Question

If \[ax + \frac{b}{x} \frac{>}{} c\] for all positive x where a,b,>0, then _______________ .

विकल्प

• `ab<c^2/4`

• `ab>=c^2/4`

• `ab>=c/4`

Answer 1

\[ \ ab \geq \frac{c^2}{4}\]

\[\text { Given }: ax + \frac{b}{x} \geq c\]

\[\text { Minimum value of} ax + \frac{b}{x} = c\]

\[\text { Now }, \]

\[f\left( x \right) = ax + \frac{b}{x}\]

\[ \Rightarrow f'\left( x \right) = a - \frac{b}{x^2}\]

\[\text { For a local maxima or a local minima, we must have}\]

\[f'\left( x \right) = 0\]

\[ \Rightarrow a - \frac{b}{x^2} = 0\]

\[ \Rightarrow a x^2 - b = 0\]

\[ \Rightarrow a x^2 = b\]

\[ \Rightarrow x^2 = \frac{b}{a}\]

\[ \Rightarrow x = \pm \frac{\sqrt{b}}{\sqrt{a}}\]

\[f''\left( x \right) = \frac{2b}{x^3}\]

\[ \Rightarrow f''\left( x \right) = \frac{2b}{\left( \frac{\sqrt{b}}{\sqrt{a}} \right)^3}\]

\[ \Rightarrow f''\left( x \right) = \frac{2b \left( a \right)^\frac{3}{2}}{\left( b \right)^\frac{3}{2}} > 0\]

\[\text { So }, x = \frac{\sqrt{b}}{\sqrt{a}} \text { is a local minima } . \]

\[ \therefore f\left( \frac{\sqrt{b}}{\sqrt{a}} \right) = a\left( \frac{\sqrt{b}}{\sqrt{a}} \right) + \frac{b}{\left( \frac{\sqrt{b}}{\sqrt{a}} \right)} \geq c\]

\[ = \sqrt{a}\sqrt{a}\left( \frac{\sqrt{b}}{\sqrt{a}} \right) + \frac{\sqrt{b}\sqrt{b}}{\left( \frac{\sqrt{b}}{\sqrt{a}} \right)} \geq c\]

\[ = \sqrt{ab} + \sqrt{ab} \geq c\]

\[ \Rightarrow 2\sqrt{ab} \geq c\]

\[ \Rightarrow \frac{c}{2} \leq \sqrt{ab}\]

\[ \Rightarrow \frac{c^2}{4} \leq ab\]