The function `f(x) = (1-sinx + cosx)/(1+sinx+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. **Evaluate \( f(\pi) \)**:
The function \( f(x) \) is not defined at \( x = \pi \). To find the value of \( f(\pi) \) that makes it continuous, we need to compute:
\[ f(\pi) = \lim_{x \to \pi} f(x) = \lim_{x \to \pi} \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x} \]

2. **Substituting \( x = \pi \)**:
First, we substitute \( x = \pi \) into the function:
- \( \sin(\pi) = 0 \)
- \( \cos(\pi) = -1 \)

Thus, we have:
\[ f(\pi) = \frac{1 - 0 - 1}{1 + 0 - 1} = \frac{0}{0} \]
This is an indeterminate form, so we will apply L'Hôpital's Rule.

3. **Applying L'Hôpital's Rule**:
Since we have an indeterminate form \( \frac{0}{0} \), we differentiate the numerator and the denominator separately:
- Numerator: \( 1 - \sin x + \cos x \)
- Derivative: \( -\cos x - \sin x \)
- Denominator: \( 1 + \sin x + \cos x \)
- Derivative: \( \cos x + \cos x = 2\cos x \)

Thus, we have:
\[ \lim_{x \to \pi} \frac{-\cos x - \sin x}{2\cos x} \]

4. **Substituting \( x = \pi \) again**:
We substitute \( x = \pi \) into the derivatives:
- \( \cos(\pi) = -1 \)
- \( \sin(\pi) = 0 \)

Therefore, we get:
\[ \lim_{x \to \pi} \frac{-(-1) - 0}{2(-1)} = \frac{1}{-2} = -\frac{1}{2} \]

5. **Conclusion**:
Thus, the value of \( f(\pi) \) that makes \( f(x) \) continuous at \( x = \pi \) is:
\[ f(\pi) = -\frac{1}{2} \]

### Final Answer:
The value of \( f(\pi) \) is \( -\frac{1}{2} \).