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

Solution :
$$\eqalign{ & f\left( x \right) = {x^2} - 3x + a + \frac{1}{a};f\left( 3 \right) = 9 - 9 + a + \frac{1}{a} > 0 \cr & \Rightarrow a + \frac{1}{a} > 0 \cr & \Rightarrow a > 0 \cr & f\left( 2 \right) = 4 - 6 + a + \frac{1}{a} \leqslant 0 \cr & \Rightarrow \frac{{{a^2} - 2a + 1}}{a} \leqslant 0 \cr & \Rightarrow \frac{{{{\left( {a - 1} \right)}^2}}}{a} \leqslant 0 \cr & \Rightarrow a = 1 \cr} $$
Therefore, $$f\left( x \right) = {x^2} - 3x + 2 = 0$$     has root 1 and 2.
$$\therefore {\alpha ^2} + {\beta ^2} = 5$$