Consider two differentiable functions `f(x),g(x)` satisfying
`6intf(x)g(x)dx=x^(6)+3x^(4)+3x^(2)+c and 2 int(g(x)dx)/(f(x))=x^(2)+c, " where " f(x) gt 0 AA x in R.`
`lim_(x to 0) (log(f(x)))/(g(x))=`

Options

A. e
B. 2
C. 1
D. 0

Explanation

To solve the problem, we will follow the steps outlined in the video transcript. Let's break down the solution step by step.

### Step 1: Differentiate the first equation
We start with the equation given in the problem:

\[ 6 \int f(x) g(x) \, dx = x^6 + 3x^4 + 3x^2 + c \]

Differentiating both sides with respect to \(x\):

\[ \frac{d}{dx} \left( 6 \int f(x) g(x) \, dx \right) = \frac{d}{dx} \left( x^6 + 3x^4 + 3x^2 + c \right) \]

Using the Fundamental Theorem of Calculus, we have:

\[ 6 f(x) g(x) = 6x^5 + 12x^3 + 6x \]

Dividing both sides by 6 gives us:

\[ f(x) g(x) = x^5 + 2x^3 + x \]

This is our **Equation (1)**.

### Step 2: Differentiate the second equation
Next, we consider the second equation given in the problem:

\[ 2 \int \frac{g(x)}{f(x)} \, dx = x^2 + c \]

Differentiating both sides with respect to \(x\):

\[ \frac{d}{dx} \left( 2 \int \frac{g(x)}{f(x)} \, dx \right) = \frac{d}{dx} \left( x^2 + c \right) \]

Again, using the Fundamental Theorem of Calculus, we have:

\[ 2 \cdot \frac{g(x)}{f(x)} = 2x \]

Dividing both sides by 2 gives us:

\[ \frac{g(x)}{f(x)} = x \]

This is our **Equation (2)**.

### Step 3: Express \(g(x)\) in terms of \(f(x)\)
From **Equation (2)**, we can express \(g(x)\) as:

\[ g(x) = x f(x) \]

### Step 4: Substitute \(g(x)\) into Equation (1)
Now we substitute \(g(x) = x f(x)\) into **Equation (1)**:

\[ f(x) (x f(x)) = x^5 + 2x^3 + x \]

This simplifies to:

\[ x f(x)^2 = x^5 + 2x^3 + x \]

Dividing both sides by \(x\) (for \(x \neq 0\)) gives us:

\[ f(x)^2 = x^4 + 2x^2 + 1 \]

### Step 5: Simplify \(f(x)\)
We can rewrite the right-hand side:

\[ f(x)^2 = (x^2 + 1)^2 \]

Taking the square root (and since \(f(x) > 0\)) gives us:

\[ f(x) = x^2 + 1 \]

### Step 6: Substitute back to find \(g(x)\)
Now substituting \(f(x)\) back into the expression for \(g(x)\):

\[ g(x) = x f(x) = x (x^2 + 1) = x^3 + x \]

### Step 7: Find the limit
We need to find:

\[ \lim_{x \to 0} \frac{\log(f(x))}{g(x)} \]

Substituting \(f(x)\) and \(g(x)\):

\[ \lim_{x \to 0} \frac{\log(x^2 + 1)}{x^3 + x} \]

### Step 8: Evaluate the limit
As \(x \to 0\):

- \(\log(x^2 + 1) \to \log(1) = 0\)
- \(x^3 + x \to 0\)

This is an indeterminate form of type \(0/0\). We can apply L'Hôpital's Rule:

Differentiating the numerator and denominator:

\[ \text{Numerator: } \frac{d}{dx}[\log(x^2 + 1)] = \frac{2x}{x^2 + 1} \]
\[ \text{Denominator: } \frac{d}{dx}[x^3 + x] = 3x^2 + 1 \]

Now we evaluate the limit:

\[ \lim_{x \to 0} \frac{\frac{2x}{x^2 + 1}}{3x^2 + 1} = \frac{0}{1} = 0 \]

### Final Answer
Thus, the final answer is:

\[ \lim_{x \to 0} \frac{\log(f(x))}{g(x)} = 0 \]