Question

If x is real and $k=\frac{x^2-x+1}{x^2+x+1}$, then
(a) k ∈ [1/3,3]
(b) k ≥ 3
(c) k ≤ 1/3
(d) none of these

Answer 1

(a) k ∈ [1/3,3]

$$\begin{gathered} k=\frac{x^2-x+1}{x^2+x+1} \\ \Rightarrow k{x}^2+kx+k=x^2-x+1 \\ \Rightarrow \left(k-1\right)x^2+\left(k+1\right)x+k-1=0 \end{gathered}$$

For real values of x, the discriminant of $\left(k-1\right)x^2+\left(k+1\right)x+k-1=0$ should be greater than or equal to zero.

$$\begin{gathered} \therefore \mathrm{if}k\ne 1 \\ {\left(k+1\right)}^2-4\left(k-1\right)\left(k-1\right)\ge 0 \\ \Rightarrow {\left(k+1\right)}^2-{\left\{2\left(k-1\right)\right\}}^2\ge 0 \\ \Rightarrow \left(k+1+2k-2\right)\left(k+1-2k+2\right)\ge 0 \\ \Rightarrow \left(3k-1\right)\left(-k+3\right)\ge 0 \\ \Rightarrow \left(3k-1\right)\left(k-3\right)\le 0 \\ \Rightarrow \frac{1}{3}\le k\le 3\mathrm{i}.\mathrm{e}.k\in \left[\frac{1}{3},3\right]-\left\{1\right\}...\left(\mathrm{i}\right) \end{gathered}$$

And if k=1, then,
x=0, which is real ...(ii)
So, from (i) and (ii), we get,
$k\in \left[\frac{1}{3},3\right]$