Question

Let \( f : \mathbb{R} \setminus \{0\} \to (-\infty, 1) \) be a polynomial of degree 2, satisfying \( f(x)f\left( \frac{1}{x} \right) = f(x) + f\left( \frac{1}{x} \right) \). If \( f(K) = -2K \), then the sum of squares of all possible values of \( K \) is:

• A. 1
• B. 7
• C. 9
• D. 6

Hint:

For quadratic equations, use the sum and product of roots to calculate expressions involving the roots.

Answer 1

The Correct Option is D

Step 1: Apply the given condition to find \( f(x) \)

Given that \( f(x) \) is a polynomial of degree 2, let \( f(x) = ax^2 + bx + c \) where \( a \neq 0 \).

Using the given condition:
\( f(x)f\left( \frac{1}{x} \right) = f(x) + f\left( \frac{1}{x} \right) \)

Substituting the polynomial expression:
\( (ax^2 + bx + c)\left(a\frac{1}{x^2} + b\frac{1}{x} + c \right) = (ax^2 + bx + c) + \left( a\frac{1}{x^2} + b\frac{1}{x} + c \right) \)

Step 2: Simplify the equation

After simplifying, the resulting equation becomes:
\( 1 - K^2 = -2K \)

Step 3: Solve for \( K \)

The equation simplifies to:
\( K^2 - 2K - 1 = 0 \)

Solving this quadratic equation:
\( K = \frac{2 \pm \sqrt{(2)^2 - 4(1)(-1)}}{2(1)} \)
\( K = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \)

Step 4: Find the Sum of Squares of the Roots

Using the identity:
\( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \)

From Vieta’s formulas:
\( \alpha + \beta = 2 \quad \text{and} \quad \alpha\beta = -1 \)

Thus,
\( \alpha^2 + \beta^2 = 2^2 - 2(-1) = 4 + 2 = 6 \)