Question

1.2 union, intersection, and complement
score: 70/100 answered: 11/15
question 11
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let ( u = {1,2,3,4,dots,10} ), ( a = {1,3,5,7} ), ( b = {1,2,4,6} ), and ( c = {3,4,6,7,9} ).
select ( (b cup a) ) from the choices below.
\\( \circ \\{2,5,6,9\\} \\)
\\( \bullet \\{1,2,3,7,10\\} \\)
\\( \circ \\{6,8,9,10\\} \\)
\\( \circ \\{4,5,8,9\\} \\)
\\( \circ \\{3,8,9,10\\} \\)
\\( \circ \emptyset \\)

Explanation:

Step 1: Find \( B \cup A \)

First, we need to find the union of sets \( B \) and \( A \). The union of two sets \( B \) and \( A \) (denoted as \( B \cup A \)) is the set of all elements that are in \( B \) or in \( A \) (or in both).

Given \( A = \{1, 3, 5, 7\} \) and \( B = \{1, 2, 3, 6\} \).

To find \( B \cup A \), we combine all the elements from both sets, removing duplicates. So:

\( B \cup A = \{1, 2, 3, 5, 6, 7\} \) (since 1 and 3 are in both sets, we only include them once)

Step 2: Find the complement of \( B \cup A \) (denoted as \( (B \cup A)' \))

The universal set \( U = \{1, 2, 3, \dots, 10\} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \)

The complement of a set \( S \) (denoted as \( S' \)) is the set of all elements in the universal set \( U \) that are not in \( S \).

So we need to find all elements in \( U \) that are not in \( B \cup A \).

From \( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \) and \( B \cup A = \{1, 2, 3, 5, 6, 7\} \), the elements in \( U \) but not in \( B \cup A \) are:

\( 4, 8, 9, 10 \)

So \( (B \cup A)' = \{4, 8, 9, 10\} \)

Answer:

\(\{4, 8, 9, 10\}\) (which corresponds to the option "4,8,9,10" in the given choices)