If `A={0,1,2,3,4,5}` and a relation R defined by (x)R(y), where `2x+y=10`. Then `R^(-1)` is

Options

A. `{(3,4),(4,2),(5,0)}`
B. `{(4,3),(4,2),(5,0)}`
C. `{(4,3),(4,2),(5,0)}`
D. `{(4,3),(2,4),(0,5)}`

Answer

To solve the problem, we need to find the inverse of the relation \( R \) defined by \( (x)R(y) \) where \( 2x + y = 10 \) for the set \( A = \{0, 1, 2, 3, 4, 5\} \).

### Step-by-Step Solution:

1. **Identify the relation \( R \)**:
We need to find pairs \( (x, y) \) such that \( 2x + y = 10 \) for \( x \) and \( y \) belonging to the set \( A \).

2. **Calculate values of \( y \) for each \( x \)**:
- For \( x = 0 \):
\[ 2(0) + y = 10 \implies y = 10 \]
- For \( x = 1 \):
\[ 2(1) + y = 10 \implies y = 10 - 2 = 8 \]
- For \( x = 2 \):
\[ 2(2) + y = 10 \implies y = 10 - 4 = 6 \]
- For \( x = 3 \):
\[ 2(3) + y = 10 \implies y = 10 - 6 = 4 \]
- For \( x = 4 \):
\[ 2(4) + y = 10 \implies y = 10 - 8 = 2 \]
- For \( x = 5 \):
\[ 2(5) + y = 10 \implies y = 10 - 10 = 0 \]

3. **List the valid pairs \( (x, y) \)**:
We only consider pairs where both \( x \) and \( y \) are in the set \( A \):
- \( (0, 10) \) - not valid since \( 10 \notin A \)
- \( (1, 8) \) - not valid since \( 8 \notin A \)
- \( (2, 6) \) - not valid since \( 6 \notin A \)
- \( (3, 4) \) - valid
- \( (4, 2) \) - valid
- \( (5, 0) \) - valid

Thus, the relation \( R \) can be written as:
\[ R = \{(3, 4), (4, 2), (5, 0)\} \]

4. **Find the inverse relation \( R^{-1} \)**:
The inverse relation \( R^{-1} \) is obtained by swapping the elements in each pair:
- From \( (3, 4) \) we get \( (4, 3) \)
- From \( (4, 2) \) we get \( (2, 4) \)
- From \( (5, 0) \) we get \( (0, 5) \)

Therefore, the inverse relation \( R^{-1} \) is:
\[ R^{-1} = \{(4, 3), (2, 4), (0, 5)\} \]

### Final Answer:
\[ R^{-1} = \{(4, 3), (2, 4), (0, 5)\} \]