Let f(x) be a polynominal one-one function such that
`f(x)f(y)+2=f(x)+f(y)+f(xy), forall x,y in R-{0}, f(1) ne 1, f'(1)=3.`
Let `g(x)=x/4(f(x)+3)-int_(0)^(x)f(x)dx,` then

Options

A. g(x)=0 has exactly one root for `x in (0,1)`
B. g(x)=0 has exactly two roots for `x in (0,1)`
C. `g(x) ne 0, x in R-{0}`
D. `g(x)=0, x in R-{0}`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given functional equation
We are given the equation:
\[ f(x)f(y) + 2 = f(x) + f(y) + f(xy) \]
for all \( x, y \in \mathbb{R} - \{0\} \).

### Step 2: Rearranging the equation
Rearranging the equation gives:
\[ f(x)f(y) - f(x) - f(y) = f(xy) - 2 \]
This can be factored as:
\[ (f(x) - 1)(f(y) - 1) = f(xy) - 1 \]

### Step 3: Define a new function
Let us define a new function:
\[ h(x) = f(x) - 1 \]
Then the equation becomes:
\[ h(x)h(y) = h(xy) \]
This form suggests that \( h(x) \) could be a power function.

### Step 4: Assume a form for \( h(x) \)
Assume:
\[ h(x) = x^m \]
Then:
\[ h(xy) = (xy)^m = x^m y^m \]
This is consistent with our earlier equation.

### Step 5: Relate back to \( f(x) \)
Since \( h(x) = f(x) - 1 \), we have:
\[ f(x) = x^m + 1 \]

### Step 6: Find \( m \) using the derivative condition
We know that \( f'(1) = 3 \). First, we compute \( f'(x) \):
\[ f'(x) = mx^{m-1} \]
Evaluating at \( x = 1 \):
\[ f'(1) = m \cdot 1^{m-1} = m \]
Setting this equal to 3 gives:
\[ m = 3 \]

### Step 7: Determine \( f(x) \)
Thus, we have:
\[ f(x) = x^3 + 1 \]

### Step 8: Define \( g(x) \)
Next, we define:
\[ g(x) = \frac{x}{4}(f(x) + 3) - \int_0^x f(t) dt \]

### Step 9: Substitute \( f(x) \)
Substituting \( f(x) \):
\[ g(x) = \frac{x}{4}((x^3 + 1) + 3) - \int_0^x (t^3 + 1) dt \]
\[ = \frac{x}{4}(x^3 + 4) - \int_0^x (t^3 + 1) dt \]

### Step 10: Evaluate the integral
Calculating the integral:
\[ \int_0^x (t^3 + 1) dt = \left[ \frac{t^4}{4} + t \right]_0^x = \frac{x^4}{4} + x \]

### Step 11: Substitute back into \( g(x) \)
Now substituting back:
\[ g(x) = \frac{x}{4}(x^3 + 4) - \left( \frac{x^4}{4} + x \right) \]
\[ = \frac{x^4 + 4x}{4} - \left( \frac{x^4}{4} + x \right) \]
\[ = \frac{x^4 + 4x - x^4 - 4x}{4} \]
\[ = 0 \]

### Conclusion
Thus, we find that:
\[ g(x) = 0 \]
for all \( x \in \mathbb{R} - \{0\} \). Therefore, \( g(x) \) is always equal to 0.

### Final Answer
The correct option is that \( g(x) \) is always equal to 0.

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