Question

Let \( A = \{(x, y) : y = e^x, x \in \mathbb{R} \} \) and \( B = \{(x, y) : y = e^{-x}, x \in \mathbb{R} \} \). Then,

• A. \( A \cap B = \emptyset \)
• B. \( A \cap B \neq \emptyset \)
• C. \( A \cup B = \mathbb{R}^2 \)
• D. None of these

Hint:

To find the intersection of sets defined by functions, equate the functions and solve for the common values of \( x \) and \( y \).

Answer 1

The Correct Option is B

Step 1: Define the sets \( A \) and \( B \).
We are given two sets: - \( A = \{(x, y) : y = e^x, x \in \mathbb{R}\} \), which represents the graph of the exponential function \( y = e^x \). - \( B = \{(x, y) : y = e^{-x}, x \in \mathbb{R}\} \), which represents the graph of the exponential function \( y = e^{-x} \).

Step 2: Find the intersection of \( A \) and \( B \).
To find \( A \cap B \), we solve the system of equations: \[ y = e^x \quad \text{and} \quad y = e^{-x} \] Equating the two expressions for \( y \), we get: \[ e^x = e^{-x} \] Taking the natural logarithm of both sides: \[ x = -x \] This gives \( x = 0 \).

Step 3: Check the value of \( y \) at \( x = 0 \).
When \( x = 0 \), both equations give: \[ y = e^0 = 1 \quad \text{and} \quad y = e^{-0} = 1 \] Thus, the point \( (0, 1) \) lies in both sets \( A \) and \( B \).

Step 4: Conclusion.
Since the point \( (0, 1) \) is in both sets, the intersection \( A \cap B \neq \emptyset \), corresponding to option (B).

Step 5: Quick Tip.
When finding the intersection of sets, equate the defining equations and solve for the common points. In this case, \( (0, 1) \) is the only common point.