Question

An AND gate is the boolean function defined by

• A. $f(x_1,x_2)=(x_1.x_2),(x_1,x_2)\in \{0,1\}$
• B. $f(x_1,x_2)=(x_1+x_2),(x_1,x_2)\in \{0,1\}$
• C. $f(x_1,x_2)=(x_1,x_{1,}{x}_2)\in \{0,1\}$
• D. $f(x_1,x_2)=(x_2.x_1,x_2)\in \{0,1\}$

Answer 1

The correct option is A

$f(x_1,x_2)=(x_1.x_2),(x_1,x_2)\in \{0,1\}$

Step 1. Given data and drawing diagram of AND gate:

The AND gate is given and drawn as follows,

[Figure $\left(a\right)$: diagram of the two-input AND gate]

Step 2. Explanation for the correct option:

1. The AND gate is defined as a logic gate having a single output for two or more inputs.
2. If both the inputs are high only then the output will be high otherwise the output will always be zero for example if $A$ and $B$ are two inputs as shown in figure $\left(a\right)$ then if either of the two is $"0"$ or both $A$and $B$ are $"0"$ then output $C$ will always be $"0"$.
3. If both $A$ and $B$ are $"1"$ then only output $C$ will be $"1"$. This can be expressed in a logical expression as,

$f(A,B)=(A.B),(A,B)\in \{0,1\}........\left(1\right)$

Solving expression $\left(1\right)$, we get

$$\begin{gathered} A.B=1.1=1 \\ A.B=1.0=0 \\ A.B=0.1=1 \\ A.B=0.0=0 \end{gathered}$$

Only option (A), $f(x_1,x_2)=(x_1.x_2),(x_1,x_2)\in \{0,1\}$ proves to be the same as the expression $\left(1\right)$ and exhibits properties of the AND gate.

Hence, 'option $\left(A\right)$' is correct.