If `f(x) = (1- cos 7(x-pi))/(x-pi), (x ne pi)` is continuous at `x =pi`, then `f(pi)` equals-

Options

A. 0
B. 1
C. `-1`
D. 7

Answer

To find the value of \( f(\pi) \) for the function

\[ f(x) = \frac{1 - \cos(7(x - \pi))}{x - \pi} \quad (x \neq \pi) \]

and ensure that it is continuous at \( x = \pi \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches \( \pi \).

### Step-by-Step Solution:

1. **Set up the limit**:
We need to find:
\[ \lim_{x \to \pi} f(x) = \lim_{x \to \pi} \frac{1 - \cos(7(x - \pi))}{x - \pi} \]

2. **Substituting \( x = \pi \)**:
If we directly substitute \( x = \pi \), we get:
\[ f(\pi) = \frac{1 - \cos(0)}{0} = \frac{1 - 1}{0} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule.

3. **Applying L'Hôpital's Rule**:
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:
- The derivative of the numerator \( 1 - \cos(7(x - \pi)) \) is \( 7 \sin(7(x - \pi)) \).
- The derivative of the denominator \( x - \pi \) is \( 1 \).

Thus, we have:
\[ \lim_{x \to \pi} \frac{1 - \cos(7(x - \pi))}{x - \pi} = \lim_{x \to \pi} \frac{7 \sin(7(x - \pi))}{1} \]

4. **Evaluating the limit**:
Now substituting \( x = \pi \) into the new limit:
\[ = 7 \sin(7(\pi - \pi)) = 7 \sin(0) = 7 \times 0 = 0 \]

5. **Conclusion**:
Since the limit exists and equals 0, we can define \( f(\pi) \) as:
\[ f(\pi) = 0 \]

Thus, the value of \( f(\pi) \) is \( \boxed{0} \).