If \(A=\{0,1,2\}\) and \(B=\{-1,0,1\}\), how many ordered pairs \((x,y)\) in \(A\times B\) satisfy \(x^2+y^2=1\)?
Answer and explanation
Correct answer: 3
Check each possible value of \(x\) from A. For \(x=0\), the equation becomes \(y^2=1\), so \(y=-1\) or \(y=1\), giving two pairs: \((0,-1)\) and \((0,1)\). For \(x=1\), \(y^2=0\), so \(y=0\), giving \((1,0)\). For \(x=2\), \(y^2=-3\), which is impossible. Thus there are three valid ordered pairs, so option B is correct.
Frequently asked questions
What is the correct answer to this question?
3
Why is this the correct answer?
Check each possible value of \(x\) from A. For \(x=0\), the equation becomes \(y^2=1\), so \(y=-1\) or \(y=1\), giving two pairs: \((0,-1)\) and \((0,1)\). For \(x=1\), \(y^2=0\), so \(y=0\), giving \((1,0)\). For \(x=2\), \(y^2=-3\), which is impossible. Thus there are three valid ordered pairs, so option B is correct.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: The Empty Set, Finite and Infinite Sets, Equal Sets.
