Practicing Success
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If $9^{x+1}+(a^2-4a-2) 3^x +1>0$ for all x ∈ R, then |
$a ∈ R$ $a ∈ R^+$ $a ∈ [1,∞)$ $a ∈ R-\{2\}$ |
$a ∈ R-\{2\}$ |
We have, $9^{x+1}+(a^2-4a-2) 3^x +1>0$ for all x ∈ R $⇒9y^2+(a^2-4a-2) y + 1 > 0$ for all $y > 0$, where $y = 3^x$ $⇒y(9y+\frac{1}{y}+a^2-4a-2)>0$ for all $y>0$ $⇒(3\sqrt{y}-\frac{1}{\sqrt{y}})^2+(a-2)^2>0$ for all $y>0$ $⇒a≠2⇒a ∈ R-\{2\}$ |
