The function `f(x) = (1-sin x + cos x)/(1+sin x + cosx)` is not defined at `x = pi`. The value of `f(pi)`, so that f(x) is continuous at `x = pi` is

Options

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

Answer

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

### Step-by-Step Solution:

1. **Identify the Function**:
The function is given as:
\[ f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x} \]

2. **Evaluate the Function at \( x = \pi \)**:
First, we need to check if \( f(\pi) \) is defined:
\[ f(\pi) = \frac{1 - \sin(\pi) + \cos(\pi)}{1 + \sin(\pi) + \cos(\pi)} = \frac{1 - 0 - 1}{1 + 0 - 1} = \frac{0}{0} \]
Since this results in an indeterminate form \( \frac{0}{0} \), we need to find the limit as \( x \) approaches \( \pi \).

3. **Finding the Limit**:
We will compute the limit:
\[ \lim_{x \to \pi} f(x) = \lim_{x \to \pi} \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x} \]
As \( x \) approaches \( \pi \), both the numerator and denominator approach \( 0 \). Therefore, we can apply L'Hôpital's Rule.

4. **Differentiate the Numerator and Denominator**:
Differentiate the numerator:
\[ \frac{d}{dx}(1 - \sin x + \cos x) = -\cos x - \sin x \]
Differentiate the denominator:
\[ \frac{d}{dx}(1 + \sin x + \cos x) = \cos x - \sin x \]

5. **Apply L'Hôpital's Rule**:
Now we apply L'Hôpital's Rule:
\[ \lim_{x \to \pi} f(x) = \lim_{x \to \pi} \frac{-\cos x - \sin x}{\cos x - \sin x} \]

6. **Evaluate the Limit at \( x = \pi \)**:
Substitute \( x = \pi \):
\[ \cos(\pi) = -1, \quad \sin(\pi) = 0 \]
Therefore,
\[ \lim_{x \to \pi} f(x) = \frac{-(-1) - 0}{-1 - 0} = \frac{1}{-1} = -1 \]

7. **Conclusion**:
To make \( f(x) \) continuous at \( x = \pi \), we set:
\[ f(\pi) = -1 \]

### Final Answer:
The value of \( f(\pi) \) so that \( f(x) \) is continuous at \( x = \pi \) is \( -1 \).