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Let $X$ be a nonempty set and let $\mathcal{P}(X)$ denote the collection of all subsets of $X.$ Define $\textit{f}:\textit{X$\times$ $\mathcal{P}$(X)}\rightarrow \mathbb{R}$ by

$f(x,A) = \begin{cases}  1 \text{ if } x  \in A & \\ 0 \text{ if } x \notin  A & \end{cases}$

Then $f\left ( x,A\cup B \right )$ equals

  1. $f(x,A)+f(x,B)$
  2. $f(x,A)+f(x,B)-1$
  3. $f(x,A)+f(x,B) - f(x,A) \cdot f(x,B)$
  4. $f(x,A)+ \mid f(x,A) – f(x,B) \mid$
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Here, $x \in X$ and $A,B \in \mathcal{P}(X)$. So, $A, B \subseteq X$.

Now,

Case 1) : If $x \in A \cup B$, So, $f(x, A \cup B) =1 $. Here, $x \in A \cup B$ means, $x$ is either present in $A$ or $B$ but if $x$ is present in both $A$ and $B$ then we have to subtract the presence of $x$ in $A \cap B$ due to double counting.

So, $f(x, A \cup B) = f(x,A) + f(x,B) - f(x, A\cap B)$.

Now, for $f(x, A\cap B)$,

Sub-Case 1) : $x \in A\cap B$. It means $f(x,A\cap B) = 1$. Here, $x \in A\cap B$ means $x \in A$ and $x \in B$ which implies $f(x,A) =1$ and $f(x,B) =1$. So, $f(x, A\cap B) = f(x, A).f(x,B)$

Sub-Case 2) : $x \notin A\cap B$. It means $x \in (A \cap B)^c \Rightarrow x \in A^c \cup B^c \Rightarrow x \in A^c\; or\;x \in B^c \Rightarrow x \notin A \; or \; x \notin B $ which means both $f(x,A) =0 $ and $f(x,B) =0 $ and since, $x \notin A\cap B$. So, $f(x, A \cap B) = 0$. So, in this case also, $f(x, A\cap B) = f(x, A).f(x,B)$.

Hence, $f(x, A\cap B) = f(x, A).f(x,B)$.

So, $f(x, A \cup B) = f(x,A) + f(x,B) - f(x, A).f(x,B)$.

Case 2) : If $x \notin A \cup B$, So, $f(x, A \cup B) =0 $. Here, $x \notin A \cup B$ means, $x \notin A$ and $x \notin B$. So, $f(x,A) =0$ and $f(x,B) =0$. So, here in this case , also,  $f(x, A \cup B) = f(x,A) + f(x,B) - f(x, A).f(x,B)$. is true.

Therefore, the option $(C)$ is correct.

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