Question

Let $f\left(x\right)=\alpha x^2-2+\frac{1}{x}\text{where}\alpha$ is a real constant. The smallest $\alpha$ for which $f\left(x\right)\ge 0$ for all $x>0$ is

• A. $\frac{2^2}{3^3}$
• B. $\frac{2^3}{3^3}$
• C. $\frac{2^4}{3^3}$
• D. $\frac{2^5}{3^3}$

Answer 1

The correct option is D $\frac{2^5}{3^3}$

$f\left(x\right)=\alpha x^2-2+\frac{1}{x}=\frac{\alpha x^3-2x+1}{x}\forall x\in (0,\infty )$
Let $\varphi \left(x\right)=\alpha x^3-2x+1$ should be positive.
${\varphi }^{\prime }\left(x\right)=3\alpha x^2-2=0$
$\Rightarrow x=\pm \sqrt{\frac{2}{3\alpha }}$
$\therefore x=+\sqrt{\frac{2}{3\alpha }}$ point of minima
$\varphi (\sqrt{\frac{2}{3\alpha }})\ge 0\Rightarrow \sqrt{\frac{2}{3\alpha }}[\alpha .\frac{2}{3\alpha }-2]+1\ge 0\Rightarrow \sqrt{\frac{2}{3\alpha }}(-\frac{4}{3})+1\ge 0\Rightarrow \sqrt{\frac{2}{3\alpha }}(\frac{4}{3})\le 1\Rightarrow \sqrt{\frac{2}{3\alpha }}\le \frac{3}{4}\Rightarrow \alpha \ge \frac{32}{27}=\frac{2^5}{3^3}$