Function `f(x) = (sin 2x)^(tan^(2)2x)` is not defined at `x=pi/4`. If f(x) is continuous at `x=pi/4`, then `f(pi/4)` is equal to:

Options

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

Answer

To solve the problem, we need to find the value of \( f\left(\frac{\pi}{4}\right) \) for the function \( f(x) = (\sin(2x))^{\tan^2(2x)} \) under the condition that \( f(x) \) is continuous at \( x = \frac{\pi}{4} \).

### Step-by-Step Solution:

1. **Identify the function**:
\[ f(x) = (\sin(2x))^{\tan^2(2x)} \]

2. **Check the value at \( x = \frac{\pi}{4} \)**:
\[ f\left(\frac{\pi}{4}\right) = (\sin(2 \cdot \frac{\pi}{4}))^{\tan^2(2 \cdot \frac{\pi}{4})} = (\sin(\frac{\pi}{2}))^{\tan^2(\frac{\pi}{2})} \]
Here, \( \sin(\frac{\pi}{2}) = 1 \) and \( \tan(\frac{\pi}{2}) \) is undefined. Thus, \( f\left(\frac{\pi}{4}\right) \) is not defined.

3. **Use the continuity condition**:
Since \( f(x) \) is continuous at \( x = \frac{\pi}{4} \), we need to find:
\[ f\left(\frac{\pi}{4}\right) = \lim_{x \to \frac{\pi}{4}} f(x) = \lim_{x \to \frac{\pi}{4}} (\sin(2x))^{\tan^2(2x)} \]

4. **Rewrite the limit**:
We can express the limit in the exponential form:
\[ f\left(\frac{\pi}{4}\right) = e^{\lim_{x \to \frac{\pi}{4}} \left(\tan^2(2x) \ln(\sin(2x))\right)} \]

5. **Evaluate the limit**:
As \( x \to \frac{\pi}{4} \):
- \( \sin(2x) \to 1 \) (since \( \sin(\frac{\pi}{2}) = 1 \))
- \( \tan^2(2x) \to \infty \) (since \( \tan(\frac{\pi}{2}) \) is undefined)

This results in an indeterminate form \( \infty \cdot 0 \). We need to manipulate it:
\[ \tan^2(2x) = \frac{\sin^2(2x)}{\cos^2(2x)} \]
Thus,
\[ \tan^2(2x) \ln(\sin(2x)) = \frac{\sin^2(2x) \ln(\sin(2x))}{\cos^2(2x)} \]

6. **Apply L'Hôpital's Rule**:
We can apply L'Hôpital's Rule to evaluate the limit:
\[ \lim_{x \to \frac{\pi}{4}} \frac{\ln(\sin(2x))}{\cot^2(2x)} \]
Differentiate the numerator and denominator:
- Derivative of \( \ln(\sin(2x)) = \frac{2\cos(2x)}{\sin(2x)} \)
- Derivative of \( \cot^2(2x) = -4\cot(2x)\csc^2(2x) \)

So we have:
\[ \lim_{x \to \frac{\pi}{4}} \frac{2\cos(2x)}{\sin(2x)} \cdot \frac{1}{-4\cot(2x)\csc^2(2x)} \]

7. **Evaluate the limit**:
After simplification, we find that the limit approaches \( -\frac{1}{2} \).

8. **Final calculation**:
Thus,
\[ f\left(\frac{\pi}{4}\right) = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \]

### Conclusion:
The value of \( f\left(\frac{\pi}{4}\right) \) is \( \frac{1}{\sqrt{e}} \).