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In the given truth table, $f(x,y)$ represent the Boolean function.

$$\begin{array}{|c|c|c|} \hline x & y & f(x,y) \\ \hline 0 & 0 & 1 \\ \hline 0 & 1 & 0 \\ \hline 1 & 0 & 0 \\ \hline 1 & 1 & 1 \\ \hline \end{array}$$

1. $x \leftrightarrow y$
2. $x \wedge y$
3. $x \vee y$
4. $x \rightarrow y$

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0

$(A)$ is the correct answer.

 $X$ $Y$ $X\Leftrightarrow Y$ T T T T F F F T F F F T

The given function is nothing but an EXNOR which is same as BICONDITIONAL.

[$\bigodot \equiv \Leftrightarrow$]

so option A is correct.

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