The minimum value of `e^((2x^(2)-2x+1)sin^(2)x)` is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To find the minimum value of the expression \( e^{(2x^2 - 2x + 1) \sin^2 x} \), we can follow these steps:

### Step 1: Define the function
Let \( f(x) = (2x^2 - 2x + 1) \sin^2 x \). We want to find the minimum value of \( e^{f(x)} \). Since the exponential function is always positive, the minimum value of \( e^{f(x)} \) will occur at the minimum value of \( f(x) \).

### Step 2: Differentiate the function
We need to find the critical points of \( f(x) \). To do this, we will differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}[(2x^2 - 2x + 1) \sin^2 x] \]
Using the product rule:
\[ f'(x) = (2x^2 - 2x + 1) \cdot \frac{d}{dx}(\sin^2 x) + \sin^2 x \cdot \frac{d}{dx}(2x^2 - 2x + 1) \]
The derivative of \( \sin^2 x \) is \( 2 \sin x \cos x \) and the derivative of \( 2x^2 - 2x + 1 \) is \( 4x - 2 \). Thus,
\[ f'(x) = (2x^2 - 2x + 1)(2 \sin x \cos x) + \sin^2 x (4x - 2) \]

### Step 3: Set the derivative to zero
We set \( f'(x) = 0 \):
\[ (2x^2 - 2x + 1)(2 \sin x \cos x) + \sin^2 x (4x - 2) = 0 \]
This equation can be simplified by factoring out \( \sin x \):
\[ \sin x \left[ (2x^2 - 2x + 1)(2 \cos x) + \sin x (4x - 2) \right] = 0 \]

### Step 4: Solve for critical points
The equation \( \sin x = 0 \) gives us:
\[ x = n\pi, \quad n \in \mathbb{Z} \]
Next, we need to analyze the term:
\[ (2x^2 - 2x + 1)(2 \cos x) + \sin x (4x - 2) = 0 \]
This term is more complex and may not yield simple solutions, but we can evaluate \( f(x) \) at the critical points we found.

### Step 5: Evaluate at critical points
Evaluate \( f(x) \) at \( x = 0 \):
\[ f(0) = 2(0)^2 - 2(0) + 1 = 1 \quad \text{and} \quad \sin^2(0) = 0 \]
Thus,
\[ f(0) = 1 \cdot 0 = 0 \]
Now, evaluate \( e^{f(0)} \):
\[ e^{f(0)} = e^0 = 1 \]

### Conclusion
The minimum value of \( e^{(2x^2 - 2x + 1) \sin^2 x} \) is:
\[ \boxed{1} \]