Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$f\left(x\right)=\frac{\alpha x^3-2x+1}{x}\ge 0\forall xϵ(0,\infty )\Rightarrow \alpha x^3-2x+1\ge 0\forall xϵ(0,\infty )\text{Now Let}\varphi \left(x\right)=\alpha x^3-2x+1{\varphi }^1\left(x\right)=3\alpha x^2-2=0x=\pm \sqrt{\frac{2}{3\alpha }}\text{So Graph of}\varphi \left(x\right)$

$\varphi \left(\sqrt{\frac{2}{3\alpha }}\right)\ge 0\sqrt{\frac{2}{3\alpha }}[\alpha \frac{2}{3\alpha }-2]+1\ge 0\sqrt{\frac{2}{3\alpha }}[-\frac{4}{3}]+1\ge 0\sqrt{\frac{2}{3\alpha }}\le \frac{3}{4}\Rightarrow \frac{2}{3\alpha }\le \frac{9}{16}=\frac{32}{27}\le \alpha$