Let `f(x) `be real valued continous funcion on R defined as `f(x)` =`x^(2)e^(-|x|)`
Number of points of inflection for `y = f(x)` is (a) 1 (b) 2 (c) 3 (d) 4

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To find the number of points of inflection for the function \( f(x) = x^2 e^{-|x|} \), we will follow these steps:

### Step 1: Define the function
The function is defined as:
\[ f(x) = \begin{cases} x^2 e^{-x} & \text{if } x \geq 0 \\ x^2 e^{x} & \text{if } x < 0 \end{cases} \]

### Step 2: Find the first derivative \( f'(x) \)
We will differentiate \( f(x) \) for both cases using the product rule.

For \( x \geq 0 \):
\[ f'(x) = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x}) = 2x e^{-x} - x^2 e^{-x} = e^{-x}(2x - x^2) \]

For \( x < 0 \):
\[ f'(x) = \frac{d}{dx}(x^2) \cdot e^{x} + x^2 \cdot \frac{d}{dx}(e^{x}) = 2x e^{x} + x^2 e^{x} = e^{x}(x^2 + 2x) \]

### Step 3: Find the second derivative \( f''(x) \)
Now we differentiate \( f'(x) \) to find \( f''(x) \).

For \( x \geq 0 \):
\[ f''(x) = \frac{d}{dx}(e^{-x}(2x - x^2)) = e^{-x}(2 - 2x) + e^{-x}(2x - x^2)(-1) = e^{-x}((2 - 2x) - (2x - x^2)) = e^{-x}(x^2 - 4x + 2) \]

For \( x < 0 \):
\[ f''(x) = \frac{d}{dx}(e^{x}(x^2 + 2x)) = e^{x}(x^2 + 2x) + e^{x}(2x + 2) = e^{x}(x^2 + 4x + 2) \]

### Step 4: Set the second derivative to zero
To find points of inflection, we set \( f''(x) = 0 \).

For \( x \geq 0 \):
\[ e^{-x}(x^2 - 4x + 2) = 0 \implies x^2 - 4x + 2 = 0 \]
Using the quadratic formula:
\[ x = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2} \]

For \( x < 0 \):
\[ e^{x}(x^2 + 4x + 2) = 0 \implies x^2 + 4x + 2 = 0 \]
Using the quadratic formula:
\[ x = \frac{-4 \pm \sqrt{16 - 8}}{2} = \frac{-4 \pm 2\sqrt{2}}{2} = -2 \pm \sqrt{2} \]

### Step 5: Count the points of inflection
From the calculations:
- For \( x \geq 0 \), we have two points of inflection: \( 2 + \sqrt{2} \) and \( 2 - \sqrt{2} \).
- For \( x < 0 \), we have two points of inflection: \( -2 + \sqrt{2} \) and \( -2 - \sqrt{2} \).

### Conclusion
Thus, the total number of points of inflection is \( 2 + 2 = 4 \).

The answer is (d) 4.

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