Function f(x) = x^x, x > 0 decreases on the interval Option 1: (0, 1/e) Option 2: (0, e) Option 3: (1/e, ∞) Option 4: (e, ∞)

Answer: (0, 1/e). Solution: A function decreases when its derivative is negative, so we need to find where f'(x) Given f(x) = x^x, take the natural logarithm of both sides: f(x) = (x^x) f(x) = x x Differentiate both sides: (f'(x))/(f(x)) = d/dx[x x] Using the product rule on the right side: (f'(x))/(f(x)) = 1 * x + x * 1/x (f'(x))/(f(x)) = x + 1 f'(x) = f(x) * ( x + 1) f'(x) = x^x( x + 1) For f'(x) 0, we have x^x > 0 always. Therefore: x + 1 Combined with the condition x > 0, the function decreases on the interval (0, 1/e).

Calculus

Application of Derivatives

Medium

common

Function $f(x) = x^x, x > 0$ decreases on the interval

$(0, \frac{1}{e})$

$(0, e)$

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

$(e, \infty)$

✅ Correct Option: 1

A function decreases when its derivative is negative, so we need to find where $f'(x) < 0$ .

Given $f(x) = x^x$ , take the natural logarithm of both sides:

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

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

Differentiate both sides:

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

Using the product rule on the right side:

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

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

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

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

For $f'(x) < 0$ :

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

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

Therefore:

$\ln x + 1 < 0$

$\ln x < -1$

$x < e^{-1}$

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

Combined with the condition $x > 0$ , the function decreases on the interval $(0, \frac{1}{e})$ .

2025: 22 May Shift 1

Match List-I with List-II

List-I	List-II
Function	Increasing on the interval
(A) $f(x) = -x^2 - 2x + 1$	(I) $(-\infty, -1)$
(B) $f(x) = x^2 + 1$	(II) $(1, \infty)$
(C) $f(x) = x^2 - 2x + 3$	(III) $(-\infty, 0)$
(D) $f(x) = -x^2$	(IV) $(0, \infty)$

Choose the correct answer from the options given below:

2025: 26 May Shift 2

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A function $f(x)=\sin x(1+\cos x), x \in [0,\pi]$ has a local maxima at $x=$

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If $f(x) = x^2 - 4x + 13, x \in \mathbb{R}$ , then which of the following are correct?

(A) $x = 2$ is a stationary point of $f(x)$ .

(B) $f(x)$ is increasing function on $(2, \infty)$

(D) $f(2) = 9$

Choose the correct answer from the options given below:

Function f(x)=xx,x>0f(x) = x^x, x > 0f(x)=xx,x>0 decreases on the interval

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Function $f(x) = x^x, x > 0$ decreases on the interval