Question

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

• A. $0$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{4}$

We have, $ax-1+\frac{1}{x}\ge 0\Rightarrow \frac{a{x}^2-x+1}{x}\ge 0$

$\Rightarrow a{x}^2-x+a\ge 0$ as $x>0$

It will hold if $a>0$ and $D\le 0$

$\Rightarrow a>0$ and $a-4a\le 0\Rightarrow \frac{1}{4}$

Therefore, the minimum value of $a$ is $\frac{1}{4}$