Let register R1 contains the binary equivalent of the last two digit of 2205099. Perform the following operations on R1 separately. Assume b is the number bits to be shifted or rotated. b=2205099 % 7 +1. a)LShiftL #b, R1b)AShiftL #b, R1c)LShiftR #b, R1d)AShiftR #b, R1e)RotateL #b, R1f)RotateLC #b, R1 if C=0g)RotateR #b, R1h)RotateRC #b, R1 if C=0

The last two digits of 2205099 are 99. The binary equivalent of 99 is 1100011.

The value of b is calculated as 2205099 % 7 + 1 = 5.

a) LShiftL #b, R1: This operation left shifts the bits in R1 by b places. So, 1100011 becomes 1000000.

b) AShiftL #b, R1: This operation is arithmetic shift left. It is the same as logical shift left for positive numbers. So, 1100011 becomes 1000000.

c) LShiftR #b, R1: This operation right shifts the bits in R1 by b places. So, 1100011 becomes 0000011.

d) AShiftR #b, R1: This operation is arithmetic shift right. It is the same as logical shift right for positive numbers. So, 1100011 becomes 0000011.

e) RotateL #b, R1: This operation rotates the bits in R1 to the left by b places. So, 1100011 becomes 0110001.

f) RotateLC #b, R1 if C=0: This operation rotates the bits in R1 to the left by b places through carry. Since C=0, it is the same as RotateL. So, 1100011 becomes 0110001.

g) RotateR #b, R1: This operation rotates the bits in R1 to the right by b places. So, 1100011 becomes 1110001.

h) RotateRC #b, R1 if C=0: This operation rotates the bits in R1 to the right by b places through carry. Since C=0, it is the same as RotateR. So, 1100011 becomes 1110001.