If x is real and k

Solution:

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

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

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

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

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,

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