Answer:
Option C
Explanation:
Here, to find the least value of $\alpha \epsilon R$ . for which
$4\alpha x^{2}+\frac{1}{x}\geq 1,$ , for all x >0
i.e. to find the minimum value of $\alpha$ when
$y=4\alpha x^{2}+\frac{1}{x};x>0$ attains minimum value of $\alpha$
$\therefore$ $\frac{\text{d}y}{\text{d}x}=8\alpha x-\frac{1}{x^{2}}$ .....(i)
Now, $\frac{d^{2}y}{dx^{2}}=8\alpha +\frac{2}{x^{3}}$ ........(ii)
when $\frac{\text{d}y}{\text{d}x}=0,$
then $8x^{3}\alpha =1$
$\frac{d^{2}y}{dx^{2}}=8\alpha+16\alpha=24\alpha ,$ Thus, y attains minimum when
$x=(\frac{1}{8\alpha})^{\frac{1}{3}}: \alpha >0.$
$\therefore$ y attains minimum when $x=(\frac{1}{8\alpha})^{\frac{1}{3}}$
i.e, $4\alpha(\frac{1}{8\alpha})^{\frac{2}{3}}+(8\alpha)^{\frac{1}{3}}\geq1$
$\Rightarrow \alpha^{\frac{1}{3}}+2\alpha^{\frac{1}{3}}\geq1\Rightarrow 3\alpha^{\frac{1}{3}}\geq1$
$\Rightarrow $ $\alpha \geq\frac{1}{27} $
Hence, the least value of $\alpha$ is $\frac{1}{27}$
