Question

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

• A. $-\frac{1}{2}$
• B. $0$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{1}{4}$

Given,
$mx-1+\frac{1}{x}⩾0$
$\Rightarrow m{x}^2-x+1⩾0\because x>0)$
(i)Co-efficient of $x^2>0$
$\Rightarrow m>0$
(ii)Discriminant $⩽0$
$\Rightarrow 1-4m⩽0$
$m⩾\frac{1}{4}$
Minimum value of $m=\frac{1}{4}$
Hence, option 'C' is correct.