Question

The least value of $a\in R$ for which $4a{x}^2+\frac{1}{x}\ge 1$, for all $x>0$, is

• A. $\frac{1}{64}$
• B. $\frac{1}{32}$
• C. $\frac{1}{27}$
• D. $\frac{1}{25}$

Answer 1

Answer: (C)

$\frac{1}{27}$

$4\alpha x^2+\frac{1}{x}\ge 1$
$\Rightarrow y=4\alpha x^2+\frac{1}{x}$
$\Rightarrow y=\frac{dy}{dx}=8\alpha x-\frac{1}{x^2}=0$
$\Rightarrow x={\left(\frac{1}{8\alpha }\right)}^{1/3}$
$\Rightarrow f\left(x\right)=\frac{4\alpha x^3+1}{x}=\frac{1/2+1}{1/(8\alpha {)}^{1/3}}$
$\Rightarrow \frac{3}{2}\times (8\alpha {)}^{1/3}\ge 1$
$\Rightarrow {\alpha }^{1/3}\ge 1/3\Rightarrow \alpha \ge \frac{1}{27}$