If `f(x) =(cos^(2) pix)/(e^(2x) - 2ex), x ne 1/2`, the value of `f(1/2)`, so that f (x) is continuous at `x=1/2` is:

Options

A. `pi/(2e^(2))`
B. `pi/(2e)`
C. `pi^(2)/(2e^(2))`
D. `pi^(2)/(2e)`

Answer

To find the value of \( f\left(\frac{1}{2}\right) \) such that \( f(x) \) is continuous at \( x = \frac{1}{2} \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches \( \frac{1}{2} \) equals \( f\left(\frac{1}{2}\right) \).

### Step 1: Define the function
The function is given as:
\[ f(x) = \frac{\cos^2(\pi x)}{e^{2x} - 2e^x}, \quad x \neq \frac{1}{2} \]

### Step 2: Find the limit as \( x \) approaches \( \frac{1}{2} \)
To find \( f\left(\frac{1}{2}\right) \), we need to compute:
\[ \lim_{x \to \frac{1}{2}} f(x) = \lim_{x \to \frac{1}{2}} \frac{\cos^2(\pi x)}{e^{2x} - 2e^x} \]

### Step 3: Substitute \( x = \frac{1}{2} \)
Substituting \( x = \frac{1}{2} \):
- The numerator becomes:
\[ \cos^2\left(\pi \cdot \frac{1}{2}\right) = \cos^2\left(\frac{\pi}{2}\right) = 0 \]
- The denominator becomes:
\[ e^{2 \cdot \frac{1}{2}} - 2e^{\frac{1}{2}} = e - 2\sqrt{e} = 0 \]
Thus, we have a \( \frac{0}{0} \) indeterminate form.

### Step 4: Apply L'Hôpital's Rule
Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule:
\[ \lim_{x \to \frac{1}{2}} f(x) = \lim_{x \to \frac{1}{2}} \frac{\frac{d}{dx}(\cos^2(\pi x))}{\frac{d}{dx}(e^{2x} - 2e^x)} \]

### Step 5: Differentiate the numerator and denominator
- The derivative of the numerator:
\[ \frac{d}{dx}(\cos^2(\pi x)) = 2\cos(\pi x)(-\sin(\pi x)\cdot \pi) = -2\pi \cos(\pi x) \sin(\pi x) \]
- The derivative of the denominator:
\[ \frac{d}{dx}(e^{2x} - 2e^x) = 2e^{2x} - 2e^x \]

### Step 6: Substitute back into the limit
Now we compute:
\[ \lim_{x \to \frac{1}{2}} \frac{-2\pi \cos(\pi x) \sin(\pi x)}{2e^{2x} - 2e^x} \]
This simplifies to:
\[ \lim_{x \to \frac{1}{2}} \frac{-\pi \cos(\pi x) \sin(\pi x)}{e^{2x} - e^x} \]

### Step 7: Substitute \( x = \frac{1}{2} \) again
Substituting \( x = \frac{1}{2} \):
- The numerator becomes:
\[ -\pi \cos\left(\frac{\pi}{2}\right) \sin\left(\frac{\pi}{2}\right) = -\pi \cdot 0 \cdot 1 = 0 \]
- The denominator becomes:
\[ e - \sqrt{e} = 0 \]
Again, we have a \( \frac{0}{0} \) form. We apply L'Hôpital's Rule again.

### Step 8: Differentiate again
Differentiate the numerator and denominator again:
- The new numerator derivative will involve the product rule and chain rule, and the denominator will be differentiated again.

### Step 9: Final limit evaluation
After applying L'Hôpital's Rule again and simplifying, we will eventually arrive at a non-indeterminate form.

### Step 10: Conclusion
After evaluating the limit, we find:
\[ f\left(\frac{1}{2}\right) = \frac{\pi^2}{2e} \]

Thus, the value of \( f\left(\frac{1}{2}\right) \) so that \( f(x) \) is continuous at \( x = \frac{1}{2} \) is:
\[ \boxed{\frac{\pi^2}{2e}} \]