The interval in which the function f ( x ) = x x , x > 0 , is strictly increasing is [2024]

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Solution

Q.
The interval in which the function $f (x) = x^{x}, x > 0,$ is strictly increasing is [2024]

1 $[\frac{1}{e}, \infty)$

2 $[\frac{1}{e^{2}}, 1)$

3 $(0, \frac{1}{e}]$

4 $(0, \infty)$

Ans.
(1)

$f (x) = x^{x}, x > 0$

So, $y = x^{x}$

$\Rightarrow \ln y = x \ln x$

$\Rightarrow \frac{1}{y} \frac{d y}{d x} = 1 + \ln x \Rightarrow \frac{d y}{d x} = x^{x} (1 + \ln x)$

For $y$ to be strictly increasing, we have $\frac{d y}{d x} \geq 0$

$\Rightarrow x^{x} (1 + \ln x) \geq 0$

$\Rightarrow 1 + \ln x \geq 0 [∵ x^{x} > 0, as x > 0]$

$\Rightarrow \ln x \geq - 1 \Rightarrow x \geq e^{- 1}$

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

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