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The number of elements in the relation R = \(x, y) : 4x2 + y2 < 52, x, y Z \ is
(4) 77
4x2 < 52 x2 < 13 x \0, 1, 2, 3\
x = 0: y2 < 52 |y| 7 ⇒ 15 values
x = 1: y2 < 48 |y| 6 ⇒ 2 13 = 26
x = 2: y2 < 36 |y| 5 ⇒ 2 11 = 22
x = 3: y2 < 16 |y| 3 ⇒ 2 7 = 14
Number of elements = 15 + 26 + 22 + 14 = 77
Answer: (4)
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