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JEE Main 2021
Mathematical Reasoning
Mathematical Reasoning
Easy

Question

Let *, ▢ \in{\wedge, \vee} be such that the Boolean expression (p * \sim q) \Rightarrow (p ▢ q) is a tautology. Then :

Options

Solution

Key Concepts and Formulas

  • Tautology: A statement that is always true, regardless of the truth values of its components.
  • Implication (→): pqp \rightarrow q is false only when pp is true and qq is false. Otherwise, it is true.
  • Logical OR (∨): pqp \vee q is true if either pp or qq or both are true. It is false only when both pp and qq are false.
  • Logical AND (∧): pqp \wedge q is true only when both pp and qq are true. Otherwise, it is false.
  • Logical NOT (∼): p\sim p is true when pp is false, and false when pp is true.

Step-by-Step Solution

Step 1: Analyze the given expression and the possible operators.

We are given the expression (pq)(pq)(p * \sim q) \Rightarrow (p \square q), where * and \square can be either \wedge (AND) or \vee (OR). We need to find the combination of * and \square that makes the expression a tautology. We will test each option using truth tables.

Step 2: Test option (A): * = \vee, \square = \vee.

The expression becomes (pq)(pq)(p \vee \sim q) \Rightarrow (p \vee q). Let's construct a truth table:

ppqqq\sim qpqp \vee \sim qpqp \vee q(pq)(pq)(p \vee \sim q) \Rightarrow (p \vee q)
TTFTTT
TFTTTT
FTFFTT
FFTTFF

Since the last column is not all T, option (A) is incorrect.

Step 3: Test option (B): * = \wedge, \square = \wedge.

The expression becomes (pq)(pq)(p \wedge \sim q) \Rightarrow (p \wedge q). Let's construct a truth table:

ppqqq\sim qpqp \wedge \sim qpqp \wedge q(pq)(pq)(p \wedge \sim q) \Rightarrow (p \wedge q)
TTFFTT
TFTTFF
FTFFFT
FFTFFT

Since the last column is not all T, option (B) is incorrect.

Step 4: Test option (C): * = \wedge, \square = \vee.

The expression becomes (pq)(pq)(p \wedge \sim q) \Rightarrow (p \vee q). Let's construct a truth table:

ppqqq\sim qpqp \wedge \sim qpqp \vee q(pq)(pq)(p \wedge \sim q) \Rightarrow (p \vee q)
TTFFTT
TFTTTT
FTFFTT
FFTFFT

Since the last column is all T, option (C) is correct.

Step 5: Test option (D): * = \vee, \square = \wedge.

The expression becomes (pq)(pq)(p \vee \sim q) \Rightarrow (p \wedge q). Let's construct a truth table:

ppqqq\sim qpqp \vee \sim qpqp \wedge q(pq)(pq)(p \vee \sim q) \Rightarrow (p \wedge q)
TTFTTT
TFTTFF
FTFFFT
FFTTFF

Since the last column is not all T, option (D) is incorrect.

Common Mistakes & Tips

  • Carefully construct the truth tables, paying close attention to the order of operations and the definitions of the logical operators. A single error in the truth table can lead to the wrong answer.
  • Remember that pqp \Rightarrow q is only false when pp is true and qq is false.
  • When testing options, start with simpler options first. If a simpler option works, it can save time.

Summary

We tested each possible combination of logical operators for * and \square in the given Boolean expression to determine which combination results in a tautology. By constructing truth tables for each option, we found that the expression (pq)(pq)(p \wedge \sim q) \Rightarrow (p \vee q) is a tautology, which corresponds to option (C).

Final Answer

The final answer is \boxed{C}, which corresponds to option (C).

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