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

Step-by-step solution

as f(x) is a polynomial of degree two let it be f(x)=a x^2+b x+c (a 0) on satisfying given conditions we get C=1 & a= 1 hence f(x)=1 x^2 also range (- , 1] hence f(x)=1-x^2 now f(k)=-2 k1- k ^2=-2 k k ^2-2 k -1=0 let roots of this equation be & then ^2+ ^2=( + )^2-2 =4-2(-1)=6