In which of the following interval the function f(x) = x^x, x > 0 is strictly increasing? Option 1: [e, ∞) Option 2: (1/e, ∞) Option 3: [1, ∞) Option 4: [0, ∞)

Answer: (1/e, ∞). Solution: To find where f(x) = x^x (for x > 0) is strictly increasing, the derivative must satisfy f'(x) > 0. Taking natural log on both sides: f(x) = (x^x) f(x) = x x Differentiating both sides with respect to x: (f'(x))/(f(x)) = d/dx[x x] (f'(x))/(f(x)) = x + 1 f'(x) = x^x( x + 1) For the function to be strictly increasing: f'(x) > 0 x^x( x + 1) > 0 Since x > 0, we have x^x > 0 always. Therefore: x + 1 > 0 x > -1 x > e^(-1) x > 1/e The function is strictly increasing in the interval (1/e, ∞).

Calculus

Application of Derivatives

Medium

applied

In which of the following interval the function $f(x) = x^x, x > 0$ is strictly increasing?

$[e, \infty)$

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

$[1, \infty)$

$[0, \infty)$

✅ Correct Option: 2

To find where $f(x) = x^x$ (for $x > 0$ ) is strictly increasing, the derivative must satisfy $f'(x) > 0$ .

Taking natural log on both sides:

$\ln f(x) = \ln(x^x)$

$\ln f(x) = x \ln x$

Differentiating both sides with respect to $x$ :

$\frac{f'(x)}{f(x)} = \frac{d}{dx}[x \ln x]$

$\frac{f'(x)}{f(x)} = \ln x + 1$

$f'(x) = x^x(\ln x + 1)$

For the function to be strictly increasing:

$f'(x) > 0$

$x^x(\ln x + 1) > 0$

Since $x > 0$ , we have $x^x > 0$ always.

Therefore:

$\ln x + 1 > 0$

$\ln x > -1$

$x > e^{-1}$

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

The function is strictly increasing in the interval $\left(\frac{1}{e}, \infty\right)$ .

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In which of the following interval the function f(x)=xx,x>0f(x) = x^x, x > 0f(x)=xx,x>0 is strictly increasing?

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

In which of the following interval the function $f(x) = x^x, x > 0$ is strictly increasing?