The maximum and minimum values of

$y=\frac{a{x}^2+2bx+c}{A{x}^2+2Bx+C}$ are those for which

Let $f\left(x\right)=\frac{a{x}^2+2bx+c}{\mathrm{\Lambda }{x}^2+2Bx+C}$

Let $P\left(x_1,y_1\right)$ be a point on the curve $y = f \left(x\right)$ where y is maximum or minimum. Then, at p

$\frac{dy}{dx}=0\text{ and }\frac{d^{2}y}{d{x}^2}\ne 0$

The equation of the tangent at $P\left(x_1,y_1\right)$ is

$y-y_1={\left(\frac{dy}{dx}\right)}_p\left(x-x_1\right)\Rightarrow y=y_1\left[\because {\left(\frac{dy}{dx}\right)}_p=0\right.$

Putting, $y=y_1\text{ in }y=\frac{a{x}^2+2bx+c}{A{x}^2+2Bx+C}$ we get the x-coordants of

P. This means that $y=\frac{a{x}^2+2bx+c}{A{x}^2+2Bx+C}$ gives only one value of x

This is possible only when. $a{x}^2+2bx+c-y\left(A{x}^2+2Bx+C\right)$ is a perfect square

Hence, option (b) is correct.