Let `A = {(x,y) in R xx R : y = 5^(x) + 3^(x)}`
`B = {(x,y) in R xx R : y = 4^(x)}`, Then

Options

A. `A cap B` is a singleton
B. `A cap B = phi`
C. `A cap B` consists of at leas 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 as follows:

- Set \( A = \{(x,y) \in \mathbb{R} \times \mathbb{R} : y = 5^x + 3^x\} \)
- Set \( B = \{(x,y) \in \mathbb{R} \times \mathbb{R} : y = 4^x\} \)

We need to determine the intersection of these two sets, \( A \cap B \).

### Step 1: Understand the functions in sets A and B
The function in set \( A \) is \( y = 5^x + 3^x \) and the function in set \( B \) is \( y = 4^x \). Both functions are exponential functions.

### Step 2: Evaluate the functions at \( x = 0 \)
For \( x = 0 \):
- In set \( A \):
\[ y = 5^0 + 3^0 = 1 + 1 = 2 \]
So, the point is \( (0, 2) \).

- In set \( B \):
\[ y = 4^0 = 1 \]
So, the point is \( (0, 1) \).

### Step 3: Evaluate the functions at \( x = 1 \)
For \( x = 1 \):
- In set \( A \):
\[ y = 5^1 + 3^1 = 5 + 3 = 8 \]
So, the point is \( (1, 8) \).

- In set \( B \):
\[ y = 4^1 = 4 \]
So, the point is \( (1, 4) \).

### Step 4: Analyze the growth of the functions
- The function \( y = 5^x + 3^x \) is increasing because both \( 5^x \) and \( 3^x \) are increasing functions.
- The function \( y = 4^x \) is also increasing.

### Step 5: Compare the values of \( y \) for different \( x \)
From our evaluations:
- At \( x = 0 \): \( A \) gives \( y = 2 \) and \( B \) gives \( y = 1 \). Thus, \( A > B \).
- At \( x = 1 \): \( A \) gives \( y = 8 \) and \( B \) gives \( y = 4 \). Thus, \( A > B \).

### Step 6: Determine the intersection
Since \( A \) is consistently greater than \( B \) for the evaluated points and both functions are increasing, we can conclude that there will be no intersection points for all \( x \in \mathbb{R} \).

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

### Final Answer
The correct option is that \( A \cap B = \emptyset \).

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