Question ID: #1041
If the set of all $a\in R-\{1\}$, for which the roots of the equation $(1-a)x^{2}+2(a-3)x+9=0$ are positive is $(-\infty,-\alpha]\cup[\beta,\gamma)$, then $2\alpha+\beta+\gamma$ is equal to
Solution:
For both roots of a quadratic equation $Ax^2+Bx+C=0$ to be positive, three conditions must be satisfied:
1. Discriminant $D \ge 0$
2. x-coordinate of the vertex $-\frac{B}{2A} > 0$
3. $A \cdot f(0) > 0$ (which is equivalent to Product of roots $> 0$)
Condition 1: $D \ge 0$
$$B^2 – 4AC \ge 0$$
$$(2(a-3))^2 – 4(1-a)(9) \ge 0$$
$$4(a^2 – 6a + 9) – 36(1-a) \ge 0$$
$$a^2 – 6a + 9 – 9(1-a) \ge 0$$
$$a^2 – 6a + 9 – 9 + 9a \ge 0$$
$$a^2 + 3a \ge 0$$
$$a(a+3) \ge 0$$
$$a \in (-\infty, -3] \cup [0, \infty) \quad \dots (i)$$
Condition 2: $-\frac{B}{2A} > 0$
$$-\frac{2(a-3)}{2(1-a)} > 0$$
$$\frac{a-3}{a-1} > 0$$
$$a \in (-\infty, 1) \cup (3, \infty) \quad \dots (ii)$$
Condition 3: Product of roots $> 0$
$$\frac{C}{A} > 0$$
$$\frac{9}{1-a} > 0$$
$$1-a > 0 \Rightarrow a < 1$$
$$a \in (-\infty, 1) \quad \dots (iii)$$
Taking the intersection of the intervals from $(i)$, $(ii)$, and $(iii)$:
$$a \in (-\infty, -3] \cup [0, 1)$$
Comparing this with the given set $(-\infty, -\alpha] \cup [\beta, \gamma)$
$$-\alpha = -3 \Rightarrow \alpha = 3$$
$$\beta = 0$$
$$\gamma = 1$$
Calculate $2\alpha+\beta+\gamma$:
$$2(3) + 0 + 1 = 6 + 1 = 7$$
Ans. (7)
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