Question

Let \(A = \{-2, -1, 0, 1, 2\}\). A relation \(R\) is defined on set \(A\) such that \(aRb \Rightarrow 1 + ab > 0\).
Statement-1: It is an equivalence relation.
Statement-2: Number of elements in \(R\) is 17.

• A. Statement 1 and 2 are false
• B. Statement 1 is true and statement 2 is false
• C. Statement 1 is false and statement 2 is true
• D. Statement 1 and 2 are true

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
An equivalence relation must be Reflexive, Symmetric, and Transitive. We check each property for the condition $1 + ab > 0$. Then, we list all pairs $(a, b)$ from $A × A$ that satisfy the condition.

Step 2: Key Formula or Approach:
1. **Reflexive:** $1 + a² > 0$. Always true for all $a \in A$. (Yes) 2. **Symmetric:** If $1 + ab > 0$, then $1 + ba > 0$. (Yes) 3. **Transitive:** If $1 + ab > 0$ and $1 + bc > 0$, does $1 + ac > 0$? Let $a = -2, b = 0, c = 2$. $1 + (-2)(0) = 1 > 0$ (True) $1 + (0)(2) = 1 > 0$ (True) $1 + (-2)(2) = 1 - 4 = -3$ (False). Thus, it is **not** transitive. **Statement 1 is false.**

Step 3: Detailed Explanation:
List pairs $(a, b)$ such that $ab > -1$: - For $a = -2$: $b$ can be $-2, -1, 0$ ($ab = 4, 2, 0$). (3 pairs) - For $a = -1$: $b$ can be $-2, -1, 0$ ($ab = 2, 1, 0$). (3 pairs) - For $a = 0$: $b$ can be $-2, -1, 0, 1, 2$ ($ab = 0$ for all). (5 pairs) - For $a = 1$: $b$ can be $0, 1, 2$ ($ab = 0, 1, 2$). (3 pairs) - For $a = 2$: $b$ can be $0, 1, 2$ ($ab = 0, 2, 4$). (3 pairs) Total pairs = $3 + 3 + 5 + 3 + 3 = 17$. **Statement 2 is true.**

Step 4: Final Answer:
Statement 1 is false and Statement 2 is true.