If x is rea, then the maximum and minimum values of expression $\frac{{\mathrm{x}}^2+14\mathrm{x}+9}{{\mathrm{x}}^2+2\mathrm{x}+3}$ will be

Let $\mathrm{y}=\frac{{\mathrm{x}}^2+14\mathrm{x}+9}{{\mathrm{x}}^2+2\mathrm{x}+3}$

$\Rightarrow \mathrm{y}({\mathrm{x}}^2+2\mathrm{x}+3)-{\mathrm{x}}^2-14\mathrm{x}-9=0$

$\Rightarrow (\mathrm{y}-1){\mathrm{x}}^2+(2\mathrm{y}-14)\mathrm{x}+3\mathrm{y}-9=0$

For real x, its discriminant $\ge 0$

i.e. $4{(\mathrm{y}-7)}^2-4(\mathrm{y}-1)3(\mathrm{y}-3)\ge 0$

$\Rightarrow {\mathrm{y}}^2+\mathrm{y}-20\le 0 \mathrm{or} (\mathrm{y}-4)(\mathrm{y}+5)\le 0$

Now, the product of two factors is negative if these are of opposite signs. So following two cases arise:

Case I: $\mathrm{y}-4\ge 0 \mathrm{or} \mathrm{y}\ge 4 \mathrm{and} \mathrm{y}+5\le 0 \mathrm{or} \mathrm{y}\le -5$

This is not possible.

Case II: $\mathrm{y}-4\le 0 \mathrm{or} \mathrm{y}\le 4 \mathrm{and} \mathrm{y}+5\ge 0 \mathrm{or} \mathrm{y}\ge -5$

Both of these are satisfied if $-5\le \mathrm{y}\le 4$

Hence maximum value of y is 4 and minimum value is -5.