Question

If $A = B = 1$ , then in terms of Boolean algebra the value of $A.B + A\;$ is not equal to
A) $B.A + B$
B) $B + A$
C) $B$
D) $\overline A .B$

Answer 1

Hint AND operator is a Boolean operator that returns a value of TRUE $(1)$ if both its operands are TRUE$(1)$ and FALSE $(0)$ otherwise. The boolean expression for AND gate, $Q=A.B$ , where Q is the output.
OR operator is a Boolean operator that returns a value of TRUE$(1)$ if any of its operands are TRUE$(1)$ and FALSE$(0)$ if both operands are FALSE $(0)$. The boolean expression for OR gate, $Q=A+B$ , where Q is the output.
NOT operator is a unary Boolean operator that returns a value of TRUE $(1)$ if the operand is FALSE $(0)$ and vice versa. The Boolean expression for NOT gate
$Q =\overline A$, where Q is the output.

Complete step by step solution:
For$A = B = 1$,
$\Rightarrow A.B + B$ Substituting values of $A$ and $B$
$= 1.1 + 1$
(The $1.1$ will give answer as $1$ , as AND operator $\to A.B$ is used)
$= 1 + 1$
($= 1 + 1$ will give answer as $1$, as OR operator $\to A + B$ is used)
$= 1$
Therefore,
(A) $B.A + B$
$= 1.1 + 1$
(The $1.1$ will give answer as $1$, as AND operator $\to A.B$ is used)
$= 1 + 1$
($1 + 1$ will give answer as $1$, as OR operator $\to A + B$ is used)
$= 1$

(B) $B + A$
Substituting values of $A$ and $B$
$= 1 + 1$
($1 + 1$Will give answer as $1$ ,as OR operator $\to A + B$is used)
$= 1$

(C) $B$
Substituting value of $B$
$= 1$

(D) $\overline {A.} B$
Substituting values of $A$ and $B$
$= \overline 1 .1$
( $1\overline {}$ can also be written as $0$ )
$= 0.1$
(The $0.1$ will give answer as $0$ , as AND operator $\to A.B$ is used)
$= 0$

Hence, option D is the correct answer.

Note: Logic gates take input in form of 0 and 1 and the algebra dealing with these numbers is known as Boolean algebra. In logic gates boolean algebra rules must be followed:
$1 + 1 = 1\left( { \ne 2} \right)$
$1.1 = 1$
$\overline 1 = 0$
$1.0 = 0$ $0.1 = 0$