JEE Main 2013Session, Shift 1MathematicsHardMathematical Reasoning115× since 2008Consider : Statement − I : (p∧∼q)∧(∼p∧q)\left( {p \wedge \sim q} \right) \wedge \left( { \sim p \wedge q} \right)(p∧∼q)∧(∼p∧q) is a fallacy. Statement − II :(p→q)↔(∼q→∼p)\left( {p \to q} \right) \leftrightarrow \left( { \sim q \to \sim p} \right)(p→q)↔(∼q→∼p) is a tautology.AOption A. Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-IBOption B. Statement -I is True; Statement -II is False.COption C. Statement -I is False; Statement -II is TrueDOption D. Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-ICheckSkip · show solution
