A function `f: R -> R` satisfy the equation `f (x)f(y) - f (xy)= x+y` for all `x, y in R` and `f(y) > 0`, then

Options

A. `f(x)f^(-1)(x)=x^(2)-4`
B. `f(x)f^(-1)(x)=x^(2)-6`
C. `f(x)f^(-1)(x)=x^(2)-1`
D. none of these

Explanation

To solve the problem, we need to find the function \( f: \mathbb{R} \to \mathbb{R} \) that satisfies the equation:

\[ f(x)f(y) - f(xy) = x + y \]

for all \( x, y \in \mathbb{R} \) with the condition that \( f(y) > 0 \).

### Step 1: Substitute \( x = 0 \) and \( y = 0 \)

Let's first substitute \( x = 0 \) and \( y = 0 \) into the given equation:

\[ f(0)f(0) - f(0) = 0 + 0 \]

This simplifies to:

\[ f(0)^2 - f(0) = 0 \]

### Step 2: Factor the equation

We can factor this equation:

\[ f(0)(f(0) - 1) = 0 \]

This gives us two possible solutions:

1. \( f(0) = 0 \)
2. \( f(0) = 1 \)

### Step 3: Use the condition \( f(y) > 0 \)

Given that \( f(y) > 0 \) for all \( y \in \mathbb{R} \), we can conclude that \( f(0) \) cannot be 0. Therefore, we have:

\[ f(0) = 1 \]

### Step 4: Substitute \( y = 0 \) in the original equation

Now, let's substitute \( y = 0 \) into the original equation:

\[ f(x)f(0) - f(0) = x + 0 \]

Substituting \( f(0) = 1 \):

\[ f(x) \cdot 1 - 1 = x \]

This simplifies to:

\[ f(x) - 1 = x \]

### Step 5: Solve for \( f(x) \)

From the equation above, we can solve for \( f(x) \):

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

### Step 6: Find the inverse function \( f^{-1}(x) \)

To find the inverse function, we set \( y = f(x) \):

\[ y = x + 1 \]

Now, solving for \( x \):

\[ x = y - 1 \]

Thus, the inverse function is:

\[ f^{-1}(y) = y - 1 \]

### Step 7: Calculate \( f(x) \cdot f^{-1}(x) \)

Now, we multiply \( f(x) \) and \( f^{-1}(x) \):

\[ f(x) \cdot f^{-1}(x) = (x + 1)(x - 1) \]

This simplifies to:

\[ f(x) \cdot f^{-1}(x) = x^2 - 1 \]

### Conclusion

Thus, the function \( f(x) \) that satisfies the given conditions is:

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

And the product \( f(x) \cdot f^{-1}(x) \) is:

\[ f(x) \cdot f^{-1}(x) = x^2 - 1 \]

### Final Answer

The correct option is:

**C. \( f(x) \cdot f^{-1}(x) = x^2 - 1 \)**

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