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 folowing intervals is (are) a subset (s) 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

Answer: (A), (D)

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

$\alpha x^2-x+\alpha =0$ has distinct real roots.
$\therefore D>0\Rightarrow 1-4{\alpha }^2>0\Rightarrow \alpha ϵ(-\frac{1}{2},\frac{1}{2})Also|x_1-x_2|<1\Rightarrow (x_1-x_2{)}^2<1\Rightarrow (x_1+x_2{)}^2-4x_1{x}_2<1\Rightarrow \frac{1}{{\alpha }^2}-4<1\Rightarrow \frac{1}{{\alpha }^2}<5or{\alpha }^2>\frac{1}{5}\Rightarrow \alpha ϵ(-\infty ,-\frac{1}{\sqrt{5}})\cup \left(\frac{1}{\sqrt{5},}\infty \right)$
Combining (i) and (ii) -
S= $(-\frac{1}{2},-\frac{1}{\sqrt{5}})\cup (\frac{1}{\sqrt{5}},\frac{1}{2})$
$\therefore$ Subsets of S can be $(-\frac{1}{2},-\frac{1}{\sqrt{5}})and(\frac{1}{\sqrt{5}},\frac{1}{2})$