Question:

Step 1: Find the value of $f(1)$ using the functional equation. We are given that $f(x)f(y) - f(xy) = x + y$ for all $x, y \in \mathbb{R}$. Substituting $x = y = 1$: $$f(1) \cdot f(1) - f(1 \cdot 1) = 1 + 1$$ $$f(1)^2 - f(1) = 2$$ $$f(1)^2 - f(1) - 2 = 0$$ Factoring the quadratic: $$(f(1) - 2)(f(1) + 1) = 0$$ This gives $f(1) = 2$ or $f(1) = -1$. Since we are given that $f(1) > 0$, we have: $$f(1) = 2$$ Step 2: Determine the explicit form of $f(x)$. Substituting $y = 1$ into the functional equation: $$f(x) \cdot f(1) - f(x \cdot 1) = x + 1$$ $$f(x) \cdot 2 - f(x) = x + 1$$ $$f(x) = x + 1$$ Step 3: Verify that $f(x) = x + 1$ satisfies the original functional equation. We check: $$f(x)f(y) - f(xy) = (x+1)(y+1) - (xy+1)$$ $$= xy + x + y + 1 - xy - 1$$ $$= x + y \quad \checkmark$$ The function satisfies the given equation. Step 4: Find the inverse function $f^{-1}(x)$. Since $f(x) = x + 1$, we solve for the inverse by setting $y = x + 1$ and solving for $x$: $$x = y - 1$$ Therefore: $$f^{-1}(x) = x - 1$$ Step 5: Calculate $f(x) \cdot f^{-1}(x)$. $$f(x) \cdot f^{-1}(x) = (x+1)(x-1)$$ Using the difference of squares formula: $$f(x) \cdot f^{-1}(x) = x^2 - 1$$ **Final Answer:** $f(x)f^{-1}(x) = x^2 - 1$ The correct option is **Option 3**.