Let `f(x)={((1+cosx)/((pi-x)^(2)).(sin^(2)x)/(ln(1+pi^(2)-2pix+x^(2))),x ne pi),(lambda, x=pi):}`

Options

A. 1
B. `-1`
C. `(1)/(2)`
D. `(1)/(4)`

Answer

To find the value of \( \lambda \) such that the function \( f(x) \) is continuous at \( x = \pi \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches \( \pi \) and set it equal to \( \lambda \).

### Step-by-Step Solution:

1. **Understanding the Function:**
The function is defined as:
\[ f(x) = \frac{(1 + \cos x) \cdot \sin^2 x}{(\pi - x)^2 \cdot \ln(1 + \pi^2 - 2\pi x + x^2)}, \quad x \neq \pi \]
and \( f(\pi) = \lambda \).

2. **Finding the Limit:**
We need to find:
\[ \lim_{x \to \pi} f(x) = \lim_{x \to \pi} \frac{(1 + \cos x) \cdot \sin^2 x}{(\pi - x)^2 \cdot \ln(1 + \pi^2 - 2\pi x + x^2)} \]
As \( x \to \pi \), both the numerator and denominator approach 0, indicating that we can apply L'Hôpital's Rule or simplify the expression.

3. **Substituting \( x = \pi - t \):**
Let \( t = \pi - x \), then as \( x \to \pi \), \( t \to 0 \). Thus:
\[ \cos x = \cos(\pi - t) = -\cos t \]
and
\[ \sin x = \sin(\pi - t) = \sin t. \]
The limit becomes:
\[ \lim_{t \to 0} \frac{(1 - \cos t) \cdot \sin^2 t}{t^2 \cdot \ln(1 + t^2)}. \]

4. **Using Standard Limits:**
We know:
- \( 1 - \cos t \sim \frac{t^2}{2} \) as \( t \to 0 \).
- \( \sin t \sim t \) as \( t \to 0 \).
- \( \ln(1 + t^2) \sim t^2 \) as \( t \to 0 \).

5. **Substituting the Limits:**
Substitute these approximations into the limit:
\[ \lim_{t \to 0} \frac{\frac{t^2}{2} \cdot t^2}{t^2 \cdot t^2} = \lim_{t \to 0} \frac{\frac{t^4}{2}}{t^4} = \frac{1}{2}. \]

6. **Setting the Limit Equal to \( \lambda \):**
Since \( f(x) \) is continuous at \( x = \pi \), we have:
\[ \lambda = \lim_{x \to \pi} f(x) = \frac{1}{2}. \]

### Conclusion:
Thus, the value of \( \lambda \) is:
\[ \lambda = \frac{1}{2}. \]