Question

If the point of minima of the function, $f\left(x\right)=1+a^{2}x-x^3$ satisfy the inequality $\frac{x^2+x+2}{x^2+5x+6}<0,$ then 'a' must lie in the interval

• A. $(-3\sqrt{3,}3\sqrt{3})$
• B. $(-2\sqrt{3,}3\sqrt{3})$
• C. $\left(2\sqrt{3,}3\sqrt{3}\right)$
• D. $(-3\sqrt{3,}-2\sqrt{3})\cup (2\sqrt{3},3\sqrt{3})$

Answer 1

The correct option is D $(-3\sqrt{3,}-2\sqrt{3})\cup (2\sqrt{3},3\sqrt{3})$

The Given inequality is $\frac{x^2+x+2}{x^2+5x+6}<0$
$\Rightarrow \frac{x^2+x+2}{(x+2)(x+2)}<0$, Numerator is positive for all real values of x
$\Rightarrow -3<x<-2$
Now $f\left(x\right)=1+a^{2}x-x^3$
$\Rightarrow f^{\prime }\left(x\right)=a^2-3x^2=(a-\sqrt{3}x)(a+\sqrt{3}x)$
For minima of $f\left(x\right)$
$f^{\prime }\left(x\right)=0\Rightarrow x=\pm \frac{a}{\sqrt{3}}$
Also $f^{\prime \prime }\left(x\right)=-6x$
If $a>0,-3<\frac{-a}{\sqrt{3}}<-2$

$\Rightarrow aϵ(2\sqrt{3},3\sqrt{3})$
If $a<0,-3<\frac{a}{\sqrt{3}}<-2$

$\Rightarrow a\in (-3\sqrt{3},-2\sqrt{3})$