6\le x^{2}-10\le6 \) \( 4\le x^{2}\le16 \) \( A=[-4,-2]\cup[2,4] \) \( |x-2|>1 \) \( B=(-\infty,1)\cup(3,\infty) \) \( A\cup B=(-\infty,1)\cup[2,\infty) \) \( A\cap B=[-4,-2]\cup(3,4] \) \( A-B=[2,3] \) \( B-A=(-\infty,-4)\cup(-2,1)\cup(4,\infty) \)
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