Question

If the expression $\left(mx-1+\frac{1}{x}\right)$ is always non-negative, then the minimum value of $m$ must be

• A. $-\frac{1}{2}$
• B. $0$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

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