The smallest value of the constant m 0 for | vedclass

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(C) Given the inequality $f(x) = 9mx - 1 + \frac{1}{x} \geq 0$ for all $x > 0$.Multiplying by $x$ (since $x > 0$),we get $9mx^2 - x + 1 \geq 0$.For a

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$\frac{1}{36}$

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(C) Given the inequality $f(x) = 9mx - 1 + \frac{1}{x} \geq 0$ for all $x > 0$.Multiplying by $x$ (since $x > 0$),we get $9mx^2 - x + 1 \geq 0$.For a quadratic expression $ax^2 + bx + c \geq 0$ to hold for all $x > 0$,we analyze the discriminant $D = b^2 - 4ac$.Here,$a = 9m$,$b = -1$,and $c = 1$.The discriminant $D = (-1)^2 - 4(9m)(1) = 1 - 36m$.For the quadratic to be non-negative for all $x$,we require $D \leq 0$.$1 - 36m \leq 0 \implies 36m \geq 1 \implies m \geq \frac{1}{36}$.Thus,the smallest value of $m$ is $\frac{1}{36}$.

The smallest value of the constant $m > 0$ for which $f(x) = 9mx - 1 + \frac{1}{x} \geq 0$ for all $x > 0$ is:

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