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4-2.Quadratic Equations and Inequations
hard
Let $f: R -\{0\} \rightarrow(-\infty, 1)$ be a polynomial of degree $2$ , satisfying $f( x ) f\left(\frac{1}{ x }\right)=f( x )+f\left(\frac{1}{ x }\right)$. If $f(K)=-2 K$, then the sum of squares of all possible values of $K$ is :
A$1$
B$6$
C$7$
D$9$
(JEE MAIN-2025)
Solution
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
$f(x)=1-x^2$
now $f(k)=-2 k$
$1-k^2=-2 k \rightarrow k^2-2 k-1=0$
let roots of this equation be $\alpha \ \beta$ then $\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta$
$=4-2(-1)=6$
$f(x)=a x^2+b x+c \quad(a \neq 0)$
on satisfying given conditions we get
$C=1 \ a= \pm 1$
hence $f(x)=1 \pm x^2$
also range $\in(-\infty, 1]$ hence
$f(x)=1-x^2$
now $f(k)=-2 k$
$1-k^2=-2 k \rightarrow k^2-2 k-1=0$
let roots of this equation be $\alpha \ \beta$ then $\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta$
$=4-2(-1)=6$
Standard 11
Mathematics
