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 \(f(\mathrm{x}) f\left(\frac{1}{\mathrm{x}}\right)=f(\mathrm{x})+f\left(\frac{1}{\mathrm{x}}\right).\) If \(f(K)=-2 K,\) then the sum of squares of all possible values of K is :

(1) 1

(2) 6

(3) 7

(4) 9

Answer 1

Correct option is (2) 6 

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

\(\mathrm{C}=1\ \&\ \mathrm{a}= \pm 1\)

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

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

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

now f(k) = -2k

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

let roots of this equation be \(\alpha\ \&\ \beta\)

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

= 4 - 2(-1) = 6