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

Question

The number of elements in the relation $R = {(x, y) : 4 x^{2} + y^{2} < 52, x, y \in ℤ}$ is [2026]

Smart Achievers · Accepted Answer

(4)

$4 x^{2} + y^{2} < 52, x, y \in ℤ$

$↓$ $↓$

$0 0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 7 \to 1 \times 15 = 15$

$\pm 1$ $0, \pm 1, \pm 2, \pm 3, \dots . . . . ., \pm 6$ $\to 2 \times 13 = 26$

$\pm 2$ $0, \pm 1, \pm 2, \pm 3, . . . . . \dots, \pm 5$ $\to 2 \times 11 = 22$

$\pm 3$ $0, \pm 1, \pm 2, \pm 3$ $\to 2 \times 7 = 14$

$Number of elements = 15 + 26 + 22 + 14 = 77$

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