If f(x) = `(sqrt(2) cos x -1) /(cot x-1) , x ne (pi)/(4)` then the value of f `((pi)/(4))` so that f(x) becomes continous at x = `(pi)/(4)` is

Options

A. `(1)/(2)`
B. `-(1)/(2)`
C. 1
D. 9

Answer

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

Given:
\[ f(x) = \frac{\sqrt{2} \cos x - 1}{\cot x - 1}, \quad x \neq \frac{\pi}{4} \]

### Step 1: Evaluate the limit as \( x \to \frac{\pi}{4} \)
We need to find:
\[ \lim_{x \to \frac{\pi}{4}} f(x) = \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \cos x - 1}{\cot x - 1} \]

### Step 2: Substitute \( x = \frac{\pi}{4} \)
Substituting \( x = \frac{\pi}{4} \):
- \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)
- \( \cot\left(\frac{\pi}{4}\right) = 1 \)

Thus, we have:
\[ \sqrt{2} \cos\left(\frac{\pi}{4}\right) - 1 = \sqrt{2} \cdot \frac{1}{\sqrt{2}} - 1 = 1 - 1 = 0 \]
\[ \cot\left(\frac{\pi}{4}\right) - 1 = 1 - 1 = 0 \]

This results in the form \( \frac{0}{0} \), which is indeterminate.

### Step 3: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
if the limit on the right side exists.

#### Differentiate the numerator and denominator:
1. **Numerator**: \( \sqrt{2} \cos x - 1 \)
- Derivative: \( -\sqrt{2} \sin x \)

2. **Denominator**: \( \cot x - 1 \)
- Derivative: \( -\csc^2 x \)

Now we can rewrite the limit:
\[ \lim_{x \to \frac{\pi}{4}} \frac{-\sqrt{2} \sin x}{-\csc^2 x} = \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \sin x \cdot \sin^2 x}{1} \]

### Step 4: Evaluate the limit
Substituting \( x = \frac{\pi}{4} \):
- \( \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)

Thus:
\[ \lim_{x \to \frac{\pi}{4}} \sqrt{2} \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right)^2 = \sqrt{2} \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{1}{2} \]

### Conclusion
The value of \( f\left(\frac{\pi}{4}\right) \) that makes \( f(x) \) continuous at \( x = \frac{\pi}{4} \) is:
\[ f\left(\frac{\pi}{4}\right) = -\frac{1}{2} \]