There are three relations `R_(1) , R_(2)` and `R_(3)` such that
`R_(1) = {(2,1),(3,1),(4,2)}`,
`R_(2)` = {(2,2),(2,4),(3,3),(4,4)} and
`R_(3)` = {(1,2),(2,3),(3,4),(4,5),(5,6),(6,7)}
then

Options

A. `R_(1) and R_(2)` are functions
B. `R_(2) and R_(3)` are functions
C. `R_(1) and R_(3)` are functions
D. Only `R_(1)` is a function

Answer

To determine which of the given relations are functions, we need to analyze each relation based on the definition of a function. A relation is a function if every element in the domain (the first element of each ordered pair) is associated with exactly one element in the codomain (the second element of each ordered pair).

Let's analyze the relations one by one.

### Step 1: Analyze Relation R1
**Given:**
\[ R_1 = \{(2,1), (3,1), (4,2)\} \]

- **Domain:** {2, 3, 4}
- **Codomain:** {1, 2}

**Check for Function:**
- For 2, the output is 1.
- For 3, the output is 1.
- For 4, the output is 2.

Each element in the domain has a unique output in the codomain. Therefore, **R1 is a function**.

### Step 2: Analyze Relation R2
**Given:**
\[ R_2 = \{(2,2), (2,4), (3,3), (4,4)\} \]

- **Domain:** {2, 3, 4}
- **Codomain:** {2, 3, 4}

**Check for Function:**
- For 2, there are two outputs: 2 and 4.
- For 3, the output is 3.
- For 4, the output is 4.

Since the element 2 in the domain has two different outputs (2 and 4), **R2 is not a function**.

### Step 3: Analyze Relation R3
**Given:**
\[ R_3 = \{(1,2), (2,3), (3,4), (4,5), (5,6), (6,7)\} \]

- **Domain:** {1, 2, 3, 4, 5, 6}
- **Codomain:** {2, 3, 4, 5, 6, 7}

**Check for Function:**
- For 1, the output is 2.
- For 2, the output is 3.
- For 3, the output is 4.
- For 4, the output is 5.
- For 5, the output is 6.
- For 6, the output is 7.

Each element in the domain has a unique output in the codomain. Therefore, **R3 is a function**.

### Conclusion
From the analysis:
- R1 is a function.
- R2 is not a function.
- R3 is a function.

Thus, the correct answer is that **R1 and R3 are functions**.

### Final Answer:
**Option C: R1 and R3 are functions.**

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