ISI2015-MMA-23

Question

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

• A. $f(x,A)+f(x,B)$
• B. $f(x,A)+f(x,B)\: – 1$
• C. $f(x,A)+f(x,B)\: – f(x,A) \cdot f(x,B)$
• D. $f(x,A)\:+ \mid f(x,A)\: – f(x,B) \mid $

Comments

https://gateoverflow.in/124234/isi-2004-miii?show=124234#q124234

Answer 1

We will eliminate options.

• A. Assume $x\in A\cap B\Rightarrow x\in A \text{ and }x\in B$. Thus, $f(x,A)+f(x,B)=1+1=2$ which is greater than 1. Hence, incorrect.
• B. Assume $x\in A$ but $x\not\in B$. This implies that $x\in A\cup B$. Thus, $f(x,A\cup B)=1$ but $f(x,A)+f(x,B)-1=1+0-1=0$. Hence, incorrect.
• C. Assume $x\in A\cap B\Rightarrow x\in A \text{ and }x\in B$. Therefore, $f(x,A)+f(x,B)-f(x,A)\cdot f(x,B)=1+1-1=1$. Further, Assume $x\in A$ but $x\not\in B$. This implies that $x\in A\cup B$. Therefore, $f(x,A)+f(x,B)-f(x,A)\cdot f(x,B)=1+0-0=1$. Similarly, $x\in B$ but $x\not\in A$. This option maybe Correct.
• D. Assume $x\in A$ but $x\not\in B$. This implies that $x\in A\cup B$. Thus, $f(x,A\cup B)=1$ but $f(x,A)+|f(x,A)-f(x,B)|=1+|1-0|=2$. Hence, incorrect.

 

Therefore, C is the correct answer.