Sets, Relations & Functions
Mathematical Reasoning / Logical Statements
Grade 11
Question:
<p><strong>Statement-1:</strong> \(\sim (p \leftrightarrow \sim q)\) is equivalent to \(p \leftrightarrow q\).<br><strong>Statement-2:</strong> \(\sim (p \leftrightarrow \sim q)\) is a tautology.</p>
<p>Statement-1 is true, Statement-2 is false.</p>
<p>Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.</p>
<p>Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.</p>
<p>Statement-1 is false, Statement-2 is true.</p>
Step-by-Step Solution
Key Concept: Recognize that p ↔ ~q is logically equivalent to ~(p ↔ q), so negating it gives ~(p ↔ ~q) ≡ p ↔ q. A biconditional is a tautology only if it's true for all truth value assignments, which p ↔ q is not.
<p><strong>Step 1: Verify Statement-1 using truth table for ~(p ↔ ~q)</strong></p><p>When p=T, q=T: ~(T ↔ F) = ~F = T; and p ↔ q = T ↔ T = T ✓</p><p>When p=T, q=F: ~(T ↔ T) = ~T = F; and p ↔ q = T ↔ F = F ✓</p><p>When p=F, q=T: ~(F ↔ F) = ~T = F; and p ↔ q = F ↔ T = F ✓</p><p>When p=F, q=F: ~(F ↔ T) = ~F = T; and p ↔ q = F ↔ F = T ✓</p><p><strong>∴ Statement-1 is TRUE: ~(p ↔ ~q) ≡ p ↔ q</strong></p><p><strong>Step 2: Check if p ↔ q is a tautology</strong></p><p>From the truth table above, p ↔ q is FALSE when p and q have different truth values (rows 2 and 3). A tautology must be true for ALL assignments.</p><p><strong>∴ Statement-2 is FALSE: p ↔ q is NOT a tautology</strong></p><p><strong>Conclusion:</strong> Statement-1 is true but Statement-2 is false.</p><p>∴ Answer: A</p>
Correct Answer: A