The maximum value of [x (x - 1) + 1 ] (1/3) , 0 ≤ 𝑥 ≤ 1 is:

(a) 0       (b) 1/2         (c) 1     (d) ∛(1/3)

 

This question is inspired from Ex 6.5,29 (MCQ) - Chapter 6 Class 12 - Application of Derivatives

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Transcript

Question 43 The maximum value of ["𝑥 (𝑥 − 1) + 1" ]^(1/3) , 0 ≤ 𝑥 ≤ 1 is: (a) 0 (b) 1/2 (c) 1 (d) ∛(1/3) Let f(𝑥)=[𝑥(𝑥−1)+1]^(1/3) Finding f’(𝒙) 𝑓(𝑥)=[𝑥[𝑥−1]+1]^(1/3) 𝑓(𝑥)=[𝑥^2−𝑥+1]^(1/3) 𝑓^′ (𝑥)=(𝑑(𝑥^2 − 𝑥 + 1)^(1/3))/𝑑𝑥 𝑓^′ (𝑥)=1/3 (𝑥^2−𝑥+1)^(1/3 − 1) . 𝑑(𝑥^2 − 𝑥 + 1)/𝑑𝑥 𝑓^′ (𝑥)=1/3 (𝑥^2−𝑥+1)^((−2)/3) (2𝑥−1) 𝑓^′ (𝑥)=1/(3(𝑥^2 − 𝑥 + 1)^(2/3) ) .(2𝑥−1) 𝑓^′ (𝑥)=(2𝑥 − 1)/(3(𝑥^2 − 𝑥 + 1)^(2/3) ) Putting f’(𝒙)=𝟎 (2𝑥−1)/(3(𝑥^2 − 𝑥 + 1)^(2/3) )=0 2𝑥−1=0 2𝑥=1 𝒙=𝟏/𝟐 Since, 0 ≤ x ≤ 1 Hence, critical points are 𝒙=𝟎 ,𝟏/𝟐 , & 1 Hence, Maximum value is 1 at 𝑥=0 , 1 So, the 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.