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

Given expression is $\frac{x^2+14x+9}{x^2+2x+3}=y$ (say)

$$\begin{gathered} \Rightarrow x^2+14x+9=y(x^2+2x+3) \\ \Rightarrow (1-y)x^2+2x(7-y)+3(3-y)=0 \end{gathered}$$

Since above equation is a quadratic equation in $\text{'}x\text{'}$.

Since $\text{'}x\text{'}$ is Real

$\Rightarrow D\ge 0$

Here $D={\left\{2(7-y)\right\}}^2-4(1-y)3(3-y)$

$$\begin{gathered} \Rightarrow D=4(49+y^2-14y)-12(3-y-3y+y^2) \\ \Rightarrow D=-8y^2-8y+160 \end{gathered}$$

Now $\Rightarrow D\ge 0$

$$\begin{gathered} \Rightarrow -8y^2-8y+160\ge 0 \\ \Rightarrow y^2+y-20\le 0 \\ \Rightarrow (y-4)(y+5)\le 0 \\ \Rightarrow y\in [-5,4] \end{gathered}$$

Therefore the minimum and maximum values of the expression will be $-5$ and 4. Hence option $\left(a\right)$ is correct.