The value of f(0) so that ` f(x) = ((4^(x)-1)^(3))/(sin(x/4)log(1+(x^(2))/3)) , x != 0 , ` is continuous everywhere in R, is

Options

A. `3(log4)^(3)`
B. `4(log4)^(3)`
C. `12(log 4)^(3)`
D. `15(log 4)^(3)`

Answer

To find the value of \( f(0) \) such that the function

\[ f(x) = \frac{(4^x - 1)^3}{\sin\left(\frac{x}{4}\right) \log\left(1 + \frac{x^2}{3}\right)} \]

is continuous everywhere in \( \mathbb{R} \), we need to ensure that

\[ \lim_{x \to 0} f(x) = f(0). \]

### Step 1: Evaluate the limit as \( x \to 0 \)

First, we need to find \( \lim_{x \to 0} f(x) \). Substituting \( x = 0 \) directly into the function gives us an indeterminate form \( \frac{0}{0} \). Therefore, we need to apply L'Hôpital's Rule or simplify the expression.

### Step 2: Simplify the numerator

Using the fact that \( 4^x - 1 \) can be approximated as \( \log(4) \cdot x \) when \( x \) is near 0, we have:

\[ 4^x - 1 \approx \log(4) \cdot x \quad \text{as } x \to 0. \]

Thus,

\[ (4^x - 1)^3 \approx (\log(4) \cdot x)^3 = \log(4)^3 \cdot x^3. \]

### Step 3: Simplify the denominator

Next, we simplify the denominator. For small \( x \):

\[ \sin\left(\frac{x}{4}\right) \approx \frac{x}{4} \quad \text{and} \quad \log\left(1 + \frac{x^2}{3}\right) \approx \frac{x^2}{3}. \]

Thus, the denominator becomes:

\[ \sin\left(\frac{x}{4}\right) \log\left(1 + \frac{x^2}{3}\right) \approx \left(\frac{x}{4}\right) \left(\frac{x^2}{3}\right) = \frac{x^3}{12}. \]

### Step 4: Substitute back into the limit

Now substituting back into the limit, we have:

\[ f(x) = \frac{\log(4)^3 \cdot x^3}{\frac{x^3}{12}} = 12 \log(4)^3. \]

### Step 5: Evaluate the limit

Taking the limit as \( x \to 0 \):

\[ \lim_{x \to 0} f(x) = 12 \log(4)^3. \]

### Step 6: Set \( f(0) \)

To ensure continuity, we set:

\[ f(0) = 12 \log(4)^3. \]

Thus, the value of \( f(0) \) that makes the function continuous everywhere in \( \mathbb{R} \) is:

\[ \boxed{12 \log(4)^3}. \]

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