Let S be the set of all non-zero real numbers 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|lt 1. Which of the following 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)

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Let S be the set of all non-zero real numbers 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|lt 1. Which of the following 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

<div class="c2h-answers">AD <div class="c2h-hint"> $\left|x_1-x_2\right|\lt 1$ ${\therefore}\left(x_1+x_2\right)^2-4x_1x_2\lt 1$ $\Rightarrow \frac 1{\alpha ^2}-4\lt 1$ $\Rightarrow 5-\frac 1{\alpha ^2}\gt 0$ $\Rightarrow 5\gt \frac{1}{\alpha ^2}$ Take the square root on both sides. $\Rightarrow\sqrt{5}\gt \left|\frac{1}{\alpha}\right|$ $\Rightarrow \sqrt{5}\gt \frac{1}{\left|\alpha \right|}$ $\Rightarrow\left|\alpha\right|\gt \frac{1}{\sqrt{5}}$ $\Rightarrow -\frac{1}{\sqrt{5}}\gt \alpha \gt \frac{1}{\sqrt{5}}$ $\Rightarrow -\frac{1}{\sqrt{5}}\gt \alpha ,\alpha \gt \frac{1}{\sqrt{5}}$Eq$1$ Also $D\gt 0$ ${\therefore}1-4\alpha ^2\gt 0$ ${\therefore}\alpha {\in}\left(-\frac 1 2,\frac 1 2\right)$Eq$2$ From ( 1 ) and ( 2 ) $\alpha \in\left(-\frac{1}{2},-\frac{1}{\sqrt{5}}\right) \cup\left(\frac{1}{\sqrt{5}}, \frac{1}{{2}}\right)$ The options A and D are correct.

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