Let A and B be two sets that `Acap X =Bcap X=phi and Acup X=Bcup X` for some set X. then

Options

A. A=B
B. A=X
C. B=X
D. `Acup b=X`

Explanation

we have ,
`A cup X=Bcup X` for some set X
`implies A cap (Acup X)=Acap(Bcup X)`
`implies A=(AcapB)cup(AcapX)[:' A cap(Acup X)=A]`
`implies A=(AcapB)cup phi[:' AcapX=phi]`
`implies A=AcapB . . . (i)`
again
`Acup X=Bcup X`
`implies Bcup (Acup X) =Bcap (Bcup X)`
`implies (B cap A)cup (BcapX)=B`
`implies (B cap A)cup phi =B`
`implies Acap B=B`
from (i) and (ii) we have
`A=B.`