If \(A=\{x\in\mathbb{Z}:-6\le x\le 6\}\) and \(B=\{x\in\mathbb{Z}:x^2<10\}\), what is \(A-B\)?
Answer and explanation
Correct answer: \(\{-6,-5,-4,4,5,6\}\)
The set \(A\) contains all integers from \(-6\) through 6. For \(B\), solve \(x^2<10\). Since \(3^2=9<10\) but \(4^2=16>10\), the integer solutions are \(-3,-2,-1,0,1,2,3\). Thus \(B=\{-3,-2,-1,0,1,2,3\}\). Removing these elements from \(A\) leaves \(\{-6,-5,-4,4,5,6\}\), so option A is correct. The endpoints \(-3\) and 3 must be removed because their squares are 9.
Frequently asked questions
What is the correct answer to this question?
\(\{-6,-5,-4,4,5,6\}\)
Why is this the correct answer?
The set \(A\) contains all integers from \(-6\) through 6. For \(B\), solve \(x^2<10\). Since \(3^2=9<10\) but \(4^2=16>10\), the integer solutions are \(-3,-2,-1,0,1,2,3\). Thus \(B=\{-3,-2,-1,0,1,2,3\}\). Removing these elements from \(A\) leaves \(\{-6,-5,-4,4,5,6\}\), so option A is correct. The endpoints \(-3\) and 3 must be removed because their squares are 9.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Operations on Sets (Union, Intersection, Difference).
