Let `A ={(x,y) in R xx R: y = 5^(x) + 12^(x)}`
`B = {(x,y) in R xx R , y = 13^(x)}`
Then

Options

A. `A cap B` is a singleton
B. `A cap B = phi`
C. `A cap B` consists of at least two points but is finite
D. `A cap B` is an infinite set

Answer

To solve the problem, we need to analyze the sets \( A \) and \( B \) defined by the equations given in the question.

### Step 1: Define the sets
The set \( A \) is defined as:
\[ A = \{(x, y) \in \mathbb{R} \times \mathbb{R} : y = 5^x + 12^x\} \]
The set \( B \) is defined as:
\[ B = \{(x, y) \in \mathbb{R} \times \mathbb{R} : y = 13^x\} \]

### Step 2: Analyze the function in set \( A \)
For set \( A \), the function \( y = 5^x + 12^x \):
- At \( x = 0 \):
\[ y = 5^0 + 12^0 = 1 + 1 = 2 \]
- As \( x \) increases, both \( 5^x \) and \( 12^x \) increase, hence \( y \) increases.
- As \( x \) decreases, both \( 5^x \) and \( 12^x \) approach 0, hence \( y \) approaches 0.

### Step 3: Analyze the function in set \( B \)
For set \( B \), the function \( y = 13^x \):
- At \( x = 0 \):
\[ y = 13^0 = 1 \]
- As \( x \) increases, \( 13^x \) increases rapidly.
- As \( x \) decreases, \( 13^x \) approaches 0.

### Step 4: Compare the two functions
Now we need to check if there are any points where the two functions intersect, i.e., if there exists \( x \) such that:
\[ 5^x + 12^x = 13^x \]

### Step 5: Check for intersection points
1. **At \( x = 0 \)**:
- From set \( A \): \( y = 2 \)
- From set \( B \): \( y = 1 \)
- No intersection at \( x = 0 \).

2. **As \( x \to \infty \)**:
- \( 5^x + 12^x \) grows slower than \( 13^x \) because \( 12 < 13 \).
- Hence, \( 5^x + 12^x < 13^x \) for large \( x \).

3. **As \( x \to -\infty \)**:
- Both functions approach 0, but \( 5^x + 12^x \) approaches 0 faster than \( 13^x \).

### Conclusion
Since there are no points where \( 5^x + 12^x = 13^x \), we conclude that:
\[ A \cap B = \emptyset \]

### Final Answer
Thus, the intersection of sets \( A \) and \( B \) is:
\[ A \cap B = \emptyset \]