The number of integral values of a for which | vedclass

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(D) For $x^2 - (a - 1)x + 3 = 0$ to have both roots positive:$1$. Discriminant $D_1 \ge 0$ $\Rightarrow (a - 1)^2 - 12 \ge 0$ $\Rightarrow (a - 1)^2 \

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$1$

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(D) For $x^2 - (a - 1)x + 3 = 0$ to have both roots positive:$1$. Discriminant $D_1 \ge 0$ $\Rightarrow (a - 1)^2 - 12 \ge 0$ $\Rightarrow (a - 1)^2 \ge 12$ $\Rightarrow a - 1 \ge 2\sqrt{3}$ or $a - 1 \le -2\sqrt{3}$. Since $2\sqrt{3} \approx 3.46$,$a \ge 4.46$ or $a \le -2.46$.$2$. Sum of roots $> 0$ $\Rightarrow a - 1 > 0$ $\Rightarrow a > 1$.$3$. Product of roots $> 0 \Rightarrow 3 > 0$ (always true).Combining these,$a \ge 5$ (for integral values).For $x^2 + 3x + 6 - a = 0$ to have both roots negative:$1$. Discriminant $D_2 \ge 0$ $\Rightarrow 3^2 - 4(6 - a) \ge 0$ $\Rightarrow 9 - 24 + 4a \ge 0$ $\Rightarrow 4a \ge 15$ $\Rightarrow a \ge 3.75$.$2$. Sum of roots $$3$. Product of roots $> 0$ $\Rightarrow 6 - a > 0$ $\Rightarrow a Combining these,$3.75 \le a Intersection of both conditions: $a \in [5, 6)$,so $a = 5$.Thus,there is only $1$ integral value of $a$.

The number of integral values of $a$ for which $x^2 - (a - 1)x + 3 = 0$ has both roots positive and $x^2 + 3x + 6 - a = 0$ has both roots negative is

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