IDNLearn.com provides a seamless experience for finding the answers you need. Our experts provide accurate and detailed responses to help you navigate any topic or issue with confidence.
Sagot :
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}
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}
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.