JEE Main 2026 Jan 22 Shift 2, Mathematics Q3: Set

JEE Main2026Jan 22, Shift 2Mathematics

The number of elements in the relation $R = {(x, y) : 4 x^{2} + y^{2} < 52, x, y \in Z}$ is

Solution

Count integer pairs $(x, y)$ with $4 x^{2} + y^{2} < 52$ . Only $∣ x ∣ \leq 3$ works since $4 (4)^{2} = 64 > 52$ .

For each $x$ , count integer $y$ with $y^{2} < 52 - 4 x^{2}$ :

$x = 0$ : $y^{2} < 52 \Rightarrow y \in {- 7, \dots, 7}$ , giving 15 values.

$x = \pm 1$ : $y^{2} < 48 \Rightarrow ∣ y ∣ \leq 6$ , giving 13 values each, total $2 \times 13 = 26$ .

$x = \pm 2$ : $y^{2} < 36 \Rightarrow ∣ y ∣ \leq 5$ , giving 11 values each, total $2 \times 11 = 22$ .

$x = \pm 3$ : $y^{2} < 16 \Rightarrow ∣ y ∣ \leq 3$ , giving 7 values each, total $2 \times 7 = 14$ .

Total $= 15 + 26 + 22 + 14 = 77$ .

Concept behind this question

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