A relation \( R = \{(a, b) : a = b - 2, b \geq 6 \} \) is defined on the set \( \mathbb{N} \). Then the correct answer will be:
The given relation is \( R = \{(a, b) : a = b - 2, b \geq 6\} \).
For each pair:
For \( (2, 4) \): \( a = 4 - 2 = 2 \) and \( b = 4 \).
Since \( b \geq 6 \) is not satisfied, \( (2, 4) \notin R \).
For \( (3, 8) \): \( a = 8 - 2 = 6 \) and \( b = 8 \). Since \( a \neq 3 \), \( (3, 8) \notin R \).
For \( (6, 8) \): \( a = 8 - 2 = 6 \) and \( b = 8 \).
Both conditions \( a = b - 2 \) and \( b \geq 6 \) are satisfied, so \( (6, 8) \in R \).
For \( (8, 7) \): \( a = 7 - 2 = 5 \) and \( b = 7 \).
Since \( a \neq 8 \), \( (8, 7) \notin R \).
Thus, the correct pair is \( (6, 8) \).
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.
(b) Order of the differential equation: $ 5x^3 \frac{d^3y}{dx^3} - 3\left(\frac{dy}{dx}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^4 + y = 0 $