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

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Aicharya · Accepted Answer

as $f(x)$ is a polynomial of degree two let it be $f(x)=a x^2+b x+c \quad(a 
eq 0)$ on satisfying given conditions we get $C=1 \& a= \pm 1$ hence $f(x)=1 \pm x^2$ also range $\in(-\infty, 1]$ hence $f(x)=1-x^2$ now $f(k)=-2 k$ $1-\mathrm{k}^2=-2 \mathrm{k} ightarrow \mathrm{k}^2-2 \mathrm{k}-1=0$ let roots of this equation be $\alpha \& \beta$ then $\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta$ $=4-2(-1)=6$

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

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