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 \( f(x) = ax e^{bx^2} \) given that it has a maximum at \( x = 2 \) and \( f(2) = 1 \).

### Step 1: Use the condition \( f(2) = 1 \)

We start by substituting \( x = 2 \) into the function:

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

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

\[ 2a e^{4b} = 1 \]

From this, we can express \( a \) in terms of \( b \):

\[ a e^{4b} = \frac{1}{2} \quad \text{(1)} \]

### Step 2: Find the derivative \( f'(x) \)

Next, we need to find the first derivative \( f'(x) \) to determine the maximum point. We apply the product rule:

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

Calculating the derivative of \( e^{bx^2} \):

\[ \frac{d}{dx}(e^{bx^2}) = e^{bx^2} \cdot (2bx) = 2bxe^{bx^2} \]

Thus, we have:

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

Factoring out \( e^{bx^2} \):

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

### Step 3: Set \( f'(2) = 0 \)

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

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

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

\[ a + 8ab = 0 \]

Factoring out \( a \):

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

Since \( a \neq 0 \), we have:

\[ 1 + 8b = 0 \implies b = -\frac{1}{8} \quad \text{(2)} \]

### Step 4: Substitute \( b \) back into equation (1)

Now we substitute \( b = -\frac{1}{8} \) back into equation (1):

\[ a e^{4(-\frac{1}{8})} = \frac{1}{2} \]

This simplifies to:

\[ a e^{- \frac{1}{2}} = \frac{1}{2} \]

Multiplying both sides by \( e^{\frac{1}{2}} \):

\[ a = \frac{1}{2} e^{\frac{1}{2}} = \frac{\sqrt{e}}{2} \quad \text{(3)} \]

### Final Values

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

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