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Q.
If the expression $\left(mx -1 +\frac{1}{x}\right)$ is always non-negative, then the minimum value of $m$ must be
Linear Inequalities
Solution:
We know that, $ax^{2} + bx + c \ge 0$, if $a > 0$ and $b^{2} - 4ac \le 0$
Now, $mx - 1+ \frac{1}{x} \ge 0 \quad\Rightarrow \quad \frac{mx^{2}-x+1}{x} \ge 0$
$\Rightarrow \quad mx^{2} - x + 1 \ge 0$ and $x > 0$
Now, $mx^{2} - x + 1 \ge 0$, 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 $\frac{1}{4}$.