If \(U=\mathbb{R}\) and \(A=\{x:x^2<16\}\), what is \(A'\)?
Answer and explanation
Correct answer: \((- infty,-4]\cup[4,\infty)\)
The inequality \(x^2<16\) is equivalent to \(|x|<4\), which means \(-4<x<4\). Therefore, \(A=(-4,4)\). Since the universal set is all real numbers, the complement consists of the two outside intervals, including the boundary points where equality holds: \(x\le-4\) or \(x\ge4\). Hence \(A'=(-\infty,-4]\cup[4,\infty)\), option A.
Frequently asked questions
What is the correct answer to this question?
\((- infty,-4]\cup[4,\infty)\)
Why is this the correct answer?
The inequality \(x^2<16\) is equivalent to \(|x|<4\), which means \(-4<x<4\). Therefore, \(A=(-4,4)\). Since the universal set is all real numbers, the complement consists of the two outside intervals, including the boundary points where equality holds: \(x\le-4\) or \(x\ge4\). Hence \(A'=(-\infty,-4]\cup[4,\infty)\), 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.
