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JEE Mathematics 2023 Question with Solution

Statement (P⇒Q)∧(R⇒Q)(P\Rightarrow Q) \wedge (R\Rightarrow Q) is logically equivalent to:

  • A

    (P∨R)⇒Q(P\vee R)\Rightarrow Q

  • B

    (P⇒R)∨(Q⇒R)(P\Rightarrow R)\vee(Q\Rightarrow R)

  • C

    (P⇒R)∧(Q⇒R)(P\Rightarrow R)\wedge(Q\Rightarrow R)

  • D

    (P∧R)⇒Q(P\wedge R)\Rightarrow Q

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The statement is (P⇒Q)∧(R⇒Q)(P\Rightarrow Q) \wedge (R\Rightarrow Q).

Find: A logically equivalent statement among the given options.

Using implication form,

P⇒Q≡¬P∨Q,R⇒Q≡¬R∨QP\Rightarrow Q \equiv \neg P \vee Q, \qquad R\Rightarrow Q \equiv \neg R \vee Q

Therefore,

(P⇒Q)∧(R⇒Q)≡(¬P∨Q)∧(¬R∨Q)(P\Rightarrow Q) \wedge (R\Rightarrow Q) \equiv (\neg P \vee Q) \wedge (\neg R \vee Q)

Now apply the distributive law:

(¬P∨Q)∧(¬R∨Q)≡(¬P∧¬R)∨Q(\neg P \vee Q) \wedge (\neg R \vee Q) \equiv (\neg P \wedge \neg R) \vee Q

Also, by De Morgan's law,

¬P∧¬R≡¬(P∨R)\neg P \wedge \neg R \equiv \neg(P \vee R)

So,

(¬P∧¬R)∨Q≡¬(P∨R)∨Q(\neg P \wedge \neg R) \vee Q \equiv \neg(P \vee R) \vee Q

Again using implication form in reverse,

¬(P∨R)∨Q≡(P∨R)⇒Q\neg(P \vee R) \vee Q \equiv (P \vee R)\Rightarrow Q

Therefore, the statement is logically equivalent to (P∨R)⇒Q(P\vee R)\Rightarrow Q. The correct option is A.

Truth Table Verification

Given: The solution table compares the truth values of the statement with candidate equivalent forms.

Find: Which option has the same truth values as the original statement.

From the provided truth table, the column corresponding to the combined condition matches the column for (P∨R)⇒Q(P\vee R)\Rightarrow Q.

Hence the option having identical truth values in all cases is A. Therefore, the logically equivalent statement is (P∨R)⇒Q(P\vee R)\Rightarrow Q.

Common mistakes

  • Confusing and with or while combining the implications. Here the given statement is (P⇒Q)∧(R⇒Q)(P\Rightarrow Q) \wedge (R\Rightarrow Q), not a disjunction. First preserve the connective, then simplify algebraically.

  • Using the wrong distributive law. From (¬P∨Q)∧(¬R∨Q)(\neg P \vee Q) \wedge (\neg R \vee Q), the correct reduction is (¬P∧¬R)∨Q(\neg P \wedge \neg R) \vee Q, not (¬P∨¬R)∧Q(\neg P \vee \neg R) \wedge Q.

  • Applying De Morgan's law incorrectly. Since (¬P∧¬R)≡¬(P∨R)(\neg P \wedge \neg R) \equiv \neg(P\vee R), writing it as ¬(P∧R)\neg(P\wedge R) is wrong. Check whether the inner connective changes from ∧\wedge to ∨\vee.

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