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Increasing and Decreasing Functions — Maths STD 12 Science — Question

Gujarat 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 $x.$
$\Rightarrow\text{x}^\text{x} > 0$
$\Rightarrow1+\log_\text{e}\text{x} < 0$
$[$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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