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

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

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$\frac{1}{4}$

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(C) Given the expression $f(x) = mx - 1 + \frac{1}{x} \geq 0$ for all $x > 0$.Multiplying by $x$ (since $x > 0$),we get $mx^2 - x + 1 \geq 0$.For a quadratic $ax^2 + bx + c \geq 0$ to hold for all $x > 0$,we analyze the conditions.If $m \leq 0$,then for large $x$,$mx^2 - x + 1$ will eventually become negative,so we must have $m > 0$.For $mx^2 - x + 1 \geq 0$ to be true for all $x > 0$,the discriminant $D$ must be less than or equal to $0$.$D = (-1)^2 - 4(m)(1) = 1 - 4m \leq 0$.$1 \leq 4m \implies m \geq \frac{1}{4}$.Thus,the minimum value of $m$ is $\frac{1}{4}$.

If the expression $\left( mx - 1 + \frac{1}{x} \right)$ is non-negative for all positive real numbers $x$,then what must be the minimum value of $m$?

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