If `f(x) = x^(2) e^(-x^(2)/a^(2)` is an increasing function then` (for a gt 0)`, x lies in the interval

Options

A. [a, 2a]
B. `(-infty, -a]cup[0, a]`
C. `(-a, 0)`
D. None of these

Answer

To determine the interval in which the function \( f(x) = x^2 e^{-x^2/a^2} \) is increasing, we need to follow these steps:

### Step 1: Find the derivative of \( f(x) \)
To find where \( f(x) \) is increasing, we first need to compute the derivative \( f'(x) \).

Using the product rule:
\[ f'(x) = \frac{d}{dx}(x^2) \cdot e^{-x^2/a^2} + x^2 \cdot \frac{d}{dx}(e^{-x^2/a^2}) \]

Calculating each part:
1. The derivative of \( x^2 \) is \( 2x \).
2. For \( e^{-x^2/a^2} \), we use the chain rule:
\[ \frac{d}{dx}(e^{-x^2/a^2}) = e^{-x^2/a^2} \cdot \frac{d}{dx}(-x^2/a^2) = e^{-x^2/a^2} \cdot \left(-\frac{2x}{a^2}\right) \]

Putting it all together:
\[ f'(x) = 2x e^{-x^2/a^2} + x^2 \left(-\frac{2x}{a^2} e^{-x^2/a^2}\right) \]
\[ = e^{-x^2/a^2} \left(2x - \frac{2x^3}{a^2}\right) \]
\[ = \frac{2x e^{-x^2/a^2}}{a^2} (a^2 - x^2) \]

### Step 2: Set the derivative greater than zero
For \( f(x) \) to be increasing, we need:
\[ f'(x) > 0 \]
This leads to:
\[ \frac{2x e^{-x^2/a^2}}{a^2} (a^2 - x^2) > 0 \]

Since \( e^{-x^2/a^2} > 0 \) for all \( x \), we can simplify our inequality to:
\[ 2x (a^2 - x^2) > 0 \]

### Step 3: Analyze the inequality
The inequality \( 2x (a^2 - x^2) > 0 \) can be broken down into two factors:
1. \( 2x > 0 \) implies \( x > 0 \)
2. \( a^2 - x^2 > 0 \) implies \( x^2 < a^2 \) or \( -a < x < a \)

Combining these conditions:
- From \( 2x > 0 \), we have \( x > 0 \).
- From \( a^2 - x^2 > 0 \), we have \( x < a \).

Thus, the combined condition is:
\[ 0 < x < a \]

### Step 4: Conclusion
The function \( f(x) \) is increasing in the interval \( (0, a) \).

### Final Answer
The interval in which \( x \) lies for \( f(x) \) to be an increasing function is \( (0, a) \).

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