Correct Answer - B
Output of `OR` gate is `A+B`. Output of `NAND` gate is `bar(A).bar(B)`.
New, `Y=(A+B).bar(A.B)=(A+B).(bar(A)+bar(B))`
If `A=1` and `B=1` then
`A+B=1`
and `bar(A)+bar(B)=0`
So, `Y=0`
If `A=0` and `B=0`, then `Y=0`
If `A=1` and `B=0`, then
`A+B` gives `1 bar(A)+bar(B)=1` gives
So, `Y=1`
If `A=1` and `B=1`, then
`Y=1` So, the given combination is `XOR` gate