The least value of `alpha in R` for which `4ax^2+(1)/(x)ge1`, for all `xgt 0`, is

Options

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

Answer

C

Explanation

Here, to find the least value of `alpha in R,` for each `4ax^(2)+(1)/(x) ge, "for all" x ge0`.
i.e., to find the minimum value of `alpha` when `y=4alpha x^(2)+(1)/(x): xgt0` attains minimum value of `alpha`.
`:. (dy)/(dx)=8 alpha x-(1)/(x^(2))" "...(i)`
Now, ` (d^(2)y)/(dx^(2))=8 alpha+(2)/(x^(3)) " ".....(ii)`
When `(dy)/(dx)= 0 "then" 8x^(3) alpha=1`
At `x=((1)/(8 alpha))^(1//3),(d^(2)y)/(dx^(2))=8 alpha+16 alpha = 24 alpha `, Thus, y attains minimum when `x=((1)/(8 alpha))^(1//3), alpha gt0`
`:.` y attains minimum when `x=((1)/(8 alpha))^(1//3)`
`i.e., 4alpha((1)/(8 alpha))^(2//3)+(8alpha)^(2//3) ge1`
`rArr a^(1//3)+2 alpha^)1//3)ge1`
`rArr 3 alpha^(1//3) ge1 rArr alpha ge(1)/(27)`
Hence, the least value of `alpha "is" (1)/(27)`