Question

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

• A. $k\in [\frac{1}{3},3]$
• B. $k\ge 3$
• C. $k\le \frac{1}{3}$
• D. None of these

Answer 1

Answer: (A)

$k\in [\frac{1}{3},3]$

We have,

$k=\frac{x^2-x+1}{x^2+x+1}$

or, $k(x^2+x+1)=(x^2-x+1)$

or, $x^2(k-1)+x(k+1)+(k-1)=0$

For the above equation $x$ will be real if the discreminant $\ge 0$.

$(k+1{)}^2-4(k-1{)}^2\ge 0$

or, $10k-3k^2-3\ge 0$

or, $3k^2-10k+3\le 0$

or, $3k^2-9k-k+3\le 0$

or, $3k(k-3)-1(k-3)\le 0$

or, $(k-3)(3k-1)\le 0$

$\Rightarrow \frac{1}{3}\le x\le 3$

or, $x\in [\frac{1}{3},3]$.