Let X be a non-empty set and let P(X) denote | vedclass

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(B) The function $f(x, A)$ is the indicator function of the set $A$,denoted as $\chi_A(x)$.By definition,$f(x, A \cup B) = 1$ if $x \in A \cup B$,and

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$f(x, A) + f(x, B) - f(x, A)f(x, B)$

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(B) The function $f(x, A)$ is the indicator function of the set $A$,denoted as $\chi_A(x)$.By definition,$f(x, A \cup B) = 1$ if $x \in A \cup B$,and $0$ otherwise.We know that $x \in A \cup B$ if and only if $x \in A$ or $x \in B$.Using the properties of indicator functions:$f(x, A \cup B) = \max(f(x, A), f(x, B))$.Alternatively,using the inclusion-exclusion principle for indicator functions:$f(x, A \cup B) = f(x, A) + f(x, B) - f(x, A \cap B)$.Since $f(x, A \cap B) = f(x, A) \cdot f(x, B)$,we have $f(x, A \cup B) = f(x, A) + f(x, B) - f(x, A)f(x, B)$.Comparing this with the given options,if we evaluate the truth table for all cases of $x \in A$ and $x \in B$,we find that the expression $f(x, A) + f(x, B) - f(x, A)f(x, B)$ correctly represents the union.

Let $X$ be a non-empty set and let $P(X)$ denote the collection of all subsets of $X$. Define $f: X \times P(X) \rightarrow R$ by $f(x, A) = \begin{cases} 1, & \text{if } x \in A \\ 0, & \text{if } x \notin A \end{cases}$. Then,$f(x, A \cup B)$ equals

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