Define relations `R_(1) and R_(2) `on set A = [2,3,5,7,10] as x`R_(1)`y is `2x=(y - 1)` and `x R_(2) ` y if x + y = 10, then the relation R given by R = ` R_(1) cap R_(2)` is

Options

A. {}
B. {3,7}
C. {(3,7),(5,5)}
D. None of these

Answer

To solve the problem, we need to find the intersection of two relations \( R_1 \) and \( R_2 \) defined on the set \( A = \{2, 3, 5, 7, 10\} \).

### Step 1: Define the relations

1. The relation \( R_1 \) is defined as \( x R_1 y \) if \( 2x = y - 1 \).
2. The relation \( R_2 \) is defined as \( x R_2 y \) if \( x + y = 10 \).

### Step 2: Find the pairs in relation \( R_1 \)

We will check each element \( x \) in set \( A \) to find corresponding \( y \):

- For \( x = 2 \):
\[ 2(2) = y - 1 \implies 4 = y - 1 \implies y = 5 \]
So, the pair \( (2, 5) \) is in \( R_1 \).

- For \( x = 3 \):
\[ 2(3) = y - 1 \implies 6 = y - 1 \implies y = 7 \]
So, the pair \( (3, 7) \) is in \( R_1 \).

- For \( x = 5 \):
\[ 2(5) = y - 1 \implies 10 = y - 1 \implies y = 11 \]
\( y = 11 \) is not in set \( A \), so no pair here.

- For \( x = 7 \):
\[ 2(7) = y - 1 \implies 14 = y - 1 \implies y = 15 \]
\( y = 15 \) is not in set \( A \), so no pair here.

- For \( x = 10 \):
\[ 2(10) = y - 1 \implies 20 = y - 1 \implies y = 21 \]
\( y = 21 \) is not in set \( A \), so no pair here.

Thus, the pairs in \( R_1 \) are:
\[ R_1 = \{(2, 5), (3, 7)\} \]

### Step 3: Find the pairs in relation \( R_2 \)

Next, we will check each element \( x \) in set \( A \):

- For \( x = 2 \):
\[ 2 + y = 10 \implies y = 8 \]
\( y = 8 \) is not in set \( A \), so no pair here.

- For \( x = 3 \):
\[ 3 + y = 10 \implies y = 7 \]
So, the pair \( (3, 7) \) is in \( R_2 \).

- For \( x = 5 \):
\[ 5 + y = 10 \implies y = 5 \]
So, the pair \( (5, 5) \) is in \( R_2 \).

- For \( x = 7 \):
\[ 7 + y = 10 \implies y = 3 \]
So, the pair \( (7, 3) \) is in \( R_2 \).

- For \( x = 10 \):
\[ 10 + y = 10 \implies y = 0 \]
\( y = 0 \) is not in set \( A \), so no pair here.

Thus, the pairs in \( R_2 \) are:
\[ R_2 = \{(3, 7), (5, 5), (7, 3)\} \]

### Step 4: Find the intersection \( R = R_1 \cap R_2 \)

Now we find the common pairs in \( R_1 \) and \( R_2 \):
- \( R_1 = \{(2, 5), (3, 7)\} \)
- \( R_2 = \{(3, 7), (5, 5), (7, 3)\} \)

The common pair is:
\[ (3, 7) \]

### Conclusion

Thus, the relation \( R \) given by \( R = R_1 \cap R_2 \) is:
\[ R = \{(3, 7)\} \]