In the combination of the following gates the output $Y$ can be written in terms of inputs $A$ and $B$ as

(1) $\overline{\mathbf{A} \cdot \mathbf{B}}+\mathbf{A} \cdot \mathbf{B}$

(2) $A \cdot \bar{B}+\bar{A} \cdot B$

(3) A·B

(4) $\mathrm{A}+\mathrm{B}$

Solution:

Answer is (2)

$\mathbf{Y}=(\mathbf{A} \cdot \overline{\mathbf{B}}+\overline{\mathbf{A}} \cdot \mathbf{B})$