Question

Let $f\left(x\right)=\alpha x^2-2+\frac{1}{x}$ 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 $\frac{2^5}{3^3}$

$f\left(x\right)=\frac{\alpha x^3-2x+1}{x}\ge 0\forall x\in (0,\theta )$

$\Rightarrow \alpha x^3-2x+1\ge 0\forall x\in (0,\theta )$

Now let $\varphi \left(x\right)=\alpha x^3-2x+1$

${\varphi }^{\prime }\left(x\right)=3\alpha x^3-2=0$ $x=\pm \sqrt{\frac{2}{3\alpha }}$

So Graph of $\varphi \left(x\right)$

$\varphi \sqrt{\frac{2}{3\alpha }}\ge 0$

$\sqrt{\frac{2}{3\alpha }}[\alpha \frac{2}{3\alpha }-2]+1\ge 0$

$\sqrt{\frac{2}{3\alpha }}[-\frac{4}{3}]+1\ge 0$

$\sqrt{\frac{2}{3\alpha }}\le \frac{3}{4}\Rightarrow \frac{2}{3\alpha }\le \frac{9}{16}$

$=\frac{32}{27}\le \alpha$