Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIncreasing and Decreasing Functions1 Mark
MCQ
The function $f(x) = x^x$ decreases on the interval:
A
$(0, e)$
B
$(0, e)$
✓
$\Big(0,\frac{1}{\text{e}}\Big)$
D
None of these
✓
Answer
Correct option: C.
$\Big(0,\frac{1}{\text{e}}\Big)$
Given, $\text{f}(\text{x})=\text{x}^{\text{x}}$
Applying log with base e on both sides, we get
$\log(\text{f}(\text{x}))=\text{x}\log_\text{e}\text{x}$
$\frac{\text{f}'(\text{x})}{\text{f}(\text{x})}=1+\log_{\text{e}}\text{x}$
$\text{f}'(\text{x})=\text{f}(\text{x})(1+\log_\text{e}\text{x})$
$=\text{x}^\text{x}(1+\log_\text{e}\text{x})$
For f(x) to be decreasing, we must have
$\text{f}'(\text{x})<0$
$\Rightarrow\text{x}^\text{x}(1+\log_\text{e}\text{x})<0$
Here, logarithmic function is defined for positive values of x.
$\Rightarrow\text{x}^\text{x}>0$
$\Rightarrow1+\log_\text{e}\text{x}<0$
$[\text{Since }\text{x}^\text{x}(1+\log_\text{e}\text{x})<0\Rightarrow1+\log_\text{e}\text{x}<0]$
$\Rightarrow\log_\text{e}\text{x}<-1$
$\Rightarrow\text{x}<\text{e}^{-1}$ $\big[\because\ \log_\text{a}\text{x}<\text{N}\Rightarrow\text{a}^\text{N}\text{ for }\text{a}>1\big]$
Here,
$\text{e}>1$
$\Rightarrow\log_\text{e}\text{x}<-1$
$\Rightarrow\text{x}<\text{e}^{-1}$
$\Rightarrow\text{x}\in\big(0,\text{e}^{-1}\big)$
So, f(x) is decreasing on $\Big(0,\frac{1}{\text{e}}\Big).$
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