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 :
Solution
<p>as $f(x)$ is a polynomial of degree two let it be</p>
<p>$f(x)=a x^2+b x+c \quad(a \neq 0)$</p>
<p>on satisfying given conditions we get</p>
<p>$C=1 \& a= \pm 1$</p>
<p>hence $f(x)=1 \pm x^2$</p>
<p>also range $\in(-\infty, 1]$ hence</p>
<p>$$\begin{gathered}
f(x)=1-x^2 \\
\text { now } f(k)=-2 k
\end{gathered}$$</p>
<p>$1-\mathrm{k}^2=-2 \mathrm{k} \rightarrow \mathrm{k}^2-2 \mathrm{k}-1=0$</p>
<p>let roots of this equation be $\alpha \& \beta$ then $\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta$</p>
<p>$=4-2(-1)=6$</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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