Let F _ 1 (A, B, C) = (A ∧ ~ B) ∨ [~ C ∧ (A ∨ B)] ∨ ~ A and F _ 2 (A, B) = (A ∨ B) ∨ (B → ~ A) be two logical expressions. Then :

Correct answer: B. F _ 1 is not a tautology but F _ 2 is a tautology F _ 1 : (A ∧ ~ B) ∨ [~ C ∧ (A ∨ B)] ∨ ~ A F _ 2 : (A ∨ B) ∨ (B → ~ A) F _ 1 : \ (A ∧ ~ B) ∨ ~ A\ ∨ [(A ∨ B) ∧ ~ C] : \ (A ∨ ~ A) ∧ (~ A ∨ ~ B)\ ∨ [(A ∨ B) ∧ ~ C] : \ t ∧ (~ A ∨ ~ B)\ ∨ [(A ∨ B) ∧ ~ C] : (~ A ∨ ~ B) ∨ [(A ∨ B) ∧ ~ C] : [(~ A ∨ ~ B) ∨ (A ∨ B)] _ ⏟ _ t ∧ [(~ A ∨ ~ B) ∧ ~ C] F _ 1 : (~ A ∨ ~ B) ∧ ~ C ≠ t (tautology) F _ 2 : (A ∨ B) ∨ (~ B ∨ ~ A) = t (tautology)

JEE Main 202126 FebruaryEvening Shift

Exam: JEE Main
Year: 2021
Date: 26 February
Shift: Evening Shift
Subject: Mathematics
Chapter: Mathematical Reasoning

JEE Main 2021 · 26 February · Evening Shift · Mathematics · Mathematical Reasoning

Let F _ 1 (A, B, C) = (A ∧ ~ B) ∨ [~ C ∧ (A ∨ B)] ∨ ~ A and F _ 2...

Let

{F}_{1}\left(A, B, C\right) = \left(A ∧ ~ B\right) ∨ \left[~ C ∧ \left(A ∨ B\right)\right] ∨ ~ A

and

{F}_{2}\left(A, B\right) = \left(A ∨ B\right) ∨ \left(B \to ~ A\right)

be two logical expressions. Then :

Options

A ${F}_{1}$ is a tautology but ${F}_{2}$ is not a tautology
B ${F}_{1}$ is not a tautology but ${F}_{2}$ is a tautology
CBoth ${F}_{1}$ and ${F}_{2}$ are not tautologies
D ${F}_{1}$ and ${F}_{2}$ both are tautologies

Correct answer

B. ${F}_{1}$ is not a tautology but ${F}_{2}$ is a tautology

Step-by-step solution

${F}_{1}: \left(A ∧ ~ B\right) ∨ \left[~ C ∧ \left(A ∨ B\right)\right] ∨ ~ A$

${F}_{2}: \left(A ∨ B\right) ∨ \left(B \to ~ A\right)$

${F}_{1}: \left\{\left(A ∧ ~ B\right) ∨ ~ A\right\} ∨ \left[\left(A ∨ B\right) ∧ ~ C\right]$

$: \left\{\left(A ∨ ~ A\right) ∧ \left(~ A ∨ ~ B\right)\right\} ∨ \left[\left(A ∨ B\right) ∧ ~ C\right]$

$: \left\{t ∧ \left(~ A ∨ ~ B\right)\right\} ∨ \left[\left(A ∨ B\right) ∧ ~ C\right]$

$: \left(~ A ∨ ~ B\right) ∨ \left[\left(A ∨ B\right) ∧ ~ C\right]$

$: \mathop{\mathop{\left[\left(~ A ∨ ~ B\right) ∨ \left(A ∨ B\right)\right]}_{⏟}}_{t} ∧ \left[\left(~ A ∨ ~ B\right) ∧ ~ C\right]$

${F}_{1}: \left(~ A ∨ ~ B\right) ∧ ~ C \ne t$ (tautology)

${F}_{2}: \left(A ∨ B\right) ∨ \left(~ B ∨ ~ A\right) = t$ (tautology)

Practice Mathematical Reasoning on Quantrex Academy →

More from Mathematical Reasoning

Negation of p ∧ (q ∧ ~ (p ∧ q)) is JEE Main 2023 The statement ( p ∧ ( ~ q ) ) ∨ ( ( ~ p ) ∧ q ) ∨ ( ( ~ p ) ∧ ( ~ q ) ) is equivalent to… JEE Main 2023 The negation of the statement ( ( A ∧ ( B ∨ C ) ) ⇒ ( A ∨ B ) ) ⇒ A is JEE Main 2023 Among the two statements ( S _ 1 ): (p ⇒ q) ∧ (p ∧ (~ q)) is a contradiction and ( S _ 2… JEE Main 2023 The converse of ((~ p) ∧ q) ⇒ r is JEE Main 2023 The number of ordered triplets of the truth values of p, q and r such that the truth… JEE Main 2023 The statement ~ [p ∨ (~ (p ∧ q))] is equivalent to JEE Main 2023 The negation of the statement (p ∨ q) ∧ (q ∨ (~ r)) is JEE Main 2023 The negation of ( p ∧ ( - q ) ) ∨ ( - p ) is equivalent to JEE Main 2023 Negation of (p → q) → (q → p) is JEE Main 2023 Among the statements (S1): (p ⇒ q) ∨ ((~ p) ∧ q) is a tautology (S2): (q ⇒ p) ⇒ ((~ p) ∧… JEE Main 2023 Statement ( P ⇒ Q ) ∧ ( R ⇒ Q ) is logically equivalent to JEE Main 2023 Which of the following statements is a tautology? JEE Main 2023 The negation of the expression q ∨ ( ( ~ q ) ∧ p ) is equivalent to JEE Main 2023 The number of values of r ∈ \ p, q, ~ p, ~ q\ for which ((p ∧ q) ⇒ (r ∨ q) ∧ ( (p ∧ r) ⇒… JEE Main 2023 Consider: (S1): (p ⇒ q) ∨ (p ∧ (~ q)) is a tautology. (S2): ((~ p) ⇒ (~ q)) ∧ ((~ p) ∨ q)… JEE Main 2023 Full Mathematical Reasoning list Mathematical Reasoning JEE Maths PYQs JEE Mathematics PYQs All JEE Main PYQs