Let be such that is a tautology. Then
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
For the expression to be a tautology, it must be true for all truth values of p and q.
Take the critical case p = T, q = F: here (p→q) = F.
If Δ = ∨, the expression becomes F ∨ (p ∇ q), which is true whenever (p ∇ q) is true. With p = T, q = F: p ∨ q = T and p ∧ q = F. So ∇ = ∨ gives (p ∇ q) = T, making F ∨ T = T ✓.
Check all rows with Δ = ∨, ∇ = ∨: the expression is (p→q) ∨ (p∨q). The only row where (p→q) = F is p=T, q=F, but then p∨q = T, so F∨T = T. All other rows have (p→q) = T, so the disjunction is always T. Hence it's a tautology.
Take the critical case p = T, q = F: here (p→q) = F.
If Δ = ∨, the expression becomes F ∨ (p ∇ q), which is true whenever (p ∇ q) is true. With p = T, q = F: p ∨ q = T and p ∧ q = F. So ∇ = ∨ gives (p ∇ q) = T, making F ∨ T = T ✓.
Check all rows with Δ = ∨, ∇ = ∨: the expression is (p→q) ∨ (p∨q). The only row where (p→q) = F is p=T, q=F, but then p∨q = T, so F∨T = T. All other rows have (p→q) = T, so the disjunction is always T. Hence it's a tautology.
[JEE Main 2023 · 25 Jan · Shift 2]