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}
