Let f:R-0 (-,1) be a polynomial of degree 2 , satisfying f(x)f((1)/(x))=f(x)+f((1)/(x)).

Question

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

Accepted Answer

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

$f(x)=a{x}^{2}+bx+c\, (a
e 0)$

on satisfying given conditions we get

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

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

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

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

$1-{k}^{2}=-2k	o {k}^{2}-2k-1=0$

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

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

Solutions · 1