Question

Find all the values of the parameter a for which the point of minimum of the function $f\left(x\right)=1+a^{2}x-x^3$ satisfy the inequality $\frac{x^2+x+2}{x^2+5x+6}<0.$

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

Answer 1

Answer: (D)

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

We first solve the inequality $\frac{x^2+x+2}{x^2+5x+6}<0$$\Rightarrow \frac{x^2+x+2}{(x+3)(x+2)}<0.$

Since $x^2+x+2={(x+\frac{1}{2})}^2+\frac{7}{4}>0$ for $x\in R,$

the inequility is satisfied if $(x+3)(x+2)<0$ when $-3<x<-2\cdots \left(1\right)$

We now find the extrema of the function $f\left(x\right)=1+a^{2}x-x^3.$

For max, or Min.,$f^{\prime }\left(x\right)=a^2-3x^2$

$\therefore x=\frac{a}{\sqrt{3}}.-\frac{a}{\sqrt{3}}$

$f^{\prime \prime }\left(x\right)=-6>0$ as $x<0$ by (1)

Hence minimum, when $a>0$ we choose $x=-\frac{a}{\sqrt{3}}$

$\therefore -3<-\frac{a}{\sqrt{3}}<-2$ by (1)

Multiplying by -ive sing and reversing and hence no change is the sign of inequality.

Thus we get $2\sqrt{3}<a<3\sqrt{3}$

Again if $a<0$ then we choose $x=\frac{a}{\sqrt{3}}$

$\therefore -3<\frac{a}{\sqrt{3}}<-2$ $\Rightarrow -3\sqrt{3}<a<-2\sqrt{3}.$

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