If the expression `[m x-1+(1//x)]` is non-negative for all positive real `x ,` then the minimum value of `m` must be `-1//2` b. `0` c. `1//4` d. `1//2`

Options

A. `-1//2`
B. 0
C. `1//4`
D. `1//2`

Answer

3

Explanation

We know that `ax^(2) + bx c ge 0, AA x in R`,
if `a gt 0 and b^(2) - 4ac le 0.` So,
`mx - 1 + 1/x ge 0` or `(mx^(2) - x + 1)/(x) ge 0`
or `mx^(2) - x + 1 ge 0` as `x gt 0`.
Now,
`mx^(2) - x + 1 ge 0` if `m gt 0` and `1 - 4m le 0`
`rArr m gt 0 and m ge 1//4`
Thus, the minimum value of m is 1/4.