If \(A=\{x\in\mathbb{R}:x^2\le 9\}\) and \(B=\{x\in\mathbb{R}:x^2<1\}\), what is \(A\setminus B\)?
Answer and explanation
Correct answer: \([-3,-1]\cup[1,3]\)
The inequality \(x^2\le 9\) is equivalent to \(-3\le x\le 3\), so \(A=[-3,3]\). Similarly, \(x^2<1\) gives \(-1<x<1\), so \(B=(-1,1)\). The difference \(A\setminus B\) removes every point strictly between −1 and 1 from A. The boundary points −1 and 1 remain because they are not members of B. Therefore, \(A\setminus B=[-3,-1]\cup[1,3]\), making option A correct.
Frequently asked questions
What is the correct answer to this question?
\([-3,-1]\cup[1,3]\)
Why is this the correct answer?
The inequality \(x^2\le 9\) is equivalent to \(-3\le x\le 3\), so \(A=[-3,3]\). Similarly, \(x^2<1\) gives \(-1<x<1\), so \(B=(-1,1)\). The difference \(A\setminus B\) removes every point strictly between −1 and 1 from A. The boundary points −1 and 1 remain because they are not members of B. Therefore, \(A\setminus B=[-3,-1]\cup[1,3]\), making option A correct.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Operations on Sets (Union, Intersection, Difference).
