Let P = Set of all integral multiples of 3
Q = set of all integral multiples of 4
R = Set of all integral multiples of 6
consider the following relations :
1 `P cup Q = R `
2. `P sub R`
`R sub (P cup Q)`
Which of the relations given above is/are correct ?

Options

A. only 1
B. only 2
C. only 3
D. 2 and 3

Answer

To solve the problem, we need to analyze the sets P, Q, and R, and then check the given relations.

### Step-by-Step Solution:

1. **Define the Sets:**
- Set \( P \): All integral multiples of 3. This can be represented as:
\[ P = \{ 3n \mid n \in \mathbb{Z} \} = \{ \ldots, -6, -3, 0, 3, 6, 9, 12, \ldots \} \]
- Set \( Q \): All integral multiples of 4. This can be represented as:
\[ Q = \{ 4m \mid m \in \mathbb{Z} \} = \{ \ldots, -8, -4, 0, 4, 8, 12, 16, \ldots \} \]
- Set \( R \): All integral multiples of 6. This can be represented as:
\[ R = \{ 6k \mid k \in \mathbb{Z} \} = \{ \ldots, -12, -6, 0, 6, 12, 18, \ldots \} \]

2. **Check the First Relation: \( P \cup Q = R \)**
- The union \( P \cup Q \) consists of all elements that are in either set \( P \) or set \( Q \).
- Elements in \( P \): \( \{ \ldots, -6, -3, 0, 3, 6, 9, 12, \ldots \} \)
- Elements in \( Q \): \( \{ \ldots, -8, -4, 0, 4, 8, 12, 16, \ldots \} \)
- The union \( P \cup Q \) includes elements like \( 3, 4, 6, 8, 9, 12 \), etc.
- Since \( R \) contains \( 6, 12, 18, \ldots \) but also includes \( 3 \) and \( 4 \) which are not in \( R \), we conclude:
\[ P \cup Q \neq R \]
- **Conclusion**: The first relation is **incorrect**.

3. **Check the Second Relation: \( P \subset R \)**
- To check if \( P \) is a subset of \( R \), we need to see if every element of \( P \) is also in \( R \).
- Elements of \( P \): \( \{ \ldots, -6, -3, 0, 3, 6, 9, 12, \ldots \} \)
- Elements of \( R \): \( \{ \ldots, -12, -6, 0, 6, 12, 18, \ldots \} \)
- The element \( 3 \) is in \( P \) but not in \( R\).
- **Conclusion**: The second relation is **incorrect**.

4. **Check the Third Relation: \( R \subset (P \cup Q) \)**
- To check if \( R \) is a subset of \( P \cup Q \), we need to see if every element of \( R \) is also in \( P \cup Q \).
- Elements of \( R \): \( \{ \ldots, -12, -6, 0, 6, 12, 18, \ldots \} \)
- Elements of \( P \cup Q \): \( \{ \ldots, -8, -6, -4, -3, 0, 3, 4, 6, 8, 9, 12, 16, \ldots \} \)
- The elements \( 6, 12 \) are in both \( R \) and \( P \cup Q \).
- Since all multiples of 6 are also multiples of either 3 or 4, we conclude:
\[ R \subset (P \cup Q) \]
- **Conclusion**: The third relation is **correct**.

### Final Answer:
- The only correct relation is: **R is a subset of (P union Q)**.