To solve this problem, we need to analyze the relation \( R \) defined on the set \( A = \{1, 2, 3, 4, 5\} \) by the condition \( xRy \) if and only if \( 4x \leq 5y \).
First, consider and list all pairs \((x, y)\) satisfying the condition \( 4x \leq 5y \):
| \( x \) | Possible \( y \) |
|---|---|
| 1 | \( (1, 1), (1, 2), (1, 3), (1, 4), (1, 5) \) |
| 2 | \( (2, 2), (2, 3), (2, 4), (2, 5) \) |
| 3 | \( (3, 3), (3, 4), (3, 5) \) |
| 4 | \( (4, 4), (4, 5) \) |
| 5 | \( (5, 5) \) |
Counting all these pairs, we have \( m = 15 \) elements in \( R \).
Next, we need to make \( R \) symmetric. A relation is symmetric if, whenever \((x, y) \in R\), then \((y, x) \in R\) as well. Our task is to make this relation symmetric by adding the minimum number of pairs.
Let's analyze:
Summing the extra pairs needed: \((3, 2), (4, 2), (5, 2), (4, 3), (5, 3), (5, 4)\), we must add \( n = 10 \) pairs to ensure all relations are symmetric.
Adding these pairs to the existing 15 elements in \( R \), we have \( m + n = 15 + 10 = 25 \).
The final answer is 25, which means:
Given: \( 4x \leq 5y \)
then
\[ R = \{(1,1), (1,2), (1,3), (1,4), (1,5), (2,2), (2,3), (2,4), (2,5), (3,3), (3,4), (3,5), (4,4), (4,5), (5,4), (5,5)\} \]
i.e., 16 elements.
i.e., \( n = 16 \)
Now to make \( R \) a symmetric relation, add:
\[ \{(2,1), (3,2), (4,3), (1,4), (2,5), (3,4), (1,5), (2,1)\} \]
i.e., \( m = 9 \)
So \( m + n = 25 \)
If the domain of the function $ f(x) = \log_7(1 - \log_4(x^2 - 9x + 18)) $ is $ (\alpha, \beta) \cup (\gamma, \delta) $, then $ \alpha + \beta + \gamma + \delta $ is equal to
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
Let $ A = \{0, 1, 2, 3, 4, 5, 6\} $ and $ R_1 = \{(x, y): \max(x, y) \in \{3, 4 \}$. Consider the two statements:
Statement 1: Total number of elements in $ R_1 $ is 18.
Statement 2: $ R $ is symmetric but not reflexive and transitive.
Let $ A $ be the set of all functions $ f: \mathbb{Z} \to \mathbb{Z} $ and $ R $ be a relation on $ A $ such that $$ R = \{ (f, g) : f(0) = g(1) \text{ and } f(1) = g(0) \} $$ Then $ R $ is: