For what values of the parameter $k$ in the inequality $\left|\frac{x^2+kx+1}{x^2+x+1}\right|<3,$ satisfied for all real values of $x ?$

We have, $\left|\frac{x^2+kx+1}{x^2+x+1}\right|<3$

$$\begin{gathered} \Rightarrow -3<\frac{x^2+kx+1}{x^2+x+1}<3 \\ Since, x^2+x+1={\left(x+\frac{1}{2}\right)}^2+\frac{3}{4}>0 \end{gathered}$$

$$\begin{gathered} -3(x^2+x+1)<x^2+kx+1<3(x^2+x+1) \\ 4x^2+(k+3)x+4>0 ...\left(i\right) \\ and 2x^2-(k-3)x+2>0 .....\left(ii\right) \\ Since 4>0 and 2>0 \end{gathered}$$

The inequality (i) will be valid, if

           $$\begin{gathered} {(k+3)}^2-4·4·4<0\Rightarrow {(k+3)}^2<64 or -8<k+3<8 \\ or -11<k<5 ...\left(iii\right) \end{gathered}$$

and the inequality (ii) will be valid, if

${(k-3)}^2-4·2·2<0 or {(k-3)}^2<16$

$$\begin{gathered} or -4<k-3<4 \\ or -1<k<7 ......\left(iv\right) \end{gathered}$$

The condition (iii) and (iv) will hold simultaneously, if 

             $-1<k<5$