The set $\left\{x \in \mathbb{R}: 4+11 x-3 x^2>0\right\}$ is the interval
The set $\left\{x \in \mathbb{R}: 4+11 x-3 x^2>0\right\}$ is the interval
- $\left(-\frac{1}{3}, 4\right)$
- $(-, 4)$
- $\left(-4, \frac{1}{3}\right)$
- $\left(-4, \frac{-1}{3}\right)$
Solution
$\begin{aligned} & \text {Given: }\left\{x \in R: 4+11 x-3 x^2>0\right\} \\ & \because 4+11 x-3 x^2>0 \\ & \Rightarrow-(x-4)(3 x-1)>0 \\ & \Rightarrow(x-4)(3 x-1) < 0\end{aligned}$
$\therefore x \in\left(-\frac{1}{3}, 4\right)$
Asked in: AP EAMCET 2023 (17 May Shift 1)
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