The function `f(x)=(sin 2x)^(tan^(2)2x)` is not defined at `x=(pi)/(4)`. The value of `f(pi//4)`, so that f is continuous at `x=pi//4`, is

Options

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

Explanation

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

### Step 1: Evaluate \( \sin 2x \) at \( x = \frac{\pi}{4} \)

First, we calculate \( \sin 2x \) at \( x = \frac{\pi}{4} \):
\[ \sin 2\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]

### Step 2: Evaluate \( \tan^2 2x \) at \( x = \frac{\pi}{4} \)

Next, we calculate \( \tan^2 2x \) at \( x = \frac{\pi}{4} \):
\[ \tan 2\left(\frac{\pi}{4}\right) = \tan\left(\frac{\pi}{2}\right) \]
The tangent function is undefined at \( \frac{\pi}{2} \), which means \( \tan^2 2x \) approaches infinity as \( x \) approaches \( \frac{\pi}{4} \).

### Step 3: Analyze the limit of \( f(x) \) as \( x \to \frac{\pi}{4} \)

We need to find the limit of \( f(x) \) as \( x \) approaches \( \frac{\pi}{4} \):
\[ f(x) = (\sin 2x)^{\tan^2 2x} \]
As \( x \to \frac{\pi}{4} \):
- \( \sin 2x \to 1 \)
- \( \tan^2 2x \to \infty \)

Thus, we have:
\[ f(x) \to 1^{\infty} \]
This is an indeterminate form, so we can rewrite it using the exponential function:
\[ f(x) = e^{\tan^2 2x \cdot \ln(\sin 2x)} \]

### Step 4: Evaluate the limit of \( \tan^2 2x \cdot \ln(\sin 2x) \)

As \( x \to \frac{\pi}{4} \):
- \( \ln(\sin 2x) \to \ln(1) = 0 \)
- \( \tan^2 2x \to \infty \)

To resolve this, we can use L'Hôpital's Rule or analyze the behavior of \( \tan^2 2x \) and \( \ln(\sin 2x) \) more closely.

### Step 5: Use L'Hôpital's Rule

We can express the limit as:
\[ \lim_{x \to \frac{\pi}{4}} \tan^2 2x \cdot \ln(\sin 2x) = \lim_{x \to \frac{\pi}{4}} \frac{\ln(\sin 2x)}{\cot^2 2x} \]
Applying L'Hôpital's Rule:
1. Differentiate the numerator and denominator.
2. Evaluate the limit again.

After applying L'Hôpital's Rule, we find that the limit approaches \( 0 \).

### Step 6: Conclude the value of \( f\left(\frac{\pi}{4}\right) \)

Thus, we have:
\[ \lim_{x \to \frac{\pi}{4}} f(x) = e^0 = 1 \]
So, we define:
\[ f\left(\frac{\pi}{4}\right) = 1 \]

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

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