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 that point, we will follow these steps:

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

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

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

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

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

Now, substituting into the sine and tangent functions:

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

We know that:

\[ \sin\left(\frac{\pi}{2}\right) = 1 \quad \text{and} \quad \tan\left(\frac{\pi}{2}\right) = \infty \]

Thus, we have:

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

This expression \( 1^{\infty} \) is indeterminate, so we need to analyze the limit as \( x \) approaches \( \frac{\pi}{4} \).

### Step 2: Rewrite the function

To resolve the indeterminate form, we can take the natural logarithm:

Let \( y = f(x) = (\sin 2x)^{\tan^2 2x} \).

Taking the natural logarithm:

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

### Step 3: Analyze the limit

We need to find the limit of \( \ln y \) as \( x \) approaches \( \frac{\pi}{4} \):

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

Substituting \( 2x \) as \( \frac{\pi}{2} \):

\[ \lim_{x \to \frac{\pi}{4}} \tan^2(2x) = \tan^2\left(\frac{\pi}{2}\right) = \infty \]

And:

\[ \lim_{x \to \frac{\pi}{4}} \ln(\sin(2x)) = \ln(\sin(\frac{\pi}{2})) = \ln(1) = 0 \]

This gives us the form \( \infty \cdot 0 \), which is still indeterminate.

### Step 4: Use L'Hôpital's Rule

To resolve this, we can rewrite it as:

\[ \lim_{x \to \frac{\pi}{4}} \frac{\ln(\sin(2x))}{\cot^2(2x)} \]

Both the numerator and denominator approach \( 0 \) as \( x \to \frac{\pi}{4} \). We can apply L'Hôpital's Rule:

Differentiate the numerator and denominator:

1. The derivative of \( \ln(\sin(2x)) \) is:

\[ \frac{2 \cos(2x)}{\sin(2x)} = 2 \cot(2x) \]

2. The derivative of \( \cot^2(2x) \) is:

\[ -4 \cot(2x) \csc^2(2x) \]

Now applying L'Hôpital's Rule:

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

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

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

Thus, we have:

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

### Step 5: Solve for \( y \)

Exponentiating both sides gives:

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

### Final Answer

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

\[ \boxed{e^{-\frac{1}{2}}} \]