The function x^x is increasing, when - Vidyadip

MCQ

The function ${x^x}$ is increasing, when

✓
$x > {1 \over e}$
B
$x < {1 \over e}$
C
$x < 0$
D
For all real $ x$

✓

Answer

Correct option: A.

$x > {1 \over e}$

a
(a) Let $y = {x^x}$ ==> $\frac{{dy}}{{dx}} = {x^x}(1 + \log x)$

For $\frac{{dy}}{{dx}} > 0$;

${x^x}(1 + \log x) > 0$ ==> $1 + \log x > 0 \Rightarrow {\log _e}x > {\log _e}\frac{1}{e}$

For this to be positive, $x$ should be greater than $\frac{1}{e}$.

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