Q110 / 160 · TS EAMCET 2016 · Mathematics

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

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

Solution

Given, $f(x)=9 m x-1+\frac{1}{x} \geq 0$ Now, $\quad 9 m x^2-x+1 \geq 0$ For $f(x) \geq 0, D$ should be less than 0 . $$ D=1-36 m $$ $$ \begin{array}{rlrl} & & 1-36 m & \leq 0 \\ \therefore & m & \geq \frac{1}{36} \end{array} $$

Hence, the correct answer is option: 'C'.

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