Consider the function `f(x)=(1-sinx)/((pi-2x)^(2))` where `xne(pi)/(2)andf((pi)/(2))=lambda`
What is the value of `lambda`. If the function is continuous at `x=(pi)/(2)`?

Options

A. `(1)/(8)`
B. `(1)/(4)`
C. `(1)/(2)`
D. 1

Answer

To find the value of \( \lambda \) such that the function \( f(x) = \frac{1 - \sin x}{(\pi - 2x)^2} \) is continuous at \( x = \frac{\pi}{2} \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches \( \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Substituting \( x = \frac{\pi}{2} \)**:
\[ f\left(\frac{\pi}{2}\right) = \frac{1 - \sin\left(\frac{\pi}{2}\right)}{(\pi - 2 \cdot \frac{\pi}{2})^2} = \frac{1 - 1}{(0)^2} = \frac{0}{0} \]
This is an indeterminate form \( \frac{0}{0} \).

**Hint**: When you get \( \frac{0}{0} \), it indicates that you can apply L'Hôpital's Rule.

2. **Applying L'Hôpital's Rule**:
Differentiate the numerator and the denominator separately:
- The derivative of the numerator \( 1 - \sin x \) is \( -\cos x \).
- The derivative of the denominator \( (\pi - 2x)^2 \) using the chain rule is \( 2(\pi - 2x)(-2) = -4(\pi - 2x) \).

Thus, we have:
\[ \lim_{x \to \frac{\pi}{2}} f(x) = \lim_{x \to \frac{\pi}{2}} \frac{-\cos x}{-4(\pi - 2x)} = \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{4(\pi - 2x)} \]

**Hint**: Remember to simplify the limit after differentiation.

3. **Substituting \( x = \frac{\pi}{2} \) again**:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{4(\pi - 2x)} = \frac{\cos\left(\frac{\pi}{2}\right)}{4(\pi - 2 \cdot \frac{\pi}{2})} = \frac{0}{0} \]
Again, we have an indeterminate form \( \frac{0}{0} \).

**Hint**: If you encounter \( \frac{0}{0} \) again, you can apply L'Hôpital's Rule once more.

4. **Applying L'Hôpital's Rule again**:
Differentiate the numerator and denominator again:
- The derivative of \( \cos x \) is \( -\sin x \).
- The derivative of \( 4(\pi - 2x) \) is \( -8 \).

Thus, we have:
\[ \lim_{x \to \frac{\pi}{2}} \frac{-\sin x}{-8} = \lim_{x \to \frac{\pi}{2}} \frac{\sin x}{8} \]

**Hint**: Always check if the limit can be evaluated after differentiation.

5. **Evaluating the limit**:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\sin x}{8} = \frac{\sin\left(\frac{\pi}{2}\right)}{8} = \frac{1}{8} \]

6. **Conclusion**:
Since \( f(x) \) is continuous at \( x = \frac{\pi}{2} \), we set \( \lambda \) equal to this limit:
\[ \lambda = \frac{1}{8} \]

### Final Answer:
\[ \lambda = \frac{1}{8} \]