Mathematics - Application of Derivative Question with Solution | TestHub

$\frac{{\mathrm{x}}^2+\mathrm{x}+2}{{\mathrm{x}}^2+5\mathrm{x}+6}<0\Rightarrow \mathrm{x}\in (-3,-2)$

Let $\mathrm{f}(\mathrm{x})=1+{\mathrm{a}}^2\mathrm{x}-{\mathrm{x}}^3$ for maximum (or) minimum,

$f^1(x)=0\Rightarrow a^2-3x^2=0\Rightarrow x=\pm \frac{a}{\sqrt{3}}$ And ${\mathrm{f}}^1(\mathrm{x})=-6\mathrm{x}$ is positive when x is negative if $\mathrm{a}>0$ then point of minimum is $\frac{-\mathrm{a}}{\sqrt{3}}$ $\begin{array}{rlr}& \Rightarrow -3<\frac{-a}{\sqrt{3}}<-2& \Rightarrow 2\sqrt{3}<a<3\sqrt{3}\end{array}$ If $\mathrm{a}<0$, the point of minimum is a $\mid \sqrt{3}$ $\begin{array}{rlr}& \Rightarrow -3<\frac{a}{\sqrt{3}}<-2\Rightarrow -3\sqrt{3}<a<-2\sqrt{3}& \Rightarrow a\in (-3\sqrt{3},-2\sqrt{3})\cup (2\sqrt{3},3\sqrt{3})\end{array}$