Verified Solution • 2025–26 Session moderate

QJEEM22Maa744: JEE Main 2022 Mathematics - Class 11 JEEM22Maa744 Solution: Let the operations . If is a tautology, then the ordered pair…

Let the operations ∗,⊙∈{∧,∨}*, \odot \in\{\wedge, \vee\} . If (p∗q)⊙(p ⊙∼q)(\mathrm{p} * \mathrm{q}) \odot(\mathrm{p}\, \odot \sim \mathrm{q}) is a tautology, then the ordered pair (∗,⊙)(*, \odot) is :

Multiple Choice Options

  • A (∨,∧)(\vee, \wedge)
  • B (∨,∨)(\vee, \vee)
  • C (∧,∧)(\wedge, \wedge)
  • D (∧,∨)(\wedge, \vee)

Step-by-Step Verified Solution

∗, ⊙∈{∧, ∨}* ,\, \odot \in \{ \wedge ,\, \vee \}

Now for (p∗q)⊙(p⊙∼q)(p * q) \odot (p \odot \sim q) is tautology

(A) (∨,∧):(p∨q)∧(p∧∼q)( \vee , \wedge ):(p \vee q) \wedge (p \wedge \sim q) not a tautology

(B) (∨,∨):(p∨q)∨(p∨∼q)( \vee , \vee ):(p \vee q) \vee (p \vee \sim q)

=P∨T= P \vee T is tautology

(C) (∧,∧):(p∧q)∧(p∧∼q)( \wedge , \wedge ):(p \wedge q) \wedge (p \wedge \sim q)

=(p∧p)∧(q∧∼q)=p∧F= (p \wedge p) \wedge (q \wedge \sim q) = p \wedge F not a tautology (Fallasy)

(D) (∧,∨):(p∧q)∨(p∨∼q)( \wedge , \vee ):(p \wedge q) \vee (p \vee \sim q) not a tautology

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