Explanation

(c)
We know that $ax^{2} + bx + c \geq 0\ if\ a > 0\ and\ b^{2} - 4ac \leq 0$
Now, $mx - 1 + \frac{1}{x} \geq 0$
$\Rightarrow \ \frac{mx^{2} - x + 1}{x} \geq 0$
$\Rightarrow mx^{2} - x + 1 \geq 0$ and $x > 0$
Now, $mx^{2} - x + 1 \geq 0$, if $m > 0$ and $1 - 4m \leq 0$ or if $m > 0$ and $m \geq \frac{1}{4}.$
Thus, the minimum value of $m\ is\ \frac{1}{4}$.