Consider `f(x) = {{:((8^(x) - 4^(x) - 2^(x) + 1)/(x^(2))",",x gt 0),(e^(x)sin x + pi x + k log 4",",x lt 0):}` Then, f(0) so that f(x) is continuous at x = 0, is

Options

A. log 4
B. log 2
C. (log 4) (log 2)
D. None of these

Answer

To find the value of \( f(0) \) such that the function \( f(x) \) is continuous at \( x = 0 \), we need to ensure that the left-hand limit and right-hand limit at \( x = 0 \) are equal to \( f(0) \).

### Step 1: Define the function
The function is given as:
\[ f(x) = \begin{cases} \frac{8^x - 4^x - 2^x + 1}{x^2} & \text{if } x > 0 \\ e^x \sin x + \pi x + k \log 4 & \text{if } x < 0 \end{cases} \]

### Step 2: Find the right-hand limit as \( x \to 0^+ \)
We calculate the right-hand limit:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{8^x - 4^x - 2^x + 1}{x^2} \]
Substituting \( x = 0 \) directly gives us \( \frac{0}{0} \), which is an indeterminate form. We will apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule
Differentiate the numerator and the denominator:
- The derivative of \( 8^x \) is \( 8^x \log 8 \)
- The derivative of \( 4^x \) is \( 4^x \log 4 \)
- The derivative of \( 2^x \) is \( 2^x \log 2 \)
- The derivative of \( 1 \) is \( 0 \)
- The derivative of \( x^2 \) is \( 2x \)

Thus, we have:
\[ \lim_{x \to 0^+} \frac{8^x \log 8 - 4^x \log 4 - 2^x \log 2}{2x} \]

### Step 4: Apply L'Hôpital's Rule again
Substituting \( x = 0 \) again gives us \( \frac{0}{0} \). We apply L'Hôpital's Rule once more:
- The derivative of the numerator is:
\[ 8^x (\log 8)^2 - 4^x (\log 4)^2 - 2^x (\log 2)^2 \]
- The derivative of the denominator is \( 2 \).

Now we have:
\[ \lim_{x \to 0^+} \frac{8^x (\log 8)^2 - 4^x (\log 4)^2 - 2^x (\log 2)^2}{2} \]

### Step 5: Evaluate the limit as \( x \to 0 \)
Substituting \( x = 0 \):
\[ = \frac{1 \cdot (\log 8)^2 - 1 \cdot (\log 4)^2 - 1 \cdot (\log 2)^2}{2} \]
This simplifies to:
\[ = \frac{(\log 8)^2 - (\log 4)^2 - (\log 2)^2}{2} \]

### Step 6: Simplify the expression
Using the properties of logarithms:
\[ \log 8 = 3 \log 2, \quad \log 4 = 2 \log 2 \]
Thus:
\[ (\log 8)^2 = (3 \log 2)^2 = 9 (\log 2)^2 \]
\[ (\log 4)^2 = (2 \log 2)^2 = 4 (\log 2)^2 \]
So we have:
\[ = \frac{9 (\log 2)^2 - 4 (\log 2)^2 - (\log 2)^2}{2} = \frac{(9 - 4 - 1)(\log 2)^2}{2} = \frac{4 (\log 2)^2}{2} = 2 (\log 2)^2 \]

### Step 7: Find \( f(0) \)
For continuity at \( x = 0 \):
\[ f(0) = \lim_{x \to 0^+} f(x) = 2 (\log 2)^2 \]

### Conclusion
Thus, the value of \( f(0) \) so that \( f(x) \) is continuous at \( x = 0 \) is:
\[ \boxed{2 (\log 2)^2} \]