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Select the correct answer.

A Boolean operator acts on two inputs: A and B. The output is 1 only if one input is the opposite of the other. Which table is the correct truth table for the operator?

A.
\begin{tabular}{|c|c|c|}
\hline A & B & Output \\
\hline 0 & 0 & 0 \\
\hline 0 & 1 & 1 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 1 \\
\hline
\end{tabular}

B.
\begin{tabular}{|c|c|c|}
\hline A & B & Output \\
\hline 0 & 0 & 0 \\
\hline 0 & 1 & 1 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 0 \\
\hline
\end{tabular}

C.
\begin{tabular}{|c|c|c|}
\hline A & B & Output \\
\hline 0 & 0 & 1 \\
\hline 0 & 1 & 1 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 1 \\
\hline
\end{tabular}

D.
\begin{tabular}{|c|c|c|}
\hline A & B & Output \\
\hline 0 & 0 & 1 \\
\hline 0 & 1 & 0 \\
\hline 1 & 0 & 0 \\
\hline 1 & 1 & 1 \\
\hline
\end{tabular}



Answer :

GinnyAnswer
Let's analyze the question and the given options step-by-step:

The Boolean operator in question produces an output of 1 only when exactly one of the inputs is different from the other. This means the output is 1 when one input is 0 and the other input is 1, and vice versa.

This behavior describes the XOR (exclusive OR) operation. The XOR operator gives an output of 1 if exactly one of the inputs is 1 (but not both).

Based on this, let's build the correct truth table for the XOR operator:

1. When [tex]\( A = 0 \)[/tex] and [tex]\( B = 0 \)[/tex], the output is [tex]\( 0 \)[/tex].
2. When [tex]\( A = 0 \)[/tex] and [tex]\( B = 1 \)[/tex], the output is [tex]\( 1 \)[/tex].
3. When [tex]\( A = 1 \)[/tex] and [tex]\( B = 0 \)[/tex], the output is [tex]\( 1 \)[/tex].
4. When [tex]\( A = 1 \)[/tex] and [tex]\( B = 1 \)[/tex], the output is [tex]\( 0 \)[/tex].

Now, let's examine the provided options and find the one that matches our table:

Option A.
\begin{tabular}{|c|c|c|}
\hline
A & B & Output \\
\hline
0 & 0 & 0 \\
\hline
0 & 1 & 1 \\
\hline
1 & 0 & 1 \\
\hline
1 & 1 & 1 \\
\hline
\end{tabular}
- This table does not match our expected output because for [tex]\( A = 1 \)[/tex] and [tex]\( B = 1 \)[/tex], it shows 1 instead of 0.

Option B.
\begin{tabular}{|c|c|c|}
\hline
A & B & Output \\
\hline
0 & 0 & 0 \\
\hline
0 & 1 & 1 \\
\hline
1 & 0 & 1 \\
\hline
1 & 1 & 0 \\
\hline
\end{tabular}
- This table matches our expected output exactly.

Option C.
\begin{tabular}{|c|c|c|}
\hline
A & B & Output \\
\hline
0 & 0 & 1 \\
\hline
0 & 1 & 1 \\
\hline
1 & 0 & 1 \\
\hline
1 & 1 & 1 \\
\hline
\end{tabular}
- This table does not match our expected output because for [tex]\( A = 0 \)[/tex] and [tex]\( B = 0 \)[/tex], it shows 1 instead of 0, and for [tex]\( A = 1 \)[/tex] and [tex]\( B = 1 \)[/tex], it shows 1 instead of 0.

Option D.
\begin{tabular}{|l|l|l|}
\hline
A & B & Output \\
\hline
0 & 0 & 1 \\
\hline
0 & 1 & 0 \\
\hline
\end{tabular}
- This table is incomplete and also does not match our expected output for [tex]\( A = 0 \)[/tex] and [tex]\( B = 1 \)[/tex].

Therefore, the correct answer is:

Option B.
\begin{tabular}{|c|c|c|}
\hline
A & B & Output \\
\hline
0 & 0 & 0 \\
\hline
0 & 1 & 1 \\
\hline
1 & 0 & 1 \\
\hline
1 & 1 & 0 \\
\hline
\end{tabular}

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