Sets

{1, 2, 4, 6, 8, 9, 10}

Step 1: First find A ∩ B = elements common to both A and B = {3, 5, 7}. Step 2: The complement of a set means all elements in U that are NOT in that set. Step 3: (A ∩ B)' = U - (A ∩ B) = {1,2,3,4,5,6,7,8,9,10} - {3,5,7}. Step 4: (A ∩ B)' = {1, 2, 4, 6, 8, 9, 10}, which has 7 elements. This can also be verified using De Morgan's law: (A∩B)' = A'∪B'.

[-3, 5)

Step 1: The set is {x ∈ R : -3 ≤ x < 5}. Identify the boundary conditions. Step 2: At x = -3, the condition is ≤ (less than or equal), meaning -3 IS included. A closed bracket [ is used for included endpoints. Step 3: At x = 5, the condition is < (strictly less than), meaning 5 is NOT included. An open bracket ( is used for excluded endpoints. Step 4: Therefore the interval notation is [-3, 5). Common mistake: Swapping open and closed brackets — remember [ ] means endpoint included, ( ) means endpoint excluded.

A ⊂ B (A is a proper subset of B)

Step 1: Solve x² - 5x + 6 = 0 to find set A. Step 2: Factor: (x-2)(x-3) = 0, giving x = 2 or x = 3. Both are integers, so A = {2, 3}. Step 3: B = {2, 3, 4}. Check: every element of A (2 and 3) is in B ✓. Step 4: But A ≠ B because 4 ∈ B but 4 ∉ A. So A is a proper subset of B, written A ⊂ B. Common mistake: Thinking A = B without checking if B has extra elements not in A.

{4, 8}

Step 1: Find A ∪ B = {1,2,3,5,6,7} (all elements in A or B). Step 2: (A∪B)' = U - (A∪B) = {1,2,3,4,5,6,7,8} - {1,2,3,5,6,7} = {4, 8}. Step 3: Verify using De Morgan's Law: A' = {3,4,7,8} and B' = {1,4,5,8}. A'∩B' = {4,8} ✓. Step 4: Both methods give the same answer {4, 8}, confirming De Morgan's Law: (A∪B)' = A'∩B'. Common mistake: Taking A'∪B' instead of A'∩B' when verifying De Morgan's Law.