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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Sets and Relations

JEE Main 2026 · 22 January · Evening Shift · Mathematics · Sets and Relations

The number of elements in the relation R=\ (x, y): 4 x^ 2 +y^ 2...

The number of elements in the relation R={(x,y):4x2+y2<52,x,y∈Z}\mathrm{R}=\left\{(x, y): 4 x^{2}+y^{2}<52, x, y \in \mathbf{Z}\right\} is

Options

  1. A77
  2. B67
  3. C86
  4. D89

Correct answer

A. 77

Step-by-step solution

We need 4x2+y2<524x^2 + y^2 < 52 with x,y∈Zx, y \in \mathbb{Z}. This requires x2<13x^2 < 13, so x∈{−3,−2,−1,0,1,2,3}x \in \{-3,-2,-1,0,1,2,3\}.
x=0x=0: y2<52⇒y∈{−7,...,7}y^2<52 \Rightarrow y \in \{-7,...,7\} → 15 values
x=±1x=\pm1: y2<48⇒∣y∣≤6y^2<48 \Rightarrow |y|\leq 6 → 13 each → 26 values
x=±2x=\pm2: y2<36⇒∣y∣≤5y^2<36 \Rightarrow |y|\leq 5 → 11 each → 22 values
x=±3x=\pm3: y2<16⇒∣y∣≤3y^2<16 \Rightarrow |y|\leq 3 → 7 each → 14 values
Total =15+26+22+14=77= 15+26+22+14 = 77
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