The Empty Set, Finite and Infinite Sets, Equal Sets

Correct answer: A. (A ∩ B) ∪ (A ∩ C)

Explanation: This is the distributive law of intersection over union: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). An element belongs to the left side when it is in A and also in at least one of B or C. That means it is either in both A and B or in both A and C, which is exactly the right side. Hence option A is correct.

Correct answer: A. \(\varnothing\)

Explanation: The power set of any set contains the empty set. For the empty set itself, the power set is \(\mathcal{P}(\varnothing)=\{\varnothing\}\), which has exactly one element. Therefore, if \(\mathcal{P}(A\cap B)=\{\varnothing\}\), the original set must be empty. Hence, \(A\cap B=\varnothing\). Notice that \(\{\varnothing\}\) is a set containing the empty set, whereas \(\varnothing\) contains no elements.

Correct answer: C. 3

Explanation: Use each possible y from B and calculate x = 2y - 1. For y = 1, x = 1, giving (1,1); for y = 2, x = 3, giving (3,2); and for y = 3, x = 5, giving (5,3). All three calculated x-values belong to A, and every y-value belongs to B. Therefore, the valid ordered pairs are (1,1), (3,2), and (5,3), so their number is 3 and option C is correct.

Correct answer: B. 3

Explanation: 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.