If \(A=\{x\in\mathbb{Z}: |x|\le 3\}\) and \(B=\{x\in\mathbb{Z}: x^2-2x\le 0\}\), find \(A\cap B\).
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Answer and explanation
Correct answer: \(\{0,1,2\}\)
First, \(|x|\le 3\) gives \(-3\le x\le 3\), so \(A=\{-3,-2,-1,0,1,2,3\}\). Next, \(x^2-2x\le0\) becomes \(x(x-2)\le0\), which is true for \(0\le x\le2\). Since \(x\) is an integer, \(B=\{0,1,2\}\). Every element of B is in A, hence \(A\cap B=\{0,1,2\}\).
Frequently asked questions
What is the correct answer to this question?
\(\{0,1,2\}\)
Why is this the correct answer?
First, \(|x|\le 3\) gives \(-3\le x\le 3\), so \(A=\{-3,-2,-1,0,1,2,3\}\). Next, \(x^2-2x\le0\) becomes \(x(x-2)\le0\), which is true for \(0\le x\le2\). Since \(x\) is an integer, \(B=\{0,1,2\}\). Every element of B is in A, hence \(A\cap B=\{0,1,2\}\).
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Operations on Sets (Union, Intersection, Difference).
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