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(A) Given that $\alpha$ and $\beta$ are roots of the equation $x^2 + \alpha x + \beta = 0$.By Vieta's formulas:Sum of roots: $\alpha + \beta = -\alpha

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inequality is satisfied by exactly two integral values of $y$

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(A) Given that $\alpha$ and $\beta$ are roots of the equation $x^2 + \alpha x + \beta = 0$.By Vieta's formulas:Sum of roots: $\alpha + \beta = -\alpha \implies 2\alpha + \beta = 0$  .......$(1)$Product of roots: $\alpha \beta = \beta \implies \beta(\alpha - 1) = 0$  .......$(2)$From $(2)$,either $\beta = 0$ or $\alpha = 1$.Case $I$: If $\beta = 0$,then from $(1)$,$2\alpha = 0 \implies \alpha = 0$. But the problem states $\alpha 
eq \beta$,so this is rejected.Case $II$: If $\alpha = 1$,then from $(1)$,$2(1) + \beta = 0 \implies \beta = -2$. Here $\alpha 
eq \beta$ holds.Now substitute $\alpha = 1$ and $\beta = -2$ into the inequality $| |y - (-2)| - 1 | $| |y + 2| - 1 | $-1 $0 This implies $|y + 2| $-2 $-4 Thus,$y \in (-4, -2) \cup (-2, 0)$.Checking option $D$: For $x^2 + x - 2 > 0$,the roots are $x = -2, 1$. The parabola opens upward,so $x^2 + x - 2 > 0$ for $x \in (-\infty, -2) \cup (1, \infty)$. This does not hold for $x \in [-1, 0]$.

Consider the equation $x^2 + \alpha x + \beta = 0$ having roots $\alpha, \beta$ such that $\alpha \neq \beta$. Also consider the inequality $| |y - \beta| - \alpha | < \alpha$,then:

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