$y = \frac{{{x^2} - \alpha \beta }}{{2x - \alpha - \beta }}$
$x \in \mathbb{R} - \left\{ {\frac{{\alpha + \beta }}{2}} \right\}$
Assuming $\beta > \alpha $ which of the following option(s) is/are correct?
(A) $y \le \alpha $
(B) $y \ge \beta $
(C) $\alpha \le y \le \beta $
(D) $y \in \mathbb{R}$
(B) $y \ge \beta $
(C) $\alpha \le y \le \beta $
(D) $y \in \mathbb{R}$
Solution
${x^2} - 2xy + [(\alpha + \beta )y - \alpha \beta ] = 0$
$\because x \in \mathbb{R},\Delta \ge 0$
$\therefore 4{y^2} - 4[(\alpha + \beta )y - \alpha \beta ] \ge 0$
$ \Rightarrow {y^2} - \alpha y - \beta y + \alpha \beta \ge 0$
$ \Rightarrow (y - \alpha )(y - \beta ) \ge 0$
The wave-curve is shown below for the expression $ (y - \alpha )(y - \beta ) $ assuming $\beta > \alpha $.