You are here
![Ex 6.3, 29 (MCQ - The max value of [x(x - 1) + 1]^1/3 is (a) (1/3)^1/3 - Ex 6.3](https://cdn.teachoo.com/2d0fcfdf-409f-4df0-b906-2083f0b6886e/slide75.jpg)
Ex 6.3,29 (MCQ) - Chapter 6 Class 12 Application of Derivatives
Last updated at August 2, 2026 by Teachoo
Chapter 6 Class 12 Application of Derivatives
Concept wise
- Finding rate of change
- To show increasing/decreasing in whole domain
- To show increasing/decreasing in intervals
- Find intervals of increasing/decreasing
- Finding slope of tangent/normal
- Finding point when tangent is parallel/ perpendicular
- Finding equation of tangent/normal when point and curve is given
- Finding equation of tangent/normal when slope and curve are given
- Finding approximate value of numbers
- Finding approximate value of function
- Finding approximate value- Statement questions
- Finding minimum and maximum values from graph
-
Local maxima and minima
- Minima/ maxima (statement questions) - Number questions
- Minima/ maxima (statement questions) - Geometry questions
- Absolute minima/maxima
Take a fresh quiz. Then take another.
Every attempt is a new AI-adaptive Teachoo quiz with 2 questions, selected from your answers, mistakes, and progress.
Transcript
Ex 6.3, 29 The maximum value of ć[š„(š„ā1)+1]ć^(1/3) 0 ā¤ x ā¤ 1 is (A) (1/3)^(1/3) (B) 1/2 (C) 1 (D) 0 š^ā² (š„)=1/3 (š„^2āš„+1)^((ā2)/3) (2š„ā1) š^ā² (š„)=1/(3(š„^2 ā š„ + 1)^(2/3) ) .(2š„ā1) š^ā² (š„)=(2š„ ā 1)/(3(š„^2 ā š„ + 1)^(2/3) ) Putting fā(š)=š (2š„ā1)/(3(š„^2 ā š„ + 1)^(2/3) )=0 2š„ā1=0 2š„=1 š„=1/2 Since, 0 ā¤ x ā¤ 1 Hence, critical points are š„=0 ,1/2 , & 1 Hence, Maximum value is 1 at š„=0 , 1 ā“ Hence, correct answer is C
