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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

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.