Sets, Relations & Functions
General
Grade 11
Question:
<p>For α ∈N, consider R = {(x, y) : 7 | (3x + αy)} on N. R is an equivalence relation if and
only if:</p>
<p>α = 14</p>
<p>α is a multiple of 4</p>
<p>4 is the remainder when α is divided by 10</p>
<p>4 is the remainder when α is divided by 7</p>
Step-by-Step Solution
Key Concept: Start with reflexivity — it immediately pins down the divisibility condition on \alpha. Symmetry and
transitivity then follow automatically.
<p><strong>Step 1</strong>: Reflexivity forces the condition. (x, x) \in R for all x \in N requires 7 | x(3 + \alpha) for all x. This holds</p><br>5<br><br>JEE Main 2019–2024 | Relations<br>Complete Solutions Booklet<br>universally only if 7 | (3 + \alpha), i.e. \alpha \equiv4 (mod 7).<p><strong>Step 2</strong>: Symmetry (when \alpha \equiv4 (mod 7)): Given 7 | (3x + \alphay). With \alpha = 4: from 3x + 4y \equiv0 (mod 7), one</p><br>checks 3y + 4x \equiv0 (mod 7). ✓<p><strong>Step 3</strong>: Transitivity: If 7 | (3x + \alphay) and 7 | (3y + \alphaz), adding gives</p><br>7 |<br>(3x + \alphay) + (3y + \alphaz)<br> = 3x + (\alpha + 3)y + \alphaz.<br>Since 7 | (\alpha + 3), the middle term vanishes, leaving 7 | (3x + \alphaz). ✓<p><strong>Step 4</strong>: Check options: \alpha = 14: 3 + 14 = 17, 17 \div 7 leaves remainder 3 ̸= 4. ✗</p><br>Only option (4) characterises all<br>valid \alpha.
Correct Answer: 4