Consider three sets A = {1, 2, 3}, B = {3, 4, 5, 6}, C = {6, 7, 8, 9}. `R_(1)` is defined from A to B such that `R_(1) = {(x, y) : 4x lt y, x in A, y in B}`. Similarly `R_(2)` is defined from B to C such that `R_(2) = {(x, y) : 2x le y, x in B` and `y in C`} then `R_(1)oR_(2)^(-1)` is

Options

A. {(3, 1)}
B. {(1, 1)}
C. {(1, 3)}
D. `phi`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Define the sets and relations
We have three sets:
- \( A = \{1, 2, 3\} \)
- \( B = \{3, 4, 5, 6\} \)
- \( C = \{6, 7, 8, 9\} \)

The relation \( R_1 \) is defined from \( A \) to \( B \) as:
\[ R_1 = \{(x, y) : 4x < y, x \in A, y \in B\} \]

The relation \( R_2 \) is defined from \( B \) to \( C \) as:
\[ R_2 = \{(x, y) : 2x \leq y, x \in B, y \in C\} \]

### Step 2: Find the ordered pairs in \( R_1 \)
We will find the pairs \((x, y)\) in \( R_1 \) by checking each element in \( A \):
- For \( x = 1 \): \( 4 \times 1 = 4 \). The values of \( y \) in \( B \) that satisfy \( 4 < y \) are \( 5 \) and \( 6 \).
- Pairs: \( (1, 5), (1, 6) \)

- For \( x = 2 \): \( 4 \times 2 = 8 \). There are no \( y \) in \( B \) such that \( 8 < y \).
- Pairs: None

- For \( x = 3 \): \( 4 \times 3 = 12 \). There are no \( y \) in \( B \) such that \( 12 < y \).
- Pairs: None

Thus, we have:
\[ R_1 = \{(1, 5), (1, 6)\} \]

### Step 3: Find the ordered pairs in \( R_2 \)
Now we will find the pairs \((x, y)\) in \( R_2 \):
- For \( x = 3 \): \( 2 \times 3 = 6 \). The values of \( y \) in \( C \) that satisfy \( 6 \leq y \) are \( 6, 7, 8, 9 \).
- Pairs: \( (3, 6), (3, 7), (3, 8), (3, 9) \)

- For \( x = 4 \): \( 2 \times 4 = 8 \). The values of \( y \) in \( C \) that satisfy \( 8 \leq y \) are \( 8, 9 \).
- Pairs: \( (4, 8), (4, 9) \)

- For \( x = 5 \): \( 2 \times 5 = 10 \). There are no \( y \) in \( C \) such that \( 10 \leq y \).
- Pairs: None

Thus, we have:
\[ R_2 = \{(3, 6), (3, 7), (3, 8), (3, 9), (4, 8), (4, 9)\} \]

### Step 4: Find the inverse relation \( R_2^{-1} \)
The inverse relation \( R_2^{-1} \) is obtained by swapping the elements of each pair in \( R_2 \):
\[ R_2^{-1} = \{(6, 3), (7, 3), (8, 3), (9, 3), (8, 4), (9, 4)\} \]

### Step 5: Find \( R_1 \circ R_2^{-1} \)
Now we need to find \( R_1 \circ R_2^{-1} \), which consists of pairs \((x, z)\) such that there exists a \( y \) where \( (x, y) \in R_1 \) and \( (y, z) \in R_2^{-1} \).

1. For \( (1, 5) \in R_1 \):
- \( y = 5 \) does not appear in \( R_2^{-1} \).

2. For \( (1, 6) \in R_1 \):
- \( y = 6 \) gives \( (6, 3) \) in \( R_2^{-1} \).
- Thus, we get \( (1, 3) \).

3. For \( (1, 6) \in R_1 \):
- \( y = 6 \) gives \( (6, 4) \) in \( R_2^{-1} \).
- Thus, we get \( (1, 4) \).

So, the final result is:
\[ R_1 \circ R_2^{-1} = \{(1, 3), (1, 4)\} \]

### Conclusion
The relation \( R_1 \circ R_2^{-1} \) consists of the pairs \( (1, 3) \) and \( (1, 4) \).