Let \(f: \mathbf{R}-\{0\} \rightarrow(-\infty, 1)\) be a polynomial of degree 2, satisfying \(f(x) f\left(\frac{1}{x}\ri…

\(f(x) f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)\)

\(\Rightarrow(f(x)-1)\left(f\left(\frac{1}{x}\right)-1\right)=1\)

Let \(g(x)=1-f(x)\)

Then \(g(x)\) is a quadratic polynomial and

\(g(x) g\left(\frac{1}{x}\right)=1 (x \neq 0)\)

Let \(g(x)=a x^2+b x+c\)

Then, \(\left(a x^2+b x+c\right)\left(\frac{a}{x^2}+\frac{b}{x}+c\right)=1\)

\(\left(a x^2+b x+c\right)\left(a+b x+c x^2\right)=x^2\)

\(a c x^4+b(a+c) x^3+\left(a^2+b^2+c^2\right) x^2+b(a+c) x+a c=x^2\)

Comparing coefficients:

\(a c=0, b(a+c)=0, a^2+b^2+c^2=1\)

Since \(g(x)\) is quadratic, \(a \neq 0\), so \(c=0\)

Then \(a b=0 \Rightarrow b=0\), and \(a^2=1 \Rightarrow a= \pm 1\)

Also \(f(x)<1 \Rightarrow g(x)=1-f(x)>0\) for all \(x \neq 0\), so \(a=1\)

Hence, \(g(x)=x^2\)

\(1-f(x)=x^2 \Rightarrow f(x)=1-x^2\)

Now, \(f(K)=-2 K \Rightarrow 1-K^2=-2 K\)

\(\Rightarrow K^2-2 K-1=0\)

\(K=1 \pm \sqrt{2}\)

Required sum of squares:

\((1+\sqrt{2})^2+(1-\sqrt{2})^2=6\)