JEE MAIN - Mathematics (2025 - 3rd April Morning Shift - No. 5)
Explanation
The relation $ R $ is defined for the set $ \mathrm{A} = \{-3, -2, -1, 0, 1, 2, 3\} $ with the condition $ 0 \leq x^2 + 2y \leq 4 $. Let's determine the pairs $(x, y)$ that satisfy this condition.
For $ y = -3 $:
Solving $ x^2 + 2(-3) \leq 4 $, we find $ x = \{3, -3\} $.
For $ y = -2 $:
Solving $ x^2 + 2(-2) \leq 4 $, we find $ x = \{-2, 2\} $.
For $ y = -1 $:
Solving $ x^2 + 2(-1) \leq 4 $, we find $ x = \{-2, 2\} $.
For $ y = 0 $:
Solving $ x^2 + 0 \leq 4 $, we find $ x = \{-2, -1, 0, 1, 2\} $.
For $ y = 1 $:
Solving $ x^2 + 2(1) \leq 4 $, we find $ x = \{-1, 0, 1\} $.
For $ y = 2 $:
Solving $ x^2 + 2(2) \leq 4 $, we find $ x = \{0\} $.
The relation $ R $ consists of the following pairs:
$ R = \{(3, -3), (-3, -3), (-2, -2), (2, -2), (-2, -1), (2, -1), (-2, 0), (-1, 0), (0, 0), (1, 0), (2, 0), (-1, 1), (0, 1), (1, 1), (0, 2)\} $
Currently, $ R $ has $ l = 15 $ elements. To make $ R $ reflexive, it must include all pairs $(x, x)$ for every $ x \in \mathrm{A} $. We identify the missing reflexive pairs $(-1, -1)$, $(2, 2)$, and $(3, 3)$, which are required to satisfy reflexivity.
Thus, $ m = 3 $ more elements are needed. Therefore, the total $ l + m = 15 + 3 = 18 $.
Comments (0)
