Let A=\ x:| x^2-10| 1\. Then

| x^2-10| 1 B=(-infinity, 1) union (3, infinity) A union B=(-infinity, 1) union [2, infinity) A intersection B=[-4,-2] union (3,4] A- B=[2,3] B- A=(-infinity,-4) union (-2,1) union (4, infinity)

Let A x x 2 10 6 and B x x 2 1 Then | null (Mathematics)

JEE Mains 2026 21st January Shift 2

Mathematics

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

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

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

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

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

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Correct Answer:

Explanation

\left|\mathrm{x}^{2}-10\right| \leq 6\\

-6 \leq \mathrm{x}^{2}-10 \leq \\

4 \leq \mathrm{x}^{2} \leq 16\\

\mathrm{A}=[-4,-2] \cup[2,4]\\

|x-2|>1\\

\mathrm{B}=(-\infty, 1) \cup(3, \infty)\\

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

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

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

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

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