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JEE Main 20246 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2024 · 6 April · Morning Shift · Mathematics · Application of Derivatives

The interval in which the function f(x)=x^x, x>0 , is strictly...

The interval in which the function f(x)=xx,x>0f(x)=x^x, x>0, is strictly increasing is

Options

  1. A(0,1e]\left(0, \frac{1}{e}\right]
  2. B(0,∞)(0, \infty)
  3. C\left.\left[\frac{1}{e}, \infty\right)\right]_V
  4. D[1e2,1)\left[\frac{1}{e^2}, 1\right)

Correct answer

C. \left.\left[\frac{1}{e}, \infty\right)\right]_V

Step-by-step solution

f(x)=xx;x>0ℓnn⁡=xℓnx1ydydx=xx+ln⁡ndydx=xx(1+ℓnx)\begin{aligned} & \mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{x}} ; \mathrm{x}>0 \\ & \ell \operatorname{nn}=\mathrm{x} \ell \mathrm{n} \mathrm{x} \\ & \frac{1}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}}{\mathrm{x}}+\ln \mathrm{n} \\ & \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}^{\mathrm{x}}(1+\ell \mathrm{n} x) \end{aligned}forstrictlyincreasingfor strictly increasingdydx≥0⇒xx(1+ℓnx)≥0⇒ℓnx≥−1\begin{aligned} & \frac{d y}{d x} \geq 0 \Rightarrow x^x(1+\ell n x) \geq 0 \\ & \Rightarrow \ell n x \geq-1 \end{aligned}$\begin{aligned} & x \geq e^{-1} \\ & x \geq \frac{1}{e} \\ & x \in\left[\frac{1}{e}, \infty\right)\end{aligned}$
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