If `f(x)=(1-sinx)/((pi-2x)^2)`,when `x!=pi/2 ` and `f(pi/2)=lambda,` the `f(x)` will be continuous function at `x=pi/2`,where `lambda=?` (a)`1/8` (b) `1/4` (c) `1/2` (d) none of these

Explanation

To determine 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 follow these steps:

### Step 1: Check the form of \( f(x) \) at \( x = \frac{\pi}{2} \)
We first evaluate \( f(x) \) as \( x \) approaches \( \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = \frac{1 - \sin\left(\frac{\pi}{2}\right)}{(\pi - 2\left(\frac{\pi}{2}\right))^2} = \frac{1 - 1}{(0)^2} = \frac{0}{0} \]
This is an indeterminate form, so we need to find the limit as \( x \) approaches \( \frac{\pi}{2} \).

### Step 2: Find the limit of \( f(x) \) as \( x \) approaches \( \frac{\pi}{2} \)
We compute:
\[ \lim_{x \to \frac{\pi}{2}} f(x) = \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{(\pi - 2x)^2} \]
Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule
Differentiate the numerator and denominator:
- The derivative of the numerator \( 1 - \sin x \) is \( -\cos x \).
- The derivative of the denominator \( (\pi - 2x)^2 \) is \( -4(\pi - 2x) \).

Now we apply L'Hôpital's Rule:
\[ \lim_{x \to \frac{\pi}{2}} \frac{-\cos x}{-4(\pi - 2x)} = \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{4(\pi - 2x)} \]

### Step 4: Evaluate the limit again
Substituting \( x = \frac{\pi}{2} \):
\[ \cos\left(\frac{\pi}{2}\right) = 0 \quad \text{and} \quad \pi - 2\left(\frac{\pi}{2}\right) = 0 \]
This again gives us a \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule again.

### Step 5: Apply L'Hôpital's Rule a second time
Differentiate again:
- The derivative of the numerator \( \cos x \) is \( -\sin x \).
- The derivative of the denominator \( 4(\pi - 2x) \) is \( -8 \).

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

### Step 6: Evaluate the limit
Substituting \( x = \frac{\pi}{2} \):
\[ \sin\left(\frac{\pi}{2}\right) = 1 \]
Thus, the limit becomes:
\[ \frac{1}{8} \]

### Step 7: Set \( \lambda \) equal to the limit
For \( f(x) \) to be continuous at \( x = \frac{\pi}{2} \), we need:
\[ \lambda = \lim_{x \to \frac{\pi}{2}} f(x) = \frac{1}{8} \]

### Conclusion
The value of \( \lambda \) that makes \( f(x) \) continuous at \( x = \frac{\pi}{2} \) is:
\[ \lambda = \frac{1}{8} \]