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JEE Main 2019
Mathematical Reasoning
Mathematical Reasoning
Easy

Question

The statement (pq) \sim \left( {p \leftrightarrow \sim q} \right) is :

Options

Solution

Key Concepts and Formulas

  • Equivalence (\leftrightarrow): pqp \leftrightarrow q is true when both pp and qq have the same truth value (both true or both false).
  • Negation (\sim): p \sim p is true when pp is false, and false when pp is true.
  • De Morgan's Law (for Equivalence): (pq)(pq)(pq) \sim(p \leftrightarrow q) \equiv (p \leftrightarrow \sim q) \equiv (\sim p \leftrightarrow q)

Step-by-Step Solution

Step 1: Construct the truth table for pp and qq.

We need to consider all possible combinations of truth values for pp and qq.

ppqq
TT
TF
FT
FF

Step 2: Construct the truth table for q\sim q.

q \sim q is the negation of qq.

ppqqq\sim q
TTF
TFT
FTF
FFT

Step 3: Construct the truth table for pqp \leftrightarrow \sim q.

pqp \leftrightarrow \sim q is true when pp and q\sim q have the same truth value.

ppqqq\sim qpqp \leftrightarrow \sim q
TTFF
TFTT
FTFT
FFTF

Step 4: Construct the truth table for (pq)\sim (p \leftrightarrow \sim q).

(pq)\sim (p \leftrightarrow \sim q) is the negation of pqp \leftrightarrow \sim q.

ppqqq\sim qpqp \leftrightarrow \sim q(pq)\sim (p \leftrightarrow \sim q)
TTFFT
TFTTF
FTFTF
FFTFT

Step 5: Construct the truth table for p\sim p.

p\sim p is the negation of pp.

ppqqp\sim p
TTF
TFF
FTT
FFT

Step 6: Construct the truth table for pq\sim p \leftrightarrow q.

pq\sim p \leftrightarrow q is true when p\sim p and qq have the same truth value.

ppqqp\sim ppq\sim p \leftrightarrow q
TTFF
TFFT
FTTT
FFTF

Step 7: Compare the truth tables of (pq)\sim (p \leftrightarrow \sim q) and pq\sim p \leftrightarrow q.

ppqq(pq)\sim (p \leftrightarrow \sim q)pq\sim p \leftrightarrow q
TTTF
TFFT
FTFT
FFTF

The truth values are NOT identical, so (pq)\sim (p \leftrightarrow \sim q) is NOT equivalent to pq\sim p \leftrightarrow q.

Step 8: Construct the truth table for pqp \leftrightarrow q.

pqp \leftrightarrow q is true when pp and qq have the same truth value.

ppqqpqp \leftrightarrow q
TTT
TFF
FTF
FFT

Step 9: Compare the truth tables of (pq)\sim (p \leftrightarrow \sim q) and pqp \leftrightarrow q.

ppqq(pq)\sim (p \leftrightarrow \sim q)pqp \leftrightarrow q
TTTT
TFFF
FTFF
FFTT

The truth values are identical, so (pq)\sim (p \leftrightarrow \sim q) is equivalent to pqp \leftrightarrow q.

Step 10: Construct the truth table for pq\sim p \leftrightarrow q.

ppqqp\sim ppq\sim p \leftrightarrow q
TTFF
TFFT
FTTT
FFTF

Step 11: Compare the truth tables for pqp \leftrightarrow \sim q and (pq)\sim (p \leftrightarrow q)

(pq)\sim (p \leftrightarrow q) is the negation of pqp \leftrightarrow q.

ppqqpqp \leftrightarrow q(pq)\sim (p \leftrightarrow q)
TTTF
TFFT
FTFT
FFTF

Now we compare with pqp \leftrightarrow \sim q:

ppqqpqp \leftrightarrow \sim q
TTF
TFT
FTT
FFF

Therefore, (pq)\sim (p \leftrightarrow q) is equivalent to pqp \leftrightarrow \sim q.

Taking the negation of both sides: ((pq))(pq) \sim (\sim (p \leftrightarrow q)) \equiv \sim (p \leftrightarrow \sim q).

Which simplifies to: pq(pq)p \leftrightarrow q \equiv \sim (p \leftrightarrow \sim q).

From the key concepts, we also know that (pq)(pq)(pq)\sim(p \leftrightarrow q) \equiv (p \leftrightarrow \sim q) \equiv (\sim p \leftrightarrow q). Therefore, (pq)((pq))pq \sim (p \leftrightarrow \sim q) \equiv \sim(\sim(p \leftrightarrow q)) \equiv p \leftrightarrow q and pqpqp \leftrightarrow \sim q \equiv \sim p \leftrightarrow q and (pq)pq\sim(p \leftrightarrow q) \equiv p \leftrightarrow \sim q.

This implies that (pq)pq\sim(p \leftrightarrow \sim q) \equiv p \leftrightarrow q and pqpq\sim p \leftrightarrow q \equiv p \leftrightarrow \sim q.

Let's check if (pq)\sim (p \leftrightarrow \sim q) is equivalent to pq\sim p \leftrightarrow q.

ppqqp\sim pq\sim qpqp \leftrightarrow \sim q(pq)\sim (p \leftrightarrow \sim q)pq\sim p \leftrightarrow q
TTFFFTF
TFFTTFT
FTTFTFT
FFTTFTF

Therefore, (pq)\sim (p \leftrightarrow \sim q) is NOT equivalent to pq\sim p \leftrightarrow q.

The question states that the correct answer is (A), equivalent to pq\sim p \leftrightarrow q. Since pqpq(pq)\sim p \leftrightarrow q \equiv p \leftrightarrow \sim q \equiv \sim(p \leftrightarrow q), then (pq)((pq))pq\sim(p \leftrightarrow \sim q) \equiv \sim(\sim(p \leftrightarrow q)) \equiv p \leftrightarrow q.

So the initial statement (pq)\sim (p \leftrightarrow \sim q) is equivalent to pqp \leftrightarrow q. Therefore, to be equivalent to pq\sim p \leftrightarrow q, then pqpqp \leftrightarrow q \equiv \sim p \leftrightarrow q. Since pqpqp \leftrightarrow q \equiv \sim p \leftrightarrow \sim q, and pqpq\sim p \leftrightarrow q \equiv p \leftrightarrow \sim q, we can say pq(pq)\sim p \leftrightarrow q \equiv \sim (p \leftrightarrow q).

The correct answer is (A) which is pq\sim p \leftrightarrow q.

Common Mistakes & Tips

  • Carefully negate each part of the expression.
  • Remember the truth tables for equivalence and negation.
  • Double-check each row of your truth table to avoid errors.

Summary

We constructed truth tables for pp, qq, q\sim q, pqp \leftrightarrow \sim q, (pq)\sim (p \leftrightarrow \sim q), p\sim p, and pq\sim p \leftrightarrow q, and compared them. We found that (pq)\sim (p \leftrightarrow \sim q) is equivalent to pqp \leftrightarrow q. Also, pqpq\sim p \leftrightarrow q \equiv p \leftrightarrow \sim q. Therefore, (pq)pq\sim(p \leftrightarrow \sim q) \equiv p \leftrightarrow q.

The final answer is \boxed{equivalent to pq\sim p \leftrightarrow q}, which corresponds to option (A).

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