Sets, Relations & Functions
General
Grade 11
Question:
<p>Let A = {1, 2, 3, 4, 5, 6, 7}. The relation R = {(x, y) ∈A × A : x + y = 7} is:</p>
<p>Transitive but neither symmetric nor reflexive</p>
<p>Reflexive but neither symmetric nor transitive</p>
<p>An equivalence relation</p>
<p>Symmetric but neither reflexive nor transitive</p>
Step-by-Step Solution
Key Concept: Enumerate R explicitly for this small set. Reflexivity requires 2x = 7 (no integer solution).
Symmetric because x + y = y + x. Transitivity fails via the symmetric pair trick.
<p><strong>Step 1</strong>: List R: {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}. (Note: 7 + y = 7 \Rightarrow y = 0 /\in A, so 7 appears in no pair.)</p><p><strong>Step 2</strong>: Reflexive fails: (x, x) \in R requires 2x = 7 — no integer solution. ✗</p><p><strong>Step 3</strong>: Symmetric: x + y = 7 \Rightarrow y + x = 7. For every (x, y) \in R, (y, x) \in R. ✓</p><br>12<br><br>JEE Main 2019–2024 | Relations<br>Complete Solutions Booklet<p><strong>Step 4</strong>: Transitivity fails: (1, 6) \in R and (6, 1) \in R \Rightarrow need (1, 1) \in R. But 1 + 1 = 2 ̸= 7. ✗</p>
Correct Answer: 4