In the case of Intel 8085, what would be the address range for a memory system where A0-A11 are the only lines used for addressing, The memory address starts from A000HQuestion 5Select one:a.A800H-AFFFFHb.A000H-A500Hc.A000H-A800Hd.A00H-A600H

The Intel 8085 uses a 16-bit address to access memory. In this case, if only A0-A11 are used for addressing, this means that 12 bits are used for addressing.

The number of unique addresses that can be generated with 12 bits is 2^12 = 4096.

Since the memory address starts from A000H (hexadecimal), the address range would end at A000H + 4096 - 1.

In hexadecimal, 4096 is 1000, so the address range would end at A000H + 1000H - 1 = AFFFH.

However, none of the options provided (A800H-AFFFFH, A000H-A500H, A000H-A800H, A00H-A600H) match this result.

Please check the question or the options again.

The Intel 8085 uses a 16-bit address to access memory. In this case, if only A0-A11 lines are used for addressing, it means that 12 bits are used for addressing.

The number of unique addresses that can be generated with 12 bits is 2^12 = 4096.

Since the memory address starts from A000H, the address range would be from A000H to (A000H + 4096 - 1).

In hexadecimal, 4096 is 1000. Therefore, the address range would be from A000H to AFFFH.

However, none of the options provided (A800H-AFFFFH, A000H-A500H, A000H-A800H, A00H-A600H) match this range. There might be a mistake in the question or the provided options.