Verified Solution • 2025–26 Session moderate

QJEEM22Ma0a27: JEE Main 2022 Mathematics - Class 11 JEEM22Ma0a27 Solution: The statement is :

The statement (∼(p⇔ ∼q))∧q(\sim(\mathrm{p} \Leftrightarrow \,\sim \mathrm{q})) \wedge \mathrm{q} is :

Multiple Choice Options

  • A a tautology
  • B a contradiction
  • C equivalent to (p⇒q)∧q(p \Rightarrow q) \wedge q
  • D equivalent to (p⇒q)∧p(p \Rightarrow q) \wedge p

Step-by-Step Verified Solution

∼\sim (p ⇔\Leftrightarrow ∼\sim q) ∧\wedge q

= (p ⇔\Leftrightarrow q) ∧\wedge q

p q p ↔\leftrightarrow q (p →\to q) ∧\wedge q (p →\to q) (p →\to q) ∧\wedge q (p →\to q) ∧\wedge p
T T T T T T T
T F F F F F F
F T F F T T F
F F T F T F F

∴\therefore ( ∼\sim (p ⇔\Leftrightarrow ∼\sim q)) ∧\wedge q is equivalent to (p ⇒\Rightarrow q) ∧\wedge p.

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