$(A)$ is the correct answer.

$X$ | $Y$ | $X\Leftrightarrow Y$ |

T | T | T |

T | F | F |

F | T | F |

F | F | T |

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0 votes

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

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

- $x \leftrightarrow y$
- $ x \wedge y$
- $x \vee y$
- $x \rightarrow y$

0 votes

The given function is nothing but an * EXNOR* which is

[$\bigodot \equiv \Leftrightarrow$]

so option A is correct.

REFER:-> https://en.wikipedia.org/wiki/Logical_biconditional

0 votes

→ The output of a digital logic Exclusive-NOR gate ONLY goes “HIGH” when its two input terminals, A and B are at the “SAME” logic level which can be either at a logic level “1” or at a logic level “0”.

→ In other words, an even number of logic “1’s” on its inputs gives a logic “1” at the output, otherwise is at logic level “0”.

→ This type of gate gives and output “1” when its inputs are “logically equal” or “equivalent” to each other, which is why an Exclusive-NOR gate is sometimes called an Equivalence Gate

→ In other words, an even number of logic “1’s” on its inputs gives a logic “1” at the output, otherwise is at logic level “0”.

→ This type of gate gives and output “1” when its inputs are “logically equal” or “equivalent” to each other, which is why an Exclusive-NOR gate is sometimes called an Equivalence Gate

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