If f(x)=tan `(pi//4 -x)//cot2x` for `x ne pi//4`. The value of `f(pi//4)` so that f is continuous at `x=pi//4` is

Options

A. `1//3`
B. `1//2`
C. `1//4`
D. `1//6`

Answer

To find the value of \( f\left(\frac{\pi}{4}\right) \) such that the function \( f(x) = \frac{\tan\left(\frac{\pi}{4} - x\right)}{\cot(2x)} \) is continuous at \( x = \frac{\pi}{4} \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches \( \frac{\pi}{4} \).

### Step-by-Step Solution:

1. **Identify the Function**:
\[ f(x) = \frac{\tan\left(\frac{\pi}{4} - x\right)}{\cot(2x)} \]

2. **Evaluate the Limit**:
We need to find:
\[ \lim_{x \to \frac{\pi}{4}} f(x) = \lim_{x \to \frac{\pi}{4}} \frac{\tan\left(\frac{\pi}{4} - x\right)}{\cot(2x)} \]

3. **Substituting \( x = \frac{\pi}{4} \)**:
\[ \tan\left(\frac{\pi}{4} - \frac{\pi}{4}\right) = \tan(0) = 0 \]
\[ \cot\left(2 \cdot \frac{\pi}{4}\right) = \cot\left(\frac{\pi}{2}\right) = 0 \]
This gives us a \( \frac{0}{0} \) form, so we can apply L'Hôpital's Rule.

4. **Differentiate Numerator and Denominator**:
- **Numerator**:
\[ \frac{d}{dx} \tan\left(\frac{\pi}{4} - x\right) = -\sec^2\left(\frac{\pi}{4} - x\right) \]
- **Denominator**:
\[ \frac{d}{dx} \cot(2x) = -2\csc^2(2x) \]

5. **Apply L'Hôpital's Rule**:
\[ \lim_{x \to \frac{\pi}{4}} \frac{-\sec^2\left(\frac{\pi}{4} - x\right)}{-2\csc^2(2x)} = \lim_{x \to \frac{\pi}{4}} \frac{\sec^2\left(\frac{\pi}{4} - x\right)}{2\csc^2(2x)} \]

6. **Evaluate the Limit**:
- As \( x \to \frac{\pi}{4} \):
\[ \sec^2\left(\frac{\pi}{4} - \frac{\pi}{4}\right) = \sec^2(0) = 1 \]
\[ \csc^2\left(2 \cdot \frac{\pi}{4}\right) = \csc^2\left(\frac{\pi}{2}\right) = 1 \]
Thus, we have:
\[ \lim_{x \to \frac{\pi}{4}} f(x) = \frac{1}{2 \cdot 1} = \frac{1}{2} \]

7. **Conclusion**:
Therefore, to make \( f \) continuous at \( x = \frac{\pi}{4} \), we set:
\[ f\left(\frac{\pi}{4}\right) = \frac{1}{2} \]

### Final Answer:
The value of \( f\left(\frac{\pi}{4}\right) \) so that \( f \) is continuous at \( x = \frac{\pi}{4} \) is \( \frac{1}{2} \).