The minimum value of x _ e x is . - Vidyadip

MCQ

The minimum value of $\frac{\text{x}}{\log_{\text{e}}\text{x}}$ is .

✓
$\text{e}$
B
$\frac{1}{\text{e}}$
C
$1$
D
None of these

✓

Answer

Correct option: A.

$\text{e}$

Given, $\text{f}(\text{x})=\frac{\text{x}}{\log_{\text{e}}\text{x}}$
$\Rightarrow \text{f}\ '(\text{x})=\frac{\log_{e}\text{x}-1}{(\log_{\text{e}}\text{x})^{2}}$
For a local maximum or a local minima, we must have $\text{f}\ '(\text{x})=0$
$\Rightarrow \frac{\log_{e}\text{x}-1}{(\log_{\text{e}}\text{x})^{2}}=0$
$\Rightarrow \log_{\text{e}}\text{x}-1=0$
$\Rightarrow \log_{\text{e}}\text{x}=1$
$\Rightarrow \text{x}=\text{e}$
Now, $\Rightarrow \text{f}\ ''(\text{x})=\frac{-1}{\text{x}(\log_{\text{e}}\text{x})^{2}}+\frac{2}{\text{x}(\log_{\text{e}}\text{x})^{3}}$
$\Rightarrow \text{f}\ ''(\text{e})=\frac{-1}{\text{e}}+\frac{2}{\text{e}}=\frac{1}{\text{e}} > 0$
So, $x = e$ is a local minima.
$\therefore$ minimum value of $\text{f}(\text{x})=\frac{\text{e}}{\log_{\text{e}}\text{e}}=\text{e}$

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