
To solve the given problem, we need to analyze the logic circuit and determine the output for the inputs \( A = 0, B = 1 \) and \( A = 1, B = 0 \), which correspond to finding \( X \) and \( Y \) in the truth table. Let's break it down step-by-step:
1. Analyzing the Circuit:
2. Calculating the Output:
Case 1: \( A = 0, B = 1 \) (Find \( X \))
Thus, for \( A = 0, B = 1 \), the output \( X = 1 \).
Case 2: \( A = 1, B = 0 \) (Find \( Y \))
Thus, for \( A = 1, B = 0 \), the output \( Y = 1 \).
Conclusion:
From the analysis, both \( X \) and \( Y \) are 1 for the respective inputs and configurations. Therefore, the correct answer is 1,1.
Let us analyze the given logic circuit and find the values of \( X \) and \( Y \).
The first AND gate has inputs \( A \) and \( B \). The output \( E \) will be:
\[ E = A \cdot B. \]
The second AND gate has inputs \( A \) and \( E \) (output from the first gate). The output \( X \) will be:
\[ X = A \cdot E = A \cdot (A \cdot B) = A^2 \cdot B. \]
The OR gate takes inputs \( A \) and \( B \) and gives the output \( Y \):
\[ Y = A + B. \]
| A | B | E | X | Y | |---|---|---|---|---| | 0 | 0 | 0 | 0 | 0 | | 0 | 1 | 0 | 0 | 1 | | 1 | 0 | 0 | 0 | 1 | | 1 | 1 | 1 | 1 | 1 |
Thus, the values of \( X \) and \( Y \) for \( A = 1 \) and \( B = 1 \) are \( X = 1 \) and \( Y = 1 \). The correct answer is Option (1).
Which logic gate is represented by the following combinations of logic gates?



The logic gate equivalent to the circuit given in the figure is
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}