Dual of(x∨y)∧(x∨1)=x∨(x∧y)∨y is
(x.y)+(x.0)=x.(x+y).y
(x+y)+(x.1)=x.(x+y).y
(x.y)(x.1)=x.(x+y).y
None of these
Replace'∨' by'+' and'∧' by'.'.
(x+y).(x+1)=x+(x.y)+y
Dual of(x+y).(x+1)=x+(x.y)+y is written by replacing'+' to'.' and'1' to'0' and vice versa
Dual of(x+y).(x+1)=x+(x.y)+y is
Hence, option(A) is the correct option.