We have,
$$
\begin{aligned}
& y=\frac{11 x^2+12 x+6}{x^2+4 x+2} \\
& \Rightarrow \quad y x^2+4 x y+2 y=11 x^2+12 x+6 \\
& \Rightarrow \quad(y-11) x^2+4 x(y-3)+2(y-3)=0 \\
& \Rightarrow \quad x=\frac{-4(y-3) \pm \sqrt{16(y-3)^2-8}}{2(y-11)} \\
&
\end{aligned}
$$
Since, $x \in$ real number
$$
\begin{array}{lr}
\therefore & D \geq 0 \\
\Rightarrow & 16(y-3)^2-8(y-11)(y-3) \geq 0 \\
\Rightarrow & 8(y-3)[2(y-3)-(y-11)] \geq 0 \\
\Rightarrow & 8(y-3)(2 y-6-y+11) \geq 0 \\
\Rightarrow & 8(y-3)(y+5) \geq 0
\end{array}
$$


$$
\begin{aligned}
& \therefore y \leq-5 \text { or } y \geq 3 \\
& \therefore a=-5 \text { and } b=3 .
\end{aligned}
$$