MCQ
The minimum value of $\text{x}\log_{\text{e}}\text{x}$ is equal to :
- A$\text{e}$
- B$\frac{1}{\text{e}}$
- ✓$\frac{-1}{\text{e}}$
- D$\text{2}{\text{e}}$
✓
Answer
Correct option: C.
$\frac{-1}{\text{e}}$
Here, $\text{f}(\text{x})=\text{x}\log_{\text{e}}\text{x}$lmplies that $\text{f}'(\text{x})=\log_{\text{e}}\text{x}+1$
For a local maxima or a local minima, we must have f'(x) = 0
lmplies that $\log_{\text{e}}\text{x}+1=0$
lmplies that $\log_\text{e}\text{x}=-1$
lmplies that $\text{x}=\text{e}^{-1}$
Now, $\text{f}''(\text{x})=\frac{1}{\text{x}}$
lmplies that $\text{f}''(\text{e}^{-1})=\text{e}>0$
Threrfore, $\text{f}(\text{e}^{-1})$ is a local minima.
Hence, the minimum value of $\text{f}(\text{x})=\text{f}(\text{e}^{-1})$
lmplies that $\text{e}^{-1}\log_\text{e}(\text{e}^{-1})=-\text{e}^{-1}=\frac{-1}{\text{e}}$
For a local maxima or a local minima, we must have f'(x) = 0
lmplies that $\log_{\text{e}}\text{x}+1=0$
lmplies that $\log_\text{e}\text{x}=-1$
lmplies that $\text{x}=\text{e}^{-1}$
Now, $\text{f}''(\text{x})=\frac{1}{\text{x}}$
lmplies that $\text{f}''(\text{e}^{-1})=\text{e}>0$
Threrfore, $\text{f}(\text{e}^{-1})$ is a local minima.
Hence, the minimum value of $\text{f}(\text{x})=\text{f}(\text{e}^{-1})$
lmplies that $\text{e}^{-1}\log_\text{e}(\text{e}^{-1})=-\text{e}^{-1}=\frac{-1}{\text{e}}$
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