If the expression left( mx - 1 + frac{1}{x} r | vedclass

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(C) Given the expression $f(x) = mx - 1 + \frac{1}{x} \ge 0$ for $x > 0$.Multiplying by $x$ (since $x > 0$),we get $mx^2 - x + 1 \ge 0$.For a quadrati

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$\frac{1}{4}$

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(C) Given the expression $f(x) = mx - 1 + \frac{1}{x} \ge 0$ for $x > 0$.Multiplying by $x$ (since $x > 0$),we get $mx^2 - x + 1 \ge 0$.For a quadratic expression $ax^2 + bx + c$ to be non-negative for all $x > 0$,we must have $a > 0$ and the discriminant $D \le 0$.Here,$a = m$,$b = -1$,and $c = 1$.Condition $1$: $a > 0 \implies m > 0$.Condition $2$: $D = b^2 - 4ac \le 0 \implies (-1)^2 - 4(m)(1) \le 0$.$1 - 4m \le 0 \implies 4m \ge 1 \implies m \ge \frac{1}{4}$.Since $m > 0$ and $m \ge \frac{1}{4}$,the minimum value of $m$ is $\frac{1}{4}$.

If the expression $\left( mx - 1 + \frac{1}{x} \right)$ is always non-negative for $x > 0$,then the minimum value of $m$ must be

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