JEE Main 2025 — 28 January, Evening Shift — Question 20

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

$\mathrm{f}(\mathrm{x})=1-\mathrm{x}^{2}$

now $f(k)=-2 k$

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

let roots of this equation be $\alpha \& \beta$

then $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$

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