$\text{ Consider }f\left(x\right)=x^2-3x+a+\frac{1}{a},a\in R-\left\{0\right\},\text{ such that }f\left(3\right)>0\text{ and }f\left(2\right)\le 0\text{ . If }\alpha \text{ and }\beta \text{ are the roots of equation }$$f\left(x\right)=0$$\text{ then the value of }{\alpha }^2+{\beta }^2\text{ is equal to }$

 

$\begin{array}{l}f\left(x\right)=x^2-3x+a+\frac{1}{a}\\ \Rightarrow f\left(3\right)=9-9+a+\frac{1}{a}>0\\ \Rightarrow a+\frac{1}{a}>0\\ \Rightarrow a>0\\ f\left(2\right)=4-6+a+\frac{1}{a}\le 0\\ \Rightarrow \frac{a^2-2a+1}{a}\le 0\\ \Rightarrow \frac{(a-1{)}^2}{a}\le 0\\ \Rightarrow a=1\\ \Rightarrow a+\frac{1}{a}=2\end{array}$

$\begin{array}{l}\text{ Therefore, }f\left(x\right)=x^2-3x+2=0\text{ has roots }1\text{ and }2\text{ . }\\ \therefore {\alpha }^2+{\beta }^2=5\end{array}$