If \(U=\{x:x\in\mathbb{Z},-8\le x\le8\}\) and \(A=\{x:x^2-4x\le0\}\), what is \(A'\)?
Answer and explanation
Correct answer: \(\{-8,-7,-6,-5,-4,-3,-2,-1,5,6,7,8\}\)
Solve the inequality by factoring: \(x^2-4x=x(x-4)\le0\). Therefore, the real solution interval is \([0,4]\). Since the universal set contains only integers from -8 through 8, \(A=\{0,1,2,3,4\}\). Removing these five integers from \(U\) leaves \(A'=\{-8,-7,-6,-5,-4,-3,-2,-1,5,6,7,8\}\), which is option A.
Frequently asked questions
What is the correct answer to this question?
\(\{-8,-7,-6,-5,-4,-3,-2,-1,5,6,7,8\}\)
Why is this the correct answer?
Solve the inequality by factoring: \(x^2-4x=x(x-4)\le0\). Therefore, the real solution interval is \([0,4]\). Since the universal set contains only integers from -8 through 8, \(A=\{0,1,2,3,4\}\). Removing these five integers from \(U\) leaves \(A'=\{-8,-7,-6,-5,-4,-3,-2,-1,5,6,7,8\}\), which is option A.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Complement of a Set and Its Properties.
