Question

The values of $k$ for which the point of minimum of the function $f\left(x\right)=4+k^{2}x-2x^3$ satisfies the inequality $\frac{x^2+2x+4}{x^2+6x+8}<0$ is

• A. $(-2\sqrt{6},\infty )$
• B. $(2\sqrt{6},\infty )$
• C. $(-2\sqrt{6},-4\sqrt{6})\cup (0,4\sqrt{6})$
• D. $(-2\sqrt{6},-4\sqrt{6})\cup (2\sqrt{6},4\sqrt{6})$

Answer 1

The correct option is D $(-2\sqrt{6},-4\sqrt{6})\cup (2\sqrt{6},4\sqrt{6})$

Ww know that,
$\frac{x^2+2x+4}{x^2+6x+8}<0\Rightarrow \frac{(x+1{)}^2+3}{(x+2)(x+4)}<0$
So,
$-4<x<-2$
Now finding the minima of the function,
$f\left(x\right)=4+k^{2}x-2x^3\Rightarrow f^{\prime }\left(x\right)=k^2-6x^2=0\Rightarrow x^2=\frac{k^2}{6}\Rightarrow x=\pm \frac{|k|}{\sqrt{6}}{x}_1=\frac{|k|}{\sqrt{6}},x_2=\frac{-|k|}{\sqrt{6}}$
$f^{\prime \prime }\left(x\right)=-12x\Rightarrow f^{\prime \prime }\left(x_2\right)>0$
So $x_2$ is the point of minima,
$\therefore -4<x_2<-2\Rightarrow -4<-\frac{k}{\sqrt{6}}<-2\Rightarrow 2\sqrt{6}<|k|<4\sqrt{6}\Rightarrow k\in (-2\sqrt{6},-4\sqrt{6})\cup (2\sqrt{6},4\sqrt{6})$