Answer:
Option C
Explanation:
Here, to find the least value of αϵR . for which
4αx2+1x≥1, , for all x >0
i.e. to find the minimum value of α when
y=4αx2+1x;x>0 attains minimum value of α
∴ \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}