Boolean Algebra MCQ Quiz - Objective Question with Answer for Boolean Algebra - Download Free PDF

Explanation:

To understand why the complement law holds, we need to analyze the truth table for the expression A + A'. A truth table lists all possible values of the variables involved and the resulting value of the expression for each combination of variable values. In this case, we are dealing with a single variable A, which can either be 0 or 1.

The correct answer is Option 3) (A + B')(C' + D).

Key Points

Given Boolean function: A'B + CD' + A'B + CD'

• First simplify the given expression:
  • A'B + A'B = A'B (redundant term)
  • CD' + CD' = CD' (redundant term)
• So, the simplified function = A'B + CD'

Now, we are asked to find the complement of this function:

Let F = A'B + CD'

Then the complement is: F' = (A'B + CD')'

Apply De Morgan’s Law:

• (A'B + CD')' = (A'B)' · (CD')'
• (A'B)' = A + B'
• (CD')' = C' + D

Therefore, F' = (A + B')(C' + D)

Hence, the correct answer is: Option 3) (A + B')(C' + D)

The correct answer is: Option 3.

Important Points 

Let's evaluate each Boolean expression to determine which ones are true:

(A). A + AB = A This expression simplifies using the Distributive Law and the Absorption Law: A + AB = A (1 + B) A (1 + B) = A (Since 1 + B = 1) Therefore, this expression is true.

(B).(A + B)' = A'B' This expression is based on De Morgan's Law: (A + B)' = A'B' Therefore, this expression is true.

(C).(A')' = A This expression is based on the Double Negation Law: (A')' = A Therefore, this expression is true.

(D).(AB)' = A' + B' This expression is also based on De Morgan's Law: (AB)' = A' + B' Therefore, this expression is true.

All the expressions (A), (B), (C), and (D) are true, So the correct answer is option 3) (A), (B), (C) and (D).

The correct answer is Both.

Key Points

• De Morgan's laws are two transformation rules that are both valid rules of inference.
• The first law states that the negation of a conjunction is the disjunction of the negations: \(\rm \overline{X \wedge Y}=\overline{X}\vee\overline Y\)
• The second law states that the negation of a disjunction is the conjunction of the negations:  \(\rm \overline{X \vee Y}=\overline{X}\wedge\overline Y\)
• Both these laws are essential in the field of Boolean algebra and digital logic design. 

Additional Information

Commutative Law

• Commutative law states that changing the sequence of the variables does not have any effect on the output of a logic circuit.
• A + B = B + A

Associative Law

• It states that the order in which the logic operations are performed is irrelevant as their effect is the same.
• (A + B) + C = A + (B + C)

Distributive Law

• A + BC = (A + B)(A + C)

The correct answer is A - II, B - IV, C - I, D - III

Key Points

FROM laws of Boolean algebra

1 + any variable = 1

Concept:

All Boolean algebra laws are shown below

Calculation:

F = X̅ Z̅ + Y̅ Z̅ + Y Z̅ + XYZ

= X̅ Z̅ + Z̅ (Y̅ + Y) + XYZ

= X̅ Z̅ + Z̅ + XYZ

= Z̅ (1 + X̅) + XYZ

= Z̅ + XYZ 

Now using Distributive Law

= (Z̅ + Z)(Z̅ + XY)

= Z̅ + XY 

• A variable is a symbol that may take on the value 0 or 1.
• A literal is the use of a variable or its complement in an expression.
• A term is an expression formed by literals and operations at one level.

For example, the following function:

F₁ = xy + xy'z + x'yz

Has 3 variables (x,y,z),

8 literals (x,y,x,y',z,x',y,z), and

4 terms (xy, xy'z, x'yz, and the OR term that combines the first level AND terms).

Given Y = A (A̅ + C) (A̅B + C) (A̅BC + C̅)

This can be written as:

Y = (A.A̅ + AC) (A̅B + C) (A̅BC + C̅)

Since A.A̅ = 0, the above expression can be written as:

Y = (AC) (A̅B + C) (A̅BC + C̅)

Y = (AC.A̅B + AC.C) (A̅BC + C̅)

With C.C = C, we can write:

Y = (AC) (A̅BC + C̅)

Y = AC.A̅BC + AC.C̅ 

Y = 0 + 0

Y = 0

De Morgan’s’ Law

\(\overline {X + Y} = \bar X.\bar Y\)

\(\overline {XY} = \bar X + \bar Y\)

Let P be the output of 1- NAND gate

\({\rm{P}} = \overline {{\rm{A}}.{\rm{B}}}\)

Let Q be the output of 1- NAND gate

\({\rm{Q}} = \overline {\bar A + B} \)

\({\rm{Y}} = \overline {{\rm{P}} + {\rm{\;Q}}} = \bar P\bar Qa\)

\({\rm{Y}} = {\rm{\;}}\overline {\overline {{\rm{A}}.{\rm{B}}} \;} .\;\overline {\overline {\bar A + B} } \;\;\)

Y = (A.B). (A̅ + B)

Y = A.B.A̅  + A.B.B

∵ A.A̅  = 0 and B.B = B

Y = AB 

\(\left( {P \ne Q} \right) \ne R = \overline {\left( {\bar P + \bar Q} \right)} + \bar R = PQ + \bar R\)

\(\\ P \ne \left( {Q \ne R} \right) = \bar P + \overline {\left( {\bar Q + \bar R} \right)} = \bar P + Q\bar R \)

∴ S1 is FALSE

But

\(Q \ne R = \bar Q + \bar R = \bar R + \bar Q = R \ne Q\)

∴ S2 is TRUE

Important Points 

𝑋 \(\ne\) 𝑌 = 𝑋′ + 𝑌′ = (𝑋.𝑌)'  

It is a NAND operation and NAND is commutative but not associative

(AB + C + DC)(AC + BC + D)

= (AB + C[1+ D])(AC + BC + D)

= (AB + C) (AC + BC + D)

= ABC + ABC + ABD + AC + BC + CD

= ABC + ABD + AC + BC + CD

= AC (1 + B) + ABD + BC + CD

= AC + BC + CD + ABD

Concept-

• The branch of algebra which deals with the values of the variables in the form of the truth values true and false is called Boolean algebra.
• True and false are usually denoted by 1 and 0 respectively.

Explanation-

If we take complement, we get negation of the variable but if we again take the complement of complemented variable, we get the same variable.

\(F\left( A \right) = \overline {\bar A} = A\)

Laws of Boolean Algebra:

Application:

f(A, B) = ∑ m(0, 1, 2, 3)

= A̅ B̅ + A̅ B + A B̅ + AB

= A̅ ( B + B̅) + A (B̅ + B)

= A̅ + A = 1