If `f(x)={{:((1-sinx)/(pi-2x),,","xne(pi)/(2)),(lambda,,","x=(pi)/(2)):},"be continuous at "x=(pi)/(2),` then value of `lambda` is

Options

A. `-1`
B. `1`
C. `0`
D. `2`

Explanation

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = \frac{\pi}{2} \). This means that the left-hand limit, right-hand limit, and the value of the function at that point must all be equal.

### Step-by-Step Solution:

1. **Identify the function**:
The function is defined as:
\[ f(x) = \begin{cases} \frac{1 - \sin x}{\pi - 2x} & \text{if } x \neq \frac{\pi}{2} \\ \lambda & \text{if } x = \frac{\pi}{2} \end{cases} \]

2. **Find the right-hand limit as \( x \) approaches \( \frac{\pi}{2} \)**:
We need to calculate:
\[ \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}^+} \frac{1 - \sin x}{\pi - 2x} \]
Substituting \( x = \frac{\pi}{2} \) gives us:
\[ \sin\left(\frac{\pi}{2}\right) = 1 \Rightarrow 1 - \sin\left(\frac{\pi}{2}\right) = 0 \]
And for the denominator:
\[ \pi - 2\left(\frac{\pi}{2}\right) = 0 \]
Thus, we have an indeterminate form \( \frac{0}{0} \).

3. **Apply L'Hôpital's Rule**:
Since we have an indeterminate form, we can apply L'Hôpital's Rule:
\[ \lim_{x \to \frac{\pi}{2}^+} \frac{1 - \sin x}{\pi - 2x} = \lim_{x \to \frac{\pi}{2}^+} \frac{-\cos x}{-2} \]
Now substituting \( x = \frac{\pi}{2} \):
\[ -\cos\left(\frac{\pi}{2}\right) = 0 \Rightarrow \text{So, the limit becomes } \frac{0}{-2} = 0 \]

4. **Find the left-hand limit**:
Since the function is symmetric around \( x = \frac{\pi}{2} \), the left-hand limit will also be:
\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = 0 \]

5. **Set the limits equal to \( \lambda \)**:
For the function to be continuous at \( x = \frac{\pi}{2} \):
\[ \lambda = \lim_{x \to \frac{\pi}{2}^+} f(x) = 0 \]

### Conclusion:
Thus, the value of \( \lambda \) is:
\[ \lambda = 0 \]