JEE Main 2023 · 29 Jan · Shift 1

JEE Main 2023 Mathematics question on Mathematical Reasoning

If pp, qq and rr are three propositions, then which of the following combination of truth values of pp, qq and rr makes the logical expression {(p∨q)∧((∼p)∨r)}→((∼q)∨r)\{(p \vee q) \wedge ((\sim p) \vee r)\} \rightarrow ((\sim q) \vee r) false?
  1. A.
    p=T, q=F, r=Fp = T,\ q = F,\ r = F
  2. B.
    p=F, q=T, r=Fp = F,\ q = T,\ r = F
  3. C.
    p=T, q=F, r=Tp = T,\ q = F,\ r = T
  4. D.
    p=T, q=T, r=Fp = T,\ q = T,\ r = F
Show answer and solution

Correct option: B

Solution

For an implication A → B to be false, we need A = T and B = F.

With p = F, q = T, r = F:

Antecedent (p ∨ q) ∧ (¬p ∨ r) = (F ∨ T) ∧ (T ∨ F) = T ∧ T = T

Consequent ¬q ∨ r = F ∨ F = F

Since antecedent is T and consequent is F, the implication is F. This is the only combination where the consequent (¬q ∨ r) is false (requiring q = T, r = F) while simultaneously the antecedent remains true (p = F makes ¬p ∨ r = T, and p ∨ q = T since q = T).

[JEE Main 2023 · 29 Jan · Shift 1]

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