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

Given that \( f(x) \) is a polynomial of degree 2, let \( f(x) = ax^2 + bx + c \) where \( a \neq 0 \). 
Step 1: Apply the given condition to find \( f(x) \). From the given condition, we have: \[ f(x)f\left( \frac{1}{x} \right) = f(x) + f\left( \frac{1}{x} \right) \] Substitute the expression for \( f(x) \) into this: \[ (ax^2 + bx + c) \left( a\frac{1}{x^2} + b\frac{1}{x} + c \right) = (ax^2 + bx + c) + \left( a\frac{1}{x^2} + b\frac{1}{x} + c \right) \] 
Step 2: Simplify the equation. From this, we simplify and solve to find the constant values. Based on the given condition \( f(K) = -2K \), the equation becomes: \[ 1 - K^2 = -2K \quad \Rightarrow \quad 1 - K^2 + 2K = 0 \] 
Step 3: Solve for \( K \). This is a quadratic equation in \( K \): \[ K^2 - 2K - 1 = 0 \] The roots of this equation are: \[ K = \alpha \quad {and} \quad K = \beta \] 
Step 4: Find the sum of squares of the roots. We use the formula for the sum of squares of the roots of a quadratic equation: \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \] From Vieta’s formulas, we know: \[ \alpha + \beta = 2 \quad {and} \quad \alpha\beta = -1 \] Thus: \[ \alpha^2 + \beta^2 = 2^2 - 2(-1) = 4 + 2 = 6 \]