Let `f(x)=(sqrt2-(cosx+sinx))/(1-sin2x), x ne pi//4`. The value `f(pi//4)` so that f is continuous is `(sqrt2=1.41)`

Answer

To find the value of \( f\left(\frac{\pi}{4}\right) \) such that the function \( f(x) \) is continuous at \( x = \frac{\pi}{4} \), we start with the given function:

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

### Step 1: Check for continuity
For \( f(x) \) to be continuous at \( x = \frac{\pi}{4} \), we need:

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

### Step 2: Evaluate the limit
First, we evaluate the limit as \( x \) approaches \( \frac{\pi}{4} \):

\[ \lim_{x \to \frac{\pi}{4}} f(x) = \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - (\cos x + \sin x)}{1 - \sin 2x} \]

### Step 3: Substitute \( x = \frac{\pi}{4} \)
Substituting \( x = \frac{\pi}{4} \):

- \( \cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)
- \( \sin\left(\frac{\pi}{2}\right) = 1 \)

Thus, we have:

\[ \cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \]

And:

\[ 1 - \sin\left(2 \cdot \frac{\pi}{4}\right) = 1 - 1 = 0 \]

So we get:

\[ \lim_{x \to \frac{\pi}{4}} f(x) = \frac{\sqrt{2} - \sqrt{2}}{0} = \frac{0}{0} \]

This is an indeterminate form, so we apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule
Differentiate the numerator and denominator:

- The derivative of the numerator \( \sqrt{2} - (\cos x + \sin x) \) is \( \sin x - \cos x \).
- The derivative of the denominator \( 1 - \sin 2x \) is \( -2 \cos 2x \).

Thus, we have:

\[ \lim_{x \to \frac{\pi}{4}} f(x) = \lim_{x \to \frac{\pi}{4}} \frac{\sin x - \cos x}{-2 \cos 2x} \]

### Step 5: Substitute \( x = \frac{\pi}{4} \) again
Now substituting \( x = \frac{\pi}{4} \):

- \( \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)
- \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)
- \( \cos\left(\frac{\pi}{2}\right) = 0 \)

Thus:

\[ \sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0 \]

So we again have \( 0 \) in the numerator. We apply L'Hôpital's Rule again.

### Step 6: Differentiate again
Differentiate again:

- The derivative of \( \sin x - \cos x \) is \( \cos x + \sin x \).
- The derivative of \( -2 \cos 2x \) is \( 4 \sin 2x \).

Thus:

\[ \lim_{x \to \frac{\pi}{4}} f(x) = \lim_{x \to \frac{\pi}{4}} \frac{\cos x + \sin x}{4 \sin 2x} \]

### Step 7: Substitute \( x = \frac{\pi}{4} \) again
Substituting \( x = \frac{\pi}{4} \):

\[ \cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \]

And:

\[ 4 \sin\left(\frac{\pi}{2}\right) = 4 \cdot 1 = 4 \]

Thus:

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

### Step 8: Set \( f\left(\frac{\pi}{4}\right) \)
To ensure continuity, we set:

\[ f\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{4} \]

### Final Answer
Therefore, the value of \( f\left(\frac{\pi}{4}\right) \) such that \( f \) is continuous is:

\[ f\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{4} \approx 0.35 \]