The interval in which the function f(x)=x^ x , x>0, is strictly i… - Vidyadip

MCQ

The interval in which the function $\mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{x}}, \mathrm{x}>0$, is strictly increasing is

A
$\left(0, \frac{1}{\mathrm{e}}\right]$
B
$\left[\frac{1}{\mathrm{e}^2}, 1\right)$
C
$(0, \infty)$
✓
$\left[\frac{1}{\mathrm{e}}, \infty\right)$

✓

Answer

Correct option: D.

$\left[\frac{1}{\mathrm{e}}, \infty\right)$

d
$ f(x)=x^x ; x>0 $

$ \ell n y=x \ell n x $

$ \frac{1}{y} \frac{d y}{d x}=\frac{x}{x}+\ell n x $

$ \frac{d y}{d x}=x^x(1+\ell n x) $

for strictly increasing

$ \frac{d y}{d x} \geq 0 \Rightarrow x^x(1+\ell n x) \geq 0 $

$ \Rightarrow \ell n x \geq-1 $

$ x \geq e^{-1} $

$ x \geq \frac{1}{e} $

$ x \in\left[\frac{1}{e}, \infty\right)$

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