Question

Which of the following is not equal to 1 in boolean algebra?

• A. A+1
• B. \(A.\bar{A}\)
• C. \(A+\bar{A}\)
• D. \(\overline{A.A}\)

Answer 1

The Correct Option is B

The correct option is (B) \(A\).\(\bar{A}\)
Option (a): In boolean algebra, if A=1→\(\bar{A}\)=0 and A=0→\(\bar{A}\)=1

Thus, A+1=1
(∵0+1=1 and 1+1=1 in boolean)

Option (b):
if A=1,A.\(\bar{A}\)=1.0=0
A=0 then A.\(\bar{A}\)=0.1=0

⇒A.\(\bar{A}\)=0 always

Option (c): Similarly, A+\(\bar{A}\)=1 always
Case 1: A=0 ⇒ A+\(\bar{A}\)=0+1=1

Case 2: A=1⇒A+\(\bar{A}\)=1+0=1

Option (d):\(\overline{A.A}\)=\(\bar{A}\) (∵A.A=A)

⇒ \(\overline{A.A}\)=0, if A = 1

⇒ \(\overline{A.A}\)=1, if A = 0

Hence, option (B) is correct.