QJEEAdv15Ma9f22: JEE Advanced 2015 Mathematics - Class 11 JEEAdv15Ma9f22 Solution: Let be the set of all non-zero real numbers such that the…

Given, x₁ and x₂ are roots of

$\alpha {x^2} - x + \alpha = 0$

$\therefore$ ${x_1} + {x_2} = {1 \over \alpha }$ and ${x_1}{x_2} = 1$

Also, \left| {{x_1} - {x_2}} \right| < 1

\Rightarrow {\left| {{x_1} - {x_2}} \right|^2} < 1 \Rightarrow {({x_1} - {x_2})^2} < 1

or, {({x_1} + {x_2})^2} - 4{x_1}{x_2} < 1

\Rightarrow {1 \over {{\alpha ^2}}} - 4 < 1 or {1 \over {{\alpha ^2}}} < 5

\Rightarrow 5{\alpha ^2} - 1 > 0

or, (\sqrt 5 \alpha - 1)(\sqrt 5 \alpha + 1) > 0

$\therefore$ $\alpha \in \left( { - \infty , - {1 \over {\sqrt 5 }}} \right) \cup \left( {{1 \over {\sqrt 5 }},\infty } \right)$ .....(i)

Also, D > 0

\Rightarrow 1 - 4{\alpha ^2} > 0 or $\alpha \in \left( { - {1 \over 2},{1 \over 2}} \right)$ ...... (ii)

$\alpha \in \left( { - {1 \over 2},{{ - 1} \over {\sqrt 5 }}} \right) \cup \left( {{1 \over {\sqrt 5 }},{1 \over 2}} \right)$