Key Concepts and Formulas

• Polynomial Representation: A general quadratic polynomial is represented as $f(x) = ax^2 + bx + c$, where $a \neq 0$.
• Functional Equations: Techniques for solving equations where the unknown is a function. Rearranging and substituting are common strategies.
• Vieta's Formulas: For a quadratic equation $Ax^2 + Bx + C = 0$ with roots $\alpha$ and $\beta$, $\alpha + \beta = -\frac{B}{A}$ and $\alpha \beta = \frac{C}{A}$.
• Range of a Function: The set of all possible output values of a function.
• Algebraic Identity: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$

Step-by-Step Solution

Step 1: Define the general form of the quadratic polynomial.

We are given that $f(x)$ is a polynomial of degree 2. Therefore, we can express it as: $f(x) = ax^2 + bx + c$ where $a$, $b$, and $c$ are constants, and $a \neq 0$. Why? This establishes the foundation for working with the given polynomial and allows us to determine the coefficients.

Step 2: Apply the given functional equation and simplify.

The given functional equation is $f(x) f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right)$. Adding 1 to both sides and factoring, we get: $f(x) f\left(\frac{1}{x}\right) - f(x) - f\left(\frac{1}{x}\right) + 1 = 1$ $(f(x) - 1)\left(f\left(\frac{1}{x}\right) - 1\right) = 1$ Why? This rearrangement simplifies the functional equation and makes it easier to work with. This form allows us to analyze the relationship between $f(x)$ and $f(1/x)$.

Step 3: Substitute the polynomial form into the simplified functional equation.

Substituting $f(x) = ax^2 + bx + c$, we have: $f\left(\frac{1}{x}\right) = a\left(\frac{1}{x}\right)^2 + b\left(\frac{1}{x}\right) + c = \frac{a}{x^2} + \frac{b}{x} + c$ Substituting these into the factored functional equation, we get: $(ax^2 + bx + c - 1)\left(\frac{a}{x^2} + \frac{b}{x} + c - 1\right) = 1$ Multiplying by $x^2$: $(ax^2 + bx + c - 1)(a + bx + (c-1)x^2) = x^2$ Why? This substitution allows us to work with the coefficients of the polynomial and relate them through the functional equation. Multiplying by $x^2$ eliminates the fraction.

Step 4: Expand and compare coefficients.

Expanding the left side, we get: $a(c-1)x^4 + (ab + b(c-1))x^3 + (a^2 + b^2 + (c-1)^2 + a + a(c-1))x^2 + (bc - b)x + a(c-1) = x^2$ Comparing coefficients of the powers of $x$ on both sides:

• $x^4$: $a(c-1) = 0$. Since $a \neq 0$, $c-1 = 0 \implies c = 1$.
• $x^3$: $ab + b(c-1) = 0 \implies ab = 0$. Since $a \neq 0$, $b = 0$.
• $x$: $b(c-1) = 0$. This is satisfied since $b=0$ and $c=1$.
• Constant term: $(c-1)a = 0$. This is satisfied since $c=1$.
• $x^2$: $a^2 + b^2 + a(c-1) + (c-1)^2 = 1 \implies a^2 = 1 \implies a = \pm 1$. Why? Comparing coefficients allows us to create a system of equations that the coefficients $a$, $b$, and $c$ must satisfy. This leads to possible values for these coefficients.

Step 5: Determine the polynomial $f(x)$ using the range restriction.

We have $f(x) = ax^2 + 1$, where $a = \pm 1$. So, $f(x) = x^2 + 1$ or $f(x) = -x^2 + 1$.

• If $f(x) = x^2 + 1$, then the range of $f(x)$ for $x \in \mathbb{R} - \{0\}$ is $(1, \infty)$, which does not satisfy the given range $(-\infty, 1)$.
• If $f(x) = -x^2 + 1$, then the range of $f(x)$ for $x \in \mathbb{R} - \{0\}$ is $(-\infty, 1)$, which satisfies the given range.

Therefore, $f(x) = -x^2 + 1$. Why? The range restriction is crucial in narrowing down the possible polynomials. Only $f(x) = -x^2 + 1$ satisfies the given range.

Step 6: Solve for K using the condition $f(K) = -2K$.

Substituting $f(x) = -x^2 + 1$ into $f(K) = -2K$, we get: $-K^2 + 1 = -2K$ $K^2 - 2K - 1 = 0$ Why? This equation provides a relationship for determining the values of $K$.

Step 7: Find the sum of squares of the possible values of K.

Let the roots of the quadratic equation $K^2 - 2K - 1 = 0$ be $\alpha$ and $\beta$. We want to find $\alpha^2 + \beta^2$. Using Vieta's formulas:

• $\alpha + \beta = -\frac{-2}{1} = 2$
• $\alpha \beta = \frac{-1}{1} = -1$ Then, $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = (2)^2 - 2(-1) = 4 + 2 = 6$ Why? Vieta's formulas provide a simple way to find the sum and product of the roots without explicitly solving for them. The algebraic identity allows us to directly calculate the sum of squares of the roots using the sum and product.

Common Mistakes & Tips

• Forgetting the condition $a \neq 0$ when defining the quadratic polynomial.
• Not considering the range restriction when determining the correct form of $f(x)$.
• Incorrectly applying or misremembering Vieta's formulas.
• Arithmetic Errors in expanding the functional equation, comparing coefficients, or calculating $\alpha^2 + \beta^2$.

Summary

We determined the polynomial $f(x)$ by using the given functional equation, the range restriction, and comparing coefficients. We found that $f(x) = -x^2 + 1$. Using the condition $f(K) = -2K$, we formed a quadratic equation for $K$ and used Vieta's formulas to find the sum of squares of its roots, which is 6.

Final Answer

The final answer is \boxed{6}, which corresponds to option (C).