Let f: mathbb{R} - {0} rightarrow (-infty, 1) | vedclass

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(B) Given $f(x)f(\frac{1}{x}) = f(x) + f(\frac{1}{x})$.Let $f(x) = ax^2 + bx + c$. Since $f(x)$ is a polynomial of degree $2$,$a 
eq 0$.Substituting

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$6$

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(B) Given $f(x)f(\frac{1}{x}) = f(x) + f(\frac{1}{x})$.Let $f(x) = ax^2 + bx + c$. Since $f(x)$ is a polynomial of degree $2$,$a 
eq 0$.Substituting $f(x)$ into the equation: $(ax^2 + bx + c)(a(\frac{1}{x})^2 + b(\frac{1}{x}) + c) = ax^2 + bx + c + a(\frac{1}{x})^2 + b(\frac{1}{x}) + c$.Comparing coefficients,we find $c = 1$ and $a = -1$ (since the range is $(-\infty, 1)$).Thus,$f(x) = 1 - x^2$.Given $f(K) = -2K$,we have $1 - K^2 = -2K$,which simplifies to $K^2 - 2K - 1 = 0$.Let the roots be $K_1$ and $K_2$. The sum of squares is $K_1^2 + K_2^2 = (K_1 + K_2)^2 - 2K_1K_2$.Using Vieta's formulas,$K_1 + K_2 = 2$ and $K_1K_2 = -1$.Therefore,$K_1^2 + K_2^2 = (2)^2 - 2(-1) = 4 + 2 = 6$.

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

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