Let `f(x) =(1- tanx)/(4x-pi), x ne pi/4, x in [0, pi/2]`. If `f(x)` is continuous in `[0,pi/2]`, then `f(pi/4)`=

Options

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

Answer

To find \( f\left(\frac{\pi}{4}\right) \) for the function \( f(x) = \frac{1 - \tan x}{4x - \pi} \) where \( x \neq \frac{\pi}{4} \) and \( x \in [0, \frac{\pi}{2}] \), we need to ensure that \( f(x) \) is continuous at \( x = \frac{\pi}{4} \).

### Step-by-Step Solution:

1. **Check for Continuity**:
For \( f(x) \) to be continuous at \( x = \frac{\pi}{4} \), we need:
\[ f\left(\frac{\pi}{4}\right) = \lim_{x \to \frac{\pi}{4}} f(x) \]

2. **Calculate the Limit**:
We need to calculate:
\[ \lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{4x - \pi} \]
First, substitute \( x = \frac{\pi}{4} \):
\[ \tan\left(\frac{\pi}{4}\right) = 1 \quad \text{and} \quad 4\left(\frac{\pi}{4}\right) - \pi = 0 \]
This gives us the form \( \frac{0}{0} \), which is indeterminate.

3. **Apply L'Hôpital's Rule**:
Since we have an indeterminate form, we can apply L'Hôpital's Rule:
\[ \lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{4x - \pi} = \lim_{x \to \frac{\pi}{4}} \frac{-\sec^2 x}{4} \]

4. **Differentiate the Numerator and Denominator**:
The derivative of the numerator \( 1 - \tan x \) is \( -\sec^2 x \) and the derivative of the denominator \( 4x - \pi \) is \( 4 \).

5. **Evaluate the Limit**:
Now we substitute \( x = \frac{\pi}{4} \) into the limit:
\[ \lim_{x \to \frac{\pi}{4}} \frac{-\sec^2 x}{4} = \frac{-\sec^2\left(\frac{\pi}{4}\right)}{4} \]
We know that \( \sec\left(\frac{\pi}{4}\right) = \sqrt{2} \), so:
\[ \sec^2\left(\frac{\pi}{4}\right) = 2 \]
Therefore:
\[ \lim_{x \to \frac{\pi}{4}} \frac{-\sec^2 x}{4} = \frac{-2}{4} = -\frac{1}{2} \]

6. **Conclusion**:
Since \( f\left(\frac{\pi}{4}\right) = \lim_{x \to \frac{\pi}{4}} f(x) \), we have:
\[ f\left(\frac{\pi}{4}\right) = -\frac{1}{2} \]

### Final Answer:
\[ f\left(\frac{\pi}{4}\right) = -\frac{1}{2} \]