CAT 2023 Slot 3 Solved Paper

It is given that ${\mathit{x}}^2+\mathit{b}\mathit{x}+\mathit{c}=0$ has two real roots. Let the roots of the equation be $\mathit{\alpha },\mathit{\beta }\cdot \left(\mathit{\alpha }>\mathit{\beta }\right)$ Then, we can say that $\frac{1}{\mathit{\alpha }}-\frac{1}{\mathit{\beta }}=\frac{1}{3}....\mathrm{Eq}\left(1\right)$
Similarly, $\frac{1}{{\mathit{\alpha }}^2}+\frac{1}{{\mathit{\beta }}^2}=\frac{5}{9}....\mathrm{Eq}\left(2\right)$
$\mathrm{Eq}\left(2\right)$ can be written as ${\left(\frac{1}{\mathit{\alpha }}-\frac{1}{\mathit{\beta }}\right)}^2+2\cdot \frac{1}{\mathit{\alpha }}\cdot \frac{1}{\mathit{\beta }}=\frac{5}{9}$
$\Rightarrow {\left(\frac{1}{3}\right)}^2+2\cdot \frac{1}{\mathit{\alpha }}\cdot \frac{1}{\mathit{\beta }}=\frac{5}{9}$
$\Rightarrow \frac{2}{\mathit{\alpha }\cdot \mathit{\beta }}=\frac{4}{9}\Rightarrow \frac{1}{\mathit{\alpha }\cdot \mathit{\beta }}=\frac{2}{9}$
$\Rightarrow \mathit{\alpha }\cdot \mathit{\beta }=\frac{9}{2}$
We know that the product of the roots is equal to $\mathrm{c}$, which implies $\mathit{c}=\frac{9}{2}$
We also know that the sum of the roots is equal to $-\mathit{b}$.
$\Rightarrow \frac{1}{{\mathit{\alpha }}^2}+\frac{1}{{\mathit{\beta }}^2}={\left(\frac{1}{\mathit{\alpha }}+\frac{1}{\mathit{\beta }}\right)}^2-\frac{2}{\mathit{\alpha }\mathit{\beta }}=\frac{5}{9}$
$\Rightarrow {\left(\frac{\mathit{\alpha }+\mathit{\beta }}{\mathit{\alpha }\mathit{\beta }}\right)}^2-\frac{4}{9}=\frac{5}{9}$
$\Rightarrow {\left(\frac{\mathit{\alpha }+\mathit{\beta }}{\mathit{\alpha }\mathit{\beta }}\right)}^2={\left(1\right)}^2$
$\Rightarrow \mathit{\alpha }+\mathit{\beta }=\pm \mathit{\alpha }\mathit{\beta }$
Hence, the maximum value of $\mathit{b}$ is $\frac{9}{2}$.
Hence, the maximum value of $\left(\mathit{b}+\mathit{c}\right)$ is 9