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 \( f(x) = x^2 e^{-\frac{x^2}{a^2}} \) is increasing, we need to find the derivative of the function and analyze where it is greater than or equal to zero.

### Step 1: Find the derivative of \( f(x) \)

The function is given by:
\[ f(x) = x^2 e^{-\frac{x^2}{a^2}} \]

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

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

Putting it all together:
\[ f'(x) = 2x e^{-\frac{x^2}{a^2}} + x^2 \left(-\frac{2x}{a^2} e^{-\frac{x^2}{a^2}}\right) \]
\[ = e^{-\frac{x^2}{a^2}} \left(2x - \frac{2x^3}{a^2}\right) \]
\[ = 2x e^{-\frac{x^2}{a^2}} \left(1 - \frac{x^2}{a^2}\right) \]

### Step 2: Set the derivative greater than or equal to zero

For the function to be increasing, we need:
\[ f'(x) \geq 0 \]
This gives us:
\[ 2x e^{-\frac{x^2}{a^2}} \left(1 - \frac{x^2}{a^2}\right) \geq 0 \]

### Step 3: Analyze the factors

1. \( e^{-\frac{x^2}{a^2}} > 0 \) for all \( x \).
2. \( 2x \geq 0 \) implies \( x \geq 0 \).
3. \( 1 - \frac{x^2}{a^2} \geq 0 \) implies \( x^2 \leq a^2 \) or \( -a \leq x \leq a \).

### Step 4: Combine the conditions

From \( 2x \geq 0 \), we have \( x \geq 0 \).
From \( 1 - \frac{x^2}{a^2} \geq 0 \), we have \( x \leq a \).

Thus, combining these conditions, we find:
\[ 0 \leq x \leq a \]

### Step 5: Final intervals

Since \( x \) must also satisfy \( x \geq 0 \), the function \( f(x) \) is increasing in the intervals:
\[ (-\infty, -a] \cup [0, a] \]

### Conclusion

The intervals where the function \( f(x) \) is increasing are:
\[ (-\infty, -a] \cup [0, a] \]