The Boolean expression
$B.(A+B)+A.(\overline{B}+A)$
can be realized using minimum number of

Question

The Boolean expression 
$B.(A+B)+A.(\overline{B}+A)$ 
can be realized using minimum number of

Accepted Answer

To understand why Option 3, which states that the Boolean expression can be realized using 1 OR gate, is correct, let's analyze the given Boolean expression:

The expression is:
$$ B.(A+B) + A.(\overline{B}+A) $$

1. **Simplification of the Expression:**

- Apply the Distributive Law:
     $$ B.(A+B) = B.A + B.B = B.A + B $$
     (since $ B.B = B $)

- Similarly, simplify the second part:
     $$ A.(\overline{B}+A) = A.\overline{B} + A.A = A.\overline{B} + A $$
     (since $ A.A = A $)

- Combine the simplified parts:
     $$ (B.A + B) + (A.\overline{B} + A) $$

- Apply the Distributive Law again:
     $$ = B.A + B + A.\overline{B} + A $$

- Apply the Absorption Law:
     $$ = B + A $$
     (since $ B.A + B = B $ and $ A.\overline{B} + A = A $)

2. **Realization using Gates:**

- The simplified expression $ B + A $ can be realized using a single OR gate. This is because the OR operation directly corresponds to the logical addition of two variables.

3. **Explanation of Other Options:**

- **Option 1: 1 AND gate**
     - An AND gate cannot realize the expression $ B + A $ because an AND gate performs multiplication (logical AND), not addition (logical OR).

- **Option 2: 2 AND gates**
     - Using two AND gates is unnecessary for the expression $ B + A $ because it requires only an OR operation, not an AND operation.

- **Option 4: 2 OR gates**
     - While two OR gates could technically be used, it is not the minimum number required. The expression $ B + A $ can be realized with just one OR gate, making the use of two OR gates redundant.

In conclusion, the expression $ B + A $ is most efficiently realized using a single OR gate, which is why Option 3 is the correct answer.