Question

The least value of the function $f(x) = ax + b/x, a > 0,b > 0,x > 0 $ is

• A. $ \sqrt{ab} $
• B. $ 2\sqrt{\frac{a}{b}} $
• C. $ 2\sqrt{\frac{b}{a}} $
• D. $ 2\sqrt{ab} $

Answer 1

Answer: (D)

Given, $f(x)=a x+\frac{b}{x}$
On differentiating w.r.t. ' $x$ ', we get
$f^{\prime}(x)=a-\frac{b}{x^{2}}$
Put $f^{\prime}(x)=0$,
$\Rightarrow a-\frac{b}{x^{2}}=0$
$\Rightarrow x=\pm \sqrt{\frac{b}{a}}$
$\therefore f^{\prime \prime}(x)=\frac{2 b}{x^{3}}$
At $ x=\sqrt{\frac{b}{a}}$,
$f^{\prime \prime}(x)=\frac{2 b}{\left(\sqrt{\frac{b}{a}}\right)^{3}}>0, \text { minima }$
$\therefore$ Minimum value of $f(x)$ is
$f\left(\sqrt{\frac{b}{a}}\right) =a \sqrt{\frac{b}{a}}+\frac{b}{\sqrt{\frac{b}{a}}}$
$=\sqrt{a b}+\sqrt{a b}=2 \sqrt{a b}$