Question:

Step 1: Rewrite the functional equation in a more useful form. We start with the given functional equation: $$f(x)f\left(\frac{1}{x}\right) - 3f(x) - 3f\left(\frac{1}{x}\right) + 5 = 0$$ Rearranging by adding and subtracting 9: $$f(x)f\left(\frac{1}{x}\right) - 3f(x) - 3f\left(\frac{1}{x}\right) + 9 = 4$$ Factoring the left side: $$\left(f(x) - 3\right)\left(f\left(\frac{1}{x}\right) - 3\right) = 4$$ Step 2: Determine the form of the polynomial. Let $g(x) = f(x) - 3$. Then we have: $$g(x) \cdot g\left(\frac{1}{x}\right) = 4$$ Since $f(x)$ is a polynomial, let us assume $f(x) = ax^n + 3$ for some constants $a$ and $n$. Then: $$g(x) = ax^n \quad \text{and} \quad g\left(\frac{1}{x}\right) = \frac{a}{x^n}$$ Step 3: Find the constants $a$ and $n$. From the condition $g(x) \cdot g\left(\frac{1}{x}\right) = 4$: $$ax^n \cdot \frac{a}{x^n} = 4$$ $$a^2 = 4$$ $$a = 2 \quad \text{(taking the positive value)}$$ So $f(x) = 2x^n + 3$. Using the condition $f(2) = 11$: $$2 \cdot 2^n + 3 = 11$$ $$2^{n+1} = 8$$ $$2^{n+1} = 2^3$$ $$n + 1 = 3$$ $$n = 2$$ Therefore, $f(x) = 2x^2 + 3$. Step 4: Verify the solution and find $f(3)$. Let us verify that $f(x) = 2x^2 + 3$ satisfies the original functional equation: $$f(x)f\left(\frac{1}{x}\right) = (2x^2 + 3)\left(\frac{2}{x^2} + 3\right) = 4 + 6x^2 + \frac{6}{x^2} + 9 = 13 + 6x^2 + \frac{6}{x^2}$$ $$3f(x) + 3f\left(\frac{1}{x}\right) = 3(2x^2 + 3) + 3\left(\frac{2}{x^2} + 3\right) = 6x^2 + 9 + \frac{6}{x^2} + 9 = 6x^2 + \frac{6}{x^2} + 18$$ $$f(x)f\left(\frac{1}{x}\right) - 3f(x) - 3f\left(\frac{1}{x}\right) + 5 = 13 + 6x^2 + \frac{6}{x^2} - 6x^2 - \frac{6}{x^2} - 18 + 5 = 0$$ ✓ Now we calculate $f(3)$: $$f(3) = 2(3)^2 + 3 = 2 \cdot 9 + 3 = 18 + 3 = 21$$ **Final Answer:** $f(3) = 21$, which corresponds to **Option 1**.