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Among the statements (S1) : $$(p \Rightarrow q) \vee((\sim p) \wedge q)$$ is a tautology (S2) : $$(q \Rightarrow p) \Rightarrow((\sim p) \wedge q)$$ is a contradiction

JEE · Math · previous-year question

  1. A.neither (S1) and (S2) is Truecorrect
  2. B.only (S2) is True
  3. C.both $$(\mathrm{S} 1)$$ and $$(\mathrm{S} 2)$$ are True
  4. D.only (S1) is True

Answer

A. neither (S1) and (S2) is True

Explanation

(S1) : $$(p \Rightarrow q) \vee((\sim p) \wedge q)$$ $$ \begin{array}{|c|c|c|c|c|c|} \hline \mathrm{P} & \mathrm{Q} & \sim p & \sim p \wedge q & p \Rightarrow q & \begin{array}{c} (p \Rightarrow q) \vee \\ (\sim p \wedge q) \end{array} \\ \hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\ \hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} \\ \hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\ \hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\ \hline \end{array} $$ Here, for every value of components compound statement (last column) are not true. Hence, $S_1$ is not a tautology. $S_2:(q \Rightarrow p) \Rightarrow((\sim p) \wedge q)$

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