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(C) Given the equation $\log_{|\alpha|} \left( \frac{x^2}{\beta^2} \right) = 1$. This implies $\frac{x^2}{\beta^2} = |\alpha|$,so $x^2 = \beta^2 |\alp

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(C) Given the equation $\log_{|\alpha|} \left( \frac{x^2}{\beta^2} \right) = 1$. This implies $\frac{x^2}{\beta^2} = |\alpha|$,so $x^2 = \beta^2 |\alpha|$. Since $\alpha, \beta$ are roots of $x^2 - |a|x - |b| = 0$,the product of roots $\alpha \beta = -|b| \le 0$. Given $|\alpha|  0$,we analyze the signs. Since $|a| Solving for $x$,we get $x = \pm |\beta| \sqrt{|\alpha|}$. The positive root is $|\beta| \sqrt{|\alpha|}$. Since $|\alpha| Since $\beta > 0$ and $\sqrt{|\alpha|} < 1$ (as $|\alpha| < |\beta|$ and the product of roots is negative),the positive root is less than $\beta$.

Let $\alpha, \beta$ be the roots of the equation $x^2 - |a|x - |b| = 0$ such that $|\alpha| < |\beta|$. If $|a| < \beta - 1$,then the positive root of $\log_{|\alpha|} \left( \frac{x^2}{\beta^2} \right) - 1 = 0$ is

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