Check sibling questions

Ex 6.5, 29 - The max value of [x(x - 1) + 1]^1/3 is (a) (1/3)^1/3

Ex 6.5,29 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5,29 - Chapter 6 Class 12 Application of Derivatives - Part 3


Transcript

Ex 6.5, 29 The maximum value of γ€–[π‘₯(π‘₯βˆ’1)+1]γ€—^(1/3) 0 ≀ x ≀ 1 is (A) (1/3)^(1/3) (B) 1/2 (C) 1 (D) 0 Let f(π‘₯)=[π‘₯(π‘₯βˆ’1)+1]^(1/3) Finding f’(𝒙) 𝑓(π‘₯)=[π‘₯[π‘₯βˆ’1]+1]^(1/3) 𝑓(π‘₯)=[π‘₯^2βˆ’π‘₯+1]^(1/3) 𝑓^β€² (π‘₯)=(𝑑(π‘₯^2 βˆ’ π‘₯ + 1)^(1/3))/𝑑π‘₯ 𝑓^β€² (π‘₯)=1/3 (π‘₯^2βˆ’π‘₯+1)^(1/3 βˆ’ 1) . 𝑑(π‘₯^2 βˆ’ π‘₯ + 1)/𝑑π‘₯ 𝑓^β€² (π‘₯)=1/3 (π‘₯^2βˆ’π‘₯+1)^((βˆ’2)/3) (2π‘₯βˆ’1) 𝑓^β€² (π‘₯)=1/(3(π‘₯^2 βˆ’ π‘₯ + 1)^(2/3) ) .(2π‘₯βˆ’1) 𝑓^β€² (π‘₯)=(2π‘₯ βˆ’ 1)/(3(π‘₯^2 βˆ’ π‘₯ + 1)^(2/3) ) Putting f’(𝒙)=𝟎 (2π‘₯βˆ’1)/(3(π‘₯^2 βˆ’ π‘₯ + 1)^(2/3) )=0 2π‘₯βˆ’1=0 2π‘₯=1 π‘₯=1/2 Since, 0 ≀ x ≀ 1 Hence, critical points are π‘₯=0 ,1/2 , & 1 Hence, Maximum value is 1 at π‘₯=0 , 1 ∴ Hence, correct answer is C

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.