JEE Main 24 June 2022 Shift 1 Solved Paper

Then (p∆∼q)∨(∼p∆q) becomes
(p∨∼q)∨(∼p∨q) which is always true, so x becomes a tautology.
Case-II
When ∆ is same as ∧
Then (p∧q)⇒(p∧∼q)∨(∼p∧q)
If p∧q is T, then (p∧∼q)∨(∼p∧q) is F
so x cannot be a tautology.
Case-III
When ∆ is same as ⇒Then (p⇒∼q)∨(∼p⇒q) is same at (∼p∨∼q)∨(p∨q), which is always true, so x becomes a tautology.
Case-IV
When ∆ is same as ⇔
Then (p⇔q)⇒(p⇔∼q)∨(∼p⇔q)
p⇔q is true when p and q have same truth values, then p⇔∼q and ∼p⇔q both are false. Hence x cannot be a tautology.