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Maxima and Minima — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMaxima and Minima1 Mark
Question
The minimum value of $\frac{\text{x}}{\log_{\text{e}}\text{x}}$ is .
  1. $\text{e}$
  2. $\frac{1}{\text{e}}$
  3. $1$
  4. None of these
✓

Answer

  1. $\text{e}$
Solution:
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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