Consider the function `f(x)=(sin 2x)^(tan^(2)2x), x in (pi)/(4)`. The value of `f((pi)/(4))` such that f is continuous at `x=(pi)/(4)` is

Options

A. `sqrte`
B. `1//sqrte`
C. 2
D. None of these

Answer

To find the value of the function \( f(x) = (\sin 2x)^{\tan^2 2x} \) at \( x = \frac{\pi}{4} \) such that \( f \) is continuous at \( x = \frac{\pi}{4} \), we can follow these steps:

### Step 1: Evaluate \( f\left(\frac{\pi}{4}\right) \)

First, we substitute \( x = \frac{\pi}{4} \) into the function:

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

Calculating \( 2 \cdot \frac{\pi}{4} \):

\[ 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} \]

Now, we find \( \sin\left(\frac{\pi}{2}\right) \) and \( \tan\left(\frac{\pi}{2}\right) \):

\[ \sin\left(\frac{\pi}{2}\right) = 1 \]
\[ \tan\left(\frac{\pi}{2}\right) = \text{undefined (approaches infinity)} \]

Thus, we have:

\[ f\left(\frac{\pi}{4}\right) = 1^{\tan^2\left(\frac{\pi}{2}\right)} = 1^{\infty} \]

This is an indeterminate form.

### Step 2: Take the logarithm

Let \( y = f(x) \), so:

\[ y = (\sin 2x)^{\tan^2 2x} \]

Taking the natural logarithm on both sides:

\[ \ln y = \tan^2 2x \cdot \ln(\sin 2x) \]

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

We need to evaluate:

\[ \lim_{x \to \frac{\pi}{4}} \ln y = \lim_{x \to \frac{\pi}{4}} \tan^2 2x \cdot \ln(\sin 2x) \]

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

\[ \tan^2 2x \to \tan^2\left(\frac{\pi}{2}\right) \to \infty \]
\[ \ln(\sin 2x) \to \ln(1) = 0 \]

This results in the indeterminate form \( \infty \cdot 0 \).

### Step 4: Rewrite the expression

We can rewrite the expression as:

\[ \ln y = \frac{\ln(\sin 2x)}{\cot^2 2x} \]

### Step 5: Apply L'Hôpital's Rule

Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule:

Differentiate the numerator and denominator:

1. Derivative of \( \ln(\sin 2x) \):
\[ \frac{d}{dx} \ln(\sin 2x) = \frac{2 \cos 2x}{\sin 2x} = 2 \cot 2x \]

2. Derivative of \( \cot^2 2x \):
\[ \frac{d}{dx} \cot^2 2x = -2 \cot 2x \cdot \csc^2 2x \cdot 2 = -4 \cot 2x \cdot \csc^2 2x \]

Now applying L'Hôpital's Rule:

\[ \lim_{x \to \frac{\pi}{4}} \frac{2 \cot 2x}{-4 \cot 2x \cdot \csc^2 2x} \]

This simplifies to:

\[ \lim_{x \to \frac{\pi}{4}} \frac{-1}{2 \csc^2 2x} \]

### Step 6: Evaluate the limit

As \( x \to \frac{\pi}{4} \):

\[ \csc^2\left(\frac{\pi}{2}\right) = 1 \]

Thus:

\[ \lim_{x \to \frac{\pi}{4}} \ln y = -\frac{1}{2} \]

### Step 7: Solve for \( y \)

Exponentiating both sides gives:

\[ y = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \]

### Conclusion

Therefore, the value of \( f\left(\frac{\pi}{4}\right) \) such that \( f \) is continuous at \( x = \frac{\pi}{4} \) is:

\[ f\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{e}} \]