If `A={(x,y):y=e^(-x)}" and "B={(x,y):y=-x}`. Then,

Options

A. `A nn B = phi`
B. `A sub B`
C. `B sub A`
D. `A nn B={(0,1),(0, 0)}`

Explanation

To solve the problem, we need to find the intersection of the two sets \( A \) and \( B \).

### Step 1: Define the sets
- Set \( A \) is defined as \( A = \{(x, y) : y = e^{-x}\} \).
- Set \( B \) is defined as \( B = \{(x, y) : y = -x\} \).

### Step 2: Understand the equations
- The equation \( y = e^{-x} \) represents an exponential decay function. As \( x \) increases, \( y \) approaches 0 but never actually reaches it. At \( x = 0 \), \( y = 1 \).
- The equation \( y = -x \) represents a straight line with a negative slope. It passes through the origin (0,0) and decreases as \( x \) increases.

### Step 3: Find the intersection
To find the intersection \( A \cap B \), we need to set the equations equal to each other:
\[ e^{-x} = -x \]

### Step 4: Analyze the intersection
1. The function \( e^{-x} \) is always positive for all real \( x \).
2. The function \( -x \) is negative for \( x > 0 \) and positive for \( x < 0 \).

Since \( e^{-x} \) is positive and \( -x \) is negative for \( x > 0 \), there can be no intersection in that region.

Now, let's check for \( x < 0 \):
- As \( x \) approaches negative infinity, \( e^{-x} \) approaches infinity, while \( -x \) approaches positive infinity.
- At \( x = 0 \), \( e^{-0} = 1 \) and \( -0 = 0 \).
- For \( x < 0 \), \( e^{-x} \) is greater than \( -x \) since \( e^{-x} \) grows rapidly while \( -x \) is linear.

### Step 5: Conclusion
Since there are no values of \( x \) for which \( e^{-x} = -x \), we conclude that:
\[ A \cap B = \emptyset \]

### Final Answer
The intersection of sets \( A \) and \( B \) is the empty set, denoted as:
\[ A \cap B = \emptyset \]

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