Let s be the set of all non-zero real numbers  $\alpha$  such that the quadratic equation $\alpha x^2-x+\alpha =0$   has two distinct real roots  $x_1$  and  $x_2$  satisfying the inequality $|x_1-x_2|<1$  , which of the following interval is a subset of s.

The quadratic equation  $\alpha x^2-x+\alpha =0$
Discriminant (D) =  $b^2-4ac={(-1)}^2-4\alpha \left(\alpha \right) =1-4{\alpha }^2$
   
For distinct real roots  $1-4{\alpha }^2>0$
  i.e,  $\alpha \in (-\frac{1}{2},\frac{1}{2})-\left(1\right)$
Now   $|x_1-x_2|<1$ 
 $\Rightarrow {|x_1-x_2|}^2<1$

$\Rightarrow {(x_1+x_2)}^2-4x_1{x}_2<1$                    $S.R=\frac{1}{\alpha }$

$\frac{1}{{\alpha }^2}-4\left(1\right)<1$                                         $P.R=1$
  $1-4{\alpha }^2<{\alpha }^2$

$$\begin{gathered} \Rightarrow 5{\alpha }^2-1>0 \\ (\sqrt{5}\alpha +1)(\sqrt{5}\alpha -1)>0 \\ \alpha \in (-\infty ,\frac{-1}{\sqrt{5}})\cup (\frac{1}{\sqrt{5}},\infty ) - \left(2\right) \\ From \left(1\right) \& \left(2\right) \Rightarrow \alpha \in (-\frac{1}{2},\frac{-1}{\sqrt{5}})\cup (\frac{1}{\sqrt{5}},\frac{1}{2}) \\ \therefore S=(-\frac{1}{2},\frac{-1}{\sqrt{5}})\cup (\frac{1}{\sqrt{5}},\frac{1}{2}) \end{gathered}$$