To solve this problem, we need to use the properties of set theory, particularly the concepts of union, intersection, and complement within the universal set [tex]\( U \)[/tex].
Given:
[tex]$
\begin{array}{l}
U = \{f, m, q, r, s\} \\
B = \{f, m, s\} \\
D = \{f, m, q\}
\end{array}
$[/tex]
Step 1: Find the complement of [tex]\( D \)[/tex] (denoted as [tex]\( D' \)[/tex])
The complement of [tex]\( D \)[/tex] consists of all elements in the universal set [tex]\( U \)[/tex] that are not in [tex]\( D \)[/tex].
Since [tex]\( D = \{f, m, q\} \)[/tex], we need to remove [tex]\( f \)[/tex], [tex]\( m \)[/tex], and [tex]\( q \)[/tex] from [tex]\( U \)[/tex].
Therefore:
[tex]$
D' = U \setminus D = \{f, m, q, r, s\} \setminus \{f, m, q\} = \{r, s\}
$[/tex]
Step 2: Find [tex]\( B \cup D' \)[/tex]
The union of [tex]\( B \)[/tex] and [tex]\( D' \)[/tex] includes all elements that are in either [tex]\( B \)[/tex] or [tex]\( D' \)[/tex] (or in both).
Given:
[tex]$
B = \{f, m, s\} \quad \text{and} \quad D' = \{r, s\}
$[/tex]
So:
[tex]$
B \cup D' = \{f, m, s\} \cup \{r, s\} = \{f, m, s, r\}
$[/tex]
Hence, the answer to part (a) is:
[tex]$
B \cup D' = \{f, m, s, r\}
$[/tex]
Step 3: Find the complement of [tex]\( B \)[/tex] (denoted as [tex]\( B' \)[/tex])
The complement of [tex]\( B \)[/tex] consists of all elements in the universal set [tex]\( U \)[/tex] that are not in [tex]\( B \)[/tex].
Since [tex]\( B = \{f, m, s\} \)[/tex], we need to remove [tex]\( f \)[/tex], [tex]\( m \)[/tex], and [tex]\( s \)[/tex] from [tex]\( U \)[/tex].
Therefore:
[tex]$
B' = U \setminus B = \{f, m, q, r, s\} \setminus \{f, m, s\} = \{q, r\}
```
Step 4: Find \( B' \cap D' \)
The intersection of \( B' \) and \( D' \) includes only the elements that are in both \( B' \) and \( D' \).
Given:
$[/tex]
B' = \{q, r\} \quad \text{and} \quad D' = \{r, s\}
[tex]$
So:
$[/tex]
B' \cap D' = \{q, r\} \cap \{r, s\} = \{r\}
```
Hence, the answer to part (b) is:
[tex]$[/tex]
B' \cap D' = \{r\}
```
