Let f:R-\0\rightarrow(-infty,1]$f:R-\{0\}\rightarrow(-\infty,1]$ be a polynomial of degree 2, satisfying f(x)fleft(frac1xright)=f(x)+fleft(frac1xright).$f(x)f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right).$ If f(K)=-2K$f(K)=-2K$ then the sum of squares of all possible values of K is:

### Related Formula Standard result for functional equation of a polynomial satisfying f(x)f(1/x) = f(x) + f(1/x)$f(x)f(1/x) = f(x) + f(1/x)$: f(x) = 1 pm x^n$$f(x) = 1 \pm x^n$$ ### Core Logic Given that f(x)$f(x)$ is a polynomial of degree 2, the identity implies: f(x) = 1 + x^2 quad textor quad f(x) = 1 - x^2$$f(x) = 1 + x^2 \quad \text{or} \quad f(x) = 1 - x^2$$ We are given the range is bounded above: (-infty, 1]$(-\infty, 1]$. - For 1 + x^2$1 + x^2$, the range is [1, infty)$[1, \infty)$. - For 1 - x^2$1 - x^2$, the range is (-infty, 1]$(-\infty, 1]$. Therefore, the correct functional form is f(x) = 1 - x^2$f(x) = 1 - x^2$. ### Step 1: Solve for K Given condition: f(K) = -2K$f(K) = -2K$ 1 - K^2 = -2K implies K^2 - 2K - 1 = 0$$1 - K^2 = -2K \implies K^2 - 2K - 1 = 0$$ Let the roots of this equation be K_1$K_1$ and K_2$K_2$. From quadratic properties (Vieta's formulas): K_1 + K_2 = 2$K_1 + K_2 = 2$ K_1 cdot K_2 = -1$$K_1 \cdot K_2 = -1$$ ### Step 2: Calculate Sum of Squares We need the sum of squares of the values of K: K_1^2 + K_2^2 = (K_1 + K_2)^2 - 2K_1K_2$$K_1^2 + K_2^2 = (K_1 + K_2)^2 - 2K_1K_2$$ K_1^2 + K_2^2 = (2)^2 - 2(-1) = 4 + 2 = 6$$K_1^2 + K_2^2 = (2)^2 - 2(-1) = 4 + 2 = 6$$ ### Pattern Recognition The functional equation f(x)f(1/x)=f(x)+f(1/x)$f(x)f(1/x)=f(x)+f(1/x)$ uniquely forces polynomials to be 1 pm x^n$1 \pm x^n$. Remembering this shortcut saves valuable time required to derive the template from general coefficients. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers and Quadratic Equations Class 12 Mathematics: Functions