Range of 8 bit 2s complement – UGC NET Question

In 8-bit 2's complement, one bit is effectively used for sign while still contributing to magnitude in a special way. The most negative value occurs when the MSB is 1 and all other bits are 0, giving -128. The most positive value occurs when the MSB is 0 and all other bits are 1, giving +127. Therefore, the representable range is from -128 to +127.

Option A:
Option A reverses the endpoints and yields an impossible range because +128 is not representable in 8-bit 2's complement. The positive side stops at +127. Thus this range is incorrect.

Option B:
Option B is correct because it matches the standard formula for an n-bit 2's complement range, which is -2^(n-1) to 2^(n-1) - 1. For n = 8, this gives -128 to +127. This range is widely used in computer architecture.

Option C:
Option C proposes a symmetric range from -255 to +255, which would require at least 9 bits for representation. It does not correspond to any standard 8-bit signed format.

Option D:
Option D gives -256 to +255, which is the range for unsigned 9-bit or similar schemes, not 8-bit 2's complement. Hence it is not applicable here.