Suppose A and B are two sets given as follows:
`A = {{x,y):x,y in R " and " y =2e^(x)}`
`B={(x,y):x,y in R " and " y=2x+1}`

Options

A. `A sube B`
B. `B sube A`
C. `A cap B = phi`
D. `A cap B = A`

Answer

To solve the problem, we need to analyze the two sets \( A \) and \( B \) defined as follows:

1. **Set A**: \( A = \{(x,y) : x,y \in \mathbb{R} \text{ and } y = 2e^{x}\} \)
2. **Set B**: \( B = \{(x,y) : x,y \in \mathbb{R} \text{ and } y = 2x + 1\} \)

### Step 1: Understand the equations of the sets

- The equation for set \( A \) is \( y = 2e^{x} \), which is an exponential function.
- The equation for set \( B \) is \( y = 2x + 1 \), which is a linear function.

### Step 2: Find the points on each graph

**For Set A**:
- When \( x = 0 \):
\[ y = 2e^{0} = 2 \quad \text{(Point: (0, 2))} \]

**For Set B**:
- When \( x = 0 \):
\[ y = 2(0) + 1 = 1 \quad \text{(Point: (0, 1))} \]
- When \( y = 0 \):
\[ 0 = 2x + 1 \implies x = -\frac{1}{2} \quad \text{(Point: (-0.5, 0))} \]

### Step 3: Sketch the graphs

- The graph of \( A \) (exponential function) will start at point \( (0, 2) \) and rise steeply as \( x \) increases.
- The graph of \( B \) (linear function) will pass through points \( (0, 1) \) and \( (-0.5, 0) \) and will have a slope of 2.

### Step 4: Analyze the intersection of the graphs

To find if there is any intersection, we need to set the equations equal to each other:
\[ 2e^{x} = 2x + 1 \]

### Step 5: Solve the equation for intersections

Rearranging gives:
\[ 2e^{x} - 2x - 1 = 0 \]

This is a transcendental equation, and we can analyze it graphically or numerically to see if they intersect.

### Step 6: Check for intersections

By observing the behavior of both functions:
- The exponential function \( 2e^{x} \) grows much faster than the linear function \( 2x + 1 \) for large \( x \).
- For negative \( x \), \( 2e^{x} \) approaches 0 while \( 2x + 1 \) becomes negative.

### Conclusion

Since the two graphs do not intersect, we conclude:
- **Intersection**: \( A \cap B = \emptyset \) (the empty set).
- Therefore, \( A \) is not a subset of \( B \) and \( B \) is not a subset of \( A \).

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

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