JEE Main 202331 Jan, Morning shiftMathematicsMediumMathematical Reasoning115× since 2008(S1) (p⇒q)∨(p∧(∼q))(\mathrm{S} 1)~(p \Rightarrow q) \vee(p \wedge(\sim q))(S1) (p⇒q)∨(p∧(∼q)) is a tautology (S2) ((∼p)⇒(∼q))∧((∼p)∨q)(\mathrm{S} 2)~((\sim p) \Rightarrow(\sim q)) \wedge((\sim p) \vee q)(S2) ((∼p)⇒(∼q))∧((∼p)∨q) is a contradiction. ThenAOption A. only (S2) is correctBOption B. both (S1) and (S2) are correctCOption C. only (S1) is correctDOption D. both (S1) and (S2) are wrongCheckSkip · show solution
