<p>If R = {(x, y) : x, y ∈Z, x2 + 3y2 ≤8}, then the domain of R−1 is:</p>
Step-by-Step Solution
Key Concept: domain(R-1) = range(R) = {y \in Z : \existsx \in Z with x2 + 3y2 \leq8}. Find all integers y for which
the constraint is satisfiable.
<p><strong>Step 1</strong>: Set x = 0 (gives maximum room for y): 3y2 \leq8 \Rightarrow y2 \leq2.67, so |y| \leq1. For y = \pm2: 3(4) = 12 > 8</p><br>even with x = 0 — impossible.<br>2<br><br>JEE Main 2019–2024 | Relations<br>Complete Solutions Booklet<p><strong>Step 2</strong>: Verify y \in {-1, 0, 1} all work:</p><br>y<br>Constraint<br>Valid?<br>0<br>x2 \leq8 \Rightarrow x \in {-2, -1, 0, 1, 2}<br>✓<br>1<br>x2 \leq5 \Rightarrow x \in {-2, -1, 0, 1, 2}<br>✓<br>-1<br>same as y = 1<br>✓<p><strong>Step 3</strong>: Distractor: domain(R) = {-2, -1, 0, 1, 2} is option (3) — designed to trap you.</p>
Correct Answer: 2