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JEE Main 202127 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
27 July
Shift
Morning Shift
Subject
Mathematics
Chapter
Mathematical Reasoning

JEE Main 2021 · 27 July · Morning Shift · Mathematics · Mathematical Reasoning

The compound statement (P ∨ Q) ∧ (~ P) ⇒ Q equivalent to:

The compound statement (P∨Q)∧( P)⇒Q\left(P ∨ Q\right) ∧ \left(~ P\right) ⇒ Q equivalent to:

Options

  1. AP∨QP ∨ Q
  2. BP∧ QP ∧ ~ Q
  3. C (P⇒Q)~ \left(P ⇒ Q\right)
  4. D (P⇒Q)⇔P∧ Q~ \left(P ⇒ Q\right) ⇔ P ∧ ~ Q

Correct answer

D.  (P⇒Q)⇔P∧ Q~ \left(P ⇒ Q\right) ⇔ P ∧ ~ Q

Step-by-step solution

PP QQ P∨QP ∨ Q  P~ P \left(P ∨ Q\right) ∧ ~ P</td><td>(P∨Q)∧ P⇒Q</td> <td>\left(P ∨ Q\right) ∧ ~ P ⇒ Q
TT TT TT FF FF TT
TT FF TT FF FF TT
FF TT TT TT TT TT
FF FF FF TT FF TT

PP QQ  Q~ Q P∧ QP ∧ ~ Q P⇒QP ⇒ Q  (P⇒Q)</td></tr><tr><td>~ \left(P ⇒ Q\right)</td> </tr> <tr> <td>T</td><td></td> <td>T</td><td></td> <td>F</td><td></td> <td>F</td><td></td> <td>T</td><td></td> <td>F</td></tr><tr><td></td> </tr> <tr> <td>T</td><td></td> <td>F</td><td></td> <td>T</td><td></td> <td>T</td><td></td> <td>F</td><td></td> <td>T</td></tr><tr><td></td> </tr> <tr> <td>F</td><td></td> <td>T</td><td></td> <td>F</td><td></td> <td>F</td><td></td> <td>T</td><td></td> <td>F</td></tr><tr><td></td> </tr> <tr> <td>F</td><td></td> <td>F</td><td></td> <td>T</td><td></td> <td>F</td><td></td> <td>T</td><td></td> <td>F</td></tr></tbody></table><p></p><tableborder="1"cellpadding="1"cellspacing="1"style="width:500px;"><tbody><tr><td></td> </tr> </tbody></table><p> </p><table border="1" cellpadding="1" cellspacing="1" style="width: 500px;"> <tbody> <tr> <td>~ \left(P ⇒ Q\right)</td><td></td> <td>P ∧ ~ Q</td><td></td> <td>~ \left(P ⇒ Q\right) ⇔ P ∧ ~ Q</td></tr><tr><td></td> </tr> <tr> <td>F</td><td></td> <td>F</td><td></td> <td>T</td></tr><tr><td></td> </tr> <tr> <td>T</td><td></td> <td>T</td><td></td> <td>T</td></tr><tr><td></td> </tr> <tr> <td>F</td><td></td> <td>F</td><td></td> <td>T</td></tr><tr><td></td> </tr> <tr> <td>F</td><td></td> <td>F</td><td></td> <td>T</td></tr></tbody></table><p>Hence,(P∨Q)∧( P)⇒Q</td> </tr> </tbody></table><p>Hence, \left(P ∨ Q\right) ∧ \left(~ P\right) ⇒ Q equivalent to  (P⇒Q)⇔P∧ Q~ \left(P ⇒ Q\right) ⇔ P ∧ ~ Q

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