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 $\left|x_{1}-x_{2}\right|<1$. Which of the following intervals is(are) a subset(s) of $S$ ?

(A) $\left(-\frac{1}{2},-\frac{1}{\sqrt{5}}\right)$; (D) $\left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)$

(B) $\left(-\frac{1}{\sqrt{5}}, 0\right)$

(C) $\left(0, \frac{1}{\sqrt{5}}\right)$

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Quadratic Equations and Inequalities

Let

S

be the set of all non-zero real numbers

α

such that the quadratic equation

α x^{2} - x + α = 0

has two distinct real roots

x_{1}

and

x_{2}

satisfying the inequality

∣ x_{1} - x_{2} ∣ < 1

. Which of the following intervals is(are) a subset(s) of

S

Answer: (A), (D)

For two distinct real roots, discriminant

Δ = 1 - 4 α^{2} > 0

, so

∣ α ∣ < \frac{1}{2}

. The roots are

x_{1, 2} = \frac{1 \pm 1 - 4 α ^{2}}{2 α}

. We have

∣ x_{1} - x_{2} ∣ = \frac{1 - 4 α ^{2}}{∣ α ∣} < 1

, which gives

1 - 4 α^{2} < α^{2}

, so

5 α^{2} > 1

, i.e.,

∣ α ∣ > \frac{1}{5}

. Combining with

∣ α ∣ < \frac{1}{2}

, we get

S = (- \frac{1}{2}, - \frac{1}{5}) \cup (\frac{1}{5}, \frac{1}{2})

. Thus options (A) and (D) are subsets of

S

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · One or more correct options