If x is real and k=x2-x+1x2+x+1, then

From k=x2-x+1x2+x+1We have x2(k-1)+x(k+1)+k-1=0As given, x is real ⇒(k+1)2-4(k-1)2≥0⇒3k2-10k+3≤0Which is possible only when the value of k lies between the roots of the equation 3k2-10k+3=0That is, when 13≤k≤3 { Since roots are 13 and 3}