Question

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(s) is/are a subset of S?

• A. $(-\frac{1}{2},-\frac{1}{\sqrt{5}})$
• B. $(-\frac{1}{\sqrt{5}},0)$
• C. $(0,\frac{1}{\sqrt{5}})$
• D. $(\frac{1}{\sqrt{5}},\frac{1}{2})$

Answer 1

The correct option is D

$(\frac{1}{\sqrt{5}},\frac{1}{2})$

Given, $x_1$ and $x_2$ are roots of $\alpha x^2-x+\alpha =0.$
$\therefore x_1+x_2=\frac{1}{\alpha }$ 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 $(x_1+x_2{)}^2-4x_1{x}_2<1$
$\Rightarrow \frac{1}{a^2}-4<1$ or $\frac{1}{{\alpha }^2}<5$
$\Rightarrow 5{\alpha }^2-1>0$ or $(\sqrt{5}\alpha -1)(\sqrt{5}\alpha +1)>0$

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