Let `C` be a curve defined by `y=e^(a+b x^2)dot` The curve `C` passes through the point `P(1,1)` and the slope of the tangent at `P` is `(-2)dot` Then the value of `2a-3b` is_____.

Answer

To solve the problem, we need to follow these steps:

### Step 1: Set up the equation of the curve
The curve is defined by the equation:
\[ y = e^{a + b x^2} \]

### Step 2: Use the point P(1, 1)
Since the curve passes through the point \( P(1, 1) \), we can substitute \( x = 1 \) and \( y = 1 \) into the equation:
\[ 1 = e^{a + b(1)^2} \]
This simplifies to:
\[ 1 = e^{a + b} \]

### Step 3: Solve for \( a + b \)
To satisfy this equation, we know that:
\[ e^{a + b} = 1 \]
This implies:
\[ a + b = 0 \]
This is our first equation.

### Step 4: Find the derivative \( \frac{dy}{dx} \)
Next, we need to find the slope of the tangent line at point \( P(1, 1) \). We differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(e^{a + b x^2}) \]
Using the chain rule, we get:
\[ \frac{dy}{dx} = e^{a + b x^2} \cdot \frac{d}{dx}(a + b x^2) = e^{a + b x^2} \cdot (2bx) \]

### Step 5: Substitute \( x = 1 \) into the derivative
Now, we substitute \( x = 1 \) into the derivative to find the slope at point \( P \):
\[ \frac{dy}{dx} \bigg|_{x=1} = e^{a + b} \cdot (2b \cdot 1) \]
Since we already established that \( e^{a + b} = 1 \), this simplifies to:
\[ \frac{dy}{dx} \bigg|_{x=1} = 1 \cdot 2b = 2b \]

### Step 6: Set the slope equal to -2
We know from the problem statement that the slope at point \( P(1, 1) \) is -2. Therefore, we set up the equation:
\[ 2b = -2 \]
Solving for \( b \):
\[ b = -1 \]

### Step 7: Find the value of \( a \)
Using the first equation \( a + b = 0 \):
\[ a + (-1) = 0 \]
This gives:
\[ a = 1 \]

### Step 8: Calculate \( 2a - 3b \)
Now we can find the value of \( 2a - 3b \):
\[ 2a - 3b = 2(1) - 3(-1) = 2 + 3 = 5 \]

### Final Answer
Thus, the value of \( 2a - 3b \) is:
\[ \boxed{5} \]