If x is real and k


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


(a) $k \in[1 / 3,3]$


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

$\Rightarrow(k-1) x^{2}+(k+1) x+k-1=0$

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

$\therefore$ if $k \neq 1$

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

$\Rightarrow(k+1)^{2}-\{2(k-1)\}^{2} \geq 0$

$\Rightarrow(k+1+2 k-2)(k+1-2 k+2) \geq 0$

$\Rightarrow(3 k-1)(-k+3) \geq 0$

$\Rightarrow(3 k-1)(k-3) \leq 0$

$\Rightarrow \frac{1}{3} \leq k \leq 3$ i. e. $k \in\left[\frac{1}{3}, 3\right]-\{1\} \quad \ldots($ i $)$

And if k=1, then,

x=0, which is real       ...(ii)

So, from (i) and (ii), we get,

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

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