What will be the foldback memory address range of the following memory chip while interfacing with 8085 microprocessor? 
Step 1: Understand the memory chip and addressing.
The memory chip in question is the 8155 static RAM, with a size of 256 x 8, which provides a total of 2048 bytes (2K).
- The 8085 microprocessor uses a 16-bit address bus, allowing it to access 64KB of memory.
- The 8205 decoder takes the higher 3 bits (A13, A14, A15) and generates the appropriate chip enable (CE) signals based on the address range specified by the memory chip.
Step 2: Determine the foldback memory range.
Given that the 8155 memory chip is 2K (2048 bytes), it occupies the range starting from the base address \( 2000H \). The range of 2K memory in hexadecimal is \( 2000H \) to \( 27FFH \).
Step 3: Conclusion.
Thus, the foldback memory address range for the memory chip is \( 2000H - 27FFH \), which corresponds to option (3).
Find the least upper bound and greatest lower bound of \( S = \{X, Y, Z\} \) if they exist, of the poset whose Hasse diagram is shown below:
Suppose \( D_1 = (S_1, \Sigma, q_1, F_1, \delta_1) \) and \( D_2 = (S_2, \Sigma, q_2, F_2, \delta_2) \) are finite automata accepting languages \( L_1 \) and \( L_2 \), respectively. Then, which of the following languages will also be accepted by the finite automata:
(A) \( L_1 \cup L_2 \)
(B) \( L_1 \cap L_2 \)
(C) \( L_1 - L_2 \)
(D) \( L_2 - L_1 \)
Choose the correct answer from the options given below: