if ax+ b x for all positive x where a, b, > 0, then. - Vidyadip

MCQ

if $\text{ax}+\frac{\text{b}}{\text{x}}\geq\text{c}$ for all positive $x$ where $a, b, > 0,$ then.

A
$\text{ab} < \frac{\text{c}^{2}}{4}$
✓
$\text{ab} > \frac{\text{c}^{2}}{4}$
C
$\text{ab} > \frac{\text{c}}{4}$
D
None of these

✓

Answer

Correct option: B.

$\text{ab} > \frac{\text{c}^{2}}{4}$

$=\text{f}(\text{x})=\text{ax}+\frac{\text{b}}{\text{x}}$
$\text{f}(\text{x})=0$
$\Rightarrow \text{a}-\frac{\text{b}}{\text{x}^{2}}=0$
$\Rightarrow\text{x}=\pm\sqrt{\frac{\text{b}}{\text{a}}}$
$\text{f}\ ''(\text{x})=\frac{2\text{b}}{\text{x}^{3}}$
$\text{f}\ ''\Big(\sqrt{\frac{\text{b}}{\text{a}}}\Big)=\frac{2\text{b}}{\Big(\sqrt{\frac{\text{b}}{\text{c}}}\Big)^{3}}>0$
$\Rightarrow\text{x}=\sqrt{\frac{\text{b}}{\text{a}}}$ has a minima.
$\text{f}\ ''\Big(\sqrt{\frac{\text{b}}{\text{a}}}\Big)=2\sqrt{\text{ab}}\geq\text{c}$
$\frac{\text{c}}{2}\le\sqrt{\text{ab}}$
$\Rightarrow\frac{\text{c}}{4}\le{\text{ab}}$

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