If : f(x) ` {: ((1-sin x)/(pi-2x)", ... " x != pi//2 ),(=lambda ", ... " x = pi//2):}` is continuous at ` x = pi//2 ` , then : `lambda = `

Options

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

Answer

To determine the value of \( \lambda \) such that the function \( f(x) \) is continuous at \( x = \frac{\pi}{2} \), we need to ensure that:

\[ \lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) \]

Given that:

\[ f(x) = \frac{1 - \sin x}{\pi - 2x} \quad \text{for } x \neq \frac{\pi}{2} \]
\[ f\left(\frac{\pi}{2}\right) = \lambda \]

### Step 1: Calculate the limit as \( x \) approaches \( \frac{\pi}{2} \)

We first compute the limit:

\[ \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\pi - 2x} \]

Substituting \( x = \frac{\pi}{2} \):

\[ 1 - \sin\left(\frac{\pi}{2}\right) = 1 - 1 = 0 \]
\[ \pi - 2\left(\frac{\pi}{2}\right) = \pi - \pi = 0 \]

Thus, we have an indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule

Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

where \( f(x) = 1 - \sin x \) and \( g(x) = \pi - 2x \).

Calculating the derivatives:

- The derivative of the numerator \( f'(x) = -\cos x \)
- The derivative of the denominator \( g'(x) = -2 \)

Now we can rewrite the limit:

\[ \lim_{x \to \frac{\pi}{2}} \frac{-\cos x}{-2} = \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{2} \]

### Step 3: Evaluate the limit

Now substituting \( x = \frac{\pi}{2} \):

\[ \cos\left(\frac{\pi}{2}\right) = 0 \]

Thus, the limit becomes:

\[ \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{2} = \frac{0}{2} = 0 \]

### Step 4: Set the limit equal to \( \lambda \)

Since we want the function to be continuous at \( x = \frac{\pi}{2} \):

\[ \lambda = \lim_{x \to \frac{\pi}{2}} f(x) = 0 \]

### Conclusion

Therefore, the value of \( \lambda \) is:

\[ \lambda = 0 \]

---