If x is real, then the value of $\frac{{\mathrm{x}}^2+34\mathrm{x}-71}{{\mathrm{x}}^2+2\mathrm{x}-7}$ does not lie between

Let $\mathrm{y}=\frac{{\mathrm{x}}^2+34\mathrm{x}-71}{{\mathrm{x}}^2+2\mathrm{x}-7}$

$\Rightarrow {\mathrm{x}}^2(\mathrm{y}-1)+2(\mathrm{y}-17)\mathrm{x}+(71-7\mathrm{y})=0$

For real values of x, its discriminant $\mathrm{D}⩾0$

$\Rightarrow 4{(\mathrm{y}-17)}^2-4(\mathrm{y}-1)(71-7\mathrm{y})⩾0$

$\Rightarrow ({\mathrm{y}}^2-3+\mathrm{y}+289)-(71\mathrm{y}-7{\mathrm{y}}^2-71+7\mathrm{y})⩾0$

$\Rightarrow {\mathrm{y}}^2-14\mathrm{y}+45⩾0\Rightarrow (\mathrm{y}-5)(\mathrm{y}-9)⩾0$

It is possible when both $\mathrm{y}-5$ and $\mathrm{y}-9$ are negative or both positive. Let $\mathrm{y}-5\le 0\Rightarrow \mathrm{y}\le 5$ and $\mathrm{y}-9⩽0\Rightarrow \mathrm{y}\le 9.$

Hence $\mathrm{y}\le 5 ...\left(\mathrm{i}\right)$

If $\mathrm{y}-5\ge 0\Rightarrow \mathrm{y}⩾5$ and $\mathrm{y}-9\ge 0\Rightarrow \mathrm{y}\ge 9.$

Hence $\mathrm{y}\ge 9.$

Therefore y does not lie between 5 and 9.