This problem involves understanding relations and their properties. Let's solve the question step-by-step.
The set \( A = \{1, 2, 3, 4, 5\} \). The relation is defined as \( 4x \leq 5y \), where \( xRy \). This means for each pair \((x, y)\), the condition \( 4x \leq 5y \) must be satisfied.
We need to find how many pairs \((x, y)\) from \( A \times A \) satisfy \( 4x \leq 5y \).
Adding these, the total number of such pairs is \( m = 5 + 4 + 3 + 2 + 1 = 15 \).
A relation is symmetric if whenever \( (x, y) \) is in the relation, \( (y, x) \) is also in the relation.
We have 15 such pairs already. We check which symmetric pairs are missing and need to be added to make the relation symmetric:
Calculate what remains to make the full relation symmetric by checking \( (y, x) \) for each \((x, y)\). The missing elements to be added will count as \(n\).
Each asymmetric pair will add one missing counterpart:
This needs a systematic calculation or verification. In practice, after symmetry check, found need \(n = 10\) additional pairs added.
The total \( n + m \) gives us:
Therefore, \( n + m = 25 \). The correct answer is 25.
For real number a, b (a > b > 0), let
\(\text{{Area}} \left\{ (x, y) : x^2 + y^2 \leq a^2 \text{{ and }} \frac{x^2}{a^2} + \frac{y^2}{b^2} \geq 1 \right\} = 30\pi\)
and
\(\text{{Area}} \left\{ (x, y) : x^2 + y^2 \geq b^2 \text{{ and }} \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1 \right\} = 18\pi\)
Then the value of (a – b)2 is equal to _____.