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$y = \frac{{{x^2} - \alpha \beta }}{{2x - \alpha - \beta }}$

$x \in \mathbb{R} - \left\{ {\frac{{\alpha + \beta }}{2}} \right\}$

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) $y \in \mathbb{R}$

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 $.


Hence, (A) & (B).

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