If the roots of sqrt{frac{1-y}{y}}+sqrt{frac{ | vedclass

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(A) Let $t = \sqrt{\frac{1-y}{y}}$. Then the equation becomes $t + \frac{1}{t} = \frac{5}{2}$.Solving for $t$,we get $2t^2 - 5t + 2 = 0$,which gives $

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$\frac{1}{2}$

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(A) Let $t = \sqrt{\frac{1-y}{y}}$. Then the equation becomes $t + \frac{1}{t} = \frac{5}{2}$.Solving for $t$,we get $2t^2 - 5t + 2 = 0$,which gives $t = 2$ or $t = \frac{1}{2}$.If $\sqrt{\frac{1-y}{y}} = 2$,then $\frac{1-y}{y} = 4$ $\Rightarrow 1-y = 4y$ $\Rightarrow y = \frac{1}{5}$.If $\sqrt{\frac{1-y}{y}} = \frac{1}{2}$,then $\frac{1-y}{y} = \frac{1}{4}$ $\Rightarrow 4-4y = y$ $\Rightarrow y = \frac{4}{5}$.Thus,$\alpha = \frac{1}{5}$ and $\beta = \frac{4}{5}$ (since $\beta > \alpha$).Then $\alpha + \beta = 1$ and $\alpha \beta = \frac{4}{25}$.The equation $(\alpha+\beta) x^4 - 25 \alpha \beta x^2 + (\gamma + \beta - \alpha) = 0$ becomes $x^4 - 4x^2 + (\gamma + \frac{3}{5}) = 0$.Let $u = x^2$. Then $u^2 - 4u + (\gamma + \frac{3}{5}) = 0$.For real roots,$u$ must be non-negative. The discriminant $D = 16 - 4(\gamma + \frac{3}{5}) \ge 0$ $\Rightarrow 4 - \gamma - \frac{3}{5} \ge 0$ $\Rightarrow \gamma \le \frac{17}{5} = 3.4$.Also,for at least one non-negative root $u$,we need the sum of roots $4 > 0$ and product $\gamma + \frac{3}{5} \ge 0 \Rightarrow \gamma \ge -0.6$.Thus,$\gamma \in [-0.6, 3.4]$. Among the options,$\frac{1}{2} = 0.5$ is in this range.

If the roots of $\sqrt{\frac{1-y}{y}}+\sqrt{\frac{y}{1-y}}=\frac{5}{2}$ are $\alpha$ and $\beta$ $(\beta > \alpha)$ and the equation $(\alpha+\beta) x^4-25 \alpha \beta x^2+(\gamma+\beta-\alpha)=0$ has real roots,then a possible value of $\gamma$ is

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