as $f(x)$ is a polynomial of degree two let it be

$$f(x)=a x^2+b x+c \quad(a \neq 0)$$

on satisfying given conditions we get

$$C=1 \& a= \pm 1$$

hence $f(x)=1 \pm x^2$

also range $\in(-\infty, 1]$ hence

$$\begin{gathered} f(x)=1-x^2 \\ \text { now } f(k)=-2 k \end{gathered}$$

$$1-\mathrm{k}^2=-2 \mathrm{k} \rightarrow \mathrm{k}^2-2 \mathrm{k}-1=0$$

let roots of this equation be $\alpha \& \beta$ then $\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta$

$$=4-2(-1)=6$$