Question

If $x$ is real and $k=\frac{x^2-x+1}{x^2+x+1}$, then

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

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 ⇒ $(k+1{)}^2-4(k-1{)}^2\ge 0$

⇒ $3k^2-10k+3\le 0$

Which is possible only when the value of k lies between the roots of the equation $3k^2-10k+3=0$

That is, when $\frac{1}{3}\le k\le 3$ {Since roots are $\frac{1}{3}$ and 3}