If x is real, the expression $\frac{{\mathrm{x}}^2+2\mathrm{x}-11}{\mathrm{x}-3}$ takes all real values except those which lie between a and b, then a and b are

$\begin{array}{lc}\text{Let}& \mathrm{y}=\frac{{\mathrm{x}}^2+2\mathrm{x}-11}{\mathrm{x}-3}\\ \Rightarrow & \mathrm{xy}-3\mathrm{y}={\mathrm{x}}^2+2\mathrm{x}-11\\ \Rightarrow & {\mathrm{x}}^2+(2-\mathrm{y})\mathrm{x}+(3\mathrm{y}-11)=0\\ \Rightarrow & (2-\mathrm{y}{)}^2-4(3\mathrm{y}-11)\ge 0\\ \Rightarrow & 4+{\mathrm{y}}^2-4\mathrm{y}-12\mathrm{y}+44\ge 0\\ \Rightarrow & {\mathrm{y}}^2-16\mathrm{y}+48\ge 0\\ \Rightarrow & {\mathrm{y}}^2-12\mathrm{y}-4\mathrm{y}+48\ge 0\\ \Rightarrow & \text{(y}-4\left)\right(\mathrm{y}-12)\ge 0\\ \Rightarrow & \mathrm{y}\le 4\text{or}\mathrm{y}\ge 12\end{array}$