• # question_answer If $x$is real and $k=\frac{{{x}^{2}}-x+1}{{{x}^{2}}+x+1},$ then [MNR 1992; RPET 1997] A) $\frac{1}{3}\le k\le 3$ B) $k\ge 5$ C) $k\le 0$ D) None of these

From $k=\frac{{{x}^{2}}-x+1}{{{x}^{2}}+x+1}$ We have ${{x}^{2}}(k-1)+x(k+1)+k-1=0$ As given, x is real Þ ${{(k+1)}^{2}}-4{{(k-1)}^{2}}\ge 0$ Þ $3{{k}^{2}}-10k+3\ge 0$ Which is possible only when the value of k lies between the roots of the equation $3{{k}^{2}}-10k+3=0$ That is, when $\frac{1}{3}\le k\le 3$ {Since roots are $\frac{1}{3}$ and 3}