Question

The set of all values of parameters $a$ for which the points of minimum of the function $=1+a^{2}x-x^3$ satisfy the inequality $\frac{x^2+x+2}{x^2+5x+6}\le 0$ is

• A. an empty set
• B. $(-3\sqrt{3},-2\sqrt{3})$
• C. $(2\sqrt{3},3\sqrt{3})$
• D. $(-3\sqrt{3},-2\sqrt{3})\cup (2\sqrt{3},3\sqrt{3})$

Answer 1

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

Given inequality is $\frac{x^2+x+2}{x^2+5x+6}\le 0$

Now, $\frac{x^2+x+2}{(x+3)(x+2)}\le 0$, Numerator is positive for all real values of x

$⟹-3<x<-2$

Let $f\left(x\right)=1+a^{2}x-x^3$

$⟹f^{\prime }\left(x\right)=a^2-3x^2$

$=(a-\sqrt{3}x)(a+\sqrt{3}x)$

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 $a>0,-3<\frac{-a}{\sqrt{3}}<-2⟹a\in (2\sqrt{3},3\sqrt{3})$

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

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