Question

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

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

Answer 1

The correct option is C $\frac{1}{27}$

$\alpha \ge \frac{x-1}{4x^3}$
$\alpha \ge \text{max}\left(\frac{x-1}{4x^3}\right)$

Let $f\left(x\right)=\frac{x-1}{4x^3}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{4}[-\frac{2}{x^3}+\frac{3}{x^4}]=0\Rightarrow x=\frac{3}{2}$
$f^{\prime \prime }\left(x\right)=\frac{1}{4}[\frac{6}{x^4}-\frac{12}{x^5}]$
$f^{\prime \prime }\left(\frac{3}{2}\right)<0$
$\Rightarrow f\left(x\right)$ is maximum at $x=\frac{3}{2}$
$f\left(\frac{3}{2}\right)=\frac{1}{27}$

$\therefore \alpha \ge \frac{1}{27}$