Question

The least value of $\alpha ϵR$ for which $4\alpha x^2+\frac{1}{x}\ge 1$, for all $x>\alpha$ is

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

Answer 1

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

The value of $f\left(x\right)=4a{x}^2+\frac{1}{x}$ appreciates infinity as $x\to 0$,

Some where in between.

The derivative $f^{\prime }\left(x\right)=8ax-\frac{1}{x^2}$

betting this to Zero,

$\begin{array}{c}8ax-\frac{1}{x^2}=0\\ \Rightarrow 8a{x}^3=1\\ x_0=\frac{1}{2a^{\frac{1}{3}}}\end{array}$

Substituting

$\begin{array}{c}f\left(x\right)=4a\frac{1}{4a^{\frac{2}{3}}}+2a^{\frac{1}{3}}=\left(1+2\right)a^{\frac{1}{3}}\\ =3a^{\frac{1}{3}}\end{array}$

To find the value where the minimum is $1$.

or, $a=\frac{1}{27}$

$\because$ the multiple of a $a$ is $4x^2$ which is always positive, clearly the value increases as ${}^{\prime }{a}^{\prime }$ increases, so

$a=\frac{1}{27}$

Then,

Option $C$ is correct answer.