Answer:
Option A,B,C
Explanation:
Given , x1 and x2 are roots of αx2−x+α=0
∴ x_{1}+x_{2}=\frac{1}{\alpha} and x_{1}x_{2}=1
Also, |x_{1}-x_{2}|<1
\Rightarrow |x_{1}-x_{2}|^{2}<1\Rightarrow(x_{1}-x_{2})^{2}<1
or (x_{1}-x_{2})^{2}-4x_{1}x_{2}<1
\Rightarrow \frac{1}{\alpha^{2}}-4<1 or \frac{1}{\alpha^{2}}<5
\Rightarrow 5\alpha^{2}-1>0
or (\sqrt{5}\alpha-1)(\sqrt{5}\alpha+1)>0

\therefore \alpha \epsilon \left(-\infty, -\frac{1}{\sqrt{5}}\right)\cup\left(\frac{1}{\sqrt{5}},\infty\right) ......(i)
Also, D>0
\Rightarrow 1-4\alpha^{2}>0 or \alpha \epsilon \left(-\frac{1}{2},\frac{1}{2}\right) .....(ii)
From Eqs. (i) and (ii) , we get
\alpha \epsilon \left(-\frac{1}{2}, -\frac{1}{\sqrt{5}}\right)\cup\left(\frac{1}{\sqrt{5}},\frac{1}{2}\right)