If `f(x)={{:(tan((pi)/(4)-x)/(cot2x)",",x ne(pi)/(4)),(k",",x=(pi)/(4)):}` is continuous at `x=(pi)/(4)` , then the value of k is

Options

A. 1
B. 2
C. `(1)/(2)`
D. none of these

Answer

To find the value of \( k \) such that the function

\[ f(x) = \begin{cases} \tan\left(\frac{\pi}{4} - x\right) / \cot(2x) & \text{if } x \neq \frac{\pi}{4} \\ k & \text{if } x = \frac{\pi}{4} \end{cases} \]

is continuous at \( x = \frac{\pi}{4} \), we need to ensure that

\[ \lim_{x \to \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}\right) = k. \]

### Step 1: Find the limit as \( x \) approaches \( \frac{\pi}{4} \)

We need to calculate

\[ \lim_{x \to \frac{\pi}{4}} \frac{\tan\left(\frac{\pi}{4} - x\right)}{\cot(2x)}. \]

### Step 2: Substitute \( x = \frac{\pi}{4} \)

Substituting \( x = \frac{\pi}{4} \):

- \( \tan\left(\frac{\pi}{4} - \frac{\pi}{4}\right) = \tan(0) = 0 \)
- \( \cot(2 \cdot \frac{\pi}{4}) = \cot\left(\frac{\pi}{2}\right) = 0 \)

This gives us the indeterminate form \( \frac{0}{0} \).

### Step 3: Apply L'Hôpital's Rule

Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:

\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]

if the limit on the right exists.

### Step 4: Differentiate the numerator and denominator

1. Differentiate the numerator \( \tan\left(\frac{\pi}{4} - x\right) \):
- The derivative is \( \sec^2\left(\frac{\pi}{4} - x\right) \cdot (-1) = -\sec^2\left(\frac{\pi}{4} - x\right) \).

2. Differentiate the denominator \( \cot(2x) \):
- The derivative is \( -\csc^2(2x) \cdot 2 = -2 \csc^2(2x) \).

### Step 5: Rewrite the limit using derivatives

Now we have:

\[ \lim_{x \to \frac{\pi}{4}} \frac{-\sec^2\left(\frac{\pi}{4} - x\right)}{-2 \csc^2(2x)} = \lim_{x \to \frac{\pi}{4}} \frac{\sec^2\left(\frac{\pi}{4} - x\right)}{2 \csc^2(2x)}. \]

### Step 6: Evaluate the limit

Now substituting \( x = \frac{\pi}{4} \):

- \( \sec^2\left(\frac{\pi}{4} - \frac{\pi}{4}\right) = \sec^2(0) = 1 \)
- \( \csc^2\left(\frac{\pi}{2}\right) = \csc^2\left(\frac{\pi}{2}\right) = 1 \)

Thus, we get:

\[ \lim_{x \to \frac{\pi}{4}} \frac{1}{2 \cdot 1} = \frac{1}{2}. \]

### Step 7: Set the limit equal to \( k \)

Since the function is continuous at \( x = \frac{\pi}{4} \), we have:

\[ k = \lim_{x \to \frac{\pi}{4}} f(x) = \frac{1}{2}. \]

### Final Answer

Thus, the value of \( k \) is

\[ \boxed{\frac{1}{2}}. \]