Let `A={1,2,3,...10}` and `R={(x, y) : x+2y=10, x, y in A}` be a relation A. Then, `R^(-1) =`

Options

A. `{(2, 4),(4, 3), (6, 2), (8, 1)}`
B. `{(4, 2),(3, 4),(2, 6),(1, 8)}`
C. `{(4, 2),(3, 4),(2, 6)}`
D. `{(y, x):2x+y=10,x,y in A}`

Explanation

To find the inverse relation \( R^{-1} \) for the relation \( R = \{(x, y) : x + 2y = 10, x, y \in A\} \) where \( A = \{1, 2, 3, \ldots, 10\} \), we will follow these steps:

### Step 1: Define the relation \( R \)
We start with the equation given for the relation:
\[ x + 2y = 10 \]
We need to find pairs \( (x, y) \) such that both \( x \) and \( y \) belong to the set \( A \).

### Step 2: Solve for \( y \)
Rearranging the equation gives:
\[ 2y = 10 - x \]
\[ y = \frac{10 - x}{2} \]

### Step 3: Determine valid values of \( x \)
Since \( x \) must be an integer in the set \( A \), we will check which values of \( x \) from 1 to 10 yield an integer \( y \).

1. If \( x = 1 \):
\[ y = \frac{10 - 1}{2} = \frac{9}{2} \quad \text{(not valid)} \]
2. If \( x = 2 \):
\[ y = \frac{10 - 2}{2} = 4 \quad \text{(valid)} \]
3. If \( x = 3 \):
\[ y = \frac{10 - 3}{2} = \frac{7}{2} \quad \text{(not valid)} \]
4. If \( x = 4 \):
\[ y = \frac{10 - 4}{2} = 3 \quad \text{(valid)} \]
5. If \( x = 5 \):
\[ y = \frac{10 - 5}{2} = \frac{5}{2} \quad \text{(not valid)} \]
6. If \( x = 6 \):
\[ y = \frac{10 - 6}{2} = 2 \quad \text{(valid)} \]
7. If \( x = 7 \):
\[ y = \frac{10 - 7}{2} = \frac{3}{2} \quad \text{(not valid)} \]
8. If \( x = 8 \):
\[ y = \frac{10 - 8}{2} = 1 \quad \text{(valid)} \]
9. If \( x = 9 \):
\[ y = \frac{10 - 9}{2} = \frac{1}{2} \quad \text{(not valid)} \]
10. If \( x = 10 \):
\[ y = \frac{10 - 10}{2} = 0 \quad \text{(not valid)} \]

### Step 4: List valid pairs in \( R \)
From the valid calculations above, we find the pairs:
- For \( x = 2, y = 4 \) → \( (2, 4) \)
- For \( x = 4, y = 3 \) → \( (4, 3) \)
- For \( x = 6, y = 2 \) → \( (6, 2) \)
- For \( x = 8, y = 1 \) → \( (8, 1) \)

Thus, the relation \( R \) is:
\[ R = \{(2, 4), (4, 3), (6, 2), (8, 1)\} \]

### Step 5: Find the inverse relation \( R^{-1} \)
The inverse relation \( R^{-1} \) consists of the pairs where the elements are swapped:
- From \( (2, 4) \) we get \( (4, 2) \)
- From \( (4, 3) \) we get \( (3, 4) \)
- From \( (6, 2) \) we get \( (2, 6) \)
- From \( (8, 1) \) we get \( (1, 8) \)

Thus, the inverse relation \( R^{-1} \) is:
\[ R^{-1} = \{(4, 2), (3, 4), (2, 6), (1, 8)\} \]

### Final Answer
The inverse relation \( R^{-1} \) is:
\[ R^{-1} = \{(4, 2), (3, 4), (2, 6), (1, 8)\} \]