If the function f(x) =`axe^(bx^(2))` has maximum value at x=2 such that f(2) =1 , then find the values of a and b

Explanation

To solve the problem, we need to find the values of \( a \) and \( b \) for the function \( f(x) = axe^{bx^2} \) given that it has a maximum at \( x = 2 \) and \( f(2) = 1 \).

### Step 1: Find \( f(2) \)
We start by substituting \( x = 2 \) into the function:

\[ f(2) = a(2)e^{b(2^2)} = 2ae^{4b} \]

Given that \( f(2) = 1 \), we have:

\[ 2ae^{4b} = 1 \]

### Step 2: Find \( f'(x) \)
Next, we need to find the derivative of \( f(x) \) to determine the critical points. We use the product rule:

\[ f'(x) = \frac{d}{dx}(axe^{bx^2}) = a e^{bx^2} + axe^{bx^2} \cdot \frac{d}{dx}(bx^2) \]

Calculating \( \frac{d}{dx}(bx^2) = 2bx \), we get:

\[ f'(x) = a e^{bx^2} + 2abx^2 e^{bx^2} = e^{bx^2}(a + 2abx^2) \]

### Step 3: Set \( f'(2) = 0 \)
Since \( f(x) \) has a maximum at \( x = 2 \), we set \( f'(2) = 0 \):

\[ f'(2) = e^{b(2^2)}(a + 2ab(2^2)) = e^{4b}(a + 8ab) = 0 \]

Since \( e^{4b} \) is never zero, we can set the expression in parentheses to zero:

\[ a + 8ab = 0 \]

### Step 4: Solve for \( b \)
From \( a + 8ab = 0 \), we can factor out \( a \):

\[ a(1 + 8b) = 0 \]

This gives us two cases: \( a = 0 \) or \( 1 + 8b = 0 \). Since \( a = 0 \) would make \( f(x) = 0 \), we discard it. Thus, we solve for \( b \):

\[ 1 + 8b = 0 \implies 8b = -1 \implies b = -\frac{1}{8} \]

### Step 5: Substitute \( b \) back to find \( a \)
Now we substitute \( b = -\frac{1}{8} \) back into the equation from Step 1:

\[ 2ae^{4(-\frac{1}{8})} = 1 \implies 2ae^{- \frac{1}{2}} = 1 \]

This simplifies to:

\[ 2a \cdot \frac{1}{\sqrt{e}} = 1 \implies 2a = \sqrt{e} \implies a = \frac{\sqrt{e}}{2} \]

### Final Values
Thus, the values of \( a \) and \( b \) are:

\[ a = \frac{\sqrt{e}}{2}, \quad b = -\frac{1}{8} \]