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 \( f(\pi) = \lambda \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \begin{cases} \frac{(1 + \cos x)}{(\pi - x)^2} \cdot \frac{\sin^2 x}{\ln(1 + \pi^2 - 2\pi x + x^2)}, & x \neq \pi \\ \lambda, & x = \pi \end{cases} \]

2. **Set up the limit**:
We need to find:
\[ \lim_{x \to \pi} f(x) = \lambda \]

3. **Substitute \( x = \pi + t \)**:
As \( x \to \pi \), let \( t = x - \pi \) so that \( x = \pi + t \). Then as \( x \to \pi \), \( t \to 0 \).

4. **Rewrite the limit**:
Substitute \( x = \pi + t \):
\[ \lim_{t \to 0} f(\pi + t) = \lim_{t \to 0} \frac{(1 + \cos(\pi + t))}{(\pi - (\pi + t))^2} \cdot \frac{\sin^2(\pi + t)}{\ln(1 + \pi^2 - 2\pi(\pi + t) + (\pi + t)^2)} \]
Simplifying the terms:
- \( \cos(\pi + t) = -\cos t \)
- \( \sin(\pi + t) = -\sin t \)

Thus, the limit becomes:
\[ \lim_{t \to 0} \frac{(1 - \cos t)}{t^2} \cdot \frac{\sin^2 t}{\ln(1 + \pi^2 - 2\pi^2 - 2\pi t + \pi^2 + 2\pi t + t^2)} = \lim_{t \to 0} \frac{(1 - \cos t)}{t^2} \cdot \frac{\sin^2 t}{\ln(1 + t^2)} \]

5. **Evaluate the limit**:
We know:
- \( 1 - \cos t \sim \frac{t^2}{2} \) as \( t \to 0 \)
- \( \sin^2 t \sim t^2 \) as \( t \to 0 \)
- \( \ln(1 + t^2) \sim t^2 \) as \( t \to 0 \)

Therefore, we can rewrite the limit:
\[ \lim_{t \to 0} \frac{\frac{t^2}{2}}{t^2} \cdot \frac{t^2}{t^2} = \lim_{t \to 0} \frac{1}{2} \cdot 1 = \frac{1}{2} \]

6. **Set the limit equal to \( \lambda \)**:
Since \( \lim_{x \to \pi} f(x) = \lambda \), we have:
\[ \lambda = \frac{1}{2} \]

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