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

Question

Let $f: \mathbf{R}-\{0\} \rightarrow(-\infty, 1)$ be a polynomial of degree 2 , satisfying $f(x) f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$. If $f(\mathrm{~K})=-2 \mathrm{~K}$, 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

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

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 $\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta$

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

Solutions · 1