JEE MAIN - Mathematics (2025 - 2nd April Evening Shift - No. 22)
If the set of all $\mathrm{a} \in \mathbf{R}-\{1\}$, for which the roots of the equation $(1-\mathrm{a}) x^2+2(\mathrm{a}-3) x+9=0$ are positive is $(-\infty,-\alpha] \cup[\beta, \gamma)$, then $2 \alpha+\beta+\gamma$ is equal to $\qquad$ .
Answer
7
Explanation
$$\begin{aligned} & f(x)=(1-a) x^2+2(a-3) x+9, f(0)=9>0 \\ & D \geq 0 \Rightarrow 4(a-3)^2 \geq 4(1-a) \cdot 9 \end{aligned}$$
$\Rightarrow a \in(-\infty,-3] \cup[0, \infty)\quad\text{..... (i)}$
_2nd_April_Evening_Shift_en_22_1.png)
$$\begin{aligned} & x_1+x_2=\frac{-2(a-3)}{1-a}, x_1 x_2=\frac{9}{1-a} \\ & x_1+x_2>0 \Rightarrow \frac{a-3}{a-1}>0 \Rightarrow a \in(-\infty, 1) \cup(3, \infty) \ldots \text { (ii) } \end{aligned}$$
$x_1 x_2>0 \Rightarrow 1-a>0 \Rightarrow a \in(-\infty, 1) \ldots($ iii $)$
$$\begin{aligned} &\Rightarrow \text { Interaction of (i), (ii) and (iii) }\\ &\begin{aligned} & a \in(-\infty,-3] \cup[0,1) \\ \Rightarrow \quad & \alpha=3, \beta=0, \gamma=1 \Rightarrow 2 \alpha+\beta+\gamma=7 \end{aligned} \end{aligned}$$
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