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JEE Main 20248 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2024 · 8 April · Morning Shift · Mathematics · Application of Derivatives

The number of critical points of the function f(x)=(x-2)^ 2 / 3 (2...

The number of critical points of the function f(x)=(x−2)2/3(2x+1)f(x)=(x-2)^{2 / 3}(2 x+1) is

Options

  1. A1
  2. B2
  3. C0
  4. D3

Correct answer

B. 2

Step-by-step solution

f(x)=(x−2)2/3(2x+1)f′(x)=23(x−2)−1/3(2x+1)+(x−2)2/3f′(x)=2×(2x+1)+(x−2)3(x−2)1/33x−1(x−2)1/3=0\begin{aligned} & \mathrm{f}(\mathrm{x})=(\mathrm{x}-2)^{2 / 3}(2 \mathrm{x}+1) \\ & \mathrm{f}^{\prime}(\mathrm{x})=\frac{2}{3}(\mathrm{x}-2)^{-1 / 3}(2 \mathrm{x}+1)+(\mathrm{x}-2)^{2 / 3} \\ & \mathrm{f}^{\prime}(\mathrm{x})=2 \times \frac{(2 \mathrm{x}+1)+(\mathrm{x}-2)}{3(\mathrm{x}-2)^{1 / 3}} \\ & \frac{3 \mathrm{x}-1}{(\mathrm{x}-2)^{1 / 3}}=0\end{aligned} Critical points x=13\mathrm{x}=\frac{1}{3} and x=2\mathrm{x}=2
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