Given, x₁ and x₂ are roots of

$$\alpha {x^2} - x + \alpha = 0$$

$$\therefore$$ $${x_1} + {x_2} = {1 \over \alpha }$$ and $${x_1}{x_2} = 1$$

Also, $$\left| {{x_1} - {x_2}} \right| < 1$$

$$ \Rightarrow {\left| {{x_1} - {x_2}} \right|^2} < 1 \Rightarrow {({x_1} - {x_2})^2} < 1$$

or, $${({x_1} + {x_2})^2} - 4{x_1}{x_2} < 1$$

$$ \Rightarrow {1 \over {{\alpha ^2}}} - 4 < 1$$ or $${1 \over {{\alpha ^2}}} < 5$$

$$ \Rightarrow 5{\alpha ^2} - 1 > 0$$

or, $$(\sqrt 5 \alpha - 1)(\sqrt 5 \alpha + 1) > 0$$

$$\therefore$$ $$\alpha \in \left( { - \infty , - {1 \over {\sqrt 5 }}} \right) \cup \left( {{1 \over {\sqrt 5 }},\infty } \right)$$ .....(i)

Also, $$D > 0$$

$$ \Rightarrow 1 - 4{\alpha ^2} > 0$$ or $$\alpha \in \left( { - {1 \over 2},{1 \over 2}} \right)$$ ...... (ii)

$$\alpha \in \left( { - {1 \over 2},{{ - 1} \over {\sqrt 5 }}} \right) \cup \left( {{1 \over {\sqrt 5 }},{1 \over 2}} \right)$$