Question

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

• A. $\frac{1}{9}$
• B. $\frac{1}{16}$
• C. $\frac{1}{36}$
• D. $\frac{1}{81}$

Answer 1

The correct option is C $\frac{1}{36}$

$f\left(x\right)=9mx-1+\frac{1}{x}$

$f\left(x\right)=\frac{9m{x}^2-x+1}{x}$

$f\left(x\right)=\frac{9m(x^2-\frac{x}{9m}+\frac{1}{9m})}{x}$

$f\left(x\right)=\frac{9m(x^2-\frac{x}{9m}+\frac{1}{(18m{)}^2}+\frac{1}{9m}-\frac{1}{(18m{)}^2})}{x}$

$f\left(x\right)=\frac{9m\left(\right(x-\frac{1}{18m}{)}^2+\frac{1}{9m}-\frac{1}{(18m{)}^2})}{x}$

$f\left(x\right)=\frac{9m(x-\frac{1}{18m}{)}^2}{x}+\frac{\frac{9m}{9m}-\frac{9m}{(18m{)}^2}}{x}$

$f\left(x\right)=\frac{9m(x-\frac{1}{18m}{)}^2}{x}+\frac{1-\frac{1}{36m}}{x}$

Since , $m>0,x>0and(x-\frac{1}{18m}{)}^2\ge 0$ , first term is alsways positive.

So, for the function to be always positive, second term must also be positive.

$⟹\frac{1-\frac{1}{36m}}{x}\ge 0$

$⟹m\ge \frac{1}{36}$

Therefore, the smallest value of m for which the function is always postive is $\frac{1}{36}$

Hence , answer is option (C).