Question 14 Marks
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 __________ .
Answer
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Both the roots are positive
$\mathrm{D} \geq 0$
$4(a-3)^{2}-4 \times 9(1-a) \geq 0$
$a^{2}-6 a+9-9+9 a \geq 0$
$a^{2}+3 a \geq 0$
$a(a+3) \geq 0$
$
a \in(-\infty,-3] \cup[0, \infty) \quad ...(i)
$
$-\frac{\mathrm{b}}{2 \mathrm{a}}>0$
$\frac{2(a-3)}{2(a-1)}>0$
$
a \in(-\infty, 1) \cup(3, \infty) \quad ...(ii)
$
$\mathrm{f}(0)=9>0$
Equation (i) $\cap$ (ii)
$a \in(-\infty,-3] \cup[0,1)$
$2 \alpha+\beta+\gamma-6+0+1=7$
Both the roots are positive
$\mathrm{D} \geq 0$
$4(a-3)^{2}-4 \times 9(1-a) \geq 0$
$a^{2}-6 a+9-9+9 a \geq 0$
$a^{2}+3 a \geq 0$
$a(a+3) \geq 0$
$
a \in(-\infty,-3] \cup[0, \infty) \quad ...(i)
$
$-\frac{\mathrm{b}}{2 \mathrm{a}}>0$
$\frac{2(a-3)}{2(a-1)}>0$
$
a \in(-\infty, 1) \cup(3, \infty) \quad ...(ii)
$
$\mathrm{f}(0)=9>0$
Equation (i) $\cap$ (ii)
$a \in(-\infty,-3] \cup[0,1)$
$2 \alpha+\beta+\gamma-6+0+1=7$
