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

• A

  $\frac{2^{2}}{3^{3}}$
• B

  $\frac{2^{3}}{3^{3}}$
• C

  $\frac{2^{4}}{3^{3}}$
• Correct answer
  D

  $\frac{2^{5}}{3^{3}}$

Solution

$\begin{array}{ll}f(x)=\frac{\alpha x^{3}-2 x+1}{x} \geq 0 & \forall x \in(0, \infty) \\ \Rightarrow \alpha x^{3}-2 x+1 \geq 0 & \forall x \in(0, \infty) \\ \text { Now Let } \phi(x)=\alpha x^{3}-2 x+1 & \end{array}$ $$ \phi^{\prime}(x)=3 \alpha x^{2}-2=0 \quad x=\pm \sqrt{\frac{2}{3 \alpha}} $$ So Graph of $\phi(\mathrm{x})$ $$ \begin{array}{l} \phi\left(\sqrt{\frac{2}{3 \alpha}}\right) \geq 0 \\ \sqrt{\frac{2}{3 \alpha}}\left[\alpha \frac{2}{3 \alpha}-2\right]+1 \geq 0 \end{array} $$ $\sqrt{\frac{2}{3 \alpha}}\left[-\frac{4}{3}\right]+1 \geq 0$ $\sqrt{\frac{2}{3 \alpha}} \leq \frac{3}{4} \Rightarrow \frac{2}{3 \alpha} \leq \frac{9}{16}$ $=\frac{32}{27} \leq \alpha$

Hence, the correct answer is option: 'D'.