Discussion
Let R be a relation in \(\mathbb{N}\) defined by \(\{(x: y): x + 3y = 10, x, y \in \mathbb{N}\}\)Then the range of R is
Answer: D
The range of R is the set \(\{1, 2, 3\}\), printed in the key's option as \(\{1, 2, 3,\}\).
Find the natural-number pairs with \(x + 3y = 10\), that is \(x = 10 - 3y \ge 1\):
- \(y = 1\): \(x = 7\)
- \(y = 2\): \(x = 4\)
- \(y = 3\): \(x = 1\)
\(y = 4\) would give \(x = -2\), which is not a natural number. So \(R = \{(7, 1), (4, 2), (1, 3)\}\) and its range, the set of second elements, is \(\{1, 2, 3\}\).
\(\{1, 2, 3, 4\}\) and \(\{2, 3, 4\}\) wrongly include 4, and \(\{1, 3, 4\}\) and \(\{1, 3\}\) leave out 2. The trailing comma in the keyed option is a printing slip; the set it lists is \(\{1, 2, 3\}\).
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