The correct answer is (d) All but Y0 are HIGH
The best explanation: In the given diagram, S0 and S1 are selection bits. So,
I/P S0 S1 O/P
D = 0 0 0 Y0
D = 0 0 1 Y1
D = 0 1 0 Y2
D = 0 1 1 Y3
Hence, inputs are S0 and S1 are Low means 0, so output is Y0 and rest all are HIGH.