The least value of `a in RR` for which `4ax^2+1/x >= 1`, for all `x > 0`, is

Options

A. `1/64`
B. `1/32`
C. `1/27`
D. `1/25`

Answer

C

Explanation

Let `f(x)=4 alpha^2+1/x,x gt 0`
Then
`f(x)=8alpha x -1/x^2=(8 alpha x^3-1)/(x^2) " and " f''(x) = 8 alpha +(2)/(x^3)`
At points of local maximum of minimum ,we must have
`f(x)=0 rArr 8 alpha x^-1=0 rArr x (1)/(8 alpha)^(1/3)`
Clearly f''(x)`=8alpha +(2)/(x^3)gt 0 " for all "x gt 0 `
So f(x) attains its minimum value at `x=((1)/(8 alpha))^(1/3)`
The minimum value at `x=(1/(8 alpha ))^(1/3)` is `f((1/(8 alpha))^(1//3))=4alpha(1/(8alpha))^(2//3)+2(8 alpha)^(1//3)`