The least value of function $f(x) = ax + \frac{b}{x}$ (where, $a > 0, b > 0, x > 0$) is:

$2\sqrt{ab}$

The correct answer is Option (2) → $2\sqrt{ab}$ ##

$f'(x) = a - \frac{b}{x^2}$ and $f'(x) = 0$

$\Rightarrow a = \frac{b}{x^2}$

$\Rightarrow x^2 = \frac{b}{a} \Rightarrow x = \pm \sqrt{\frac{b}{a}}$

Now, $f''(x) = -b \cdot \frac{(-2)}{x^3} = + \frac{2b}{x^3}$

At $x = \sqrt{\frac{b}{a}}$, $\quad f''(x) = + \frac{2b}{\left(\sqrt{\frac{b}{a}}\right)^3} = \frac{+2b \cdot a^{3/2}}{b^{3/2}}$

$= +2b^{-1/2} \cdot a^{3/2} = +2 \sqrt{\frac{a^3}{b}} > 0 \quad [∵ a, b > 0]$

$∴$ Least value of $f(x)$, $f\left(\sqrt{\frac{b}{a}}\right) = a \cdot \sqrt{\frac{b}{a}} + \frac{b}{\sqrt{\frac{b}{a}}}$

$= a \cdot a^{-1/2} \cdot b^{1/2} + b \cdot b^{-1/2} \cdot a^{1/2}$

$= \sqrt{ab} + \sqrt{ab} = 2\sqrt{ab}$