If x is real, then the maximum and minimum values of the expression $\frac{{\mathrm{x}}^2-3\mathrm{x}+4}{{\mathrm{x}}^2+3\mathrm{x}+4}$ will be

Let $\mathrm{y}=\frac{{\mathrm{x}}^2-3\mathrm{x}+4}{{\mathrm{x}}^2+3\mathrm{x}+4}$

$\Rightarrow (\mathrm{y}-1){\mathrm{x}}^2+3(\mathrm{y}+1)\mathrm{x}+4(\mathrm{y}-1)=0$

For x is real $\mathrm{D}\ge 0$

$$\begin{gathered} \Rightarrow 9{(\mathrm{y}+1)}^2-16{(\mathrm{y}-1)}^2\ge 0 \Rightarrow -7{\mathrm{y}}^2+50\mathrm{y}-7⩾0\Rightarrow 7{\mathrm{y}}^2-50\mathrm{y}+7\le 0 \\ \Rightarrow (\mathrm{y}-7)(7\mathrm{y}-1)⩽0 \end{gathered}$$

Now, the product of two factors is negative if one in $-\mathrm{ve}$ and one in $+\mathrm{ve}$.

Case I: $$\begin{gathered} (\mathrm{y}-7)\ge 0 \mathrm{and} (7\mathrm{y}-1)\le 0 \\ \Rightarrow \mathrm{y}\ge 7 \mathrm{and} \mathrm{y}\le \frac{1}{7} \end{gathered}$$

But it is impossible

Case II: $(\mathrm{y}-7)\le 0 \mathrm{and} (7\mathrm{y}-1)\ge 0$

$\Rightarrow \mathrm{y}\le 7 \mathrm{and} \mathrm{y}\ge \frac{1}{7}\Rightarrow \frac{1}{7}\le \mathrm{y}\le 7$

Hence maximum value is 7 and minimum value is $\frac{1}{7}$