It is given, for all \(x \in R\)
\(\begin{aligned}
&\left|\frac{x^2+k x+1}{x^2+x+1}\right| < 3 \\
& \Rightarrow \quad-3 < \frac{x^2+k x+1}{x^2+x+1} < 3 \\
& \Rightarrow \quad-3 x^2-3 x-3 < x^2+k x+1 < 3 x^2+3 x+3 \\
& {\left[\because x^2+x+1 > 0, \forall x \in R\right] }
\end{aligned}\)
So, \(4 x^2+(k+3) x+4 > 0, \forall x \in R\)
Then \(D < 0 \Rightarrow(k+3)^2-4(4)(4) < 0\)
\(\begin{array}{ll}
\Rightarrow & (k+3)^2-8^2 < 0 \\
\Rightarrow & (k+3-8)(k+3+8) < 0 \\
\Rightarrow & k \in(-11,5) \quad \ldots (i)
\end{array}\)
and \(2 x^2+(3-k) x+2 > 0, \forall x \in R\)
Then \(D < 0 \Rightarrow(3-k)^2-16 < 0\)
\(\Rightarrow \quad(k-3-4)(k-3+4) < 0\)
\(\Rightarrow \quad k \in(-1,7) \quad \ldots (ii)\)
From intervals (i) and (ii), we get
\(k \in(-1,5)\)
Hence, option (b) is correct.