Let `R_(1)` be a relation defined by `R_(1)={(ab)|ageb,a,bepsilonR}`. Then `R_(1)` is

Options

A. an equivalence relation `R`
B. Reflexive, transitive but not symmetric
C. Symmetric, Transitive but not reflexive
D. Neither transitive nor reflexive but symmetric

Explanation

To determine the properties of the relation \( R_1 = \{ (a, b) | a \geq b, a, b \in \mathbb{R} \} \), we will check if it is reflexive, symmetric, and transitive.

### Step 1: Check for Reflexivity
A relation is reflexive if for every element \( x \) in the set, the pair \( (x, x) \) is in the relation.

- For any \( x \in \mathbb{R} \), we need to check if \( x \geq x \).
- Since \( x \) is always equal to itself, \( x \geq x \) holds true.

Thus, \( (x, x) \in R_1 \) for all \( x \in \mathbb{R} \). Therefore, \( R_1 \) is reflexive.

### Step 2: Check for Symmetry
A relation is symmetric if for every pair \( (a, b) \in R \), the pair \( (b, a) \) is also in the relation.

- Suppose \( (a, b) \in R_1 \), which means \( a \geq b \).
- For symmetry, we need to check if \( b \geq a \) also holds.
- If \( a > b \), then \( b \) cannot be greater than or equal to \( a \). Hence, \( (b, a) \notin R_1 \).

Thus, \( R_1 \) is not symmetric.

### Step 3: Check for Transitivity
A relation is transitive if whenever \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \) must also be in \( R \).

- Assume \( (a, b) \in R_1 \) and \( (b, c) \in R_1 \).
- This means \( a \geq b \) and \( b \geq c \).
- From these inequalities, we can conclude \( a \geq c \) (because if \( a \) is greater than or equal to \( b \) and \( b \) is greater than or equal to \( c \), then \( a \) must be greater than or equal to \( c \)).

Thus, \( (a, c) \in R_1 \), and therefore, \( R_1 \) is transitive.

### Conclusion
The relation \( R_1 \) is reflexive and transitive but not symmetric. Therefore, \( R_1 \) is a partial order relation.

### Final Answer
The relation \( R_1 \) is reflexive and transitive but not symmetric.

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