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PSPCL JE EE 2019 Shift 2 Previous Year Paper

Option 2 : X.(X̅ + X̅ Y) = 1

ST 1: DC Circuits

2405

20 Questions
20 Marks
25 Mins

__Theory:__

- In mathematics and mathematical logic, Boolean algebra is the branch of algebra in which values of the variables are the truth values which are true and false, usually denoted by 1 and 0 respectively.
- Boolean algebra is used to analyze and simplify the digital logic circuits.
- The three basic Boolean operates are AND, OR, and NOT.
- Boolean algebra is a system of mathematics based on the logic that has its own set of rules or laws which are used to define and reduce Boolean expression

**Laws of Boolean Algebra:**

**Identify Law:**

A + 0 = A

A ⋅ 1 = A

**Idempotent law:**

A + A = A

A ⋅ A = A

**Annulment law:**

A ⋅ 0 = 0

A + 1 = 1

**Complement law:**

A ⋅ A̅ = 0

A + A̅ = 1

**Commutative law:**

A ⋅ B = B ⋅ A

A + B = B + A

**Double Negative law:**

A̅̅ = A

**De-Morgans theorem :**

\(\overline {A + B} = \bar A \cdot \bar B\)

\(\overline {A \cdot B} = \bar A + \bar B\)

**Distributive law:**

A ⋅ (B + C) = A ⋅ B + A ⋅ C

A + (B ⋅ C) = (A + B) ⋅ (A + C)

**Absorptive law:**

A + (AB) = A

A (A + B) = A

**Associative law:**

A + (B + C) = (A + B) + C = A + B + C

A ⋅ (B ⋅ C) = (A ⋅ B) ⋅ C = A ⋅ B ⋅ C

__Calculation:__

**Option: i)**

X ⋅ (X̅ + 1) = X

L.H.S ⇒ X ⋅ (X̅ + 1)

L.H.S = X ⋅ (1) [∵ Annulment law ⇒ A + 1 = 1]

L.H.S = X [∵ Identify law ⇒ A ⋅ 1 = A]

∴ L.H.S = R.H.S

**Option: ii)**

X ⋅ (X̅ + X̅ Y) = 1

L.H.S. ⇒ X ⋅ (X̅ + X̅ Y)

L.H.S = X ⋅ [X̅] [∵ Absorptive law ⇒ A + (A ⋅ B) = A]

L.H.S = 0 [∵ Complement law ⇒ A ⋅ A̅ = 0]

L.H.S = 0 ≠ 1

L.H.S ≠ R.H.S

∴ X ⋅ (X̅ + X̅ Y) = 1 **(False statement)**

**Option: iii)**

X + X̅ = 1

L.H.S ⇒ X + X̅

L.H.S=1 [∵ Complement law ⇒ A + A̅ = 1]

∴ L.H.S = R.H.S

**Option: iv)**

1 + X = 1

L.H.S ⇒ 1 + X

L.H.S = 1 [∵Annulment law ⇒ A + 1 = 1]

**∴ L.H.S = R.H.S**