Sets

{x : x² = 4 and x is odd}

Step 1: Check each option to see which has NO elements. Step 2: Option A: x < 5 and x > 2 gives x = 3, 4 (natural numbers) — NOT empty. Step 3: Option B: x² = 4 gives x = 2 or x = -2. Neither 2 nor -2 is odd, so this set has NO elements — EMPTY SET. Step 4: Option C: x is an even prime — the number 2 is an even prime, so this set = {2} — NOT empty. Step 5: Option D: x + 3 = 3 gives x = 0, so the set = {0} — NOT empty (note: 0 ≠ ∅). Final Step: Only Option B satisfies the condition of having no elements, making it the empty set.

A ⊂ B and A ≠ B

Step 1: Check if A ⊂ B: every element of A (1, 2, 3) must be in B. Step 2: 1 ∈ B ✓, 2 ∈ B ✓, 3 ∈ B ✓ — so A ⊂ B. Step 3: Check if A = B: B has elements 4 and 5 which are not in A, so A ≠ B. Step 4: Since A ⊂ B and A ≠ B, A is a PROPER subset of B. Final Step: The correct statement is A ⊂ B and A ≠ B. Disjoint sets share no common elements — but A and B share 1, 2, 3, so they are not disjoint.

{x : x = n², n ∈ N}

Step 1: Observe the pattern in the set: 1, 4, 9, 16, 25, ... Step 2: Check: 1 = 1², 4 = 2², 9 = 3², 16 = 4², 25 = 5². Step 3: These are perfect squares of natural numbers. Step 4: Verify other options: 2n gives even numbers (2,4,6,...); n+1 gives (2,3,4,...); n³ gives (1,8,27,...) — none match. Final Step: The set-builder form is {x : x = n², where n ∈ N}. This is the set of all perfect squares of natural numbers.

{1, 6}

Step 1: First find A ∪ B = all elements in A or B = {2, 3} ∪ {3, 4, 5} = {2, 3, 4, 5}. Step 2: Now find (A ∪ B)' = all elements in U that are NOT in A ∪ B. Step 3: U = {1, 2, 3, 4, 5, 6}, A ∪ B = {2, 3, 4, 5}. Step 4: Remove {2, 3, 4, 5} from U: remaining = {1, 6}. Final Step: (A ∪ B)' = {1, 6}. This can also be verified using De Morgan's law: A' ∩ B' = {1,4,5,6} ∩ {1,2,6} = {1,6} ✓